What is the difference between ΔG and ΔrG?

Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference.

The Gibbs free energy change, $\Delta G$

You are quite right in saying that $\Delta G$ represents a change in a system over a time interval. The notation $\Delta G$ itself implies that it is a difference in $G$ between two things: an initial state and a final state. Let's use a real example in order to make things clearer. Consider the thermal decomposition of ammonium nitrate:

$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$

If someone were to ask you "what is $\Delta G$ for this reaction?", you should really, technically, be telling them: it is not well-defined. That is because of two things. Firstly, the reaction conditions, e.g. temperature and pressure, are not specified.

But more importantly, a balanced equation does not tell you exactly how much ammonium nitrate is reacting. In your beaker, you could have $1~\mathrm{mol}$ of $\ce{NH4NO3}$ reacting according to the above equation, but you could equally well have $2~\mathrm{mol}$, or you could have $5000~\mathrm{mol}$ (I must say that's one huge beaker though), or you could have $0.001~\mathrm{mmol}$.

The stoichiometric coefficients in the equation do not indicate the amounts of compounds reacting.

So, a more proper question would read like this:

  $\pu{2 mol}$ of ammonium nitrate fully decompose according to the balanced equation $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. Calculate $\Delta G$ for this process.

Great. Because the decomposition is complete, we now know that our final state is $2~\mathrm{mol}~\ce{N2O(g)} + 4~\mathrm{mol}~\ce{H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$, and our initial state is $2~\mathrm{mol}~\ce{NH4NO3(s)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. So,

$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot G_\mathrm{m}(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{NH4NO3(s)})]
\end{align}$$

We can't quite find the absolute molar Gibbs free energies, so the best we can do is to use Gibbs free energies of formation.

$$\begin{array}{c|c}
\text{Compound} & \Delta_\mathrm{f}G^\circ / \mathrm{kJ~mol^{-1}}\text{ (at }298~\mathrm{K}\text{)} \\ \hline
\ce{NH4NO3(s)} & -183.87 \\
\ce{N2O(g)} & +104.20 \\
\ce{H2O(g)} & -228.57
\end{array}$$
$$\scriptsize \text{(data from Atkins & de Paula, }\textit{Physical Chemistry}\text{ 10th ed., pp 975-7)}$$

(Note that the use of standard formation Gibbs free energies is only because of the conditions specified in the question, which conveniently corresponds to the standard state. If we specify different conditions, we can still find $\Delta G$, but we would have to use different data.) So:

$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{NH4NO3(s)})] \\
&= [(2~\mathrm{mol})(+104.20~\mathrm{kJ~mol^{-1}})] + [(4~\mathrm{mol})(-228.57~\mathrm{kJ~mol^{-1}})] - [(2~\mathrm{mol})(-183.87~\mathrm{kJ~mol^{-1}})] \\
&= -1073.62~\mathrm{kJ}
\end{align}$$

Note that we have units of kJ. Since $\Delta G$ is the difference between the Gibbs free energy of one state and another, $\Delta G$ has to have the same units as $G$, which is units of energy.

Now, this does not necessarily mean that $\Delta G = G_\text{products} - G_\text{reactants}$. For example, if I changed my question to be:

  In the Haber process, 100 moles of $\ce{N2}$ and 300 moles of $\ce{H2}$ are reacted at $800~\mathrm{K}$ and $200~\mathrm{bar}$ according to the equation $\ce{N2 + 3H2 -> 2NH3}$. Only 10% of the starting materials are converted under these conditions. Calculate $\Delta G$ for the process. (These numbers are made up.)

then, your final state would not be the pure products. Your final state is not 200 moles of $\ce{NH3}$. Your final state is $90~\mathrm{mol}~\ce{N2}$, $270~\mathrm{mol}~\ce{H2}$, and $20~\mathrm{mol}~\ce{NH3}$.

In general, one could write, for a chemical reaction,

$$\Delta G = \sum_i (\Delta n_i) G_{\mathrm{m},i}$$

where $\Delta n_i$ is the change in the amount of compound $i$ (in moles), and $G_{\mathrm{m},i}$ is the molar Gibbs free energy of the pure compound $i$, under the $T$ and $p$ conditions specified. Going back to our ammonium nitrate example, we would have $\Delta n_{\ce{NH4NO3}} = -2~\mathrm{mol}$, $\Delta n_{\ce{H2O}} = +2~\mathrm{mol}$, and $\Delta n_{\ce{H2O}} = +4~\mathrm{mol}$.

The Gibbs free energy change of reaction, $\Delta_\mathrm{r} G$

As you have correctly stated, $\Delta_\mathrm{r}G$ is the slope of a graph of $G_\mathrm{syst}$ against the extent of reaction, commonly denoted $\xi$. This is the easiest way of interpreting $\Delta_\mathrm{r}G$. This question contains a slightly fuller derivation and explanation of what $\Delta_\mathrm{r}G$ means.

However, $\Delta_\mathrm{r}G$ does indeed, somewhat, refer to the instantaneous difference between the "molar Gibbs free energies" of the products and reactants. This is different from $\Delta G$ in three main ways.

First, $\Delta G$ is the difference between the Gibbs energy of the entire system at two points in time. Here, we are interpreting $\Delta_\mathrm{r}G$ as the difference between the "molar Gibbs free energies" of two components of the system: reactants and products. In other words, the system generally contains both reactants and products, and $\Delta_\mathrm{r}G$ may be thought of as the difference between the Gibbs energies of the product part of the system, and the reactant part of the system, even though you cannot separate them in the laboratory.

Secondly, it is the difference between the chemical potentials, not the molar Gibbs free energies. The molar Gibbs free energy is simply defined by $G_i/n_i$; the chemical potential is a partial derivative and is defined by $\mu_i = (\partial G/\partial n_i)_{T,p,n_j}$. In the case where there are no other species present (i.e. species $i$ is pure), then the chemical potential is identical to the molar Gibbs free energy.

Lastly, it is weighted by the stoichiometric coefficients $\nu_i$ instead of the change in the amount $\Delta n_i$. The stoichiometric coefficient is a dimensionless quantity, which is negative for reactants and positive for products. So, in the ammonium nitrate decomposition as written at the very top, we have

$$\nu_{\ce{NH4NO3}} = -1; \qquad \nu_{\ce{N2O}} = +1; \qquad \nu_{\ce{H2O}} = +2$$

and our expression for $\Delta_\mathrm{r} G$ is

$$\begin{align}
\Delta_\mathrm{r} G &= \sum_i \nu_i \mu_i \\
&= \nu_{\ce{N2O}}\mu_{\ce{N2O}} + \nu_{\ce{H2O}}\mu_{\ce{H2O}} - \nu_{\ce{NH4NO3}}\mu_{\ce{NH4NO3}} \\
&= \mu_{\ce{N2O}} + 2\mu_{\ce{H2O}} - \mu_{\ce{NH4NO3}}
\end{align}$$

Now, note the units again. The chemical potential is a partial derivative of the Gibbs free energy (units $\mathrm{kJ}$) with respect to the amount of $i$ (units $\mathrm{mol}$), and so it must have units $\mathrm{kJ~mol^{-1}}$. And from our above expression, $\Delta_\mathrm{r}G$ must also have units of $\mathbf{kJ~mol^{-1}}$, since the stoichiometric coefficients are dimensionless.

Does it matter if you start with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ of ammonium nitrate? The answer is now, no. The amount of starting material does not affect $\mu_i$, nor does it affect $\nu_i$. Therefore, $\Delta_\mathrm{r}G$ is independent of the amount of starting material.

Why is $\Delta_\mathrm{r} G$ an instantaneous difference between the chemical potentials? Well, that is because the chemical potentials of the reactants and products, $\mu_i$, are changing continuously as the reaction occurs. Therefore, if we want to calculate $\Delta_\mathrm{r} G$, we have to take a "snapshot" of the reaction vessel: otherwise it makes absolutely no sense to speak of $\mu_i$ because we wouldn't know which value of $\mu_i$ to use.

Compare this with $\Delta G$ above: the quantities of $G_{\mathrm{m},i}$ are constants that do not vary depending on the extent of reaction. Therefore, we do not need to specify a particular extent of reaction to calculate $\Delta G$.

A diagram to sum up

Note that, to adequately define what $\Delta G$ is, you need a starting point and an ending point. I showed two possibilities for $\Delta G$; there are infinitely many more.
Let's say the reaction goes to completion. If you double your starting material and double your product, the difference between $G_\text{products}$ and $G_\text{reactants}$ will also be doubled. So, it's important to specify!
Likewise, to adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$.
If you double your starting material and double your product, the curve is stretched by a factor of 2 along the y-axis, but it is also stretched by a factor of 2 along the x-axis because $\xi$ is also doubled. (For a mathematical explanation see the definition of $\xi$ given in the linked question earlier.) So, while the difference $\Delta G$ is doubled, the gradient $\Delta_\mathrm{r} G$ remains unchanged.
The units should also be clear from this. Since $\Delta G$ is a difference between two values of $G$, it has to have units of $\mathrm{kJ}$. On the other hand, $\Delta_\mathrm{r}G$ is a gradient and therefore has units of $\mathrm{kJ/mol}$.

A caveat

Unfortunately, the notation $\Delta G$ is often loosely used and treated as being synonymous with $\Delta_\mathrm{r}G$. You will therefore see people give $\Delta G$ units of $\mathrm{kJ~mol^{-1}}$. For more information refer to Levine, Physical Chemistry 6th ed., p 343. I would personally recommend making a distinction between the two.


How can the standard deviation for a set of experimental value be interpreted in terms of precision?

But what standard deviation is small enough to be considered precise?

"Sufficiently precise," ironically enough, is something that can't really be defined precisely in general terms.
In other words, a level of precision that is perfectly fine in one context might be unacceptably low in others.
For example, if I'm trying to estimate how much table salt I need to buy to supply my catering business for the next six months, I might be perfectly happy with a not-very-precise calculation at $\pm\,30\%$, since salt is cheap and will keep just fine if I buy too much.
On the other hand, if I'm a pharmacist compounding a prescription containing a drug where the level in the body leading to nasty side effects is not all that much higher than the level that provides the desired pharmacological activity, I probably want to use (and am probably required by law to use!) methods and apparatus that'll give me, say, $\pm\,3\%$ precision on my measurements.

Is there a better method for quantifying precision?

Not really.  Pretty much every metric I've seen for quantifying precision involves the standard deviation in some way.  Depending on the situation, one might use:

$s$, the standard deviation by itself

$s/\bar x$, the standard deviation divided by the mean, also called the relative standard deviation or coefficient of variation (CV)

$s/\!\left(\bar x \sqrt n\right)$, the relative standard deviation divided by the square root of the number of measurements

This is equal to the standard error $\left(s/\!\sqrt n\right)$ divided by the mean

The "Comparison to standard deviation" subsection of the Wikipedia article on the coefficient of variation lists some situations where these different quantities would be useful relative to others.

From this standard deviation, would this experiment's results be considered precise?
If not, why not, and how would I interpret it?

In my opinion: Yes, these results are precise. In this case, since the mean value is far from zero and there's only one dataset being examined, I would use the coefficient of variation (CV) as my metric of precision.  With these values, I get a CV of $1.88\%$.  On the whole, I think this is an entirely reasonable CV for a titration experiment. The data could exhibit more precision, obviously, but that low of a CV implies very good repeatability to me.
As porphyrin noted, that $\pu{17.1 mL}$ measurement you obtained is perfectly reasonable in a statistical sense, as it's $25\%$ of a small dataset, residing $1.7$ standard deviations away from the mean. This puts it well line with the expected $32\%$ occurrence of values $x$ where $|x-\bar x|/s > 1.0$.
In other words, I disagree with Mithoron: I don't think that value is an outlier at all.  The usual metric I'm familiar with for identifying outliers is where $|x-\bar x|/s > 3.0$, which should occur only $\approx 0.34\%$ of the time for normally-distributed data.

The uncertainty of the burette is $\pm\,\pu{0.05 mL}$.

A brief note: I believe the small uncertainty on the burette as compared to the standard deviation in your data $(0.05 / 0.38 \approx 13\%)$ indicates that the majority of the spread in your measurements comes from elements of the experimental apparatus/procedure other than the burette.

Further reading: This article at Inorganic Ventures is a good writeup of the mean and standard deviation, and how they relate (or not) to accuracy and precision.


Synthesis of Aspirin, made unknown fluid

Often, acetylizations are performed with acetic anhydride being deployed both as reagent, as well as solvent.  Consequently, acetic anhydride is used in excess when referring to the number of $\ce{-OH}$ to react.

The addition of water you describe in step 3) intends to hydrolyse this excess of acetic anhydride to form acetic acid.  Apart from acetic acid equally representing a good solvent for many organic products, it takes time to complete this hydrolysis of acetic anhydride. You may accelerate the hydrolysis of the acetic anhydride if the beaker is immersed into an ultrasound bath, or just continue stirring the mixture at room temperature as usual.

The second point is, albeit its high melting point (136 $\mathrm{^\circ{}C}$), the crystallization of aspirine equally takes some time, too. Crystallization of organic materials from water may demand some more attempts prior to success than from organic solvents, too.  Cooling the mixture often will lower the solubility of your product in the solvent, however, if cooled too quickly to too low temperatures you may obtain either just a glassy gel / syrup instead of a solid easy to filter off, or a solid that has to be recrystallized to purify it further.  Scratching the inner walls of the beaker with a glass rod intends to create fresh sites where crystallization may start.

For peracetylating glucose with acetic acid, for example, it is not too strange to stirr the quenched reaction mixture with an excess of water over night at room temperature to convert the initially "honey-like" product into a good solid.


How to balance this redox reaction (using half reaction method):$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2}$$

You really can't balance this equation for two reasons.

(1) There is hydrogen on the left hand side, but none on the right.

(2) There are two competing redox reactions. One produces $\ce{NO}$ and the other produces $\ce{NO_2}$. The overall reactions are:$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  ->  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$  $$\ce{3Cu_{(s)}  +  8HNO3_{(aq)}  ->  3Cu(NO3)2_{(aq)}  +  2NO_{(g)}  +  2H2O_{(l)}}$$

Thus the ration of $\ce{NO2_{(g)}}$ to $\ce{NO_{(g)}}$ is a kinetics problem. In concentrated acid the top reaction predominates and you get mostly the first reaction. In dilute acid the second reaction predominates and you get mostly the second reaction.

Now as an example let's look at the redox reaction of the first reaction.

Looking at the copper first we have: $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$

Just looking at $\ce{NO3^{-} -> NO2 + e^{-} }$ the charge balances, and the $\ce{N}$ atoms balance, but the $\ce{O}$ atoms do not balance and hence this is not a valid half cell reaction. The valid reaction is: $$\ce{e^{-} + 2HNO3_{(aq)}  -> NO^{2-}_{3} + NO2_{(g)}  +  H2O_{(l)}}$$

Using "redox" I can simply balance the electrons. So I need the Cu reaction as is, but I need to multiple the nitrite acid reaction by two. $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$ $$\ce{2e^{-} + 4HNO3_{(aq)}  -> 2NO^{2-}_{3} + 2NO2_{(g)}  +  2H2O_{(l)}}$$

Now that the electrons match, I can add the two equations:
$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  ->  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$


What does ³D mean in energy level diagram for helium?

The index 1 or 3 is used for saying that the spins of the two electrons are parallel or anti-parallel. One spin is up or +1/2, and the other one is down, or -1/2. This is the case in the ground state : the 2 electrons of Helium have two antiparallel spins (one +1/2 and one -1/2). The atom has no total spin, as the two electrons cancel the magnetic effect of their spins. The total spin of the atom is then 1/2 - 1/2 = 0. In such a configuration, the He atom is called para-helium. This remain true if one electron is excited and goes to an upper orbital like 2s, 3s, etc. provided that the two electrons maintain their spins (one up and one down). The total spin is zero, and the atom is an excited state of para-helium.

It is difficult, but not impossible, to change the spin of one electron, so that both electrons have the same spin. In this configuration, the helium is called ortho-helium. The total spin for the whole atom is 1/2 + 1/2 = 1. In that case, there are three possibilities for the projection of the total spin in an external magnetic field. +1 in the sense of the field, -1 in the opposite sens, and zero, if the total spin is perpendicular to the magnetic field. This number 3 is reported in upper index. The differences of the three projections must be an integer.


Is axis of symmetry considered a true symmetry?

A molecule is chiral (1, 2) if it does not have an improper axis of rotation. Since your drawing does not seem to show the meso form of 2,3-dichlorobutane, it is optically active.


Estimation of CO2 by titrimetric method using phenolphthalein indicator

But when we calculate the concentration of $\ce{CO2}$ we consider only reaction 2 which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in Step 3 and the CO2 (the $\pu{10\%}$) that was directly converted into $\ce{H2CO3}$ (eq 1) are not taken into account.

Almost $\ce{H2CO3}$ in the sample will be converted to $\ce{HCO3-}$ as you proceed with the phenolpthalein endpoint. Search for "Speciation of Carbonate as function of pH" (Sorry I do want to immediately link the graph but it is only a http and stackexchange won't accept it).

At ~8.3 or so, all carbonate species is at the form of bicarbonate. All carbonic acid in the solution will be transformed as bicarbonate. The amount of carbonate ion is almost nonexistent. The $\ce{H+}$ released will be neutralized by NaOH, so aside from reacting directly with $\ce{CO2}$, it will also react with the $\ce{H+}$ released by $\ce{H2CO3}$, so (1) is accounted on the amount of NaOH that will be used.

Also, the sum of all equations (2) and (3) is just

$$\ce{2 NaOH + 2CO2 -> 2NaHCO3}$$

So in essence, for 1 mole of NaOH you react with one mole of $\ce{CO2}$ to produce one mole sodium bicarbonate.

In totality, for every one mole $\ce{H2CO3}$, you react with one mole NaOH. For every direct reaction of NaOH with $\ce{CO2}$, the ratio is $1:1$.

So I guess that's all reactions accounted for.


Numerical related to crystallisation of water below its freezing point

In the question Crystallization begins after you add a small piece of ice
.
 So it has to begin that's been given in the question.

I don't know thermodynamically whether it's plausible or not, but it's a given in the question a logical alternative way to approach the question (according to me) is

 I formatted it as a spoiler just in case you want to think a bit more or anybody else does

   The temperature has increased because latent heat of fusion is being released i.e ice being formed from water is  releasing heat.

PS: I may be wrong


Acidic conditions, high heat, and return to carboxylic acids

In the cyanohydrin reaction aldehydes or ketones can react with HCN to form a cyanohydrin.  Inorganic cynanides and acid can be used in place of HCN, to generate the required cyanide anion.   The reaction proceeds by typical addition of the nucleophilic cyanide anion to the carbonyl carbon to form the tetrahedral intermediate, which can ultimately be protonated to yield the corresponding cyanohydrin.

Once formed, the nitrile group of the cyanohydrin can undergo acidic (or basic) hydrolysis by the same mechanism as any other common nitrile.  The only difference in the case of a cyanohydrin is that the "R" group contains a hydroxyl substituent on its alpha carbon.

A useful variant of cyanohydrin formation is the Strecker amino acid synthesis.  Instead of forming a cyanohydrin, an aminonitrile is formed and the nitrile is hydrolyzed to the corresponding alpha-amino acid.  A detailed mechanism can be found in the above link.


How to determine the volume of water necessary to produce a solution with 4.9 g/l sulfuric acid?

We have talked about it in chat and maybe the question is a bit ambiguous. I would interpret the last line as a mass concentration. So what you are actually looking to create is a $\gamma(\ce{H2SO4}) = \pu{4.9 g/l}$ solution. Then the calculation becomes straight forward.
Keep in mind that water is your solvent, so you can't actually dilute it; also writing down the concentration is a bit pointless here.

You can easily convert the mass concentration to an amount concentration. The relative molar mass $M$ is given by the mass $m$ of the substance divided by the amount of substance $n$:
$$M = \frac{m}{n}\tag1\label{molarmass}$$

The mass concentration is defined as
$$\gamma = \frac{m}{V}.\tag2\label{massconc}$$
The amount concentration is defined as
$$c = \frac{n}{V}.\tag3\label{amountconc}$$
You can use \eqref{molarmass} in \eqref{amountconc} and obtain
$$c = \frac{m}{M} \times \frac{1}{V}
=\frac{m}{V} \times \frac{1}{M}.\tag4$$
Substitute \eqref{massconc} back in and you have
$$c = \frac{\gamma}{M}.\tag5\label{amounttomass}$$

Now in the question you have given the following:
\begin{align}
M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\
c_\mathrm{i}(\ce{H2SO4}) &= \pu{0.1 mol/l}\\
V_\mathrm{i}(\ce{H2SO4}) &= \pu{100 ml}\\
\gamma_\mathrm{f}(\ce{H2SO4}) &= \pu{4.9 g/l}
\end{align}
I have given the initial values the subscript $\mathrm{i}$ and the final values the subscript $\mathrm{f}$ (instead of concentrated and diluted).

You are now looking for the volume of solvent you have to add, which is
$$\Delta V = V_\mathrm{f} - V_\mathrm{i}.\tag5\label{volume}$$

Then you are right, you need the following relation
$$c_\mathrm{i} \times V_\mathrm{i} = c_\mathrm{f} \times V_\mathrm{f},
\tag6$$
rearranged and using \eqref{amounttomass}
\begin{align}
c_\mathrm{i} \times V_\mathrm{i} &=
\frac{\gamma_\mathrm{f}}{M} \times V_\mathrm{f},\\
V_\mathrm{f} &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},
\end{align}
and using \eqref{volume}
\begin{align}
V_\mathrm{i} + \Delta V &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},\\
\Delta V &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i} - V_\mathrm{i}\\
&= \left(\frac{M \times c_\mathrm{i}}{\gamma_\mathrm{f}} - 1\right)V_\mathrm{i}.
\tag7\label{dilute}
\end{align}

Now simply plug in your values and get the solution (hover over box to reveal):

   $\displaystyle \Delta V = \left(\frac{\pu{98.1 g//mol} \times \pu{0.1 mol//l}}{\pu{4.9 g//l}} - 1\right)\times \pu{0.100 l} = \pu{0.100 l}.$


How can a compound having a plane of symmetry be Chiral?

I think you are getting confused between chirality and E-Z isomerism in cyclic structures. So let's clear that up first.
According to IUPAC:

chirality
The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

Chiral molecules are always dissymmetric (lacking $S_n$) but not always asymmetric (lacking all symmetry elements except the trivial identity). However, asymmetric molecules are necessarily always chiral.
For a basic E-Z isomerism definition:

A cycloalkane has two distinct faces, and any substituent on a ring lies toward one of two faces. When two substituents on a ring point to the same face, they are cis. When the two substituents point to opposite faces, they are trans.

Now, I'll address your second question first:

[OP]How is the compound as a whole a Chiral compound because afaik Compounds having a plane of symmetry(Which passes through the two chiral centers in this compound) are supposed to be achiral.

You're right. This compound does possess a plane of symmetry. However, that only means it can't form enantiomers. It can very well form diastereomers (in this case a pair of E-Z isomerism).
Some further examples to illustrate the point:

The trans isomer is achiral, meaning this structure doesn't have enantiomers. However, it still has a cis diastereomer which can show chirality.
An example closer home:
Image courtesy
I'm sure you must've figured out by now, that the carbon on top isn't a capable of showing optical isomerism. However, due to the dissimilar substituents present ($\ce{H\text{ and } CH3}$), the compound can show cis-trans isomerism.
Further reading:


Find the volume of the carbon dioxide product

You need to determine the limiting reactant. Use the equation and the number of moles of $\ce{CH_3CH_2OH}$ and $\ce{O_2}$.
In the reaction, you need 1 mol of $\ce{CH_3CH_2OH}$ and 3 mol of $\ce{O_2}$ to get 2 mol of $\ce{CO_2}$. If you have less than 1 mol of $\ce{CH_3CH_2OH}$ but 3 mol of $\ce{O_2}$, you cannot get 2 mol of $\ce{CO_2}$. In such a case, $\ce{CH_3CH_2OH}$ is the limiting reactant. The amount of $\ce{CO_2}$ will thus be lower than 2 mol.

Using the number of moles of the limiting reactant, you should be able to calculate the number of moles of $\ce{CO_2}$ produced. Use proportions!

The last step will involve using the ideal gas equation, by assuming that $\ce{CO_2}$ behaves like an ideal gas. $$PV=nRT$$
You have the pressure $P$, the temperature $T$, $n$ (the number of moles of the $\ce{CO_2}$), $R$ is a constant. Find the volume $V$.


The rules for numbering of bicyclic compounds such as azulene are given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

First, you need to generate a "preferred orientation" of the molecule:

  P-25.3.2.3.2 Polycyclic fused ring systems are oriented in accordance with the following criteria considered in order until a decision is reached:
  (a) maximum number of rings in a horizontal row;

Because of this, the two rings need to be horizontally aligned with each other in the preferred orientations. Note that this rule alone does not distinguish between the two orientations on the left.

Then, the numbering proceeds as follows:

  P-25.3.3.1.1 The numbering of peripheral items in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter 'a', 'b', 'c', 'd', etc.

This leads to two possible numbering systems for the two orientations above. Which one is preferred is given by a further rule

  P-25.3.3.1.2 If alternative numberings for a ring system remain [...] the following criteria are applied in order until a decision is reached: [...]
  (c) low locants are assigned to fusion carbon atoms;

The ketones in your molecule have greater seniority over the thiols (P-41), and hence the ketones are expressed as the suffix (leading to a name ending in "dione") instead of the thiols (a name ending in "dithiol"). Note that if the thiols are deprotonated this seniority ranking would be reversed, as anions have greater seniority over ketones (hence you would have a name ending in "bis(thiolate)").

The quinone derived from azulene is named as azulene-x,y-dione, where x and y are the locants of the two carbonyl carbons (P-64.2.2.2.3). The asymmetry in the quinone leads to ambiguity in the numbering, which is resolved by the next rule:

  P-14.4 When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority: [...]
  (c) principal characteristic groups and free valences (suffixes);

Finally, the –SH groups are inserted as prefixes into the name, as they can no longer be suffixes. The appropriate prefix for –SH is 'sulfanyl'. The multiplier used is "bis" and not "di" to avoid ambiguity:

  P-16.3.6 The prefixes 'bis', 'tris', 'tetrakis', etc. are also used to avoid ambiguity:
  (a) before a mononuclear subset of a polynuclear acyclic structure;
  Examples: bis(sulfanyl) (preferred prefix; defines two –SH groups, see P-63.1.5; whereas disulfanyl defines the –SSH group; see P-63.3.1)

As you can see, the manual generation of such names is in no way trivial. Most people use software to automatically generate names, which is generally sufficiently clear for the chemical community to understand them. However, it should be mentioned that these automatically generated names are not always the preferred IUPAC names: for example, ChemDraw 16 gives "2,6-dimercaptoazulene-1,5-dione" as the name, even though the prefix "mercapto" is no longer recommended (P-63.1.5).


I already made a comment about some of what I am about to say but I will provide a partial answer. I say partially because I could not find any mechanism for the second product. However, from literature, what I found was that in acidic conditions, $\ce{KMnO4}$ will oxidize naphthalene into the first product, phthalic acid1. In basic/alkaline conditions, $\ce{KMnO4}$ will oxidize naphthalene into Phthalonic Acid2. The results also agree from other papers. The proposed mechanism for the acidic solution reaction is here (taken from the paper in the first link):

It is essentially the same mechanism for the cleavage of a double bond but it happens in two positions in naphthalene. Note this is the mechanism for $\ce{KMnO4}$ oxidative cleavage:

Also note that the mechanism in the first image was based on a 2017 paper studying the kinetics of the reaction and then formulating a mechanism. So this reaction is still being studied. Along the same lines, I could not find the mechanism for the reaction of $\ce{KMnO4}$ and naphthalene in alkaline conditions. I would assume that its mechanism is still being researched.
References

https://www.jchps.com/issues/Volume%2010_Issue%201/40-0581116.pdf
The Oxidation of Naphthalene to Phthalonic Acid by Alkaline Solutions of Permanganate, R. Arthur Daly, The Journal of Physical Chemistry 1907 11 (2), 93-106, DOI: 10.1021/j150083a001
http://www.orgsyn.org/demo.aspx?prep=cv2p0523
https://www.acros.com/_rainbow/pdf/oxidation_brochure_manga.pdf
http://studymaterial.unipune.ac.in:8080/jspui/bitstream/123456789/8059/1/POLYNUCLEAR%20HYDROCARBONS.pdf


Are there any significant uses of the compound formed by magnesium and anthracene?

It seems like an idea of using magnesium anthracene systems for the $\ce{MgH2}$ production persisted since 1980s [1] till late 2000s, when new more efficient method with better scalability for industrial use was established.

One of the recent reviews in hydrogen-storage applications [2, p. 220] compares the older two-step process of $\ce{MgH2}$ synthesis:

  First, metallic magnesium is dissolved in anthracene to form Mg-anthracene solution. Then, the hydrogenation of the Mg-anthracene solution produces $\ce{MgH2}$ with $\ce{CrCl3}$ or $\ce{TiCl4}$ catalyst.

with the most recent direct synthesis from Mg powders and tablets [3]. Their experimental trial involved the following:

  Powder and tablet Mg were placed on trays inside a 50 kg hydrogenation furnace. After purging with Ar, the furnace was ﬁlled with 99.9999% hydrogen with a pressure of less than 2 MPa. The temperature of the process was controlled so as to remain near the boundary between the regions $\ce{\{MgH2\}}$ and $\ce{\{Mg + H2\}}$ in the thermal equilibrium diagram, and maintained for 30–45 h. The hydrogen consumption behavior was monitored by a pressure gauge. Afterward, the furnace was gradually cooled down to room temperature. The hydrogenation yield (%) was calculated via the weight difference of Mg samples before/after the process.

$\ce{MgH2}$ obtained with this method showed a hydrogen yield of around 96% with great reproducibility and scalability, and was proved as industry-ready.

But still, magnesium-anthracene · 3THF is used at a lab-scale in fine organic, organometallic, and inorganic syntheses [4, pp. 300-307]:

as a source of soluble zerovalent magnesium;
in preparation of finely-divided metal powders and transition metal complexes;
catalytic synthesis of Grignard compounds.

Bibliography

Bogdanovic, B. Acc. Chem. Res. 1988, 21 (7), 261–267. DOI 10.1021/ar00151a002.
Sasaki, K.; Li, H.-W.; Hayashi, A.; Yamabe, J.; Ogura, T.; Lyth, S. M. Hydrogen Energy Engineering: A Japanese Perspective; Springer Japan: Tokyo, 2016. ISBN 978-4-431-56042-5.
Uesugi, H.; Sugiyama, T.; Nii, H.; Ito, T.; Nakatsugawa, I. Journal of Alloys and Compounds 2011, 509, S650–S653. DOI 10.1016/j.jallcom.2010.11.047.
Fürstner, A. Active metals: preparation, characterization, applications; VCH: Weinheim; New York, 1996. ISBN 978-3-527-61517-9.


Is there an explanation to the atomisation energy trend of transition metals?

The hump is apparent because of the dip. To be more clear, it is good to know that all these attributes follow the same trends:

Metallic character
Strength of metallic bonding
Enthalpy of atomisation
Melting point
Density

If you plot a graph with any of these attributes on the $y$ axis, you'll notice that more or less in each of these cases, the graph rises first reaches a maxima just before the middle of the series, undergoes a dip in the middle again rises and then finally decreases till the end of the series.

It can be somewhat explained (these observations are made on an experimental basis) by the strength of metallic bonding. As the number of unpaired electrons increase, the metallic bonding becomes stronger. Note that $\ce{Cr}$ has the highest number of unpaired $e^-$s: $6$. This explains the hump.

$\ce{Mn}$ on the other hand has a half-filled electronic configuration. Thus, it is more stable. Stability order on the grounds of electronic configuration follows the order:

  Fully filled > Half filled > Partially filled

A lesser strength of metallic bonding, plus the stability collectively account for the dip.

Also, as a quick roundup of densities, density of $\ce{Os}$ is the highest: $\pu{22.59 gm cm^{-3}}$ and $\ce{Ir}$ stands quite close $\pu{22.56 gm cm^{-3}}$. Another fun fact, gold and mercury are also quite dense: $\pu{\rho_{Au} = 19.32 gm cm^{-3}}$, $\pu{\rho_{Hg} = 13.56 gm cm^{-3}}$. As a comparison (as to how dense I actually mean): $\pu{\rho_{Pb} = 11.34 gm cm^{-3}}$.

The low melting points of $\ce{Mn}$ and $\ce{Tc}$ are also mapped to half-filled stable configurations.


$\Delta G_m$ is the free energy barrier for the motion of atom from one site to another. The equation overall looks like a transition state approach to this activated process, and thus $\nu$ is the frequency at which this motion is attempted, i.e. a $\nu$ is a characteristic lattice frequency. In such models, it is often assumed to equal the Debye frequency.

Finally, $\omega$ is the rate of transition from one site to another, or something proportional to it (because the proportionality symbol is used rather than an equality). $\omega$ can be thus be the ionic conductivity, or a diffusion constant.

You can find the same expression and graph in this chapter of undergrad lecture notes, for confirmation.


What elements don't observe the octet rule?

Octet rule is more advisory than a rule; It's usually obeyed by the main group elements.
Exception are paramagnetic compounds (obvious reason); first for elements in periodic system (He-configuration), 3-valent boron compounds; depending on counting system boron clusters (f.e. dodecaborane with 6-coordianted boron); transition metal complexes (these usually follow 18e rule, however there are plenty of examples with 14, 16 or even 20 electron complexes).


Which acids are created by the combination of a gas and water?

I don't think there is a specific name for these acids. But I can think of at least 4 more. The gases are $HCl,\ SO_2,\ SO_3,$ and $NO_2$.


Identifying unknown solution with indicators. why is one pKa value ignored and how to treat negatives values?

As you noted, the $\mathrm{pH}$ is fairly close to 9.

It's probably quite different from the $\mathrm{p}K_{\mathrm{a}}$ value of $\alpha$-naphtholphthanlein because if you have any of the yellow acidic form around, the solution will take on a greenish tinge. One order of magnitude seems sufficient so, $\mathrm{pH} \geq 9$

The sample with thymolphtalein follows the same logic in that since we don't see any blue, there must be little of the base form, so the $\mathrm{pH} \leq 9$.

9 is very close to 8.8, the $\mathrm{pH}$ where we would expect equal forms of thymol blue. Indeed, $10^{0.2}\approx 1.05$, so both forms are present in solution in roughly equal amounts.

Further assuming that intensity of the two colors is similar, you would have a mix of yellow and blue, also known as green.


How could household vinegar create a salt on car carpet?

I once had a humidifier and I ran it with tap water.  I soon noticed a build up of white dust on the carpet right in front of the humidifier.  I suspect your situation might be related to what caused my white dust.  Tap water contains salts (just like table salt), the harder the water the more salts it contains.  They are present in low concentration so you don't normally see them.  In my case, as more and more tap water evaporated it led to a visible build up of these salts.  In your case, as the water on the carpet dried, it left behind the same salts.  Maybe you really soaked your carpet with a lot of water, maybe your water is very hard, and finally, maybe salts were actually put into the carpet during their manufacturing process - and maybe washing them with vinegar (dilute acetic acid) solubilized the salts and brought them to the surface, making them visible upon drying.


Separation of alkylated 1,2,4-triazole in solution

Joule and Mills Heterocyclic Chemistry [1] confirms that 1,2,4-triazoles alkylate with ~9:1 selectivity for N1 over N3, citing a paper in Tetrahedron Letters  from an industrial group at Merck as the source of the data[2].

The ratios reported were based on 1H-NMR of the crude product.[#] In this case, all of the compounds formed are known compounds, and as such it was easy to compare the data for 1-alkyl vs 3 alkyl. If the compounds formed were completely novel, it may be harder to tell which is which, though more complex NMR methods such as H-N or C-N correlation are often used in the pharmaceutical industry where this issue of regioselectivity often arises.

As an aside, the paper points out that the N1 and N3 alkylated products are easily separable, in this case by distillation, but in normal cases, silica gel chromatography (or HPLC if needed) can usually be used to separate the different regioisomers, with sufficient effort!

[1]: Mills, K.; Joule, J. A. Hetrocyclic Chemistry; Wiley:Chichester- Sussex, 2010

[2]: Bulger, P. G.; Cottrell, I. F.; Cowden, C. J.; Davies, A. J., Dolling, U-H. Tetrahrdron Lett., 2000, 41, 1297

[#]: The reaction was worked up, but not purified prior to working out the ratios. This is good practice, as it assures that you haven’t accidentally columned away some of the minor product, as is often the case


Say conductivity of the electrolyte is represented by $\kappa$, molar conductivity by $G$, resistance by $R$ and resistivity of the electrolyte by $\rho$
I'm assuming you know the definitions to the above since you've asked this question.
Now, resistance $R$ = $\frac{\rho\space l}{A}$ where $l$ is the length of the conductor and $A$ is the area of cross section.
For an electrolyte, $l$ can be taken equal to the distance between the electrodes and $A$ can be taken as the surface area of the electrodes.
Since conductivity is the reciprocal of resistivity ($\kappa$ = $\frac{1}{\rho}$),
$R$ = $\frac{\space l}{\kappa\space A}$
Reciprocating both sides,
$\frac{1}{R}$ = $\frac{\kappa\space A}{l}$
Since molar conductitivy is defined as the conductance of the entire solution having 1 mole of electrolyte dissolved in it and conductance is $\frac{1}{R}$,
$G$ = $\frac{\kappa\space A}{l}$
where $G$ is molar conductivity


Reaction kinetics, relating half life to reaction rate

Recall the meaning of the half life $t_{1/2}$: At $t_{1/2}$, a concentration (or partial pressure) is decreased to the half of its initial value.

It is crucial to realize that this represents exactly the $\mathbf{-\frac{dp_A}{dt}}$ in your rate equation!

With other words, $-\frac{dp_A}{dt} = k\cdot p_A\cdot p_B^2$ becomes $-\frac{54}{2\cdot100} = k\cdot54\cdot(1.0)^2$ for your first set of data.

Solve for $k$ and do the same for the other data sets. What is the result?


The short answer is yes. Machine learning, data mining, AI and other techniques are highly useful in chemistry.

I completely agree with Fred's answer that lots of machine learning, expert systems and statistical analysis in chemistry goes back a long time. This is particularly true in analytical chemistry - match a mass spec or NMR or IR against a library of known compounds.

Now I saw your tag "computational chemistry" and there are some newer applications here. Basically the question is whether you can predict properties (e.g., the heat of formation, dipole moment, etc.) that would normally require quantum mechanics.

There are a few groups attempting this, but one of the more successful attempts so far has come from Anatole von Lilienfeld.

  The researchers found that with a landscape of more than 5000 molecules, the error for predicting atomisation energies of new molecules drops below 10kcal/mol, approaching the 5kcal/mol accuracy of hybrid DFT. 'Calculating a molecule's atomisation energy using hybrid DFT would take on average one hour on a single CPU,' says von Lilienfeld. 'With machine learning, it's milliseconds.'

There are more conventional approaches to use machine learning, evolutionary algorithms and the like to perform parameterization of force fields, semiempirical quantum mechanics, etc.

Basically, if you have a lot of data, machine learning techniques can be effective ways to analyze the data and use it for other purposes.


Keep in mind, that the reason reactions occur spontaneously is because there's an overall driving force: $\Delta G = \Delta H - T\Delta S$. Since there are fewer molecules on the right hand, we can guess that $\Delta S$ isn't driving this reaction. ($G$ is Gibbs free energy, $S$ is entropy, $H$ is enthalpy, and $T$ is temperature.)

So then your question boils down to "why is this reaction favorable for $\Delta H$?"

Even before I look up the bond enthalpies for each of the bonds involved, I have an intuition. Yes, we're breaking the double bond in cyclohexene. We're also breaking the $\ce{Br-O}$ bond in $\ce{Br-O-H}$. So that's two bonds that are broken.

You're right -- that costs energy, so there's clearly something more energetically favorable on the other side. Well, in the product, I'm forming a $\ce{C-Br}$ and a $\ce{C-OH}$ bond.

So the net reaction breaks two bonds, and forms two bonds. As you say, the general intuition is that the "double" part of a double bond isn't quite as strong as the standard $\sigma$ single bond.

Let's look up the bond enthalpies also through Wikipedia:

$\ce{C=C}$ (vs. $\ce{C-C}$ which remains in the product): ~146 kcal/mol vs. ~84 kcal/mol = ~62 kcal/mol
$\ce{Br-O}$ = ~56 kcal/mol (I had to look this one up, so it's a bit iffy)
$\ce{C-Br}$ = ~69 kcal/mol
$\ce{C-O}$ = ~85 kcal/mol

Net $\Delta H \approx$ -36 kcal/mol.

Even if some of the bond enthalpies are off a bit, it's clearly energetically favorable. It has nothing to do with how the species are mixed, or really the temperature. (Well, my guess is that $\Delta S$ is not favorable here, since two molecules form one, so high temperature might make the reaction less favorable from a $\Delta G$ standpoint.) Nor do you need UV light or anything like that -- the difference in bond strengths will push this reaction.


Which Oxygen atom in HCOOH (formic/methanoic acid) does Carbon donate its electrons to, to obtain a partial positive charge?

Electronic effects are cumulative. For example, the central carbon atom in an orthoester like HC(OEt)3 is surrounded by 3 oxygen atoms which add up.

In the case of formic acid (which is what you described in your question), yes the carbon atom is donating its electron density to BOTH oxygen atoms. However, the oxygen atom involved in the C=O bond will attract more electrons, because the double-bond is shorter.


Reference request - Nomenclature of organophosphorus compounds?

The usual substitutive nomenclature for organic compounds was extended to various other elements including phosphorus. The current rules for name construction of organic compounds containing phosphorus are included in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). These recommendations cover typical organophosphorus compounds that are used as pesticides and chemical warfare agents. However, they do not include the nomenclature for organometallic compounds, coordination compounds, polymers, and some natural products.


Caution: This answer is exclusively from my perspective.

Moments after uploading this question I realized the answer myself. For those who strike upon this question. I 'think' the answer goes like this:

The combined gas law states that :

$\frac{PV}{T}=nR$ ----------  [or $PV=nRT$ ideal gas equation] (where 'n' is the number of moles of the gas in consideration)

Now therefore:

$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}$ = $n_1R+n_2R$ = $R\left(n_1+n_2\right)$

Now

$n_1+n_2=n_3$ (NOTE: $n_3$ is not the number of moles of a single gas but the gases $G_1$ and $G_2$ combined)

This can be said as the volume gets added up. Therefore further referring to Avogadro's Law (of equal volume) we the latter equation yields the equation in question.

PLEASE DO CORRECT THIS ANSWER IF IT IS FAULTY.


How to determine sulfate ions in a solution by precipitation?

Acquaint yourself with the so-called solubility rules. Depending on which cation sulfate is paired with, the sulfate compound may or may not be soluble. Generally, compounds containing sulfate ($\ce{SO_4^2-}$) compounds are soluble. (For essentially the same reason that sulfuric acid $\ce{H2SO4}$ is a strong acid, if you're interested.)

Well-known insoluble sulfate compounds are $\ce{PbSO4, SrSO4, CaSO4}$, and $\ce{BaSO4}$. Note that all of these cations have a somewhat high (+2) charge and are large.

If sulfate anions are present in your aqueous solution and you a a soluble compound containing $\ce{Pb^2+, Sr^2+}$, etc, then those cations will form an insoluble precipitate with the sulfate.


Equilibrium between barium carbonate, and barium oxide and carbon dioxide

Your conclusion is correct.

In the expression for equilibrium, we consider the activities for the solids to be 1. This ultimately leads to:$$K_p=p_{\ce{CO2}}$$

[Note that $K_P=f_{\ce{CO2}}$, where $f_{\ce{CO2}}$ is the fugacity of $\ce{CO2}$. Under "ideal" conditions, $f_{\ce{CO2}}=p_{\ce{CO2}}$, where $p_{\ce{CO_2}}$ is the partial pressure due to $\ce{CO2}$.]

In other words, the equilibrium partial pressure of $\ce{CO2}$ will be constant. One may compare this to the aqueous tension in case of water VLE.

Increasing the volume would cause more $\ce{BaCO3}$ to dissociate, producing more $\ce{CO2}$. The ideal gas law tells us that: $$n_{\ce{CO2}}\propto p_{\ce{CO2}}\cdot V$$

Your equilibrium pressure due to $\ce{CO2}$ does not change.


Why is there carbon lining on the steel vessel in Hall Heroults process?

An obvious answer is to why the presence of carbon, per a source Chapter 4: The Role of Carbon in Metallurgy on the historical extraction of iron from its ores, for example, to quote:

Carbon plays a crucial role in the chemical preparation of metals, particularly iron. ... This chapter discusses carbon as an intermediary product used in very large quantities to reduce and insulate metals and semi‐conductors from their natural oxides.

In a few words, carbon is a reducing agent, an insulator and a good oxygen scavenger.
Also, given the electrolysis nature of the Aluminum process, it is a cheap and very noble cathode. The employment of other noble metal cathodes would be more expensive and likely add impurities.


Why does boron add to the less substituted carbon in the hydroboration of an alkene?

For the hydroboration of isobutene

there is a steric aspect:

And though sterics may play some part, it doesn't explain near 100% regioselectivity. More important in determining regiochemistry, I would say, are the electronics of the transition state:

Boron is electrophilic, and as such will draw the electron density of the alkene π-bond towards itself. This C–B bond formation occurs earlier than the C–H bond formation, which in turn imparts a partial positive charge to the other carbon of the double bond in the transition state of the reaction.

This somewhat resembles a carbocation, which you might remember are stabilized by additional substituents via hyperconjugation. Thus, hydroboration results in an anti-Markovnikov product for mostly the same reason that acid catalyzed hydration results in a Markovnikov product — stabilization of positive charge.


How does one prove Avogadro's law from Gay-Lussac's?

His name was Amedeo Avogadro, not Amidio nor Amadeo nor Avegadro nor Avagadro. The full name was Lorenzo Romano Amedeo Carlo Avogadro, Count of Quaregna and Cerreto.
Avogadro's law states that

equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

More exactly, it also includes single atoms, e.g. for a case of noble gases, due the way how IUPAC defines a molecule.
But remember the Avogadro law implies ideal gases, similarly as the Gay-Lussac's law,Boyle's law and Charles' law. For ideal gases, sizes of molecules play no role. Both the size and volume of ideal gas molecules are considered negligible.
Real gases have deviations from this ideal behavior. See  van der Waals equation as the most used real gas state equation.

$$\left(p+a \frac{n^2}{V^2}\right)(V-n b)=nRT$$
$$\left(p+ \frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m} - b)=RT$$

where parameters $a$ (addressing cohesive forces ) and $b$ (addresing non zero molecular volume) are specific for the particular gas.
Then the molar amount of a real gas for given $V$, $p$ and $T$ is the solution of the cubic equation for  variable $n$ :
$$pV - n ( pb + RT ) + a \frac{n^2}{V} - ab \frac{n^3}{V^2}  = 0 $$
It clearly shows that the same volumes of different real gases, at the same temperature and pressure, contain ( usually slightly ) different counts of molecules/atoms.


What is the pressure point where deviations from ideal gas law occur?

It depends on the gas. If the molecules have strong interactions with each other you can see measurable deviations at pressures around 5 atm. But if you have a "more ideal" gas you may have to pressurize it to much higher pressures. There is no set value.


As with all the periodic trends concerning differences inside a single group, it will have something to do with the size of the atoms involved.

You already noted that fluorine is the smallest of the halogen atoms and iodine is the largest. This is due to there being only two core electrons in fluorine while iodine has 46. While adding an electron to any of these will increase their size, the size differences between them will remain the same due to said core electrons; hence, fluoride will be the smallest and iodide will be the largest.

This means that fluoride has a high charge-per-volume ratio while iodide’s is much lower (compare that to Pearson’s HSAB concept — it’s the same idea). A higher negative charge density means that it draws cations towards it better and thus stabilises the crystal structure. A secondary effect is that the smaller size of fluoride means that it must have a lower coordination number in the crystal structure, thus having less positively charged neighbours, thus attracting them more.

Together, these influences mean that it is just harder to break up the high ordering of the crystal structure to melt the salts the higher up you are in the group.

A higher degree of covalent bonding for iodine may play a role, but I would deem it neglectable.


Reaction between phenylamine and halogens

Chlorination of aniline with chlorine is rapid and gives 2,4,6-trichloroaniline according to the references cited in this paper here. The earliest reference cited is 1845 by AW Hofmann so it has been around a long time. More modern approaches use N-Chlorosuccinimide as the reaction is more controllable


Why does a mixture of siloxene and cerium(IV) sulfate luminesce?

The idea of chemiluminescence is pretty simple, you need an oxidizer and a substance that actually gives a glow.

According to Wikipedia cerium sulfate is rather an oxidizer especially under acidic condition, which actually explains why we need sulfuric acid, so basically, whatever is siloxene (I suspect it is a silicon oxide kind of structure) it is an oxidant that when oxidizes the difference in energy level causes a glow in the visible spectrum.

Unless more clarification about siloxene, this could be the only possible answer.

sources:

NurRage explanation about chemiluminescence
Wikipedia for properties about cerium sulfate

EDIT: what was meant by potential sources is that the method of using words was mine but the information is from the sources sorry for inconvenience


Does a weak acid undergo it's weak ionization upon the addition of any kind of base?

A weak acid does not (contrary to what you show in your equation) predominantly ionize before the base is added. But keep in mind that proton transfer reactions (at least those between oxygen and/or nitrogen atoms) occur almost instantaneously in both directions, so the equilibrium you illustrated is established immediately (for a weak acid, with the big arrow pointing left). So when (strong) base is added, H3O+ is consumed immediately, then more acid is ionized to re establish the equilibrium--and this goes on as you add more base until you have consumed almost all of the weak acid. These reactions are controlled by the relative energies of the reactants and products, and not by reaction rates.


What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree?

Is simply a dissolution no reaction occur and as said Niculau the pH change, if occur, is absolutely negligible.

Salt in general affect taste lowering the bitterness of the food enhancing the positive sensory attributes of foods. In fact the reason of this behavior is the sodium ion $Na^+$ (even litium ion works) and his interaction with the receptors in taste buds. So sodium acetate therethically works as well, Gery Beuchamp has studied many of this salt, quoting from NCBI:

  various sodium-containing ingredients have been known to reduce the
  bitterness of certain compounds found in foods, including quinine
  hydrochloride, caffeine, magnesium sulfate, and potassium chloride

This is an article about the suppression of bitterness by sodium. From here I'm afraid is more a question of physiology.

I read this the first time here:

This, Herve. Molecular gastronomy: exploring the science of flavor. Columbia University Press, 2006 p. 94


Why does multipying a scalar in both sides of a balanced chemical equation, change related quantities?

Maybe because the equilibrium constant, Gibbs energy, enthalpy, etc. are not the fundamental characteristics you think they are.

Really, what does enthalpy mean, for example? Say, a reaction $\ce{A + B -> C + D}$ has certain enthalpy. This means that running a reaction with 1 mole of A and 1 mole of B gives us certain amount of heat. Now let's multiply everything (including coefficients and enthalpy) by 2. That would be $\ce{2A + 2B -> 2C + 2D}$. In other terms, running a reaction with 2 moles of A and 2 moles of B gives us twice as much heat, just like we would expect.

Now think of the equilibrium constant. Say, we have a reaction $\ce{2NO2<->N2O4}$ and the constant just happens to be 4. This means ${\ce{[NO2]^2\over[N2O4]}}=4$. OK, and what if we write this reaction in a different way, like $\ce{4NO2<->2N2O4}$? Obviously, the equilibrium concentrations are still the same, but the expression for the equilibrium constant has changed to ${\ce{[NO2]^4\over[N2O4]^2}}$. That's a square of the previous expression, hence its numeric value would be $4^2=16$. So what? It's not like we can just up and measure this constant with some kind of "constantometer", anyway. It's the real concentrations that matter; and if we calculate the concentrations using either version of the constant, they will turn out the same.

This is also in accordance with $\ln K = {\Delta G\over RT}$. When you square the constant, its logarithm doubles, and so does $\Delta G$.


Use of geometry index for the determination of coordination environment

For your first question, the original paper (J. Chem. Soc., Dalton Trans. 1984, 1349–1356) that described the geometry index $\tau_5$ defined it as an "index of trigonality". For example, they write for a compound with $\tau_5=0.48$

  By this criterion, the irregular co-ordination geometry of $\ce{[Cu(bmdhp)(OH2)]2+}$ in the solid state is described as being $48\%$ along the pathway of distortion from square pyramidal toward trigonal bipyramidal.

For your second question, while I haven't been able to find a $\tau_7$ or $\tau_8$ used in the literature, it seems possible to define such parameters under the right conditions. To devise a $\tau_8$, we can see that for a regular cube $\ce{MX_8}$, there can only be bond angles of $70.5^\circ$ (between adjacent $\ce{X}$ in the same square) and $109.5^\circ$ (between opposite corner $\ce{X}$ of the same square or between corner $\ce{X}$ of different squares). However, an antiprism instead has an angle of $99.6^\circ$ separating the $\ce{X}$ of different squares.

(Image obtained from Inorganic Chemistry by Miessler and Tarr)

This suggests using a formula reminiscent of $\tau_5$ to define $\tau_8$ as the antiprismatic distortion index. One possibility is $$\tau_8=\frac{\beta-\alpha}{9.9^\circ}$$ where $\beta > \alpha$ are the two largest valence angles and $9.9^\circ$ is a normalization factor to make it between $0$ and $1$. So when $\alpha=\beta=109.5^\circ \to \tau_8=0 \to$ cubic geometry and $\alpha=109.5^\circ$ $\beta=99.6^\circ \to \tau_8=1 \to$ antiprismatic geometry.

This will only work if the structure is a regular antiprism (i.e an anticube). The same is true for defining $\tau_7$ between a pentagonal bipyramid and a monocapped trigonal prismatic. This is because the angles for these will vary if all the attached groups are not the same and so a consistent scheme based on the angles would not suffice.

I also imagine that $\tau_7$ would be harder to define in this way because I don't think there is pair of angles that on its own could describe the distortion between the two geometries.


Why the bond strength of HCl is more than HBr?

You can think of the overlapping volume between the electron clouds, but a simpler form to show why this happens is when you consider the electrical potential between the atoms. Even tough this isn't an ionic compound, it has many characteristics that are similar.

If I remember correctly, this is the expression for the potential energy between charged particles: $E_p = k\frac{Q_1 \cdot Q_2}{d}$

Since the radius of $\ce{Br}$ is greater than the radius of $\ce{Cl}$, the potential energy involved in $\ce{H-Br}$ bond will be smaller than $\ce{H-Cl}$ bond.

Another thing that came to my mind, and I'm not sure, is that as the 4p cloud of $\ce{Br}$ is bigger than that of $\ce{Cl}$, the relative amount of the volume that is overlapped with the 1s cloud of $\ce{H}$ gets smaller with $\ce{Br}$ than with $\ce{Cl}$, which means the connection is weaker.


Why is the entropy of container A the greatest?

An accurate summary of the relative molar heat capacities of the different species is given in another answer and recapitulated in the following table together with the total heat capacity for each gas:

$$\begin{array}{|case|Cv|} \hline \textrm{molecule} & C_{vm}/R & NC_{vm}/R \\ \hline  \mathrm{A} & 3/2 & 1.5 \\ \mathrm{B} & 5/2 & 1.25  \\\mathrm{C} & 7/2 & 1.75 \\ \hline \end{array}$$

where $N$ is the total amount of gas, which based on the problem described in the OP is either $\pu{1 mol}$ of atoms ($\mathrm{A}$) or $\pu{\frac{1}{2} mol}$ of homonuclear diatomic molecules ($\mathrm{B}$ and $\mathrm{C}$).

Since the entropy change for the isochoric temperature change is $\Delta S = C_v \log\left( \frac{T_2}{T_1} \right)$, it follows that $\Delta S_C > \Delta S_A > \Delta S_B$.


Yield of alkyl chloride in the Hunsdiecker reaction

The Hunsdiecker reaction is a halogenation-decarboxylation reaction of silver salts of carboxylic acids.

To understand what causes the yield to change, first we have to consider the mechanism of the reaction.

As seen from the mechanism the homolytic cleavage of the $\ce{O-X}$ bond is the slowest step in the reaction and thus is the rate determining step. So as you predicted it all comes upon the stability of halogen radical. It's commonly known that the bromine radical is fairly stable than the chlorine radical and thus the Hunsdiecker reaction gives better yield with bromine rather than with chlorine.


What the balanced chemical equation for the conversion of starch to dextrin by dry heating?

Let's take a look at the structure of both molecules. First, starch:

Now, dextrin:

Do you see the difference? They are the same molecule (both are glucose polymers), the only difference is the molecular weight, or the length of the polymer chain.

Starch, by convention, is 300-600 monomers long. Dextrin is something less than that.

Since the number of dextrin molecules you get will depend on the polydispersity (size distribution), average chain length, and extent of the reaction, there is no way to write a balanced reaction equation for a real reaction.

You could assume that a starch molecule of a given length breaks into dextrin molecules of a given length, and write an equation. It would look like this:

$\ce{C6H11O5[C6H10O5]_{m}C6H11O6 + (\frac{$m-n$}{2})H2O->} \frac{(m+2)}{(n+2)}\ce{C6H11O5[C6H10O5]_{n}C6H11O6}$

This would only work for one length of starch and dextrin (in reality there could be 10 - 1000) and would also only be true if $m > n$, and $m$ is evenly divisible by $n$.

Also, in the question you specify dry heat, but there is no way to balance the equation without including water (unless you use hydrogen and oxygen gas). I suspect that even under dry heat conditions, there is moisture on the bread, and it is consumed in the reaction. This could be part of the reason that toasting drys bread.


Why doesn't limestone thermally decompose instead of turning to marble?

Short answer:

It's not hot enough, and the pressure is too high.

Long answer:

You say:

  intense heat and pressure

The key word is pressure rather than heat. Let's look at the phase diagram of the CaO and CO2 compositional system:

(Calculated and plotted using Perple_X)

Pure limestone and marble are composed of calcium carbonate: CaCO3. In air, the temperature required for calcium carbonate decomposition is 898 °C (considering simple equilibrium and ignoring any vapour pressure effects). As pressure increases, the temperature requirement for this equilibrium decomposition increases very rapidly. It increases so fast that in fact, at above 80 bar, the calcium carbonate will melt into a liquid (call it calcium carbonate magma or lava) instead of decomposing to lime and carbon dioxide.

While 80 bar may seem a lot for us humans (it is higher than the pressure exerted by a 45 kg woman wearing stiletto heels), it is nothing compared to the pressure below us, deep in the Earth. Marble forms at a wide pressure range, but for this discussion let's say that it's somewhere between 2 and 10 kbar. That's 2000 to 10000 bar! Way off the scale of the phase diagram. Also, marble forms in the temperature range of 300 to 800 °C. Again, it's not even enough to melt the calcium carbonate, let alone decompose it to gas.


Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:
$$\ce{A + B <=> C + D}$$
We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$
Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:
$$\Delta G^0=-RT\ln K$$
(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)
Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:
$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$
For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium  constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:
$$\ce{A + B -> C + D}$$
Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.


Confusion in half filled or full filled electronic configuration

Having a half-filled or filled subshell is stabilising, as you say. But for Cl, the difference in energy between 3s and 3p is greater than the additional stabilisation from filling the 3p subshell. 3d and 4s are closer in energy, so for Cr it is favourable to promote a 4s electron and have a 4s1 configuration.


The flow rate of the heating fluid $F_\mathrm j$ probably refers to the mass flow rate (e.g. expressed in $\mathrm{kg/s}$) since the indicated flow rates into the heat exchanger and out of the heat exchanger are identical. (The corresponding values for the volume flow rate (e.g. expressed in $\mathrm{m^3/s}$ would be different since the change in temperature causes a difference in density.)

$$F_\mathrm j=\frac mt\tag{1}$$

The change in temperature of the heating fluid $\Delta T$ is given by the different temperatures of heating fluid before and after the heat exchanger:

$$\Delta T=T_\mathrm j-T_\mathrm{ji}\tag{2}$$

The ratio of the amount of heat energy $Q$ (e.g. expressed in $\mathrm{J}$) that is transferred to a system at constant pressure and the resulting increase in temperature $\Delta T$ is given by the heat capacity at constant pressure $C_p$ (e.g. expressed in $\mathrm{J/K}$):

$$C_p=\frac Q{\Delta T}\tag{3}$$

You may know the value of the specific heat capacity at constant pressure $c_p$ (e.g. expressed in $\mathrm{J/(kg\cdot K)}$) for the heating fluid at the average temperature of the heat exchanger, which is defined as heat capacity at constant pressure divided by mass $m$:

$$c_p=\frac{C_p}m\tag{4}$$

Now you can rearrange the definition of specific heat capacity $\text{(3)}$ to solve for the heat $Q$:

$$Q=C_p\cdot\Delta T\tag{5}$$

insert the result for $C_p$ from equation $\text{(4)}$

$$Q=c_p\cdot m\cdot\Delta T\tag{6}$$

the result for $\Delta T$ from equation $\text{(2)}$

$$Q=c_p\cdot m\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{7}$$

and the result for $m$ from equation $\text{(1)}$

$$Q=c_p\cdot F_\mathrm j\cdot t\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{8}$$

Therefore, the thermal power $P$ of the heat exchanger (e.g. expressed in $\mathrm{W=J/s}$) is

$$\begin{align}P&=\frac Qt\\[6pt]
&=c_p\cdot F_\mathrm j\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{9}\end{align}$$

The sign of the values for $Q$ and $P$ depends on whether heat is transferred from the fluid to the reactor or from the reactor to the fluid.

Note that the specific heat capacity $c_p$ generally depends on temperature; therefore, the specific heat capacity of the fluid changes during the passage of the heat exchanger. That is why you should use an appropriate average value. A precise engineering calculation would better be based on the specific enthalpy $h$ of the fluid at the given temperature and pressure before and after the heat exchanger.

Example

The characteristic data of a typical nuclear power plant with a pressurized water reactor are

Coolant mass flow rate $F=18\,800\ \mathrm{kg\ s^{-1}}$
Coolant pressure $p=158\ \mathrm{bar}=15.8\ \mathrm{MPa}$
Reactor pressure vessel inlet temperature $T_\text{in}=291\ \mathrm{^\circ C}$
Reactor pressure vessel outlet temperature $T_\text{out}=326\ \mathrm{^\circ C}$

The specific heat capacity of water at the inlet temperature of $T_\text{in}=291\ \mathrm{^\circ C}$ and a pressure of $p=15.8\ \mathrm{MPa}$ is $c_{p,\text{in}}=5.2553\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$. The corresponding specific heat capacity at the outlet temperature of $T_\text{out}=326\ \mathrm{^\circ C}$ is $c_{p,\text{out}}=6.4388\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$. For simplicity’s sake, we may want to use the specific heat capacity for the average temperature $\tfrac12\left(T_\text{in}+T_\text{out}\right)=308.5\ \mathrm{^\circ C}$, which is $c_p=5.6787\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$.

Using equation $\text(9)$, the thermal power of the reactor can be calculated as

$$\begin{align}P&=c_p\cdot F\cdot\left(T_\text{out}-T_\text{in}\right)\\[6pt]
&=5.6787\ \mathrm{kJ\ kg^{-1}\ K^{-1}}\times18\,800\ \mathrm{kg\ s^{-1}}\times\left(326\ \mathrm{^\circ C}-291\ \mathrm{^\circ C}\right)\\[6pt]
&=3736584.6\ \mathrm{kJ\ s^{-1}}\\[6pt]
&\approx3737\ \mathrm{MW}
\end{align}$$

By way of comparison, the literature value for the reactor core thermal power is $3765\ \mathrm{MW}$.

Note that, in this example, heat is transferred from the reactor to the fluid.


Why does copper not react with nitrogen?

Sketching with broad strokes: the first key point is that oxygen and nitrogen are major components of the atmosphere in the form of $\ce{O2}$ and $\ce{N2}$ molecules, and not as ions. $\ce{N2}$ is a very stable compound because the two nitrogen atoms in its molecules are united by a particularly strong triple bond. $\ce{O2}$ is overall more reactive than $\ce{N2}$, largely as a reflex of the bond between the atoms not being as strong. Specifically concerning oxidation reactions (of which those you suggest are typical examples), oxygen atoms having more affinity for electrons than nitrogen atoms contributes towards making $\ce{O2}$ being a better oxidiser than $\ce{N2}$.


Why is the carbon-carbon bond in hexachloroethane shorter than in ethane?

One reasoning would be applying Bent's rule:

In ethane the $\ce{C-H}$ bonds are approximately composed from $\ce{61\% sp^{3.3} C}$ and $\ce{39\% s H}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{2.3} C}$ orbitals.
In hexachloroethane the $\ce{C-Cl}$ bonds are composed from $\ce{48\% sp^{3.7} C}$ and $\ce{52\% sp^{4.8} Cl}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{1.8} C}$ orbitals.

Higher $\ce{s}$-character in bonds usually means that the optimal overlap of these orbitals is at shorter inter-nuclear distances. Since the $\ce{s}$-character in the $\ce{C-C}$ bond in hexachloroethane is much higher than in ethane, the bond length is expected to be shorter.

(Values obtained from an Natural Bond Orbital Analysis on the DF-BP86/def2-SVP level of theory.)


Two issues arise immediately: 1) dissolving a variable thickness of iron carbonate (probably also of varying composition) and 2) not affecting the fossilized bone (which also probably has varying composition). A third issue might be that the iron carbonate seems to be either very tightly bonded to the bone, or perhaps even part of it by being part of the first substitution layer.

If the fossil is known to be mostly silica, acetic acid (vinegar) could work. Best method would be to scrub the fossil under the surface of the solution with frequent examination to see where additional scrubbing is appropriate, and where not.

If the fossil is largely calcium carbonate, recourse could be to oxidize the ferrous to ferric using a) calcium nitrite solution or b) hydrogen peroxide (H2O2). Calcium nitrite is a commercial corrosion inhibitor for rebar steel in concrete, and functions to oxidize ferrous to ferric at pH 11-12. The product at high pH is ferric oxide (hydrated). The aim in this process is to chemically attack the iron carbonate and turn it into something else that is not attached to the fossil. Scrubbing under the liquid surface (using rubber gloves!!) could turn the iron carbonate to a powdery product that is not attached to the fossil, at high enough pH to avoid dissolving calcium compounds. The chemical reaction is:
2 Fe++  +  2 NO2-  +  2 OH-  -->  Fe2O3  +  H2O  +  2 NO.
The NO will oxidize in air to NO2 which forms nitric and nitrous acids in water, and is toxic, but so is the nitrite, so you have to be very careful not to breathe the vapors or leave the oxidizing solution on your skin.

Perhaps a safer method would be to use hydrogen peroxide. This reaction is well-known, but is usually done in acid; a pH of 5 should be safe enough for the fossil, even if it is carbonate-based. Doing the reaction in acid keeps the ferric iron soluble, but hydrogen peroxide is also reactive in alkaline solution and a high pH might be more suitable for carbonate substrates. The reaction is:
H2O2  + 2 H+  +  2 Fe++  -->  2 H2O  +  2 Fe+++.
There is the added advantage that using excess peroxide causes foaming which could help remove particles from the fossil. Foaming causes mist which should not be breathed, and rubber gloves should be worn because H2O2 will eat your skin if you leave it on you for any significant period.


In the case of strong acid-strong base titration at equivalence point with the addition of very small amount of titrant (like a drop) the pH change is very large. That means the range over which indicator changes its color is likely to be passed in a drop of titrant, resulting in a sharp end point.

Whereas in the other cases the change of pH at equivalence point is less dramatic and there will be a smaller change in pH at equivalence point. So, the indicator will change its color slowly resulting in a wider end point.


What is the diagram of the roto-inversion of $\bar 6$?

The image below gives some example of different rotation inversion operations. For the $\bar 6$ operation rotate by 60 degrees (black dot to open circle in the direction towards 3) then invert through the centre of coordinates to the opposite side. Inversion changes a point at (x, y, z) to (-x, -y, -z) thus, in the diagram point 1 becomes point 2 after the whole operation, then 2 becomes 3 and so on. Two $\bar 6$ operations moves 1 to 3 and 6 operations places it back where it started.

[The image is from 'Crystallography, An introduction ' by W. Borchardt-Ott, 3rd ed publ Springer.]


Terminology for describing nuclei participating in metallic bonds

Both your statements mean the same thing, however it is convenient to think of metals as kernel + sea of electrons, as it greatly simplifies visualization, conceptualization and thus calculations. From a purely numeric viewpoint, i prefer the 'sea of free electrons' model or that cations are surrounded by free electrons.


Product of primary amine and acid anhydride

It should be very clear that a primary amine is a nucleophile and an acid anhydride an electrophile. In the series of carbonyl and carboxyl derivatives, an acid anhydride is classified as quite reactive: more so than aldehydes or ketones, much more than esters or amides but less than acid chlorides. Thus, we should all agree that the first step is the nucleophilic attack on the $\ce{C=O}$ π system via a tetrahedral intermediate to give N-methyl succinic monoamide.

Scheme 1: Mechanism of the monoamide formation.

With that, we can consider whether further reactions will take place. Where we originally had an anime and an acid anhydride, we now have a carboxylate (the $\ce{C=O}$ bond that wasn’t attacked) and a secondary amide. The secondary amide still has a lone pair we might consider to be nucleophilic — but that lone pair is taking part in amide resonance and is thus not available for nucleophilic attack (at least not well). And to remove the amide’s proton, we would need a strong base which we don't have. Thus, the amide side of the equation is far less reactive.

What about the carboxylate? Thus highly symmetric structure is also very unreactive towards nucleophilic attack in the same way as an amide. Due to the negatively charged oxygen donating one of its lone pairs to the π system, the carbon atom is much less electrophilic than in an acid anhydride. Thus, in this case also there is no desire to react.

Since we have two functional groups of low reactivity, the reaction ends here. However, with enough heat (supplying activation energy), it is possible for the sluggish intramolecular reaction between the the amide and the carboxylate to occur. In principle, the mechanism is the same except that you must transfer a few protons and remove water.

Scheme 2: potential onwards reaction to succinimide under harsh conditions.


What is a good method to measure the redox potential of a cellular system?

There are many complexes of d-elements, that can reversibly lose/gain one electron, changing their color in process. They are known as Redox Indicators. Some of them are stable enough to be used in vivo.


If you write out the electron configuration of Fe, you will find it has $3\text{d}^6 4\text{s}^2$ in its valence level, meaning it can theoretically take on any oxidation state from +1 to +8 (and even some negative states, for that matter).

In practice it is extremely uncommon to find any oxidation states other than +2 or +3 for iron - which explains why a simple periodic table containing only the most common naturally occurring states will omit them - but they are possible, just not very stable or long-lived. The polyatomic ion you have observed is called ferrate.


How do I find mass equivalence of a photon?

$$E = \frac{h\cdot c}{\lambda}$$

$$1\ \mathrm J = 6.24150934 \times 10^{18}\ \mathrm{eV}$$

then you have
$$E = mc^2 = \frac{h\cdot c}{\lambda}$$


What does being an "elementary substance" mean?

From LATINTOS:

  An elementary substance is a pure chemical substance that consists of atoms belonging to a single chemical element.

Now you should be able to solve your question.


Here are some links to the mechanism of the degradation of spermidine and some analogues in living systems.

reference 1
reference 2
reference 3
reference 4

Based on these reactions I would guess that an oxidative pathway is followed here as well, once the sample has been exposed to air.  Instead of a nice enzyme catalyzed oxidation, here we would have a free radical oxidation. Here is a general outline of such a process.  The first step would involve generation

of the more stable secondary nitrogen radical (as opposed to the primary terminal nitrogens) and formation of the nitrogen hydroperoxide.  Elimination of hydrogen peroxide could produce the two possible imines which could be further oxidized by oxygen to produce both possible diamines and both possible amino-aldehydes.  You could test your material for the presence of hydrogen peroxide, but the smell of putrescine ought to be a "dead" giveaway.


Can I somehow 'reuse' the chemicals in glow sticks?

Here is a generalized version of what's going on chemically.  The key steps are dioxetane formation (the highly strained 4-membered ring) and its subsequent decomposition and energy transfer to a fluorescent dye, which then decays back to its ground state by emitting light.  By using different fluorescers, you can change the wavelength (color) of the light emitted.

Each company that manufactures glowsticks has their own patent(s) and proprietary process, hence the variation from product to product.

  So I was wondering if I could (pick one or more):
  1.Recycle the chemicals in it to make it glow again
  2.Use the chemicals for other interesting (and maybe even useful) experiment

1) Since the starting oxalate ester and concentrated hydrogen peroxide are consumed in the reaction, to make it glow again you would need to put in a new batch of both the oxalate ester and concentrated hydrogen peroxide.  You probably wouldn't want to open the glowstick because:

there might be some concentrated hydrogen peroxide remaining (if I were the manufacturer, I would want this cheap reagent present in a slight excess to make sure all of the ester is consumed)
many of the commonly used fluorescers are irritants, both externally and internally
phthalate esters are commonly used as solvents and they pose health risks

2) Even after all of the starting materials are used up, the fluorescer still remains in solution.  Exposing the glowstick to a black light will still cause it to fluoresce.


There is a famous saying that paper never refused an ink, which means anyone can write anything on the paper and the paper won't say anything back to the writer. Internet has made this situation worse. One should develop a healthy level of skepticism when reading website and forums.

  Why is copper the only transition element to give a coloured flame? Most other flame tests are given by alkali metals and alkaline earth metals, which is understandable and is a common trend across the group. Copper stands out as an exception.

This is not true. Boron compounds, when burned in alcohol, burn with a similar green color. Now here is the fundamental point to understand. When we see colors in the flames, usually it is due to very simple molecules of those metallic compounds in the flame. It is a molecular emission. If you can buy a pocket spectroscope, you will just a broad bands of color. So a flame spectrum, specially a low temperature flame, is not equal to an atomic emission spectrum. If I introduce copper into a very hot flame (>2000 Kelvin), you will not see anything, because copper atoms emit light mainly in the UV range.

Therefore all the common flame tests with calcium, barium, strontium show the flame emission spectrum of simple diatomic molecules of Ca/Ba/Sr in the flame. Copper is no exception. You may also wonder why a hydrocarbon flame is blue, there is no transition element there?

Vanadium is yellowish green

Cerium makes the flame yellow

Lead is blue white

Yttrium is beautiful crimson, which often appears on textbook covers on inductively coupled plasma spectroscopy (very high temperature atomic emission). Wikipedia has a big list.

Atomic spectroscopists use Grotrian diagrams to assign the transition lines the emission spectrum. Nobody uses VBT to explain the spectrum. Predicting the correct emission spectrum even of a diatomic molecule in the flame, will perhaps require the knowledge level of a PhD in physics.


Is it accurate to describe these radiation measurement units in terms of where-it's-measured and relationship-to-time?

Maybe some numerical values may help.

Radioactivity. The activity is the number of disintegrations occurring per second in a given source. The unit is the Becquerel : $1 Bq = 1 s^{-1}$. For example, the experience shows that $3.7·10^7$ nucleus are decomposed per second in $1$ g $\ce{Ra-226}$. The activity of $1$ g $\ce{Ra-226}$ is $3.7·10^7 Bq$ which was previously called $1$ Curie ($1$ $Ci$). One Curie is a huge activity for usual measurements in the laboratory.

Exposure dose. If a gamma-ray radioactive source is in air, the gamma rays ionize the air around. As a consequence, they deposited some energy in a small volume of air situated at a distance from the source. If the small air sample is far away from the source, the amount of energy is smaller than if it is near the source. The amount of energy so absorbed by the air sample is called the dose (or exposure dose) and is determined with dosimeters, and it is measured in $J/kg$ air. The unit of dose is röntgen ($1~r$), equal to $0.01 J/kg$.

The experience shows that $1 Ci$ of a gamma-ray emitter delivers a dose of $1$ röntgen per hour ($1 r/h$) in air at $1 m$ distance from the source. This value is  somewhat dependent on the nature of the radioactive source. Let's forget about this difficulty for the present purpose. Most important, the dose decreases with the square of the distance. So, to be consistent with the units, the dose delivered by a gamma-ray source is $1 r·m^2 Ci^{-1} h^{-1}$. A small volume of air (some milliliters) situated at $1 m$ distance from a $1 Ci$ gamma source, absorbs in one day ($24 h$) a dose equal to $(1~r·m^2·Ci^{-1} h^{-1})·1 Ci·24 h/(1 m)^2 = 24~ r = 0.24 J/kg $. This energy is finally transformed into heat.

If the irradiated matter is not air, but another substance, like water, the amount of energy is called absorbed dose and is expressed in Gray, abbr. $Gy$ ($1 Gy = 1 J/kg$). The experience shows that an exposure dose of $1$ röntgen delivers an absorbed dose of $0.01 Gy$ in water, like in air.

Living tissues are mostly made of water. But they don't react the same way to a given absorbed dose. Some molecules like ADN are much more sensitive to a given adsorbed dose than water. To compare the biological effects of a given dose expressed in Gray, it is necessary to define a new parameter, the effective dose, which is expressed in Sievert, abbr. $Sv$. $1 Sv = 1 Gy$ in water. But $1 Sv$ = up to $20~Gy$ in sensitive tissues.

The present explanation is just a first approach of the domain of dosimetry and radioprotection. As quickly mentioned in the text, the energy deposited in a target depends also on the energy of the gamma ray, which was forgot here. And we have not spoken about alpha and beta rays, whose dosimetry is not similar to gamma rays. Dealing with these problems should take hours, and cannot be done in the present message.


How does hyperconjugation explain stability of partial positive charges?

The solution is referring to the partial positive charge on the carbon opposite the developing C-B bond. It is stabilized by substituents similar to how a full carbocation is stabilized by substituents, e.g., in the tert-butyl carbocation. This doesn't need to be a completely empty p-orbital—electron density will be dynamically influenced by any imbalance in charge distribution—but if it's easier to conceptualize, you could imagine it as the pi-bond electrons being fully transferred to the C-B bond, leaving an empty p-orbital on the opposite carbon, which is stabilized by hyperconjugation.

The reality should be closer to a dynamic process at both sites, though. The pi-bond transfers electron density into the C-B bond, while the substituents stabilize the rising positive charge on the other carbon. As the C-B bond develops, the B-H bond simultaneously weakens, and begins moving its electron density toward the partial-positive carbon.


Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid?

The hydrated forms might be an azeotrope which means there's a content of water present that is difficult to remove via distillation techniques which may be how the P-toluene sulfonic acid was prepared.

The anhydrous form would require further processing.

The moisture present in the hydrate forms may affect the melting point if it were acting as an azeotrope with the P-toluene sulfonic acid, also it wouldn't be as high purity as the anhydrous form which would display the true melting point.


Equilibrium of more than two inter reacting species


When you say:

 When I draw the structure of it in such a way so that the 2 CH3 groups are
 adjacent to each other (on paper ) and take its mirror image I see that I got a
 mirror image which is not superimposable on the original structure.

and then:

On the other hand when I draw the structure such that the 2 CH3 groups are on
opposite sides of the central carbon and then take a mirror image of it ,
rotate it by 180° in the plane of the paper I get the original structure.

This tells me you are drawing structures where the bonds are at right angles to each other, like this:

     F                              CH3
     |                              |
 CH3-C-CH3        and           CH3-C-H
     |                              |
     H                              F

This will make the seeing of the structures much harder, you can't quite see the 'tetrahedral'.  In fact both of the drawings you have indicated are the same, only the connectivity matters -- that is the central carbon is bonded to a fluorine, a hydrogen and two methyl groups.

This is a good place to use models.  And if you don't want to buy tinker toys, tooth picks and colored marshmallows (or cheeses) will work just as well.

To resolve your conundrum, expand requirements such that if there exists any arrangement of the groups in your drawing that can form a superimposable mirror image, the molecule is not chiral.

Also consider using dashes and wedges, such as:


Has anyone encountered the use of Omega to represent equilibrium?

I would guess $Ω^m$ notation is a personal invention and I couldn't find any sources that would standardize it.

The choice of these symbols for abbreviating "equilibrium" may be justified that the glyph "Libra" ♎ originates from the Greek letter Omega Ω.
Further, from Merriam–Webster Online:

  Equilibrium contains a root from the Latin libra, meaning "weight" or "balance". As a constellation, zodiac symbol, and astrological sign, Libra is usually pictured as a set of balance scales, often held by the blindfolded goddess of justice, which symbolizes fairness, equality, and justice. Equilibrium has special meanings in biology, chemistry, physics, and economics, but in all of them it refers to the balance of competing influences.

Uppercase letter "m" is probably because $Ω^m$ is a shortened form (contruction) of a single word that ends in the same letter: equilibrium.

$$\underset{Ω}{\underbrace{\text{equi}\underset{♎}{\underbrace{\text{libr}}}\text{iu}}}\underset{^m}{\underbrace{\text{m}}}$$


What is an alternate way of preparing a sodium hydrogen phosphate/sodium dihydrogen phosphate buffer through titrating an acid with a strong base?

This picture says more than a thousand words:

You want a mixture of $\ce{HPO4^2-}$ and $\ce{H2PO4^-}$. You can start with any species (in a solution with a certain pH) and add another species whose pH is "on the other side" of the desired pH. If the pH of your starting solution is lower than the desired pH, you can also titrate with NaOH.
What you actually do (of the six options) is determined by solubility of the solution and by availability. If you titrate two phosphate solutions of the same concentration, there will be no dilution effect. If you titrate with NaOH, you would used fairly concentrated NaOH to avoid dilution, but would have to add water to a defined volume (or measure the volume after reaching your desired pH) to figure out the concentration of your buffer.
You should not titrate with HCl because then you end up with extra sodium choride in your buffer.


To begin with, dissociating molecules to a plasma at the atomic level will not only create ions but also neutral atoms. These will indeed reassociate on cooling, and the question is how to predict what the products will be. In the extreme case of a simple system such as pure hydrogen gas (dissociating, then recombining to H2) or, say chlorine plus bromine dissociating/recombining to give BrCl, it should be easy to predict the products formed and perhaps more about the behavior of the system as a function of temperature.

If you have a more complex system, including (for the sake of argument) atoms such as carbon that can form complex ring and chain structures, predicting the outcome will be impossible - since (1) the different ways to combine atoms may lead to a huge number of possible intermediates, many of whose energies/reactivity are not known; (2) even if you could predict these intermediate structures, the number of further possible reactions between them becomes astronomical, and (3) kinetic barriers (activation energies) for  reactions are generally harder to predict that ground state energies. As is often the case when theory comes up short, you may have to go into the laboratory and do some experiments. Possible approach: aim the plasma (along with a stream of inert gas such as argon) at a supercooled probe in a vacuum and spectrally characterize the species in the resulting matrix. I imagine that you could end up with some pretty amazing chemical species that way.

I would urge those among my colleagues who are familiar with the current state of theoretical chemistry to add to this discussion. I love a question that makes you go back to basics and rethink everything from the ground up.


Many different carbon ions exist in the gas phase.

In Electronic spectra of linear carbon anions, spectra of $\ce{C_2^-}$, $\ce{C_3^-}$... $\ce{C_10^-}$ are discussed.

Dianions such as $\ce{C_7^{2-}}$ have also been observed.

In the solid phase, aluminum carbide and berylium carbide have more than 50% ionic character according to the Handbook of Refractory carbides and nitrides, and correspond to $\ce{C^{4-}}$.  Calcium carbide, which could be considered a salt of acetylene, corresponds to  $\ce{C2^{2-}}$  rather than $\ce{C^{2-}}$


Plasticisers and other impurities from polyethylene laboratory bottles?

Polyethylene (polythene, to some) may have plasticizers, such as adipates, terephthalates, or phthalates, but they are not inherently needed in all cases. Hopefully, the supplier of those alcohols does not bottle them in cntainers with plasticizers, but a wash bottle, purchased separately, might very well have them added to the plastic, particularly because a squeeze bottle must be very flexible.
It is also possible that the polymer itself degraded over time, and with exposure to light, yielding ethylene monomer and other scission products. Compare the odor of the alcohol with that of ethylene, a faint sweet and musky odor (or "odour" if it's ethene).
As to which plasticizer is used, if any, you'd need to consult the bottle manufacturer. You could substitute a glass wash bottle, e.g. a gas wash bottle with aspirator tube, or a PTFE pipette pump, if you're concerned about contamination, e.g. because your measuring phthalates.
However, if your concerned about personal health, this is not the correct forum.


Why do metallic elements have a very small band gap while nonmetallic elements have a large band gap?

The energy of the orbitals may be the first point to explain it at this level. You can use the example of a simple 1-D LCAO to form the bands(check page 723 of Atkins physical chemistry 8th edition). For instance, the energy difference between the bands formed by s orbitals and p orbitals will be small or large depending on the atom.


What is the parent chain in a hydrocarbon with two different rings?

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

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

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

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

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

So far, we have three possible names:

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

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

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

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

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

This leaves us with two possible preferred IUPAC names:

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

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

  P-44.2 SENIORITY ORDER ONLY FOR RINGS AND RING SYSTEMS

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

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