Thermodynamic vs kinetic basicity I'm writing something related to basicity and came across few papers that distinguish thermodynamic and kinetic basicity.

"In particular, the study of hydrogen bonding (H-bonding) within proton sponges has provided rigorous knowledge into the inter-related roles of symmetry, resonance, and aromaticity in H-bonding, as well as a better understanding of kinetic versus thermodynamic basicity and the emergence of a more reﬁned view of the entropic and enthalpic factors contributing to H-bond strength." ( DOI:10.1021/acs.joc.5b01743 )

"There is a tendency, however, that sterically demanding proton sponges with high thermodynamic basicity with strongly protected protonation sites often have a low kinetic basicity." ( DOI:10.1021/ja052647v )

"Thus, this compound class has been the subject of considerable scientific interest challenging chemists from differentfieldstofindsuitablemethodsfortheirsynthesisand handling, evaluate their basicity strength, and understand the principles underlying the thermodynamic and kinetic foundation of their basicity."( DOI:10.1002/anie.201612446)

"The thermodynamic basicity of all newproton pincers was measured by nmr titration experiments with corresponding neutral organic super bases of known pK>25 in CD3CN, while their kinetic basicity was measured by dynamic nmr proton exchange measurements." (Activating Unreactive Substrates: The Role of Secondary Interactions, Edited by Carsten Bolm and F. Ekkehardt Hahn)

So, my questions are:

1) What is the exact distinction between these two terms?

2) Is thermodynamic basicity the one mentioned in common basicity defintions? (arrhenius, lewis, bronsted-lowry)

3) Why are they measured differently?

4) Are they denoted diferrently, do they have different units and is there some general mathematical formula for each that would help me get deeper understanding and distinction between these two terms?

5) Also, how are these two mesaures of basicity related to gas phase proton affinity and intrinsic basicity and pKb?

I have worked on a computational project where I had to study the sites of deprotonation in this molecule:

When deciding between mechanisms I had to consider both thermodynamic and kinetic. For example, I had this case: deprotonations of two different hydrogens(bond with N Blu atoms in the picture)were similarly thermodynamically favored (similar free Gibbs energy of reaction) but one H was towards the inside of the ring and the other pointed towards the outside, therefore, because kinetically favored, the mechanism proposed was the one with the external H removed.

Your questions:

  What is the exact distinction between these two terms?

A Thermodynamically favored base means that the Gibbs energy of the rection where an H+ binds to the protonation site(eg.lone pair of the N) is very negative.
However, if this protonation site is difficult to access, then this reaction is very slow or even impossible and a less thermodynamically favored protonation reaction may occur instead(if the molecule has more than one protonation sites or other bases are present).

  Is thermodynamic basicity the one mentioned in common basicity defintions? (arrhenius, lewis, bronsted-lowry)

yes, and the kinetic effect is evident only in limited numbers of cases.

  Why are they measured differently?

To make it easy and avoid competitive reaction considerations, suppose you have only one type of protonation site. If you want to know how easy is for the proton to access that site, you have to determine the speed of the reaction(dynamic experiment). If you want to know only about thermodynamic(energies involved) then you let the reaction go to equilibrium and just measure the pKa. If the deprotonation site is very inaccessible then this thermodynamic experiment may take very long because you have to wait it reaches the equilibrium. In this case, if it takes more than your experiment time scales than you will measure an apparent pKa that is lower than expected because your reaction has not reached the equilibrium yet. This apparent lower pKa is due to the kinetic effect.

  Are they denoted differently, do they have different units and is there some general mathematical formula for each that would help me get deeper understanding and distinction between these two terms?

I don't know if there are other ways to express this other than kinetic and thermodynamic acid/base.
Thermodynamic: you wait for the reaction to reach the equilibrium and using the $\Delta G$ of reaction determine the pKa.
$ΔG = -RT\ln K_\mathrm{a}$
 $pKa = -log(K_\mathrm{a})$
 In case of a difficult to access protonation site you may not wait for enough and measure a lower pKa but this is an apparent pKa because the definition of it is purely thermodynamic.

  Also, how are these two mesaures of basicity related to gas phase proton affinity and intrinsic basicity and pKb?

Not expert of these techniques, I worked only theoretically but as usual, the solvent affects the value of the pka. The most stable conformation in the gas phase may not be the most stable in solution anymore and there's the solvation energy contribution to consider when you determine the Gibbs energy of the protonation reaction. Therefore, the measurements in solution may not be comparable with the gas phase.

Finally, I can give you some example of very evident kinetic effect:
 In the molecule below the lone pair of the nitrogen is difficult to access for an H3O+ molecule and the pKa measured(not the thermodynamic pKa) will be different depending on how much you wait during the titration experiment. I don't know the timescales for this molecule.

or you can have a base trapped in a fullerene molecule:

This is the most extreme case with an infinite equilibrium time. The pKa of the molecule trapped is still the same but you will experimentally measure the pKa of your solvent and conclude it is not basic.


Does the amplitude of the RF pulse affect the NMR signal? I know that the NMR signal depends on the static magnetic field strength $ B_0 $ (by Boltzmann equation) and its gyromagnetic factor. While NMR occurs when the RF pulse frequency matches the equation:
$$ \nu = \frac{g}{h}\mu_{n}B_{0} $$
what about its amplitude? Does the amplitude of the RF pulse affect the NMR signal? For example, does it affect the signal strength?

V. approx explanation. The rf pulse is at 90$^\text{o}$ to the static field $B_0$ and has the effect of moving the magnetisation (sum of spin vectors) away from alignment along $B_0$ (z-axis normally) into the xy plane where it rotates and cuts through detector coils and induces FID signal. The FID can only be detected with a magnetisation component in the xy plane. The size of the rf pulse's magnetic field $B_I$ is proprotional to angle $\theta$ moved away from z-axis as $B_I\sim \tan(\theta) $ and so via magnetisation's component in xy plane (arrow on axis in sketch) to the size of the FID.


Predicting the spontaneity of non-isothermal reactions We use free energy equations -- Helmholtz or Gibbs -- to predict whether or not a reaction is spontaneous. These equations depend on constant temperature.

This forum post describes a scenario where volumes of water at different temperatures are mixed. The $\Delta G$ appears to be positive. (The poster's textbook gives $\Delta G = 99.5 \, \mathrm{Cal}$, a result I've been able to duplicate.) This seems to show that $\Delta G$ doesn't predict spontaneity for non-isothermal reactions.

So what does?

For an adiabatic spontaneous process in a closed system (like the one in your example), according to the Clausius Inequality, the change in entropy is greater than zero.  If you have a closed system in which a chemical reaction can occur, if the system is adiabatic, the same criterion applies.

The criterion that, at constant temperature, if a reaction is spontaneous, $\Delta G^0$ in going from pure reactants to pure products at 1 bar will be negative is just a rule of thumb which, over the years, has confused students to no end.  If you start out with pure reactants and mix them, spontaneous reaction will always occur to some extent.  It's just that, if $\Delta G^0$ is positive, the reaction will tend to proceed  to a lesser extent when equilibrating, and if $\Delta G^0$ is negative, it will tend to proceed to a greater extent when equilibrating.  So, in this sense, every reaction is spontaneous.


How to calculate concentrations in a diluted mixture of multiple solutions? I have a homework problem as follows:

  A solution was prepared by mixing $25.0\ \mathrm{mL}$ of $0.208\ \mathrm{M}$ $\ce{Fe(NO3)3}$ and $4.50\ \mathrm{mL}$ of $0.00190\ \mathrm{M}$ $\ce{KSCN}$ and diluting to $50.00\ \mathrm{mL}$ with solvent, $0.10\ \mathrm{M}$ $\ce{HNO3}$. Calculate the initial concentrations of the thiocyanate ion and the iron(III) ion before any reaction takes place.

Now, how do I proceed with the question? At first I thought that it should just be the same concentration as the solutions, $0.208\ \mathrm{M}$ and $0.00190\ \mathrm{M}$, because there is only one thiocyanate and one iron(III) ion in the solutions. But the fact that we have an $\ce{HNO3}$ solution confuses me. Am I somehow supposed to use $M_1 V_1 = M_2 V_2$?

Am I somehow supposed to use $M_1V_1=M_2V_2$?

Yep!
The problem has given you three different solutions:

Iron(III) nitrate
Potassium thiocyanate
Nitric acid

You know the concentrations of each, and you know (or can calculate) the volumes of each that are mixed to yield the final solution.  What you'll actually be using in the calculation are three different equations:

$M_{1i}V_1 = M_{1f}V_f$
$M_{2i}V_2 = M_{2f}V_f$
$M_{3i}V_3 = M_{3f}V_f$

$M_{\#i}$ is the initial concentration of each of the three chemicals in their separate starting solutions, and $M_{\#f}$ is the final concentration of each after mixing, that the problem is asking for (except for the $\ce{HNO3}$).


Total number of stereoisomers of truxillic acid Total number of stereoisomers of the compound will be:

This was a question asked in our mock test. I've tried by considering pseudo-chirality on the carbon atoms. But I don't know where to start.
I need help finding the different stereoisomers. It would be appreciated if you could draw the different isomers.

At first glance, not only does it seem like there are no stereogenic centres in truxillic acid, it is also highly symmetrical. However, if we consider different conformations of the chemically identical units (H, COOH or Ph), which give rise to different isomers, some carbons at the 4-membered ring can possibly become stereogenic.

Stereogenic centres are labelled with an asterisk: ε-truxillic acid and peri-truxillic acid do not have stereogenic carbon centres as their 4-membered ring carbons have chemically and conformationally identical neighbours.
It took me a while to get this.Thanks to @andselisk and wikipedia.


Can reaction of gastric acid with swallowed things be dangerously exothermic? Okay, okay. I know that swallowing a large enough amount of any substance would be considered dangerous. That isn't the point of this question, however.

As you probably know, the stomach has hydrochloric acid in it to help digest food, and kill off some microorganisms. I suspect the swallowing aluminum foil (or a decent-sized ingot if aluminum) would be dangerous, because it reacts with hydrochloric acid exothermically to form aluminum chloride and hydrogen gas.

$$\ce{2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)}$$

Here is a video of aluminum's exothermic and vigorous reaction with hydrochloric acid:
https://youtu.be/LaaORZ0ofCY?t=1m12s

Now imagine that going on inside your stomach. I believe that the hydrogen gas may be able to escape through the mouth, but the intense heat might evaporate water (thus creating more gas that must escape), and would burn and kill cells inside the stomach. Plus, there's always the possibility that the hydrogen gas that is escaping could ignite, and burn through the mouth (where it escaped from) and into the body.

So why isn't everyone worried about accidentally swallowing aluminum foil? Wouldn't it be considered very dangerous to do so? Is there something I'm missing or otherwise not thinking of?

For an average healthy adult person maximum concentration of hydrochloric acid
in gastric juice is $C = \pu{0.16 M}$, and on average there is $V = \pu{2 L}$ of gastric acid in the stomach, assuming it's daily secretion of roughly $\pu{1.5 L}$ and some leftovers. Considering that the acid hasn't been used for the purpose yet, and the following reaction

$$\ce{Al + 3HCl -> AlCl3 +1.5H2},\tag{1}$$

one can make an assumption that the mass of aluminium metal that can possibly be digested is

$$m_\ce{Al} = \frac{1}{3} CVM_\ce{Al} = \frac{1}{3} \times \pu{0.16 mol L^{-1}} \times \pu{2 L} \times \pu{26.98 g mol^{-1}} = \pu{2.85 g}.$$

In reality the outcome depends on the health conditions and biometrics of the given individual. People suffering from acid deficiency won't be able to digest even this amount. Larger or overweight specimen can have larger/stretched stomach, so the amount of acid and its pH may vary quite a lot.

On the other hand, having certain decease such as gastric ulce may amplify the effect of digestive system reacting to the produced $\ce{AlCl3}$ and heat. Also, aluminium(III) chloride is considered to be a neurotoxin. Acute oral
toxicity ($\ce{LD_{50}}$, mouse) of $\ce{AlCl3}$ is $\pu{3805 mg/kg}$. Reaction (1) produces approx. $\pu{14 g}$ of $\ce{AlCl3}$, which is not all that bad.

As @MaxW noticed in the comments, swallowing any metal is potentially dangerous and may or may not cause unwanted effects.


Decreasing magnitude of peaks in cyclic voltammogram? I'm in an instrumentation course and we covered cyclic voltammetry earlier in the semester and will be doing an experiment in CV next week.

My question is this:

If you have a reaction and sweep the potential, will the new peaks be entirely within the previous peak? (Meaning: lower magnitude in the subsequent peaks?) It seems that with each sweep, it should have lower electrical potential.

Or would the peaks trace over the previous ones, nicely?

Pardon me for not fully understanding (or if my question is wrong.)

Enjoy a poorly drawn image of what I mean:

Perhaps your observation is an indication of irreversible reaction conditions.  More specifically, a tight deposit on the electrodes hampering the advancement of the electrochemical reactions. This happens sometimes with some less soluble organic dyes used organic electronics, too.  Take a look if the electrodes are still shiny, maybe they need to be cleaned again.  If this is the case, try out a different solvent/electrolyte system.

One aiding companion to trouble shoot is this.


How do acids prevent ferrous sulfate from getting oxidised? If $\ce{FeSO4.7H2O}$ is dissolved in an acidic environment, for example in 0.05 M aqueous solutions of methanoic or ethanoic acid, would that prevent it from getting oxidized if an oxidizing agent like bromine water was added to it? If so, how? Since this is not examining oxidation with respect to air or oxygen.

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water.
Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily:
Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$       +0.770 V and
O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$    +0.401 V
(The half cell voltages are only indicative of what's going on, because we are not at standard concentrations of 1 M.)
The light-colored solution quickly becomes yellowish and consumes all the oxygen in the vessel, and then sucks in more air and consumes that oxygen, and then things slow down a little, but diffusion still lets a bit more oxygen get in, and the "ferrous" becomes somewhat ferric, and the liquid becomes a muddy color and you wish you could preserve the nice blue-green color somehow.
You could preserve the ferrous if you could somehow complex it and keep it from oxidizing to ferric. But nature prefers to complex the ferric ion: smaller ion, bigger charge. Typical ferric stability constants are larger than ferrous by ten orders of magnitude.
It is instinctive to suppose that metals will oxidize in the presence of O$_2$, and if higher valences are possible, well, then, the metal will go as far as possible. But the oxidation of ferrous to ferric is unfavored (+0.770 V) and Fe$^2$$^+$ needs help from O$_2$. The reduction of water is also unfavored, but is driven forward by the consumption of OH$^-$ ions by precipitation by Fe$^+$$^3$, which also keeps the Fe$^+$$^3$ low.
LeChatelier predicted as much: by increasing the acidity, you reduce the availability of electrons from OH$^-$, and O$_2$ is more able to take electrons from Fe$^+$$^2$.
So you could say that adding acid stabilizes the ferrous ion, but what it really does is inhibit the oxygen molecule. Bromine is totally unaffected.


Stripeless cleaning of windows Last week I was discussing with a friend how we thought the stripeless cleaning of windows is achieved when using a cleaner like Windex Glass Cleaner as opposed to just using hot water, in which case you do get stripes left behind.

At first we thought that the stripeless nature was caused by surfactants in the cleaner that allow the (polar) dirt on the windows to be collected in the water through micellation. However, when we use normal cleaner not specifically designed to clean glass/windows, we also get the stripes even though we are sure that there are also surfactants in the normal cleaner.

The question is: what is the difference between the glass cleaner and the regular cleaner that makes sure that windows become stripeless clean?

One thought we considered but couldn't really work out was that the stripe formation would be something caused by a combination of the marangoni effect and evaporation, similar to the coffee-stain effect which can be overcome by specific tuning of the surfactants as shown here.

Could someone explain how the stripeless cleaning works? Can fluid dynamics explain what happens or is it really something chemical going on?

When searching I also ran into a solution that (at least the manufacturers claim) will also give stripeless clean windows. The interesting thing is that they use pure $\ce{H2O}$ so without any surfactants but also without any dissolved salts like regular tap water has.

Cross-post: https://physics.stackexchange.com/q/56800/7433

After quick digging I found, that glass cleaner solutions usually contain alcohols and surfactants. This additions reduce surface tension and allows the solution to interact with glass more easily, distributing the solution over the entire surface.

The white stripes visible after water dried are usually solid salts. If the water does not contain any salts, but only compounds, that can quickly evaporate, it will not leave stripes. But it is not addition, but absence of additions that do the trick. For example, most bleacher/toilet cleaners contain hypoclorites, salts of unstalble $\ce{HOCl}$ acid and leave $\ce{NaCl}$ after evaporation

When tap water is allowed to dry, it usually leaves white stripes, but they are not visible on most surfaces

The only ideas I can get why the salts don't deposit if using a glass cleaner is that salts are removed (and if you have a cleaner not meant to be diluted, the water in it may be deionized) or to trap salt ions into organic ligands so resulted salts were not crystalline (possible, but I do not like the idea).


What does the plus-minus sign with a value in parentheses mean? I always see the plus-minus sign with a value in parentheses after a physical property is described such as melting and boiling point, like this:

$$\text{Melting point }(\pm 1~^\circ\mathrm{C})$$

What does it mean?

It's simply describing the absolute error of the relevant quantities.

So if this precedes a table of melting points, you can expect the true value to vary by 1 degree, e.g.

$\ce{H2O}: 0.0 \pm 1~^{\circ}\mathrm{C}$

means the melting point is between $-1$ and $+1~^{\circ}\mathrm{C}$.


How do I get MarvinSketch to use the ACS style of drawing? I have a Wikipedia account and the people there have a tendency for getting upset over chemical structures that I created using MarvinSketch not conforming to the ACS drawing style.

Is there any way to make MarvinSketch use the ACS drawing style?

You can find some information on JACS style here.

If you are using Marvin Sketch 5.1, first clean your structure, then got to edit then format. Click the Apply changes for all the atoms / bonds and then Load Style. Pick J. Am. Chem. Soc and OK.

In Marvin Sketch 6 open File then Document Style and continue as above

For more information see the documentation here.

That should work


Confusing definition of the term precision According to the textbook, we say a measurement is precise when series of measures are close to each other.

But in real use, I feel like the term is used in a different meaning.

For example, scale A weighs to cube of lead and displays 10.5g while scale B displays 10.4977g.
In this case, one say the measurement made by scale B is more precise than that of A. That is, scale B can distinguish more fine weights than A.

Isn't this usage of the word precise different from textbook definition? Please explain clearly the real usage of the term precise.

In colloquial English, precision is synonymous with accuracy. If you look it up on Google, the first definition is:

  the quality, condition, or fact of being exact and accurate.

In the field of science, however, it takes on another meaning:

  the degree to which repeated measurements under unchanged conditions show the same results

There is also measurement resolution:

  the smallest change [an instrument] can detect in the quantity that it is measuring

Which is similar to the definition precision takes on in numerical analysis:

  the resolution of the representation, typically defined by the number of decimal or binary digit

In short, precision can take on different meanings depending on how you choose to apply it. If you talk about the precision in a set of measured data, you will be using the scientific definition; if you talk about the precision of an instrument, you will be using the numerical analysis definition.


Reaction of carboxylic acid with phosphorus trichloride The reaction of carboxylic acid $\ce{RCOOH}$ with $\ce{PCl5}$/ $\ce{SOCl2}$ / $\ce{PCl3}$ yields an acyl chloride $\ce{RCOCl}$. With $\ce{PCl5}$ and $\ce{SOCl2}$ the reaction is quite simple utilizing one mole reactants and yielding products.
But in case of $\ce{PCl3}$ the balanced reaction is as follows:
$$\ce{3RCOOH + PCl3 -> 3RCOCl + H3PO3}$$
First i thought the reaction would proceed with PCl2− attacking the oxygen from which the oxygen-hydrogen bond cleavage happened but then I realized that it would form product OPCl which is not a real compound i suppose so i am stuck how would the reaction proceed
So what is the mechanism for this reaction?

A plausible mechamism of this reaction is discussed in a paper by Xiao et. al1 where they tried to react benzoic acid with $\ce{PCl3}$ to get acyl chloride. They noted the following mechanism:

Although the detailed mechanism was not clear, a possible mechanism is
proposed [...] the first step should be a reaction of the acid with
$\ce{PCl3}$, probably via a four-center transition state A, with
carbonyl oxygen attacks the electron-deficient $\ce{P(Cl)}$ while $\ce{Cl}$
attacking the electronically deficient carbonyl carbon. Formation of a
$\ce{Cl-C}$ bond and $\ce{P-O}$ bond (A’), further $\ce{C-O}$ bond cleavage and
hydrogen transfer took place to give the products B and C.
Product C then reacts further with the acid to give the
corresponding chlorides and $\ce{HP(O)(OH)2}$

This product C will again react with carboxylic acid two times so as to replace all the chlorine atoms and ultimately forming phosphorous acid (actually phosphorous acid technically exist as $\ce{HPO(OH)2}$ called phosphonic acid) So, the final reaction would be three moles of carboxylic acid reacting with $\ce{PCl3}$ to form three moles of acyl chloride and phosphonic acid thus fully utilizing the three chlorine atoms in $\ce{PCl3}$. The major advantage of this reaction is that there is only one side product i.e phosphonic acid which is non-toxic and easily removed by filtration whereas reaction with $\ce{PCl5/SOCl2}$ gives two side products making the reaction more volatile.
Reference

Atom-efficient chlorination of benzoic acids with PCl3 generating acyl chlorides, Jing Xiao and Li-Biao Han, Journal of Chemical Research, 2019, Vol. 43(5-6) 205–210 (PDF)


Effect of oxidation aluminium reflectivity Aluminium is a fairly good reflector, but oxidation will reduce it's optical reflective properties.

Aluminium is used for reflectors either being deposited on a substrate, or polished and protected by a protective layer to avoid oxidation.

How will the oxide layer of a several year exposed polished aluminium (indoor) will impact the reflectivity.

Will it be considerable or minimal ?

The lifetime of aluminum depends on the environment. If it's near an ocean, where air is humid and often contains salt particles, it may become pitted or look a milk white, rather than shiny, in just months. Anodizing can protect it for years, though.

Because the surface of anodized aluminum is porous (until sealed), it may have a somewhat more or less matte finish depending on treatment. Bright dip anodizing, depending on the aluminum alloy, offers a more shiny finish.

The Alzak process is used for lighting reflectors, and there is information on its specular reflectivity.


Creating Boric Acid from Borax and Hydrochloric Acid I'm working on an experiment in school and I need to create 200 mL of 2 mol/L boric acid. According to a formula I found on wikipedia, boric acid can be created with a reaction with hydrochloric acid.

  $$\ce{Na2B4O7·10H2O + 2 HCl -> 4 H3BO3 + 2 NaCl + 5 H2O}$$

So to calculate the amount of moles of boric acid needed, I multiplied the volume with the concentration to get 0.4 mol. After using the molar ratios in the equation, the molar mass of borax decahydrate and the concentration of HCl, I calculated that I would need 38.14 g of borax and 33.56 mL of 5.96 mol/L HCl in order for a complete reaction.

After mixing these together, I ended up with about 40 mL of a really thick solution, which is not the intended 200 mL. What did I do wrong and how do I actually create boric acid of this volume and concentration? Am I missing steps in between?

Assuming that boric acid is completely soluble in water, you did everything correctly so far, you simply missed the last step: dilution.

First you calculated the amount of substance of boric acid you would need to produce a two molar solution. With that you basically assume, that you dissolve solid boric acid in water.
You then calculated the amount of borax you need for complete reaction with hydrochloric acid. Since this is already hydrogen chloride dissolved in water, you bring some of the water you assumed before. But this is as your calculation says only about 40 mL.
Still assuming that it is completely soluble, you would have produced a solution that is about ten molar. You still need to dilute it to the final volume, to have the desired concentration.

Now let's go one step further. Boric acid is not very soluble in water under standard conditions, see Wikipedia. As A.K. pointed out, solubility is 47.2 g/L at 20 °C. That translates to 0.76 mol/L. As you can see solubility increases with temperature, in boiling water the maximum concentration is about 4.5 mol/L.
So what you are trying to attempt is not possible, at least not in pure water at room temperature. Try heating it up a bit, somewhere around 70 °C should be enough.

Alternatively PubChem has kindly provided us with a little more statistics about solubility.
In methanol you can obtain the desired two molar solution at room temperature, as solubility is 173.9 g/L at 25 °C.


Does heavy water taste sweet? In this YouTube video from Cody's Lab, Cody claims that heavy water tastes sweet.

He does some fairly convincing comparisons but still expresses a little doubt that the effect is real.

Has this been studied by others? Is the effect verified and if so, what possible explanations exist?

According to H.C. Urey and G. Failla, Science  15 Mar 1935,
Vol. 81, Issue 2098, pp. 273, there's no difference in the taste of ordinary and heavy water.


How to determine the pH of a mixture of two weak acids? We have two solutions:

Solution 1 is $\ce{HCOOH}$, its concentration is $c_1 = \pu{10^-2 mol/l}$, its volume is $V_1 = \pu{50 ml}$, and its $\mathrm{pH}_1 = 2.9$.
Solution 2 is $\ce{CH3COOH}$, its concentration is $c_2 = \pu{10^-2 mol/l}$, its volume is $V_2 = \pu{50 ml}$, and its $\mathrm{pH}_2 = 3.4$.

How would be the equation and the ICE table, and what is the $\mathrm{pH}$ of the mixture of these two solutions?

I used numbers to differentiate the solutions, and to understand how it works, but for no other reason.

I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work.

Below a more general approach.

Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:

Reactions

\begin{align}
  \ce{HA + H2O &<=> H3O+ + A-} &
  K_{\mathrm{a},(\ce{HA})}
    &= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
  \tag{1}\label{eq:KAcidHA}\\
  \ce{HB + H2O &<=> H3O+ + B-} &
  K_{\mathrm{a},(\ce{HB})}
    &=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
  \tag{2}\label{eq:KAcidHB}\\
  \ce{2 H2O &<=> H3O+ + OH-} &
  K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}

Mass balance

\begin{align}
  C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
  C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}

Charge balance

$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$

Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
  \frac{K_\mathrm{w}}{\ce{[H3O+]}}
  +\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HA})}}
  + \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$

or as polynomial

\begin{align}
\begin{split}
 &\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
  - K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}

This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version[1].

The \eqref{eq:GeneralPol} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H3O+]}^4$:

\begin{align}
\begin{split}
 &\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
  - K_\mathrm{w} \right]\\
-&\left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  \right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}

The \eqref{eq:GeneralPolSimp1} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll \ce{[H3O+]}^3$:

\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}

The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.

$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
      + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$

Or using the initial concentrations

$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4\left(%\displaystyle
      \frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
        }{V_\ce{HA} + V_\ce{HB}}
      + \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
        }{V_\ce{HA} + V_\ce{HB}}
    +K_\mathrm{w}\right)}
}{2}$$

Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid[3], we have
$$\ce{pH}=3.05.$$

Addendum

@MaxW made some warnings about the assumptions which simplificate ($\ref{eq:GeneralPol}$)(see comment) and ($\ref{eq:GeneralPolSimp1}$)(see comment) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.

As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.

Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.

Acknowledgement

Thanks to @MaxW for draw attention to limitations of assumptions.

Reference

Edited based on first @MaxW's comment.
Edited based on second @MaxW's comment.
Haynes, W. M.;  Lide, D. R.; Bruno, T. J., CRC Handbook of Chemistry and Physics. 97 ed.; CRC Press: 2017; pp. 5–88.


Diels Alder reaction with maleic anhydride. Reactivity of different dienes I don't understand question 6, although I took notes and the professor mentioned an explanation in class.

Can someone please help using easier terms?

a) The compound can rotate into a trans configuration that is not suitable for Diels Alder

b) The compound is locked into a cis configuration that is favorable. Furthermore, a carbon group is slightly electron donating

c) The compound has an electron withdrawing group.

d) What compound is that? It's compounds labelled b) and c)

Comment: Normal demand Diels Alder is enhanced by an electron withdrawing group in the dienophile and an electron donating group in the diene. Look at reaction mechanism tab (http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction)

I am not sure as to why that is. I have forgotten.


How to conveniently separate Cd impurities from ZnO without contaminating it? How could I separate Cd impurities from ZnO without contaminating it and without using overly expensive equipment or extremely dangerous solvents?

I don't want to contaminate it by introducing other heavy metals or being left with more than trace levels of undesired solvents.

I'd be fine losing some amount (say max 10-20%) of my ZnO in the process, if necessary.

Please ask me for any further clarity I could bring to this question.

I know nothing about this so I'd be happy with any contributions on how to easily separate any reasonable portion of $\ce{Cd}$ salts from $\ce{ZnO}$ in general (precipitating it out or solving it away from $\ce{ZnO}$ for instance). Cycling the zinc through a series of reaction and then back to ZnO would be fine also (if not leaving trace levels of anything biologically harmful).

Since you actually considered the use of a calutron, I assume that you are interested in separation of relatively small amounts of pure ZnO - on the order of a few grams.

Cadmium and zinc are similar in many respects, which makes me think of rare earths. One of the separation techniques used there is ion exchange chromatography. Resins with special affinity for many ions (either to catch it or reject it) are available. Some contaminants are not easy to remove by conventional ion exchange resins. In many cases, very specific resins have been developed for these contaminants. Selective resins from Rhom and Haas, www.lenntech.com, are available today for the removal of:
• Boron
• Cadmium, mercury and other heavy metals
• Chromate
• Lead
• Nickel
• Nitrate
• Perchlorate
and some other contaminants.

On the more researchy side, the stability constants of the tetra ammonium complexes of zinc and cadmium differ by a large factor, unlike their complexes with EDTA and similar chelators. For Zn(NH3)4++, the dissociation constant is 9.8 x 10^-10; for Cd(NH3)4++, it is 2.5 x 10^-7. (Reference Book of Inorganic Chemistry, Latimer and Hildebrand, 1951, 3rd ed., pp 135-6). The impure ZnO could be dissolved in NH4Cl (which will also dissolve CdO) and the solution run thru a column. Here's the research: a column of what? Perhaps a column of the impure ZnO. Since the zinc complex is more stable (more soluble), it should concentrate in the first eluate, leaving CdO behind. Solid ZnO would rip NH3 off the cadmium complex and become soluble, so the ZnO will turn into CdO. I have no idea how useful this scheme could be, but the big difference in stability constants is an eye-catcher.


How do I compare the C=C bond lengths in this cumulene? How do I arrange these chemical bonds in an increasing/decreasing order based upon their lengths?

I understand that the bond (a) will be longer than (b) because (a) is a bond between sp² hybrid and sp hybrid carbon atoms, as opposed to (b) which is a bond between two sp hybrid carbon atoms. But how do we compare bond (c)?

Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).

At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).

We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).


Mesh size of the silica gel What does the Mesh in the mesh size of silica gel that is used for various chromatography techniques mean?

Mesh is a unit representing the grain-size distribution of a material. So it represents either the opening of the mesh, or the number of grain per surface unit depending on what you are considering.

You can find here the conversion in grain size and surface unit related to the "Mesh".


Calculating the partition function based on molar heat capacity Given the molar heat capacity of a partition function as a function of temperature, how would one determine the partition function?

Say the molar heat capacity = $\alpha T^2$

All you have to do is to use some of the well known results from Statistical Thermodynamics:
The molar heat capacity at constant volume is

\begin{equation}
  \bar{c}_V = \left( \frac{\partial \bar{U}}{ \partial T} \right)_V
\end{equation}

where $\bar{U}$ is the inner energy, $T$ is the temperature, and $V$ is the volume.
The molar inner energy in terms of the partition function $Z$ and the temperature is given by:

\begin{equation}
  \bar{U} = \frac{k_{\text{B}} T^{2}}{n} \frac{\partial \ln Z}{ \partial T} \ .
\end{equation}

where $n$ is the amount of substance.

Now, you simply have to plug the equation for $\bar{U}$ into the equation for $\bar{c}_V$ and after using the chain rule of differentiation you get the following differential equation:

\begin{equation}
  \bar{c}_V = 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} + \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} \ .
\end{equation}

Finally, you just have to use your formula for the molar heat capacity, $\bar{c}_V = \alpha T$, and solve the resulting differential equation

\begin{align}
  \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} - \alpha T^2 &= 0  \\
 \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{1}{T} \frac{\partial \ln Z}{ \partial T} - \frac{n \alpha}{k_{\text{B}}} &= 0
\end{align}

for $\ln Z$.
If solving differential equations by hand is not your thing, you can simply charge Wolfram Alpha with the task and get:

\begin{align}
  \ln Z &= \frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \\
  Z &= \exp\left(\frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \right) \ ,
\end{align}

where $C_1$ and $C_2$ are constants of integration which can be fixed by enforcing boundary conditions for the temperature.


Why is styrene an aromatic compound? Why is styrene aromatic? Doesn't it have 8 pi electrons, thereby violating Hückel's rule? Can someone offer an alternate, more through explanation?

As Wikpedia Article says ,the criteria for a molecule to be aromatic are --

the molecule must follow Hückel's rule, having 4n+2 electrons in the delocalized, conjugated p-orbital cloud;
the molecule must be able to be planar;
the molecule must be cyclic; and,
every atom in the ring must be able to participate in delocalizing the electrons by having a p-orbital or an unshared pair of electrons.

When we are talking about aromaticity , we consider the electrons inside the ring. You are right that the styrene has 8 pi electrons but when checking for Huckel's rule, we only consider the electrons inside the ring. The alkenyl group is just an attachment to the stable aromatic compound which might take part in resonance but wouldn't affect the compound's aromaticity.


Mass percent and mole percent of a composition The Problem
A gas mixture contains hydrogen, carbon monoxide, and carbon dioxide. Calculate:
a) The mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}$;
b) The molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both 0.35.
My Attempt
I first calculated the molar masses:
molar_mass$\ce{H2}$ = 2.01588 g
molar_mass$\ce{CO2}$ = 44.0095 g
molar_mass$\ce{CO}$ = 28.0101 g

a) The mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}$;

Since the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}$, then the mol% of $\ce{H2}$ must be 85.
I set the mixture equal to 10 mols. This means that there are 1.5 moles $\ce{CO}$ and 8.5 moles $\ce{H2}$. I then converted these values to grams.
1.5 mol $\ce{CO}$ = 42.015 g $\ce{CO}$
8.5 mol $\ce{H2}$ = 17.135 g $\ce{H2}$
The total grams is 59.15 g.
The percent of $\ce{H2}$ would then be 29% since 17.135 / 59.15 = 0.29.

b) The molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both 0.35.

Since the mass fractions of $\ce{CO}$ and $\ce{CO2}$ are both 0.35, the mass fraction of $\ce{H2}$ must be 0.3.
I set the mixture equal to 100 g. This means that there would be 35 g $\ce{CO}$, 35 g $\ce{CO2}$, and 30 g $\ce{H2}$.
Converting these to mols:
35 g $\ce{CO}$ = 1.25 mol $\ce{CO}$
35 g $\ce{CO2}$ = 0.795 mol $\ce{CO2}$
30 g $\ce{H2}$ = 14.88 mol $\ce{H2}$
This gives us a molar composition of
0.88 $\ce{H2}$;
0.07 $\ce{CO}$;
and 0.05 $\ce{CO2}$.
My answers matched up with the solution key, but did I solve this question in the 'right' way? Or is there a better/quicker alternative?

Your approach in general looks OK except for

terminology (e.g. "total grams" should be total mass);
units (e.g. values provided for molar masses should have dimensions $\pu{g mol-1},$ not $\pu{g});$
excessive precision (you are given fractions with two significant figures, but there are mass values reported in your intermediate calculations having five significant digits; also, there is no need to calculate molar masses up to six significant figures — two would suffice);
math notations ("textual formulas" including chunks such as "molar_mass" or "1.5 mol CO" are not acceptable; use standardized symbols instead, e.g. $M$ or $n(\ce{CO}) = \pu{1.5 mol},$ correspondingly).

I was, however, taught to solve problems differently.
Solve the problem algebraically first, then plug in the physical quantities.
This way your solution becomes modular, generalized, and errors are easier to notice.
No imaginary "set" or "fixed" masses, volumes or other quantities introduced out of nowhere.
There is no point in intermediate calculations and they make tracking significant symbols harder than it should be.
Finally, assigning numerical indices to the compounds in the mixture saves you some time and paper, so let's do it now:
$$\underset{1}{\ce{H2}}\quad\underset{2}{\ce{CO}_\vphantom{}}\quad\underset{3}{\ce{CO2}}$$
and move on to the questions:

a) Calculate the mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}.$

By definition, mass fraction of $i$-th component in the mixture is
$$ω_i = \frac{m_i}{\sum_j m_j}.\tag{1.1}$$
For a binary system with given mole fraction of the second component $x_2:$
$$ω_1 = \frac{m_1}{m_1 + m_2} = \frac{1}{1 + \frac{m_2}{m_1}} = \frac{1}{1 + \frac{x_2M_2}{(1 - x_2)M_1}} = \frac{1}{1 + \frac{0.15 × \pu{44 g mol-1}}{(1 - 0.15) × \pu{2.0 g mol-1}}} = 0.29.\tag{1.2}$$

b) Calculate the molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both $0.35.$

By definition, mole fraction of $i$-th component in the mixture is
$$x_i = \frac{n_i}{\sum_j n_j} = \frac{m_i}{M_i\sum_j\frac{m_j}{M_j}} = \frac{ω_im}{M_i\sum_j\frac{ω_jm}{M_j}} = \frac{ω_i}{M_i\sum_j\frac{ω_j}{M_j}}.\tag{2.1}$$
Plugging in the corresponding quantities:
$$x_1 = \frac{0.30}{\pu{2.0 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.88 \tag{2.2.1};$$
$$x_2 = \frac{0.35}{\pu{28 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.07 \tag{2.2.2};$$
$$x_3 = \frac{0.35}{\pu{44 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.05 \tag{2.2.3}.$$
As you can see, using this approach takes less space and in my opinion is much more presentable.
Given you know elementary algebra and use standardized notations, you can greatly improve literacy of your answers and make them more concise.


Adding hydrogen ion concentration while calculating pH I have just started learning how to calculate pH and I have noticed that adding [$\ce{H+}$] makes a negligible difference in few cases whereas in other cases it makes a major change in the final value. Why is that?

pH is calculated by a logarithmic formula given by:

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

Lets say, the initial hydrogen ion concentration is $\ce{[H+]_0}$, and change in concentration is $d\ce{[H+]}$. The change in pH would be:

$$d\text{pH}=-\log{(\ce{[H+]_0}+d\ce{[H+]})}$$

This can otherwise be written as:

$$d\text{pH}=-\log{\ce{[H+]_0}}×\frac{d\ce{[H+]}}{\ce{[H+]_0}}$$

You can see that the change in pH depends on the initial hydrogen ion concentration. It's in fact inversely proportional to the initial concentration. Adding a drop of concentrated acid to a base would give a much bigger change in pH compared to adding a drop to pure water, which in turn is bigger compared to adding a drop to an acid.


Lowest freezing point room temperature liquid? What substance that is liquid at STP has the lowest freezing/melting point? I think it's either 2-methylpentane or 3-methylpentane, but I get different numbers from different sources.

After a bit of random searching through small molecules, I found this very large dataset with 28,645 experimentally measured melting points. Sorting them by increasing melting point and checking the first few entries individually, it seems that N,N-diethylmethylamine holds the record, with a melting point of -196°C (!) and a boiling point of 63-65°C.

I also mentioned another possibility for very low melting liquid substances at room temperature, namely ionic liquids. Even though the species present are much larger and have ionic interactions, presumably the ions pack so extremely poorly in the solid that freezing is very unfavourable.  Aldrich already offers multigram quantities of ionic liquids melting at -88°C, and ongoing research offers even lower melting alternatives, such as the ones in this article which purportedly do not crystallize until -150°C. This limit may well be improved upon in the next few years.

If mixtures are allowed as well as pure substances, then a combination of small organic molecules or ionic liquids will very likely reach even lower temperatures than the examples above without freezing.


Mixing two containers of each water and ethanol to varying degrees of purity In the following question, the water/wine mixing problem was recast in chemical terms. However, partial molar volumes can make this a much more challenging problem than it appears on the surface.

https://chemistry.stackexchange.com/questions/33579/wine-water-mixing-problem

For example, consider a container of $\pu{10 L}$ of ethanol and another container of $\pu{10 L}$ of water. To make a point, let's transfer $\pu{5 L}$ of ethanol to the water container and then transfer $\pu{5 L}$ of the water/ethanol solution back to the ethanol container. What is the purity in each container at the end?

We need to consider that mixtures of water and ethanol have less volume than the two do separately:

For example, after the first transfer, the water container contains $\pu{10.000 kg}$ of water and $\pu{3.945 kg}$ of ethanol for a total mass of $\pu{13.945 kg}$. However, we need to compute mole fraction to determine the volume of the solution:

$$\begin{array}{c|c|c}
\ & \text{water} & \text{ethanol}\\ \hline
\text{mass in kg} & 10.000 & 3.945\\ \hline
\text{moles} & 555 & 85.6\\ \hline
\text{mole fraction} & 0.866 & 0.134
\end{array}$$

From our graph, it looks like we have an excess volume of $\pu{-0.75 mL/mol}$. Since we have total moles $=640.6$, we have an excess volume of $\pu{-0.75 mL/mol}\times \pu{640.6 mol}\times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$. Thus the actual volume is not $\pu{15 L}$ but more like $\pu{14.5 L}$.

If we transfer $\pu{5 L}$ back, we are left with less than $\pu{10 L}$ in the water container. We will also have less than $\pu{10 L}$ in the ethanol container!

  What volume $V_1$, as a fraction of the original volume $V_0$, needs to be transferred in order for both containers to have the same final purity?

$$\begin{array}{c|c|c}
\ & \text{water} & \text{ethanol}\\ \hline
\text{Volume in concern, in L} & 10.00 & 5.00 \\ \hline
\text{Density in kg/L} & 1.000 & 0.789 \\ \hline
\text{Molar mass in g/mol} & 18.015 & 46.069\\ \hline
\text{mass in kg} & 10.000 & 3.945\\ \hline
\text{moles} & 555 & 85.63\\ \hline
\text{mole fraction} & \frac{555}{555+85.63} = 0.866 & \frac{85.6}{555+85.63} = 0.134
\end{array}$$

From the given graph, it looks like we have an excess volume of $\pu{-0.75 mL/mol}$ (at $\chi_\text{EtOH}=0.134$). Since we have total amount of water and ethanol, $555+85.63=\pu{640.63 mol}$, we can calculate the excess volume of the container:

$$\pu{-0.75 mL/mol} \times \pu{640.63 mol} \times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$$
Thus the actual volume is not $\pu{15 L}$, but $\pu{(10.00+5.00-0.480) L}=\pu{14.52 L}$

Suppose, we transfer $V_1$ of water/ethanol mixture back to the container-2 containing the remaining of $\pu{5.00 L}$ of ethanol. That $V_1$ contains:
$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 38.22V_1 \ \pu{mol}\text{ water}$$
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 5.90V_1 \ \pu{mol}\text{ EtOH}$$

Thus container-2 has $38.22V_1 \ \pu{mol}$ of water and $\left(5.90V_1 + \pu{85.63 mol}\right)$ of ethanol after the addition of $V_1 \ \pu{L}$ of water/ethanol mixture from container-1.

Suppose we have made both containers have the same purity (by $\%(w/w)$) by this action. Let's calculate the masses of water and ethanol ($m_{w2}$ and $m_{et2}$, respectively) in container-2:

$$m_{w2} = 38.22V_1 \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= 688.5 V_1 \ \pu{g}\text{ water}$$

Similarly,

$$m_{et2} = \left(5.90V_1 + 85.63\right) \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(271.8V_1 + 3945\right) \ \pu{g} \text{EtOH}$$

Now, we'll calculate the masses of water and ethanol ($m_{w1}$ and $m_{et1}$, respectively) in container-1. Amounts of water and ethanol in this container is:

$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}$$
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}$$

Thus,

$$m_{w1} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= \left(555 - \frac{555V_1}{14.52}\right)\times   \pu{18.015 g}\text{ water}$$

Similarly,

$$m_{et1} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(85.63 - \frac{85.63V_1}{14.52}\right)\times   \pu{46.069 g}\text{ EtOH}$$

Theoretically, the condition for same purity by $w/w$ is:

$$\frac{m_{w1}}{m_{w1}+m_{et1}} = \frac{m_{w2}}{m_{w2}+m_{et2}} $$

or

$$\frac{m_{et1}}{m_{w1}+m_{et1}} = \frac{m_{et2}}{m_{w2}+m_{et2}} $$


Regioselectivity of bromination of alkenes Recently, in my chemistry lesson, we were discussing the mechanism of the bromination of alkenes. The teacher brought up the involvement of the cyclic bromonium intermediate. She also mentioned that the nucleophile attacks the more substituted carbon atom in the bromonium ion. However, this raised many questions among my classmates. Firstly, based on steric reasons, the nucleophile is more likely to attack the less-substituted carbon since there is less hindrance. Hearing this, the teacher then said that the reason for the preference is based on electronic factors. However, this makes little sense as well, considering that the presence of more alkyl groups will decrease the partial positive charge on the more substituted carbon, making it less attractive to the nucleophile (be it $\ce {Br^-}$ or $\ce {H2O}$).

I went to seek clarification by looking at more authoritative texts, such as Advanced Organic Chemistry: Part A: Structure and Mechanisms by Carey and Sundberg, as well as Guidebook to Mechanism in Organic Chemistry by Peter Sykes. However, they were citing very weird reasons that were not very intuitive. Here are some extracts from relevant chapters of the two texts:

From the former,

  Unsymmetrical alkenes nevertheless follow the Markovnikov rule because the partial positive charge that develops is located predominantly at the carbon that is better able to accommodate an electron deficiency...

From the latter,

  With an unsymmetrical alkene, e.g. 2-methylpropene (32), the more heavily alkyl-substituted carbon will therefore be preferentially attacked by the residual nucleophile, $\ce {Cl^-}$.

Why do the alkyl groups result in greater cationic character when they are in fact, electron releasing?

With bromination, the first step is the formation of an adduct complex, which can be explained that the π-bond polarises the dibromide. The next step is the formation of the bromonium ion, which carries an overall positive charge. A symmetry-changing induction like substituents will always result in asymmetric carbon-bromine bond lengths in the bromonium ion. Since it is a cation, similar rules for the stability/instability as for carbocations apply. Therefore the higher substituted carbon will have the longer $\ce{C-Br}$ bond length, and the higher charge (more positive/less negative). This makes it a better target for the residual nucleophile $\ce{Br-}$ to attack there.

Here is the bromonium ion for the reaction
$$
\ce{\underset{alkene}{H3C-CH=CH2} + Br2 ->
[\underset{bromonium ion}{H3C-CH(Br)CH2}]+ + Br- ->
\underset{$trans-$alkane}{H3C-CHBr-CH2Br}}
$$
calculated at the DF-b97d3/def2-SVP level of theory, charges with NBO6, to illustrate the concept:

Additionally the HOMO has the higher coefficient at the lesser substituted carbon, while the LUMO has the higher coefficient at the higher substituted carbon.

From this, you can see that both quotes from the textbooks are correct.


Is ionic charge just a label? We say that when an atom loses an electron it becomes positively charged ion, and similarly when an atom gains electrons it becomes a negatively charged ion. My question is that are 'positive' and 'negative' just labels indicating more protons than electrons or vice versa, or do the ions actually display an increased interaction towards other charged particles?

These ionic charges actually do behave as you would expect from electric charges: cations will be drawn towards the negative pole in an electrolytic cell while anions will be drawn towards the positive pole. The same feature can be observed with α and β radiation (which are $\ce{^4_2He^2+}$ and electrons, respectively); these are deflected in an electric field as expected.

On a more philosophical level, the entire concept of one charge being ‘positive’ while the other is ‘negative’ is of course meaningless. We might as well have called them ‘East’ and ‘West’ charges akin to the North and South poles of a compass. The only physically observable is that the two act as opposites; two particles of like charge will repel, two of opposite will attract. Having positive and negative charges makes calculations much easier though.


Limiting Reactants without a given mass This is a question I have for Chemistry homework, but I don't understand how to solve it without a given mass. I narrowed the choices down (see below _) I know a limiting reactant well limits the reaction (so if there are 2 reactants) and the react left is the excess reactant.

I thought Propane would be the limiting reactant because the oxygen is in excess (though that was incorrect).

Then, I though Oxygen would be the limiting reactant (naturally) since its a combustion reaction (though that is incorrect).

Determine whether each reaction depends on a limiting reactant. Explain why or why not and identify the limiting reactant. (Select all that apply.)

(c) Propane (C3H8) burns in excess oxygen to produce carbon dioxide and water.

Yes, one reactant is naturally limiting.
Yes, there are two reactants.
No, all reactants are in excess.
No, there is only one reactant.
propane
oxygen
carbon dioxide
water
Not enough information to identify which is the limiting reactant.

Yes, one reactant is naturally limiting.
Yes, there are two reactants.
No, all reactants are in excess.
propane
oxygen
Not enough information to identify which is the limiting reactant.

It seems that you would need information about how much of each reactant you have when conducting this. edit: "Excess" is the key word here; so it hints that Oxygen is NOT the limiting reagent. Considering how combustion reactions are structured, the answer seems to be the other reactant here.


Delocalised electron in graphite I have a question regarding the delocalised electron in graphite. There has been conflicting information between books.

Some books said the delocalised electron in graphite exists $\bf{between}$ layers.
Some said the delocalised electron in graphite exists $\bf{within}$ the hexagonal layer.

I support the second statement, because carbon has electronic configuration of 1s2 2s2 2p2. Graphite has sp2, therefore leaves 1 p orbital. And the giant pi bond should be extended over the layer, instead of between.

I am not sure my evidence is sound.
Any advice would be greatly appreciated. Thank you in advance.

It's just a matter of definition. If a "layer" is understood as an infinitely thin plane passing through the centers of carbon atoms, then the electron density of the huge delocalized $\pi$ orbital is located above and below that plane, i.e., technically between layers. If a layer is defined so as to include whole atoms, then the said orbital is within layer.

In any case, the delocalized electrons are more or less "free" to move along the planes and not perpendicular to them, which explains the high anisotropy in electrical conductance and other properties of graphite.


How do I visualise bicyclo[4.4.1]undeca-1,3,5,7,9-pentaene? While reading my organic chem book, I came across this compound:

Since this is an exercise question, I do not know the name of this compound.
I am unable to visualize this compound (how it looks out of the paper) as I have never come across two rings joined by a bridge-like structure, my book says it is a methylene bridge.
I appreciate your assistance in determining the structure of this compound.

This compound is 1,6-methano[10]annulene, and a classic example of an unusual compound displaying aromaticity.

Aromaticity requires the pi-system to lie in a plane. One might speculate that [10]annulene (shown below) would be aromatic, because it contains a cyclic pi-system of 10 electrons, which appears to be planar. However, because the hydrogens indicated are trying to occupy the same space, the compound is forced out of planarity, and the compound it non-aromatic.

Bianchi and coworkers reported the crystal structure of 1,6-methano[10]annulene. In this compound, the hydrogens that clash in [10]annulene are replaced by a methylene (CH2) group. From directly above, the compound looks basically like naphthalene, with a little extra space in the middle. More interestingly, from the side, it's clear that the atoms of the 10-membered ring are nearly planar, with that methylene bridge oriented perpendicular to the plane.

Reference: Structure of 1,6-methano[10]annulene, R. Bianchi, T. Pilati and M. Simonetta, Acta Cryst. (1980). B36, 3146-3148, https://doi.org/10.1107/S0567740880011089


Can an acid have a high pH? I am studying anesthetics and came across thiopental which is considered a weak acid. However it has a pH of 10.5. I always learned that greater than 7 is alkaline. I’m looking for some guru to help me outs thanks!

No, acids can't have pH greater than 7. By definition, at 25 °C, solutions with a pH less than 7 are acidic and solutions with a pH greater than 7 are basic.

In your question you say thiopental and pH = 10.5. Actually this is sodium thiopental with pH range of 10.2-11.2 for a 8% solution SOURCE. The solution is alkaline because of hydrolysis.

The pH value however can be less than 0 or greater than 14 for very strong acids and bases respectively. (Lim, Kieran F. (2006). "Negative pH Does Exist". Journal of Chemical Education. 83 (10): 1465.)


Catalyst for Formic acid decomposition I wanted to decompose formic acid into $\ce{CO2}$ and $\ce{H2}$ for a school experiment. I was wondering what would be a suitable catalyst to use?

In A Novel Example of Molecular Hydrogen Generation from Formic Acid at
Visible-Light-Responsive Photocatalyst, the authors describe the formation of carbon dioxide and hydrogen from formic acid using visible light from a halogen lamp and freshly prepared $\ce{Cu2O}$ (or platinum-covered $\ce{Cu2O}$) as a catalyst.

The preparation of the $\ce{Cu2O}$ is described as follows:

  Ten cubic centimeters of $\mathrm{1.0~mol~dm^{-3}}$
  $\ce{CuSO4*9H2O}$ aqueous solution was dropped
  into $\mathrm{100~cm^3}$ of $\mathrm{1.0~mol~dm^{-3}}$ $\ce{Na2S2O3}$ aqueous solution, and
  subsequently, the mixture solution was dropped into $\mathrm{200~cm^3}$
  of $\mathrm{1.0~mol~dm^{-3}}$ $\ce{NaOH}$ solution at 80 °C.

  The resulting reddish-brown precipitate ($\ce{Cu2O}$) was separated by filtration and washed with deionized water, after which the precipitation was dried at 40 °C in an $\ce{Ar}$ atmosphere.

The conversion of formic acid is described as follows:

  The photocatalytic experiment was conducted in a glass vessel (volume: $\mathrm{20~cm^3}$) containing $\mathrm{10~cm^3}$ of $\mathrm{2.5~mol~dm^{-3}}$ $\ce{HCOOH}$ and a known amount of photocatalyst (10 mg (ca. 70 μmol)) [...] in
  $\ce{Ar}$ atmosphere with stirring. The pH of the $\ce{HCOOH}$ solution was adjusted to 5.0 by the addition of $\ce{NaOH}$ solution.

  A halogen lamp was used as the light source (light intensity: $\mathrm{280~mW~cm^{-2}}$ ) along with a sharp-cut filter (S76-L42, Suruga Seiki Co. Ltd.), using which the samples were irradiated with light of wavelength $\lambda > 420~\mathrm{nm}$. [...]

In summary, it seems possible to perform the experiment at school. UV filters are available from different manufacturers, a Schott N-WG320 might work.


How to calculate the amount of sodium sulfate needed to lower the silver ion concentration of a saturated silver sulfate solution? How many moles of $\ce{Na2SO4}$ we have to add into a saturated $\pu{0.5 L}$ solution of $\ce{Ag2SO4}$ so that the concentration of $\ce{Ag}$ is $\pu{4.0 \times 10^-3 M}$? ($K_\mathrm{sp} = 1.4 \times 10^{-5}$)

I found out first $\ce{[SO4^2-]}$ in a normal saturated solution:
\begin{align}
K_\mathrm{sp} &= [\ce{Ag+}]^2[\ce{SO4^2-}],\\
1.4 \times 10^{-5} &= 4s^3,\\
s &= 0.015
\end{align}
So we have $\pu{0.015 M}$ of $\ce{SO4^2-}$ in this solution.
Then, I find out $\ce{[SO4^2-]}$ needed for a saturated solution of $\ce{Ag2SO4}$ but with $[\ce{Ag}] = \pu{4.0 \times 10^-3 M}$:
\begin{align}
1.4 \times 10^{-5} &= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}],\\
[\ce{SO4^2-}] &= 0.875
\end{align}
So we need $\pu{0.875 M}$ of $\ce{SO4^2-}$ in such saturated solution.
So the difference of $\ce{SO4^2-}$ is $\pu{0.875 M} - \pu{0.015 M} = \pu{0.86 M}$. So we need to add $\pu{0.86 M}$ of $\ce{SO4^2-}$, that is $\pu{0.43 mol}$ of $\ce{SO4^2-}$ ($\pu{0.86 M} \times \pu{0.5 L} = \pu{0.43 mol}$).
Eventually, we obtained we need to add $\pu{0.43 mol}$ of $\ce{Na2SO4}$.

The above are my calculations, can anyone tell me if my steps can be justified?

Your calculations are true to a certain extend. However, you neglected the amount of $\ce{SO4^{2-}}$ ions precipitated as $\ce{Ag2SO4}$. And also, since we are calculating in the means of concentration terms, we have to assume that addition of $\ce{Na2SO4}$ solid did not change the original $\pu{0.5 L}$-volume.

You calculated correctly the concentration of $\ce{[SO4^2-]}$ ions ($s$) in the original saturated solution, which is equal to the solubility of $\ce{Ag2SO4}$ ($s$):

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (2s)^2 \times s= 4s^3= 1.4 \times 10^{-5} $$
$$\therefore \; s = \left(\frac{1.4 \times 10^{-5}}{4}\right)^\frac{1}{3} = 1.52 \times 10^{-2}$$

You also calculated correctly the concentration of $\ce{[SO4^2-]}$ ions in the solution after certain amount of $\ce{Na2SO4}$ solid was added (suppose it is $m$ mols) to make  the concentration of $\ce{Ag+}$ ions is equal to $\pu{4.0 \times 10^{-3} M}$:

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}] = 1.4 \times 10^{-5}$$
$$\therefore \; [\ce{SO4^2-}] = \left(\frac{1.4 \times 10^{-5}}{16.0 \times 10^{-6}}\right) = \left(\frac{1.4}{1.6}\right) = \pu{0.875 M}$$

Thus, you have $\pu{0.875 M}$ of $\ce{SO4^2-}$ ions in this solution. However, note that some of $\ce{SO4^{2-}}$ ions would be precipitated as solid $\ce{Ag2SO4}$ when you keep adding $\ce{Na2SO4}$ solid (vide supra) in order to maintain the equilibrium (Le Chatelier's principle). This amount is equal to the half of $[\ce{Ag+}]$ ions precipitated with it:

$$\therefore \; [\ce{SO4^2-}]_{ppt} = \frac{1}{2}\left(2 \times 1.52 \times 10^{-2} - 4.0 \times 10^{-3} \right)= \pu{1.32 \times 10^{-2} M}$$

Suppose you add $m$ amount (in moles) of $\ce{Na2SO4}$ to the original solution. which already had $\pu{1.52 \times 10^{-2} M}$ of $\ce{SO4^{2-}}$ ions. Thus, $[\ce{SO4^{2-}}]$ remain in the solution:

$$[\ce{SO4^{2-}}]_{remain} = \pu{0.875 M} = m + 1.52 \times 10^{-2} - 1.32 \times 10^{-2} $$

$$\therefore \; m = 0.875 - 0.20 \times 10^{-2} \approx \pu{0.873 M}$$

Thus, you need to add $\pu{0.873 mol/L}\times \pu{0.5 L} \approx \pu{0.437 mol}$ of $\ce{Na2SO4}$ to the original solution.


N(delta)-acetyl... vs N(alpha)-acetyl Can anyone tell me what  the difference between these two compounds is:

N(alpha)-acetyl-ornithine

N(delta)-acetyl-ornithine

The only difference I can think of is the chiral properties of the nitrogen atom, i.e. nitrogen inversion.

If that is the case, does it make sense to differ between these to compounds, or will they be constantly interconverted?

The difference does not lie in stereochemistry. Rather, N$^\alpha$-acetylornithine and N$^\delta$-acetylornithine are regioisomers of one another.

The superscript Greek letter denotes which nitrogen is substituted with an acetyl group. The lettering is relative to the carboxyl group as shown in the image below:

Note that:

The carboxyl group’s carbon does not get the $\alpha$, because it is considered part as the functional group from with the labelling begins
The $\alpha$ carbon is asymmetric. Both L- and D-compounds (or (S) and (R) according to CIP nomenclature) are possible for both N$^\alpha$- and N$^\delta$-acetylornithine.


Octet rule violation in Period 2 elements My textbook states,

  Period 2 elements typically obey the octet rule. Period 3 and later elements can expand their valence shells. Elements in Periods 5 and 6 of the p-block show variable valence (the inert-pair effect).

I was surprised that it said period 2 elements typically obey the octet rule. What are some circumstances in which period 2 elements break the octet rule? I though they couldn't because they don't have available d orbitals.

Typically the octet rule becomes less likely to work, regardless of the period, when we move to a group where there are too few valence electrons to go around.  Prominent examples in Period 2 include compounds of beryllium in Group 2 (such as the fluoride, especially when gaseous or liquid) and boron in Group 13 (as in alkyl boranes like this one).  Carbon can also form species that are most stable with fewer electrons than the octet rule would predict, a "violation" most prominently seen in the cyclopropenyl cation, where another "rule" that is much more reliable when properly used comes into play.


How many electrons can an orbital of type f hold? I was taking a chemistry test and I encountered the following question:

  How many electrons can an orbital of type f hold?

  A. 6
  B. 10
  C. 2
  D. 14
  E. 1

Since there can be [-ℓ, ℓ] orientations and since the orbital type f has ℓ = 3, we should have 7 possible orientations with 2 spins, so ${7 \times 2 = 14}$, so I thought the correct answer was D (14).

However, I got it wrong and the correct answer is marked as C (2). Is it an error in the test, or am I missing something?

You're correct that there are seven possible spatial orientations for an f-type orbital, and hence seven possible orbitals in one f-type sub-shell. However, the question specifically asks for the maximum number of electrons in one such orbital, and any single atomic orbital, regardless of the sub-shell type specified by $l$, can only hold two electrons. This is by virtue of the Pauli exclusion principle. Fourteen would be the maximum number of electrons across an entire f-type sub-shell, but the question only asks about one orbital.


How is a phase equilibrium defined for a one-component system? A question on this site asked whether a one-component system is at equilibrium when melting or boiling, and the disparate answers were somewhat dependent on the definition of phase equilibrium. Another question asks whether two phases can ever exist in equilibrium at the boiling point. The term "phase equilibrium" seems to imply that under certain conditions, a one-component system with two phases is indeed at equilibrium.

Searching for the definition of phase equilibrium, I found these two:

Definition 1

  Phase equilibrium is the study of the equilibrium which exists between or within different states of matter namely solid, liquid and gas. Equilibrium is defined as a stage when chemical potential of any component present in the system stays steady with time.

Source: https://www.wattco.com/casestudy/phase-equilibrium

Definition 2

  Phase equilibrium is the state of thermodynamic system, in which the different phases of the substance having common boundary surfaces do not vary quantitatively.

Source: https://www.elsevier.com/books/computational-materials-science/ovrutsky/978-0-12-420143-9

Implications of the distinct definitions

According to definition 1, water present as two phases at the melting point would be at equilibrium even if slowly melting isothermally. According to definition 2, the same situation would not be called equilibrium because the amount of ice is decreasing over time (and thermodynamic parameters such as entropy and inner energy are changing over time).

Phase rule

For the mentioned system, the phase rule specifies one independent degree of freedom (you can change one intensive parameter, say temperature, and all other intensive parameters, in this case just pressure, will be determined, given that the system is supposed to be at equilibrium). Because it is a one-component system, the concentrations (or activities) are constant and are not a degree of freedom. If you are used applying the equilibrium concept to solution chemistry, this is somewhat unusual.

Question: What is the official definition of phase equilibrium, and how does it apply to a one-component system where the chemical potentials are independent of the mole ratios?

Melting is a process. A process is not a state. Since only states can be "at equilibrium", it doesn't make sense to speak of a real melting substance as being in an equilibrium state. However, melting is deemed reversible if performed infinitely slowly, and during such a transition the system passes through a series of equilibrium (or "quasi-equilibrium") states. But such a process is an idealization.  In reality, you can't "melt ice isothermally". Real transitions are therefore irreversible because system and surroundings are not actually in thermal equilibrium during heat transfer$^\dagger$.
Note that if you place in contact two of the intermediate states sampled during the transition neither will change, that is, each remains in its original equilibrium state.
The second definition encompasses the first one (equality of chemical potential across phases). The second definition therefore better captures what is meant by phase equilibrium. The conditions for phase equilibrium are those used to derive the phase rule, namely mechanical, chemical and thermal balance between phases. A phase is a region of uniform composition, temperature and pressure.

$^\dagger$ We say that it is possible to melt a pure substance using an ideal reversible process that leads the system infinitely slowly through a series of equilibrium states (each intermediate stage in that reversible process is an equilibrium state).  In order to move between intermediate states, an infinitesimally small amount of heat has to be transferred from the infinitesimally warmer surroundings (exploiting, say, something like a Maxwell demon). In what is a logical sleight of hand, we say that the temperature difference is so small however, that the system and surroundings are effectively isothermal. This does not reflect what happens in a real process.


Redox properties of methyl radical The methyl radical, $\ce{.CH3}$, is a strong oxidant and a strong reductant according to Wikipedia.

Why is this so? Any additional reasons to the ones below?

I know that an oxidant itself is reduced; an oxidant helps something else lose electrons, so the oxidant must be the one gaining electrons. Given that the methyl radical has an unpaired electron, it is quite reasonable that it would be amenable to having an extra electron to pair up with its lone electron.

I know that a reductant itself is oxidized; a reductant helps something else gain electrons. The carbon in the methyl radical has an oxidation state of -3, and I know that carbon can adopt a variety of higher oxidation states, such as +3 in acetic acid.

Why is this so?

Is this so?

Following is a tabulation of the ionization potentials (IP) and electron affinities (EA) for some alkali metals and halogens (kJ/mol):

\begin{array}{|c|c|c|c|c|c|c|}\hline
& IP & EA & && IP & EA          \\ \hline
F & 1681 & 331 && Li & 520 & 60  \\ \hline
Cl & 1251 & 352 && Na & 496 & 53 \\ \hline
Br & 1140 & 328 && K & 419 & 49  \\ \hline
I & 1008 & 298 && Rb & 403 & 47  \\ \hline
\end{array}

A good reducing agent readily donates electrons and would be expected to have a low IP.  Sodium would be expected to act as a good reducing agent and chlorine as a poor one, and, indeed the IP of sodium is much lower than that of chlorine.

A good oxidizing agent readily accepts electrons and would be expected to have a large EA.  Chlorine would be expected to act as a good oxidizing agent and sodium as a poor one, and, indeed, the EA of chlorine is much higher than that of chlorine.

This NIST webpage tabulates experimental IP's and EA's for the methyl radical.  The IP is given as (9.84 eV) 945 kJ/mol and the EA appears to be less than (0.6 eV) 58 kJ/mol.

Let's compare the IP and EA of the methyl radical to values in the above table for strong oxidizing and reducing agents.  The IP of the methyl radical is significantly larger than those reported for the alkali metals and a bit less, but relatively close to the IP's for all of the halogens except fluorine. This comparison suggests that methyl radical is, at best, a weak reducing agent. The EA of methyl radical is significantly less than those reported for the halogens and right around those reported for the alkali metals. This comparison suggests that methyl radical is, at best, a very weak oxidizing agent.


What is this fluorinated organic substance? The corresponding author of the paper where this formula was published as appendix passed away. Can someone help me with identifying what is the name of this compound and why have strange zig-zag between Nitrogens

The name of the compound is 1-chloromethyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane bis(tetrafluoroborate) (CAS #: 140681-55-6), which is commonly known as Selectfluor, a trademark of Air Products and Chemicals (see Waylander's comment elsewhere). Different view of the compound is given below (to you to understand the zig-zag feature):

Introduced in 1992, this organic salt is used as a fluorine donor in organic synthesis (Ref.1). For example of using Selectfluor as a source of fluorine, see Ref.2:

References:

R. Eric Banks, Suad N. Mohialdin-Khaffaf, G. Sankar Lal, Iqbal Sharif, Robert G. Syvret, "1-Alkyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane salts: a novel family of electrophilic fluorinating agents," J. the Chem. Soc., Chem. Commun. 1992, (8), 595-596 (DOI: 10.1039/C39920000595).
Timothy J. Barker, Dale L. Boger, "$\ce{Fe(III)}$/$\ce{NaBH4}$-Mediated Free Radical Hydrofluorination of Unactivated Alkenes," J. Am. Chem. Soc. 2012, 134(33), 13588-13591 (https://doi.org/10.1021/ja3063716).


Mechanism for reaction of ketene with water In the Arndt-Eistert synthesis, a ketene is the intermediate. The ketene is then hydrolysed in the presence of $\ce{Ag2O}$. What is the mechanism for the hydrolysis of the ketene?

Is an oxonium ion formed or will there be nucleophilic attack of $\ce{H2O}$ on the carbonyl carbon?

The silver salt, which can also be $\ce{AgCF3COO}$, is added to the mixture to cleave off $\ce{N2}$, creating a singlet carbene which induces the 1,2-shift to generate the ketene. From then on, it has no further purpose.

Our solvent is water, which could technically act as an acid, a base or a nucleophile. Reacting as a base does not forward our mechanism, so we need to consider whether water is stronger as an acid (left-hand mechanism in scheme 1) or as a nucleophile (right-hand mechanism).

Scheme 1: Two possible mechanisms for ketene hydrolysis.

A few thoughts on the two mechanisms, first. In whichever way water reacts, it will react with the most reactive site, first. In case of a nucleophilic attack, the most electrophilic carbon is clearly the ketene carbon. Water has the possibility of cleaving either the $\ce{C=C}$ double bond or the $\ce{C=O}$ double bond. The $\ce{C=O}$ bond is preferred because it usually represents the molecule’s LUMO and the resulting carbanion would be rather unfavourable and probably immediately flip to an oxyanion. The final [1,3]-proton shift cannot occur in a concerted manner for electronic reasons. It is also highly unlikely that a single water molecule is both acceptor and donor as I drew it for brevity. I would assume a pair of molecules acting from both sides in a stepwise manner.

In case of the water-acidic mechanism, the most basic position in the molecule would be the oxygen atom so one would expect protonation there first. Again, I would assume the [1,3]-proton shift to be non-concerted which could provide major flaws to the mechanism. I also do not see the carbon position having sufficient basicity to be protonated in any noteworthy manner in rather neutral water. But let’s just assume it could happen for the sake of the following argument.

When it comes to rating water’s properties, we should note that it’s a mediocre base, a rather bad nucleophile but an even worse acid. So while one can save the nucleophilic mechanism by saying ‘yeah, water’s not a good nucleophile, but it’s the best we have floating around and the ketene position is rather reactive’ we cannot save the acidic mechanism in any similar way. Ketones are, in good approximation, unprotonated in aquaeous solution, so protonating an even worse position is all the more unlikely.

Of course, it is conceivable that both mechanisms happen to some extent. By playing around with the solution’s pH, you could probably favour either: an acidic solution will possibly facilitate the acidic mechanism while a basic solution will definitely facilitate the nucleophilic one — hydroxide is a so much better base than water.


Observing crystal of water By just freezing water and using an optical microscope, can I observe water crystals?

Or do I have to use a special light to see?

No special light is required, although you will want to keep the light source distant to avoid melting your ice. You will also want to experiment with light angles to get the best resolution of the different angled facets of the ice crystal.

Without the right equipment and experience it will be difficult to get an image quite as spectacular as the crystal you've shown. The keys to getting the best image are avoiding heat, from a lamp or the surrounding air, which causes sintering of the finest faceted crystals.

A good way to view ice under a microscope is to place crushed dry ice under a dark covering (depending on your equipment, getting something that fits and still focuses may be your greatest challenge). Whether you use glass or plastic or whatever else isn't very important. The dry ice is cold enough that ice will nucleate on the surface of your choice and grow from there. Be prepared to watch the whole show, from freezing to melting!


Enthalpy and Internal Energy for Isothermal Expansion Considering a system with constant atmospheric pressure , i.e a massless piston sitting in a cylinder containing water. At constant temperature say $\rm 100\,^{\circ} C$, the water turns into vapour and pushes the piston up.
Now shouldnt the $\Delta U=0$ as $Q = -p(V_2 - V_1)$, considering the Ideal Gas Behaviour
And then as $\Delta H = \Delta U + \Delta n\cdot R\cdot T$, then shouldn't $\Delta H = \Delta n\cdot R\cdot T$?

Then according to the question in image below, why isn't $\Delta U = 0$?

You are right that an in in ideal gas, internal energy is a function of temperature only, and that in this problem, temperature is not changing.  However, I think you are confused about how broadly the ideal gas law applies to this problem.

The question states that the ideal gas law applies to the water vapor.  But the question is about a phase change of water.  Let's break down some of the components of the problem.  In the question we have:

Liquid water.  The ideal gas law does not apply to liquid water.
Water vapor.  The ideal gas law does apply.
A phase change of liquid water to water vapor.  $\ce{H2O(l) <=> H2O(g)}$  The ideal gas law does not apply to the process of the phase change, simply because processes are not gases and cannot be modeled by the ideal gas law.

Thus only one of three "components" of the problem is an ideal gas.

As a look at any reasonable steam table will tell you, the internal energy of water vapor is higher than the internal energy of liquid water.  This difference is the heat of vaporization (at constant volume).


Testing thermite for chemical properties I'm writing a movie in which a bunch of military explosive ordnance experts destroy a turbine with thermite and the local fire chief has to work out whether it is plain $\ce{Fe2O3}$ and aluminium powder thermite or military-grade thermite with other stuff in it.

How would he test for these ingredients and better be able to discern between the two?
And what sort of language would he use to translate his finding?

Typical military-grade thermite mixtures (e.g. used in incendiary hand grenades) contain a large amount of barium nitrate (e.g. 29.0 % in Thermate-TH3, see Patent No. US 6766744). It should be possible to detect the barium oxide that is left over after the fire, maybe first with a simple flame test on site followed by AAS (atomic absorption spectroscopy) in a laboratory.


What is the lattice structure of manganese? A transition element is defined as the one which has incompletely filled d orbital in its ground state or in any one of its oxidation state. Zinc , cadmium and mercury are not typical transition elements because they have fill filled d orbital in their ground state as well as in their common oxidation state. These elements do not show none of the characteristic properties of transition metals like the lattice structure as shown below in picture. They probably do not have proper metal-metal bonding, a characteristic feature of transition metals. But, still they are studied along with the chemistry of transition metals. But what about manganese? It is a core and important element in transition series. Here, X is unknown lattice structure.

Ah, good old Mn. This is one of the uglier ones in terms of phases.
$\alpha$-Mn, the room temperature phase, is CBCC (A12 family), fairly unusual. Near 1000K, Mn transitions to $\beta$-Mn, with a simple cubic (A13 family) crystal structure. $\gamma$-Mn is FCC, face-centered cubic (A1), and $\delta$-Mn is BCC, body-centered cubic (A2).

One excellent reference is A.T. Dinsdale, SGTE Data for Pure Elements, CALPHAD 15(4) 317-425 (1991). There is a readily available PDF of an updated version of that paper that can be found (Dinsdale). This paper gives the accepted Gibbs free energy functions for the elements in their various stable, and some unstable, forms.

Editing to add additional information on the $\alpha$- and $\beta$-Mn crystal phases. Both of these phases are pretty weird, and each is the first observed crystal with their structure (hence they are the 'prototype' in crystallography terms).

$\alpha$-Mn was described by A.J. Bradley and J. Thewlis in 'The Crystal Structure of $\alpha$-Manganese', Proc. Royal Society 115, 456-471 (1927). The unit cell is based on body-centered cubic, but contains 58 atoms representing 4 distinct positions. It can be thought of as a bcc Bravais lattice with a 29 atom basis. Mn is the only element that exhibits this crystal structure.

$\beta$-Mn was described by G.D. Preston in Phil. Mag. 5(33) 1207-1212 (1928). The unit cell is simple cubic, containing 20 atoms in 2 groups. Preston's unit cell drawing is significantly worse than that of Bradley and Thewlis, although representations of the unit cell can be easily found on the web these days. Again, Mn is the only element that exhibits this crystal structure.

The $\delta$-Mn and $\gamma$-Mn are normal, boring bcc and fcc crystals, so no need to point to references for them.

I would provide links to the papers if I could - my institute does not have direct access to them.

As yet further info on the crystal structure references.  The A12, A13, ... notation is the Strukturebericht notation used by crystallographers. Unfortunately, there are a number of other notations used as well, including Schonenflies and Hermann-Mauguin. The Strukturebericht is pretty basic, mostly a catalog of crystal types, while the others convey the actual symmetries of the crystal structure. Some of this confusion of notations seems to date from the early days of crystallography when it was done in many places and languages. There are seven crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups to be described to some level by the various notations. For this answer I stuck to the Strukturebericht because that is the notation used in the Dinsdale reference on Gibbs free energies of the elemental phases.


How much worse is methane than CO2 as greenhouse gas? I can't find a coherent answer to this question. The numbers are all over the place.

25 times greater
84 times more potent

The reason I'm looking is because I can't find a reason in physics why this would be so*:

  There is nothing “special” about methane that makes it much powerful than $\ce{CO2}$ (link to source)

What makes $\ce{CH4}$ absorb more infra-red than $\ce{CO2}$? What experiments have shown this?

Methane absorbs infra-red at different bandwidths than $\ce{CO2}$ (see here). Does the earth radiate more heat in those bandwidths?

Let's take a look at this graph just so we have a reference.

This shows the IR absorption bands of various gases found in the atmosphere. For this answer, let's focus in on the absorption bands of $\ce{H2O}, {CO_2}$, and $\ce{CH_4}$.

First, the answer to why methane is considered a worse greenhouse gas than carbon dioxide is largely because methane absorbs in regions where carbon dioxide does not absorb at all. That is, methane absorbs different wavelengths of light than carbon dioxide.

Another thing to note is that methane's absorption peaks overlap with some of the absorption peaks of water vapor. This is an interesting thing to note because the strength of methane's absorptions in those overlapping regions are much stronger than the absorptions of water.

So, we have two things which contribute to why methane is sometimes called a worse greenhouse gas than carbon dioxide. First, it absorbs in areas where $\ce{CO2}$ does not absorb. Second, these absorptions are very strong.

Something you might be interested in, however, is the fact that there are some people who make the argument that methane essentially doesn't matter as a greenhouse gas. This would be due to the overlap in absorption between water vapor and methane. For instance, methane has an average concentration of 1.7 ppm (parts per million) in earth's atmosphere. Compare this with the fact that the earth's atmosphere can be composed of 1% water vapor at sea level and .4% across the whole atmosphere on average.

Thus, the amount of water vapor in the atmosphere is much more than that of methane at any given time. So, it can be argued that we do not need care much about methane as a greenhouse has because water vapor will do all the IR absorption that methane would have done.

I don't know enough about atmospheric chemistry to really say if I agree or disagree with those claims. One thing I do know, however, which is interesting is the fact that the idea of global warming would seem to present more of a problem with water vapor than methane or carbon dioxide because as temperature increases, the ability of the atmosphere to hold water vapor increases which would accelerate the warming from water vapor, which would increase the amount of water vapor the atmosphere can hold...

Also, I believe that the metric which says methane is more dangerous than carbon dioxide is a comparison which does not account for other things which might absorb that light. In other words, the figure you quote do not account for the fact that water vapor might make methane irrelevant as a greenhouse gas.

Interesting Read: http://www.nasa.gov/topics/earth/features/vapor_warming.html


How is drift due to static electricity eliminated from an analytical balance? I've recently encountered a problem when using an analytical (0.1 mg accuracy) scale to determine the mass of a sample of ferrocene powder. The reading in the balance seems to drift either upwards or downwards when I drop some of the powder in an erlenmeyer flask. The reading stabilizes when I add water to form a solution.

I've heard that static electricity might cause this kinds of problems when measuring powders so I suspect this is the case here, because the problem only appears when I add the ferrocene powder. I've try reaserching some ways to get rid of the problem, a static gun seems to work in some cases (no idea if it would work with a powder though) and I've heard that air ionisers also work, but don't have access to either of these.

Although this is the first time I've encountered this situation, I've heard its pretty common, so what do people usually do to get rid of the problem in a cheap manner? Is there an easy solution?

I'm using gloves to operate the scale, which I've been told to do in order to keep the flasks clean of oil from my fingers. Can the gloves be worsening the problem? What can I do about that?

Static charge can affect the measurements made by an analytical balance, as can magnetic materials because the mechanism in modern analytical balances involves a feedback loop between the load that is measured and an electromagnet. Insulating weighing vessels (plastic or glass) tend to accumulate static charge, especially when handled with another insulator (gloves) by the triboelectric effect.

You should first make sure that the body of the balance is grounded properly according to the manual for your model.

As a short-term solution to static charged glass, you might try arranging the sample so that it is father away from the balance. One way might be to place the sample on top of an empty sample bottle on the balance. As you mentioned, a more robust solution for discharging static build-up is to have a commercially available air ionizer nearby. The device will either produce an equal amount of positive and negative ions in the air at high voltage AC or DC electrodes, or with an ionizing radioactive material such as is found in anti-static brushes used for cleaning vinyl audio records.

If you are frequently measuring magnetic materials, many analytical balances can be configured in a bottom-loading arrangement. In this setup a platform is hooked beneath the balance, and the sample is supported sufficiently far away from the force-restoring motor so that it does not influence the measurement.


What is Markovnikov's rule? What is Markovnikov's rule? How can we use it to predict the products of additions to alkenes? And is it always obeyed, or are there exceptions to this rule?

Markovnikov's rule is used to predict how (for example) HX, where X is a halide, adds to an unsymmetrical olefin.  As the diagram below shows, Markovnikov addition follows the rule that the "H" will add to the least substituted carbon while the "X" will add to the most substituted carbon.  This is due to the fact that these reactions proceed by protonaton of the olefin to yield a carbocation.  Protonation at the least substituted carbon yields the most stable carbocation (in this case a secondary carbocation).  Finally the carbocation reacts with $\ce{X^-}$ to yield the final product.

In some cases the Markovnikov rule is not followed.  Such additions to olefins are termed "anti-Markovnikov".  These reactions typically proceed through a radical intermediate rather than a carbocation.  For example, addition to the olefin might be initiated by the addition of a bromine radical to generate the most stable radical intermediate, which would place the radical center at the most highly substituted carbon. Addition of a hydrogen radical completes the process and generates the final product which has the reverse addition pattern as compared to the Markovnikov addition product.


Are there any free phosphorus-31 NMR spectral libraries? I know that there are some sites like Sigma Aldrich and AIST for $\ce{^1H}$ NMR and $\ce{^13C}$ NMR, but was wondering if there were any free sites to compare $\ce{^31P}$ NMR spectra?

There are quite a few it seems, a couple of examples are below:

University of Wisconsin Phosphorus NMR data, starts with a shifts overview.

Stanford's Magnetic Resonance Laboratory site P NMR at the SMRL, provides comparisons and a protocol for testing.

A general NMR site is NMR Wiki - has many links to many sources.