A cereal sample was analysed for Calcium content as follows: A cereal sample was analysed for Calcium content as follows: A 5g cereal sample was digested with HCl and the resulting mixture made up to 50ml in a volumetric flask. The absorbance of this solution was found to be 0.351. A 7ppm calcium solution gave an absorbance of 0.254. Assuming linearity between absorbance and concentration, calculate the calcium content of the cereal in mg Ca/100g cereal. (8 marks)

Seems so simple but is confusing me as when I take into account calcium content of the cereal in mg Ca/100g cereal.

A 5g cereal sample was digested with HCl and the resulting mixture made up to 50ml in a volumetric flask.

The absorbance of this solution was found to be 0.351.

A 7ppm calcium solution gave an absorbance of 0.254

7 ppm Ca solution/ absorbance 1 x absorbance 2 =

step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

step 2: 78.5158/50 ml  (of initial volume sample)= 1.57 ppm ca in the first solution.

Supposing the density of the first solution to be near that of water:

step 3: (1000g) x (1.57 x 10^-6) = 1564 g Ca

step 4: (1564 g ca / 5g) = 312.8 g/100 g

answer: 3.128x10^5 mg/100 g would be the initial answer I receive.

My second attempt looking at it a different way...

feeling that I have done step 1 correctly.  Starting at step 2

78.5158/(50 ml x 2) (multiplying initial volume by 2 to achieve 100 ml)
= .785158 ppm ca in 100 ml.

step 3: (1000g x 7.85x10^-4g)=7846 g ca /100 ml

7846000 mg ca / 100 ml

answer: 7.846 x 10^6 mg ca/ 100 ml

third attempt:

Seeing that the initial sample is 5 g and the question is looking for calcium content in a 100 g sample...

5g sample x 20 = 100 g

therefore other components are multiplied by 20

V= 50ml x 20 = 1000 ml

step 1: step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

step 2: 78.5158/1000 ml  (of initial volume sample x20)= 0.0785 ppm ca in the first solution.

step 3: (1000g) x (.0785 x 10^-6g) = 72.5 g Ca

step 4: (72.5 g ca / 100g) = .725 g Ca

answer: 725 mg Ca content/ 100 g cerial

I feel that I am missing a step in calculations.  I would prefer to read a resource of working this out myself than finding the answer.
I feel more confident in my third attempt of finding a correct answer.

This question is from a past paper.

The absorbance of this solution was found to be 0.351.
A 7ppm calcium solution gave an absorbance of 0.254

You have slightly (~1.5 times) higher absorbance (.35) than a solution of 7 ppm calcium (0.24)

  step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

This is not correct. You should have proportion
7/0.254 = x/0.351   leading to x = 7*0.351/0.254

You get 10 times higher value. This is the first error. You should have done 7*(0.351/0.254) = 9.67 ppm.

In 50 mL it is A, but when it was in 5 mL it was 50/5 = 10 times higher.
7*(0.351/0.254)*(50/5)=96.7 ppm.

This means you have 96.7 mg calcium per 1 kG of cereal (ppm is one millionth, and thus, mg per kG). Then one tenth of that amount of cereal (0.1 kg/1 kg) contains 1/10 of the calcium. that is 9.67 mg.

Once again, you should have had these steps:
So, you should have these steps:

(1) Since the final sample absorbs .35/.254 ~1.38 times more it is 1.38 times more concentrated than 7 ppm. That is 9.67 ppm.

(2) It was diluted 50/5= 10 times, so it was 10 times more concentrated before. That is 96.7 ppm.

(3) For 100 g of cereal: 100 000 mg * 96.7 /1 000 000 = 96.7 mg /10 = 9.67 mg.


Are there any negatively charged electrophiles, or simply neutral and positive? When reading up on electrophilic aromatic substitution, most resources refer to electrophiles as positively-charged species, "in general" or "mostly".

What are non-positively charged electrophiles? Are they simply neutral atoms/molecules more willing to gain electrons due to having less grasp on their own electrons? Are there any, as strange as it sounds, negatively charged electrophiles?

Yes, there are many neutral electrophiles.

One large class are the compounds with leaving groups. Some that you have likely encountered are: acids (HCl, HBr, sulfuric acid, etc) which are electrophilic at the hydrogen; unhindered alkyl halides/sulfonates (e.g., methyl iodide for SN2 reaction); dihalides (e.g., Br2 for electrophilic addition to an alkene).

A second large class are compounds with pi bonds to an electronegative atom. Most importantly this class includes the carbonyls (aldehydes, ketones, esters, etc). Michael acceptors fall into this class as well, which have an extended pi system containing an electronegative atom.

Since you are covering electrophilic aromatic substitution, soon you may learn about nucleophilic aromatic substitution, where a highly electron poor aromatic ring (e.g., having multiple nitro groups) bearing a leaving group can be attacked by a strong nucleophile.

In all of these cases, although the compounds are neutral, there is a partial positive charge at the electrophilic site explained by a polarized bond (acids, alkyl halides), resonance (carbonyl), or spontaneously (dihalides).

Neutral electrophiles are less reactive than their positively charged counterparts, which means that the nucleophiles are typically stronger for a reaction to occur.

As for negatively charged electrophiles, I can think of only a couple of cases that could be considered that way. For example, in the reduction of a carboxylic acid with lithium aluminum hydride, the first step is deprotonation of the acid to give a  carboxylate. LiAlH4 is very reactive and adds a hydride to the carboxylate. That could be seen as nucleophilic attack on a negatively charged electrophile, and many people will draw it that way. In reality though, there is something stabilizing the negative charge, either the lithium or aluminum. So it's not really a negatively charged carboxylate that's being attacked. I don't think that there are really any negatively charged electrophiles, but someone may have an example I haven't seen or thought of.


Products of the reaction between ethoxyethene and HI Problem

The products formed in the reaction
$$\ce{CH2=CHOCH2CH3 + HI ->}$$
are
(a) $\ce{CH2=CHI}$ and $\ce{CH3CH2I}$
(b) $\ce{CH2=CHI}$ and $\ce{CH3CH2OH}$
(c) $\ce{CH3CHO}$ and $\ce{CH3CH2I}$
(d) $\ce{CH3CH(I)OCH2CH3}$

Answer

 (c) $\ce{CH3CHO}$ and $\ce{CH3CH2I}$

Question
Does ether cleavage occur or electrophilic addition of $\ce{HI}$ to the double bond take place or both? I'm equally inclined towards both (c) and (d).

Here is a mechanism for the cleavage. After proton transfer, the carbocation has a resonance contributor. Nucleophilic attack can occur on the ethyl group because there is a good leaving group.

The electrophilic addition can happen, with iodide attacking the carbocation. However, it is reversible. Iodide is a good leaving group, and the carbocation is resonance stabilized. The cleavage reaction is much less reversible.


Prop-2-yn-1-ol NMR spectrum differences Looking at prop-2-yn-1-ol spectrum that was taken from:

Spectrometric identification of organic compounds. Robert M. Silverstein and G. Clayton Bassler. J. Chem. Educ., 1962, 39 (11), p 546. DOI: 10.1021/ed039p546.

Why OH peak is so sharp? And why it is located around 1 ppm? The book does not specify the solvent used.

At the same time if one takes a look at 1H NMR spectrum of the same compound taken on 90 MHz machine in CDCl3. Source: Spectral Database AIST with SDBS No. 74HSP-00-252

Why the OH peak shifted so much? But it is still as sharp as in other spectrum.  Please someone explain such pattern.

The chemical shift of protons in polar groups can vary a lot between solvents of different polarity, because the average electron density around the proton changes. Hydrogen bonding etc. A nice overview can be gotten from the article NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist.

For example water shifts from 4.79ppm in water($D_2O$) down to 0.4ppm in d6-benzene. Or the OH group of ethanol from 0.5 in benzene to 1.3 in chloroform to 4.6(!) in DMSO. The methyl and methene groups otoh just shift by 0.3 ppm max. between solvents.

The signals in your spectra are sharp, because there is either a fast exchange, meaning the peak comes at the average CS of both positions of the proton, or no exchange. Slow exchange (either statistically or because the reaction is slow) leads to broadening.

(Slow = exchange happens on the NMR timescale, a millisecond plus/minus a few decades, depending on your $T_{2}$ etc.)


Calculating the pH of a ammonium hydrogen sulfate solution I am attempting to complete a question in which I need to use an ICE table to calculate the pH of a $\pu{2.61 mol L-1}$ $\ce{NH4HSO4}$ (ammonium hydrogen sulfate) solution.

When $\ce{NH4HSO4}$ dissociates I know that it splits into $\ce{NH4+}$ and $\ce{HSO4-}$ ions.

According to my textbook, $\ce{HSO4-}$ is actually an acid and not a base, so I set up my equilibrium expression like such:

$$\ce{NH4+ + HSO4- + 2 H2O <=> 2 H3O+ + SO4^2- + NH3}$$

$\ce{NH4+}$ has a $K_\mathrm{a}$ of $\pu{5.56e-10}$. $\ce{HSO4-}$ has a $K_\mathrm{a}$ of $\pu{1.2e-2}$.

According to my ICE table:

$\ce{NH4+}$ and $\ce{HSO4-}$ have starting concentrations of $\pu{2.61 mol L-1}$.

Both lose $x~\pu{mol L-1}$, and have equilibrium concentrations of $2.61 - x$.

$\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$.

To calculate the final answer I am using the $K_\mathrm{a}$ of $\ce{NH4+}$ and plugging in the values from my ICE table for the equilibrium concentrations of the products and reactant.

Since $K_\mathrm{a}$ of $\ce{NH4+}$ is so small, I am disregarding the changes and substituting $2.61$ instead of $2.61 - x$.

The final answer I am getting is not the same as the answer in the back of the book which is approximately $\mathrm{pH} = 0.76$.

What is wrong with my process?

The hydrogen sulphate ion is a much stronger acid that the ammonium ion. Thus almost every hydronium ion in the solution comes from the dissociation of a hydrogen sulphate ion.

This said, the assumption

  $\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$.

is wrong.

When you solve the quadratic equation

$$K_a = \frac{[\ce{H3O+}] [\ce{SO4^{2-}}]}{[\ce{HSO4^{-}}]} = \frac{x^2}{c - x}$$

you will get a pH near your text book's value.


How does boiling water extract dyes from vegetables? Why does boiling water always seem to release and become the colors of whatever vegetables I am boiling?

For example, beet root and red cabbage both vividly color hot water.

I'd assume it's something to do with collisions with water particles breaking down cell walls and releasing a pigment, but how would this work on a physical level?

These dyes (NOT pigments$^*$) are highly water soluble. As you pointed out, boiling causes cell walls to rupture, releasing the cell plasma into the water. Boiling of the plasma fluid causes pressure build up, something the cell walls aren't designed to withstand.

You can also extract them with cold water by mashing up the raw material in a food blender or grinding it in a mortar and pestle. This has the same effect of rupturing the cell walls.

An interesting fact is that many of these dyes are pH sensitive: their colour is different in acid or alkaline conditions. Notably red cabbage extract changes from wine red to greenish blue when you add some sodium bicarbonate (baking powder) to it, because 'Bicar' is a (weak) alkali. Then adding a lot of vinegar (a weak acid) will change it back to its original colour.

$^*$ pigments are inorganic substances (many are refined minerals) and mostly insoluble in water.


Why isn't ammonium chromate called diammonium chromate? I have to name $\ce{(NH_4)_2CrO_4}$. I know the answer is ammonium chromate. But I don't understand why it isn't called diammonium chromate.

You don't need the prefix because you know (or rather are assumed to know) the charges on the constituent ions of an ionic compound, which this is. The ammonium ($\ce{NH4^+}$) cation has a charge of +1 and the chromate ($\ce{CrO4^{2-}}$) anion has a charge of -2. Since the formula unit has to have a net charge of zero, it follows that "ammonium chromate" must have the empirical formula of ($\ce{{(NH_4^+)}_2CrO4^{2-}}$). Similarly "ammonium nitrate" must have an empirical formula of $\ce{NH4NO3}$ because the nitrate anion has a charge of -1.  In both cases, you can deduce the formula from the name and your pre-existing knowledge of the properties of the ions.

The prefixes are used much more often in simple molecular compounds because the elements in periods 2 and 3 can often be in multiple oxidation states, and you need the prefixes to figure out the actual molecular formula, the names of the elements are not enough. The most famous example is probably nitrogen and oxygen, which form at least half a dozen neutral compounds. You need prefixes so you can distinguish between $\ce{NO2}$ (nitrogen dioxide) and $\ce{N2O}$ (dinitrogen monoxide).  "Nitrogen oxide" wouldn't cut it, because you wouldn't know the oxidation state of $\ce{N}$ and $\ce{O}$ and could not deduce the formula from the name.

Even when the constituents of ionic compounds can form more than one ion, e.g. for transition metals, we usually choose to write the oxidation state of the varying ion instead of using prefixes, e.g. $\ce{FeO}$ is "iron(II) oxide" instead of "iron monoxide" and $\ce{Fe2O3}$ is "iron(III) oxide" instead of "diiron trioxide."

You may observe that this is all shot through with logical inconsistency and redundancy. That is a reflection of the chaotic history of chemical nomenclature and the need to balance systematic naming conventions to make it easier for the newcomer with the ability to use the vast existing knowledge base that uses older and more ad-hoc or individually varying schemes.


How can dish-soap make glass less misty? Warm glass that face a cold temperature becomes misty - but becomes less misty if sprayed with dish-soap. Moreover, dish-soap sprayed on glass makes the glass less misty, if it face a drastic change in temperature - at least makes it harder for the glass to become misty.

Why is it so?

The fogging is caused by tiny droplets of water. These condense onto the glass when hot humid air comes in contact with a cold surface . The dish-soap acts as surfactant. The dish soap prevents droplets from forming. It forces any water molecules that condense onto the surface to be evenly spread across the surface. Thus, the water molecules form a thin film instead. This film is usually thin enough to not impede vision.


Selection of parent carbon chain for IUPAC naming How to select the parent carbon chain for IUPAC naming of following compound?

I think the red numbering is correct, but in the answer key, they have used the blue numbering.
Which is correct?
My question is regarding selection of parent chain in an alcohol. While selecting the parent carbon chain (base name) to write the IUPAC name of an alcohol, do we have to give preference to the one which gives the lowest locant to the $\ce{–OH}$ group (viz. primary functional group here), or the one which has the maximum number of carbon atoms?

When assigning numbers to carbon atoms in an aliphatic skeleton, your first priority must be the longest carbon chain. The chain with red numbers is one carbon atom smaller than the one with blue numbers. Blue - 9, red - 8.
It's also helpful to know that when you have two carbon chains of the same length, the one that's the most branched must be assigned the highest priority. Here we have exactly this case, where two chains of the same length can be found. One of them has 4 branches (2 ethyl radicals, 1 alcohol and 1 bromide) and the other one has 3 branches (1 hydroxyl, 1 ethyl and 1-(1-bromoethyl)).


Why aren't methanal and ethanal also ketones? Why aren't the aldehydes methanal/formaldehyde and ethanal/acetaldehyde also ketones? Ketones are defined by replacing any H - C - H in an alkane with C=O, which is true for both of these: methanal replaces two hydrogen atoms in methane with C=O, while ethanal does it with ethane.

Aldehydes deserve a different name because they have different properties, for example being easily oxidized to acids, as in the Benedict test which distinguishes "reducing" sugars like glucose (an aldehyde) from non-reducing ones like fructose (a ketone).  The relation between aldehydes and ketones is like the relation between alcohols and ethers, both of which can be described as R-O-R, but the properties are significantly different when one of the R's is a hydrogen.


To predict if a substance is deliquescent FeCl3,FeSO4 and CuSO4 all are deliquescent solids.

  Is there any way to predict this without knowing this fact?
  I am looking for a way to predict if a substance is deliquescent for my upcoming exams.

There is no general rule, just partial empirical rules for particular anions, which can be memorized from textbooks.

There is theoretical reasoning why salts  are soluble or not, concerning the energy of the crystal lattice, energy of ion hydration, entropy change etc. But easier way is to go for empirical rules and "chemical sense".


Safe explosion demonstrations As an end of high school treat, my class is trying to convince our chemistry teacher to make an explosion as a demonstration. What sort of reactions are beautifully explosive but safe enough to do in a school lab environment, when performed by a teacher who knows what they're doing?

From our friends at Periodic Videos, the following come to mind:

Gun cotton, more precisely known as nitrocellulose, which is a very simple nitration that you could show prior to the pyrotechnics.

Sodium and Potassium are also good, though you may not always get an explosion, depending upon the available hydrogen above the metal.  Which suggests...

Hydrogen! This is the classic explosion in oxygen. In fact, on a small scale, the electrolysis of water into hydrogen and oxygen can be used to show that the mixture of these gases is very explosive. But keep it small!

All of these experiments avoid outright explosives, which seems prudent in a high school setting.


Why is the concentration halved at the intersection of Hg²⁺/Hg₂²⁺? For the couple $\ce{Hg^2+}/\ce{Hg2^2+}$: $E^\circ = \pu{0.91V}$.
  At the border the concentration is $c = \pu{0.10 mol \cdot L-1}$ for all ions. Hence build the stability diagram of mercury.

I have:
$$\ce{2Hg^2+ +2e- <=> Hg2^2+}$$

Then by the Nernst relation I have:
$$E = E^\circ + \pu{0.03 V} \times \log\left( \frac{\left[\ce{Hg^2+}\right]^2}{\left[\ce{Hg2^2+}\right]} \right)$$

In the solution of the exercise they write:

  At the border $\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}$.

I don't understand why it is $\frac{c}{2}$.

The total concentration of all ions is $c = \pu{0.10 mol L-1}$. The conservation equation of element mercury requires that
$$\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}.$$

Generally, it's a convention to make the concentration of ions at the border equal. On the other hand, matter conservation implies that the concentration of each ion is $\frac{c}{2}$.


Can Hückel's rule be applied to inorganic compounds? Hückel's rule helps us know which molecular orbital, bonding or antibonding contains more electrons, and it is derived from the same. And on the basis of this, we can find out the stability of the compounds.

But, this rule was specifically introduced for cyclic organic compounds to know if they are aromatic or not. But again, the reasoning remains the same across all compounds, that is, if more electrons are in BMO than in ABMO – stable; otherwise not.

So, I was wondering why we couldn't use the same rule for cyclic inorganic compounds? I don't know, but I might be missing out somewhere. Any help would be appreciated.

Also: This pdf here, agrees that Hückel's rule can be applied to inorganic cyclic compounds, but with a 'maybe'.

  Whereas organic
  compounds need to be planar to be aromatic, no such limitation applies to inorganics. An inorganic compound may be expected to be aromatic if it satisfies a “counting rule”; the familiar
  Hückel rule applies to planar aromatics and the Hirsch rule works for spherical analogs.

Huckel's rule can be applied to "inorganic" cyclic molecules but only if they have a delocalised pi system. They also have to have flat (or close to flat) rings.

Recently some N5 rings which 6 pi electrons have been made, these are "inorganic" aromatic rings.

Also borazine has been the subject of considerable debate, it has been discussed on stack exchange before. It is the subject of a interesting paper: Islas, R.; Chamorro, E.; Robles, J.; Heine, T.; Santos, J. C.; Merino, G. Borazine: to be or not to be aromatic. Struct Chem 2007, 18 (6), 833–839. DOI: 10.1007/s11224-007-9229-z.


What is the difference of molar volume between dissolved 35Cl- and 37Cl-? Im am geochemist, dealing with isotopes such as 35Cl and 37Cl.

For some model I'm trying to make I need to be able to estimate what is the difference of partial molar volume between dissolved 35Cl and 37Cl in water (as chloride Cl-).

Huge literature exists concerning the partial molar volume of Cl- and other ions, but I can't find anything on specific studies on variations of volume on different isotopes. As in isotopes geochemistry we're often looking a permil variations of ratios, any tiny difference in molar volume of 35Cl and 37Cl would have an importance for me.

I don't have an exact answer for chloride specifically, but consider the theoretical equations in the following two articles:

Isotope Effects on Molar Volume and Surface Tension: Simple Theoretical Model
and Experimental Data for Hydrocarbons J. Chem. Phys. 44, 457 (1966)

  We assume the molecules are held together in a
  condensed phase by van der Waals "bonds" or covalent
  bonds represented by 6-12 or Morse potential-energy
  functions, V(x). For monatomic molecules x is taken
  to represent the distance between molecular centers.

and

Lattice Constants of Separated Lithium Isotopes J. Chem. Phys. 27, 1030 (1957)

which finds atoms of Lithium-6 $4 \times 10^{-4}$ times more separated than those of Lithium-7 and also cited to a reference which considers lithium-6 versus lithium-7 fluoride-19.


Analytical Chemistry Textbook Recommendation Could anyone recommend a good book for self-learning that provide a clear, step by step guide on solving such chemical equilibria questions.

It should cover the following topics: weak acid and strong base titration, buffers, redox reactions, electrochemical reactions, complexation reactions, potentiometric titrations.

Preferably, I would like it if the textbook also provide work solutions to the practice questions it gives. I heard that Fundamentals of Analytical Chemistry by Skoog is good and covers the topics that I have listed above but I don't think it provides any worked solutions to its questions.

Any help will be much appreciated.

Fundamentals of Analytical Chemistry is a fine book and it does have an student solution.


What is maximum energy level of molecules? I have been reading about the relation of molecular graph spectrum, that is, the set of eigenvalues of the Laplacian matrix of the molecular graph, and molecular properties. Recently, I came across a paper that finds an upper bond for maximum eigenvalue (DOI: 10.1109/ICNC.2012.6234727). To motivate their research, authors state that

  The eigenvalues of adjacency matrix of the graph is used to present
  the energy level of specific electronic. Particularly, spectral radius
  of graphs is the maximum energy level of molecules.

I was wondering if anyone can elaborate on what maximum energy level of molecules is. Is it the maximum energy level among all the atoms of the molecule?

In the painting "Dream Caused by the Flight of a Bee" by Salvador Dali, among other strange items, we may see a fish flying in midair. There is also one in the drawing "Big Fish Eat Little Fish" by Bruegel the Elder. One may conclude that there is a flying fish in every picture out there. One may even publish a paper to that effect. But I believe you see the shortcomings of this way of reasoning.

Same thing here. Molecules don't have the maximum energy level. You may think of the energy required to break a  molecule, but that's not a level; rather, it is a lower bound of the continuous spectrum. You may think of the energy levels of its constituent atoms, but those are quite irrelevant to the energy levels of the molecule. Moreover, atoms don't have the maximum energy level either.

At this point you may wonder: what were the authors thinking? Were they (together with the editors) just lunatics? Not at all! Then what did they have in mind? Why, they said it loud and clear: the spectra of graphs. How are those related to the spectra of molecules? Is a molecule the same as a graph? Of course it isn't... unless we are looking at it within certain simplistic approximation where it is.

Yes, this is the old good Hückel method once again. You take a conjugated $\pi$ system, draw a graph of the molecule, solve it for eigenvalues, and get the $\pi$ energy levels of the molecule (only upside down). Naturally, there is but a finite number of those, and there inevitably is a maximum. That's what they mean when talking about the "maximum energy level of a molecule".

Now, these $\pi$ levels typically include the frontier orbitals (HOMO and LUMO), and those next to them, and answer pretty much any chemical question about the molecule. All in all, it is quite natural and easy to forget altogether about all other orbitals and energy levels of the molecule ($\sigma$ bonds and stuff). But they are still there, and what I said before is still true.

So it goes.


Why is twice as much hydrogen produced than chlorine in the electrolysis of brine? I have seen a great many number of diagrams which show the electrolysis of brine in a diaphragm cell, but show twice the volume of hydrogen produced. For example:

On the other hand, other diagrams show that the same volume of hydrogen and chlorine is produced. For example:

So who's right and why?

It appears, as the others have said that the same number of moles of gas of chlorine and hydrogen is produced.

Then why are there different levels of solution?

Since the solution is taken out in the cathode chamber, the level of solution present is lower than the anode chamber where sodium chloride solution is being added in.


How to Run GAMESS and Avogadro on Command Line? I always use GAMESS and Avogadro on my own laptop. Recently I installed them on our university supercomputer and started using them by remote logging-in.

On the laptop everything was super easy but unfortunately I can not find any documentation for using them from Terminal Command Line ( I mean sending a job and receiving the result all from Terminal Command Line ).

It would be great if someone knows a good documentation or tutorial for this subject.

Contact the HPC helpdesk at your university or search for some instructions compiled by them.

We cannot know whether your local cluster uses PBS (TORQUE/MAUI), SLURM or anything else to reserve cores, nodes, memory and cpu time to run jobs.


how to use di-tert-butyl dicarbonate (diBOc) as a solvent? how should we use di-tert-butyl dicarbonate(diBOc) as solvent?

In reference, diBOc was in the form of a liquid but when I checked the bottle it was in solid state.
diBOc should be stored at 8-10 ℃. So how is it possible for it to be used as liquid?
Is it OK to melt it at RT? Wouldn't melting, make diBOc have structural differences?

Di-tert-butyl dicarbonate melts around room temperature (m.p.=23°C).  It does not decompose at this or even slightly higher temperatures.  For example, it is typically purified by distillation under reduced pressure at temperatures up to around 65°C.  At higher temperatures it will decompose to isobutene, t-butyl alcohol and carbon dioxide.

In any case it is OK to melt di-tert-butyl dicarbonate and use it between room temperature and up to around 40-60°C.  Just make sure that when you open the bottle and use it that you do so in a dry environment as water will decompose di-tert-butyl dicarbonate.


Why a particular enantiomer of a diastereomer shows affinity towards another? A particular enantiomer can show affinity towards another enantiomer of another diasteromer eg. body receptors with chiral drugs, chiral stationary phase of chromatography etc.

But what's the reason behind this affinity? A broad answer will be encouraged.

Let's set aside compounds for the moment.

Your hands are chiral objects, that is, they are non-superimposable mirror images of each other [see more here]. If you pick up something that is not chiral, like a cup, there is no difference whether you use the left hand or right hand. The "energy" of those two interactions (right hand + cup or left hand + cup) is the same.

Now, let's say you get a pair of gloves made to fit your hands. If you put your left hand into the right-hand glove, it feels different than if you put your left hand into the left-hand glove. You could think of it as those two interactions (left hand + left glove vs left hand + right glove) having different energies.

The reason for this is that enantiomers of chiral objects interact with non-chiral objects with the same energy, but interact with enantiomers of chiral objects with different energy (ignoring coincidental cases, as @Zhe suggests).

The way one enantiomer of a compound "fits" with enantiomers of other compounds (objects) like proteins or chromatography instruments is difficult to predict. Sometimes it is useful to actually model the interactions, but often, like in chromatography, it's enough to know that there should be a difference in energetics of the enantiomers of a compound with a single enantiomer of another chiral object.


Why does Mg react vigorously in NaCl solution and less so in water? When you put $\ce{Mg}$ into water a few $\ce{H2}$ bubbles appear. But when you put $\ce{Mg}$ into a $\ce{NaCl}$ solution there is a vigorous release of $\ce{H2}$, why is this and what reactions are taking place?

It would still be the same reaction.

\begin{align}
\ce{Mg (s) &-> Mg^2+ + 2 e-} & E^\circ&=\pu{2.36 V}\\
\ce{2 e- + 2H2O (l) &-> H2 (g) + 2OH-} & E^\circ&=\pu{-0.828 V} \times 2
\end{align}

The combined reaction would be spontaneous ($E > 0$), but $\ce{Mg(OH)2 (s)}$ is highly insoluble,
\begin{align}
\ce{Mg(OH)2 (s) &<=> Mg^2+ (aq)  + 2OH- (aq)}, &
K_\mathrm{sp}=7.1 \times 10^{-12},
\end{align}
leaving the solution insufficient ions to conduct the electrons necessary to fast complete the reaction (despite of the escaping hydrogen gas).
$$\ce{Mg (s) + 2H2O (l) -> H2 (g) + Mg(OH)2 (s)}$$

This is like an electrochemical cell with no salt bridge (or bridge with only de-ionized water). Hence a little salt ions would accelerate the reaction forward.

In contrast, calcium solid, having comparable standard reduction potential, will undergo a similar redox reaction in water and needs no help for additional ions because $\ce{Ca(OH)2 (s)}$ is about a million times more soluble ($K_\mathrm{sp}=6.5 \times 10^{-6}$) than magnesium hydroxide, providing sufficient free ions in the solution to drive the vigorous release of hydrogen gas. (There are a couple of YouTube video demos.)


How to remove pyridine from Reaction mixture containing chiral molecule with diacetal protection group? While working with chiral molecule, we have pyridine in our reaction mixture. Molecule has four chiral centre and protection group-diacetal and acetyl groups which may break in harsh conditions. Please suggest work up methods to remove pyridine.

Extract the organic soln of the product with dilute aq. Cu(II) until the characteristic blue of the Cu-pyridine complex is absent from the aq. phase.


Elemental Potassium as both an oxidizer and a reducer On this table I realized that elemental potassium could adopt oxidation states of -1, 0, and +1.

http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements

Does this mean that elemental potassium can function as both an oxidizer and a reducer? I understand that oxidizers are reduced (gain electrons), and that reducers are oxidized (lose electrons). How can potassium anion be formed? The potassium anion must be extremely unstable due to potassium's extremely low electronegativity, correct?

Also in what compound does potassium actually have a negative oxidation state?

Compounds having potassium anions are referred to as potassides and are rare, usually involving K- within a cyrpt such as:

Crystalline Salts of Na- and K- (Alkalides) that Are Stable at Room Temperature J. Am. Chem. Soc. vol. 121, pages 10666-10667 describes preparation of K+(aza222)K- , where "aza222" is the compound in the above figure, but with the 6 H's replaced with methyl groups.  The compound is formed by dissolving potassium metal in methyl amine solution of the methylated aza222.


"Middle row anomaly" of the periodic table I was studying about the periodic table recently, and was reading a topic associated with oxides of halogens, and came across the following line

  The bromine oxides, $\ce{Br2O}$, $\ce{BrO2}$, $\ce{BrO3}$ are the least stable halogen oxides (Middle row anomaly) and exist only at low temperatures. They are very powerful oxidizing agents.

So, I went to search this on Google, and found these lines

  It refers to the instability of oxides of bromine as compared to relative stability of oxides of chlorine and Iodine at room temperature, the former being stable only at low temperatures.

But the line above is simply restating what the book had already told.
So, is their any specific reason why this happens, or is this only due to the experimental data we have gathered?

This is due to the transition metal contraction.

Bromine has the electron configuration $\ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. $The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation states of bromine (I, IV, VI for the examples you give) because the ionization energies are higher than you might expect from a simplistic approach to periodic trends.


Why does Comfrey Break Down Easily? comfrey (Symphytum officinale) is a well known plant to organic growers. One of its features is that the leaves break down very quickly, either in water or in a container closed from UV light. There must be some chemical explanation for this, can you explain to me what that is ?

Thanks Erin

Erin, the main the problem with the question is apparently this part "One of its features is that the leaves break down very quickly, either in water or in a container closed from UV light." You also mention the compound comfrey and wonder if this is responsible for a quick decomposition? I don't think so.

The key question is that Comfrey plant decomposes quickly, as you observed, but compared to which plants? I don't think it is unique that it decomposes rather quickly in soil, water etc. Don't other plants like spinach leaves decompose or spoil quickly? You can easily eliminate UV because UV cannot penetrate compost. It must be bacterial / fungal action.

You can use Google Scholar to fetch scientific papers only and one of the scientific papers says "Comfrey on a dry weight basis is very high in ash, averaging 18 percent at Rosemount and ranging from 13 to 42 percent in other parts of the world (4). A partial
elemental composition of comfrey forage is reported in Table 4. The forage is particularly high in potassium and is higher than many other forage crops in calcium,
phosphorus, iron, and copper. Carbon, hydrogen, oxygen, nitrogen, and sulfur are the major non-mineral elements. The combined high nitrogen and high mineral concentrations make comfrey forage an unusually good material for composting, mulching, and organic fertilization of crops"

From Comfrey: A Controversial Crop https://conservancy.umn.edu/handle/11299/141442

The high nitrogen content could be reason for its quick decomposition by bacteria or fungi, however it is hard to assign a single compound "comfrey" as being responsible for it.


Molecular orbital energies prediction with ML algorithms In the recent years, computational chemistry community has focused on Machine Learning algorithms to predict molecular properties. Unfortunately, many of the authors of such papers are not chemists, and hence, there are little explanation on chemical aspects of this type of works.

One well known problem is to predict different properties of QM9 dataset given their atom features. The properties include HOMO/LUMO orbital energies, atomization energies, etc. The inputs include type of heavy atoms in the molecule, type of their bonds, type of atom hybridization, etc. You can see a complete list in the following tables

The question is how the inputs are related to the outputs. For example, I understand that HOMO\LUMO energies are solutions of the electronic Schrödinger equation. But which one of the inputs features affects these? I appreciate it if someone can expand on which ones affect frontier molecular energies. Thanks!

As several comments have mentioned, interpreting a machine learning method can often be difficult. Outside the question of machine learning, I think your question centers around "what would I want to know to roughly predict HOMO / LUMO energies?"

Unfortunately QM9 isn't very diverse chemically, but if we think from a general qualitative molecular orbital picture (e.g., building up diatomic molecules), you want to know:

Number of valence electrons (i.e., which orbitals are filled)
Atom type (which tells you valence electrons and atomic orbital energies)
Type of bond (e.g., single sigma bonds clearly will yield different MO energies than pi bonds)
Conjugation - conjugated and aromatic systems will have different MO energies than plain sigma-bonded species
Bond distances (from the Coulomb repulsion which encodes a distance $1/r$)

Given the above information, it's fairly easy to sketch out MO diagrams for $\ce{O2}$ or $\ce{HOH}$ so one can imagine that they're important for a machine learning algorithm too.

Now how do you know if they actually matter to the ML model?

One easy way is to drop one of the descriptors and retrain the model. Does it get worse? How much?

There are statistical techniques to give you some insight as well. For example, one can use LASSO to drop factors from the input representation that don't have much effect. You can also look at simple statistical correlation between input factors. I'd guess that "Atom Type" and "Atomic Number" are probably highly correlated with each other and only one is truly necessary.. but that's just a guess.


Which is more basic: 1,2‐dihydropyridine or 1,4‐dihydropyridine? Which compound is more basic of the two?

My thinking: In compound A, the lone pair of nitrogen are in resonance with two other double bonds. The resonance chain here seems to be longer as compared to in compound B. However, in compound B, the lone pairs of nitrogen seem to be more 'busy' because of the two double bonds on its either side,when compared to compound A (@YUSUFHASAN provided this second logic).
So, which of the two compounds should be more basic? (The main idea is that, the lesser 'busy' lone pair electrons of nitrogen means more 'free' electrons to pick up $\ce{H^+},$ which is what bases do.)
Do note that this question is supposed to be answered qualitatively. (By this I mean, not looking at tabular data from research papers or books!)

In compound $\ce{A}$ , lone pair on $\ce{N}$ is in conjugation with two $\ce pi$ bonds (as shown below). Therefore this lone pair is delocalized (resonance structures in the figure below) to a greater extent.

In  compound $\ce{B}$ , lone pair on $\ce{N}$ is in conjugation with a single $\ce pi$ bond , while the other pi bond is not in conjugation or is cross conjugated.

Since lone pair on $\ce{N}$ in compound $\ce{B}$ is less delocalized then in compound $\ce{A}$ , compound $\ce{B}$ is more basic then compound $\ce{A}$.


DIY Brass Etching methods (safer etchants, waste, and disposal of) I want to etch brass for making name plates etc. (Also for copper PCBs but brass is probably more complicated given the other metals).

Several etchants commonly used, and I'm sure many more are possible; I'm not fussed what is used. Even vinegar, although something a bit faster would be great.

I want a method easy for someone with minimal chemistry knowledge and equipment to safely perform, and dispose of byproducts etc as household waste, or down a drain.

(Since AFAIK for example copper compounds for example are hazardous to the environment, and etchants can be dangerous/toxic/corrosive), and in the UK for example, getting it disposed of professionally is prohibitively expensive for responsible DIYers on low budgets)

What would be a good choice of etchant? And what would the process be for using and disposing of the waste / bi-products? In general, the less hazardous the better.

PCB resists and techniques would be used, so need an etchant compatible with any of those.

An easy step-up from vinegar would be hydrochloric acid, $\ce{HCl}$. It's known as muriatic acid and is sold in bulk at pool-supply and home-improvement stores. The concentration from such sources is typically ~8.9M, or 32% (by mass). Such is what I measured from home; it's typically advertised as 30%-32%. There are rumors (which I haven't been able to confirm but suspect are true) that $\ce{HCl}$ from these sources often has iron contamination, but that is generally a minor concern for etching processes.

$\ce{HCl}$ has been widely used for etching aluminum hardware and paint-thinning in construction etc. It is highly reactive with (unoxidized) aluminum but can still etch other metals with more effort / more concentrated solution. You probably already know, but it's best practice to clean and polish a surface before etching, to remove impurities like oil and dirt, but also to expose the pure metal by removing protective oxides on the surface. The rougher the better (you can always polish-up later).

Brass is a simple alloy of $\ce{Cu}$ and $\ce{Zn}$. The reaction enthalpy of $\ce{Al + HCl}$ is -1,062.74 kJ, whereas that with $\ce{Zn}$ is only -153.90 kJ. Copper is hardly reactive with $\ce{HCl}$ alone, so it follows that $\ce{HCl}$ is a much stronger etchant for aluminum than for brass, although it is still possible to use it. There are many types of brass -- generally, the more $\ce{Zn}$, the better $\ce{HCl}$ will etch the material:

It is fine to dispose of $\ce{HCl}$ down the drain etc., but only do so after diluting it (carefully, slowly and substantially with water).

You can boost the performance of $\ce{HCl}$ as an enchant by introducing hydrogen peroxide, $\ce{H2O2}$. This combination has very strong oxidative effect and can dissolve metals like copper with ease. If you want to be environmentally friendly, this would be a good stopping point. The most powerful etchant will result from higher concentration of $\ce{H2O2}$. It's typically sold at 3% in stores but food-grade 30% $\ce{H2O2}$ can be bought online or made at home via this method.

I have had great success etching copper (and nickel) using a home-made solution of copper II chloride $\ce{CuCl2}$. Copper (I/II) chloride is very hazardous to aquatic life and (dissolved in solution or not) human life, but with careful handling it's a fine thing to use. It is quite easy to produce and handle, and does not corrode other (more precious) metals, like gold, so you can get selective etching if you have a certain alloy or something like RAM cards from PC parts. (I have isolated gold metal from these using $\ce{CuCl2}$ solution). To make it, simply add copper metal to a solution of concentrated $\ce{HCl}$ and very slowly add concentrated $\ce{H2O2}$. The copper metal will dissolve/oxidize into $\ce{Cu^{+2}}$, and if you dry the resulting blue-green solution (you don't have to) completely, you'll be left with a brown $\ce{CuCl2}$ powder. It's hygroscopic so if you leave it out it will form a blue hydrate. So I keep mine in a tightly closed plastic bottle. To use it as an etchant, just take some of the solid and dissolve it in water (again). The chief downside, as you mentioned, is the toxicity of copper chloride -- I have never thrown mine away. You can re-use it many times.

If I ever had to dispose of it, to do so properly I would use electrolysis. Attaching a DC power supply with the (-) terminal on a piece of copper metal (in dissolved $\ce{CuCl2}$ solution) and the (+) terminal in the solution (or attached to an iron nail) would do the trick, though you'd have to wait some time for the copper in solution to reduce -- and to push equilibrium you'd need a moderate/high voltage. Once you've reduced most or all of the aqueous copper to solid metal, the solution leftover is relatively harmless.


Is the boundary of an atom well defined? That question might seem weird, but that's what I have been imagining lately; going through Bohr's theory on the hydrogen atom. He clearly defines the radius of an orbit with an exact mathematical expression. But what do they mean when they say any atom (say $\ce{Na}$) has a radius $r$. Are we to consider atoms as solid, rigid spheres? If that's the case, what about water?

If all the atoms are defined as solid, rigid, spherical balls, how come an aggregation of so many solid objects be a liquid? How can it flow? So, in a nutshell what is the boundary of an atom made up of, or even this: does it have a defined boundary?

This troubles me a lot. I think maybe the modern atomic model has dealt with this problem. I haven't read it in detail, and I would be highly thankful to any references.

No, the boundary of an atom is not well defined. According Quantum mechanics the electron density around an atoms decreases exponentially, thus getting asymptotically close to zero.
Considering atoms as rigid spheres is only an approximation. The criteria used to decide on the actual value of the atomic radius is arbitrary. You can for example choose the expectation value $\langle\hat r\rangle$ of the electronic wavefunction. For the Hydrogren atom, this results in the bohr radius $a_0$. But you might as well choose the maximum of the electron density ($\frac{3}{2}a_0$ for $\ce{H}$), or something based on experiments.
Now the aggregate state is something completely different and not directly  related to atomic radius and whether they are rigid sphere or not. Solid means the molecules/atoms of a substance are fixed in their positions. They can vibrate, but do not change their average position with respect to each other. In gases molecules may move freely. And a liquid is something in between, you have some order if you look at a small group of molecules, but not on a larger scale as for solids.


Do you assume undrawn hydrogens are pointing away from you when determining R/S? When determining R/S of stereoisomers, the group with the smallest atomic number needs to be in the back (dashed line). But if no group attached to a chiral center is drawn in dashed line, do we assume that the undrawn hydogren is dashed? I have attached an example below, the chiral center is circled in red. Thanks

When no bond from the chiral center has up or down graphical indication, it can mean any or unknown stereochemical configuration. Which might be confusing, because it can also mean, that the stereochemical configuration on this atom was mistakenly omitted.

To explicitly indicate that any or unknown configuration, wavy line bond can be used:

(Hydrogen atom with the dashed bond on the (other) chiral center can be can be omitted at all, the stereochemical meaning is unambiguous, much more if the bond wedge was filled with black.)


Calculating Kp in two different ways The degree of dissociation of $\ce{NH3}$ at $\pu{1 atm}$ is 20 % as follows:
     $$\ce{2NH3 <=> N2 + 3H2} $$

I follow two ways but end up with two different answers.

Here, I assume that initial amount of reactant mole is 2 and that of product is 0. Then I take the amount of ammonia in equilibrium to be $2-2\alpha$, nitrogen to be $\alpha$ and hydrogen to be $3\alpha$. Then, I find $K_\mathrm{p}$ (where $\alpha=0.2$).

Here, as dissociation is 20 %, I assume that amount of ammonia in equilibrium will be 0.8 and that of nitrogen and hydrogen will be 0.2 and 0.6 respectively. Then also, I find $K_\mathrm{p}$, but it differs from the initial one.

Where am I mistaken?

First solution is perfect.

The issue is in the second solution.

If we start with 1 mole of $\ce{NH3}$ then with 20% dissociation we are left with 0.8 moles as 0.2 moles react these 0.2 moles give o.1 mole of $\ce{N2}$ and 0.3 moles of $\ce{H2}$

Hence $$K_\mathrm p = \frac{[\ce{H2}]^3[\ce{N2}]}{[\ce{NH3}]^2} = \frac{[\ce{0.3}]^3[\ce{0.1}]}{[\ce{0.8}]^2}$$


Percent composition by volume of the reaction H2 + Br2  2HBr For the reaction $\ce{H2(g) + Br2(g) <=> 2HBr(g)}$ at a certain temperature, $K=1$. Determine the composition (in per cent by volume) of the equilibrium reaction mixture if the initial mixture consisted of $3$ moles of $\ce{H2}$ and $2$ moles of $\ce{Br2}$.

The answer given is $49.6\%$ of $\ce{H2}$, $29.6\%$ $\ce{Br2}$, $20.8\%$ $\ce{HBr}$.

I made an ICE table with following data:

$$
\begin{array}{c|ccc}
   *        & \mathbf I   & \mathbf C   & \mathbf E \\
\hline
 \ce{H2}    & 3   & -x  & 3-x \\
 \ce{Br2}   & 2   & -x  & 2-x \\
 \ce{HBr}   & 0   & 2x  & 2x \\
\hline
\end{array}
$$

Using this data I made the equation:

$$
1=\frac{4x^2}{(3-x)(2-x)}
$$

My work in solving it was:

$$
(3-x)(2-x)=4x^2 \\
3x^2+5x-6=0 \\
x=\frac{-5+9.85}{6}\text{ or } x=\frac{-5-9.85}{6}
$$

I used the answer $0.808$ for $x$. I calculated the volume percents as:

$$
\%~\ce{H2} = {3-x\over 5} = 43.84\% \\
\%~\ce{Br2} = {2-x\over 5} = 23.84\% \\
\%~\ce{HBr} = {2x\over 5} = 32.33\%
$$

But, these don't match the answer given. Where have I gone wrong?

Your equation and your math appear correct to me: I also get $x=0.808$ from the quadratic equation, and calculate the same percentages.

As long as you are supposed to assume ideal gas behavior in the question, it appears to me that the answer key is incorrect.


Why should epoxides be more reactive at a position where positive charge is stabilised? In the above reaction, I am told that the nucleophilic chloride anion attacks the trisubstituted carbon because it has the most partial positive character due to its ability to stabilize positive charge.

How does that statement make any sense?

I'm not following the train of "logic," which is as follows:

The nucleophile will want to attack the location with the greatest partial positive charge. Okay, Coulomb's law. I'm good with this statement.
The carbon with the greatest partial positive charge is the one that can best stabilize positive charge. (?)
Alkyl groups are slightly electron-donating; the tertiary carbon will be able to better stabilize partial positive charge.
Thus, the chloride anion attacks the tertiary rather than the secondary carbon. I don't see how this follows from the previous assertions. Doesn't stability correspond negatively with reactivity? If so, what is the real reason the chloride anion attacks the tertiary carbon rather than the secondary carbon?

To the question mark in (2): this implication is simple:  if the charge were on a neighboring carbon you'd have a state of higher energy.  "Can better stabilize" is a synonym for "gives lower overall energy".

To the objection in (4): the stability ranking refers to various electronic distributions in the possible transition states. In the transition state of an SN2 reaction, and when there is a positively charged leaving group, the carbon undergoing nucleophilic substitution will bear a partial positive charge in the transition state. So, the degree to which this positive charge is stabilised settles the matter of which carbon will be attacked preferentially.

The reactivity ranking refers to the fitness of each atom in the intermediate as a site of attack by the nucleophile. The most likely target is the one which leads to the lower-energy transition state, i.e. has a lower activation energy barrier.

So there is no contradiction.  The carbon most able to stabilize a positive charge is also the one most likely to be the target of a nucleophilic attack (other things being equal; of course there could be factors, like sterics, changing the ranking).


chemical equilibrium reaction for burned gases in engine I do not understand how equation(3) occurs.(taken from the book Internal Combustion Engines by John B.Heywood)
$$ $$
Consider a reactive mixture of ideal gases. The reactant species $M_a,M_b$ etc.and the product species $M_l,M_m$ etc. are related by the general reaction whose stoichiometry is given by:
$$\upsilon _aM_a+\upsilon _bM_b+...=\upsilon _lM_l+\upsilon _mM_m+...$$
$$ $$

This can be otherwise written as:
$$ \sum_{i}\upsilon _iM_i=0$$
where the $\upsilon _i$ are the stoichiometric coefficients and by convention are positive for the product species and negative for the reactant species.

Let $\delta n_a$ of $M_a$ react with $\delta n_b$ of $M_b$,etc. and produce $\delta n_l$ of $M_l$,$\delta n_m$ of $M_m$,etc..

These amounts are in proportion,given by the equation (3):

(3) $$\delta n_i=\upsilon _i\delta n$$

1.Does $\delta n$ here signify 'extent of reaction'?

2.I would eventually like to use the number of moles of each species in expressing the chemical potential,with the gibbs free energy already known. But,the part I don't understant is equation 3. What does the author mean by proportional? Could you give a example?

Gibbs free energy is given as:
$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\delta n_i$$

which by equation(3) can be re-written as(WHY??):

$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\nu _i\delta n$$

It's simpler than it might look. As a random example:

$$\ce{3A + 2B->2C + 4D}$$

If I were to ask you how much of $\ce{D}$ would be produced after $5\  mol$ of $\ce{B}$ were consumed, then you would immediately answer $10\  mol$ "due to the stoichiometry". In more elaborate terms, for a $\Delta n_B=n_B^{after}-n_B^{before}=-5\ mol$, one has $\Delta n_D=n_D^{after}-n_D^{before}=+10\ mol$. Any infinitesimal change in quantity of any species $i$ ($\delta n_i$) is accompanied by a shift in the quantities of all other species precisely determined by the stoichiometry of the reaction (the stoichiometric coefficients, $v_i$), such that $\frac{\delta n_i}{v_i}$ is constant for all species $i$. For the above reaction, $-\frac{\delta n_A}{3}=-\frac{\delta n_B}{2}=\frac{\delta n_C}{2}=\frac{\delta n_D}{4}$. The ratio $\frac{\delta n_i}{v_i}$ is defined as $\delta n$, and you can think of it as a counter for "how many times the reaction happened" ($\delta n>0$ if left to right, $\delta n<0$ if right to left).

$\delta n$ is a general change in amount of substances, but the particular case where changes are calculated relative to the equilibrium condition ($\Delta n=n^{equilibrium}-n^{before}$), then it represents the extent of reaction.


Balancing chemical systems - searching and sorting preliminary information I was hoping I could get help finding ways to sort information in the problem. What kind of information does "the conversion of toluene to benzene..." give us that is relevant towards solving the problem for values $R_G$ and $R_L$?

From your diagram, $R_G$ is the recycled gas - the hydrogen that has not been consumed plus the methane that has been produced. $R_L$ is the recycled liquid - the unreacted toluene. Both go back into the mixer with fresh toluene added at the rate given and fresh hydrogen added to bring the mole ratio of hydrogen to toluene (and hydrogen to methane) to what is given.

$R_L$ is easier. The problem tells you that 80% of the toluene is converted into benzene and 8% of the toluene is converted into biphenyl. What percentage of toluene is left over? The original feed flow of toluene is 3450 lb/hr, but the amount of toluene entering the gross feed to the reactor is higher (includes $R_L$).

$$G_{\text{tol}}=F+R_L$$
$$R_L = 0.12\times G_{\text{tol}}$$

For $R_G$, you need to know how much hydrogen you have going into the mixer. You are told that the mole ratio of hydrogen to toluene in the gross feed is 4:1. Now that you have $R_L$, you can figure out the total toluene in the gross feed, and then do the stoichiometry calculation to get to total hydrogen in the gross feed. Since you have the balanced equations, you can determine how much hydrogen is consumed. If 80% of the toluene is consumed in the generation of benzene, how much hydrogen is consumed? How much methane is produced? Do the same with the conversion of toluene to biphenyl.

Once you are able to obtain the molar flow rates of all molecules out from the reactor, you should be able to calculate the flow rate of the gas exiting from the separator. This gas stream (Let's call it $K$) contains only $\ce{CH4}$ and $\ce{H2}$ of mol fractions $y$ and $1-y$ respectively. The mol fractions can be calculated considering that all $\ce{CH4}$ and $\ce{H2}$ that exists the reactor leaves the separator in $K$.

The molar composition of the $R_G$ and $P$ streams that leave $K$ is the same. This means that you are now able to use a molar flow rate balance on the mixing point between $M$, $R_G$ and the stream to the mixer (Lets's call this $Y$).

$$Y = M + R_G$$

By doing a balance on methane, you can find $R_G$.

$$(0.5)*(Y) = (1-y)(R_G) $$

Since you know $Y$ and $y$, you will then be able to find $R_G$.


What is the white powder that I have extracted from cinnamon? In Advanced higher chemistry, I am extracting cinnamaldehyde using this procedure- https://www.flinnsci.ca/api/library/Download/ba6ab218508047ea843bb0d9e580be16
. After evaporating the hexane, instead of the cinnamaldehyde (a liquid), a solid white power is formed. We have no clue what it is and think it may be either something that is naturally extracted from the cinnamon bark or something that was added during 'production' (got it from Tesco's).

"6 grams of cinnamon can contain from to 42 to 189 mg cinnamaldehyde." http://herbpedia.wikidot.com/cinnamaldehyde  That's not very much.

"Cinnamon consists of a variety of resinous compounds, including cinnamaldehyde, cinnamate, cinnamic acid, and numerous essential oils." https://www.hindawi.com/journals/ecam/2014/642942/

The extraction method starts with 10 grams of cinnamon and salt is used to salt out the cinnamaldehyde from the steam distillate. Could the solid be salt carried over from the separation funnel? If the solid you obtained is more than about 400 mg, it might well be contaminated with NaCl. You could add a milliliter or two of H2O to see if you get a (saturated) salt solution plus the cinnamaldehyde oil on top. If the cinnamon you started with was very old or had very little cinnamaldehyde to begin with, the amount of oil you obtain might well be overshadowed by a little NaCl contamination.


Can you combine elemental gold and boron? Is it possible to combine gold and boron to form compounds? A simple example being:

$$
\ce{Au + B -> AuB}
$$

Yes, gold and boron can bond with each other to form alloy clusters. They are very important in semiconductor industry and used in fabrication process. Here(1) is a short note on gold-boron alloy:

  Photoelectron spectroscopy and density-functional theory are combined to investigate the electronic and structural properties of a
  series of $\ce{B-Au}$ alloy clusters: $\ce{B6Au_{n}^{−}}$ and
  $\ce{B6Au_{n}}$ ($\ce{n = 1−3}$). Rich spectral features are
  observed for each species, and vibrational structures are resolved for
  numerous detachment transitions of $\ce{B6Au^−}$ and
  $\ce{B6Au^{2−}}$. Electron affinities of $\ce{B6Au_{n} (n = 1−3)}$ are
  evaluated to be 2.70 ± 0.03, 2.91 ± 0.02, and 3.21 ± 0.05 eV,
  respectively. Global structural searches are performed for both the
  anions and their neutrals. The calculated electronic binding energies
  are compared with experimental measurements to establish the anion
  global-minimum structures and their corresponding neutral states.
  The ground-state structures of these clusters are shown to be planar
  or quasi-planar with an elongated B6core, to which the first and
  second Au atoms are bonded terminally and the third Au in a bridging
  position. All three anion clusters are π antiaromatic.
  Natural bond orbital analyses show that the $\ce{B-Au}$ bonding is
  highly covalent, providing new examples for the $\ce{Au/H}$ analogy
  in Au alloy clusters.

You can find more information about gold-boron alloy clusters in the reference below.

Reference

The Journal of Chemical Physics 138, 084306 (2016); doi: http://dx.doi.org/10.1063/1.4792501
https://www.google.com/patents/US3211550
https://www.ncbi.nlm.nih.gov/pubmed/23901981
http://aip.scitation.org/doi/10.1063/1.4816010
https://www.researchgate.net/publication/235886590_On_the_structures_and_bonding_in_boron-gold_alloy_clusters_B6Aun-_and_B6Aun_n_1-3
http://www.google.co.in/patents/US3137595


Short Cut Method to Calculating pH of Polyprotic Acid? I am doing the following question:

  Calculate the pH of a $4.00\ \mathrm{mol\ L^{-1}}$ solution of citric acid.
  $\mathrm{pK_{a1}} = 3.09~~~~~ \mathrm{pK_{a2}} = 4.75~~~~~ \mathrm{pK_{a3}} = 6.40$

Usually for polyprotic acids, we can assume that the second, third and other dissociations  of $\ce{H+}$ ions are negligible when compared to the first dissociation. However I believe that this is only valid when the pKa values are separated by around 4. So for citric acid, this would be an invalid assumption as the pKa values are pretty close to each other.

Now, I can find the pH of the solution by doing the long way by first finding the amount of $\ce{H+}$ ions formed by the first dissociation, then using that to find the amount of $\ce{H+}$ ions formed by the second dissociation and so on.

However the room that they have given for this question is pretty small and has only has enough room for a couple of lines of working out and I definitely wouldn't be able to fit all my working out in that box.

So am I missing some short cut method to finding the pH in cases like this, or is the above assumption that I have outline at the start still correct for citric acid?

The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid.

You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps.

(For the sake of simplicity I write $c(\ce{H+})=x$ and $k_{a,n}=k_n$.)

Exact: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$
Neglecting $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$
Neglecting $k_2$ and $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{k_1}{x+k_1}$$

Solving those equations give the following pH values (with super unrealistic high precision):

Exact:               1.2469291050971560
w/o $k_3$:           1.2469291065371300
w/o $k_2$ and $k_3$: 1.2470654318802320

As a pH can only be measured precisely within something like $\pm \left(10^{-2}\ldots10^{-3}\right)~\mathrm{pH}$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected.

So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days.


Law of Mass action VS Rate law The Law of mass action has 2 parts - the equilibrium rate part & the reaction rate part.
The "reaction rate part" states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to it's stoichiometric coefficient. The rate law states that rate of a reaction is directly proportional to the product of active masses of the reactants with each  reactant term raised to some power which may or may not be equal to stoichiometric coefficient of the reactant.

Why are the two laws different when in fact we are talking about the powers to which active masses of reactants should be raised in both the cases ? And which law is correct?

First of all, you need to understand the basic difference between the two.

Consider the reaction $$\ce{A + B -> C}$$ Law of mass action states that in a rate equation the powers that the active masses of components raised are always stoichiometric factors.

For a reaction they are always fixed. However, the rate of  reaction differs with the physical conditions. This is applicable to both equilibrium reactions and moderate reactions that are unidirectional.

This makes law of mass action kind of theoretical in nature. In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.

Both the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.


How can the effect of a pulse sequence on a strongly coupled spin system be analysed? In introductory NMR theory there are two models which are commonly used to analyse the effect of a pulse sequence: namely, these are the vector model and the product operator formalism.

The vector model serves as a useful pictorial representation, but is generally considered too simplistic to be used when the system contains coupled spins. Usually at this point the product operator approach is introduced, based on the Liouville–von Neumann equation

$$\frac{\mathrm d\hat{\rho}}{\mathrm dt} = -\mathrm{i}[\hat{H}, \hat{\rho}]$$

and its solution $\hat{\rho}(t) = \exp(-\mathrm{i}\hat{H}t)\hat{\rho}(0)\exp(\mathrm{i}\hat{H}t)$. Then with a knowledge of the initial density matrix $\hat{\rho}(0)$ and the relevant Hamiltonians $\hat{H}$, the density matrix after a pulse sequence $\hat{\rho}(t)$ can be calculated.

Yet, it is also stated that the product operator approach "only applies to weakly coupled spin systems" (Keeler, Understanding NMR Spectroscopy 2nd ed., section 7.1.3). "Weakly coupled", of course, means that the coupling constant $J_\mathrm{IS}$ is much smaller than the difference in resonance frequencies $|\nu_\mathrm I - \nu_\mathrm S|$. This raises two questions:

Why does the product operator formalism not work for strongly coupled systems?
How can the effect of a pulse sequence on such systems be analysed?

The product operator formalism still applies to strongly coupled systems.  The simple "magic formula" for calculating a closed-form solution to the step-wise time evolution only applies when the terms of the Hamiltonian commute with each other, which is not the case in a strongly-coupled system.  Maybe more accurately stated: in a weakly-coupled system, the non-commuting terms can be neglected.  The other assumption that allows the removal of non-commuting terms is that the pulses are very short compared to other evolution times, so during pulse periods the free-evolution terms are omitted from the Hamiltonian.

The effects of a pulse sequence without these on such a system can still be found by computing the exponential of the matrix rather than using the shortcut.

Now a shameless plug: the SpinDrops tool (http://spindrops.org) can be used to simulate and visualize the effects of pulse sequences with arbitrary Hamiltonians. It shows an exact 1-1 mapping of coupled-spin operators to the visualization which can not be done with the vector model.  The setting 'Pure Pulses' can be disabled to include the free-evolution Hamiltonian during pulses, but incorporating strong coupling would currently need to be done by defining the Hamiltonian explicitly.


Why can't you achieve the equilibrium exactly in the middle (K=1) when you add enough heat/energy? My reasoning would be that given the high amount of energy added, both the activation energies of the reaction from left to right and right to left are excessively surpassed thus both directions are as easily achieved. Please correct my reasoning if this is wrong. Thanks!

There are indeed reactions where the equilibrium constant passes through unity as a function of temperature.  The water-gas shift reaction

$\ce{CO + H2O <=> CO2 + H2}$

has an equilibrium constant that gradually decreases with increasing temperature (it's slightly exothermic as written), and the constant is close to $1$ at $\pu{800°C}$.


Is it always the case that when a reaction's products are higher in energy than its reactant, then the reaction is not favored? If this is the case, can you give a brief explanation why is this always the case?

This question brings up the broader point of what it means for a reaction to be favored. Your question describes thermodynamic favorability and that is just dependent on the relative energy of the products and the reactants. An analogy to this is to think of energy levels as literal heights subjected to gravity. If you have a ball on the ground (reactants) and you want to put it on a shelf (products), you have to put in energy. But, once it is in the product state, gravity will want to drive the ball back to the reactant state because it is lower in energy.

Another type of favorability is kinetic favorability; this relates to the rate at which the forward and reverse reactions can occur. There are cases where a thermodynamically favorable reaction doesn't occur (or rather occurs at an incredibly slow rate) due it being kinetically unfavorable. An example of this is diamond and graphite; it would be thermodynamically favorable for diamond to convert to graphite, however the reaction to do so is so kinetically unfavorable that you are unlikely to see it occur.


Molecular similarity using Tanimoto coefficient I'm trying to use a method for evaluating similarity of molecules. I have seen many publications in the machine learning community have used Tanimoto coefficient. However, when I read about it, it states that this is a measure of similarity for molecular fingerprint representation.

I was wondering if this means that the measure of distance is not for the molecule itself but for its representation?
If so, is there a measure that compares two molecules based off of their inherent structure rather than the particular representation of choice?

Also, any other suggestions on the choice of measure would be appreciated.

I was wondering if this means that the measure of distance is not for
  the molecule itself but for its representation?

Inherently, you must decide on a representation of a molecule in cheminformatics. In most cases, the representation is a "2D" valence-bond connection table (e.g., SMILES, SD file, ChemDraw, etc.) familiar to anyone who has taken an organic chemistry class.

So yes, a similarity measure depends on the representation. Tanimoto coefficients are usually based on comparing binary fingerprints.

  If so, is there a measure that compares two molecules based off of
  their inherent structure rather than the particular representation of
  choice?

I'm not sure how to answer this question since a representation forces choices about an "inherent structure" of a molecule. If I pick SMILES, I have no coordinates, implied hydrogen atoms, and choices about implied bond orders aromaticity, etc.

I think perhaps your question is about fingerprints - that "Is is there a similarity measure between a molecular graph - without generating fingerprints as an intermediate step?"

In that case, I think graph similarity measures are what you want, e.g. from Peter Willett, "RASCAL: Calculation of Graph Similarity using Maximum Common Edge Subgraphs" The Computer Journal (2002) pp 631–644

The reason people use fingerprint similarity measures is that they're easy to implement, and have worked fairly well for a wide variety of tasks.


Fundamental physical forces behind osmosis I wonder which fundamental forces act behind osmosis process? Sum of these forces oppose gravity force, like in this figure:

So I wonder what are these forces?

If you dribble a basketball, the force of your hand on the ball, that keeps it bouncing, is not (1) gravity; (2) the weak force; or (3) the strong force. It must be electromagnetic force--the electrons in the molecules of your skin repelling the electrons in the molecules on the surface of the ball. Also, your skin and the ball remain (largely) intact because of the electromagnetic force that hold them together, respectively. Same for osmotic "force". The motions of the solute and solvent molecules are not governed meaningfully by anything other than electromagnetic force.


Are ions actually charged particles? For instance, if an atom loses an electron and becomes a cation, such as sodium, how does that make the sodium atom positively charged? The sodium's atomic radii decreases due to less electron-electron repulsions which in effect makes the sodium atom more electron dense.

The common response I get from TA's is that positive and negative charges cancel each other out so essentially one proton isn't getting canceled out which results in a net positive charge. However, protons and electrons don't cancel each other out in an atom (effective nuclear charge), so I know that response is incorrect since the electrons closest to nucleus of an atom feel a greater positive charge than the valance electrons.

Yes, ions are actually charged particles and neutral atoms are not. The TA's version is correct: the positive and negative charges cancel each other out. In a positive ion, we have one extra proton that gives the whole thing a net positive charge. In a neutral atom, electrons and protons cancel each other out exactly. How can they not to? After all, their electric charges are precisely identical, just with opposite signs.

Now when we speak about cancelling out like that, we only mean the net charge. Take a mole of neutral atoms and a mole of positive ions (just theoretically, of course, otherwise the consequences might be catastrophic), and even a crude electroscope will tell you the difference between the two right away.

If you specifically intend to ask about one ion, let's look at it from a different perspective. A neutral atom or molecule is, well, neutral, but that doesn't mean it would not participate in electrostatic interactions: there are those small positive and negative charges inside it. They seem to be moving, or at least not being all in one place, and you can feel tiny electric fields resulting from that. The corresponding electric potential decays with distance as $\frac1{r^2}$ for a dipole, and even faster than that for a quadrupole or other configurations of charges. The same applies to an ion, but it also has a net charge, and its potential behaves like $\frac1r$. When looking from a far enough distance, you will only feel the pull of the net charge. Look at a neutral atom from the same distance, and you will feel nothing.

Sure enough, if you look very closely, or even inside an atom (or an ion), you will see a lot of things going on around you, and the net charge will not be all that important.

So it goes.


Is there a gauche interaction between C-3 and C-6 in cyclohexane? To my understanding, a gauche interaction occurs between two R groups when staggered by 60 degrees in a Newman Projection. When looking down the C-1 to C-2 bond of a cyclohexane, the Newman Projection shows that C-3 and C-6 are staggered by 60 degrees. Would this still be a gauche interaction, and if so, why does it not result in an energy increase in the chair conformation? I was told this is because cyclohexane in the chair conformation acts as a long chain carbon, but I did not understand how that results in no energy increase.

Yes, you're exactly right. Not only is the axial methyl gauche with respect to the methylene unit that you point out, but there are also gauche interactions of the two methylenes that are drawn in the "bridging" positions of your Newman projections.

So, why is the chair conformation so stable?

For one, gauche interactions are really not that destabilizing. You're talking about maybe a few kJ mol-1. Realistically, the axial interactions between that methyl group and other axial hydrogens is going to be more energetically unfavorable. But this is easily solved by a ring flip, which maintains the chair conformation.

Whoever mentioned that chair cyclohexane "acts like a long carbon chain" was probably referring to the fact that the bond angles are close to 109.5 degrees, which means that it is far less strained than rings that are fewer than 5 carbons or greater than 7 carbons. However, that's true for boat cyclohexane as well, so it's not that relevant to the question at hand.

For reference, boat cyclohexane is about 25 kJ mol-1 higher in energy than chair cyclohexane, so the gauche interactions you're talking about are a very minor contribution the the energy of the molecule.


Osmium tetroxide and meta-chloroperoxybenzoic acid Can these be used to induce acetal ring closures? For example, when dealing with a molecule that has both a C=O functional group and an alkene as part of its structure and there is the possibility of creating a 5 or 6 membered ring through intramolecular -OH attack on the C=O ... can we just use the $\ce{OsO4}$ to create the diol and expect it to provide us sufficient acidity to catalyze ring closure?

Same question goes for meta-chloroperoxybenzoic acid.

With osmium tetroxide, no it does not appear that dihydroxylation and acetalization can be done simultaneously. Usually dihydroxylation is carried out with catalytic osmium tetroxide with a stoichiometric amount of N-methylmorpholine N-oxide (NMO). The latter regenerates the Os(VIII) species necessary to carry out the dihydroxylation and results in the formation of N-methylmorpholine, which is a weak base. Without an acid available for catalysis, the acetalization does not occur. The sequence can be accomplished in "one-pot" however. In practice, you can monitor the progress of the dihydroxylation reaction. When it is deemed complete, add catalytic acid directly to the same reaction mixture, which allows the acetal formation.

Reference: 10.1016/j.tetlet.2012.02.045

Similarly, the overall transformation has been accomplished with mCPBA, but it's not clear if happens without any further intervention. In the referenced paper below, it's not entirely clear to me when the epoxide opens, because the procedure specifies that the epoxidation step is carried out with exclusion of water. It's likely that the epoxide is stable until addition of water, which opens the epoxide to the diol and then the acetalization can occur, catalyzed by the meta-chlorobenzoic acid byproduct from the epoxidation. In some cases, they simply isolate the epoxyketone as well.

Reference: 10.1002/ejoc.201101087


What's the name of the problem of finding equilibrium concentrations? Is there any specific name of the problem of finding equilibrium concentrations with known equilibrium constants and initial concentrations?

Would it be appropriate to call it "direct equilibrium problem", by analogy with the similar problem in chemical kinetics--finding concentrations as function of time for known rate constants--that is called "direct kinetic problem"?

I don't think the phrase "direct equilibrium problem" is used in this context.

"Multiple-equilibrium problem" would be appropriate.

As in the phrase "It involves solution of the multiple equilibrium problem" J. Molecular Biology vol. 247 page 774.

Or "chemical equilibrium problem" if you don't want to imply that there is more than one equilibrium.


Determining actual safety from a vague material safety data sheet My desktop computer system employs a liquid cooling loop featuring a 6-litre external reservoir and radiator combination "cooling tower" unit. It isn't a cooling tower in the strict sense as it does not achieve temperature reduction through any evaporative method - only by radiating heat through the metal structure. However, the reservoir is not sealed and I have noticed a reduction in the level of liquid over the past few months since I started using it - I must assume that I'm breathing-in this liquid in some amount.

The coolant used describes itself as:

  XSPC EC6 is a high performance, eco-friendly coolant for PC water-cooling. It’s based on a blend of refined vegetable extracts with non-toxic corrosion inhibitors and non-toxic dyes. It offers superb protection for copper, brass, steel, nickel, aluminium and has been tested with a range of acetal and acrylic plastics

Never one to trust marketing materials, I requested a material safety data sheet from the manufacturers, and details of the ingredients. I only received an MSDS back without an ingredients listing. The MSDS reports the following:

  COMPOSITION / INFORMATION ON PRODUCT COMPONENTS
  Hazardous Components for EC
  Low oral toxicity proprietary antifreeze formulation based on refined vegetable extracts and non-toxic corrosion and scale inhibitors.
  Classified as GRAS (Generally Regarded as Safe) as a food additive and not classified as harmful for handling or transport purposes.
  However we recommend good working practices are employed and personal protection gear is worn when handling any chemical.

  [...]

  EXPOSURE CONTROL/PERSONAL PROTECTION
  Respiratory Protection: Respiratory protection if there is a risk of exposure to high vapour concentrations.

  [...]

  TOXICOLOGICAL INFORMATION
  Acute Toxicity: LD 50/oral/rat: >20000 mg/kg

  [...]

  ECOLOGICAL INFORMATION
  Mobility: The product will dissolve rapidly in water.
  Ecotoxicity: The product is rated as:
  Algae EC50 (72h):>100mg/l.
  Fish : LC50 (96h):.100mg/l, Oncorhynchus mykiss
  Bacteria: >1000 mg/l Daphnids (acute) EC50 (48h):>100mg/l

My guess is that it's some glycol-based solution - I'm disappointed they don't reveal the actual ingredients.

Is this substance anything I should be concerned about? Is there any  carcinogenic potential?

Also, how can I determine the specific heat capacity of the coolant? As it is used in a reservoir system instead of an active heat-exchange system the heat-capacity is important - I would like to be able to calculate exactly what TDP (Thermal Design Power) computer components I can use and how long I can run them for at-load without fear of overloading the cooling system.

General common sense should always remind you not to drink the liquid stick your finger in it, rub it all over your skin or breathe its vapours for extended periods of time. Paracelsus is best known in Germany for this quote:

  All things are poison and nothing is without poison; only the dose allows that a thing is not a poison.

If you are now scared out of your wits, then please relax and consider that preface a general disclaimer. Because all the toxicologic data I can see points towards the liquid being entirely harmless. Here’s why:

  Acute Toxicity: LD 50/oral/rat: >20000 mg/kg

That’s more than $20~\mathrm{g/kg}$ — an incredible number for $\mathrm{LD_{50}}$ values. Usually, the highest $\mathrm{LD_{50}}$ values I see are $2~\mathrm{g/kg}$. And the greater than basically means ‘we couldn’t kill the rats with that oral dose so we gave up.’

  Low oral toxicity proprietary antifreeze formulation based on refined vegetable extracts and non-toxic corrosion and scale inhibitors. Classified as GRAS (Generally Regarded as Safe) as a food additive and not classified as harmful for handling or transport purposes. However we recommend good working practices are employed and personal protection gear is worn when handling any chemical.

(Emphases mine) As you can see, there are numerous references to ‘nope, this thing really won’t harm you.’ The GRAS classification actually seems to be concerning food additives. No danger here.

  Respiratory Protection: Respiratory protection if there is a risk of exposure to high vapour concentrations.

I’ll admit, this sounds a tad unsafe at first glance. But if you know what MSDS write for truly unsafe compounds, you’ll realise that this is saying ‘If we write “safe”, somebody will stick their nose in, suffocate and then sue us, so we have to put some kind of caveat here.’

The only thing that actually hints at a potentially harmful substance is the thing about high vapour concentrations. But remember that this is a cooling liquid; you don’t want a cooling liquid evaporating away, that would reduce the cooling efficacy. The minute amounts you are losing will dilute well in the air in your room. Open the window once a day or every two days and you’ll have removed anything that can potentially harm you. The liquid is as safe as it gets.

However, don’t sue me if you decide to turn off common sense and start drinking the liquid or stop venting your room or whatnot.


Vapour pressure and equilibrium constant Consider a case I where I have a container halfway filled with a pure volatile solvent. Now we know that liquid molecules will start getting converted into vapour phase, and then dynamic equilibrium will be reached. The equilibrium constant here is given by $K= [P_0]$, where $P_0$ is the partial pressure of the vapours.  ($[P]$ due to pure liquids is considered as 1, so denominator is 1 and hence $K = [P_0]$).

Now we know $K$ only depends on temperature and is independent of any impurities being added, right?

Now, consider the case if I add some impurities to the pure solvent. We know there will be a lowering of vapour pressure and now if I write the equilibrium constant equation, $K = [P_1]$ , where $P_1$ denotes the new vapour pressure when impurities are present, and also $P_1 < P_0$.

So the problem is: $K$ has changed ! I mean previously $K$ was $P_0$ and now $K$ is $P_1$.

So $K$ value has changed by adding impurities. But my book says impurities has no effect on equilibrium constant. Where am I wrong?

Equilibrium constants contain activities of all components involved. For gases, these are proportional to partial pressures, so we use those instead. For liquids and solids, they are constant, so we tend to disregard them altogether. But the activity of a liquid ceases to be constant as you start adding impurities. All of a sudden, you have to use $K=\dfrac{P_{gas}}{a_{liquid}}$.


Do acids without hydrogen taste sour? I have heard that acids taste sour because of hydrogen (I don't really know exactly).  According to the definition of acids:

  An acid (from the Latin acidus/acēre meaning sour) is a chemical substance whose aqueous solutions are characterized by a sour taste, the ability to turn blue litmus red, and the ability to react with bases and certain metals (like calcium) to form salts

Some acids may not contain hydrogen.  Would they also taste sour?

Look at the Wikipedia article on sourness for an explanation of that. But you also really need to really read a basic introduction to acid/base chemistry, because the question does not make much sense… Acids are, by definition, substances that increase the concentration of H+ ions in an aqueous solution.


Are radioactive isotopes of silver dangerous? I know we all think of things like uranium as being radioactive in the nasty radiation poisoning sort of way. But do elements like silver have similar effects in their radioactive isotopes?

I'm an author doing some research for a book I'm working on. Would there be a particular quantity or exposure time that would be necessary to make it toxic?

Say we did know the amount to make it toxic. I know regular silver can cause Argyria which is mostly a cosmetic illness, but would the side effects of the radioactive silver be similar to that of other radioactive materials?

A glance through the table of the isotopes in the venerable CRC handbook reveals that the longest lived radioactive isotope of Ag is $Ag^{108m}$, made by neutron capture by $Ag^{107}$.  this has a listed half life of "> 5y", but the capture cross section is only $35\pm5$ barns, pretty small for thermal neutrons.  So, it would be really hard to get a lot of radioactive silver.

Next, all isotopes listed decay by $\beta, \beta^{+}$, EC, or IT (electron capture of internal transitions for the metastable nuclei), with decay energies ranging from 93 keV ($Ag^{107m} \rightarrow Ag^{107}$) to 6MeV ($Ag^{116}$).  Those are ionizing radiation, which would increase your probability of getting cancer.

The 'nasty radiation poisoning' things are actually nasty primarily because of being heavy metals, the radiation comes along for the ride, so to speak.  The normal Uranium isotopes are very long lived, so emit very little radiation per unit time.  Polonium, the scary poison of choice in stories and perhaps by the KGB, in contrast, has a short lifetime (138 days for $Po^{210}$), and spews lots of ~5MeV $\alpha$ particles, heavily damaging tissue (much more so than gammas).

Given the difficulty in making the stuff, assuming you would try $Ag^{108m}$, (small cross sections, can't chemically separate from the stable $Ag^{107}$, etc., your best bet would just be acute heavy metal poisoning from Ag salts (which may or may not be highly toxic). Or, using a silver candlestick to bludgeon the poor victim to death.


Can you get oxygen form silica or alumina? If I had a lot of silica or alumina, like on the moon, would I be able to split the silica or alumina into aluminium and oxygen/ silicon and oxygen? It doesn't matter how unpractical it is. If there is a way, how can it be done?

If you don't care too much about practicality, ionic compounds (i.e. salts) (which most minerals are) can be separated by melting them and then electrolysing. The only problem is that most salts have very high melting points. The needed temperature can be lowered by dissolving the salt in another lower melting point liquid, as happens in the Hall-Heroult process for aluminium production, but that only works if the dissolving liquid is more resistant to electrolysis than the stuff you're trying to electrolyse.

An even more general method of separating basically any compound is by heating it into a hot enough plasma to atomise it. The atoms can be separated using a calutron (which works similarly to a mass spectrometer). In reality such separation is only used with very small amounts of materials for analysis and historically for uranium enrichment, but there is no reason in principle why this could not be scaled up to an industrial scale.


What determines the atom or molecule that receives electrons during oxidation? I'm no chemist, so bear with me.

From my understanding, the reason metal oxidation occurs is because the metal is losing electrons to the oxygen ions in the air; and that an ion is an atom or molecule which either has more or less electrons than the "standard" of that atom/molecule; which is why it's "taking" electrons from something else.

What prompts my question has to do with soldering. I'm learning that one reason for the use of flux is to prevent this oxidation and protect the soldering tip. So according to what I said before, a soldering iron becomes oxidized because it has lost electrons to the air, and the high temperatures hasten this process.

Now to get to the detailed question: I understand the very basic oxidation process. But, once the metal has lost electrons, doesn't that now make it composed of ions? If so, Why does the metal continue to corrode? Why isn't the metal now taking electrons and virtually "restoring" itself? What is it about oxygen ions that make them the ones that take electrons from metals such as copper? Why isn't copper the one taking electrons from the air? Does it have to do with valence electrons or electron shells or something?

When a metal like iron is oxidized, each iron atom loses 3 electrons. These electrons are kept by oxygen atoms from the air. The iron atom becomes an ion charged 3+, and the oxygen atom becomes an ion charged 2-. The positive ions still attract negative ions, and they will never separate. Two iron ions and three oxygen ions are necessary to make a new compound which must be not charged. This assembly is a new substance called rust, which has a brown color. Once such a transformation has happened, the next iron atoms can repeat this operation : the amount of rust will slowly increase.

All metals can be oxidized by more or less the same way. But it could be quicker or longer.  For copper, this reaction is not so frequent. Copper makes a copper oxide which is black, but you may wait for months to see it.


How do I remove free biotin after biotin-protein conjugation? I want to quantify a biotin-protein conjugate after chemical biotinylation. After biotinylation, the solution contains both biotinylated protein and free biotin.

I intend to use a HABA-avidin biotin quantitation kit like this one from Pierce. Since the biotinylated protein and the free biotin will both displace HABA and decrease the absorbance, I need to remove the free biotin so that I only quantify the biotinylated protein.

How do I separate the free biotin from the solution?

I did some bioconjugation with NHS and maleimide biotins to proteins, peptides, and oligosaccharides. I removed the free biotin using spin filters or dialysis. Just be sure to use a MWCO well below the mass of your protein. I even used prep scale HPLC to remove biotin from the oligosaccharide. The HABA kit works well, I think I used that same one to quantify peptide - avidin linking. But try to keep your biotin protected from light and oxygen, it oxidizes fairly easily, I was getting +16 and +32 masses on LC-MS and couldn't explain it until I found some 50 year old paper on biotin sulfoxide. The oxidized biotin doesn't bind avidin nearly as well as normal biotin.


Hydrogen bonding in solid bicarbonates Why is there two different kinds of hydrogen bonding in solid bicarbonates $\ce{NaHCO3}$ and $\ce{KHCO3}$? In both cases we consider anions of each for hydrogen bonding .

The anions involved in hydrogen bonding are the same for $\ce{NaHCO3}$ and $\ce{KHCO3}$. But, why is there a difference in hydrogen bonding?

I found something interesting in Acta Crystallogr. 1952, 5 (2), 292–292, which is attached below.

As someone rightly pointed out in the comments, we'll know this only if we know the crystal structure of the compound in question.

What's interesting to note is that hydrogen bonding, and the arrangement of atoms in space under influence of the same varies from compound to compound, and is dependent on the crystal structure of the same.


What is the nomenclature of this alkane? In this alkane, the blue line is parent chain. It is 12 carbon based.
I wonder why red line cannot be parent chain. It does not compete with blue line about the position of substitution.

And also, i understand 5-isopropyl part but i cannot understand octan-4-yl part. As i learned, when we take substituent part of already substituted part, we take carbon number one on the branch point. So i think this molecule has nomenclature about 6-(1-propylpentyl) but it is wrong. Is there any nomenclature rule that i have missed?

The nomenclature rule that you have missed is given in Section P-46 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) (explaining that compound acyclic substituents consisting of a principal chain and one or more acyclic substituents can be named in two ways which can be found in Subsection P-29.2) and in Subsection P-29.3.2.2, which clarifies that the preferred names for simple substituent prefixes derived from saturated parent hydrides with a single free valence that is not at the end of the longest chain are formed by using alkanyl substituents according to Method (2) of Subsection P-29.2:

  (2) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

For the substituent that is mentioned in the question, the preferred alkanyl-type prefix according to Method (2) is ‘octan-4-yl’.

Note that, according to current IUPAC nomenclature, the prefix ‘isopropyl’ is only retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’, which is also formed according to Method (2).

Therefore, the complete name for the compound that is given in the question is 6-(octan-4-yl)-5-(propan-2-yl)dodecane.