text:
What is the definition of a "Simple enzyme" exactly? Is it an enzyme that does not require a co-factor?
I'm having a hard time making a distinction between simple and complex enzymes. the definitions I've found so far all say this:

Simple enzymes are enzymes that are only made out of protein, and complex enzymes are ones that, in addition to the protein portion, have a non-protein portion as well, which is called a co-enzyme/prosthetic group (I've seen both terms being used)

Does this mean that any protein that utilizes a co-factor, even if that co-factor is simply a co-enzyme or an activating metal, i.e., not an actual part of the enzyme's structure, is considered a complex enzyme? Or are the two terms restricted to the structure of the enzyme (i.e., whether they have prosthetic groups, be they organic or inorganic, or not)?

Having searched for the concept, it seems some people refer to an "enzyme-substrate complex" as a "complex enzyme". Your definition is something I haven't heard before. Maybe you can provide more context on where you first encountered the definition.

The concept's explained in our college's lectures. I'm not sure what references the lectures are based on though.

What is the definition of a “Simple enzyme” exactly?

This is not a term that is commonly used. There are commonly used terms for an enzyme lacking its cofactor (apoenzyme) and having its cofactor (holoenzyme). Also, the term "enzyme complex" is used for the enzyme bound (non-covalently) to other molecules.

Does this mean that any protein that utilizes a co-factor, even if that co-factor is simply a co-enzyme or an activating metal, i.e., not an actual part of the enzyme's structure, is considered a complex enzyme? Or are the two terms restricted to the structure of the enzyme (i.e., whether they have prosthetic groups, be they organic or inorganic, or not)?

Yes, just apply the definition you were given. If the metal/co-enzyme/prosthetic factor is absent, the enzyme does not work.


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Why do phase changes occur?
So I was doing a Gizmo for my Science class and I came across this question: "Why do phase changes occur?" I could not find anything about it in my resources. Does anybody here know the answer?

They occur because that's just how physics works. There isn't anything profound going on. Let's take water as an example. Water boils because enough energy has been introduced into the system such that molecules are able to break away from the interactions keeping them in the liquid state.

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Radical addition and radical cleavage of cyclohexene and cyclohexane

Using reaction mechanism, why radical bromination of cyclohexene is faster than cyclohexane? Draw the stereochemistry of most stable product(s) of both cyclohexene and cyclohexane.

Here are what I get so far

I'm not sure about the kinetics of the reaction but I worry the kinetics vs thermodynamics factor (and I can't find any reaction coordinate that can show energy of possible radical position in cyclohexene and cyclohexane in one diagram).
Radical bromination of cyclohexene can undergo two competing reactions, radical addition and radical cleavage then followed by SN2 mechanism. I'm not sure if cyclohexane-bromonium ion can be formed in radical addition.
Drawing stereochemistry, should I prove it using chair conformation for both cyclohexene or cyclohexane to show the most stable product? I really don't know how to show the bromonium ion on that chair conformation and how SN2 attacks the ion (I thought it should be attack on backside but I can't prove why backside). Maybe some people will just throw "check this reference" but how I even can connect the conformation to resonance-stabilized cyclohexene product.

Can someone help me to connect the dots and what I miss from interpreting the question? Thank you.

How much does a double bond stabilise a radical in the allyl position?

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General Oxidant - Reductant Battery
In general, is it true that if I connect an aqueous oxidant to an aqueous reductant via a wire then a current flows along the wire? In particular, does current flow in the following scenario?
Imagine putting one end of a wire in hydrogen peroxide (a common oxidizer) and the other end inside water containing formic acid or pure carbon (reductants).
I understand that standard Energizer alkaline batteries are composed of separated $\ce{Zn}$ and $\ce{MnO2}$. Connecting the terminals allows electrons to flow from the $\ce{Zn}$ to the $\ce{MnO2}$, oxidizing/reducing them to $\ce{ZnO}$ and $\ce{MnOOH}$.

is it true that if I connect an aqueous oxidant to an aqueous
  reductant via a wire, will a current flow along the wire? In
  particular, does current flow in the following scenario?

No true. Nothing will happen as the circuit is incomplete. The oxidant or the reductant species have to see each other "face to face". Guess what is the most powerful reducing and oxidizing agent in the world? It is the cathode or the anode respectively. Consider an electrolytic cell which is calm and unstirred. The anode and the cathode are dipping in the solution. The moment you turn on the current, only the species which is next to the electrode gets oxidized or reduced. Nothing happens to the bulk right away, so physical promixity is important in electrochemistry.


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Organometallic Synthesis: Cp2Ti(I)Me from TiCl4
I've been working on this synthesis question for a couple days but I'm a little stuck. The question is:

Using only $\ce{TiCl_{4}}$, $\ce{NaCp}$, $\ce{PhLi}$, $\ce{MeI}$, and $\ce{EtMgBr}$, synthesize $\ce{Cp_{2}Ti(I)Me}$. (should take 4 to 6 steps)

But it seems to me the synthesis can be done in 2 steps like so:
$\ce{TiCl_{4} + 2NaCp -> Cp_{2}TiCl_{2}}$
$\ce{Cp_{2}TiCl_{2} + MeI -> Cp_{2}Ti(I)Me}$
Which doesn't match the question's "4-6 steps" note, so I'm not sure what I'm missing.

You need a reductive step to go from Ti(IV) to Ti(III)

@waylander Cp2Ti(Me)I is titanium IV.

My bad, not enough coffee


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Chemical kinetics

Consider a reversible reaction $\ce{A}$ converts to $\ce{B}$ and $\ce{B}$ converts to $\ce{A}$ with forward and backward rate constants $k_\mathrm{fwd} = k_\mathrm{rev }= \pu{1 s-1}.$ Suppose we start with a 1 molar solution of $\ce{A}$. How long will the concentration of $\ce{A}$ take to reach 0.75 molar?

Here's what I have done:
$$\ce{A <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] B}$$
$$k_\mathrm{fwd} = k_\mathrm{rev} = \pu{1 s-1}$$
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k_\mathrm{fwd}[\ce{A}]^n - k_\mathrm{rev}[\ce{B}]^m$$
For the first order reaction $n = m =1$:
$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= k_\mathrm{rev}[\ce{B}] - k_\mathrm{fwd}[\ce{A}] \\
  &= k_\mathrm{rev}([\ce{A}]_0 - [\ce{A}]) - k_\mathrm{fwd}[\ce{A}] &\qquad \{[\ce{B}] = [\ce{A}]_0 - [\ce{A}]\} \\
  &= k_\mathrm{rev}[\ce{A}]_0 - k_\mathrm{rev}[\ce{A}] - k_\mathrm{fwd}[\ce{A}] \\
  &= k_\mathrm{rev}[\ce{A}]_0 - [\ce{A}](k_\mathrm{rev} + k_\mathrm{fwd}) \\
  &= [\ce{A}]_0 -2[\ce{A}] &\qquad \{k_\mathrm{rev} = k_\mathrm{fwd} = 1\}
\end{align}
$$
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = [\ce{A}]_0 -2[\ce{A}]$$
$$\int_{[\ce{A}]_0}^{[\ce{A}]_t}\frac{\mathrm d[\ce{A}]}{[\ce{A}]_0 - 2[\ce{A}]} = \int_0^t\mathrm dt$$
$$\frac 1 2 \ln{\left(\frac{0.5 - 0.75}{0.5 - 1}\right)} = t$$
$$t = \pu{-0.35 s}$$
My answer is coming in negative. I am a biology student and not good at calculus. Is something wrong with the math or the whole process is wrong?

You have made an error in the final integration:
$$\begin{align}
t &= \int_{[\ce{A}]_0}^{[\ce{A}]_t}\frac{\mathrm d[\ce{A}]}{[\ce{A}]_0 - 2[\ce{A}]} \\
&= \frac{-1}{2} {\ln|{A_{0}-2[A]}|_{A_0}}^{A_t} \\
&= \frac{-1}{2} \ln\left|\frac{1-2\cdot 0.75}{1-2\cdot 1}\right| \\
&= \frac{-1}{2}\ln(0.5) \\
&= \ln(2)/2 \\
&= \pu{0.35 s}
\end{align}$$

How do you get -1/2

Make a substitution say $x=[A]_0 - 2[A], dx = -2d[A]$ and change the limits of integration accordingly. In general, $\int \frac{dx}{a+bx} = \frac{1}{b} \ln|a+bx| + C$


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Molar ratios not leading to complete reaction in Schweizer's reagent production
In my home lab, I've reacted $\ce{Cu(OH)2}$ with aqueous ammonia to produce Schweizer's reagent.  However, I'm surprised by the amount of ammonia required to react with this amount of $\ce{Cu(OH)2}$. Have I mistaken some calculation or formula?
Background/procedures:
8.89 g of copper (II) hydroxide was prepared by combining 24.98 g of $\ce{CuSO4 (aq)}$  with excess aqueous ammonia, to which was added 8.01 g $\ce{NaOH (aq)}$, precipitating a blue-green solid, which was filtered and allowed to dry, producing a blue-green powder with no visible hint of black copper oxide. This is believed to be reasonably pure copper (II) hydroxide.
This 8.89 g of copper (II) hydroxide was combined with aqueous ammonia (assayed at "28% to 30%"), gradually, in amounts of 30 mL, 30 mL, 70 mL, and 70 mL, a total of 200 mL. Also, 80 mL of distilled water was added, but I assume this is merely diluting the reagent and not affecting the reaction.
At each addition of aqueous ammonia, more of the copper hydroxide reacted to form a clear, deep blue liquid, consistent with the expected appearance of Schweizer's reagent. However, a sizable amount of green-blue solid remains:

Assumptions and calculations:

8.89 g $\ce{Cu(OH)2}$, with a molar mass of 97.56 g/mol, equals 0.0911 mol $\ce{Cu(OH)2}$.
200 mL of "28-30%" aqueous ammonia, at roughly 14 mol/L, equals 2.8 mol $\ce{NH3}$.
Balanced equation for this reaction is $\ce{Cu(OH)2 + 4 NH3 -> [Cu(NH3)4](OH)2}$ ; this indicates a ratio of 1:4 for copper hydroxide to ammonia.
However, my ratio so far is closer to 1:28, and it's still not enough.

Question:
If my calculations are correct, I've added about 7 times the amount of ammonia needed to fully react with this amount of copper hydroxide, but it appears not all of it has reacted. Where have I miscalculated?

I remember notes the proper procedure is precipitation Cu(OH)2 by NaOH, decanting and dissolving it in ammonia. There was warning the S. reagent prepared by other way does not dissolve cellulose.


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How does resonance fail in approximating chemical structures?
In the book "Concise Inorganic Chemistry" by Prof. JD Lee, it says here:

These contributing structures do not actually exist. The $\ce{CO3^2-}$ does not consist of a mixture of these structures, nor is there an equilibrium between them. The true structure is somewhere in between and is called a resonance hybrid. Resonance was widely accepted in the 1950s but is now regarded at best as clumsy and inadequate, and at worst as misleading or wrong!

I want to know where exactly does resonance fail in approximating the structures?

Related: What is resonance, and are resonance structures real?

The fact that the resonant structure aren't real at most clean up a misunderstanding of the theory not really why it might not be a very good model (which is case is, there is no need to overemphasises pros and cons). MO theory might be conceptually more elegant and correct by principle. However computational chemists still working on the wavefunctions of hydrogenoid systems such as HH+  !

Resonance structures are one model that explains bonding, but not a very good one
Resonance structures are widely misunderstood. Individual resonance structures are not real compounds that exist, but an approximation that assumes all structures are made from bonds consisting of two electrons. The true structure of a compound is thought to be an average mixture of the various possible resonance structures consisting of simple bonds.
The archetypal example is benzene. If we model benzene as a structure consisting of "normal" 2-electron bonds, we get a ring consisting of alternating single and double bonds. This contradicts the observed structure where all the bonds are the same (and are somewhere between the character of single and double carbon-carbon bonds). The resonance model of benzene consists of two structures with the normal double and single bonds swapped, giving an average overall structure where all the bonds are an average of single and double bonds.
For structures with more complexity than benzene, the resonance structures are more complex and frequently don't make equal contributions, nor do they do a good job of predicting the real structures. Molecular orbital based approaches (which recognise the difference between pi-bonds and sigma bonds and do a much better job of recognising the fractional character of many real bonds) are far better.
That is why resonance structures "fail". They make drawing simple structures easier, but they don't do a good job of predicting real structures.


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When pH > pKa and when pH < pKa and the effectiveness of buffer solutions
I understand that when $\mathrm{pH}= \mathrm{p}K_\mathrm{a}$, the buffer solution will be at its maximum capacity, and there will be equal concentrations of the acid/conjugate acid and the base/conjugate base.
However, when $\mathrm{pH} > \mathrm{p}K_\mathrm{a}$, why is it that $\ce{[A^-] > [HA]}$? Shouldn't it be the other what around since the Henderson-Hasselbalch equation is:
$$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A^-]}}{\ce{[HA]}}$$
So $\frac{\ce{[A^-]}}{\ce{[HA]}}$ should be less than $0$, so $\ce{[HA]}$ must be greater.
Why is it that when $\mathrm{pH} > \mathrm{p}K_\mathrm{a}$, the concentration of the conjugate base ($\ce{[A-]}$) is greater than the concentration of the acid ($\ce{[HA]}$)?

Have a look at the method described in this answer https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566

It's pKa, not kPa.

...as pKa = - log Ka ...

To calculate $\mathrm{pH}$ of buffer solutions, Henderson-Hasselbalch equation is a very good assert:
$$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A^-]}}{\ce{[HA]}}  \tag{1}$$
Also, you can rewrite the equation as:
$$\mathrm{pH} - \mathrm{p}K_\mathrm{a} = \log \frac{\ce{[A^-]}}{\ce{[HA]}} \tag{2}$$
Accordingly, as you correctly suggested, for monobasic acid, when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, the equation $(2)$ becomes:
$$\mathrm{pH} - \mathrm{p}K_\mathrm{a} = \log \frac{\ce{[A^-]}}{\ce{[HA]}} = 0 = \log 1$$
Thus, $\ce{[A^-]} = \ce{[HA]}$. Now, look at the equation $(2)$ again for the condition of $\mathrm{pH} > \mathrm{p}K_\mathrm{a}$: The term $\left(\mathrm{pH} - \mathrm{p}K_\mathrm{a}\right)$ should be positive, and hence $\log \frac{\ce{[A^-]}}{\ce{[HA]}} > 0$. Mathematically, the only way this could happen is if and only if $\ce{[A^-]} > \ce{[HA]}$.
Similarly, if the condition is $\mathrm{pH} < \mathrm{p}K_\mathrm{a}$, the term $\left(\mathrm{pH} - \mathrm{p}K_\mathrm{a}\right)$ should be negative, and hence $\log \frac{\ce{[A^-]}}{\ce{[HA]}} < 0$. Again, in mathematics, the only way this could happen is if and only if $\ce{[A^-]} < \ce{[HA]}$.


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Chain with double bond or triple bond

In this hydrocarbon there can be two main chains: one with two double bonds and one with a double and a triple bond
Both main chains contain seven carbon atoms.
Which will be considered correct and why?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the first relevant criterion to be considered in choosing a principal chain is the length of the chain.
When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached:

P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)

Since both possible chains in this case have two multiple bonds, Rule (a) is not enough to make a choice.
According to Rule (b), the principal chain is the hepta-1,6-diene because it has the greater number of double bonds.
Therefore, the preferred IUPAC name (PIN) is 4-(prop-2-yn-1-yl)hepta-1,6-diene.


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Diastereoselectivity in the zinc-mediated Barbier allylation
Penumati & Kommu (2014) report the stereoselective synthesis of (+)-$\alpha$-conhydrine from R-(+)-glyceraldehyde. In their synthesis, the imine 6 underwent a zinc-mediated Barbier allylation to produce secondary amine 7.

They explained the preferential formation of the major product 7a over 7b via the following Felkin-Ahn model, claiming that the anionic allyl nucleophile approaches the less hindered face. However, it appears to me that there is no significant difference in the steric hindrance posed by each face of the imine group. Hence, I am skeptical if application of the Felkin-Ahn model here is appropriate and whether it does give the correct predicted major product. I thought of using the chelation model, since certainly there is a possibility $\ce {Zn^2+}$ can be involved, but I realised that it doesn't seem to be appropriate as well, or at least that is what appears to be the case for me.

Reference
Penumati, N. R.; Kommu, N. Stereoselective Synthesis of (+)-$\alpha$-Conhydrine from R-(+)-Glyceraldehyde. Organic Chemistry International, 2014. doi: 10.1155/2014/982716

The Felkin-Anh model isn't about which "face" is less hindered (true, the nucleophile always attacks opposite $R_L$, but you can just rotate the bond until $R_L$ is on the other side of the carbonyl). It's the positions of $R_M$ and $R_S$ that dictate the stereoselectivity, and when the nucleophile attacks via the Burgi-Dunitz angle, you want $R_S$ to be in its way and not $R_M$. The chelation model would give you the wrong diastereomer. FWIW, that journal doesn't look particularly reliable...

@orthocresol Actually, I don't even think that the above diagram depicts the Felkin-Ahn model... It does not match my understanding, i.e. what I've read from Claydens. Indeed, it is rather dubious... How would you rationalise such stereoselectivity then?

Polar Felkin-Anh might be appropriate here: the C-O bond takes the place of the 'large' substituent, which would give you the correct stereochemistry. That would be my best guess, but no guarantees...

@orthocresol I have tried the polar Felkin-Ahn approach and it does not seem to predict a convincing result that is consistent with the above stereochemistry. Perhaps, you could illustrate what you meant in an answer.

@orthocresol I have recently read about the evolution of stereochemical models used to predict the stereochemical outcomes of particular reactions. It seems to me that no one model suggested can be universally applied... Do you have any thoughts as to which is the closest model to one that can be universally applied? (I would consider this so-called "polar Felkin-Ahn" and the Felkin-Ahn based on sterics as the same model)


text:
Why does the Heisenberg uncertainty principle work in an atom?
Where is my logic wrong?
An electron can only assume quantized energies in an atom.
If an electron is localized in space in a hydrogen atom, its radius is known. If its radius is known, its electrostatic potential energy is known. Then, if the electron assumes quantized energies, its kinetic energies are quantized by $E_\mathrm{total} - E_\mathrm{potential}$ for all quantized possible total energies of the electron. This gives us quantized kinetic energy and thus momentum, giving us quantized frequencies. If you sum all of the waves of those frequencies, will it give the localized position wave-function? They can't right due to the fact that they're quantized.
This is probably very wrong; I just don't understand how.

This doesn't work because you don't know the radius of the electron about the nucleus. When you see values of atomic radii, these are average values inferred from the distance between atoms in molecules or the spacing of ions in crystals. Electrons have a probability distribution of their position and this distribution doesn't have a max cutoff distance. Technically, the electron could be arbitrarily far away from the nucleus, though the probability for any significant distance is vanishingly small.

You are referring to the Bohr model of the atom, which was an early model that ultimately proved somewhat insufficient to describe reality.

Recall that the uncertainty principle only applies to knowing at the same instant in time the position and momentum. So it is possible to measure the energy  (momentum squared/mass) exactly, but therefore, not where the electron is.

@Zhe "somewhat insufficient" means it is unable to reproduce anything quantum in nature, including Heisenberg uncertainty principle

@Greg Sorry, is that a suggestion for an edit?

If the Heisenberg Ubcertainty Principle didn't work for atoms, they and everything made of them, including us, would collapse into black holes.  And God didn't want that.

@Zhe Your comment, it is up to you. In the given context that model is completely incorrect from a conceptual point of view (due to its classical nature), which has lead the OP to incorrect assumptions and incorrect physical picture.

Electrons don't have a well-defined orbital radius even in a hydrogen atom
An electron "orbiting" in a hydrogen atom does not have a well-defined position or even just a well-defined radius just because it has a well-defined energy.
The orbital is more like a 3D probability density function covering a cloudy space of a defined shape (spherical in the case of the lowest energy orbital but not of a single radius and extending with a finite probability from the nucleus to a vaguely defined boundary we often regard as the radius of the atom).
Different orbitals have different, increasingly complex, shapes defined by solutions of the Schrodinger equation. Heisenberg's uncertainly relationship limits our ability to define the location and momentum of the electron at the same time. The fuzzy probability clouds defined by Schrodinger's solutions are the best we can do.


text:
Selection of parent chain (IUPAC nomenclature)

In the given compound, there are (apparently) two possibilities of choosing the parent (main) chain for its nomenclature in accordance with IUPAC rules:
(A) being the one containing 4 carbon atoms and the triple bond in the parent chain;
(B) being the one containing 6 carbon atoms and the hydroxyl group as a substituent in the parent chain.
Which one shall I choose while naming this compound? My book claims it to be (A), but I do not understand. (B) is clearly the longer chain and contains the hydroxyl group, which has a higher priority in nomenclature as compared to the triple bond. Where am I going wrong?

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. Apparently, you have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the aldehyde group ($\ce{-CHO}$).
The next rule (P-44.3.2) is looking at the longest chain. Therefore, the correct answer is B; the principal chain contains six carbon atoms and the hydroxyl group. Thus, the complete name is 2-ethynyl-5-hydroxyhexanal.

The next rule (P-44.4.1.1) would be looking at the multiple bond, but it is not needed since we already found the principal chain.
Note, however, that the IUPAC recommendations on this matter have been changed. In the current version, the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed.

(…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

If your book is still using old nomenclature rules, it would expect the other answer A to be correct.


text:
Ratio of uncharged to charged side chains

The $\mathrm{p}K_\mathrm{a}$ of the side chain imidazole group of histidine is $6.0.$ What is the ratio of uncharged to charged side chains at $\mathrm{pH}~7?$

Here's my attempt to the solution:
$$7 = 6 + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$
$$1 = \log\frac{[\ce{A-}]}{[\ce{HA}]}$$
$$10[\ce{HA}] = [\ce{A-}]$$
$$\underset{\text{uncharged}}{1} : \underset{\text{charged}}{10}$$
I've used the Henderson-Hasselbalch equation to the solve this, considering imidazole group to be the acid $\ce{HA}$. However, the correct answer appears to be 10:1 uncharged to charged ratio. If I use $\ce{HA^+}$ to be the acid, the answer turns out correct.
Can anyone tell me what have I done wrong? If I was wrong in assuming $\ce{HA}$ please tell me why. Let me know if some clarification is needed.

It's not acidic, but basic - it's pKa of conjugated acid, not neutral base.

What Mith said. You have the $\mathrm pK_\mathrm a$ of histidinium, not histidine. A spoken and written inacurracy that is way too common.

I think you are simply misunderstanding the "uncharged to charged" part of the question. Imidazole is a base, so it gains a proton to become charged when the pH < pKa. That's why you got the right answer when you used $\ce{HA+}$ as the acid, as you put it. In this case, unlike with most general chemistry problems of this nature, the acid is charged while the conjugate base is uncharged.
Below pH 6, imidazole exists largely as a protonated imidazolium ($\ce{HA+}$) species. Above pH 6, it is primarily in the neutral ($\ce{A}$) imidazole form. When imidazole is protonated (i.e., in acid or $\ce{HA+}$ form) it is charged. When it is deprotonated (i.e., in base or $\ce{A}$) form, it is uncharged.
So the proper expression is $$7 = 6 + \log\frac{[\ce{A}]}{[\ce{HA+}]}$$ which should lead you to the correct answer:
$$1 = \log\frac{[\ce{A}]}{[\ce{HA+}]}$$
$$10 = \frac{[\ce{A}]}{[\ce{HA+}]}$$
So, the ratio of uncharged:charged species is $\frac{[\ce{A}]}{[\ce{HA+}]}$, or 10:1.


text:
Combustion and pyrolysis of sulfides and organosulfurs
Last week there was a huge fire in french chemical factory Lubrizol, we now know the main products that burnt, to simplify the discussion I extracted a few compounds:

isobutyl and isobutene polysulfide 500 tons,
dithiophosphoric acid and corresponding zinc salt (R = isopropyl) 200 tons,
various hydrocarbons (mainly mineral oil) 2000 tons
as well as  calcium sulfonate and calcium salt of methylphenol sulfide (100 tons), amides/esters of phosphoric acid, polymers, aromatic amines & alcohols, various reaction products of $\ce{P2S5}$ with isopropanol 400 tons

What kind of sulfur compounds should we expect to be emitted from pyrolysis and combustion of such organosulfur compounds?

Note the authorities claim the smoke analysis didn't show any $\ce{H2S}$ and that the $\ce{SO2}$ level wasn't higher than everyday measurements in cities.
There is a paper on the pyrolysis of polyethylenesulfide showing cyclization leads to a bunch of products, I wonder if it is applicable here  (also I would like to know why no $\ce{H2S}$ is formed?)

feel free to add any thoughts on combustion pyrolysis principles

I'd expect $\ce{H2O, CO, CO2}$, and $\ce{SO2}$ to be created during the combustion of these compounds. I'd wager a decent amount of $\ce{SO2}$ was created, but was diluted in the atmosphere.
Regarding $\ce{H2S}$, maybe it was produced, but then reacted according to the following:
$$\ce{H2S + O2 -> SO2 + H2O}$$
Additionally, there were probably a wide variety of odd sulfur compounds, but their existence was transient as they reacted to form $\ce{SO2}$ for instance. Furthermore, the authorities may not even be looking for these more complex molecules.

I think it is contradictory with their analysis, also $SO_2$ is 2 times heavier than air even if the density is lowered by the high temperature of the fumes. In one of their analysis there is organosulfur $<10\mu g/m^3$, same for $SO_2$. I can't make it consistent with those hundred of tons of sulfur that burnt. Where should I search for references showing $SO_2$ is always formed when burning organosulfurs ?

Also I was wondering if, due to the huge amount of fuels consuming all the $O_2$ in air, it would be more pyrolysis than combustion (it burnt for 12 hours so the temperature was extreme)


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More use of pentane(+hexane+..) when compared with methane?
Methane is one of most cleaner of vehicle fuels when compared with other hydrocarbons. It can be easily and continuously produced in form of biogas in biodigestors working on farm and animal wastes. What are the reasons and/or roadblocks that are keeping us from un tapping and harnessing such fuel instead of going for fossil fuels?

It looks like Wikipedia got this well covered. Also, a "(+hexane+..)" notation looks like a title for a witch house mix.

We’re using it and it’s hard to change. 2) hexane is a liquid while methane is a gas; go figure.

It looks like you tried to completely replace your question with another one, rendering the existing answer irrelevant. Please avoid doing so, and ask a new question instead.

Using methane as fuel in vehicles needs a method to deliver it to vehicles
Methane is widely used as a fuel. It is the major component of natural gas, which is used to power domestic cookers and central heating boilers. It is also used to power gas turbines that generate electricity.
But all these application need a way to get the methane to the location where it is used. And that means pipelines. Needless to say, a car permanently connected to a gas pipe is a little inconvenient if you want to drive more than a few meters.
There are, however, many gas-powered vehicles. But they require special infrastructure to deliver the gas and special tanks to store it under pressure. The pressure tanks in cars are heavy and expensive both of which lower the attractiveness compared to diesel or petrol which can simply be poured into an unpressurised tank as liquids. These extra factors lower the net attractiveness of methane (or, for similar reasons, hydrogen).
So methane may be cleaner and (often) cheaper than liquid fuels, but the overall costs of infrastructure and storage offset that advantage for most users.

Cant we reversably adsorb methane over some porous metal surface so it can be safely transported and made available as fuel. i ve heard that hydrogen one of the cleanest fuel can be adsorbed over Palladium metal.

@user240988 In principle, yes. In practice the weight and storage density is nothing like good enough (for methane or hydrogen). Methane can be transported as a liquid but this requires extreme refrigeration and is only viable in large ships.


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If a beta particle is just an electron, what causes it to be radioactive/dangerous?
Like what the title says, my question is if a beta particle is just an electron, what causes it to be dangerous?
I understand that a beta particle is not only an electron but it is also one with a lot of energy.
I wish to understand what happens when this electron hits a living cell, the effects, and why it is dangerous.

Its energy.  Anything is dangerous given enough energy, including ordinary nitrogen and oxygen molecules.

@OscarLanzi What does Tornado Wikipedia page have to do with beta-particles?

"Anything" (not just electrons) is dangerous given enough energy, in that case air.

@QuesoPez It looks like you've already answered the question from the title, and the remaining part regarding interaction with living tissue looks like a homework question.

Sorry, but I'm a 9th grader taking chemistry. I'm learning stuff on my own, and wikipedia pages don't help because I'm not smart enough to understand what they're talking about :)

Not really the same but one can as well ask " if a book [thrown towards you] is just a bunch of paper sheets, what causes it to be a blunt body/dangerous? ". This is not to make fun of you, but rather to let you think and likely understand the analogy.

We should be able to answer questions at the level of the questioner.
A beta particle is described as an electron with a lot of energy. We generally use this term to describe electrons emitted by a nuclear transformation. Different atoms emit electrons with different energy: the energy of the electrons emitted by tritium are so low (5.7 keV) that it is difficult to observe them with a Geiger counter (they won't penetrate the walls of the counter) https://en.wikipedia.org/wiki/Common_beta_emitters , but some nuclei emit very high energy electrons: Cesium-137 emits particles with an energy of 661 keV, so high that the energetic electron can eject more electrons from matter - that is, they can ionize lots of molecules in the human body and cause major changes in DNA - and that could lead to serious effects, like cancer  https://en.wikipedia.org/wiki/Beta_particle .
For a reference, x-rays (photons, not electrons) have energies between 100 eV and 100 keV https://en.wikipedia.org/wiki/X-ray . While the energies of x-rays can be high enough to cause damage, they are used at low doses and for such a short time - just enough to be caught on film or CCD devices - such that the damage to the body is small enough to be repairable. Radioactive elements might be in contact with the body for a much longer time and have a much larger total effect.

Danger is a relative term.  In general for beta-emitters it's not dangerous to be in the same room.  Even a relatively hard beta, like P32 (around 1.7 million electron volts) will not penetrate its container. If ingested and absorbed into the body, however, the beta-emitter will be in close proximity to the other molecules of the body.  So its energy can cause disruption of those molecules.  Even a relatively soft beta, like tritium's (around 15 thousand electron volts) can cause problems if enough is ingested.  One damage mechanism is mutations.  Another is formation of poisonous compounds such as free radicals.
Nucleic acids (RNA and especially DNA) carry information regulating the functioning of the cell.  Radiation-incduced mutation of the DNA could cause the cell to behave badly, such as multiplying without control, forming a cancer.


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Confusion about hybridization terminology
I was just introduced to the concept of orbital hybridization. I believe I understand the idea behind it, but there is some accompanying terminology that prevents me from achieving a greater understanding.
In the following image:

I see that one of the electrons in the $2s$ orbital of carbon is promoted to one of the $2p$ orbitals. I start to get confused when hybridization occurs, simply because there are four $\ce{sp^3}$ hybrid orbitals, as also shown in this diagram:

I learned that the name $\ce{sp^3}$ comes from the combination of one s-orbital and three p-orbitals ($\ce{s + p + p + p}$), and this makes sense to me. What doesn't make sense is why each individual hybrid orbital is called an $\ce{sp^3}$ orbital. When the $\ce{2s}$ orbital, for example, becomes an $\ce{sp^3}$ orbital, why is it referred to by its combination with p-orbitals ($\ce{sp^3}$) when it was just an s-orbital before?
Could someone please offer a simple explanation that would help clear this up? Why is each individual hybrid orbital called an $\ce{sp^3}$ orbital? It seems like all four orbitals together should be called $\ce{sp^3}$, because they all make up the combination defined by $\ce{sp^3}$ ($\ce{s + p + p + p}$).

You didn't understand the idea behind it. As far number of orbitals is concerned, the picture you've posted is pretty clear, isn't?

@Alchimista I meant I understood the idea behind hybridization. And yes, the picture was clear. It's the basis of my question. I see that there are four orbitals. I wanted to know why.

Think of it as mixing a substance (say a glass of juice with three of water), where in the end the total amount of the substance has to be conserved. The "mixed" four orbitals have to contain what you started out with.

@Kman3 Buck Thorn's example is the right way to think about it. You go from having one s orbital and three p orbitals to four mixed orbitals that are one-quarter s-type and three-quarters p-type.

@BuckThorn Thanks for your help! So, using your example, does each $\ce{sp^3}$ orbital have $\ce{s}$ and $\ce{p}$ characteristics, just like how four glasses of the juice-water mixture would have both water and juice in them?

Yes, the orbitals are sometimes shown as "balloons" and when mixed then the balloons have intermediate shapes  with properties from all sources.

The number of hybridized orbitals is equal to the number of original orbitals. The name of the hybridized orbitals (e.g., $sp^{3}$) means just that the ratio of the contributions of $s$ to $p$ orbitals is $1:3$. We do not write $s^{\frac{1}{4}}p^{\frac{3}{4}}$.


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Selective Esterification of Amino Acid
I am going to prepare tert-butyl protected threonine. I was going for the classic esterification methodology which is acid catalyzed (sulfuric acid) esterification under reflux of tert butyl alcohol + threonine. My worry is that the -OH r group of threonine will also be esterified. My lab has very few reagents. We don't have TMSCI, silicone, amberlyst or carbodiimide. Does any of you have any suggested methods using tert butyl alcohol?
Thank you!

T-BuOH plus sulfuric will just give a mess. You need to protect the amino group first then the -OH. Here is how it is done with serine  https://link.springer.com/article/10.1007/BF02975241


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How does solidification occur?
What does it mean that "When vapor pressure of solid is greater than liquid, there is a net conversion from solid to liquid".
What exactly occurs in the process of solidification of liquids?

You can think of the amount of vapor in equilibrium with a particular condensed phase as being indicative of the relative stability of the condensed phase. A more stable condensed phase results in a lower vapor pressure.
More formally, stability of a pure substance in a particular phase is usually described in terms of its chemical potential $\mu$ in that phase. A lower (more negative) chemical potential implies a higher stability. Equilibrium between two phases means that the chemical potential in both is the same. In addition, the chemical potential of a vapor (assumed to behave like an ideal gas) can be related to the value at some reference pressure $p^\circ$ using the following equation:  $$\mu(g) = \mu^\circ + RT\log \left( \frac{p}{p^\circ} \right)$$ This means that at equilibrium with a condensed phase $$\mu(\text{cond}) = \mu^\circ + RT\log \left( \frac{p}{p^\circ} \right)$$ and that the relative chemical potential of condensed phases i and j can be related via their respective vapor pressures as $$\mu_i-\mu_j = RT\log \left( \frac{p_i}{p_j} \right)$$ This means that if the vapor pressure of phase $i$ is higher than that of $j$, the chemical potential of phase $i$ is greater than that of $j$ at that temperature and pressure of the system. And only the most stable condensed phase survives.


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When will the Lassaigne's method for detection of nitrogen and Kjeldahl's method for quantitative estimation of nitrogen fail?
Lassaigne's method:
The organic compound is fused with dry $Na$ in a fusion tube. The fused mass after extraction with water is boiled and filtered. This filtrate is called sodium extract (S.E.), which is used to detect elements other than $\ce{C}$ and $\ce{H}$ in organic compound.
If the organic compound contains nitrogen then the S.E. will form $\ce{NaCN}$. Further reactions with $\ce{FeSO4 + NaOH}$ and $\ce{FeCl3 + HCl}$ gives blue or green colour, confirming the presence of nitrogen.
This test fails in case of diazo compounds because sodium metal can not extract nitrogen from them. What about the compounds containing nitrogen in the ring or having nitro groups?
Kjeldahl's method:
Nitrogen containing organic compound is heated with conc. $\ce{H2SO4}$. $\ce{N}$ in the compound converted to $\ce{(NH4)2SO4}$. The mixture is then heated with excess of $\ce{NaOH}$. The liberated $\ce{NH3}$ gas is absorbed in excess of standard solution of $\ce{H2SO4}$. From the amount of acid reacted with $\ce{NH3}$, % of $\ce{N}$ can be calculated.
This method is not applicable to diazo, nitro and compounds having nitrogen in the ring. This is because nitrogen of these compounds does not change to ammonium sulphate upon reacting with conc. sulphuric acid.

You are right about the limitations of century old Lassaigne's method. Why are they still teaching it? Modern day CHNSO elemental analysis is done with gas chromatograpy systems. If you are quoting from a book, provide a reference.
This is an excerpt from a 1953 work Sulfur in Organic Compounds, by a Modified. Lassaigne's Method ACTA CHEMICA SCANDINAVICA 7 (1953) 1395

Objections to Lassaigne's method are: a) low boiling substances evaporate
  before reaction with the sodium, b) diazo-, azo and polynitro compounds evolve nitrogen c) some heterocyclic compounds do not react.

OK... So I am rephrasing it: "The sodium extract can not react with the nitrogen present in nitro, azo, diazo and nitrogen containing heterocyclic aromatic compounds.


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Identifying Structure based off 1H-NMR & Mass Spectroscopy Data
I have almost the whole night trying to identify this structure with no avail as there is always one part that does not fit. The determined formula for this structure was $\ce{C13H17BrO}$ from the $m/z=268$ (M-100%), $m/z=269$ (M+1-14.7%), and $m/z=270$ (M+2-98.6%).
The $\ce{^1H}$-NMR data is as follows with integration in parentheses:
a. $\pu{7.30-7.27 ppm}$: multiplet (5) - arene with 5 hydrogens
b. $\pu{3.20 ppm}$: singlet (2) - R-CH2-Br attached to tetra-substituted carbon?
c. $\pu{2.45 ppm}$: triplet (2) - R-CH2-CH2-CO-CH3
d. $\pu{2.17 ppm}$: singlet (3) - R-CH2-CO-CH3
e. $\pu{1.68 ppm}$: sextet (2) - R-CH2-CH2-CH3
f. $\pu{0.90 ppm}$: triplet (3) - R-CH2-CH3
And this is the structure I come up with:

which is the only structure that fits all the hydrogen integrations, but does not support the chemical shift at all.
I believe this is part of the molecule

because it takes care of shifts d,c,e,f. However, it is impossible to attach that structure to the remaining Arene, CH2-Br, and tetra substituted carbon. Actually spent over 6 hours on this problem and I'm losing my mind please help me.

You are just concerned with the shift of 2.45 right? I think your proposed structure does take care of that shift. The inductive effect of the carbonyl and the bromine atom may just be sufficient to cause such a deshielding effect.

c. 2.45 ppm triplet (2)- R-CH2-CH2-CO-CH3? $$$$
Did you mean R-CH2-CH2-CH3?

Did you deduce molecular formula by MS data or it is given? If so, I believe you may need  additional decimal point on parent peak and others to do so. For example, ChemDraw is given M+1 peak as 14.1%  instead of 14.7% on your data for $\ce{^{79}Br}$ containing peak.

Also, the part of molecule you are suggesting is a complete molecule. So, it can't be a part.

TanYongBoon Yeah, the shift at 2.45 doesn't really make sense according to this chart that I was given since a shift a 2.40 correlates to the CH2 attached to a carbonyl, but I too think its somewhat plausible that the 2.45 shift could be cause by deshielding thanks!. Karsten According to my proposed structure it is Ch2-ch2-ch3, but the ppm shift and splitting is more in line with the ch2 attached to a carbonyl, sorry I might have worded it strangely. Mathew I determined the formula from (M+1)/13 and 1:1 ratio of M to M+2 for 79 Bromine, but it didn't match up to the exact values!


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Ingredients for cleaning optical elements
I have recently purchased a cleaning solution for precision optics which has the following ingredients (amounts were not specified):

Ethyl Alcohol
Methylethylketon
Aqua
Diethyl Ether
Isopropyl

Typically reagent grade isopropyl alcohol alone is used for cleaning optical elements. So, does the mixture above provide any advantages over using pure isopropyl? (Possible advantages may include a faster dry time, better removal of oils/smudges, and less toxicity)
Any explanations would be greatly appreciated.

It might possibly be intended to dissolve an adherent organic film, such as that left by evaporation of plasticizers from foam packaging.
However, it also might well partially dissolve paints and cements used in the lens assembly, leaving yet harder-to-remove dirt on the lens. Unless you have a specific need for those solvents, use distilled water and/or isopropanol, according to B&H.

According to the manufacturer, the solution is meant to be non-invasive to the thin film coatings on the lenses, so it is suitable for camera lenses as well as laser optics. The performance appears to be similar to pure Isopropyl. So, I wasn't sure if the admixtures really offer any benefits...

Butanone, also known as methyl ethyl ketone (MEK) (see Wikipedia) is an effective solvent for paints and varnishes. If a bit got on a painted ring around an optical element, it would deposit a film of paint on the edges of the lens. This might not be a problem with black anodized lens rings, but why take a chance?

Thank you so much for the explanation. I see. Do you know of any advantages or disadvantages associated with the other two ingredients (namely, ether and ethyl alcohol)?

About the same as butanone, though less likely to dissolve paint or cement.


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Why does the dipole moment of molecules go from positive to negative?
In physics the dipole moment is said to go from the negative to the positive pole, check for instance https://en.wikipedia.org/wiki/Electric_dipole_moment (Elementary definition).
This makes sense to me since in an electric field (which is directed from positive to negative) the dipole moment aligns with the electric field, such that in the aforementioned definition the positive pole points towards the negative pole of the electric field and vice versa.
(also see: https://physics.stackexchange.com/questions/89157/motion-of-a-dipole-in-an-electric-field)
Now, in all representations of molecular dipole moments I could find online the dipole moment is directed from the positive to the negative pole.
Check for instance: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Dipole_Moments
or just have a look at google images for the search term 'dipole moment molecule'.
This doesn't make any sense to me, is there any good reason for this discrepancy?

You have assessed the problem correctly. Physicists and chemists use just the opposite directions. Keep in mind that both are arbitrary choices, however the physicist's convention is consistent with electrostatics. All you have to do is to be aware of this direction in the chemistry literature. I have rarely seen any student ever thinking about it which is rather sad.
If you want more details read this article Misconceptions in Sign Conventions: Flipping the Electric Dipole Moment

"Most chemists draw the direction of an electric dipole backwards. This inconsistency in sign convention is more than a trivial oversight or matter of convenience. We are introducing an incorrect convention that is central to our student’s understanding of fundamental thermodynamics; which way is up? As a student matures from general chemistry to organic through analytical, physical, inorganic, and biochemistry they are reintroduced to similar, powerful concepts that hold our discipline together. However, this reconnection of chemical principles comes with each subdiscipline’s own idiosyncrasies and a highly contextual framework. Reexamination of a central concept from the perspective of a new subdiscipline should not introduce misconceptions about that concept. When misconceptions introduced through chemical language can be avoided, we should change the way we speak."

This doesn't make any sense to me, is there any good reason for this discrepancy?

Provided you remember the convention you are using it shouldn't matter. You could argue that either way "makes sense". If you draw it the way that you suggest is logical, then you are right, the low energy configuration will see the dipole moment vector aligned along external (applied) electric field lines, pointing from negative to positive charge. In addition, if you need to draw the electric field lines generated by the dipole moment and assume it is a point dipole then the electric field lines seem to stream through the dipole vector (look again at the animated figure in the wikipedia page). This is illustrated by the following frame from the wikipedia animation:

The lines loop around the vector from the positive end to the negative end. This is a physically appealing picture and consistent with the effect of an electric dipole aligned in an external field in a low-energy configuration, which is that its own field opposes the external one.
But if you don't consider the molecule to be a point dipole (which it most decidedly isn't) then the internal field lines between positive and negative charges inside the molecule "look all wrong" (they will oppose the direction of the moment vector if drawn to point from negative to positive charges in the molecule). Again, this is illustrated by one of the frames from the wikipedia animation:

On the other hand, if you draw the moment vector "according to a chemist", then the internal field lines do stream along in the same direction as the vector, which might be helpful to visualize the relative charge distribution inside the molecule. It is also easier (imho) to think of a vector pointing from more positive to more negative values. So perhaps it is that chemists, in drawing the dipole moment, consider individual charges in the molecule, and look for a shorthand to understand this internal distribution, whereas physicists consider an abstract mathematical point dipole. Mathematically however it is a potentially serious source of confusion and mistakes to draw it the "chemists way". It does leave you wondering what they were thinking.
In answers to the linked question of which yours is a duplicate you will see the argument that the direction of the dipole vector "follows naturally" from the mathematical definition of the moment, but I don't find the argument convincing. For instance, you may write for a dipole generated by two equal but opposite charges
$$\mu=q_+(x_+ -x_-)$$
which suggests the dipole points from the negative to the positive charge, but you could just as well write
$$\mu=q_-(x_- -x_+)$$
which suggests the opposite. So what's right?
I haven't read the article cited by @M.Farooq but it probably presents a conclusive perspective on the matter.

Note a recent example of chemists doing things "the right way":
What can we say about the dipole moment of terephthalic acid?

"But if you don't consider the molecule to be a point dipole (which it most decidedly isn't) then the internal field lines between positive and negative charges inside the molecule "look all wrong" Are there any figures which show this? I don't really understand this statement tbh.

@Stikke I added a couple of frames from the wikipedia animation to illustrate my point


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The structure of exploding buckminsterfullerene-nitrous oxide compounds
Recently, looking into structures of buckminsterfullerenes, I have found several articles claiming that attaching 12, or “several” nitrous oxide molecules to the buckminsterfullerene will cause it to degrade exothermically, heating the surroundings to several thousand degrees in “picoseconds”. The degradation would produce more nitrous oxide, further carrying on the reaction.
As I remember, they didn’t have any cited sources, or give any creditable evidence, or give the structure of the nitrous oxide-buckminsterfullerene compound. So I’m wondering, where are the NOs attached on the C60?, and what is the structure of this compound?(preferably a diagram?)
I searched it on many places, but to no avail.
Thanks,
Max0815

Downvoters, mind I ask you why?

NOs are not yet attached to C60, so it is a little early to ask "where".

@IvanNeretin I don’t understand. Could you clarify a bit more?

What's to understand? In effect, you are asking what is the height of the spire on a particular building in a land far away. We know very little of that land. We are not even quite sure that the said building exists, let alone whether it has a spire.

@IvanNeretin So the buckybomb has not been built yet you mean?

Well, there were some reports of making something of which we don't even know the exact composition, much less the structure. In terms of my analogy, some people have been to that land and claim to have seen the building, but the solid evidence (like a photo) is yet to come. Then again, pretty much all organics with enough nitro groups is explosive, so that's hardly exciting news anyway.

Just look for some of the most common functionalization of fullerene, it can be addition and so on. But then look at the last comment of Ivan Neretin. "So what? " is appropriate out of a context.

@IvanNeretin https://phys.org/news/2015-03-buckyballs-bucky-bombs.html which says “To build the miniature explosives, Prezhdo and his colleagues attached 12 nitrous oxide molecules to a single Bucky-Ball and then heated it. Within picoseconds, the Bucky-Ball disintegrated—increasing temperature by thousands of degrees in a controlled explosion.
The source of the explosion's power is the breaking of powerful carbon bonds, which snap apart to bond with oxygen from the nitrous oxide, resulting in the creation of carbon dioxide, Prezhdo said.”


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ml/min to mol/min
After completing a continuous distillation experiment, i am trying to convert my feed flow which i set to 30 ml/min and into mol/min to determine a mass balance equation.
If i use the equation below to determine my molar feed flow for the liquid. I am not too sure if i would use atmospheric temperature and pressure or use a specific measurement from the experiment.
PV= nRT
n/time = RT/PV

Use T/P from the experiment and remember to use appropriate units (K units for T).

@Emm You cannot use the above equations for liquids. And the 2nd one is wrong !

Note that the equation $$pV=nRT$$ is valid for a gas only ( more exactly an ideal gas), not for a liquid .
Liquid
$$\frac nt = \frac {V}{t} \cdot \frac{\rho }{ M}$$
Beware of the proper units. Instead of standard ones, these are more suitable for liquids:
$$\mathrm{[mol / s] = [ mL / s ] \cdot \frac {[g/mL]}{[g/mol]}}$$
Gas
$$\frac nt = \frac {V}{t} \cdot \frac{p  }{ RT}$$  For $p$ and $T$, the actual gas conditions apply, including implicit atmospheric pressure.
So eventual pressure correction of the pressure gauge depends on if it measures the absolute pressure, or relative pressure wrt atmospheric one. For the latter, the atmospheric pressure must be added.
Beware of the proper units:
$$\mathrm{[mol / s] = [ m^3 / s ] \cdot \frac {[Pa]}{[J/K/mol][K]}}$$


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Balanced equation for Schweizer's reagent production?
In Thompson's Illustrated Guide to Home Chemistry Experiments the chemical formula for Schweizer's reagent is represented as $\ce{[Cu(NH3)4](OH)2}$ (although Wikipedia represents it as $\ce{[Cu(NH3)4(H2O)2](OH)2}$ ).
Using Thompson's formula, it seems a balanced equation for producing Schweizer's reagent by disolving copper (II) hydroxide in aqueous ammonia would be:
$\ce{Cu(OH)2 (s) + 4 NH3 (aq) -> [Cu(NH3)4](OH)2 (l)}$
This indicates a ratio of 1:4 for copper hydroxide to ammonia.
Is the above balanced equation, and the indicated 1:4 ratio, correct? If not, what might be a correct representation?
(Note: This is a more specific question related to a broader question about Schweizer's reagent production.)

Yes, but as you can see from what you wrote earlier, it's not like stoichiometric ratio has to be enough, rather contrary.

Mithoron, can you elaborate? If the balanced equation is correct and I use a 1:5 ratio, shouldn't all of the copper hydroxide be dissolved?

Stoichiometric ratio is 1:4 but it doesn't mean using more will complete reaction.

Ahah, you mean because of chemical equilibrium a particular reaction may or may not result in complete consumption of all reactants into products?


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How much time does sulfur dioxide take to oxidize to trioxide in air?
I saw that $\ce{NO2}$ is oxidized really fast, but if I put $\ce{SO2}$ in a jar with air at STP, will it take hours, days or years to react into $\ce{SO3}?$

I'd say it will take days if you add some water, or even longer otherwise.

In the reaction $\ce{O + SO2 +M \to SO3 + M }$ in the gas phase and where $M$ is an inert gas such as He the rate constant is $3\cdot 10^{14},\mathrm{cm^6,mol^{-2},s^{-1}}$.  See Westernberg & deHaas J. Chem. Phys. 63, 5411 (1975); https://doi.org/10.1063/1.431347. Usually in the manufacture of sulphuric acid the oxidation is done with a catalyst such as $\ce{V2O5}$.


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What can we say about the dipole moment of terephthalic acid?
The dipole moment of quinol is not zero because it can exist in two different energetically favourable conformations. The weighted average is calculated to arrive a dipole moment of 1.6 D.
According to Wikipedia, the dipole moment of terephthalic acid is zero.
But few sources have a different opinion.
Can we say that terephthalic acid can also exist in these two conformations, leading to a high dipole moment?

Yes it has dipole moment. Only in one configuration, it's dipole moment is 0. But overall due to C-C bond rotation, it has non zero dipole moment.

At room temperature the COOH groups are free to rotate (in sub nanoseconds ) so this will make the molecule symmetric and so zero dipole moment. In a crystal, just possibly, packing may cause a dipole to exist although this seems unlikely.

It isn't even obvious to me that the Wikipedia value for quinol is right. The reference is from 1945 and the method is unclear.

The direction of the marked dipole moments seem to be incorrect.

@harry Different people follow different conventions, i.e., one towards more electronegative atom and the other is towards more electropositive atom.

You are indeed correct, a study by Karthikeyan et. al.[1] suggests that the Dipole moment of  terephthalic acid is indeed around ~2.6D (in air), the same as your latter source.
From the paper,

The calculated dipole moment of
the title compound is 2.57 Debye in gas and 3.14 in solvent

The same paper later mentions,

The calculated value of dipole moment is found to be 2.733
Debye. The highest value of dipole moment is observed in the component of µZ which is -2.733
D since the dipole vector is oriented negative Z direction. The dipole moment is zero in µX and
µY component.

This is expected since terephthalic acid has only two stable rotamers with practically equal energies, both of which are planar.[2]
.
Now even though the paper does not mention this clearly anywhere, what this suggests to me is 2.5704 D is the average Dipole moment of the 3a rotamer (varying from 2.4078 to 2.733 due to COOH twisting vibrations? unclear from the paper), since the Dipole moment of 3b should definitely be zero.
Since both the rotamers exist in nearly equal concentrations, This could suggest that the average dipole moment of should be written as ~2.6/2 D? It's unclear.
References:

Karthikeyan, N.; Joseph Prince, J.; Ramalingam, S.; Periandy, S. Electronic [UV–Visible] and vibrational [FT-IR, FT-Raman] investigation and NMR–mass spectroscopic analysis of terephthalic acid using quantum Gaussian calculations. Spectrochim. Acta, Part A 2015, 139, 229–242. DOI: 10.1016/j.saa.2014.11.112.

Zoran, Marković; Dalibor, Bajduk; Ivan, Gutman ; Geometry and conformation of benzenecarboxylic acids, Journal of the Serbian Chemical Society 2004 69, 877-882. DOI: 10.2298/JSC0411877M

Welcome to Chem SE. I cannot access the journal. Can you summarize the reason in your answer? Is it because of two different conformers; one out of which has non-zero dipole moment (may be = 2 x 2.6D)?

@Apurvium I added the summary from the paper, but further reading revealed several inconsistencies in the paper that weren't entirely explained. I added what I think might be the case. See I may have erred somewhere.


text:
What is the largest one step radical cyclization ever achieved?
When I was an undergrad, roughly around 2005, I took a grad level class that covered various topics in advanced organic synthesis. I don't remember much clearly from the class, but one thing that I can't forget was this one reaction where a single chain compound was, in one step, cyclized to a multi-ring system (maybe 4 rings) via radical cyclization. This cyclization was precipitated by the release of N_2 from an azido group.
It was similar to the cyclization of squalene into lanosterol, but it was not enzymatically catalyzed. I can remember being absolutely astounded that nature had figured out how to do this, and then to later learn that some crazy bastard had done something similar without protein scaffolding was extremely humbling.
I unfortunately can't find my notes from that class, and my google searching hasn't helped me, either. I will try to reach out the original professor, but I think that's a long shot. And it's of course possible, in all honestly, that I could be misremembering it. But, I still have a fuzzy photo in my memory of the lower left hand side of a page of a journal article where this reaction was drawn out. If only I had enough detail to draw it!
So, I guess I'll just ask a more general question to the community. What is the largest single step cyclization that you are aware of? By "largest", I mean "number of atoms that end up as part of a ring that were not originally part of the ring". Very large diameter, single bond/ring radical formations are entropically unlikely, so that should be a fair metric.


text:
Glyoxylic acid synthesis
In my quest for a non-electrochemical glyoxylic acid synthesis, I've found a few that seem simple and pretty OTC. The first is the reduction of oxalic acid with magnesium metal, which if I'm not mistaken forms a magnesium salt which is unwanted. The second is a mixture of nitric acid and ethyl alcohol, which I couldn't find any kind of yield and it was only briefly mentioned in a patent. The third is the oxidation of a glyoxal solution with nitric acid and again I couldn't find any info on reaction time, yields, and whatnot. Has anyone tried any of these methods with reasonable results?

My natural inclination would be towards ozonolysis of fumaric or maleic acid.


text:
How is it possible to melt diamonds?
On another thread on this website it mentioned how it was possible to melt diamonds, but what would be the micro-level occur for that to happen. This question applies to Silicon Oxide aswell. They both have infinite structures. Which bonds would break or even weaken for them to transition states? For example, when a regular covalent solid is heated, the intermolecular forces of attraction weaken and the solid melts. How would that work for a structure like that of diamond?

See, all bonds in diamond are exactly the same. Just how can we tell which of them would break? (Some would, that's for sure.) What does "which" even mean in this context?

you might consult a phase diagram : http://phycomp.technion.ac.il/~anastasy/teza/teza/node5.html

see also: https://chemistry.stackexchange.com/questions/6068/what-is-known-about-liquid-carbon

Why do you say "when a regular covalent solid is heated, the intermolecular forces". To zeroth order that's not true, an appreciable proportion of the atoms just start having enough thermal energy to overcome them. And given it is not all atoms that have sufficient think how that might answer your question.

The comments in this question come pretty close to providing the answer. If necessary, I can organise the concepts a bit more in a new answer here.


text:
Do companies recover carbon dioxide and water from the purification step of cryogenic air separation, and how?
In the 3rd step of cryogenic air separation, filtered air is purified to remove water, $\ce{CO2},$ and hydrocarbons from the $\ce{N2}$-$\ce{O2}$-$\ce{Ar}$ stream. Im interested in knowing what happens with that $\ce{CO2}$ and water. Can both resources be recovered in a useful form? Can the $\ce{CO2}$ be purified? Are Praxair or others currently doing this?
Context: Im doing research on a theoretical industrial process which uses pure $\ce{CO2},$ water, and $\ce{N2}.$ I want to know if all these inputs can be obtained from a standard air separation facility (of course $\ce{N2}$ can). Im particularly interested in $\ce{CO2}$ recovery.
Please note I'm not interested in buying gases. I'm working on a publication about a theoretical system to produce protein from air and solar power. In your answer it would be helpful to provide links to papers or websites which give descriptions of the recovery process (energy costs etc).

Please show that you have understood how much CO2 and oxygen and nitrogen there are in normal air, and how much C, N, O and H a simple protein contains.

You mean you are going to feed the CO2 into a greenhouse?

@Karl Your point is that the ratio of CO2 0.04% to N2 80% will be very low from air. True. Good aprox of cells is c4h7o2n. protein is about 16% N. The system also uses direct air capture (DAC) of CO2 to supplement missing C. However I want to recover CO2 from the N2 production if possible.

@BuckThorn No, it does not use plants. You can look at Solar Foods company to get an idea of a related technology.

Well, yea. My point is that to produce proteins from air, you need to process a huge excess of nitrogen and oxygen. If you can economically do DAC, recovering the CO2 from air separation is just a very tiny bonus.

@Karl, I never replied. Your comments were spot on. It was obv after, but glad you pointed it out!

It is unlikely that this is an economically feasible $\ce{CO2}$ recovery method, which means commercial gas producers probably don't bother to isolate $\ce{CO2}$ this way (surely not water). The removal of $\ce{CO2}$ and water is performed with a pre-purification unit (PPU) that contains an adsorbing material such as alumina or molecular sieve. When the material nears saturation with impurities, the PPU needs to be taken off-line for regeneration. This is done by flowing warm $\ce{N2}$ over the adsorbent to heat it. It certainly seems possible to attempt capture of the $\ce{CO2}$ at this stage. Water is likely to be released from the adsorbent after the $\ce{CO2}$, allowing in principle a stream containing CO2 and N2 to be isolated. However isolation of $\ce{CO2}$ from this stream is expensive. As explained in a brochure from the CO2 Capture Project:

$\ce{CO2}$ can be separated from other gases by cooling and condensation. Cryogenic separation is widely used commercially for streams that already have high  $\ce{CO2}$ concentrations (typically >90%) but it is not used for more dilute  $\ce{CO2}$ streams.
A major disadvantage of cryogenic separation of  $\ce{CO2}$ is the amount of energy required to provide the refrigeration  necessary  for  the  process,  particularly  for  dilute  gas  streams. Another  disadvantage  is that some components, such as water, have to be removed before the gas stream is cooled, to avoid blockages.
Cryogenic separation has the advantage that it enables direct production of liquid  $\ce{CO2}$, which is needed for certain transport options, such as transport by ship. Cryogenics would normally only be applied to high concentration, high pressure gases, such as in pre-combustion capture processes or oxygen fired combustion.

This is complemented by information provided by Universal Industrial Gases Inc.:

Producing carbon dioxide as a commercial product requires that it be recovered and purified from a relatively high-volume, CO2-rich gas stream, generally a stream which is created as an unavoidable byproduct of a large-scale chemical production process or some form of biological process.
In almost all cases, carbon dioxide which is captured and purified for commercial applications would be vented to the atmosphere at the production point if it was not recovered for transport and beneficial use at other locations.
The most common operations from which commercially-produced carbon dioxide is recovered are industrial plants which produce hydrogen or ammonia from natural gas, coal, or other hydrocarbon feedstock, and large-volume fermentation operations in which plant products are made into ethanol for human consumption, automotive fuel, or industrial use.  Breweries producing beer from various grain products are a traditional source.  Corn-to-ethanol plants have been the most rapidly growing source of feed gas for CO2 recovery. CO2-rich natural gas reservoirs found in underground formations found primarily in the western United States and in Canada are another source of recoverable carbon dioxide.


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What does it mean to calculate "weight of SiO2 in the form of Si"?
This is with reference to the typical mass balance problems in iron making. Whenever we are asked to mole balance $\ce{SiO_2}$ it is treated this way:
$W_{\ce{SiO_2}}$ from iron ore + $W_{\ce{SiO_2}}$ from coke = $W_{\ce{SiO_2}}$ in the form of $W_{\ce{Si}}$ in hot metal+$W_{\ce{SiO_2}}$ in slag.
Please explain the meaning of the highlighted part. What is the meaning of "weight in the form of"?

The sum expression looks weird. What kind of work $W$ silicon dioxide is supposed to perform?

@andselisk That is weight not work..LOL

Weight is denoted with a small letter $w$ (see e.g. Wikipedia) and measured in Newtons which also makes no sense in this context. In reality, what you probably have here is mass $m$. Not many reasons to LOL, actually, as it's all pretty sad.

When the silicon in $\ce{SiO2}$ is converted it of course ceases to be present as $\ce{SiO2}$, but you can "pretend" that it still is. It is algebraically simpler to compute a weight of $\ce{SiO2}$, equivalent to the reduced weight $W_{\ce{Si}}$. All you have to do is subtract the remaining weight of $\ce{SiO2}$ from the starting amount. This is a bit simpler then to report $W_{\ce{Si}}$, since the latter calculation requires conversions using molecular and atomic weights for $\ce{SiO2}$ and $\ce{Si}$, respectively.
It also makes it particularly easy to compute a quantity like %yield as follows:
$ \% yield = 100 \times W_{\ce{SiO_2}}$ in the form of $W_{\ce{Si}}$ in hot metal/$(W_{\ce{SiO_2}}$from iron ore+ $W_{\ce{SiO_2}}$ from coke)
Here %yield refers to the molar yield of reduced Si relative to total starting amount of Si.


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Can a universal indicator turn colorless?
I learned that litmus paper could be bleached by chlorine to turn colorless. I’m wondering if any chemical substances can react with all the components of a universal indicator such that it turns colorless, or stops being a pH indicator?

I would suspect the same chlorine bleach.  Universal indicator consists of organic dyes.

@Oscar Lanzi really? I heard that chlorine just turns it purple or red

A fun demo is to make red cabbage soup, which goes from deep red at very low pH to deep violet at high pH, and then add a bit of chlorine bleach to turn it clear.


text:
Not a chemist, needs help with a calculation regarding addition of HCl to sodium formate solution
As stated I am not a chemist, but I need help on a question I have.
Assume I have a Formic acid 75% and I add between 5-10 kg to 1.000 liter of water. If I wanted to add sodium formate and HCl 37% to water and achieve a similar result  with regards to pH and "pure" undissociated formic acid how much of each product would I need to add?
I have calculated, probaly wrong, that I would need 8 ml of HCl to 1.000 L to lower pH to around 4 which is close to pKa of FA (3,75), but I do not know if this will give me the same amount of undissociated FA as the original scenario.
For my application idea pH is less important than undissociated (antibacterial) FA.
Hope someone takes up the (over my head) calculation challenge

When you mean "75%", do you mean weight by weight? That means, in your 5-10 kg mass of the substance you add, 75% of that mass is the mass of formic acid? Similarly, you may wish to specify what is meant by "$\ce {HCl}$ 37%".

At $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=3.75$ , there will be 50% of non-dissociated formic acid and 50% of formiate.
You need 50 % of stoichiometric amount of hydrochloric acid to convert 50% of the formiate to the acid.
If we consider the percentage as (w/w) %,  then $\pu{1 kg}$ of 75% formic acid ($M=\pu{46
g/mol}$) contains $\pu{16.3 mol}$ of formic acid.
For usage of $\pu{5 kg}$ to $\pu{1000 L}$, $\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a}-\log(c)\right)\\
=\frac12\cdot \left(3.75-\log{\frac{5\cdot 16.3}{1000}}\right)\\
=\frac12\cdot \left(3.75-\log{0.0815}\right)=2.42$
For $\pu{10 kg}$, it is 2.27.
$\pu{1 kg}$ of the above formic acid  is equivalent to $\pu{ 16.3 mol}$ of sodium formiate, i.e. $\pu{ 1108 g}$
For 50% conversion, you need $\pu{ 8.15 mol}$ of $\ce{HCl}$, i.e. $\pu{ 8.15\cdot 36.5 = 297 g}$ $\ce{HCl}$, what is equivalent $\pu{ 804 g}$ of 37% $\ce{HCl}$


text:
Isoelectric point defintion
My professor said that in order to have PI, both + and - charge have to exist, and therefore phosphoric acid cannot have a Pi. I am confused because I can't find any definition that states both + and - charge need to exist on text or online. Online, the definitions of PI indicates something similar to "at which the molecule carries no electrical charge." So why can't H3PO4 be PI if it the molecule has 0 charge?

Because there is also present $\ce{H2PO4-}$ It could have an IP if its dissiciation was compensated by existence of the same amount of $\ce{H4PO4+}$ in strong acid.

You can't have an isoelectric point with $\ce{H2PO4^-}$ and $\ce{H4PO4^+}$. The positive and negative charges need to be on the same molecule. Isoelectric points are most typically encountered for amino acids and proteins.

https://en.wikipedia.org/wiki/Isoelectric_point  "The isoelectric point (pI, pH(I), IEP), is the pH at which a molecule carries no net electrical charge or is electrically neutral in the statistical mean."   I do agree the IEP is used mostly ( not exclusively ) for amino acids and proteins.

That means the nature of IEP is the molecule is statistically neutrall, with balanced charged forms. That applies to both amino acids and phosphoric acid in strong acid.(at some point ) Being a zwitterion is just one of ways, where all 3 forms are balanced.


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What are D_b and D_a in the Generalized Fugacity Coefficients table?
I've been referring the book: Chemical Process Principles: Part-II Thermodynamics. I found a table that gives me fugacity coefficients in a tabular form corresponding to various reduced pressures and reduced temperatures. This is it:

I dont know what the $D_b$ and $D_a$ in the table. In the bottom of the table, I see:
$$\left(\frac{f}{p}\right)^{'} = \left(\frac{f}{p}\right)(10^{D({z_c}-0.27)})$$
Is the $D$ in this equation related to the $D_b$ and $D_a$ in the table?

Have you checked the text in the book?

I wasn't sure what $D$ stands for. But, I assume $a$ and $b$ are van der Waal coefficients in real gas equation.

See page 574 in your book.

$D_b$ and $D_a$ are deviation terms which must be used in the place of $D$ to obtain the fugacity coefficients for fluids whose critical compressibility factor is not 0.27. Replace $D$ with $D_a$ if $z_c > 0.27$ and with $D_b$ if $z_c < 0.27$.


text:
d-Menthol vs dl-menthol: Does an enantiomer and its racemic mixture have different melting points?
I am checking out Wikipedia and it shows a different melting point (m.p.) for l-menthol $(\pu{42 ^\circ C}$ to $\pu{45 ^\circ C})$ vs dl-menthol $(\pu{36 ^\circ C}$ to $\pu{38 ^\circ C}).$
How is this possible? d-menthol and l-menthol being enantiomers, shouldn't they have identical physical properties including the m.p.?

The two enantiomers of a molecule possess the same physical properties (e.g., melting point, boiling point, polarity etc.) and so behave identically to each other.

If so, why would the racemic mixture have any different m.p.? I.e. you can't just physically  mix two pure materials with the same m.p. and end up with a material with a different m.p., right?
I checked out Common Fragrance and Flavor Materials by Surburg and Panten [1, p. 55] just to be sure and the data agrees. The m.p. of the d-menthol and dl-menthol again seem to be different:

Other physical constants of commercially available levorotatory and racemic
  menthols are: (−)-menthol [2216-51-5], mp $\pu{43 °C},$ $n_\mathrm{D}^{20}~1.4600,$ $[α]_\mathrm{D}^{20}~–50^\circ;$ (±)-
  menthol [15356-70-4], mp $\pu{38 °C},$ $n_\mathrm{D}^{20}~1.4615.$

References

Surburg, H.; Panten, J. Common Fragrance and Flavor Materials: Preparation, Properties and Uses, 5th, Completely Revised and Enlarged edition; Wiley-VCH: Weinheim, 2006. ISBN 978-3-527-31315-0.

In stereochemical vocabulary, a racemic mixture (racemate) is one that has equal amounts (50:50) of left- and right-handed ($d$- and $l$-) enantiomers of a chiral molecule. According to Wikipedia: Racemic Mixture, a racemic mixture can be crystallized in four ways:

Conglomerate (sometimes racemic conglomerate): If the molecules of the substance have a much greater affinity for the same enantiomer than for the opposite one, a mechanical mixture of enantiomerically pure crystals will result. The melting point of the racemic conglomerate is always lower than that of the pure enantiomer. Addition of a small amount of one enantiomer to the conglomerate increases the melting point. Roughly 10% of racemic chiral compounds crystallize as conglomerates.
Racemic compound (sometimes true racemate): If molecules have a greater affinity for the opposite enantiomer than for the same enantiomer, the substance forms a single crystalline phase in which the two enantiomers are present in an ordered 1:1 ratio in the elementary cell. Adding a small amount of one enantiomer to the racemic compound decreases the melting point. But the pure enantiomer can have a higher or lower melting point than the compound. A special case of racemic compounds are kryptoracemates, in which the crystal itself has handedness (is enantiomorphic), despite containing both enantiomorphs in a 1:1 ratio.
Pseudoracemate (sometimes racemic solid solution): When there is no big difference in affinity between the same and opposite enantiomers, then in contrast to the racemic compound and the conglomerate, the two enantiomers will coexist in an unordered manner in the crystal lattice. Addition of a small amount of one enantiomer changes the melting point just little bit or not at all.
Quasiracemate: A quasiracemate is a co-crystal of two similar but distinct compounds, one of which is left-handed and the other right-handed. Although chemically different, they are sterically similar (isosteric) and are still able to form a racemic crystalline phase. One of the first such racemates studied, by Pasteur in 1853, forms from a 1:2 mixture of the bis ammonium salt of (+)-tartaric acid and the bis ammonium salt of (−)-malic acid in water. Re-investigated in 2008,[6] the crystals formed are dumbbell-shape with the central part consisting of ammonium (+)-bitartrate, whereas the outer parts are a quasiracemic mixture of ammonium (+)-bitartrate and ammonium (−)-bimalate.

The most common type of racemate is the racemic compound (2nd type). The crystal of a racemic compound is composed of enantiomer pairs, distributed in an organized array, and representing a true molecular compound. Thus, these racemic compounds have unique physical properties, including melting points, solubility, density, and solid state spectra, different from that of either crystalline enantiomer (Crystallization: MSU.edu). Specifically, melting points of this kind of recemates can be either higher, identical, or lower than that of their corresponding pure enantiomers. The reference 1 states that the thermodynamic and crystallographic approaches have unambiguously demonstrated that racemic-menthol is a racemate. Hence, showing unique physical properties different from that of either crystalline enantiomer such as melting point being lower than that of either pure enanthiomer can be expected.
To address your statement "would it be fair to say that the boiling points would always be the same for the enantiomers as well as the racemic mixture, however the melting points may vary?": In most of the cases yes (see Table below). Also be aware that some other physical properties such as solubility can be different as well. For example, solubility of racemic-tataric acid (m.p.: $\pu{206 ^\circ C}$) is $\pu{25 g}$/$\pu{100 mL}$ while that of either enanthiomers of tataric acid (m.p.: $\pu{172 ^\circ C}$) is $\pu{135 g}$/$\pu{100 mL}$ (Crystallization: MSU.edu).
The following table emphasizes this trend (Ref.2):
$$
\begin{array}{c|lcr}
\text{Stereoisomer} & \text{m.p., } \pu{^\circ C} & \text{b.p., } \pu{^\circ C} & [\alpha]^{20}_D\text{ (in 20% EtOH)} \\
\hline
l\text{-Menthol} & 43.0 & 216.5 & -50.0 \\
dl\text{-Menthol} & 38.0 & 216.5 & 0.0 \\
d\text{-}neo\text{-Menthol} & -15.0 & 211.7 & +20.0 \\
dl\text{-}neo\text{-Menthol} & 52.0 & 211.7 & 0.0 \\
d\text{-}iso\text{-Menthol} & 82.5 & 218.6 & +26.0 \\
dl\text{-}iso\text{-Menthol} & 53.5 & 218.6 & 0.0 \\
d\text{-}neo\text{-}iso\text{-Menthol} & -8.0 & 214.6 & +2.0 \\
dl\text{-}neo\text{-}iso\text{-Menthol} & 13.5 & 214.6 & 0.0 \\
\end{array}
$$
References:

Yohann Corvis, Philippe Négrier, Stéphane Massip, Jean-Michel Leger, Philippe Espeau, "Insights into the crystal structure, polymorphism and thermal behavior of menthol optical isomers and racemates," CrystEngComm 2012, 14(20), 7055-7064 (DOI: 10.1039/C2CE26025E).
Kenneth L. Waters, George D. Beal, "Some physical and chemical properties of commercial racemic menthol," Journal of the American Pharmaceutical Association 1945, 34(2), 52-56 (https://doi.org/10.1002/jps.3030340208).

Great answer! Nitpicking follow up question: Since you said "most" and not "all" are you aware of any corner cases where the BP varies among  a pair of enantiomers?

@curious_cat: Not yet. ;-)

If solid menthol were just amorphous, then you would expect the racemate and each enantiomer to exhibit identical melting points. However, if you read further on the Wikipedia page you reference, you'll see that racemic vs. enantiopure menthol crystallizes differently:

The two crystal forms for racemic menthol have melting points of 28 °C and 38 °C. Pure (−)-menthol has four crystal forms, of which the most stable is the α form, the familiar broad needles.

(This information is trustworthy and corroborated in the primary literature; here is one example)
The melting point of a crystalline solid has to do with how well the molecules can pack together. In other words, on a molecular level, there is an orientation dependence. The better the molecular packing, the stronger the intermolecular forces, and the higher the melting point - since it takes more energy to separate the molecules. This is why the enantiopure menthol has a higher melting point.

Thanks! So would it be fair  to say that the BPs would always be the same for the enantiomers as well as the racemic mixture however the MPs may vary?


text:
Explain me concept of vapour pressure like I am five
I have browsed internet to understand the concept of vapour pressure but something isn't linking for me. How do we relate vapour pressure to temperature for liquid only and any simplified definition of vapour pressure would be more than welcome! Can we relate vapour pressure to critical temperature in any way? Thanks in advance.....

Molecules of liquid, similarly as molecules of gas, have a particular statistical distribution of energy.
Molecules of liquid bond each other by intermolecular bonds, that deny them to escape from liquid to gas and become free.
Well, it is denied for most of them. At any temperature, there always exists a fraction of molecules with high enough energy to break bonds and escape.
This fraction increases with temperature of liquid, approximately exponentially.
If you put liquid to a closed container with vacuum, the escaping molecules start forming a gas with growing pressure. That is the vapour pressure.
When a molecule of such a gas hits the liquid, it is trapped by the bonds and stays in the liquid. The rate of this process is proportional to the vapour pressure.
At the particular vapour pressure, dependent on temperature, the rate of escaping liquid molecules ( evaporation rate ) equals to rate of trapping of molecules ( condensation rate ) and the liquid  evaporation ( and condensation ) rate is zero.
This vapour pressure is called the saturated vapour pressure ( for given liquid at given temperature ).
This saturated vapour pressure does not depend on presence of other gases, like air, where it is called the saturated vapour partial pressure.(with partial often omitted)
The partial pressure is the contribution of a particular gas compound to the total gas pressure, proportional to its molecular fraction within the gas.


text:
Equilibrium pressure

Ammonium carbamate dissociates as follows:
$$\ce{NH2COONH4(s)<=> 2NH3(g) + CO2(g)}$$
The value of $K_p$ for this reaction is found to be equal to
  $\pu{2.92 \times 10^-5 atm^3}$. If one mole of ammonium carbamate is heated in a sealed container, the total pressure developed in the container is:

(A) $\pu{0.0194 atm}$
(B) $\pu{0.0388 atm}$
(C) $\pu{0.0582 atm}$
(D) $\pu{0.0667 atm}$

I first tried considering initial moles as $1$ and taking the moles used at equilibrium as $x$. I was able to form an equation with equilibrium constant, $x$, total pressure(with the help of fact that partial pressure = mole fraction * total  pressure; and also that equilibrium cost = (product of partial pressure of products raised to their stoichiometric coefficients) / (partial pressure of reactants raised to their coefficients).
But I am getting only one equation while two unknowns: $x$ and total pressure.

This problem is relatively easy to solve if you remember that the vapor pressure of the products is independent of the amount of solid, provided there is enough solid for it to be in equilibrium with the gases. Therefore:
$$K_p=p_{\ce{NH3}}^2p_{\ce{CO2}}$$
Since $p_{\ce{NH3}}=2p_{\ce{CO2}}$
$$K_p=4p_{\ce{CO2}}^3$$
and $p_{\ce{CO2}}=\pu{0.0194 atm}$. This is one third of the total pressure, which must be $\pu{0.0582 atm}$.
Answer: C


text:
How can I find out a substance's name from its structural formula?
When visiting a botanical garden in Stockholm, I came across the following mosaic tile:

I tried using MolView, but it does not seem to identify it. Also, I have tried their advanced search (similarity), but that provided no results.
I have also tried searching for it based on SMILES formula on Chemspider.
I am wondering whether there any other tools that can help me to find out what substance it is.
Extra context information: it was next to a DNA structure artwork, so I guess it is somewhat related to evolution, DNA etc.
Question: How can I find out a substance's name based on its structural formula?

That's betanin, from which beets derive much of their deep purple color, and which may have neuro-protective (Ref. 1) features. The following is the structure retrieved from the wikipedia (on 2019-10-5), which closely matches that in the mosaic:

I used the editor available on the Sigma-Aldritch website to generate a structure and submit a search:

The program identified betanin as the top hit despite my sloppy structure.
The SMILES code is: C(OC(C1O)OC(C(=CC2[N+]3=CC=C(CC(N4)C(=O)O)C=C4C(=O)O)O)=CC=2CC3C([O-])=O)(C(C1O)O)CO
The IUPAC name (retrieved from the wikipedia) is (2S)-1-{2-[(2S)-2,6-dicarboxy-2,3-dihydropyridin-4(1H)-ylidene]ethylidene}-5-(β-d-glucopyranosyloxy)-6-hydroxy-2,3-dihydro-1H-indol-1-ium-2-carboxylate
As noted in a comment, the stereochemistry in the wikipedia image above and that in the mosaic does not match that of the structure in the Sigma-Aldrich catalogue. I used ACD/ChemSketch freeware to generate the following image that does match the one in the catalogue:

American Chemical Society. "Vegetable compound could have a key role in 'beeting' Alzheimer's disease." ScienceDaily, 20 March 2018.

As an aside, it's interesting to note that work on L-dopa, a biosynthetic precursor of betanin, led to a Nobel prize award for Swedish scientist Arvid Carlsson. Also, quoting from the wikipedia:

The 2001 Nobel Prize in Chemistry was also related to l-DOPA: the Nobel Committee awarded one-quarter of the prize to William S. Knowles for his work on chirally catalysed hydrogenation reactions, the most noted example of which was used for the synthesis of l-DOPA.

The structure-drawing engine (and presumably also search engine) provided through the Sigma-Aldrich structure search web service is JSDraw2 (JSDraw V3.2.3) by Scilligence.

I am quite disappointed by the actual result (a food dye), but the method is useful. +1. Thanks.

Well, I'm guessing that it colors some of the flowers and tubers at the garden you visited, but there might be more to it than that. The family of compunds is pretty interesting if you are willing to look into it.

Oh, it makes perfect sense then. An interesting puzzle for a botanical garden.

https://www.sciencedaily.com/releases/2018/03/180320084414.htm Didn't your mom ever tell you to "eat your beet soup"?

BTW I believe the ACD software, like a number of competitors, can perform structure searches, but the freeware can't.

The first image (the one from Wikipedia) is wrong. Its creator mirrored the structure of another image without mirroring the wedged/dashed bonds, inverting the stereochemistry. I've replaced it on the Wikipedia page with an older, correct image. (Also, sadly, the mosaic is wrong for the same reason.)

@LegionMammal978 Good catch, I didn't bother to compare structures. In fact I picked up the original image from the wikipedia (not ACD software-generated). I updated and hopefully ironed out any errors.


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What is the purpose of phase matching when drawing a MO diagram?
During the lecture, we were learning how to draw the MO diagram for $\mathrm{BF_3}$.
I understand that only orbitals with the same symmetry labels can interact and form molecular orbitals. My professor always elaborates a lot on "phase matching," and he would draw the orbitals out.
However, I do not understand how these orbital drawings are supposed to assist me when drawing the MO diagram. How would it affect my process for drawing the MO diagram?

If I interpret correctly phase matching is guaranteed when the orbitals have matching symmetry. You are then allowed to combine them in the MO diagram.

Phase matching can matter for determining whether you are forming a bonding vs an antibonding orbital.


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Can the presence of radon (Rn-222) be determined by testing dust?
If a relatively unventilated area, like a basement, were to be exposed to radon gas (Rn-222), would the dust in the area possess a detectable amount of lead? Considering lead 214 is a decay product of radon 222, will a room exposed to radon gas possess dust with a detectable and differentiable amount of lead relative to a room that is unexposed?

Chemically not, as amount of 214Pb is unrelated to total amount of lead.

The dust, or rather walls, is better to be tested for 226Ra. And the best is usage to passive long term scanners of 222Rn, detecting alpha decays of gaseous radon.

The 26 minute half life of Pb-214 would seem to be a problem...