text:
Impurities in household baking soda
For a long time, I have been using store bought baking soda (with sodium bicarbonate as the sole ingredient listed) as the sodium bicarbonate in my (qualitative) home experiments, but recently I noticed that a 100mL solution with 3.6g baking soda developed light-yellow precipitates after air exposure for two days. I was pretty sure the baking soda had all dissolved initially.
Also, I started to worry about this because of the weird smell associated with baking soda when heated.
Anyways, what are the main impurities in baking soda? I especially want to know about the source of the yellow precipitate and smell, but a full list or some articles on this topic would be most appreciated.
Searching in search engines and online databases using terms like "baking soda smell" gives results about using baking soda as a deodorant; "impurity in baking soda" does not yield an answer; "baking soda composition" only shows sodium bicarbonate. The vendors I asked persisted that their product contained nothing but sodium bicarbonate, "100% pure with no other substance." (It's unlikely that they admit the presence of impurities, as they think I would refuse to buy "impure" baking soda!)

Why not to ask the vendor?  The third parties cannot know what minor impurities, possibly intentional additives for baking context, are present or added. //  The SE site expects you  search, read and think  about the topic. Then, if you still need to ask, provide the explicit summary of partial answers/ideas/thoughts you have got.  That prevents us to tell you what you already know or what you can easily find yourself. Effort not shown may be considered as effort not done and the question may get closed.
How do I ask a good question?.

Thanks for commenting. I have asked a few vendors just now, and will edit the question and show my personal search results.

Do you have older, well behaving lot ( or from different vendor) of baking soda to compare, to eliminate possibility it is because of the used water?

What was the smell like ? Rather organic or inorganic ? Weird is rather vaque.

What was the source of water and how did you store the solution?

I have been making solutions with the same brand of baking soda for a long time, but I noticed the ppt for the first time yesterday. I was using a new bag of baking soda too. 2. The smell was a bit like ammonia, not as pungent, but definitely unpleasant and not something I'd associate with food. 3. I used tap water and stored the solution in a clear glass jar.

The appearance of light brown precipitates over time in tap water is a sign of (dissolved) iron impurities in a solution. Iron can come anywhere, iron pipes, water, baking soda itself.
For example, the now famous Hamaliyan pink salt, which comes from Khewra mines from Pakistan rather is slightly pink. If you dissolve the salt in water and let water sit for a few minutes, you will see rust colored soilds sitting at bottom. That is iron oxides.
Simply change the baking soda brand, and if you still see yellow ppt in water aftter letting it sit for days. Make a conc. solution to accelerate precipitation.
Heating food-grade baking soda should not generate any smell. The so-called weird smell must be coming from heating the metallic vessel and volatilization of adsorbed organic stuff. Most likely the vessel was a kitchen item (?).
If your baking soda was sitting for a long time in a fridge or outside, it does tend to adsorb smells.

That's a great point. Ferric ions do precipitate under alkaline conditions, and the smell could the baking soda releasing previously adsorbed chemicals in the air. I was using a beaker for the heating though, so I felt the smell should still be from the baking soda.       But after all, has nobody done a composition test for common baking soda?

Just a sanity check: You are not using baking powder, right? Food grade baking soda from a reliable company should be pure.

No. I'm sure it was baking soda; it has "sodium bicarbonate" as the sole ingredient.

@anemia0 with respect to "nobody has done a composition test for common baking soda" there is no such thing as "common baking soda" of fixed composition. The impurities will vary depending on the method and facility (and even batch) of preparation. For each grade (ie food, industrial, etc), there are limits for specified impurities, and every batch will have an associated certificate of analysis (such as this one here: https://ishtarcompany.com/wp-content/uploads/2018/12/Sodium-Bicarbonate-Spec-1014x1024.jpg), but those generally aren't available to household consumers


text:
How to get tin(II) oxide from tin(IV) oxide?
How to get $\ce{SnO}$ from the deposited $\ce{SnO2}$? Any method is ok.
If just heat the $\ce{SnO2}$, can it change to $\ce{SnO}$?

No, it won't react at all - SnO2 is thermally stable.

As noted in the comments and the other answer that $\ce{SnO2}$ is thermally stable and it is in fact $\ce{SnO}$ that thermally disproportionates to $\ce{SnO2}$ and most of the synthesis involves tin(II) salts e.g. tin(II) oxyhydroxide. However, there is a way to synthesize it from a tin(IV) salt. If you hydrolyze tin(IV) iodide, it forms tin(IV) oxide and hydroiodic acid which will then react with each other to form tin(II) oxide and iodine.
$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$
$$\ce{SnO2 + 2HI -> I2 + SnO + H2O}$$
(taken from this answer. Upvote that answer also if you like this answer)

Common ways of preparing tin monoxide ($\ce{SnO}$) all involve using another tin(II) compound as a precursors, such as reacting a tin(II) salt solution with sodium hydroxide. Not only does attempting to heat $\ce{SnO2}$ to decomposition fail; it is actually the $\ce{SnO}$ that decomposes with heating, disproportionating to give $\ce{SnO2}$. See https://en.wikipedia.org/wiki/Tin(II)_oxide.

The answer is apparently a possible yes, but not precisely "just heating", as correctly noted by my esteem colleagues.
Per this 2008 source: "Experimental study of SnO2/SnO/Sn thermochemical systems for solar production of hydrogen, to quote:

The thermal reduction of tin(IV) oxide occurs in the temperature range 1400–1600°C following a zero order kinetic law of Arrhenius with an activation energy of 394.8 kJ mol−1... The operating conditions that prevent gaseous stannous oxide (SnO) from recombining with O2 are defined. The effect of a quenching device (water-cooled finger) is negligible whereas operation at low total pressure or low O2 and SnO partial pressures leads to nearly pure SnO product.

So, a high temperature treatment of tin(IV) oxide to stannous oxide is apparently possible under select operating conditions. In particular, low oxygen concentration and low $\ce{SnO}$ partial pressures may be able to produce a relatively pure $\ce{SnO}$ product.

source: Stack Exchange

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

sort: votes


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Can I degrade PLA into something usable?
I have a lot of 3D printer waste and don't have a recycling machine for it.
As a side project, I was wondering if there is any way to degrade/compost/recycle it meaningfully.
I have my Uni's chemistry lab available to me so I could play around with a lot of things.
I am only an amateur chemist, but the teacher likes me and is interested in the project too.
And just for fun, if we could make something out of it like glucose or alcohol or something, I read you can do that with organic molecules, that would be double the fun.
I will add some papers I have found:

http://greensxm.com/composting-pla-plastic/

https://onlinelibrary.wiley.com/doi/epdf/10.1002/pen.25511#:~:text=4.4%20Enzymatic%20degradation,lipase%2C%20esterase%2C%20and%20alcalase.

https://sci-hub.st/10.1007/s12010-010-9117-7

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


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Notation for orbital coefficient matrix in RHF vs UHF
My question stems from Table 1 of this paper:
"Generalized Hartree–Fock Description of Molecular Dissociation, by
Carlos A. Jiménez-Hoyos, Thomas M. Henderson, and Gustavo E. Scuseria"
https://pubs.acs.org/doi/10.1021/ct200345a
In particular, I am confused about the notation for the subscript of the elements in the orbital coefficient matrix. For a real RHF ansatz for the two-electron/two-orbital system, for example, we should have
\begin{gather}
  |\psi\rangle
  = C_{11}^{\sigma\sigma}|\phi_1\alpha,\,
    \phi_1\beta\rangle + C_{22}^{\sigma\sigma}|\phi_2\alpha,\,
    \phi_2\beta\rangle,
\end{gather}
(where I have changed notation slightly from the original paper by putting the sigma's in the superscript) while for a real UHF, we have
\begin{gather}
  |\psi\rangle
  = C_{11}^{\sigma\sigma}|\phi_1\alpha,\,
    \phi_1\beta\rangle + C_{22}^{\sigma'\sigma'}|\phi_2\alpha,\,
    \phi_2\beta\rangle,
\end{gather}
where the $\phi_1$ and $\phi_2$ are spatial orbitals and $\alpha$ and $\beta$ are the spin orbitals.
Does the superscript $\sigma\sigma$ vs $\sigma'\sigma'$ signify the variational coefficients? For example, does $C_{11}^{\sigma\sigma}$ correspond to a variation of the alpha and beta spin coefficients with the same variational parameter, while a term say $C_{12}^{\sigma\sigma'}$ corresponds to varying alpha and beta with different parameters? I don't believe this notation is explained in the paper.
I also find similar notation here (unfortunately, no online pdf). Now, RHF, UHF, GHF, etc. are defined by the matrix-structure of the orbital coefficient matrix. It looks like the block-diagonal terms correspond to how we vary the spatial component in the HF ansatz with respect to the different spin sectors (correct?) Do off-diagonal terms in the orbital coefficient matrix correspond to variation of the spin sectors themselves? Is there a more accessible reference that discusses this reasoning in more detail?

They are spin-blocking the canonical MO coefficients $\mathbf{C}$, in other words treating the spin orbitals as two-component spin vectors. I think this spin-structure is easier to see with one-body operators like the Hamiltonian (Fock operator) or density matrix, but we can still apply the spin-blocking in this case.
Say we have an atomic spin orbital
\begin{gather}
  |\Phi\rangle
    = |\phi\rangle|\sigma\rangle
    = \pmatrix{\phi_\alpha\phi_\beta},
\end{gather}
where $|\phi\rangle$ is the spatial component and $|\sigma\rangle \in \{|\alpha\rangle, |\beta\rangle\}$ (spin-up and spin-down, respectively).
Then the general canonical spin orbital $|\Psi\rangle$ can be obtained from the coefficient matrix $\mathbf{C}$ as
\begin{gather}
  |\Psi\rangle
    = \mathbf{C} |\Phi\rangle
    = \pmatrix{
        C_{\alpha\alpha} & C_{\alpha\beta} \\
        C_{\beta\alpha} & C_{\beta\beta}
      }
      \pmatrix{
        \phi_\alpha \\
        \phi_\beta
      }.
\end{gather}
In the RHF case, $C_{\alpha\alpha} = C_{\beta\beta}$, and so
\begin{gather}
  |\Psi\rangle
    = \mathbf{C} |\Phi\rangle
    = \pmatrix{
        C_{\alpha\alpha} & 0 \\
        0 & C_{\alpha\alpha}
      }
      \pmatrix{
        \phi_\alpha \\
        \phi_\beta
      }
    = C_{\alpha\alpha}
      \pmatrix{
        \phi_\alpha \\ \phi_\beta
      },
\end{gather}
which describes two spin-paired electrons in a single spatial orbital.
In the UHF case, $C_{\alpha\alpha} \neq C_{\beta\beta}$, and so
\begin{gather}
  |\Psi\rangle
  = \mathbf{C} |\Phi\rangle
  = \pmatrix{
      C_{\alpha\alpha} & 0 \\
      0 & C_{\beta\beta}
    }
    \pmatrix{
      \phi_\alpha \\
      \phi_\beta
    }
    = \pmatrix{
        C_{\alpha\alpha} \phi_\alpha \\
        C_{\beta\beta}\phi_\beta
      },
\end{gather}
which describes two opposite spin electrons in a separate spatial orbitals.
In the GHF case, where we relax the requirement for electrons to be eigenfunctions of $\hat{S}_z$, we can write the general case as
\begin{gather}
  |\Psi\rangle
  = \mathbf{C} |\Phi\rangle
  = \pmatrix{
      C_{\alpha\alpha} & C_{\alpha\beta} \\
      C_{\beta\alpha} & C_{\beta\beta}
    }
    \pmatrix{
      \phi_\alpha \\
      \phi_\beta
    }
  = \pmatrix{
      C_{\alpha\alpha} \phi_\alpha + C_{\alpha\beta} \phi_\beta\\
      C_{\beta\alpha} \phi_\alpha + C_{\beta\beta}\phi_\beta
    }.
\end{gather}
All the $C_{\sigma\sigma'}$ coefficients are independent variational parameters, subject to orthonormality constraints. Moreover, coefficients may be real or complex-valued.

Does the alpha/beta subscript on the spatial $\phi$ denote the spin orbital or the spatial orbital?

I have usually seen the different spatial orbitals denoted by A and B, as opposed to $\alpha$ or $\beta$

Alpha/beta on spatial $\phi$ denotes the spatial orbital corresponding to that particular spin block. It's not common to work with the full spin blocked matrix, since off-diagonal spin elements are generally zero, with GHF being the exception.

I believe I understand. So when writing a 2-electron Slater determinant for some GHF state, say, would it be correct to say that each element mixes the $\alpha$ and $\beta$ sectors; i.e., each element would be of the form $\psi_\nu(r,,s)=\alpha(s)\sum_\mu c_{\nu \mu}\phi_\mu(r)+\beta(s) \sum_\mu c_{\nu \mu}'\phi_\mu(r)$? (here, the c's are the elements of your C matrices). I am looking in Szabo's QC book, but they don't seem to consider the case of constructing Slater determinants with spin orbital mixing a la GHF


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Comparing stability of halide ions
I have to compare stability of halide ions : $\ce{F-},\ce{Cl-},\ce{Br-},\ce{I-}$
I am confused between two factors : As we go down the group , the size of given species increases which spreads the negative charge more ( as incoming electron goes into a larger orbital and hence feels less repulsion) , however the electron gain enthalpy factor conveys a different result.
According to the first factor $\ce{I-}$ must be most stable but according to second factor $\ce{Cl-}$ must be most stable as chlorine's electron gain enthalpy is most negative.
I think here one factor should dominate over other but can't seem to figure out.
I searched various online  resources and each one contradicts the other.
Can you please guide me?

I'm pretty sure this depends on context; e.g. fluoride is most stable in aprotic, soft solvents but iodide is most stable instead in protic, hard solvents.


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How can a 90 degree pulse cause precession perpendicular to the applied field in proton NMR
From my understanding, a proton's magnetic moment in an NMR experiment can undergo precession in only two discrete states: either parallel to the applied magnetic field, or antiparallel to it. Absorbing light of propper frequency allows it to go from the lower energy parallel configuration, to the higher energy antiparellel one.
This framework seems to conflict with all explanations I've found regarding pulsed (proton) NMR. Regarding 90 degree pulse experiments, I've read that they work by supplying a broadband radio frequency pulse perpendicular to the applied magnetic field, causing all protons to to absorb the light and enter an excited state. This then somehow causes them to change the precession of their magnetic moments to be along a plane perpendicular to the applied field. This precession then gradually moves back to the ground state, where it is parallel to the applied field.
I'm confused as to how the magnetic moment can ever move to be perpendicular like this, as that would mean it's in between being parallel and antiparallel to the applied field. This conflicts with my initial understanding, in which the moments can only exist in the 2 discrete states. Where am I going wrong?

Well, you're right: spins don't exist only in spin-up or spin-down. This simplification is very sadly propagated in typical organic chemistry books, probably because the truth is too complicated.

How does it really work then?

An answer is possible, but it would be rather long, so in the meantime can I suggest reading a proper NMR book (see https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry and Ctrl-F for Keeler or Levitt). Note, also, that if the magnetisation is parallel to the external field, it cannot precess about it. This is because precession is essentially a rotation about that axis; anything lying on the axis of rotation doesn't rotate.

Simply speaking, the pulse brings a small fraction (equal to the difference in occupation between the two states in equillibrium) of the spins to precess in phase. That turns the static net magnetisation in $B_0$ direction into a rf ("precessing", i.e. measureable via induction) magnetisation orthogonal to $B_0$. If your book talks about "light" and "all spins", I recommend you find a nice dumpster for it.

The 90° pulse lets 50% of the spins change state. Or at least that's what the result looks like, in our simple picture.

https://mriquestions.com/phase-coherence.html

Have a look at the Bloch sphere. This is a representation of a two-level system such as a spin 1/2 system. If a vector points to one of the poles of the sphere, you have a pure spin up or spin down state. A 90 degree pulse rotates this vector to the equatorial plane representing a superposition of spin up and spin down states. Note that the vector precesses around the vertical axis with an angular velocity depending on the difference between the eigen frequency and the frequency of the pulse.

Spins don't exist only as spin-up or spin-down.
I make no apology for making this big and bold. It is so disappointing to see this misconception being propagated in so many textbooks.
Spin-up and spin-down are two of the infinite possible quantum states. In general, a spin-1/2 nucleus has a state
$$\psi = c_\alpha \psi_\alpha + c_\beta \psi_\beta$$
where $c_\alpha, c_\beta$ are complex numbers and $\psi_\alpha$ and  $\psi_\beta$ represent the 'spin up' and 'spin down' states respectively. In order for the state to be 'physically sensible', we require that $|c_\alpha|^2 + |c_\beta|^2 = 1$.
A pure spin-up state would have $c_\alpha = 1$ and $c_\beta = 0$; and a pure spin-down state would have $c_\alpha = 0$ and $c_\beta = 1$.* However, there is no stipulation that these are the only possible values. It is perfectly sensible to have a linear combination, or a superposition, of spin-up and spin-down:
$$\psi = \frac{1}{\sqrt{2}} \psi_\alpha + \frac{1}{\sqrt{2}} \psi_\beta$$
It is easy to verify that $|c_\alpha|^2 + |c_\beta^2| = 1/2 + 1/2 = 1$, as required. This is a perfectly valid state for a spin-1/2 nucleus.
So, if we identify the two special states $\psi_\alpha$ and $\psi_\beta$ with "along the positive $z$-axis" and "along the negative $z$-axis" respectively, then it logically follows that all the other states can be identified with pointing in a different direction.
Formally, the way to do this is to calculate the expectation value of $x$-, $y$-, and $z$-angular momentum from each state.† I'll not go through the maths here, but essentially you will get three values out of this, often denoted $(I_x, I_y, I_z)$. Then you can plot this on what is known as the Bloch sphere. For example, if you calculate this for the spin-up state, you will get $(I_x, I_y, I_z) = (0, 0, 1)$,‡ which essentially points straight up along the $z$-axis. It follows, that different combinations of $c_\alpha$ and $c_\beta$ lead to different values of $I_x, I_y, I_z$ which in turn lead to different 'directions' on the Bloch sphere.
In general, you can convert freely between these two representations of the spin states: you can either use the $(c_\alpha, c_\beta)$ representation, or you can use the $(I_x, I_y, I_z)$ representation.§

NMR on one spin
If we start with a single spin pointing along the $+z$-axis, i.e. spin-up, then the effect of a 90° pulse would be to tilt it to somewhere in the $xy$-plane. It could be along the $x$-axis (in which case we'd have $I_x = 1$ and $I_y = I_z = 0$), or it could be along the $y$-axis, etc. etc. In general, such states are superpositions of $\psi_\alpha$ and $\psi_\beta$. The exact state you get depend on the phase of the pulse, which isn't super important for now.
Let's say, for argument's sake, that you get a state along the $x$-axis. It turns out that this state is not actually 'stable'; if you leave it to be, it will evolve in time. How does it evolve? It turns out that if you work through the mathematics of time-dependent quantum mechanics, this state rotates about the $z$-axis with a characteristic frequency $\omega$. We can express this in terms of the coefficients:
$$\begin{pmatrix}I_x \\ I_y \\ I_z \end{pmatrix} = \begin{pmatrix}\cos(\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix}$$
You can verify that if we plug in $t = 0$, we get back $(I_x, I_y, I_z) = (1, 0, 0)$, which corresponds to "immediately after the pulse".
Fundamentally, NMR is based on detecting these oscillations. The electronics are set up to detect terms which behave like $\cos(\omega t)$ as a function of time; Fourier transformation of these terms allows us to extract the frequencies $\omega$ and hence plot an NMR spectrum.
Now here comes the issue when discussing NMR on single spins: we can't actually continually measure $I_x$ or $I_y$ on a single spin. This is not only technologically impossible, but also highly problematic considering that the act of quantum measurement leads to 'wavefunction collapse'.
This is the point at which we need to consider many spins, or macroscopic properties.

NMR on many spins
We now consider an ensemble of many (identical) spins. For example, we could have a sample of water, in which all of the $\ce{^1H}$ spins are identical.¶
Are all of the spins going to point up? Well, if the energy difference between spin-up and spin-down was large enough, then maybe they would. However, NMR energy differences are extremely tiny! As a result, there isn't enough bias towards the lower-energy states. What you get is that all the spins point in essentially random directions, with a slight overall bias towards the $+z$-axis. We call this a polarisation, in that the spins generally lean towards the $+z$-axis.
The 90° pulse acts on each of these spins differently. If one of those spins happens to start off along the $+z$-axis, it will be tilted into the $xy$-plane correctly. If one of the spins happens to already be in the $xy$-plane, then it may end up somewhere else entirely.
What can be shown (using a formalism called density matrices) is that the overall polarisation, which starts off as $+z$, mimics the single-spins case. That is to say, the 90° pulse brings the net polarisation into the $xy$-plane, after which it starts evolving (or precessing) at its characteristic frequency $\omega$. At this point we tend to call these "coherences" instead of "polarisations", which is a slightly complicated term to define properly, but essentially it revolves around the idea that: even though each individual spin is doing its own thing, collectively, they act in tandem to provide a net effect that can be detected.
If you have different types of spins, like in a more complicated organic molecule, then you would get have multiple types of polarisation which precess at different rates. Again, this can be Fourier transformed to extract the individual frequencies of each component.
Now, the fact that we are dealing with many spins helps us to avoid the traps of measurement previously mentioned. The first thing to note is that we have many spins, so the signal that we detect is likely to be rather bigger than just that from a single spin.# The second thing to note is that we no longer care about each individual spin's state: we're not measuring each spin's wavefunction individually, but rather we are measuring how the entire ensemble is biased or polarised. The net bias of the ensemble can be constructed in many ways; by measuring it, we don't actually enforce any requirement on the individual spin's wavefunctions. This means that we don't run into awkward issues with wavefunction collapse.

Further reading
The NMR books by Keeler (Understanding NMR Spectroscopy, 2nd ed., Wiley) and Levitt (Spin Dynamics, 2nd ed., Wiley) explain this more thoroughly than I can here.

Footnotes
* More precisely, one is equal to $0$ and the other is equal to some arbitrary complex number with magnitude $1$, which can be represented as $\mathrm{e}^{\mathrm{i}\theta}$ for any angle $\theta$.
† Note that the Heisenberg uncertainty principle (HUP) doesn't stop you from doing this. The HUP doesn't tell you anything about the actual expectation value itself, it only tells you about the uncertainties in them. For example, consider position and momentum: the HUP says that you cannot measure both simultaneously and with perfect accuracy. It doesn't say that you cannot measure these simultaneously, nor does it say that you cannot measure them at all: it is only the combination of simultaneously and perfect accuracy that is forbidden.
‡ There is a constant factor of $\hbar/2$ which I've neglected. It's not hugely important. The important part is that $I_x = I_y = 0$ and $I_z \neq 0$.
§ Going deeper, this is a result of a homomorphism between the SU(2) and SO(3) groups, which has quite a bit of significance in spin physics.
¶ Yes, technically, there's also going to be $\ce{H3O+}$ and $\ce{OH-}$.
# Because the energy gap is small, though, the net polarisation is still generally pretty small: NMR is considered an insensitive technique.

I will "cheat" by circumventing a detailed attempt to describe how a macroscopic magnetization arises from an ensemble of spins in the presence of a static external magnetic field, or how such an ensemble responds to perturbations in the form of applied RF fields.
Consider instead that when we probe a spin and say that it points up or down relative to a magnetic field, what we are saying is that one specific component of the spin angular momentum - that along the field - points up or down, and that the magnitude of that component (the projection of the spin angular momentum onto the coordinate axis defined by the field) is fixed (it is a quantum property). There will be an additional component in the plane perpendicular to the field, and which according to the Uncertainty Principle may point anywhere on that plane, the direction being indeterminate, provided the total angular momentum vector retains the right magnitude.
Next, it is worth distinguishing between quantum and macroscopic properties. Spin angular momentum is a quantum property and when attempting to describe it the Uncertainty Principle must be kept in mind at all times. Precession on the other hand is a macroscopic (classical E&M) concept that can be used to describe the motion of a very large collection of spins about an external magnetic field. This ensemble property can be dealt with classically without giving the Uncertainty Principle much thought.
One way to bridge the two perspectives is to consider that an applied RF field perturbs the quantum state of a spin and can result in a mixed state in which it is no longer up or down, it is in a superposition of those. An ensemble of spins in such a state is referred to in NMR as a "quantum coherence", to imply that all of the spins are behaving the same way (they are coherent), but so as not to forget the QM nature of the individual spins.

From my understanding, a proton's magnetic moment in an NMR experiment can undergo precession in only two discrete states: either parallel to the applied magnetic field, or antiparallel to it.

You have to distinguish bulk magnetization, which you can treat classically, with the magnetic moment of a single proton, which is governed by quantum mechanics. The precession picture is for the classic view.

Absorbing light of propper frequency allows it to go from the lower energy parallel configuration, to the higher energy antiparellel one.

When you look at the classic NMR spectrometer (scanning), switching spin states is a good picture. If you scan too slowly, you reach saturation (same number of spin up and spin down) and lose the signal. You would have to wait (with kinetics determined by the so-called T1 parameter, several seconds for a 600 Mhz instrument) until you are back at the equilibrium situation, where a small excess of spins is in the lower energy state.

This framework seems to conflict with all explanations I've found regarding pulsed (proton) NMR.

In pulsed NMR, you start things off with a 90 degree pulse, resulting in half the spins up and half down. The reason you still get a signal is that there is a coherence (or spin polarization). This coherence goes away after a while (the protons from different molecules experience a slighlty different field because of tumbling and conformational differences), with kinetics determined by the so-called T2 parameter. T2 is typically much shorter than T1, on the order of tens of milliseconds. So once you lose coherence, you have to wait out T1 before you can run the next scan.

Regarding 90 degree pulse experiments, I've read that they work by supplying a broadband radio frequency pulse perpendicular to the applied magnetic field, causing all protons to to absorb the light and enter an excited state.

No, the pulse results in a coherent state for a 90 degree pulse. For a 180 degree pulse applied to the equilibrium state, you would get what you describe. However, most pulse sequences start with a 90 degree pulse (because that is how you get a signal).

This then somehow causes them to change the precession of their magnetic moments to be along a plane perpendicular to the applied field. This precession then gradually moves back to the ground state, where it is parallel to the applied field.

For this picture to be almost correct, you need two things: First, you have to look at bulk magnetization rather than single protons. Second, you have to be in a rotating frame. In the laboratory frame, the magnetization always precesses around the B0 field (the constant, much stronger field) and it would be difficult to describe what the pulses do.

I'm confused as to how the magnetic moment can ever move to be perpendicular like this, as that would mean it's in between being parallel and antiparallel to the applied field. This conflicts with my initial understanding, in which the moments can only exist in the 2 discrete states. Where am I going wrong?

As orthocresol states in the comments, you can have a superposition of states (up and down) if you want to explain the magnetization perpendicular to the B0 field at the level of single protons. This is explained handwaivingly here (with a pinch of quantum mechanics if you click on the advanced explanation).


text:
Rate of Reaction - Potassium Permanganate and Hydrogen Peroxide
I'm interested in an aerospace project that requires I know the reaction rate ($[\pu{M/s}]$) of,
$$\ce{2 KMnO4 (aq) + 3 H2O2 (aq) -> 3 O2 (g) + 2 MnO2 (s) + 2 KOH (aq) + 2H2O (l)}.$$
I'm just curious if there is an analytic method for finding it?
To that end, I also feel I should double-check. To predict the heat ($[\pu{kJ}]$) released I only need to do the following calculation:
$$\Delta H_\mathrm{reaction} = \sum\Delta H_\mathrm{products} - \sum\Delta H_\mathrm{reactants}$$
Is this correct?
My goal is to predict the thrust generated by a hypothetical propulsion system, using this reaction as opposed to combustion.
My thought is if I can derive a theoretical reaction rate then I can calculate a theoretical value for $\mathrm{d}\Delta H_\mathrm{reaction}/\mathrm{d}t$, then thanks to the ideal gas law and thermal physics I should be able to calculate a theoretical max thrust.

You use mhchem \ce{} by a weird way.
You can write $$\ce{KMnO4(aq) + 3 H2O2(aq) -> 3 O2(g) +2 MnO2(s) +2KOH(aq) +2H2O(l)}$$ to get
$$\ce{KMnO4(aq) + 3 H2O2(aq) -> 3 O2(g) +2 MnO2(s) +2KOH(aq) +2H2O(l)}$$
Note that phase states are recommended not to be written as subscripts.
See  https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

BTW, it is permanganate, manganate is $\ce{K2MnO4}$. // I doubt the rate can be calculated from basic principles, so unless it is published, it must be measured. TD preference and kinetic rate are independent.

The rate of catalyzed decomposition of H2O2 will depend on the state of division of the catalyst. That will certainly change over the course of the reaction as the MnO2 is formed, then gets reacted upon.
H2O2 has been used since before World War II for propulsion of rockets and torpedoes. https://en.wikipedia.org/wiki/High-test_peroxide
Although it is not stated explicitly in the Wikipedia article, it seems that the rate of decomposition of H2O2 is determined simply by the rate at which it is pumped into the propulsion chamber (with the catalyst separately pumped in). This assumes essentially instantaneous decomposition, which is apparently an adequate approximation.
The reaction of H2O2 with MnO2 or KMnO4 would be similar to hypergolic fuels. The first hypergolic fuel actually used H2O2 along with hydrazine hydrate. "Hypergolically-fueled rocket engines are usually simple and reliable because they need no ignition system. Although larger hypergolic engines in some launch vehicles use turbopumps, most hypergolic engines are pressure-fed." The reaction is said to be spontaneous, but more than that, it is fast, immediate, explosively fast. It is kept from exploding by limiting the amount that gets to react. You could probably call it a limited explosion; the astronauts have essentially declared that their trips are like riding an explosion. https://en.wikipedia.org/wiki/Hypergolic_propellant


text:
Does Markovnikov's Rule apply to reactions like Dehydration of Alcohols?
I understand Markovnikov's rule essentially tells us the major product in addition reactions when a smaller polar molecule is added onto an unsymmetrical larger compound, however does it also apply to the dehydration of alcohols ie not addition reactions?
Or perhaps is there another rule, regardless can you determine the major/minor product?

You’re looking for Zaitsev's rule


text:
Why does the stratosphere have a lot of ozone, but not the Earth's surface?
Ozone ($\ce{O3}$) and dioxygen ($\ce{O2}$) are both allotropes of oxygen. However, nearly 90% of all ozone is found in the stratosphere, whereas near the earth's surface (or overall in the troposphere) there is hardly any naturally-formed $\ce{O3}$.
Why is this so? Is this because the natural formation of $\ce{O3}$ requires special atmospheric conditions, such as the presence of some catalyst or reactant that is only in the stratosphere?

Because the stratosphere also has a lot of hard UV radiation.

related: Ozone is above, nitrogen is below

The relative AND total amount of ozone in that thin layer of the stratosphere is TINY.

The relative amount of ozone in the stratosphere is so tiny that its mole fraction is about $10^{-5}$ - $10^{-6}$

https://chemistry.stackexchange.com/questions/13573/endothermicity-of-reaction https://chemistry.stackexchange.com/questions/69767/why-is-it-that-only-ozone-absorbs-uv-radiation https://chemistry.stackexchange.com/questions/84032/determining-the-rate-of-stratospheric-ozone-destruction

https://chemistry.stackexchange.com/questions/28485/how-much-ozone-is-produced-by-ionization-of-air-and-how-turn-ozone-into-oxygen

Ozone is produced naturally when highly energetic UV radiation breaks down oxygen molecules and cause them to split apart in a process called photolysis. If a free oxygen atom collides with another oxygen moleule, it joins up, forming ozone.
$$\ce{O2 + UV -> 2 O}$$
$$\ce{O + O2 -> O3}$$

So, why most of the ozone resides in stratosphere?

The equilibrium concentration of ozone depends on the altitude and is maximum at 25km which is just above troposphere.

Only a small amount of ozone accumulates in the troposphere because other UV photons break down ozone molecules, turning them back into oxygen. When ozone is being broken down as fast as it is formed, its concentration remains constant. The exact height where the ozone concentration is maximum is 25km.
Ozone absorbs many of the high-energy UV photons, converting their energy into heat. This energy heats the air, and the ozone rich air rises above the troposphere. This higher temperature differentiates the stratosphere from the troposphere.
Most of the high-energy UV photons is used up by stratospheric oxygen and no significant amount of these photons reach in the lower part of the atmosphere.

There are certainly other factors that plays a role. You can find more information in this question post from EarthScience.SE.

I would replace "rises" by "stays". And you should perhaps mention that the max. ozone concentration in the stratosphere is naturally well below 10 ppm.


text:
Finding the reaction order from a given plot of chemical kinetics
What will be the order of the reaction for a chemical change having $\log t_{1/2}$ VS $\log a$ Where $a=$ Initial concentration of reactant and $t_{1/2} =$ Half Life?

0 order
1st order
2nd order
None of These

Actually the answer I found by searching in internet is option 1 but following my calculations, I am getting the answer as option 4.
My calculations are as follows:
For 0 order reaction:
$$t_{1/2} = \frac{a}{2K}$$
Taking $\log$ on both sides:
$$\log t_{1/2} = \log{\frac{a}{2K}}$$
$$\log t_{1/2} =  \log a - \log 2K \tag1$$
This is a Straight line equation of type $y= mx - C $.
MY DOUBT IS
If you notice in graph you will get to know that the intercept is positive while in the equation $(1)$, intercept is negative so how is it possible?
My Background- actually I didn't have kept Mathematics as my major subject in senior secondary high school so I have little knowledge with respect to graph so please forgive me if I am wrong at formulating (1).

Whoever indicated the precise angle of 45 degrees in that graph should have known that it can only be true for a particular combination of time units (for $t_{1/2}$) and for concentration for ($a$).

At least in the image you have shown, the location of zero of the y-axis is not indicated. How do you know the intercept is positive?

The equation that you have derived is correct. For a zeroth-order reaction, $$\log t_{1/2}=\log a-\log2K$$
By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the $K$ value is very small. In case of any reaction having $K\lt0.5$, $2K$ becomes less than $1$ and hence $\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than $0.5$).

There are a lot of units problems/assumptions implicit in your answer.  For example, you could just change the time units you use to measure the rate constant.  If it has an inconvenient value like 3 mol per L per s, you can just decide to use a time unit of nanoseconds, so then $k$ will be $3\times10^{-9}$ mol per L per nanosecond.  And that is much less than 0.5.

The generic slope-intercept equation for a line is:
$$y=mx+c$$
When given angle $\theta$ for a straight line, the slope $m$ is calculated by:
$$m=tan\;\theta$$
In this case:
$$m=tan\;(45°)=1$$
Substituting $m$ into the generic slope-intercept equation for a line, and comparing it with your derived equation (1):
$$y=x+c$$
$$logt_{1/2}=log\;a-log\;(2K)$$
We can see that:
$$y=log\;t_{1/2}$$
$$m=1$$
$$x=log\;a$$
$$c=-log\;(2K)$$
The slope $m$ is not negative, since it's equal to 1.
As far as the intercept $c$ is concerned, as explained by other answers, the value of $K$ for this particular reaction rate is such that taking the base 10 logarithm of $2K$ value will itself result in a negative number, which will then become a positive number once mutiplied by the $-$ sign next to the logarithm function.
In other words, $log(2K)$ is negative and $-log(2K)$ is positive.
In conclusion, both $m$ and $c$ are positive in this problem.


text:
How exactly are unit cells arranged in a crystal?
Does a certain type of compound only contain a single type of unit cell (eg.bcc,fcc,etc) or different type of units cells combine to form a crystal of any compound?

One compound means one crystal structure mean one unit cell. You don't combine different units cells together.

Whoever taught you about crystallography has willfully omitted the most basic principle. Can you explain how this happened? It ought to be exposed.

The unit cell repeats in all three dimensions. In a perfect crystal, you get exact copies of the contents of the unit cell. The unit cell edges a, b, and c are the translation vectors used to translate the contents of the unit cell in three direction. The unit cell is a construct, the atom arrangement it describes is real.
The content of the unit cell can be simple (just one atom) or complicated (complex structure with thousands of atoms, e.g. a protein).
If the unit cell does not repeat in three dimensions, you don't have a 3D crystal. Instead, you might have an amorphous solid, a quasicrystal, a crystal with defects, or some other non-perfectly ordered structure.

Ok thank you!. So do the arrangement of these unit cells give rise to voids? for example in NaCl , Na occupy all octahedral voids, How exactly do these voids arise?

At the beginning, there was the void. Once you place chloride ions in a regular arrangement, only some of the void is occupied by chloride. Connecting the chloride ions in your mind with lines shows octahedrons, the centers of which are maximally distant from the chloride ions and are large enough for sodium ions to fit in. This is an artificial order of making the crystal, of course. Without the sodium ions, the chloride ions would blow apart.


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How to determine positions of groups on compound based on H-NMR spectroscopy (C9H13NO)
Based on the below $\pu{200 MHz}$ $\ce{^1H}$-$\mathrm{NMR}$ spectroscopy, I have to determine the molecular structure of compound $\bf{B}$, which oxidises to a compound $\bf{A}$ with molecular formula $\ce{C9H11NO}$.

I am pretty sure compound $\bf{B}$ will then be $\ce{C9H13NO}$.
I know how to look at peaks and thereby neighbours, integral and thereby hydrogens on group in total.
From the shift, peaks, and integral I believe the group furthest downfield to be an aromatic (benzene) ring.
From this I am down to the following 3 possible compounds (ignoring chirality), but I'm unsure of how to determine with the shift by hand (no simulating software).

The following experiments have been done on compound A:
2,4-DNPH -> Red sediment -> carbonyl-group
Fehling -> no sediment -> not aldehyde, but ketone
Br(aq) -> no decolouration -> no C=C
Polaroid -> light shines through -> presence of asymmetric C-atom

None of these structures matched the spectrum. For example, how do you explain the two-proton sharp singlet at  around $\pu{2.0 ppm}$?

Also, it would be helpful if you calculated number of unsaturations first since you have the molecular formula.

I believe that singlet to be -NH2. From experiment with Br we find that there should be no C=C bonds.

Can't be the third one because A would be an amide, and 2,4 DNPH would not react with it.

You have found the presence of carbonyl group. but, where is the carbonyl group in your structures?

The third compound is not isolable but an intermediate.

The carbonyl group would be on compound A which is oxidated compound B. So I expect B to instead have a hydroxy group.


text:
Aren't electrophilic addition reactions of alkenes also nucleophilic addition reactions?
And if they are nucleophilic substitution reactions as well, then why do we not call them "electrophilic and nucleophilic addition reactions of alkenes"?
Consider the reaction between ethene and HBr. The HBr bond undergoes heterolytic fission resulting in a bromide anion and hydrogen cation. The hydrogen cation quickly forms a dative bond with one of the carbon atoms (the hydrogen atom is the electrophile and it got added to ethene).
After that, the bromide anion (which is a nucleophile) forms a dative bond with the other carbon atom (acting like a nucleophile) and is "added" to the structure.
So why is such a reaction called "electrophilic addition" even though nucleophilic addition is also taking place? Why isn't it called "electrophilic & nucleophilic addition"?

Substrate is alkene. Alkene get saturated so it is an addition reaction. first group to add is proton (in your example) which is a electrophile. Therefore, it is an electriphilic addition. The rest is to complete reaction. Think about redox reaction. There can't be a reduction without oxidation. Calling the reaction is with respect to one molecule. Here it is alkene.


text:
How to remove graphite substrate of boron carbide coating?
I have a boron carbide coating (the thickness is either 90 microns or 220 microns) on a graphite substrate (the thickness is 0.5 inch - about 13 mm, the other dimensions are 2x2 inch$^2$ - about 5x5 cm$^2$). I need to remove the graphite substrate to (hopefully) have thin free-standing boron carbide plates. Any suggestions? Chemical or mechanical methods? And who could perform that? In particular, I considered burning graphite in air or oxygen, but I have not found clear information on what conditions I need to burn graphite without destroying boron carbide.

If it were a thin substrate, I'd suggest flipping it substrate up in a vacuum chamber and using an oxygen plasma to etch through the carbon. However, a 13 mm etch would take impractically long, and likely would not be uniform after going through so much material.
If you can start with a much thinner carbon substrate, or if you can mechanically remove most of it while keeping the surface parallel to the $\ce{B4C}$ surface, perhaps by cleaving the graphite, plasma etching might still be a good choice. However, mechanically reducing thickness likely will be difficult.

Thank you. So what kind of etching rate should I expect? And will boron carbide be immune to such etching?

Rate: slow; check manufacturer's literature, for example, or contact them. As for attacking the boron carbide, keep that side shielded, against a protective surface, perhaps adding a layer of photo-resit atop,

But it will not protect the boron carbide from the inner side, unless the substrate is removed incompletely.

True, but since the removal is slow, you should be able to gauge when it's done.

Well, hopefully, I will be able to find a more selective approach.


text:
Could an LPG gas ever become lighter than air?
Both of the LPG gases, propane ($\ce{C3H8}$) and butane ($\ce{C4H10}$) are heavier than air. Propane is $1.55$ times heavier and butane is $2.08$ times heavier. Is same-temperature air always lighter than the LPG gases? If not, at what temperature(s) would a change(s) take place?
The reason I am asking this is because I recently opened a barbecue gas canister to drain it empty. But I stopped the draining not too long after, when I read on the net that LPG gas was heavier than air and would therefore not raise into the sky. The thought then came to me that maybe there is a difference between summer and winter here. That the gas may diffuse easier during the warmer part of the year.

This is why many indoor garages forbid the entrance of LPG-powered vehicles.

Every useful word avoided in the initial question formulation is paid later by several words posted by responders and the question poster during clarification. :-)

Density is not the relevant concept for your actual question. Gases mix eventually, so the propane and butane would escape the open container no matter what their density (within reason, i.e. given the densities and the gravity we have - the situation would be different in an ultracentrifuge or on the surface of the sun).

At any but the same temperature and pressure, both propane and butane are always denser than air.
As gas density (with ideal gas approximation) is $$\rho = \frac {pM}{RT},$$
where
$p$ is pressure,
$M$ is molar mass (mean one for air),
$R$ is the universal gas constant $\pu{8.314 J K-1 mol-1}$,
$T$ is absolute temperature.
Therefore, for the same temperature and pressure, density is proportional to molar mass.
$$ \rho \propto M$$
As the molar masses of both gases are bigger than the mean air molar mass (44 resp. 58 versus 28.8 g/mol), the gases are always denser than air.

To address the recent question edits, eventually better dissipation of LPG gas in summer is not because changes of relative density, but due faster diffusion and possibly better thermal or forced convection.

At ambient pressure, the gas phase of a compound is of lesser density than their corresponding liquid (or solid) phase.  The two latter are sometimes referred as condensed, too; molecules are much closer to each other than in the gaseous state and by consequence, there is a greater mass per unit of volume (i.e., an increase of density).
Thus, if the density of compound X as gas exceeds the density of air, then the density of compound X as liquid equally exceeds the density of air.


text:
Does it matter, and if so, how should "dAMP" (meaning "deoxyadenosine monophosphate") be pronounced?
When an expert says "dAMP" (meaning "deoxyadenosine monophosphate") is it pronounced "dee ay em pee", "lowercase dee ay em pee", "damp", "dee amp", or something else (and does it matter)?

Does it matter? Laser is also an acronym but we pronounce it as a word instead of individual letters.

@nilayghosh Good point. I have edited the question accordingly.

https://chemistry.stackexchange.com/questions/162571/why-is-deoxyadenosine-monophosphate-called-damp-and-not-damp?rq=1

Pronounce it as dee-ay-em-pee. That sounds the most natural. "damp" means something else, so you can't use it.

@SRMaiti The fact that DAMP means something already makes it easier to learn. A sort of mnemonic in effect. Not sure that in chemistry you can't pronounce dAMP as "damp" because the latter means something else - would you care to expand on that?

Of course it matters: wrongly pronounced, people either don't understand or immediately identify the speaker as a beginner. Otoh of course it doesn't matter: you'll find out quickly when you start talking to the pros.

@MatthewChristopherBartsh Well, generally people avoid words that mean something else already. For example, if you mix tea and coffee, it is called "cofftea", not "toffee" because there is already a thing called toffee! Since people prononunce AMP as ay-em-pee, it's makes sense to pronounce the d in front of it. Besides how would you pronounce dGMP? It can't be pronounced in go anyway, so you are forced to pronounce each letter separately.

@SRMaiti I'd pronounce "dGMP" as "dee jee em pee". But that doesn't mean I can't pronounce "dAMP" as "damp" or "dee amp".

@NilayGhosh When you write: "Laser is also an acronym but [emphasis mine] we pronounce it as a word instead of individual letters", that's essentially saying: "even though it's an acronym, we still pronounce it as a word."  In fact, laser is an acronym precisely because we pronounce it as a word.  That's what acronym means:  An abbreviation pronounced as a word.  So your comment would have been better phrased as: "Laser is considered an acronym because we pronounce it as a word, rather than as individual letters".

@theorist I agree I could have phrased it that way but it's not applicable for all cases. Take an acronym that cannot be  pronounced as a word e.g. RDX then we have to pronounce as individual letters.

@NilayGhosh  I'm afraid that's not correct.  Abbreviations that aren't pronounced as words, like "RDX" "CPU", or "HIV" are not acronyms.  That's what I've been trying to explain.  Instead, they're initialisms.  Quoting from my Mac's dictionary:  "Acronym. An abbreviation formed from the initial letters of other words and pronounced as a word (e.g. ASCII, NASA)." "Initialism.  An abbreviation consisting of initial letters pronounced separately (e.g., CPU)."  For more info., see: https://www.proofreadnow.com/blog/bid/65823/Acronyms-Initialisms-and-Abbreviations

I spent years of my life - and have the publications to show for it - studying how motor proteins work. Thus, a good portion of my working vocabulary at the time was comprised of the word "MgATP," which all parties pronounced "em gee ay tee pee." I hereby cast my vote for "dee ay em pee" in the case we're talking about here.


text:
Number of stereoisomers of 1,3,5-trimethylcyclohexane
When I searched the stereoisomers of 1,3,5-trimethylcyclohexane, I got four structures given below.

I don't get how structures 1 and 2, 3, and 4 are not the same. Could anyone explain?

1 and 2 are the same all right.

@IvanNeretin what about 3 and 4 how are they different if we just turn the ring upside down?

They aren't different either.

1,3,5-Trimethylcyclohexane doe not show optical isomerism as it has no chiral carbon atoms. Therefore, we are left with counting the total number of cis–trans isomers.
1,3,5-Trimethylcyclohexane has two cis–trans isomers which are given by structures 1 and 3 in your diagram.
1 and 2 are identical because if you look from behind the plane of the screen, 1 will turn into 2. Similar argument holds for 3 and 4. Hence we get two distinct cis–trans isomers.

They are conformational isomers e,e,e and a,a,a  and  a,e,e and e,a,a. So it depends on temperature. Molecules are 3-dimensional.


text:
Why does BH3/THF reduce the carbonyl group to a methylene rather than an alcohol?

This reaction is taken from: Pintér, Á.; Sud, A.; Sureshkumar, D.; Klussmann, M. Autoxidative Carbon-Carbon Bond Formation from Carbon-Hydrogen Bonds. Angew. Chem., Int. Ed. 2010, 49 (29), 5004–5007. DOI: 10.1002/anie.201000711.
Normally carbonyl groups are reduced to alcohols, why this reaction can go directly to methylene?

This is more like an amide than a ketone.

Thanks for the edit, that's a meaningful idea you mentioned. THX a lot.

If you got your answer, feel free to self answer it. Self answer are always appreciated and it also help future readers.

There is a hint in Wikipedia that the substrate, rather than the borane, may be the main factor. In this case the substrate is aromatic:

Aromatic carbonyls are more readily reduced to their respective alkanes than aliphatic compounds.[1] For example, ketones are reduced to their respective alkyl benzenes by catalytic hydrogenation[2,3] or by Birch reduction[4] under mild conditions.

While this passage does not mention specifically reduction by borane, the same effect would likely apply. If we suppose that the carbonyl compound is first reduced to an alcohol, removal of the hydroxyl group would require formation of a radical or ion, which in an aromatic compound such as the OP's substrate would be stabilized by conjugation with the aromatic ring(s). Such a species would then be reduced by accepting a hydrogen in place of the hydroxyl group.
With borane acting as a Lewis acid on the OP's substrate, an aromatic carbocation (isoelectronic with anthracene) would form and could then accept a hydride ion to give the methylene product.
Cited References
1.
Nishimura, Shigeo (2001). Handbook of Heterogeneous Catalytic Hydrogenation for Organic Synthesis (1st ed.). New York: Wiley-Interscience. p. 583. ISBN 9780471396987.
2.
Zaccheria, Federica; Ravasio, Nicoletta; Ercoli, Mauro; Allegrini, Pietro (2005). "Heterogeneous Cu-catalysts for the reductive deoxygenation of aromatic ketones without additives". Tetrahedron Letters. 46 (45): 7743–7745. https://doi.org/10.1016/j.tetlet.2005.09.041.
3.
Walker, Gordon (1956). "Reduction of Enols. New Synthesis of Certain Methoxybenzsuberenes via Hydrogenation of Dehydroacetic Acids". Journal of the American Chemical Society. 78 (13): 3201–3205. https://doi.org/10.1021/ja01594a062.
4.
Hall, Stan; Lipsky, Sharon; McEnroe, Frank; Bartels, Anne (1971). "Lithium-ammonia reduction of aromatic ketones to aromatic hydrocarbons". The Journal of Organic Chemistry. 36 (18): 2588–2591. https://doi.org/10.1021/jo00817a004.


text:
Why d(z^2-x^2) and d(y^2-z^2) join to form dz^2 orbital?
The possible combinations are xy,yz,zx ,d(x^2-y^2) , d(y^2-z^2),d(z^2-x^2) 7 total for d , but i heard that from Scrodinger equation one concludes that only 5 are possible but i dont see how intuitevly or via some reasoning one gets this may someone provide a brief reason for this?

I personally never really understood that argument either. I think the straightforward answer is to actually solve the Schrodinger equation; you don't get 6 or 7 solutions from it, you can only get 5, because $m_l$ is constrained to be $\in {-2, -1, 0, 1, 2}$. One way of putting it would be that even if you made a 6th or 7th wavefunction, it would not be linearly independent, i.e. it would not be a genuinely new wavefunction but rather a combination of the existing five.

That said, I'm not hugely keen on writing an answer unless you explain what you were told more precisely. It's not clear what "the possible combinations" means - who came up with these combinations, and how were they generated?

Sir by possible combinations i mean x,y,z possible orbitals which might form 3C2 ways of choosing and 3C2 ways of x^2.. ones that i was telling . I just did a bit combinatorics in this nothing  else.

May you show how they are linearly independent or some stuff Sir it might give me clue and some text references which is easier to understand ?


text:
Comparing dipole moments of hydrogen halides
How can we compare dipole moments of halide ions?
As we go down the group, the induced charged hydrogen decreases as electronegativity difference between hydrogen and corresponding halogen decreases but the sepeartion between the poles of dipoles increases and since dipole moment is given by, $$\mu=qd$$
the $q$ factor and $d$ factor contradicts each other, so how can we do comparisons in such cases?

Go and measure $q$ and $d$.

Or look up the data on the compounds in Wikipedia. Hint: the maximum is where you probably think it is.


text:
Diels-Alder and Synthesis Problem: Confused about how to add carbon chain and double bonded oxygen

I am working on this synthesis problem. My first step was converting the dienophile given into a diene via E2 elimination. From the product given, I concluded that the dienophile will carry the double bonded oxygen. I think I understand how to draw the Diels Alder reaction.
However, I am stuck on how to generate the carbon chain from the two wedges.
I appreciate any help or hints.

This is not just one step Diels Alder reaction. It is a multi-steps synthetic problem. Al least, you can approach by retro-synthetic steps to find out the diene and dienophile.

@MathewMahindaratne Yes i know that. But please read my description to see what Im confused about.

Your text is not very clear. You conclude that the dienophile will have the doubly bonded oxygen --- which? There are four doubly bonded oxygens in the product. What E2 elimination are you talking about? There isn't any functional group that can be eliminated from the starting molecule. It's not possible to deduce this from the description you gave. If you can draw a diagram to show your suggested reaction scheme, that will help greatly.

Conceptually, you may think about the right hand hemisphere of the target molecule as derived from butane-1,4-diol:

You would use its bis-enolate as substrate for a Diels-Alder reaction with a (Z)-configurated dienophile.  This aims to install R' and R'' on the same side of the cyclohexene built (cis relationship).  There would be multiple functional group interconversions (FGI); reduction of the double bond in the cyclohexene, cleavage of the ether groups, oxidation of the then revealed 1,4-diol into a 1,4-dione (without tapping into an analogue of the redox couple of hydroquinone / benzoquinone).
The diene exhibits a pattern of six carbon atoms between the two carbonyl-C, thus on paper one way to disconnect this one were an ozonolyis (reductive workup):

A bibliographic search should query for examples where the triple substituted diene reacts faster than the double substituted.  One plausible access to the diene (to perform the ozonolysis) could be the Birch reduction of toluene.


text:
Alumina solubility in cryolite (Hall-Heroult process)
For the Hall-Héroult process, the solubility of alumina in flux seems to be important.  But why should I care about the persence of alumina in the bath? I can use feeder to keep the concentration constant because electrolysis decomposes $\ce{Al2O3}$. It can be 0.1% or 10% what would be the difference? Concentrations does not affect decomposition voltages of the components, so the voltage would only decompose $\ce{Al2O3}$ and would not touch the flux.

I have found the answer!
The limitation is in the current you can apply (limiting current)
If you exceed it - the anode effect may occur and NaFAlF3 will start to decompose instead of alumina.
So your cell will make you rich faster if the solubility is higher!

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.  And credit the source of the illustration in case it were not your own work.

Most of the bath is cryolite ($\ce{Na3AlF6}$) and not alumina ($\ce{Al2O3}$) because for electrolytic reduction, we do not use an aqueous solution, we have to use melt alumina. But alumina melts at a very high temperature ($\pu{2072 ^\circ{}C}$).
So we use cryolite which serves two purposes:

Cryolite acts as a solvent for alumina and dissolves it.
It increases the conductivity of the melt and decreases the melting point of alumina to about $\approx \pu{950 ^\circ{}C}$

Thus a suitable composition of the bath is required to make the process possible and feasible.

I do know that cryolite bath should dissolve alumina, but what would be the advantage of having say 15% solubility instead of 1% solubility of alumina in the bath? Whenever the flux is untouched by the applied voltage we only need to dissolve SOME fraction of Al2O3 and we can run the cell while adding new Al2O3 from the feeder to keep this fraction constant. Why should I want to dissolve more % of alumina? What is the effect of alumina concentration in the bath?


text:
What is an efficient way to generate CO2 electrochemically?
I would like to build a small carbon dioxide generator for experimental purposes. There are two requirements that are set in stone:

It has to be electrochemical. (Easy to switch on/off with electricity.)
It has to generate only one gaseous product, which is carbon dioxide. (No need to separate the gases, which might prove difficult.)

I have thought and tried using oxalic acid in a cell with a proton exchange membrane, which did not work. (And I have not figured out why so far.) I kept generating oxygen instead of carbon dioxide. Also the byproduct was hydrogen, which is problematic in my application, and the PEM is very fragile, so I could not build up any pressure. There are a few requirements, that are debatable. For example if a reactant or product is highly toxic or extremely expensive, that would possibly be a dealbreaker. Also it would be nice to be able to enclose the cell and build up a bit of pressure with carbon dioxide (only 1-2 bar).
Unfortunately researching this is rather difficult, since the main topic of research is binding carbon dioxide, which are 99.9% of the publications that pop up with the keywords I have searched for. Are there other things I could try? I have not tried a lot and I am already running out of ideas, but please bear with me, I am a theoretician and have not been in the lab for quite a long time.

Maybe not to your spec but you can readily buy dry ice - could you not keep some of it around in a deep freeze and use as-needed? It'll still sublimate but might be enough of a solution to get you on your way.

Carbon dioxide is easily liquefied at common temperatures. The most efficient way to release it from a cylinder is with a solenoid valve, which could use far less electricity than that required to produce the same amount by electrolysis. A solenoid toggle valve, e.g., https://www.smcpneumatics.com/MTV-3-M5.html , would use no power when static, i.e., open or closed, so would be more efficient, yet. Also, the raw material is pure CO2, so there is no waste to dispose, later.

@ToddMinehardt Unfortunately that is not at all possible in my setup. While it would be relatively easy to get, storage really is the main problem there. I do not think the deep freeze would be sufficient for the timescales I had in mind and Dewars are really expensive (not even sure a Dewar would be much better than a styrofoam box in the deep freeze anyways).

@DrMoishePippik I am well aware of solenoid valves and CO2 in bottles. The ease of a solenoid valve comes with additional costs though. Pressure reducer, cylinder and all the other stuff (hell throw in a MFC for good measure, if I want to have any chance of accurate electronic control over the flow) quickly add up to an enormous cost. This question was not about how to get CO2 from a bottle, but how to generate it electrochemically.

You might want to provide additional constraining specs (budget etc). You will want to monitor how much CO2 you are releasing for instance, how are you doing this?

Buy a pellet gun that uses CO2 cartridges. Do not load pellets. Stick a tube into the nozzle of the gun (pistol) and from there into your apparatus. 12 gram CO2 cartridges (800 psi) are available. A gun shop can probably help you. Or you could put a balloon onto the pistol - bang, bang - detach the balloon and attach to your apparatus (not much over 1 atm then, tho.

The Kolbe electrolysis may be helpful to you. It is an electrochemical decarboxylation process with free radicals as intermediates, with applications in syntheses of symmetric alkanes.

$\ce{H2}$ is generated as a side product on the cathode, but using aprotic solvents or a substance with a higher reduction potential than that of $\ce{H2}$, like $\ce{CuSO4}$, should fix this. Alternatively, as suggested by Poutnik, consider separating the compartments with a porous glass.

There is always possible to arrange  cathode + anode divided by salt bridge, diaphragm or porous glass to have both gases separated.


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Why 1-methylethanol is not preferred IUPAC name for isopropyl alcohol?
For the compound $\ce{CH3-CH(OH)-CH3}$ I propose two names: propan-2-ol and 1-methylethanol. Which is the correct name as per the IUPAC guidelines?
According to Organic Chemistry by Morrison and Boyd, the name is given as propan-2-ol. Why is 1-methylethanol not accepted?

I am in prefer of the latter name to be correct by giving more priority to the functional group rather than number of carbons .

The Wikipedia does a pretty reasonable job of explaining this: https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry#Alcohols and https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry#Order_of_precedence_of_group

The name is derived from the longest carbon chain - in this case propyl. 1-Methyl ethanol is incorrect.

As side note, chemistry.se's page resources for learning chemistry includes a dedicated section «References about Nomenclature», too.  You might want to start reading the 4-page summary, or the brief guide; both illustrated and published open access as .pdf.

"Confusion regarding Nomenclature" is a horrible title (it's not even a question — it's a complaint), and please don't use math mode where it makes no sense, e.g. for plain text. When citing third-party sources, make sure your citation is complete and unambiguous so that others don't have to guess your textbook's edition, year and page number your are referring to.

@buttonwood As another side note, links to the guides can also be found when you click on the "nomenclature" tag and then on "Lern more …". I don't know if that's easier, though.

@Long Indeed, this is new for me. Perhaps because I did not use tag «nomenclature» as often as possible (homework for me). Briefly, I compared the two .pdf documents the tag points to (2015  vs 2020) with diffpdf and leave it with «their organization to present the material is different» assuming the rules by 2015 are those in the 2020 document. Because your name is seen frequently when answering these questions. possibly you know better. For a quick look on the 4-page summaries and access to all Color Books, I liked the link to the resource page.

@Waylander , yeah that would be correct but there is also a rule that no matter of the longest carbon chain , the numbering should start from the carbon which is nearest to functional group . So we should start naming it from mid carbon as it's nearest to alcohol group .

Please reference this rule you to which you allude, Get_Maths.  Perhaps you mean that 2-hexanol should not be named 5-hexanol. If so, then the lower number is secondary to the chain length. @Waylander

No carbon chain can ever be 1-substituted with a carbon-based substituent, because that means the chain itself is one carbon longer than you originally said.


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Organic chemistry aromatic compounds

How can we determine which of the above conpounds are aromatic . My teacher taught me to remove a hydrogen and shift the electrons but  I did not understand it  . Could someone help me so that i can learn and understand this subject better.

If you're supposed to apply the 4n+2 rule, you can't for the fourth or fifth molecule (2-pyrone, cyclopentadienone). The 4n+2 rule does not apply to systems with pendant pi bonds, and the people who posed this problem should have known that.

@OscarLanzi could you please check if my answers are correct as  i am totally confused . ( the first one is aromatic (4n+2 = 6 ) the second and third aren't . the fourth one is aromatic and the fifth isn't )

How many of the following are aromatic?

Among these compounds first four are aromatic where fifth compound is not aromatic.

Coming to the first compound we have three π-bonds and a vacant orbital on B. So there are 6 electrons in conjugation and it obeys Hückel's rule. Hence, it is aromatic.
Here we have two π-bonds and a lone pair on N. So there are six electrons in conjugation and hence, it is aromatic.
Note: Lone pair on right nitrogen is not involved in conjugation.
Similar to 1 and 2, we have a vacant orbital on B, a lone pair on N and two π-bonds. So totally there are six electrons in conjugation and hence, it is aromatic.
Form the following image you can understand why 4th compound is aromatic. Since there are six electrons in delocalisation where four electrons are from two π-bonds and two electrons from oxygen's lone pair.

I hope you can understand why 5th compound is not aromatic as it does not obey Hückel's rule.

This is probably what they want in the textbook, but not really the whole story for molecules 4 and 5. See my comment to the question.


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Doesn't the eutectic diagram violate Gibbs's phase rule?
EDIT: As the question hasn't attracted any answers, I tried to put in more detail.
Gibbs's phase rule for a 2 component system at constant pressure says that $F=3-P$, i.e., that the number of degrees of freedom is three minus the number of phases present. This should mean that when $P=3$, there should be no degrees of freedom and three phases should only be in equilibrium at a single point.
However, looking at the eutectic phase diagram, this seems not to be true. Even visually - four lines intersect at the eutectic point, not three.

And my understanding is that the three phases, $\alpha$, $\beta$ and $L$ can be in equilibrium along a line at the eutectic temperature, not only at a single point.
This can be seen by looking at the Gibbs free energy of the three phases at the eutectic temperature, which should look like this:

By the common tangent construction, the three phases are at equilibrium along a range of compositions, and thus $F=1$, not $0$.
Is there a mistake in my understanding, or is it a 'violation' of the phase rule?

At constant pressure, F = C - P +1 and there is only one point where all three phases coexist. You can't vary the composition away from the eutectic point and keep all 3 in equilibrium.

To add on to what Jon Custer has mentioned, any system that lies along this "eutectic line" will instead segregate into two phases, one eutectic and the other α/β. This can be understood from the common tangent construction, which shows you that, for a given composition, there are two local minima in free energy corresponding to the two phases. Minimizing free energy involves choosing the proportion of the two phases subject to the constraint of the total composition of your system.

The three-phase (eutectic) equilibrium phase compositions are not along a line in the phase diagram. Each of the three phases is instead at a single point (specifically, where the isotherm touches the field having that phase), which dovetails with zero degrees of freedom. The full isothermal line is not a range of equilibrium compositions but a boundary between different two-phase regions above and below that temperature. In contrast, other lines actually do represent a range of equilibrium compositions when the necessary degree of freedom (shown in the diagram as the temperature being allowed to vary as a function of composition) exists.

Indeed, seems plausible that is the root of their misunderstanding.


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Determination of geometrical isomers for chains with rings and double bonds

How can we determine the geometrical isomers for this compound . It contains a ring and 3  double bonds . I am confused because i was told to consider 1 ring = 1 double bond. When i tried it i got it as $2^4$ but its not the right answer . Kindly explain how to find the geometrical isomers for the compound by taking the above compound as an example.

The correct answer is 2⁴ .

You could consider 1,4-disubstituted cyclohexane = 1 double bond, if the two substituents are different. However, the 1.3-disubstituted cyclopentane is less symmetric, so you are off by one.

The molecule contains three $\ce{C=C}$ double bonds which may be in either (E), or in (Z) configuration, a total of $2^3$ variations if taken alone (blue rectangles).  The molecule contains two stereogenic centres, which may be either in (R), or in (S) configuration, a total of $2^2$ variations if taken alone (red rectangles):

Since you want to run the permutations on double bonds and stereogenic centers as independent of each other, there are $2^5 = 32$ stereo isomers possible (five parameters, two levels each).


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Exceptional electronic configuration of Chromium
I understand that the reason for the exceptional electronic configuration of Cr is the increased stability of half filled set of orbitals
But here's an excerpt from my textbook

…consider the case of Cr, which has $3d^54s^1$ configuration instead of $3d^44s^2$; the energy gap between the two sets ($3d$ and $4s$) of orbitals is small enough to prevent electron entering the $3d$ orbitals.

What do they mean by prevent the electron from entering $3d$ orbital? Didn't one electron from the $4s$ orbital actually go to the $3d$ orbital?
Is it likely that it's a typo to write "prevent" instead of "permit"?

Yes I believe that's a typo, it should say "permit." The reason we see these Aufbau's principle exceptions in transition metals is because the $4s$ and $3d$ orbitals are very similar in energy.
In chromium, having a $4s^2$ $3d^4$ configuration results in electron-electron repulsion due to the two electrons in the $4s$ orbital. For this reason, chromium adopts a $4s^1$ $3d^5$ configuration, in which each electron occupies its own orbital. Recall that Hund's rule essentially states that you fill each orbital once before going back with the second electron in regards to orbitals of the same energy (in the same subshell). In the case of chromium, we are dealing with orbitals that are almost in the energy ($3d$ and $4s$) so you can essentially view this as a special case of Hund's rule that extends to orbitals of nearly the same energy.
For what it's worth, the other notable exception is copper, which adopts a $4s^1$ $3d^9$ configuration over $4s^2$ $3d^8$. For reasons that go beyond the scope of this answer, before you hit the transition metal elements ($Z \le 21$), $3d$ is higher in energy than $4s$, so we fill $4s$ before $3d$. But beyond this point, $3d$ becomes lower in energy than $4s$, so you lose $4s$ electrons first when we're talking about ionization. Knowing this, it is energetically more favorable to have $4s^1$ $3d^9$ because the higher energy orbital only has one unpaired electrons and the lower energy orbitals have paired ones.
The important thing to note from these two examples is that there is no special added stability form merely having a half full electron shell (i.e., something is not magically more stable because of having a half filled subshell). Adopting this form of configuration has consequences that are perfectly explainable to students of intro chemistry—do not brush it off as mere "magic."

I had mentioned half filled set of orbitals. I was referring to the stability given by exchange energy


text:
How does one obtain a CuCl2 (solid or in solution with water) from a solution with HCl
I have a solution of CuCl2 and with HCl, and I want to seperate CuCl2 from the HCl. I cannot boil or distill the solution, since I don't have a fume hood or access to a safe enough space outdoors. So I need to find a way to get the CuCl2 out of solution, or at least get rid of the HCl in order to boil off the water safely. I can filter out percipitates, if that helps in any way.
Any ideas?
Edit: I reacted small amounts with H2O2 and bubbling air through, and an excess of copper. This worked, but takes far too long. I have access to copper oxides (likely aged), but there was a discussion in the comments about solubility.

If you remember my correct answer (use copper metal in the presence of air and even weak acid like dilute HCl). it will work being electrochemcial in nature. This sources confirms that CuO will NOT dissolve in dilute HCl which is what you likely have. To quote:  pubs.acs.org/doi/10.1021/acsreagents.4106 to quote: "The monograph also details the following specifications and corresponding tests for verifying that a substance meets ACS Reagent Grade specifications including: Assay, Insoluble in Dilute Hydrochloric Acid,." . So, if the HCl is dilute stirring every hour is more like hourly forever!

If CuO, as the source claims, becomes insoluble at some HCl concentration level, how can there not be an associated HCl concentration issue? Note, as the electro-chemistry proceeds regardless of concentration as long as there is some HCl,  I would argue that such an approach is the only possible path.  But, I have further refined my answer and have presented it for input/review elsewhere. Upon completion of that process, I may provide a search phrase here for those actually seeking a workable solution and interested in advancing their range of chemistry (important for me, working on patents).

@AJOKER I did react a small batch with copper metal and bubbling air through the solution, and yielded the desired product. A byproduct of this reaction was insoluble CuCl, which I am able to reuse in future runs. However, the reaction is very slow. Is there a way I can accelerate the reaction? And could I ask you to quote the relevant parts of your source? I am unable to access it, since I am am a physics student, and our department has no license. Thank you. I would love to accept your answer, but I am unable to find it.

@AJKOER Please be careful when accusing other participants of particular conduct, here regarding their voting, which I remind you is strictly anonymous unless they leave a comment directed to you indicating that they have voted in a particular way. Note that accusations on your part could be construed as harassment. Others are at liberty to vote as they please on other posts, as are you.

There is a reputed assertion in my deleted answer that it is 'speculative'. My answer actually explains the chemistry around the prep of basic copper chloride per an existing patent and cited, with reaction equations no less, in Wikipedia here https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide  as a 'commercial process'.  The only speculation exists for those not acquainted  with the associated science.

Apparently your solution contain $\ce{CuCl2}$ and $\ce{HCl}$. If you want to save only $\ce{CuCl2}$ you should first add an excess of black copper oxide $\ce{CuO}$ and stir at least one hour. The following reaction will consume unwanted $\ce{HCl}$ : $$\ce{CuO + 2 HCl -> CuCl2 + H2O}$$ Then filtrate the solution to get rid of the $\ce{CuO}$ excess. The obtained blue solution contains no $\ce{HCl}$ any more. You may evaporate it to $≤ 100°$C. You should obtain by cooling to room temperature a blue-green hydrated copper chloride $\ce{CuCl2·2H2O}$. But this hydrated copper chloride is not easy to use, because it is deliquescent : it will soon get dissolved in the water attracted from the humidity of the air. It is much better to heat it to a temperature > $110$°C. Heated to between $110$°C and $498$°C, $\ce{CuCl2}$ solutions produce, after evaporation, an anhydrous copper chloride $\ce{CuCl2}$, which is a brownish yellow powder. And this powder is easier to use, because it is only slightly hygroscopic. It slowly attracts water but will not produce a viscous liquid like $\ce{CuCl2· 2H2O}$.
These two copper chlorides are both very soluble in water. At $0$° C, $100$ mL water dissolves $\ce{71 g CuCl2}$ and $\ce{110 g CuCl2·2H2O}$.
$\ce{CuCl2}$ melts at $498$°C, and is decomposed into $\ce{CuCl + Cl2}$ at $993$°C.

Maurice: Do you really expect, given the limited resources outlined, access to Cupric Oxide ? If this is, indeed, correct, this answer is not useful. Please do not downgrade my both useful, correct and interesting answer.

Neutralization of CuCl2 by NaOH and boiling solution with Cu(OH)2 provides CuO easily.

No, do not add NaOH, as it can destroy all (or part) of your CuCl2 leaving an insoluble Cu(OH)2 and even result in a clear solution of NaCl  and water. Any amount of NaOH also introduces a non-removal Sodium ion presence. And why use NaOH as one more likely has Baking Soda (NaHCO3) ? My answer (edited for a clear reference to Patina) introduces no contaminants and occurs naturally.

@AJKOER It was - thought obviously - meant to just a part of the solution, not the whole one. Sure, NaHCO3 would work. // Another way is adding  metallic copper, as you similarly pointed out, and carefully adding hydrogen peroxide. //  Your answer  diverges a lot - let others to evalute its qualities.

Thanks, I have re-arranged my answer for more clarity as I progress in levels of chemistry around copper.

Maurice: Have you actually done this? Per this source https://pubs.acs.org/doi/10.1021/acsreagents.4106 to quote: "The monograph also details the following specifications and corresponding tests for verifying that a substance meets ACS Reagent Grade specifications including: Assay, Insoluble in Dilute Hydrochloric Acid,..." . So, if the HCl here is dilute, stirring CuO every hour is not going to work, especially if the CuO has been subject to an aging effect.

@AJKOER. I have done this synthesis once. But I have not checked the purity of the obtained product. And I have not downgraded your answer. I never downgrade.

do have access to copper oxides (black, red and Grey...). I am however certain, that the copper oxides I can obtain, were not freshly synthesized, so aging effects will have occurred, I guess. I am not planing on adding NaOH, as I don't want contamination in the product. So, what method is preferable given my situation? I want to obtain a dry or dihydrate powder of CuCl2 (Copper(II) chloride, not 100%sure if that is the correct formula). I have a solution of CuCl2 and CuCl (I and II oxidation states) with HCl, and do not own a fume hood.

I have used black copper oxide bought on the market. Merck sells two qualities, one $96$% and one >$99$%. Adding any one of these oxides will neutralize the remaining $\ce{HCl}$ present in solution. So that the result will be pure $\ce{CuCl2}$ in solution. A simple evaporation and heating to $120°$C will give pure $\ce{CuCl2}$


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Experimental setup
I would like to know your opinion regarding the setup of a system for oxygen purging with nitrogen. Before ordering the necessary materials, I'd like to be sure if this setup is feasible or if it can be enhanced or even simplified, so any advice or new ideas are more than welcome.
To summarize, the experiment I'm going to perform is as follows:
-Oxidative weathering in solution of various sulfide minerals will be studied under different chemicals conditions. Some of the experiments must be performed in anoxic conditions, so all dissolved oxygen must be purged with nitrogen.

It is necessary to control the temperature throughout the experiment (40-42 ºC). The duration of the experiments will be 72h (non-stop). Stirring is also required. My first question: is it advisable to use a hot stirring plate together with the use of a silicon bath (or other suitable heating bath oil)? Is there any safety recommendation on the use of conventional hot plates for a prolonged period?
About the materials needed. As container, I've thought of using normal Erlenmeyer flasks (125 ml would be enough) but septum-sealed. The setup I have in mind is to use a small diameter Pasteur pipette inserted into a Tefzel tubing connected directly to the nitrogen tank. Then, just pass the pipette through the septum (nitrogen inlet). For the outlet and positive pressure, maybe I could use a needle tip (wide diameter). For better visualization, I include a very basic drawing with the assembly scheme I propose:

How can I perform these experiments in series? That is, three flasks with different solutions purging, being stirred inside heating baths at the same time, but using a single tank of nitrogen.
Any advice on the assembly? Should I use another type of glassware instead of an Erlenmeyer? Are there better alternatives instead of using a Tefzel tubing (connectors, nitrogen tank adapters…)?
Do not hesitate to give your opinion. Thank you in advance.

If you can, join a collaboration with a group active in organic or inorganic chemistry, or catalysis possessing a Schlenk line.  Its typical use in these lab aims to perform reactions without interference of oxygen and water.  If you have to perform experiments of this type multiple times, consider to purchase such an installation.
The Schlenk line basically is an advanced two-tube manifold with connection to a vacuum pump, and a protection gas:

(image credit to https://schlenklinesurvivalguide.com/)
The principal advantage is that you evacuate the reaction flasks (round bottom flasks, but not Erlenmeyer flasks) and thus effectively replace the atmosphere in these containers by a dry inert gas e.g., nitrogen. A bit more expensive per liter is argon (the protection gas used e.g., for welding); in comparison to nitrogen, argon is of higher density than air and thus yields a good «blanket» covering your samples.  Using a Schlenk line equally offers the advantage to connect to multiple (three to five) reaction vessels in parallel, which may be immersed in a bath (e.g., silicon) on top of a heating plate.
Request an on-site hands-on training session by a more senior chemist how to perform a pump-and-thaw cycle (cycling in said survival guide, one of multiple youtube video providing a demonstration) and the maneuvers you might need which are known as Schlenk techniques.

If you can't find a group to borrow access to a Schlenk line, a very much down-scaled approach would be to use multiple test tubes in lecture scale to create small compartments of controlled atmosphere.  It is cheaper than a Schlenk line, but does not offer pump-thaw cycles.  These test tubes may closed by a rubber stopper intentionally pierced twice for gas-in and gas-out, and are large enough to accommodate a rugby-shaped stirring bar.  Though initially not set up for this purpose, you may alter the set up described recently by Johnson and West in the Journal of Chemical Education:

(image credit to Johnson et al.)
The very first and very last tubes in such a series would be filled partially by silicon oil to close the vessels against normal atmosphere.  The inner tubes in this series (marked red) could be used to expose your hot material to oxygen with slight over pressure.  (The sequence how the smaller tubes reach into the test tubes does matter as does the level of filling the liquids to prevent an accidental transfer of solution/liquid into an adjacent test tube.)  The ensemble has a height of about $25\dots\pu{30 cm}$ depending on the rack used

(image credit to Johnson et al.)
but it equally is possible to hold the tubes housing an intentional reaction with a clamp immersed in a bath.
Reference:
Johnson, S. E.; West. J. K. A Simple, Inexpensive, Mercury-Free Bubbler Apparatus. J. Chem. Educ. 2021, 98, 2482–2485; doi 10.1021/acs.jchemed.1c00224.

Thank you very much for your reply. The proposed Schlenk line system seems very suitable. Unfortunately, it is not possible to spend funds on the purchase of this equipment, because we are not going to perform these experiments under anoxic conditions many times. However, I will ask an organic group from the other department if they have this system. In any case, I would like to know more alternatives, how I can get a good oxygen purge with a more "lab-made" configuration.

This would be a pity if a university active in chemistry has neither permanently installed Schlenk lines, nor some to borrow from the inventory.  Some chemistry departments actually have a «shop» to borrow machinery (like heating plates), glass ware (like round bottom flasks) and chemicals, so it might be necessary to give the colleagues a telephone call to learn when this counter is open.

The proposed new setup looks very promising. I suppose I could adapt it to my experimental purpose. Even if it's a down-scaled approach compared with the Schlenk lines, In my opinion I could get good results, I'll test the performance of the system with previous trials. Thank you so much for your explanation, the reference and for your feedback.


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When oxygen is connected only to carbon, are then the oxygen and carbon connected with a double bond?
My teacher showed us three molecules: a carbohydrate molecule, a protein molecule, and a lauric acid molecule. I noticed that, when an oxygen atom is connected to one carbon atom and isn't connected to any other atom, then the oxygen and carbon atoms are connected with a double bond. Is my hypothesis correct, and has anyone ever found it before?

Yes-ish and yes

Carbon monoxide $\ce{C#O}$

@ron So it seems my hypothesis is wrong.

@KamalSaleh, not necessarily wrong. Things are just a little more complicated than they at first appear. 4 is the normal valency of Carbon, but that can sometimes be changed. In the case of CO (carbon monoxide), an electron jumps from the oxygen to the carbon atom, changing the valency of carbon from 4 to 3, and that of oxygen from 2 to 3. They then join with a triple bond.  And some elements prefer ionic bonds, which are different again.  But start with the simple concepts, and work your way up to the next level after you are confident enough.

@RayButterworth The valence of carbon is by definition always four, nothing will change that. It is definitely not as simple as you make it out to be. It has also nothing to do with bonding.

@Martin-マーチン, I'm not making it simple, I'm trying to keep it simple.  Read the original question.

@RayButterworth I don't know that much about Chemistry, so I should follow your advice.

@RayButterworth By writing what you wrote, you're not keeping things simple, but you are oversimplifying and you are considering two subjects. And in my opinion, this is making it wrong, and it is not useful. Especially not to someone starting out in chemistry.

@Martin-マーチン, if you have a more suitable answer to this question, one that will help rather than confuse the OP, please post it, and I'll upvote it.

Over 150 years ago August Kekulé had a dream about how atoms might form molecules.  You might find it interesting to read about it.
Chemistry is largely about valence, the number of connections that each atom is able to make with other atoms.
In the cases you are talking about, with co-valent bonds, you can imagine each carbon atom as having 4 arms reaching out, each oxygen having two arms, and each hydrogen having 1 arm.
If an oxygen atom is connected to only one other atom, then that connection must use both arms, holding hands with 2 of the carbon atom's arms (just as you noticed), forming a double bond.
The carbon's other two arms could be connected to another oxygen atom, forming CO2 (carbon dioxide), or to 2 Hydrogen atoms, forming CH2O (formaldehyde).
Or it could double bond to another carbon atom, which in turn has two more arms to connect to something else, such as C2H5OH (drinking alcohol).
Molecules based on carbon are so common and potentially so complicated, that there is a whole field (organic chemistry) based on it.
If you are at all interested in chemistry, the concept of valence will be essential to your understanding.

There are counter examples that are commonly rationalized with a negative formal charge on oxygen (as in hydroxide ions, $\ce{OH-}$, or deprotonated alcohols, $\ce{R-O-}$,)or a positive formal charge (as in the $\ce{C#O}$ @Ron mentioned in a comment to the question or in the hydronium ion, $\ce{H3O+}$).

I realize you are trying to keep it simple, but there is the possibility of forming a radical too: https://pubchem.ncbi.nlm.nih.gov/compound/Ethoxy-radical