text:
Why do bonding and anti-bonding states split in energy (mathematically)?
I am a physicist, and when I hear about the splitting of energy levels for MO antibonding and bonding states, I have a rough intuition that tells me this is reasonable to expect, but I can't think of (nor find) a good formal explanation.
My best shot at understanding this is borrowing from Avoided Crossing and saying that if I have a 2 atom system with a total Hamiltonian $ H = H_1 + H_2 $ that is the sum of each individual atom Hamiltonians, then I can take the Eigenstates for each individual Hamiltonian as a basis.
$$\begin{align}
H_1 | \psi_1 \rangle &= E_1 | \psi_1 \rangle \\
H_2 | \psi_2 \rangle &= E_2 | \psi_2 \rangle
\end{align}$$
Then the total hamiltonian in this basis becomes,
$$\begin{align}
H &= \left(\begin{matrix}
E_1 & \langle \psi_2 | H_1 | \psi_2 \rangle \\
\langle \psi_1 | H_2 | \psi_1 \rangle & E_2
\end{matrix}\right)
\end{align}$$
In the limit of the atoms being really far apart, I would assume the off diagonal terms would vanish, and then the eigenstates are just the atomic orbitals. But if the atoms come closer together, then we get off diagonal terms which causes the splitting as described in avoided crossing.
Is this correct?

You are (essentially) correct and the off-diagonal terms are related to so-called "overlap" between atomic orbitals (AOs). It is common to talk of the AOs "overlapping" to form molecular orbitals (MOs), and as you rightly said, the off-diagonal terms (or "overlap") goes to zero in the limit of infinite bond length. The term "avoided crossing" is not commonly seen in introductory MO theory, but I think it is fair to describe it as such.
Of course there are some slight subtleties. Let's stick to a one-electron system such as $\ce{H2+}$ for now. $\psi_i$ refers to the 1s orbital on hydrogen atom $i$ and $E$ is the energy of a 1s orbital on hydrogen. The minor issue is that $\hat{H} \neq \hat{H}_1 + \hat{H}_2$ (following your notation where $\hat{H}$ is the molecular Hamiltonian and $\hat{H}_i$ is the Hamiltonian for hydrogen atom $i$), because:
$$\begin{align}
\hat{H} &= -\frac{\nabla^2}{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{R} \\
\hat{H}_1 &= -\frac{\nabla^2}{2} - \frac{1}{r_1} \\
\hat{H}_2 &= -\frac{\nabla^2}{2} - \frac{1}{r_2} \\
\end{align}$$
(I used atomic units. $r_i$ is the distance between the electron and nucleus $i$, and $R$ is the distance between the two nuclei). So actually, it is more like
$$\hat{H} = \hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}$$
and therefore the first term in the matrix
$$\begin{align}
\mathbf{H}_{11} &= \left<\psi_1\middle|\hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}\middle|\psi_1\right> \\
&= E + \left<\psi_1\middle|-\frac{1}{r_2}\middle|\psi_1\right> + \frac{1}{R} \neq E
\end{align}$$
(since $\langle\psi_1|\hat{H}_1|\psi_1\rangle = E$, and also $\langle\psi_1|(1/R)|\psi_1\rangle = (1/R)\langle\psi_1|\psi_1\rangle = 1/R$).
For more information I refer you to Atkins' Molecular Quantum Mechanics 5th ed., pp 262–266. The terminology can be a bit confusing and also Atkins does not use atomic units (in his text he uses $j_0 = e^2/4\pi\varepsilon_0$ which is simply equal to $1$ in atomic units). Adjusting for this, the final quoted formulae for the matrix elements are:
$$\alpha = \mathbf{H}_{11} = \mathbf{H}_{22} = E - j' + \frac{1}{R}; \qquad j' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_1\right>$$
(as found earlier), and
$$\beta = \mathbf{H}_{12} = \mathbf{H}_{12} = \left(E + \frac{1}{R}\right)S - k'; \qquad S = \langle\psi_1|\psi_2\rangle;\quad k' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_2\right>$$
If you now solve the generalised eigenvalue equation $\mathbf{Hc} = E\mathbf{Sc}$ (the simple $\mathbf{Hc} = E\mathbf{c}$ only works when your basis set is orthonormal, but in this case it is not) you obtain the two eigenvalues
$$E_\pm = E + \frac{1}{R} - \frac{j' \pm k'}{1 \pm S}$$
Note that when the atoms are infinitely far apart both $S$ and $k'$ vanish, so (1) the off-diagonal element $\beta$ vanishes and (2) the "bonding" and "antibonding" orbitals are of the same energy, i.e. $E_+ = E_-$.


text:
Structure of PH4Br
What is the structure of $\ce{PH4Br}$? I would like to know which atom is bonded to which. Are all five atoms directly bonded to phosphorus?

No, not in the solid state. The most recent crystal structure analysis suggests there are nearly ideal tetrahedra of phosphonium cations $\ce{PH4+}$ with bromine anions $\ce{Br-}$ in between bound via H-bonds [1, ICSD#23691]:

Reference

Schroeder, L. W.; Rush, J. J. The Journal of Chemical Physics 1971, 54 (5), 1968–1973. DOI 10.1063/1.1675127.

andelisk, does PH4Br exist as a liquid or gas?

@HolgerFiedler $\ce{PH4Br}$ sublimes at approx. 30 °C.

andelisk So what should prevent me to imagine that phospine and hydrogen bromide still exist in the solide phase and they simply sorted in a crystalline order? (I’m really a newbie in chemistry, so sorry for the naive question.)

@HolgerFiedler Well, you can imagine anything you want, but neutron diffraction suggests otherwise (BTW have you checked the paper cited?).

Behind a paywall? Doesn’t the sublimation of you HBr2 point to a very week bonding. Or more to a very homogeneous mixture?

@HolgerFiedler Oh, I'm very sorry, I thought you had access to the paper. You are right, the compound is rather unstable at ambient conditions and readily recombines to $\ce{PH3}$ and $\ce{HBr}$ upon sublimation.

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text:
Polarity of Acetic Acid, Ethanol and 1-Propanol
In order to compare acetic acid, ethanol and 1-propanol in terms of solubility in hexane, I would like to order them in terms of (a)polarity.
By looking up the dipole moments, I found that these molecules can be ordered in terms of increasing polarity like this.
1-Propanol (1.68D) < Ethanol (1.69D) < Acetic acid (1.74D)
Without looking up the values for the dipole moments, how can I predict which one of these molecules will be the most/least polar?

Dipole moment should not be called "polarity".  It has only tangential connection with real polarity of substances which is  connected with solubility. Also all of these are miscible with hexane which is hardly deducible from dipole moments.

Alright so based on what should I order these molecules in terms of miscibility with hexane?

Miscible means it dissolves in all proportions - completely soluble.

Isn't it possible to compare these 3 molecules?

Of course it is, but not on basis of solubility in n-hexane in room temperature. There are various measures of properties of solvents, like dielectric constant, donor number.

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text:
Calculating Average Rate

A $\pu{3.45 g}$ piece of marble ($\ce{CaCO3}$) is weighed and dropped into a beaker containing $\pu{1.00 L}$ of hydrochloric acid. The marble is completely gone $\pu{4.50 min}$ later. Calculate the average rate of reaction of $\ce{HCl}$ in $\pu{mol/L/s}$. Note that the volume of the system remains at $\pu{1.00 L}$ through the entire reaction.

I'm not very sure why the units are in $\pu{mol/L/s}$ instead of just $\pu{mol/s}$. Here is what I did:
$$\text{Rate} = \frac{\pu{3.45 g}~\ce{(CaCO3)}}{\pu{4.5 min}} = \pu{0.0128 g/s}$$
$$\pu{0.128 g}~\ce{CaCO3} = \pu{1.28e-3 mol}$$
$$n(\ce{HCl}) = \pu{2.58e-3 mol}$$
$$\text{Rate} = \pu{2.58e-3 mol/L/s}$$
However, the answer given is $\pu{2.55e-4 mol/L/s}$. What am I doing wrong?

You haven't accounted for the stoichiometry of the reaction, and I suppose you wrongly converted minutes to seconds. Always start solving problems like this with writing down the chemical reaction:
$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$
By definition rate of consumption of hydrochloric acid over time $\Delta t$ is:
$$r=\frac{\Delta c(\ce{HCl})}{\Delta t}$$
Since all calcium carbonate reacted completely:
$$\Delta c(\ce{HCl}) = \frac{\Delta n(\ce{HCl})}{V} = \frac{2n(\ce{CaCO3})}{V} = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})}$$
where $m$ is mass, $M$ - molar mass, $V$ - volume.
And the average rate is:
$$r = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})\Delta t} = \frac{2\cdot\pu{3.45 g}}{\pu{1 L}\cdot\pu{100.09 g mol-1}\cdot\pu{4.50 min}\cdot\pu{60 s min-1}} = \pu{2.55e-4 mol L-1 s-1}$$
Also, be careful with notations. Use proper capitalization, and don't equate moles to grams! This is not tolerable in natural sciences.


text:
What are the CIP rules for cyclic substituents?
Here's what I read on Wikipedia (section "Cycles"):

To handle a molecule containing one or more cycles, one must first expand it into a tree (called a hierarchical digraph by the authors) by traversing bonds in all possible paths starting at the stereocenter. When the traversal encounters an atom through which the current path has already passed, a ghost atom is generated in order to keep the tree finite. A single atom of the original molecule may appear in many places (some as ghosts, some not) in the tree.

I don't understand how to assign CIP priorities to cyclic systems, for example, consider this:

How do we decide the CIP priorities? What does that paragraph I found on Wikipedia really mean? Could someone please give a detailed explanation?

I presume you want to know the E/Z geometry of the double bond. You need to determine on each end of the double bond which ring has priority. Intuitively one may argue that cyclohexane is larger than cyclopentane and cyclobutane is larger than cyclopropane. In this simple case, this is true and the double bond has the Z-configuration. However, digraphs are used in more complex cases. Your structure below has four colored dots. Each attached ring bond is "cut" and stretched out until one returns to the original atom. The three red atoms from the cyclohexane ring are identical. The same is true of the other three rings. Now treat the digraph as you would any acyclic alkene. The digraph has the Z-configuration.
Addendum: For two more complex cases, go here and here.

Is the ghost atom counted or not by that I mean if we were to compare cyclo hexane and meth-Di(pentane)yl which would have a higher priority?

@AvatarShiny The ghost atom is counted as an atom of the same atomic number as its non-ghost counterpart but connected to nothing. Taking the fourth atom of one of the green chains and comparing it to the fourth atom of one of the red ones, both have C, H, H. We need to dive down another layer to find a difference. The red C (fifth in the chain) is connected to C, H, H. The green C is a ghost connected to nothing. Since $\ce{C}>\text{nothing}$, the red chain wins.

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text:
Hexanoic acid and copper/brass - toxicity?
We have a sensor made of brass through which we are flowing organic solvent vapours mixed with air, typically hexane and various alcohols. After a year or so the interior is covered in a greenish-black slime. The smell is definitely that of hexanoic acid, so I am assuming much of the deposit will be the copper salts of the various acids.
Is there likely to be any significant toxicity with the probable compounds that may require special cleaning procedures?


text:
Can naphthenic acid be ionized in n-dodecane?
I am not very clear about the difference between ionization and dissolution. I know that naphthenic acid can dissolve in n-dodecane, but I do not know if it is ionized in n-dodecane.
Thank you!

No, it isn't. Moreover, chances are it is dimerized.

It would not ionize (though the correct term is that it won't dissociate, ie. the O-H bond in the acid won't cleave giving carboxylate anion).
It's because n-dodecane has no lone electron pairs, so it can't complex the hydrogen cation. Unless we're talking about extreme vacuum or plasma, there are no free $H^+$, it always exist in a form of a complex cation with a molecule of solvent or the like. But such solvent must have some electron donor: lone pair, double bond, etc. to form the complex.
The only other possible way here would be self-dissociation, where one of the substance molecules accept the hydrogen carion from the other one, but here it is also borderline impossible as the additionally protonated acid molecule would be very unstable and immidetely such auto-dissotiation would reverse.
The fact that the naphtalenic acid dissolves in n-dodecane is due to low polarity of both, that is acid molecules don't bind strongly to each other and n-dodecane molecules don't bind strongly to each other.

Thank you for your detailed reply！Now，I understand this question.

Therefore, naphthenic acid may ionize in the presence of water, right?

Yes. Although it probably wouldn't dissolve in pure water, so it would need to be some appropriate solvent mixture.


text:
Tip of my tongue: what is the name of this process?
I remember back in Chemistry class, we did this thing where we would drop some liquid into a solvent. The moment it touched the solvent, it would turn into a gel sphere with the solvent inside it, and you could take these little spheres out and they'd essentially be capsules of solvent. What is the name of this process? The closest thing it reminds me of is Prince Rupert's Drop.

The general term for the process would be "encapsulation".
The Wikipedia article on micro-encapsulation lists different processes by which such encapsulation is achieved. Two which sound similar to your described process are interfacial polycondensation and interfacial cross-linking:

In interfacial polycondensation, the two reactants in a polycondensation meet at an interface and react rapidly. ... Condensed polymer walls form instantaneously at the interface of the emulsion droplets.

Alternatively, the process your class did might have been ionotropic gelation:

Ionotropic gelation occurs when units of uric acid in the chains of the polymer alginate crosslink with multivalent cations.


text:
Using cryptands to make a negatively-charged solution?
In our discussion group, my friend mentioned that metal cations can be separated from other cations in solution using cryptands. After addition of the cryptands to the solution of the cations, chelation of the cations takes place and since the complexes formed are more soluble in organic solvent, addition of an organic solvent to this resultant solution would allow these complexes to be migrate to the organic layer. The organic layer can then be decanted and the metal ions have now been separated successfully from the aqueous solution. This was my friend's proposal.
From hearing this, I quickly thought of the possibility of creating a negatively-charged solution. By applying the method of using cryptands which my friend has proposed to separating sodium ions from a solution of sodium chloride, wouldn't the remaining chloride ions in solution give the resultant solution an overall negative charge? This led me to think that there must be something wrong with his idea as this should not be able to occur, based on my instinct. However, I do not know exactly how to refute my friend, to say his proposal does not work. Thus, I would like to ask if anyone knows the reason why my friend's method would not work.

If cations move anions are drawn with them. If anions can't do that then cations stay close to them.

Never, ever. The initial charge separation immediately leads to a small electrical potential, against which no further separation is possible. (The largest possible potential in any electrochemical system is in the range of 5V) If you think about a quantitative separation, that voltage difference would easily go into the megavolts, even for a rather dilute system.

For some reason this question reminds me of Maxwell's demon paradox. You don't really separate anions and cations here; in reality the resulting metal ($\ce{M+}$) cryptand ($\ce{L}$) complex is going to be associated with anionic part ($\ce{X-}$) in organic solvent ($\ce{s}$) after it's being extracted from aqueous phase ($\ce{w}$). For the detailed process, refer to [1], for example.
There are two main processes to consider:

$$
\begin{align}
\ce{L_\mathrm{s} + M^+_\mathrm{s} &<=> LM^+_\mathrm{s}}\label{rxn:1}\tag{1}\\
\ce{\overline{\ce{L}} + M^+_\mathrm{w} + X^-_\mathrm{w} &<=> \overline{\ce{LMX}}}\tag{2}
\end{align}$$

You would be correct if only \eqref{rxn:1} occurs, however, this is not the case.
Reference

Fyles, T. M. Can. J. Chem. 1987, 65 (4), 884–891. DOI 10.1139/v87-149 (Open Access).


text:
Molar mass and molar volume of monochloroethylene

Monochloroethylene gas is used to make polyvinylchloride (PVC). It has a density of $\pu{2.56 g L-1}$ at $\pu{22.8 ^\circ C}$ and $\pu{101 kPa}$. What is the molar mass of monochloroethylene? What is the molar volume under these conditions?

$$PM=dRT$$
$$M=\frac{\pu{2.56 g L-1}\cdot\pu{295.8 K}\cdot\pu{0.0821 atm L mol-1 K-1}}{\pu{0.9968 atm}} = \pu{62.37 g mol-1}$$
After this I can't find molar volume. Which formula should I use?
I know $PV=nRT$, but without knowing the gas mass I can't use this formula, I believe.

You need to start with PV=nRT. Let n=1 and solve for V. Then using density you can calculate the molar mass.

Yes, I tried to do your way and got it. thanks . But i realized even if i do this my way plugging in the numbers to this equation $d=\frac MV$ i get the same result

You might get the same number, but take a look at the units. The units $\mathrm{g}\cdot\mathrm{L}^{-1}$ are not the units for molecular mass $\mathrm{g}\cdot\mathrm{mol}^{-1}$

Right totally. İt was a typo my bad

You don't need to know the volume to determine the molar volume $V_\mathrm{m}$ of a substance. In your notations:
$$V_\mathrm{m} = \frac{M}{d} = \frac{\pu{62.37 g mol-1}}{\pu{2.56 g L-1}} = \pu{24.36 L mol-1}$$
Alternatively, you could've just used ideal gas law:
$$V_\mathrm{m} = \frac{V}{n} = \frac{RT}{P} = \frac{\pu{0.0821 atm L mol-1 K-1}\cdot\pu{295.8 K}}{\pu{0.9968 atm}} = \pu{24.36 L mol-1}$$
P. S. You determined molecular mass correctly, but, as MaxW pointed out, the units are wrong: you probably made a typo and put $\pu{g L-1}$ instead of $\pu{g mol-1}$.

@MaxW No, this is incorrect. Molar volume has SI units of $\pu{m3 mol-1}$, which are also equally good converted to $\pu{L mol-1}$ (see IUPAC Green Book, p. 47). There is a good reason it's called molar volume. You must divide both part by $n$ in your formula.

you're right. I deleted my previous comment which was confusing.

@MaxW No prob, I also tend to overthink quite often:)

Thank you guys so much. Even though this might be an easy question, it feels good to have everything clear


text:
Dicumyl synthesis

2,3-Dimethyl-2,3-diphenylbutane (Dicumyl) (63). A solution of 2-bromo-2-phenylpropane (10.0 g, 50 mmol) in anhydrous diethyl ether (25 mL) was stirred with magnesium turnings (0.60 g, 0.025 g atom) overnight. The reaction mixture was poured into aqueous $\ce{NH4Cl}$ solution (100 mL, 5%) and extracted with dichloromethane (100 mL). The solvent was evaporated, and the solid residue was recrystallized from 95%
  ethanol to give 2,3-dimethyl-2,3-diphenylbutane  (2.4 g, 47%).

Does anyone have any intuition as to how this reaction is working mechanistically? Is it a Grignard formation and then homocoupling event? Can anyone see a reason for why an ortho methyl substituent on the phenyl ring might interfere with this reaction?

Clarification: am I correct that they don't care about the atmosphere, i.e. no argon/nitrogen atmosphere is implied in the reaction?

The ammount of magnesium was 25 mmol, so only half of the 2-bromo-2-phenylpropane was turned into magnesium bromide.
After a molecule of magnesium bromide formed, I would guess it undergo a elimination of $\ce{MgBr^+}$ moity since it is on a tertiary (and also benzyl) carbon and form a relatively stable carboanion. The carboanion then attacks one of the 2-bromo-2-phenylpropane molecules forming the product. But I also wouldn't rule out $\ce{S_{N}2}$ type substitution.

I don't know what the exact mechanism is, but I would be very surprised to see a naked carbanion without any sort of metal. I've never seen or heard of such a thing before, except in highly simplified mechanisms (which incidentally give a horribly wrong impression that organolithiums react almost instantaneously). Aromatic rings don't even stabilise negative charges all that much; the pKa of toluene is still very high, ~40. With two electron-donating methyl groups, one would expect 2-phenylpropane to be even less acidic than toluene.

Well, that's interesting as nothing is added here. Today Palladium-catalyzed cross-couplings are quite common, but on a historical side, they took their idea from the very first reactions of sodium and organo-chlorides (Wurtz-coupling) and the magnesium mediated formation of biaryls. Back then transition metal salts were added which lead to the idea to use $\ce{Pd}$ which is way more carbophilic and can easily enter and leave the catalytic cycle. I have heard that the idea for the cross-coupling derived from the Schlenk-equilibrium, so your Grignard-reagent dimerizes and then equilibrates to from a $\ce{MgBr2}$ and a $\ce{MgR2}$. I'd love to say now that this simply reductively eliminates to form $\ce{Mg}$ and your product, but I have my doubts (although I have heard people talk about it before).
So in your case, we probably have to go the Wurtz path and say that the negatively polarized carbon on your Grignard-reagent nucleophilically attacks the $\ce{C-Br}$ bond in an $\ce{S_{N}2}$ reaction. If you used sodium the bond would be much more ionic and you'd probably also end up in simply deprotonating the second equivalent of starting material forming a 1-propene derivate.

negatively polarized carbon on your Grignard-reagent nucleophilically attacks the C-Br bond || Not going to happen with tertiary carbon.


text:
Meaning of carbon hybridization
What do we mean when we talk about carbon hybridization?
I'm trying to understand how it works, what is the difference between non-hybridized and hybridized carbon?

Carbon is not hybridised. In a too simple model, carbon has s and p valence orbitals, like those in the hydrogen atom. The geometry of those is not useful to make bonds. Instead, the orbitals that form bonds have different properties, either 4 sp3 or 3 sp2 and 1 pi or 2 sp and 2 pi orbitals. A not too bad description of those can be reached by mathematically "mixing" the properties of s and p orbitals. Hence their name "hybrid". To simplify this description, we just say C is hybridised sp3 or sp2 or sp. Nonhybridised does not occur normally.

Actually hybridization refers to the electron orbitals around the atom not the atom itself. It actually follows the model called VSPER of electron distribution. So it's not limited to carbon, it applies to all atoms.
Atoms depending on the number of electrons they have in their elemental form, has a set number of orbitals. And in order they are as follows, S orbital 2e-, P orbital 6e-, and some other you don't really encounter in intro ochem.
The hybridization is how the orbitals combine to create bonds. You should really read a book about this, I really recommend organic chemistry by Klein chapter 1.


text:
Why must polymers have a repeating unit?
In organic chemistry, we learned that small molecules can form a polymer via a process called polymerization. For example, $\ce{CH2=CH-Br}$ molecules can form the polymer
\begin{align}
\ce{nCH2=CH-Br-> -[-CH2 -}&\ce{CH -]_n -}.\\
&\;\ce{|}\\
&\ce{Br}
\end{align}
So the polymer is a periodic chain $\ce{-CH2-CHBr-CH2-CHBr-\cdots}$. But since every monomer can have two orientations ($\ce{-CH2-CHBr -}$ or $\ce{-CHBr-CH2 -}\!$), there is no requirement that all monomers must be in the same orientation, does the polymer have to be a peroidic chain? Can it be a random chain with a structure that looks like
$$\ce{-CH2-CHBr-CHBr-CH2-CHBr-CH2-CH2-CHBr-CH2-\cdots}\,?$$
Most textbooks emphasize that $n$ is random, but still assume that the unit repeats.

I recommend you look at the roots of ‘polymer’ poly means many, mer means segment, mono means one. Monomer - one segment. Polymer - many segments.

@JavaScriptCoder, the discussion is periodic vs random. I'm not sure if a random chain is actually possible, or it is simply energetically not favorable on the scale of thermal fluctuations $k_BT$.

honestly I can’t see why it couldn’t be random, I mean DNA could be random and still considered a polymer

Random chains are very much possible. Then again, this particular one is probably repeating.

@IvanNeretin, entropy favors randomness, but energy favors order. For example, having two adjacent $\ce{-CHBr -}$ segments can increase the repulsion of negative charges on bromine. If entropy can overcome this, then we will have a random chain. Otherwise the chain will want to order itself.

This sounds like a pretty comprehensive answer to your own question.

That's how things would work out in theory. I'm still expecting an experienced answer that tells me what happens in practice.

Well, that's what happens in practice, too. Related reading: https://en.wikipedia.org/wiki/Tacticity.

There is two things: The repeating unit, and its orientation. The former is always the same in your case, the latter not necessarily. Of course you can also mix monomers (doesn't always work however.

Entropy would favour a random orientation of each monomer unit when building up the chain, however whenever you have two different substituents at the ethylene unit, then electrostatics and sterical hindrance will make one side be preferred to make the connection to the growing chain end.
Polymers that do grow via chain growth therefore have a regular structure, but there can be defects. The defect density depends on how strong the preference for one side is during the polymerisation reaction. You can have as little as one in ten thousand. That's actually a bit of a problem, because the regularity often allows polymers to crystallise, which is important for the technical application. You want the number of defects always to be the same, because too high crystalinity makes your material brittle, too low makes it soft.
Polymers that make step growth (any two molecules randomly connect with their end groups, until you have a long chain) could be different, but also there often a regular structure occurs, because of A-B + A-B , either the A+B like to react, or A+A and B+B.

I realize this problem is related to a 1D Ising model with nearest neighbor interactions. Since the Ising model in 1D does not have a ferromagnetic phase transition at nonzero temperatures, defects are inevitable once the chain grows long (to approach the thermodynamic limit).

@ZhuoranHe No, that doesn't really apply afaik. We are not talking about a 1D crystal. The chain is flexible (rotation around sp3-bonds, ), so there is a lot of room for entropy in it. Isotactic polypropylene for example can be made practically defect-free, much less than one defect per molecule at several thousand monomers per molecule, and you could also (not only theoretically) sort out chains with a defect.

I know the chain is not straight.  In a simplified model that only cares whether adjacent monomers have the same or opposite orientations, the rotational degrees of freedom of the sp3 bonds would be uncoupled to the ordering of the orientations. To make the mean defect-free length long, we need to reduce $\exp(-\Delta E/k_BT)$ by increasing $\Delta E$, the energy cost of having a defect. If $\Delta E=10k_BT$, then the defect-free length could be $\sim 10^4$.

@ZhuoranHe The orientation of repeating units cannot reach thermodynamic equillibrium, because the activation energy to flip one unit cannot be reached without completely destroying the chain. The ends of a long gaussian chain have no way of knowing they belong the same chain. A fixed, known defect imo is not even a defect in the way Ising is looking at it. (?)

I think they still can reach equilibrium during the reaction. You see the number $n$ can randomly change. What this means is that the bonds between monomers are forming and breaking all the time. To preserve the defectless chains sifted from the reaction system, the temperature would then have to be lowered.

$n$ can only grow during most polymerisation reactions, because we are very far below the so called ceiling temperature. Once a covalent bond is made, it stays. I am very sure Ising, Peierls et al. only apply to systems with translational periodicity. A stiff chain like C=C=C=C=C=C=C would indeed be sth different, that's not a polymer like all others, and would just break when it comes close to some critical lenght.


text:
How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution?
Below was question 34 in the USNCO 2017 exam:

If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?
(A) Roughly 50%
  (B) Roughly 75%
  (C) Roughly 95%
  (D) Over 99%

My solution is as follows:

Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K_\mathrm{sp}$.
Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$
Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$
Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation:
$$
(0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$
Find percentage of calcium hydroxide precipitated:
$$\frac{0.019}{0.02} \times 100\% = 95\%,$$
hence (C)

I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).
Is there a simpler method?

A less analytic aproach:
Initial concentrations:
$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$
$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.
Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.
The concentrations that remain in solution are:
$\ce{[Ca^{2+}}]=0.0126(1-p)$
$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$
Since the solubility product remains the same:
$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$
Now, since there are only four scenarios, try them out substituting every value of $p$ in  the equation and take the closest one to $8\cdot10^{-6}$:
\begin{array}{|c|c|}\hline
    p&K_\mathrm{sp}\\\hline
    0.50&\pu{7.99E-05}\\
    0.75&\pu{3.56E-05}\\
    0.95&\pu{6.46E-06}\\
    0.99&\pu{1.27E-06}\\\hline
\end{array}
The closest one is for roughly 95% precipitation, without solving any equation.

When you run into a cubic in evaluating concentrations, one approach you can use to solve it is the method of successive approximations.
$$(0.0126-x)(0.1252-2x)^2=8\cdot10^{-6}$$
$$(0.0126-x_0)(0.1252)^2=8\cdot10^{-6}$$
$$(0.0126-x_1)(0.1252-2x_0)^2=8\cdot10^{-6}$$
For example, moving from the first line to the 2nd line, I guess that $2x=0$ and solve for $x_0$. This first guess is not great, but I can use the solution from that approximate equation to generate a better guess. So in the 3rd line, I approximate $2x=x_0$ and solve for $x_1$. You can continue this procedure until $x_{n+1}\approx x_n$, which in this case occurs very quickly. I obtain $x_0=0.01209$, $x_1=0.0119747$, and $x_2=0.011976$ after which the value doesn't change. Checking the ratio, we see that it gives just about $95\ \%$. $$\frac{0.011976}{0.0126}\times100=95.047\ \%$$
An important side note for this case is why I approximated $2x$ rather than $x$. This is because approximating $x$ will cause you to converge towards the two complex solutions to the cubic equation.

Note what the question actually asks you, "roughly". Approximation is an important tool in ionic equilibrium calculations.
Here, you have $0.0126M \text{ }\ce{Ca}^{+2}$ and $0.0252M\text{ } \ce{OH-}$ in the original solution. By addition of $0.1M \text{ }\ce{NaOH}$, which is a strong electrolyte, you'll not only add $0.1M\text{ } \ce{OH-}$ to the solution, but also cause $Q$ to exceed $K_{sp}$, hence, the salt will be precipitated out.
Now, the concentration of our ions before precipitation is $0.0126M \text{ }\ce{Ca}^{+2}$ and  $0.1252M\text{ } \ce{OH-}$. Let $y$ be the fraction of existing ions that are not precipated out. So, we'll be left with $0.0126 \cdot y M \text{ }\ce{Ca}^{+2}$ and $0.1252 \cdot (2y-1) M\text{ } \ce{OH-}$ ions. Equating their ionic product to the salt's $K_{sp}$, we get:
$$0.0126 \cdot y\cdot(0.1252\cdot (2y-1))^2=8\times10^{-6}$$
$$y\cdot(2y-1)^2\approx0.0405$$
$$y\cdot(1+4y^2-4y)\approx0.0405$$
Magic trick$^\text{TM}$: Neglect $y^3$ at this step. If $y$ comes out be $\le5\%$, we'll assume this neglection to be correct.
$$4y^2-y+0.0405\approx0$$
$$y\approx0.050838$$
That means nearly $(100-5)\% = 95\%$ of the salt was precipitated out. Hence, your answer. Without a graphing calculator.
PS: If you actually solve the original equation, you'll get $y=0.05$. Hence, our approximate answer is very reasonably close.

How did I get $(2y-1)$ as fraction of final concentration of $\ce{OH-}$ ions?
Fairly easy. Observe that $y$ is the fraction of concentration of ions not precipitated out, while $x$ is the fraction of concentration of ions precipitated out. Then $x+y=1$. So, $1-x=y=\text{ concentration of }\ce{Ca}^{+2}$ ions. And $1-2x=1-2\cdot(1-y)=2y-1=\text{ concentration of }\ce{OH-}$ ions.
But why did I take $y$ anyway? What was wrong with the original $x$?
Our aim was to reduce this problem into an equation which can be solved without a calculator, i.e., by approximation. Taking $y$ ensured that in the end we could neglect higher powers of $y$ (note that we cannot neglect higher powers of $x$) and solve the question easily.

Crux of my answer: if while solving an ionic equilibrium question you're stuck in a tough calculation/unsolveable equation, consider modifying your approach to allow the use of neglection.
Hope it helps!

Ok, I admit, it wasn't any magic trick :P But just a clever observation that if $x\le 0.05$, then $x^3\le0.000125$, which is too small to affect our final result significantly.


text:
How does dry air get saturated in the Ostwald-Walker Method?
This question was already asked and this was the answer that was posted:
"So basically, before you start the experiment the vapors have already formed, meaning that thermodynamic equilibrium has been achieved. Then you start bubbling through dry gas, which then mixes with the vapor gas already in the bubbles. If you set the gas flow appropriately, you get a near-equilibrium situation (see the graphic below) and the gas that exits the bulb assemply has exactly the same amount of vapor in it as the vapor "layer" (which can not be thought of as a layer anymore because of the bubbling) above the liquid surface.
We do not have an equilibrium situation, which is why so much care has to be taken that we get as close as possible to the equilibrium situation. (Also, we do not have a closed system, so everything starts to get slightly problematic at this point. But to a first degree of approximation, this will hold up fine.)"
(For more, visit: How does Ostwald-Walker method work? )
My questions are:
1) What is this 'near' equilibrium situation that is created?
2) If dry air gets saturated when it passes through the solution bulbs, then how is it able to absorb more solvent vapours when it is passed through the solvent bulbs?

The answer to your second question asked is because of the pressure difference between Solvent and solution's vapour pressure


text:
IUPAC naming of 1,2-disubstituted cyclohexane derivative
I'm facing the challenge of naming this molecule:

We have to name it as an alkyl-substituted cycloalkane, since the ring has more carbons than the longest substituent (isopropyl). My questions:

When numbering carbons on the ring, we can start counting from either the methyl group (1-methyl...), or the isopropyl group (1-isopropyl...). Which one should I start with? Or do I need to start numbering with the substituent which name comes first, alphabetically?
I set the carbon number 1 on the methyl substituent, giving me the following name: 1-methyl-2-isopropylcyclohexane. However, my book says the correct name is: 1-isopropyl-2-methylcyclohexane. So the solution managed to sort the name both numerically and alphabetically. Is this what I need to do? Is the name I gave incorrect?
Which prefixes count towards alphabetically-naming a compound? Which prefixes don't? All I know is that the numeric prefixes like di- or tri- don't count. Are there any others which don't count either?

Related: Nomenclature of an organic compound

You may want to give it a try yourself; I already did. The book doesn't answer any of those 3 questions, at least not in the chapter it's supposed to: Chapter 4 (Cycloalkanes). So, unless the truth is hidden in any of the 1350 pages remaining (which would clearly be John's fault), the book gives the reader no clue on how to proceed in this case.

Simple prefixes (simple substituent groups such as methyl and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘methyl’ is considered to begin with ‘m’; ‘isopropyl’ is considered to begin with ‘i’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Thus, the correct alphanumerical order in your example corresponds to the name x-isopropyl-y-methylcyclohexane.
Furthermore,

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g; however, Rule f does not permit a decision to be reached between ‘1-isopropyl-2-methylcyclohexane’ and ‘2-isopropyl-1-methylcyclohexane’ since both names correspond to the locant set ‘1,2’.
According to Rule g, your example is named as 1-isopropyl-2-methylcyclohexane rather than 2-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.
Note, however, that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.
Therefore, the preferred IUPAC name for your example is actually 1-methyl-2-(propan-2-yl)cyclohexane because ‘methyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).

Let me first quote some rules given by IUPAC (1) and then come to your question.
ref.1
http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm#a_2__25
Acyclic Hydrocarbons
Rule A-2. Saturated Branched-chain Compounds and Univalent Radicals
2.25 - Univalent branched radicals derived from alkanes are named by prefixing the designation of the side chains to the name of the unbranched alkyl radical possessing the longest possible chain starting from the carbon atom with the free valence, the said atom being numbered as 1.
Examples to Rule A-2.25

The following names may be used for the unsubstituted radicals only:

2.3 - If two or more side chains of different nature are present, they are cited in alphabetical order .
The alphabetical order is decided as follows:
(i) The names of simple radicals are first alphabetized and the multiplying prefixes are then inserted.

(ii) The name of a complex radical is considered to begin with the first letter of its complete name.

2.4 - If two or more side chains are in equivalent positions, the one to be assigned the lower number is that cited first in the name.

For arranging substituents  , which are treated as prefixes  the following rule
is followed.
R-0.1.8 Order of prefixes
(a) Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplying affixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Now comming to your question , it has two substituents methyl and isopropyl at equivalent positions. Using rule 2.4 and R-0.1.8  , numbering is started from isopropyl end and named.

References:
1.http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm#a_2__25
2.http://www.acdlabs.com/iupac/nomenclature/93/r93_93.htm

Note that the references correspond to obsolete IUPAC recommendations from 1979 and 1993, respectively.

@Loong , i have  since checked for 2004 recommendations  at the following url ,  www.iupac.org/reports/provisional/abstract04/favre_310305.html  ,but the site seems to be down.I will most certainly recheck.Thankyou

Thank you for this thorough answer. One question though, you mentioned "equivalent positions" twice, but I still don't know what you mean by that. If you could give me further clarification on this, I'll be very grateful!


text:
How can organocuprates generated with catalytic Cu(I) be competitive in 1,4-addition to an enone?
Organocopper compounds (e.g. $\ce{LiCuR_2}$) are often prepared from an organolithium or Grignard reagent when a soften nucleophile is needed, such as for a conjugate addition to an $\alpha,\beta$-unsaturated carbonyl. Clayden et al. (p.499) recommend generating it in situ catalytically via transmetallation with 1% $\ce{CuI}$.
My question: if at the start, there is 1% of organocopper reagent and 99% Grignard reagent, why don't those 99% do direct addition? Why is the organocopper reagent so much more reactive that conjugate addition is the main outcome of the reaction?

My guess would be the greater nucleophilicity of the cuprate over the $\ce{R-Mg-X}$

@EashaanGodbole Thanks! But why is it more nucleophilic?

The formation of -ate complexes (e.g. boronate, aluminate (as in $\ce{LiAlH4}$) results in the formation of high-energy $\sigma$-bonds. There's the extra reduction in effective electronegativity due to the negative charge too...

@EashaanGodbole But isn’t the C-Mg bond higher in energy than C-Cu?

Would it be higher in energy than $\ce{C-Cu^{-}}$?

You assume the rates of reaction are nearly equal. Since 1,4-addition of the Cu-reagent is highly efficient relative to 1,2-addition of a Grignard reagent (say, ~10:1) AND you have only 1% Cu, then the formation of the Cu-reagent and conjugate addition is at least 1000 times faster than 1,2-Grignard addition.

Sorry could you please explain: (1) why the 1,4 addition is highly efficient, (2) how you get those 1000. Thanks!

@EashaanGodbole Do you mean that $\ce{C-Cu^{-}}$ is higher in energy than $\ce{C-Mg^{+}}$? It might be! But that would also imply that the C is more $\delta-$. Isn’t the whole point of using Cu is that it’s more electronegative and so can reduce the electron density around C, making it less hard to disfavor direct addition?

Related: Hard-Soft Acid-Base (HSAB) Theory

@EashaanGodbole Could you please clarify what you mean exactly? Like the whole of your argument? Sorry, I still don’t follow


text:
What facts are being used in this atomic structure question?
I wish to understand how to solve this atomic structure question:

For the ground state, the electron in the H-atom has an angular
  momentum = $\hslash$ , according to the simple Bohr model. Angular
  momentum is a vector and hence there will be infinitely many
  orbits with the vector pointing in all possible directions. In
  actuality, this is not true, because:

Bohr model gives incorrect values of angular
momentum.
only one of these would have a minimum energy.
angular momentum must be in the direction of spin of electron.
electrons go around only in horizontal orbits.

Actually, we have been taught Bohr orbits a lot in class, with formulae for everything (atomic radius/electron orbital speed/various energy levels/etc.)  and also basics of the quantum mechanical model. However, we didn't go into such a great depth of Bohr orbits as asked in the above question.
Specifically, I don't know of any reasonable arguments against points 1,2,3. Point 4 is evidently wrong based on the quantum mech model.

How do I develop any claim for/against the points 1,2,3? What are the facts involved here?

The only correct answer is #1.
In actuality, the ground state of the hydrogen atom has zero angular momentum.
Regarding #2 and #4, there is no preferred plane in which one point should orbit another point.  By symmetry, they are all equivalent.
Regarding #3, there is no angular momentum in the ground state, and in a state whether there is angular momentum, it doesn't need to be in the same direction as the electron spin angular momentum.

Thank you! Can you please add some detail as to why it has zero angular momentum? And why are points 2 and 3 wrong?

Oh, I just noticed your elaboration (next time please comment if you update your answer :)) I understand your logic for point 2, 3 and 4. For why there is zero angular momentum, I came across an old answer by you (coincidence ;))? Does that answer also apply in this question?


text:
What is the driving force for imine formation?
I know this that when a primary amine is reacted with a carbonyl compound, an imine is formed:

I'm wondering why the position of this equilibrium lies to the product side / right-hand side. Why is the formation of the C=N double bond favoured?

similarly to https://chemistry.stackexchange.com/questions/10722/what-makes-c-o-more-stable-that-coh%e2%82%82/10739?s=7|27.7077#10739

No, i didn't get it!

If we simply compare the relative reactivities of the product and reactant sides,  we can see that the attack of the more nucleophilic lone pair of the nitrogen would attack faster than that of the oxygen.  The ratio of the rate constants gives the equilibrium constant.  Voila

In practice, a drying agent is sometimes added to the reaction: I've used molecular sieves before, but a cursory database search brings up a few other options, e.g. magnesium sulfate, silica gel, or Montmorillonite (this list is probably not exhaustive). These remove water from the system and pull the equilibrium position over to the right-hand side.
That's one possibility, but I don't think that's all there is to it - I think a lot of imine formations proceed perfectly ok even without dehydrating agents being added to them. However, you'll have to hear about these from somebody else, as I don't know what would drive those reactions. It could well be that simply adding an excess of either amine or carbonyl drives the reaction to completion.
Interestingly, a 2009 paper by Saggiomo and Lüning1 describes their investigations into supposed imine formation in water. Apparently, both the carbonyl compound (benzaldehyde or salicylaldehyde) and amine (aniline) do not dissolve in water - not particularly surprising. The imine did not form appreciably in water and only started to form when water was removed during workup and purification.
P/S This is just the top hit I found on Google, it's hardly a complete literature search.

Saggiomo, V.; Lüning, U. On the formation of imines in water—a comparison. Tetrahedron Lett. 2009, 50 (32), 4663–4665. DOI: 10.1016/j.tetlet.2009.05.117.

See,We compare any reaction thermodanimaclly and kinetically
Let me put the thermodynamical aspect first which is that the formation of water is the very good sign of thermal stability for a reaction and a energy of 572 KJ/mol is realesed and hence helps to meet with the enrgy demands for formation of carbon and nitrogen bond
And suprisingly carbon and nitrogen bond is stable due to presence of lone pair on nitrogen and high density of electrons around it.
Now coming back to thermodynamics the carbon nitrogen energy is comparatively very high 614KJ/mol for this carbon and nitrogen bond. Also important thing note oxygen is more electronegative but yet as wr see oxygen is ready to leave with hydrogen to form carbon nitrogen bond double bond.
Now Kinetics
The product hence formed is by first order Kinetic nad hence also it is easy for hydrogen to migrate to alchol group and hence give water and with it leave it as to provide a negative charge to nitrogen and henceforth the water is left to make the way for carbon and nitrogen bond to form.Also notice  there will be a positive charge on oxygen so would leave easilyand hence there will a negative charge on "NZ" leaving formation of the double bond.
See this for data  https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies and
http://witcombe.sbc.edu/water/chemistryelectrolysis.html


text:
How do I heat my borosilicate reflux column?
I am gonna use a ceramic tape heater, but cannot wrap it around evenly. Is the glass column gonna be okay?

I am running some tests on plastic pyrolysis. I am using the reflux condenser to make the heavier hydrocarbons flow back into the boiling flask. I am also trying out various catalyst in the reflux condenser which will facilitate catalytic cracking.
I have decided to maintain the temperature at top of the reflux column to be of 150C and it will allow the lighter fraction to be removed off and the heavier fraction to be condensed back into the boiling flask.
The boiling flask is made up of quartz and the rest of the glassware is borosilicate.
I am gonna use the glass setup to test catalysts. If results are satisfactory, will get a microreactor made up of steel for further testing. I can better monitor and control temperature in a metal setup I think. Gonna try on only PE and PP types of plastic.
I am a structural engineer by profession and am new to organic chemistry. Thanks for you help in advance.

I take it "get a different reflux column" (e.g. one that will work with your heating tape) is out of the question? -- A big concern I would have is: if the shape of the reflux column is important, inadequate or uneven application of heat may yield results other than what you desire. Perhaps going into more details about why you're using that column and why you need to apply heat to the column may provide improved answers.

Thanks R.M. The heating tape is not flexible as I want it to be. I am running some tests on plastic pyrolysis. I am using the reflux condenser to make the heavier hydrocarbons flow back into the boiling flask. I am also trying out various catalyst in the reflux condenser which will facilitate catalytic cracking.

I have placed an order for a heat gun which can heat upto. Planning to wrap the column in aluminum foil and use the heat gun on it. Gonna use the IR thermometer which I have to measure temperatures.

I am gonna use the glass setup to test catalysts. If results are satisfactory, will get a microreactor made up of steel for further testing. I can better monitor and control temperature in a metal setup I think.

That should pose no issue regarding the glass
Without images to see how unevenly you will apply the heat, I have to modify my instinctual reaction "That cannot possibly be an issue" to "Most likely it will be OK". I'd be much more worried about variation in the distilling if you are at some point drawing a distillate. If you cannot control the heat flux, the yield and temperature profile will be offset from the theoretical and can (if the process is delicate) lead to variation in the results.
But borosilicate glass is sturdy, you can probably heat it with a naked propane flame - it is designed to have a low thermal expansion coefficient. Directly from that, it will have a high thermal shock resistance.

I am hoping to maintain the temperature with a range of 150C. (appox)

@Vanadium This looks like some heavy duty industrial research... Well... First things first, at 150°C I have no doubt the column will do fine from a temperature point of view. But there is room for improvement in your setup. Add some clips, or in a pinch some steel wire to secure the wrap around the column, then add aluminium foil to reduce the stresses and heat loss. There are standard solutions for this, but wire and foil will get you far. Also, I can't see a cooling unit, is that on purpose? Or is it just hidden from view, behind the mounting rack?

Just some tests for pyrolysis. Will do. I was thinking of using aluminium foil around the column. That would help to better distribute the heat I guess. Thanks for your answer Stian.

Yep, The condenser is just behind mounting rack.

I am gonna use the glass setup to test catalysts. If results are satisfactory, will get a microreactor made up of steel for further testing. I can better monitor and control temperature in a metal setup I think.

@Vanadium It should, but at 150°C you have plenty of options. Just make sure you keep some leftover money to get good temperature measurements. =)

Tried covering the set up with aluminum foil. Works great.  The column got up to 180C without external heating. Thanks Stian.

The first answer is great and the borosilicate glass can indeed survive harsh conditions. Here I'd like to suggest a DIY-project to improve the efficiency of the heater and make the column be warmed more evenly. Take a sheet of fiberglass cloth, sew the heating element in a serpentine pattern with the fiberglass threads, then wrap the crafted heating mantle around the column. The exact size can be estimated based on the height and the diameter of the wide part of the column. I sketched an approximate scheme:

Alternatively, you can leave a line of column uncovered (1-2 cm) for visual inspection.  Fiberglass can withstand the same temperatures as your heating mantle, but I suppose you've already discovered it the hard way.

Hey andselisk, Thanks for the answer and great illustration.

I am also thinking of wrapping the reflux column with aluminium foil and then use the ceramic tape heater to wrap over the aluminium foil. I hope this will lead to a better heat distribution.

@Vanadium No, don't... Aluminium would block the heat radiation, which (arguably) is the primary mechanism for transferring heat in such a setup. You want to enclose the heat inside, not shield the equipment from the heater.

Oh thanks. I assumed primary heat transfer is through conduction. And so wrongly thought it would help in even distribution via conduction.

That explains a lot of things. Thanks a lot @Stian

@Vanadium No problem. Usually you do not even try to heat the whole column, you heat the liquid, which boils, then transfers heat up the column. you wrap the column to limit heat loss. If you need very precise temperature control or exactly the same temperature in the entire column you make a setup like the good mr Andelisk has so nicely drawn, and then wrap with a reflective material outside of that. Aluminium reflects well, and thus doesn't emit well, so it traps heat by not being able to emit it, in combination with high internal reflection.


text:
Any substance too poisonous to measure an LD50?
A recent question on poisons was answered in part by bringing up the concept of $\pu{LD_{50}}$ and animal testing and so on; none of that was new to me, but it did bring to mind a question I don’t know the answer to:
Has there ever been a substance that scientists have actively tried to determine $\pu{LD_{50}}$ for, and could not because they were unable to determine a dosage low enough to kill “just” half of the subjects?
Please note that this question is not restricted, strictly, to any technical definition of “poison.” Rather, answers must meet these requirements:

There must be research (considering that I am looking for something that might be a negative result and those do not get published as much as they should be, I won’t demand peer-reviewed publication, but the goal should have been to attempt peer-review publication if it had worked out—effectively, I mean serious work), and
that research must use the term “$\pu{LD_{50}}$” to characterize the substance.

Basically, I don’t want to get into debates about what is, or is not, a poison here. If a researcher is willing to call something $\pu{LD_{50}}$, then I am willing to accept it as a “poison” for the purposes of this question.
I also don’t want speculation, or for a user here to characterize something as $\pu{LD_{50}}$ when the underlying research doesn’t call it that. It is not enough to say, for example, “well I’m sure even a single atom of antimatter inside your body would be pretty bad,” you need to cite a particular researcher who has performed experiments with the goal of determining what they themselves called $\pu{LD_{50}}$ for that substance.
I suspect the answer is no, but I have no idea how to research something like this.

Comments are not for extended discussion; this conversation has been moved to chat.

Of course no. Botulotoxin is probably the strongest known, and still its $\rm LD_{50}$ is counted in nanograms per kilogram, which is pretty manageable. Sure, working with such tiny amounts requires some special measures, but still, it is way greater than one molecule. You can divide it again, and again, and again.
So it goes.

Thanks; I’d heard botulotoxin referred to as the strongest, but I wasn’t sure that there might not be caveats to that, strongest with the $\pu{LD_{50}}$ actually measured or something.

"one prion molecule makes two - those two go on and make four, those four make eight, and so on, until the whole brain is just one liquifiied spongy mass. " https://blogs.scientificamerican.com/a-blog-around-the-clock/bio101-protein-synthesis-transcription-and-translation/ I wouldn't rule out one prion molecule delivered in a particular way killing (slowly) more then 50% of the people.

@DavePhD It's not like you have LD50 for them, also you vastly overestimate their power. Organisms have ways to deal with misfolded proteins. BTW this is supposed to be about poisons and blogs on sci. amer. aren't good sources.

@Mithoron Technically, the only requirement I have for a substance here is that someone tried to determine an $\pu{LD_{50}}$ for it. If it’s not technically a poison per se, but harmful in some other way (e.g. these prions, or Bill K’s suggestion of uranium), it counts as long as someone’s tried to get an $\pu{LD_{50}}$. Let me know (or feel free to edit yourself) if you think the question needs more clarification on that point.

@Mithoron I'm with DavePhD on this one... One single virion or prion can be easily enough to kill an adult human, if the immune system is compromised or the virus/prion is "new" and/or "smart" enough for the immune system to overlook it for a couple of days. Some viruses do just that - they replicate slowly so that the immune system's defense reaction is not triggered shortly after the infection (as that triggering  usually requires a rapid increase in pathological activity in short time), and when the immune system "notices" the virus, it is compromised by the virus already (HIV virus).

@vaxquis that kind of "if" does not an LD50 make.

Seeing as I’m the one who asked the question, I think I’m entitled to determine what it’s asking about, not you. Yes, I understand that $\pu{LD_{50}}$ is typically used for toxicants, would not generally be used for pathogens, etc. But the definition of $\pu{LD_{50}}$ really is compatible with anything that might, at some dosage, kill 50% of subjects. And while scientists would almost never use it that way, almost never is not the same as never. If a researcher decided to use $\pu{LD_{50}}$ as a measure of a substance’s danger, that is good enough for me here.

(But it does have to be an actual research effort made to determine “the LD50LD50\pu{LD_{50}} of X,” not just some speculative answer here suggesting that if someone were to try to do so with some dangerous substance, it would meet the criteria of the question. I am using the “research effort made” standard as the bar to meet to qualify as counting as an “LD50LD50\pu{LD_{50}},” as a valid answer to the question.)

It's not even that easy to draw a line between prions and toxins, considering that prions are just proteins, just like e.g. botulotoxin.

Botulotoxin doesn't reproduce. I think that's quite a solid line.

@IvanNeretin Just so you know, I did update the question to clarify what I meant, in case that changes your answer. I suspect it doesn’t, and I suspect that this is still the correct answer, but since there had been some confusion  (and because another answer asserted that $\pu{LD_{50}}$ actually has been used for e.g. prions, though it’s difficult for me to confirm that this is the case), I have un-accepted this answer for now. In all likelihood I will end up coming back and accepting this answer again, but thought you deserved the head’s up.

Well, that makes it quite a bit harder to answer. I think I'd rather not change my answer, even though prions might now be a viable candidate.

Right, that’s why I didn’t just look up the deadliest toxin, see it had a documented $\pu{LD_{50}}$, and consider my question answered. Anyway, thanks again for the answer in any event.

Folks, please take these side discussions to the chatroom that was automatically made for the comments under the question.  This is really not what comments are made for.  Thanks!

Otherwise said no homeopathic poison exidt :)

Wouldn't prions be more of a biological than chemical hazard, just like a virus?

Some options that do not work, but have been suggested:

Prions: looks like LD50 has been successfully determined for the most common Sc237 and 263K prions: Prions: A Challenge for Science, Medicine, and the Public Health System. Though it is also referred to as "Infectious dose" as opposed to "Lethal dose".
Polonium: While some sources claim that polonium cannot be characterized by LD50, this is not due to high lethality but just because the effects vary depending on what kind of cancers or acute radiation disease the radiation causes. Actual LD50 is around 90 ng for humans for acute radiation disease.
Antimatter: looks like around 1 ng would be plenty to kill a person. LD50 would be much lower, but probably easy enough to measure anyway. Real problem in the experiment would be delivering the antimatter without premature annihilation.

But in reality, it just depends on the means available to the scientist trying to determine the LD50. For example in this study, "For the remaining solvents LD50 could not be determined due to volume
limitations, and have been given a value of <1.0 ml/kg."
The limit of current science would have to be a poison that would kill with a single molecule, as it is quite possible nowadays to manipulate and separate single molecules. Or otherwise single indivisible unit, such as a single cell.

Please limit your answer to only those substances that have had actual research into determining their “$\pu{LD_{50}}$.” So that book on prions might work (though it’s unclear to me since only one page appears to use the term and I don’t get to see enough context to see if they’re actually discussing the $\pu{LD_{50}}$ of the prions), maybe polonium. Also, things that do have an $\pu{LD_{50}}$ aren’t actually answers to the question; they could be useful if you can prove they are as deadly as things can possibly get (suggesting that the answer to the question is no).

@KRyan, well, if you want to be nitpicky like that, I'll just point to the second last paragraph that fits the nitpicky interpretation of your question by having a group of scientists who were actively trying to determine LD50 and couldn't.

I don’t consider it nit-picky at all; a question open-ended enough to allow random speculation like antimatter here should have been closed as being overly-broad and/or primarily opinion-based. Anyway, yes, the second paragraph could work, except I have absolutely no way of confirming your assertions here. A link to the work would improve matters tremendously. Note the updates to the question clarifying matters.

Yeah, paywalls suck. If you want, searching for that phrase on google scholar should (I think) show that it does exist there even if you don't have access to the full text.

This is about #1 again? Yeah, I tried that, only got one hit (and one false-positive for a citation abbreviated as LD), which did talk about $\pu{LD_{50}}$ but it wasn’t clear what substance was being discussed in that section, and I couldn’t get at the surrounding context. Certainly not enough to see that they attempted to determine an $\pu{LD_{50}}$ but could not due to being unable to isolate a small enough dosage.

Still would like to see the “other” sources for #2 (the ones that aren’t the “some” that say you cannot characterize polonium with $\pu{LD_{50}}$) that do talk about it having an $\pu{LD_{50}}$, though again if they were able to actually determine a dosage than it isn’t quite an answer to the question. Though again, if you want to answer no and then use prions and polonium as evidence of even ridiculously dangerous things that nonetheless we could get an $\pu{LD_{50}}$ for, that’s fine—but you have to actually spell out that this is your answer in your answer to do that.

(As above)
Folks, please take these side discussions to the chatroom that was automatically made for the comments under the question. This is really not what comments are made for. Thanks!

@jonsca Actually for what it’s worth, in my opinion, all of these comments are directly pertinent to attempting to improve the answer, not side discussions. If jba is not going to act on them, they may be obsolete and worthy of migrating to chat or deleting, but at least in my opinion it has not been long enough for that (but then I am not very familiar with this Stack, and certainly haven’t seen what flags have been raised here). On the other answer, I agree.

@KRyan I think the initial confusion was about my list. I did clarify that to "Options that have been suggested but do not work:". As for what is closest to a real answer I have, is the study with "For the remaining solvents LD50 could not be determined.." but the full text to that is unfortunately paywalled.

About antimatter. Shouldn't ng of antimatter simply destroy ng of matter?  Would initiate a  cascade?

@Alchimista It would destroy a nanogram of matter, yes, and release a lot of energy in the process. 180 kJ for a nanogram, enough to heat 0.7 kg of surrounding tissue to boiling point.

Forgot that!  :))


text:
Looking for Desiccants
I am an engineer and not the best at chemistry, I am horrible at it, but I was wondering if it is safe to mix sodium hydroxide and calcium oxide to make a desiccant that also captures carbon monoxide?
I am asking because I am trying to make a high quality zero air generator and need to lower the humidity and carbon monoxide and this seemed like a way to do it.

Why would you want to mix NaOH with CaO? Yes, it is safe, but also useless.

Sodium hydroxide and calcium oxide would remove water and carbon dioxide, but they would not remove carbon monoxide. For any significant flow,  you would also have to figure out how to prevent the airflow from either clogging the "filter" or creating "channels" through the filter.

Can't you just buy sodalime to remove the CO2 and water? Won't do anything for the CO though.

I suspect there might be a typo and OP meant carbon dioxide $\ce{CO2}$, not carbon monoxide $\ce{CO}$, since both NaOH and CaO are suited for $\ce{CO2}$ adsorption. Also, you are right that this mix can help to get rid of trace amounts of water, however hydration is quite exothermic (e.g. your adsorber is going to get warm).

You will also need activated carbon to remove VOCs

I have other drying components and use both activated charcoal and Purafil. I am just looking to remove the small amount of moisture that remains, I am currently getting a dewpoint of -34 C but I need to get to -40C or lower. I also need to remove CO and CO2 and this just seemed like a way to go. I am still looking for a second CO scrubber but that was not this request, I must of typed monoxide by mistake; sorry. I have a CO scrubber that converts CO into CO2 I am just going to slow the air flow through to increase affect but I wanted a second CO scrubber. That's my next question for here.


text:
Redox potential of a lead–acid battery
In the German Wikipedia there are two reactions on the poles of the battery shown with the following potentials:

$$
\begin{align}
\ce{Pb + SO4^2- &-> PbSO4 + 2 e-} &|\pu{-0.36 V}\\
\ce{PbO2 + SO4^2- + 4 H+ + 2 e- &-> PbSO4 + 2 H2O} &|\pu{+1.68 V}
\end{align}
$$
$$E_\mathrm{Ges}^0 = \pu{1.68 V} - (\pu{-0.36 V}) = \pu{2.04 V}$$

I do understand the potential for the second $\pu{1.68 V}$ since for the second reaction the underlying redox pair is $\ce{Pb^4+ + 2 e- -> Pb^2+} $. For this redox pair the electrochemical standard potential is $\pu{1.69 V}$.
But for the first reaction I think that the underlying redox pair has to be $\ce{Pb^2+ + 2 e- -> Pb}$. This redox pair has standard potential of $\pu{-0.1263 V}$.
This result in a voltage of $\approx\pu{1.55 V}$. But Wikipedia and a book of mine tell the the voltage of this battery type is $\pu{2.04 V}$.
What the reason for the $\pu{-0.36 V}$?
Source: This is from the German Wikipedia article on lead-acid batteries. Unfortunately the English version doesn't contain the calculation of the voltage. I took the standard potentials from the book Elektrochemie by Hamann.

The potentials depend on the form of the compounds.
It is true that in solution
$\ce{Pb^{2+}  (aq) + 2e^- -> Pb (s)} $
is -0.126 V.
But in the case of a battery we have:
$\ce{PbSO4 (s) + 2e^- -> Pb (s) + SO4^{2-} (aq)}$
And in this case the $\ce{Pb^{2+}}$ is in solid form and the potential is -0.356 V.
In a battery the sulphate is insoluble and it is required that it sticks to the electrode, otherwise the reverse reaction can not occur.
A table of potentials can be found here

The underlying redox pair is, as you say, $\ce{Pb^2+ +2e− -> Pb}$ and has standard potential of $\pu{−0.1263 V}$.
But standard potential is for $\pu{1 M}$ concentration. If you look at the "underlying" reaction, you must correct for the reduced concentration of $\ce{Pb^2+}$ due to its insolubility. The correction is made by using the Nernst equation and the solubility (or solubility product) of $\ce{PbSO4}$, and the half-reaction potential increases because the solubility of $\ce{Pb^2+}$ is low.


text:
What would affect the rate of recharge of a lead-acid battery?
I am trying to perform an investigation regarding lead-acid batteries and would like to specifically look into the recharging mechanism. However, I realised that this might be quite difficult since the recharge equation is
$$\ce{2PbSO_4 +2H_2O->4H^+ +2SO_4^2- + Pb + PbO_2}$$
Would anything except for the voltage supplied affect the rate in which the charging takes place (or the current through the electrodes)? For example, would changing the concentration of the electrolyte affect the rate of recharge?
EDIT: I was just looking at the equation and thought that if I increased the concentration of the electrolyte, which is the sulfuric acid, it would actually cause the equilibrium to shift left, meaning that the current would be lower with higher concentrations of sulfuric acid. Is this correct?


text:
etymology of supernatant
Why do we call a solution that has been centrifuged a supernatant? It seems to me that a "natant" should be something that floats (from Latin "natare" meaning "to swim"), and a "supernatant" should be something which lies above the natant, not below it. Shouldn't the liquid be called a subnatant? Or maybe a superprecipitate?
Edit: fixed Latin (see below)

It means "the liquid that swims above", from the Latin terms super = above and natare = to swim.
So it is the liquid that remains on top.
The subnatant would be the liquid at the bottom, in case you should centrifuge two liquid phases.
The term you name, nato, means son in Latin.

OK, so it is "swimming above" the precipitate and not "above" something that is "swimming", such as a floating pellet. That makes more sense!


text:
Does BET theory apply to type H3 and H4 isotherm for calculating surface area?

BET applies to type 2 and 4. That is basically unanimous.

You should elaborate and put some links.

This is enough.If you know the field, you will know what I am talking about.

The sample that shows loop of H3 type have (very likely) no micropores (small uptake in low p/p0 range). In this case you can apply BET theory in common p/p0 range (0.05-0.3). The sample that shows H4 hysteresis have micropores (large uptake in low p/p0 range). The BET theory can be applied here, but in a much lower p/p0 range. For activated carbon, for example, one would look for a linear(!) range between p/p0 = 0.02-0.1. Generally, the BET theory is not suitable for microporous materials.


text:
Nomenclature of amoxicillin
I am trying to understand the nomenclature of amoxicillin:

The IUPAC name that Wikipedia has given is (2S,5R,6R)-6-{[(2R)-2-amino-2-(4-hydroxyphenyl)-acetyl]amino}-3,3-dimethyl-7-oxo-4-thia-1-azabicyclo[3.2.0]heptane-24-carboxylic acid.
I think I understand most of it but I have two problems:

Why is there a 24 before the carboxylic acid?
There is a $\ce{C=O}$ group between two nitrogens that doesn't appear in the name. I was expecting an amide, rather than amino or oxo.

To preface this, I'm not sure if this is indeed the correct IUPAC name as per the latest 2013 recommendations. It might well be, but I don't know. However, that's not crucial to your question, which is more about understanding the name that is given on Wikipedia. Here's the structure of amoxicillin:

The amide C=O is in "acetyl", which refers to the $\ce{CH3C(=O)}$ group (see e.g. acetyl chloride, $\ce{CH3COCl}$):

The '24' is probably a typo and should instead be '2'.

Following this numbering scheme you can see that the lactam C=O is in '-7-oxo'.

Thanks for your help.  It does indeed appear to be a typo as on the pubchem website (https://pubchem.ncbi.nlm.nih.gov/compound/amoxicillin) it has the IUPAC name as (2S,5R,6R)-6-[[(2R)-2-amino-2-(4-hydroxyphenyl)acetyl]amino]-3,3-dimethyl-7-oxo-4-thia-1-azabicyclo[3.2.0]heptane-2-carboxylic acid.  Apologies as I have should of course checked somewhere else first.

The answer given by orthocresol explains (and also partly corrects) the name that was suggested by Wikipedia and was mentioned in the question; however, it deliberately does not discuss the correct name according to current (2013) IUPAC recommendations.
The resulting name ‘6-{[2-amino-2-(4-hydroxyphenyl)-acetyl]amino}-3,3-dimethyl-7-oxo-4-thia-1-azabicyclo[3.2.0]heptane-2-carboxylic acid’ (ignoring stereodescriptors for simplicity’s sake) is actually not wrong; however it is not the preferred IUPAC name.
The difference concerns the ‘[2-amino-2-(4-hydroxyphenyl)-acetyl]amino’ substituent group. For the preferred IUPAC name, the locants ‘2’ for the amino and the 4-hydroxyphenyl substituents would be omitted since all substitutable hydrogen atoms of an acetyl group ($\ce{-CO-CH3}$) have the same locant, resulting in the name ‘[amino(4-hydroxyphenyl)acetyl]amino’. However, the preferred suffix for a substituent of the type $\ce{-NH-CO-R}$ is ‘carboxamido’ (i.e. ‘acetamido’ for $\ce{-NH-CO-CH3}$) rather than ‘acylamino’ (i.e. ‘acetylamino’ for $\ce{-NH-CO-CH3}$). Thus, the preferred name for the complete substituent group is 2-amino-2-(4-hydroxyphenyl)acetamido. (Note that the ocants ‘2’ for the amino and the 4-hydroxyphenyl substituents can no longer be omitted since acetamido has substitutable hydrogen atoms with different locants.) Therefore, the preferred name for the complete compound is 6-[2-amino-2-(4-hydroxyphenyl)acetamido]-3,3-dimethyl-7-oxo-4-thia-1-azabicyclo[3.2.0]heptane-2-carboxylic acid.

To understand this name, we first have to see that the carbocylic acid group ($\ce{-COOH}$) is the principal characteristic group in this case. Therefore, this group is chosen for citation at the end of a name by means of a suffix. Accordingly, the senior parent structure of this compound is bicyclo[3.2.0]heptane-2-carboxylic acid.

The name of the parent structure is modified according to the positions of the hetereoatoms to 4-thia-1-azabicyclo[3.2.0]heptane-2-carboxylic acid.

Since the principal characteristic group is expressed by means of a suffix at the end of a name, the other substituent groups have to be expressed as prefixes. Considering only the groups that are directly attached to the parent structure leads to the name 6-acetamido-3,3-dimethyl-7-oxo-4-thia-1-azabicyclo[3.2.0]heptane-2-carboxylic acid.

As mentioned above, however, the acetamido group has two further substituents: (a simple substituent group ‘amino’ and a compound substituent group ‘4-hydroxyphenyl’), which changes the acetamido group to the complex substituent group ‘2-amino-2-(4-hydroxyphenyl)acetamido’.


text:
Carbon atoms which carbocation and carboanion are BOTH less stable then arene/vinyl
I'm trying to come up with a compound class in which iodine atom would be a bad leaving group both whether it leaves or takes charge with it (I know that $\ce{I^+}$ on its own would be horrible leaving group, but I'm thinking of a reaction where it oxidizes and so might leave forming a carbanion).
Aryl and vinyl carbons are both good candidates, but I'm wondering if some other compounds might work better or just as well.
So the question is about finding a compound $\ce{RR'(R'')C-I}$ such that both $\ce{RR'(R'')C^+}$ and $\ce{RR'(R'')C^-}$ are not stable. Or in other way, a compound with both $\mathrm pK_\mathrm a$ higher and $\mathrm pK_\mathrm b$ lower than that of aryl or vinyl.


text:
Magnesium oxide lab question
For my magnesium oxide lab, one of the question asks why I need to include the cover of the crucible when weighing it with the magnesium before the reaction, then again after with magnesium oxide, and in the beginning when its empty. Does it change if I don't include the cover? anyone have any hints or tips as to why I need to include the cover?

When you burn magnesium it sputters. Without a cover bits would be thrown out. With the cover bits stick to the inside of the cover.
(1) Weight(crucible+lid+magnesium) - Weight(crucible+lid) = Weight(magnesium)
(2) Weight(crucible+lid+magnesium oxide) - Weight(crucible+lid+magnesium) = Weight(oxygen)

The question is asking why you want to cover when weighing it. Not during the experiment

Think... if there is magnesium oxide on the inside of the cover from burning, what happens if you ignore the cover in the initial and final weighings?

Magnesium is going to be covered with a thin layer of $\ce{MgO}$ and is not really an issue here, but for calcined $\ce{MgO}$ you need to avoid both physical absorption and chemisorption of water vapors and $\ce{CO2}$ from the air:
$$
\begin{align}
\ce{MgO + H2O &-> Mg(OH)2}\\
\ce{MgO + CO2 &-> MgCO3}
\end{align}
$$
Both magnesium hydroxide and carbonate are also hygroscopic and form a wide range of hydrates when kept in humid environment. It is necessary either to weigh the sample very quickly, or use the crucible with the lid tightly closed.
Ideally for such measurements you want to use a weighing bottle specifically designed for precise mass determination of hygroscopic and reactive solids:

(Source)
In both cases make sure to tare the balance with both masses of the vessel and the lid beforehand in order to subtract those values from the experimental data. Of course, if you label the vessels, this also should be done prior to weighting when the crucible is cooled down to the room temperature.

How does the water vapor and CO2 cause the need to weigh the crucible with the lid closed? Does it not work if  you dont measure it with the lid?

Baloney. The crucible lid is not air tight.

@ChemPsycho I added some more details. If you want to make a good use of analytical balance, you got to weight air-sensitive substances with the lid closed.

@MaxW It's not airtight, but it a) minimizes the contact of the substance with the atmosphere of the entire room; b) develops a habit to properly handle air-sensitive substances. I guess this is a teaching lab, where stuff get shattered quite often, and crucibles are generally less expensive than weighing bottles.

Why would u want this to be air tight. Oxygen has to go in to mix with magnesium

@ChemPsycho When you weigh substances, you do not want stuff to react with anything.

What would it react to if there was no lid? co2?

@ChemPsycho your question is unclear. The way I interpret it is that you are trying to weigh the MgO product which you already obtained from a reaction previously. However, your comments here make no sense in that context, so I guess I was wrong? The problem is that nobody can accurately guess at what you mean by "my magnesium oxide lab", as nobody has your lab book with them. Would you care to elaborate by editing your question?

Edited my question to make it more answerable.

@ChemPsycho I edited the answer a bit to address new edit; however, in the nutshell it's still the same idea.

Im in a 10th grade honors chem class. Don't understand u

@ChemPsycho What exactly you don't understand?

I still dont understand why u need to cover the crucible when weighing it before you put in the magnesium ribbon, then after you put in the magnesium ribbon and weigh again, then weight again of magnesium oxide after heating

@ChemPsycho Again, you want the masses of pure substances only. To prevent the unwanted side reactions, you limit the contact with the atmosphere by putting the substance in a closed vessel. What the side processes are is quite explicitly explained in the answer. If you still have no idea what's going on, you might want to do some additional readings on basic chemistry, or talk to you technical assistant.

Again baloney. (1) Weight(crucible+lid+magnesium) -Weight(crucible+lid) = Weight(magnesium) (2) Weight(crucible+lid+magnesium oxide) -Weight(crucible+lid+magnesium) = Weight(oxygen) // As I said, you need the lid to contain the magnesium oxide within the crucible and lid. Without the lid you'd lose some of the magnesium oxide.


text:
Naming Convention: 2,4-dichlorophenylacetic acid

Why is this the correct name for this compound? I understand the 2,4-dichloro part, but...

Why is it a phenol if there is no hydroxyl group directly bonded to the  hydrocarbon ring?
Why is it an acetic acid if it has a CH2COOH instead of a CH3COOH branch on it?

For reasons passing understanding, a cyclic, aromatic, $\ce{C6H5}$ radical fragment (so a benzene minus one $\ce{H}$) is called "phenyl" for the purpose of nomenclature.
I doubt that "2,4-dichlorophenylacetic acid" is a truly IUPAC conforming name, so I'll outline the evolution of this name instead. First, there is acetic acid. You substitute one of the methyl $\ce{H}$s by phenyl: phenylacetic acid (tough to get nowadays, thanks to meth cookers). Now you substitute chlorine atoms for $\ce{H}$s on phenyl, voila.
Note that substitution here is conceptual, not chemical - one would synthesize this very differently.

I only had a brief flip through the Blue Book, but if I am not mistaken the PIN for this compound is probably (2,4-dichlorophenyl)acetic acid, as parentheses are required around compound substituents.


text:
how can i synthesis unsaturated ester via wittig reaction?
What is the best pocedure to synthesis ethyl cinnamate using benzaldehyde and (carbethoxymethylene)triphenylphosphorane (wittig reagent) and chloroform as solvent via wittig reaction?

Why are you using chloroform as solvent? It precludes the use of strong bases.

A good system for this is using DBU as the base in refluxing chloroform.  Bull. Chem. Soc. Japan (2003) 76 1675


text:
Why can’t organozinc reagent react with alkylalkoxysilane？
we know that Grignard reagents can react with alylalkoxysilane ,such as：
 RMgCl + MeSi(OMe)3 → MeRSi(OMe)2 + MeOMgCl  (R=phenyl & benzyl)。
But the organozinc reagent such as BnMgBr can’t react with alkylalkoxysilane under the same conditions，even if  the additive TMEDA（N,N,N',N'-tetramethyl-ethane-1,2-diamine）  or LiCl was used as catalyst 。

Compare electronegativities of magnesium and zinc and carbon. The difference for magnesium and carbon is $\pu{1.24}$, while for zinc it is $\pu{0.9}$ in Pauling scale. Also, the energy neccessary to oxidize $\ce{M^+}$ to $\ce{M^2+}$ (difference between second and first ionization energy) is about $\pu{715 kJ/mol}$ for magnesium compared to about $\pu{825 kJ/mol}$ for zinc. These informations both tell you that there is much larger negative charge (as in partial charge) on the carbon atom in organomagnesium compounds than in organozinc compounds. Magnesium is also more prone to leaving as a $\ce{MgX^+}$ moiety than zinc as $\ce{ZnX^+}$. In other words, the $\ce{C-Mg}$ bond breaks much more easily than $\ce{C-Zn}$ bond.
In conclusion, magnesium has lower electronegativity (stronger metallic character if you wish) than zinc and that makes organomagnesium compounds much more reactive as nucleophiles than organozinc compounds.

Thank you for your warm reply!
I assume that TMEDA plays a catalytic role. TMEDA was added to form a complex with Zn to increase the activity of the BnZnBr. But there is no effect 。
.
What is the way to make the reaction going smoothly?


text:
What makes dimethyl compounds so extreme?
I've noticed that dimethyl compounds tend to have extreme reactive, including bioreactive, characteristics.
Am I being misled by anecdotal selection bias, or is this generally true.  And if generally true, are there clear theoretical explanations as to why?
Examples that come to mind:

Dimethyl sulfoxide (DMSO) readily penetrates skin, and can carry lots of things across the skin barrier along with itself.
Miniscule amounts of dimethyl mercury will not only penetrate skin and many common PPE but are also irreversibly lethal.
Many dimethyl-metals are pyrophoric (i.e., they auto-ignite in a standard atmosphere).

I have a feeling that it may be partly an anecdotal selection basis because I don't think the properties you have mentioned are unique to dimethyl compounds.

This website may provide some useful information regarding the point on absorption of chemicals through the skin: https://www.ursamajorvt.com/blogs/the-blog-cabin/37974081-how-do-toxics-enter-our-bodies

This is because of two trivial facts, one from chemistry and another from math: (1) methyl is the smallest organic radical, and (2) it is not uncommon for a function to reach an extremum at the border of its domain.
Arguably, this is not even true in any significant way. Not sure about DMSO, but I firmly believe that organic compounds of mercury are all about as deadly as dimethylmercury. Well, those with bigger organic radicals might be slightly less dangerous because they are not quite as good at penetrating the cell barriers, or lower solubility, or because they contain less mercury per gram. The maximum is probably reached at methyl; but the maximum is not sharp.
All that applies to pyrophoric metalorganics as well.
So it goes.


text:
How would I use significant figures to find the answer of these dimensional analysis questions?
In chemistry class, I have been given a set of dimensional analysis questions to answer. I'm not sure if the units factor into whether or not significant figures will need to be applied. I know that when multiplying and dividing, the number in the equation with the least significant figures determines the number of significant figures in the problem. However, I'm confused how to use significant figures to answer these questions.

Question 1: At $\pu{4{^\circ}C}$, water has a density of exactly $\pu{1 g/mL}$. What mass would one gallon of water have at $\pu{4{^\circ}C}$? (There are roughly 3.78541 liters in a gallon.)

Here is my work:
$$
\begin{align}
&1 \frac{\pu{g}}{\pu{mL}} * \frac{\pu{1000 mL}}{\pu{1 L}} * \frac{\pu{3.78541 L}}{\pu{1 gal}} * \pu{1 gallon} \\
&= \frac{\pu{1000 g}}{\pu{1 L}} * \frac{\pu{3.78541 L}}{\pu{1 gal}} * \pu{1 gallon} \\
&= \pu{3785.41 g}
\end{align}
$$
I'm not sure if I'm supposed to round 3785.41 to the one sig fig (making it $\pu{4000 g}$), five sig figs (making it 3785.4), or just leave my answer as it is.

Question 2: The currency used in much of Europe is called the "euro." If one euro is worth 1.2951 U.S. dollars, how much would one million U.S. dollars be in euros?

Here is my work:
$$
\begin{align}
&\$1,000,000 * \frac{1 €}{1.2951} \\
&= 772141.1474 \\
&= \pu{700000 €}
\end{align}
$$
In this question, I assumed that either 1 or 1,000,000 determines the significant figures which is why I rounded the number to the first significant figure. I don't think the 1.2951 determines the significant figures because it contains the most significant figures.

Question 3: How many seconds are in 1.0000 weeks?

Here is my work:
$$
\begin{align}
&\pu{1.000 weeks} * \frac{\pu{7 days}}{\pu{1 week}} * \frac{\pu{24 hours}}{\pu{day}} * \frac{\pu{60 min}}{\pu{1 hour}} * \frac{\pu{60 seconds}}{\pu{1 minute}} \\
&= \pu{604,800} \\
&= \pu{600,000 seconds}
\end{align}
$$
In this question, I believed that there would be only one significant figure due to the unit of days (7) having only one significant figure. Thus, the answer would be rounded to the first significant figure making the final answer 600,000.

Do not go into banking unless you want a vacation in Leavenworth. If I write you a check for 1M dollars and each dollar is 1/1.2951Euros How many Euros would you want? Significant figures are used only to accommodate uncertainty in Measurements. Those questions are somewhat inappropriate since they involve definitions not measurements. I love the rough definition of a gallon. So your job is to identify the measurement that determines the significant figures.  BTW a week is a week there can't be 1.00000weeks. How would 1.1 weeks be measured?

Question 1
It is reasonable to assume the 4°C, and the density of water are exact. The problem specifies that there 3.78541 liters in a gallon, which is six significant figures. So the answer should have six significant figures.
Question 2
The million dollars is an exact number of dollars, but the conversion rate is only specified to five significant figures. So the number of euros should be expressed to 5 significant
figures.
Question 3
The number of weeks is given as 1.000, so the value has 4 significant figures. There are exactly 7 days in a week, exactly 24 hours in a day, exactly 60 minutes in an hour, and exactly 60 seconds in a minute. So the number of seconds has to be expressed to 4 significant figures.

Adding to what @MaxW has written, as a check it might be helpful to think about the errors in each of your cases.
In (1) the uncertainty in the number of litres is $\pm 0.000005 $ (just less than the number of places quoted assuming that the number given is subject to random error) and use this to determine the uncertainty in your final answer.
In (2) there can be no uncertainty in the exchange rate to euros as it is not subject to experimental error so the number could be argued to be 1.2951000$\cdots$. And so it could also be argued that any number of decimal places is ok, except that the currency is only divided into 1/100 of a dollar so this ultimately limits number of decimal places. (If you consider this to be too contrived deal with it as in (1).)
In case (3) this is essentially the same as (1) but is somewhat artificial in that we must assume that the week has an error of 0.00005 which gives an uncertainty of about 30 secs in the answer.

I agree that using significant figures is a rudimentary method of error propagation. // I think you've over analyzed (2). It is a toy problem.  // You're just wrong that the week has an error in the number of days of 0.00005. The error of 0.00005 is in the number of weeks.

@MaxW, possibly over egged it and yes my error, a typo .I have corrected it.

In (3) the error is 0.005%. For 604,800 seconds that is about 30 seconds, not 3.


text:
In biochemistry, when do rate constant units matter in calculating free energy changes?
In my experience, it is always been the ratio of products to reactants that mattered, so as long as your units are consistent; what they are doesn't matter. However, an in-class problem today required that the units be changed to molar (M) from millimolar (mM). Can you summarize when/why this matters in some cases, but not in others?

$K$ should be a unitless value. Otherwise, you'd get a nonsensical unit when you try to take the log of the unit. Each component of the product $K$ is usually an activity, which is frequently similar in value to the molar concentration, or a fugacity, which is frequently similar in value to the partial pressure of a gas.

IUPAC permits the "equilibrium constant" to have units https://goldbook.iupac.org/html/E/E02177.html , but not the "standard equilibrium constant"  https://goldbook.iupac.org/html/S/S05915.html

Good clarification to have. Though converting to a free energy with units is ¯\(ツ)/¯

If the equilibrium constant is defined as a ratio of quantities having units (concentrations, partial pressures, molalities), then unless the sum of the exponents in the numerator happens to equal the sum of the exponents in the denominator, the units matter.


text:
Heat capacity determination
Since heat capacities are determined in a calorimeter taking advantage of the known heat capacity of water, how was the heat capacity of water originally determined?

https://physics.stackexchange.com/questions/297887/how-was-the-specific-heat-capacity-of-water-determined-experimentally

Actually used calorimetric tools like DSC calorimeters have nothing to do with heat capacity of water.

It looks to me like DSC calorimeters require a reference sample with known heat capacity.  Whether the reference is water or some other substance, the problem is the same.

The heat capacity of water was initially determined by doing mechanical work (moving a turbine) in the water until the temperature rose.
They then made the equivalency of mechanical and thermal energy from that.

Thanks.  That makes sense.  But I imagine that those early determinations weren't terribly precise.  So I'm guessing that heat capacities were gradually refined in some way that didn't depend solely on a well-defined reference heat capacity such as water or any other.  Perhaps the Dulong-Petit law was an early subsequent step.  Likely there were many refinements along the way.


text:
1H NMR impurity peaks don't align across samples
I have a series of four 1H NMR spectra for a polymer I'm synthesizing under different reaction conditions. However, the acetone and chloroform peaks are shifted relative to each other across the samples - see the attached images, where if the spectra are calibrated to the chloroform peak, the acetone peaks do not overlap.
I'm relatively a novice at NMR, so I apologize if the answer is obvious, but what could be causing something like this? Is it a common observation?

Can your polymer have some interaction with acetone? Through hydrogen bonding or so?

Looks like your peaks at 7.2 are moving also. The acetone peaks appear to be  quite different concentrations - are these plotted to the same y-scale? The top acetone trace looks to be a much larger integral than the bottom trace. Concentration influences chemical shift significantly.

Could it just be that the NMR did not get recalibrated to the standard and you measured these samples on different days?

That is why TMS is (was) recommended

@RaoulKessels - yes, there are hydroxyl groups on the end of the polymer.

@long - these are not on the same y-scale. The acetone is also from washing the tubes, so I wouldn't expect the concentrations to be well-controlled. I've since re-done these spectra with cleaner tubes and the acetone peaks are gone, so this isn't of much concern beyond calibrating the spectrum. The CHCl3 peak is from the solvent, which is CDCl3.

@jheindel these were all measured consecutively. If there was a 'hardware' calibration error, wouldn't we expect the peaks to shift the same amount, such that using the CHCl3 peak as a reference would also align the acetone peaks?


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Lead-Acid Battery: Standard Electrode Potential
I'm making a lead-acid battery for an investigation. Since I only had sulfuric acid and lead electrodes, I decided to firstly run 2 lead electrodes through electrolysis in a sulfuric acid electrolyte. This would create the following half equations:
$$\ce{4H+ + 4e- -> 2H2}$$
$$\ce{Pb +2H2O->PbO2 +4H+ +4e-}$$
At this stage, I still need to perform some theoretical calculations of the required potential to run this electrolysis. Obviously, the standard reduction potential of the hydrogen ion is 0V. But I can't find the potential for the 2nd reaction. Could someone provide me with a source for it?

Your reactions are wrong.
$$\ce{Pb(s) + HSO^−_4(aq) -> PbSO4(s) + H+(aq) + 2e−}$$
$$\ce{PbO2(s) + HSO^−4(aq) + 3H+(aq) + 2e− -> PbSO_4(s) + 2H2O(l)}$$
Lead acid batteries are fairly reversible, hence their use as rechargable batteries. Anything that produces a gas, like in your reduction reaction, is unlikely to be reversible.

You can find the electrode potentials for a lot of reactions on Wikipedia:
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)
I also think your second reaction isn't entirely correct. Lead atoms don't react with water to create lead oxide. Instead it's lead and sulfate ions reacting with water to create lead oxide and sulfuric acid.