Would the mixture of vinegar and baking soda make a cleaning agent? I've seen it suggested countless times that combining vinegar and baking soda is a great home cleaning solution, but this always seemed like nonsense to me.

I recognize that baking soda, a base, can eat away at certain kinds of grime and that vinegar, an acid, can operate similarly on different kinds of grime.

Mixed together, of course, the two create a neat foaming reaction and I think this is where the alleged myth would originate - people see the violent foaming and assume cleaning is occurring.

Isn't it true that the two are primarily reacting with each other and yielding only a relatively inert and neutralized sludge?  That the mix reacts with any grime less than they would if both were applied in isolation against the grime?  I assume that the only plausible benefit from this mix would be in any pockets of non-neutralized components floating on the foam and physically reaching new areas of grime that the individual volumes would not cover.

But I guess there could be more to cleaning effectiveness besides pH!

Am I right or are they?

Well, cleaning is a very broad subject. This is a question which could make a great Science Fair Project. Because vinegar is just a dilute solution of acetic acid (ignoring trace compounds present), and because both baking soda and sodium acetate are pretty soluble, it wouldn't probably be practical to try to make a saturated acidic solution of BS. That is, any precipitates will be made from alkaline solution, under normal circumstances.

There's two questions which would need to be answered:

Does fresh V+BS slurry act the same way as old V+BS slurry?
Are either better than water + BS?

Well, hmm. To be honest, IDK, but I'd guess "maybe" for both (which is why it would be interesting to see what happens). Let me explain:

Acetic acid dissolves soap scum (mostly K, & Ca salts of fatty acids) and dissolves water hardness (mostly Ca & Mg carbonates) So starting off getting rid of some or all of that junk could definitely be a good thing (and acetic acid is only going to be around for a very short time if BS is present in excess). But we know that acids attack metals, so leaving vinegar on metal isn't a good idea...and guess what? Once the BS fully reacts, the acid will be gone! I can't think of any other plausible reason that V+BS would be better than W+BS.

One other potential reason would be if the vinegar changed the abrasiveness of the BS, making it more effective (which is another thing you could test for), but I doubt it would. (possible but not quite plausible, if you know what I mean). Anyway, what you've got left is W+BS after the reaction, so what's the worst it can do?

I do know for sure that most cleaners (with the exception of many metal cleaners) are alkaline because higher pH tends to dissolve (and/or lift-off (which means form a colloid of or weaken the adhesion to the surface-to-be-cleaned of)) organics better than acidic solutions. Cleanser, which is usually just mostly silica, certainly is more abrasive than BS, but that also means it will tend to scratch (or fog) more surfaces. I don't know, but I'll bet most cleansers are alkaline for the reason mentioned above. (alkaline solutions are NOT more slippery, but they form soaps out of your skin, and those soaps are more slippery. So, the slipperiness you feel when you touch an alkaline (basic) solution is just your skin dissolving :)


How to classify the hydride shift in the following diazacyclopenta[cd]pentalene system? How is this hydride shift reaction type called?

I think it's not one of following types:

cycloaddition (Diels Alder)
electrocyclic ring opening
sigmatropic rearrangement
ene reaction
cheleotropic reaction

Therefore I'm thinking what else can been the name of the observed reaction.

The ring opening reaction with hydride shift has been taken from Heterocycles 2004, 64 (1), 447–461 (DOI: 10.3987/COM-04-S(P)48).

It took me a while to see what was actually happening. Initially, I was focused on a [1,5] suprafacial sigmatropic proton shift followed by a diaza-electrocyclic ring opening reaction. However, after drawing the corresponding orbitals I realised that I would have a eight-electron antiaromatic π system, in which a proton shift would need to occur antarafacially — not possible.[1]

It still is clear that it is some kind of pericyclic reaction, and going through the list we can exclude everything except for a (retro-) group transfer reaction for which the (retro-) ene reaction is the most well-known example. (We are transforming one σ bond into one π bond which excludes all the others.)

It took me a while to actually see the retro-ene system in the substrate, but finally I did, and I took the liberty of highlighting it in red in the scheme below. Note that you have to ignore the carbon-nitrogen bond which belongs to the lowest five-membered ring from the ene-system. It is not taking part, if anything it is scaffolding.

Scheme 1: Retro-ene reaction of the title compound with the pericyclic six-membered ring system highlighted. Dashed bonds represent those that are broken/formed within the reaction.

Counting electrons, I realised that my initial assumption of a proton shift was wrong and that it is indeed a hydride shift.

Notes:

[1] [1,n]-rearrangements are classified as suprafacial or antarafacial depending on whether the group is on the same side or the other side of a π system after migration. Technically, there is a further difference between retention and inversion, but hydrogen cannot rearrange under inversion so that is moot. The general rule for proton transfers is: If the number of bonds is $4n + 1$, the mechanism is suprafacial, for $4n + 3$ it is antarafacial.


Usage of TFA in Pfitzner-Moffat oxidation and Collins vs Jones oxidation Is the addition of TFA in Pfitzner-Moffat oxidation to protonate the DCC to form a positive carbocation for the alcohol to attack? Or is there more to it? Why is TFA used and not other acids? Are other acids to weak for this?

Lastly for the Collins vs Jones oxidation, the only discernible differences are the solvent: $\ce{H2SO4}$ vs DCM, and that the Collins oxidation makes use of a pyridin chlorochromate, does this complex dampen the oxidative effect of chromium? So much so that the Collins oxidation can not directly oxidise a alcohol to a carboxylic acid, whereas Jones reagent with $\ce{H2SO4}$ and $\ce{CrO3}$ can?

Moffat & Pfitzner have accidentally found their famous oxidation while working on mono-nucleotides (acually 5'-phosphates). Breifly, when pyromidium thymidine-5'-O-phosphate was reacted with dicyclohexylcaebodiimide  in anhydrous DMSO at room temperature resulted a violent reaction to give 90% of 5'-aldehyde product (Ref.1). They further pursued this finding and published their finding as specific reaction (Ref.2). The acid they have used in first finding was anhydrous orthophosphoric acid, which have given 90% yield. Hence, it is not necessary to have TFA as the acid. Following is the suggested mechanism for Pfitzner-Moffat oxidation:

About Collins vs Jones oxidation, both of them used chromium-based oxidants. Oxidations with chromium oxidants proceed through a similar mechanism. Different between Collins and Jones are conditions: Anhydrous (in pyridine/$\ce{CH2Cl2}$) vs aqueous (use of $\ce{CH2Cl2}$). Following is the suggested mechanism for chromium-based oxidants:

After first oxidation to aldehyde, chromium oxidant needs $\ce{CH-OH}$ part available to further oxidize to carboxylic acid (see in both mechanism, how oxidation initiated and finished). In anhydrous conditions, this feature is lacked. So, oxidation stops at aldehyde moiety. However, trace of water make aldehyde in equlibrium with its gem-diol moiety, which gives the necessary requirement of $\ce{CH-OH}$ feature. So, in Jones oxidation condition, oxidation will further continue to give carboxylic acids.

References:

K. E. Pfitzner, J. G. Moffatt, “The Synthesis of Nucleoside-5' Aldehydes,” J. Am. Chem. Soc. 1963, 85(19), 3027-3027 (https://doi.org/10.1021/ja00902a035).
K. E. Pfitzner, J. G. Moffatt, “A New and Selective Oxidation of Alcohols,” J. Am. Chem. Soc. 1963, 85(19), 3027-3028 (https://doi.org/10.1021/ja00902a036).


Reaction of 1,1-diphenylethanol on Montmorillonite K10 under microwave radiation What is the product of 1,1-diphenylethanol + Montmorillonite K10 Clay undergoing microwave radiation? What is the mechanism for it?

My effort:

1,1-diphenylethanol + Montmorillonite K10 Clay catalyst used in a microwave-assisted synthesis of pyrazoles from enaminones.

Used for:

Support for catalysts
Preparation of clay nanopaper composites of nacre-like structure

Catalyst for:

Conversion of triose sugars with alcohols to alkyl lactates
Deprotection of ketone dimethylhydrazones

Is my answer correct?

Is my answer correct?

No! Your "answer" is just a random collection of facts and does not address the question.

Actually, Ben Norris has already kindly provided all the necessary information in his comment:

Montmorillonite K10 is a solid, acidic catalyst.
Microwave irradiation means that you are applying heat to the reaction.

Under the given conditions, 1,1-diphenylethanol (1) is protonated, eliminates water (which might be driven out since heat is applied) and yields 1,1-diphenylethene (2).

Given that you're still applying heat in the presence of an acidic catalyst, the reaction will not stop here.

Protonation of 2 again results in the known bisbenzylic cation, which will add to a second molecule of 1,1-diphenylethene in Markovnikov fashion.

The resulting intermediate is will undergo an intramolecular Friedel-Crafts alkylation which furnishes the indane derivative 3.


No chlorate esters? Why? Why are there no chlorate esters? Or are they just undocumented or too unstable? I've seen perchlorate esters, but google "chlorate esters" and only perchlorate esters come up.

Your initial statement is wrong. There are organic chlorates. When you want to search compounds, try chemical search engine like PubChem or ChemSpider instead of google. Even google lead you to something if you read carefully. It is no reason that chlorate ester can not exist and they do exist. Below is the link to the PubChem entry of Choric Acid, Methyl Ester, which is Methyl Chlorate.

https://pubchem.ncbi.nlm.nih.gov/compound/18695386#section=Top


Why is a buffer solution best when pH = pKa i.e. when A-/HA=1 Buffers work best when  $\mathrm{pH}$=$\mathrm{p}K_\mathrm{a}$

From the Henderson-Hasselbalch equation,

$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log_{10}\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$

If $\mathrm{pH}$=$\mathrm{p}K_\mathrm{a} \implies\log_{10} \ce{\frac{[A-]}{[HA]}} = 0 \implies \frac{[\ce{A-}]}{[\ce{HA}]}= 1$

But why does having $\frac{[\ce{A-}]}{[\ce{HA}]}= 1$  make it a good buffer solution?

When I look at a $\log_{10}$  graph, it looks like the further to the right $x$ i.e. $\ce{\frac{[A-]}{[HA]}}$ is, the less effect any change in $\ce{\frac{[A-]}{[HA]}}$ has on $\log_{10}\frac{[\ce{A-}]}{[\ce{HA}]}$ (and hence effect on $\mathrm{pH}$ as well).

So wouldn't this indicate that a good buffer would have a large $\log_{10} \ce{\frac{[A-]}{[HA]}}$ (as e.g. going from $\ce{\frac{[A-]}{[HA]}} = 10$   to $\ce{\frac{[A-]}{[HA]}}= 11$  will have a smaller change (0.04 in this example) in $\log_{10}\ce{\frac{[A-]}{[HA]}}$ than going from $v= 1/1$   to $\ce{\frac{[A-]}{[HA]}}= 2/1$) (0.3 in this example). Where both of these changes show addition of 1 M of base.

Note: 2 similar questions where asked before: Why is buffer capacity at a maximum when the ratio of its components is 1? and Why does a buffer work best at the pH closest to its pKa?

The questions didn't show what exactly they didn't understand which I am attempting to do here. Also the answers were not useful as one said said the buffer is best when X (which I know), but not didn't explain why this is the case. The other questions answer provided a link which I read but was beyond me.

Preliminaries

There are many ways in which you can go about defining a "Buffer Capacity". A good intuitive way to start would be to say, it is tha maximum amount of acid (or base) the solution can tolerate before it starts showing a significant change in pH. While this is not bad, it obviously begs the question what is deemed "significant"?

However, let us start at the very beginning by examining (in very basic terms) how a buffer works.

Say, you have some amount of $\ce{A^-}$ in the solution and some amount of $\ce{HA}$. You add a strong acid to the solution, and this leads to an influx of $\ce{H^+}$ ions; the buffer responds by forming more $\ce{HA}$ through the combination of $\ce{A^-}$ and $\ce{H^+}$. Similarly, you add a strong base to the solution, and you have a lot of $\ce{OH^-}$ floating around, and the soultion responds appropriately, by favouring the disassociation of $\ce{HA}$, the $\ce{H^+}$ so released combines with the $\ce{OH^-}$ and neutralises it.

The equilibrium we are interested in is $$\ce{HA <=> A^- + H^+}$$

an increase in acidity causes a shift to the left, and an increase in basicity causes a shift to the right.

So, in this way a buffer acts a proton sink, absorbing/releasing protons as appropriate to maintain the pH of the solution.

The buffer is said to be broken when we add to much acid/base, and this consumes all of the $\ce{A-}$ / $\ce{HA}$ respectively.

A solution with more weak base, $\ce{A-}$, has a higher buffer capacity for addition of strong acid. A solution with more weak acid, $\ce{HA}$, has a higher buffer capacity for addition of strong base.So although the pH of a buffer is determined by only the ratio conjugate base and acid, the ability of the buffer to absorb strong acid or base is determined by the individual concentrations of the conjugate base and acid.

So it might seem intuitive to have the optimum at equal concentrations of conjugate base and acid.

Intuitive definitions

$$\ce{HA + OH^- <=> A^- + H2O }$$
$$\ce{A^- + H3O^+ <=> HA + H2O}$$
$$pH + pOH = 14$$

The following discussion is largely inspired by this paper (free to read)

Namely, if you you have $$\mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{A^-}{HA}\right)$$

Since, $\mathrm{pK_a}$ is a fixed number, an changes that can take place to the pH value are caused by the $\log \left( \frac{A^-}{HA}\right)$ term.

so, we can say $$ \mathrm{pH} = \mathrm{pK_a} + \delta \\ \text{where}\ \delta = \log \left( \frac{A^-}{HA}\right)$$

make $\delta = 0$ and you are done. You can call that your ideal case, an ideal buffer.

Now, you can consider a buffer's resistance to base and acid separately, and you can define an acid and base buffer capacity ($\beta_a$ and $\beta_b$), for very simple cases (read caveats).

The stoichiometric ratio of of conjugate base and acid is 1:1. $\Delta$ denotes an arbitrary change in $\delta$ after $\alpha$ mol of $\ce{HA}$ has reacted (after addition of a strong base)

$$\delta + \Delta = \log \left( \frac{[A^-] + \alpha}{[HA] - \alpha}\right)$$

One then notes that bases attempt to increase the pH of a solution so $\Delta = 1$, and also let us denote $\beta_b$ as the buffer base capacity. Also, note that $\alpha = \beta_b$. We are interested in the maximum amount of base a 1:1 buffer solution can tolerate before the pH increases by one unit. We are assuming a 1:1 stoichiometry between the the weak acid in our buffer and the strong base that is being added to it.

$$10^{\delta + 1 } = \left( \frac{[A^-] + \beta_b}{[HA] - \beta_b}\right) $$

We solve for beta and substitute for $\delta$ to ultimately get,

$$ \beta_b = \frac{9[\ce{HA}][\ce{A^-}]}{10[\ce{HA}] + [\ce{A^-}]}$$

Now, let us consider that we are adding a strong monoprotic acid. The approach is similar to the one described above. A few key differences is one lets $\Delta = -1$ and let $\alpha = -\beta_a$

I am omitting the algebra involved. You can attempt as an exercise it if you'd like, and ask for clarifications in the comments if you run into trouble. We obtain a similar relation

$$ \beta_a = \frac{9[\ce{HA}][\ce{A^-}]}{10[\ce{HA}] + [\ce{A^-}]}$$

$\beta_a = \beta_b$ if and only if $\delta = 0$ Otherwise, the buffer either displays stronger resistance towards acids or bases.

Caveats

This is a good exercise to develop a good basic understanding, however doesn't have a lot of practical uses. Here, only a simple buffer was considered, i.e one without multiple weak bases and acids. Additionally, polyprotic acids haven't been accounted for.

Formal defintion of buffer capacity

The formal definition of buffer is a little bit intimidating, but I will throw it in here anyway. It obviously makes little sense to have two different kinds of capacities, one for acids and one for bases in everyday life. So we need something more general, which still incorporates the intuition we have developed so far (although it might not seem obvious)

Let $n$ be number of equivalents of added strong base (per 1 L of the solution). Note that addition of $dn$ moles of acid will change pH by exactly the same value but in opposite direction. I will try to derive a formula connecting buffer capacity with pH, pKa and buffer concentration- all numbers we can obtain easily.

For simplicity, I assume that the strong base is monoprotic, and we have volume of 1 (which will allows me to use concentration and number of moles interchangeably)

$$\beta = \frac{\mathrm{d}n}{\mathrm{d}pH} \tag{1}$$

Charge balance of the solution is given by the equation
$$[\ce{A^-}]+[\ce{OH^-}] = [\ce{B^+}] + [\ce{H^+}] \tag{2}$$

$[\ce{B+}]$ is nothing but concentration of the strong base present, or the number of moles present since we assume a volume of 1 L(:D ), so it is simply $n$ in the solution.

$$C_{b} = [\ce{HA}]+[\ce{A-}] \tag{3}$$

and $$[\ce{HA}] = \frac{[\ce{A-}][\ce{H+}]}{K_a} \tag{4}$$

from 3 and 4,

$$ [\ce{A-}] = \frac{C_bK_a}{K_a+ [\ce{H+}]} \tag{5} $$

Using the definition of $K_w$, equation 2 and 5

$$n = \frac{K_w}{[\ce{H+}]} - [\ce{H+}]+ \frac{C_bK_a}{K_a+ [\ce{H+}]} \tag{6}$$

$$\beta = \frac{\mathrm{d}n}{\mathrm{d}pH} = \frac{\mathrm{d}n \mathrm{d}[\ce{H+}]}{\mathrm{d}[\ce{H+}]\mathrm{dpH}} \tag{7}$$

Now, you can evaluate the derivative given above, and after some alebraic manipulations you get
$$\beta = (2.303) \left( \frac{K_w}{[\ce{H+}]} + [\ce{H+}]+ \frac{C_bK_a}{(K_a+ [\ce{H+}])^2} \right )$$

Generalising this for solutions containing several buffers

$$\beta = (2.303) \left( \frac{K_w}{[\ce{H+}]} + [\ce{H+}]+ \sum_{}^{} \frac{C_bK_a}{(K_a+ [\ce{H+}])^2} \right )$$

I'd like to point out that the first two terms in the buffer capacity formula are not dependent on the buffer presence in the solution. Why? They just tell us solutions of high (or low) pH are resistant to pH changes. Simple.

Above plot shows how the buffer capacity changes for the 0.1M solution of acetic buffer. As expected buffer exhibits the highest resistance to acid and base addition for the equimolar solution (when pH=pKa). From the plot it is also obvious that buffer capacity has reasonably high values only for pH close to pKa value. The further from the optimal value, the lower buffer capacity of the solution. Solution containing only conjugate base (pH 8-10) has buffer capacity of zero, for the higher pH presence of the strong base starts to play an important role. In the case of pure acetic acid solution (pH below 3) pH is already low enough to be resistant to changes due to the high concentration of $\ce{H+}$.


Using TBAI (tetra-butylammonium iodide) as a catalyst. Ideas on the mechanism or the driving force? for my etherification of an alcoxide with alkylbromide I used TBAI as a catalyst. The catalyst transfers the alkylbromide into an alkyliodide which is more reactive.

I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide? Is there a mechanism to explain that?

I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide?

It's not so much a driving force as much as it is a dynamic equilibrium of sN2/sN1 reactions. Halides can exchange each other in equilibrium if one is not removed thus processes that remove the unwanted product such as the Finkelstein reaction are use to achieve complete substitution (think Le'Chatlier's Principle).

$$\ce{R-Br +I- <=> R-I + Br-}\qquad\quad\ \ \  \text{(Equillibrium)}$$
$$\ce{R-Br +I- +Na+ <=> R-I + NaBr (s)}\qquad \text{(Finkelstein Reaction)}$$

  Is there a mechanism to explain that?

Given the information you provided, Your reaction is proceeding as follows:

$$\ce{R-Br + I- -> R-I + Br-}\tag{1a}$$

$$\ce{R-I + R'-O- -> R-O-R' +I-}\tag{1b}$$
Where as the reaction:
$$\ce{R-Br + R'-O- -> R-O-R' +Br-}\tag{2}$$

cannot proceed or does not proceed at a fast enough rate and thus the catalyst can improve the reaction rate by first substituting the less reactive halide. The Wikipedia article on Williamson ether synthesis agrees with this:

  However, if an unreactive alkylating agent is used (e.g. an alkyl chloride) then the rate of reaction can be greatly improved by the addition of a catalytic quantity of a soluble iodide salt (which undergoes halide exchange with the chloride to yield a much more reactive iodide, a variant of the Finkelstein reaction).


Thermal decomposition products of polyurea? Are the thermal decomposition products of polyurea polymers known, especially of the urea group, in the absence of water?

Looking at the formula

R-NH-C=O
     |
    HN-R

there appears to be no obvious candidate for a decomposition product - not enough oxygen for CO2, not enough hydrogen for Ammonia.

Any ideas or, even better, references?

Here's one abstract:

  The primary fragmentation processes in the thermal decomposition of polymers were studied in detail on a series of structurally related polyureas by direct pyrolysis with a mass spectrometer. Our results indicate that polyureas I–III undergo a quantitative depolycondensation process analogous to that observed for N-monosubstituted polyurethanes. The thermal decomposition of polyureas IV–VI proceeds by intramolecular hydrogen transfer processes that occur at higher temperatures with respect to depolycondensation. Polycarboxypiperazine VI is decomposed by a single-stage decomposition mechanism that leads to fragments with amino end groups and carbon oxide.

I took a look at the article.  Depolycondensation is accurate: most of the polyureas examined decomposed to the "parent" isocyanates and amines.  The temperature range examined was 150°C to 350°C.

If you're interested in what happens at higher temperatures, I would guess a decomposition to mixtures of CO, ammonia, water, and possibly methane and/or hydrogen.  But I didn't find any references that would support me after a quick look


Transition metal or transition element? With reference to the periodic table’s d-block, are the terms “transition metals” and “transition elements” interchangeable when describing these elements?
Are there any situations where it would be incorrect to use one phrase instead of the other?

Properly speaking, a "metal" is any substance with certain specific properties; chiefly, that it presents the charge transport behaviour of an electrical conductor. Since these properties are phase-dependent, it'd be more accurate to talk about "metallic phases" rather than "metallic substances" - as a given substance can present both metallic and non-metallic phases (to cite a famous, even if still not completely settled, example, hydrogen is thought to present a metallic phases, but not all hydrogen phases are metallic).

There are other ways to conceptualise metallic properties (such as a focus on band structure, or in terms of bonding) but these are pretty much equivalent to the definition that focuses on electrical conductor behaviour, as these are the underlying mechanisms that explain electrical conductors.

In any case, is it correct to identify certain elements as "metals", beyond talking about a specific metallic phase?

The answer is that no, it's not formally correct (IUPAC Gold Book prefers the term "transition elements"), but it is a well-established term, so when we refer to "metals" there is usually little ambiguity. Note that the term metal is also used in accepted definitions that focus on the properties and not on the elements (such as "molecular metal").


What is a rigorous definition of gas volume, and how is the Van der Waals equation derived? I am confused about the justification for the corrections to the ideal gas law in the Van der Waals equation: $$p=\frac{nRT}{V-nb}-a\left(\frac{n}{V}\right)^2$$

I understand that the equation attempts to correct for attractive and repulsive forces between molecules, and that at high volumes the corrections are negligible, at intermediate volumes the $V-nb$ correction dominates (since attractions are dominant and this shrinks the volume, as should occur for attractions), and at low volumes the $a(n/V)^2$ correction dominates, since repulsions are dominant.  However, the 7th edition of Atkins' and de Paula's Physical Chemistry justifies part of this equation as follows:

  The repulsive interactions are taken into account by supposing that they cause the molecules to behave as small but impenetrable spheres.  The nonzero volume of the molecules implies that instead of moving in a volume $V$ they are restricted to a smaller volume $V-nb$.  This argument suggests that the perfect gas law $p=nRT/V$ should be replaced by $$p=\frac{nRT}{V-nb}$$ when repulsions are significant.

I am confused by this argument-- what exactly is the definition of 'volume' here?  They seem to be saying that it is the empty space around the gas molecules, but it seems to me that volume should be defined as the space 'taken up' by the gas.  Why should the space taken up by the particles themselves be subtracted?  This isn't done for solids or liquids as far as I know.

These thoughts have also made me realize that I'm not quite sure what it means for a gas to 'take up' space.  Does anyone have a rigorous definition of gas volume?

EDIT: The exchange with Chris in comments has raised further questions.  It now seems to me that the $V-nb$ correction actually accounts for repulsions rather than attractions.  I think I was incorrect in thinking that the $V-nb$ correction dominated at intermediate volumes and $a(n/V)^2$ at low volumes.  If $V-nb$ is for repulsions, then it should dominate at low volumes, but I can't tell from the equation which correction will actually be dominant.  Also I am now wondering whether it even makes sense to connect one correction with repulsions and one with attractions.

EDIT: Follow-up questions for F'x:
- I thought that 'density' meant mass/volume.  Is the use of it to represent the inverse of molar volume (as you have used it) common?
- Where is the phase-transition in the red van der Waals curve?
- I am still a little unclear on gas volume.  Are the 'volume available to the gas' and 'volume of the gas' the same?  You say it's the same as the shape of the container, but aren't we subtracting the volume of the actual gas particles from that?  The $V$ term represents the volume of the gas, correct?  And $V-nb$ represents the 'volume available to the gas'.

I edited the first van der Waals equations in your question, because it was incorrect.

First, the volume available to the gas is pretty much was you think: it's the space left for it to occupy, i.e. the volume delimited by the container. If you think of a gas tank, it's the interior volume of the tank. For systems of macroscopic dimensions, there is no real ambiguity as to its definition, which can be purely geometrical. For nanoscale systems, such as accessible volume in porous media, the definition is much harder: it is typically done by constructing a Gibbs dividing surface between the two phases considered (here, a solid and a gas).

Second, importance of both terms in a real-life example. Let's write the van der Waals equation in terms of density, rather than volume and quantity:

$$ \left(P+a\rho^2\right)\left(\frac{1}{\rho}-b\right)=RT $$

Let's take a practical example of a gas with a large quadrupolar moment, CO2. We have $a$ and $b$ parameters for it from this source, and we can plot $\rho(P)$ at $T=300$ K:

(black: ideal gas; red: van der Waals gas; green: only $b$; blue: only $a$). We see that excluded-volume correction by itself is not much of a gain, but it's actually crucial once you include the attraction, because otherwise the phase transition (the cusp in the blue curve) occurs at much too low pressure.

Finally, regarding the derivation of the van der Waals equation, this is typically a topic of statistical thermodynamics and there are many such demonstrations. My favorite is the classical one by Hill, but it may not use the notations in modern textbooks. I'm also partial to that in McQuarrie’s book. All statistical physics derivation introduce the excluded volume in a very natural fashion, but they show more variation on the way they arrive at the $a$ term (so of them find the exact analytical form of the equation, while in some derivations you find a series expansion whose first term is the van der Waals equation).


Analysis of cations using sodium thiosulphate A question in an exam was as follows:[1]

How many of the following salts are first soluble in $\ce{Na2S2O3}$ solution, followed by the formation of a white precipitate and finally the precipitate turns black on standing
$\ce{AgNO3}$, $\ce{Pb(CH3COO)2}$, $\ce{Hg(NO3)2}$, $\ce{BiCl3}$, $\ce{CuSO4}$

The answer given is:

 $5$, which means that all of them give this test.

I understand that this is a test for cations since the reagent is sodium thiosulphate.
A pre-emptive guess on what the reactions would be that the first reaction forms the thiosulphate of the cation followed by the conversion of thiosulphates into sulphides (from the color change given).
However, I was unable to find any text referencing this test and so haven't been able to verify the actual reactions that take place.[1]
What cations does this test give a positive result for? What are the reactions that take place?

[1]: The previous version of the question asked for a test using sodium dithionite which was a typo (due to reasons provided in the answer below), confirmed after M.Farooq's suggestion to ask the professor who set the exam.

I do not think the question had sodium dithionite. This is a mistake. It must be sodium thiosulfate, because sodium dithionite is such a strong reducing agent that it will not solubilize these metals. The properties mentioned in the answer match with thiosulfate.
Sodium dithionite is such a strong reducing agent that it will reduce copper (II) to copper.
With this correction...
Interesting question but requires some knowledge of colors of sulfides. Sometimes you would have to use intuition in such exam questions. The hint is "...finally turn black on standing". Recall that silver tarnish is brown black, so are lead, copper and bismuth sulfides. Mercury also forms black sulfide. So in a state of exams, a safe bet would be all cations.
Thiosulfate has complex properties (a) it is a strong reducing agent (b) it forms complexes and (c) it decomposes under various conditions to pure sulfur and sulfite ion in acidic medium.
From Jander and Blasius, Textbook of Analytical and Preparative Inorganic Chemistry (in German)

Von den Salzen sind nur $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ und $\ce{PbS2O3}$ schwer löslich. Sie werden jedoch durch Sodalösung quantitativ in das lösliche Natriumsalz überführt. Mit vielen Schwermetallionen (z.B. $\ce{Ag(I)}$, $\ce{Fe(II)}$ und $\ce{Cu (I)}$) bildet Thiosulfat im Überschuss lösliche Komplexe.

An english translation of this would be

Of the salts only $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ and $\ce{PbS2O3}$ are poorly soluble. However, they are quantitatively converted into the soluble sodium salt by soda solution. With many heavy metal ions (e.g. $\ce{Ag(I)}$, $\ce{Fe(II)}$ and $\ce{Cu(I)}$) an excess of thiosulfate forms soluble complexes.

In the book the equations of conversion to sulfide are given as. Ask your teacher if this a typo in the exam.

4. Nachweis als $\ce{AgS2O3/Ag2S}$
\begin{align}
\ce{S2O3^2- +2Ag+ &-> Ag2S2O3\downarrow} \\
\ce{Ag2S2O3 + 3S2O3^2- &-> 2[Ag(S2O3)2]^3-} \\
\ce{Ag2S2O3 + H2O &-> Ag2S + H2SO4} \\
\end{align}

That is all you have to remember that thiosulfate will precipitate $\ce{Ba}$, $\ce{Ag}$ and $\ce{Pb}$ initially but excess would dissolve it.
I would be interested to know otherwise.


Tert-butoxide in SN2 reactions Can Tert-butoxide displace bromide via SN2 reaction? I read that Tert-butoxide is bulky and has steric hindrance, even with primary substrates. How is this reaction (picture below) possible?

Source: Levy, Arrow pushing in Organic Chemistry

This seems interesting.
The $(CH_3)_3CO^-$ is a strong nucleophile and we observe here methyl bromide is being attacked and there is no point of elimination. There has to substitution in this case.
The bulkier the nucleophile is the more probable is the elimination product due to steric hindrance. In majority of the cases Tert-butoxide undergoes elimination reaction.
The further factor includes temperature at which the reaction is being carried out.

We can show this relation with $$\delta G  = \delta H - T \delta S$$
Hope it was helpful.


Quantum mechanics electron probability Schrödinger's equation yields us the solutions of 90% probability of presence of electrons in orbitals. Hence, the electrons in theory should be able to exist elsewhere in space as well.
In that case, they could exist in another atoms' vicinity, possibly in its orbital as well. So, what would be the quantum numbers in this case? Shouldn't they be affected by possible interference, hence changin quantum properties like spin, etc.
P.S. I have very limited knowledge on quantum mechanics, so please excuse me if any statement of mine is incorrect or sounds rudimentary.

The issue here is that Schrödinger's equation doesn't yield us a "solution of 90% probability". It doesn't yield a probability at all. It yields a wave function, $\psi(x,y,z,t)$, which has units $\frac{1}{\sqrt{m^3s}}$. To get a probability density, we square its norm: $|\psi|^2=\psi^*\psi$. Note: even this is not a probability. A probability is dimensionless, whereas this has units $\frac{1}{{m^3s}}$. We can represent the probability of the particle described by $\psi$ being in the range (x, y, z, t) and (x+dx, y+dy, z+dz, t+dt) by multiplying the norm squared by dxdydz*dt: $P(x,y,z,t) = |\psi(x,y,z,t)|^2dxdydzdt$. Of course, these increments are infinitesimal, so this alone would evaluate to zero. However, this approach allows you to integrate over a range of x,y,z,and t:
$P(x+\Delta{x},y+\Delta{y},z+\Delta{z},t+\Delta{t}) = \int{dx\int{dy\int{dz\int{dt|\psi(x,y,z,t)|^2}}}}$.
A better understanding of this will help you understand the issues with your second question. However, we can also rely on the Pauli exclusion principle to assert that there cannot be two Fermions that share the same set of quantum numbers (note that some of the quantum numbers may be shared, but not all simultaneously.


What's the least common element in the universe? What's the least common element in the universe? And what is its frequency measured in Hertz?

(edited based on @JonCuster's comment) The rarest known elements are most probably the man-made heavy elements. These isotopes (elements) are very unstable. They most probably also formed in supernovas, but their natural formation is minimal, and due to their short lifetime, they do not accumulate either.
As Jon added, most probably heavier elements are also formed in supernovas, however, I am not aware of any of them observed. The reason for that is 1) again, short lifetime prevents any large accumulation, 2) under the formation conditions they are not atoms, but bare nuclei /plasma, so one cannot observe their atomic spectrum.


Why does selenophene not undergo aromatic substitution? Other derivatives like thiophene, pyrrole give electrophilic substitution but why not selenophene? Is it because selenophene does not have enough electron donating effect and not having its electrons in conjugation, making the ring anti-aromatic?

The Se atom has such diffuse valence orbitals compared with carbon that there is not good pi overlap, so we don't really have a conjugated cycle.  By definition when there isn't a conjugated cycle, it's "not aromatic", not anti-aromatic or non-aromatic as they both require cyclic conjugation. Words involving 'aromatic' are tricky.


Comparing the zero point vibrational energies of DCl and HCl (a) Compare the zero point energy (i.e. $E_0)$ in joules for $\ce{H^{35}Cl}$ and $\ce{D^{35}Cl}$ (assuming both are anharmonic oscillators).
(b) Explain the effect this will this have on their chemistry (you may wish to use a diagram to help here).

I have calculated a value for the vibrational frequency and have a value of the anharmonicity constant for $\ce{H^{35}Cl}$ (but not the anharmonicity constant for $\ce{D^{35}Cl}).$ I am unsure on how to proceed with these as all I know is an equation linking vibrational energy to the vibrational frequency and anharmonicity constant. I would be so grateful for any insight into this.

I don't know which equations you are using, but this is the one I was taught:
$$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$
Here, $\tilde{\nu}$ is the wavenumber of the vibration; $v$ is the quantum number that represents the vibrational levels $(v= 0, 1, 2,\ldots);$ $\chi$ is the anharmonicity constant:
$$\chi=\frac{\tilde{\nu}}{4\,D_\mathrm{e}},\tag{2}$$
where $D_\mathrm{e}$ is the depth of the lowest point of the potential well (not the depth of the $v = 0$ vibrational level!).
The zero-point energy is the energy when $v=0$ i.e. the molecule is in the lowest vibrational state. So you have to substitute that in the energy equation.
Now, we can assume that the well depth $(D_\mathrm{e})$ of the $\ce{H-Cl}$ and $\ce{D-Cl}$ bonds are same.
The only difference between $\ce{H-Cl}$ and $\ce{D-Cl}$ would be in the value of the vibrational wavenumber $(\tilde{\nu}),$ which you can get from the vibrational frequencies that you already have, as you wrote in the question. Once you substitute that in the equation, you will get your answer.


Why are paramagnetic materials attracted to ferrimagnetic materials(such as iron, nickel, and cobalt)? Diamagnetism exists in every molecule or ionic compound. It is the ability to be repelled by a magnetic field.

So you might think from this, that liquid oxygen would either be repelled by the magnet or not affected at all, yet it is attracted to it.

$\ce{O2}$ is paramagnetic, but how can a paramagnetic material like oxygen be attracted to a ferrimagnetic material like Iron other than the fact, that neutral attracts both positive and negative charges and $\ce{O2}$ is neutral.

Is it just this, or is there another part to it, like charged iron atoms inducing a dipole in $\ce{O2}$ or something?

First of all, in your question i feel a confusion about the nature of magnetism and paramagnetism. These magnetic interactions have nothing to do with attraction/repulsion of electric charges, and magnets do not attract or repel Coulomb charges (when we discuss the magnetic part of the interaction). In short static, magnetic and electric properties are two different stories.

Second, iron is ferrOmagnetic and not ferrImagnetic (capitalisation is by me). Forget ferrimagnetic materials for a second, and let's just talk about ferromagnetic, paramagnetic and diamagnetic materials:

Diamagnetic materials have no free electronic spin in the system, therefore they do not interact with magnetic field as a first approximation. There is a very small repulsive interaction coming from the polarisation of paired spins, but it is very small compared to any other interactions. Most organic compounds are in this category.
Paramagnetic materials have free electronic spins, which can flip / rotate. When put the sample in magnetic field, this field polarises the spins which turns into an attractive interaction. This is analogous with the electric field induced dipole moments, but this is a magnetic interaction, not an electronic interaction. The alignment of spins are constantly changing, as simple thermal excitation can rotate them. Therefore paramagnetic materials has no permanent magnetic moments like magnets. Typical examples are molecular oxygen, transition metal complexes, etc.
In ferromagnetic materials the free electronic spins are having a strong interaction with each other, and aligned parallel in strict order. This alignment is so stable that the magnetic moment of spins adds up, and there can be a macroscopical magnetic field around the sample (i.e. it is a magnet).

When you see the attraction between paramagnetic and magnetic materials, you see the polarisation of the paramagnetic materials in the magnetic field of ferromagnetic materials, which is an attractive itneraction.


Are these organo-metallic compounds? My understanding says organo-metallic = organic + metallic compunds.

One already known to me is Grignard Reagent ($\ce{RMgX}$).

But would you consider $\ce{NaOMe}$ or $\ce{NaCCH}$ as an organometallic compound? Would you consider $\ce{LiOAc}$, $\ce{LiOMe}$, $\ce{LiN(Me)2}$ or $\ce{LiMe}$?

To my understanding, only $\ce{NaC#CH}$ and $\ce{LiCH3}$ are organometallic compounds.

According to the IUPAC definition of organometallic compounds, these must contain

  at least one metal-carbon bond between an organic molecule, ion, or radical and a metal.


In spectroscopy, is it possible for the sample to be excited multiple times? The context of my question is rotational spectroscopy (using microwave radiation), where there is the $J = \pm 1$ selection rule. The way I understand spectroscopy is that the sample is irradiated with a range of frequencies. Is it possible, during the irradiation of the sample (say CO), for the molecule to be excited by 2 levels by 2 successive photons of the appropriate frequencies?

Now this does not affect the locations of the peaks per se, because the relaxation of course has to happen one level at a time. I am specifically wondering about the distribution of peaks, which is supposed to follow the Boltzmann distribution. Let us say, for example, we have a sample at absolute zero, where all the molecules are at the ground state $J = 0$. Is it possible for a molecule to be excited twice (or more), leading to the rotational spectrum showing more than one peak?

Thank you in advance.

Very technically? Yes.

Realistically? The probability is small enough that even if it does happen, the peaks for the multiple transitions are going to be small enough that we cannot really observe them on the spectrum. The lifetime of a given excited state is so small compared to the analogous time in the ground state that it can basically be considered zero for the situation you're talking about. I'm sure it has happened, but compared to the one at a time transitions, your spectrum isn't really going to show it.


Is it possible to displace cerium from cerium sulphate using aluminium to get pure cerium metal? I'm unable to find cerium any reactivity series, hence the question..

No
The std. electrode potential for Ce3+/Ce is -2.33V, the std. electrode potential for Al3+/Al is -1.66V. This shows that Al is not reactive enough to reduce Ce(III), it is reactive enough to take Ce(IV) to Ce(III) so I think you would get a mixed salt of Ce/Al sulfate.
This reference here states that calcium metal is used with cerium fluoride as a preparative method.
Note: you can't do this in aqueous solution. Your reactive metal will react with the water preferentially.


Can the compounds NOF3 and POF3 both exist?Why? Today I came across a question (in one of my books) like

  Can the compounds $\ce{NOF3}$ and $\ce{POF3}$ both exist? Why?

After some googling I found that both of them do indeed exist. I can understand that $\ce{P}$ can extend its octet due to empty d orbitals. But $\ce{N}$ can't do so. So how can it possibly exist?

Both molecules can be considered the acid (hydrate) trifluoride of the respective acid (phosphoric acid or nitric acid). In the case of phosphoric acid, this is immediately obvious: Just replace the three hydroxide groups with a fluorine atom, each. For the nitrogen compound, you initially need to (formally) add water before you can do the substitution:

$$\ce{O=N+(OH)-O- + H2O -> [(HO)3-N+-O- ] -> F3N+-O-}$$

(Note: this is not a proper chemical reaction, merely an illustration of the thought processes.)

You could also approach this compound from the ammonium cation $\ce{NH4+}$ by replacing three hydrogen atoms with fluorine and the fourth with oxygen. Or you start off with $\ce{NF3}$ (isoelectronic to $\ce{PF3}$ and then oxidise the nitrogen by adding a single oxygen atom much in the same way you would oxidise from $\ce{HNO2}$ to $\ce{HNO3}$.

In fact, for many of the compounds traditionally explained with d-orbital participation for main-group elements, an explanation adhering to the octet rule (no more than eight valence electrons per main-group atom; two for hydrogen/helium) is often closer to the truth. This includes, but is not limited to $\ce{OPF3, SO4^2-}$, sulphonic acids and many more.


No current in homemade galvanic cell I built a galvanic cell using 0.01 M solutions of copper sulfate and zinc sulfate. I used copper and zinc electrodes and immersed them into the proper solutions (copper into copper sulfate, zinc into zinc sulfate).

I used approximately 30 mL of each solution and put them into separate beakers. I connected the solutions using a salt bridge that was filled with 0.01 M potassium nitrate solution. I used cotton to cover the ends of the salt bridge (a plastic tube).

I measured the voltage across the copper and zinc electrodes and I got around 1 V. However, there is no current being produced according to an ammeter. The galvanic cell cannot light up a small light bulb for instance.

Furthermore, I took voltage measurements over a 50 minute period and the the voltage looked relatively constant throughout the period, though very slightly lower at the end.

Your small copper-zinc cell may not light up a bulb as often shown in textbooks. Check your ammeter settings, it must show some current. How are you connecting it in series? voltmeter is connected in parallel. Now an interesting question. Imagine you join 12 (e.g., Duracells) in series to generate about 12 Volts. Can one start a car with 12 Duracells? Why do you need a large car lead-acid battery? The answer lies in current. Same potential difference does not mean same amount of current supply. You may think 1 V is close to an ordinary cell, your single copper-zinc cell cannot supply enough current to even light up a bulb. Cell potential is a thermomdynamic concept. Current supply is a kinetic issue-how fast the electrons transfer to or from electrodes. Now you can think of the reason why there is no commercial battery made from zinc and copper. See Daniell cell for an interesting history. Check the size of electrodes in Daniell cells.


Negative Molar Specific Heat of a Gas It is known that the specific heat of a gas is process dependent. So it must be theoretically possible to have a negative value for a gas according to the following equation (for polytropic process, $PV^n = \mathrm{const}$):

$$C = \frac R{γ - 1} + \frac R{1 - n}$$

where $C$ is molar specific heat and $γ$ is adiabatic exponent.

Supposing $γ$ is $\frac{5}{3}$ and $n$ is $\frac{4}{3}$, $C$ comes out to be negative. Is it practically possible and if so what would it signify? As you provide more heat to a gas in such a process, would it lose temperature?
Please clarify.

I'm going to continue to resist calling C the heat capacity because, as used here, it is not path-independent.  Instead, I will take it to merely represent the derivative of the heat added per mole of ideal gas with respect to temperature for a polytropic process path.

With this caveat, here is my interpretation of what is happening mathematically when C takes on negative values:  If the gas is expanding, a negative value of C means that we added heat to the gas during the expansion, but not enough to offset the expansion cooling that would occur in an adiabatic reversible expansion, so its temperature decreases while we add heat.  If the gas is being compressed, a negative value of C means that we removed heat from the gas during the compression, but not enough to offset the compression heating that would occur in an adiabatic reversible compression, so its temperature increases while we remove heat.  It's as simple as that.


Why do different gases have the same molar volume? Molar volume:

  The molar volume, symbol $V_\mathrm m$ is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

At S.T.P conditions the molar volume of any ideal gas is $\pu{22.711 dm^3}$. My question is: how can two gases with giant difference in molecular size (say $\ce{He}$ and $\ce{Fe}$) have the same volume, considering that the size difference between them is huge, shouldn't the volume they occupy also be different?

Thanks to @MaxW for pointing out a fault.

You have the right thought process; if two real gases have different sized molecules, they won't have the same molar volume. The issue is that your statement about gases at STP refers to ideal gases which by assumption are just point particles that only interact via elastic collisions. To get a better sense of the difference that considering the finite size of molecules and their electrostatic interactions can make, you will want to read up on the various real gas models used, such as the Van der Waals, Redlich-Kwong, and Peng-Robinson models.

As a small side note, MaxW is correct that the definition of STP you are using is out of date so the molar volume at STP with the new definition is slightly higher. Its still quite common to see people accidentally continuing to use the old definition, so be on the lookout for that.


Is it possible to achieve a level of "truly zero" concentration? If we take some aqueous solution and dilute it further and further, will the concentration of the solution ever get to zero? I would say no, simply because total dilution implies that all the molecules of the solute have literally disappeared. But, the fact that I am unable to figure out where the molecules have gone doesn't make my argument compelling at all. I am led to believe that there is a far better answer and/or explanation to my questionUPDATE: While it is true that this question closely resembles the linked one, the answer provided in this question is a lot better as it gives a much deeper insight into the dilution process.

It depends how you dilute it. If you take an aqueous solution of A and just add pure water (absolutely 100% water), the concentration of A will never quite be null. In this case however, you will reach a point where the concentration of A is so small that it can be considered null for your applications.
If, however, you dilute the solution, take a sample, then dilute that sample (and so on), you could reach a concentration of exactly 0M. Imagine you have diluted the solution enough so that it contains exactly 1 molecule of A. When you take your sample for the next dilution, if this molecule isn't in the sample, the concentration will be exactly null. If it does happen to be in the sample, it could be left behind when you draw the next sample, or the next, and so on.
In practice though, the water you use for the dilution will likely contain impurities. You will maybe not achieve exactly 0M, but the concentration could be so small that it is undetectable and have no measurable consequence.


How can the size of atom be the cause of octet expansion? Lately, I was reading The Lewis Theory of Covalent Bonding by Peter Atkins in Appendix 4 of 'Elements of Physical Chemistry'. There he was talking about expansion of octet . As he wrote:

  Many molecules cannot be written in a way that conforms to the octet rule. Those classified as hypervalent molecules requires an expansion of octet. Although it is often stated that octet expansion requires the involvement of d-orbitals, and is therefore confined to Period 3 and subsequent elements, there is good evidence to suggest that octet expansion is a consequence of an atom's size, not its intrinsic orbital structure. [...]

I had always bore in mind that octet expansion was always due to d orbitals.

But size of atoms? How can it be related to octet expansion?

Your revised version of Atkins’ statement is correct, no hypervalent main group compounds require the involvement of d-orbitals. Rather, all of those molecules formerly explained by d-orbital participation can just as well be explained with 4-electron-3-centre bonds (including resonance structures) and charge separation (e.g. in sulphate; no need for resonance), giving rise to all-octet Lewis structures. In fact, calculations have shown the d-orbital participation to be minimal and neglegible at best.

A much more relevant question is whether more than $x$ substituents actually fit around the central atom. This is especially notable for the difference between the second and third periods: The reason why nitric acid is $\ce{HNO3}$ while phosphoric acid is $\ce{H3PO4}$ is likely due to the much smaller size of nitrogen, not allowing for tetracoordination by oxygens in the $\mathrm{+V}$ oxidation state while the larger phosphorus atom can support four oxygen atoms. If I remember correctly, that was also discussed as a reason why $\ce{NF5}$ hadn’t been found at the time of Klapötke’s article while $\ce{PF5}$ has been known for quite a long time.

It gets more pronounced when you go even lower in the periodic table, i.e. to higher periods. Antimony is known to be coordinated by six oxygens in the $\mathrm{+V}$ oxidation state ($\ce{H7SbO6}$, if you wish, although $\ce{H[Sb(OH)6]}$ is a more accurate description) — something phosphorus could technically do ($\ce{PF6-}$ with small fluorine atoms is known) but usually doesn’t. It is mainly the greater atom size of antimony that allows the coordination of two additional oxygen atoms.


What is the purpose of the Carothers equation? I don't really understand the equation. I feel like it is simply demonstrating the kind of exponentiality characteristic for the step polymerization reactions, but it cannot be useful for actual calculations. I mean - according to the equation a system which has executed, let's say, a polymerization reaction with 99% conversion will have an average degree of polymerization of 100 even if the system has fewer than 100 monomers. Am I missing something?

It is an equation useful for describing the behaviour of a large system, not predicting the fate of a few monomers.
Polymerization is a stochastic process—if we only have 80 monomers, the resulting polymer molecules will vary wildly in length if we perform the experiment several times (and logically, could not ever have an average degree of polymerization of 100), and the equation could not provide much insight into the results.
However, 80 monomers is a very tiny quantity.
If we take one of the monomers used to make PET, terephthalic acid, at 166.13 g/mol works out to around $2\times 10^{-23}\ \textrm{kg}$ for 80 molecules, a quantity that is almost impossible to actually measure out so it doesn't really matter that the equation doesn't work.
At a more practical scale, with even 1 g of the monomer, we have ~$10^{21}$ monomers and we will see the average behaviour of the system that the equation describes.


Basicity of substituted pyridines Which of the following pyridines are most basic?

I would've considered D, since chlorine is the most electronegative, and would therefore have the largest partial negative charge, making it more basic than the nitrogen substituent. The answer given, however, is B. How can I reason this for myself?

Basicity has everything to do with electron density on the electron-pair donor.  Electronegative subsituents actually decrease basicity because they draw electrons towards themselves, though nitrogen and chlorine have comparable electronegativities (3.04 and 3.16 respectively), so I'm not sure you can use that argument here. Instead of inductive effects, let's consider the resonance structures of our molecules:

$\hspace{5.6cm}$

A positive charge on a nitrogen is always more favorable than a positive charge on a chlorine. From this we know nitrogen is donating its electrons into the ring to a much greater extent than chlorine. Because the electron density is much greater on the heterocyclic nitrogen in B, it will therefore be the more basic of the two.


Can hydrogen on platinum reduce carboxylic acids and esters? Can $\ce{H2/Pt}$ reduce carboxylic acids and esters?

I am confused about the reducing nature of hydrogen in presence of platinum. In some books, it reduces carboxylic acid and ester while in others, not. I have checked Finar and Boyd books for it.

This is based off my memory, as far as I can remember what my teacher taught me.

Note, instead of $\ce{Pt}$, you can go for $\ce{Ni}$ (or Raney Nickel) or even $\ce{Pd}$ as a catalyst.
Functional groups such as acid halide, cyanide, nitro, aldehyde, ketone, alkene, and alkyne are easily reduced by $\ce{H2/Ni}$.  Some other groups can be reduced with $\ce{H2/Ni}$ as well, but if someone's asking you a question in an exam, they want you to refer to the list of groups which are easily reduced (more in the reactivity order below).  So the answer in your case would be a no.
Next, here's a reactivity order:
First, the groups which are easily reduced:
$$\begin{align} \text{acid halide} > \ce{R-C#C-R} > \ce{R-C=C-R} > \ce{R-CHO} > \\ &\text{ketone}>  \ce{R-CN} > \ce{R-NO2} > \ldots \end{align}$$
Then the groups that are reduced only with difficulty:
$$ \ldots > \ce{R-X} > \text{benzene} > \ce{R-COOH} > \text{amide} > \text{ester} > \text{acid anhydride}$$
$\ce{R-X}$ would require temperatures around $\pu{50^\circ C}$, benzene would require a temperature of about $\pu{100^\circ C}$ and groups from carboxylic acid to acid anhydride would require temperatures around $200-\pu{250^\circ C}$.


What does it mean to shift equilibrium? What does it mean to shift a chemical equilibrium? For example,

  the equilibrium shifts to the left …

I don't understand that.

Probably you are having problems with Le Chatelier's Principle.

Suppose you have an equilibrium established between four substances $\ce{A}$, $\ce{B}$, $\ce{C}$ and $\ce{D}$.

  What would happen if you changed the conditions by increasing the
  concentration of $\ce{A}$?

According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of $\ce{A}$ decreases again - by reacting it with $\ce{B}$ and turning it into $\ce{C + D}$. The position of equilibrium moves to the right. Of course, this assumes that there is still some amount of $\ce{B}$ left in the reaction vessel.

For better understanding, refer this.


pH calculation - adding distilled water after titration Will someone help me figure out how can I solve the following question?

We are given 110 ml of acetic acid (pKa=4.76), with concentration of 1M, to which we add 10 ml of NaOH with concentration of 1M.

Afterwards, we add the solution to 880 ml of distilled water.

What would be the pH of the solution, after the addition of water?

Will you please help me figure this out?

I obviously need to use Henderson-Hasselbalch here, but how does the addition of water change things?

Thanks a lot in advance

So I presume you can work out the initial H-H equation. I got pH 5.8? might be worth checking.
Then use the equation $$c_1 \cdot V_1 = c_2 \cdot V_2$$
Where $c_1 = [\ce{H+}]$

$[\ce{H+}] = 10^{-\mathrm{pH}}$

so I got $[\ce{H+}] =10^{-5.8} = 1.58489 \times 10^{-6}$

Now we can rearrange our original equation: $(c_1 \cdot V_1)/V_2 = c_2$

$c_1=1.585\times10^{-6}$

$V_1=120\ \mathrm{ml}=0.12\ \mathrm L$

$V_2=(120\ \mathrm{ml}+880\ \mathrm{ml})= 1\ \mathrm L$

$c_2=1.902\times10^{-7}$

$\mathrm{pH}=-\log[\ce{H+}]$

$\mathrm{pH}= -\log(1.902\times10^{-7})$

$\mathrm{pH}= 6.72$

Hope this helps, Its been a long time since I did anything like this so let me know how it goes and if this is correct. cheers.


Questions about air-quality testing in high school The Pinelands Regional High School in Ocean County New Jersey was closed for four days last week for air quality concerns after rooftop construction. Yesterday was the students‘ first day back. There have been conflicting reports of health issues, but it generally seems that the first day back is going OK.
There were two rounds of air-quality testing conducted last week. The first found unacceptable levels of multiple chemical compounds, the second, conducted two days later (after being cleaned), found no dangerous levels of any chemical.
Full results document of each test

Test 1: October 3
Test 2: October 5.

All of the unacceptable results found in round one (Indicated by red flags):
Sample: Room 164

Chemical | Actual ug/m3 | Times greater | NIOSH Res uh/m3

Ethelbenzene | 2.7 | 2.45 | 1.1
1,2,4-Trimethylbenzene | 15 | 2.05 | 7.3

Sample: Upper level - Media

Chemical | Actual ug/m3 | Times greater | NIOSH Res uh/m3

Benzene | 8.9 | 24.72 | 0.360
Ethelbenzene | 37 | 33.64 | 1.1
Xylene (p,m) | 110 | 1.1 | 100
1,2,4-Trimethylbenzene | 180 | 24.86 | 7.3
Naphthalene | 11 | 132.52 | 0.083

The clean results of round two

Chemical | Actual ug/m3 | NIOSH Res ppm

Benzene | 8.9 | 0.360
Ethelbenzene | 37 | 1.1
Xylene (p,m) | 110 | 100
1,2,4-Trimethylbenzene | 180| 7.3
Naphthalene | 11 | 0.083

This report concludes: "TTI has evaluated the results and can confirm that none of the target compounds exceeded limits for either exposure or notification at both the state and federal levels."
Confusion
There are a number of things I find confusing about this.

First, the  first round of tests analyze specific areas in the school, but the second does not identify any area.
Both rounds seem to have measured almost exactly the same levels.
The recommended NIOSH residential levels are measured in ug/m3 in round one testing, but ppm in round two. In round one, for example, the recommended maximum for benzene is 0.360, which according to this tool, translates to 0.00036 ppm. However, on the second test, has a recommended maximum of 0.1 ppm. Also, the same calculator translates benzene’s ug/m3 of 8.9 to 0.0089 ppm, but in the Test two results, it is translated to 0.0028.
Finally, parents are simply unfamiliar with these chemicals and what the levels imply, and how they affect teenagers. It is hard to fight what you can’t see or smell.

Help understanding and interpreting these results, and especially in giving parents some reassurance if it is warranted, would be appreciated. Thanks.

First off the acceptance levels the first time are way lower then the second time, you should ask why there is a large difference. (I have no clue, it's quite important since the second time those limits are NOT met!)

I found NO acceptance level for the chemicals (long term exposure.)
THE most accepted guidelines are usually set by the world health organisation and even they do not dare to set limits since it is difficult see http://www.euro.who.int/en/health-topics/environment-and-health/Housing-and-health/publications/2010/who-guidelines-for-indoor-air-quality-selected-pollutants
For common benzene they wrote a small book.

But to put it in perspective, I think that you inhale way more of those chemicals during filling your car. Then staying in the classroom for a week ;) [Citation needed!]
Edit: the concentrations are actually all more elevated then the averages mentioned by literature. Can we please have a copy of the raport? Next time it might be better to measure air quality according to the WHO guidelines since we simply do not know what we are looking at.

"First, the first round of tests analyze specific areas in the school, but the second does not identify any area.
Both rounds seem to have measured almost exactly the same levels."
I agree, this is weird. You should ask where and how this is sampled. But there are regulations for the sampling, let's assume they are professionals and based on previous results this is a worst case measurement.

"The recommended NIOSH residential levels are measured in ug/m3 in round one testing, but ppm in round two. In round one, for example, the recommended maximum for benzene is 0.360, which according to this tool, translates to 0.00036 ppm. However, on the second test, has a recommended maximum of 0.1 ppm. Also, the same calculator translates benzene’s if/m3 of 8.9 to 0.0089 ppm, but in the Test two results, it is translated to 0.0028."
PPM is a horrible unit without actual dimensions just like percentages. Are we talking about volumes, weights , molars or a mix? Anyway, just use the ug/m3 since that's what really Is measured. (I you like you can see the right conversions inside the WHO document http://www.euro.who.int/en/health-topics/environment-and-health/Housing-and-health/publications/2010/who-guidelines-for-indoor-air-quality-selected-pollutants but that's only if you like math ;)

"Finally, parents are simply unfamiliar with these chemicals and what the levels imply, and how they affect teenagers. It is hard to fight with you can’t see or smell."
I totally agree! But don't underestimate the nose ;)
I GUESSE, looking at the numbers there is no extreme dangerous exposure. The second report concludes the the same. The world health organisation does not exactly know what levels are really exceptional dangerous, therefore it's by "law save" as long as there are no observed health problems. But just ask the professionals ;)
This is where my field ends, it's up to you to tell the right people the right information and ask the right questions, (you are probably way better at that then I am ;) good luck!!


Specific volume of an air parcel So I have an unsaturated air parcel in the atmosphere that ascends adiabatically from the $1000\:\mathrm{mbar}$ level to the $700\:\mathrm{mbar}$ level with its initial temperature being $10\:°\mathrm{C}$.

I have to find the initial specific volume in $\mathrm{m^3}\:\mathrm{kg^{-1}}$.

I know that the specific volume formula is $v=1/\rho$ and that the equation of the state of an ideal gas can be written as $p=\rho RT$ where $R$ is $8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}}$

So I have $1000\:\mathrm{mbar}=\rho(8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}})(283.15\:\mathrm{K})$ and solve then for $\rho$ and then get $0.42477$ and plug that for $v=1/0.42477$ and get $2.354\:\mathrm{m^3}\:\mathrm{kg^{-1}}$ as the initial specific volume. Am I doing this right?

The ideal gas law states
$$ pV = nRT \Leftrightarrow \frac{n}{V} = \frac{p}{RT}\ .$$

For the density $\rho = m/V$ and the mass $m = M\cdot n$ with the molar mass $M$, we get a new formula for the density:
$$ \rho = M\frac{n}{V} = M\frac{p}{RT}$$

With the definition of the specific volume we get
$$ v = \frac{RT}{Mp}\ . $$

So unless you know the molar mass of your parcel of air, you will not get out a numerical value for the specific volume.


Anion-exchanged compound - smaller 1H NMR integrations I anion exchanged one of my compounds (Br- to PF6-) and the NMR integrations changed as shown on the two NMRs below. The aliphatic region peaks integrate to smaller values in the second one. Are there any known cases where PF6- somehow influences the relaxation time?

The other possibility is that this product decomposed to a similar compound, which would slightly throw off the integrations...

On face value, there is nothing here that necessarily suggests a change in relaxation times. Before thinking too hard about what chemistry is happening, you should assure us that your data is reliable and consistent. You should provide details of your experimental conditions; are you using a calibrated π/2 pulse for excitation or a shorter pulse approximation? What relaxation delay are you using? Have you measured T1 relaxations for these nuclei? You could consider repeating your experiment using a much greater relaxation rate between scans to see whether the integrations change.

These are more interesting phenomena in the spectra that need rationalising; loss of fine structure for downfield peaks for instance.

As for your question though, there are lots of examples of counterions changing relaxation times. This is a major tool used for probing outer sphere complexation, and can be used to determine how anions and cations interact around the ligand. It is used widely in protein structure (complexation with paramagnetic cations) and investigations with ionic liquids. If this is a counterion-induced relaxation change, you should be able to account for this by repeating your NMR experiment using appropriate parameters.


Precipitation reaction stoichiometry? I have a question where I do not understand the final step to solve...here it is:

What will be the Strontium ion concentration remaining after 30.0 mL of 0.10 M Na2SO4 solution are added to 70.0 mL of 0.20 M Sr(NO3)2 solution?

First, I wrote the equation: Sr + SO4 -> SrSo4 (I know this is excluding ions, because I am primarily focused on the stoichiometry part)

I figured moles of Na2So4 and Sr(NO3)2, which correspond to moles of strontium. These are .003 moles Na2SO4 and 0.014 moles Sr(NO3)2. This is where I got stuck, looked to my book's answer explanation, and got more confused. The book states
"0.0030 mol of sulfate ion will combine with 0.0030 mol of strontium ion, leaving 0.011 mol of strontium in a total volume of 100.0 mL."

Why is the reaction leaving 0.011 moles of strontium? What was the calculation that obtained this number?

Thank you!

Sr(NO3)2 + Na2SO4 = SrSO4 + 2NaNO3

So, here we have, 1 mol Sr(NO3)2 reacts with 1 mol of Na2SO4 by unitary method, you can say, 0.003 mol Sr(NO3)2 reacts with 0.003 mol of Na2SO4.
But checking the reactants we have 0.003 moles Na2SO4 and 0.014 moles Sr(NO3)2

So only 0.003 moles of each reacts with the other to leave out any excess in either of the reactants, as in this case, Sr(NO3)2.

When 0.003 mol reacts of Sr(NO3)2, we have $$0.014 (initial) - 0.003 (used) = 0.011mol$$ of Sr(NO3)2 left.


How many watt-hours are theoretically necessary to produce 2kg of methane from sea water and air? Given the composition of sea water in mol:
\begin{array}{lr}
  \ce{H2O}    &:& \pu{53.6 mol}\\
  \ce{Cl-}    &:& \pu{0.54 mol}\\
  \ce{Na+}    &:& \pu{0.469 mol}\\
  \ce{Mg^2+}  &:& \pu{0.053 mol}\\
  \ce{SO4^2-} &:& \pu{0.028 mol}\\
  \ce{Ca^2+}  &:& \pu{0.010 mol}\\
  \ce{K+}     &:& \pu{0.010 mol}\\
  \ce{Ct}~(\ce{CO2 + HCO3- + CO3^2-})
              &:& \pu{0.002 mol}\\
\end{array}

And the composition of air:
\begin{array}{lr}
  \ce{N2}    &:& \pu{78.08 mol}\\
  \ce{O2}    &:& \pu{20.95 mol}\\
  \ce{Ar}    &:& \pu{0.934 mol}\\
  \ce{CO2}   &:& \pu{0.04 mol}\\
\end{array}

My general problem: considering that these concentration do not change during the reaction (excess of reactive) and that the product ($\ce{CH4}$) is taken away immediately, how much energy is theoretically needed to produce $\pu{2 kg}$ of $\ce{CH4}$ from the optimal mix of these two phases?

My specific questions:

How can I be sure of which ions/molecule to definitely "get out" of the equation ?
How can I be sure which reactions are the best ones for my case ?
How do I do a dynamic equilibrium with more than 1 phase ?
How do I do a dynamic equilibrium with added energy ?

Too much for making it of any practical interest

This question cannot be answered easily, there are too many factors. Bytheway it would not be practically convenient making such a process. With $\ce{CO2}$ as your only source of carbon, you would need an extremely energy dispendious air separation apparatus to get some $\ce{CO2}$. You can get then hydrogen from water electrolysis and use  Sabatier reaction (you need a Nickel catalyst) to get finally methane.


Chemistry Spreadsheet Software? Are there any chemistry spreadsheet-like software that:

Generates 2D structure for the entered compounds (based on name or SMILES)
Easy to search and create customized reports (such as similarity search)

that preferably runs on linux and is not too expensive?

I'm aware of ICM Chemist and JChem for Office (Windows), and I'm wondering if there are any other alternatives available.

I can't comment on spreadsheet-like tools.

But I'm frequently using Open Babel, which is free (as in free speech and free beer).

If you're not so much into using command-line tools or using programmes via their bindings to programming languages, a GUI for Open Babel is available too, but I've never used it.

  Generates 2D structure for the entered compounds (based on name or SMILES)

Names might be tough in general, but SMILES (Simplified Molecular Input Line Entry System) can be processed with Open Babel when the identifier is read from a file. Output to large number of chemical formats and image formats is possible.

obabel -ismi stuff.smi -omol -O stuff.mol

would convert a SMILES string to a mol file.

Converting a collection of SMILES to MOL and writing them to a SDF file is not a big thing:

#!/usr/bin/env python

import os
import pybel

# dictionary of SMILES taken from chemspider.com
insecticides = {
    'parathion' : 'CCOP(=S)(OCC)OC1=CC=C(C=C1)[N+](=O)[O-]',
    'disulfaton' : 'CCOP(=S)(OCC)SCCSCC',
    'permethrin' : 'C1(C(C1C(=O)OCC2=CC(=CC=C2)OC3=CC=CC=C3)C=C(Cl)Cl)C',
    'furathiocarb' : 'CCCCOC(=O)N(C)SN(C)C(=O)OC1=CC=CC2=C1OC(C2)(C)C',
    'methomyl' : 'CC(=NOC(=O)NC)SC'
}

# write dict items to multi-component SDF file
outdir = '/home/klaus/workspace/chemistry/stackexchange/similarity/'
outname = 'insecticides.sdf'
outpath = os.path.join(outdir, outname)
outfmt = 'sdf'
datafile = pybel.Outputfile(outfmt, outpath, overwrite = True)

for name, smiles in insecticides.iteritems():
    mol = pybel.readstring('smi', smiles)
    mol.title = name
    datafile.write(mol)

datafile.close()

EDIT

  Easy to search and create customized reports (such as similarity search)

For the substructure search, I used the Drug-likeness NCI subset from ligand.info. The unpacked SDF file (445.2 MB) has 193,323 entries on 14,757,262 lines (wc -l).

From the SDF file, I first generated a binary fastsearch file using

obabel data.sdf -ofs -O data.fs

Searches are based on SMILES and SMARTS patterns.

I thought that 2-aminobenzothiazole might be an interesting fragment to search for and with a bit of fiddling, I wrote the suitable SMARTS pattern: c12ccccc1nc(N)s2

If in doubt about the correct representation, it is always a good idea to convert the pattern to an image file here: obabel -:"c12ccccc1nc(N)s2" -opng -O pattern.png.

The search was performed using:

obabel -ifs data.fs -s"c12ccccc1nc(N)s2" -osvg -O results.svg -xd -xc5 -xA -xC

and generated a SVG table with 233 entries.


How do I determine if a process is reversible without numeric information? How do I determine if a process is reversible without numeric information?

It would be nice to know how to determine whether a process is reversible in order to solve problems LIKE this one:

A weather balloon is filled with Helium gas and released from the ground. It goes up 18km and achieves a diameter of 15m. Determine if the following values are greater than zero, less than zero, or equal to zero: ΔV, ΔP, ΔT, ΔU, ΔH, S_sys, S_surr, S_tot.

Here is my attempt at solving this problem. I believe at least half of my error comes from not knowing how to determine if this process, or any other process merely described, is reversible.

Obviously, ΔV increases because the balloon expands as the diameter gets to be large compared to what it presumably was at the time of release, even though the diameter at that time isn't mentioned.
ΔV > 0

ΔP < 0 because pressure decreases as altitude increases.

ΔT < 0 because temperature increases as altitude increases.

ΔU < 0 because ΔT < 0
ΔH < 0 because ΔT < 0

It seems that I have assumed that the external properties of T and P apply to the inside of the balloon, so I guess the system is reversible.
Stot = 0

Ssys = -Ssurr

I don't really know where to go from here...

I want to argue that because the balloon actually does manage to overcome gravity, it's change in volume must be pretty substantial (making the w = -PdV be what causes the balloon to go up). Maybe it's more substantial than the change in temperature, giving a large positive term due to volume in the entropy state equation listed above.
Ssys > 0

However, if the system is being looked at as the balloon itself rather than the air inside the balloon, then when the balloon isn't inflated, it has more freedom of movement than after it's fully inflated and therefore more entropy, meaning that the final entropy change of the system would be negative.

If I am correct in guessing that this process is reversible AND in guessing that Ssys > 0, then Ssurr < 0.

I would rather not have all of my answers to physical chemistry problems be guesses. How can I solve this problem without guessing? How can I know if my guesses are correct or not? How can I better approach this problem? It's possible that the correct answer to this problem is due to something that I have completely overlooked. How do I not overlook considerations while doing these types of problems?

Any process is irreversible if it involves:

Friction
Heat transfer across a finite temperature difference
Expansion when there is a finite pressure difference
Mixing of difference substances under any real circumstances
Anything else where work is not being maximized

So air friction against the balloon rising at a non-zero velocity makes the process irreversible.

Heat transfer from the balloon when the inside temperature is different from the outside temperature makes the process irreversible.

For more information see: http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html


Synthesis of anhydrous sodium tungstate I am interested in preparing an anhydrous sample of sodium tungstate (Na2WO4), which is normally found as the dihydrate.  Finding an example of this in literature is proving difficult, I am interested if someone can suggest a preparation method based on prior experience or difficulties I should encounter.

Hoermann [1] prepared anhydrous sodium tungstate by melting the 1:1 mix of sodium carbonate and tungsten(VI) oxide (and, subsequently, growing single crystals); alternatively, he proposed prolonged drying of a dihydrate at 100 °C:

$\ce{Na2MoO3}$ und $\ce{Na2WO3}$ sind wasserfrei durch Zusammenschmelzen von 1 Mol $\ce{Na2CO3}$ mit 1 Mol $\ce{MoO3},$ bzw. $\ce{WO3},$ oder durch volliges Entwassern der Hydrate bei 100° zu erhalten. Die Salze zeigen ansgepragte Polymorphie.

Busey and Keller [2] obtained anhydrous sodium tungstate (reported water content 0.12%) by drying a dihydrate at 200 °C in vacuum. Purity has also been confirmed by Raman spectroscopy and powder x-ray diffraction:

The $\ce{Na2WO4}$ was prepared by heating the dihydrate at 200° in a vacuum.

References

Hoermann, F. Beitrag zur Kenntnis der Molybdate und Wolframate. Die binären Systeme: $\ce{Li3MoO4-MoO3},$ $\ce{Na2MoO4-MoO3},$ $\ce{K3MoO4-MoO3},$ $\ce{Li2WO4-WO3},$ $\ce{Na2WO4-WO3},$ $\ce{K2WO4-WO3},$ $\ce{Li2MoO4-Na2MoO4},$ $\ce{Li2WO4-Na2WO4},$ $\ce{Li2MoO4-K2MoO4}.$ Z. Anorg. Allg. Chem. 1929, 177 (1), 145–186. DOI: 10.1002/zaac.19291770117.
Busey, R. H.; Keller, O. L. Structure of the Aqueous Pertechnetate Ion by Raman and Infrared Spectroscopy. Raman and Infrared Spectra of Crystalline $\ce{KTcO4},$ $\ce{KReO4},$ $\ce{Na2MoO4},$ $\ce{Na2WO4},$ $\ce{Na2MoO4 · 2 H2O},$ and $\ce{Na2WO4 · 2 H2O}.$ The Journal of Chemical Physics 1964, 41 (1), 215–225. DOI: 10.1063/1.1725625.


Is it possible to dissolve Raspberry ketone in warm water? I poured a small amount of RK in cold water and it clumped together so I warmed it in the microwave and it disappeared in the water meaning it dissolved, but i was under the assumption that RK is insoluble in water, so what happened ?

According to PubChem, 4-(4-hydroxyphenyl)-2-butanone (Raspberry ketone) is estimated to be partially soluble in water, ca. 25.1 g/L @ 25 celcius.

https://pubchem.ncbi.nlm.nih.gov/compound/4-4-Hydroxyphenyl-2-butanone#section=Experimental-Properties

My initial thought is what would one need with Raspberry ketones? Their health effects seem to lack any clinical evidence.


What is the meaning of "charge separation" in resonance? I've read that the resonance structures in which there is lesser 'charge separation' contribute more to the resonance hybrid than the ones with greater charge separation.

But I am unable to understand why it is so.

I mean isn't it correct that the presence of two charges on adjacent atoms leads to "less stability" due to bond stretching or/and compression?

But apparently, the actual observation is contrary to this argument.

I guess I'm actually confused about the meaning of 'charge separation'? Does it refer to the distance between the location of the charges or the difference between their amounts?

There are two ways to think of charge separation and both make resonance structures less stable:

Moving electrons from a bond to a single atom to generate a cation/anion pair. For example:
$$\ce{R2C=CH-CH=CR2 <-> R2C=CH-\overset{+}{C}H-\overset{-}{C}R2}$$

Moving double bonds to separate opposite charges. For example:
$$\ce{R2C=CH-\overset{+}{C}H-\overset{-}{C}R2 <-> R2\overset{+}{C}-CH=CH-\overset{-}{C}R2}$$

Usually, however, the first way is meant. The general idea behind required a minimal charge separation is that in general (as long as we do not form super-octet structures) double bonds are lower in energy than separate charges on adjacent atoms as you correctly mention. Therefore, if we minimise the number of formal charges the idea is that we are concurrently maximising the number of double bonds, thereby generating a better resonance structure which contributes more to the actual composition.
On the other hand, even in the second case it is better to keep the charges together (in the absence of other effects such as strongly electronegative or electropositive atoms). This is because the charges attract each other and they do this better if they are close.


Does the Fermi level correspond to the highest occupied energy level? I know that, in the band structure of a solid, the Fermi level is the energy level at which there is a 50% probability of finding an electron (at thermodynamic equilibrium), and that it doesn't have to correspond to an actual energy level (ie it's in the band gap in an insulator).
But Atkins and Shriver's Inorganic Chemistry doesn't mention this and instead defines it as the highest occupied energy level at 0 K. (page 103, 5th edition)
I was wondering if these definitions are actually equivalent. If so, can you explain how? Given the fact that in an insulator the Fermi level is in the band gap, it seems like at best the definition in Atkins is oversimplified.

Be aware of Fermi level versus Fermi energy, as it seems to me the book speaks about the latter.
The quotes below are from both Wikipedia links.

In band structure theory, used in solid state physics to analyze the energy levels in a solid, the Fermi level can be considered to be a hypothetical energy level of an electron, such that at thermodynamic equilibrium this energy level would have a 50% probability of being occupied at any given time.

The Fermi energy is a concept in quantum mechanics usually referring to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature. In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy, whereas in a metal, the lowest occupied state is typically taken to mean the bottom of the conduction band.

The Fermi energy is only defined at absolute zero, while the Fermi level is defined for any temperature.

The Fermi energy is an energy difference (usually corresponding to a kinetic energy), whereas the Fermi level is a total energy level including kinetic energy and potential energy.

The Fermi energy can only be defined for non-interacting fermions (where the potential energy or band edge is a static, well defined quantity), whereas the Fermi level remains well defined even in complex interacting systems, at thermodynamic equilibrium.

Since the Fermi level in a metal at absolute zero is the energy of the highest occupied single particle state, then the Fermi energy in a metal is the energy difference between the Fermi level and lowest unoccupied single-particle state, at zero-temperature.

A collection of non interacting fermions, often called the Fermi gas is a simplified model useful in the solid state physics. There is supposed the considered fermions do not interact, so the energies of these fermion quantum states do not depend on the distribution of  these fermion quantum states.
A typical case would be the classical hydrogen atom model, applied on multielectron atoms. As electrons in the same atom interact quite a lot, it is not a good model.
For solid states, electrons and electron vacancies are often taken as  quasi fermions following Fermi-Dirac statistics for the Fermi gas,
as they are separated enough to be considered as non interacting.


What does stepped Ni(111) surface mean? I think (111) surface refers to a flat surface of atoms that intersect with x,y,z axis at 1.

Stepped surfaces generally refer to surfaces (of a particular index) which have 'steps' like a flight of stairs. A cursory search on google images can yield some useful examples, such as this one:

As you can see from the example, a stepped surface will be a surface which includes steps which go from one layer to the next.


Lewis base strength and the hydrogen proton My prof told me that we can compare Lewis base strength based on electron donation to the $\ce{H+}$ proton - i.e sigma bond formation with the $\ce{H+}$ proton - and proceeded to conclude that $\ce{S^{2-}}$ is a weaker Lewis base than $\ce{O^{2-}}$, since the former is a weaker base than the latter. He then proceeded to generalize that the oxide ion is always a stronger Lewis base than the sulfide ion.

Is this a valid method of comparing the strength of Lewis bases? I have two issues:

Isn't Lewis acid/base strength quantified using kinetics, not thermodynamics? Don't stronger LAs react faster, and weaker LAs react slower?
Even if this comparison were valid, would the comparison still be valid outside of water solution?

Does this mean that this question is invalid? No medium is specified.

I think he knows what he's talking about. In explaining the much greater stability of silver sulfide ($K_{\mathrm{sp}} \approx 6 \cdot 10^{-51}$) vs. $\ce{Ag2O}$ ($K_{\mathrm{sp}} = 3.5 \cdot 10^{-16}$), he notes that charge density does not explain this disparity. In fact, charge density would imply that the oxide salt should be more stable.

  For instance, although $\ce{O^2-}$ is a stronger B/L base than $\ce{S^2-}$, $\ce{S^2-}$ can be (and usually is) a stronger Lewis base than $\ce{O^2-}$ since S has a lower EN than O. Furthermore, recall that Lewis base strength depends on the Lewis acid with with the Lewis base interacts.

The text goes on to mention other factors such as having a metal cation with a surfeit of valence d electrons, and how that can populate the empty valence 3d orbitals of $\ce{S^2-}$. (S has 3d orbitals?)

This makes sense; a good nucleophile has less stabilized electrons - i.e. electrons that are further from the nucleus. It still seems that the questions posted above put B/L acidity ahead as the main deciding factor for sigma-donor strength.

One of the keys in the definition of Lewis acids and Lewis bases is that they react to form a Lewis adduct.  To compare the relative strengths of different Lewis bases, you would allow them to react with the same Lewis acid, form the required adduct and measure the enthalpy of formation of the adduct. The adduct with the largest enthalpy of formation would contain the strongest Lewis base.

1) So Lewis acid and base strength is about thermodynamics.

2) You could run the reaction in any solvent (or gas phase) as long as all of the Lewis acids or bases you want to compare are able to form an adduct in that solvent.  There is no reason why the order of a series of enthalpies of bond formation has to remain constant when you move from solvent to solvent, so your relative strengths could well change from one solvent to another.  The same applies to oxide \ sulphide relative Lewis base strengths if you substitute some other Lewis acid in place of the proton.  Again the relative ordering might well change.


Are all NO3- salts soluble in water? If so, why? All the examples of $\ce{NO3-}$ salts are soluble in water (all that I know about).
Is it always so or there is some salt which doesn't dissolve in water?

If so what is the reason behind it?

No, the correct way of putting it is
$$\mathrm{Almost~all~of~the~\mathbf{inorganic}~nitrate~salts~are~soluble~in~water. }$$

The families of organic nitrate salts are typically nitrates of azoles and imidazoles. Some bright examples are (R) & (S)-miconazole nitrates, isoconazole nitrate and econazol nitrate. Econazol nitrate (Other names: Spectazole, EN) is the most common organic nitrate salt, and is believed to be very slightly soluble in water, less than $0.1 \mathrm{\frac{g}{100~g}}$ (1, 2, 3)

In inorganic nitrate salts, with a bit of patience, I've noticed that barium nitrate, mercury(I) nitrate, and cobalt(III) nitrate are the least soluble of nitrate salts with a monoatomic cation. They are in the brink of insolubility, but I still wouldn't call them insoluble, since

Their solubility isn't that low at STP; somewhere around $\frac{5~\mathrm{g}}{100~\mathrm{g}}$. ($\ce{Hg2(NO3)2.2H2O}$ decomposes in water, so as RSC says)
Their solubility increases greatly with temperature.

The reason, you say?

Well, let's think of why one compound is soluble in water while the other isn't. When an ionic compound is formed, energy is released. This is called Lattice enthalpy in constant pressure. The process of formation of lattice from separate ions is usually highly exothermic. Therefore, when an ionic compound dissolves in water, and when this lattice "breaks up", the reaction would be highly endothermic.

So, to dissolve in water, the ions have to 'overcome' the lattice energy. How? Of course, the net resulting attractions between water molecules and the ions have to be stronger. This attraction is kind of a formation of a very very weak bond (and its weakness is the reason it's not categorized as such) and thus, is exothermic. This energy release when the dissolution occurs is called enthalpy of hydration, provided that the ions be in the gaseous state.

So, nitrate is a really big anion, with a single charge. The lesser concentration of negative charge results in relatively lesser lattice enthalpies. Also, the possibility of the creation of hydrogen bonds with water enhances solubility by increasing enthalpy of hydration.

That's why almost all of nitrates are soluble.


Carbocation Rearrangements in SN1 reactions Would one get 1-(1-chloroethyl)-1-methylcyclobutane?

I am not sure if the driving force identification is correct? Is my going from a 2° to a 3° carbocation correct? Does anyone have any tips for knowing when to look out for rearrangements?

I think the solvolysis would involve rearrangement to the cyclopentyl carbocation as indicated in the figure below.  The driving force for the rearrangement is two-fold

without rearrangement, the cyclobutyl compound would generate a secondary carbocation; with rearrangement a more stable tertiary carbocation is generated instead
release of strain as the 4-membered ring opens to a 5-membered ring

The tertiary cyclopentyl carbocation ultimately captures a solvent molecule as indicated.


Is there a better way balancing reaction equations than systematic trial and error? I was introduced to the concept of balancing chemical equations in school and I did some practice examples, which were easy and took me about 2 minutes per question. I used the systematic trial and error method and I don't really like it.

I'm worried that my teacher will give us a more difficult question, so I'm looking for a method to solve chemical equations such as
$$\ce{C3H8 + O2 -> CO2 + H2O}.$$

I have tried matrices, but I don't really understand them, as I am not in grade 12 yet. If you know matrices, can you explain how I can balance the chemical equation with aforementioned method (thoroughly)? I would also be grateful if you have any other useful and quick methods which work for pretty much any high school chemical equation balancing problem.

When using matrices, you are really just solving a system of equations. You can use any technique you know to solve the system of equations that a chemical equation represents. Using matrices is only one technique that may be useful for complicated equations or if you have a calculator/MATLAB on hand. In the given example, it would be easier to solve it directly with guess and check.
Writing the system of equations
In your example
$$\ce{C3H8 + O2 -> CO2 + H2O},$$
write in some variables as coefficients:
$$\ce{a C3H8 + b O2 -> c CO2 + d H2O}.$$
Now, we can write out a system of equations based on the stoichiometry of each element on either side of the equation. In other words, write equations that express the number of each element on the left hand side to the number of each element on the right hand side:
\begin{align}
3a &= c\\
8a &= 2d &\Longleftrightarrow&& 4a &= d\\
2b &= 2c + d
\end{align}
Finally, we have to choose a value for one variable to substitute in. This will be the fourth and final equation. We want integer coefficients here, so we will choose a substitution that gives us all integer coefficients. Otherwise, just multiply your coefficients at the end so that they are all integers. Here, I'll let $a = 1$.
Solving the system
Substitution
Substituting the first two equations into the third gives:
\begin{align}
2b &= 6a + 4a = 10a &\Longleftrightarrow&& b &= 5a
\end{align}
\begin{align}
a &= 1\\
b &= 5\\
c &= 3\\
d &= 4
\end{align}
And the balanced equation is
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$
Using matrices
To answer the question though, let's do it with matrices. I won't go through the whole explanation of solving the system but I'll show you how to build the matrices and get an answer. Remember, we are just representing the same system of equations from before, but with matrices now!
We need to build two matrices, $A$ and $b$. Reorganize the equations we wrote previously so that all of the variables are on the left hand side and a constant is on the right hand side. Also, notice I added in our equation let $a = 1$:
\begin{align}
1a + 0b + 0c + 0d &= 1\\
3a + 0b -1c + 0d &= 0\\
0a + 2b -2c - 1d &= 0\\
4a + 0b + 0c - 1d &= 0
\end{align}
These equations are written like this:
\begin{align}
A &=
\begin{bmatrix}
1 & 0 & 0 & 0\\
3 & 0 & -1 & 0\\
0 & 2 & -2 & -1\\
4 & 0 & 0 & -1\\
\end{bmatrix}\\
b &=
\begin{bmatrix}
1\\
0\\
0\\
0\\
\end{bmatrix}
\end{align}
Notice that $b$ is just a column of the constants on the right hand sides of the equations, in the same order as they were given. The order is important here!
There are a few ways to solve this manually or in a calculator/MATLAB. If you have a TI-8x calculator or similar, you could use its rref function (this transforms an augmented matrix into reduced row echelon form, I'm sure google could help you out) or you could multiply $A^{-1}b = x$. In MATLAB, just do: A\b. This gave me:
$$
x =
\begin{bmatrix}
1 \\ 5 \\ 3 \\ 4
\end{bmatrix}
$$
Each row in this final matrix $x$ is the value for $a$, $b$, $c$, $d$ in that order, giving the balanced chemical equation
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$


Can atoms delocalize in molecules? Electrons can delocalize over multiple equally probable bonding structures. Can atoms delocalize over multiple equally probably bonding structures in certain cases?

It seems to me that any such effect would be most pronounced with very small atoms such as hydrogen.

Furthermore, I would expect such effects to be mostly to occur with hydrogen bonds and cation-pi interactions.

One possibility I am imagining sticking the ends of a bunch of linear molecules that end in OH groups or fluorine atoms together.

Can atoms delocalize over multiple equally probably bonding structures in certain cases?

If it happens at all, it exceedingly rare. Through my 10-years work in chemistry, I didn't once found a case where it may be relevant. However, what does and indeed affect chemistry is reversible rearrangements.

The simplest case is rotation of molecules around single bonds. Another one is nitrogen inversion. More complex cases also exist. Such things sometimes make interpretation of NMR quite difficult. NMR gives a peak for each non-equal nucleus in the molecule. When internal rearrangement is involved, it sometimes makes two atoms 'equal', i.e. exchanging places too fast for NMR to catch the difference and producing a single peak instead. This, however, is 'quasi-delocalisation', since there is no need in resorting to delocalisation to explain said phenomena.

Another quantum phenomena often observed is tunneling. It affects slightly effective reaction rates/barriers and vibrational spectra.

Anyway, de Broglie wave length for proton is thousand times smaller than for electrons, so quantum effects in nucleus movement usually overshadowed by quantum effects in electrons movement.


What is an easy way to reduce a boronic acid to borane? I have a selection of boronic acids that I want to reduce to borane forms. Some of the boronic acids in question are phenylboronic acid (C6H5B(OH)2), tolylboronic acid (C7H7B(OH)2), and p-phenylenediborinic acid (C6H4(B(OH)2)2). I want to convert them into phenylborane (C6H5BH2), tolylborane (C7H7BH2), and p-phenylenediborane (C6H4(BH2)2).

I think I can do this by dissolving the boronic acid in ether and adding Lithium Aluminum Hydride (LAH). Giving a reaction that looks something like:

However, I am afraid that, given how aggressive a reducing agent LAH is I might get something like:

So I guess I have two questions:

Will LAH attack the boron-carbon bond shown in the second reaction diagram?
Can the reduction of the boronic acid be accomplished via a less aggressive reducing agent, perhaps Sodium BoroHydride?

Just a note: I am well aware that the boranes produced will be oxygen sensitive, either converting back to boric acid on exposure to oxygen, or catching fire.

LAH will attack the boron-oxygen bonds without going after the boron-carbon bond, see here (Eq. Z, p. 142).

But it also deprotonates the hydroxyl groups as a strong base like a Grignard reagent. You need to prevent that.  Convert your acid to an ester (OH -> OR), remove the water and then apply the LAH.