What is the molecular interpretation of Raoult's law? I was reading about Raoult's law which states:

It states that the partial vapour pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Regarding the molecular origin of this law, Peter Atkins in his book writes:

The molecular origin of Raoult's law is the effect of the solute on the entropy of the solution. In the pure solvent, the molecules have a certain disorder & a corresponding entropy; the vapour pressure then represents the tendency of the system & its surroundings to reach a higher entropy.
[...] Because the entropy of the solution is higher than the pure solvent, the solution has a lower tendency to acquire an even high entropy by the solvent vaporising. In other words, the vapour pressure of the solvent in the solution is lower than that of the pure solvent.

I've not understood his explanation especially the bold lines above.
$\bullet$ Firstly, how could vapour pressure represent the tendency of the system & its surroundings to reach a higher entropy?
$\bullet$ Why does the solution has a lower tendency to acquire an even high entropy by the solvent vaporising ? After-all every system has greater tendency to acquire high entropy, isn't it? Is not this statement contradictory to the Second law by saying lower tendency to acquire high entropy?
Could anyone please explain that to me?

I believe what the quote is saying is that once equilibrium is reached between the vapor-liquid mixture, then there will be no tendency to increase entropy. This is because entropy is not the only factor to consider. In any system, one must always consider the interplay between both entropy and energy. For instance, at equilibrium, $$\Delta G=\Delta H-T\Delta S=0$$ which means that at equilibrium, $$\Delta H=T\Delta S$$So, while you raise a good point that in most cases, a system will tend to maximize entropy, in a closed system, one must consider the interplay between entropy and energy which in this case are exactly equal (technically they differ by a scaling factor which is the absolute temperature).

Additionally, what is meant by "vapour pressure represents the tendency of the system and its surroundings to reach a higher entropy" is that one can make an inference about the energetic and entropic properties of a solution based on the vapor pressure of that solution. For instance, if a system has a very low vapor pressure (consider a solid as an extreme example) then the solution is very energetically stable and thus it has a very low tendency to reach higher entropy. But, if you think about acetone, which has a vapor pressure of $30.6 ~\text{kPa}$ at $25^{\circ}~\text{C}$ (ten times that of water at the same temperature), one could say that this then has a strong tendency to increase its own entropy and the entropy of the surroundings.


Galvanic cell involving a hydrogen electrode and copper electrode in sodium hydroxide A galvanic cell consists of a standard hydrogen electrode and a copper
  electrode.  Suppose that the copper electrode is immersed in a
  solution that is $0.100\ \mathrm{M}$ in $\ce{NaOH}$ and that is saturated
  with $\ce{Cu(OH)2}$. Find cell potential

  $\begin{alignat}{2}\ce{2 H+ (aq) -> H2 (g)}\quad E_0 = 0.00\ \mathrm{V}\\
\ce{Cu^2+ (aq) + 2 e- -> Cu (s)}\quad E_0 = 0.34\ \mathrm{V}\end{alignat}$

  $K_\text{sp} = 1.6 \times 10^{-19}$ for $\ce{Cu(OH)2}$

I started by doing writing off the Nernst equation and realized I do not have $\ce{Cu^2+}$ concentration, and since I’m given $K_\text{sp}$ of copper, I assumed I should use it to find copper (after doing ice table for $\ce{Cu(OH)2 -> Cu^2+ + 2 OH-}$). But then I stumbled on $0.100\ \mathrm{M}$ of $\ce{NaOH}$, what should I do with it? and what does saturated mean in the given context of the problem?

In this case, "saturated" means that there is enough dissolved copper (II) hydroxide such that no more will dissolve, i.e. the equilibrium represented by the $K_\mathrm{sp}$, $\ce{Cu(OH)2_{\ (s)}<=>Cu^2+_{(aq)} + 2OH-_{(aq)}}$ holds true. So first, let's figure out the concentration of copper (II) hydroxide that will dissolve in 0.1 M NaOH (making the assumption, as mentioned in the comments, that $[\ce{OH^-}] = \mathrm{C_\ce{NaOH}})$:

The $K_\mathrm{sp}$ is really really small, so we can actually ignore
its contribution to the $[\ce{OH^-}]$, but for fun, you can do the
whole ICE table thing and end up with:
$$\begin{align}
     [\ce{OH^-}] &= .1 + 2x \\
     [\ce{Cu^2+}] &= x \\
     K_\mathrm{sp} &= [\ce{OH^-}]^2[\ce{Cu^2+}] \\
     1.6\times 10^{-19}  &= (.1 + 2x)^2(x) \\
     x = [\ce{Cu^2+}] &= 1.6 \times 10^{-17}\ \mathrm{M}
     \end{align}
   $$

Next, using the Nernst equation, we can calculate the copper half-cell reduction potential (Assuming T = 298.15 K):

$$\begin{align}
    E &= E^\circ - \frac{RT}{zF}\ln\frac{1}{[\ce{Cu^2+}]} \\
      &= 0.34\ \mathrm{V} - \frac{RT}{2F}\ln\frac{1}{1.6 \times 10^{-17}} \\
      &= -0.16\ \mathrm{V}
    \end{align}
    $$

Finally, the whole cell potential:

First, negate the calculated reduction potential to get the oxidation potential:
$$\begin{align}
    E_\mathrm{ox} &= -E_\mathrm{red} \\
      &= 0.16\ \mathrm{V}
    \end{align}
    $$
Then, because the SHE's reduction potential is defined as 0 V, the whole cell potential is the same as the oxidation half-cell potential:

$$\begin{align}
    E_\mathrm{ox} &= E_{(\ce{H})\mathrm{red}} + E_{(\ce{Cu})\mathrm{ox}} \\
                  &= 0\ \mathrm{V} + 0.16\ \mathrm{V} \\
                  &= 0.16\ \mathrm{V}
    \end{align}
    $$


Is it possible to have negative bond orders? In the MO theory, is it possible to have a molecule with bond order less than 0? The least bond order I have come across as of now is 0 (for di-atomic noble gases), indicating that the molecule cannot exist and the atoms get pulled apart.

Is it possible to create a scenario (at least theoretically) where we deal with a negative bond order by manipulating the system's conditions or any of that sort?

No, it's not possible, theoretically either.

If you see the Molecular orbital theory, maximum number of anti bonding electrons = number of bonding electrons.

And bond order is (Number of bonding electrons - number of anti bonding electron)/2

So, the minimum value that you can get is 0 and not negative, as anti bonding electrons can't exceed the number of bonding electrons.


How to determine the reactivity of a set of compounds that may undergo nucleophilic substitution? Which of the following compounds is most reactive under $\mathrm{S_N1}$ conditions?

  Arrange the following compounds in order of decreasing rate of $\mathrm{S_N1}$ hydrolysis:

  A. II > III > IV > I
  B. IV > III > II > I
  C. III > IV > II > I
  D. I > II > III > IV

Both these questions are based upon the rate of reactivity of $\mathrm{S_N1}$ reaction.

1) In this I thought the answer should be B because the rate of $\mathrm{S_N1}$ reaction depends on the ease of formation of carbocation. Being more electronegative, the electron density between carbon and chlorine bond is quite shifted towards chlorine, so it should be easier for chlorine to leave, hence forming the carbocation. Options A, B, and C allow the carbocation to be resonance stabilized so D cannot be the answer.

But the answer given is A.

Where am I wrong with this logic? I guess it might be because of more polarizing power of iodine, if it is this, then how are we supposed to figure out which one will be the dominating factor?

2) I think we check this on the basis of carbocation stability as well. Carbocation will be formed as $\ce{Br}$ leaves. Alkyl chain is electron releasing, so it must increase the electron density on the benzene ring which should be better for the stability of carbocation (as there is more electron density on the benzene ring to stabilize the positive charge). Hence the answer should be I > II > III > IV

But the answer is A.

Where am I wrong with my logic in both?

In question 1), you are wrong about the tendency of leaving groups. While you are correct that the carbon-chlorine bond is more polarised towards chlorine, iodide is still a better leaving group. This is because it is much larger and can thus stabilise the resulting negative charge much better.

You can observe this in $\mathrm{S_N2}$ reactions, where iodides are much more reactive than bromides which are more reactive than chlorides (and fluorides are basically unreactive) even though the same polarisation argument could have applied.

In question 2), the stabilisation that alkyl groups give carbocations, or the electron density they donate into a phenylic ring actually comes from $\ce{C-H}$ bonds. Thus, the para-xylene derivative can stabilise a cation better since there are three $\ce{C-H}$ bonds it can use for stabilisation. No matter which way the methyl is oriented, it can stabilise. para-ethylbenzyl bromide can only be stabilised in two out of three orientations (when a $\ce{C-H}$ bond is orthogonal to the phenyl ring) and para-isopropylbenzyl bromide only in one. Thus, the rate of hydrolysis is accordingly.


What is difference between aluminium mercury alloy and gallium aluminium alloy? Aluminum mercury alloy is used as a chemical reagent to reduce compounds, such as the reduction of imines to amines,in this case mercury destroy aluminium oxide layer which protect aluminium to react with water.

Alloy of aluminium and gallium which similar to above alloy make the aluminium more reactive by preventing it from forming an oxide layer.

In both cases mercury and gallium plays a same role of destroying oxide layer on aluminium to react with water but why can't gallium aluminium alloy reduce imines to amines like mercury aluminium alloy ?

We can rationalize the differences in the reactivity as follows.  Gallium has a higher ionization energy compared to mercury.  During a redox reaction the metal is oxidized and it loses an electron, therefore how "easy" it is for a metal to lose its electon is iimportant. Since gallium is harder to inonze than mercurey, it implies that it is "easier" for mercury to lose its electron compared to Gallium, making mercury a better stronger reducing agent.

I may come back and add some quantitative information later, but i'm going to go eat now. I think looking in a table of standard reduction potentials would be reasonable.


Why is brine solution more effective than granular salt in melting ice on a surface? Why is a brine solution consisting of a small fraction of salt in water, more effective in melting ice on a surface, than just pure salt?

This is because the brine solution is already in solution, whereas granular salt has to dissolve first. If ice is present instead of water, this is difficult, as there is no liquid water for the salt to dissolve in.
Secondly, salt dissolution is an endothermic process, so it will further reduce the temperature of the surroundings. For example, if you have an ice and water solution and add salt crystals you can bring the temperature of the ice/water combination to a much lower temperature (even below $\pu{-10^\circ C}$!).


Free radicals vs Hund's rule contradiction? How do we reconcile these two statements?

Formation of electron pairs is energetically favorable.  Therefore, free radicals are particularly reactive
Hund's rule - it takes energy to pair electrons, so orbitals are typically filled singly first

The first statement in the question is not correct. Formation of electron pairs is not favorable. That would lead to putting two electrons in the same orbital and there is some effect from charge-charge repulsion.

Electrons end up paired in bonding or stabilized orbitals because there is space for a second electron, and the overall stabilization that would result from putting a second electron in is greater than the destabilization of getting that electron from somewhere else.


Why are higher acids solid while lower acids are liquid? I was just going through acids on Wikipedia in my free time, and noticed this neat trend:

Formic acid, Ethanoic acid, uptil Nonanoic acid, are all liquids. The first few are colourless, while the latter are yellowish and also "oily" (more viscous).
Decanoic acid and all its elder brothers are white crystals (or "powders")
Aromatic acids like benzoic acid or picric acid are solids.

I am unsure if there could be a reasonable logic for the colours. But, I am sure there could be a good logic for the physical state of these acids at room temperature.

To me, all these acids have hydrogen bonding common, so we can rule out that factor. Apparently, the physical state seems to be related to the "size" of the acid. The heavier acids  are solids, while the lower ones are liquid. My question is: why is it so?

PS: List of all acids at one place is at the bottom of this Wikipedia page.

Acids with a larger size have greater intermolecular forces than 'smaller' acids, meaning it would take more heat energy to break those bonds in larger acids than in smaller ones. Hence at room temperature the heavier acids would be solid.


How does one separate ethanol from acetic acid using aqueous extraction? Would you add in $\ce{NaOH}$, deprotonating the acid, and then wash the mixture in water, thereby removing the $\ce{Na^+}$ and $\ce{CH3O^-}$ ions in an aqueous layer? But then wouldn't the ethanol be dissolved in the water as well, as it is polar? In general, how do you separate two polar substances from each other using extraction?

If you don't use extraction, how else would you separate the two, aside from distillation?

All answers are appreciated.

Well, distillation is almost certainly the best method. However, since you've explicitly ruled that out, there are only a few alternatives I can think of:

Ethanol can be separated from water with reasonable efficiency (at least on the microscale) by a process of "salting out." The ethanol/water mixture is saturated with $\ce{NaCl}$. The solvation of the salt ions is a highly thermodynamically favorable process, and as that occurs the molecules of water become increasingly less able and available to form hydrogen bonds with ethanol. I've never tried this with acetic acid specifically, but I can't think of any reason why it wouldn't work.
I would imagine with the right choice of eluent, separation by column chromatography using silica or alumina should be feasible.
Fractional freezing may be possible in theory, although I have no idea as to its efficacy for this specific case.

I'm also assuming here that you're looking strictly for methods of physical separation, not chemical (so, strictly as a general example: adding something like calcium oxide to consume water in the generation of insoluble calcium hydroxide would be out of the question).


Proof of Markovnikov's Rule Is there a proof for Markovnikov's Rule? Specifically, why does the H bond to the carbon with the most hydrogen atoms already bonded to it?

In most reactions that follow markovnikov's rule, a carbocation is formed and rearranged such that it is stabilised to the maximum extent .

The stability usually comes from the inductive effect, that is the electron releasing methyl groups that lead to a better charge distributed system. More is the number of such methyl groups surrounding the carbocation, more will be the stability and more such intermediates will be formed. These stable intermediates will be attacked by the negative reagent. This justifies the Markonikov's rule (at least in cases where carbocations are formed).

Note: This only applies to reactions such as acid catalysed hydration, action of halogen acids on alkenes, etc. In some reactions such as oxymercuration-demercuration no carbocation is formed but it still follows Markonikov's rule and uses a different reaction mechanism.


How to dissolve composite In2O3 with SnO2 without affecting stainless steel? I found this article, but it was not helpful. The $\ce{In2O3}$ is probably amorphous. $\ce{HNO3}$ and $\ce{HCl}$ do not work. Any ideas?

In case your question is about the recovery of indium from ITO, the following might be helpful:

In Recovery of indium from indium tin oxide by solvent extraction, published in Hydrometallurgy, 2011, 107, 56-61 (DOI), the abstract seems to suggest that $\ce{H2SO4}$ is more suitable than $\ce{HNO3}$. Furthermore, a solvent extraction scheme for indium, using di-(2-ethylhexyl)phosphoric acid (DEHPA) is proposed.

In Recovery of indium from used indium–tin oxide (ITO) targets, published in Hydrometallurgy, 2011, 105, 207-212 (DOI), the authors agree on using $\ce{H2SO4}$ (at 90 °C) and decrease the ymount of tin in solution by sulfide precipitation.


Formulas of synthetic polymers I have a question regarding synthetics. $\ce{(C8H8)}_n$ is the formula for polystyrene. What exactly does the $n$ stand for?

Thanks

n is the number of molecular units in a chain, also called the degree of polymerization. This varies based on the molecular weight of the polymer. The molecular weight is the degree of polymerization times the molecular weight of a monomer.


How do stoichiometric coefficients effect the half-life of a reaction? $$\ce{2A -> 3B + C + 2D}$$

This is a first order reaction. Rate constant $k$ for reaction is given to be (say) $\pu{1.386E-2 min^{-1}}$. From this we get the value of $t_{\frac{1}{2}}$ of reaction as $\frac{\ln(2)}{k} = \pu{50 min}$. Now this $t_{\frac{1}{2}}$ should mean half life time of reaction (and not for $\ce{A}$). However, in my textbook it is written that if concentration of $\ce{A}$ is $\pu{200 M}$ at $t=0$ then at $t = \pu{100 min}$ the conc. of $A$ would be $\pu{50 M}$ as $100$ minutes is two times of the half life (and thus the conc. of $A$ should reduce by $2^2$ times).

I think this is wrong as the half life of $A$ is half of the half-life of reaction (since rate of consumption of $A$ is double the rate of reaction as can be seen from the stoichiometric coefficients). According to me after $100$ minutes the concentration of $A$ should be $\pu{25 M}$. Is my logic correct? Or am I missing something?

Consider what happens to your argument if I rewrite the equation in the following equivalent form:
$$\ce{A -> \frac{3}{2}B +\frac{1}{2}C +D}$$
That is, you write

  ... since rate of consumption of A is double the rate of reaction as can be seen from the stoichiometric coefficients

In contrast, the rate of the reaction, in this case, is defined to be the rate of decay of $\ce{A}$,
$$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$
When you look at it like this, it is more clear that the stoichiometry should not have an effect on the decay process as you claimed.

Furthermore, if we integrate the equation above after separating by variables,
$$\int_{[\ce{A}_0]}^{[\ce{A_t}]}\frac{1}{[\ce{A}]}\mathrm{d}[\ce{A}] = \int_0^t-k\mathrm{d}t$$
we get,
$$\ln\left(\frac{[\ce{A_t}]}{[\ce{A}_0]}\right)=-kt.$$

If we try to find the time at which the original concentration has halved, we get the formula you have written above which is,
$$t_{\frac12}=\frac{\ln(2)}{k}.$$

The important point here, and this isn't true for other reaction-orders, is that the half-life is independent of initial concentration. This is not what one would intuitively expect, and indeed it's not true in general, but it is characteristic of all exponential-decay processes.

Indeed, this is a very deep result which doesn't tend to get emphasized enough in my opinion.

It is true that sometimes you can say that a reaction will be second order based on stoichiometric coefficients, but you are told this is first order, so that is probably the source of your confusion.

So, your textbook is right.


Determine mechanism: E1 vs. E2 I am asked to indicate what is the product of this reaction, and the mechanism by which it is formed:
3-Bromo-3-ethylpentane + sodium hydroxide in methanol, 50 ºC
I identify it as an elimination reaction. I would say that it is a bimolecular elimination, E2, because a tertiary alkyl halide and a strong base (NaOH) are used. However, my teacher told me that it was unimolecular, E1, and I don't understand why.

3-Bromo-3-ethylpentane is so bulky that an E2 elimination seems to be unlikely as the $\ce{OH}^{-}$, despite being a strong base, has a very low chance of directly attacking the beta $\ce{H}$. Moreover, proceeding by the E1 mechanism, we have a tertiary carbocation which is quite stable compared to primary and secondary carbocations.
However, as @Zhe has pointed out, looking at the experimental data gives the best answer. A general rule of thumb (although exceptions may exist) is that tertiary alkyl halides tend to follow E1 mechanism rather than E2 even under strong bases.


How to pronounce "potassium manganate(VII)"? The formal names for such inorganic compounds having metals showing of variable oxidation states involve use of Roman numerals, for example "copper(II) sulphate" or "potassium manganate(VII)" or "hexaaquacopper(II)". Of course, common names like "copper sulphate" or "potassium permanganate" do exist, but if I wanted to use the formal ones, how do I correctly pronounce them?

Simple: Just read the roman numerals as a number.

Copper-two sulphate
Potassium manganate-seven
Hexaaquacopper-two.

For negative oxidation numbers use a minus:

Tetracarbonylferrate-minus-two ($\ce{[Fe(CO)4]^2-}$)


What is the parent chain in a hydrocarbon with two different rings? There is cyclopentyl attached to one terminal carbon atom and cyclohexyl attached to the other terminal carbon atom and there are three carbon atoms linking them. Since the cyclohexyl ring has the greatest number of carbon atoms, it should be the parent chain. So the name should be 3-cyclopentylpropyl cyclohexane.
But my book says that the middle propyl chain should be the parent chain, the cyclohexyl and cyclopentyl groups should be the substituents, and the correct answer is 1-cyclohexyl-3-cyclopently propane.
What is the correct name. And if my book is correct, why is it so?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

So far, we have three possible names:

Method (1): “(3-cyclopentylpropyl)cyclohexane” or “(3-cyclohexylpropyl)cyclopentane” (ring preferred to chain)

Method (2): “1-cyclohexyl-3-cyclopentylpropane” (chain preferred to ring)

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

This leaves us with two possible preferred IUPAC names:

“(3-cyclopentylpropyl)cyclohexane” or “(3-cyclohexylpropyl)cyclopentane”

  P-44.1.3 Seniority order only for rings and ring systems. Criteria that apply only when the choice for parent structure is between two or more rings or ring systems are given in P-44.2.

  P-44.2 SENIORITY ORDER ONLY FOR RINGS AND RING SYSTEMS

  P-44.2.1 If application of P-44.1 does not effect a choice, general criteria for determining ring seniority given below are applied successively until there are no alternatives remaining. (…)
  The senior ring or ring system:
  (…)
  (e) has the greater number of skeletal atoms;
  (…)

Therefore, the preferred IUPAC name is “(3-cyclopentylpropyl)cyclohexane” (not “(3-cyclohexylpropyl)cyclopentane”) since cyclohexane has more skeletal atoms than cyclopentane.


Nucleophile in a SN2 reaction In a $\mathrm{S_N2}$ reaction, a good nucleophile is considered to be willing to donate electrons. So how do we generalize this fact across the periodic table? Does that mean the more electronegative an atom is, the better the nucleophile?

As electronegativity increases the holding capacity of electron by the atom increases. So it is less likely to donate electrons. So we can generalise that access a period in periodic table nucleophilicity decreases.


Does caesium or francium have a lower ionization energy? $\ce{Cs}$ and $\ce{Fr}$ are in the same group of lowest first ionization energy, but which element has the lowest ionization energy, and why?

which element has the lowest ionization energy?

The ionization energy for Cs is lower than that of Fr.

  why?

The Wikipedia article on Francium states:

  Francium has a slightly higher ionization energy than caesium, 392.811(4) kJ/mol as opposed to 375.7041(2) kJ/mol for caesium, as would be expected from relativistic effects, and this would imply that caesium is the less electronegative of the two.

That is, relativistic effects are more pronounced for francium than for cesium. A greater positive charge density in the nucleus for the larger atom (and higher atomic number) means higher velocity for (here) the 7s electron, which has a higher relativistic mass and thus stays closer to the nucleus than in the case of Cs (where we're dealing with the 6s electron).

The relativistic effect manifests in a greater effective electronic mass which in turn decreases the effective atomic radius, which makes it harder to pull off that electron (hence the higher ionization energy). See this Wikipedia article for insights on atoms with a high atomic number and associated electrons with a low principal quantum number, my source for almost all of the answer contained in this post.


Is there also a leveling effect for weak acids in water? The leveling effect says that any acid stronger than the hydronium ion ($\ce{H3O+}$) will completely dissociate in water. Hence no differentiation of acid strength is possible between eg HBr and HCl in water.

I was wondering why no one ever talks about the leveling effect for acids weaker than water. Isn't it so that these acids would not dissociate at all and hence no differentiation is possible on this side either?

Also, what is the reasoning behind the fact that acids stronger than the hydronium ion will completely dissociate in water? Why can't it be an equilibrium (say if it's only slightly stronger than hydronium)?

To answer your first question, weak acids don't get levelled in water. Each solvent has their acid-base equilibrium window, and water has one that is between 0-14, meaning that any solvent with a pKa within that range can be distinguished in water. There are other solvents with a ore narrow acid-base discrimination Window, such as methanol can acid, with a range between -10 to about 3, which will be unable to discriminate between weak acids. The reasoning behind the acid-base discrimination window is a bit long winded (and maybe a little confusing), so I will try to break it down as much as I can.

Say you have an acid in water, then the pKa + pKb = pKw
This is true for any acid or base in water. If pKa<0 or pKb<0 then then most of the compound will be dissociated in water, as an acid or base, respectively. This is undesirable, especially if the pKa/pKb<<0. As you have noted in your comments, when the pKa of a compound is less than 2 units compared to water, then in water only 1% of it will remain in the undissociated form. If pKa>pKw, you will obtain a negative pKb which will behave more like a base in water. It follows that compounds with a 0< pKa< pKw  will fulfill the condition where both pKa and pKb> 0. Now guess what pKw is? 14. Hopefully this might explain to you why water is able to distinguish between weak acids with a pKa between 0 to 14.


Finding pH of a buffer solution on addition of hydrogen iodide to it Question:

  A buffer with $\mathrm{pH} = 4.88$ was prepared by dissolving $\pu{0.10 mol}$ of benzoic acid ($K_\mathrm{a} = \pu{6.3E-5}$) and $\pu{0.50 mol}$ of sodium benzoate in sufficient pure water to form $\pu{1.00 L}$ solution. To a $\pu{70.0  mL}$ aliquot of this solution was added $\pu{2.00 mL}$ of $\pu{2.00 M}$ $\ce{HI}$ solution. What was the $\mathrm{pH}$ of the new $\pu{72.0 mL}$ solution?

My solution:

Using the Henderson–Hasselbalch equation

$$4.88 = -\log(6.3 \cdot 10^{-5}) + \log \left( \frac{0.10-x}{x} \right),$$

$$x = \pu{0.0173 mol}$$

Amount of conjugate base in $\pu{70.0 mL}$ is

$$0.0173 \times 70/1000 = \pu{1.211e-3 mol}$$

Amount of $\ce{HI}$ added is $\pu{4e-3 mol}$.

I'm stuck here having more $\ce{H+}$ than my base.

Amount of acid added:
$$0.002\ \mathrm L \times 0.2\ \mathrm{mol/L} = 0.004\ \mathrm{mol}$$

This will neutralize $0.004\ \mathrm{mol}$ of the benzoate: $$0.50\ \mathrm{mol} - 0.004\ \mathrm{mol} = 0.496\ \mathrm{mol}$$ of benzoate

This will increase benzoic acid by $0.004\ \mathrm{mol}$: $$0.10\ \mathrm{mol} + 0.004\ \mathrm{mol} = 0.104\ \mathrm{mol}$$ of benzoic acid

$$[\ce{H+}] = K_\mathrm a\left(\frac{n(\ce{HA})}{n(\ce{A-})}\right)$$ (yes, it's supposed to be $[\ce{HA}]/[\ce{A-}]$, but volume cancels out leaving amount of substance so long as your numbers leave you inside the titration curve between the starting $\mathrm{pH}$ and the equivalence point $\mathrm{pH}$)

$$[\ce{H+}] = 6.3 \times 10^{-5} \left(\frac{0.104\ \mathrm{mol}}{0.496\ \mathrm{mol}}\right) = 1.32 \times 10^{-5}$$

$$-\log\left(1.32 \times 10^{-5}\right) = \mathrm{pH} = 4.88$$
(actually $4.8794$, so you have a small change in $\mathrm{pH}$, as you would expect with a buffer with so much of the conjugate base in solution, but not enough to make a significant difference.)

I'm not at all clear why people are married to the Henderson–Hasselbalch equation for buffer problems. The equilibrium expression covers just about any situation I can think of, there's no formula to remember if you know how to do an equilibrium expression.


Determine the mole fraction Under certain conditions, the well-known fertilizer $\ce{NH4NO3 (s)}$ explodes according to the following balanced reaction:

$\ce{2NH4NO3 (s) -> 2N2 (g) + 4H2O (g) + O2 (g)}$

Exactly 8 g of $\ce{NH4NO3}$ is placed in a 1.18 L constant-volume container containing dry air at 56.01 kPa and 15 ºC. The explosive reaction goes to completion and the temperature rises to 527 ºC. The container remains intact after the explosion. Determine the mole fraction of N2 in the gas phase after the explosion.

Data: Assume that the composition of dry air is 79% N2 and 21% O2 by volume.

So I know that $\ce{N2}$ takes 0.9322 L of volume. I calculated it by 0.79*1.18 = 0.09322

The only thing that I don't know is the number of moles, so do I just use the formula of ideal gas law which is $PV = nRT$,
 P = 0.5528 atm( 56.01kPA) V = 0.09322

However, which temperature should I use?
And am I using the right approach for this problem?

You're asked to find the mole fraction of $\ce{N2}$ after the reaction. That is defined by:

$$x_{\ce{N_2},\text{f}} = \frac{n_{\ce{N_2},\text{f}}}{n_{\text{tot},\text{f}}}$$

where $x$ indicates a mole fraction, $n$ the number of moles and the subscript $\text{f}$ means "final". Furthermore, since the container is sealed, no gases can escape the container. This means that:

$$\begin{align}
n_{\ce{N_2},\text{f}} &= n_{\ce{N_2},\text{i}} + \Delta n_{\ce{N_2}} \\
n_{\text{tot},\text{f}} &= n_{\text{tot},\text{i}} + \Delta n_{\text{tot}}
\end{align}$$

where the symbol $\text{i}$ indicates "initial" and $\Delta n$ indicates the change in the number of moles caused by the decomposition of $\ce{NH4NO3}$. The four quantities on the right-hand side of these two equations are the quantities that you will have to find.

$n_{\text{tot},\text{i}}$ - This can be found by using the ideal gas law, $pV = nRT$ with the data given before the explosion. Substituting $p = 56.01\mathrm{~kPa}$, $V = 1.18\mathrm{~L}$, and $T = 286{~K}$, you will find that $n_{\text{tot},\text{i}} = 0.0276\mathrm{~mol}$.
$n_{\ce{N_2},\text{i}}$ - Since 79% of pre-explosion air is $\ce{N_2}$, $n_{\ce{N_2},\text{i}} = (0.79)(0.0276\mathrm{~mol}) = 0.0218\mathrm{~mol}$.
$\Delta n_{\ce{N_2}}$ - This is related stoichiometrically to the number of moles of $\ce{NH4NO3}$. In fact, it is numerically equal, as the coefficients of $\ce{NH4NO3}$ and $\ce{N2}$ are the same. The molar mass of $\ce{NH4NO3}$ is $80.04\mathrm{~g~mol^{-1}}$, so $\Delta n_{\ce{N_2}} = (8\mathrm{~g})/(80.04\mathrm{~g~mol^{-1}}) = 0.100\mathrm{~mol}$.
$\Delta n_{\text{tot}}$ - This is also related stoichiometrically to the number of moles of $\ce{NH4NO3}$. Each mole of $\ce{NH4NO3}$ leads to the formation of $(2+4+1)/2 = 3.5\mathrm{~mol}$ of gaseous products (observe that water will be gaseous after the explosion due to the rise in temperature). Therefore, $\Delta n_{\text{tot}} = (3.5)(0.100\mathrm{~mol}) = 0.35\mathrm{~mol}$.

Putting it all together, we have:

$$x_{\ce{N2},\text{f}} = \frac{0.0218 + 0.100}{0.0276 + 0.35} = 0.323$$

Note that the mole fraction is dimensionless. The exact final temperature is also of no use; its only purpose is to tell you that water will exist as a gas (and therefore contribute to the total number of moles). It could be any other number above $100^\circ\mathrm{C}$ and the final answer would be the same, as the mole fraction is independent of temperature.

I hope I didn't mess up the arithmetic.


How to determine the specific heat of aluminum? I have a question from my AP preparation book that I got wrong and do not understand the explanation:

  A student designs an experiment to determine the specific heat of aluminum. The student heats a piece of aluminum with a mass of $\pu{5.86 g}$ to various temperatures, then drops it into a calorimeter containing $\pu{25 mL}$ of water. The following data is gathered during one of the trials:

  Initial Temperature of $\ce{Al}$: 109.1
  Initial Temperature of $\ce{H2O}$: 23.2
  Final Temperature of $\ce{Al + H2O}$: 26.8

I correctly calculated the heat gained by the water ($\pu{376.2 J}$), and the specific heat of aluminum ($\pu{0.78 J/g*C}$). However, how do I calculate the enthalpy change for the cooling of aluminum in water? The book's procedure is as follows:

$$\frac{\pu{5.86g}~\ce{Al}}{\pu{26.98g}~\ce{Al}} = \pu{0.217 mol}~\ce{Al}$$

$$\frac{\pu{376 J}}{\pu{0.217 mol}} = \pu{1730 J//mol} = \pu{1.73 kJ//mol}$$

I see that one may calculate specific heat by dividing the heat gained by water by moles of aluminum and converting to kilo-joules per mole in this scenario. However, why is that? What is the reasoning behind this procedure? How would you phrase this procedure to apply to this general type of question?

If the water and aluminum are regarded as incompressible, then the internal energy is a function only of temperature.  In addition, the heat capacity at constant pressure is equal to the heat capacity at constant volume.  If the pressure is constant, then the change in enthalpy will be equal to the change in internal energy.  So the change in enthalpy of the aluminum is equal to its heat capacity times its temperature change.  The same goes for the water.  So the total change in internal energy and/or enthalpy for the combination of water and aluminum is zero.


Crystal structures of heaviest nuclides' compounds What is the heaviest element of the periodic table for which there is a compound whose crystal structure has been completely resolved by X-ray diffraction?

Since there were no further suggestions, I decided to use the references from my comment in an answer, which, I believe, is element 99, einsteinium.

Both single crystal x-ray diffraction (SCXRD) and powder x-ray diffraction (PXRD) have certain sample requirements/limitations:

Mass/amount

PXRD: lower limit is in micrograms region, ideally at least a couple of milligrams of polycrystalline sample;
SCXRD: a single crystal of a size of fractions of a millimeter.

Time

PXRD: solid polycrystalline sample must be homogeneous which sometimes requires prolonged synthesis time (minutes, hours) due to diffusion barrier (solid-solid, solid-liquid or solid-gas). A measurement itself on average only takes minutes;
SCXRD: in addition to the synthesis, growing a suitable single crystal for analysis normally takes days, maybe hours if you are lucky. A measurement itself on average takes hours, sometimes days.

Looking both at the end of periodic table of elements and these two constrains, I suspect it is unlikely that current techniques can be applied for transfermium elements and fermium itself as we neither have the necessary amounts nor time.

Among transuranium elements with $Z < 100$ plutonium $\ce{_{94}Pu}$ compounds appear to be of the greatest interest and probably the greatest number of published structures, too.
However, einsteinium $\ce{_{99}Es}$ exists as several relatively stable isotopes with half-lives of the magnitude of months and days and has a somewhat developed chemistry to speak of.

In 1968 Fujita et al. [1] having 3 μg of $\ce{^{253}Es}$ synthesized and reported crystal structures of einsteinium(III) chloride $\ce{EsCl3}$ and einsteinium(III) oxochloride $\ce{EsOCl},$ both determined via powder x-ray diffraction.
Reading the full text is encouraged, mainly to admire authors' ingenuity in overcoming technical difficulties (self-contamination of the isotope, film darkening by radiation, measurement of lattice parameters in temperature range from the room temperature to over 400 °C and so on) and overall concise writing style of the report.

$\ce{Es^3+}$ in both structures has C.N. 9 and coordinates in a single-capped square antiprism ligand environment:

Figure 1. Crystal structure of $\ce{EsCl3}$ (ICSD-109002). Color code: $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#B31FD4}{\Large\bullet}~\ce{Es}$.

Figure 2. Crystal structure of $\ce{EsOCl}$ (ICSD-109003). Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#B31FD4}{\Large\bullet}~\ce{Es}$.

A few years later Haire and Baybarz [2] determined crystal structure of einsteinium(III) oxide $\ce{Es2O3}$ via electron diffraction, but this is another story.

References

Fujita, D. K.; Cunningham, B. B.; Parsons, T. C. Crystal Structures and Lattice Parameters of Einsteinium Trichloride and Einsteinium Oxychloride. Inorganic and Nuclear Chemistry Letters 1969, 5 (4), 307–313. https://doi.org/10/bxzkjh. (PDF)
Haire, R. G.; Baybarz, R. D. Identification and Analysis of Einsteinium Sesquioxide by Electron Diffraction. Journal of Inorganic and Nuclear Chemistry 1973, 35 (2), 489–496. https://doi.org/10/bbfgp4.


What legal guidelines exist for completeness of a Materials Safety Data Sheet? I know that the answer to this question will vary by country, but some aspects must be similar amongst those countries that have authorities overseeing chemical and industrial hygiene. The Wikipedia article on the topic is not so good. My understanding for the US is that a MSDS must be correct and complete to the best knowledge of the preparer of the sheet.

My question is thus:

  Is there are legal requirement to the correctness and completeness (to the best of the preparer's knowledge) for a MSDS? If not, then why bother with MSDS at all?

This question was inspired by the commercial for this product from 3M. You can view the commercial by clicking on the videos link at the bottom of the site. Never mind how much nonsense their claim of "botanically pure ingredients" is, I was intrigued by the MSDS link at the top of the page. I wanted to know what the "botanically pure" substances were.

There are two versions of this product, with different scents. The MSDS for each is linked: Breeze and Lemongrass. Section 2: Ingredients lists the same four items for both products:

Water 65-80%
Cloth 20-30%
Sodium lauryl sulfate <3% (surfactant, not a naturally occurring substance)
Thyme Oil <0.1% (antibacterial, at least thymol is, and botanical in origin, but certainly not a pure substance)

No odorant is listed in the ingredients, but Section 3 clearly indicated they have different smells. 3M knows what odorant was added, and if legal requirements exist for completeness, then 3M is earning a finger-shaking and a slap on the wrist. So, again,

  Is there are legal requirement to the correctness and completeness (to the best of the preparer's knowledge) for a MSDS? If not, then why bother with MSDS at all?

OSHA's FAQ says that chemicals that make up less than 1% (0.1% if it is a carcinogen) of a product do not need to be listed on the MSDS. So it looks like 3M could be in compliance if the oderants are present in trace amounts (in that case, I have no idea why they feel the need to include the thyme oil).

The only relevant thing OSHA seems to say about consumer products specifically is that employers are not required to provide an MSDS for consumer products used in the workplace, as long as the consumer product is used in the same manner as a consumer would.

The Consumer Product Safety Commission specifically disclaims jurisdiction over chemical products, and the Chemical Safety Board doesn't appear to have anything related to consumer chemical products.


On diagonal terms in the Coulomb matrix I have seen many machine learning algorithms for prediction of quantum chemistry properties that use Coulomb matrix as their input. Coulomb matrix is defined as,

$$\boldsymbol{M}_{i j}^{\mathrm{Coulomb}}=\left\{\begin{array}{cc}{0.5 Z_{i}^{2.4}} & {\text { for } i=j} \\ {\frac{Z_{i} Z_{j}}{R_{i j}}} & {\text { for } i \neq j}\end{array}\right.,$$

where the diagonal terms:

  ... can be seen as the interaction of an atom with itself and are
  essentially a polynomial fit of the atomic energies to the nuclear
  charge.

I was trying to read more about these diagonal terms and how exactly this equation is fitted. Essentially, something that says: "... and from there we get $0.5Z^{2.4}$." If these energies can be extended to bonds as well, as the off-diagonal term in Coulomb matrix represents repulsion forces rather than potentials, any references would be appreciated as well.

In terms of how they got the relation for the diagonal elements, I believe it is relatively straightforward. Given a list of atomic energies (energy of atom relative to separated nucleus and electrons), one can try to fit these values to a function of nuclear charge. If you plot a few values, it suggests trying an exponential fit (I don't know why all the papers say polynomial fit, as this could not lead to a noninteger exponent). So, you can try to fit the values to a function of the form $aZ^b$ using, for example, least-squares regression. You should wind up with something pretty close to the equation given in your question, with some variation because I don't know what atomic energies they used to do the fit. (I'm also not sure why the function isn't negative since the atomic energies are negative. My best guess is the sign winds up not being important as a descriptor in the network).

As to why they choose to use these values along the diagonal, they describe the system in terms of the energies of the individual atoms, along with the energy of interactions between atoms given by the off-diagonal terms. The original Rupp et al. paper compared two Coulomb matrices by looking at the difference between their eigenvalues, with small differences implying similar molecules. These eigenvalues can be thought of as effective atomic energies within the molecule, that is, energies of the individual atoms with the influence of the surrounding atoms included. Similarly, the eigenvectors would describe fragments of the molecule or an effective atom which account for the influence of neighboring atoms.


Aromaticity of fused rings I understood how to find aromatic nature of single aromatic rings (like benzene) or that of combination of rings which are independently aromatic, such that the electrons of one ring do not enter into another ring (like biphenyl).
However I don't have much clarity on finding the aromatic nature if electrons from 2 or more rings can delocalise into each other.
Like biphenylene

(Image Source:pubchem)
If I only look at the central ring it seems to be antiaromatic. However there can be resonance throughout the entire molecule that can cause it to be more anti aromatic. Also the top and bottom rings might have aromatic character on resonance in individual rings too. These are just my guesses so I don't know what is the actual nature of the compound, but I didn't understand why this compound would be stable if it is associated with some antiaromatic character? Please elaborate on this aspect

Antiaromaticity doesn't exist in states that are observable. Its extended presence is a contradiction to its definition. Instead the double bond system exists as an independent system instead of being conjugated. This effect is known as the Jahn-Teller Distortion. A very famous example of said effect is cyclo-butadiene which exists in the form of a rectangular compound as explained in this answer
Proof that this is also the case in biphenylene is as follows:
Taken from the wikipedia page on biphenylene, (emphasis mine)

[...] Questions of bond alternation and ring currents have been investigated repeatedly. Both X-ray diffraction and electron diffraction studies show a considerable alternation of bond lengths, with the bridging bonds between the benzenoid rings having the unusually great length of $\pu{1.524 Å}$. The separation of the rings is also reflected by the absence of the transmission of NMR substituent effects through the central [4n] ring. However, more sensitive NMR evidence, and particularly the shifting of proton resonances to high field, does indicate the existence of electron delocalization in the central [4n] ring. [...]

The reference stated in the Wikipedia article is Tetrahedron 1964, 20 (5), 1179–1184, which states:

Taking into consideration the fact that the molecule is planar and that the two six-membered rings are known to be $\pu{1.52 A}$ apart and assuming that the charge density around the molecule is uniform, we calculated the chemical shifts between the ortho and meta hydrogens due only to ring currents in two cases, i.e., two ring currents (0.13 ppm) and three ring currents (0.34 ppm). The chemical shift for the two ring-current case is in good agreement with the experimentally determined value (0.105 ppm), but it should be borne in mind that ample physical chemical evidence exists to show that the six-membered rings are not entirely independent electronic systems.


I (might) have accidentally combined these mold-killers at home: Any adverse reactions? A trailer is a small enclosed space. About 2 months ago we had some mold and I applied some mold killer to a few spots that I could not remediate. Below I've listed 2 products I used. (I definitely used product #1, and it's possible I applied product #2 to the same 1 x 2 foot area.)

Air testing shows the mold is gone, but there's still something in the air that one can feel isn't good to breath in. It tickles your throat, tightens up throat and sinuses, headache, and sometimes makes you cough. (Again, this isn't a medical question. I don't live in the trailer, it's sitting.) The only way to find out what that is though, is to disect what 2 products were sprayed. Would mixing these 2 products create something super nasty? (Like formaldehyde, Chlorine, Chloroform, etc., the list goes on.)

We did a Volatile Organic Compound (VOC) test a few weeks ago and it turned out to be pretty much negative. It did pick up some very low levels of toluene, hexane, ethyl benzene, and xylene, but far from anything alarming. The levels were far below the EPA's acceptable exposure levels, but the reason I mention these chemicals is in case they are valuable information and tip you off on anything.

Also, in case this is a tip for anyone; All my fabrics smell and they really make me cough (etc.). After I moved out of the trailer to remediate it, I noticed that all of my cloths, towels, bedding, etc. (i.e. all fabrics) had a weird chemical smell until they were washed. It sort of smelled like a smokey chemical smell... but not really smokey. Hard to explain. But what's really interesting is; I had a pile of clothes I'd left in a closet. The other day I put them in my car to drive to my new house, and I started coughing in the car, the headache set in again, irritated throat... all the same symptoms, only now its some kind of gassing coming off my clothes that were in the trailer. Weird! It makes me wonder if you're aware of a specific chemical that can stick to clothes like that?

I have 2 questions:

QUESTION 1: Is/was it dangerous to mix these 2 products? Did I create an even worse chemical in doing so? (I've listed their chemical composition at the bottom of this post)

QUESTION 2: Is there a chemical method to 'pull' chemicals like these out of the trailer? For example, I've heard about people putting ammonia in a bowl and leaving in their trailer to pull out formaldehyde. Not sure if that is a valid chemical method?

Product #1: Mold Armor Mold Remover & Disinfectant

Ingredients:

Ethanol is the main ingredient. These ingredients below are the small/active ingredients

Quaternary ammonium compounds, benzyl-C12-16-alkyldimethyl, chlorides (< 0.1 %)
Decanaminium, N,N-Dimethyl-N-octyl-chloride {Decyldimethyloctylammonium chloride} (< 0.1 %)
Octanaminium, N,N-Dimethyl-N-octyl-chloride (< 0.1 %)
Decanaminium, N-Decyl-N,N-dimethyl-chloride (< 0.1 %)

See product documentation (SDS), here.

Product #2: Concrobium Mold control

Ingredients:

Water is the main ingredient. These 3 ingredients below are the small/active ingredients

Trisodium phosphate
Sodium carbonate (.95%)
Sodium bicarbonate

See product documentation (SDS), here.

Edit

I reworded my question. Again, not a medical question. This is a chemistry question about mixing 2 products, and their reactions.

I wouldn't be too worried about any chemicals that came from either of the products, or from reactions between the products. Any of those would have been volatile enough that they're long gone by now.

From your description, it sounds like you may be having a reaction to either residual mold or another allergen present in the trailer.

Mold is very difficult to fully eradicate. Even if you cleaned it off of all visible surfaces, it could very easily be growing behind something and you'd never know it without ripping out pretty much everything that's not glass or metal. Also, even if you've managed to kill all the most that was originally present, there could still be residual mold spores or proteins that get kicked up into the air very easily. If you're sensitive to mold, it doesn't take much for your immune system to react, giving you eye and upper respiratory tract irritation.

If this is the case, it may be easier to get rid of the trailer. To fully decontaminate, you'd want to go in and do a full tear out of everything that could have mold on it. After you're down to bear metal, you'd need to wet-wipe everything with a good biocide, making sure that you don't spread possible contamination with your wiping cloth. The point isn't only to kill the mold - it's to get rid off every possible protein fragment that may have been generated by the mold or mold spores.

While it's true that some people are sensitive to mold generated mycotoxins, more common is that people have immune reactions to the mold proteins, and it is not a small job to fully get rid of them.

Now, before you go full bore, keep in mind that it could also be something other than mold. Dust mites and roaches are other common allergens that come to mind. It may be worth going to an allergist to get tested for these and other common environmental allergens. You may find out that you're actually allergic to some kind of pollen (like poppies!) and it just happens that the trailer was driven through a poppy field at some point...


What constituents in soft drinks are the cause their strong pungent smell? Wikipedia and PubChem tells us that $\ce{CO2}$ is odorless in lower concentrations but can have a mildly pungent acidic smell at higher concentrations.

However, there is no pungent smell from chemical reactions such as the combustion of cooking gas, effervescence from carbonate upon the addition of an acid, and the sublimation of a piece of dry ice.

But many soft drinks and soda waters (including Sprite, 7 up, Pepsi, Coca-Cola, and Thums Up) emit, open opening, a strong burning smell that may cause severe irritation to eye and nose. The same burning is experienced during the burping (or belching) that comes out after consuming a glass of such drink

So, I want to know, is this due to $\ce{CO2}$? Maybe the manufacturers pump too much $\ce{CO2}$ into their drinks that it reacts in a different way than $\ce{CO2}$ from LPG-oven flames or carbonate/acid reactions? Or maybe that 'burning effect' is due to some mysterious additive?

Yes, that strong smell is mostly $\ce{CO2}$, possibly with a faint tint of volatile aromatizers from the beverage.

Our sense of smell works in surprisingly non-straightforward ways. Indeed, the smells of the same compound in high and low concentrations are sometimes perceived as quite different, to the point of being totally unrecognizable. A classic example is skatole, which may produce either a pleasant flowery aroma or a fecal stench, depending on the concentration alone. Looks like $\ce{CO2}$ is just a somewhat less extreme example of the same effect.

Air with 5% of $\ce{CO2}$ doesn't smell much of anything (not that it is OK to breathe it for long), which leads most people to conclude that $\ce{CO2}$ is odorless. This is quite wrong. If you try to inhale some pure $\ce{CO2}$ (which I do not recommend or endorse), that's when the smell kicks in. Your breathing just stops instinctively, as if you were trying to inhale water. By that moment, though, some $\ce{CO2}$ reaches your olfactory nerves. I am at a loss of words as to what it feels like. Being slapped across your face with a shovel seems like a moderately adequate approximation. Besides, you feel that familiar refreshing sharp odor of an open Coca-cola bottle.

So no, this is not due to some mysterious additive.


Solubility Of Benzoic Acid Why is benzoic acid well soluble in benzene (and other unpolar solvents, see e.g. wikipedia). What causes the hydrogen bonds in the solid to break? Something (i.e. polar and ionic substances) is soluble in water because of the hydration enthalpy with water, but i can see no similar cause for a solubility in benzene!

I would guess that benzoic acid forms dimers or trimers connected by hydrogen bonds in nonpolar solvents. To the outside, these only show the phenyl rings.

The colligative properties of those solutions would clearly show how many individual particles they contain. I´ll leave the literature research for you. ;)


Why does contaminated brake fluid conduct electricity? Using a voltmeter to test for water contamination in brake fluid Contaminated brake fluid conducts electricity. In the automotive repair world, brake fluid is considered contaminated by two ways:

Brake fluid is hygroscopic. Higher water content in brake fluid means lower boiling point. Brake pads/rotors get hot which can boil the water near near the pads/rotors. Boiled water is steam which is a compressible gas. Having a gas in the brake lines leads to a spongy brake pedal. So, high water content absorbed by brake fluid is considered contaminated.
Brake lines contain copper. When microscopic bits copper get in the brake fluid, they can build up in the ABS unit causing failure of the ABS system. Thus, high amounts of copper mean contaminated brake fluid.

Since pure water and copper are both nonelectrolytes, how can brake fluid conduct electricity? Is there some other contaminant I am missing here that is an electrolyte? Is the water being absorbed, somehow, by the brake fluid not strictly H2O?
I am asking because I want to know if I can use a voltmeter to test if there is too much water contamination in brake fluid.

Even pure water is somewhat ionized. If water is more conductive than pure brake fluid, the test is valid. Why don't you try this as an experiment? To sample(s) of brake fluid fresh from a new, unopened, can, add distilled water, drop by drop, stirring each time and look for a change in conductivity. Measure the proportions, and you can graph the effect of moisture on conductivity.
N.B. There are two types of automotive hydraulic brake fluid in common use: glycol-based and silicone-based. You should test the conductivity of each, and the effect of moisture on each.
BTW, unless so much metal was suspended in the fluid to make a thick slurry, that would not affect conductivity. It is possible some metals, such as aluminum, would react with the fluid and ionize, but those metals should not be part of the brake system.


Bonding in Square Antiprismatic Compounds Both $\ce{IF_8^-}$ and $\ce{XeF_8^{2-}}$ have (distorted) square antiprismatic geometries.

I've been wondering how to explain the bonding in such compounds.  Hypercoordinate bonding (as used to explain the bonding in $\ce{PCl_5}$, for example) doesn't seem to work in these cases.  Nor does the resonance explanation used in the case of $\ce{IF7^-}$ (which would involve something like $\ce{XeF_8^{2-} <-> XeF6 + 2F-}$ in this case) seem very strong.  Can the bonding and geometry in the above octacoordinated compounds be explained using  s and p orbitals?

From what research I've done, there isn't a definitive answer, though people have ideas about what causes this phenomenon. Both sources here cited note the lack of adherence to the VSEPR model, and hypothesize that the steric effects of the eight fluorine atoms prevent the central unbonded lone pair from existing in a definite space, thereby distorting the antiprismatic shape.
To paraphrase an inorganic chemistry textbook:

$\ce{[XeF8]^{2-}}$ is difficult to interpret using VSEPR theory. There is a lone pair on the central xenon atom, which in the VSEPR model would be predicted to occupy a definite position about the central atom as do lone pairs in $\ce{NH3}$, $\ce{SF4}$, and $\ce{IF5}$. In $\ce{[XeF8]^{2-}}$, however, there is no definite location for this central lone pair.
One explanation is based on the degree of crowding around xenon—with eight fluorines attached, repulsions between the xenon-fluorine bonds are too strong to enable a lone pair to occupy a well-defined position. This gives $\ce{[XeF8]^{2-}}$ a very slightly distorted square anti prism structure, with one face being slightly larger than the opposite face. Although this distortion may be a result of the way in which these ions pack in a crystal structure, it is also possible that the distortion is caused by a lone pair exerting some influence on the size of the larger face.

From a piece of literature specifically analyzing the antiprismatic structure of xenon(VI):

We believe that the larger $\ce{Xe-F}$ separations are due to a weak interaction between some fluorine atoms and the neighboring cations, thus causing a slight but significant distortion of the antiprism. However, the distortion might also be explained as a manifestation of some steric activity of the xenon lone pair. Since the eightfold xenon coordination observed here provides no clearly defined ninth coordination position for the lone pair, it appears to offer little support for electron repulsion models which have been used with considerable success in rationalizing the structures of other noble gas compounds.


Benzophenone Ketyl color origin I was recently preparing the benzophenone ketyl radical by reacting it with sodium in toluene. Although this compound is really well known and it seems somewhat obvious and accepted in literature that the radical causes a color I am still unable to find anything on the actual transition. There are some luminescence experiments but at least to my research there doesn't seem to be a good explanation on the actual color.

My suggestion would be that perhaps occupying the former LUMO of the benzophenone to generate a SOMO causes the orbital to shift in energy. I always confuse when something is stabilized or destabilized but since it seems to be in interaction with the two benzene rings I'd suggest that by occupying the orbital you can also lower it in energy a bit and thus promote the excitation of a former HOMO electron into the singly occupied SOMO with visible light.

According to what I found there are various mechanisms depending on the environment around the radical orbital how it can stabilize itself and thus lower the gap between the two orbitals (former HOMO and new SOMO).

TL;DR: The primary transition responsible for color is apparently the ${\pi{-}\pi}^*$ transition of the conjugated ketyl radical anion. This absorbs usually around the green to orange portions of the visible spectrum, leaving behind the purple-blue, and sometimes green color.

Original answer (background info):
According to Wikipedia/Benzophenone, the ketyl radical is formed by first, transition of the ketone from a singlet (double bond) to a triplet (diradical) state, and then abstraction of an electron from a donor to form the ketyl radical anion.
As I understand, only the radical anion is colorful. Both ref 1 and Wikipedia/Ketyl indicate that color disappears if the radical anion is allowed to oxidize (back to the ketone, which should be in equilibrium with the diradical). So my guess is, when an extra electron is added to the triplet state, this enables new electronic transitions that produce color.
These two sources also indicate both solvent and substituent effects: In the Wikipedia article, the solvent is toluene, and color is deep blue for benzophenone. In the paper, the solvent is isopropyl alcohol, and color is greenish-blue for benzophenone, and varies from blue to green depending on substituents. This also suggests that conjugation plays an effect, since the substituents can directly affect only the phenyls. But this, in turn, affects the energy levels of the double bond.

Edit:
A Google search for "benzophenone ketyl spectrum" (including the image search) gives several results of varying relevance, such as refs 2-4. In particular, from ref 2 (dealing with uranium ketyl radicals):

The electronic absorption spectrum of 2 in toluene features an intense absorption band ($\lambda_\text{max} = \pu{562 nm}$, $\epsilon = \pu{5680 M^{-1} cm^{-1}}$) in the visible region (Figure 8, top). This absorption gives rise to the ketyl complex’s distinctive and intense purple color and is assigned to the ${\pi{-}\pi}^*$ transition of a highly conjugated ketyl system. The transition energy of $\pu{562 nm}$ is blue-shifted from that of the sodium benzophenone ketyl radical, which exhibits the ${\pi{-}\pi}^*$ transition at $\pu{618 nm}$.

Ref 3 talks about $D_1 \leftarrow D_0$ and subsequent $D_n \leftarrow D_1$ transitions, and provides an energy level diagram. I assume $D_n$ refers to the doublet states that form upon abstraction of a hydrogen by (or addition of an electron to) the triplet diradical.
Finally, ref 4 adds:

Bzp is known to have a high ISC quantum yield following UV excitation, but until recently the mechanism of population of the lowest triplet excited state ($T_1$) after photoexciting to its $S_1$ (${nπ}^*$) state remained uncertain. ... These investigations indicate that the dominant pathway involves fast ISC from $S_1$ to $T_2$ (${\pi\pi}^*$) as an intermediate state, with the subsequent rate of population of the lowest vibrational levels of $T_1$ controlled by the time scale for dissipation of excess vibrational energy to the solvent bath after internal conversion (IC) from $T_2$.

(Bzp = benzophenone, ISC = intersystem crossing). It then proceeds to discuss these transitions in more detail.

J. Am. Chem. Soc. 1933, 55 (1), 391–395; doi:10.1021/ja01328a057
J. Am. Chem. Soc. 2008, 130 (20), 6567–6576; doi:10.1021/ja801007q
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Electron interference Assuming that electrons behave as both waves and particles, and that waves(like in double slit experiment) can interfere with themselves:

Do electrons actually move in any direction, or do they just 'teleport'(similar to how when making a Quantum Leap, they instantaneously appear at a different shell, rather than actually move continuously).
Do electrons 'subtract' with other electrons for the probability function of an electron being present in that location. i.e. if there is a covalent bond, will the electrons interfere with the other probably function aside from repelling it?.
Do electrons only have positive probabilities. Is it more like how gravity never repels, only attracts, or more similar to how waves of light have peaks and troughs. Do electrons ever cancel out with others? If so, how?

Traditional understanding of trajectory is of no use for electrons. Granted, on macroscale you may detect something close to movement and trajectory, (electron beams and electron cannons take their names from this) but it still far from what we have in classical mechanics. Ideally you should abandon any attempts to relate quantum and classical mechanics and view quantum mechanics as an abstract mathematical construct.
Incorrect question. If you have a system of two or more particles, you should consider the wavefunction of combined system, not separate wavefunctions for separate particles. Still, for electrons being fermions, there is a strict rule that wavefunction must be zero for positions in phase-space where two electrons of same spin are at same location and since electrons are of same charge, probability of having two electrons of different sping close to each other is usually lower than to have them at some distance.
Yes, only positive probabilities. Though, this does no apply to wavefunction itself, it may be negative at some point (actually, its value can be a complex number).


Why are trans fats worse than saturated fats? Saturated fat molecules have no double-bonded carbons, so they are long and straight, which means they stack easier and tend to form solids at room temperature, and solids are better at forming plaques in your arteries and interacting with cholesterol in the bloodstream.

Moving onward to hydrogenation, to my current understanding, if you hydrogenate vegetable oil all the way, you get saturated fats (no double bonds anywhere because you've shoved in so much hydrogen to occupy all the bonds). But for the purposes of making trans fats, it's usually partial hydrogenation.

I assume that most vegetable oils that get partially hydrogenated must be polyunsaturated fats to begin with (meaning more than one double-bond) because partial means you aren't replacing all the double bonds, so there must be at least two double-bonds present.

The result of partial hydrogenation means you remove these double bonds and start straightening out the chains, which makes them easier to stack (which is why trans fats tend to be solid or semisolid), like saturated fats.

My question:

Why are trans fats worse than saturated fats? It looks as though partial hydrogenation is simply taking a polyunsaturated fat and bringing it closer and closer to the status of a saturated fat, but not all the way. If the goal is to get a fat that is solid and lasts longer, why not just use a saturated fat?

And for whatever reason saturated fats aren't the answer: Isn't the end result of partial hydrogenation technically just another unsaturated fat? I thought unsaturated fats were considered healthy? So how are trans fats worse than fully-stackable saturated fats?

Or upon reflection maybe my question is, why are trans fats considered so much worse than other unsaturated fats? If all hydrogenation does is break up double bonds and insert hydrogen, straightening out any cis-kinks in the chain, doesn't this just generate the same acid with fewer double bonds and straighter chains? How is this any different than going out in nature and finding the same unsaturated fat, rather than going to the trouble of converting?

What's causing the difference, here? Where is my understanding breaking down?

Someone may very well give a more thorough answer but from what I understand, most naturally occurring unsaturated fats have their double bonds in the cis conformation which is generally higher in energy than the trans conformation.

Thus, the idea is that when you ingest trans fats (i.e. fats with a trans double bond) your body has to somehow hydrogenate this double bond before it can break it down further. This is more difficult for your body to do because it must either isomerize to a cis bond and then hydrogenate, or just hydrogenate the trans bond which is not easy for your body to do effectively. In some cases hydrogenation of trans fats can't actually happen because nature basically makes cis double bonds in fats (which sounds weird but is true) so your body has ways of breaking these fats down but not trans fats. So, the fat just gets stored or excreted without doing anything to it.


How to calculate the amount of gas in a compressed cylinder I am trying to calculate a cost per hour use of a gas we buy in a cylinder. The details I have been working with are:

Gas used CP-grade N2 material number: 110628-L can

Gas: N, density = 1.251 g/L @ 0.101 MPa M (molar mas) = 14.007
Gas cylinder: Pressure = 200 bar or 20 MPa Cylinder volume 9.45 m3 or 9450 L
Operating pressure of gas from cylinder = 4.5 bar or 0.45 MPa and flow rate = 60 cL/min

I have been applying the ideal gas law to PV=nRT to figure out how many total litres of N are in the can at 4.5 bar pressure to then get a rough estimate of how many hours of flowing gas I get. However I seem to be getting unreasonable numbers, potentially from me using the equation or reasoning incorrectly:

When I apply the gas law under isothermal V2=V1/(P2/P1) conditions to see how much the internal gas volume expands from the 20 MPa pressure, as a sanity check I wanted to see how many L at 1 atm came out. The result was 1.89 ML which makes little sense as the conversion of 1.89 ML N2 to mass = 2.364 Mg.

Similarly it equates to 420 kL @ 0.42 MPa.

Where am I going wrong?

Maybe the specifications mean that its 9.45m3 or 9450 L of compressed gas to 200 bar?

Turns out my assumptions were wrong about the stated $9\,450\ \mathrm L$ of $\ce{N2}$ gas. The $9\,450\ \mathrm L$ is the total volume of standard pressure gas compressed into the cylinder to $20\ \mathrm{MPa}$.

Using $V_2=V_1/(p_2/p_1)$ having $p_1=0.101\ \mathrm{MPa}$, $p_2=20\ \mathrm{MPa}$, $V_1=9\,450\ \mathrm L$. The total volume of gas within the cylinder is about $47\ \mathrm L$ a reasonably fitting size for the cylinder dimensions.

Hence to answer my own question, at the pressure of usage ($0.45\ \mathrm{MPa}$) the volume of gas (using same equation as above) is $2\,121\ \mathrm L$ which at $60\ \mathrm{cL/min}$ equates to $59\ \mathrm h$ of flow roughly.


Detection of magnesium in a buffer? I have a PolyA tailing kit for putting PolyA tails on the 3' end of RNA. The kit comes with a proprietary 5X buffer and a 25mM MnCl2 solution, and in the standard set up I use 20uL of 5X buffer and 10uL of 25mM MnCl2. When trying to force the enzyme to accept an alternative substrate I did the reaction without the MnCl2 and it worked. Further research revealed that the enzyme also works with Mg++ ions, even though the rate is faster with Mn++, and some old papers on the enzyme mention using both Mg and Mn in the buffer. This leads me to believe that the buffer contains Mg++, but I don't know for sure because the manufacturer won't tell me.

I immediately thought I could use atomic absorption spectroscopy, but our department does not have one. I looked up colorimetric assays online, but the kits cost about $300, which is more than I want to spend on this idea. What I'm asking is if anyone knows a relatively simple method to detect magnesium in a solution? I have access to a UV-Vis spectrophotometer and fluorimeter.

EDIT: After digging for other phone numbers for the manufacturer I found someone who would tell me a little about the buffer, and it does contain magnesium. He couldn't me how much though. Spectro-Phone-otemotry.

$\ce{Mg^2+}$ forms complexes with 4-(2-Pyridylazo)resorcinol that can be detected spectrophotometrically: Simultaneous spectrophotometric determination of calcium and magnesium in water, described by E. Gómez, J.M. Estela, V. Cerdà in Analytica Chim. Acta 1991, 249, 513-518.

(DOI: 10.1016/S0003-2670(00)83027-5)


Rules of naming ionic compounds While brushing up on my Chemistry basics, I came across a topic about naming ionic compounds. It is a question regarding Roman numerals.
It says "Roman numerals should not be included in the name when the metal can only form one cation."

The cation of $\ce{Zn^{2+}}$ has +2 charge, and $\ce{NO3-}$ anion has -1 charge.

which results in having $\ce{Zn(NO3)2}$

If Roman numerals are used for transitional metals and given that zinc is part of the transitional metals, why is this ionic compound named as zinc nitrate instead of zinc(II) nitrate?

Thanks.

Zinc and cadmium are generally considered to be "almost invariably" in the oxidation state +2 (cfr. the introduction to this paper about one of the few exceptions), therefore the indication of the charge is unnecessary.
Zinc is usually listed as a "Type-I cation" (see Monoatomic ion), i.e. a cation that only appears in one oxidation state, or its case is emphasised as an exception among the transition metals:

Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion.
-- Chemistry LibreTexts

(For the sake of completeness, the IUPAC Red Book section IR-5.4.2.2 "Use of charge and oxidation numbers" neither mentions "Type I/II cations" nor treats the zinc case separately)


What is the relationship between the two definitions of the difference between basicity and nucleophilicity? (Clarification: this is a question about terminology, not so much about when to use different concepts.  I think I do understand which characteristics of a species to use when trying to determine something about a reaction in some circumstances.  I just don't know what word to call it.)

When someone on the internet (here or elsewhere) is asked the difference between "basicity" and "nucleophilicity", they typically give one of the following two answers:

Basicity is about how well a species donates electrons to a proton, while nucleophilicity is about how well a species donates electrons to a more general atom or ion (especially carbon).
Basicity is about thermodynamics (i.e. equilibrium constant) while nucleophilicity is about kinetics (i.e. activation energy).

While these two answers may be related, they do not seem identical!  So I am wondering which one is actually correct, or whether there is some way that they are both correct.  In particular, I'm wondering what happens when you have a situation that puts those two descriptions at odds with each other.

We could have a reversible reaction involving carbon (though I know that's not the usual situation) that does in fact have some equilibrium.  When we're talking about that reaction's equilibrium, do we still use the word "basicity"?
In a reaction with a proton rather than a carbon, when we're talking about kinetics, do we still use the word "nucleophilicity"?

I hope I've gotten my question across OK.

The second set of definitions is the correct one.

Basicity is a thermodynamic property. In essence, the basicity is measured by determining the equilibrium constant of the following (generic) reaction:
$$\ce{B- + H2O <=> B-H + OH-}\tag{1}$$

The proton transfer that is the gist of this is an extremely fast process — faster than diffusion control due to the Grotthuss mechanism. You may have witnessed this during acid-base titrations: equilibrium is reached extremely rapidly. Another way to observe this is by adding carbonate bases to an acidic solution; $\ce{CO2}$ evolution is basically instantaneous. Therefore, there is no way to observe just one step (forward or reverse); you basically always have them both.

Nucleophilicity is a property that is related to the rapidness of the following, again generic, reaction:
$$\ce{Nu- + R-LG -> Nu-R + LG-}\tag{2}$$

As similar as this may seem to the above, the lack of a reverse arrow may already show that this is tied to only one directiong; that it is not an equilibrium process but rather a kinetic one. Most of the time, there will be one species that is clearly a good nucleophile ($\ce{Nu-}$ in the scheme) and one that is only a good leaving group ($\ce{LG-}$ in the scheme). The reaction will be driven to completion and stay there.

To exemplify the difference, consider what would happen if you added hydroxide to the acid-base equilibrium $(1)$ versus additional leaving group to the nucleophilic substitution $(2)$. In $(1)$, the equilibrium would readjust; if the base is less basic than hydroxide it will be deprotonated, i.e. the reaction will be shifted back to the reactants. In $(2)$, typically nothing will happen, e.g. if you add $\ce{OPPh3}$ to a Mitsuobu reaction; showing the kinetic nature of the process.

Another way to look at the problem — and to separate the parameter basicity from the parameter nucleophilicity — is to look at the influence of steric effects. For example, consider ethanolate, propanolate and deprotonated isobutyl alcohol. In all three cases, the negative charge is unambiguously located on the oxygen atom. Thus, we would expect a thermodynamic property to be relatively equal for the threeof them. At the same time, there is a clear increase in steric hindrance from ethanolate to the isobutyl alcohol. This steric effect should manifest itself in reaction kinetics and lead to the latter being less nucleophilic than the former. Indeed, sterically hindered bases are often described as non-nucleophilic while the same structure with less hindrance often works fine as a nucleophile.

Finally, let’s work at debunking the claims laid out in the first set of definitions. First, basicity. After the Arrhenius and Brønsted-Lowry theories of acids and bases came the Lewis theory. In this, the reaction of a Lewis base with a Lewis acid — not necessarily a proton — is an acid-base reaction. One of many possible examples would be the formation of a transition metal complex such as $\ce{[Cu(NH3)4(H2O)2]^2+}$. Here, the ammine ligand is acting as a base, but it is donating to a non-hydrogen atom (namely copper). The claim that basicity only goes towards hydrogen is obviously false.

Unfortunately, it is not possible to fully debunk the second part since a nucleophilic attack on a proton is not really possible (that would indeed be the thermodynamic acid-base reaction). However, we may remember that most of the non-metals can act as the recipient of a nucleophilic attack; for example sulphur in a Swern oxidation, phosphorus in a Horner-Wadsworth-Emmons reaction or iodine in a Dess-Martin oxidation.


Effect of solvent in the reaction of alkyl halide with nitrite ion $$\ce{RX + NaNO2 -> ?}$$

Is there a difference in the outcome of the reaction if the solvent is changed from an alcohol (e.g. EtOH) to a polar aprotic solvent such as DMSO or DMF?

I have two books - one book mentions an alcoholic solvent gives $\ce{RONO}$ as product; the other mentions $\ce{RNO_2}$ is the product in DMSO/DMF. Can anyone please confirm?

Let me first clear out the difference between reaction of alkyl halide with $\ce{AgNO2}$ and $\ce{NaNO2/KNO2}$

The synthesis of nitro compounds from reaction of alkyl halides with alkali metal nitrites using DMSO/DMF as medium is known as the Victor meyer reation($\ce{AgNO2}$) or Kornblum reaction($\ce{NaNO2/KNO2}$). The products of these reaction are generally mixture of nitroalkanes and alkyl nitrites. But the main difference between them is that the % yield of nitroalkane is higher on reaction with $\ce{AgNO2}$ than that of $\ce{NaNO2/KNO2}$.Whereas % yield of alkyl nitrite is higher on reaction with $\ce{NaNO2/KNO2}$ than that of $\ce{AgNO2}$. The required mechanisms for the reactions are..

Now as for the reaction of $\ce{NaNO2/KNO2}$ with alcohol, there is no involvement of alkyl halide. Alkyl nitrite can also be prepared by adding concentrated $\ce{H2SO4}$ to the aqueous solution of $\ce{NaNO2/KNO2}$ and alcohol. The mechanism of this reaction is

Source books:

The Nitro Group in Organic Synthesis By Noboru Ono
Organic Chemistry of Explosives By Jai Prakash Agrawal, Robert Hodgson
Sourcebook of Advanced Organic Laboratory Preparations By Stanley R. Sandler, Wolf Karo


How are alkalide ions synthesized? How are alkalide molecules synthesized, for example, I have read in a paper that the reaction $$\ce{2Na -> Na+ + Na-}$$ is exothermic with a $\Delta H = \pu{-438 kJ/mol}$.

I have also read that in the presence of a certain type of crown ether, under specific conditions, $\ce{Na+}$ crystallizes at around $\pu{-20 ^\circ C}$.

Can someone walk me through the mechanism through which sodide, or any alkalide, in general, is actually synthesized?

The reaction energetics

Using the method of the superposition of configurations, the electron affinity of $\ce {Na}$ was theoretically determined to be $\ce {+0.54 eV}$ $\ce {^1}$, that is, around $\ce {-52.1 kJ/mol}$. The gas-phase process

$\ce {2Na (g) -> Na^+ (g) + Na^- (g)}$

has been determined to be endothermic by $\ce {4.54 eV}$ $\ce {^2}$ while the solid-state process

$\ce {2Na (s) -> Na^+ .Na^- (s)}$

has also been estimated to be endothermic by $\ce {0.8 eV}$ $\ce {^2}$.Thus, your assertion regarding the large exothermicity of the reaction is questionable. However, it is worthy to note that the $\ce { \Delta H_f}$ and $\ce {\Delta G_f}$ for $\ce {Na^+.Cry Na^-}$, where $\ce {Cry}$ = [2.2.2] cryptand, are $\ce {-10 kJ/mol}$ and $\ce {+28 kJ/mol}$ respectively $\ce {^3}$. Your large exothermic enthalpy may likely refer to the lattice energy, i.e. for the process

$\ce {M^+.Cry (g) + M^- (g) -> M^+.Cry M^- (s)}$.

For $\ce {M = Na}$, the $\ce {\Delta H}$ and $\ce {\Delta G}$ for the above process are $\ce {-323 kJ/mol}$ and $\ce {-258 kJ/mol}$ respectively $\ce {^3}$.

Preparation of the alkalide

$\ce {Na^-}$, $\ce {K^-}$, $\ce {Rb^-}$, and $\ce {Cs^-}$ anions are stable both in suitable solvents and in crystalline solids$\ce {^3}$. The latter can be prepared either by cooling a saturated solution $\ce {^4}$ or by rapid solvent evaporation.

The principal difficulty in preparation of crystalline salts containing alkalide ions by the method of cooling a saturated solution is the low solubility of these alkali metals in the amine and ether solutions $\ce {^3}$. Without a sufficiently large concentration of the metal dissolved in solution, precipitation of the solid upon cooling would be insignificant. This difficulty was resolved by the use of crown-ether and cryptand complexes, such as those of [18]crown-6 and [2.2.2] cryptand] $\ce {^3}$. The complexating agent complexes with $\ce {M^+}$ , shifting the equilibrium (1) far to the right, significantly increasing the concentrations of the dissolved metal ions.

(1) $\ce { 2M (s) -> M^+ (sol) + M^- (sol)}$

(2) $\ce { M^+ (sol) + Cry (sol) -> M^+.Cry}$

This technique of using complexating agents was also what Dye et al. used in their synthesis in 1973 $\ce {^4}$. As reported by Dye et al., a sufficiently concentrated solution of sodium metal (in excess) dissolved in ethylamine with [2.2.2] cryptand was first prepared. The solution is then cooled to dry ice temperatures, giving a gold-coloured crystalline solid precipitate. Through thorough analysis, this precipitate was then determined to be $\ce {Na^+.Cry Na^- (s)}$ with $\ce {Cry}$ being the [2.2.2] cryptand.

References

Weiss, A. W. Theoretical Electron Affinities for Some of the Alkali and
Alkaline-Earth Elements. Phys. Rev., 1968, 166 (1), 70-74
Tehan, F. J.; Barnett, B. L.; Dye, J. L. Alkali anions. Preparation and Crystal Structure of a Compound which contains the Cryptated Sodium Cation and the Sodium Anion. J. Am. Chem. Soc., 1974, 96 (23), 7203–7208
Dye, J. L. Compounds of Alkali Metal Anions. Angew. Chem., 1979, 18 (8), 587-598
Dye, J. L.; Ceraso, J. M.; Lok, M. T.; Barnett, B. L.; Tehan, F. J. A Crystalline Salt of the Sodium Anion (Na-). J. Am. Chem. Soc., 1974, 96 (2), 608-609


Molar mass and molar volume of monochloroethylene Monochloroethylene gas is used to make polyvinylchloride (PVC). It has a density of $\pu{2.56 g L-1}$ at $\pu{22.8 ^\circ C}$ and $\pu{101 kPa}$. What is the molar mass of monochloroethylene? What is the molar volume under these conditions?

$$PM=dRT$$

$$M=\frac{\pu{2.56 g L-1}\cdot\pu{295.8 K}\cdot\pu{0.0821 atm L mol-1 K-1}}{\pu{0.9968 atm}} = \pu{62.37 g mol-1}$$

After this I can't find molar volume. Which formula should I use?

I know $PV=nRT$, but without knowing the gas mass I can't use this formula, I believe.

You don't need to know the volume to determine the molar volume $V_\mathrm{m}$ of a substance. In your notations:

$$V_\mathrm{m} = \frac{M}{d} = \frac{\pu{62.37 g mol-1}}{\pu{2.56 g L-1}} = \pu{24.36 L mol-1}$$

Alternatively, you could've just used ideal gas law:

$$V_\mathrm{m} = \frac{V}{n} = \frac{RT}{P} = \frac{\pu{0.0821 atm L mol-1 K-1}\cdot\pu{295.8 K}}{\pu{0.9968 atm}} = \pu{24.36 L mol-1}$$

P. S. You determined molecular mass correctly, but, as MaxW pointed out, the units are wrong: you probably made a typo and put $\pu{g L-1}$ instead of $\pu{g mol-1}$.


Volume expansion from phosphate buffer solution I want to mix a phosphate buffer at $\mathrm{pH\ 7.2}$ at $20~\mathrm{^\circ C}$. I do this by dissolving $0.0036\ \mathrm{mol}$ of sodium dihydrogen phosphate monohydrate ($\ce{NaH2PO4*H2O}$) and $0.0063\ \mathrm{mol}$ of di-sodium hydrogen phosphate dihydrate ($\ce{Na2HPO4*2H2O}$) in water per liter of buffer. I add $0.15~\mathrm{M}\ \ce{NaCl}$. How much water do I add if I want $1~\mathrm{L}$ of buffer? Do I just use $1~\mathrm{L}$ of water — will the salts not expand the volume? Is it insignificant and at what point do I need take into account the volume expansion of the solution?

Notes:

  $$M(\ce{NaH2PO4 . H2O}) = 137.99~\mathrm{g/mol} \Longrightarrow 137.99~\mathrm{g/mol} \times 0.0036~\mathrm{mol} = 0.496764~\mathrm{g}$$

  $$M(\ce{Na2HPO4 . 2 H2O}) = 177.99~\mathrm{g/mol} \Longrightarrow 177.99~\mathrm{g/mol} \times 0.0063~\mathrm{mol} = 1.121337~\mathrm{g}$$

  $$M(\ce{NaCl}) = 58.44\ \mathrm{g/mol} \Longrightarrow 58.44\ \mathrm{g/mol} \times 0.15~\mathrm{mol/L} = 8.766~\mathrm{g/L}$$

There is an exact way of doing this and there is a practical way of doing this.

The exact way involves a volumetric $1~\mathrm{l}$ flask. You would first fill the salts into the volumetric flash, add some water to dissolve them and then carefully add more water until the flask contains exactly $1~\mathrm{l}$ of solution. That removes the error of the salts increasing the overall volume.

However, it is also part of the exact way to add a pH-monitoring device before you have finised to make sure the pH is exactly at $7.2$ — and to carefully add phosphoric acid or sodium hydrogen phosphate if the pH is off. (A better solution would be to make sure the pH is over the desired value so phosphoric acid can simply be added; as it is a liquid it can be added and dissolved easier.) For most applications outside of thermodynamics, this is an overkill.

The practical way involves one of the following:

using an Erlenmeyer flask, beaker or similar with a $1~\mathrm{l}$ mark; filling the salts first and then the solution up to the mark
simply measuring $1~\mathrm{l}$ of water and adding that.

Why are the practical solutions okay? In most cases, it does not matter whether the molar concentration of, say $\ce{NaCl}$ is exactly $0.15~\mathrm{M}$ or whether it is $0.14~\mathrm{M}$. This rather unimportant experimental error is simply accepted. It also does not really matter whether the pH is exactly $7.2$ or whether it is, say $7.5$. Therefore, one often decides to go the not-so-exact but easier and faster way.


Can a reaction have carbocation resonance followed by a hydride shift? Consider 5-methylcyclopentadiene:

If HX attacks at carbon 1, we have an option of 1,2 or 1,4 (carbocation resonance) addition. In the case of 1,4 addition, since the carbocation (secondary) will be one bond away from a tertiary carbon, will there be a hydride shift, thus yielding in 1,5 addition?

In essence, can you have both carbocation resonance and a hydride shift in one reaction? Or is it that once a carbocation is resonance stabilized, there is no need for stabilization via hydride shift?

Substituted cyclopentadienes are an unique molecules, which are not stable at room temperature (recall dimerization). For example, 1-methylcyclopent-1,3-diene (1) or 2-methylcyclopent-1,3-diene (2) or 5-methylcyclopent-1,3-diene (3; diene in the question) would rearranges to a mixture of all 3 positional isomers if leave at room temperature for 1-2 days by hydrogen migration (Ref.1 & 2). The abstract of Ref.1 states that:

  A number of substituted cyclopentadienes display characteristic properties which are representate of all members in the series: (a) the endocyclic double bond isomers are in the thermodynamic equilibrium; the process of establishing the equilibrium being retarded; (b) the transformation of isomeric cyclopentadienes proceeds by steps, i.e. a proton from the methylene group (5th position) passes mainly to the adjacent C-atom (1st position). The proton-cyclopentadienyl anions are probably transitional states in the transformation of isomers. It is assumed that this transformation is a first order reaction. As a result of this investigation it has been shown that the structure of the substituted cyclopentadiene (or the composition of the isomer mixtures) depends mainly on the character of the substituent and not on the method of synthesis.

The ratio of 1 : 2 : 3 would be 45:54:1. Kinetically, the process is first order and activation energy is $\pu{20.4 kcal/mol}$ over temperature range of $\pu{298-313 K}$ (Ref.3). The kinetic studies by NMR has been suggested that rearrangement is due to 1,2-hydrogen shift (Ref.4).

Yet, these cyclopentadienes are also suseptible to protonation by strong acids (Ref.5). Thus, it is very possible that protonation at $\ce{C1}$ 1,5-addition after 1,2-hydride migration. However, as explained above, 5-methylcyclopent-1,3-diene (3) is relatively the least stable positional isomer. At room tempeature, it would stay mostly as 1-methylcyclopent-1,3-diene (1) and 2-methylcyclopent-1,3-diene (2) with $\approx 1:1$ ratio. Therefore, addition would be either 1,4- and 1,2-.

References:

V. A. Mironov, E. V. Sobolev, A. N. Elizarova, "Some general characteristic properties of substituted cyclopentadienes," Tetrahedron 1963, 19(12), 1939-1958 (https://doi.org/10.1016/0040-4020(63)85008-5).
Sigmund M. Csicsery, "Methylcyclopentadiene Isomers," J. Org. Chem. 1960, 25(4), 518-521 (https://doi.org/10.1021/jo01074a009).
Raymond K. Crandall, “Investigation of the electronic and thermal rearrangement properties of spiro[2,4]hepta-4,6-dienes,” M.S. Thesis, Rochester Institute of Technology, Rochester, NY, 1982.
S. McLean, P.Haynes, "Hydrogen migration in cyclopentadienes," Tetrahedron 1965, 21(9), 2329-2342 (https://doi.org/10.1016/S0040-4020(01)93887-6).
Ronald F. Childs, Maung Zeya, “Diprotonation of 5-Acyl-1,2,3,4,5-pentamethylcyclopentadienes,” Can. J. Chem. 1975, 53(22), 3425-3430 (https://doi.org/10.1139/v75-490)


Can you push a liquid and a gas into a container at different pressures? I have a gas ($\ce{CO2}$ for example) and I want to infuse it into a liquid ($\ce{H2O}$ for example). I need a higher pressure of $\ce{CO2}$ to do this. So I push in water into a sealed container (at 7 psi) and add $\ce{CO2}$ (at 14 psi).

Is this possible? Will it back up? What if I use a check valve? Can I use a non-electrical actuator of some sort to let liquid in, and then add gas 'automatically'? What about something that functions like a carburetor?

I need to let the new solution out after it is mixed and also allow a consistent flow if I leave the exit open. If I have a liquid at 7 psi and a gas at 14 psi and I am trying to push them both into a sealed container, how is this done? Letting it out? Doing it on the fly?

I know you do NOT carbonate water at these pressures. Just an example with real gas/liquid.

Pressure is due to the random impact of molecules on the surface, regardless of whether water or carbon dioxide or anything else. Once equilibrium is reached, all components are at the same pressure.

That said, there's no problem using a moving stream of $\ce{CO2}$ to propel and to mix with water at much lower pressure, using Bernoulli's principle. The automotive carburetor and spray atomizer operate on this basis.

The alternative, of course, is to add the water to the container through a check-valve, then add $\ce{CO2}$ at a higher pressure.

The choice is a continuous process vs. a batch process.


How does a space group determine crystal structure? If I have an intermetallic and want to know the specific arrangement of atoms in the lattice, how can I determine that? What I mean is the Cartesian (xyz) position, or the (abc) positions in the lattice. In my case, I am working with Ni3Sn4, which has a space group C2/m. The space group diagram can be found here.

To build lattice you need the space group, the three unit cell dimensions, the three unit cell angles, each atom's fractional coordinates (e.g., a *.cif file overall),

Acta Cryst. B38, 598 (1982)  doi:10.1107/S056774088200346X
Ni3Sn4, C2/m, a=12.214, b=4.060, c=5.219 A, beta=105.0 degrees (others 90 degrees by default), V=250 A^3 (verify), Z=2 (formula units/unit cell)

plus

http://www.chem.gla.ac.uk/~louis/software/ortep3/ or

http://www.shapesoftware.com/00_Website_Homepage/ or

http://www.iucr.org/resources/other-directories/software?result_42405_result_page=A
the big kahuna.

For 3D visualization, if you are not practiced,
http://www.berezin.com/3d/3dprism.htm


Synthesis of mandelic acid from benzaldehyde For an upcoming exam, I have to describe the synthesis of 2-hydroxy-2-phenylacetic acid (mandelic acid) from benzaldehyde. The provided answers describe quite a long process, consisting of 8 stages. However, I've been suggested a much simpler method, consisting of:

$\ce{C6H5CHO + HCN -> C6H5CH(OH)CN}$
$\ce{C6H5CH(OH)CN + HCl + 2H2O -> C6H5CH(OH)COOH + NH4Cl}$

This is even simpler than the 3-stage process described in the Wikipedia article. Could it be a viable answer?

The direct reaction with HCN seems to be how the mandelic acid was synthesized in the 19th century. If you check how the reaction with sodium bisulfite is actually performed (Organic Syntheses, Coll. Vol. 1, p.336 (1941); Vol. 6, p.58 (1926).) you'll see that this is not more complicated than the direct HCN route as those first two steps are performed in one reaction step.
From this preparation, several preparations via the HCN route are cited:

Naquet and Louguinine, Ann. 139, 299 (1866); Wallach, Ann. 193, 38
(1878); Spiegel, Ber. 14, 239 (1881); Ultee, Rec. trav. chim. 28, 254
(1909); Rule, J. Chem. Soc. 113, 12 (1918).

I looked at the first page of the oldest of those papers from Naquet and Louguinine from 1866 and it states:

Bekanntlich bildet sich diese Säure bei der wechselseitigen Einwirkung des Benzoesäure-Aldehyds und der Cyanwasserstoffsäure bei Anwesenheit von verdünnter Chlorwasserstoffsäure gemäß der Gleichung:
$$\ce{C7H6O + CNH + 2 H2O + HCl -> NH4Cl + C8H8O3}$$

Translated, this means that you can synthesize mandelic acid from benzaldehyde, HCN and HCl.
I'm assuming you're not planning to actually synthesize it using this route yourself, I would guess that avoiding the use of HCN directly was one of the motivations behind the synthesis route using sodium bisulfite.
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Compounds that adopt the zincblende structure Zinc sulfide, copper iodide and copper bromide all adopt the zincblende structure. My lecture notes suggest that this structure is adopted preferentially over the wurtzite structure if the bonding is more covalent (why is this?); thus, it’s adopted where there is a polarizing cation and a polarizable anion.

However, surely a polarizing cation should be small and highly charged (i.e. not ions like $\ce{Cu+}$ surely? It has a low charge and large-ish radius). What makes this polarizing? I’ve also heard this about $\ce{Ag+}$.

Generally when you get a preference between two structures due to directional bonding, the first order environment of the cation is different. In this case however both are tetrahedrally coordinated. The Wurtzite structure has a marginally greater Madelung constant (1.641 versus 1.638).

Of the top of my head, I think the dominant factor in favour of the blende structure is stronger directional cation-cation and anion-anion interactions. In the case of Cu(I), the d-orbitals are going to be fairly diffuse due to the low charge, potentially allowing some covalent interaction between adjacent sites. A much better example would be GaAs, where both cation and anion are realtively soft/polarizable, which adopts blende, in contrast to AlN, with hard ionic interactions, which adopts wurzite. I don't think it's really as simple as this because the difference in energy between the two structures is often very small; you can see this in the relative stability of both ZnS polymorphs.