Can carbanion rearrange to attain more stability?

Sooner or later anionic rearrangement name reactions should turn up in your organic chemistry classes. I will present two in this post.

The first and arguably most well-known are the Wittig rearrangements; these can proceed as [1,2] or as [2,3] rearrangements.[1] The Kürti/Csakó states that:[1]

  The [1,2]-Wittig rearrangement proceeds via a radical-pair dissociation-recombination mechanism, while the [2,3]-Wittig rearrangement is a concerted, thermally allowed sigmatropic process proceeding via an envelop-like transition state in which the substituents are pseudo-equatorial.

This means that strictly speaking only the [2,3]-Wittig rearrangement could be considered an actual carbanion rearrangement as per your question but many chemists may also consider the [1,2]-Wittig rearrangement one since it effectively does generate a rearranged and thereby more stable anion. Both reactions are shown in scheme 1.

Scheme 1: the [1,2] (top) and [2,3] (bottom) Wittig rearrangements.

The second example has recently gained significant applications in synthetic organic chemistry; it is the same general reaction except it employs a silyl ether instead of an alkyl ether. Here, the stability of both sides of the reaction is similar so both the forward and backward direction are known; these are the Brook and retro-Brook rearrangements (see scheme 2).

Scheme 2: The retro-Brook rearrangement (top) and Brook rearrangement (bottom).[2,3]

West et al. who reported the reactions in scheme 2 note the interesting effect that the amount of butyllithium added influences the equilibrium. If substoichiometric or catalytic amounts are added, the normal Brook rearrangement takes place with the silyl ether being the more stable species. Under excess butyllithium conditions, the retro-Brook rearrangement takes place indicating that under those conditions the alcoholate is the more stable species.[3]

I mentioned that arguably the Brook rearrangement is more common in organic synthesis; this is due to Amos B. Smith’s anion relay chemistry or ARC.[4] Smith uses priviledged synthetic intermediates that can give alcoholates upon an initial reaction. The previously introduced silyl group migrates towards the alcoholate, liberating another carbanion which can subsequently participate in another nucleophilic attack. This allows multiple anion-dependent transformations to occur in a single reaction vessel.

References:

[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, p. 490.

[2]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, p. 64.

[3]: R. West, R. Lowe, H. F. Stewart, A. Wright, J. Am. Chem. Soc., 1971, 93, 282–283. DOI: 10.1021/ja00730a065.

[4]: A. B. Smith III, M. Xian, J. Am. Chem. Soc., 2006, 128, 66–67. DOI: 10.1021/ja057059w.


Solute mole fraction for hydrated magnesium sulfate dissolved in water

Alright, let’s take this one step at a time.

First of all, you want to determine a molecular formula, and among your known data you have a mole fraction. This suggests it’s a good idea to work with moles to solve the problem. As such, the first step is to convert the amount of water added and magnesium sulfate hydrate used. Calculating the number of moles of water added is easy:

$$90\ \mathrm{g}\ \text{water added} \times \frac{1\ \mathrm{mol}\ \ce{H2O}}{18\ \mathrm{g} \ \ce{H2O}} = 5\ \mathrm{mol}\ \ce{H2O}$$

The number of moles of $\ce{MgSO4.$x\,$H2O}$ is unknown because even though you know how much mass was dissolved, you don’t know its formula completely. However, $\ce{MgSO4}$ contributes $120\ \mathrm{g\ mol^{-1}}$ to the molar mass, and each additional molecule of water in the hydrate stoichiometry adds a further $18\ \mathrm{g\ mol^{-1}}$, so that the molar mass of $\ce{MgSO4.$x\,$H2O}$ is $120\ \mathrm{g\ mol^{-1}} + x \times 18\ \mathrm{g\ mol^{-1}} = (120+18x)\ \mathrm{g\ mol^{-1}}$. With this molar mass, the number of moles of $\ce{MgSO4.$x\,$H2O}$ is:

$$70\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O} \times \frac{1\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}}{(120+18x)\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O}} = \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}$$

From the definition of mole fraction:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{total}}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}$$

Now, one is tempted to write the following and solve for $x$:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}\ }{\frac{70}{120+18x}+5 \ }=0.039$$

However, this equation is incorrect. The problem is that the following process was not taken into account:

$$\ce{MgSO4.$x\,$H2O(s) \xrightarrow{water} MgSO4(aq) + $x$\ H2O(l)}$$

In other words, when the magnesium sulfate hydrate dissolves in water, the water molecules contained in the solid are released and become an indistinguishable part of the solvent; the final mixture is a simply a solution of the solute $\ce{MgSO4}$ in water. Therefore, the number of moles of solvent is not simply $5\ \mathrm{mol}$ as calculated at the start, but rather a larger value, which should include the number of moles of water released when the solid dissolves.

From the dissolution equation, we see that $1\ \mathrm{mol}$ of $\ce{MgSO4.$x\,$H2O}$ generates $1\ \mathrm{mol}$ of $\ce{MgSO4}$ and $x\ \mathrm{mol}$ of $\ce{H2O}$ from crystallized water. However, we know the amount of moles of $\ce{MgSO4.$x\,$H2O}$. This means the following:

$$n_\text{solute}=n_{\ce{MgSO4}}=1 \times n_{\ce{MgSO4.$x\,$H2O}}= \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4}$$

$$n_\text{solvent} = n_\text{added water} + n_\text{crystallized water} = n_\text{added water} + x \times n_{\ce{MgSO4.$x\,$H2O}} = \bigg(5 + \frac{70x}{120+18x}\bigg)\ \mathrm{mol}\ \ce{H2O}$$

Now we can finally apply these values to the mole fraction equation:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}}{\frac{70}{120+18x} + \bigg(5 + \frac{70x}{120+18x}\bigg)}=0.039$$

Solving for $x$ provides the answer $x=7.03$, which means there are seven moles of water per mole of $\ce{MgSO4}$, and thus the formula for the solid is $\ce{MgSO4.7H2O}$, representing magnesium sulfate heptahydrate.


The C-13 chemical shift of acyl chlorides vs ketones

Halogens $(\ce{Cl},$ $\ce{Br},$ $\ce{I})$ also act as conjugative donors. Not very good ones but you still have to take that into account especially because, as you said, conjugative effect is larger than inductive effect (and it is very different in $\ce{^1H}$- or $\ce{^13C}$-NMR).
The $\ce{^13C}$ chemical shift in $\ce{CH2Cl2}$ is $\pu{53 ppm}$ but the one for $\ce{CH2I2}$ is $\pu{–62 ppm}$ (yes, with a minus sign).


Electrolysis rust removal - side products - identify lead

Add the information regarding the composition regarding the electrodes. However lead is very heavy it would not like to float on the top. Secondly, lead compounds are mostly white with an exception of oxides (yellow/ red) and iodide (yellowish). Most likely the black floating scum is carbon in iron. Wear gloves, and wash your object with a mild abrasive such as a baking soda, and it should be fine. After all it was meant as a hand tool. It will not become "toxic" after a brief electrolysis.


Finding work done for a chemical reaction at non-constant pressure and temperature

There are a series of five assumptions you need to make to get from the general expression for work, to the expression you wrote in your OP.  With each assumption, you are applying successively more constraints to your system.  In the post below I give the assumptions, and the equation that results from each, so that you can see how the form of the equation evolves as successively more constraints are applied.
We start with the general expression for work:
$$\text{đ}w = \vec{F} \cdot d \vec{x}$$
[We use the form $\text{đ}w$ rather than $dw$ because work is path-dependent, and its differential is thus inexact; the dyet glyph, $\text{đ}$, is used to indicate this.]
Now we integrate and express the dot product in terms of scalar quantities:
$$w = \int_{path} \vec{F} \cdot d \vec{x} = \int_{path} |F| \,d |x| \cos(\theta)$$
Because work is path-dependent, it is necessary to integrate over the actual path, as indicated in the above integrals, rather than merely from the initial state to the final state.
Dropping the explict scalar notation, we can write it as:
$$w =  \int_{path} F d x \cos(\theta)$$
Now let's add some assumptions:
Assumption 1: The force is applied in the direction of motion.
As the angle between the direction of motion and the direction of force is $0$, and $\cos(0) = 1$, we have:
$$w =  \int_{path}F dx \cos(\theta) = \int_{path}F dx$$
Assumption 2:  The work is $\bf{pV\text{– work.}}$
For $pV\text{– work}$, we can take work done on the system to be either positive or negative.  I'll use the former convention. And we can only reliably use the external pressure to determine the work; the internal pressure is not always defined, particularly for rapid expansions or compressions.  To indicate this explicitly, we write "$p$" as $p_{ext}$. Then (using "$A$" for area):
$$p_{ext} = F/A;\,\,dV = -A dx\,\, \text{(negative sign is because as $x$ increases, $V$ decreases)}$$
$$\Rightarrow  \int_{path} F dx =  -\int_{path}(p_{ext}A) (dV/A) = -\int_{path}p_{ext} dV$$
Thus:
$$w = -\int_{path}p_{ext}dV$$
This is the general expression for $pV\text{– work}$.  It applies regardless of the conditions.
Assumption 3: The system is in mechanical equilibrium with its surroundings.
In this case, $p_{ext} = p_{sys}$.  By convention, we designate $p_{sys}$ simply as "$p$".
Note that this assumption restricts us to paths in which the system is continously in mechanical equilibrium with its surroundings.  This limits us to paths whose work depends only on the difference between the initial and the final state.  Thus we do not need to concern ourselves with doing an explicit path integral. I.e., because the path restriction is now implicit, we can simply integrate between intitial and final states (which I've designated as $i$ and $f$):
$$w =  -\int_{i}^{f}pdV$$
Assumption 3a: The system is at constant pressure.
Since $p$ is a constant, we can take it out of the integral:
$$w =  -\int_{i}^{f}pdV=  -p\int_{i}^{f}dV = -p\Delta V$$
I labeled this assumption as "3a" rather than 4 because we didn't need to both specify the system is at constant pressure and in mechanical equilibrium with its surroundings, since if it is at constant pressure it is also in mechanical equilibrium with its surroundings.
But I broke the steps up, since I wanted to give the expression following assumption 3 for completeness.
Assumption 4: The system is an ideal gas.
Substituting $V=nRT/p$, and remembering that the pressure is constant, gives:
$$w =  -p\int_{i}^{f}dV =  -p\int_{i}^{f}d(n R T/p)=-R\int_{i}^{f}d(n T)=-R\int_{i}^{f}ndT-R\int_{i}^{f}Tdn$$
Assumption 5: The temperature is constant.
Since $T$ is a constant, we can take it out of the integral:
$$w =-R\int_{i}^{f}d(n T)=-RT\int_{i}^{f}dn = -RT \Delta n$$
Or, equivalently:
$$w =-R\int_{i}^{f}ndT-R\int_{i}^{f}Tdn = -R\int_{i}^{f}Tdn = -RT\int_{i}^{f}dn = -RT \Delta n$$


Solvent effect on gold nanoparticle and silver nanoparticle in UV spectroscopy?

Attenuation of light related to the properties of material is governed by beer-lambert law Many molecules absorb ultraviolet or visible light. The absorbance of a solution increases as attenuation of the beam increases. Absorbance is directly proportional to the path length, b, and the concentration, c, of the absorbing species. Beer's Law states that
A = ebc, where e is a constant of proportionality, called the absorbtivity.

Different molecules absorb radiation of different wavelengths. An absorption spectrum will show a number of absorption bands corresponding to structural groups within the molecule. For example, the absorption that is observed in the UV region for the carbonyl group in acetone is of the same wavelength as the absorption from the carbonyl group in diethyl ketone.

here is beer law description:http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm


When you write a chemical equation, you don't care whether the reaction has gone to completion or not. On a deeper level, none of them do. There is always an equilibrium, and there is always some starting compound left. To find out how much is that "some", you need to know the equilibrium constant and make certain calculations.

The written equation does not tell you that each and every molecule in the sample should react exactly like that. This is simply not what it is all about. It does, though, tell you another thing: if you got such-and-such amounts of products, you must have lost such-and-such amounts of the starting compounds. This is simply the conservation of mass, and there is no going around it.


How to determine the percentage of iron(II) and iron(III) in a solution?

You’re already very close. You have realised that:

You can titrate the $\ce{Fe^{+II}}$ content with a potassium dichromate standard solution; and
You can turn $\ce{Fe^{+III}}$ into $\ce{Fe^{+II}}$ by adding metallic zinc.

Note that in the first bullet point of mine, your second reaction, you are creating $\ce{Fe^{+III}}$, hence iron(III) that was present before you started won’t be modified. Similarly for the other reaction and iron(II). That should lead us to following methodology, hidden in a spoiler tag:

   1. Titrate the iron(II) content in an aliquot of your solution. That gives you the content of iron(II) and only iron(II).

 2. Take another aliquot and reduce all iron(III) to iron(II). That will give you the total iron content.

 3. Substract the value from 1. from the value from 2. to get the content of iron(III).


How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination?

The following three products wil be formed:

Your book may be giving the answer to the number of monochloro products formed, as there are only two distinct monochloro products formed on chlorination of isobutane.


What is the difference between polyethene and polyethane?

The reason it is called polyethene (or polyethylene) is because it was produced from the monomer of ethylene. To produce polyethylene, a radical initiator is added to ethylene, which attacks one carbon and opens up the double bond leaving a radical at the end. This radical attacks another ethylene molecule, etc. The chain stops forming when it encounters another radical. So, from ethylene, we get polyethylene (even though it looks like polymethylene).


Comparing the pKb of piperazine and methylpiperazine

In terms of the logic employed by your answer key

Whether an inductive effect is stabilizing depends on the nature of the molecule and reaction you're thinking about.

You're right to think of the methyl group as inductively release electron density into the ring. However, this inductive release from the methyl group is actually increasing the electron density in the piperazine ring, which in turn makes the N lone pair higher in energy and more reactive, thus making it a stronger base. However!

What is actually the case in real life

I was able to find a detailed study of $\mathrm{p}K_\mathrm{a}$ values of protonated piperazines and methylpiperazines, and assuming $\mathrm{p}K_\mathrm{b} = 14 - \mathrm{p}K_\mathrm{a}$.

Then at 298 K...

$\mathrm{p}K_\mathrm{b}\text{(piperazine)} = 4.27$
$\mathrm{p}K_\mathrm{b}\text{(N-methylpiperazine)} = 4.86$
$\mathrm{p}K_\mathrm{b}\text{(N,N'-dimethylpiperazine)} = 5.62$
$\mathrm{p}K_\mathrm{b}\text{(2-methylpiperazine)} = 4.43$

This tells us that your answer key is actually wrong, and that in real life sterics may be playing a role in the basicity of the N atom. This data is in line with the fact that secondary bases tend to be slightly stronger than tertiary ones. For example, the

$\mathrm{p}K_\mathrm{b}(\text{diethylamine)} = 3.02$
$\mathrm{p}K_\mathrm{b}(\text{triethylamine)} = 3.25$


Do We change sign of electrode potential when reversing an equation?

The reduction potentials are a special case of standard reaction enthalpy, because you can transform $\Delta E$ into $\Delta H$ just by calculating with a few constants.

Standard reaction enthalpy and all other thermodynamic values related to reactions change sign when the direction of reaction is reversed.


I have read your question several times to figure out what your problem is.
Below, I will try to give you a simple explanation without any fancy mathematics or complex equations.

Now, suppose we dissolve the weak acid "A" in water. The acid will react with water and form hydrogen ions and the corresponding base "B" of the weak acid. We further suppose that the pH of this solution is 3 at equilibrium.

We have the following equilibrium:

$$\ce{A + H2O <--> H3O+ + B}$$

The concentrations are supposed to be low, so we can state an equilibrium constant:

$K = [\ce{H3O+}]*[\text B]/[\text A]$

Now, we add a strong acid to this solution. What will happen?
The pH of the solution will drop. Suppose it will drop to pH = 2. We will get an immediate response from the equilibrium reaction of the weak acid system. Since the amount of hydrogen ions have increased in the solution by the addition of the strong acid, the equilibrium will be pushed to the left.

Pushed to the left? What does it mean?

Well, the new situation implies that the weak acid cannot give off  the same amount of hydrogen ions as it did at pH = 3. pH is now 2 and we have a completely new situation.

Now I will target on what I think is your main problem.

When you add a strong acid to the weak acid solution, the pH will drop.
The equilibrium of the weak acid will then reset into a new equilibrium determined by the actual pH and the equilibrium constant of the system.

At pH = 3, hydrogen ions were formed from the weak acid. These (or other) hydrogen ions will be forced to react with the corresponding base B, giving more A at pH = 2.
Thus, the net effect is that the hydrogen ion concentration related to the strong acid will be in principle unchanged.


Does halogenation by phosphorus tribromide lead to an inverted product?

Yes, the haloalkane product will have the inverse stereochemistry of the alcohol that it was produced form. $\ce{PBr3}$ does this by converting the $\ce{-OH}$ into a good leaving group and providing a nucleophile all in one. This is can be seen in the two step mechanism:

$\hspace{2.4cm}$

When $\ce{Br-}$ ejects $\ce{PBr2OH}$ in an $\mathrm{S_N2}$ reaction, the stereochemistry is inverted:

$\hspace{3.1cm}$


What is the shiny structure under a metal coating?

Likely an intermetallic , I think there are a couple different ones. I am more familiar with "spray and diffuse"  or vapor deposition ( Alon). Alonizing is very long ( > 24 hr) but gives a high integrity coating , can't remember the thickness.  As I remember the "diffuse" cycle takes a couple hours. I expect your process produced a thinner coating but still with some intermetallics.


Grasping the concept of Electronic Spin, Effective Spin and Fictitious Spin

Here's what I believe they're trying to say.
First, note that the discussion is mostly limited to high spin complexes (3 unpaired electrons out of the 7 d electrons), so the figure you reproduced is only representative of high spin states.
Second, recall that when D > 0 (ie left side of figure), the five d orbital energy levels split so that we have three t2g orbitals that are lower in energy than two e2 orbitals because of crystal field splitting. When D < 0 (right side of figure), the reverse is true, two e2 orbitals are lower in energy than three t2g orbitals.
In the absence of a crystal field, there is no difference in energy across the five orbitals, so for the case of 7 d electrons, we have a 4A state, which is the pre-split state in both cases in the figure.
In typical simple crystal field models, D>0 is associated with octahedral (i.e. OH) coordination and D<0 with tetrahedral (TH) or cubic coordination, but the authors give examples of TH with both D>0 and D<0, so the sign of D should not be assumed based on geometry here (nor should geometry be assumed based on sign).
Looking at the figure, then, we can start with the right side (because it's easier to understand) and see the transition from 4A to a state of D<0. That is the application of the crystal field. Since D<0, we know that the two e2 orbitals are lower in energy than the three t2g orbitals. Since we have high-spin d7, our electronic state is e24t2g3, with all three t2g electrons unpaired.
What we see in the figure is that there are two pairs of energy levels, with the states with spin $\pm \frac32$ at lower energy than those with spin $\pm \frac12$. The spin $\pm \frac32$ and $\pm \frac12$ values refer to the sum of the spins of the three unpaired electrons. The $\pm \frac32$ states are those in which all three t2g electrons have the same spin (either all $+\frac12$ or all $-\frac12$), while spin $\pm \frac12$ means that one of the three has a different spin than the other two. We know from Hund's rule that unpaired electrons in degenerate orbitals have the lowest energy when they all have the same spin direction, due to the exchange energy term. That is the source of the splitting in energy between the two possible total spin states.
We now get to the "fictitious spin" part. This is actually relatively simple. It has nothing to do with actual spin and simply indicates the number of experimentally relevant spin states, but is expressed in $\frac 12$ increments to have the familiar form of a spin value. Thus, fictitious spin of $\frac12$ just means two possible states, as one would observe for an electron with possible spins $\pm \frac12$. A fictitious spin of 1 would mean three possible states, as we would have for the case of two unpaired electrons.
In the figure, the fictitious spin is $\frac12$ because under the experimental conditions, only the two lower energy levels (ie actual spin of $\pm \frac32$) are observed. The two states with spin $\pm \frac12$ are higher enough in energy that they do not occur.
The left side can be interpreted similarly, except that here, when D>0, the three t2g orbitals are lower in energy than the two e2 orbitals, and the electron configuration is t2g5e22, with one of the t2g and both of the e2 electrons unpaired. Now, since the unpaired electrons are not in three degenerate orbitals, the lower energy states have the unpaired t2g opposite spin to the unpaired e2 electrons (which by Hund's rule again are the same spin).
The fictitious spin is interpreted in the same way on this side of the figure, simply meaning that only two states are accessible under the conditions of interest.


What happens in an exothermic reaction at the atomic level?

In a chemical reaction, bonds are broken and bonds are formed. For example, when elemental oxygen and hydrogen form water, we are breaking bonds in the elements and making heteroatomic bonds connecting hydrogen and oxygen atoms:
$$\ce{O2 + 2H2-> 2H2O}$$

To emphasize the bonds, we could also write:

$$\ce{O=O + H-H + H-H -> H-O-H + H-O-H}$$

In a thought experiment, we could first separate the elements into "isolated" atoms, and then recombine them to form water. Bond dissociation energies estimate how much energy it takes to separate bound atoms into "isolated" atoms. In the case of the example reaction, it takes less energy to break all the bonds of the reactants (one double bond connecting oxygen atoms and two single bonds connecting hydrogen atoms) than to break all the bonds of the products (four single bonds between oxygen and hydrogen atoms). As a result, there is an excess of energy that becomes available, resulting in an exothermic reaction.

  What affects the amount of heat energy released?

The reaction follows a more subtle path than completely ripping molecules apart into atoms and then making new molecules. However, because enthalpy is a state function, the result in terms of reaction enthalpy is the same. So the only thing that affects the reaction enthalpy is the set of reactants and products.

The amount of heat depends on how you set up the reaction. If the temperature before and after the reaction is the same, if the pressure is constant and there is no non-PV work, it is equal to the reaction enthalpy. If the reaction is used to do work (electrochemical cell, steam engine etc), if it is a photochemical reaction, if it gives off light, or if any other of the conditions given in the previous sentence are not met, that will affect the amount of heat transferred.

  Which form of energy is converted to heat energy?

The energy "comes from" the different energy of electronic states in the reactants and products. You can call this chemical energy, or a combination of kinetic and potential energy of the electrons. We see that electrons going from a higher energy state to a lower energy state give off energy when considering atomic spectra (sodium giving off a yellow light in the Bunsen burner, for example). In the case of reactions, electronic states are different between reactants and products not because they are in excited states, but because of the different configuration of nuclei giving rise to different electronic states.

  How does this happen?

Unless the reaction gives off energy directly (by emitting electromagnetic waves - rather rare), the available energy gets transformed into kinetic energy of the nuclei (e.g. molecular vibration), and then is transferred to the surrounding by the common ways of heat transfer until the system is at thermal equilibrium again. If the reaction is endothermic, some of the activation energy (provided by molecular collisions) is not "returned to the pool" but instead used to get the electrons into their higher energy states in the products (compared to the reactants), leading to a net heat transfer from surroundings to the chemical reaction.


$\ce{HF}$ reacts with glass ($\ce{SiO2}$).[1]

$$\ce{SiO2 + 4HF -> SiF4 + 2H2O}$$

$\ce{SiF4}$ is not a solid that consists of vertex-connected tetrahedra like $\ce{SiO2}$ but is a gas at room temperature. Technically, $\ce{HF}$ is not a solvent since in this case it reacts with the glass vessel.

According to Spierings:[2]

  $\ce{HF2-}$ ions are adsorbed on surface silanol groups, the $\ce{HF}$ molecules on vicinal silanol groups and $\ce{H+}$ on surface bridging oxygens in siloxane units. [...] These are transformed into surface groups such as $\ce{\bond{#}Si-F}$ and $\ce{\bond{#}Si-O-SiF3}$. The adsorption of $\ce{HF}$ and $\ce{HF2-}$ increases the electronic density on the bridging oxygen in the siloxane unit. This in turn makes these oxygens more basic, so more $\ce{H+}$ ions are adsorbed, which leads to more siloxane bonds being broken per time unit, i.e. a kind of catalytic effect. [...] The catalytic action of $\ce{H+}$ ions on breaking siloxane bonds also occurs in the dissolution of glasses in acidic and weakly alkaline solutions.

Hollemann, A. F.; Wiberg, E. Hollemann–Wiberg: Lehrbuch der Anorganischen
Chemie, 34th ed.; de Gruyter: Berlin, 2007; Vol. 102.
Spierings, G. A. C. M. Wet chemical etching of silicate glasses in hydrofluoric acid based solutions. J. Mater. Sci. 1993, 28 (23), 6261–6273. DOI: 10.1007/BF01352182.


Why is arsenic trichloride more likely to be polar than aluminum trichloride?

It is because $\ce{AlCl3}$ has a trigonal planar geometry (leading to no permanent dipole moment) and $\ce{AsCl3}$ has a pyramidal geometry (leading to a permanent dipole moment).


It turns out somebody already answered the question in a different post:

  Xanthan Gum can handle hydrochloric / muriatic acid (HCl) for a while,
  depending on how concentrated the acid is. Here is how I made an HCl
  gel to clean an old toilet that was heavily coated with lime scale...

  I first dissolved a level teaspoon of xanthan gum into 1/4 cup of
  rapidly stirring water. After a few hours of letting it sit, with
  occasional stirring, I had a very thick, viscous goo. To make the HCl
  gel, I mixed one part of this goo with 1.5 parts of 31% hydrocholoric
  acid, by volume. This gave a gel of a suitable viscosity to get the
  lime dissolved off the toilet. I don't know how long this gel would
  last before the HCl broke down the gum, but my gel lasted several
  hours, which was enough to get the toilet clean.


How many protons does phenolphthalein lose in basic solutions?

ChemSpider phenolphthalein's $\mathrm{p}K_\mathrm{a}$ value: 9.7 (at room temperature). Here they've is only given one $\mathrm{p}K_\mathrm{a}$ value, but it is still reasonable to assume that there are other forms in other pH conditions, like in this image:

(source: Wikipedia)

In other sources like academic ones, it is always shown with 2 forms, and these are associated with the supposedly (especially back titration from pink to colourless is very hard to catch) sharp color change in pH 8-10 range. If one is to use an indicator for that pH region, phenolphthalein is chosen but not for the other regions so no one is interested in possible other forms.

It is also understandable if people actually don't even know the other forms, or try to use them also since pH meter(also possibly UV-Vis spectrometry) is used for a long time to track the pH, I have only seen these indicators in student laboratories for teaching purposes, not in any of research laboratory.


How can the 18 electron rule be used to explain that Fe(η5-C5H5)2 doesn't react with hydrogen but Ni(η5-C5H5)2 does?

Hydrogenation of NiCp2 takes place on the ligand, not the metal, to reduce donor electron count, giving a low-energy closed-shell 18 electron count.

http://en.wikipedia.org/wiki/Nickelocene
http://en.wikipedia.org/wiki/Ferrocene
http://en.wikipedia.org/wiki/Hapticity
http://www.chem.mun.ca/homes/cmkhome/Organometallics_The%20Basics.pdf
   p. 9, 10.  "η" is "hapticity," the number of ligand atoms contributing pi-electron orbitals to the metal center.

$\ce{Cp^-}$ is aromatic (draw the Frost diagram) when coordinated $\ce{η^5}$ on-face, or  $\ce{η^1}$ non-aromatic as a simple anion coordinated on-vertex.  Saturating one localized double bond leaves cyclic delocalized allyl anion coordinated $\ce{η^3}$.

In the case of $\ce{TiCp4}$, two Cp anions are face pentahapto, two are vertex monohapto.  If we take a proton or C-13 NMR, will we see discrete protons or carbons for the $\ce{η^1}$-Cp, or will allowed 1,5-shifts scramble all five protons?  All five carbons?  Will it be temperature dependent if it happens?  If it happens, will H and C have different kinetics?


Is the t-butyl carbocation more stable than the benzyl carbocation?

I am using a very simplistic quantum chemical approach of the following isodesmic reaction:

I have used Gaussian 16 Rev. A.03 and the DF-B97D3/def2-TZVPP level of theory. The summaries of the calculations are included below.

On this level of theory the depicted reaction has an energy change of $\Delta G = \pu{- 37.1 kJ mol-1}.$ Therefore one could assume that the 2-methylpropan-2-ylium cation is more stable than the phenylmethylium cation. These values were estimated at $T = \pu{298.15 K}$ and $p = \pu{1 atm}.$

One certainly can do more calculations, but it is, however, a start.

Calculation summaries

INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch2_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -270.5777462860 hartree
zero-point corr.             (ZPE):            +0.114824 hartree/particle
thermal corr.                  (U):            +0.120571 hartree/particle
ther. corr. enthalpy           (H):            +0.121515 hartree/particle
ther. corr. Gibbs en.          (G):            +0.085614 hartree/particle
entropy (total)            (S tot):              +75.561 cal/(mol K)
heat capacity (total)       (Cv t):              +22.578 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch3_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -271.4854244270 hartree
zero-point corr.             (ZPE):            +0.125049 hartree/particle
thermal corr.                  (U):            +0.131428 hartree/particle
ther. corr. enthalpy           (H):            +0.132372 hartree/particle
ther. corr. Gibbs en.          (G):            +0.093359 hartree/particle
entropy (total)            (S tot):              +82.110 cal/(mol K)
heat capacity (total)       (Cv t):              +23.831 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h10_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -158.4201368210 hartree
zero-point corr.             (ZPE):            +0.128849 hartree/particle
thermal corr.                  (U):            +0.134606 hartree/particle
ther. corr. enthalpy           (H):            +0.135550 hartree/particle
ther. corr. Gibbs en.          (G):            +0.101200 hartree/particle
entropy (total)            (S tot):              +72.297 cal/(mol K)
heat capacity (total)       (Cv t):              +20.521 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h9_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -157.5167928240 hartree
zero-point corr.             (ZPE):            +0.114033 hartree/particle
thermal corr.                  (U):            +0.120557 hartree/particle
ther. corr. enthalpy           (H):            +0.121501 hartree/particle
ther. corr. Gibbs en.          (G):            +0.083653 hartree/particle
entropy (total)            (S tot):              +79.659 cal/(mol K)
heat capacity (total)       (Cv t):              +20.965 cal/(mol K)

(Compiled with tools-for-g09.bash, which somewhat surprisingly works for g16.)

Optimised geometries

14
phch2_q1/b97d3tzvpp.freq.xyz
C          0.08663       -0.19347        0.09384
C          0.05691       -0.11475        1.46208
H          1.02702       -0.19311       -0.45239
H         -0.82919       -0.25883       -0.48899
C         -1.20323       -0.11721        2.16356
C         -1.22180       -0.03857        3.53614
C         -0.00341        0.04387        4.24119
C          1.24453        0.04873        3.58467
C          1.28552       -0.02918        2.21254
H         -2.12409       -0.18167        1.59151
H         -2.16067       -0.03932        4.07940
H         -0.02695        0.10566        5.32621
H          2.15902        0.11360        4.16439
H          2.23039       -0.02773        1.67720

15
phch3_q0/b97d3tzvpp.freq.xyz
C          0.00007        0.00000       -0.04694
C          0.00682       -0.00000        1.46143
H          1.01682       -0.00000       -0.44951
H         -0.51968       -0.88199       -0.43936
H         -0.51968        0.88199       -0.43936
C         -1.19805       -0.00000        2.17854
C         -1.20438       -0.00000        3.57251
C          0.00015        0.00000        4.27958
C          1.20614       -0.00000        3.57927
C          1.20606       -0.00000        2.18275
H         -2.14093        0.00000        1.63572
H         -2.14998        0.00000        4.10835
H         -0.00257        0.00000        5.36594
H          2.14940        0.00000        4.11910
H          2.15082        0.00000        1.64397

14
t-c4h10_q0/b97d3tzvpp.freq.xyz
C          0.00369       -0.00664        0.00284
C         -0.01764        0.03020        1.53527
C          1.44139        0.03012       -0.52791
C         -0.74698       -1.23338       -0.52790
H         -1.04360        0.03406        1.92005
H          0.49077       -0.85020        1.94859
H          0.49182        0.92075        1.92008
H          1.46219        0.03417       -1.62347
H          1.97408        0.92055       -0.17569
H          2.00042       -0.85041       -0.18649
H         -0.76068       -1.24961       -1.62346
H         -0.26410       -2.15775       -0.18616
H         -1.78452       -1.24933       -0.17586
H         -0.51485        0.89146       -0.36388

13
t-c4h9_q1/b97d3tzvpp.freq.xyz
C         -0.00184       -0.00056        0.03904
C          0.01681        1.45949        0.05366
H         -0.46367        1.80346       -0.88071
H          1.00698        1.90257        0.15089
H         -0.66770        1.83073        0.83269
C          1.25179       -0.74789        0.02785
H          1.76765       -0.48991       -0.91708
H          1.93393       -0.35706        0.79738
H          1.13840       -1.82916        0.09679
C         -1.27605       -0.71190        0.03180
H         -1.24998       -1.55839       -0.66820
H         -1.35257       -1.20345        1.02233
H         -2.15046       -0.07921       -0.11824

(g09.chk2xyz does not work for g16.)


Calculating Ratio of Stability Constants

You used all the information from the question except the line that would have helped you solve it

  The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$

From your first cell you find that

$$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 103.97$$

which means that in your second cell

$$\frac{[\ce{Fe(CN)6^4-}]}{[\ce{Fe(CN)6^3-}]} = 103.97$$

and $E^\circ$ for the cell with cyanide is $+0.361~\mathrm{V}$. (You have a typo in your equation, the 4- should be in the numerator and 3- should be in the denominator. Also your equation is missing units but I won't bother with that.) This is quite far off from the accepted literature value (around $+0.430~\mathrm{V}$) but let's move on.

  Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above

No, that is wrong. The ratio of the stability constants refers to that expression you have written, when all of the compounds are placed into the same solution. You cannot simply use the two equations earlier because those describe two separate solutions.

You have to use the difference in $E^\circ$ to work out the ratio of $\beta$.

The way to do this would be to realise that if you take the difference between the $E^\circ$ you get

$$\begin{align}
\ce{Fe^3+ + Fe(CN)6^4-} &\longrightarrow \ce{Fe^2+ + Fe(CN)6^3-} & E^\circ &= +0.410~\mathrm{V}
\end{align}$$

The equilibrium constant for this reaction is the quantity that you are looking for

$$K = \frac{[\ce{Fe^2+}][\ce{Fe(CN)6^3-}]}{[\ce{Fe^3+}][\ce{Fe(CN)6^4-}]}$$

and so

$$K = \exp\left(\frac{-\Delta_\mathrm{r} G^\circ}{RT}\right) = \exp\left(\frac{-nFE^\circ}{RT}\right) = 8.589 \times 10^6$$


"how will you make a cell from one electrode?".

There is no need to, because there is nothing which prohibits you from writing a Nernst equation for a half cell.* Remember the rule no.1 in electrochemistry, whenever you mention half-cell potential, your next automatic question should be "with respect to what?". If the textbook is quiet and you are required to look up tables, it is understood that it is with reference to the hydrogen electrode under standard condition which is set at 0 V (exactly).

Now at pH 10, your standard conditions have changed and the activity or hydrogen ion concentration differs. It is no longer zero. So proceed with the Nernst equation for the half cell and work it out.

*(Yes, there is no single electrode galvanic cell, but for fun and general knowledge, people have done single electrode reduction using electrostatics (Bard and others). Faraday did it centuries ago. It was published in a very reputable magazine.)


Van't Hoff Equation with changing Enthalpy

There is. Suppose that at 298 K $$\Delta G^○_{298} = -RT\ln{K_{p\,298}} = \Delta H^○_{298} - T\Delta S^○_{298},$$ and if both $\Delta H^○$ and $\Delta S^○$ depend on $T$, $$-RT\ln{K_p} = \Delta H^○_{298} + \int_{298}^T \Delta C_pdT - T\Delta S^○_{298} - T\int_{298}^T \frac{\Delta C_p}{T}dT.$$ Then both sides of it can be divided by $T$ and knowing that $\int udv = uv - \int vdu$, where in our case $v = 1/T$ and $u = \int_{298}^T \Delta C_pdT$, we can "simplify" the last equation: $$-R\ln{K_p} = \frac{\Delta H^○_{298}}{T} - \Delta S^○_{298} - \int_{298}^T \frac{\int_{298}^T \Delta C_pdT}{T^2} dT.$$ I'll not endeavor to differentiate that with respect to $T$, but now you can probably figure out the rest. Since $C_p$ is usually a not-that-complex polynomial of $T$, the math should be rather straightforward.


Role of 10-camphor sulfonic acid as a catalyst in protecting group chemistry

For a given protecting group, there is no single best way of introducing it, just as there is no single best way of oxidising an alcohol – the conditions you use depends upon the substrate you have and the tolerance of it towards things such as acid, base, and heat.

In your question you ask specifically about conditions to introduce a benzylidine acetal, namely why aqueous acids aren't used – the tl;dr answer is that they are!

Below is an excerpt from the section in Greene's Protecting Groups in Organic Synthesis that deals with formation of such a group:

  PhCHO, ZnCl2, 28°C, 4 h.
  PhCHO, DMSO, concd. H2SO4, 25°C, 4 h.
  PhCHO, TsOH, reflux,  H2O, 72% yield.
  Sulfuric acid on silica gel, PhCH(OMe)2, 75–93% yield.
  PhCH(OCH3)2, I2, CH3CN, 60–89% yield.

  Taken from: Greene's Protecting Groups in Organic Synthesis

What all of these conditions have in common is that some kind of acid catalyst is required. Mineral acids (sulfuric), Lewis acids (zinc chloride or iodine) and organic acids (tosic acid) have all been shown to work in the literature, and indeed most people involved in complex molecule synthesis will have tried most of them.

Which condition is most appropriate is a matter of context (and often a lot of screening). Solid organic acids (tosic acid, camphor sulfonic acid, pyridine para-toluene sufonate) are often used in late-stage synthesis as they're able to be dried, this allows the reactions to be conducted under (almost) anhydrous conditions. Use of sulfuric acid dictates that water (quite a lot of it) is present, which can knock off other delicate protecting groups/functionality in a molecule.


Pressure-Volume work is a type of mechanical work .
There are also non-mechanical forms of work where in pressure and volume terms are not involved .
The most famous example is electrical work . The work done to move electrical charges ( like motion of electrons in a conductor) comes under this category .


Why do Gas Laws only use Kelvin Temperature?

First you must know how was Charles' discovered his law?
See on wikipedia Jacques Charles

Here is a quote from Wikipedia

  Around 1787 Charles did an experiment where he filled 5 sites to the same volume with different adds. He then raised the temperature of the sites to 80 °C and noticed that they all increased in volume by the same amount. This experiment was referenced by Gay-Lussac in 1802 when he published a paper on the precise relationship between the volume and temperature of a gas. Charles' law states that under constant pressure, an ideal gas' volume is proportional to its absolute temperature. The volume of a gas at constant pressure increases linearly with the absolute temperature of the gas. The formula he created was $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

As all statements of experimental laws are completely based on the results of experiment performed and cannot be altered. Charles'law is also an experimental law, and it was discovered from the experiment performed by Jacques Charles and he found that the Volume of a ideal (close to Ideal) gas is directly proportional to absolute temperature.

You are saying that

  Why volume can't be proportional to temperature in Celcius?

But how it can be if it is not the result of the experiment.

Suppose a cricketer just scored six and you are asking why can't we write his score as four instead of six. In the same way Charles discovered that Volume is directly proportional to absolute temperature and you are saying why it is not temperature in Celcius.

If you want to know why volume can only be proportional to absolute temperature - experiment is the proof.


Elements interact with the rest of the world through their electrons. How those electrons interact with other atoms (or electromagnetic radiation) determines how that atom behaves.

You see this with isotopes. Different elements are distinguished from each other by the number of protons in their nucleus. Changing the number of neutrons has little effect on their properties. Why is this? It's not the protons themselves which change the properties, instead it's how the change in total positive charge of the nucleus changes how many electrons the atom can hold and how they behave that changes the properties. Different numbers of electrons - and specifically different numbers of electrons with respect to the net charge on the atom - change the properties of the atom. Changing the number of neutrons changes the nucleus, but has little effect on how the electrons behave, so has little effect on the chemistry of the atom.

So what does that have to do with "periodicity" of the periodic table? Well, there's a structure  to how the electrons are "stored" in an atom. These are orbitals. For a bunch of quantum mechanical reasons that are rather advanced for this question, there are repeating patterns in these orbitals. Most particularly, in the l quantum number which determines the shape of the orbital. Each period corresponds (roughly) to the n quantum number. For each n, there are a set of ls which go along. (Because of tradition, orbitals with different l quantum numbers get letter labels: s, p, d, f, ...)

This is why you see periodic trends. Having exactly filled s and p orbitals is the most stable, and each n has its own set of s and p. So each time you get a filled s and p, you get an inert noble gas. For halogens (like chlorine and bromine), you need just one extra electron to fill the s and p orbitals, so they really like taking a single electron from other elements. In contrast, the alkali metals (like sodium and potassium) can most easily get to a completely filled s and p orbital state by losing a single electron, resulting in their propensity for a positive charge. It's how close the atoms are to filled s and p orbitals - regardless of their n quantum number - and what it takes to get there which provides the bulk of the "periodicity" of the periodic table.

It turns out that the s and p orbitals are the most "exposed", so what's going on there contributes the majority of the behavioral change between elements. Elements that differ only in how the electrons in their d orbitals are situated have much less property differences (this is why most transition metals have similar properties). Elements which differ only in f orbitals (like the Lanthanides) are even more alike. There certainly are differences here, but they tend to rely on more subtle effects than those that cause the alkali/chalcogen/halogen/noble gas differences.

Note that this sort of orbital-based periodicity breaks down as you get lower in the periodic table. That's because you actually start to get relativistic effects, and how the orbitals behave starts to change. Periodic trends are just that - trends - and aren't absolute.


I can tell you how to find something that will make tracing paper transparent, but I don't have an example to hand.

Tracing paper is partially opaque because it is "rough" and contains many small fibres of cellulose that scatter light. These scatter light effectively because of their size and because their refractive index is very different from air. If you can find a liquid which has a similar refractive index, that will make the paper much more transparent than in its natural state.

I'm sure their are some, but I don't know any right now.


How long does a linear hydrocarbon chain have to be before we call it polyethylene?

Actually, polyethylene is still an alkane as it fits IUPAC definition perfectly fine [1, p. 1313]:

  alkanes:
  Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms.

Smaller alkane molecules you are referring to, are oligomers. IUPAC defines the prefix oligo- as follows [1, p. 1353]:

  oligo:
  A prefix meaning 'a few', and used for compounds with a number of repeating units intermediate between those in monomers and those in high polymers. The limits are not precisely defined, and in practice vary with the type of structure being considered, but are generally from 3 to 10.

At the same time removing numerical restraints when referring to oligomer molecules [2, p. 2289]:

  1.2 oligomer molecule
  A molecule of intermediate relative molecular mass, the structure of which essentially comprises a small plurality of units derived, actually or conceptually, from molecules of lower relative molecular mass.

Obviously, the boundary between oligomers and polymers is blurred. Usually for the emergence of "polymer" sufficient degree of polymerization $N$ starts with $10^2 - 10^4$ for synthetic polymers and about $100 - 200$ for globular proteins.

A significant feature of any polymer as compared to its molecular weight counterpart is that when one or several units are added or removed into the polymer molecule, the properties of the polymer material remain almost unchanged. Properties of oligomers substantially depend on molecular weight.

As a typical example, melting points of low molecular weight alkanes gradually increases with polymerization degree (Source):

reaching a plateau of approx. $\pu{400 K}$ for linear low density polyethylene.

References

Moss, G. P.; Smith, P. A. S.; Tavernier, D. Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995). Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375. https://doi.org/10.1351/pac199567081307.
Jenkins, A. D.; Kratochvíl, P.; Stepto, R. F. T.; Suter, U. W. Glossary of Basic Terms in Polymer Science (IUPAC Recommendations 1996). Pure and Applied Chemistry 1996, 68 (12), 2287–2311. https://doi.org/10.1351/pac199668122287.


Visualizing rotation about a double bond

Adding one more element to help you long-term: I think the underlying reason you are having a problem with this is that the visualization you are using is a little misleading.

You're selecting an atom with your red dot.  You can't rotate around an atom. You can rotate around the bonds between the atoms, though, assuming that the constraints of that bond allow it.

So in the first example, there are two possible carbon-carbon bonds you can rotate around that would both correspond to that red dot. Your dot should be on the sigma bond, based on your diagram.

In the second, you have three carbon-carbon bonds connected to your red dot. The one you are rotating is the double bond, not either of the bonds in the ring.  So the dot should instead be on that bond.

If you change the way you're marking the connection in your head, it could help you think about the nature of that connection, and that could help you with the rest of the information everyone is providing.


Nomenclature of ester with cyclic substituents

The names of esters are usually formed by placing the alcoholic component in front of the name of the acid component as a separate word. The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-65.6.1 General methodology
Neutral salts and esters are both named using the name of the anion derived from the name of the acid. Anion names are formed by changing an ‘-ic acid’ ending of an acid name to ‘-ate’ and an ‘-ous acid’ ending of an acid name to ‘-ite’. Then, salts are named using the names of cations, and esters the names of organyl groups, cited as separate words in front of the name of the anion.

Since the acid component of the ester that is given in the question is cyclobutanecarboxylic acid, the corresponding ester is named as … cyclobutanecarboxylate.
However, that does not completely answer your question why the extra carbon atom is named using the ‘carboxy’ part.
Simple carboxylic acids are named substitutively using the suffix ‘oic acid’ as follows.

P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

Thus, the ‘carboxy’ part is not used in such simple cases. For example, pentanoic acid is not named ‘butanecarboxylic acid’.
However, the acid component of the ester that is given in the question is not such a simple case since the $\ce{-COOH}$ group does not terminate the chain of the parent structure (the cyclobutene ring). In such cases, the suffix ‘carboxylic acid’ is used instead.

P-65.1.2.2 The suffix ‘carboxylic acid’ is used for all carboxylic acids not covered by P-65.1.2.1, except for benzoic acid, a retained name (see P-65.1.1.1). The carboxy group can be attached to any atom, carbon or heteroatom, of any parent hydride; (…).


I think the key feature is the surface tension of mercury-air which is extremely high (about 7 times water-air which is already high), combined with the chemical nature of the sponge, which is typically cellulose.

Since cellulose is a big polysaccharide it has a reasonably favorable interaction with both water (due to the OH groups) and alcohols and alkanes (due to the carbon chain). This causes the contact angle of water, alcohol and alkanes with the sponge to be low (at least lower than 90 degrees) thus allowing capillary action to suck the liquid into the sponges pores. The metallic nature of mercury has an unfavorable interaction with the sponge thus resulting in a contact angle above 90 degrees. This will not allow capillary action because the cosine of angles above 90 degrees is negative, therefore the sponge cannot soak up the mercury.

A sponge made out of metal (silver is common I believe) would be able to soak up the mercury because it has a favorable interaction in terms of surface energy.


What explains the case hardening color effect?

Case hardening (carburizing or induction or flame) does not color steel. You are seeing “temper” colors which are unrelated to case hardening. It is caused by different thickness of very thin layers of iron oxide. These colors were developed during tempering heat-treatment after case hardening. Starting with yellow, then blue and purple in a continuous transition. Faint yellow starts about 300 °F, longer times give more color development of color.
The same colors also develop on stainless steels and titanium at different temperatures and times. A stainless motorcycle exhaust pipe will show the range of colors as temperature changes along its length. If a titanium weld shows yellow color it has been exposed to air at elevated temperature (and is likely defective because of absorbed oxygen and nitrogen).


Why is carbon dioxide bent in its first excited state?

This effect is the so-called Renner–Teller effect and is a consequence of a coupling between vibrational and electronic motion and thus a breakdown of the Born–Oppenheimer approximation. You might be familiar with the Jahn–Teller effect which is related to the Renner–Teller effect. In degenerate states of linear molecules with three or more atoms, the total energy of the molecule can be reduced by bending the molecule. Consider a three atomic linear molecule in a degenerate state (such as $\ce{CO2}$ in its excited $\Pi$ state). When the molecule bents, the degeneracy is lifted as it can bend in the $x$ or $y$ direction. Because of symmetry requirements, the potential has to be an even function of the bending coordinate. Ignoring the lifting of the degeneracy we may approximate the zero-order potential with a Taylor expansion as
$$
V^0=ar^2+br^4+\cdots,
$$
where $r$ is the bending coordinate and $a$ and $b$ are constants describing the potential ignoring the vibronic interaction. When we do consider the vibronic interaction, the degenerate potential splits in two, but the splitting must still be an even function of $r$.
$$
\Delta V=\alpha r^2 + \beta r^4 + \cdots
$$
If the interaction is so large that $a<\alpha/2$, the potential will not have its minimum at $r=0$, but will have two minima at $\pm r_e$ and the molecule has a lower energy when it is bent.
See also Herzberg Vol 3, p 26.


Is there any way to determine if the compound is oxide or peroxide or superoxide?

generally the most universal way to prove this is X-ray crystallography. It allows to obtain crystal structure of target compound directly. It has, however, an inherent flaw: it can't reliably detect position of hydrogen atoms, so in some cases some uncertainty may remain.

In case obtaining crystal is impossible, the second option is NMR, that allows to get various info about atom's neighborhood, but interpreting NMR data is really hard. In case of superoxide, EPR may prove direct evidence.

Generally there is no way to detect such things from brutto-formula. For example, compound $\ce{CH2O2}$ may be either dioxirane (methylene peroxide) or formic acid. In practice, however, peroxides are really unstable, so in most cases if compound is not overloaded with oxygen (like, $\ce{H2SO5}$) and is reasonably stable towards thermal decomposition, it is most likely oxide and not peroxide. Peroxides are usually formed either from another peroxide or electrochemically, or (in rare cases) via cold oxidation with molecular oxygen. And there are exactly three common superoxides.


Entropy is a state function, which means its value does not depend on how it gets from state A to state B, rather it only depends on the final state B itself.  In this case, we can imagine a process which takes the system from state A to state B via a reversible process, and your expression for the change in entropy would be correct:
$$\Delta S = \frac{q_\text{rev}}{T_\text{sys}} - \frac{q_\text{rev}}{T_\text{surr}}$$
where I have explicitly labelled the heats as reversible heat flows (this quantity does depend on the path taken).
Even if in reality such a reversible path would not be taken, the final change in entropy would still have the same value, so it is still valid to use that expression.


Are intermolecular forces a type of chemical bond?

The IUPAC definition of "chemical bond" is:

  When forces acting between two atoms or groups of atoms lead to the formation of a stable independent molecular entity, a chemical bond is considered to exist between these atoms or groups. The principal characteristic of a bond in a molecule is the existence of a region between the nuclei of constant potential contours that allows the potential energy to improve substantially by atomic contraction at the expense of only a small increase in kinetic energy. Not only directed covalent bonds characteristic of organic compounds, but also bonds such as those existing between sodium cations and chloride anions in a crystal of sodium chloride or the bonds binding aluminium to six molecules of water in its environment, and even weak bonds that link two molecules of $\ce{O_2}$ into $\ce{O_4}$, are to be attributed to chemical bonds.

So the answer is "yes" in some cases.

See also the IUPAC definition of "molecule":

  ... must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state

So, for example, a water-water dimer, held together by hydrogen bonding, has a monomer-monomer potential energy surface that is deep enough to confine at least one vibrational state, and it would be appropriate to refer to the hydrogen bond as a chemical bond.


Doubts Regarding Ceric Ammonium Nitrate Test for Alcohols

The question does not mention in what the ammonium cerium nitrate is dissolved in, but I guess it is water. If so, one expects that sufficiently large compounds that one wishes to test are not sufficiently soluble in water, such as decanol. Obviously, this is not entirely correct because disaccharides such as lactose do are soluble in water.

I expect a sufficient solubility to be important for the test such that it finishes in a timely manner and does not become inconclusive due to prolonged reactions with air, water etc.


AKM: In your response to MM the self-condensation of diacetyl 1 occurs stepwise and not with the double enol of diacetyl 1 (biacetyl; butan-2,3-dione). Although your question relates to the acid-catalyzed condensation of diacetyl, historical condensations of diacetyl were conducted under alkaline conditions. The products formed from the base or acid-catalyzed are numerous and dependent on reaction conditions.
Quinone 2 was prepared under basic conditions by von Pechmann.1 Diels and coworkers2 isolated from an alkaline reaction a dimer, C8H12O4, which was assigned lactonic structure 3 with no spectroscopic evidence.3 The structures are color-coded to display the four carbon moiety of diacetyl. Clearly, lactone 3 shows a branched red carbon framework. An undefined "benzylic acid rearrangement" was suggested without details. The structure was later revised to diketone 4 based upon spectroscopic evidence and a more reasonable normal carbon chain.4
Trimer 5, C12H18O6, has also been isolated under alkaline conditions and the stereochemistry confirmed by NMR and IR.5
An acid-catalyzed condensation of diacetyl in dilute aqueous acetic acid (70oC, 13 days) leads to C10 structure 10.6 The structure was confirmed spectroscopically and by periodate cleavage with loss of acetic acid with concommitant formation of diketone 11. Structure 10 does not form a hemiketal as does structure 5 owing to the resonance stabilizing effect of the α, β-unsaturated ketone. The mechanism proposed is described in the sequence of steps 4 --> 10.
As can be seen from the examples provided --- and that's not all of them --- the self-condensation of diacetyl is not straightforward.

1) Pechmann, H. v., Berichte, 1888, 21, 1417; Otte, R. and Pechmann, H. v., 1889, 22, 2115.
2) Diels, O., Blanchard, W. M., Heyden, H. v. d., Berichte, 1914, 47, 2355.
3) Machell, G., J. Chem. Soc., 1960, 683.
4) Birch, A. J. and Moye, C. J., J. Chem. Soc., 1957, 412.
5) Cresswell, R. M., Smith, W. R. D. and Wood, H. C. S., J. Chem. Soc., 1961, 4882.
6) Shapiro, R., Hachman, J., Wahl, R., J. Org. Chem., 1966, 31, 2710.


Quantify the effectiveness of baseline correction in Raman spectra

Could you also mention the correction method? Your auto-fluorescence background is really huge and your signal to noise ratio is low. One of the most cited methods in automated background correction is indeed based on polynomial fitting, albeit little but more sophisticated than just least square fitting. This article is worth reading.

Automated Method for Subtraction of Fluorescence from Biological Raman Spectra
https://journals.sagepub.com/doi/abs/10.1366/000370203322554518

Another simple option is that you can choose a spectral range where to apply a polynomial baseline. Say from 0 to 1200 you apply a polynomial of degree 5, but 1200 onward you choose polynomial of degree 6. However, you have thousands of spectra and it is not clear whether your background shape is similar in each case or not.

Addendum:  I am afraid there is no solid way to compare the baseline correction approaches. The reason is that you do not know the "true" model of the autofluorescence. This is also stated by Ed in comments. He is a leading spectroscopist. In such cases, you should simulate a baseline like these authors and do a comparison. Make a synthetic baseline by polynomials with say n=5, make it more complicated by mixing certain regions with polynomial  n=6, 3, 2, and then try your approaches. Don't add any peaks to the baseline, and just check which technique is showing least residuals. Is there a trend in the residuals? The trend is also important. The residuals should be scattered. What technique did you use in your case?

Alternatively, you can try a known fluorescent compound, Rhodamine G, and collect a fluorescence spectrum. In this case you will know the true shape of the spectrum. Use that as a true model and try your baseline corrections.

One general advice given to us by a professor, who was an author of Raman spectroscopy monograph, was that a better and better fit does not indicate reality. The model may be completely wrong but may have a correlation coefficient close to unity. Thus fitting a polynomial or any experimental baseline correction is just a utilitarian approach, which is changing the aesthetics and cosmetic appearance of a Raman spectrum. There is nothing fundamental going on.


HEC works ok, but it's not that easy to use. You have to hydrate in pure water for a few hours or more, ideally with stirring, and then incorporate the alcohol a little at a time with high shear mixing. You can use a blender, though it will incorporate a lot of air. If you use too much HEC, the "mother" gel (the HEC and water that will become 30% of your final product) will be very firm and it will be difficult to avoid lumps throughout, and the line is very thin between this and ending up with an unusually thin product. When I'm using this, I err on the thin side and use about 0.8% w/w HEC, so the "mother" is 2.67%.

Some other polymers are much easier to work with if you can get hold of them, notably Sepimax Zen aka. GelMaker PH (Polyacrylate Crosspolymer-6). These can be added directly to a 70% isopropanol or ethanol mixture and need no further ingredients to activate or correct the pH afterwards.

I have not tried aloe vera but I think it would be difficult to get a gel with 70% alcohol. I know lots of YouTubers and bloggers are recommending it. Success may depend on what other thickeners have been used in the aloe vera product. I've seen some videos where the result was the extremely thin liquid with a bit of stickiness that you would expect from the recipe.


Have we found stable, super-heavy elements?

There already is an element, heavier  (higher atomic and mass numbers) than all naturally occurring ones, that is stable enough for everyday use.  You are probably using it, well, everyday, in your smoke detector.

It's americium, element 95, with three isotopes having 100+ year half-lives.  Americium-241 has a half-life of over 400 years, for americium-243 it's over 7000 years and there is even an odd-odd isomer, with mass number 242, having a half-life over 100 years.  (https://en.wikipedia.org/wiki/Americium).

This is not perfect (and whether it's called "superheavy" depends on definition), but the americium is out-lasting other components of smoke detectors, or industrial gauges where americium is also used.  In that sense we can indeed call the most long-lasting americium isotopes stable.


Why doesn't hydrogen peroxide decompose into hydrogen and oxygen?

We have two theoretical decomposition pathways:

$$\ce{H2O2(l) -> H2O(l) + 1/2 O2(g)} \tag{1}$$

$$\ce{H2O2(l) -> H2(g) + O2(g)} \tag{2}$$

The required data (all at $298~\mathrm{K}$) is taken from the appendix of Atkins' Physical Chemistry, 9th ed:

$$\begin{array}{cccc}
\hline
\text{Species} & \Delta_\mathrm{f} H^\circ\mathrm{~/~kJ~mol^{-1}} & S^\circ_\mathrm{m} \mathrm{~/~J~K^{-1}~mol^{-1}} & \Delta_\mathrm{f} G^\circ\mathrm{~/~kJ~mol^{-1}} \\
\hline
\ce{H2O2(l)} & -187.78 & 109.6 & -120.35 \\
\ce{H2O(l)} & -285.83 & 69.91 & -237.13 \\
\ce{H2(g)} & 0 & 130.684 & 0 \\
\ce{O2(g)} & 0 & 205.138 & 0 \\
\hline
\end{array}$$

The short way

The standard Gibbs free energy change of a reaction ($\Delta_\mathrm{r} G^\circ$) can be calculated from the individual standard Gibbs free energies of formation exactly analogously to the enthalpy change:

$$\Delta_\mathrm{r} G^\circ = \sum_J \nu_J[\Delta_\mathrm{f} G^\circ(\ce{J})]$$

So, for reaction 1,

$$\begin{align}
\Delta G^\circ_1 &= [1(-237.13) + \frac{1}{2}(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\
&= -116.78~\mathrm{kJ~mol^{-1}}
\end{align}$$

For reaction 2,

$$\begin{align}
\Delta G^\circ_2 &= [1(0) + 1(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\
&= +120.35~\mathrm{kJ~mol^{-1}}
\end{align}$$

Clearly reaction 1 is favourable and reaction 2 isn't.

The long way

There's absolutely no need to do this, but you could use the "extra" data given above to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for both reactions. You'd get:

$$\begin{align}
\Delta H^\circ_1 &= [1(-285.83) + \frac{1}{2}(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\
&= -98.05 ~\mathrm{kJ~mol^{-1}} \\
\Delta H^\circ_2 &= [1(0) + 1(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\
&= +187.78 ~\mathrm{kJ~mol^{-1}} \\
\Delta S^\circ_1 &= [1(69.91) + \frac{1}{2}(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\
&= +62.879 ~\mathrm{J~K^{-1}~mol^{-1}} \\
\Delta S^\circ_2 &= [1(130.684) + 1(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\
&= +226.222 ~\mathrm{J~K^{-1}~mol^{-1}}
\end{align}$$

If you're only interested in the final result, you don't need to do all this at all. But it does give you some insight into the difference between the two reactions: reaction 1 is enthalpically favoured whereas reaction 2 is entropically favoured.

Generally speaking, enthalpy usually plays a larger role in determining $\Delta G^\circ$. However, it is not a very good idea to generalise this statement and compare reactions solely on the basis of their $\Delta H^\circ$.

Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, and taking $T = 298~\mathrm{K}$,

$$\begin{align}
\Delta G^\circ_1 &= -98.05 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+62.879 ~\mathrm{J~K^{-1}~mol^{-1}}) \\
&= -116.79~\mathrm{kJ~mol^{-1}} \\
\Delta G^\circ_2 &= +187.78 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+226.222 ~\mathrm{J~K^{-1}~mol^{-1}}) \\
&= +120.37~\mathrm{kJ~mol^{-1}} \\
\end{align}$$

consistent with our earlier findings.


Why are acid-base reactions so exothermic?

One can think of it as simply because of electrostatic attraction. Basically, an acid produces $\ce{H+}$ ions in water while a base produces $\ce{OH-}$ ions in water. So when you mix an acid and a base (e.g. sulfuric acid and sodium hydroxide) in a beaker, you get a solution containing $\ce{H+}$ and $\ce{OH-}$ ions. So, when these ions are far apart, there exists potential energy between them (because opposite charges attract each and want to come closer together). So when these ions come closer together, the attraction between them becomes strong enough that a bond forms between the ions to form $\ce{H2O}$. During this process, the initial potential energy between the ions decreases and is converted to heat energy. Hence this is why it is exothermic.

So generally speaking, energy is released when a bond is formed, while energy is required (energy is absorbed) to break a bond. So in the case of acid and base reactions, no bond is being broken, bonds are only being formed. So it must be an exothermic reaction then.


Comparision of ONO Bond angle in NO2 and N2O4

The bond angle in NO$_2$ is $\approx 134^\mathrm{o}$ and the NO bond length 120 pm, in $\ce{N2O4}$ the N-N bond is very long 175 pm and the angle $\approx 126^\mathrm{o}$.

Nitrogen dioxide is paramagnetic with the odd electron in the $\sigma$ bonding sp$^2$ orbital in the 'third' lobe as it were with the O atoms occupying the other two. Additionally there are 4 electrons in the $\pi$ system, two bonding and two antibonding. NO$_2$ also has several low lying excited states which greatly complicates matters.

Joining up two NO$_2$ naturally leads to $\ce{N2O4}$ which is planar and has a very long single $\sigma$ N-N bond. Why the molecule is planar is not clear as it might be expected to twist to remove oxygen-oxygen repulsion. Intuitively one would expect an MO to extend over the whole molecule and so prevent twisting, but the NN bond is very long and single.

Hopefully someone has more up to date data, otherwise you will have to search the literature for MO based (i.e. numerical quantum/chemistry) approaches to describe the bonding.


Rate of change of reaction rate of 0th, 1st, 2nd order reaction?

The logarithm of the concentration of one of the components will vary linearly. However, the concentration does not vary linearly with time, only its logarithm does.

Reaction Rate $= -\frac{d[A]}{dt} = kA = $ constant * [concentration] for a first order reaction.

If you integrate this using separation of variables you end up with

$\ln \frac{[A]_t}{[A]_0} = -kt$

if you plot $\ln[A]_t$ verse time the slope will be constant.

If you plot concentration as a function of time it will not be constant because we can rearrange the above equation to get

$[A]_t = [A]_0 \exp(-kt)$

which is clearly exponential and not linear.

So yes, for a first order reaction, the reaction rate is constant because it depends linearly on the reaction rate constant which is... constant because it is the slope of the logarithm of concentration verse time. Here is a useful link:

www.chem.libretexts.org/reaction_rates

Here is a short answer to a similar question that is useful: chemistry.stackexchange.com


Finding the mass of NaNO3 in original mixture from preciptate

The dried precipitate is $\ce{Ag2S}$.
The number of moles of $\ce{Ag2S}$ is:
$$n=\frac{m_{\ce{Ag2S}}}{M_{\ce{Ag2S}}}=\frac{0.44}{32+108\times 2}=\frac{0.44}{248}=0.00177\,\mathrm{mol}$$
The number of moles of ion $\ce{Ag(I)}$ is$$n'=2n=0.00355\,\mathrm{mol}$$

On the other hand, the percentage by mass of $\ce{AgNO3}$ in the original mixture is: $$\frac{m_{\ce{AgNO3}}}{m_{\ce{(AgNO3 + NaNO3)}}}\times 100=\frac{n'\times M_{\ce{AgNO3}}}{m_{\ce{(AgNO3 +NaNO3)}}}\times 100=\frac{0.00355 \times (108+14+16 \times 3)}{1.61}\times 100=37.48\%$$

So, the percentage by mass of $\ce{NaNO3}$ is: $100-37.48=62.52\%$


Copper reduced from solution is readily oxidized by air.

When I cared, I made a moderately fine copper powder from citrate. Threat copper salt solution with slight molar excess of sodium (potassium) citrate solution (can be made on site from corresponding hydroxide and easily available citric acid), shake and let the solution stay for a while. It immediately turns deep blue, but the precipitate may appear several minutes later. Filter off the precipitate and then heat on a sand bath at ~200-300 Celsium in a vial plugged with cotton wool. I didn't check if evolved gases are toxic, so use a fume hood or good ventilation. The citrate goes through several changes in color, but eventually produces metallic copper powder. Cool it with wool in. The powder seems to be mildly (inconsistently) pirophoric, so store in a small air-tight container.

Hydrogen on it's own certainly can reduce copper, but the process is neither fast, nor pragmatic.


How to determine fire hazard of heating ethylene glycol to dissolve borax and boric acid

Chlorates, perchlorates, permanganates and other strong oxidants conflagrate even at room temperature with ethylene glycol.

Boric acid and borax, however, are not considered strong oxidants and should probably be safe to dissolve below the boiling point of water. That said, I would test a small batch in a safe place, to be sure there is not a large gain in temperature on dissolving.

BTW, you might be interested in reading of the preservation of the Swedish warship Vasa in polyethylene glycol, and long-term issues with acidification. N.B. Polyethylene glycol is generally considered innocuous, but ethylene glycol is toxic if ingested.


It seems this name should not be in the blue book. It is an error.

The following instruction is what Corrections to Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013 has given for your problem:

  Page 82, P-15.3.2.4.1, example 4 on this page.
  Delete 1,1′-(2-ethenyl-1,2,3-trimethyldisilazane-1,3-diyl)bis(ethan-1-one).

As Igors Mihailovs (OP) pointed out in comment section to the question, the Recommendations 2013 are not error-free. You may see how nany errors have been corrected in this web page. For example, page 75 has 4 errors in that page alone!


How can you build a model of tetrahedral coordination from objects found at home?

Inflate balloons, and tie them «at their stem» like a bouquet of flowers.  If you take four of them, not too much inflated, you well demonstrate a situation close to $sp^3$ hybridization. These models equally work well in larger lecture halls by the way, and intentionally using different colors allows many options.

(source)

(screen photo

You need some worked examples?  See videos like this or this.

You need a scientific paper? Well, there are as well, e.g., this (open access) expanding the picture to extended $\pi$-systems:


Urban air pollution and the smell of freshly cut lawn grass?

The information page from the Compound Interest page of the same name describes it well (below).

A summary of the main points are that grass already emits small amounts of VOCs (volatile organic compounds) - mostly comprising of 6-carbon atom based aldehydes and alcohols.

When the lawn is cut, the main smell is from (Z)-3-hexenal, at which only 0.25 parts per billion can be sensed (smelled) by humans - this compound is short lived, hence why the smell is a freshly cut smell.

As a little twist, in research presented in the paper Air Pollution and the Smell of Cut Grass, the authors suggest that a significant proportion of aerosol VOC's in urban air pollution actually originate from the release of these chemicals when lawns are cut - and given the sheer number of manicured lawns in an urban centre, this starts to make sense.

Some of these released chemicals are highly photoreactive. The flux of VOCs is seasonal as well in many places, contributing to seasonal variations of VOC-based photochemical smog.

Within the article, VOC flux measurements taken in the urban area of Melbourne Australia indicate that 6% of the flux arises from uncut grass; however, this flux doubles when the lawn is cut. The combined flux is comparable to industrial sources.

It is suggested in the article, that

  within an
  urban airshed, during the growing season, about one third of the emissions of
  photochemically reactive VOCs come from grass and cut grass.


Why is pyridinium chlorochromate selective compared to chromate, dichromate etc.?

Though pyridinium chlrochromate (PCC) is a salt, the pyridnium cation is easily dissolved by a wide variety of organic solvents. As such, the chlorochromate anion then becomes dissolved in the organic solvent, and can oxidize $1º$ and $2º$ alcohols present in solution to carbonyls. Oxidation of carbonyls to carboxylic does not occur when using PCC for the very reason you said yourself—no water is present to hydrate the carbonyl species to its geminal diol to allow for further reaction.

Other benefits of PCC include that it is not particularly hygroscopic, is commercially available, and can be stored for some time, though other methods (Swern oxidaiton, Moffatt oxidation, Dess-Martin Periodinane) are generally preferred as the reaction workup for PCC can be tedious and the chromium(IV) species produced is toxic.


How can one optimise the formation of copper acetate?

Your question and title do not correspond, IMHO. I'll answer one of both:

Vinegar is mostly water, with some acetic acid ($5\%$). A classic experiment, performed by a lot of kids at school, is to mix vinegar with baking soda. The baking soda neutralizes the acid and produces carbon dioxide. The carbon dioxide is what you're looking for: it's the sign that the reaction is taking place.

Pour a pinch of baking soda inside your solution. If it makes bubbles on contact, your solution still contains a lot of vinegar. If you don't see any bubbles, even when stirring, most of the acid has been consumed.


Electroplating copper yields rough surface?

There are few things that you can improve.

First, beads are forming because your copper anodes are not pure copper. Beads are forming around impurities. You can try to remove (filter) impurities by firmly wrapping anodes in paper tissue or better with water filtration polypropylene felt filter so only Cu ions can pass into solution. You can after that collect impurities already present in solution by electroplating few hours on some scrap piece of wire.

Second, holes should be drilled cleanly. You should invest into carbide drill bits and good drill stand.

Third, you should also think about adding Cl ions to solution, NaCl or HCl in small quantities. Plated surface will be shinier...

Best regards,
Pavle


What is the density of nitrogen at standard conditions?

One possibility for differences in measured density of an element is isotopic variation. For example, sulfur isotope abundance may vary, depending if the element is derived from soil, from sulfide minerals or as the native element from a volcanic vent. Nitrogen has two common stable isotopes: 14N and 15N. It is possible that the measurement of density varied because of natural variation.

However, it is possible the density varied because the definition of "Standard Temperature and Pressure" (STP) has changed over time (e.g., 101.325 kPa pre-1982, 100 kPa afterwards).

It is possible the density measurements varied because the definition of mass has changed over time.

It is possible the density measurements varied because of experimental error.

So if the discrepancies are a cause for concern in your work, check the source given for each measurement to help explain the differences.


Why is the formation free energy of carbon dioxide almost constant in Ellingham diagrams?

Let's subdivide the CO2 standard entropy change reversible process into 3 parts:

Isothermal expansion of one mole of oxygen from 1 bar to the equilibrium partial pressure of oxygen in a reactor operating at equilibrium at a total pressure of 1 bar (a Van't Hoff equilibrium box)

Injection of one mole of oxygen at its equilibrium partial pressure into the equilibrium box reactor through a semipermeable membrane  while removal of one mole of CO2 through a semipermeable membrane at its equilibrium partial pressure in the reactor

Isothermal compression of one mole of CO2 from its equilibrium partial pressure in the equilibrium box reactor to a pressure of 1 bar.

The overall changes (standard changes) in G, H, and S for this 3-step process are -394359 J/mole, -393509 J/mole, and 2.8508 J/mole-K.  Now, let's break this down to the changes in the three steps.  Based on the standard change in G for the overall process, the equilibrium partial pressures of oxygen and CO2 in the equilibrium box are 8.08E-70 bars and essentially 1 bar.
In step 2, since the reaction is operating at equilibrium, the change in G is zero, while the changes in H and S are $\Delta H_2=\Delta H^0=-393509$ J/mole and $\Delta S_2=\Delta H_2/298.15 = -1319.8$ J/mole-K.
In steps 1 and 3, the expansion of oxygen and the compression of the CO2 are isothermal, so $\Delta H_1=\Delta H_3=0$.  In step 1, $\Delta G_1=R(298.15)\ln{8.08\times 10^{-70}}=-394359$ J/mole and $\Delta S_1=-R\ln{8.08\times 10^{-70}}=1322.7$ J/mole-K.  In step 3, the initial and final CO2 pressures are essentially 1 bar, so $\Delta G_3=\Delta S_3=0$
So, it seems that, merely by coincidence, the changes in entropy in step 2 compared to the sum in steps 1 and 3 are of nearly equal magnitude and opposite in sign (-1319.8 vs +1322.7).  This means that their difference is only 2.85, which represents the overall entropy change.  It is coincidental because the mechanistic changes in step 2 compared to steps 1 and 3 are very different.


Initially the idea was that -ate suffix is reserved for a mononuclear acid with highest oxidation state, while -ite suffix was reserved for a mononuclear acid with second highest oxidation state. The per-...-ate construct was reserved for anion with 'extra' oxygen, typically a peroxocompound and hypo-...-ite construct was reserved for acid extra low on oxygen.

However, since for halogens $\ce{XO4^-}$ anions were obtained rather lately, perhalogenates has this formula and simple halognates are $\ce{XO3^-}$. Mn subgroup (Mn, Tc, Re) inherited this in the sense that highest oxidation state corresponds to per- ... -ate construct

Similarly, ortho- prefix was reserved for mononuclear acids with highest number of hydroxogroups possible, while meta- prefix was reserved for an acid with one water molecule less (typically, polymer/oligomer)

Putting it all together using phosphorous as an example

Perphosphate $\ce{PO5^{3-}}$
Ortophosphate - $\ce{PO4^{3-}}$
Metaphosphate - $\ce{(PO3^-)_n}$
Phosphite $\ce{HPO3^{2-}}$ (since $\ce{H3PO3}$ is actually $\ce{H2[HPO3]}$ for some arcane reasons) Note: depending on the chemistry course you are taking (less advanced courses mainly), some will use $\ce{PO3^{3-}}$ as phosphite
Hypophosphite - $\ce{H2PO2^-}$ (since $\ce{H3PO2}$ is actually $\ce{H[H2PO2]}$

And for chlorine (see above about perhalogenates)

Perchlorate $\ce{ClO4-}$
Chlorate $\ce{ClO3-}$
Chlorite $\ce{ClO2-}$
Hypochlorite $\ce{ClO-}$

Not every degree needs to be present, for example only simple carbonates and percarbonates exist, and sometimes extra stages (usually polynuclear) occur, such as dithionites. This system is widespread in areas heavy on traditions, such as medicine and heavy industry.

There is a gentle pressure from IUPAC to move to much less confusing systematic naming, but the success is limited, especially considering widespread chemicals, which are familiar anyway. It is similar to situation with RJ plugs: everyone recognises phone and Ethernet connectors, but very few care about their full formal name.


Are all alkene geometrical isomers achiral?

While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.

Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.


Why should fire extinguishers containing carbon dioxide not be used on magnesium fires?

The reaction of magnesium metal (Mg) with carbon dioxide is extremely exothermic (heat evolving). In other words, carbon dioxide makes magnesium metal burn.

$$\ce{2Mg(s) + CO2(g) -> 2MgO(s) + C(s)}$$


Why aqueous hydroxide is slippery (beyond just saponification)

Glass itself is soluble in strong alkalis, forming sodium silicate solution, which is a syrupy, somewhat slippery-feeling substance. You can use sodium silicate to make a fire-resistant coating or high-temperature cement.

As you state, it's also possible the glove itself is being attacked. If the glass feels slippery in $\ce{NaOH}$ and $\ce{KOH}$ solutions, but not in $\ce{NH4OH}$, then it is most likely a silicate forming, though.


Elements other than carbon that can form many different compounds by bonding to themselves?

Is carbon the only element that can do this?

No, carbon is not the only element with such characteristics.

  If not, then what are the other elements can also do this?

There is a whole number of elements such as silicon, arsenic, germanium.

  Is there a term to describe such elements?

At least I'm unaware of such a term, which might be furnished by our far wiser community.

  What is the chemical characteristic that allows this to occur?

Catenation.

More information:

According to the Molecular Orbital Theory, the condition for a compound to exist is that it should have more electrons in the bonding orbitals than in the anti-bonding orbitals. So, as long as you have the bonding orbitals filled more, you can have pretty anything, more than just chains of atoms.

Thus, the existence of a compound also depends on the precise conditions in which the compound is kept, for example sodium forms different types of chlorides under different conditions and that as pointed out by Poutnik in the comments, $\ce{He2^1+}$ and a ton of others are discovered and still more awaiting discovery.


Your teacher gave you a poor answer.

You are observing three different things.

(1) The solution isn't instantly homogeneous when HCl is added. It takes a couple of seconds of stirring to mix it thoroughly.

(2) The calcium hydroxide solution initially has considerable buffer capacity. As the solution gets less basic then a small amount of HCl can change the pH by a relatively large amount. So a drop of HCl in 50 ml of 0.1 molar calcium hydroxide won't change the pH much, but a drop of HCl in a  neutral solution at pH 7 would change the pH by quite a bit.

(3) The electrode and meter have a response time. So if you could move the pH electrode instantly from an acid solution to a basic solution it would take the electrode and meter themselves a few seconds to equilibrate. (The pH electrode has a hydrated silica layer that forms the boundary on the surface which takes time to equilibrate. The pH meter itself will have some RC time constant so that electrical noise is minimized.)


Which has a lower boiling point, Ozone or CO2? Why?

The primary thing to realize is that, although the ozone molecule has a bent structure, it is a homonuclear molecule.
This means that between the individual oxygens, the bond moments being formed are quite less, as inherently, both the oxygens have the same Pauling electronegativity values. Only a slight difference in the same will occur, due to the asymmetrical bonding scheme in ozone.

It's worth noting that bond moments in carbon dioxide are much more pronounced than those in ozone, even though they end up being canceled vectorially (like you said).

For comparing the boiling points, we need to see the extent of intermolecular attraction as a measure for the same. Due to the difference in electronegativities between $\ce{C}$ and $\ce{O}$, $\ce{CO2}$ ends up having more partial charges on its constituent atoms as compared to ozone. Hence, the boiling point of carbon dioxide is more than ozone


Why does a complete discharge destroy a Li-Ion-Battery?

In a Lithium ion cell, the anode material can dissolve in the electrolyte, and then on recharge, precipitate in the midst of the electrolyte and insulating membrane, short-circuiting the cell. Further, the cathode material can release oxygen, which migrates away and does not get reincorporated on charging.

Another problem with most secondary (storage) cells, Pb-acid as well as Li, is physical destruction of the structure on over-charging and under-charging. During the charge-discharge cycle, there is some change in volume of the electrodes, which causes them to spall, shedding pieces of the active material. This leads to loss of capacity and sometimes a short-circuit.

Even a fully charged cell deteriorates gradually... but increased discharge increases the growth of precipitates. Li-ion cells self-discharge about 2%/mo, (half-life of ~2 years), so a charged cell is quite stable, with no perceptible growth of whiskers or precipitate for years.


The oxidation of aldehydes and alpha-diketones with peroxy compounds

The Baeyer-Villiger oxidation is conducted under acidic conditions such as an alkyl peroxide in the presence of a mineral acid (e.g.; H2SO4) or a peracid (e.g.; peracetic acid or m-chloroperbenzoic acid). In the reactions described in the diagram, the reactions of aldehyde 1 or α-diketone 7 are initiated by proton of the carbonyl oxygen followed by addition of peroxide (an alkylperoxide is illustrated). Ultimately, the oxygen of the RO-moiety is protonated and is the effective leaving group (ROH) in the rearrangement. In both sequences, the red route is favored; the blue route disfavored. Cation 3 is a more stable species than cation 5 because of hyperconjugation with the alkyl group or resonance stabilization if R'=aryl. Cation 3 affords a carboxylic acid while cation 4 would lead to formate ester 6.
In the case of peroxyhemiketal 8, acyl migration leads to cation 9 which is more stable than species 11, which has a positive charge adjacent to a carbonyl group, which is destabilizing. Thus, anhydride 10 is formed in preference to α-ketoester 12.


For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount?

From what I remember from Fick's laws, I would say that the factor that limits the rate of dissolution is the concentration of the already-dissolved compound. At first, you don't have any caffeine dissolved so the rate of dissolution is fastest. As time goes by, more and more caffeine becomes solvated and that hinders the dissolution of remaining caffeine from the tea leaves. At any point of time until saturation, the 300 ml cup will have faster dissolution rate because it has more "free space" to solvate the caffeine.

Edit: However, after thinking this for a while, I have to add that in practice the difference might be too small to be detected. The solubility (~ saturation point) of caffeine in water is 20 g/l at 20°C while the concentration of caffeine in tea is 0.1 - 0.4 g/l – huge difference. It might be that from caffeine's point of view, the "free space" is more or less the same in both cups.


Is there a difference between ‘ammonium hydroxide’ and ‘aqueous ammonia’?

Ammonia is a weak base, ammonium is a weak acid. Water is a very weak acid and hydroxide is a very strong base. Taken together, this tells us that equilibrium $(1)$ is pushed towards the reactant side.

$$\ce{NH3 (aq) + H2O (l) <<=> NH4+ (aq) + OH- (aq)}\tag{1}$$

Thus, it is practically impossible to form stoichiometric ammonium hydroxide; the $\mathrm{p}K_\mathrm{a}$ values tell us that ammonium ions and hydroxide ions put together will generate water and ammonia.

However, it is also true that equation $(1)$ is an equilibrium — everybody has seen a solution of ammonia in water display basic properties so hydroxide ions (and thus ammonium ions) are unquestionably there. The predominant species is dissolved ammonia, though.

$$\ce{NH3 (g) + H2O (l) <=>> NH3 (aq) + H2O (l)}\tag{2}$$

It is correct that the most important process when dissolving ammonia in water is $(2)$ — $(1)$ plays a more minor role. Thus, it is also more correct to speak of solutions of ammonia — but at the same time ammonium hydroxide is commonly understood even if not formally correct.

By the way beware: occasionally students are sent off to fetch ‘solid ammonium hydroxide’ as a joke — similar to students getting sent off to fetch ‘vacuum gas’, ‘three phenol rings’ etc.


chemical possibility in oil - 8th graders have questions for a project

A chemical that can be added to crude oil so that the oil solidifies upon contact with water to either seal leaks or to make the oil recoverable sounds like a very good idea. Unfortunately, such a chemical is exceedingly unlikely to exist.

When processing crude oil, you typically want the oil itself and not any additives. That means, the chemical in question would have to be added in concentrations as low as possible. Therefore, it’s reaction with water must be very strong. But this in turn means that the traces of water present everywhere (yes, even inside your sample of crude oil) would almost be enough to cause the reaction. Therefore, outside of water-free atmospheres (used in laboratories under inert gas to perform water and air-sensitive reactions) this chemical would hardly be usable.

A second problem is getting the chemical reasonably dissolved. Crude oil is a very viscous mixture unless heated. To use the chemical effectively, it would need to be added almost immediately and distributed well in the sample of crude oil — which would mean immediately heating the crude oil to such temperatures that efficient mixing is possible (likely under pressure to not lose the more valuable but also more volatile petrol components) right after getting it out of the ground. Especially on oil rigs, I see major difficulties with this process; and it is basically on oil rigs that you would want this chemical the most.

Finally, there is the ‘how would it work’ problem. I have extreme difficulty imagining a reaction with water that would lead to solidification. The only idea I can come up with is a compound that would be liquid at standard conditions but adds water to become a solid which crystallises out of the crude oil phase. However, this solid would likely not do you the favour of crystallising where you need it and you would need so much of it that my first argument becomes invalidated. Therefore, I conclude that such a compound cannot exist.

If such a compound exists, I’m sure we would have heard of it or it would be in use (or have been in use if there are significant and severe environmental issues with it that are worse than the issues with crude oil itself.


Why can temperature changes have units of either degrees Celsius or kelvin?

The temperature in kelvin is equal to the temperature in degrees Celsius plus 273.

So, $\Delta T$ will be the same whether the temperature was reported in $\mathrm{K}$ or $^\circ\mathrm{C}$.  As an example if $T_\mathrm{initial} = 0\ ^\circ\mathrm{C}\ (273\ \mathrm{K})$ and $T_\mathrm{final} = 100\ ^\circ\mathrm{C}\ (373\ \mathrm{K})$, $\Delta T$ is 100 degrees no matter whether you used $^\circ\mathrm{C}$ or $\mathrm{K}$ in your computation.