Hydrolysis of transition metals' halides? So, it's a relatively common known solubility rule that any hydroxide with a cation not in the first two groups is basically insoluble. So supposing we have a transition or post-transition metal $M$ and say it forms a charge of $n+$ (so it's cation is $\ce{M^{n+}}$) why would $\ce{MX_n}$ (where $\ce{X}$ is a halide) remain in solution? Why doesn't the following reaction occur?
$$
\ce{MX_n(aq)} + n\ \ce{H_2O(l)} \rightarrow \ce{M(OH)_n(s)} + n\ \ce{HX(aq)}
$$
My thought was that perhaps this reaction does occur, but just very slowly and not fast enough for the $\ce{M(OH)_n}$ to precipitate in noticeable amounts. However, I don't really know and this is just a guess. Could someone explain to me why this doesn't happen?

$\ce{MCl_n(aq)} + n\ \ce{H2O(l)} \rightarrow \ce{M(OH)n(s)} + n\ \ce{HCl(aq)}$ does tend to go to the right. "Basically insoluble" is not quantitative enough to generalize. If the hydroxide is very insoluble, the acid produced by hydrolysis may not keep it in solution. But if the hydroxide is even only slightly soluble, it may not precipitate out. The acidity of the solution is a subtle indicator that the reaction has shifted a little bit to the right. The reaction isn't slower, just not always so extreme.

If you put $\ce{SiCl4}$ in water, it gives $\ce{HCl}$ + silica gel because $\ce{SiO2}$ (or "$\ce{Si(OH)4}$") is so insoluble.

If you put $\ce{MgCl2}$ in water, no precipitate appears because $\ce{Mg(OH)2}$ is slightly soluble (~2 mg/L), and the $\ce{HCl}$ produced by the hydrolysis reaction is acidic enough so that if a significant amount of $\ce{Mg(OH)2}$ were produced, it would just redissolve. $\ce{Mg(OH)2}$ gives a pH of about 10 in water, so at pH's lower than 10, $\ce{Mg(OH)2}$ doesn't precipitate out. BTW, $\ce{MgCl2}$ is slightly acidic because of the hydrolysis reaction.

$\ce{AlCl3}$ is complicated because of ionic complexes, but in water it hydrolyzes to give a very acidic solution and some aluminum hydroxide complexes or gels which are of varying solubility. $\ce{AlCl3}$ has been used as a water treatment because it produces flocculent precipitates that carry down impurities.

$\ce{NiCl2}$ does not precipitate in water because $\ce{Ni(OH)2}$ is even more soluble than $\ce{Mg(OH)2}$.

$\ce{FeCl3}$ is quite acidic in water. You might expect the commercial 40% solution to be a cloudy product because of the production of acid, but it is a clear, deep red color. Oh, but when it is diluted and the pH rises because the acid is diluted, it precipitates in flocs like $\ce{AlCl3}$. It is used to clarify water by precipitating dirt and dust.


Why does aluminum have a lower first ionization energy than magnesium? I used to use the explanation that $s$ orbitals penetrate better than $p$ orbitals, however, could the explanation be that $3s$ is shielding the $3p$ in aluminum?

The easiest way to explain it is that $\ce{Al}$ has one unpaired electron in it's highest energy orbital ($\mathrm{3p}$), and $\ce{Mg}$'s highest energy orbital ($\mathrm{3s}$) the electrons are paired. It is energetically favorable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Here's what their orbital pairings look like:

and if you look at the general trend you can see that this occurs whenever all the valence electrons are paired:

Sources:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html
http://en.wikipedia.org/wiki/Electron_pair


Why do these reactions not occur? The following reactions do not occur:
$$\ce{HCl(aq)}\ +\ \ce{Pb(NO3)4(aq)} \to$$
$$\ce{HCl(aq)}\ +\ \ce{Cu(NO3)2(aq)} \to$$

Is this because in both cases both reactants are entirely dissolved as their components ion?
$\ce{NO3}$ is a conjugate base of a strong acid  so it is a very weak base so it doesn't participate in reactions, right?

When put in water a mixture such as $\ce{HCl}$ and $\ce{Pb(NO3)4}$ goes to complete ionic dissociation. Lead nitrate is soluble in water and hydrochloric acid is a very strong acid. Nitric acid is also strong, which means those dissolved nitrate ions are not reversing back to the molecular form.

Their ions will be fairly stable in solution and will only combine in case a more stable compound is formed. You could come up with reactions in which some compounds will go out of solution (like $\ce{Cl2}$), such as $\ce{ 2HCl + HNO3 \to HNO2 + Cl2 + H2O} $. Lead could also act as an oxidizing agent. But I would argue that if you consider the change in free energy or the net redox potential of those reactions, you'd find that they are quite unfavourable. If you want to calculate it itself, you could start from a reduction potential table like this one.


Aromatisation of a ring which includes nitrogen Question

Complete the following reaction sequence:

Answer

My solution
I was able to complete all the steps except for the last one. This is what I ended up with:

What I'm not getting is how the $\ce{-OH}$ groups are removed and the ring is aromatised. If anyone could provide me with a mechanism for the same, it would be really helpful.

This would be the mechanism I propose for the reaction:

I have taken the $\ce{pH}$ of the medium to be $4.5$. That should explain the protonation of the $\ce{-OH}$ groups.


Description of a van't Hoff apparatus A "van't Hoff apparatus" is mentioned as a means of performing a reaction involving gases in a reversible manner, for instance here.

Unfortunately I have been unable to find a description of such a device using a web search (including on this site). It may be that the device has other names. I intuit from the post above that the apparatus is similar to the device used to measure the Joule-Thompson coefficient of a gas.

Could someone provide a description of the nature and operation of such an apparatus (even if it is only a theoretical device)?

The Van't Hoff equilibrium box is a chamber containing a mixture of reactants and products at equilibrium.  It is connected to an array of cylinders, each containing one of the pure reactants or products.  The chamber itself is operated at constant total pressure and temperature.  The cylinders are each connected to the equilibrium box through a semipermeable membrane that only passes that species into the chamber.  Pure species are introduced in stoichiometric proportions to the chamber through these membranes at the same partial pressures as in the reaction mixture, and pure products are removed in stoichiometric proportions at the same partial pressures as in the reaction mixture.  So the operation of the equilibrium box is reversible, and the delta G in going from pure reactants in to pure products out is zero.


What is the percentage of d-tartaric acid in a mixture of d and l isomers? Specific rotation for pure tartaric acid is $-12 ^\circ$. A mixture of d-l isomers is prepared which has a specific rotation of $+6^\circ$. What is the percentage of d-tartaric acid in a mixture of $d$ and $l$ isomers?

The pure tartaric acid must have been $l$-tartaric acid.

My attempt:

$$\frac{l-d}{l+d}=\frac{+6}{-12}=\frac{1}{-2}$$ So $d=3l$. Which implies $$\frac{d}{d+l}=\frac{3}{3+1}=0.75 $$

However the answer given in my textbook is $66.67 \%$. Most probably I didn't do any calculation error. So where did I go wrong?

Your calculation is correct.

You can check your result since you know that pure (+)-tartaric acid has a specific rotation of $+12^\circ$ and that pure (−)-tartaric acid has a specific rotation of $-12^\circ$. Therefore, a mixture of $75\ \%$ (+)-tartaric acid and $25\ \%$ (−)-tartaric acid has a specific rotation of
$$0.75\times(+12^\circ)+0.25\times(-12^\circ)=6^\circ$$

A similar exercise can be found in Hart, Hadad, Craine, Hart Organic Chemistry – A Short Course: Google Books

The corresponding solution can be found here: Google Books


Conceptual understanding of solution / solute in a question 6.853g of sodium cyanide was dissolved in water and brought to volume in a 100ml volumetric flask. The final mass of the solution was found to be 107.166g

My understanding -

A solution is understood when I put a solute into a solvent (water) and dissolves it.

So in this case, if I wanted to find the composition of solution expressed as moles,
I must take mass multiply by the molecular weight of Sodium Cyanide, do I use the mass of the solution or the mass of solute ?

With this then I can go on to find molarity, normality as well as weight percent (mass of solute/mass of total solution) and weight/volume percent (mass of solute/volume of total solution).

Mole fraction is the ratio of the number of mole of a component to that of the number of moles of the total mixture. In your case, 6.853g of NaCN equals 6.853/49.002 = 0.139 moles.

The rest is water i.e. 107.166 - 6.853 = 100.313g. this corresponds to (100.313/18.015) = 5.568 moles.

Now, total moles = 0.1398 + 5.568 = 5.708moles.

The mole fraction of NaCN = 0.139/5.708 = 0.024.


Reaction between sodium carbonate and water In this Wikipedia article, it mentions:

  As the concentration of sodium carbonate increases, it undergoes
  hydrolysis to form sodium hydroxide.
  $$\ce{Na2CO3 + H2O → 2NaOH + CO2}$$

Because this occurs in boilers, I can assume that it requires a heat of no more than 100 degrees Celsius. Why is this process not used when commercially producing $\ce{NaOH}$, rather than the chloralkali process?

Possible reasons:

Not enough concentration of sodium hydroxide is formed
The sodium hydroxide is contaminated with significant amounts of sodium carbonate
The reaction is too slow

The three reasons you propose are technically true, but they are not the reason why this is not the process used to produce sodium hydroxide.
But let me first clarify some points (and rant about a bit).

Hydrolysis is a reaction where a compound is split whilst incorporating a molecule of water. The reaction :
$$\mathrm{Na_2CO_3 + H_2O \rightarrow 2 NaOH + CO_2}$$
is not a hydrolysis. It is classified as an acido-basic reaction ($\mathrm{H^{+}}$ exchange) followed by a dehydration reaction. A dehydration is kind of the opposite of an hydrolysis (the true opposite would be called hydration), because is involves the loss of water. The transformation decomposes according to (forgetting about sodium ions):
$$\mathrm{{CO_3}^{2-} + H_2O \rightarrow {HCO_3}^- + HO^-} \\
\mathrm{{HCO_3}^{-} + H_2O \rightarrow {H_2CO_3} + HO^-} \\
\mathrm{H_2CO_3 \rightarrow CO_2 + H_2O }$$
$\mathrm{H_2CO_3}$ is carbonic acid. Carbon dioxide is its anhydride.
The overall transformation actually consumes two molecules of water as acid, and one molecule of water is produced upon dehydration of carbonic acid. But this does not appear in the simplified equation above.

This reaction is actually not thermodynamically favorable ; the reverse transformation is the one that is spontaneous. What drives it here is the fact that the carbon dioxide is gaseous and is driven off the water, into the atmosphere. Unless you work under a $\mathrm{CO_2}$-free atmosphere, it would not be possible to produce sodium hydroxide not contaminated with carbonate.

Now, why is this not used industrially ? The answer lies in the cost of the processes : investments, availability of reactants, price of energy, etc.

Calcinating sodium carbonate at more than 100°C actually gives off sodium oxide ($\mathrm{Na_2O}$), but it also releases $\mathrm{CO_2}$ in the atmosphere, and you have to obtain sodium carbonate first. This is no commonplace mineral. Most of it is actually produced by the Solvay process, which converts two plentiful materials, namely calcium carbonate and sodium chloride into calcium chloride and sodium carbonate :
$$\mathrm{2\,NaCl + CaCO_3 \rightarrow Na_2CO_3 + CaCl_2}$$

The chloralkali process produces sodium hydroxide (and chlorine) directly from sea salt, and the  products are hardly contaminated.


Understanding a concentration term I am having a problem wrapping my mind around the idea that in $230\:\mathrm{mL}$ of $0.615\:\mathrm{M}$ $\ce{KCl}$ there is $0.141\:\mathrm{mol}$ of $\ce{Cl^-}$ ions. I guess I still do not understand moles.

Is this calculation saying that there $0.141\:\mathrm{mol}$ of $\ce{K}$ and $\ce{Cl}$ in one mole of $\ce{KCl}$?

I know this may be a silly question, but I just want to make sure I am following the correct way of thinking when it comes to this concept.

Your concentration is 0.615 M. The M stands for molar concentration, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution.
A mole is just like any other unit for amount, except that it is built on a very large number. Since, for example, there at $6.02\times 10^{23}$ atoms of chlorine in 35.45 grams of chlorine, we generally do not discuss numbers of atoms (or molecules) as a useful measurable amount. How much would a dozen atoms of chlorine weigh? Much less than any of our balances or scales could measure!
One mole of stuff is an amount of stuff equal to the number of carbon atoms in 12 grams of $\ce{^{12}C}$, most common isotope of carbon. This number happens to be Avogadro's Number - $6.02\times 10^{23}$, which is really large! This is also the number of atoms or molecules present in a number of grams equal to one formula weight of that substance, so it is a convenient number. We know that when we weigh out 12.01 grams of carbon, 1.008 grams of hydrogen, 15.99 grams of oxygen, etc., we know that we always have the same number of atoms in each sample, because each is one mole. We tend not to worry too much about the large number hiding in the definition of one mole.
Now, if you have one mole of a compound, like $\ce{KCl}$, then you know that there is one mole of the $\ce{KCl}$ formula unit in that sample. Since each $\ce{KCl}$ formula unit has one $\ce{K+}$ and one $\ce{Cl-}$, then each mole of $\ce{KCl}$ has one mole of $\ce{K+}$ ions and one mole of $\ce{Cl-}$ ions.
Now, you have your 0.615 M solution, and you only have 230 mL of it. How much $\ce{KCl}$, $\ce{K+}$, and $\ce{Cl-}$ is in that sample? Well, you would have 0.615 moles of each in 1 liter, but you do not have one liter. You have 230 mL. Since there are 1000 mL in a liter, we can do a little maths:
$$230\text{ mL}\times \dfrac{1 \text{ L}}{1000\text{ mL}}\times\dfrac{0.615\text{ mol KCl}}{1\text{ L}}=0.141 \text{ mol KCl}\equiv 0.141 \text{ mol}\,\ce{K+}\equiv 0.141 \text{ mol}\,\ce{Cl-}$$


How mass spec formate and acetate adducts masses are calculated? If I look at any tables or calculators, I find that negative ion formate adduct has a monoisotopic mass of +44.9982 units. If I try to calculate myself, I get slightly different result:

H + C + O + O =
1.007825 + 12.0107 + 15.9994 + 15.9994 =
45.017325

The difference is 0.019, what is already not indifferent, as usually I calculate with a tolerance of +- 0.02 units, which means this can cause me miss many hits. I could not find out what may cause this difference.

Later had similar experiences with the acetate, as in tables +59.013851 can be found, but if I calculate:

3 x H + 2 x C + 2 X O =
3 x 1.007825 + 2 x 12.0107 + 2 x 15.9994 =
59.043748

Which means even more, 0.0298 difference.

My question is what causes the difference, and what is the correct wat to calculate these masses?

In your calculation, you are using the average molecular mass of the structure, where atomic masses are based on the natural abundance of all isotopes of the element.

H:  1.00794 u
C: 12.01070 u
O: 15.99940 u
O: 15.99940 u
─────────────
   45.01744 u

However, for the exact molecular mass (or monoisotopic mass) of the structure, the atomic masses of each atom are based on the most common isotope for the element.

H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
────────────────
      44.99765 u

Furthermore, you are missing the mass of the extra electron of the formate ion. The electron mass is $m_\mathrm e=0.000548579909070(16)\ \mathrm u$. Thus,

H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
e−:    0.00055 u
────────────────
      44.99820 u


Shielding effects and atomic size The atomic size on going from aluminum to gallium decreases because of poor shielding effect of the $(n-1)d$ electrons, but on going from copper to zinc, the size increases due to the same shielding effect. Why ?

Aluminum and gallium are on different periods. When an additional shell is (fictitiously) added to aluminum, the size is seen to decrease. This is, as correctly mentioned, due to the poor shielding offered by the $(n-1)d$ orbitals to the incoming electrons in the ultimate $p$ subshell.

However, between copper and zinc, the matter seems to be different–their electronic configurations are different.

While copper has a $4s^13d^{10}$ arrangement, zinc has a $4s^23d^{10}$. I would presume the fully filled spherical $4s$ orbital to considerably repel (also read as: shield) the d-electrons,  and hence the effect appears as a minute increase in size.

There can be deeper discourses, but I think this should suffice for now.


Which platonic solids have been built as carbon skeletons? There are 5 platonic solids: the tetrahedron, the octahedron, the cube, the dodecahedron and the icosahedron.

How many of these have been realised as carbon skeletons by chemical synthesis?

The previous answer and comments have already listed out the known platonic hydrocarbons, however, since the OP had expressed an interest in the synthesis of these platonic hydrocarbons in one of his comments, I have provided some commentary to that effect along with links to the original papers that first reported these compounds. Also, of interest would be this paper
Tetrahedrane
Technically speaking, Tetrahedrane ($\ce{C4H4}$) still remains a hypothetical compound, as an unsubstituted version hasn't been synthesised yet.

Two stable derivatives, whose synthesis has been achieved include tetra-tert-butyltetrahedrane (prepared in 1978 by Günther Maier) and tetra(trimethylsilyl)tetrahedrane (reported in 2002).
Tetra-tert-butyltetrahedrane
The original paper discussing the synthesis can be found here. The synthesis has been summarised below. The authors rationalise the stability of this compound as follows:

If  all  the  corners of the tetrahedrane unit are occupied by bulky groups then on lengthening  one bond of the  ring  skeleton , two  tert-butyl groups  tending  to increase  their  mutual  separation  will  be forced  back  together  by  the  other ones  ("corset  effect"); a tetrahedral  structure permits  the  four  substituents  to adopt a regular spherical distribution and hence a maximum mutual separation.

In simpler terms, trying to break a bond forces the bulky substituents together increasing strain.

Tetrakis(trimethylsilyl)tetrahedrane

Again, the synthetic route is given below. The original paper is also by Günther Maier.
Reaction with methyllithium has yielded the stable tetrahedranyllithium derivative. These coupling reactions with this lithium compound have allowed access to more derivatives, for example the dimer in the last step of the reaction scheme given above.
This compound is far more stable than the tert-butyl analogue. On account of the silicon–carbon bond being longer than a carbon–carbon bond, the corset effect is reduced. $\sigma$-donor character of the trimethylsilyl explains the increased stabilisation of the tetrahedrane.
The tetrahedrane skeleton is made up of banana bonds, and hence the carbon atoms are high in s-orbital character
Cubane

Cubane was first synthetically realised by Eaton and Cole in 1964. The synthetic scheme from the original paper is given below:

Note: In the original synthesis, the starting material for this, 2-bromocyclopentadienone, was obtained from 2-cyclopentenone via allylic bromination with N-bromosuccinimide in carbon tetrachloride followed by addition of molecular bromine to the alkene gives a 2,3,4-tribromocyclopentanone. After treatment of this compound with diethylamine in diethyl ether, which causes elimination of two equivalents of hydrogen bromide, we get the desired diene product.
Again, the cubane is kinetically stable because of lack of readily accessible decomposition paths.
Dodecahedrane

Dodecahedrane was synthesised by Leo Paquette in 1982, and later by Prinzbach (via pagodane) in 1987.
The pagodane synthetic route has been given below.

Paquette's 1982 organic synthesis takes about 29 steps, and starts from cyclopentadiene, and for the sake of brevity, I have not presented it here.


Reaction of copper (II) ion reacting with sodium hydroxide Based on an experiment, a solution that contained $\ce{Cu^2+}$ formed a precipitate with $\ce{NaOH}$. Is this how you write the equation?

$$\ce{2NaOH + Cu^{2+} -> Cu(OH)2 + 2Na+}$$

On the experiment: solution A is mixed with NaOH (solution turned white). Using a flame test solution A is color green. I figured solution A has $\ce{Cu^{2+}}$ ion.

Yes, the reaction is correct however it may be preferable to either include all or no spectator ions.
Use either a net ionic equations (omit the Na+), molecular equation (include the copper compound full) or complete ionic equation.

For example, if the compound is copper sulphate the equations would be (without states)

Net ionic
$$\ce{Cu^2+}+\ce{2OH^-->Cu(OH)_2}$$
Molecular equation
$$\ce{CuSO_4}+\ce{2NaOH->Cu(OH)_2}+\ce{Na_2SO_4}$$
Complete Ionic Equation
$$\ce{Cu^2+ +SO_4^{2-}}+\ce{2Na^+ +2OH^-->Cu(OH)_2}+\ce{2Na^+ +SO_4^2-}$$

Edit

In the case where you don't know what copper compound it is, use a net ionic equation. As a side note, copper hydroxide is not white. You might want to check your calculations.

If you're wondering what it is, check what other compounds will emit green.


Is the neutral di-µ-oxo-bridged manganese dimer with eight ammonia ligands a reasonable structure? In its guide to model antiferromagnetic couplig, Gaussian uses $\ce{Mn2O2(NH3)8}$ as a model compound.[1] It argues that manganese is in II oxidation high spin state; in other words, in both manganese are all d orbitals are singly occupied. Furthermore it creates a structure, where the metal centres are octahedral coordinated. The level of theory used is U-B3LYP/3-21G, which is in my opinion not enough to discuss these complexes in the first case.

I could not find any evidence of a structure proposed by Gaussian, the closest I have gotten is given as a cation $\ce{[Mn2O2(NH3)8]^z+}, (z\in2,3,4)$.[2]

I have reason to doubt the structure given in the tutorial,[3] as in the $\frac{5}{2}$ high spin case there are only four orbitals in manganese left which could accept bonding partners. Given that two of them are used by the bridging oxo ligands, only two ammonia should bond to the manganese centres. This would lead to the formula $\ce{Mn2O2(NH3)4}$. The situation changes once you start removing electrons or pair them up. Then the question changes to whether ammonia is a strong enough ligand to induce lower spin states.

Summarising: Is $\ce{Mn2O2(NH3)8}$ a reasonable structure?

For convenience, here is the structure as proposed by Gaussian (they do create a different one within the tutorial though):

References & Notes:

Gaussian Website: http://gaussian.com/afc/
Mcgrady, J. E.; Stranger, R. Redox-Induced Changes in the Geometry and Electronic Structure of Di-μ-oxo-Bridged Manganese Dimers. J. Am. Chem. Soc. 1997, 119 (36), 8512–8522 DOI: 10.1021/ja964360r.
In my optimisations on a different level of theory I was unable to reproduce the structure.

The structure is extremely reasonable to the point where I would ask why any other di-oxido-bridged manganese(II) ammine complex would not show this structure if it were proposed. To describe why, I am going to be lazy and just copy my molecular orbital scheme of a general octahedral 3d-metal complex that I have used many times before.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

As you can see, the four empty orbitals ($\mathrm{a_{1g}}$ and $\mathrm{t_{1u}}$) are indeed present and part of the bond-forming orbitals, but the fact that the $\mathrm{e_g}$ orbitals are also half-filled does not prevent them from participating, too. In fact, these orbitals turn into $\mathrm{e_g^*}$ orbitals in the resulting complex, formally antibonding. This, however, doesn’t matter, because there are still many more electrons in bonding orbitals ($\mathrm{t_{1u}, a_{1g}}$ and $\mathrm{e_g}$) that this small amount of lost energy is more than offset. Typically, for a high-spin $\mathrm{d^5}$ complex, the bond order would be somewhere around 5/6. Still well within the overall bonding range.

If one looks at the kinetic stability of various complexes, one will find that $\mathrm{d^3}$ and low-spin $\mathrm{d^6}$ often give very inert octahedral complexes as no antibonding orbitals (of the simplified scheme) are populated—but even a $\mathrm{d^{10}}$ or $\mathrm{d^0}$ metal will form these complexes, their structures just won’t be as long-lasting. (Zinc has been shown to equilibrate between tetra, penta and hexacoordinated complexes in aquaeous solution as I was told in my coordination chemistry lecture.)

So yes, this octahedral complex is perfectly fine.


How does NaOH modify pyranodioxins? I am given the molecule in black which is treated with $\ce{NaOH}$. The product is in blue.
Could someone explain the mechanism? All I have so far is that the $\ce{-OH}$ group on the bottom right carbon has been deprotonated to $\ce{-O-},$ but am not sure how the rest of the reaction should proceed.

The molecule in black consists of a cyclic ketal and a hemiacetal. The hemiacetal is known to be in equilibrium with its aldehyde form, which is the the molecule in blue, in aqueous solutions even without $\ce{NaOH}$. Recall the reaction of glucose (a cyclic hemiacetal) with dilute sodium hydroxide, which is known as Lobry de Bruyn-van Ekentein rearrangement (Ref.1).
Nonetheless, a hemiacetals can be decomposed to the relevant carbonyl and the alcohol under basic conditions as explained here (a reliable mechanism is given). In your case both carbonyl and the alcohol groups are within the same molecule (the molecule in blue):

All acetals and ketals are stable to alkaline conditions since they don't carry free hydroxyl group.

References:

Roy L. Whistler, J. N. BeMiller, "Alkaline Degradation of Polysaccharides," Advances in Carbohydrate Chemistry 1958, 13, 289-329 (DOI: https://doi.org/10.1016/S0096-5332(08)60359-8).


Calculating the concentration of copper(II) from a non-standard cell potential A galvanic cell consists of a standard hydrogen electrode and a copper electrode.
  \begin{align}
  \ce{2 H+ (aq) + 2e- &-> H2 (g)}&
  E^\circ &= 0.00\ \mathrm{V}\\
  \ce{Cu^2+ (aq) + 2 e- &-> Cu (s)}&
  E^\circ &= 0.34\ \mathrm{V}
\end{align}
  Suppose the copper electrode is placed in a solution of unknown copper (II) concentration. The cell potential is measured to be $\pu{0.195 V}$. Determine the concentration of $\ce{Cu^2+}$.

Since there's no indication that the $\ce{Cu^2+}$ is in the standard state, I started off by using the formula for non-standard cell potential:
$$E=E^\circ-\frac{RT}{nF}\ln Q$$

However, I can’t seem to get the right answer. I was wondering if I set up the problem incorrectly or how I go about getting the concentration of copper.

You are on the right track with that equation. There is a problem with the question, though: you do not know which direction the reaction is proceeding in. Therefore, you get two possible answers.

One of the two possible ways to write the reaction that's occurring is

$$\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+ (aq)}$$

For this reaction, the value of $E_\mathrm{cell}^\circ$ is given by $E^\circ(\ce{Cu^2+}/\ce{Cu}) - E^\circ(\ce{H+}/\ce{H2}) = \pu{+0.34 V}$. This seems to be the more logical direction for the reaction to proceed in, as we have a positive value of $E_\mathrm{cell}^\circ$. Consequently, to achieve the smaller potential difference of $\pu{0.195 V}$, all we need to do is to reduce the concentration of $\ce{Cu^2+}$ from the standard of $\pu{1 mol dm-3}$, so that the forward reaction becomes slightly less favourable. The corresponding value of $n$ is $2$. The reaction quotient can be approximated by

$$Q = \frac{([\ce{H+}]/c^\circ)^2}{([\ce{Cu^2+}]/c^\circ) \cdot (p(\ce{H2})/p^\circ)} = \frac{c^\circ}{[\ce{Cu^2+}]}$$

since the $\ce{H+}/\ce{H2}$ electrode is operating under standard conditions (and where $c^\circ = \pu{1 mol dm-3}$). The temperature is not specified, but we could assume it to be $\pu{298 K}$.

Therefore:

$$\begin{align}
E_\mathrm{cell} &= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &= \pu{0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{c^\circ}{[\ce{Cu^2+}]}\right) \\
[\ce{Cu^2+}] &= \pu{1.246 \times 10^-5 mol dm-3}
\end{align}$$

If the reaction is actually running in the reverse direction, then you have the reaction

$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$

with $E^\circ_\mathrm{cell} = \pu{-0.34 V}$. Going through exactly the same steps as above (but taking care to invert the expression for $Q$, since the reactants and products are swapped), you get to

$$\begin{align}
E_\mathrm{cell} &= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &= \pu{-0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{[\ce{Cu^2+}]}{c^\circ}\right) \\
[\ce{Cu^2+}] &= \pu{8.019 \times 10^-19 mol dm-3}
\end{align}$$

What's happening here? It turns out that as you reduce the concentration of $\ce{Cu^2+}$ further, $E_\mathrm{cell}$ drops more and more, until it actually becomes thermodynamically favourable for the reverse reaction to occur. This concentration isn't particularly sensible, but you can certainly achieve it by serial dilution.

How do we resolve this dilemma? The most straightforward way is to find out which electrode is the anode and which is the cathode, which you can do with the voltmeter. For voltmeters to display a positive reading, as is suggested by the question, you need to have electrons flowing from the black wire to the red wire through the voltmeter. This means that the black wire must be connected to the electrode which provides it with electrons, i.e. the anode, since the anode is the site of oxidation (the chemical species gives up its electrons to the electrode and is itself oxidised in the process).

So, if your voltmeter displays a positive reading and the black wire is connected to the standard hydrogen electrode, it means that $\ce{H2}$ is being oxidised to $\ce{H+}$. That means that we are in the first situation above, where the forward reaction $\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+ (aq)}$ is occurring.


Observability of Orbitals and Orbital Energies This question comes from some thoughts I had after reading this question.

First of all, is an orbital an observable? I know the answer to this question is no because there is no "single-orbital operator" or whatever you'd like to call this. I brig this up because exactly what this means should be a part of a thorough answer. Plus an orbital is a state and we don't observe the eigenvectors but the eigenvalues.

Second, is the energy of an orbital an observable? It's more tempting to think that the answer to this is yes, but if it's true that there is no "single-orbital operator" then there must be no corresponding eigenvalue which represents the energy of this orbital. As I understand it, if one runs a HF calculation and sums the energies calculated for the individual orbitals, this is the total electronic energy. In the answer linked above, however, it is said that there is an infinite number of unitary transformations for a given set of orbitals to another set. Is there any correspondence between the energies of these transformed orbitals and the original set of orbitals? By that I mean, the orbitals themselves change, but would the optimized energies of each orbital change and it is the only the wavefunction and total energy which stay constant?

Finally, how do the answers to the above questions connect with Koopmans' theorem and photoelectron spectroscopy? That is to say, Koopmans' theorem says that the energy of the HOMO as calculated from Hartree-Fock corresponds to the first ionization energy of the system. Additionally, in photoelectron spectroscopy, it sure seems like people are observing the energies of individual orbitals, but I don't really know anything about this technique in detail so that may just be a misunderstanding.

Additionally, is Koopmans' theorem still true once we perform a unitary transformation on the Hartree-Fock wavefunction?

The reason I lump all these questions together is because I think in order to sufficiently answer the question of whether or not an orbital energy is observable, all of these points should be addressed, so I just asked them all at once. I hope it's not overkill.

Excluding the trivial case of a one electron system, where the wavefunction for the system is an orbital (1 electron wavefunction) and so could be measured via the electron density, orbitals are not observables. There is no known Hermitian Operator that returns orbitals (or their energies) from its expectation value. So is that the end of the discussion? Not necessarily.

In a 2006 paper, Shabazian and Zahedi$^1$ argue that what constitutes an observable depends on the conventions we establish for describing a system. They give as an example the Quantum Theory of Atoms in Molecules (QTAIM) developed by Richard Bader$^2$. Strictly speaking, the normal formulation of Quantum Mechanics doesn't allow us to describe subsystems within a molecule. We form molecules from atoms, but once the molecule is formed, there are no longer atoms; rather there is just a single wavefunction which describes the whole system. This is horribly inconvenient from a classical chemists perspective, as much of our understanding of chemistry comes from subdividing molecules into important functional groups that give the major properties of the molecule. To resolve this, Bader proposed a definition of an atom in a molecule by defining boundary conditions for a atom based on flux of the electron density. This gives an unambiguous, experimentally measurable way of determining observables like atomic charge, which without this set of external conditions is not considered an observable at all. While this has as of yet not been done, Shabazian and Zahedi argue that orbitals could be made observables by establishing a similar set of external conditions.

So while orbitals are not currently observables, what is and isn't an observable can change as new theoretical models develop.

Shahbazian, S. & Zahedi, M. Found Chem (2006) 8: 37. https://doi.org/10.1007/s10698-005-8247-4
Bader, R. Chem. Rev., 1991, 91 (5), pp 893–928
DOI: 10.1021/cr00005a013


Formation of Acetal from Hemiacetal? In the reaction mechanism of formation of Acetal from Hemiacetal, why doesn't H+ attack the Oxygen of OR Group instead of the OH?

Since O in OR group has +I effect from two alkyl groups, shouldn't it donate its lp more easily?

The acetal reaction is performed in $\ce{H+/ROH}$ condition (e.g., $\ce{H+/CH3OH}$). The general mechanism can be found in any Organic Chemistry textbook. The reaction is reversible. The first step is protonation of aldehyde (or ketone) carbonyl oxygen by an acid ($\ce{H+}$):

The protonation makes the carbonyl carbon more susceptible to nucleophilic attack by available alcohol (e.g., methanol in the scheme). This gives the tetrahedral protonated hemiactal intermediate, which undergoes deprotonation to give stable hemiacetal. All of these steps are reversible, but forward reaction is favored due to highly available alcohol molecules (Le Chatelier's Principle).

The hemiketal has two oxygen molecules, $\ce{OH}$ and $\ce{OR}$, both of which are susceptible to protonate. However, if $\ce{OR}$ is protonate, reaction would go backward. Of cause, it protonates because the reaction is reversible (see the scheme). Yet, more favored is $\ce{OH}$, which would subsequently eliminate as a $\ce{H2O}$ molecule. The resultant positively charged carbonyl compound undergoes second nucleophilic attack by another $\ce{CH3OH}$ molecule to give protonated acetal, which proceeds to give expected acetal by deprotonation. Since we are using the alcohol as the reaction solvent, forward reaction is favored by multiple times.

OP's suggestion of protonation of $\ce{OH}$ group is legitimate, but reaction goes backward for a moment, but forward reaction favorably proceed to the end (Recall Fischer esterification). If you want back your aldehyde (or ketone), you can make backward reaction favored by using water as your solvent ($\ce{H+/H2O}$).


Do minerals and/or metals found in water tend to settle to the bottom? I was reading an article that suggested that if one lets water sit for a day in a bottle or pot, then some of the metals in the water, such as lead, will settle to the bottom.

Does anyone know if this theory has been tested?

The reasoning seems somewhat logical, since particles in water tend to sink to the bottom. However, at the molecular level I know that ionic interactions may cause unintuitive behaviors.

I do have a TDS and pH meter. Perhaps I could let water sit for a day, then check the TDS of the water on top. Then compare that with the TDS of water that has been thoroughly mixed. Does anyone see any flaws with this idea?

  Excessive toxic metals.

  Many city water supplies, especially wells,
  contain some excessive amounts toxic metals or toxic amounts or forms
  of other minerals.  Common toxic metals include arsenic and lead,
  which are found in many pesticides and herbicides and find their way
  into the wells and other water supplies.  At times, these are
  difficult to remove from large amounts of water, so they are just left
  there even if the water violates environmental standards, especially
  in poorer nations.

  Other minerals that often contaminate drinking
  water include too much iron, manganese, boron and copper depending on
  the location. One way to remove some of these easily is to let a
  pitcher or pot full of tap water sit for a day and some of the metals
  will settle to the bottom of the pot.

  You can then pour off the
  water, leaving most of the minerals in the bottom of the pot.  You may
  lose some beneficial minerals this way, but it is one way to remove
  some toxic metals, as well.

-Dr. Lawrence Wilson

http://www.drlwilson.com/ARTICLES/WATER.htm#INTRO

I have experimented with this phenomenon and found that allowing water to sit for a day may have a non-trivial effect on the TDS gradient in the water. This effect could be due to various reasons, however.

Materials: Glass pitcher, plastic & porcelain & glass containers; tap water from near Chicago, IL, USA; TDS meter

Methods: I poured tap water into various containers and allowed the water to stand for about 30 hours. I then measured the TDS at the top of the water. Then I stirred the water vigorously and let the water sit for about a minute. I took a TDS reading again at the top of the water. I repeated the experiment a total of 4 times.

Results (Updated 12/23/2016):
$$\begin{array}{ccc}
 & \text{TDS Before Stirring} & \text{TDS After Stirring} \\ \hline
\text{Glass} & 138 & 141 \\
\text{Plastic} & 150 & 146 \\
\text{Porcelain} & 153 & 152 \\
\text{Glass} & 147 & 149 \\
\end{array}$$

Conclusion: The results are interesting. I will discuss a few possible reasons behind the results, but I'm not certain.

TDS (after) was higher both times for glass. Maybe this is because when the water sits, a small number of ions adsorb to the glass surface.
For plastic and porcelain, TDA (after) was lower for both plastic and porcelain. This may be due to dust that settled and dissolved into the water, as I did not cover the containers (looking back, covering the containers would probably have made the experiment more controlled). In addition, maybe ion adsorption happens less with porcelain and plastic.


What is electrolytic protection? My highschool textbook tells me that sacrificial protection (more technically-cathodic protection) and electrolytic protection are different. I have searched high and low for an explanation of electrolytic protection, but I got nothing.

It describes the electrolytic protection, stating that the steel/iron structure is hooked up to an electrolytic (not electrochemical!) cell as the negative terminal (cathode) and the anode is an inert material (titanium, in the example).

Why does this work, if it does at all?

I understand why a HS text might call it "electrolytic" , but in the real world it is called "cathodic protection" or CP. That is what to look for on the net. It is a multibillion dollar industry. Essentially all pipelines ( oil, gas, products, water , etc.) use it . A large majority off offshore steel structures ( oil  platforms, ships, etc) use it. And many steel products are galvanized . There are two basic types ; sacrificial coatings and impressed current. Sacrificial coatings are represented by galvanized steel , also aluminum and cadmium coatings. Impressed current, CP is also  two categories : Sacrificial anodes , usually aluminum ( which do not use rectifiers) , and inert anodes , usually titanium or graphite. These are buried in soils made more conductive with graphite powder (typically) and electrically connected to rectifiers  which are then connected to the steel to be protected . The voltage is adjusted and monitored along the protected steel , eg. pipeline. Pipelines are coated with an electrically insulating coating to reduce the amount of current required to maintain a protective voltage , normally lower than - 0.65 V ( Cu - Cu SO4 half cell) Too low a voltage can cause hydrogen gas bubbles under the coating and hydrogen embrittlement of certain steels.There is also anodic protection but it has very specific limited applications. So, there is such a thing as "electrolytic" protection.


Electrophilicities of carboxylic acid derivatives I would like to ask about the rate of hydrolysis for 4-hydroxyphenyl acetate (A), N-(4-hydroxyphenyl)acetamide (B), 4-hydroxyphenyl chloroacetate (C), and methyl 4-hydroxybenzoate (D). I know that hydrolysis depends on the positive charge of the carbonyl carbon and also the substituent around it (e.g. electron withdrawing).

Therefore, from the picture above I can see that compound C is the most reactive since it has a phenol and a chlorine as its substituents. Compound B is the least reactive due to nitrogen being the weakest electronegatively as compared to oxygen (I think?).

I am confused between compounds A and D because they are isomers of each other. The oxygen and phenol group are on the same substituent in compound A and are in two different substituent in compound D. Does this affect anything? or can I say that the strength of the electron withdrawing groups are additive, meaning compound A and D are similar in terms of reactivity?

Background

When a carbonyl compound reacts with a nucleophile we form a tetrahedral intermediate.  Since the carbonyl carbon becomes $\ce{sp^3}$ hybridized in the tetrahedral intermediate there is no significant resonance stabilization - the various tetrahedral intermediates have (approximately) the same energy.  However, strong resonance effects can occur in the starting carbonyl compound, potentially stabilizing the carbonyl compound.

In other words, while a substituent attached to the carbonyl carbon can stabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

If we place the carbonyl compound and the tetrahedral intermediate on a reaction coordinate we can see that anything that stabilizes the carbonyl compound will increase the activation energy for reaction, slowing down the reaction rate.

Below are the resonance structures we can draw for our carbonyl compound.

If structures II and  III contribute significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structures II and III contribute less.

Answer

Three of your compounds are esters, one is an amide.  We know that resonance structure III is more important in an amide than in an ester (due to electronegativity differences, positive charge on nitrogen is better than positive charge on oxygen).  Therefore the amide (B) will react the slowest.

Now to separate the three esters.  You are right that ester C will be the most reactive.  This is because the inductively electron withdrawing chlorine substituent destabilizes resonance structure II.

In compound D we can draw stabilizing resonance structures like the following;  the para-hydroxy substituted aromatic ring is very effective at moving electron density towards the electron deficient carbonyl carbon thereby stabilizing it.

The methyl group in compound A also donates electron density to the carbonyl carbon, but not nearly as effectively as the aromatic ring.

Therefore our overall order of carbonyl reactivity is C > A > D > B


Role of THF in oxymercuration of alkenes I am trying to self learn solomon's organic chemistry book, but there is no explanation over why we should use THF in oxymercuration. After a quick search I found out that THF help stabilize in hydroboration, but is it the same for oxymercuration? and I still don't understand why and how THF can help stabilize in hydroboration.

$\ce{THF}$ is a polar aprotic solvent, that is, it can stabilize positive charges but not negative ones since it won't form hydrogen bonds with the help of any of its own hydrogen atoms.

Thus the carbocationic intermediate, that is the mercurinium ion is still stabilized, allowing the reaction to proceed.

And in Hydroboration, $\ce{THF}$ acts as the solvent as well but plays the additional role of shifting the equilibrium between diborane and borane to the side of the latter. As you might know, the Boron atom of $\ce{BH3}$ is electron deficient, which plays a key role in the formation of the trialkylborane intermediate of the Hydroboration reaction. Formation of diborane quenches this electron deficiency of borane by the formation of two 3-centred, 2-electron bonds in the dimer.


Do both iodine and potassium iodide turn dark in the presence of starch? I will do a titration to standardise a solution of sodium thiosulfate with potassium permanganate solution and potassium iodide solution.
My book says that when the potassium permanganate solution begins to lighten, add a few drops of starch indicator and titrate until the solution turns clear.
This seems to suggest to me that starch turns $\ce{I_2}$ dark but not $\ce{KI}$ as some of the $\ce{KI}$ solution would be still in the flask. Is this the case?

Starch is a long ribbon or a long filament made of a great number of glucose units attached to one another like the wagons in a train. But this filament is wound in a helicoidal way. And the inner part of this helix is a sort of long hole. And the dimension of this hole is just big enough to allow the iodine molecule $\ce{I2}$ to enter and stay there, maintained by van der Wals forces. But when the iodine molecule is inserted in the starch helix, it produces a "starch-iodine" complex which has an intense blue color.


Predicting relative solubilites of salts without Ksp values Is there a way of comparing which salt is more soluble without knowing their $ K_{sp}$ data or performing experiments?

For example, comparing $\ce{Hg2Cl2}$ and $\ce{Cr2(SO4)3}$.

I am not sure but can I compare lattice and hydration energy relatively or the ion size and estimate?

Experiments will give concrete results but I was wondering if there is another method just to gauge the solubility.

(The relevant chat discussion)

To a certain extent, you can predict the relative solubilities of two salts, especially when there is a large difference in solubility. The existing answer mentions Fajans' rules and obviously it is quite easy to predict which of $\ce{NaCl}$ and $\ce{AgI}$ is more soluble.

But I'm inclined to be pessimistic and say no, this is not generally applicable.

You're perhaps on the right track. The dissolution of the salt arises from a couple of measurable quantities: the lattice enthalpy of the salt, and hydration enthalpies of the ions. You could measure both, and try to come up with a trend.

However that's often far easier said than done. The trends are sometimes quite subtle, and even if you have all the relevant data and limit yourself to a "simple" system like the alkali metal halides, you'll find that the interplay between two factors leads to a very delicate balance, and the resulting trend is quite complex. I don't think it's something that we can predict solely in our heads without doing any measurements.

For example, if you simply compare $\ce{NaF}$ and $\ce{NaI}$, $\ce{NaI}$ has a smaller lattice enthalpy (favours dissolution) but the $\ce{I-}$ ion also has a smaller hydration enthalpy (disfavours dissolution). So which factor wins out? There's no simple heuristic or rule of thumb that applies here, not any that I know of, at least.

How about two obviously more different salts? Let's try $\ce{NaF}$ and $\ce{MgF2}$. $\ce{NaF}$ has a smaller lattice enthalpy (favours dissolution) but also the sum of the hydration enthalpies is smaller than $\ce{MgF2}$ (more ions and $\ce{Mg^2+}$ is more charged than $\ce{Na+}$ - which disfavours dissolution of $\ce{NaF}$). Again we have the same dilemma.

In fact, that's just the enthalpic changes, so only half the story. There's also entropy to care about, which can either be positive or negative (and it's extremely difficult to predict this!).


Why does soap reduce surface tension of water One explanation I read is that soap's long non-polar end will seek the air at the interface between water and air, and that the ionic end will form ionic-dipole bonds with the water. In the process of moving to the surface, the soap molecules force apart the water molecules, and hence the water molecules no longer have hydrogen bonds with each other, and the surface tension is weakened. However, wouldn't the ionic heads of the soap molecules just form stronger ionic-dipole bonds with the water, increasing inter molecular forces, and therefore increasing surface tension? What is the true explanation for why soap reduces surface tension of water?

Apoorv did not give the right explanation, because this would imply, that surface tension goes especially down, when micelles are formed, but the opposite is the case. Compare e.g.: first chapter of The Colloidal Domain by D.Fennell Evans and Håkan Wennerström.

First thing to note is that pure water has a higher surface tension than hydrocarbon chains which is due to the strong dipole dipole interaction.

If there is a very dilute solution of surfactant in water no micelles are formed but you have monomers "floating around". In contrast to ions in water they are not homogenously distributed, but they concentrate especially on the surface where their polar head is directed to the water and their hydrophobic chain is directed to the air. In the end you substituted water molecules with hydrocarbon chains which leads to a lower surface tension.

If you increase the concentration of surfactant, they start to form micelles in the water (and not on the surface). So after reaching this critical micelle concentration (called CMC) the surface tension stays quite constant. This constant behaviour is used experimentally to estimate the CMC.

See the figure below Equation 11:
http://infohost.nmt.edu/~jaltig/SurfaceTension.pdf

Please note that Apoorv described some kind of reversed micelle. The normal way is that they stick their polar head to the water and have their hydrocarbon chain inside the micelle.

In the picture the white group is the polar head:
https://en.wikipedia.org/wiki/Micelle#/media/File:Phospholipids_aqueous_solution_structures.svg


Determination of +/- reflections in sigma molecular term symbols This page, which depicts the molecular term symbols for the $\ce{O2}$ and $\ce{O2-}$ (Edit: Mistaken on $\ce{O2-}$) molecules, perhaps best summarizes the full scope of my questions. In general, I don't understand how a specific electron configuration relates to either +,- or both reflections.

Why does the linked page switch reflections for different spin multiplicities, ie. why does $^3\Sigma^-$ have a minus reflection but $^1\Sigma^+$ a positive reflection?
For $\ce{O2-}$, the left configuration leads to $^3\Sigma^- + ^1\Sigma^+$. Why are two sigma terms needed to describe this configuration?
As a more general question, how accurate is it to use the molecular orbital representation for deriving term symbols? Most sources I have seen describe electron configurations as three degenerate p-orbitals or 5 degenerate d-orbitals and never invoke MO diagrams. However, it seems necessary for deriving both parity and reflections. Is the MO approach always correct? Can the term symbols be derived without using MO diagrams, specifically for the parity and reflections?

Thank you for any advice.

The $ \pm $ superscripts refer to symmetry operations defined in the $D_{\infty h}$ point group.  The operation on the molecular orbital is reflection in a plane containing the nuclei. If the orbital in unchanged then the sign is $+$, if changed then $-$.

The electron configuration for ground state oxygen is
$(1\sigma _g)^2(1\sigma _u)^2(2\sigma _g)^2(2 \sigma _u)^2(3\sigma _g)^2(1\pi _u)^2(1\pi _g)^2(1\pi _g)^1(1\pi _g)^1$

However, the only electrons we need to consider are the two in the $\pi_g$ orbitals. Adding up the individual orbital and then spin angular momenta separately (equivalent to Russell-Saunders coupling in atoms) then combining both gives the term symbols $^3\Sigma, ^1\Delta ,^1\Sigma $. The calculation is done in a similar way to that for atoms and is shown below in the notes.

The subscript g is found because there is an even number of $\pi_g$ electrons.
The superscript $ \pm $ is because any state ( singlet or triplet) must be asymmetrical overall to exchange and as the triplet has a symmetrical spin wavefunction the spatial part is asymmetrical, and vice-versa for the singlet, the superscript for the triplet is $-$. Thus the superscript gives the term symbol $^3\Sigma_g^-$. Hund’s rule dictate that this is the lowest energy level, followed by $^1\Delta$ at $7918 \pu{cm^{-1}}$ and $^1\Sigma_g^+$ at $13195.22 \pu{cm^{-1}}$.

The $\ce{O_2^-}$ has one unpaired electron in a $\pi$ orbital (as does $\ce{O_2^+}$) so the term symbol is $^2\Pi_g$.

As to your third question the overall symmetry should not change when you go from atomic to MO's.

Notes

The individual $m_l$ and $m_s$ are listed below. The values restricted by the Pauli principle are removed.
The $m_l$ values are $\pm 1$ and the $m_s \pm 1/2$. The the table the spins are labelled as $\pm$ only. The total angular momentum is given as $\Omega = \Lambda + \Sigma$ by analogy with $J=L+S$ in atoms.

$$\begin{matrix}
m_{l1} & m_{l2} & m_{s1} & m_{s2} & \Lambda & \Sigma \\
1 & 1 &  + & + &\text{etc.  Pauli forbid'n}\\
\\
1 & 1 &  + & - & 2 & 0 & ^1\Delta\\
1 & 1 &  - & + & 2 & 0 &  .. \\
-1 & -1 &  + & - & -2 & 0 & ..\\
-1 & -1 &  - & + & -2 & 0 & ..\\
\\
1 & -1 &  + & +  & 0 & 1 & ^3\Sigma\\
-1 & 1 &  + & +  & 0 & 1 & ..\\
1 & -1 &  - & -  & 0 & -1 & ..\\
-1 & 1 &  - & -  & 0 & -1 & ..\\
\\
1 & -1 &  + & - & 0 & 0 & ^1\Sigma\\
1 & -1 &  - & + & 0 & 0 & ..\\
\end{matrix}$$


What do locants with prime symbols (e.g. 4’) mean? I was reading an article about acetaminophen and it says its chemical name is 4'-hydroxyacetanilide.

What does the 4 signify and most importantly, what does the apostrophe mean? First semester chem student so if you can answer in layman terms that would be great.

4 means the fourth atom and the prime symbol specifies the part of the molecule (one ' means the secondary structure), so 4'-hydroxyacetanilide means you have acetanilide with a hydroxide group on the fourth atom of the secondary structure.
Naming organic compounds works like this:

  Find the longest continueous chain of carbon atoms. Called the parent chain this chain establishes the base alkane name for the compound.
  Number the carbon atoms in the parent chain beginning with the end closest to any substituents.
  Using the correct substituent name (methyl, ethyl, chloro,etc.), and the carbon atom number on the parent chain.
  If a particular substituent occurs more than once, use a prefix (di-, tri-, tetra-) to indicate the number of those substituents.
  List the alkyl substituents in alphabetical order.

Source: http://intro.chem.okstate.edu/1014/lecture/organic.html

Here are the IUPAC guidelines about the nomenclature of organic compounds:
http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter1.pdf


How to calculate the theoretical yield of a Diels-Alder reaction? A Diels-Alder reaction happens between 9-Anthracenemethanol (208.26 g/mol) and N-Methylmaleimide (111.10 g/mol).

I'm using 0.069 g 9-Anthracenemethanol (0.033 mol) and 0.11g N-Methylemaleimide (0.0009 mol, limiting reagent).

Since DA reactions have 100% atom economy, the product has the molar mass of both products combined, 319.36 g/mol.

Multiplying that by the moles of limiting reagent, we get 0.28 g product.
However, the mass of both starting materials combined (0.069g + 0.110g) together is not even 0.28g. What am I doing wrong?

You are missing a  couple of zeros in the number of moles of your 9-anthracenemethanol.  I calculate 0.00033 mol of that reagent, which therefore becomes your limiting reagant, and I calculate a total yield of 0.105 g of product, with about 0.73 g of  N-Methylemaleimide left over.


Which karat gold is harder, 24 kt or 18 kt? $\pu{24 kt}$ gold indicates that $24/24$ part of it, i.e. $100\%$ of it is gold. $\pu{18 kt}$ indicates that $18/24$ part of it is gold. As gold is a soft element, I reckoned that $\pu{18 kt}$ gold would be harder than $\pu{24 kt}$ gold.

I recently saw on a website that $\pu{18 kt}$ gold is harder than $\pu{9 kt}$ gold, which surprised and confused me. Now I want to know which karat gold is harder of $\pu{18 kt}$ and $\pu{24 kt}$.

18 kt gold is generically harder than 24 kt gold.  Both the site you linked and another site agree on that.  Pure gold really is quite soft, and when you add things to it, it will get harder.

Now, one key to your problem with understanding the hardness of gold and its alloys is that multiple different alloys can be called "18 kt gold".  Different alloys will have different properties and hardness. Interestingly, there is a titanium/gold alloy that is ~4× harder than titanium normally is. The paper - Science Advances 2016, 2 (7), e1600319 - on it looks interesting.

Back to the topic on hand, note in the table below that there are different values for hardness for each carat of gold based on how it has been physically worked and how it has been alloyed.  Also note that all the values listed for 18 kt gold are significantly above that of 24 kt gold.

Source: Total Materia; Gold and Gold Alloys; Table 3: Mechanical Properties of Typical Gold Alloys

It might be worth a separate question of why 9 kt gold might be listed as softer than 18 kt gold, if it has more in the way of impurities.


Electrochemistry (Spontaneous Reactions) Determine which of the following pairs of reactants will result in a spontaneous reaction at 25 celsius?

Pb2+(aq) + Cu(s)

Ag+(aq) + Br-(aq)

Li+(aq) + Cr(s)

Fe3+(aq) + Cd(s)

None of the Above.

I've tried using using tables like this http://ch302.cm.utexas.edu/images302/Electrochemistry_Reduction_Potentials.jpg to use the equation Ecell = Ecath - Eanode.

I'm hoping anyone could answer the question, and explain why each of them could be, or could not be spontaneous. (I know that an Ecell being positive results in spontaneity).

Here's my work:

Pb2+(aq) + Cu(s)

.34-(-.13) = .47

Ag+(aq) + Br-(aq)

.80 - (.76) = .04

Li+(aq) + Cr(s)

-.90-(-3.05) = 2.15

Fe3+(aq) + Cd(s)

-.04 - (-.81) = 77

None of the Above.

As you can see, they're all spontaneous. I'm wondering if I grabbed the wrong values from the above table. The question asks about ONE spontaneous reaction, from my calculations, they're all spontaneous.

@MichaelD.M.Dryden

For $\ce{Pb^2+_{(aq)} + Cu_{(s)}<=> Pb_{(s)} + Cu^2+_{(aq)}}$, your potentials are the wrong way around. The cathode is the electrode at which the reduction occurs so it's lead minus copper.

For $\ce{2 Ag+_{(aq)} + 2 Br^{-}_{(aq)}<=> Ag_{(s)} + Br2_{(l)}}$, I don't know where you got .76 V for bromide, but that's not the right number. Also, consider the $K_{sp}$ of $\ce{AgBr}$ and think about whether a non-redox reaction is occurring.

For $\ce{3Li^+_{(aq)} + Cr_{(s)}<=> 3Li_{(s)} + Cr^3+_{(aq)}}$, your potentials are backwards again and the potential for $\ce{Cr}$ is wrong. Consider how easily elemental lithium is oxidized; just putting it in water will evolve hydrogen and oxidize the lithium, so anything that reduces the lithium cation would have to be a phenomenal reducing agent.

For $\ce{2Fe^3+_{(aq)} + 3Cd_{(s)}<=> 2Fe_{(s)} + 3Cd^2+_{(aq)}}$, you have the right order, but the wrong potential for $\ce{Cd}$.

If you're having trouble keeping the order straight, it may help to think of it this way:
$$E_{\mathrm{cell}}=E_{\mathrm{reduction}} + E_{\mathrm{oxidation}}$$ where $E_{\mathrm{oxidation}}=-E_{\mathrm{reduction}}$. It's the same equation, but instead of keeping track of anode and cathode, the half-cell where reduction is happening will match what the table says and you use the potential as-is (since it's a table of reduction potentials), and the half-cell where oxidation is happening will be backwards from that in the table, so you negate the potential.


Are geminal disubstituted alkenes more stable than their cis/trans isomers? I know for a fact that cis alkenes are less stable than trans alkenes due to steric repulsions.

But what bugs me is the fact that why systems such as 2-methylprop-1-ene, where I believe steric repulsions are greater than that in cis-but-2-ene, are more stable than its cis and trans counterparts.

I saw this being given in the stability order of alkenes in Solomons and Frhyle.
I looked up Morrison and Boyd and it said the same thing. Paula Bruice said otherwise stating trans to be more stable.

Which is the correct order, and why?

Paula Bruice

Solomons and Frhyle

For reference, free energies of formation:

$$
\begin{array}{c|c} \text{Compound} & \Delta G^{\circ}_{f}\text{ in kJ mol}^{-1} \\\hline \text{2-methylpropene} & 58.1 \\ E\text{-2-butene} & 63.0 \\ Z\text{-2-butene} & 65.9  \\\hline \end{array}
$$

This is a very good question and if popular books give conflicting answers, then it must be reasoned out. Unfortunately, Paula Bruice has given the wrong answer while the other two books have given no explanation for this comparison.

For the answer I assume that you know about hyperconjugation and the various contributing structures it involves.  This gives a basic idea of hyperconjugation:

Source: tutorcircle.com

Note that on hyperconjugation, the other carbon of the alkene gets a negative charge on it.

If the hyperconjugation structures of 2-methylprop-1-ene and trans-but-2-ene are drawn:

you will notice that:

$ \ce{C_1} $ of the former has a negative charge on it
$ \ce{C_2} $ of the latter has a negative charge on it
The latter's methyl group makes the negatve charge less stable due to +I (inductive effect)
The former has no such destablizing effect as the carbon containing the negative charge is just connected to hydrogens
As all the hyperconjugation structure have this effect applying on them, the net effect cant be ignored for the overall stability

Hence the overall stability of  2-methylprop-1-ene > trans-but-2-ene


What does it mean to define entropy as a measure of a systems internal energy per unit temperature which is unavailable to do work? Entropy was defined in one of my findings as "the measure of the internal energy of a system per unit temperature which is unavailable to do work". Can someone help me understand this definition better cause I have always defined entropy as the degree of disorder in a system.

The answer to this question goes back to the Gibbs equations according to which $$\mathrm{d}U=T\,\mathrm{d}S-p\,\mathrm{d}V$$ and for a reversible process $$đq = T\,\mathrm{d}S$$ So, if you have a change in internal energy $\mathrm{d}U$ and it does work $p\,\mathrm{d}V$ then it has also absorbed $T\,\mathrm{d}S$. The verbiage you are quoting is very awkward and is not very useful, nevertheless for your future studies you will be better off if you will never think of entropy as "disorder" in any sense. Entropy is not disorder and such concept will never help you solve any real problem. Also, never think of a body containing "heat"; a body has energy depending on its state but independent of external constraints; and it has entropy that depends both on its state and its external constraints. For example, an ideal gas after it has expanded into a larger volume has the same energy as before it has done no work and absorbed no heat but its entropy is larger than it was in a smaller available volume. The entropy is the same in all possible molecular configuration consistent with the volume, so it is the same whether the molecules occupy uniformly the available volume or because of a large fluctuation spontaneously collapse into a smaller volume form the larger one. So it is the constraint (the available) volume that determines the entropy not how the molecules are distributed in it in any given instant of time. No disorder.


Does K2CO3 react in minute amounts with glassware at higher temperatures? At a temperature of $\pu{750 ^\circ C}$, $\ce{K2CO3}$ and $\ce{SiO2}$ react with each other in substantial proportions, forming $\ce{K2SiO3}$ and $\ce{CO2}$ in the process$\ce{^{[1]}}$.
We all know as well that potassium hydroxide reacts with glass. So, I was wondering if $\ce{K2CO3}$ was reacting with pyrex and/or normal glass in minute amounts, either in solution at say >$\pu{150 ^\circ C}$ or when being dried under vacuum in an oven.
One putative indirect mechanism in my wild imagination would be if $\ce{CO2}$ could leave the solution (maybe only at higher temperatures) and if potassium hydroxide at all existed in solution, upon the dissolution of $\ce{K2CO3}$ in water?
References:

Shikha Agarwal, In Engineering Chemistry: Fundamentals and Applications;  Cambridge University Press: Delhi, India, 2016 (ISBN: 978-1-107-47641-7).

In agriculture, potassium silicate ($\ce{K2SiO3}$) is used as foliar fertilizer ($\ce{K2SiO3}$ is water soluble). The major synthetic method of the preparation of $\ce{K2SiO3}$ is heating $\ce{SiO2}$ and $\ce{K2CO3}$ at high temperatures $(\pu{600-850 ^\circ C})$ in various mole ratios (e.g., Ref.1):
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2} \tag1$$
According to Ref.1, alkali metal carbonates such as $\ce{Na2CO3, K2SiO3, Rb2CO3,}$ and $\ce{Cs2CO3}$, all give corresponding metal silicates in this reaction (e.g., silica to metal carbonate mole ratios used in these preparations are $1:2$ to $3:1$).
Keep in mind that the temperatures used in these preparations are below the melting points of both reactants, $\ce{SiO2}$ ($\pu{1713 ^\circ C}$) and $\ce{K2CO3}$ ($\pu{891 ^\circ C}$), thus, the reaction is essentially a solid state one.
Potassium silicate fertilizer grades were successfully produced by direct fusion of silica ($\ce{SiO2}$) and potassium salt ($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of the mixtures (melt temperatures are as high as $\pu{1350 ^\circ C}$). The range of the weight ratio of silicon dioxide/potasium solid used in this method is $1:1$ to $5:1$. The reaction between $\ce{SiO2}$ and $\ce{KOH}$ is being:
$$\ce{SiO2 + 2KOH ->[\Delta] K2SiO3 + H2O} \tag2$$
According to the Ref.1, the other salts (e.g., halides and sulfates) of the metal carbonate catalyze the reaction. For example, in preparing sodium silicate, 6.12 parts by weight sodium carbonate, 3.47 parts by weight silica of minus 325 mesh particle size, and 0.41 part by weight sodium sulfate were ground together (which is $\ce{SiO2 + Na2CO3 + 0.05Na2SO4}$). The ground mixture was heated at a temperature of approximately $\pu{700 ^\circ C}$. for $\pu{4.5 h}$, after which time, the reaction was found to be 51% complete (c.f., without catalyst under same condition, completion was ~$23\%$).
Based on these date, it is safe to say that when silica made contact with an alkali carbonate or hydroxide at elevated temperatures, we can expect a reaction between two chemicals, regardless of they are been in solid form or melt. Once washed, the formed metal silicate would be washed away by water, leaving silica with damaged surface. According to Wikipedia:

The most familiar, and historically the oldest, types of manufactured glass are "silicate glasses" based on the chemical compound silica (silicon dioxide, or quartz), the primary constituent of sand. Soda-lime glass, containing around 70% silica, account for around 90% of manufactured glass.

Therefore, it is no wonder when you applied alkali hydroxide solution (e.g., $\ce{NaOH}$ or $\ce{KOH}$) solution to glass surface for a long time and found the surface become roughed. The alkali carbonated would do the same in higher temperatures.
Note: $\ce{Li2SiO3}$ is only partially soluble in water. However, its $\ce{Li4SiO4}$ version has an advantage of being $\ce{CO2}$ absorber (Ref.3) according to the reaction:
$$\ce{Li4SiO4 + CO2 ->[\Delta] Li2SiO3 + Li2CO3} \tag3$$
Unlike other carbonates, two $\ce{Li2CO3}$ molecules react with one $\ce{SiO32}$ molecule to give $\ce{Li4SiO4}$:
$$\ce{2Li2CO3 + SiO2 ->[\Delta] Li4SiO4 + 2CO2} \tag4$$
References:

Isadore Mockrin, “Production of Silicates,” US Patent 1958, 2,823,098 (PDF).
Srie Muljani, Bambang Wahyudi, Ketut Sumada, Suprihatin, “Potassium silicate foliar fertilizer grade from geothermal sludge and pyrophylite,” MATEC Web of Conferences 2016, 58, 01021 (DOI: 10.1051/matecconf/20165801021) (Proceedings of the 3rd Bali International Seminar on Science & Technology)(PDF).
Xianyao Yan, Yingjie Li, Xiaotong Ma, Jianli Zhao, Zeyan Wang, “Performance of $\ce{Li4SiO4}$ Material for CO2 Capture: A Review,” Int. J. Mol. Sci. 2019, 20, 928 (22 pages) (DOI: 10.3390/ijms20040928).


Coordination number of carbon in methyllithium tetramer According to J. D. Lee's Concise Inorganic Chemistry, the coordination number of carbon in $\ce{Li4(CH3)4}$ is $7$. But the structure of the tetramer is

Doesn't carbon have coordination number $6$ here? Am I missing something?

TL; DR: coordination number of carbon atom in methyllithium tetramer is indeed 7, arising from 6 intramolecular interactions, as you suggested, and an extra bond with the lithium atom of the next tetramer.

The key point is that you need to look beyond the tetramer. I'm sorry to inform you that the image from Wikipedia of the standalone cubane $\ce{Li4(CH3)4}$ is useless for the determination of coordination number (C.N.) of carbon. In fact, there is a 3D network of interlinked clusters, and the description given in J. D. Lee's Concise Inorganic Chemistry is in fact very precise [1, pp 304]:

  The structure of the $\ce{(LiCH3)4}$ cluster is unusual. The four $\ce{Li}$ atoms occupy the corners of a tetrahedron. Each methyl $\ce{C}$ atom is above a face of the tetrahedron, and forms a triple bridge to the three $\ce{Li}$ atoms that make up the face of the tetrahedron. The intramolecular $\ce{Li-C}$ distance is
  $2.31~\mathrm{Å}$. The $\ce{C}$ is bonded to the three $\ce{H}$ atoms in the methyl group. The $\ce{C}$ is also bonded to a $\ce{Li}$ atom in another tetrahedron (with an intermolecular $\ce{Li-C}$ distance of $2.36~\mathrm{Å}$). The coordination number for the $\ce{C}$ atom is therefore $7$.

The problem I stumbled upon is that I didn't find any ready-to-use CIFs for methyl lithium with 3D coordinates (there are plenty for the derivatives and some powder XRD of questionable quality). I ended up creating my own CIF based on data published by Weiss et al. [2] with VESTA program and visualized the content of the unit cell with OLEX2:

Figure 1. Unit cell of the methyl lithium tetramer with the central cubane core $\ce{Li4(CH3)4}$ linked via four crystallographically equivalent $\ce{C}$ atoms with 4 neighboring cubanes displayed as tetrahedra $\ce{Li4}$ for clarity. Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#CC80FF}{\Large\bullet}~\ce{Li}$; $\color{#909090}{\Large\bullet}~\ce{C}$.

Figure 2. Coordination environment of the carbon atom and interatomic distances (Å).

In addition, I run the test for C.N. with Dirichlet subroutine of the ToposPro, which also suggests the C.N. 7:

Central atom:C1 0.320 0.320 0.320 Rsd:0.900
 D(CP):0.624  ( 0.3698 0.3698 0.3698 )
 D(VDP):0.364  ( 0.3490 0.3490 0.3490 )
Atom:0.980 < r < 2.370  <r>=1.747   Top: 1.229 < R < 1.556  <R>=1.357
CN=7:0:0 NV=10 V=3.052/6.126 S=13.376 Cpac=0.161 Ccov=5.175
G3=0.114818789
Face distribution: {3/1 4/3 5/3 }
Vertex distribution: {3/10 }

  N  Atom   x      y      z     Dist.   D1      D2     SAng.
  1  H_1  0.3510  0.1920  0.3510  0.97957  0.89982  0.07975  27.31224
  2  H_1  0.1920  0.3510  0.3510  0.97957  0.89982  0.07975  27.31224
  3  H_1  0.3510  0.3510  0.1920  0.97957  0.89982  0.07975  27.31224
  4  Li1  0.3690  0.3690  0.6310  2.30686  0.89982  1.40704   5.76860
  5  Li1  0.3690  0.6310  0.3690  2.30686  0.89982  1.40704   5.76860
  6  Li1  0.6310  0.3690  0.3690  2.30686  0.89982  1.40704   5.76860
  7  Li1  0.1310  0.1310  0.1310  2.37007  0.89982  1.47025   0.75748

Figure 3. Voronoi-Dirichlet coordination polyhedron for the carbon atom.

Content of the CIF file:

#======================================================================

# CRYSTAL DATA

#----------------------------------------------------------------------

data_VESTA_phase_1

_chemical_name_common                  'Methyl lithium tetramer'
_cell_length_a                         7.24000
_cell_length_b                         7.24000
_cell_length_c                         7.24000
_cell_angle_alpha                      90
_cell_angle_beta                       90
_cell_angle_gamma                      90
_space_group_name_H-M_alt              'I -4 3 m'
_space_group_IT_number                 217

loop_
_space_group_symop_operation_xyz
   'x, y, z'
   '-x, -y, z'
   '-x, y, -z'
   'x, -y, -z'
   'z, x, y'
   'z, -x, -y'
   '-z, -x, y'
   '-z, x, -y'
   'y, z, x'
   '-y, z, -x'
   'y, -z, -x'
   '-y, -z, x'
   'y, x, z'
   '-y, -x, z'
   'y, -x, -z'
   '-y, x, -z'
   'x, z, y'
   '-x, z, -y'
   '-x, -z, y'
   'x, -z, -y'
   'z, y, x'
   'z, -y, -x'
   '-z, y, -x'
   '-z, -y, x'
   'x+1/2, y+1/2, z+1/2'
   '-x+1/2, -y+1/2, z+1/2'
   '-x+1/2, y+1/2, -z+1/2'
   'x+1/2, -y+1/2, -z+1/2'
   'z+1/2, x+1/2, y+1/2'
   'z+1/2, -x+1/2, -y+1/2'
   '-z+1/2, -x+1/2, y+1/2'
   '-z+1/2, x+1/2, -y+1/2'
   'y+1/2, z+1/2, x+1/2'
   '-y+1/2, z+1/2, -x+1/2'
   'y+1/2, -z+1/2, -x+1/2'
   '-y+1/2, -z+1/2, x+1/2'
   'y+1/2, x+1/2, z+1/2'
   '-y+1/2, -x+1/2, z+1/2'
   'y+1/2, -x+1/2, -z+1/2'
   '-y+1/2, x+1/2, -z+1/2'
   'x+1/2, z+1/2, y+1/2'
   '-x+1/2, z+1/2, -y+1/2'
   '-x+1/2, -z+1/2, y+1/2'
   'x+1/2, -z+1/2, -y+1/2'
   'z+1/2, y+1/2, x+1/2'
   'z+1/2, -y+1/2, -x+1/2'
   '-z+1/2, y+1/2, -x+1/2'
   '-z+1/2, -y+1/2, x+1/2'

loop_
   _atom_site_label
   _atom_site_occupancy
   _atom_site_fract_x
   _atom_site_fract_y
   _atom_site_fract_z
   _atom_site_adp_type
   _atom_site_B_iso_or_equiv
   _atom_site_type_symbol
   Li1        1.0     0.131000      0.131000      0.131000     Biso  1.000000 Li
   C1         1.0     0.320000      0.320000      0.320000     Biso  1.000000 C
   H1         1.0     0.351000      0.351000      0.192000     Biso  1.000000 H

References

Lee, J. D. Concise Inorganic Chemistry, 4th ed.; Chapman & Hall: London ; New York, 1991. ISBN 978-0-412-40290-6.
Weiss, E.; Hencken, G. Über Metall-Alkyl-Verbindungen: XII. Verfeinerung der Kristallstruktur des Methyllithiums. Journal of Organometallic Chemistry 1970, 21 (2), 265–268. DOI: 10.1016/S0022-328X(00)83621-9. (In German)


Is there any replacement for ethyl mercaptan in LPG? Ethanethiol, commonly known as ethyl mercaptan, is a colorless gas added to LPG in order to detect its leakage.

Apart from ethyl mercaptan, is there any gas or liquid that may be used as a replacement to this?

Why would one want to replace it?

The choice of ethanethiol is a reasonable compromise between its toxicity and the extremely low detection limit around 1 ppb (1mg/1000 kg).


Ammonium chloride in aqueous solution When Ammonium chloride is in an aqueous medium its cation part ammonium undergo hydrolysis. Since ammonium is conjugate acid of a weak base it readily release one $\ce{H+}$ and converts back to $\ce{NH3}$. But its given in my textbook that hydrolysis of Ammonium chloride gives ammonium hydroxide. Why Ammonium hydroxide is formed instead of staying as ammonia? Because according to my understanding ammonium hydroxide is formed when ammonium ion and hydroxide ion attract each other to have an ionic bond, and since ammonium ion converts back to ammonia how is this possible?

When ammonium chloride is hydrolyzing, there is established equilibrium

$$\ce{NH4+  + H2O <<=> NH3 + H2O+}$$

with $$\mathrm{pH}=7-\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$

$$\mathrm{pOH}=14 - \mathrm{pH}=7+\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$

Concentration of ammonia is several orders higher than concentration of hydroxide ions, with origin of water dissociation, so no ammonia hydroxide there.

When ammonia is dissolved, there is established equilibrium

$$\ce{NH3  + H2O <<=> NH4+ + OH-}$$

with $$\mathrm{pH}=14-\frac 12 ( 4.75 - \log {c_{\ce{NH3}}} )$$

Equimolar concentrations of $\ce{NH4+ + OH-}$ are sometimes formally called ammonium hydroxide, but using this term is deprecated and it cannot be isolated.


Why are some organic reactions run at elevated temperatures? We are using a water bath at $70\ ^\circ\mathrm{C}$ for different reactions to identify oxygen-containing organic compounds, such as the Tollens reaction, the iodoform test and hydroxamic acid test.

What is the function of heat in these reactions? Does the heat act as a catalyst?

Though they can produce a similar effect (faster reaction rate), heat doesn't act as a catalyst.

Reaction rate is dependent on how often reagents can overcome the activation energy of a reaction. Therefore, a lower activation energy, $E_\mathrm{a}$, means a faster reaction.

A catalyst lowers the activation energy of a reaction by producing lower energy transition states. This is how catalysts speed up reactions. A generic version of this is shown in the image below. Transition states occur at the local maxima of the graph in this image.

Heat, on the other hand, increases the average kinetic energy of the molecules within a system (i.e. most molecules are moving more quickly). By doing this, the chances of two particles colliding and having enough energy to react are higher because the particles will move faster. Therefore, they will collide more often, and when they do, they will have more "momentum" (so to speak) to get over this energy hill. By making it so more of the reactant particles can overcome this activation energy, heat also speeds up reactions.


Replicating an illustration for the crystal structure of ferrihydrite I came across this image of crystal structure in an article The Structure of Ferrihydrite, a Nanocrystalline Material [1]:

Does someone know the software in which it was produced? I have been looking for a way to visualize crystal structures of some minerals but I am stuck looking for a piece of software capable of doing it. I am aware of Jmol, but it does not show crystal structures too well.

References

Michel, F. M.; Ehm, L.; Antao, S. M.; Lee, P. L.; Chupas, P. J.; Liu, G.; Strongin, D. R.; Schoonen, M. A. A.; Phillips, B. L.; Parise, J. B. The Structure of Ferrihydrite, a Nanocrystalline Material. Science 2007, 316 (5832), 1726–1729. https://doi.org/10.1126/science.1142525.

I think the software used could be Diamond (http://www.crystalimpact.com/diamond/)


Equilibrium constant for a solid-solid equilibrium Suppose there are two equations:

$$
\begin{align}
\ce{A(s) &<=> B(g) + C(s)} &\quad &K_1 \tag{R1}\\
\ce{B(g) &<=> D(s) + E(s)} &\quad &K_2 \tag{R2}
\end{align}$$

These reactions have equilibrium constants $K_1$ and $K_2$, respectively. Both of these reactions are occurring simultaneously inside a container.

When I add these equations, I get

$$\ce{A(s) <=> C(s) + D(s) + E(s)} \tag{R3}$$

Since I added the reactions, I must multiply the equilibrium constants to get the equilibrium constant for above reaction which essentially is the net of the two reactions.

I have two questions:

How do I even write equilibrium constant for solid solid equilibrium?
Since $\ce{B}$ doesn't appear in equilibrium constant, is it safe to say that the concentration of $\ce{B}$ doesn't affect the equilibrium concentrations of any other species?

How do I even write equilibrium constant for solid solid equilibrium?

First, let's write the expression for the reaction quotient:

$$Q = 1$$

Q is a constant, always one, as long as all reactants and products are present. If you run out of one, it is zero or infinity.

Now, there are two cases. If the reaction is at equilibrium, K is also 1. If not, you can't measure the equilibrium concentrations (the reaction will go to completion, that is not an equilibrium). If you want to obtain the equilibrium constant, you would have to determine the Gibbs energy at standard state, and calculate K from that.

This is certainly quite confusing when we contrast it with other reactions (like R1 and R2), which will reach equilibrium. It gets less confusing if we rewrite R3 as:

$$\ce{A(s) -> C(s) +D(s) +E(s)}$$

An example you might be familiar with is ice turning into water. The reaction quotient is always one. Depending on the temperature, the equilibrium constant is smaller than one (all water freezes), is larger than one (all ice melts) or is exactly one (water and ice coexist in an equilibrium). You might think it is unlikely that we adjust the temperature exactly right to hit the melting point, but in fact it self-adjusts because melting ice is endothermic.

  Since B doesn't appear in equilibrium constant, is it safe to say that the concentration of B doesn't affect the equilibrium concentrations of any other species?

It is best to un-ask the question, for two reasons.

There is no equilibrium
The concentration of the other species are constant, so they are not affected by the reaction at all (unless the species is used up)

For reactions where at least one species is a solute or a gas, i.e. a species whose concentration changes continuously and where there is some entropy of mixing, the Gibbs energy will have a minimum somewhere along the way between all reactants and all products. In this special case, everything is a bit different.


What is the name of cyclopropane with a double bond? I was drawing out all the isomers of $\ce{C5H8}$ and came upon two structures of cyclopropane attached with double bonds. The group is neither a methylene nor an ethyl... what is it? What are the names of these two molecules?

Ethylidenecyclopropane

 Vinylcyclopropane (or Ethenylcyclopropane)


How does this method of calculating the water equivalent of a calorimeter work? I am doing a AP Chemistry lab in which we determine the heat capacity of a calorimeter (a concept that seems like it should depend on a lot more than just the calorimeter.) We have to calorimeters with water of different temperatures (which is tracked the whole time) but equal mass. We then mix them. We then track temperature again. We then extrapolate to the time of mixing (with a linear regression for some reason.) We look at how much energy was lost by the hot water and gained by the cold water based on all 3 lines (each calorimeter before, and the mixed calorimeter after.) Next we take the difference of the energy gained and energy lost at a single instant divide by the temperature change of the cold calorimeter (I have no idea why the cold one) and this is the calorimeters heat capacity.

I am having trouble understanding this, since at a single instance, the energy gained by the hot water and lost by the cold should be equal, and any difference would be error. It also seems completely random to divide by the temperature of the cold water and not the hot. What is going on here?

P.S. Later in the lab, extrapolating to the time of mixing is used to find the energy change due to a reaction, which makes more sense.

P.S.S One Calorimeter was fully mixed into the other (I forget which one, but either way does not make sense, and calorimeters themselves (like the material it was made from) had negligible heat capacity.))

P.S.S.S It is the teacher's first year doing this lab I think.

So you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is.

First, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and heat absorbed by the calorimeter. If energy is conserved, $$q_{hot} + q_{cold} + q_{calorimeter} = 0$$

One thing that's confusing you is that last term, $q_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!

Now, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. It also gives the approximate temperature change for the calorimeter if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.

The heat capacity of the calorimeter is $$C = \frac{q_{calorimeter}}{\Delta T_{calorimeter}}$$
You know $\Delta T_{calorimeter}$ from your time vs temperature graph. You know $q_{calorimeter}$ from the energy conservation equation above; $q_{calorimeter} = -q_{hot}-q_{cold}$. Therefore you know $C$. You should know how to get $q_{hot}$ and $q_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?


What is 'equality' of chemical bonds? Why are all the $\ce{C-O}$ bonds in $\ce{CO_3^2-}$ equal but not those in $\ce{H2CO3}$?

For $\ce{CO_3^2-}$ you have the following possible resonance structures:

The structure bellow, is the one which represents better the real ionic molecule, in which all the bonds are equal. You can delocalize the negative charge through all the ion.

For $\ce{H2CO3}$ the structure is:

As you can see, in this case, bonds are fixed, and there is no possible resonance delocalization, so you have two single bonds (less strong) C-O and one double bond C=O (stronger). For $\ce{CO_3^2-}$ you have that all bond lenths the same showing an intermediate situation between single and double bonds.


Why is thallium(III) oxide stable while thallium(III) chloride unstable Why is thallium(III) oxide stable while thallium(III) chloride unstable

Both the compounds have +3 oxidation state of thallium and the electronegativity difference between chlorine and oxygen is not very large. Then why is there such a large difference between the stability of the 2 compounds?

The stability of thallium(III) oxide and thallium(III) chloride can be compared by observing two properties:

Size of corresponding ions

Size of $\ce{Tl^3+}$ is 0.95 Å, size of $\ce{O^2-}$ is 1.44  Å and that of $\ce{Cl-}$ is 1.8  Å. So, it pretty seems that difference in size between thallium and oxygen atoms is less than that of difference in size between thallium and chlorine atoms. So, thallium(III) oxide is stable than that of thallium(III) chloride. (data)

Difference in electronegativity

Electronegativity of thallium is 2.04, oxygen is 3.44 and that of chlorine is 3.16. So, difference in electronegativity of thallium-oxygen is more than that of thallium-chlorine. So, thallium(III) oxide is stable than that of thallium(III) chloride. (data)


Can one turn a nickel blue? I've heard of artificial coin toning, but haven't been able to find a recipe for creating a blue (or blue-ish) nickel coin.

While not a metallurgist, I work with one. She tried nitric acid, which looked more like a coating and it rinsed off -- hoping for a permanent color change.

I've read that nickel is not a reactive metal, so not sure if this is possible.

I did it accidentally, I had coins in the bottom of my school bag and a packet or paracetamol and ibuprofen leaked without me knowing, the tablets got crushed into a powder and were left in the bottom of my school bag with the coins for about 6 months. They're now green, and it doesn't wash off with water and it doesn't look like a coating.


Carbonyl/Nitrosonium Complex Stability Why is the nickel complex exhibiting the strongest pi-bonding?

From what I understand, charge density plays a role. Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts.

Also I note that some have positive overall charges; could this play a role? Metal cation <-> NO+ repulsion? So we might want a metal anion?

Also would size play a role? Size goes hand in hand with charge density. And what about "backbonding"?

Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts.

Ahem. Nope. $\pi$-bonding in $\ce{CO/NO+}$ complexes is from back-donation, from occupied metal d-orbitals onto unoccupied $\pi^*$ orbitals of the ligand.

Looks like homework question, so no further details provided.


Is a beaker/reaction vessel like the one pictured available commercially? We're doing photocatalysis with the catalyst fixed to a 2" Si wafer, washing an aqueous solution (methylene blue, the concentration varies but in the picture I believe it's 0.1 mMol/l) over the top in beakers that look like this

(the lid is the heatsink for the 365 nm UV LED that drives the reaction; that part is pretty much optimised).

The few we have are custom made, and we need some more.  I can't find a source of anything comparable, but I'm only a physicist who doesn't necessarily know what's available or what terms to search for.

It needs two hose barbs as near the bottom as possible, and a flat base wide enough to lift a 2" wafer out (i.e. about a 60 mm minimum internal diameter).

They're also visible in this timelapse video of the reaction.

If such a thing exists, what's it called? Or is there something we could repurpose? Both the cost and the lead time of custom glassware are a bit high.

A beaker / reactor of such type would be rather expensive as well. If design and elegance do not matter you can easily adapt every suitably sized beaker by holding two pipes at the levels at which the hose barbs are supposed to be. Perhaps you choose a beaker a bit wider so there will be no turbulence or differences of sort as compared to what you are using.

Any flask or other container, combined with two small tubes and a couple of stands will effectively let you save money and time. Perhaps you must get space for the inlet and outlet through the led holder lid.


How to calculate dew point of air-water system I want to calculate the maximum water holding capacity of air at a given temperature (as in $\frac{max\ mass\  of\   water}{mass\  of\  dry\  air} $ vs temperature)
I looked up online and there are various models to do such calculations. However for most of them the temperature range is not more than 100 C. Are there any relations which cover a higher temperature range, upto 300-400 C?
If not, is not possible to determine saturation point above the boiling point of water?

Edit: I am considering a constant atmospheric pressure. Thus, I want to check variation only with temperature.

How to calculate dew point of air-water system? I want to calculate the maximum water holding capacity of air ... However for most of them the temperature range is not more than 100 C. Are there any relations which cover a higher temperature range, up to 300-400° C? If not, is it not possible to determine saturation point above the boiling point of water?

At very high temperatures I could only find the Mollier enthalpy-entropy chart (for steam) here's one for up to 1931.67 degrees rankine (800 degrees celsius) and another for up to 2200 degrees rankine (949.072 degrees celsius):

Source: "Mollier H-S Diagram"

Source: "Mollier enthalpy-entropy chart for steam, US units"
Source: Wikipedia - Enthalpy–entropy chart

On the diagram, lines of constant pressure, constant temperature and volume are plotted, so in a two-phase region, the lines of constant pressure and temperature coincide. Thus, coordinates on the diagram represent entropy and heat.
The Mollier diagram coordinates are enthalpy $h$ and humidity ratio $x$. The enthalpy coordinate is skewed and the constant enthalpy lines are parallel and evenly spaced.

See the Wikipedia webpage for more information, or assistance reading the diagram.
For restricted ranges (of which you complained, but are useful for comparison and to view the bottom of the chart) see Wikipedia's webpage for "Psychrometric charts":

Source: "Psychrometric Chart for Sea-level pressure using SI units"
Wikipedia's webpage for "dew point temperature" shows some calculations and means to make simple estimates, but it cuts you off at 100°C.


Concentration of a solution of NaOH I'm about to try to make my own soap for the first time. In order to create my recipe, I'm using this calculator.
As I'm using a lye solution, I have to fill the box %w/w of the solution. The calculator says:

  If you are using a ready made lye solution, check the box above and
  enter the enter the strength of your solution. The solution strength
  needs to be measured as a percentage of lye weight to water weight (%
  w/w).

I'm a bit confused here, I would have thought the w/w ratio was the ratio of the mass of NaOH in the solution to the total mass of the solution (and not the water mass used to disolve NaOH). So my first question is: what is really that w/w ratio?

My second question is how to interpret what I read on the lye solution (1L) I've bought. It's written:

  Sodium hydroxyde - N°CAS: 1310-73-2: 30%

What does this 30% refer to? Is this a ratio of volumes? Of mass? Is this the w/w ratio I want to know?

Thanks for helping me clarify this!

Whatever that lye calculator is doing it is either wrong by design or internally correct and using a very obscure definition that no other chemist would understand.

Percentage solutions are always (barring calculators programmed by laymen) defined as weight of the solute divided by weight of the solution times 100. So a $\pu{30\%}$ solution of $\ce{NaOH}$ means $\pu{30g}\ \ce{NaOH}/\pu{100g}\ \text{solution}$.


How is Al2Cl6 covalent and Al2O3 ionic? I encountered a question where the type of bonding for various oxides are prompted. Two of which are $\ce{Al2Cl6}$ and $\ce{Al2O3}$.

I thought that chlorine atoms are more electronegative than oxygen atoms, so if one of them is ionic and the other is covalent, surely chlorine would be ionic, because it may be able to "tear off" the electrons from aluminium atoms.

This is evidently not the case according to the answer key of this question. I would like to know an explanation for the ionic/covalent properties of these bonds.

You are right that the difference in electronegativity is responsible for the ionic bonds in aluminum and oxygen, and normally that would be the case with chlorine too, but $\ce{Al2Cl6}$ is a special molecule.

The structure of $\ce{Al2Cl6}$ can be viewed as two aluminum atoms covalently bonded to four chlorine atoms each. Two of the chlorine atoms bridge between the two aluminum atoms. The following is a possible representation:

A special type of bonding called 3 center 4 electron bonding is present in the structure. These bonds result from a combination of a filled p orbital and two half filled p orbitals. This causes there to be a filled bonding and a filled non-bonding orbital. The bond orders between each bridging chlorine and aluminum atom are 0.5; the structure of 3 center 4 electron bonds makes it so that the bonding orbital is delocalized over both the bonds. $\ce{Al2Cl6}$ has two of these bonds and they are responsible for the covalent nature of the molecule.