Gaussian16 Raman calculation from IR checkpoint file

Raman intensities require polarizabilities to be calculated, IR intensities only dipole moments. Your checkpoint file may therefore not have the properties you require, in which case you'll need to do more calculations. Using the polar=raman keyword would apparently save you a little bit of time, because the program then already knows your force constants and normal modes from the preceding freq calculation, but there will still be computational effort involved in calculating the polarizabilities.
Note that there are some cases where Gaussian also give you the polarizabilities "for free", as the manual states on the polar=raman page:

Note that Polar is done by default when second derivatives are computed analytically.

You can use the chkchk tool to see what's really stored in the checkpoint file.


What you are doing is right. The value to use depends on the definition of the standard state. The values of $K_c$ and $K_p$ are indeed different, but what you are not accounting for is the fact that $\Delta G^o$ also changes with the equilibrium constant used.

The naught over $G$, representing "standard state" establishes some rules. For homogeneous gaseous reaction, the standard $\Delta G$ corresponds to the standard state of all gases, which is defined in terms of pressure $P=\pu{1atm}$, and hence $K_p$ needs to be used with the quoted value of $\Delta G^o$.
For reactions involving solutes and pure liquids, the standard states are defined differently, in terms of their concentrations, and hence $K_c$ is to be used. Since the definition of standard state affects the value of $\Delta G$, we must choose that form of the equilibrium constant which uses the activity of the reacting species in terms of the quantity used to define the standard state of that species. This maintains uniformity.


Surely you are aware that the calculation of this sort is but a very rough estimate, yet still I feel the urge to stress that once more. Having said that, let's look at the double bond in the first cycle: isn't it exocyclic with respect to the other cycle?


I have used the image you used $2nd$, see carefully the markings.
The $C-O$ bond with benzene ring is stronger than the $C-O$ bond of ethyl, as there is resonance with benzene making it partially a double bond. So, that $C-O$ is difficult to break compared to the $C-O$ bond in ethyl branch.


Expressing solubility in binary solvents with non-additive volumes

For solvent mixtures, especially in mixture or solubility design of experiments,  I work in mass for the solvents.  It's unaffected by temperature,  more accurate (very advantageous if small values are being used) and should get around your problem. You can quote your initial results as %w/w


Why does lead(II) acetate does not ionize appreciably?

A source in my native language says that lgK of PbAc2 is 4.08. It seems that lead acts as a central atom with two acetates as two-dentant ligands and two molecules of water acting as ligands too.
That means that lead acetate has its lattice ruined during dissolving, but it still forms a coordination compound.
 I am not really good in coordination chemistry of lead, but I just found out that lead wants to have octahedral geometry around it, so I would assume the structure of that would be

All bonds from lead ion are not really covalent rather than coordination bonds. Mind the charges!
P. S. aq=water
P. P. S. People in the comments are saying lead is too large for acetate to be bidentate,thus I would assume acetate just forms coordination compound with lead just like a simple monodentate ligand.


Why is the nitrile group not hydrolyzed as well during hydrolysis of imine in Thorpe reaction?

It has to do with hybridization. The nitrogen in a nitrile is sp hybridized; imine sp2 and ammonia sp3. The more s-character in hybrid orbitals, the less basic the electron pair (protonated nitrile pKa ~ -10; pKb ~24). Ammonia has the lowest percent s-character (25%, NH4+ pKa 9.25). Imines lie in between although most examples you will find are measured in DMSO and are loaded with biasing phenyl groups.
Compare the carbon series: acetylene (sp, pKa ~25), ethylene (sp2, ~44), methane (sp3, pKa ~50). The closer the electron density is to carbon, the more acidic the C-H bond. If electrons are held closer to carbon, then it is more difficult for homolytic cleavage of a C-H bond. That is, acetylene is willing to give up a proton but not a hydrogen atom. Thus, Bond Dissociation Energy runs in the opposite order as acidity.

The harder it is to protonate, the harder it is to hydrolyze.

There were no conditions provided by user 481311 for the hydrolysis. Imines, such as the acetaldehyde methylimine (CH3CH=NCH3), will hydrolyze in water. As the User illustrates, imines hydrolyze readily in acid medium. These hydrolyses may be conducted at ambient temperature or at mild temperatures. Nitriles, owing to their low basicity, require more vigorous conditions like refluxing aqueous acid. Its all about conditions. I hope this helps.


Simultaneous Equilibrium Reactions, One Goes to Completion

Initially

Suppose that initially, before any reaction, these are the species and amounts present:

$$\ce{CH4}: m \textrm{ moles} $$
$$\ce{H2O}: w \textrm{ moles} $$
$$\ce{C2H6}: e \textrm{ moles} $$
$$\ce{H2}: h \textrm{ moles} $$
$$\ce{CO}: c \textrm{ moles} $$

After the second reaction completes

Let's assume that the second reaction is limited by the amount of ethane present, not the amount of water.  (This must be true, otherwise the first reaction could not be at "equilibrium" because one of the equilibrating species would  head to a concentration of zero.)  If the reaction is limited by ethane, then when it completes, $2e$ moles of carbon monoxide and $5e$ moles of hydrogen are formed.

$$\ce{CH4}: m \textrm{ moles} $$
$$\ce{H2O}: w - 2 e \textrm{ moles} $$
$$\ce{C2H6}: 0 \textrm{ moles} $$
$$\ce{H2}: h + 5e\textrm{ moles} $$
$$\ce{CO}: c + 2e\textrm{ moles} $$

After the first reaction equilibrates

The first reaction doesn't go to completion.  Instead, let's assume the extent of the reaction from our current state to the equilibrated state is $\xi$ moles.  That means that the species inventory becomes:

$$\ce{CH4}: m - \xi \textrm{ moles} $$
$$\ce{H2O}: w - 2 e - \xi \textrm{ moles} $$
$$\ce{C2H6}: 0 \textrm{ moles} $$
$$\ce{H2}: h + 5e + 3 \xi \textrm{ moles} $$
$$\ce{CO}: c + 2e + \xi \textrm{ moles} $$

These values can now be plugged into an equilibrium equation:

$$ K = \frac{[\ce{H2}]^3 [\ce{CO}]}{[\ce{CH4}][\ce{H2O}]}$$

$$ K = \frac{(h + 5e + 3 \xi)^3(c + 2e + \xi)}{(m-\xi)(w - 2 e - \xi)} $$

Now "all" you have to do is solve that equation for $\xi$.


After boiling water in an electric kettle, should I empty it?

Your boss is somewhat right. Probably not wrong enough to argue with him.

There are several types of scaling that can be caused by mineral content in a water supply. One of the most common is called "temporary hardness" or "carbonate hardness." At high or neutral pH, calcium and magnesium carbonates have quite low solubility in water: 139 ppm for magnesium carbonate, and a mere 6.7 ppm for calcium carbonate, both at 25°C. (NOTE: Until starting this reply, I hadn't realised how little agreement there is on the solubility of calcium carbonate. Every source that I checked agrees that it is quite low; but no two sources agreed to even one sig. fig.!)

However ordinary, clean rain water often has a pH as low as 5.6, through absorption of atmospheric carbon dioxide. (Actual acid rain is even lower!) When such water runs through deposits of these very common minerals, some of them dissolve through the reaction:

$\ce{CaCO_3 + H_2O +CO_2 <-> Ca^{2+} + 2HCO_3^{-} }$

and the similar one for magnesium. Calcium bicarbonate has an aqueous solubility of 166 g/L at 20°C -- some 13,000 times more soluble than the carbonate. Of course, the dissolution process from hard rock is slow, and most runoff will not actually be saturated.

The problem is that our reaction, above, is an equilibrium; anything which tends to drive it back to the left will cause precipitation of the insoluble carbonate. One of those things is just atmospheric outgassing; this is how stalactites and stalagmites form. But a much faster method is boiling the water. That decomposes the bicarbonate ion, and rapidly drives off the freed carbon dioxide gas. The resulting precipitate of calcium carbonate is called "temporary hardness" because, as you have observed, it is easy to remove with any mild acid. Some other forms of scaling are much less forgiving.

Now, if your water hardness problem is severe, a lot of carbonate will start to precipitate as soon as the kettle is heating up. This is why another commentator suggested that it is too late to do anything after the kettle has boiled: that the scale deposits have already grown. However, there are two complications. Firstly, the supersaturated calcium carbonate solution doesn't necessarily all precipitate at once; the scale crystals may continue to grow for some time. I can't say just how long, but a process whereby large, hard crystals tend to cannibalise small, soft ones can continue for weeks.

Secondly, you will not see hardness that bad from mnay common water sources. Well water, maybe, but most municipal supplies will already be softened to the greatest degree that it is economical. If the water hardness problem is mild, some of the calcium carbonate will precipitate immediately, but some (more than 6.7 ppm!) will stay dissolved in the very hot water, until the solubility falls with falling temperature and more of it precipitates.

Long story short, your boss's concerns are not unreasonable. On the other hand, as you note, temporary hardness is easily cleaned off, and only temporary hardness is related to boiling. On the third hand, it's his kettle and he's the boss, so do as you're told!


The aluminum may not hydrolyze all the way to a neutral hydroxide.  It could form hydroxo complexes that remain in solution, especially if the solution is made acidic by the same hydrolysis.

Consider a half-reaction in which the Al is combined with only one or two hydroxide ions.

Some other tips:

1)  Since you have an acidic solution don't put hydroxide ions on the left.  Put water and balance with hydrogen ions on the right.

2)  Sulfate ion is a weak base.  It will combine with hydrogen ions to make bisulfate.  So add sulfate on the left and incorporate the hydrogen ions noted above into bisulfate ions.


Effect of pressure on solubility of salts

The answer is "(b) Decreases."  The reason is that the saturated solution occupies more volume than the starting reagents.  At higher pressures, creating this extra volume requires more $P dV$ work be done by the dissolving reagents.

  According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?

I find it easier to think in terms of volumes, not densities.  High pressure should favor low volumes.  Imagine a balloon filled with saturated solution.  If we squeeze the balloon by raising the pressure on the outside of it, the balloon will "want to" shrink.  In the example given, the balloon can shrink only by saturated solution converting back into solid salt and liquid water, so that is what would be thermodynamically favored.

Update: Based on discussion in the comments, I wanted to add a point about compressibility and thermodynamic stability.  The degree to which a system opposes a change in pressure is called the compressibility and is related to dV/dP for a given system, which you can find from the equation of state. In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value. That is, the volume must shrink as pressure rises for mechanically stable thermodynamic systems.


You are right.In case of benzene and PMB, the delocalization is not good enough. So a strong acid is required.

As for the case of OMe, a strong acid as well as strong nucleophile will be needed. A weaker acid can't do the job. Let me explain why.
Note that carbon oxygen bond is quite strong.
So what we need is a strong nucleophile. But in case of OMe as a protecting group, even a stronger nucleophile will not work if the oxygen is not heavily protonated due to two reasons:-
1. Strength of C-O bond.
2. SN1 can't occur as methyl carbocation not stable.

Here acid is a stoichiometric reagent, not simply a catalyst. If a weaker acid is used, it will just establish a protonation-deprotonation equilibrium. However, a stronger acid will shift the equilibrium towards protonation.
And haloacids are best suited.

Conclusion being that a weaker acid will render nucleophile incapable to cleave ether in case of OMe.


How can I carry out a titration to identify mercury in wastewater?

The potentiometric titration of mercury with dithiooxamide has been successfully applied after preliminary separation of mercury by reduction (Comprehensive Water Analysis, Volume 2, pp.186-187, Elsevier).

Selective identification of mercury in waste water can be accomplished by fluorescence enhancement chemosensors (SENSOR LETTERS 2010, 8, 1–6). In this recent paper, the ﬂuorescent emission intensity of the probe is enhanced upon binding to mercury ions. The ﬂuorescence enhancement is attributed to a 1:1 complex formation between the probe and Hg(II) ion. The chemosensor can be applied for the quantiﬁcation of mercury; it shows a high selectivity toward Hg(II) ions in comparison with common metal ions.


How does a reference electrode work in a three electrode system?

The reference electrode lets you control the potential difference so you don't apply too much driving force to the working electrode and end up reducing water etc. No current passes through it.

You have the basic idea but if absolutely no current passed through the reference electrode then you couldn't measure any voltage!

The gist is that you can't measure the "voltage" of a single electrode. You need a complete cell to get a voltage. The reference electrode is used with a very very low current to measure the voltage of the electrolytic cell formed with the working electrode.


Remember first that the integrated rate laws that we are familiar with for zeroth, first, and second order reactions are based on systems involving a single reactant in their rate determining step (or those that have been manipulated to give very large excesses of one reactant such that its concentration remains essentially fixed during the reaction process.)

With that in mind, there are only two variables involved, the concentration of our reactant of interest, $[A]$, and time, $t$.  We can't directly integrate the differential rate law; it's a differential equation and must be solved as such.

For an example, consider the first order differential rate law:
$rate = k[A]$

The meaning of  $rate$ here is the rate of change of the concentration of reactant A with respect to time.  Since we want the rate to be positive, we use the opposite of that quantity.

$-\frac{d[A]}{dt} = k[A]$

This is a differential equation, specifically one that is separable (and thus fairly easy to solve.)  Multiply both sides by $-dt$ and we get:

$\frac{d[A]}{[A]} = -k\ dt$

$\int{\frac{1}{[A]}d[A]} = \int{-k\ dt}$

Integrating both sides and applying initial condition to find the constant gives the integrated first order rate law:

$ln[A]=-kt+ln[A_0]$

The other integrated rate laws are found in similar ways.


Sulphuric Acid in Esterification Reaction

The esterification is not the only thing catalysed by sulphuric acid. Basically, it’s catalytic activity is protonation of whatever feels happy enough to be protonated. In your desired esterification mechanism, we want to protonate the acid:

$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=\overset{+}{O}H)-OH + HSO4-}\tag{1}$$

This will ease the nucleophilic attack of the alcohol onto the carboxy carbon. However, we also have other atoms that can be protonated.  For example, the hydrogen-carrying oxygen may also accept the proton:

$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=O)-\overset{+}{O}H2 + HSO4-}\tag{2.1}$$

This could then lead to the removal of water and the formation of a very reactive acyl cation:

$$\ce{R-C(=O)-\overset{+}{O}H2 -> R-C#\overset{+}{O} + H2O}\tag{2.2}$$

This in itself isn’t bad as it could also react with the alcohol in the desired fashion. However, the alcohol also can be protonated:

$$\ce{R-CH2-CH2-OH + H2SO4 <=> R-CH2-CH2-\overset{+}{O}H2 + HSO4-}\tag{3.1}$$

And depending on the type of alcohol you have, a number of side reactions may occur. For instance, you could simply eliminate $\ce{H3O+}$ from the protonated alcohol to give an alkene:

$$\ce{R-CH2-CH2-\overset{+}{O}H2 -> R-CH=CH2 + H2O + H+}\tag{3.2}$$

If you have a suitable neighbouring group that can stabilise positive charges, semi-pinacol-type rearrangements could occur. I won’t draw that with the simple primary alcohol I drew because it would be unlikely in e.g. propanol. It could be a problem in other cases, though.

Finally, the intermediates I drew up can react even further and unpredictably with sulphuric acid still present.

If you add sulphuric acid slowly, the local acid concentration will remain rather low and the reaction mixture is rapidly homogenised. This means that side reactions caused by sulphuric acid will be suppressed. If, however, you add it quickly, there will be areas of your solution with very high local sulphuric acid concentrations. In these, these side reactions are accelerated. Furthermore, concentrated sulphuric acid acts as a dehydrating agent which further speeds up the reactions we don’t want (they all include losing a water molecule).

These side reactions — especially the dehydrating ability of sulphuric acid — tend to end up in ‘black tar’ which would give a more or less uniform black solution. What exactly happened is hard to say except that some polymerisation probably took place.


Hardness and softness are terms used to describe which aspect of acidic behaviour?

The terms softness and hardness are used in HSAB theory to distinguish Lewis acids and bases by charge density, polarizability, electronegativity, and, in the case of molecular compounds, HOMO and LUMO energies.

Charge density is inversely proportional to atomic/ionic radius and proportional to oxidation state (in case of a neutral atom) or charge (in case of an ion). Hence charge density increases with decreasing radius and increasing oxidation state.

Polarizability is the tendency of the electron shell of an atom/molecule to become deformed in an electric field, which can originate e.g. from the charge of an adjacent ion or the partial charge of a dipole. An electron cloud of an atom or ion is more easily deformed with increasing volume, as it is more diffuse, and its outer electrons tend to be more loosely bound. Therefore, larger atoms and molecules tend to be more polarizable, and higher polarizability is often accompanied by lower charge density or oxidation state.

In the case of molecules, when a Lewis acid reacts with a Lewis base, electrons from the highest occupied molecular orbital (HOMO) of the Lewis base are transferred to the lowest unoccupied molecular orbital (LUMO) of the Lewis acid. How well this transfer of electron density works depends on the relative energies of the acid LUMO and the base HOMO. The stability of the reaction product will increase with the energy difference between its HOMO and LUMO (source).

The HSAB theory predicts that hard Lewis bases react preferentially with hard acids, and that likewise soft bases form stronger bonds with soft acids. For example, $\ce{Li+}$ and $\ce{BF3}$ are hard acids and $\ce{F-}$ and $\ce{NH3}$ are hard bases due to their small size and high charge density/low polarizability of the ions. $\ce{BF3}$ therefore readily forms Lewis acid-base adducts with fluoride and ammonia, and $\ce{LiF}$ is less soluble in water than $\ce{LiCl}$ and $\ce{LiBr}$. $\ce{Ag+}$ is an example for a soft acid (large ion with low charge), and the halide ions become increasingly softer bases with increasing ionic radius. Silver therefore forms stronger bonds with heavier halides like $\ce{I-}$ and $\ce{Br-}$, and this also explains the decreasing solubility of the silver halides in water, which is classified as a hard solvent. In the case of soft acids and bases, the difference of electronegativity tends to be smaller, and the bonds have more covalent character (e.g. silver iodide) than those between hard acids and bases (e.g. lithium fluoride) with larger difference in electronegativity (source).

The shift of reaction equilibria can also be predicted with this concept, like in the following example (reference):

$$\ce{HgO + H2S \rightleftharpoons HgS + H2O}$$

This equilibrium will be shifted to the right side, because $\ce{Hg^2+}$ as a soft acid prefers to bond to the soft base $\ce{S^2-}$. Likewise, oxygen as a hard base forms stronger bonds with hydrogen (hard acid).


Difference between S and I in the context of magnetic resonance experiments

The nomenclature you mean has nothing to do with the electron spin S, it is something completely different. In heteronuclear experiments with two spins it is common to call them $I$ and $S$, with $I$ being the more sensitive nucleus. They're still both spin 1/2 nuclei.

$I$ is one spin, and $S$ is a different spin in this nomenclature, with a scalar coupling between them.

This nomenclature is usually introduced with the INEPT transfer and experiments like HSQC or HMQC.


How come water would not affect the results of a titration?

The volume and concentration of base equals the number of moles of base. This is use to determine the number of moles of acid in the flask. Since you are interested in the number of moles of acid in the flask, why would adding water (which doesn't have any acid) change the number of moles of acid?


Why can't sulphate ion be discharged in the anode of electrolysis?

In aqueous solution the anode will oxidize the water instead, as this reaction can be done at much lower potential than oxidizing the sulfate ion.

Sulfates can be oxidized if you take the water away.  Potassium peroxydisulfate can be prepared by electrolyzing a solution of potassium bisulfate ($\ce{KHSO4}$) in sulfuric acid solvent.


Intrinsic Reaction Coordinate - What does 'Connectivity' really mean?

Various criteria may be applied in order to ascertain the connectivity of structures. A common solution is a path generated by requiring it to be a Newton path on the potential energy surface (E):
$$
(\nabla\nabla^\ast E(x))^{-1}\nabla E(x) = -\frac{dx}{dt}
$$
Two points $x_0$ and $x_1$ are connected only if $\exists x:\mathbb{R}\to\Omega,\tau_0, \tau_1\in \mathbb{R}$ such that $x(t)$ is a Newton path and $x(\tau_i)=x_i$, i.e., starting and ending structures, where $\Omega$ is your space of structures (usually centered 3D coordinates). $x$ is then a parameterization of that path on $[\tau_0,\tau_1]\subset\mathbb{R}$.

Sometimes it is additionally required that there is at most 1 maximum on $E(x)$ between $\tau_0$ and $\tau_1$.

You may be interested in reading this page. In a literature search, you may want to read up on the work by Vanden-Eijnden, Bofill, Quapp, and Henkelman. This topic is too rich to discuss in this forum.

To address the specific examples, assuming the starting point is in the convex attraction basin of the nearest optimum and the used optimizer converges in such instances, the optimum is connected in the above sense. Most optimizers used in geometry optimizations are some variant of (quasi-) Newton-Raphson optimizer, so this criterion is given. There are several reasons why the IRC may not converge to the same optimum as the subsequently used geometry optimizer, including changes of basis sets, convergence tolerances, or in case of quasi-Newton methods, the Hessian guess.

The fact that the optimizer brings the water closer to the other molecule is no indication of lack of connectedness. Instead the IRC run did not end at an optimum. It is actually expected that the energy of two (neutral) molecules decreases as they approach each other from infinity even in the complete basis set limit (van der Waals forces). You just found the vdW/BSSE-bound pre/post-complex.


Thermal conductivity and enthalphy data for nitrogen gas

By way of comparison, using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I have got a specific enthalpy of $h=306.26\ \mathrm{kJ/kg}$ and thermal conductivity of $\lambda=25.601\ \mathrm{mW/(K\ m)}$ for nitrogen at a temperature of $T=295.15\ \mathrm K$ and a pressure of $p=0.5\ \mathrm{bar}$.

Note that the absolute value of enthalpy at a single state point is meaningless; it is only the difference between two different state points that matters.


The way to get the information you seek is to obtain a safety data sheet (SDS) for either your product or each of its ingredients.
Safety data sheets have slightly different content by jurisdiction, but most follow a common format, especially those that follow the Globally Harmonized system (GHS).
In a SDS that follows the GHS, Section II identifies the hazards of the substance using tightly defined common phrases. If the product or one of its compounds is toxic in some way or corrosive, Section II will have that information.
Section II contains "Hazard Statements" and "Precautionary Statements". Here is a SDS for acetic acid. There are three hazard statements:

H226 - Flammable liquid and vapor
H314 - Causes severe skin burns and eye damage
H402 - Harmful to aquatic life

There are numerous precautionary statements. The Hazard statements provide the information you want. Here is a list of GHS hazard statements.


Why are there only less than 120 elements out there ?

They haven't stopped at 118. The researchers at CERN and Dubna are looking into heavier elements. Such as element 120 and 122 etc. It's all about the "island of stability". https://en.wikipedia.org/wiki/Island_of_stability
Where the right amount of protons and neutrons increase the stability of the nucleus. It's going to be very difficult to get there due to the increased mass of the target nuclei and increased mass of colliding nuclei. So really the only limit to the elements scientists can create, is the tools they use


Where to find compounds IC50 on an enzyme?

To my knowledge, there are no data bases accumulating  half maximal inhibitory concentration values on a enzyme ($\mathrm{IC_{50}}$). According to Wikipedia:

  Half maximal inhibitory concentration ($\mathrm{IC_{50}}$) is a measure of the potency of a substance in inhibiting a specific biological or biochemical function (i.e. an enzyme, cell, cell receptor, microorganism, etc.).

$\mathrm{IC_{50}}$ is commonly used as a measure of antagonist drug potency in pharmacological research. According to the FDA, $\mathrm{IC_{50}}$ represents the concentration of a drug that is required for 50% inhibition in vitro. It is comparable to an $\mathrm{EC_{50}}$ for agonist drugs.

The major reason for not to accumulate $\mathrm{IC_{50}}$ values is their abundance. For example, for given specific enzyme, there may be vast number of inhibitors present. All of these inhibitors have their specific $\mathrm{IC_{50}}$ value. Thus, if you need to find a $\mathrm{IC_{50}}$ value of specific inhibitor on a particular protein (or enzyme) or a receptor or other biological system, you should search using both names of the inhibitor and biological system as key words. Thus, you would find any literature, which may have $\mathrm{IC_{50}}$ values for sought combination.

For example, Galectin-3 is  a protein that is a member of the lectin family (carbohydrate-binding proteins). Galectin-3 is approximately $\pu{30 kDa}$, and contains a carbohydrate-recognition-binding domain with about 130 amino acids that enable the specific binding of $\beta$-galactosides, such as N-acetyllactosamine (14; the parent substrate; $\mathrm{IC_{50}} = \pu{158 \mu M}$ under the conditions; see insert red box). Several derivatives of N-acetyllactosamine (15 - 18) have displayed inhibitory activities with different $\mathrm{IC_{50}}$ values ranging from $13$ to $\pu{31 \mu M}$ (See the plot below):

On the other hand one inhibitor can be active against more than one biological systems with different $\mathrm{IC_{50}}$ values.

Relevant References:

P. Sörme, B. Kahl-Knutsson, U. Wellmar, B.-G. Magnusson, H. Leffler, U. J. Nilsson, “[12] Design and Synthesis of Galectin Inhibitors,” In Methods in Enzymology, Volume 363 – Recognition of Carbohydrates in Biological Systems, Part B: Special Applications;   Y. C. Lee and R. C. Lee, Eds.; Academic Press: Amsterdam, The Netherlands, 2003, pp. 157–169.
P. Sörme, Y. Qian, Per‐G. Nyholm,  H. Leffler, U. J. Nilsson, “Low micromolar inhibitors of galectin-3 based on 3'-derivatization of N-acetyllactosamine,” ChemBio Chem 2002, 3(2-3), 183–189 (https://doi.org/10.1002/1439-7633(20020301)3:2/3<183::AID-CBIC183>3.0.CO;2-%23).
J. Hynes, Jr., S.  K. Nair, “Chapter Nine - Advances in the Discovery of Small-Molecule IRAK4 Inhibitors,” Annual Reports in Medicinal Chemistry 2014, 49(1), 117–133 (https://doi.org/10.1016/B978-0-12-800167-7.00009-2).


Is there an open Chemical database that includes GHS/CLP classifications?

Since you're a Dutch neighbour, the European Chemicals Agency might be an option.


Using argon is, in effect, a costly physical path to remove oxygen, hence your interest in using nitrogen.
However, I suspect, a more efficient, cost-effective, and likely safe chemical path is also possible. For example, assuming a control structure (so as to mitigate potential deaths from O2 depletion, as has been reported, for example, among sewer workers working in a recently drained iron pipe), oxygen scavenging can result from a limited exposure to iron surface previously scrubbed with an acid. Here, lemon juice appears to be a friendly, pleasant, and effective choice.
Note, iron powder and ferrous salts have been used in the food industry to remove problematic oxygen (see, for example, THE USE OF OXYGEN SCAVENGERS AND ACTIVE PACKAGING TO REDUCE OXYGEN WITHIN INTERNAL PACKAGE ENVIRONMENTS).
One should be able to stoichiometrically compute, and experimentally validate, a safe amount of Fe (or its surface area) to employ. This is to limit a continuing unsafe scavenging action if, for example, subsequent maintenance work is required.
You can also employ commercially available products containing Calcium Chloride, for example, to remove water vapor.


Have you ever seen a picture of graphite, to begin with? It shares the same feature, i.e., that each carbon is bonded to exactly 3 other carbons. You may call it "alternating double bonds", but this is not quite correct either, because in fact all bonds are equal. Think of it as a resonance structure, much like benzene. This is the reason for graphite conductivity, too.

The idea about that lone pair is plain wrong. Carbon has but 4 valence electrons; once 3 of them are spent for sigma bonds, how could it have a lone pair?

Pure buckminsterfullerene does not conduct electricity (strictly speaking, it is a semiconductor, not a conductor like any metal). Some of its derivatives do, but that's another story.


What value should I use for P in the ideal gas equation for an unknown molar quantity?

Strictly speaking, the pressure inside must be certainly higher inside due to the restoring forces generated by the stretching of the plastic bag. It is this force that causes the gases to gush out when punctured or opened.

Bottom line: the assumption of the pressure inside being equal to atmospheric pressure can be considered correct with a considerably large error margin, provided we are talking about a ziploc bag and not the tyre of a tractor.


What is the thermal state of a reaction at equilibrium?

You're right, as far as I can see. If we look at a particular transformation $A \leftrightharpoons B$ there is a forward reaction and a backwards reaction. They are associated with enthalpy change of different signs and the same magnitude. At equilibrium the rates are the same, so the forwards and backwards reactions release and absorb the same amount of heat and it appears as if nothing is happening.

There is no separate "reaction at equilibrium" with its own enthalpy change. It's not really clear to me what "thermal state of a reaction" means, as systems have states and reactions are pathways between systems.


Acidity or acid strength can be expressed either in terms of the proton affinity (PA) of the conjugate base, e.g., PA($\ce{A-}$), or in terms of the acid dissociation constant, e.g., $\mathrm{p}K_{\mathrm{a}(\ce{HA})}$. The former expression is often favored among gas-phase chemists, whereas the latter expression is most familiar to solution chemists.

In aqueous solution, $\mathrm{p}K_{\mathrm{a}(\ce{CH3CO2H})} = 4.76$ while  $\mathrm{p}K_{\mathrm{a}(\ce{CH3CH2CO2H})} = 4.87$, thus acidity of acetic acid is greater than that of propanoic acid. Keep in mind that solvent plays a huge role in $\mathrm{p}K_{\mathrm{a}}$ measurements. For example, compare $\mathrm{p}K_{\mathrm{a}}$ measurements in $\ce{H2O}$ and in $\ce{DMSO}$:

$$
\begin{array}{c|cc}
\hline
\ce{HA} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{H2O} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{DMSO}  \\
\hline
\ce{CH3CO2H} & 4.76 & 12.3  \\
\ce{H2O} & 15.7 & 31.2 \\
\ce{CH3OH} & 15.5 & 27.9 \\
\ce{C6H5OH} & 9.95 & 18.0 \\
\hline
\end{array}
$$

However, the proton affinity (PA) measurements of the conjugate bases of these acids in gas phase tell a different story. Proton affinity is a thermodynamic measurement conducted in the gas phase. These measurements give a pure view of base/acid strength, without complications from solvent effects such as salvation, thus take the solvent out of the equation.

Cardwell, et al. (Ref.1) has measured the proton affinities of various $\ce{R-CO2H}$ acids, including ethanoic and propanoic acids ($\ce{R}=\ce{CH3}$ and $\ce{CH3CH2}$, respectively). They conclude that:

  The acidity increment (with the size of $\ce{R}$-group) is relatively small and can be attributed to the stabilizing effect of the polarizability of the alkyl group on the negative ion.

References:

Gary Caldwell, Richard Renneboog, Paul Kebarle, “Gas phase acidities of aliphatic carboxylic acids, based on measurements of proton transfer equilibria,” Canadian Journal of Chemistry 1989, 67(9), 611-618 (https://doi.org/10.1139/v89-092).
Also read: Diethard K. Bohme, Edward Lee-Ruff, L. Brewster Young, “Acidity order of selected Brønsted acids in the gas phase of $\pu{300  K}$,” Journal of American Chemical Society 1972, 94(15), 5153-5159 (https://doi.org/10.1021/ja00770a002) and references there in.


questions about the synthesis of iodobenzene from aniline

This procedure looks basically OK. The key thing here is keeping the reaction mixture below 5C. It is good practice to cool your KI soln to 5C, and add it slowly so that the temperature of the reaction mixture stays below 10C until addition is complete. Using very concentrated KI soln may be an issue as the reaction is exothermic, better to have it more dilute and cold.

I would use an organic solvent (EtOAc or Et2O) at stage 9 to extract the product simply because you are doing this on a relatively small scale. On this scale chromatography is a better option than distillation for final purification.


Retrosynthesis of a cyclopropane type molecule

How about this? Cyclopropane from carbene + maleic anhydride. Add 2 eq methyl Grignard and trap with methyl iodide.

Possibly, you could do the exhaustive addition of Grignard first, then cyclopropanate and methylate in one shot with diazomethane. Thanks to Beerhunter for this suggestion in comments.


The 'white colour' is due to the fact that the product does not absorb light in the visible part of the spectrum but forms as many fine crystals which in themselves scatter room lights. So the 'whiteness' is due to the scattering of light back towards your eyes after the light is repeatedly reflected of the many facets of the many crystals in the ppt. If you use red light to illuminate then the 'white colour' will become red. If the ppt is coloured then absorption of light occurs at some wavelengths as well as scattering at all wavelengths.

At the surface of a crystal some small fraction of light is reflected, but with millions of randomly oriented crystals numerous reflections occur. The amount of reflection at each surface depends on the difference in refractive index between the crystal and air for the wavelength of light used. If you could add a solvent say dichloromethane which has  a high refractive index and will probably not dissolve the ppt then the crystals can be made to almost disappear as scattering is now far less.  You can do the same thing with frosted glass beads which, when in air, scatter a lot but in water become virtually invisible.


What happens in an acid and metal reaction where the acid is twice as concentrated and the metal twice as much in quantity?

Without providing more detailed specifications of the metal such as: state of the metal - powder or solid, shape of the solid form etc., the rate of the reaction will be the same, but the second reaction will be a bit faster because of the bigger temperature change (as it affects reaction rate coefficient).

I assume that you meant 3 and 6 $cm^3$ and not $cm$.
The same metal reacts in both reactions. So we can assume that the density of both metals is also the same ($\rho_1=\rho_2=\rho$). We can further assume that metals were dissolved completely and that the temperature change is actually temperature increase.

$\rho=\frac{m_1}{V_1}$, $\rho=\frac{m_2}{V_2}$

$\frac{m_1}{V_1}=\frac{m_2}{V_2}$

$\frac{m_1}{3}=\frac{m_2}{6}$

$\frac{m_1}{1}=\frac{m_2}{2}$

The bigger piece of metal weights 2 times more than the smaller metal. We can convert weights to moles (molar mass of both metals is the same):

$n=\frac{m}{M}$

$\frac{n_1}{1}=\frac{n_2}{2}$
so we can say $2*n_1=n_2$

The 1st reaction:

$n_1$ reacts with 0,5M of acid.

The 2nd reaction:

2*$n_1$ react with 1M solution and it is the same as $n_1$ reacts with 0,5M

The volume of the solution is the same and the specific heat capacity will be nearly the same. The reaction enthalpy corresponds to the amount of reacted material. Because two times more material reacts in the second reaction,  the temperature change will be nearly two times bigger (nearly - because of the change in specific heat capacity).


Homemade "milk of magnesia" (suspension of magnesium hydroxide in water)

Milk of magnesia is typically just a suspension of about $\pu{8\%}$ wt/v (i.e. $\pu{8g / 100 mL}$) magnesium hydroxide in water. To get a good suspension you will want to have very fine magnesium hydroxide. Ideally, it should feel like flour when rubbed between you fingers. Regardless, there will be some settling and you will need to shake it thoroughly prior to each usage. No other preparation should be necessary.


Briefly, since ethyl alcohol expands more rapidly with increasing temperature than does water, then measuring the mix by volume at a warmer temperature would give you less alcohol and more water, because the alcohol would be less by weight for the same volume.

Thermal Coefficient of Expansion (approximate, measured at room temperature)

Alcohol, ethyl (ethanol):   0.00109 / K

Water:                            0.00021 / K

Using these constants, you can figure out the answer to your question. But note the wording: given a mixture. If the EtOH/H2O is already mixed, changing the temperature does not change the concentration or proof, and you would have to separate the ingredients to measure their respective volumes. BTW the volume of the mix is slightly less than the sum of the volumes of its ingredients.

Moral: if your measuring your drinks by shot-glass rather than with a scale, you'll get a weaker drink in a warm bar ;-)


Currently, the definition of Avogadro constant depends on the definition of the kilogram, and thus has an uncertainty. The recommended value[1][2] is

$$N_\pu{A} = \pu{6.022140857(74) \cdot 10^{23} mol-1}.$$

It will probably be fixed, however, coming May 20th, 2019 on World Metrology Day to exactly[3][4]

$$N_\pu{A}^{2019} = \pu{6.02214076 \cdot 10^{23} mol-1}.$$

See the references for a thorough review of the history, reasons for redefinition, how some relations become exact and others inexact, and to which degree, etc.

References

[1] Marquardt, R.; Meija, J.; Mester, Z.; Towns, M.; Weir, R.; Davis, R.; Stohner, J. 'A Critical Review of the Proposed Definitions of Fundamental Chemical Quantities and Their Impact on Chemical Communities (IUPAC Technical Report)'. Pure and Applied Chemistry, 2017, 89 (7), 951–981. DOI: 10.1515/pac-2016-0808.

[2] Mohr, P. J.; Newell, D. B.; Taylor, B. N. 'CODATA Recommended Values of the Fundamental Physical Constants: 2014'. Rev. Mod. Phys., 2016, 88 (3), 035009. DOI: 10.1103/RevModPhys.88.035009.

[3] Marquardt, R.; Meija, J.; Mester, Z.; Towns, M.; Weir, R.; Davis, R.; Stohner, J. 'Definišion of the Mole (IUPAC Recommendation 2017)'. Pure and Applied Chemistry, 2018, 90 (1), 175–180. DOI: 10.1515/pac-2017-0106.

[4] 'A new definition of the mole has arrived'. IUPAC press release. archived link: goo.gl/5haPwJ


Why must the movement of electrons through the wire be balanced by a movement of ions in the solution?

Karl Ratzch is basically correct, but I will try to provide a little more detail.

Electric charges, when separated, exhibit a very large tendency to try to get back together again.  This is represented in basic Physics by Coulombs Law and the large value of the constant in that relationship.

I presume we are talking about a battery here. If the flow of electrons out of the wire is different from the rate of ions arriving at the plate, the net charge on the plate will change.  Depending on which flow rate is greater, the voltage of the plate will instantly go up or down.  This will then change the rate of both charge flows such that the two almost instantly get back to the same value.

More detail: Assume that the flow of electrons out of the plate through the wire is larger than the flow of anions into the plate from solution.  This reduces the net negative charge on the plate.  This reduces the "force" that is pushing electrons down the wire.  The actual name of that "force" is voltage.  The voltage will change such that it doesn't push electrons out the wire as fast and will repel anions in solution less aggressively allowing more of them to reach the plate (per second).  The opposite is also true.  As a result, the flow in and out of the plate will always remain almost exactly the same, with small differences existing for very brief periods of time when the voltage changes.


Why does zinc sulfide glow when hit by electrons?

This is an example of phosphorescence. The Wikipedia page of zinc sulfide and phosphorescence can explain it better than I do, but in short, when zinc sulfide get hit by electrons, electrons transfer some of its energy to zinc sulfide and excites its electron. Zinc sulfide stores the energy for a while, then emit light when the electron goes back to its ground level.


Molecule with a quadrupole moment in an electric field

To determine the force on an arbitrary multipole moment, we first expand the E-field in a Taylor-series around the point $\vec{r}=0$:

$$\vec{E}(\vec{r})=\vec{E}_0+\vec{r}\cdot(\nabla\vec{E})_0+\frac{1}{2}\vec{r}\vec{r}:(\nabla\nabla\vec{E})_0+...$$

where the number of dots in the product denotes how many indices to contract over. We can then determine the force using this expansion, since it is just the charge multiplied by the E-field:

$$\vec{F}=q\vec{E}_0+\vec{\mu}\cdot(\nabla\vec{E})_0+\frac{1}{2}\mathbf{\Theta}:(\nabla\nabla\vec{E})_0+...$$
Typically, people use the traceless quadrupole, which would add an additional factor of $\frac{1}{3}$ in the third term, but I'm going to work with the basic definition throughout. So this shows that a quadrupole subject to a large electric field double gradient (not sure of the correct terminology for this) will experience a force even if it is charge neutral and has no net dipole.

For completeness, we can also write the energy since the force is just the gradient of the energy:

$$W=q\phi-\mu\cdot\vec{E}-\frac{1}{2}\mathbf{\Theta}:(\nabla\vec{E})+...$$
where we see, as TryHard said, that the E-field gradient interacts with the quadrupole.

As an aside, one can also obtain the torque (cross product of force and r) on an arbitrary multipole:
$$\vec{T}=\vec{\mu}\times\vec{E}_0+\frac{1}{2}\mathbf{\Theta}\dot{\times}(\nabla\vec{E})_0+...$$

(I'm using the nonstandard notation $A\dot{\times}B$ to mean $\sum_{ijkl}\epsilon_{jkl}e_iA_{jl}B_{lk}$, where $\epsilon_{jkl}$ is the Levi-Civita operator and the $e_i$ are a set of orthonormal unit vectors). This suggests that a quadrupole will experience a torque when interacting with a field gradient.

For more information on this, I would read chapter 3 of Jeanne McHale's Molecular Spectroscopy. There are some typos and notational quirks(e.g. in the equation for torque, McHale just writes a cross product for the 2nd term, which isn't defined between matrices), but its overall a good book to give background in classical electrostatics and a quantum mechanical description of spectroscopy.


An explosion is a type of spontaneous chemical reaction that, once initiated, is driven by both a large exothermic change (great release of heat) and a large positive entropy change (great quantities of gases are released)   $\ce{^1}$ from  reactants to products (a thermodynamically favorable process).

The energetic stability of the gaseous products and hence their generation comes from the formation of strongly bonded species like carbon monoxide, carbon dioxide, and (di)nitrogen, which contain strong double and triple bonds having bond strengths of nearly 1 MJ/mole.

Consequently, most commercial explosives are organic compounds containing $\ce{-NO2, -ONO2}$ and $\ce{-NHNO2}$ groups that, when detonated, release gases like the aforementioned (e.g., nitroglycerin, TNT, HMX, PETN, nitrocellulose).

In general, Detonation is used to describe an explosive phenomenon whereby the decomposition is propagated by an explosive shock wave traversing the explosive material at speeds greater than the speed of sound within the substance.

The following factors affect the stability of an explosive: (1) Chemical constitution.
(2) Temperature of storage. (3) Exposure to sunlight (4) Electrical discharge.

Certain groups like nitro  (–NO2), nitrate (–ONO2), and azide (–N3), are intrinsically labile or Kinetically there exists a low activation barrier to the decomposition reaction which makes compounds exhibit high sensitivity to flame or mechanical shock.

Nitroglycerin $\ce{^2}$ is a primary explosive  that is extremely sensitive to stimuli such as impact, friction, heat, static electricity, or electromagnetic radiation ultimately leading to detonation.

You can see that 4 moles of nitroglycerin produces 35 moles of hot gases.The concentrated gases are under very high pressure, so they expand rapidly. The heat speeds up the individual gas particles, boosting the pressure even higher. The gas expands faster than the speed of sound, it generates a powerful shock wave.

As a rule of thumb, explosives with an activation energy of decomposition
higher than 170 kJ/mol are stable for thousands of years at room temperature, whereas for values below 155 kJ/mol, chemical stability is limited.$\ce{^3}$

As can be seen from Table below $\ce{^4}$, the commonly used explosives from the classes of aromatic and aliphatic nitro compounds, secondary nitramines, and organic azides are very stable, whereas aliphatic nitrate esters suffer from much lower chemical stability.

TNT is a secondary explosive less sensitive than a primary explosive and requires substantially more energy to be initiated.

2,4,6-trinitrotoluene (TNT) is a molecular explosive that exhibits chemical stability due to its high performance, comparatively low shock sensitivity, low melting temperature of TM = 353–354 K (80–81 °C)$\ce{^5}$ $\ce{^6}$

References

https://en.wikipedia.org/wiki/Explosive#Stability
http://www.ch.ic.ac.uk/rzepa/mim/environmental/html/nitroglyc_text.htm
Chimia 58 (2004) 401 -408
https://www.ingentaconnect.com/content/scs/chimia/2004/00000058/00000006/art00010?crawler=true
M. L. Hobbs and M. J. Kaneshige, in Proceedings of the Society for Experimental Mechanics Annual Conference, Albuquerque, NM, 2009
APPLIED PHYSICS LETTERS 104, 021911 (2014) ,
" Chemical stability of molten 2,4,6-trinitrotoluene at high pressure "
Dana M. Dattelbaum, Raja S. Chellappa, Patrick R. Bowden,Joshua D. Coe,
and Madeline A. Margevicius


Quick-and-Dirty Molecular Dynamics by Mass-Weighted Atom Translations?

The random displacements in your suggested method are uncorrelated from step to step.  What you are suggesting is essentially heavily damped molecular dynamics at the Brownian limit.  The limitations are obvious, you won't sample correlated, large-scale motions efficiently.  (That might be a student project right there: turn up the Langevin parameter, see when you lose rotations of phenyl rings, and ask yourself what this means.)


Will the dextrorotatory and laevorotatory enantiomers always have a set of stereocenters with opposite absolute configuration?

Precisely, these are based on the configurations proposed by German chemist Emil Fischer, who established the structures of many sugars.

  Emil Fischer elucidated these configurations for the aldohexoses in
  1896. According to the Fischer convention, D sugars have the same absolute configuration at the asymmetric center farthest removed from
  their carbonyl group as does D-glyceraldehyde. The L sugars, in
  accordance with this convention, are mirror images of their D
  counterparts,

Similary,the  Fischer projection for glucose  is shown below.

It is very important not to confuse these enantiomers with epimers   (two sugars that differ only in the configuration around one carbon atom e.g D-glucose and D-mannose, which differ only in
the stereochemistry at C-2)

References

Voet and Voet Biochemistry
Lehninnger Principles of Biochemistry
Biochemistry (Campbell and Farrel)


Which of the compounds give the same SN1 and SN2 products?

For starters, none of your products show chirality in any way, so it does not matter whether stereoinformation is lost or whether you convert a racemic mixture into a racemic mixture where every single molecule if you had labelled it pre-reaction now had the opposite stereochemistry — in both cases, both reactants and products are racemic mixtures. So you cannot take the presence or absence of chiraliy to distinguish here.

What we can say, is that every compound can take part in an $\mathrm{S_N2}$ reaction (some faster, some slower) but not every one will easily take part in an $\mathrm{S_N1}$ reaction. Specificly, (a) and (d) would give a secondary carbocation intermediate while (b) would even give a primary carbocation. These cations are very unstable and would immediately undergo Wagner-Meerwein type rearrangements towards more stable — read tertiary or resonance stabilised — carbocations. Or, to quote one of my professors from uni:

  Exactly one single unstabilised secondary carbocation exists, and that is the prop-2-ylium cation. (Because it cannot rearrange to a more stable one.)

Thus, only (c) will form a stable carbocation without any rearrangement happening: It would be an allyl cation which is well resonance stabilised. But now we have to dig a step deeper, because while all other reactants will certainly not give the same product in both cases, it could be that the correct answer is ‘none of these’, i.e. that somehow we can distinguish between products — because once we have the allyl cation, we could let the nucleophile attack on carbon 1 or carbon 3. Thankfully, though, both attacks give the same product here, because there are no other asymmetries present in the molecule, so we can transform one into the other by rotation.

And, never forget, the $\mathrm{S_N2}$ reaction could also follow the $\mathrm{S_N2'}$ mechanism — this, too, would switch the identities of the two carbons, so we can truly say that both mechanisms will lead to the same product.


How do you draw (E)-1-phenyl-1,3-butadiene?

ManishEarth put together the pieces to understand the connection of the atoms to form the molecule.  Yet since the molecular structure requested in the original post is about (E)-1-phenyl-1,3-butadiene, the answer actually is opposite configuration:

By convention, alkenes posses a reference plane orthogonal to the plane of drawing.  In case the alkene is substituted twice by groups attached at the termini of the alkene -- in the upper example a phenyl and an alkyl chain -- these may be either at the same or the opposite side of this reference plane.

For the ease of convenience, two drawings simplifying the substitutents by "R" demonstrate this. In the first example, they are opposing each other, so their relationship is described either by (E) or (trans). The second example shows them at the same side of the reference plane, therefore, this alkene has a (Z)- or (cis)-configuration.

You might ask about the origin of (E) and (Z).  They are derived from German entgegen and zusammen; similar like the Latin (trans) for across, and (cis) for nearby.  Both ways to describe these configurational isomers are found in the literature for one and the same molecule, yet in an article or report, you should stick to either the (E)/(Z) or (cis)/(trans). (And yes, in italics.)


How can you qualitatively tell when the hydrolysis of aspirin is over?

First I was thinking that maybe a chemical indicator could be used, however the pKa values of all three components are relatively close to each other (especially aspyrin and salicylic acid). It might still be a bit of a help to stick a pH electrode in the system and observe the change - however, this is not exactly something you can see easily.
Another thing that came to mind is that salicylic acid (as all phenols) will form a colorful complex with Fe(III). However, the appearance of the red-ish color will only tell you when the reaction has started, since it's borderline impossbile to observe that the solution is not getting any darker anymore. Again, if you have some machinery, you could stick your solution into a cuvette and put it in a photometer.
Apart from the two ideas above, I don't really see an easy way of observing anything.


Lewis base strength and the hydrogen proton

One of the keys in the definition of Lewis acids and Lewis bases is that they react to form a Lewis adduct.  To compare the relative strengths of different Lewis bases, you would allow them to react with the same Lewis acid, form the required adduct and measure the enthalpy of formation of the adduct. The adduct with the largest enthalpy of formation would contain the strongest Lewis base.

1) So Lewis acid and base strength is about thermodynamics.

2) You could run the reaction in any solvent (or gas phase) as long as all of the Lewis acids or bases you want to compare are able to form an adduct in that solvent.  There is no reason why the order of a series of enthalpies of bond formation has to remain constant when you move from solvent to solvent, so your relative strengths could well change from one solvent to another.  The same applies to oxide \ sulphide relative Lewis base strengths if you substitute some other Lewis acid in place of the proton.  Again the relative ordering might well change.


Can I solve the partial derivative (dG/dH)s ? Does it make sense?

$$dH=TdS+VdP$$
At constant S, $$dH=VdP$$
$$dG=-SdT+VdP$$
If S is constant, then $$\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP=0$$But,$$\left(\frac{\partial S}{\partial T}\right)_P=\frac{C_p}{T}$$and$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$So, $$dT=\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_PdP$$and$$dG=\left[V-S\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$Dividing by dH then gives:
$$\left(\frac{\partial G}{\partial H}\right)_S=1-\frac{\alpha T}{C_p}S=1+\frac{(G-H)\alpha}{C_p}=1+\frac{\alpha T}{C_p}\left(\frac{\partial G}{\partial T}\right)_P$$


Potassium permanganate (VII) + glucose + sodium hydroxide =?

This is a redox reaction in which the permanganate ion is reduced and the glucose is oxidised. Potassium permanganate is usually used in acid solution and under these conditions is a very powerful oxidising agent in which the manganese is reduced from the oxidation state of +7 to +2 with a colour change from purple to colourless (actually extremely pale pink). In alkaline solution manganese is only reduced from +7 to +6 changing colour from purple to green. (The blue colour you mention is due to a mixture of purple and green). The green manganate(VI) ion is unstable and slowly disproportionates to manganate(VII) (purple) and manganese(IV) oxide which is brown. The manganate(VII) goes on to react as before until the brown manganese(IV) oxide is all that remains. In alkaline solution this tends to form a colloidal suspension which (if fairly dilute) can appear orange. The oxidation of the glucose forms no coloured productss.


Effect of noble gas configuration on polarising power

Recall the fact that the basic difference between an ionic bond and a covalent bond, is basically just charge separation. In ionic bonds, charges are well separated, and the bond arises due to opposite charge attractions. Covalent bonds, on the other hand, have the electrons shared in between them, and it is these electrons that hold the nuclei of the two bonded atoms together.

But sometimes, a covalent bond isn't completely covalent, nor is an ionic bond completely ionic. Even an ionic bond may have some degree of "covalentness" in them. You can compare the covalentness of two ionic bonds with the help of a few rules formulated by Fajans. A few of these rules would be:

Rule 1: Smaller the cation and greater the charge, more the covalent character.

Why is this true? If you consider a cation to be similar to a uniformly charged sphere, the electric field generated at it's surface would be:

$$E=\frac{1}{4\pi \epsilon}\frac{Q}{R^2}$$

Where $Q$ is the charge on the cation, and $R$ is it's radius. It's quite clear now that highly positive cations and smaller cations can generate higher electric fields, and can strongly pull the electrons of the anion into the space between the ions, which gives rise to a partial covalent character.

In your example, it was given that "cation with pseudo-inert gas configuration can polarize better than a one with inert gas configuration". This is very much true, and can be explained using screening effects of orbitals. In the case of $\ce{Na+}$, it's valence electrons are in the $p$ orbitals, while in the case of $\ce{Cu+}$, they are in the $d$ orbitals. It's a known fact that $d$ orbitals shield the electrons from nuclear attractions poorly compared to the $p$ orbitals (see Why do screening effect decreases due to d-orbital? for more details). As a consequence, the nucleus attracts the valence electrons much more strongly in case of $\ce{Cu+}$, leading it to have a smaller size, and more polarizing power.

Rule 2: Greater the size of the anion, more is it's polarizability and greater is the covalent character.

Now the first question that would arise is, what is polarizability and why does it affect covalent character? Polarizability refers to the ability of the anion to get polarized, which is charge separation within the ion. Larger anions allow more room for the electron clouds within them to move around, which causes the cation to polarize it better. See the picture below for better clarity:


Limits on contaminant size in filtration for extraction, and standard alternatives?

There are two approaches regarding the size selectivity of solid to be filtered. Most common is some sort of "tortuous path", that is a fine weaving of some type of fiber having some average size cutoff regarding what gets filtered out and what passes through.  Secondly there are filters that are actually grids of a specific size of hole or square to restrict the solids from passing through. These look like a screen sieve through a microscope and I've seen them as small as 1 micron, though this is extremely small and usually used for sterilization.

In reality, you are more likely to have problems with your filter becoming clogged than with too much solids getting through. Your idea of centrifugation is good, but as a pre-filter step to help prevent clogging the filter.  In this case you could either decant or suction off the bulk of the liquid before the finer filter finish. The requirement for final filtration size is application dependent.


One definition of entropy is that the number of microstates that makes up a macrostate influences the probability of observing the macrostate.  So your wavefunction needs to be multi-configurational and include both $\mathrm{H}_2^+$ and $\mathrm{H}^+ + \mathrm{H}$. The usual approach is to invoke the Born-Oppenheimer approximation, which makes the wavefunction depend on a single static set of nuclear coordinates, i.e. single-configurational. This approach will just give you the energy and you must compute reaction probability using statistical mechanics.

Multi-configurational SCF does include entropy in the following sense.  Let's say you have a molecule with a HOMO-LUMO gap that is sufficiently small to allow for significant mixing. You run a CASSCF(2,2) to determine the LUMO contribution and is 5%.  Now suppose you have the same system, same HOMO-LUMO gap, but the LUMO is doubly degenerate.  Your CASSCF(2,3) would now have a larger LUMO contribution, though I am not sure it would be 10%.


How can the vapor pressure increase until it reaches a value close enough to the atmospheric pressure to sublimate without reaching its melting point?

Compounds can sublimate even if the vapor pressure is much lower than the atmospheric pressure. This is how we can smell solids. If you put the solid substance in a closed container, the partial pressure of the substance in the gas phase will increase until it reaches the vapor pressure, and you will be at equilibrium. In an open system, it will continue to sublimate, with a concentration gradient in the gas phase (that is how the sense of smell is used to locate smelly objects).

  they asked us to heat a beaker that contained camphor and to cover it with a watch glass with ice on it.

In your experimental setup, you never reach an equilibrium. The vapor pressure of the camphor in the beaker is higher than the vapor pressure of the camphor deposited on the cold watch glass. Accordingly, camphor will continue to sublimate from the solid on the bottom and continue to deposit on the camphor on the watch glass.

  This seems weird. Does this happen to other substances?

Yes, you can see the snow sublimate in the winter when temperatures are below the freezing point of water. A dusting of snow on a drive way (or ice on a windshield) will "disappear" (sublimate) without liquid water present at any time. When food is freeze-dried, the same process is used. The key for this to work is that the air has to be dry (i.e. the partial pressure of water in the air has to be lower than the vapor pressure of ice).


What does the value of the pH meter mean when there aren't any hydronium ions in the substance being tested?

Nowadays you have to differentiate between pH measurements with non-glass sensors and with glass sensor. In the typical pH meter with a glass sensor, there could be many things that could go wrong resulting in biased results. In general, some electrodes may be sensitive to interfering ions such as $\ce{Ag+}$. So some potential could be generated also by other ions not only hydronium ions. However, the most common thing is that you are actually reading biased measurements because you are perturbing the status of your instrument.

Interaction of the solvent with the materials of your pHmeter

First thing you have to consider is the interaction of the solvent with the materials of your pH meter. You can easily damage some part of it or dehydrate the glass membrane of your electrode.

Solubility of the electrolyte in the junction

Then you have to bear in mind that in your glass pHmeter there is a junction that leaks an electrolyte for forming the saline bridge if this electrolyte is not soluble in your solvent the "circuit" will not close and you will have a wrong measurement.


From Greenwood and Earnshaw's excellent book Chemistry of the Elements:

  When the metals are burned in a free supply of air the predominant product depends on the mtal: $\ce{Li}$ forms the oxide $\ce{Li2O}$ (plus some $\ce{Li2O2}$), $\ce{Na}$ forms the peroxide $\ce{Na2O2}$ (plus some $\ce{Na2O}$) [...]

In general, formation of inorganic peroxide salts in this fashion doesn't appear to be especially favorable for other metals, at least by comparison to the variety of other oxides possible. Sodium's peculiar preference seems anomalous to me, especially given that the lattice energy for the oxide is higher (albeit slightly). This may be a function of the availability of oxygen, which is roughly $21\%$ in air, and is certainly temperature-dependent as well.

As for organic substances, formation of peroxide linkages is possible from a variety of functional groups and using various reagents. Ordinary $\ce{O2}$ will form peroxide bonds with various substrates via radical mechanisms, especially under photochemical conditions. Carboxylic acids react with $\ce{H2O2}$ to form peroxyacids. These are just two examples that come to mind, and numerous others exist.


Why Does 0.10 M Dihydrogen Phosphate Dissociate Rather than Hydrolyse in a Neutral Solution of Water?

The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species $\ce{H2PO4-, HPO4^{-2} , PO4^{-3} ,H3PO4}$  in solution, that ($\ce{H2PO4-}$)   is amphoteric that behaves both:

as an acidreleasing $\ce{H+}$ in aqueous solution and( forming $\ce{HPO4^{-2}}$ ions which can produce very few of $\ce{PO4^{-3}}$ ions)

$$\text{
Ionization:}\ce{H2PO4- <=> HPO4^{-2} +\color{red}{H^+} \quad\left(K_\mathrm{a}=\color{red}{6.2\times10^{-8}}\right)}$$
OR

as a base(can undergo hydrolysis forming ($\ce{H3PO4}$) acid
$$\text{Hydrolysis:}\ce{H2PO4-  +  H2O <=> H3PO4 + \color{blue}{OH-} \quad\left(K_\mathrm{b}=\color{blue}{1.3\times10^{-12}}\right)}$$

So ,$\color{red}{K\mathrm{a}}>\color{blue}{K\mathrm{b}}$ ,thus the first reaction is the dominant,so ,$\pu{pH}<7 $ and the solution is$~\color{red}{\text{acidic}}$.


Comparing acid strength of oxoacids derived from oxides

The question quoted in the OP is designed to test a periodic table trend concept.

Particularly, within a given row of the periodic table, excluding noble gases, the element oxides form increasingly acidic aqueous solutions from left to right across the row.

See the lecture by Dr. Pinkington The Acid-Base Character of Oxides and
Hydroxides in Aqueous Solution for additional information.

Only answer (A) is increasingly acidic left to right across the periodic table.


How can water exist in three states in freezing point?

If you only have a single component, water, present, then the liquid, vapor, and solid can only exist at the triple point.  If there is a second component present, say air, the phase rule tells us that there is an additional degree of freedom, pressure.  So, at a total pressure of 1 atm, the freezing point is 0 C (slightly lower than the triple point temperature), and you have liquid water, solid water, and, in the gas phase, water vapor mixed with air.


Why is quinone mono oxime more stable than 4-nitrosophenol?

Two factors come into play: (1) the inherent difference in resonance energy is not as great as you might think, and (2) the greater tendency for a dissociable proton to bind more tightly to an oxime function rather than a phenol function appears to counterbalance the reduced $\pi$-electron bonding difference.
Less than meets the eye
Suppose you were to use the Hückel model to estimate the pi-electron binding energy for several eight-carbon conjugated systems.  By that method I find the following results:

For the conjugated straight chain $\ce{CH2=(CH=)_6CH2}$ the total $\pi$-electron binding energy works out to $9.52\beta$, where in this theory $2\beta$ represents one isolated $\pi$ bond.

For styrene, with a $\pi$-electron pattern matching the nitroso-phenol, the result is $10.53\beta$.  A difference of about one $\beta$ unit is typical when an aromatic ring is compared with the opened ring.

For para-xylylene whose quinonoid $\pi$-electron structure matches the oxime tautomer described here, we get $9.92\beta$.

Of course the aromatic structure is most stable, but not all non-aromatic structures are created equal.  The Hückel-model calculations indicate that even though the quinonoid structure is typically called non-aromatic, in fact it captures a significant part (about $40$% based on the numbers) of the extra $\pi$-electron stabilization we usually associate with an aromatic ring.  The energetic difference between the two competing $\pi$-electron configurations is thereby less than the "aromatic" and "non-aromatic" labels would seem to imply.
Swimming upstream
Now let's see where the dissociable proton is more stable.  To do that we look up the $pK_a$ values typically associated with each prospective acidic functiin:

Phenol gives $pK_a=9.95$.

Acetoxime, $\ce{(CH3)_2C=NOH}$, gives $pK_a=12.2$.

The oxime function is clearly holding the proton more tightly; placing the acidic proton there instead of on the phenol function saves $2.2$ powers of ten in the equilibrium constant.
We cannot directly compare this with the Hückel-model calculations from the previous section, because that model is only semi quantitative and a more complex calculation would be needed to capture the effects of the heteroatoms in the conjugated structure.  But the results do suggest that the combination of the quinonoid structure capturing part of what is usually aromatic $\pi$-electron stabilization with the preference of a dissociable proton to bind to the oxime function leads the oxime tautomer to win out over the nitroso-phenol.


How to calculate the percentage of calcium carbonate in sea shells in an experiment

$\ce{CaCO3}$ is basic so it will neutralize the acid.

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O +CO2}$$

In your experiment you need to devise a way of finding out what the end point of the reaction is (using a $\mathrm{pH}$ indicator to find when the mixture is neutral). You need to be able to measure the amount of hydrochloric acid used in moles  (you should measure the volume and you know the concentration. The number of moles of water is twice the amount of moles of the calcium carbonate (from the chemical equation above).

You should measure the mass of the amount of shell you use at the start. You will be able to figure out the mass of the calcium carbonate in that sample of shells that you use (since you now know the number of moles of calcium carbonate). Then:

$$(\frac{\mathrm{Mass_{\ce{CaCO3}}}}{\mathrm{Mass_{shells}}}) \times 100\%=\mathrm{Percent}~\ce{CaCO3}$$

You should be able to do it with just one concentration but your teacher probably wants to talk to you about reliability and experimental errors.


Do carboxylic acids form hydrates like aldehydes and ketones?

Let me go against the previous answers, and say that they "do" form hydrates.
What you should consider is that the equilibrium towards such hydrates is extremely unfavorable, making them virtually non-existent.

There have been, anyway, clever ways of "blocking" those forms as orthoesters. If I recall correctly, Corey used a protecting group of the OBO family in the synthesis of a natural product.

A more recent paper, which describes a protecting group of that nature, is this:
https://pubs.acs.org/doi/abs/10.1021/ol047493n?journalCode=orlef7

Anyway, in normal mechanism-writing, that hydrate is not considered, being highly unfavored and unlikely to form.


Why doesn't self condensation occur in acetoacetic ester synthesis

Self-condensation is typically not an issue with 1,3-dicarbonyl compounds, such as the one in your question. Why? Due to their acidity ($\mathrm{p}K_\mathrm{a}$ around 9 to 11), most 1,3-dicarbonyls exist in their enolate form in solution. Therefore, there is very little of the original electrophilic 1,3-dicarbonyl which the enolate can attack. Hence, the amount of self-condensation taking place is relatively little and usually not a concern.


Understanding lattice matching of graphene on silver (111) crystal surface and meaning of the √3 x √3 lattice?

"Root 3" or "$\sqrt{3}$" or "$\sqrt{3} \times \sqrt{3}$" are slang for the Wood's notation (1, 2) $ \left( \sqrt{3} \times \sqrt{3}\right)R30$, a commonly occurring configuration of one hexagonal 2D lattice with respect another.

In this example the substrate is the (111) face of silver. Since it is face-centered cubic with lattice constant of about 4.08 Angstroms the (111) surface is hexagonal with a lattice constant 4.08 / $\sqrt{2}$ = 2.88 Angstroms.

The figure caption says "2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice."

The annotation in the OP's drawing copied below shows solid arrows for the Ag(111) lattice vectors the dashed arrows are rotated by 30 degrees and $\sqrt{3}$ longer, and they match the periodicity of a $2 \times 2$ graphene cell.

The lattice constant of free graphene is about 2.46 Angstroms and double that is 4.92. Compared to $\sqrt{3}$ times 2.88 = 4.99 Angstroms it's a fairly good lattice match.


Why are the ethyl ester methylene protons not diastereotopic in ethyl (+-)-2-methylbutyrate?

You are correct to conclude that they are diastereotopic. There appears to be a bit of splitting evident if you look closely at the NMR, but it might be expected to be very small given the distance between the ethyl group methylene protons and the chiral center. But even if the distance were greater and the resulting splitting too small to be observed, the protons would still be technically diastereotopic.


Can someone induce cancer in another person

[…] So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?

Yes, in principle.

Information on known cancerogenic compounds can be found in CMR (Canceogenic, Mutagenic and toxic to Reproduction) lists published by national organizations responsible for workplace safety.

In Germany, this would be the IFA (Institut für Arbeitsschutz der Deutschen Gesetzlichen Unfallversicherung).

List with old nomenclature

List with new nomenclature

I have added these lists just for the sake of completeness!

If you suspect any of the activities that you described, contact the local authorities.


To begin with, I want to present you approximate representations of the four cycloalkenes in question. Unfortunately though, these representations cannot, in this case, be used to fully replace building the molecules; I suggest you do so with the molecular modelling kit of your choice.

I’ll start off with the six-membered ring. This — as is often correctly pointed out — is essentially strainless and the chair conformation displayed is the most stable one. While frequent interconversion between different chair forms can occur, switching axial and equatorial substituents. We can thus expect that a reaction will, in principle, always take place in this chair form.

If we examine the dihedral angles, we notice that an axial substituent is anti-periplanar to the adjacent two carbons’ hydrogens. This means, it can easily anti-eliminate in both directions. No substituent whether equatorial or axial has any ecliptic hydrogen in either direction. This means, that a syn-specific elimination, which requires an ecliptic configuration, cannot occur out of the chair conformation. The $4~\%$ syn-elimination the table references must occur from other, minor configurations such as the boat.

Cyclopentane is also hardly strained. However, the configuration of the ring is much more volatile. Furthermore, even in the shown configuration there are possibilities of syn-eliminations; for example see the two carbons opposite to the upwards-pointing one. Still, anti-eliminations are also well possible, especially from the other three carbons that form the upwards arch, and account for most of the elimination.

A similar case can be made for cycloheptane, which is even more strained. There is not much for me to say here that I haven’t said for cyclopentane other than that you should build it.

And finally cyclobutane. This is the most strained of the systems considered and probably the one with the most fluctuating configuration. Here, it looks like an anti-elimination is possible; but this requires the substituent to occupy an axial position. This configuration is strongly sterically hindered. Syn-eliminations can occur from any position by only moderate shifts from the geometry shown. This may explain why anti-eliminations are strongly disfavoured here. I would, however, once again stress how helpful it is (especially in this case) to build the model.


One way to approach this problem is think about what kind of material would best promote crystallization, or ice-nucleation, and then essentially use a material that does not have those characteristics.

Consider cloud seeding for example. The whole premise of this is that there is a dearth of ice-freezing-nuclei in the atmosphere, so in the presence of supercooled water droplets, you introduce a compound with a similar crystal structure to ice, like silver iodide for example. Even crystalline compounds with different structures like sodium chloride and dry ice are also effective. So, since we're trying to do the opposite of ice-nucleation, we want opposite characteristics of the surfaces in contact with the supercooled water; nothing crystal-like in nature, something as smooth and clean as possible on a very small scale.

Teflon is a highly hydrophobic synthetic polymer, meaning it's similarity to both liquid and solid water is minimal. A smooth, clean, unscratched Teflon surface is probably the ideal choice. A cheaper option using the same concept would be other organic polymers like polyethylene (probably what the bottles in your video link were made of), polypropylene, HDPE, etc.

Another option that is not crystal-like by definition is any type of consumer glass container. Again, any nano-scale defects or contamination will degrade it's ability to inhibit ice formation from supercooled water.


How to find maximum and minimum oxidation number of an element

The rule you quoted is generally true for main group elements — but only if you count the groups in the older main-group/transition-metals formality. The newer IUPAC terminology counts the s, p and d element groups one after the other, so that sulphur would be in group 16. For that nomenclature, you need to subtract 10 from the group number for the maximum oxidation state.

Example: Phosphorus (group 15 by current terminology; group V by older terminology). Maximum oxidation state: $\mathrm{+V}$. Minimum oxidation state $\mathrm{-III}$.[1]

This does have a general quantum chemical explanation. These oxidation states add up to eight, which is exactly the number of electrons that typically make up the outer (valence) shell — where chemistry happens.

If we start at phosphorus($\mathrm{V}$) and want to remove another electron, we would have to remove this electron from the core orbitals. These are already pretty stable (i.e. low in energy) in the ground state, but for every electron removed they get stablised further. Thus, they end up in a very stable state and removing gets all the much harder.

Likewise, every electron added to a neutral compound will destabilise all orbitals of that atom. If three are already added to phosphous, giving phosphorus($\mathrm{-III}$), the next electron would need to be added to a distant orbital, which again is not something easy. This is why there is typically a range of eight for the chemically accessable oxidation states of main group metals.

However, I am not willing to bet anything that the s-elements of periods 4 and higher (potassium and below and calcium and below) stick to that rule; for the reason see and interpret below. So far, only $\mathrm{+I}$ and $\mathrm{-I}$ are known for alkaline metals (group 1) and only $\mathrm{+II}$ and $\mathrm{+I}$ for alkaline earth metals are known (save the elemental oxidation state $\pm 0$).

Transition metals are a lot harder. In theory, you could assume a range of either ten (d-electrons only), twelve (d and s-electrons) or eighteen (d, s and p-electrons). In fact, experimentally a range of ten has been established for chromium, manganese and iron while a range of twelve has been established for osmium and iridium. (Source: Wikipedia)

So far, the highest oxidation state has been found for iridium ($\mathrm{+IX}$). Platinum($\mathrm{X}$) has been predicted. (Source: also Wikipedia) Note that these states are well within the ‘s and d’ idea I alluded to earlier. To the best of my knowledge, there are no pieces of evidence pointing towards the ‘s, p and d’ idea.

There are a large number of unknown oxidation states for transition metals, e.g. while chromium($\mathrm{-IV}$) and ($\mathrm{-II}$) are known, chromium($\mathrm{-III}$) is not.

The notion that metals could only have positive oxidation states is incorrect. As an example, $\ce{[Fe(CO)4]^2-}$ with an iron oxidation state of $\mathrm{-II}$ is known.

All of this complicates the analysis strongly. Until much more research has been performed, you should probably not attempt to predict maximum and minimum oxidation states of these elements.


Why do sealed plastic bottles of pop collapse as the contents get old?

It is most likely the CO2, H2O and air in the pop bottle headspace diffused (leaked) out of the pop bottle.
You are correct: the pressure has reached ambient (not dropped below ambient). Because the pressure in a freshly bottle pop bottle was originally higher than the pressure in the room (normally atmospheric pressure), and the bottle's plastic walls were smooth and strong, any decrease in pressure inside the pop bottle would cause the pop bottle walls to collapse.
If the bottles were left longer, the CO2 and water would continue to evaporate from the liquid phase until the headspace contained approximately the same pressure as the air in the room. If somehow all the water evaporated, you'd be left with sugar and a bunch of other solids (that were dissolved). I doubt full evaporation of water would occur.

Also, if the bottle is made from PETE (polyethylene terephthalate), prolonged contact with water (including pop) will result in many by-products. Two of them are:

a. UV exposure, air and water (together) promote peroxide formation on polymer end-groups (OH, COOH). Another way of saying this is that the bottle walls degrade.

b. The PETE plastic could leach some trace antimony (if that was used in the polymerization process) and/or phthalates, suspected to be endocrine disruptors.

IF water permeates through the bottle walls, it can allow for microoganisms to diffuse in/out of the bottle. You might get some mould in there!

Those are just some things that could happen.


Yes, in The Characterization of Linear Polyethylene SRM 1475. II. Determination of Total Methyl Content by Infrared Spectrophotometry and references cited therein, it is confirmed that the polymer chains terminate with methyl groups (the repeating unit plus an additional hydrogen as you phrase it).


Oxidation numbers of atoms in Ti(SeO3)2?

The selenite anion ($\ce{SeO3^{2-}}$) has a net charge of -2. Titanium's oxidation state should be obvious given this information. As for selenium's oxidation state, since we know each oxygen has a charge of -2, and since the entire polyatomic anion has a net charge of -2, the following equation must balance:

$$\ce{Se}+3 \cdot (-2)=-2$$


How can I calculate the pH of the solution created by reacting sodium with water?

You answer is right, but your reasoning is not.

Amount of $\ce{Na} = \frac{0.023}{23} = 0.001$ mole

$[\ce{NaOH}]= \frac{0.001}{0.100} (\mathrm{moleL^-1}) = 0.01$ M

$\downarrow$

$p\ce{OH} = 2 \to p\ce{H} = 14 - 2 = 12$


How many of different types of chemical species are in a certain volume of aqueous solution of an acid

The following is an approximate calculation, since among other things it makes the approximation that the activities of each substance are equal to their molarities, which is not true in such concentrated solutions. Also I take some shortcuts to make the calculation quicker.

Let's look at a volume of $1\ L$. The mass would be $1323\ g$, 98% of which ($1296.54\ g$) is due to the acid, and 2% ($26.46\ g$) due to water. This initially translates into $10.36\ mol$ of acid, and $1.47\ mol$ of water. However, some of the water and acid are consumed to create hydronium and conjugate base.

To calculate the dissociation of the acid, you may want to apply $K_{a}=\frac{[H_{3}O^{+}] [A^{-}]}{[HA]}$ (the Henderson-Hasselbach equation is derived from this by taking the log on both sides, so they're equivalent), but this is actually a very bad approximation, since you need to take into account the concentration of water in non-dilute solutions. We'll do it anyway and I worked it out to be $[H_{3}O^{+}]=[A^-]=0.704\ mol.L^{-1}$. Since the volume we're working with is $1\ L$, that means $n_{H_{3}O^{+}} = n_{A^-} = 0.704\ mol$. Subtracting those amounts from the initial amounts of water and acid, we reach $$ n_{H_{2}O}=0.77\ mol$$ $$ n_{HA}= 9.66\ mol$$ $$n_{H_{3}O^{+}}=n_{A^-}=0.704\ mol$$
To find the actual number of each entity, just multiply by the Avogadro number $6.022\times 10^{23}$ and you have the amount of each in a litre of solution.


A question about geometry of benzenetetrol

The melting point of 1,2,4,5-benzenetetrol is 210C (https://www.chemicalbook.com/ProductChemicalPropertiesCB41307665_EN.htm), so there is very probably significant intermolecular attraction between molecules, probably in three dimensions, which would involve the hydrogens pointing up and/or down, and not in-plane.

For a crystal structure, https://pubchem.ncbi.nlm.nih.gov/compound/1_2_4_5-Benzenetetrol suggests that 3 hydroxyls are planar while one is as out of plane as you would expect for cyclohexanol. Believe it or not. (I don't believe it.)

The (predicted) NMR spectrum shows two peaks, one very sharp for the aromatic protons, the other fairly sharp for the hydroxyls, but no experimental data seems to be available. The sharpness of the predicted spectrum suggests that the prediction is that free rotation is likely in solution.


The difference isn't that stark. However, you have to bear in mind the Configuration of electrons on the central nitrogen. We often will draw ammonia with the electron pair atop the nitrogen. That lone pair will readily accept a proton, making it a base.

To put it another way, ammonia is far more ready to accept a proton than to release one thanks to the lone pair.

Of course, it also depends on the environment it's in. But in general, you'll usually see ammonia (or just nitrogens in compounds) referred to as basic


Here is some data on other well-soluble salts:

Comparing the values, both the anion and the cation contribute to magnitude of enthalpy and entropy. The larger cations seem to contribute to a larger entropy of dissolution, as do the larger anions. The entropy is concentration-dependent, so I assume the data is for a standard state of about 1 mol/L dissolved salt.

There is a nice discussion about the dissolution of ammonium chloride here. The enthalpy of dissolution (14.4 kJ/mol) and the entropy of dissolution (75.3 J/(K mol), source) are close to that of KCl, so there is nothing super-special about the ammonium ion.

  [...] since their ΔH values are usually much larger than the TΔS values they would have, especially at room temperature.

At the beginning of the dissolution process, the entropy dominates. At the solubility limit, entropy and enthalpy contributions are equal and add up to zero.

  What makes ammonium nitrate soluble?

I did not find a data table that includes entropy and enthalpy of dissolution for ammonium nitrate, but you could easily calculate these from entropy and enthalpy of formation of the reactants and products. On second view, I am having trouble finding the data for aqueous nitrate, but this textbook has some values. There is a report that nitrate comes in two forms in aqueous solution, perhaps complicating things.

Some people seem to think that the dissolution of nitrates is special in terms of enthalpy (compare to e.g. carbonate) and others argue that the entropic contribution is special. It would be easier to answer the question, for example, "what makes ammonium nitrate more soluble than ammonium carbonate".


For a cathode ray tube (old-style TV), you want the electrons traveling collision-free in a straight line from the cathode to the screen (anode), where they generate light hitting a phosphorescent layer. For a neon light, you want the electron hitting gas particles so that the entire tube glows.

This is further explained here: Why does the cathode ray tube only start glowing at low pressures?


Elimination of bromide and acetate under reducing conditions

What reaction can lead to this elimination when it seems that no
  proton is leaving the molecule (instead of bromide) and acetate is
  leaving?

This is a reductive elimination reaction.  The classic textbook example is the reductive elimination of trans-1,2-dibromocyclohexane to cyclohexene in the presence of zinc dust and acetic acid.

It is basically a redox reaction where the zinc metal is oxidized and the dibromocyclohexane is reduced.  As with many reactions involving electron transfer from metal surfaces, the mechanism of this reaction is not fully understood.  Perhaps the zinc inserts itself into the carbon-bromine bond as a first step, but in any case, electron flow proceeds in the direction shown above.

The reaction has found many uses and has been widely modified to include cases such as yours where a bromine and another group are eliminated to form a double bond.  Here is a link to a paper where one of the bromines is replaced by an alkoxy group and the same zinc-mediated reductive elimination occurs.  Here is a link to another paper where zinc is used to perform a reductive elimination in a series of compounds where both bromines have been replaced, one by acetate and the other by hydroxyl.


How to calculate the molar mass of a protein from its osmotic pressure?

Use osmotic pressure $Π=iMRT$
From this get your concentration and calculate molarity of solution.
By the way if you are looking more on this topic then you must consider the chapter on solutions and their colligative properties in your chemistry textbooks


How to tell which species has the highest ionization energy?

Like most chemical conundrums you need to simplify the problem using some knowledge of chemistry.

First there are two electron configurations. $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$ all have the configuration ${1s^2 2s^22p^6}$. $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$ all have the configuration ${1s^2 2s^22p^6 3s^2 3p^6}$.

For $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$, $\ce{Mg}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Mg^{2+}}$.

For $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$, $\ce{Ca}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Ca^{2+}}$.

Now it is just necessary to compare the ionization of $\ce{Mg^{2+}}$ to $\ce{Ca^{2+}}$. Since the $2p$ orbitals are closer to the nucleus than the $3p$ orbitals, it will be harder to ionize $\ce{Mg^{2+}}$ than $\ce{Ca^{2+}}$.


If you are trying to observe corrosion in a specific area, agar is a convenient way to prevent convection currents from obscuring what caused the corrosion and for localizing the indicator.


Is there an A value for the carbaldehyde and vinyl groups?

Comparing the axial conformations of vinylcyclohexane and
  cyclohexancarbaldehyde: Which molecule is more strained?

As you point out, "A-values" can be used to get some idea of the relative "size" of a substituent in the cyclohexane system.  More specifically, the A-value reflects both the steric size of the substituent and the length of the bond connecting the substituent to the cyclohexane ring (the longer the bond, the further away the substituent and steric interactions are therefore reduced).

The A-value is just the free energy difference between the axial and equatorial conformers. Since destabilizing steric interactions can occur in the axial conformation, the larger the substituent, less axial and more equatorial conformer will be present.

(image source)

As a consequence, the "larger" the substituent the larger the A-value. The A-value for a methyl group is 1.8, using

$$\mathrm{\Delta G = -RT*ln(K)}$$

we can solve for K and find that the axial/equatorial ratio for methylcyclohexane is ~ 5/95 (as expected, the less sterically strained equatorial conformer predominates) at room temperature.

(image source)

Here is a link to a nice table of A-values.  The A-value for a vinyl group ($\ce{-HC=CH2}$) is 1.6, while it is only 0.7 for the aldehyde group ($\ce{-CHO}$). Hence the aldehyde group is the smaller of the two. Or said differently, axial vinyl-cyclohexane would be more strained than axial cyclohexane-carbaldehyde


You applied the POAC incorrectly . Even your stoichiometric answer is incorrect.

First Balance the reaction

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

We get that one mole of $\ce{O2}$ requires 2 moles of $\ce{H2}$.
Thus 10 moles $\ce{O2}$ need 20 moles of $\ce{H2}$.


Does the boiling boint of miscible liquids follow a linear relationship?

Except if molecular composition of your two liquids are similar, the boiling point of the mixture will not proceeds linearly but like this for instance :

With the lower blue curve the boiling curve. In between the two curve you'll then have a mixture of vapor and liquid and only vapor on the upper part the binary diagram.

So you should find out an already drawn binary diagram for your mixture or draw it out by yourself first.


Yes, you've described a property of amines. When protonated by an acid (e.g. $\ce{HCl}$), an amine is converted into into a water soluble salt. This is actually a thing to do in pharmaceuticals to increase the amount of a drug that can be dissolved in water and be taken up by the body. The next time you have the chance take a close look at the chemical names on a drug label and theres a good chance you might see "hydrochloride."

Just as something can be protonated, it can be deprotonated. For the hydrochloride salt of an amine, $\ce{NaHCO3}$ should be basic enough to deprotonate it, and cause it to precipitate out of solution, as it would no longer be a water-soluble salt.


Reasoning retention factor of benzoic acid

The answer is the dimer formation of the benzoic acid. This is why higher concentrations lead to more extended dimer formation, thus higher $\mathrm{R_f}$ values as the carboxyl group "gets shielded."

One thing of note is that $\mathrm{R_f}$ values are not directly dependent on the polarity. They also depend for example on the size of the molecule (Oswald radius). And the relation between properties of the molecules (as stated for example in the form of molecular descriptors like $\log P$) and the resulting $\mathrm{R_f}$ is very complicated indeed. This is also illustrated by the fact that there are predictors for IR spectra, NMR spectra, solubility and lots of molecular descriptors but I have never heard of a good $\mathrm{R_f}$ value predictor.

This is why I would always be careful to talk about it like "the polarity" of the compound, especially as "the polarity" is not a clearly defined quantity.


What is the CNC bond angle in methyl isothiocyanate?

I have just answered What is the CNC bond angle in methyl isocyanate? Pretty much everything written there also applies to methyl isothiocyanate, just a bit more extreme.
Here is the structure calculated on the RI-BP86/def2-SVP level of theory:

The bond angle is $\angle(\ce{CNC}) = 156^\circ$, so we expect a much larger contribution of the right side in the following resonance description:
$$\ce{H3C-N=C=S <-> H3C-N^+\bond{3}C-S^-}$$
Reasons for this are not trivial, but that would exceed the scope of this answer.
TL;DR From VSEPR we would expect a trigonal coordination and a $120^\circ$ angle. Refining with resonance, we'll expect a widening of the bond angle. Everything beyond that  needs much more involved models.


How to find the concentration of barium ions in a mixture of barium sulfate and barium thiosulfate?


Can anyone please help me how do you determine kinetically the fastest product of sulphonation of $\beta$ napthol?

Electrophile can substitute 2-naphthol ring on 3 possible positions: 1-position through intermediate I, 6-position through intermediate II, and 3-position through intermediate III (See following diagram). Among these three intermediates, intermediate I is the most stable (You must visit Mechanism of formation of 2-naphthol red dye to get excellent explanation why substitution at 1-position is the preferable one over 3-position, although both are ortho to the phenolic $\ce{OH}$ group). Thus, the kinetically fastest product of sulphonation of 2-naphthol is 2-hydroxynaphthalene-1-sulfonic acid (Oxy Tobias acid). There will be no 3-hydroxynaphthalene-2-sulfonic acid (Ref.1).

The best evidence for this preference has been displayed during specific reaction using Claisen rearrangement (See above diagram in the box). After first rearrangement free phenolic $\ce{OH}$ group was allylated again and subjected to heat. For our surprise, the rearragement is proceeded to give 100% 1,1-diallyl derivative as shown. No 1,3-diallyl-2-naphthol was detected (according to NMR evidence).

Based on Ref.1, some 2-hydroxynaphthalene-6-sulfonic acid (Schaeffer's acid) has also been produced under kinetic control. Yet, longer heating time allowed to give disubstituted product (2-hydroxynaphthalene-1,6-disulfonic acid) preferably, thus thermodynamic product cannot be determined by these evidence.

Reference:

Paolo Beltrame, Giorgio Bottaccio, Paolo Carniti, Giuseppina Felicioli, “Sulfonation of 2-naphthol by sulfuric acid: rate measurements and kinetic model,” Ind. Eng. Chem. Res. 1992, 31(3), 787-791 (https://doi.org/10.1021/ie00003a022).