Entropy paradox and Confusion with Gibbs energy Define $G=H-TS_\text{sys}$. Then,
\begin{align}
\mathrm dG &=\mathrm dH-S_\text{sys}\,\mathrm dT-T\,\mathrm dS_\text{sys}\\[3pt]
&=-T\left(\frac{-\mathrm dH}T+\mathrm dS_\text{sys}+S_\text{sys}\,\frac{\mathrm dT}T\right)
\end{align}
At conditions of constant temperature and pressure, $\frac{-\mathrm dH}T=\mathrm dS_\text{surr}$ and $\mathrm dT=0$. Thus, we end up with:
$$\mathrm dG=-T(\mathrm dS_\text{sys}+\mathrm dS_\text{surr})=-T(\mathrm dS_\text{universe})\tag{E-01}$$
Using $H=U+pV$, and by using the first law of thermodynamics, it can be shown that:
$$\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$$
At constant $p,T$, this equation reduces to $\mathrm dG=0$. Using this in $\text{E-01}$ yields $\mathrm dG=\mathrm dS_\text{universe}=0$, at constant $T$ and $p$.
So how exactly do we obtain the so-called "spontaneous processes", which have a negative value of $\mathrm dG$ at constant $T$ and $p$?
I can somewhat see why the result makes some sense: The conditions assumed in the derivation resemble a reversible thermodynamic process, for which it's known that $\mathrm dS_\text{universe}$ is zero. The question in bold though is still unanswered. Maybe I am confused regarding what is actually meant by $\Delta G$.

The essence of what's going on is this:
Your equation, $dG = Vdp -SdT$, is correct, but has restrictions you don't realize. Specifically, it applies only to systems that are (a) closed (so no adding or subtracting material), (b) have only a single component* (so no mixing, phase changes, or chemical reactions) and (c) that can do $pV\ \hspace{-.45em}$-work only.
[*It also applies to multi-component systems where the composition is fixed.]
To understand why, under such conditions, $dT = 0 \text{ and } dp = 0 \Rightarrow dG = 0$, let's apply the Gibbs phase rule to such a system.  The phase rule says:
$$F = C − P + 2,$$
where C is the number of components, P is the number of phases at equilibrium (not to be confused with $p$, pressure), and F is the number of degrees of freedom.
Since $C = 1 \text{ and } P = 1$, we obtain $F = 2$.  That means we only have two degrees of freedom, i.e., two independent ways in which we can vary the intensive properites of the system. If the only type of work we can do is $pV\ \hspace{-.45em}$-work, the only ways we can adjust the system's intensive properties are by changing its temperature or its pressure.
Hence, if we apply the restrictions $\boldsymbol{dT=0 \textbf{ and } dp =0}$ to a closed single-component system where we only allow $\boldsymbol{pV}$-work, the system can't change!  And if the system can't change, then of course $\boldsymbol{dG = 0}$!
But, you might protest, $dG$ is not generally zero, even at fixed  $T\text{ and } p$.  So how do we reconcile this with what we wrote above?  Well, we need a more general expression for $dG$ that allows for non-$pV\ \hspace{-.45em}$-work, addition and subtraction of material, and changes in composition:
$$dU = \text{đ}q + \text{đ}w +\sum_i \mu_i dn_i
= \text{đ}q + \text{đ}w(pV) + \text{đ}w (non\text{-}pV) + \sum_i \mu_i dn_i$$
Since we can calculate $dU$ using any path, let's use a reversible path:
$$dU = TdS - pdV + \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
And since:
$$G = U+pV-TS \Rightarrow dG = dU + pdV +Vdp - TdS - SdT$$
$$\Rightarrow  dG =  VdP -SdT+ \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
Here, $\sum_i \mu_i dn_i$ is the sum of the chemical potential of each species $i$  $(\mu_i)$ times the change in the amount of species $i$ $(dn_i)$. This accounts for the change in $U$, and thus in $G$, as we change the composition
Hence, even if $dT = 0 \text{ and } dp = 0$, if we have non-$pV\ \hspace{-.45em}$-work and/or a change in composition (e.g., mixing, a change in phase, gain or loss of material, or a chemical reaction), it will not be the case that $dG$ is restricted to be zero.


What happens when the amount given doesn't allow for equilbrium to be achieved in a reaction? The reaction $$\ce{A.3H2O(s) <=> A.H2O(s) + 2H2O (g)}$$ has a $K_p = \pu{400 mmHg2}$
One millimole each of $\ce{A.3H2O}$ and $\ce{A.H2O}$ are taken in a $\pu{1 liter}$ container at $\pu{300 K}$
Relative humidity is $80 \%$ and saturation water vapor pressure is $\pu{85 mm Hg}$ at $\pu{300 K}$. Find the final composition of the container

My attempt
\begin{align}
K_p &=(P_\ce{H2O})^2 = 400 \\
\therefore P_\ce{H2O} &= \pu{20 mm Hg}
\end{align}
Now,
\begin{array}{lc} \hline
\ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline
i &1\times10^{-3} && 1 \times10^{-3} && \pu{68 mm Hg} \\
c &+x && -x && -P \\
e &2\times10^{-3} && 0 && \pu{30 mm Hg} \\ \hline
\end{array}
$\pu{38 mm Hg}$ drops because $\pu{2 millimoles}$ of water vapor was consumed
$$ P=\frac{1}{12}\frac{300\times 2\times 10^{-3}}{1} = \pu{38 mm Hg}$$
The answer should have been $\pu{20 mm Hg}$ pressure of water vapor but I am afraid there isn't enough reactant to make that much vapor pressure.
I asked a friend and he suggested that we can take $\pu{2 millimoles}$ each of the reactants as they are solid and don't influence the $K_p$ expressions while also not influencing the ratio of the added moles.
I do not get why this can be done but this allows for the vapor pressure at equilibrium to become $\pu{20 mm Hg}$ as required

This takes the assumption that the water vapour present acts as an ideal gas. Therefore $PV=nRT$ is valid.

Flaws/Misconceptions:
(1) You directly used pressure as you have in the RICE table.

Using pressure in a RICE table is not very useful since the RICE table deals with the amount of reactants and products and not pressure.

(2) The reaction doesn't reach equilibrium

The reaction reaches equilibrium as shown below. This means that the question in title doesn't need to be considered.
However, in such a case where equilibrium isn't reached, then the reaction stops due to lack of reagents to go forward. When you add more reagent it keeps moving forward until equilibrium is reached after which addition of solid reactants wouldn't cause a reaction to take place.

Now, as OP has stated, the $P_\ce{H2O}= \pu{20 mm Hg}$ at equilibrium. and it is given that relative humidity is $80\%$ for a saturation vapor pressure of $\pu{85 mm Hg}$ and so that means the intial pressure exerted by $\ce{H2O(g)}$ is:
$$P_\ce{H2O}= \frac{8}{10}\cdot 85=\pu{68 mm Hg}$$
Now using the ideal gas law, for the initial condition, we get:
\begin{align}
n_i(\ce{H2O})&= \frac{68}{76}\cdot\frac{1}{0.0821 \times 300} \\
&\approx\pu{0.00036 mol}
\end{align}
Similarly for the final case, we get:
\begin{align}
n_f(\ce{H2O})&= \frac{20}{76}\cdot\frac{1}{0.0821 \times 300} \\
&\approx\pu{0.00011 mol}
\end{align}
Therefore, $\pu{0.00025 mol}$ of $\ce{H2O}$ reacts for the condition to reach equilibrium. Now drawing the RICE table for the reaction we get:
\begin{array}{lc} \hline
\ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline
i &1\times10^{-3} && 1 \times10^{-3} && 3.6 \times 10^{-4} \\
c &1.25\times10^{-4} && -1.25\times10^{-4} && -2.5\times10^{-4} \\
e &1.125\times10^{-3} && 8.75\times 10^{-4} && 1.1\times10^{-4} \\ \hline
\end{array}
So the final composition is:
\begin{array}{lc} \hline
\text{Component} &\text{Amount} \\ \hline
\ce{A.3H2O} & \pu{1.125 \times10^{-3} mol} \\
\ce{A.H2O} & \pu{0.875 \times10^{-3} mol} \\
\ce{H2O} & \pu{1.1 \times10^{-4} mol} \\ \hline
\end{array}


Can I make Cobalt Violet with these materials? I'm given $\ce{CoCl2*6H2O}$ as well as $\ce{Na3PO4*12 H2O}$

The chemical formula for Cobalt Violet is $\ce{Co3(PO4)2}$. I haven't done stoichiometry in a long time but I think that's creatable, right?

Some website says that I need $\ce{NaH2PO4·10H2O}$, instead of the Sodium phosphate dodecahydrate which is what I have.

The dihydrogen phosphate anion is actually a weak acid and will dissociate to make both $\ce{HPO4^{2-}}$ and $\ce{PO4^{3-}}$ when dissolved in solution. If it works with the dihydrogen phosphate salt, it should work with the hydrogen phosphate and phosphate salts, too. (After all, it's the phosphate anion that precipitates out with the cobalt, if your formula is correct.)


Is histidine drawn wrong in these mechanisms for chymotrypsin? While trying to understand the mechanism of chypotrypsin, I have been looking at a few mechanisms, which are found in my lecture notes, and one found in my textbook. In both cases, it looks like the histidine has the wrong connectivity with the rest of the amino acid chain. Is there a reason for this, or is it an oversight on behalf of the publishers?

Under aqueous conditions you can tautomerise the double bond, so it doesn't matter. The two nitrogens are, for all intents and purposes, equivalent.


Is there a T-V or P-V chart relating LiBr temperatures for recovery? I'm working on a project and would like to consider a LiBr solution to be used as a desiccant. I don't know however the temperature of air vs pressure to recover it. I was thinking of using a nozzle/diffuser and room temperature air to recover it's desiccant properties, but I'm not sure that this is possible.

Without some fancy equipment to measure the salt concentration in the solution,  a simple saturated solution of lithium bromide would have some relative humidity which would be a function of temperature.  I looked quickly for some relative humidity table of lithium bromide and found the following reference which has an equation that can be used for calculating the relative humidity ($RH_{sat}$) over a saturated solution of lithium bromide as a function of the temperature in Celsius ($T_C$).

  "Humidity Fixed Points of Binary Saturated Aqueous Solutions" JOURNAL OF RESEARCH of the National Bureau of Standards, Vol. 81A, No.1, January-February 1977, Lewis Greenspan (pdf link)

$RH_{sat} = 7.75437 - 0.0654994T_C + 0.420737\times10^{-3}T_C^2 $

Some values for $RH_{sat}$ are:

7.75 % at 0 C
5.41 % at 100 C

So as long as some solid lithium bromide remained then the solution would be saturated. If all the lithium bromide dissolves, then you'd simply boil the solution to regenerate it. Lithium bromide is about twice as soluble in a boiling solution (somewhat over 100 C) as at room temp (20 C). So boil down to some starting volume of solution or until you see solid lithium bromide in the boiling solution. When the solution cools to room temperature about half the lithium bromide will crystallize out regenerating the saturated desiccant solution.

If you do want to boil this to regenerate then be sure and use some boiling stones. The solution would seem likely to bump.


What should be the pKa and IUPAC name of this compound? Note that this is not squaric acid as it has no double bond between the carbons holding the hydroxy groups.

I know this isn't exactly an answer, but I think it might help in reaching a qualitative understanding. These are the geometry optimized structures of the protonated and singly deprotonated form of the molecule (I could not obtain a theoretical pKa; it is a much more involved calculation than I anticipated). I would expect the pKa to be fairly low, as the deprotonated structure would seem to be fairly stable due to hydrogen bonding.

(Images were obtained using GaussView. Optimized structures were from Gaussian Optimization calculation using APFD, Basis set 6-311G+(2d,p)).


Why do people commonly refer to Na2HPO4 as Sodium Hydrogen Phosphate, rather than Disodium Hydrogen Phosphate? First of all, I am aware of this question.  The purpose of my question is to understand why people seem to break the rules that IUPAC has set (See example five on page 76 of the book), especially since this is not a compound that has a historic name, like $\ce{NH3}$ being ammonia or $\ce{N3-}$ being azide.  This compound has two species of cations, with sodium having a 2:1 ratio to hydrogen.  According to example five on page 76, compounds that have multiple cations are supposed to give prefixes to every compound in an ion that has a quantity greater than one.  Why do we not do this for sodium hydrogen phosphate?

Well, turns out that I needed to look at this a little differently.  I looked at calcium hydrogen carbonate and then realized that hydrogen is part of the polyatomic ion.  It's a bit harder to recognize when there are no parenthesis, like in the case of sodium hydrogen phosphate, but the same thing applies there.  Therefore, the name, sodium hydrogen phosphate is correct because there are still only two distinct ions in the compound.


Genuine confusion about the equilibrium constant I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured. But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.

How then can K = 1 imply that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured?

Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured.

There is nothing special about K = 1. At equilibrium, no matter the value of K, there is no net reaction. Forward and backward reactions proceed at the same speed.

  But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.

Yes, even a reaction with K = 1 can react. Just imagine you start with some reactant and no product. This situation is not at equilibrium. The reaction will proceed in the forward direction until Q = K. Then, you can disturb the system, changing Q, and there will be a net reaction again, changing concentrations until Q reaches K again.

  Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

You said "please", so I will try for a short, simple and comprehensive answer. The value of K allows you to calculate equilibrium concentrations, no matter which concentrations you started with. For a lower K, the product equilibrium concentrations will be lower than for a higher K (same initial concentrations). The opposite is true for reactant equilibrium concentrations.

It should be clear that the equilibrium concentrations for a given K are not always the same. The simplest example is doubling initial concentrations (while keeping the volume constant). If the equilibrium concentrations were equal for both cases (initial concentrations doubled or not), it would go against the conservation of matter. (We doubled the amount of species present initially, so we can't have the "old" amount of equilibrium species).


Hydrolysis of an ester An organic compound X of formula $\ce{C5H10O2}$ forms two compounds when boiled in aqueous sodium hydroxide solution. One of these compounds forms a ketone when treated with acidified potassium dichromate.

  Compound X could be

  \begin{array}{ll}
\textbf{A} \quad \ce{CH3COOCH(CH3)2}. \qquad & \textbf{C} \quad \ce{CH3CH2COOCH2CH3}. \\
\textbf{B} \quad \ce{CH3CH2CH2CH2COOH}. \qquad & \textbf{D} \quad \ce{(CH3)2CH2COOCH3}.
\end{array}

Can someone confirm that the correct answer is option A instead of C?

Since one of the compounds (alcohol) forms a ketone, therefore it must be secondary alcohol. This means the ester must be branched, so it would be A?

Oxidation of an alcohol (with potassium dichromate) gives a ketone when the alcohol is secondary, like isopropanol.  So, which ester out of the given choices has a secondary alcohol component?

You see that A is correct as claimed.


Standard hydrogen electrode, why hydrogen? Why is the hydrogen electrode used as the reference to set electrode potentials? Could any other element be used instead?

There are two separate issues here. The first one is defining a zero for potential. In theoretical physics, it is the energy of two charges separated by infinity distance.

In electrochemistry, this is chosen to be the potential of the hydrogen electrode since that is something that can be made reproducibly and hydrogen was the de facto 'zero' for a number of other quantities.

These days, there are much more practical and easy to use reference electrodes like the silver/silver chloride (as chipbuster mentioned) or the saturated calomel or the mercury/mercurous sulfate. Here is a good list of them from a manufacturer of such electrodes: http://www.koslow.com/select_reference_probe

Full disclosure: I used to work for a company that resells electrodes from Koslow.


Does aluminum oxide react with rubidium? I have rubidium vapor inside of the vacuum chamber. Inside the vacuum chamber, there are two flat stainless steel coated with aluminum oxide at temperature ~100 °C.

Since rubidium is active, I am worrying that it will interact with the aluminum oxide $\ce{Al2O3}$ coating. Any ideas?

As suggested in the comments, rubidium aluminates do exist.
Aluminates such as $\ce{RbAlO2}$ and $\ce{Rb6Al2O6}$ are prepared at higher temperatures (above 550 °C [1]) or even from the melt.

However, alkali metal vapors at elevated temperatures act as reducing agent, so I wouldn't expect aluminates anyway.
On the other hand, aluminides such as $\ce{RbAl}$ (Zintl phase) are formed from the melt at much higher temperatures (above melting point of aluminium) and pressures exceeding atmospheric.

Since the corundum-coated plate is supposed to perform in vacuum under mild temperatures, I think it shouldn't be affected by rubidium vapors.
There is a possibility of coating embrittlement over time as rubidium diffusion occurs, but I wouldn't expect it to be an issue.

References

Schläger, M.; Hoppe, R. Darstellung und Kristallstruktur von $\ce{K6[Al2O6]}$ und $\ce{Rb6[Al2O6]}$. Zeitschrift für anorganische und allgemeine Chemie 1994, 620 (5), 882–887. https://doi.org/10.1002/zaac.19946200522.


Why does dimethyl sulfoxide have higher boiling point than acetone? They have very similar structure except for a central atom, but the difference in boiling points is very large $(\pu{186 ^\circ C}$ for DMSO and $\pu{56 ^\circ C}$ for acetone).
How is central atom impacting the boiling point?

Firstly, sulfur has a larger atomic radius than carbon, so we would expect DMSO (Ddimethyl sulfoxide) to have a much larger and thus polarisable electron cloud than acetone. The London dispersion forces between DMSO are thus stronger, causing DMSO to have a higher boiling point.
Secondly, if we compare the dipole moments of the $\ce{S=O}$ and $\ce{C=O}$ bonds, the $\ce{S=O}$ bond is much more polarised. This is because the $\mathrm{3p-2p}$ overlap between $\ce{S}$ and $\ce{O}$ is much less favourable than the $\mathrm{2p-2p}$ overlap between $\ce{C}$ and $\ce{O}$. As such, the resonance structure on the right is actually the more significant resonance structure. This makes the $\ce{S=O}$ bond much more polarised as compared to the $\ce{C=O}$ bond, causing the permanent dipole-permanent dipole interactions between DMSO to be much stronger than the same interactions between acetone molecules.

Lastly, we would realise that DMSO has a trigonal pyramidal shape while acetone is planar. Due to this, the net dipole moment in DMSO is likely to be stronger due to it being a more asymmetric molecule.


How is the crystal direction important for etching silicon surfaces? I am a student of masters study with a background only in electrical engineering, but I have studied Miller indices and crystallographic planes and directions, personally, to help myself understand the process of silicon etching which is required in my current course of study. But, I am unable to understand the following written in my text book:

  "When Silicon is etched anisotropically, i.e. depending on crystal direction, and a {1 0 0} surface direction is used, cavities will result which have the same {1 0 0} orientation at the bottom and are determined laterally by {1 1 1} surfaces. The reason is that {1 1 1} surfaces are etched off using a minimal etch rate and this way becomes the most resistant"

Could someone please help me understand this concept.

Certain types of wet etching agents have differential reactivity on different crystal surfaces. Consider that the Si-Si distance and coordination environment will be quite different.

Let's start with an isotropic etch - it just eats silicon. Well, if you started with a pinprick (i.e., just a point and not a gap in the mask) you'd etch out a hemisphere. Normally, we etch through gaps in the mask, so you'll get rounded sides.

With an anisotropic wet etch you can get flat, angular side-walls defined by crystal planes.

  For instance, potassium hydroxide (KOH) displays an etch rate selectivity 400 times higher in <100> crystal directions than in <111> directions.

OK, so let's consider that etch through a gap in a mask into a <100> surface. As we go down 400 atomic layers of Si, we'll clearly have some etching along the side-walls (e.g., an "undercut"). But also consider that depending on the size of the feature for the etch, you'll go down into the <100> but not completely uniformly. Different wet anisotropic etches will have different selectivity rates between <100> and <111> and thus you get different aspect ratios.

Incidentally, dry etching (e.g., RIE, plasma) can give more vertical side-walls.


Does syn-dihydroxylation form racemic mixture? I have been asked to determine the nature of the product for the following reaction:

And as I recognized this reaction to be a case of syn-dihydroxylation, my obvious response was that the product formed will be meso.
But, in the solution to the problem, it says it makes a racemic mixture.

Shouldn't both the $\ce{-OH}$ be on the same side and form a meso-compound? And what does the $(sym)$ stand for?

Oxidation  of alkene to 1,2-syn-diol by cold basic $\ce{KMnO4}$ is a common classification test for alkene in undergraduate organic chemistry laboratory courses. OP had correctly recognized that this reaction is a case of syn-dihydroxylation, and OP's conclusion of the product to be meso is also correct since the alkene in hand is a cis-but-2-ene. The schematic representation of the proposed mechanism is given below:

Source of the scheme
Since but-2-ene structure is planer and a prochiral molecule, it has two faces (Si- or Re-face; above or below the plane) to get attacked by a reactant, here it is $\ce{KMnO4}$. The above mechanism displays the attack by only one face. The oxidation gives butan-2,3-diol by syn-attack on that face. The product has two stereocenters and the stereochemistry of that product is $(2R,3S)$. As indicated in the diagram, the product also has a plane of symmetry, and hence, the $\ce{C}$2 stereocenter is the mirror image of the $\ce{C}$3 stereocenter. Therefore, the product is a meso-compound.
If the $\ce{KMnO4}$ attack is done on the opposite plane, the product would be $(2S,3R)$-butan-2,3-diol, if numbering has maintained the same. This is essentially the same meso-product, as we discussed above.
If the starting substrate is trans-but-2-ene, the $\ce{KMnO4}$ attack from one face gives you $(2S,3S)$-butan-2,3-diol, while the $\ce{KMnO4}$ attack from opposite face gives you $(2R,3R)$-butan-2,3-diol. Since possibility of attacking from the Si-fase and/or Re-face is $0.5$ (or $50\%$), you get essentially a racemic mixture.
Based on above facts, OP's textbook description of the reaction is erroneous, not only on stereochemistry, but also on the notation of the reaction (it is syn- not sym-addition).


What reactions occur when mixing copper sulfate and sodium hydroxide? I tried reacting copper sulfate with sodium hydroxide to get copper hydroxide, which should precipitate, according to the following equation:
$$\ce{2NaOH + CuSO4 -> Cu(OH)2(s) + Na2SO4}$$
I looked on Google Images, and the color of copper hydroxide is light-blue, but something interesting happened when I mixed these two solutions: the precipitate formed - and was originally light-blue, as expected - however, it turned into this black-green sludge within a few seconds. I'm thinking a complex ion might have formed between the $\ce{Na2SO4}$, and the $\ce{Cu(OH)2}$. I just saw a video on this, posted by NileRed, but he doesn't go into detail. Does anyone have an idea of what's going on?

As noted in your question and in one of the comments, copper forms many different complexes having a variety of colors from red to green to blue to black and probably more.  Of course the situation here is even more "complex" (sorry) as you can end up with mixtures of the different copper complexes, as appears to be the case over the course of your experiment.

The formation of some cupric oxide ($\ce{CuO}$) would account for the appearance of an insoluble black precipitate.  This could happen from the decomposition of some of the $\ce{Cu(OH)2}$.

$\ce{Cu(OH)2}$ itself should be a greenish to pale-blue insoluble precipitate.

It is possible that the original $\ce{Cu(OH)2}$ precipitate appeared more blue than green due to the blue $\ce{CuSO4}$ solution.  As the reaction proceeded, $\ce{CuO}$ and more $\ce{Cu(OH)2}$ precipitated to give the black-green precipitate you observed.


How can I calculate the charge distribution of a water molecule? I would like to obtain the 3D charge density of a water molecule for plotting purposes (i.e. not necessarily super-accurate). I know this is a tough problem, but I'd like to give it a try. Since I'm currently using python, I've identified PyQuante and PySCF, and gather that I should learn how to run density functional theory (DFT) calculations. Before I do that, I'd like to find out if there is something else out there I should use instead. Multiple suggestions are welcome.

I've installed PyQuante, which according to it's README.txt may be more accessible to understanding.

For example, the color plots in this answer give a hint of what I'm talking about, though I'll ultimately plot the field strength contours and some electric field lines.

I'd like to obtain a large table of values (or expressions/expansions?), so I can create and work with surfaces and field line plots myself.

Additionally is it possible that there exist supplementary data in published articles. Where I might find that? Although this is about existing data, I actually do plan on putting in the time to have at least a minimal understanding of the calculation itself. If DFT is it, then I will be spending time studying that.

Two suggestions: MolCalc: Calculate Properties > Polarity and Solvation gives you a 3D surface plot with the electrostatic potential superimposed.

Another suggestion is Avogadro/GAMESS/MacMolPlt. See for example here


Error propagation Titration I hope I am right in this section.

I am unsure with error propagation. When calculation the error in a titration, many errors has to be taken into account:

Error in Glassware/ Error in Balance/ Error in Burette etc.

I learned that the absolute and relative error have only 1 significant figure and that the total amount is rounded to the decimal place of the error.

Therefore   5.34532g ± 0.001428g  would be 5.345g ± 0.001g

The relative error 0.001g/5.345g = 0.00018709 = 0.0002

If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot?
Wouldn't be rounding the error just in the end make more sense?

Many thanks in advance

I learned that the absolute and relative error have only 1 significant figure

That really isn't  true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 could all be significant.

You then simplified to 5.345g ± 0.001g which is fine. My personal preference would be more like 5.345g ± 0.0014g since ± 0.00067g would round to ± 0.001g as well. So I'd prefer one more digit in the ± "precision" than the measurement itself.

  If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot?

Yes absolutely. Carry "extra" significant figures until the final step.

  Wouldn't be rounding the error just in the end make more sense?

That is the preferred method.

We typically use a calculator now to do the calculations. You can set the calculator to show 3 significant figures but it carries more internally.

There are various statistical techniques which allow all the measurements used in the calculations to have random errors injected and then the  calculation is simulated many times to see how the errors propagate through the calculations.


Fixed cell parameters, TINKER I am using TINKER package to calculate the external reorganization energy for pentacene crystal. But I have problems with that - I've generated .xyz file of 5 molecules with VESTA and I need to perform the calculation on those five molecules (image below) with periodic boundary conditions (PBC).

How can I find the proper size of cell and keep it fixed for calculations? When I use keywords A-axis, B-axis, C-axis and alpha, beta, gamma and assign lattice vectors and angles to them after optimization or minimization for the charged system like that apparently cell gets broken and molecules are distorted out of box:

It's a little hard to tell from your second visualization, but to me, it simply seems as if the molecules have shifted somewhat with respect to the periodic box.

A unit cell or periodic boundary "box" is simply a frame of reference. There's no walls. The molecules don't "see" it.

It's merely an indication that there's translational symmetry. In short, in the left-hand view of the second image, the molecule "breaking the box" is also wrapped around into the box by translational symmetry.

I don't know what visualization tool you're using, but if you display more than one unit cell, you'll easily see that the molecules "wrap" around.


Do any substances in a liquid state work as well as water to put out fires? Is there a liquid substance or compound that works as well to put out fire as water does?

Carbon tetrachloride was once used for this purpose as it has practically no flammability at lower temperatures .


How can the formation of an alkene be explained during the hydrolysis of an ester when the alkoxy moiety is tertiary? $$ \ce{RCOOR' + H2O -> RCOOH + R'OH} $$
But when $\ce{R'}=3^\circ$ it may result in the formation of alkene. But by the normal mechanism given in books and in various sites I am not able to explain its formation. I think that somewhere $\ce{R'^{+}}$ must be formed for alkene to be the product.

By the normal mechanism given in books and in various sites I am not able to explain its formation

That's right, and it means that there must be a different mechanism predominating.

This is the "standard" mechanism for ester hydrolysis in acid:

If you look closely at the colour-coding, you will notice that it is the blue bond in the ester which connects the $\ce{C=O}$ to the $\ce{OR'}$ that is broken. This is called the acyl bond. The rate-determining step involves two molecules - it is a bimolecular reaction. Therefore, this ester hydrolysis mechanism is called $\mathrm{A_{AC}2}$: $\mathrm{A}$ for acid, $_\mathrm{AC}$ for acyl bond, and $2$ for bimolecular.

Let's use $\ce{R'} = \ce{C(CH3)3}$, the prototypical example of a tertiary alkyl group. The alternative mechanism goes as follows:

Notice here it is the green bond, between the $\ce{O}$ and the $\ce{R'}$, that is broken. This is called the alkyl bond. The slow step is unimolecular. So, following the same logic as before, this hydrolysis is called an $\mathrm{A_{AL}1}$ mechanism. (It doesn't matter which of the red and blue bonds ends up as the carbonyl group and which one ends up as the $\ce{OH}$ group - they are interchangeable via a simple proton transfer.)

As you correctly suspected, the formation of the $\ce{C(CH3)3}$ carbocation occurs. Depending on the reaction conditions, it could be attacked by a nucleophile (e.g. if you use ethanol as the solvent, you might get some ether byproduct), or it could just simply lose a proton to give isobutene.

What factors may work to favour the $\mathrm{A_{AL}1}$ pathway? Well, the most obvious thing is the formation of the relatively stable tertiary carbocation. Another possibility is steric hindrance arising from the bulky $\ce{R'}$ group, which disfavours nucleophilic attack of water in the "normal" pathway. In fact, t-butyl esters like the one shown above are so resistant to nucleophilic attack (under basic conditions) that they are sometimes used as protecting groups for carboxylic acids!


Why does the temperature remain constant during freezing? I know it's the most silly question one could ever ask but why is that so?

I am aware of what latent heat means and I also know it has something to do with freezing too.

Say I have with me, a mug of hot water and I put it to freeze. And so the temperature reduces first to 273 K and then remains constant to convert into ice and decreases once again.
 But why does the temperature remain constant though? Does this has something to do in molecular level?

Creating an ordered solid from a disordered liquid releases energy

At the molecular level this is what is happening when you cool something like water.

At first the cooling reduces the average kinetic energy of the water molecules so they slow down. Basically, temperature is the average kinetic energy so removing energy lowers the temperature.

At some point, though, water starts to form ice where the molecules are not wandering about randomly and get rigidly stuck together in ice crystals. But this process releases energy: the ordered ice crystal has a significantly lower energy than water at the same temperature. Converting liquid water to ice releases that energy.

What this means is that when extracting energy from warm water the result is simply a lower temperature. But one you get to 0°C the extra energy extracted is used to convert the liquid to the solid rather than reducing the temperature. Once all the water has solidified, further removing energy reduces the temperature again.

The key is that the process of freezing itself involves energy changes so instead of reducing the temperature, the energy comes from the conversion of liquid too solid.


I have 1 litre of water at 45 °C. How long would it take to evaporate when ambient temperature is 25 °C? As a fun experiment, I'm trying to evaporate water in my kitchen. I've heated the mixture to 45 °C, and I'm wondering how long it might take to completely evaporate.
Assume that the water is 2 cm deep in a flat pan (if that is relevant).

The evaporation rate $E_r$ can be approximated by:
$$E_r = C \times A \times (x_s - x) = 0.013$$ kg/h

$C$... evaporation coefficient $ = 25 + 19 \times v$

$A$... water surface $= 0.03 $ m$^2$

$x_s$... saturation humidity $= 0.027$ kg/kg (at 30 °C)

$x$... absolute humidity $= 0.011 $ kg/kg (at 30 °C and 40 % rel. humidity)

$v$... air speed above water surface $= 0.08$ m/s

If you mesure the air temperature above the water surface - which I approximated with 30 °C - you can get a way better approximation for your case. Also having the rel. humidity you can then find the values for $x_s$ and $x$ in a Mollier diagram (h,x-diagram).

PS: You will waste a lot of energy when you want to obtain the water temperature at 45 °C. Heating the air instead should be more effective.

Reference: Link (in German)


Why don't alkynes undergo preferential anti bromination to the degree that alkenes do? Alkenes, for the most part (unless there is a phenyl group or highly ionising solvent) undergo almost exclusively anti addition of bromine. However, under similar conditions alkynes, although undergo preferential anti-addition, it is not as exclusive as alkenes. Even with electron withdrawing substituents like two $\ce{COOH}$, only 70% is anti. (Peter Sykes, page 182)
Why?

From what I know of the mechanism, Addition of Br2 takes place in the following mechanism

For an alkyne, the intermediate would not be stable due to the ring strain on 3 member ring with a double bond in it, thus reducing the rate of reaction.

Mechanism referred from: https://en.wikipedia.org/wiki/Halogen_addition_reaction


Confusion regarding the Shape of XeF4 I found on internet that the shape of $\ce{XeF4}$ is square planar but what if one place $2$ of $\ce{F}$ on the equitorial position and rest two on axial position and the two lone pair on the leftout axial position then its shape will not be square planar but will change to something else

Can someone please explain this to me.

The shape of $\ce{XeF4}$ is indeed square planar as you mentioned it is so because:

The lone pairs are kept in anti position to minimise repulsion between them and also from the electrons in fluorine atoms. Keeping the lone pair anti results in 4 empty positions which are filled with fluorine, giving rise to a square planar shape and octahedral geometry.

Please fell free to comment any doubt, and give the tick if you are satisfied.


Calculate the transition dipole moments between the 1s inner shell and the high-lying states of a molecule I am trying to calculate the transition dipole moments between the 1s inner shell and the  high-lying states of N2, namely, $<\psi_{1s}|{\bf r}|\psi_{high-lying}>$.

For this, I installed GAMESS (US) and calculated molecular orbitals of N2.
However, I only got HOMO-6 to LUMO+2.
I want LUMO+3, 4, 5... and if possible, 1s.

How can I do this in GAMESS? Or do I need other softwares?

Might have found the option you are looking for. GAMESS controls MO printing with the NPREO keyword. If you can specify ranges of orbitals either by their number (e.g. NPREO(1)=1,10 prints the first 10 orbitals) or relative to the HOMO (e.g. NPREO(1)=-3,-5 prints from HOMO-3 to LUMO+4). You can also give 3rd and 4th numbers to list the range of orbital energies to print.

These print options are described on p.23 of the input descriptions document on the Gordon research group page.


Removal of water from sulfuric acid Suppose I have a dehydration reaction of methanol to produce methoxymethane:
$$\ce{2 CH3OH(l) ->[H_2SO_4] C2H6O(g) + H2O(l)}$$
Usually, people will use sulfuric acid to dehydrate the methanol. Assuming all the methanol was reacted, how do I get rid of the water in the sulfuric acid? Will simply heating the sulfuric acid be sufficient?

You should be adding stoichiometric amounts of diluted acid. Then distilation to remove the methanol by destilation. Alternatively can also use molecular sieves, or add sodium sulfate to dry the methanol


Why do no known atoms have electrons in the g or h subshells? I'm learning about orbital quantum numbers. While checking several elements on the periodic table I noticed that there aren't any atoms that have electrons in the g or h subshells. Why is this?

Well, the first available sub-shell for "g" electrons would be 5g (i.e., 1s, 2p, 3d, 4f, so 5g). Based on current trends, we'd need row 8 of the periodic table. We just haven't found those elements yet.

Let's look at it this way.

After row 1 (1s electrons only), there are 2 rows of the s and p-block elements before "d" opens up with $\ce{3d^1 4s^2}$
After two rows of s- p- and d-block elements, the "f" block opens up with cerium: $\ce{4f^1 5d^1 6s^2}$

So we'd need two more rows of the f-block (i.e., the lanthanides and actinides) and then complete the row.

Right now, we have found element 118, so if we can synthesize a few more, we can open up the "g" block.

Now, there are elements that have g electrons in excited states, so g and h-orbitals are relevant to some chemistry. But the ground-state elements haven't been discovered yet.


How can the bulky octadecyltriethoxysilane molecule promote single-crystal growth? Octadecyltriethoxysilane is very bulky (source),

yet it was used to promote single crystal growth (source). How can such a bulky molecule induce crystal growth? (For example epitaxial growth happens only when both materials have similar unit cell dimensions.)

The function of the octadecyltriethoxysilane monolayer in this case is not to promote epitaxial growth but to reduce the surface energy of the dielectric layer (SiO2), on top of which the semiconductor (pentacene/rubrene/etc.) is about to be deposited. The reduced surface energy allows the crystal to grow "comfortably".

Organosilanes are typical compounds for modifying oxide layers (Pernstich et al., 2004):


What substance has the lowest solubility product? What substance has the lowest $K_\mathrm{sp}$ and what is its value? The lowest I could find is $2.6\cdot 10^{-124}$ for cobalt(III) sulfide $\ce{Co2S3}$.

The $K_\mathrm{sp}(\ce{Co2S3})$ value of magnitude of $10^{-124}$ appears in paper by Goates et al. [1].
The refined "thermodynamic" value of $\pu{2.6E-124}$ that you've listed and is used by numerous textbook up to these days has been proposed by Waggoner [2].
However, thirty years later (late 1980s) there's been another study by Licht [3], which showed significant deviation from the previous studies due to two factors:

A new value of the free energy of sulfide ion formation for aqueous solutions  has been used:
$$Δ_\mathrm{f}G^\circ(\ce{S^2-(aq)}) = \pu{(111 ± 2) kJ mol-1}$$
Previous value [1] was
$$Δ_\mathrm{f}G^\circ(\ce{S^2-(aq)}) = \pu{20.64 kcal mol-1} \approx \pu{83.68 kJ mol-1}$$

Free sulfide activity $a(\ce{S^2-})$ used for $K_\mathrm{sp}$ determination

$$
\begin{align}
\ce{M_xS_y &<=> x M^{$2y/x$+} + y S^2-} &\qquad K_\mathrm{sp} &= a(\ce{M^{$2y/x$+}})^x \cdot a(\ce{S^2-})^y \\
\ce{HS- &<=> H+ + S^2-} &\qquad K_2 &= \frac{a(\ce{H+})\cdot a(\ce{S^2-})}{a(\ce{HS-})}
\end{align}
$$

has also been misinterpreted in early studies by acidification of hydroxyl $\ce{OH-}$ instead, erroneously substituting $K_2$ by $K_\mathrm{w}.$

Table I from [3] lists more recent $\mathrm{p}K_\mathrm{sp}$ values, among which the one for $\ce{Co2S3}$ that has increased significantly $(\mathrm{p}K_\mathrm{sp}(\ce{Co2S3}) = 49.9,$ $K_\mathrm{sp}(\ce{Co2S3}) \approx \pu{1.26E-50}).$
Five least soluble sulfides from that table are:
$$
\begin{array}{lrc}
\hline
\ce{M_xS_y} & \mathrm{p}K_\mathrm{sp} & K_\mathrm{sp} \\
\hline
\ce{Ir2S3} & 196.3 & \pu{5.0E-197} \\
\ce{Bi2S3} & 115.1 & \pu{7.9E-116} \\
\ce{Mo2S3} & 107.8 & \pu{1.6E-108} \\
\ce{Ni3S4} & 104.5 & \pu{3.2E-105} \\
\ce{In2S3} & 96.3 & \pu{5.0E-97} \\
\hline
\end{array}
$$
So that it seems like the new champion (at least among sulfides) is iridium(III) sulfide $\ce{Ir2S3}$ with $K_\mathrm{sp} = \pu{5.0E-197}$.
References

Goates, J. R.; Gordon, M. B.; Faux, N. D. Calculated Values for the Solubility Product Constants of the Metallic Sulfides. Journal of the American Chemical Society 1952, 74 (3), 835–836. DOI: 10.1021/ja01123a510.
Waggoner, W. H. Textbook Errors: Guest Column. The Solubility Product Constants of the Metallic Sulfides. Journal of Chemical Education 1958, 35 (7), 339. DOI: 10.1021/ed035p339.
Licht, S. Aqueous Solubilities, Solubility Products and Standard Oxidation-Reduction Potentials of the Metal Sulfides. Journal of The Electrochemical Society 1988, 135 (12), 2971. DOI: 10.1149/1.2095471.


Formal nomenclature for fulvene I am a little bit confused trying to name the compound fulvene:

I'm not sure whether its name is 5-methylene-1,3-cyclopentadiene or                        3-methylene-1,4-cyclopentadiene.

Fulvene consists of a cyclopentadiene ring and a $\ce{{}=CH2}$ substituent group. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefix ‘methylene’ is no longer recommended to designate the $\ce{{}=CH2}$ group. The preferred prefix is ‘methylidene’; thus, the systematic name of fulvene without locants is methylidenecyclopentadiene.
Concerning the numbering of locants, the current IUPAC recommendations read as follows.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, low locants are given first to the double bonds in the ring according to (e) and then to the prefix ‘methylidene’ according to (f). Thus, the correct systematic name is 5-methylidenecyclopenta-1,3-diene rather than 3-methylidenecyclopenta-1,4-diene since the locant set ‘1,3’ for the double bonds is lower than ‘1,4’.

Note that according to IUPAC nomenclature since 1993, locants are placed immediately before that part of the name to which they relate (e.g. cyclopenta-1,3-diene rather than 1,3-cyclopentadiene).


Electron withdrawing groups and acidity I know that a strong electron withdrawing group reduces electron density and makes a molecule more acidic. Also, a meta substituent will have less effect than ortho and para substituence and the para isomer will be more acidic than the ortho one.
The point that I wonder: Does the acidity always increase with the number of withdrawing group as well?

Some group, let's say, phenol has one strong electron withdrawing group like nitro, carboxyl, chloro etc. If phenol would have two more of that same withdrawing group attached, would it be simply more acidic? Or, should another factors be considered to determine the acidity? I thought that maybe it would be more reliable to look $\mathrm{p}K_\mathrm{a}$ values but couldn't find any molecule that fits that situation.

Yes and no, with more electron withdrawing groups at ortho and para positions would certainly increase acidity but other factors do play a role. For example,
In trinitrobenzoic acid

here the two $\ce{NO2}$ at ortho positions are very bulky and due to steric repulsion rotate the $\ce{COOH}$ group thus making the compound non planar due to which there will be no more resonance thus, reducing acidity drastically. This is the most important factor that plays role in reducing the acidity.


Does a stoichiometric coefficient represent molecules or moles of a substance? I understand that the one mole of atoms is equal to $6.02\times10^{23}$ atoms. However, when I used various resources to learn about stoichiometry, they showed me a balanced equation:

$$\ce{2H2O->2H2 + O2}$$

and said "2 moles of water" decomposed to form "2 moles of molecular hydrogen" and "1 mole of molecular oxygen". It is my understanding that there are only 2 molecules of water, not $2 \times 6.02 \times 10^{23}$ molecules of water and so on with the other parts of the equation as well. Could someone please clear up my confusion?

Balanced equations represent ratios of reacting chemical species, not absolute quantities. What we mean when we say $$\ce{2H2 +O2->2H2O}$$ is not literally that two molecules of hydrogen react with one of oxygen, but rather that two units of hydrogen react. These units could be moles or molecules or something else entirely.


How many electrons can have the quantum number set n=5, l=3, ml=-1? How many electrons can have the quantum number set $n=6,\ l=3,\ m_l=-1$?

Also, please explain why. I know that n describes the number of shells in an atom but what do n, l, and ml have to do with electrons?

This post answers the question of the of what the other quantum numbers mean:

What do the quantum numbers actually signify?

As for how many electrons can have $n=5,\ l=3,\ m_l=-1$...

The Pauli Exclusion Principle states that two identical fermions (an eletron is a type of fermion) cannot occupy the same quantum state.

  In other words, two electrons (in the same atom) cannot have identical quantum numbers.

For each $n$, the values of $l=0,1,2,...,n$
For each $l$, the values of $m_l=-l,...,-1,0,+1,...,+l$

For each $m_l$ there are only two allowed values of $m_s$, namely $m_s = + \frac{1}{2}$ and $m_s = -\frac{1}{2}$. Thus, each set of $\{n,l,m_l\}$ can only describe two electrons. In your case, we have two electrons with $n=5,\ l=3,\ m_l=-1$:

$\{n,l,m_l,m_s\}=\{5,3,-1,+\frac{1}{2}\}$
$\{n,l,m_l,m_s\}=\{5,3,-1,-\frac{1}{2}\}$


Electrochemical half cell without redox couple In a book about electrochemistry the situation of a cell is described of which one half cell has an "undefined thermodynamic potential" as no redox couple exists:

$$
Pt/Cu^{2+}\ (0.01\ M),\ Cd^{2+}\ (0.01\ M),\ H_2SO_4\ (1\ M)
$$

Estimating the potential where reduction of Cu sets in by thermodynamic data when a potential is applied,

$$
Cu^{2+} + 2e \rightleftharpoons Cu,\ +\ 0.340\ V\ vs. NHE\\
Cd^{2+} + 2e \rightleftharpoons Cu,\ -\ 0.4025\ V\ vs. NHE\\
O_{2+} + 4H^+ + 4e \rightleftharpoons H_2O,\ +\ 1.229\ V\ vs.\ NHE,
$$

they argue that there is a rest potential range $0.340\ V ... 1.229\ V\ vs. NHE$ where no current is to be expected. Beyond these values current flow is expected. I don't understand this, because I understand $0.340\ V\ vs. NHE$ to be the standard potential of copper reduction under standard conditions, but the concentration of 0.01 M is not "standard" (= 1 M), right?

On the other hand, I'm afraid I have a general lack of understanding a system where one species of a redox pair is missing, because argueing in terms of Nernst's equation for such systems always leads to infinitely large (or small) potentials. In the case of copper above Nernst is not applicable in my opinion (correct ?), but what does the general relation

$$
 \lim_{[R] \rightarrow 0} (E_0 +  \frac{RT}{nF} \ln\frac{[O]}{[R]})=\infty
$$
for the reaction $O+ne\rightleftharpoons R$ mean? And how can one figure out the rest potential range for a half cell on thermodynamic basis (neglecting kinetics)? What does the situation for copper looks like where [R] does not appear in Nernst's equation? When the solution in the book is right, why?

Nernst's equation

The Nernst Equation can be derived from the Gibbs energy $\Delta G = nFE $. The free energy for a cell reaction:

$$\Delta G = \Delta G^\circ - RT \ln Q$$

can be written as:

$$E = E^\circ - \frac {RT}{nF} \ln Q$$

for reaction

$$\ce{O + $n$e -> R}$$

$$E = E^\circ -  \frac{RT}{nF} \ln\frac{a_\mathrm R}{a_\mathrm O}$$

where $a_i$ is activity of species $i$.

Activity
In a solution, the activity of $i$ is defined: $a_i=\gamma_i \frac {C_i}{C^0}$ where $C_i$ is the concentration of solute, $C^\circ$ is standard concentration ($1\ \mathrm M$), and $\gamma_i$ is unitless activity coefficient.

The equation can be simplified in:

$$E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]}{[\ce{O}]}$$

But for the reaction

$$\ce{Cu^2+ + 2e- -> Cu}$$

the Nernst's equation will be:

$$
E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]=1}{[\ce{O}]}
$$

because activity of solid (deposited copper) is 1.

If you try to calculate the potential of $0.01\ \mathrm M$ solution you will obtain the difference between standard concentration and the potential at $0.01\ \mathrm M$.

$$\begin{align}
E &= E^{\circ'} -  \frac{0.02559\ \mathrm V}{2} \ln\frac{1}{0.63}\\[6pt]
&= 0.34\ \mathrm V -  0.0059\ \mathrm V
\end{align}$$

The computed potential ($0.0059\ \mathrm V$) is lower than the standard potential and they are probably neglecting it.


How would this reaction look like in a potential energy diagram? (SN1 Reaction) My task is to explain, why molecule A reacts in a SN1-reaction faster than molecule B. (I would do this by comparing their potential energy diagrams)

My idea is to have a closer look at the reaction mechanism at first:

TS1 is here the intermediate with the carbo-cation. That's the step, where the reaction is slow, where one needs a lot of energy to remove the Bromine.
Maybe I guess one does not need this lot of energy in molecule A compared to B because the Bromine is in the background (trans- isomer) there and not in the foreground? Furthermore, in SN1 reactions the reaction rate is determined by the substrate (the first step).
So, the activation energy of molecule A is smaller then the activation energy needed in B and that's the reason, why the reaction ist faster?

An interesting question is how (or if) one really knows if both reactions proceed via bonafide SN1 type mechanism. Under the assumption that both reactions proceed via bonafide SN1 mechanism, we must infer that the carbocation (carbenium ion) intermediate 1 in both cases must be the same. While the previous answer correctly assumes that this intermediate may be described well as planar (at the carbon centre), the attack by water is the faster step and hence should not affect the overall rate of the reaction. Since both A and B must proceed through the same intermediate, (although not rate determining!), both must react at the same rate with the incoming nucleophile.

A better explanation can be deduced by looking at the most preferred conformations in the two starting materials. These are designated 2 and 3 below, for A and B respectively. In both cases, the bulky tertiary butyl group will likely prefer to be sat at an equatorial position to minimise 1,3 diaxial interactions. Once tertiary butyl group is "locked" at the equatorial position, we have to place the bromine axially (up) in A (2), and equatorially (down) in B (3).It is possible to argue (although still very qualitative and hand-wavy) that 3 is likely more stable than 2, since both groups are equatorial in 3.

Now, the rate of a reaction depends on the activation free energy barrier, which is the difference between the free energy of the reactant and that of the transition state. At this stage, without computation it is impossible to predict which transition state will be at a higher or lower energy. However, speculatively, to a first approximation the energies of the transition states would be very similar. Under this assumption, the activation energy barrier for A (higher energy conformation) should be less than that for B, which should explain the observed reaction rates.

Having said that, I would not necessarily think that energy diagrams are the best way to explain this difference in reactivity, or indeed if there is a single, simple explanation.


Intermolecular and Intramolecular Forces (Vapor pressure of n-butylamine and methanol) What are the different properties that intermolecular forces and Intramolecular forces attribute to? i.e. intermolecular forces contribute to BPs, vapor pressures, etc.

Specific problem: why does n-butylamine have lower vapor pressure at room temperature than methanol? Explain in regards to IMFs.

What are intermolecular molecular forces?
The forces(electrostatic force) between molecules of a compound. These forces are responsible for intermolecular distances and how tightly the molecules are held together. If the intermolecular forces(van der Waals force) are large, it means the molecules are tightly held together in a group and they are fixed in their position.

Now what is boiling point? It's a temperature at which liquid state of a compound changes into its gas state. What's the difference between gaseous and liquid phase? Gaseous phase has very large intermolecular distance when compared to that in liquid phase. If a large force exists between the molecules and they are quite fixed in their position, then large amount energy or heat will be required to move them away from each other. Large heat means greater temperature and hence greater boiling point. So the lesson is that, the stronger the intermolecular forces, the higher the boiling point.
The similar concept in applicable in melting point. the stronger the intermolecular forces, the higher will be the melting point.

You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably(In case you may be wondering, how are they related. Click here). If the intermolecular forces are large, then molecule in liquid phase are not able to leave to liquid easily, hence lowering the escaping tendency and therefore lower vapour pressure.

Note: This is a very basic explanation.


Étard Reaction of Ethylbenzene What is the major product when the following compound undergoes Étard reaction?

Is it acetophenone or a terminal aldehyde.
The mechanism discussed in Mechanism for an Étard reaction (chromium complex) does not give any conclusive prediction.

Ethylbenzene yields acetophenone when it undergoes Étard reaction.
The mechanism is as follows:


Does the benzene ring in styrene show a negative inductive effect on the ethenyl group? I know that a group exerts an inductive effect on another one if there is a difference in electronegativity. In the case of styrene (1) all the atoms are sp2-hybridized. So, how could the phenyl ring show a negative inductive effect on the ethenyl group?

This can be explained by looking at the molecular orbitals. The molecular LUMO of benzene (which accepts electrons) is closer to the HOMO of ethylene (which given electrons) as compared to the large gap between HOMO of benzene and LUMO of ethylene.

Surely, the HOMO, LUMO don't exist separately and only exist as a molecular orbital. But the separation makes for easy transfer of electrons from ethylene to benzene unlike the opposite.
Because I have involved pi orbitals into the solution, I believe mesomeric effect is more apt here. To talk about inductive effect we have to involve a more complex idea (group electronegativities). Resources for such data are not easily accessible. I would appreciate if someone can find the group electronegativities of benzene and ethylene and post them here for the analysis of inductive effect as well.
Surely, if present, a soft nucleophile will it will attack the 2-carbon on ethylene side-group.


Weighing by difference When preparing a known mass of something (e.g. when producing a standard solution) it is often that in my school lab that we first use a weighing boat to measure the mass and then once transferred into a beaker we reweigh the boat and calculate the mass transferred.

Do we this even though we zero the scale anyway?

Now to my knowledge would this not increase the uncertainty as instead of just using the beaker on the scale and adding the mass directly which would lead to one uncertainty we instead weigh it twice which would thus have 2x the uncertainty in the scale.

Could somebody explain where my reasoning is wrong and why this method is more accurate?

Although there are several sources of weighing error that would be minimized by using the weighing by difference method (as discussed in the comments), I want to illustrate what I see as the limiting problem, that of the relative uncertainty of the measurement, which roughly scales (no pun intended, really) with the mass being measured.

Lets use a bit of an exaggerated situation for illustration. In particular, we'll say that we have one balance that can measure down to $\pu{0.001g}$, and can handle up to $\pu{100g}$ total mass. In reality you would likely have a high-mass balance for weighing directly into the beaker, or a low-mass balance for using the weighing boat. We are just using one really good balance to simplify this illustration, but the concept is the same either way.

Lets say our beaker has a mass of $\pu{50g}$, our weighing boat is $\pu{1g}$, and we are to measure out $\pu{1g}$ of material.

The beaker plus the material to be measured will then have a measured mass of ~$\pu{51.00 +/- 0.05g}$. This means that the relative uncertainty in the measured mass of the material is no better than $\pu{5\%}$($\pu{0.05g}$ out of $\pu{1g}$).

The weighing boat plus the material to be measured out will have a measured mass of ~$\pu{2.000 +/- 0.002g}$. This means that the relative uncertainty in the measured mass of the material is $\pu{0.2\%}$ ($\pu{0.002g}$ out of $\pu{1g}$).

So, even with two measurements rather than one, using the weighing boat method results in much less total uncertainty in the measured mass of your material as compared to weighing a small quantity of material directly into a large beaker.


Reaction of nitro compounds with strong alkali? Why does a primary and a secondary nitro compound behave as an acid in the presence of strong alkali whereas tertiary nitro compounds do not?
Can anyone explain it with reactions?

In order for a compound to act as an acid, it must have a proton which can be abstracted. As you correctly pointed out, tertiary nitro compounds have no $\alpha$ hydrogens and therefore they cannot be deprotonated at the $\alpha$ position.

Of course, they could be deprotonated elsewhere, but if the other groups are just alkyl groups then this isn't really feasible as they have such a high $\mathrm{p}K_\mathrm{a}$.


Given a forward reaction with positive enthalpy, is its reverse reaction be endothermic or exothermic? I'm pretty sure the sign changes.

But, I want to be sure if it becomes endothermic or exothermic.

Yes, in a reverse reaction the sign changes. That is because Enthalpy is a state function, one that depends on only the current state of the function. That means the energy required to push a reaction to its products would be directly reversed to push it back to its reactants.

A key use of this knowledge is in Hess's law, and Born Haber Cycle.


Is there a liquid that looks like water but boils at a low temperature? I just watched a movie where someone put their hand into boiling water. Disregarding how this was actually done, I wondered whether there was a liquid which could be used for an illusion of someone "putting their hand into boiling water".

For this it would have to look like water, boil at a comfortable temperature (< 50 degree celsius) and should not be so toxic that you can't put your hand into it. Smell doesn't matter.

Is there some liquid that fulfils these requirements?

The following list gives some boiling points

I've added the suggestions of Ben Norris and Todd Minehardt.

Common organic solvents

Pentane: 36 °C.
Diethylether: 35 °C.
MTBE: 55 °C.
Acetone: 56 °C.

Please read the Material Safety Data Sheets (MSDS), pay attention to the hazard statements and decide whether you (still) want to put your bare hands into these solvents!

Refrigerants

Dibromodifluoromethane: 23 °C
Trichlorofluoromethane: 24 °C
2,2-Dichloro-1,1,1-trifluoroethane: 28 °C
1,1-Dichloro-1-fluoroethane: 32 °C


Why is benzoic acid a stronger acid than 3-aminobenzoic acid? Why is benzoic acid a stronger acid than 3-aminobenzoic acid? Obviously at meta position, only negative inductive effect will operate. But, how does that make it a weaker acid than benzoic acid?

The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.2$. The $\mathrm{p}K_\mathrm{a}$ of aniline (aniline's conjugate acid really) is $4.8$.

The two macroscopic $\mathrm{p}K_\mathrm{a}$s of 3-aminobenzoic acid (aniline-3-carboxylic acid) are $3.1$ and $4.8$.

When the microscopic $\mathrm{p}K_\mathrm{a}$s of two groups in a molecule are similar, the macroscopic $\mathrm{p}K_\mathrm{a}$s are not reflective of one particular group protonating/deprotonating. The neutral form will include both a zwitterionic and an non-zwitterionic form in the case of 3-aminobenzoic acid.

When the groups are close in space within the molecule, there will be electrostatic interaction. If the amino group is protonated, it facilitates deprotonation of the carboxylic acid group. If the carboxylic acid group is deprotonated, it facilitates protonation of the amino group.

Addition: There is a very relavent article: Carbon-13 NMR determination of acid-base tautomerization equilibriums J. Org. Chem., 1984, 49 (4), pp 691–696.  It reports that the neutral species is 59% zwitterion and 41% molecular (non-zwitterion).


Difference of dipole moments of dichloromethane and trichloromethane? Why is the dipole moment of $\ce{CH2Cl2}$ ($1.60 ~\mathrm{D}$) greater than that of $\ce{CHCl3}$ ($1.08~\mathrm{D}$)?

Based on my knowledge of vectors, I feel it should be the other way around.

Just to add some quantification to Ben Norris's answer. Consider each $\ce{C-Cl}$ bond, which has a bond dipole moment of magnitude $A$. The contribution from $\ce{C-H}$ is neglected here to simplify the calculations.

Now consider dichloromethane. The resultant of the two $\ce{C-Cl}$ dipoles will have a magnitude of $$2A\cos\frac\theta2$$
where $\theta\approx109.5^\circ$, the angle between the vectors from the ends of tetrahedron to the centre. This turns out to be $1.154A$.

For tricholoromethane, the three $\ce{C-Cl}$ dipoles will be at an angle $\theta$ such that $\cos\theta=1/3$ with a straight line symmetrically passing through its centre. Thus the resultant in this case will be $$3A\cos\theta$$
which will be $1A$, lesser than in the case of dichloromethane.

To visualise how can you get those angles and cosines, see this page.


Is terylene a copolymer? Terylene is a polyester made from terephthallic acid and ethylene glycol. Even though we use two reagents, they undergo esterification to give bis(2-hydroxyethyl) terephthalate (monomer) which then undergo polymerization.

So, is it correct to consider terylene as a copolymer? My reasoning tells that it should be a homopolymer. Same line of reasoning gives nylon, phenol-formaldehyde resin, bakelite — all as homopolymer.

But majority of questions which I go through considered the above polymers as copolymers. What is correct? Are they homopolymers or copolymers?

No, PET it is definitely not a copolymer. Neither is e.g. Nylon 6,11 (or any other polyamide with two numbers). The monomers alone would not polymerise, or at least only by a very different reaction (e.g. ethylene gylcol gives PEG).

What you can have are alternating copolymers, but there the sequence is a statistical property. It comes when a reactive chain end B* has a higher probability to add an A (to give BA*), and a chain end A* to add B (to give AB*). That probability is sensitive to monomer composition, the temperature, pressure, catalyst, etc. If you want to read more about that, "copolymerisation parameters" is the topic to look up. The monomers you see there are afaik invariably olefines, i.e. they have a double bond that undergoes polymerisation.


Formyl chloride naming How is the naming of formyl chloride $\ce{CHOCl}$ justified? What is the functional group in it?

Why can't I name it as methanoyl chloride as acid halide is given more preference than aldehyde?

Welcome to parallel worlds of trivial and systematic names. We also use formic acid, acetic acid, acetyl chloride.

You can name them all systematically as methanoyl chloride, methanoic acid,  ethanoic acid and  ethanoyl chloride. But in the real lab life, nobody call them like that.

Formyl chloride has  its name from
formyl-  as acyl derived from formic acid, the same as acetyl- from acetic acid.

Technically, formyl chloride is an aldehyde as well, as formyl is de facto an aldehyde group, being formally a member of the serie :

$$\ce{Cl-CH2OH <> Cl-CHO <> Cl-COOH}$$

But by naming conventions, both trivial and systematic, being a derivate of an acid replacing $\ce{OH}$ by $\ce{Cl}$ takes precedence to naming it as an aldehyde.

As a sidenote, it is rather funny aldehyde names are derived from the respective acids and not alcohols.  Because the aldehyde etymology says aldehyde="alcohol dehydrogenatus", so it could be as well methylaldehyde and ethylaldehyde, instead of formaldehyde and acetaldehyde.


Why does CO2 have higher boiling point than CO? The intermolecular forces between $\ce{CO2}$ molecules are dispersion forces, while the forces between $\ce{CO}$ molecules are mostly dipole-dipole attraction forces. So, why does $\ce{CO2}$ have a higher boiling point than that of $\ce{CO}$?

CO2 has more electrons than CO. This means that it has a much larger electron cloud as compared to CO, so its more easily polarised and thus, the ease of forming instataneous dipole-induced dipole bonds increases. Even though CO is a polar molecule and it forms permanent dipole-permanent dipole bonds, in this case the id-id bonds are stronger.

P.s. Just a 17 year old A Level Chem student here, I might be wrong


Resonance condition for absorption of energy In NMR, for example, why it is not possible to excite nuclei from one spin state to another using energy of arbitrary frequency?

This is indeed the basis of modern Fourier transform NMR (the technique that all modern spectrometers use).

The molecule is irradiated with a range of frequencies and the response detected. Application of some clever maths (the Fourier transform) then allows the individual signals to be observed.

You can think of this a bit like bashing every key on a piano and recording the noise. Application of Fourier transform would then allow you to get back the sound for the individual keys.


In Flame Emission Spectroscopy, what part of the flame is used to measure the concentration of the analyte? Flame Emission Spectroscopy is an analytical measuring device that relates the intensity of the emitted light relative to to the concentration of the sample.

As I am reading an article about FES, I noticed how the author purposely inserted the different parts of the flame and their activity.

The above image was from an article in Chemistry LibreTexts about FES.

The author said "The Primary Combustion Zone is where the initial decomposition occurs and molecular fragments are observed. The Interzonal Region is the hottest part of the flame and atomic fragment are observed. The Secondary Combustion Zone is cooler overall and a conversion is seen from atoms back to stable molecules and oxides."

I have read the article but it was never mentioned what part of the flame does the transducer or sensor was placed on.

I have read the article but it was never mentioned what part of the flame does the transducer or sensor was placed on.

It was never mentioned because no sensor or transducer is actually placed on or in the flame. What spectroscopist do is that they focus the light (coming from the flame) into a device called the monochromator. The job of the monochromator is to separate the wavelengths of the light from the flame. The desired wavelength is then focussed on the detector. Thus the detector is really far from the actual flame.

Now the key question is which part of the flame is useful. There is no single correct answer. The criterion is we scan the height of the flame and check which part of the flame is showing the maximum intensity of the desired signal. This is done by changing the height of the burner. Most spectrometers do it automatically but old lovely instruments had knobs for manual adjustment.

Normally for most elements one would like to see the hottest part of the flame, where the concentration of the atomic emission is the highest.


What is the relation between volume strength and normality of H2O2? I have been told that:
$$\text{equivalent weight of }\ce{H2O2} =\text{molecular weight of }  \ce{H2O2}$$
as it's $n$-factor is $1$ (from balanced equation).
But in my book it's given that:

Molarity $= \frac {V }{11.2}$
Normality $= \frac {V} {5.6}$

Why are they different if equivalent and molecular weights are equal?

The n- factor must be obviously wrong because two electrons are involved in the reduction of peroxide to oxygen. Write the half cell yourself. Indian chemistry textbooks should have pity on the world and stop teaching these equivalents to young students. This is a thing of the 18th century. Those who invented this concept moved on.


On the assignment of axial and equatorial to substiuents of polyhexacyclics Must the assignment of axial and equatorial substituents of polyhexacyclics always be with respect to each six-membered ring or is there a shorthand?

For example, how would X, Y, and Z be assigned axial/equatorial in the following substituted adamantane?

The structure you have drawn where X = Y = Z = H is known as adamantane, C10H16. It has 3 degrees of unsaturation but visually appears to have 4 rings. Once 3 rings have been defined (rings A, B and C), then all C-C bonds have been accounted for. Put another way, you only need to erase three C-C bonds and add hydrogen to obtain a saturated, acyclic hydrocarbon, C10H22.
In structure 1 ring A is highlighted in red with X and Y equatorial to ring A. In structure 2 substituents X and Z are both axial to ring B. Structure 3 has the configuration of the Z substituent changed to equatorial relative to ring B. Ring C bears the same relationship to ring A as ring B does to ring A. Therefore in structure 4, substituent Y is axial to ring C. Now locate ring "D" and define equatorial/axial relationships of the Z substituent to ring "D" in structures 1 - 4.


Does hydrochloric acid change in concentration over time? I have new bottles of 1 N hydrochloric acid from Fisher. The acid comes standardized from the company:

  Standardized at 25 °C to pH 8.2 against sodium hydroxide that has been standardized against NIST™ potassium acid phthalate.

Will it remain standardized after opening the bottle, or do we have to standardize it regularly?

First of all, check the expiry date. In general HCl solutions are secondary standards. 1 N is just a approximation. You need to standardize it every time with a primary standard such as sodium carbonate. On the other hand, constant boiling HCl is indeed a primary standard. It is not very well known, perhaps because modern analytical chemists have become lazier.

Constant-boiling hydrochloric acid
John L. Sheldon
J. Chem. Educ., 1935, 12 (8), p 377
https://pubs.acs.org/doi/abs/10.1021/ed012p377


Alkaline Ester Hydrolysis Reaction In an ester hydrolysis reaction using a dilute base what is the pH change before and after the reaction occurs. In this reaction I want to react methyl ethanoate with dilute sodium hydroxide. I want to know if the pH of the solution will change after I add the ester to the base.
If it doesn't, is there a reaction I can do with esters that will result in a pH change?

Alkaline hydrolysis of ester:

  $\ce{RCOOR' + OH- -> RCOO- + R'OH}$

Almost all organic acids are weak acids:

  $\ce{RCOO- + H2O <<=> RCOOH + OH-}$

Therefore, a lot of $\ce{OH-}$ is consumed during the esterification, with the product $\ce{RCOO-}$ generating little $\ce{OH-}$.

There is a net consumption of $\ce{OH-}$, making the solution more acidic, lowering the pH.


Why does zinc plate copper in silver penny lab In class, we performed a lab that involved heating a solution of zinc sulfate with zinc strips at the bottom of the beaker. A penny was placed on top of the zinc strips and soon after the penny became covered in a layer of zinc. What process causes the zinc layer to develop?

The explanation was that the zinc metal transferred electrons to copper which in turn reduced the zinc ions in solution, but this does not seem to make any sense as the zinc is being oxidized to just be reduced again. Why does this happen.  Is a thin zinc layer more stable?

Here are the instructions for the experiment

Electricity is involved: the copper (well, copper plated zinc in the US) penny is in direct contact with the zinc, forming a short-circuited connection. Forcing the plating action "backwards" (copper being more electronegative than zinc) requires current.

Try the same experiment but with a separator, such as a piece of filter paper, between zinc and copper; there should be no plating.


Why is chloroethene said to be a conjugated system? Why is chloroethene said to be a conjugated system?Is it because of p orbitals present in chlorine or something?

Conjugation can be between pi-bond and another pi-bond, lone pair of electrons and a pi-bond(as in your case) and pi-bond and a vacant orbital(p or d).


How does aniline react with HNO3 in the absence of sulfuric acid? How does aniline react with $\ce{HNO3}$ in the absence of sulfuric acid? I read that nitration occurs at slow rate, but it didn't explain how nitronium ion is formed.

In the absence of sulfuric acid, aniline does react with nitric acid to form anilinium nitrate. It is observed as an intermediate product during the production of diazobenzene nitrate (benzenediazonium nitrate). But the species is too unstable for mass production and cannot be stored for long hours. Here is the description:

  20 g of aniline are placed in a beaker, well cooled, and “boiled-out”
  nitric acid, previously diluted with half its volume of water,
  carefully added till the mixture sets to a thick crystalline paste –
  aniline nitrate.[...]

Caution: The synthesis is not to be carried out due to high explosive nature of diazobenzene nitrate. Explode in dry state.


What happens when mixing elemental mercury with sodium hypochlorite solution Say you spilled elemental mercury from a thermometer and used household bleach on it, what would be the reaction if any?

There are several redox reactions can be taken place when you add household beach on elemental mercury metal. However, this question is more like a homework question. Thus, I give you insight to the answer and you may read a bit and find out what's happening. This reaction is studies and results have been published (Reference 1), the abstract of which states that:

  The absorption of elemental $\ce{Hg}$ vapor into aqueous hypochlorite was measured in a stirred tank reactor at $25$ and $\pu{55 ^{\circ}C}$. $\ce{NaOCl}$ strongly absorbs $\ce{Hg}$ even at high pH. Low $\mathrm{pH}$, high $\ce{Cl-}$ and high-temperature favor mercury absorption.  Aqueous free $\ce{Cl2}$ was the active species that reacted with mercury. However, chlorine desorption was evident at high $\ce{Cl-}$ and $\mathrm{pH}<9$. Overall second-order reaction was observed between mercury and chlorine with an apparent second-order rate constant of $\pu{1.7\times 10^{15} M^{-1} s^{-1}}$ at $\pu{25 ^{\circ}C}$ and $\pu{1.4\times 10^{17} M^{-1} s^{-1}}$ at $\pu{55 ^{\circ}C}$. Gas-phase reaction was observed between $\ce{Hg}$  and $\ce{Cl2}$ on apparatus surfaces. Strong mercury absorption in water was also detected with $\ce{Cl2}$ present. Results indicate that the chlorine concentration, moisture, and surface area contribute positively to mercury removal.

I hope you will figured it out when you consider the standard reduction potentials of $\ce{Hg}$ and $\ce{OCl-}$ in any electrochemical series. For example, see Reference 2.

References:

L. L. Zhao, G. T. Rochelle, “Mercury absorption in aqueous hypochlorite,” Chemical Engineering Science 1999, 54(5), 655-662 (https://doi.org/10.1016/S0009-2509(98)00263-2).
S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).


How to reverse damage done by nail polish remover without acetone to a laptop keyboard? Today I was trying to clean my laptop with a nail polish remover (ingredients: ethyl acetate, peg-40, hydrogenated castor oil, panthenol, parfum, d-limonene, geraniol, ci 26100, ci 60725). Not only it didn't get clean, but some white thing remained on the keys.
I tried boiling water and a piece of cloth to make it warm and easier to remove without any success.
What can be done to reverse the damage?

The solvent appears not only to have removed paint on the keys but also to have caused solvent crazing in the plastic itself. Regrettably, there is probably not much that can be done to actually remove the crazing, which may have penetrated too deeply to remove by buffing, but you might be able to hide it with a substance such as an oil that has the same index of refraction as the plastic.

If the appearance is unacceptable, you might replace the keyboard, use a plug-in keyboard, or get a keyboard cover, such as the Kuzy Russian Keyboard Cover, specific to the Mac Pro. Individual key stickers would probably not adhere well.


Why is methyl chloride a primary alkyl halide? In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.
  Chemguide.co.uk

For $\ce{CH3Cl}$, the carbon which carries the halogen atom has no other alkyl groups attached, yet my class notes say "the alkyl halide has a primary centre". So is a primary alkyl group one with is attached to 0 or 1 other alkyl group and not just 1?

Yes, you are correct.

The IUPAC defines alkyl groups in the following manner:

  Univalent groups derived from alkanes by removal of a hydrogen atom from any carbon atom $\ce{–C_{$n$}H_{$2n+1$}}$. The groups derived by removal of a hydrogen atom from a terminal carbon atom of unbranched alkanes form a subclass of normal alkyl (n-alkyl) groups $\ce{H(CH2)_{$n$}}$. The groups $\ce{RCH2}$, $\ce{R2CH~(R \neq H)}$, and $\ce{R3C~(R \neq H)}$ are primary, secondary and tertiary alkyl groups, respectively.

Note that for primary alkyl groups hydrogen may be a remainder, i.e. $\ce{R $=$ H, Alkyl, Aryl}$, therefore any halogenoalkane with the structure $\ce{RH2C-X}$ is primary.


Are fractional oxidation states possible? To get the oxidation state of carbon (denoted $x$) in octane, $\ce{C8H18}$, I used the fact that hydrogen has an oxidation state of $+1$:

$$8x + 18\cdot 1 = 0 \qquad \Longrightarrow \qquad x = -\frac{18}{8} = -2.25$$

But in that way, I get an oxidation number that is fractional, or not an integer. Is that normal?

The "oxidation number" is a theoretical value used to do electron bookkeeping and is one way of comparing the number of electrons "owned" by an atom in a molecule or ion versus how many valence electrons present in the atom as depicted on the periodic table.

Your calculation is assuming that all of the carbon atoms in octane have the same oxidation state, which is incorrect. The two terminal carbons have an oxidation state of -3 while the inner 6 have oxidation states of -2. Taking the average of these values gives you an average oxidation state, which in your case is a fractional number.

It is doubtful that the oxidation state calculated in this manner is of any use, since it is based on the assumption that all the carbons in the molecule are the same.


Educational practice problems/systems/reactions in computational chemistry I am not sure if this question is suitable for Chemistry SE, but I think fairly objective answers can be given, and not just I will benefit from them

I have almost completed the first year of a MSc program in theoretical (and computational) chemistry, and have gone from knowing absolutely nothing about computational chemistry, to having heard of some of the stuff I need to master. I just took the step of installing GAMESS on my personal computer, and I intend to start doing test calculations on my own.

While courses provide theoretical foundations, my experiences are that actual computations are quite limited. I took the initiative to engage in a student research project which started February this year, and during this time I have gained a lot of practical experience with calculations (exploring the PES, finding and optimizing transition state structures, calculating MEP's, looking at molecular orbitals). And to me this kind of experience is extremely valuable.

So I would like to practice with well-studied systems/reactions/whatever. In short, I want a diverse set of systems, some of which may require "manual labour" (e.g. CASSCF). With well-known systems I can find experimental and theoretical values to compare with, and likely also discussions about how certain methods perform for each system. These systems should be small enough so that they can be comfortably run on a Macbook Pro (2,9 GHz Intel Core i7, 8GB RAM, SSD) without taking too long.

So question basically is this: What is a good set of educational systems/reactions, suitable for an interested student in computational chemistry?

There's a free software quantum chemistry package called Psi4. If you're a GNU/Linux user, under Ubuntu or any Debian-based distro, you can install it with a simple terminal command.

sudo apt-get install psi4

They have a well documented site, including a section dedicated to education with simple projects and friendly message boards. As of today, they list the following projects in their education section:

What is the radius of an atom?
What makes a molecule polar or non-polar?
How do we calculate the most accurate energies for a boron atom?
Can we visually predict binding energies?
When does water become $\ce{OH + H}$? A study in molecular orbitals.
What happens to the orbitals as a one-dimensional box gets longer? A study in molecular orbitals.
Is $\ce{C3H+}$ present in the Horsehead nebula?
What is symmetry in chemistry?

The version available in Ubuntu repository is functional, but quite outdated, 1.1.5, as of Mar 2020. If one is serious about learning it, my advice is to download it directlly from their site. For me the only issue with it is that Avogadro, my output viewer of choice, seemingly doesn't provide native support to view Psi4 output files. But I count that as a missing feature in Avogadro, not Psi4.


Why can't diamagnetic, and paramagnetic magnetize? Ferromagnetic materials can be made into permanent magnets.  How come diamagnetic, and paramagnetic materials be made into magents?  Does it have to do with the valence electrons, and how they are arranged, and their spin?

The concept of Ferromagnetism, Paramagnetism and Diamagnetism is completely different, though they all show magnetic properties.

FERROMAGNETISM: This is constituted of domain and each domain contains a large number of molecules (the number varies from material to material). These domains tends to align themshelves in a certain fashion in presence of magnetic field in order to achieve lowest energy and hence ferromagnetic materials show attraction to presence of magnets. Now these domains are not loose, but the arrangement can be made permanent in presence of some prolonged magnetic field. The amount of magnetism something can hold can be explained by hysteresis loop (NCERT Class 12 Physics Text Book). The arrangement of domains are shown below:
PARAMAGNETISM and DIAMAGNETISM depends on the number of unpaired electrons present in a system. The presence of unpaired electron to the system gives it a magnetic moment (given  by $\mu=\sqrt{n(n+1)}$, in Bohr Magenton , where n is the number of unpaired electrons). Hence the number of unpaired electrons, i.e. the excess number of electrons in the same spin, determines the magnetic moment. A material is called diamagnetic if the value of $\mu = 0$.

Thus paramagnetic materials are permanent magnets by their intrinsic property, but the magnetic moment is too weak to detect it physically. The increase or decrease in magnetic moment, or, the number of unpaired electrons in paramagnetic materials, is impossible by physical methods like keeping them in touch with magnet, or rubbing a magnet, or keeping it in strong magnetic field as done with ferromagnetic materials.

However, this is possible through chemical methods. For example, $\ce{[Fe(CO)6]^{3+}}$ is diamagnetic while $\ce{[Fe(H2O)6]^{3+}}$ is paramagnetic. However, this is not magnetizing, it is just changing what you are working with.


What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers? The question is as follows:

  What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: $n=4, m_{l}=+3$

Since $m_{l}=+3$ I know that $l=3$ and then I wrote them all:
$$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}$$
which means that the maximum number of electrons is $70$, but the actual configuration in the answer was:
$$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 5d^1, 4f^{14}$$
which means that the maximum number of electrons is 71. What didn't I count for?

Here are the electron configurations of the relevant elements:

Source: https://chemistrytheperiodictable.files.wordpress.com/2017/11/electronic-config-lanthanides.png?w=474&h=399

The first $f$ electrons get filled in the 6th period of the table of elements. First, the $6s$ get filled, then (for lanthanum) the $5d$ gets filled for the first time, and then the $4f$ subshell. As you go across the lanthanides, the $5d$ is sometimes filled with one electron and sometimes empty.

  What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: n=4,ml=+3

The answer depend on how you interpret "last electron". Going from Thulium to Ytterbium, you could argue the last electron is going into $4f$, whereas from Ytterbium to Lutetium, the last electron is going to $5d$. On the other hand, the $5d$ is filled before $5f$ going from Barium to Lanthanum. Comparing Lanthanum to Lutetium, all 14 electrons seem to be going into the $4f$ subshell, supporting the notion that the answer to the question is 71.

In the end, it is not the best question - it does not have a clear answer, and you don't really learn anything from the answer other than that the energies of $4f$ and $5d$ are close to each other.


Calculating molar conductance of CaCl2 Ok, so the following question is given in my text book.

  Of the molar conductance value of $\ce{Ca^2+}$ and $\ce{Cl^-}$ at infinite dilution are respectively $118.88\times10^{-4}$ and $77.33\times10^{-4}$ then that of $\ce{CaCl2}$ is  (all have same unit)

So what I did was I added the molar conductivity of $\ce{Ca^2+}$ and added $2\times$ molar conductivity of  $\ce{Cl^-}$. I got the answer. It was one of the options. So I basically wanted to know: Did I do it correctly and is my process correct? And if I did it wrong then what is the correct process?

Yes, the process is correct, but do you understand why?


Can nitrate and sulfate ions be oxidised during electrolysis of their molten compounds? $$\begin{align}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{S2O8^2- + 2e- \;&<=> 2SO4^2-}\quad\quad &&E^\circ = +2.01\ \mathrm{V}
\end{align}$$

From the above, I know that an aqueous solution of a sulfate salt undergoing electrolysis will not oxidise to peroxydisulfate ions over water being oxidised to oxygen at the anode unless it is at extremely high concentrations.

However, what if the solution is molten? Will the product at the anode be peroxydisulfate?

Also, can nitrate ions be oxidised? Perhaps to peroxynitrate or peroxynitrite?

Not only they can, but they must get oxidized. There is no other option. This is the only way the electric current can pass between the electrode and the solution.

Whether or not the products will be some peroxyanions is another question. At high temperatures typical of molten salts, I don't think they will survive. Hence I would expect $\ce{O2}$ + disulfate in the first case and $\ce{O2 + NO2}$ in the second.


How did Michael Faraday melt Sodium Chloride? Michael Faraday melted sodium chloride in a silver spoon and ran an electric current through the molten salt, which released chlorine as a gas and left pure sodium behind.

Sodium chloride melts at $800\ \pu{^\circ C}$. How did Michael Faraday melt the salt?

*I know he invented the bunsen burner, but did he use that?

The answer is given in Faraday’s own work, Experimental Researches in Electricity, Volume 1, freely available thanks to The Project Gutenberg:

  399. Other substances, which could not be melted on glass, were fused by the lamp and blowpipe on platina connected with one pole of the battery, and then a wire, connected with the other, dipped into them. In this way chloride of sodium, sulphate of soda, protoxide of lead, mixed carbonates of potash and soda, &c. &c., exhibited exactly the same phenomena as those already described: whilst liquid, they conducted and were decomposed; whilst solid, though very hot, they insulated the battery current even when four troughs were used.

The text above shows that the lamp in question is the spirit lamp, and the whole aparatus is described here:

The crux of it is that an open flame might not have enough heat, so a conventional flame is simply contained into a more restricted space.


What is the IUPAC systematic name for methane? The 1993 IUPAC Blue Book claims that "methane" is a semisystematic name:

  Semisystematic names also exist, such as "methane", "propanol", and "benzoic acid", which are so familiar that few chemists realize that they are not fully systematic. They are retained, and indeed, in some cases there are no better systematic alternatives.

Propan-1-ol and phenylmethanoic acid are obvious systematic versions of propanol and benzoic acid. (Though I disagree that they are "better".)

What is the "fully systematic" name for methane—or is this a case where there is no "better systematic alternative"?

The latest edition of the Blue Book, the 2013 Revision, clarifies this:

  Methane is a retained name (see P-12.3) that is preferred to the systematic name ‘carbane’, a name never recommended to replace methane, but used to derive the names ‘carbene’ and carbyne for the radicals $\ce{H2C^2-}$ and $\ce{HC^3-}$, respectively.

So it seems to be the case that the systematic IUPAC name is only there for consistency with the ions, and is not recommended (and not used) to describe methane. Similarly, "ethane", "propane", and "butane" are preferred to their systematic equivalents "dicarbane", "tricarbane", and "tetracarbane".


What are some ways to measure sugar concentration of fruits? I've tried to do titrations using the quantitative Benedict's Solution and it was difficult to use as I do not have a way of keeping a constant temperature. I am currently using a Polarimeter to find the angle of the various concentrations of sugar in my solutions but it does not seem very clear or efficient.

The solute will alter the density of the solvent, which is used for example to determine the concentration of sulfuric acids by a hydrometer and in particular by help of a saccharometer.  Depending on your equipment you will need at least about 20 mL of analyte.
A popular alternative among winemakers is to rely on the change of the refractive index of a solution (refractometry) instead.  With portable devices as on the picture, you basically determine the critical angle of total reflection of the juice of few grapes placed on a prism.  The analysis is quick (seconds) and needs little (mL or less), the reading typically yields refractive index and content of sugars present as expressed for example in Oechsle, Baumé, Brix, or mass%.

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Note, however, that these techniques are not specific to sugar(s).  The presence of other compounds soluted equally may affect the refractive index, as well as the wavelength of light used and temperature during the recording.  To prevent adulteration of wine, for example, some more advanced refractometers already determine the refractive index at multiple wavelengths simultaneously, too.