Resonance structures of some aromatic compounds Do compounds like tropolone, furan, and thiophene (which are aromatic) show resonance?

If yes then what are the resonance structures that can be drawn? Also, do all aromatic compounds show resonance?

Resonance structures can be drawn for all aromatic compounds.  Tropolone, furan and thiophene are considered to be aromatic compounds because

we can draw resonance structures
and they are cyclic compounds with 4n+2 (n=1) pi electrons

Here are the resonance structures we can draw for tropolone, the positive charge can be delocalized to every carbon in the ring.

Just like in benzene, where instead of drawing 2 resonance structures (top of following figure) we usually just draw the one resonance structure shown at the bottom of the figure;

instead of drawing all 8 resonance structures for tropolone, we usually represent them by just drawing the one resonance structure shown at the bottom of the tropolone figure.

and here are the resonance structures for furan

and for thiophene, they are the same as furan, just replace the oxygen with a sulfur atom in the resonance structures.


In the blue bottle experiment, will it eventually be "used up" after repeated use? The blue bottle experiment is a bottle half full of a colorless solution (dextrose, KOH, and methylene blue, in the version that I saw) and air that, when shaken, turns blue through the methylene blue oxidizing. Over time, the blue color fades while the methylene blue is reduced and the solution goes back to colorless. This can be repeated.

My question is, can this be repeated indefinitely, or is there some chemical in the solution that is being eaten up? Is the mechanical energy of shaking enough to sustain the reaction indefinitely? If it can be used up, what exactly is being used up?

The blue bottle experiment, and other similar reactions like the "chemical traffic light", are not infinitely repeatable.  They are driven by the oxidization of glucose (or some other suitable reducing agent, such as ascorbic acid) by atmospheric oxygen, and will eventually stop working once the bottle runs out of either glucose or oxygen.  However, each cycle only consumes a rather small amount of glucose and oxygen, so the reaction can be repeated quite many times.

Also, the blue indicator dye (methylene blue) itself is not used up in the reaction.  Even if the reaction does "run out of fuel", it can be made to work again just by adding more glucose to the solution and/or simply by opening the bottle to let more oxygen in.  However, since the glucose is not getting fully oxidized all the way to water and carbon dioxide in this reaction, the various incomplete oxidization products will gradually accumulate in the solution.

I'd expect that, if you kept adding more and more glucose to the solution, eventually you'd end up with a sticky sugary alkaline goo full of partly oxidized glucose products that was too viscous to be swirled around the flask any more.  At that point, you'd really have no choice but to dispose of the experiment and start over with fresh ingredients.


Back titration of excess HCl used for dissolving limestone $15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $200~\mathrm{mL}$ with water. A $20.0~\mathrm{mL}$ portion was then neutralised by $8.50~\mathrm{mL}$ of a $0.1~\mathrm{M}\ \ce{NaOH}$ solution.
Calculate:

Amount of excess $\ce{HCl}$ in the $20.0~\mathrm{mL}$ portion
Amount of excess $\ce{HCl}$ in the $200~\mathrm{mL}$ portion
Amount of $\ce{HCl}$ which reacted with $\ce{CaCO3}$

My attempt:
When the limestone is dissolved,
$$\ce{2 HCl(aq) + CaCO3(s) -> CaCl2(aq) + CO2(g) + H2O(l)}$$
$$n(\ce{HCl}) = c \times V = (15 \times 10^{-3}~\mathrm{L}) (1.4~\mathrm{M}) = 0.021~\mathrm{mol}$$
$$n(\ce{CaCO3}) = x~\mathrm{mol}$$
When the solution is diluted to $200~\mathrm{mL}$ the molarity of $\ce{HCl}$ can be found by,
\begin{align}
n_\mathrm i &= n_\mathrm f\\
c_\mathrm i V_\mathrm i &= c_\mathrm f V_\mathrm f\\
c_\mathrm f &= \frac{c_\mathrm i V_\mathrm i}{V_\mathrm f}\\
\therefore c_\mathrm f (\ce{HCl}) &= \frac{(15\times 10^{-3}~\mathrm{L})(1.4~\mathrm{M})}{(200\times 10^{-3}~\mathrm{L})} = 0.105~\mathrm{M}
\end{align}
I'm having doubts about my working up until this stage. Am I correct so far?

It is generally a good idea to approach problems in the order that sub-questions are asked in. There is often a paedagogic reasoning in that order that can help you approach the answer.

In this case, the key point is that the limestone is impure $\ce{CaCO3}$, and that thus we do not know how much of the acid was initially neutralised with $\ce{CO2}$ evolution. Approaching the question from this side, as you did, is not very helpful, since we are missing an important value to calculate from.

What we do know, however, is the final end of the experiment: that $8.50~\mathrm{ml}$ of $0.1~\mathrm{M}\ \ce{NaOH}$ were able to neutralise a certain portion of the remaining hydrochloric acid. You should think of this as being a titration.

Once you have realised that, it makes sense why the first question asks for the amount of substance in the $20.0~\mathrm{ml}$ aliquot — it is the one you can actually calculate quickly, and it will lead you to the entire solution.

Thus, your plan of action is:

find the concentration or amount of $\ce{HCl}$ in the $20.0~\mathrm{ml}$ portion by known formulae used for titration problems (your work so far shows you can do this);

   We have $\ce{HCl + NaOH -> NaCl + H2O}$, so
 $$\begin{align} n_\ce{HCl} = n_\ce{NaOH} &= (\pu{8.50 mL})(\pu{0.1 M}) \\ &= \pu{0.00085 mol} \end{align}$$

thencefrom calculate the concentration or amount of $\ce{HCl}$ in the entire $200~\mathrm{ml}$ sample from similar formulae;

   A $\pu{20 mL}$ portion of the $\ce{HCl}$ simply has one-tenth of the original amount of $\ce{HCl}$, so in the entire $\pu{200 mL}$, we have:
 $$n_\ce{HCl} = (\pu{0.00085 mol}) \times \frac{\pu{200 mL}}{\pu{20 mL}} = \pu{0.0085 mol}$$

Next, decide how much hydrochloric acid was lost in the dissolution of the limestone.

   At the start, we had a grand total of
 $$\begin{align} n_{\ce{HCl},\mathrm{tot}} &= (\pu{15.0 mL})(\pu{1.4 M}) \\ &= \pu{0.021 mol} \end{align}$$
 which you already calculated. After reacting with the limestone, we were left with only $\pu{0.0085 mol}$ of that. So, the amount that reacted is simply:
 $$n_{\ce{HCl},\mathrm{reacted}} = \pu{0.021 mol} - \pu{0.0085 mol} = \pu{0.0125 mol}.$$


What exactly is happening when sodium, potassium, or other alkali metal explodes in water? There are lots of videos on YouTube showing sodium, potassium, etc. exploding when dropped into water (this, for example).

I understand that when an alkali metal is exposed to water, a violent exothermic reaction occurs where a hydroxide and hydrogen gas are produced, but why and how does the sample of metal end up detonating and fragmenting?  Physically speaking, how can a block of metal seemingly blast apart from the inside when the reaction occurs on the surface of the sample?

The Wikipedia article on alkali metals explains this, but I still don't seem to understand how this would result in an explosion.  Diagrams would be very helpful.

The key of the question is how does the release of hydrogen gas and energy result in an explosion?  Is there thermal runaway, and is the metal physically destabilized in some way during the reaction?

Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the surface of the droplet increases in area rapidly forming a spiny, porcupine like shape as fingers of the molten metal are shot into the liquid at astonishing speed. The larger the surface area gets the faster the reaction occurs, leading to a runaway effect. The study was published in Nature Chem. (Ref.1).
References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, "Coulomb explosion during the early stages of the reaction of alkali metals with water," Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).


Conversion of acetaldehyde to butane-1,4-diol How to convert acetaldehyde (ethanal) to butane-1,4-diol. I just tried doing aldol, but it leaves an alkyl group at the end. Also it cannot be done by oxidizing ethane with basic permanganate.

Cross Aldol and add H+

Oxidise Aldehyde to Carboxylic Acid

use any of the reagents

3.Perform Arndt–Eistert reaction

Use LiAlH4 for reduction of acid to alcohol

And finally you get 1,4-dihydroxybutane.

Image courtesy Wkipedia, SigmaAldrich


Book Recommendation for Molecular Orbital Theory I posting my question here since I have seen similar questions but I haven't found the specific answer I am searching for. I am seeking textbook recommendations on Molecular Orbital Theory with comprehensive sections and explanations on diatomic molecules. I want to learn how to construct the molecular orbital energy-level diagrams and to write the electronic configuration for diatomic molecules.
I have seen several books and papers such as Mulliken's papers, Spectra of Diatomic Molecules book by Herzberg, Molecular Quantum Mechanics by Atkins, and other modern books, etc but unfortunately, I haven't found sections that treat other than simple diatomic molecules. To be more specific, homonuclear diatomic molecules and Hydrides (i.e. LiH, BH, CH, ...) in the case of heteronuclear diatomic molecules. So I am looking for books that treat molecular orbital theory more than simple diatomic molecules (moderate and heavy heteronuclear diatomic molecules) and I haven't found any!
Any help in this regard is much appreciated.

If you are interested in an applications, look at MO theory in the context or organic reactions, which is a part of chemistry where the theory works well.  Check out Fleming's Frontier Orbitals and Organic Chemical Reactions:
https://onlinelibrary.wiley.com/doi/book/10.1002/9780470689493 .


What do you mean by "periodic" in the periodic table? Modern periodic law states:

  “The physical and chemical properties of the elements are periodic
  functions of their atomic numbers”.

But I don't think this is so! (Forgive me for my stupidity. But please try to understand what I am saying and if you find any mistakes then please let me know).

If I consider the property of Li then I find that after 7 element there comes an element Na which have similar properties like Li. Now according to periodic law I must get another element which has similar properties to that of Li after 14 elements. And indeed I found one named K . Now again the same thing must happen (no?) ie, there must be an element which have similar properties to Li and is 21 elements away(Mn). But I don't find such element! Why?

The properties of Element is periodic so why then this "periodicity" doesn't hold in case of Li?

First you need to understand what a periodic function is in mathematics. For example, if you are standing at any point P on a circle, you start walking on the circumference and count the degrees you completed. Once you complete 360 degrees, and you will find yourself on exactly the same point P. Sines, cosines and many other functions are periodic. This term existed before the chemists started using the term periodic law in the early 1800s or perhaps well before that. If $f(x)$ is a function with period $T$, then $f(x+T)= f(x)$ for a periodic function in a defined range of $[a,b]$. The period $T$ may change to $T'$, when x is in a different range say, $[c,d]$. This is  an example of a discontinuous periodic function. As you will see, periodic properties of the elements form a discontinuous periodic pattern in a loose sense.

Now come to the periodic law: "The law that the chemical elements, when listed in order of their atomic numbers (originally, atomic weights), fall into recurring groups, so that elements with similar properties occur at regular intervals." [unabridged OED].

Instead of $x$ for a mathematical function, the independent variable is the atomic numbers Z (Zahl for a number in German) for the chemists. So this periodic law has a very loose connection with mathematical periodic function, because none of the values are actually repeated. Each element is unique, but in a loose way, their properties or even a better word is trends "repeat" themselves after a certain number of elements. See this chart for example. Atomic radii vs. Z, do you see a loose periodic pattern?

Once you have got the idea of periodic function in a loose way for the chemical elements, get ready to the see the physicists style of periodic table. It explains why some properties loosely repeat themselves.

Each element in the column, corresponds to filling of a certain orbital, for example look at the column $s^1$. All elements falling in that column correspond to the filling of $s^1$ electrons. Then move to the next column $s^2$, all elements listed there corresponding to filling of $s^2$. Can you see a pattern? In this sense, one calls it periodic function in a modern way, that certain elements repeatedly fall in $s^1$ column, some elements repeatedly fall in $s^2$ column.


What is the product of a reaction between an acyl chloride and a secondary alcohol? For the following two reactants, I want to predict the major product.

I would have thought the second reactant $\ce{CH3CH(OH)CH3}$ splits into $\ce{CH3CHCH+}$ and $\ce{OH-}$ , and the $\ce{OH-}$ would replace the $\ce{Cl-}$ in the first reactant $\ce{Phenyl-COCl}$, giving B.  My main reason for thinking this would occur is that I always see e.g. water form $\ce{H+ and OH-}$.

However, the answer is C. The $\ce{CH3CH(OH)CH3}$ gives up only its  $\ce{H+}$, and the rest of it joins $\ce{Phenyl-COCl}$ to form C.

How do I predict which atom(s) will leave/join? And for this particular question, how do I know the bond in the 2nd reactant cleaves at O-H not at C-OH?

It's a bit tricky because there's probably a reagent that's missing. You might add DMAP or triethylamine to this reaction which would further activate the acyl chloride. But the alcohol oxygen is the most nucleophilic moiety in this reaction. The most electrophilic moiety is the carbonyl group, especially since it's activated by the electron withdrawing chlorine.

Thus we would predict that a standard acylation (a carbonyl reaction) will take place where we add a nucleophile to the carbon of a carbonyl group. After which, reformation of the carbonyl group will result in kicking out a chloride anion. Finally, a very fast proton transfer will take place that breaks the O-H bond you are referring to. Notice that this is really an after thought and not really the key point of this reaction.

Take a look at the mechanism for the reaction in your textbook (or here).


Oxidation of 2-Methyl, 2-Phenyl Propane Considering oxidation mechanism of side-chains in arenes, it is known that side-chains get oxidized to carboxylic group ($-COOH$).

Below is a picture of 2-Methyl, 2-Phenyl Propane. When it gets refluxed with acidified $KMnO_4$ or acidified $K_2Cr_2O_7$, it should be oxidized to give benzoic acid.

But according to the provided source, it seems 2-Methyl, 2-Phenyl Propane doesn't react with the above oxidizing agents.

My question is,

Why does this compound disobey the general oxidation of benzyl derivatives?

Note, the source has a OH on position 4 of the ring, that makes it difficult for me to name it.

Source: https://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_s13_qp_41.pdf
Qno:5 c) i)

Or is the OH that makes the difference in this question?

2-Methyl,2-Phenyl Propane:

For oxidation of aromatic side chains, you must have at least one Benzylic hydrogen. tert-butyl Benzene (which is actually 2-methyl,2-phenyl propane) doesn't have any benzylic hydrogen.

-By @Soumik Das


How do antioxidants work? A friend and I are doing a project that involves the process behind manufacturing Jet A-1 and specifically the specifications highlighted in DEF 91-91. I was interested to see the information at the bottom of page 8.

Specifically:

Why are 2,6-ditertiary-butyl-phenol, 2,6 ditertiary-butyl-4-methyl-phenol and 2,4-dimethyl-6-tertiary-butyl-phenol used rather than "more common" antioxidants like ascorbic acid?
How do these antioxidants function and do they work any differently
to biochemical antioxidants?
What factors determine the effectiveness of an antioxidant?
Why are a mixture of chemicals used at different ratio's rather than just one. Do they work more/less effectively at different temperatures for example?

1) solubility. Typical liquid fuels consist entirely of hydrocarbons and are non-polar and thus cannot dissolve polar molecules like ascorbic acid.

2) The molecules mentioned in the provided quote can form stable radicals on loss of $\ce{OH}$ group hydrogen, i.e. act as a radical trap, transforming active radicals into non-active particles. It is a typical mechanism for antioxidants to work.

3) Strongly depends on the definition of effectiveness employed. Differen applications demand different properties.

4) Because in industry people typically do not bother with isolation of individual products if a mixture can work just as well. Sure, one can isolate individual products of phenol alkylation, but it is a tiresome work, so why bother, if the mixture works just fine?


What are the half-equations of manganese and oxygen in the following redox reaction? The equation is $$\ce{2Mn^2+(aq) + 4OH^-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)}$$
I know that the Manganese (II) gets oxidized to Manganese (IV), but I'm not sure about the other half. I know oxygen gets reduced, but using simply $\ce{O2 +2e^-->2O-}$ doesn't work because the equations won't balance out then.

By the way, this chemical equaltion is used in the Winkler titration.

The half-equations should look like this:

$$\ce{2Mn^2+(aq) \to 2Mn^4+(aq) + 4e^-}$$

and

$$\ce{4OH^-(aq) + O2(aq) + 4e^- -> 4O- + 2H2O(l)}$$


Why does the iron rust with less salt For my school project I was going to measure the rate that salt has on the rusting of iron.

To do this I set up an experiment where I had 15 beakers containing 200 milliliters of distilled water. I poured into the beakers varying moles of salt $\ce{(NaCl)}$. I had $\pu{0mol}$, $\pu{0.25mol}$, $\pu{0.5mol}$, $\pu{0.75mol}$ and $\pu{1mol}$ of salt. So now there are three beakers for every concentration. I then put a iron nail in every beaker, put plastic wrap over them to stop some of the oxygen flow and left it be for 10 days.

What I was surprised to find was that the beakers with no salt in them had more rust than the one ones that had. It was exactly the opposite of what I expected would happen, and I can't find a website that explains how this occurs. Does anyone have an explanation?

Corrosion is a balancing game. In this case, by covering the beakers, you have limited the driving force for oxygen dissolution by limiting the amount of oxygen available to dissolve into solution and then the higher salt concentrated solution is able to further minimize the amount of oxygen that can be dissolved and participate in the redox reaction (as explained in the linked question's answer).


Effect of mixing acids on solubility Please help me clarify the title if you come up with a better way to put it (you certainly will).
My question is the following:
If you make a mix of equal parts of various acids (Bicarbonate, Hydrochloric, Nitric, Sulfuric, etc.) and throw some metal in there that will ionize and wait for a while until you have an equilibrium, what will that resulting equilibrium look like vs. the known solubilities of the expected salts? Where and in what form will the metal ions most likely end up? Will the metal's solubility tend towards an average of the expected solubilities? Or tend towards the solubility of the least soluble metal salt? Or of the most soluble metal salt? Or exceed the solubility of the most soluble salt?
This is assuming that no acid is / would have been limiting in reacting with the metal in the first place.
My assumption is that most metal ions would end up precipitating as the least soluble salt but I'm very much unsure of myself.
Upon Poutnik's advice, here are two examples I provided in the comments:
Case No 1: I drop 1 mole Cadmium metal in a solution with 5 moles of sulfuric acid and a great excess of water beyond the solubility limit of the product (I assume it would dissolve entirely), I then add 5 moles of phosphoric acid. Would that (a) precipitate most of the nickel (almost as if only phosphoric acid was present), or will (b) the solubility be somewhere between that of the sulfate and phosphate salts, or (c) most of the nickel remain in solution as if only the sulfate was present?
Case No 2: I drop 1 mole of lithium and 1 mole of cadmium metal into 2 moles of sulfuric acid with enough water for both to be entirely dissolved. I then add one mole of phosphoric acid, leave the solution in a sealed container with heavy stirring and come back 1000 years later. I then filter out and analyze the precipitate: Will it be all cadmium / mainly cadmium / an even mix of both cadmium and lithium phosphate salt? (knowing both cadmium and lithium form insoluble phosphate salts, but the solubility of the lithium salt is several orders of magnitude above that of the cadmium salt). Let's assume there is no double salt formed for the sake of simplicity.

All strong acids dissociate in water: $\ce{HX + H2O -> H3O+ + X-}$.
Metals reacts with hydronium ions, not with a particular acid, e.g. like $\ce{M(s) + 2 H3O+ -> M^2+ + H2(g) + H2O}$.
( With nitric acid, it is more complicated due its reduction to $\ce{NO}$ or $\ce{NO2}$, but it is not important now. )
Metallic cations and acid anions are not paired with a particular counter ion, until they start precipitating. They all form together a "ionic soup".
As a rule of thumb, the guidance is the salt solubility. But in case of mixed anions and/or cations and non stoichiometric cation/anion ratios, more important is solubility product
$$K_\mathrm{sp}=[\ce{M^n+}]^{m} \cdot [\ce{X^m-}]^{n}$$
The salts that would reach their solubility product would start precipitate,
but then hard-to-control kinetic factors of crystallization initiation and propagation take part.
The excess of the counter ion decreases solubility of the salt, compared to salt solubility in water. It is similar effect as gravimetric precipitation of very little soluble salts, where excess of the counter ion causes much lower ion residual concentration, compared to being just in water.
Other ions may alternate salt solubility by side reactions  ( if applies ) with the considered ions, or by affecting activity coeeficients - see below.

For the case 1, sulphates would have the minor effect of decreasing somewhat the activity conefficients that therefore somewhat increased solubility of the phosphate. But it is general aspect of ions presence.
For the case 2: It depends on if there is formed a double salt like LiCdPO4 or not.

In particular specific cases, there is possibility of forming double salts, where precipitate consists of integrated more than 1 cation or anion. E.g. crystallization from solution containing $\ce{Fe^2+, NH4+, SO4^2-}$ may lead to crystallization of "Mohr salt" $\ce{(NH4)2Fe^{II}(SO4)2 \cdot 6 H2O}$. Similarly for alums with $\ce{K}$ and $\ce{Al/Cr/Fe^{III}}$ as  $\ce{M^{I}M^{III}(SO4)2 \cdot 12 H2O}$.

During rapid crystallization there occurs also coprecipitation:

There are three main mechanisms of coprecipitation: inclusion, occlusion, and adsorption. An inclusion occurs when the impurity occupies a lattice site in the crystal structure of the carrier, resulting in a crystallographic defect; this can happen when the ionic radius and charge of the impurity are similar to those of the carrier. An adsorbate is an impurity that is weakly bound (adsorbed) to the surface of the precipitate. An occlusion occurs when an adsorbed impurity gets physically trapped inside the crystal as it grows.

In the case of rather concentrated solutions of well soluble salts, there is complication we should consider the ion activities  instead:
$$K_\mathrm{sp}=a_{\ce{M^n+}}^{m} \cdot a_{\ce{X^m-}}^{n}$$
Acticity coefficients for diluted solutions can be calculated by Debye-Huckel theory, but for concentrated solutions it is hard to theoretically predict them. We can use concentrations for rough predictions, but the numbers may not fit well.


E/Z configuration of the central double bond in a highly branched poly-ene What would be the configuration of the marked double bond in the attached molecule?

For simplicity, I shall assign numbers to the double bonds so that it is easier to reference them. This numbering is chosen solely after their position in space and does not correspond to the number of carbons in a PIN or otherwise systematic name.

1                 10
  > 5         8 <
2    \       /    11
       > 7 <
3    /       \    12
  > 6         9 <
4                 13

The terminal bonds, 1–4 and 10–13, are easy since they are all disubstituted: (1Z,2Z,3E,4E,10E,11Z,12Z,13E)

1Z                 10E
   > 5         8 <
2Z    \       /    11Z
        > 7 <
3E    /       \    12Z
   > 6         9 <
4E                 13E

Next up are the intermediate double bonds. If we simply follow the atoms, neither side will ever have priority, thus the somewhat less common rule that Z has priority over E takes effect. This allows us to label the double bonds (8Z) and (9Z).

1Z                  10E
   > 5         8Z <
2Z    \       /     11Z
        > 7 <
3E    /       \     12Z
   > 6         9Z <
4E                  13E

However, we cannot determine a stereodescriptor for bonds 5 and 6: both 1 and 2 have the same geometry (Z) as do both 3 and 4 (E). Thus, neither of these double bonds is asymmetric as they only feature three different substituents with the two identical ones being on the same side.

1Z                   10E
   > 5Ø         8Z <
2Z     \       /     11Z
         > 7 <
3E     /       \     12Z
   > 6Ø         9Z <
4E                   13E

Moving on to the final double bond 7, we face the same problem. The substituents 8 and 9 are identical, both feature an E double bond attache to E and Z. Thus again, we are unable to determine a stereodescriptor since the double bond is not asymmetric.

1Z                    10E
   > 5Ø          8Z <
2Z     \        /     11Z
         > 7Ø <
3E     /        \     12Z
   > 6Ø          9Z <
4E                    13E

Note that we can assign priorities to the left-hand side of double bond 7: the bottom half (3,4,6) has a lower priority than the top half (1,2,5) since the latter features Z double bonds where the former features E. But assigning a descriptor to 5, 6 and 7 is impossible due to their symmetric other side.


How do erasable pens work? I understand the principle behind an eraseable pencil. Graphite is highly slippery, you rub it onto the paper, and an eraser can rub it off again.

I've had friends at school with 'eraseable ink pens'. The ink was 'special' and the eraser was 'special' and it could erase the pen in the same way a regular eraser could erase pencil.

My question is - how do eraseable pens work? (Ie what substances are involved and what are their properties?)

Instead of a typical oil/dye ink,  erasable pens use a compound closer to rubber cement, which acts like a liquid when a bit of pressure is exerted. 1
A bit of history and a link to patents can be found here
EDIT: Be sure to check out the other answer here about thermochromic ink too.


What are compounds with the same mass called? It may be that the answer is just "no" and that's why I can't find it, but is there any common name for compounds with the same mass?

I'm thinking in the context of mass spectrometry where these compounds require additional steps to resolve.

Isomers are the obvious example, but I was wondering if there was a more general term.

In mass spectrometry, ions with the same nominal mass (but possibly differing exact masses) are called isobaric.


How to perform a relaxed scan in GAMESS? I have the following input Gaussian input for my system. The coordinates are given in Cartesian format and it is scanning through 8 steps, with a bond distance between 1 and 5 having a difference of 0.1 Å in each step, and the bond distance between 1 and 2 is fixed.

#hf/6-31g** opt=modredundant

OPT

1 1
 O     0.086655     0.133386    -0.150088
 O    -1.636963     0.213966    -1.809353
 H    -0.149749     0.260762     0.763793
 H     0.802607    -0.484925    -0.261350
 H    -0.423645     0.140377    -0.629197
 H    -2.313930    -0.430164    -1.970372
 H    -1.720592     0.938467    -2.416119

B 1 5 S 8 0.1
1 2 F

What is the analogue of this in GAMESS?

Unfortunately a feature like ModRedundant in Gaussian, which one can use to perform relaxed scans, is not available in GAMESS. Even the possibility to perform a rigid scan seems a bit complicated. You can request a calculation with RUNTYP=surface and you have to specify a $SURF group. See example 29 of the tests, which you should find in path/to/gamess/tests/standard/exam29.inp.

A relaxed scan is a bit harder to perform, and only straight forward for varying distances. You have to perform a series of single point calculations, with partially frozen coordinates. I describe the general procedure of doing that in How do I perform a partial optimisation in GAMESS?

Here I will outline it for your problem, based on what you can read there.

Before we do that however, you should be aware, that there is a python script by Andrew Hoos on github. I did not have the time to test it. It is quite old, and I cannot predict how the development of GAMESS will influence the functioning of rscan.

I assume you have read and understood the linked post, so in an effort to safe some space, I use a more automated approach here.

First, set up the general file, which you want to use for a check run. I have used your geometry.

 $CONTRL
   EXETYP=CHECK
   SCFTYP=RHF RUNTYP=ENERGY
   COORD=UNIQUE   ICHARG=1   MULT=1   NZVAR=15   $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   $END
 $GUESS
   GUESS=HUCKEL    $END
 $DATA
   H5O2
   C1
   OXYGEN      8.0      0.086655000         0.133386000        -0.150088000
   OXYGEN      8.0     -1.636963000         0.213966000        -1.809353000
   HYDROGEN    1.0     -0.149749000         0.260762000         0.763793000
   HYDROGEN    1.0      0.802607000        -0.484925000        -0.261350000
   HYDROGEN    1.0     -0.423645000         0.140377000        -0.629197000
   HYDROGEN    1.0     -2.313930000        -0.430164000        -1.970372000
   HYDROGEN    1.0     -1.720592000         0.938467000        -2.416119000
   $END

It is again essential that you include NONVDW descriptions. Since the proton will move in between the two water molecules, I have included both pairs. Note that the 1-2 distance will not be included in the list of internal coordinates (even if you specify it directly). You can use NONVDW(1)=1,2 instead though.

Now that you have checked everything you can run the first optimisation:

 $CONTRL
   SCFTYP=RHF   RUNTYP=OPTIMIZE    COORD=UNIQUE
   ICHARG=1   MULT=1   NZVAR=15     $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   IFZMAT(1)=1,1,2,     1,1,5
   FValue(1)=2.3938489, 0.700
   $END
 $GUESS
   GUESS=HUCKEL   $END
 $DATA
   H5O2
   C1
   OXYGEN      8.0      0.086655000         0.133386000        -0.150088000
   OXYGEN      8.0     -1.636963000         0.213966000        -1.809353000
   HYDROGEN    1.0     -0.149749000         0.260762000         0.763793000
   HYDROGEN    1.0      0.802607000        -0.484925000        -0.261350000
   HYDROGEN    1.0     -0.423645000         0.140377000        -0.629197000
   HYDROGEN    1.0     -2.313930000        -0.430164000        -1.970372000
   HYDROGEN    1.0     -1.720592000         0.938467000        -2.416119000
   $END

Note that I am keeping the automagic creation of the coordinates this time. To be on the safer side, you should probably follow more closely the other post. In any case, we are now ready to continue. Well use the optimised geometry of the last run, as well as the orbitals, which can be found in the auxiliary output.

 $CONTRL
   SCFTYP=RHF    RUNTYP=OPTIMIZE    COORD=UNIQUE
   ICHARG=1   MULT=1   NZVAR=15     $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   IFZMAT(1)=1,1,2,     1,1,5
   FValue(1)=2.3938489, 0.800
   $END
 $GUESS
   GUESS=MOREAD
   NORB=15
   $END

! ----- RESULTS FROM SUCCESSFUL RHF      GEOMETRY SEARCH -----
! ----- COORDS, ORBS, GRADIENT, AND APPROX. HESSIAN -----
!  COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
!      ATOM  CHARGE       X              Y              Z
! ------------------------------------------------------------
 $DATA
   O2H5
   C1
   OXYGEN      8.0   0.0758752612   0.1029703596  -0.1380388056
   OXYGEN      8.0  -1.6329486914   0.2211188064  -1.8103049937
   HYDROGEN    1.0  -0.0725810430   0.4747088980   0.7504261439
   HYDROGEN    1.0   0.9330501410  -0.3174832331  -0.3330916306
   HYDROGEN    1.0  -0.4237766906   0.1376065810  -0.6270666776
   HYDROGEN    1.0  -2.3424944360  -0.4268887357  -2.0060894609
   HYDROGEN    1.0  -1.7921347121   0.9257714517  -2.4735439582
   $END

! --- OPTIMIZED RHF      MO-S --- GENERATED AT Fri Dec 15 00:48:25 2017
! E=     -150.1620919294, E(NUC)=   47.9559336673
 $VEC
 1  1 9.93613099E-01 3.20702947E-02-2.24634863E-03 1.56904217E-04-2.19957003E-03
 1  2 3.13313084E-04 9.04941985E-04 6.59436960E-04-4.52587984E-05 6.45901704E-04
 1  3-4.70970670E-03-4.70987324E-03-9.86284464E-03 1.88174472E-04 1.88109591E-04
 2  1-5.80880573E-04 1.71794788E-03-8.47977388E-04 5.87852394E-05-8.29926100E-04
 2  2 9.94111746E-01 2.64448946E-02-2.47981701E-03 1.59721737E-04-2.45660777E-03
 2  3 1.64987868E-04 1.65066081E-04-3.05915593E-03-5.57763329E-03-5.57725393E-03
 3  1 2.41238581E-01-8.54557950E-01 8.35552605E-02-5.83631489E-03 8.18285646E-02
 3  2 7.39936297E-03-2.26953665E-02-1.91551065E-03 1.44103019E-04-1.84324895E-03
 3  3-7.30179599E-02-7.30300125E-02-1.26560023E-01-2.80500756E-05-2.84225764E-05
 4  1-1.47611593E-02 6.57491207E-02 2.15954260E-02-1.48570875E-03 2.11365513E-02
 4  2 2.35829657E-01-8.53953087E-01 6.86900515E-02-4.36681579E-03 6.81986435E-02
 4  3 1.56173985E-02 1.56167812E-02-3.51168589E-02-1.45207104E-01-1.45193326E-01
 5  1 2.65621341E-02-1.38559271E-01-5.11829955E-01 3.54136523E-02-5.00838599E-01
 5  2 1.64839489E-02-6.35695788E-02 4.44897012E-02-3.02940337E-03 4.36454714E-02
 5  3-2.12387862E-01-2.12451288E-01 3.07944883E-01-3.22077160E-02-3.22120362E-02
 6  1-6.42384605E-06 3.19756766E-05-4.30678730E-01 3.39316521E-01 4.64142719E-01
 6  2-8.23546629E-06 3.50155053E-05-4.64034770E-03 3.65246668E-03 4.98286518E-03
 6  3 3.26546905E-01-3.26522265E-01-2.41992633E-05-2.15894395E-05 1.43171984E-05
 7  1 3.03726648E-05-1.46010812E-04 3.61316021E-01 8.78725612E-01-3.07129752E-01
 7  2-9.63734293E-07 4.12738444E-06 1.19803044E-02 2.92946887E-02-1.01831260E-02
 7  3 3.50564742E-05 4.37481046E-05 1.83831688E-05-2.02321039E-02 2.02282616E-02
 8  1 3.91710932E-06-1.95703405E-05 1.97311245E-02 4.80455996E-02-1.67388085E-02
 8  2 5.95251643E-06-3.08145320E-05-2.27533068E-01-5.59223458E-01 1.93232191E-01
 8  3 3.64914685E-05-2.06253221E-05 8.95712681E-08 4.24180458E-01-4.24177941E-01
 9  1-2.23062153E-02 1.54500963E-01-1.24786038E-01 8.66527729E-03-1.22090919E-01
 9  2 8.76656449E-02-4.57576664E-01-5.50893665E-01 3.63552632E-02-5.43466463E-01
 9  3-3.91196174E-02-3.91748912E-02-1.19065907E-01 2.64674392E-01 2.64677791E-01
10  1 1.09047481E-04-7.57403559E-04-4.23084857E-03 3.61227555E-03 5.66486586E-03
10  2 1.69891575E-04-8.14724604E-04 6.04758690E-01-4.68585751E-01-6.43938341E-01
10  3 5.24314248E-03-4.87589514E-03 5.88487922E-04 2.55282495E-04 2.68854270E-04
11  1 1.54833774E-01-1.06847947E+00-2.73466865E-01 1.87755601E-02-2.67446840E-01
11  2 2.46273283E-02-1.43176172E-01-9.48992300E-03 8.45045780E-04-8.70879842E-03
11  3 7.99564096E-01 7.99664119E-01 2.53819180E-01 7.71090350E-02 7.71448689E-02
12  1 5.72763567E-06-2.78066490E-05 5.58361863E-01-4.39877926E-01-6.01681240E-01
12  2-9.45283650E-06 6.04268124E-05-9.09590832E-05 2.17244079E-05 3.25012306E-05
12  3 8.69427097E-01-8.69318789E-01-3.23352070E-05-6.93390818E-05-3.27231202E-05
13  1 4.13030403E-03-7.72647532E-02 1.79424923E-01-1.24368684E-02 1.75491985E-01
13  2 1.34789464E-01-8.74151452E-01 4.71611018E-01-3.07542286E-02 4.66703957E-01
13  3-1.13602815E-01-1.13734341E-01 2.65565976E-01 7.69534468E-01 7.69759188E-01
14  1-1.26314109E-06 1.52361151E-05-7.19364333E-04-1.76212662E-03 5.75477468E-04
14  2 2.07150162E-05-1.49811355E-04 3.43071888E-01 8.42940764E-01-2.91210891E-01
14  3 2.75400846E-05-1.38879856E-05-3.40876355E-05 8.31340062E-01-8.31078482E-01
15  1-8.63113810E-02 9.44284502E-01-7.40065924E-01 5.12577143E-02-7.24284153E-01
15  2-1.65007257E-02 6.66544153E-02 2.89063586E-01-1.96228027E-02 2.83830925E-01
15  3 2.15524124E-01 2.15552519E-01-1.69321183E+00 1.62283414E-01 1.62340378E-01
 $END

And you'll have to do that for every step of you scan. If you are a bit talented with scripting, then you can surely automate the process.

I did not run the whole scan, but here are the energies of the ones I completed:

OH = 0.7   #1       NSERCH= 13     ENERGY=    -150.1620919
OH = 0.8   #2       NSERCH=  4     ENERGY=    -150.2905386
OH = 0.9   #3       NSERCH=  4     ENERGY=    -150.3504096
OH = 1.0   #4       NSERCH=  5     ENERGY=    -150.3755229
OH = 1.1   #5       NSERCH= 60     ENERGY=    -150.3875894
OH = 1.2   #6       NSERCH=  9     ENERGY=    -150.3898837
OH = 1.3   #7       NSERCH= 20     ENERGY=    -150.3872606
OH = 1.4   #8       NSERCH= 36     ENERGY=    -150.3774125


Acid-base titration: Unknown base in burette or flask? While finding the concentration of $\ce{NaOH}$ via titration with $\pu{0.05M}$ Oxalic acid solution (in our lab) we put NaOH in the burette whereas added phenolphthalein to oxalic acid in titration flask. But in this video the demonstrator does just opposite ( i.e., $\ce{NaOH}$ in flak with phenolphthalein). I am confused in deciding which one might be correct. My view is that either would do the job but I need some clarification.

The base should be in the burette, and the acid in the flask. With this technique, the point of equivalence is obtained with a rather high degree of accuracy.


Extraction of Sn from SnO2 In the extraction of $\ce{Sn}$ from $\ce{SnO2}$ by carbon reduction method, why is  $\ce{CO}$ formed instead of $\ce{CO2}$ as per the following reaction?

$\ce{SnO2 + C -> Sn + CO}$

Above a $1000\ \mathrm K$, the $\Delta_\mathrm fG$ of $\ce {CO}$ from $\ce C$ is more negative than that of $\ce{CO2}$ formation from $\ce C$. Therefore, during smelting, when coke ($\ce C$) reacts with $\ce {SnO2}$, the formation of $\ce {CO}$ rather than $\ce {CO2}$ is thermodynamically preferable.

Below that temperature, the reduction with carbon is not possible since $\Delta_\mathrm fG(\ce{SnO2})\lt\Delta_\mathrm fG(\ce{CO2})\lt\Delta_\mathrm fG(\ce{CO})$. Therefore at all temperatures where this reduction is possible, thermodynamically carbon monoxide formation would be more favourable as compared to carbon dioxide formation. To see for yourself, the variation of $\Delta_\mathrm fG$ of various compounds, have a look at Ellingham Diagrams, a link to a comprehensive set of which is Given Here.

$\Delta_\mathrm fG$ is the Gibbs free energy of formation.


Library / function for computation of molecular mass / molecular weight in C We are looking for a function to be used for computing the molecular mass / molecular weight / formula weight in C, either in a library or standalone.  So far we have found the following implementations:

R language: shown here (limited implementation)
Matlab: shown here (full implementation)
Python: shown here (issue is discussed, preliminary implementation)
Perl: shown here (for protein molecular weights)

Ideally, the function would be able to handle a string input which could be either an element or compound.  In the latter case, the total number of each element forming the compound would be determined prior to atomic mass look-up.


Sodium Sulfate or Sodium Carbonate for electrolysis in water I am not a chemist but I have a question regarding electrolytes in electrolysis.
I have done a lot of reading but cannot find the answer I'm looking for regarding Sodium Sulfate (or Sulfide)
I intend to clean a film off a Stainless Steel plate, so that will be my cathode.
For my Anode I will be using a Platinum coated Titanium plate.
At a small scale I have been using Sodium Carbonate (Na2CO3) and it has worked fine, however I intend to scale it up to ~1m^2 plates for the Anode and Cathode, as well as 150A @ 15v DC for ~3min in Deionised water with many repeats, so temperature rise of the water should also be accounted for.
I would like to make this as safe/clean as possible and am considering switching out the electrolyte at this stage to something more stable. I have read that Sodium Carbonate can precipitate directly onto the electrodes at high currents. I believe this is not an issue with Sodium Hydroxide but want to avoid that for obvious reasons. I have stumbled upon Sodium Sulfate as an alternative but very little information on its breakdown and potential pollutants. Is there any other that I am missing/suggestions?
I will be using a fume hood, however Sodium Chloride is definitely not an option.
Thanks for your time,
Tea
Edit:
So I have done some reading and have mostly worked out the half equations of Sodium Sulfate (Na2SO4):
Cathode:
(2)H+ + (2)e- -> H2(g)
Anode:
(2)O2- -> O2(g) + (4)e-
Remaining in aqueous solution:
NaOH (aq) + S (?)
As an update then, can I ask what, if anything, happens to the NaOH and Sulfur?
Also, what form will the Sulfur remain in? I assume it will try to re-bond with oxygen, which will be abundant in the system? Will either be outgassed?
Assuming the NaOH is constantly building up, at what point does it become overwhelming in the system and need to be removed?
Thanks so much for reading this far.
tea

If you electrolyze a solution of sodium carbonate, it will be fine at the beginning. Later on, some problem may happen around the anode where $\ce{O2}$ is released. Indeed some acid ($\ce{H+}$) is produced in solution according to the half-equation : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ and these acidic ions $\ce{H+}$  react with the carbonate ions $\ce{CO3^{2-}}$according to $$\ce{CO3^{2-} + H+ -> HCO3^- + H2O}$$ In presence of the ion sodium $\ce{Na+}$, the ion hydrogenocarbonate $\ce{HCO3^-}$ reacts to produce sodium hydrogenocarbonate $\ce{NaHCO3}$ which is not very soluble in water, and may make a white deposit around the anode $$\ce{Na+ + HCO3^- -> NaHCO3(s)}$$ Furthermore if the electrolysis is continued a long time, a supplement of acid ions will react with these $\ce{HCO3^-}$ ions producing some $\ce{CO2}$ by the reaction  $$\ce{H+ + HCO3- -> CO2 + H2O}$$
$\ce{CO2}$ is a gas, and it is relatively soluble in water. In the beginning the $\ce{CO2}$ produced will stay dissolved in the liquid. But on the long run, $\ce{CO2}$ will get out of the solution and get mixed with the oxygen produced at this electrode. So instead of getting pure $\ce{O2}$ at the anode, you will obtain a mixture $\ce{O2 + CO2}$.
So you better use a solution of sodium sulfate $\ce{Na2SO4}$ instead of sodium carbonate. The above mentioned drawbacks do not occur with sodium sulfate.
In a $\ce{Na2SO4}$ solution, the ions $\ce{Na+}$ are moving towards the negative electrode (cathode), and the ions $\ce{SO4^{2-}}$ are moving toward the positive electrode (anode). They move but they don't touch their electrodes and they are not discharged. Instead that is water that either give or absorb electrons.
At the negative electrode (cathode) where $\ce{Na+}$ ions are approaching, the following reaction occurs :  $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ and the $\ce{OH-}$ ions neutralize the approaching $\ce{Na+}$ ions. Of course some $\ce{H2}$ bubbles are appearing on the cathode.
At the positive electrode (anode), the following reaction happens $$\ce{2 H2O ->  O2 + 4 H+ + 4 e-}$$ and the $\ce{H+}$ ions neutralize the approaching $\ce{SO4^{2-}}$ ions. Of course some $\ce{O2}$ bubbles are produced on the anode. Half as much as $\ce{H2}$, because $4$ e- produce $1$ $\ce{O2}$ and $2 \ce{H2}$
To summarize, electrolyzing a $\ce{Na2SO4}$ aqueous solution will produce $\ce{H2}$ and a solution of $\ce{NaOH}$ at the cathode. And it will produce $\ce{O2}$ plus a solution of $\ce{H2SO4}$ at the cathode.
Now if the anodic solution is stirred and mixed with the cathodic solution, the following reaction happens $$\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O}$$ As a consequence, $\ce{Na2SO4}$ is regenerated, and the only chemical that is consumed in the whole process is water. The only product getting out of the whole process is the formation of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode.


How to compute the number of ATP molecules that can be produced from breaking down an unsaturated free fatty acid? From studying $\beta$-oxidation of free fatty acids, I've learned how to compute for the ATP equivalence of even-chain and odd-chain free fatty acids.

What really bothers me now is how to compute for the ATP equivalence of an unsaturated fatty acid. Since one of the major difference of the process is in unsaturated fatty acids, normal $\beta$-oxidation rounds will occur until the unsaturation is encountered. The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of FADH2.

Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

Definitely subtract a molecule of $\ce{FADH2}$. After all, you can skip the first of the the four steps in each cycle of removing two carbons (formation of double bond, adding water across the double bond, oxidizing the double bond to a carbonyl group, oxidative decarboxylation). You would also have to check whether any of the steps are coupled to ATP hydrolysis (e.g. the formation of the link to the acyl carrier protein, or transport steps) if you want to get an accurate number of ATP molecules synthesized per fatty acid turned into carbon dioxide.

The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of $\ce{FADH2}$.

Most unsaturated fatty acids in unprocessed food are in the cis-form. To turn them into the requisite $\ce{\Delta^2}$ trans-double bond, you need one or two additional enzymes, see https://www.ncbi.nlm.nih.gov/books/NBK22387/. For turning cis- into trans- and, if necessary, moving the double bond into the correct position, an isomerase is used. If there are two double bonds, one is reduced first by a reductase.


Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties? The equation:

$$E^{。}_{cell}= \frac{RT}{nF}\ln K_{eq}$$

We all know cell potential is intensive, not affected by the amount, Because: $volt=\frac{joule}{coulomb}$. Both joule and coulomb will be doubled altogether.

But as seen from the equation, cell potential is affected by the number of electrons transferred.

No, there is no violation of "intensivity".  The reason is that $K_{eq}$ depends on $n$, and changes in one cancel the other out.

For example, consider the electrolysis of water:

$$\ce{2H2O_{(l)} -> 2H2_{(g)} + O2_{(g)}}$$

The equilibrium constant for this reaction is $K_{1}=\frac{[\ce{H2}]^2 [\ce{O2}]}{1}$ and if you wrote out each electrochemical half reaction separately, $n$ would be 4.

Now consider this reaction:

$$\ce{4H2O_{(l)} -> 4H2_{(g)} + 2O2_{(g)}}$$

The equilibrium constant is now $K_{2}=\frac{[\ce{H2}]^4 [\ce{O2}]^2}{1}=(K_1)^2$.  If you wrote out the half-reactions for this reaction, $n$ would be 8, twice as big.  But the $\ln K_{eq}$ term would also be twice as big, since $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.


Elements which do not form oxides I had thought that this question would be easy to answer but my searching has been unsuccessful in finding a clear answer.

For the metals, I expect that they all form oxides.  Some very readily, e.g. sodium and magnesium, and some reluctantly, e.g. gold and platinum, but I cannot find a metal which does not form an oxide.

The metalloids also all appear to form oxides.

The noble gases are obvious candidates for exceptions but xenon has a known oxide.  I think that it is expected for radon but I don't know whether that is confirmed.  A fluoride is known for krypton but no oxide yet.  No oxide is known or expected (?) for any of the lighter noble gases.

This leaves the other non-metal elements which are sufficiently few to check easily.  I find no exception except for the possible pedantic case of fluorine.  It forms a binary compound with oxygen but since it is more electronegative the compound is called oxygen fluoride rather than fluorine oxide.

Some elements are too radioactive to study e.g. francium and astatine but I expect that both would form oxides if we could manage to study them.

Edit: a good point that came up in the comments is oxygen itself.  A pedantic exception together with fluorine.

Is that correct?

The first thing that came to mind is oxygen itself, as the term "oxide" suggests one other element in its chemical formula. Merriam-Webster suggests a more strict version: "a binary compound of oxygen with a more electropositive element or group". So technically pure oxygen is not an oxide.

Putting oxygen aside, helium $\ce{He}$, neon $\ce{Ne}$ and argon $\ce{Ar}$ are probably the only three conventional elements that come to mind are not yet proven to form oxides. I find it rather tricky to back up this statement properly as the lack of knowledge about an object doesn't prove its non-existence; ideally one might want to do a stability simulation at extreme pressures and temperatures to prove the existence of $\ce{(He|Ne|Ar)O_x}$ or lack thereof. There are algorithms like USPEX that are capable of that – by the way, with the help of USPEX a family of krypton oxides has been recently predicted.


Why does hydrogen form on such a long time scale? If we were trying to figure out the time scale for a gas-phase reaction between two hydrogen atoms in a molecular cloud (which has density $~\pu{10^4cm^-3}$), apparently the reaction would happen on a time scale proportional to the inverse of the density multiplied by $10^{15}$ years.

Aside from the cloud not being dense and the probability that a collision will surpass the activation energy is small, is the time elongated because you need to induce a dipole moment between two hydrogen atoms to actually bind them together?

I recently read somewhere that even if the reactants would go into the unbound state of $\ce{H2}$ they can't emit a photon that would let it go into the bound state of $\ce{H2}$. I guess they are talking about quantum mechanics, but it is not obvious to me what the bound and unbound states of molecular hydrogen are. (If someone could clarify that too)

Why is the time-scale so long?

The time-scale is so long because the density is not just low, it's virtually nonexistent.

The particle density of any gas can be determined at any temperature and pressure using the ideal gas law ($PV=nRT$). Let's use air as an example. At $\pu{25 o^\circ C}$ ($\pu{298 K}$), and 1 atmosphere, the particle density of air ($n/V$) is:

$$ \frac{n}V =\frac{RT}P $$
$$ \frac{n}V =\frac{(\pu{0.08026 L atm (K mol)^{-1})(\pu{298  K})}}{{\pu{1 atm}}} = \pu{23.9 mol L^-1} $$

Converting to molecules per $\pu{cm^3}$:

$$23.9 \frac{\mathrm{mol}}{\mathrm{L}} \times 6.022\times10^{23}\frac{\text{molecules}}{\text{mol}}\times\frac{1 \text{ L}}{1000 \text{ cm}^3}=1.44\times10^{22}\frac{\text{molecules}}{\text{cm}^3}$$

The density of gasses we are used to dealing with is 18 orders of magnitude higher than the density you quoted. If we have a gas of hydrogen atoms at air density, then the timescale for formation of H2 is:

$$ 10^{15} \frac{\text{years}\times\text{cm}^3}{\text{particles}} \times  \frac{1 \text{ cm}^3}{1.44\times10^{22}\text{ molecules}}=6.94\times 10^{-8}\text{ years} $$

$$6.94\times 10^{-8}\text{ years}= 2.54\times 10^{-5}\text{ days}=6.08\times 10^{-4} \text{ hours}=2.19 \text{ seconds}$$

At "normal" densities, the timescale is much shorter.

The second part of your question deals with the quantum mechanics of the activation barrier to this reaction, which I cannot do justice to at this time.


Why is 1-bromo-3-methylbut-2-ene named like that and not as 2-(bromomethyl)but-2-ene? I had this confusion related to nomenclature of halo substituted compounds.

Why is it that 1-bromo-3-methylbut-2-ene is named like that and not as 2-(1-bromomethyl)but-2-ene? why do we prefer the carbon chain including the carbon attached to the bromine and not the chain including the carbon of the methyl group?

Which IUPAC rule was used in making this preference between those two carbon chains of equal length?

Here's 1-bromo-3-methylbut-2-ene's structure:

You have only two possible chains to choose from: a 3-carbon chain and a 4-carbon chain (doesn't matter which methyl group would be part of that chain, the other one will become a substituent on the chain).

You won't choose the 3-C chain for two reasons:

It is not the longest chain.
It does not contain the double bond in the molecule.

Thus, you are left with a 4-C chain, which necessarily includes the bromine atom.


Rate of reaction increases exponentially with concentration In our chemistry practical we dropped 0.07g of Magnesium into various concentrations of an acid (sulfuric or hydrochloric - only one was used, not sure which), and measured the rate of reaction by the rate of the H2 produced (used a gas syringe). We then collected data at a suitable interval of the culmulative amount of H2 produced.
We drew the graph (with multiple concentrations - 0.25, 05, 0.75, 1, 2). Then we measured the steepest point (fastest rate) and took the gradient.
Plotting rate against concentrations (0.25, 05, 0.75, 1, 2), we came to the conclusion that rate varies exponentially with concentration.
But the rate law of a typical reaction is:

r = k * [A] * [B]

, which would mean doubling [B] would double the rate law, not have an exponential effect on the reaction. Why did we come to the conclusion that concentration has an exponential effect?

In our case the rate law, I would think, would be elementary, but we are not dealing with concentrations of both Mg and HCl, just HCl, as the Mg was solid - therefore no concentration of Mg in solution.

My question is: In a reaction between two liquids, is the rate of reaction linearly dependent on the concentration?
AND: What is the rate law of our reaction? - Why did we come to the conclusion we did?

Thanks for the help!! :)

Rate equation is experimentally determined. So, you can't directly write $r=k[A][B]$ (where r is rate, k is rate constant). Probably you might have performed first order reaction. For which $r=k[A]=\frac{2.303}{t}log\frac{[R]}{[R_0]}=\frac{2.303}{t}log{e^{kt}}$.

The above equation suggests that rate is proportional to exponential function.


Is hydrogen bonding generally defined to include only three period two elements? Traditionally, hydrogen bonding has been defined to only include interactions between a positively polarized hydrogen and three period-two elements: nitrogen, oxygen, and fluorine.

Why was this definition proposed? A common question is why can't chlorine, by the aforementioned definition, hydrogen bond?

Does this have anything to do with hard and soft acid-base theory? Hydrogen is a hard acid, and chlorine is a comparatively soft base ... and the period two elements are all relatively hard?

I don't think there is any such traditional definition requiring $\ce{N}$, $\ce{O}$ or $\ce{F}$.

For example, in table 7 and the discussion thereof in Hydrogen Bonding Annual Review of Physical Chemistry Vol. 22: 347-385 the hydrogen bonding in the following species are discussed:

$\ce{ClHCl^-}$
$\ce{BrHBr^-}$
$\ce{IHI^-}$
$\ce{BrHCl^-}$

as well as related neutral radicals.

Hydrogen bonds in $\ce{S-H}$ containing compounds are discussed.  These are just examples, others that do not involve $\ce{F}$, $\ce{O}$ or $\ce{N}$ are also discussed.

The following is copied (footnotes omitted) from Definition of the Hydrogen Bond (IUPAC Recommendations 2011) Pure and Applied Chemistry Volume 83, Issue 8 (Aug 2011)

  2. DEFINITION

  The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a
  molecular fragment $\ce{X-H}$ in which $\ce{X}$ is more electronegative than $\ce{H}$, and an atom or a group of
  atoms in the same or a different molecule, in which there is evidence of bond formation.

  A typical hydrogen bond may be depicted as $\ce{X-H\bond{...}Y-Z}$, where the three dots denote the bond.  $\ce{X–H}$ represents the hydrogen bond donor. The acceptor may be an atom or an anion $\ce{Y}$, or a fragment or a molecule $\ce{Y-Z}$, where $\ce{Y}$ is bonded to $\ce{Z}$. In some cases, $\ce{X}$ and $\ce{Y}$ are the same. In more specific cases, $\ce{X}$ and $\ce{Y}$ are the same and $\ce{X-H}$ and $\ce{Y-H}$ distances are the same as well leading to symmetric hydrogen bonds. In any event, the acceptor is an electron rich region such as, but not limited to, a lone pair of $\ce{Y}$ or π-bonded pair of $\ce{Y-Z}$.

  The evidence for hydrogen bond formation may be experimental or theoretical, or ideally, a combination of both. Some criteria useful as evidence and some typical characteristics for hydrogen bonding, not necessarily exclusive, are listed below, numbered E# and C#, respectively. The greater the number of criteria satisfied, the more reliable is the characterization as a hydrogen bond.

  2.1 List of criteria

  For a hydrogen bond $\ce{X-H\bond{...}Y-Z}$:

  (E1) The forces involved in the formation of a hydrogen bond include those of an electrostatic origin, those arising from charge transfer between the donor and acceptor leading to partial covalent bond formation between $\ce{H}$ and $\ce{Y}$, and those originating from dispersion.
  (E2) The atoms $\ce{X}$ and $\ce{H}$ are covalently bonded to one another and the $\ce{X-H}$ bond is polarized, the $\ce{H\bond{...}Y}$ bond strength increasing with the increase in electronegativity of $\ce{X}$.
  (E3) The $\ce{X-H\bond{...}Y}$ angle is usually linear (180º) and the closer the angle is to 180º, the stronger is the hydrogen bond and the shorter is the $\ce{H\bond{...}Y}$ distance.
  (E4) The length of the $\ce{X-H}$ bond usually increases on hydrogen bond formation leading to a red shift in the infrared $\ce{X-H}$ stretching frequency and an increase in the infrared absorption cross-section for the $\ce{X-H}$ stretching vibration. The greater the lengthening of the $\ce{X-H}$ bond in $\ce{X-H\bond{...}Y}$, the stronger is the $\ce{H\bond{...}Y}$ bond. Simultaneously, new vibrational modes associated with the formation of the $\ce{H\bond{...}Y}$ bond are generated.
  (E5) The $\ce{X-H\bond{...}Y-Z}$ hydrogen bond leads to characteristic NMR signatures that typically include pronounced proton deshielding for $\ce{H}$ in $\ce{X-H}$, through hydrogen bond spin–spin couplings between $\ce{X}$ and $\ce{Y}$, and nuclear Overhauser enhancements.
  (E6) The Gibbs energy of formation for the hydrogen bond should be greater than the thermal energy of the system for the hydrogen bond to be detected experimentally.

  2.2 Some characteristics of hydrogen bonds

  (C1) The $\mathrm{p}K_\mathrm{a}$ of $\ce{X-H}$ and $\mathrm{p}K_\mathrm{b}$ of $\ce{Y-Z}$ in a given solvent correlate strongly with the energy of the hydrogen bond formed between them.
  (C2) Hydrogen bonds are involved in proton-transfer reactions ($\ce{X-H\bond{...}Y -> X\bond{...}H-Y}$) and may be considered the partially activated precursors to such reactions.
  (C3) Networks of hydrogen bonds can show the phenomenon of co-operativity, leading to deviations from pair-wise additivity in hydrogen bond properties.
  (C4) Hydrogen bonds show directional preferences and influence packing modes in crystal structures.
  (C5) Estimates of charge transfer in hydrogen bonds show that the interaction energy correlates well with the extent of charge transfer between the donor and the acceptor.
  (C6) Analysis of the electron density topology of hydrogen-bonded systems usually shows a bond path connecting $\ce{H}$ and $\ce{Y}$ and a (3,–1) bond critical point between $\ce{H}$ and $\ce{Y}$.


On the activity of liquids and solvents in equilibium constants This is a confusion I've had for quite sometimes, first when learning about equilibria (due to contradicting teachers), then when trying to teach it (due to contradicting sources, in books and on the Internet).

When facing an equilibrium (let's be original and take $\ce{a A + b B <=> c C + d D}$ as an example), the equilibrium constant is
$$K = \frac{a(\ce{C})^c \cdot a(\ce{D})^d}{a(\ce{A})^a \cdot a(\ce{B})^b}$$
where $a(\ce{X})$ is the activity of X.

For pure solids, the activity is 1; therefore solids "do not appear" in the expression of the constant.
For dilute solutions, the activity of the solute can be taken as its molar concentration. Thus, supposing A, B, C and D are not solids, the equilibrium constant can be written as

$$ K_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

But what happens when facing liquids (e. g. water)? In that case, I have seen two things :

Just putting the "concentration" of the liquid into the equilibrium constant.
A pure (or nearly pure, such as the solvant in a dilute solution) liquid has an activity of (or nearly of) 1. Therefore, the liquid should not appear in the constant.

Let's apply both cases to acidity constants, for the reaction $\ce{HA + H2O <=> A- + H3O+}$.

As $[\ce{H2O}] \approx 55.55$ mol/l in dilute solutions,
$$K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}][\ce{H2O}]}
\textrm{ and }
K_a = \frac{K_c}{[\ce{H2O}]} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
As $a(\ce{H2O}) \approx 1$ in dilute solutions,
$$K_a = K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

Which way is correct? I'd be more inclined to follow the option #2, but I've been stared at wide-eyed like a heretic when talking about that.

I concluded on this question after discussion with the physical chemistry professor at my home university, and reading some physical chemistry books.
Although many references include the concentration of water in the expression of Ka, this is incorrect: the activity of pure water is 1. In dilute solutions, $a(\ce{H2O}) \approx 1$.
Therefore, the second option is correct:
$$K_a=K_c=\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
This has a direct consequence on the values of pKa for the two couples of water: $pKa(\ce{H3O+}/\ce{H2O})=0$ and $pKa(\ce{H2O}/\ce{OH-})=14$, instead of the values $-1.7$ and $15.7$ that are often found in textbooks (which result from including $[\ce{H2O}]=55.55$ mol/l).
A good article on this matter can be found here: What is the pKa of water?.
References:

B. Leyh, Department of Chemistry, University of Liège, Belgium.
Atkins, P., & De Paula, J. (2006). Atkins’ physical chemistry (8th ed.). London, England: Oxford University Press.
Reid, P., & Engel, T. (2006). Physical Chemistry. Philadelphia, PA: Pearson Education.


What causes electrons to move from zinc to copper? Here is something that no video in youtube about electrochemistry can explain me about galvanic cells.

Suppose we have $\ce{Zn}$ metal immerse in $\ce{Zn^{2+}}$ in one side,  then $\ce{Cu}$ metal immerse in $\ce{Cu^{2+}}$ in the other side. Then, if we link the $\ce{Zn}$ metal to $\ce{Cu}$ metal with some conductor wire, electrons will flow from the $\ce{Zn}$ metal to the copper wire, but this will stop quickly without a salt bridge.

My question is why do the electrons move from $\ce{Zn}$ metal through the wire to the $\ce{Cu}$ metal in the first place? What causes that?

There is no reaction since the electrons will only flow one way in the wire. When you connect the zinc metal to the copper metal with a wire, there is a voltage potential generated between them, just like in a thermocouple.

From Wikipedia:

  Any junction of dissimilar metals will produce an electric potential related to temperature.

In this case, we don't care about the temperature effects.

When a salt bridge is placed between the solution, then the electrons are carried by the anions in the solution from one electrode to the other electrode.

So the zinc metal releases electrons when it dissolves, which travel through the wire easily, to the copper. The $\ce {Cu^{2+}}$ in solution grabs the electrons from the copper electrode, adding copper metal to the electrode. Since there is now more negative ions in this solution, the anions move across the bridge to the zinc electrode. The anions attack the zinc metal, causing the zinc to loose electrons when it dissolves.

I'm not sure what starts the reaction, nor why the zinc dissolves easily.


Why is condensation in dyed beverages colorless? Does food coloring somehow evaporate and turn into nothing?   Or does the beverage itself being much more color intensive?

I had a bottle of Gatorade, colored Lemon-Lime (and flavored as such). I noticed that after leaving the bottle in the car during a hot day and returned to the car, that condensation had formed on the top parts of the bottle.

But the condensation is colorless; does this mean it is solely water that is evaporating from the drink itself as condensation occurs, leaving food coloring or the color of the beverage behind?

Given that the majority of beverages such as Gatorade is water, as stated in the comments, the condensate is predominantly water, as can be seen in the Patent Electrolyte blends providing reduced salty taste (which includes Gatorade), that water amounts to well over 99% by composition, of course this will vary. The dyes themselves don't make up much of the volume.

A common food dye used, according to the Indiana University online experiment webpage How Much Red Dye is in My Gatorade®? , is the red dye, also known as 'Allura Red'

its properties according to the Wikipedia page are stated as:

  It usually comes as a sodium salt, but can also be used as both calcium and potassium salts. It is soluble in water; Its melting point is above 300 °C (572 °F).

It is likely that the majority of the food dye based salts would remain in solution.

Any salts (such as the example above), transported with the evaporating/condensing water would be too small of a concentration to cause any visible effects.

A similar question and answer is Does salt vaporise?


NCl₃ + H₂O → NH₃ + HOCl , the resulting solution is acidic, basic or neutral? When $\ce{NCl3}$ is dissolved in water, $\ce{NH3}$ and $\ce{HOCl}$ is formed.

Can you please explain me whether formed solution is acid or basic or neutral? thank you!

Here's our initial reaction, $$\ce{NCl3 + 3H2O -> NH3 + 3HOCl}$$ $\ce{NH3}$ is a weak base and $\ce{HOCl}$ is a weak acid, but there is more of the acid, so at this point the solution would be mildly acidic. However $\ce{NH3}$ and $\ce{HOCl}$ are known to react as follows $$\ce{NH3 + HOCl->NH2Cl + H2O}$$ and in aqueous solution the resultant chloramine can decompose as shown in the following equation $$\ce{3 NH2Cl ⟶ N2 + NH4Cl + 2 HCl}$$  Clearly, an acid is generated in this last reaction.  So although the initial products are neutral, eventually the solution will be strongly acidic.


Is the final temperature of a solution affected by the order of mixing components? For example, I have two beakers each containing water. One is at 90 $^{\circ}$C and one is at 10 ${^\circ}$C.

Will the final temperature of the solution I make from mixing the two components be different if I slowly poured the 90 ${^\circ}$C water into the 10 ${^\circ}$C water and vice versa?

Temperature is a so-called state function:

  In thermodynamics, a state function or function of state is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium state of the system. State functions do not depend on the path by which the system arrived at its present state. A state function describes the equilibrium state of a system.

The final temperature of your system does not depend on how the system got there (order of mixing, for example).


What is the difference between physical and chemical bonds? If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds chemical and physical?

From the freedictionary.com, chemical bond:

  Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.

What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these "several forces" excactly?

In the answer I would like to see the lists of chemical and physical bonds.

In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is the sharing of electrons between two or more nuclei. In contrast, other interactions are often said to be intermolecular (which is somewhat more specific than the term “physical”.

In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).

Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In The Nature of the Chemical Bond (1960), he gives the following definition:

  The Chemical Bond Defined.—We shall say that there is a chemical bond between two atoms or groups of atoms in case that the forces acting between them are such as to lead to the formation of an aggregate with sufficient stability to make it convenient for the chemist to consider it as an independent molecular species.

A bond is what links atoms into molecules, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…
The complete opposite: consider all interactions as chemical bonds, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like this one:

  A chemical bond is the physical phenomenon of chemical species being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons and is a phenomenon that is fully described by the laws of quantum electrodynamics. In general, strong chemical bonds are found in molecules, crystals or in solid metal and they organize the atoms in ordered structures. Weak chemical bonds are classically explained to be effects of polarity between molecules which contain strong polar bonds. Some very weak bond-like interactions also result from induced polarity London forces between the electron clouds of atoms, or molecules. Such forces allow the liquification and solidification of inert gases. At the very lowest strengths of such interactions, there is no good operational definition of what constitutes a proper definitional "bond".

This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).
You can classify interactions by energy: decide that chemical bonds are those interactions that have an energy higher than a certain threshold, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.
Finally, what I believe is the most common description is to look at the nature of the interaction, and classify it following a certain convention. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.
I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the New Oxford American Dictionary as well, which I quite like:

  chemical bond a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.

Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is sharing (or transfer) of electrons. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).

Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…


How do I run MP2 and/or CCSD-level hyperpolarizability calculations with GAMESS? I have used GAMESS before for TDHF (RPA) / TDDFT level first hyperpolarizability ($\beta$) calculations. However, I cannot figure out how to calculate $\beta$ with MP2 and CCSD for comparative purposes. Do I have to use the FFIELD keyword for the finite field approach? If that is the case, what is the general workflow necessary to calculate the time-varying hyperpolarizability?

As you guessed, to calculate $\beta$ within GAMESS for MP2 and CCSD, you would need to perform first- and second-order finite difference, respectively, as there is no coupled cluster gradient needed, so no true CCSD dipole moment.

However, there is a more fundamental problem: it is impossible to calculate time-dependent response using a finite field approach. Take the general form of 1-dimensional 2nd-order central difference,
$$
f''(x) \approx \lim_{h \to 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2},
$$

which can directly translate to a diagonal element of the polarizability tensor formed from energies at finite electric fields:

$$
\alpha_{zz}(x) \approx \lim_{h \to 0} \frac{E(x+h)-2E(x)+E(x-h)}{h^2}
$$

where $h$ is a finite electric dipole field of some strength applied along the $z$-direction. What is $x$? Is it as simple as converting the desired frequency $\omega$ to units of electric field strength? No; that still corresponds to a static field rather than a time-varying field. How are the different perturbation directions handled? $x$ must be set to 0.

From a molecular properties review by Jurgen Gauss, page 5:

  (...) disadvantages of the numerical differentiation scheme are

  a) that there is no straightforward extension to the computation of frequency-dependent properties (...)

Page 37:

  While analytic derivative theory is sufficient for the theoretical treatment of time-independent (static) properties, the underlying theory needs to be extended for the calculation of time-dependent (dynamical) properties. In particular, the fact that there is -- unlike for the static case -- in the time-dependent case no well-defined energy explains why the simple derivative theory discussed so far is not applicable.

Page 39 (with some manipulation by me):

  It can be shown that the linear, quadratic, etc. response functions $\left<\left<A;B\right>\right>_{w_{b}}$, $\left<\left<A;B,C\right>\right>_{w_{b},w_{c}}$, etc. and thus the frequency-dependent properties of interest can be determined as derivatives of the so-called time-averaged quasi energy

  $$
Q(t) = \left<\tilde{\psi}\left|\left(H - i\frac{\partial}{\partial t}\right)\right|\tilde{\psi}\right>
$$

  with the phase-isolated wavefunction

  $$
\tilde{\psi}(t) = e^{iF(t)} \psi(t)
$$

where $F$ is a time-varying external field, and $\psi(t)$ is the usual wavefunction where MO coefficients and thus the density are explicitly time-dependent.

The only numerical (rather than analytic) method for calculating dynamic properties I am aware of is direct time-propagation of the wavefunction. If you want to calculate frequency-dependent CCSD hyperpolarizabilities, your best (free) bet is DALTON. Note that their coupled cluster module is closed-shell only.


Different names for Fidaxomicin (OPT-80)? Is OPT in OPT-80 an abbreviation? If so, what is the full name?

Fidaxomicin is the International nonproprietary name (INN) of a pharmaceutical, which was/is distributed by different companies under different trade names.

To my knowledge, OPT-80 was a provisional name/signature used by Optimer Pharmaceuticals Inc.. The drug was later sold as Dificid.


Combustion without hydrogen Is there any fuels without hydrogen that will combust. My friend believes there might be and do to my experience I have never heard of one. Please explain and if u can show a balanced equation of this combustion.

A combustion reaction is classified as a reaction between a fuel and oxygen to produce energy. The reaction: 2CO(g) + O2(g) = 2CO2(g) fulfills this definition, is exothermic, and does not use hydrogen.


Calculating temperature after adding heat to ice If $53.2~\mathrm{kJ}$ of heat are added to $15.5~\mathrm{g}$ of ice at $-5^\circ\mathrm{C}$, what will be the resultant state of matter in which water is present and also calculate its final temperature.

I have done this:

Step 1: Energy required to make ice at 0 Celsius
Step 2:Energy required to make  water at 0 Celsius
Step 3: Energy required to make steam at 0 Celsius
Step 4: Energy required to make steam at 100 Celsius
Step 5:Energy required to make steam at 'x-100' Celsius

Please leave a comment stating if any errors are present. The answer I got is x = 808 Celsius. The correct answer is 303 Celsius.

It is an enthalpy balance that requires knowledge of the enthalpy change at each step: (The steps you have listed are not correct.)

Solid heat capacity of ice of the form e.g. $C_p^\mathrm{solid}(T)$. This allows you to calculate the amount of heat (or energy) required to warm the sub-cooled ice from $-5~^{\circ}\mathrm{C}$ to $0~^{\circ}\mathrm{C}$.
(Latent) heat of fusion, $\Delta H_\mathrm{fus}$ then must be added to transform the solid ice to liquid water at $0~^{\circ}\mathrm{C}$.
Next we must warm the liquid water from $0~^{\circ}\mathrm{C}$ to its boiling point at $100~^{\circ}\mathrm{C}$, which requires knowledge of the liquid heat capacity of water e.g. $C_p^\mathrm{liquid}(T)$.
Now at the boiling point, we must account for the required amount of energy to transform the liquid water into its vapor state at $100~^{\circ}\mathrm{C}$ using the (latent) heat of vaporization of water $\Delta H_\mathrm{vap}$.
The final step then involves super-heating the vapor from $100~^{\circ}\text{C}$ to $T_\mathrm{final} = ?~^{\circ} \mathrm{C}$, which requires knowledge of the vapor heat capacity of steam e.g. $C_p^\mathrm{vapor}(T)$.

You know the overall enthalpy change ($53.2~\mathrm{ kJ}$) and you know the starting temperature ($-5~^{\circ}\mathrm{C}$), so all you have to do is add the results from each of the steps and solve for $T_\mathrm{final}$ e.g. $303~^{\circ}\mathrm{C}$.


Can caustic soda (sodium hydroxide) be used to dissolve urine odour? I have an area of artificial grass at home in the back garden and during the summer months, the area emits a smell which I can only think is from the dog's urine when he pees all year long. The smell isn't a problem in the winter or during wet days.

I read a lot of posts online about using white vinegar and baking/bicarbonate of soda

I don't have any of that at home, but I do have caustic soda from unblocking drains/pipes that I use once in a while. I wonder if caustic soda can be used in a Pump Sprayer with Water or some other cleaning agent to help break down the smell from urine?

I'm a complete noob with chemistry but I thought this is the right forum to ask the question.

Well, it would actually make things a little awkward.
Why do Vinegar and baking soda work?
The reason for the usage of vinegar or sodium bicarbonate must first be understood.

Vinegar ($\ce{CH3COOH}$) is an acid, and hence protonates the carbonyl oxygen present in urea[1].

$\ce{NaHCO3}$ (baking soda) is amphoteric, meaning it can react with both acids and bases. Urea is basic, and hence forms a salt. This answer on our site talks exclusively about the deodorizing property of baking soda.

Both actions are quite much the same, and prevent the hydrolysis of urea (which happens to release ammonia ($\ce{NH3}$) which is the "foul" smelling substance)

Why wouldn't sodium hydroxide work?
Not that $\ce{NaOH}$ wouldn't react, but because it would accelerate the hydrolysis of urea[2].

So if you're looking to stop the hydrolysis of urea (and the inevitable generation of ammonia) $\ce{NaOH}$ might not be the right guy.
What I wouldn't advise
But, hey, if you're looking to remove all the urea once and for all, you could pour a few buckets of $\ce{NaOH}$ and let the ammonia create a riot in your unsuspecting neighborhood.
What I would advise
Urea hydrolysis is catalysed by the enzyme urease (so it is dead slow without it)
From reference [3] (emphasis mine)

Urea is a stable compound with a decomposition half-life in aqueous media of 3.6 years. However, hydrolysis of urea is 104 times faster when the urease enzyme is present.

If we could inhibhit its action, we'd have got what we needed.
If you're wondering where urease comes from and why the dog urine stench is greatly enhanced in summer months, you're looking at the same answer: microorganisms.
It has been published that Metals and fluorides could be used to inhibhit urease. I quote further from source [3].

2.3. Inhibiting chemicals
Seven different chemicals, silver nitrate, zinc nitrate, sodium fluoride, glacial acetic acid, vinegar, citric acid, and sulfuric acid, were used to inhibit urea hydrolysis.

The list is by no means exhaustive.
From source 4, different classes of compounds which might be used:

The classes of urease inhibitors include: amides and esters of phosphoric acid, thiols, hydroxamic acids, phosphinic and thiophosphinic acids, boric and boronic acids, phosphate, heavy metal ions, bismuth compounds, quinones, and fluoride.

If you're not keen on purchasing baking soda, I'm sure you'll find any one of the above an interesting buy.

References:

Interaction of urea with weak acids and water, Prabhat K. Das Gupta and S. P. Moulik, J. Phys. Chem. 1987, 91, 5826-5832 DOI:10.1021/j100306a061

What is the reaction between NaOH and urea?

Characterization of urea hydrolysis in fresh human urine and inhibition by chemical addition, Hannah Ray, Daniella Saetta, Treavor H. Boyer, Environ. Sci.: Water Res. Technol., 2018, 4, 87-98 DOI: 10.1039/C7EW00271H

A combined temperature-pH study of urease kinetics. Assigning pKa values to ionizable groups of the active site involved in the catalytic reaction, Barbara Krajewska, J. Mol. Cat. B: Enzymatic Volume 124, Feb. 2016, 70-76 DOI: 10.1016/j.molcatb.2015.11.021

EDIT
There has been some discussion on if the stench is truly ammoniacal or is just bacterial odor. In that case, I would suggest sprinkling of significantly higher concentration of acid. Extreme pH is known to disrupt bacterial life forms and the acid will also prevent to a great extent the possibility of urea hydrolysis, thus preventing the odor anyway.


IR spectra and hydrogen bonds Normally we see IR signals that correspond to the vibration of covalent bonds. Can a (strong) hydrogen bond (or a vibration thereof, to be precise) correspond to an IR signal as well? How about other non-covalent forces? What are examples were hydrogen bonds can be seen in the IR spectrum?

Your question implies that hydrogen bonds give rise to discrete peaks in the vibrational spectrum. It's better to say that hydrogen bonding affects the position of already existing peaks, in particular those vibrational modes coming from the $\ce{X-H}$. A hydrogen bonding network may also broaden low-frequency peaks coming from librations and hindered rotations.

For example, take the $\ce{O-H}$ stretching region in water.

The vapor spectrum consists of water molecules effectively in the gas phase that aren't interacting with each other, or are doing so rarely enough that the molecular configurations are "countable", shown by the discrete peaks between ~3600-3900 wavenumbers. A lack of water molecules interacting means hydrogen bonds aren't forming.

Moving to a bulk phase, where the water molecules are now interacting with each other all the time, hydrogen bonds are being formed and broken much more often. This results in a continuous peak, as more configurations are being sampled. In the case of water, charge transfer into the $\ce{O-H}$ bond causes a red-shift (lowering) of its frequency, moving the center of the distribution to 3400 wavenumbers. The distribution begins to narrow again in the solid phase, as the number of configurations is still large but dynamical switching between them slows.

It is also possible to experimentally measure spectra of small clusters through more advanced techniques such as mass spectrometry-detected rare gas tagging. Note the appearance of peaks around 3400 wavenumbers as the size of the cluster grows. (The presence of a proton doesn't affect these spectra in the hydrogen bonding region.)

A similar effect can be seen in molecular dynamics simulations. See how the peaks broaden and shift to lower energies when going from 1 water to the dimer, then 3 in a line/triangle, and finally 4 in a square.

First image taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html. Experimental cluster spectra taken from http://pubs.acs.org/doi/abs/10.1021/acs.jpca.5b04355. BOMD plot my own creation (unnormalized dipole ACF spectra from HF/STO-3G, 0.36 fs time step, 10000 steps).


Stereochemistry of product in Diels-Alder reaction I have a few questions on this problem I solved:

1) Why does carbon "b" on methyl vinyl ketone attack carbon "5" on the lactone? Why not carbon "2"?

2) The answer key says that the end product should have "trans" stereochemistry. The internal ester group should be opposite the carbonyl group as shown in my picture. Why is this the case? Does this have to do with the endo rule?

1) The regiochemistry of the product can be predicted by drawing resonance structures of the diene and dienophile, allowing us to assign partial charges to the reactive centers. Doing that shows us that 2 and a are partially positive while 5 and b are partially negative. We'd then predict that 5 attacks a and b attacks 2. This answer has a figure showing prediction of regiochemistry this way.

2) We would predict that the product is the endo- product, which would have the bridging ester on the opposite side of the cyclohexene relative to the methyl ketone. This has to do with the endo-rule, which is classically explained by secondary orbital overlap between the the lobe of the HOMO on carbon 4 interacting with the lobe of the LUMO on carbon c. Keep in mind that this trend does exist, but mixtures of endo and exo are common.


Which of sodium or lithium will be more reactive with oxygen? I do not understand how you can work out which element will react more readily with another element.

For example, I have lithium (Li) and sodium (Na) and I am trying to work out which element will be more reactive with oxygen. How do I manage to work that out?

Reactivity of Group 1 and 2 elements increases as you go down the periodic table.
So sodium is more reactive than lithium. Sodium will react with oxygen forming $\ce{Na2O}$ (sodium oxide). Lithium forms lithium oxide $\ce{(Li2O)}$.

Basically, there are general trends (reactivity with acids, air, etc.) down a group or along a period.


What causes metallic objects to ring? If I use a iron rod to hit a 3" × 12" iron plate there is the distinct sound of a metallic ring.

What is the cause of this distinct sound? Is it the crystal structure of iron or that iron is a metal and most metallic objects will ring when struck?

Whether something rings when struck depends on the mechanical properties of the bulk material. Ringing depends on the ability of the bulk material to recover from a mechanical impact elastically with little loss of energy. That way the block of material can vibrate for a significant period of time and we hear this because of the interaction of these vibrations with air, creating sound.

The bulk properties of a material depend on both the nature of the material and the crystal structure of the bulk solid. Ringing doesn't happen if the material has ways of dissipating mechanical energy quickly rather than by vibrating elastically. This can be because of a low elasticity but also because the crystal boundaries may create ways to dissipate energy.

So some metals don't ring. Lead makes very bad bells (as does mercury when solid as demonstrated in this amusing video). Many polymers don't ring as they have many internal ways to dissipate vibrational energy quickly. But many elastic materials like brass, iron or glass do ring because they don't have any easy internal ways to absorb vibrational energy.


If absorbance is 0, what is the equilibrium constant? If the absorbance of something is 0, what is the equilibrium constant? In a reaction, the absorbance is 0, so the concentration of the product is 0. The concentration of one of the reactants is also 0. What does this say about the equilibrium constant? You would have to divide by 0, suggesting that keq doesnt exist for this reaction.

Are we talking about uv-vis absorption?

Supposed that you are monitoring the right wavelength and that the extinction coefficients are not ridiculously small, you are outside the limits of experimental error.

If you still do not see any absorption, this means that the species isn't there (anymore).

If it isn't there anymore, it can't particiate in the equilibrium any longer - there is no measurable equilibrium anymore.

If you still need to measure something in the case of an irreversible and fast reaction, your experimental setup needs to change:

You might need to perform time-resolved  uv spectroscopy and use a stopped-flow setup for the controlled mixing of your reactants.

If you don't have anything but a uv-vis spectrometer, a cuvette and a microliter syringe, do it the old way using the standard addition procedure

Prepare the cuvette with a solution of one reactant with a decent concentration
Prepare a solution of the second reactant. This might be highly concentrated it you want to avoid dilution effects. If the reaction is fast, you can only measure the absorption of the first reactant and the product anyway
Measure the initial absorption of the first reactant
Add some volume of the second solution to your cuvette
Measure the uv absorption of your first reactant. Do you see any change?
Can you measure the absorption of the product too?

Repeat steps 4-6.


Liquifying metal and toxic fume exposure I just read a very entertaining Youtube comment thread on a video where TheBackyardScientist melts tin on the stove in his (enclosed) kitchen, the proceeds to cook a steak on it.

According to the mass of commentaries (and votes), what he did was idiotic due to the toxicity of the fumes released when melting the tin, and the fact that he was in an enclosed space without a respirator.

Reading about tin and organotin compounds a bit, nothing is mentioned about them being 'released' upon melting, but rather that they are a particular subset of compounds of tin, and some happen to be toxic.

Regarding 'metal fume fever' and its causes, it seems that it is a small subset of metals that when oxidized could cause nervous system damage, in which tin does not take part.

Aside from burning yourself horrendously, what dangers are associated with the melting of (we can assume nearly pure) tin, particularly in a closed space? Is what he did really that dangerous?

You ask:

  Aside from burning yourself horrendously, what dangers are associated with the melting of (we can assume nearly pure) tin, particularly in a closed space? Is what he did really dangerous?

According to the SDS for tin metal from Sigma-Aldrich, tin metal (at the purity level for the product mentioned here, which is <= 100% and may not be that of what was used in the video you mentioned) is not a hazardous substance or mixture.

There is some danger specifically relating to inhalation of tin dust or aerosols (see sections 6.1 and 7.1 in the linked document):

  6.1 Personal precautions, protective equipment and emergency procedures
  Use personal protective equipment. Avoid dust formation. Avoid breathing vapours, mist or gas. Avoid breathing dust.

  7.1 Precautions for safe handling
  Avoid formation of dust and aerosols. Further processing of solid materials may result in the formation of combustible dusts. The potential for combustible dust formation should be taken into consideration before additional processing occurs. Provide appropriate exhaust ventilation at places where dust is formed.

In section 8.2 (Exposure controls) it is noted that:

  Respiratory protection
  Respiratory protection is not required. Where protection from nuisance levels of dusts are desired, use type N95 (US) or type P1 (EN 143) dust masks. Use respirators and components tested and approved under appropriate government standards such as NIOSH (US) or CEN (EU).

In section 11.1 (Information on toxicological effects, a subsection of Toxicological Information), information pertaining to toxicity can be found. Inhalation here will refer to tin dust or aerosols that can be inhaled:

  Acute toxicity

  LD50 Oral - Rat - male and female - > 2,000 mg/kg (OECD Test Guideline 423)

  LC50 Inhalation - Rat - male and female - 4 h - > 4.75 mg/l (OECD Test Guideline 403)

  Dermal: No data available

If the tin was/is at the near 100% purity level, the main risks associated with handling it are associated with the formation of dust (workplace inhalation limits are given in section 8). The HMIS (Hazardous Materials Identification System) ratings in section 16 are 1 for health hazard, 0 for flammability and physical hazard, and no value assigned for chronic health hazard.

Conclusion: Assuming the person in the video did not inhale or ingest tin dust (a proposition that largely becomes moot after it has melted, in any case), there appears to be little to no danger associated with the activity with respect to melting tin in a closed space (especially such a small amount of it).


What is the possible side product when mixing citric acid, NaCl, Cu, Al, H2O2, and an unknown photoresist? When etching circuit boards, a somewhat commonly used substitute for (expensive and messy) ferric chloride is a mixture of an acid, a source of chloride ions, and hydrogen peroxide. If I understand the chemistry correctly, the acidic environment allows the copper to react with the chlorine ions to make $\ce{CuCl2}$, which will more slowly react with the copper again to make $\ce{2 CuCl}$. Adding oxygen to the solution (from the peroxide or just by aerating it) makes $\ce{CuO}$ and $\ce{CuCl2}$ again, allowing the process to repeat.

I meant to mix $\ce{H2O2}$, acetic acid and table salt, but accidentally used citric acid instead of acetic acid. After realizing my mistake, I added scraps of aluminum foil to the mixture to displace the copper, and render the mixture safer to dispose of. I went to filter out the copper and dispose of the rest, when I found that there was a layer of some kind of gelatinous substance coating the exposed surfaces in my container. Does anyone have any idea of what it could be, and how I might safely dispose it?

1) You should not use acetic acid (or any organic acid) in this setup. It potentially may form some (even if minuscule) amount of chloracetic acid which is really hazardous. Furthermore, copper acetate (and other carboxilates) has low solubility. Go for either genuine $\ce{HCl}$ or for $\ce{H2SO4}$ used in lead accumulators.

2) Most likely it is aluminium hydroxide. If the substance is bluish, it may be also copper hydroxide.