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Density calculation with cubic equation of state
I want to calculate the density of CO2 using a cubic equation of state in order to plot the P-v diagram. From theory I know that I get one real root when the substance is in single-phase region and two when it is in two-phase region. In the latter case the smaller volume is associated to liquid while the other to vapor. I'm using Matlab to solve the cubic eos, incrementing the pressure and the temperature at fixed intervals.  How can I merge the two solutions to represent the actual Pv plot?

First off, in the two-phase region you have three roots not two.
Smallest volume = liquid
Largest volume = gas
Middle volume = an unstable state you can't really reach
To determine whether a given temperature and pressure  gives the liquid or the gas you must compare the Gibbs free energy, $G$, between the liquid and gas roots as determined from solving the cubic equation.
The change in Gibbs free energy from liquid to gas is given by
$\Delta G = \int_{liquid}^{gas}{V dP}$
where $P$ and $V$ are pressure and volume along a path of constant temperature.  Note that the need for constant temperature; you can't just hold pressure constant and change the volume because that is not constant temperature.
The proper way to do it is to render
$dP=(dP/dV) dV$
where the derivative $dP/dV$ is obtained as a function of temperature and volume by differentiating the equation of state at constant temperature.  Then you have
$\Delta G = \int_{liquid}^{gas}{V(dP/dV) dV}$
with $dP/dV$ known as a function of volume at constant temperature.  You carry out the integral, which requires knowing how to integrate rational functions, and check whether $\Delta G$ is positive, negative, or zero:
$\Delta G>0$ means liquid has lower free energy, so liquid is more stable.  You plot the point for the liquid phase root.
$\Delta G<0$ means gas has lower free energy, so gas is more stable.  You plot the point for the gas phase root.
$\Delta G=0$ means both phases have the same free energy, you hit the equilibrium between liquid and gas. You plot both points and connect them with a line segment corresponding to both phases in equilibrium.  Jackpot!


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Explanation for solubility of group 2 halides
I understand that the solubility (in terms of moles/volume) of group 2 halides increase with increase in anion size, i.e.
$$\ce{MF2 < MCl2 < MBr2 < MI2},$$
where $\ce{M = Mg, Ca, Sr, Ba},\dots$ due to large decreases in lattice enthalpy.
But what is the explanation for the following discrepancies?

$\ce{BeF2 > MgF2 = CaF2 < SrF2 < BaF2}$
$\ce{BeCl2 < MgCl2 < CaCl2 > SrCl2 > BaCl2}$

In other words, how can I determine the solubility order of $\ce{BX2, MgX2, CaX2, SrX2, BaX2}$ (where $\ce{X = F, Cl, Br, I}$)? The given orders of solubility have been verified with data.

Think about what influences lattice enthalpy.

What do you mean by solubility? Mass per volume/mass, moles per volume/moles? What about hydrates?

My data was based on mass per volume. But your right, when comparing solubility, the solubility should be in terms of moles per volume. I am seeing if some change is required for the  solubility orders. No i'm not talking about hydrates.

No, the order remains the same.

To me it does not make too much sense to try to compare and explain the solubility (mass concentration) of anhydrous group 2 chlorides.
First of all, mass concentration does not seem to be the adequate physical quantity for comparison. Some of the chlorides have such high solubility that their saturated solution contains significant less water per volume than others. Molar fractions of solute and water would be a better choice.
Second, all of the chlorides precipitate as the hydrated salt from water. At normal temperature these are

$\ce{BeCl2\cdot 4 H2O}$
$\ce{MgCl2\cdot 6 H2O}$
$\ce{CaCl2\cdot 6 H2O}$
$\ce{SrCl2\cdot 6 H2O}$
$\ce{BaCl2\cdot 2 H2O}$
$\ce{RaCl2\cdot 2 H2O}$

Besides differences in structure their different stoichiometry makes a comparison difficult. And, most important, the formation of hydrates shows that we probably should not discuss the solubility of anhydrous chlorides at all.
Third, some anhydrous halides are more covalent than ionic compounds ($\ce{BeF2, BeCl2}$). This makes it difficult to argue with lattice energies and ionic radii.

So, because of their significant differences in structure and properties (ionic or covalent), lattice enthalpy and hydration enthalpy are not adequate/appropriate to explain the solubility of anhydrous group 2 halides. Am I right?

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Which base accepts the proton abstracted from water in propyne hydration?
I am enrolled in Organic Chemistry 1 at a junior college and I need some assistance in understanding this mechanism. I have previously taken chemistry in high school with no problems, this would be my first time not comprehending the subject. We are covering the hydration of alkynes.
In the general mechanism, the triple bond of propyne attacks $\ce{H2SO4}$. The hydrogen, the electrophile, goes to the least substituted carbon. We now have a carbocation so water comes in as a nucleophile and attacks. But, now we have a positive charge on the oxygen, so a base de-protonates the molecule. My question is, what is this base?
After the base attacks the lone pairs from oxygen travels down to the carbon and the pi bond then attacks $\ce{H3O+}$. This is the keto-enol tautomerizatiom. My second question is, where did we get this $\ce{H3O+}$? Our list of reagents is $\ce{H2O, H2SO4}$ and $\ce{HgSO4}$. We get a product called acetone. I'm just very fuzzy on the in-between steps.

You're in water. How about just water?

Protons can basically jump on and off at any time if there is sufficient incentive for them to do so. In your reaction, you not only have the hydrogensulphate anion $\ce{HSO4-}$ that can accept a proton to go back to sulphuric acid but also water itself which can be protonated to $\ce{H3O+}$. The only important aspect is to identify the least weak base and use that in your mechanism.
Furthermore, general chemistry should have taught you that a solution containing sulphuric acid and water will contain $\ce{H3O+}$ ions by deprotonation immediately.

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Decimal places vs significant figures
I was given an online assignment problem today. The answer was to be given to 2 decimal places. The answer I gave was $$P_2=0.06$$
This to me seems like 2 decimal places. However the answer was incorrect. I then wrote $$P_2=0.064$$
It was correct. I know I'm being knit-picky but I genuinely just want to know your thoughts on this.
Would you call this 3 decimal places and 2 significant figures, or 2 decimal places and 2 significant figures. In the context of the assignment it doesn't matter a whole lot but I still think he should have written 2 significant figures.
What's your 2 pence on the matter?

The confusion might arise because the answer might be expected in scientific notation.

It's important to have both concepts, decimal places and significant digits, because logarithms convert between them.  For example, if you know hydrogen ion activity to 2 significant digits, then you know pH to 2 decimal places, or vice versa.

My verdict on $0.064$ is three decimal places, two significant figures. The underlying philosophy is: how many figures would you give if you converted this to "scientific" notation - which would be $6.4 \cdot 10^{-2}$. Since the results must be independent of representation and nobody would argue about the number of significant figures or decimal places here, this is the way to go for me.

I received some chemistry education both in the US and Germany. Unfortunately, my experience is that in Germany these two concepts are often confused, even on university level and by lecturers. All high school level books on Chemistry I looked at in the US contained a section on this and were clear and adamant.

Except that we shouldn't be talking about decimal places in scientific notation. Or at the very least it's ambiguous unless specified that the answer should be in scientific notation with a single digits in front of the decimal (which is the default, but other representations aren't de facto incorrect). I think it's kind of silly to talk about decimal places in scientific notation though.

There's no room for my two-pence on the matter. "Significant figures" and "decimal places" mean exactly what you think they do, and the online assignment is wrong. Pick out a random sampling of textbooks on Google Books, or even the Wikipedia article, and you can find the distinction clear as day, e.g. this math book from Cambridge or the Olmsted and Williams chemistry book.
If we're being generous, maybe the problem was originally written to have the answer 0.64, and a factor of 10 was introduced later without the question being appropriately updated.

I have since a long time a non-professional interest in numerics and the answer is unequivocal: You and TAR86 are right, the given answer is wrong.
Decimal places are defined by the numbers after the decimal point:

0.1/3456.8/2e-1    =   1 decimal place
0.01/234.90/3e-2   = 2 decimal places
0.001/12.912/4e-3  = 3 decimal places

Significant figures/digits are the effective number of digits (without leading zero)

0.1/3/1e3/2e-1   = 1 significant figure/digit
0.23/3.2e6/12    = 2 significant figures/digits
78.9/8.12e-3      = 3 significant figures/digits


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The function of sulfuric acid in a reaction involving halide ions
The purpose of the following reactions is to make halogen gas:

\begin{align}
\ce{2 NaCl (s) + MnO2 (s) + \boxed{\ce{2 H2SO4 (l)}} &-> Na2SO4 (s) + MnSO4 (s) + 2 H2O + Cl2 (g)} \\
\ce{10 KI (s) + 2 KMnO4 (s) + \boxed{\ce{8 H2SO4 (l)}} &-> 6 K2SO4 (s) + 2 MnSO4 (s) + 8 H2O + 5 I2 (g)} \\
\ce{6 KBr (s) + K2Cr2O7 (s) + \boxed{\ce{7 H2SO4 (l)}} &-> 4 K2SO4 (s) + Cr2(SO4)3 (s) + 7 H2O + 3 Br2 (g)}
\end{align}

And so we need to use a strong oxidizing agent (in these examples: $\ce{MnO2}$, $\ce{KMnO4}$, $\ce{K2Cr2O7}$) to oxidize halide ions out of their compound.
What I don't understand is the reason why we would need to use sulfuric acid in the reaction if there's another oxidizing agent available and why sulfuric acid is commonly used in this specific redox reaction. Can we use another strong acid instead?

Yes, it's just acid here.


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Molar Gibbs Energy for an ideal gas
I'm given a problem where an ideal gas is expanded isothermally. I'm given the molar volumes from the start and the end and basically asked to find the change in Gibbs Energy.
I started with the equation
$$G=H -TS$$
We can write the general form $dG$
$$\begin{align}
dG & = dH -SdT - TdS \\
    & = (dE + PdV + VdP)- SdT -TdS \\
    & = (TdS - PdV) + PdV + VdP - SdT - TdS \\
    & = VdP -SdT
\end{align}$$
It's an isothermal process $\to$ (const.) $T$. Hence $SdT \to0$
Now we are left with $dG=VdP$. We need the Gibb's energy in terms of moles and molar volumes. Hence entropy (which is zero anyway) and volume need to change to molar volume. Our new equation is in terms of moles
$$dG_m=V_mdP$$
if we assume $V_m$ to be constant which it isn't but for argument sake we can get the following relation
$$\Delta G_m = V_m\Delta P=RT \ln\Big{(}\dfrac{P_f}{P_i}\Big{)}$$
since $V_m = \frac{V}{n}=\frac{RT}{P}$ for an ideal gas.
My problem is - I would like to know if I can assume the pressure to be constant and have
$$\Delta G_m = RT \ln \Big{(}\dfrac{V_f}{V_i}\Big{)} $$
Is this equation valid for finding the molar Gibbs energy, knowing the molar volumes (initial and final) and at constant temperature.
I think it's wrong because $dG_m = V_mdP$ is valid for constant $T$ but the pressure cannot be constant. I've just been struggling with trying to find a relationship with gibbs energy and molar volume and not knowing pressures.

You want to know if there is a way to expand an ideal gas where pressure and temperature are both constant?

You are aware that, for an ideal gas at constant temperature and number of moles, PV= const., correct?

$P_i V_i = P_f V_f$ yep - thanks chester. I'll be honest though, I didn't see for a while :)

Your final result is almost correct, although your derivation is somewhat questionable, and, in particular, this equation is wrong: $$\Delta G_m = V_m\Delta P = RT\ln\left(\frac{P_f}{P_i}\right).$$ For one, as you've mentioned, $V_m$ isn't constant, so we can't use the finite difference implied by $\Delta$, and I don't understand how you would have gotten the second equality without having integrated.
Instead, we should have $$\Delta G_m = \int_{G_{m,i}}^{G_{m,f}}\mathrm{d}G_m = \int_{P_i}^{P_f}V_m\,\mathrm{d}P = RT\int_{P_i}^{P_f}\frac{\mathrm{d}P}{P} = RT\ln\left(\frac{P_f}{P_i}\right) = RT\ln\left(\frac{V_i}{V_f}\right).$$ The third equality follows from $V_m = RT/P$, as you note, and the final equality follows from substituting $P = nRT/V$, the ideal gas law, and cancelling out terms.


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Role of 10-camphor sulfonic acid as a catalyst in protecting group chemistry
I recently did a reaction in the lab where 10-camphor sulfonic acid was used as the catalyst instead of something more conventional like hydrochloric acid or sulfuric acid. Dry acetontrile was used as a solvent. The scheme is shown below:

I'm not too sure why it is used. My initial thoughts are it's due to the solubility of 10-CSA in water/organic solvents. Alongside this using a conventional acid such as hydrochloric acid, the water content of the acid may affect the equilibrium. Although 10-CSA still has water in it, it would not be comparable to hydrochloric or sulfuric acid. 10-CSA also is quite bulky, which could provide some selectivity on which positions are deprotonated, although I don't see this as being likely due to its conjugate base having such a high pKa. Aside from this what else could be plausible explanation for the usage of 10-CSA?

For a given protecting group, there is no single best way of introducing it, just as there is no single best way of oxidising an alcohol – the conditions you use depends upon the substrate you have and the tolerance of it towards things such as acid, base, and heat.
In your question you ask specifically about conditions to introduce a benzylidine acetal, namely why aqueous acids aren't used – the tl;dr answer is that they are!
Below is an excerpt from the section in Greene's Protecting Groups in Organic Synthesis that deals with formation of such a group:

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

Taken from: Greene's Protecting Groups in Organic Synthesis

What all of these conditions have in common is that some kind of acid catalyst is required. Mineral acids (sulfuric), Lewis acids (zinc chloride or iodine) and organic acids (tosic acid) have all been shown to work in the literature, and indeed most people involved in complex molecule synthesis will have tried most of them.
Which condition is most appropriate is a matter of context (and often a lot of screening). Solid organic acids (tosic acid, camphor sulfonic acid, pyridine para-toluene sufonate) are often used in late-stage synthesis as they're able to be dried, this allows the reactions to be conducted under (almost) anhydrous conditions. Use of sulfuric acid dictates that water (quite a lot of it) is present, which can knock off other delicate protecting groups/functionality in a molecule.


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Comparison of acidity between angelic and tiglic acid
Which among angelic acid and tiglic acid (cis and trans isomers of 2-methylbut-2-enoic acid respectively) more acidic and why?
It is given in a book(A guidebook to mechanism in organic chemistry for the JEE by Peter Sykes and B. H. Kishan 6ed. p.3.29 Q.1)that the cis(Z-) form is more acidic but I think the trans(E-) form will be more acidic due to interaction of oxygen in conjugate base with hydrogen on C3

I agree with you

user32369 I am a student who is just getting introduced to BH kishan s book.Is the book good or bad?

@user113223 the book is pretty good. It is not your conventional guide; I use it more as a reference book and would suggest you do the same. The book would be better if it had a proper index. Whatever you do, don’t learn from it. There are better books for learning from for JEE.

can you suggest me a proper book for learning what organic chemistry actually is?

I have some thoughts of my own. However, I am not certain if they are correct.
The key to the question certainly cannot lie in the electronic factor because the cis- and trans- isomers do not differ in this aspect, except for the difference in the overall dipole moment. Thus, I believe we can only rationalise this with steric considerations. Let's first take a look at the chemical structures.

Fig. 1: Structure of tiglic acid

Fig. 2: Structure of angelic acid
The steric interaction I would like to focus on is the repulsion between the methyl groups and the negatively-charged oxygen atom of the carboxylate ion. Due to resonance, the negative charge would be delocalised over the two oxygen atoms of the carboxylate ion conjugate base. In the trans-isomer (angelic acid), the methyl group is located closer to one of the oxygen atoms (i.e. the oxygen atom of the carbonyl group). Thus, there would be a stronger repulsive interaction between the two. In the cis-isomer (tiglic acid), the methyl group is located further away from that oxygen atom, thus the intramolecular repulsion would be lessened, thus it is more stable.
Since a more stable conjugate base implies that the acid is stronger, the cis-isomer (tiglic acid) is more acidic than the trans-isomer (angelic acid). This argument is in full agreement with the information presented in your reference book.

No, this doesn’t really make sense since an oxygen anion is not really any larger than a carbonyl oxygen.

@Jan Does the size of the anion matter? I didn't assume that in my answer did I?

@Jan All I assumed in my answer is that there is significant repulsion between the electron cloud of the methyl group and the negatively-charged oxygen atom of the carboxylate ion.

I asked my teacher and he seems to have a similar idea as you(repulsive dipole-dipole interaction in cis form leading to slight instability in the acid), although he did mention a couple of times that the difference will be extremely small.

@user32369 I agree that the difference should be small.


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Mathematical explanation behind dq = TdS
Can somebody mathematically prove why $$\mathrm dq = T\,\mathrm dS$$ for a reversible process?
I realize I am supposed to provide some research but I can't find anything. The best I got is $\mathrm dE = T\,\mathrm dS$ which is basically $\mathrm dG = \mathrm dH - T\,\mathrm dS$ and it seems to be a circular argument (since for derivation of $G$ most proofs use the equation $\mathrm dq=T\,\mathrm dS$).

$\def\d{\mathrm{d}}\def\c{_\mathrm{C}}$Firstly, note that a more correct version of your equation is
$$\delta q_\mathrm{rev}=T\d S.\tag1$$
We use a Greek delta to signify that heat is not, in general, a state function. Its value during a change does not simply depend on the beginning and end points but also on the way which the change is traversed. Subscript '$\mathrm{rev}$' is for a reversible process.
Secondly, I should warn that the proofs I supply are in no way rigorous but should suffice for a standard thermodynamics course.
Thirdly, I have used a small $q$ for heat, reserving capital $Q$ for a partition function.
Classical thermodynamics
The classical definition of entropy in thermodynamics is due to German scientist Rudolf Clausius. He showed that, for a cyclic process, the following inequality holds:
$$\oint\frac{\delta q}{T} \leq 0.\tag{2}$$

Lemma. For a Carnot cycle, the following line integral over a circular path equals zero: $$\oint\c \frac{\delta q_\mathrm{rev}}{T} = 0.\tag{L}$$

Proof
Recall and remember that in a Carnot cycle efficacy $\eta_\mathrm{C}$ of a system may be expressed in two equivalent ways, by a heat quotient and a temperature quotient.
$$\eta\c=1+\frac{q_2}{q_1}\tag{3a}$$
$$\eta\c=1-\frac{T_2}{T_1}\tag{3b}$$
Substrating one from the other yields
$$\frac{q_2}{q_1} + \frac{T_2}{T_1}=0\tag{4}$$
which, through algebraic manipulation, is the same as
$$\frac{q_1}{T_1} + \frac{q_2}{T_2}=0\tag{5}.$$ Hence for a Carnot cycle, reversible by definition, we optain
$$\oint\c \frac{\delta q_\mathrm{rev}}{T} = 0.\tag{L}$$
I will omit the subscript $_C$ for Carnot in the integral when I mention in the text or it is obvious by the same subscript on some other quantity.
Proof of Clausius integral inequality
Start with the first law of thermodynamics.
$$\d U = \delta q + \delta w \tag6$$
Therefore, for a Carnot engine,
$$w\c = -\oint \delta q_\mathrm{rev}.\tag7$$
Let us take some other engine, called engine $2$. Again from the first law of thermodynamics,
$$w_2 = -\oint \delta q_2.\tag8$$
Assume that, contrary to the theorem,
$$\oint \frac{\delta q_2}{T}>0.\tag9$$
Let us join these two cycles to form a composite engine. The work done by this system will be
$$w_\mathrm{total} = w\c + w_2 = -\oint(\delta q_\mathrm{rev} + \delta q_2) = -\oint\delta q_\mathrm{total}. \tag{10}$$
By our lemma $\mathrm{(L)}$, assumption $(9)$ and result $(10)$, we see that
$$\oint\frac{\delta q_\mathrm{rev} + \delta q_2}{T} \stackrel{(10)}{=} \oint \frac{\delta q_\mathrm{total}}{T} \stackrel{\mathrm{(L)},(9)}{>} 0 \tag{11}.$$
Now change the sign or direction of the Carnot engine, and its size so that no work is done in the composite cycle with engine $2$. (This should always be possible via Carnot's theorem which I will not prove here.) Then
$$w_\mathrm{total} = \oint\delta q_\mathrm{total} = 0. \tag{12}$$
If we break cyclic line integrals apart into $y-1$ smaller integrals that add up to the cyclic integral, we achieve
$$W_\mathrm{total} = \sum_{i<y} \int_{b_i}^{b_{i+1}} \delta q_i = \sum_{i<y} q_i(b_j) \stackrel{(12)}{=} 0,\tag{13a}$$
$$\oint \frac{\delta q_\mathrm{total}}{T} = \sum_{i<y} \int_{b_i}^{b_{i+1}} \frac{\delta q_i}{T_i} = \sum_{i<y} \frac{q_i(b_j)}{T_i} \stackrel{(11)}{>} 0.\tag{13b}$$
The terms $q_i(b_j)$ may be both positive and negative but in equation $(13a)$ cancel out to give zero. In equation $(13b)$ the terms are weighted by temperatures in such a way that positivity wins. Therefore, on the whole, negative $q_i(b_j)$ terms have been divided by larger temperatures whereas positive terms possess smaller temperature denominators. Physically, this implies that heat is transferred from a lower temperature reservoir to a higher temperature bath without doing any net work. This is in violation of the second law of thermodynamics (Clausius statement).
Therefore, our assumption $(9)$ that
$$\oint \frac{\delta q_2}{T}>0$$
is actually false. The result
$$\oint \frac{\delta q_2}{T}\leq 0$$
immediately follows.

Corollary. For any reversible process, not just a Carnot cycle, $$\oint \frac{\delta q_\mathrm{rev}}{T} = 0.$$ In other words, temperature becomes an integrating factor in a reversible process.

Proof
By Clausius integral inequality $(1)$, we know immediately that
$$\oint \frac{\delta q_\mathrm{rev}}{T} \leq 0. \tag{14}$$
Assume again to the contrary that
$$\oint \frac{\delta q_\mathrm{rev}}{T} < 0. \tag{15}$$
Since this is a reversible process, we may change the direction of the engine. Only the sign of the integral will alter.
$$\oint \frac{\delta q_\mathrm{rev}}{T} > 0. \tag{16}$$
If one proceeds analagously to the proof of the Clausius integral inequality, this will lead to a contradiction. Thus, assumption $(15)$ is not valid, and by $(14)$ we are left with an equality.
Why bother with all of this?
We have now proven that for any reversible process,
$$\oint \frac{\delta q_\mathrm{rev}}{T} = 0. \tag{17}$$
This means that we may define a change in a new state function $S$ which we call entropy such that
$$\Delta S= \int_{S_1}^{S_2} \frac{\delta q_\mathrm{rev}}{T}\tag{18},$$
or in differential form,
$$\d S = \frac{\delta q_\mathrm{rev}}{T}. \tag{19}$$
This is the definition of a differential of entropy. There is nothing else to derive; what we showed above is that we may indeed define such a state function. To achieve your equation, multiply both sides of equation $(19)$ by temperature.
$$T\d S = \delta q_\mathrm{rev}. \tag{CT}$$
Statistical thermodynamics
Another way to go would be to define entropy in statistical thermodynamics. Then the formula $\mathrm{(CT)}$ may be derived.
Statistics in general
For a discrete random variable $X$, entropy is defined in statistics to be the function $\mathrm{H}$ such that
$$\mathrm{H}(X)= -\sum_k p_k\ln p_k \tag{GS}$$
where small $p$ denotes probability.
Boltzmann statistical thermodynamics
Say we have a large number $\mathcal{A}$ of isolated systems which we together call an ensemble. Suppose the system may be described as
\begin{align}\begin{split} \text{energies}\ \ &E_1\ \ &E_2\ \ldots &E_n\ \ldots\\ \text{number of such systems}\ \ \ \  &a_1\ \ &a_2\ \ldots\ &a_n\ \ldots\end{split}\end{align}
where
$$\mathcal{A} = \sum_i a_i.\tag{20}$$
The systems of different energies in this ensemble are distinguishable. The systems that have the same energy, i.e., are degenerate, are indistinguishable. If we supplied each system regardless of energy with an index of energy, the number of ways to order these indices would be
$$\mathcal{A}!.\tag{21a}$$
But as we said, the systems of equal energy are indistinguishable. The number of ways of ordering each of these is
$$a_j!.\tag{21b}$$
So to avoid overcounting we must divide:
$$K(a_1,a_2,\ldots,a_n,\ldots) = \frac{\mathcal{A!}}{\prod_i a_i!}.\tag{21c}$$
You may think of this $K$ as the total number of anagrams of words with $\mathcal{A}$ letters where each letter repeats itself $a_j$ times. Boltzmann used the big letter $W$ instead of $K$ and defined entropy to be
$$S = k_\mathrm{B}\ln W.\tag{BD}$$
The natural logarithm guarantees, among other things, that entropy is not a huge quantity, and that entropy for different (independent) systems is additive. In other words, if
$$W_\mathrm{total} = W_\mathrm{A}W_\mathrm{B}$$
then
$$S_\mathrm{total} = S_\mathrm{A} + S_\mathrm{B}.$$
Boltzmann entropy as a special case of statistical entropy
Begin with the statistical definition.
$$\mathrm{H}(X)= -\sum_k p_k\ln p_k \tag{GS}$$
Say each and every state in $W$ has equal probability. Then, since the sum of probabilities must equal $1$, we optain
$$p_j = \frac{1}{W}.\tag{22}$$
(You may prove that this constraint actually maximises entropy. This is what we expect from a measure of 'information' in a system where 'knowledge of the system' is smallest in the sense that each state has equal probability. See 'Intuitive explanation of entropy?' on math.stackexchange.com by An.Ditlev.)
Substituting
$$\mathrm{H}(X)= -\sum_k \frac{1}{W}\ln \frac{1}{W} = -W \frac{1}{W}\ln \frac{1}{W} = -\ln \frac{1}{W} = \ln W.\tag{23}$$
This shows that Boltzmann constant is actually unnecessary and a figment of historical definitions. We shall see this in one other way down below. We conclude that Boltzmann's definition of entropy is a special case of a more general entropy when probabilities are equal. In thermodynamics, this more general entropy is called Gibbs entropy.
$$S\equiv -k_\mathrm{B}\sum_k p_k\ln p_k\tag{GD}$$
This is our new definition of entropy. Here we have an absolute quantity; in classical thermodynamics (without the third law) we defined the differential of entropy.
Equivalence of Gibbs entropy and Clausius entropy
A partition function $Q$
Intuitively, we assume that the relative populations of system $a_m$ and $a_n$ are given by some function depending on energy.
$$\frac{a_m}{a_n} = f(E_m - E_n)\tag{24}$$
The difference in energy is a natural choice because any arbitrary zero for an energy state should cancel. Furthermore, the function $f$ ought to be transitive.
$$\frac{a_x}{a_n} = \frac{a_m}{a_n}\frac{a_x}{a_m} \iff f(E_x - E_n) = f(E_m - E_n)f(E_x - E_m)\tag{25}$$
The most obvious function with these properties is the exponential function. Thus we define
$$a_j = C\mathrm{e}^{-\beta E_j}, \ \ C\in\mathbb{R},\tag{26}$$
where $\beta$ is positive everywhere by convention. Its form is to be determined. We start by tackling the constant $C$. Sum both sides of equation $(26)$.
$$\mathcal{A}=\sum_i a_i = C\sum_i\mathrm{e}^{-\beta E_i} \tag{27}$$
Divide equation $(26)$ by equation $(27)$.
$$\frac{a_j}{\mathcal{A}} = \frac{\mathrm{e}^{-\beta E_j}}{\sum_i\mathrm{e}^{-\beta E_i}} \tag{28a}$$
In the limit, the left-hand side becomes probability of being in a state $a_j$.
$$p_j = \frac{\mathrm{e}^{-\beta E_j}}{\sum_i\mathrm{e}^{-\beta E_i}} \tag{28b}$$
The denominator is defined to be a partition function $Q$. Assumuming that $E=E(N,V)$ (number of particles and volume, respectively) we see immidiately
$$p_j(N,V,\beta) = \frac{\mathrm{e}^{-\beta E_j(N,V)}}{Q(N,V,\beta)}. \tag{29}$$
The first law of thermodynamics with partition functions
From the definition of expectation value for a discrete variable $X$,
$$\mathrm{E}X = \sum_ip_ix_i.$$
Let us identify internal energy with an expectation value of energy. Then
$$U = \sum_i p_i(N,V,\beta)E_i(N;V).\tag{30}$$
Take a differential of both sides.
$$\d U = \sum_i p_i(N,V,\beta)\d E_i(N;V) + \sum_i E_i(N;V)\d p_i(N,V,\beta)\tag{31}$$
This expression for probability and energy holds for a reversible process. If we associate the first term with work, the second term will be heat. Doing so will allow us to prove the equivalence but it will also lead to correct results for ideal gases if you do the maths. For brevity (ha ha), I will gloss over such discussions but in my opinion the rigorous explanation (if there is one) deserves its own question. (I am not aware of the rigorous proof.) Hence,
$$\delta q_\mathrm{rev} = \sum_i E_i(N;V)\d p_i(N,V,\beta).\tag{32}$$
Proof of equivalence
Recall the definition of Gibbs entropy.
$$S\equiv -k_\mathrm{B}\sum_k p_k\ln p_k\tag{GD}$$
Take the differential of both sides.
$$\d S = -k_\mathrm{B}\sum_k (1 +\ln p_k)\d p_k\tag{33}$$
The term $\sum_k \d p_k$ will equal zero by linearity and because probabilities sum to $1$.
$$\sum_k \d p_k = \d\sum_k p_k = 0\tag{34}$$
Thus,
$$\d S = -k_\mathrm{B}\sum_k \ln p_k\d p_k.\tag{35}$$
Substituting equation $(29)$ into the natural logarithm term in equation $(35)$ yields
$$\d S = k_\mathrm{B}\sum_k \left[\beta E_k(N,V) + \ln Q(N,V,\beta)\right]\d p_k.\tag{36}$$
Similarly to equation $(34)$ the final logarithm term cancels.
$$\sum_k \ln Q(N,V,\beta) \d p_k = \ln Q(N,V,\beta)\d \sum_k p_k = 0\tag{37}$$
We are left with
$$\d S = k_\mathrm{B}\beta\sum_k E_k(N,V) \d p_k(N,V,\beta)\tag{38}$$
which by equation $(32)$ translates to
$$\d S = k_\mathrm{B}\beta\delta q_\mathrm{rev}.\tag{39}$$
Note that $k_\mathrm{B}\beta$ is also an intgrating factor. From our corollary before and to complete the equivalence, we are forced to choose
$$\beta = \frac{1}{k_\mathrm{B}T},\tag{40}$$
and finally,
$$\d S = \frac{\delta q_\mathrm{rev}}{T} \implies T\d S = q_\mathrm{rev}.\tag{19,CT}$$
In conclusion, we have shown that Gibbs entropy is equivalent to Clausius entropy up to an additive constant. This is because a differential of a constant is zero. The choice of $\beta$ again demonstrates how superfluous Boltzmann's constant really is. We introduce it in the Gibbs entropy definiton, only to be forced to get rid of it. For more information on Boltzmann's constant, and how it relates to temperature and the zeroth law, see 'Is the Boltzmann constant really that important?' on physics.stackexchange.com by DanielSank. And so,
$$S_\mathrm{Gibbs} = S_\mathrm{Clausius} + r,\ \ r\in \mathbb{R}.\tag{41}$$
If the most classic form (zero entropy at absolute zero) of the third law of thermodynamics holds, and it always does not,
$$r = 0. \tag{42}$$
TL; DR
The reason why
$$\delta q_\mathrm{rev}=T\d S\tag1$$
holds depends on your starting point.

There are different forms of the first and second laws. Proving that a state function $S$ exists varies based on one's axioms.
Entropy may be defined in various ways. If a classical definition is used, there is only to prove that we may indeed define such a state function. If a statistical definition is applied, we must first prove the classical definition as a special case.
Note that the second law of thermodynamics only defines a difference or differential in entropy. Together with the third law, one may establish an absolute scale. However, the statistical definition is already absolute to begin with.
Boltzmann entropy is a special case of Gibbs entropy, not the other way around. Gibbs entropy agrees with Clausius entropy (in reversible thermodynamics) up to an additive constant.

$$S_\mathrm{Gibbs} = S_\mathrm{Clausius} + r,\ \ r\in \mathbb{R}\tag{41}$$
If the most classical form of the third law holds, $r = 0$.

But $$dU=TdS-pdV=δQ+δW \rightarrow δQ=TdS$$
What I am missing?

$dU = \delta q + \delta w$   First Law of Thermodynamics
$dS = \frac{\delta q}{T}$     Second Law of Thermodynamics (reversible reaction)
From these two equations we can derive by substitution of the second equation into the first:
$dU = TdS + \delta w$
Now we calculate the work using the definition
Work = Force $\times$ Distance
$\delta w = -F dr$ and, by dividing the work by an area and multiplying displacement by an area,
$\delta w = -\frac{F}{A} d(rA)$ or $\delta w = -P dV$
Substituting this for work gives:
$dQ = TdS$ :)
Source: NCSU

I was dusting off my thermodynamics knowledge there, sorry for any errors

This answer is alright; but the bit about work and the first law of thermodynamics is entirely unnecessary. If you look at your second law of thermodynamics and multiply both sides by temperature, you have the relatioship given by OP.


text:
How to calculate the energy of a single proton?
Using e.g. the VWN5 functional, what is a simple way to calculate the energy of a single proton? Of course, it is not possible to perform a calculation of a proton in isolation, as it bears no electrons. So I first attempted to use the experimentally determined value (see e.g. M. Dewar and K. Dieter, J. Am. Chem. Soc. 1986, 108, 8075.), but this leads to highly erratic proton affinities.
Could it be that I need to determine the proton affinity for any given functional + basis set (e.g. VWN5, 6-31+G** in my case)? What would be an effective way to get a good rough estimate of the energy of a proton?

What "energy of single" proton? It can have any kinetic energy you can imagine.

A good rough estimate of the energy of a proton is $0$.

Energy of a proton? That doesn't make any sense. Referenced to what?

I do not have access to the paper referenced. If you wish to calculate gasphase proton affinities (which would be simple energies), you assign the energy zero for the proton. Thus, two calculations and you're done. If you look for pKb values, you need a solvation treatment and then the functional/basis set come into play.

I think looking at a single proton takes it out of the realm of chemistry and into physics.

An isolated proton will have energy given by the Schrodinger equation (just like any molecular system):
$$
-\frac{\hbar^2}{2\mu}\nabla^2\psi + V\psi = E\psi
$$
($\mu$ is reduced mass; $\hbar$ is $h/2\pi$; $V$ is potential energy; $\psi$ is the wavefunction)
The potential of this single-proton system is zero; the kinetic energy will depend on the momentum of the particle.
If you're using a generic quantum-chemistry software, the Born-Oppenheimer approximation is probably being applied, i.e. the proton ($\ce{H+}$) will be fixed in space with zero momentum. Thus its total energy will be zero, unless you deactivate BO-approximation or introduce another particle.
It's not typical to consider the energy of an isolated proton for obvious reasons (try finding one in nature!). Nonetheless it is calculable if the wavefunction is known.
Another way to consider its kinetic energy is by the classical equation $K = \frac{1}{2} mv^2$; if you consider an approximation of the lonely proton's speed as roughly that of atoms in liquid water, 1 angstrom per picosecond, you obtain ~ $5.03$ kJ/mol. This is similar to the thermal energy available at room temperature, $k_B T$, ~ 2.48 kJ/mol.
You're better off calculating energy of $\ce{H+}$ by getting energy for a molecule, and subtracting the energy of its deprotonated form. In that way you can evaluate proton affinity for a given molecule.

I think I have figured it out now: Dewars method (see above reference) is based on AM1 which gives different energy values than ab initio methods. Therefore, he has to consider the heat of formation of a single proton, which is less stable than hydrogen and hence corresponds to a positive heat of formation. However, ab initio methods yield the electronic energy, which is obviously zero for a proton. Therefore, PA = Edeprot - Eprot.

don't forget the charge will change for the deprotonated form, and you have to indicate the charge of the system in QM software to compute the wavefunction/energy/etc.

Thanks I know this much ; ) Anyway, multiplicity wouldn't add up for the wrong charge meaning it will fail if the charge state is wrong making it pretty "idiot-safe" at least in the ORCA implementation...


text:
In an experiment testing catalyst influence on reaction, is NaOH a reactant or a catalyst?
I have recently completed an experiment with hydrogen peroxide (5%), 0.1M KI and 1M NaOH.
Is the sodium hydroxide a reactant or a catalyst in this experiment?

In basic solution the half reactions are:
Reduction:
$\ce{H2O2 + 2 e- -> 2 OH-}$
Oxidation:
$\ce{I- + 6 OH- -> IO3- + 3 H2O + 6 e-}$
So the overall reaction is
$\ce{3 H2O2 + I- -> IO3- + 3 H2O}$
thus $\ce{OH-}$ is overall neither a reactant nor a product. Hence it is either a catalyst, an inhibitor, or a spectator ion.


text:
Windex and Red wine
The other day I was cleaning my counter tops with windex. There was a couple red wine (cabernet sauvignon if it makes a difference) stains from the night before and the stains turned an gray/blue color upon being sprayed with the windex. What is the cause of the color change? Is it simply the blue windex mixing with the red wine?

Arguably, most organic colors are pH indicators.

Red cabbage soup goes through similar color changes (though I'm not sure if it is best served with meat or with fish).


text:
Orbital diagram in which the Aufbau principle is violated

The orbital diagram in which the Aufbau principle is violated is:

The answer given for this question is 3, but according to me if asked for Hund's rule violation then it would be 1, as electrons are not distributed among the subshell in such a way to give maximum number of unpaired electron with parallel spin.

Ok. So what is your question?

That's the ground state for carbon. The pairing energy is lower than the difference between $2s$ and $2p$, so you definitely pair in $2s$ before moving up to $2p$.

But you’re not being asked for Hund’s rule violation …?

None of these violate Hund's rule. Aufbau is violated in 3 because $s$ was not filled before moving on to $p$.

The answer is 3 because according to Aufbau principle electrons are distributed according to energy levels that means S orbital should be filled completely before starting P . This doesn't happen in case 3  ,hence it is the answer.


text:
In-vivo racemization of thalidomide
In 1961, thalidomide was first introduced as an over-the-counter cure for morning sickness. Pregnant women across the country began to take the drug, which worked well, but shortly after, babies began to be born with severe birth defects. Experts linked this to thalidomide and later discovered that the r-enantiomer, which was the cure for morning sickness, underwent a certain reaction (or reactions) within the body that could change its configuration to that of the s-enantiomer. Does anyone know the steps of this reaction or at least what it is called?

The stereocentre has an acidic proton attached (the H). This can undergo enolisation (due to the presence of the carbonyl). As the enolate form is planar (and hence chiral), the stereochemical information is lost (there is no chiral memory). When the enolate re-converts to the carbonyl form, it does so to form a racemic mixture. The process is usually called epimirisation and is common to all stereocentres directly adjacent to a carbonyl (under the right conditions, some are more prone to epimirisation than others-  in this case the epimirisation happens in vivo).

With all that text at the beginning, this sounds like a homework assignment with only the last sentence fixed :D


text:
Cationic Detergent with no odor
I am looking for a cationic detergent for a project.  Unfortunately most ( or all?) cationic detergents are quaternary ammonium that smell like rotten fish or even worse
The product I am working on can not have odor. Is there such cationic detergent out there or an odor neutralizer ?

sigh I guess you didn't check your misunderstanding in practice. These are quaternary ammonium salts and only neutral amines not their salts may have bad smell.

@Mithoron thank you. That is a good point.  What I am looking for is something with foaming capacity and not just disinfectant like most of them. If you know any QA salt that can make foam and bubbles ( like soap ) that would be the answer and I highly appreciate that.

Also you are right , edited the question to fix ammonium


text:
Concentration of Carbon dioxide in Blood
Let's say we're able to decrease concentration of carbon dioxide in blood, how will the pH of the blood be affected?
My Reasoning : Since blood has approximately 92% water, it is safe to assume the carbon dioxide being acidic forms weakly acidic Carbolic acid thereby decreasing the concentration of Protic water resulting in an increased pH, therefore decreasing concentration of carbon dioxide should decrease pH of blood.
However this is not the answer, pH of blood would significantly increase if concentration of carbon dioxide decreases, where am I going wrong? (perhaps hemoglobin has a role?)

This is a well understood process: https://en.wikipedia.org/wiki/Respiratory_alkalosis#Mechanism

You’re wrong at that point where you consider $\ce{H2CO3}$ to be the final species and forget the subsequent deprotonation equilibrium.

Blood has a buffering capacity to maintain a relative stable pH.

Google homeostasis.

Carbonic acid will increase acidity leading to a decrease in pH, not as you have it. It doesn't decrease 'protic water,' but rather it will form an equilibrium with H2O and form carbonic acid, which weakly donates a proton to increase the hydronium ion concentration, thus decreasing pH.
For example, rain is naturally 10X more acidic (pH 6) than neutral water due to the 0.04% CO2 in air.
The pH of blood in theory would increase if we decreased the amount of dissolved CO2 but in reality the human body has an incredibly complex set of mechanisms to maintain the slight BASICITY of blood.


text:
Is a carbon chiral it two of its groups are cis - trans isomers?
For example:
In this image, the two groups sticking to a carbon are cis-trans isomers. So does that make the carbon chiral, exhibiting optical properties?
What if the two groups were both having the same configuration(both cis or both trans), would it then be achiral?

Thanks for any help :D

Related: Which alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol?

So it'll exhibit optical properties and will have enantiomers? Thanks.

The other posters replied to your main question. I will add a couple of observations. 1) chirality is a property of the whole molecule, not of any of its parts in isolation; thus there are no 'chiral centers' or 'chiral' carbon atoms, there are stereocenters and stereogenic carbon atoms, which may be found in chiral molecules . 2) having stereoisomers, or enantiomers in this case, is not the same as, or automatically implying, displaying optical properties; it is likely to observe distinct optical properties in opposite enantiomers of the same molecule, but not a universal rule.

Any time all four groups bonded to carbon are distinct, including any sort of isomeric groups, you have a chiral center.  Comments associated with the related question referenced by @Loong (since deleted, unfortunately) indicate that the cis vs trans scenario is even included in the $R$ vs. $S$ naming convention.  This case is an $S$ configuration as the cis group takes second place and is counterclockwise from the hydroxyl group in the appropriate view.

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text:
Resonance of carbonyl

Carbonyl has two resonance structures, where the $\pi$-bond between C and O breaks and the electrons "move" to the oxygen atom. This gives an empty p-orbital in carbon and a p-orbital in oxygen with two electrons, leading to a positive charge on carbon, and a negative charge on oxygen. To my understanding, oxygen gets the lone electron pair because it has a higher electronegativity than oxygen.
My question: is it possible that the two electrons "move" to carbon instead of oxygen, leading to a third resonance structure (I know that these resonance structures don't really exist, and that the real structure is somewhere in between), so that carbon has a lone electron pair? It would probably contribute little to the resonance energy compared to the other structures. Or is it not possible at all?

You can do that

Yes, you can write it and it's not significant contributor.

Thank you both! I could find anything about it, but it seemed logical to me.

Yes, resonance should never stop at the most likely resonance structure. If you’re really willing to play the game, you can take it even further and consider singlet diradical structures where the bond has been broken into a spin up and a spin down electron, each one of which resides in a different orbital. And more, and more.
(Side note: this is why it is useless to ‘count’ resonance structures. The only correct answer to ‘How many resonance structures does $\ce{X}$ have?’ is ‘Always one more.’.)


text:
Do electronic balance weighings still need to be corrected for buoyancy error?
Buoyancy error is a weighing error due to the difference in the buoyant force exerted by a medium (commonly air) on the object and on the standard masses. The difference in buoyant force occurs when the object and the standard masses have standard masses (Skoog et al., 2014).
Knowing this, I am confused as to whether electronic macroanalytical balances need to be corrected for buoyancy error because of the following factors:

Electronic macroanalytical balances still get calibrated by weighing standard masses.
Closing the doors of the balance does not make the system inside of it evacuated. It still has air which can exert buoyant force on the object and on the standard masses.

Given the factors I mentioned, do electronic balance weighings stil need to be corrected for buoyancy error?
References:

Fundamentals of Analytical Chemistry, Skoog, et al., 9e

Well, if imagine trying to weigh a helium-filled balloon on an analytical balance by taping it to the weighing pan, I think the result will be negative.  But I don't think the balloon has negative mass, so ...

A balance cannot determine the density of the weighted material and air automatically. For this reason a buoyancy error correction must be applied for high precision measurements.
Suppose you weigh $\pu{100.0000 g}$ water and have $\pu{100.0000 g}$ reference weights with a density of $\pu{8.0000 g cm^{-3}}$. Then the static buoyancy in air is equivalent to about $\pu{120 mg}$ and $\pu{15 mg}$ respectively, accounting for a difference of about $\pu{105 mg}$ in the reading of the balance.
Besides the static buoyancy there may also be dynamic buoyancy due to air flow caused by temperature differences between the weighted material and the surrounding air.
Dynamic buoyancy can be avoided be weighing in thermal equilibrium. Static buoyancy can be corrected by formulae involving the density of air, the density of the weighted material and the density of the reference weights.

So would you have to accurately measure the pressure and temperature too. The temperature to determine the thermal expansion and so additional air displacement. Or just take the measurement in a vacuum...

Not to forget air humidity. Together with pressure and temperature you could calculate air density quite accurately.

I think they must as the density of the material being weighed will change the correction. eg. The buoyancy of lead is much different from that of feathers.

This is the correct answer. For analytical measurements, the buoyancy error should be accounted for. In general, what you're weighing will have a different density than the standard masses. Hence, calibration will fail to account for this effect (electronic balance or otherwise). SWHC also provide references (DOI:  10.1021/ed061p51, DOI: 10.1021/ac50021a062) which are very pertinent. I recommend to add a bit more detail to your answer to make it better received.


text:
Calculation of entropy for an isothermal irreversible expansion
How do I calculate the entropy change for an irreversible isothermal expansion against $p_\text{ext}=p_0$?
For a reversible isothermal expansion we know $\delta S_\text{uni}=0$ so $\delta S_\text{sys}=-\delta S_\text{surr}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$
However I can't figure out why $\delta S_\text{sys}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$ remains for irreversible expansion.
I also found out an equation for irreversible expansion
$\delta S_\text{uni}=q_\text{rev}/T-q_\text{irrev}/T$
Is it correct? If yes? Can you explain how do we get it?
Edit
Is this correct ?
$\delta S_\text{uni}=\delta S_\text{sys}+\delta_\text{surr}$
$\delta S_\text{sys}=\frac{q_\text{rev}}{T}$
$\delta S_\text{surr}=-\frac{q_\text{irrev}}{T}=p_\text{ext}\delta V/T$

Here are the steps to determining the change in entropy for an irreversible process on a closed system:

Use the first law of thermodynamics to determine the final thermodynamic equilibrium state of the system for the irreversible path.
Totally forget about the irreversible path.  It is of no further use.  Focus only on the initial equilibrium state of the system and the final equilibrium state.
Devise a reversible path for the system that takes it from the initial equilibrium state to the final equilibrium state.  This reversible path does not have to bear any resemblance whatsoever to the real irreversible path, other than it must pass through the same initial and final end points.  (Entropy is a function of state).
Calculate the integral of dq/T for the reversible path that you have devised.  This will give you the change in entropy of the system for the irreversible path as well as for the reversible path.

For your problem, this procedure will give you the equation that you have written.  Note that, for the irreversible path, the system temperature matches the surroundings temperature (isothermal) only at the interface with the surroundings, but not throughout the interior of the system.  For an irreversible expansion, this is what we define as "isothermal."

So is then change in entropy of universe 0 for isothermal irreversible process? Since by 4th point qrev and qirrev are same.

Btw can you also include how to deal when expansion is against a constant pressure?

Are you saying that you don't know how to find the final equilibrium state for this case using the first law of thermodynamics?

I have never used the concept of initial and final states. I was taught to work with changes ( which I know is basically the same thing). Can you please show how to solve this question (I am not asking it for homework. I am asking because it would help me understand better). Determine the entropy change in sys, surr, uni, when a sample of helium has of mass M grams at 298K and 1 bar doubles its volume in isothermal irreversible expansion against external pressure =p

Well, if you know the mass M, then you know the number of moles.  And you know the initial temperature and pressure, so, from the ideal gas law, you know the initial volume.  What is that equal to?  At time zero, you suddenly drop the pressure to p and hold it at this value until the system reequilibrates. at 298 and p.  What is the final volume?  What is the ratio of the volumes?

If initial volume is x then final is 2x where x=PV/RT

Oops.  I didn't notice that the expansion is stopped at 2x before the system has  had a chance to equilibrate at pressure p.  So even if p is less than 0.5 bar, the final volume is double the initial volume, and the final temperature is the same as the initial temperature.  So you know the final state:  volume = 2x, temperature = 298, pressure = 0.5 bar.  So the initial and final states are now established.  How would execute step 3 of the procedure?

Let us continue this discussion in chat.

I think by thermodynamical definition of entropy you can calculate entropy change exactly only in reversible changes only. As Clausius inequality refers we can not calculate exactly the entropy change in an irreversible process. We can only estimate its limiting value. Statistically we can determin it by using Boltzmann theorem in terms of most probable thermodynamic probability of the state.


text:
Other molecules like fullerenes?
When I first heard about buckminsterfullerene, I thought it was a very fascinating molecule, with its interesting complex ring structure and pseudo-symmetry.  I wonder though, can you get similar structures with other atoms?  Maybe neighbors of similar size, like boron, nitrogen or oxygen?  Or maybe silicon, one row down?   What would their chemistry be like?

This looks kinda too broad. There are various similar molecules.

Check out https://en.wikipedia.org/wiki/Borospherene

https://en.wikipedia.org/wiki/Boranes, maybe?

Simple P4O10 has a hollow, if not spherical structure. https://en.wikibooks.org/wiki/Introduction_to_Inorganic_Chemistry/Review_of_Chemical_Bonding#/media/File:Phosphorus-pentoxide-3D-balls.png

So does $\ce{P4O6}$.  Hexamethylenetetramine is isoelectronic and isostructural with $\ce{P4O6}$ apart from the outer hydrogen atoms.

There are plenty heteroatomic molecular clusters with even more complex structures. Or do you want a single element assembly only?

@andselisk, I'm looking for single-element cases with similar roughly spherical shapes; the Borospherene suggested by Mithoron is a great example.

Cool question! Always stay curious!
A while ago, I had a similar question that started it's journey with silicon, being of similar bonding capability to carbon. However, over the course of my research, I realized that silicon was just too low on the periodic table (and therefore much heavier) to be so extremely stable in such a wide variety of forms such as fullerenes and long, complex hydrocarbons.
However, my search was not in vain. I stumbled upon Boron Nitride, an isoelectronic analog to carbon graphite. Many different compounds have been synthesized from boron nitride, consisting of a wide array of fullerene analogs, such as Boron Nitride Nanotubes, or BNNTs, which by themselves have plenty of applications. For example, paper products made from these nanotubes are highly flame retardant, as is shown in the image below:

The trick is that $\ce{B-N}$ has very similar properties to $\ce{C=C}$, with very little difference in molecular weight, and so many analogous compounds can be built using either monomer. However, boron nitride is different enough to have unique applications in certain areas. Cool, right?

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text:
What is the reaction mechanism of nitrogen leaving from diazo compounds?
When a diazo compound is formed, and then a nucleophile is added to the mixture. Will nitrogen gas leave first or will the nucleophile have to "do" a substitution reaction? What I mean is will the carbocation be formed due to nitrogen leaving or will it be an SN2 reaction? Or will it depend on the carbon that the diazo group is attached to, (like primary, secondary, etc.) depending on the stability of carbocation formed? In short, I mean to ask if it's SN1 or SN2.
Is it a thumb rule that when an unstable diazo compound is formed, that the nitrogen leaves without any nucleophile intervention?

Because when I asked it to my professor, he said its a MUST Carbocation forming reaction (he didn't specify sn1... even though i asked him that if it is sn1...), however he is not someone i can trust THAT much.. Thanks:D

Probably depends a bit on the reaction. A real chemist would probably check kinetics of the reaction to see if it's first order in the nucleophile or not.

@Zhe Maybe a real physical chemist would. A real organic chemist would run with whatever they get out.]

@Jan fair point and important missing qualifier..

The reaction is $\ce{S_N1}$

Among the evidence for the $\ce{S_N1}$ mechanism with aryl cations as
  intermediates, is the following:

The reaction rate is first order in diazonium salt and independent of the concentration of Y.
When high concentrations of halide salts are added, the product is an aryl halide but the rate is independent of the concentration of the
  added salts.
The effects of ring substituents on the rate are consistent with a unimolecular rate-determining cleavage.
When reactions were run with substrate deuterated in the ortho position, isotope effects of $\sim 1.22$ were obtained. It is
  difficult to account for such high secondary isotope effects in any
  other way except that an incipient phenyl cation is stabilized by
  hyperconjugation, which is reduced when hydrogen is replaced by
  deuterium.

That the first step is reversible cleavage was demonstrated by the observation that when  $\ce{Ar^{15}N+ #N}$ was the reaction species,
  recovered starting material contained not only $\ce{Ar^{15}N+ #N}$ but
  also $\ce{ArN^{+} #N}$This could arise only If the nitrogen breaks
  away from the ring and then returns. Additional evidence was obtained
  by treating $\ce{PhN+ #N^{15}}$ with unlabeled $\ce{N2}$ at various
  pressures. At $\pu{300 atm}$, the recovered product had lost $3 \%$ of
  the labeled nitrogen, indicating that $\ce{PhN2+}$ was exchanging with
  atmospheric $\ce{N2}$

Source: March's Advanced Organic Chemistry: Aromatic Substitution: Nucleophilic and Organometallic

What if it was a aliphatic diazonium compound ( i am aware that its unstable but assume it to be kept at a certain temp where its stable). Will it still remain SN1 or will it depend on the substrate?

@Cinnamaldehyde yes, you are right they are unstable hence, immediately nitrogen gas is given off and a planar carbocation is formed which reacts via $\ce{E1}$ or $\ce{SN1}$


text:
Ionic Compounds
Do elements share chemical or physical properties in an ionic compound?
I mean if a metal would have an ionic compound with a nonmetal would the compound have physical properties of the metal and chemical properties of the nonmetal or the opposite?
P.s. I said "traits" in my comment... I meant properties

No.$\mathstrut$

Could you please elaborate a little bit what you are meaning? E.g. take table salt which is a ionic compound of sodium (a metal, solid) and chlorine (nonmetal, a gas).

Hello and welcome to Chemistry.SE. Your question as written is very hard to follow and I (and at least one person in the comments) cannot make sense of it. Could you try expanding the question, maybe giving some examples? As it stands, I believe it will be closed. Best of luck!

@aventurin Well let's say a metalloid would do a compound with a metal. Metalloids have the physical traits of a metal (by physical traits I mean like conductivity) and chemical traits of a nonmetal (by chemical traits I mean if it would lose electrons in a compound or get one)
If it does a compound with a metal would the compound have physical traits of a metal or a nonmetal? And what about chemical traits?
P.s. Both would be solid

What you talkin' 'bout, Willis?

@IvanNeretin No? Would you explain a bit please?

Ions do not have properties of the neutral atom, so the comparison of properties that the OP wants here doesn't make sense.

sigh  Asking in the way you do now doesn't make real sense. Compounds have  little to nothing in common with the elements from which are made. One could say, the more complicated they are, the less, but even this isn't true.

No. When elements form compounds, the physical and chemical properties of this new compound are unique to this compound; explicitly, they are different from all elements a compound is made up of and there is no way to derive one from the other.
Note that sometimes there are different compounds made from the same elements such as iron(II) chloride and iron(III) chloride. Or maybe even more obvious: tin(IV) chloride (a liquid) and tin(II) chloride (an ionic solid). And finally note that even elements have different modifications with different physical and chemical properties such as phosphorus (white, red, purple, black) or carbon (graphite or diamond).


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Where is the mirror plane in 1-bromo-1-chloroethene?
I'm studying point groups in inorganic chemistry and having a difficult time understanding some of these concepts.
In my book, Inorganic Chemistry: Fifth Edition by Miessler, Fischer, & Tarr, I am presented with a table showing that $\ce{CHFClBr}$ is in the $C_1$ point group and has no symmetry other than the identity operation.

$\ce{H2C=CClBr}$ is in the $C_\mathrm s$ point group and its symmetry is listed as having only one mirror plane.

I'm having difficulty seeing this mirror plane. I would have listed this compound as being in the $C_1$ point group. Does this have something to do with the molecule being flat?

The molecule is planar, so the mirror plane is the mirror plane of the molecule

I would suggest to "play" a bit with different molecules and discover their the symmetry operations and point groups:  http://symmetry.otterbein.edu/challenge/index.html. I also noticed that students with a background in FPS games use to comprehend symmetry operations and elements better:)

@andselisk Yeah, I'm way more than halfway through my homework and just now grasping it. Any reason behind the fps game thing? I play fps games and am wondering how it would help understand symmetry.

@MelanieShebel I would surmise that those people are more practiced at certain types of spatial reasoning skills. But there are other ways to develop those as well.

@MelanieShebel It's rather subjective, but I suspect that old-school non-linear labyrinth-type maps (such as good old Doom, Quake and System Shock) train spatial thinking and visual memory, and help to grasp flat screen representation of 3D entities faster because you are constantly under pressure and have to react/adapt swiftly. It also might just be my excuse for still playing those games as well, so don't take it seriously:)

For all symmetry operations, you should assume the atoms reduced to a point size. Then, if you have a perfectly flat molecule (like in the case of your disubstituted ethene), all atoms will be in the same mathematical plane meaning there is nothing above or below the plane.
Therefore, the plane is an element of symmetry that puts the nothing above the plane on the nothing below the plane and leaves the atoms themselves (in the plane of symmetry) untouched.

Is it really necessary to assume that atoms reduce to point size? I do not think so. Models as ball and sticks work well for symmetry consideration. Moreover not reducing to point size does convey the message that there are not things such as egg shaped atoms. Perhaps I am rough on this description but it should be clear what I mean. I prefere a picture in which the plane does cut the molecule in its upper and lower halves....

@Alchimista No, it’s not necessary but I like the picture because points will always be in the plane. I’m sorry, but there are too many negations ins your sentence about egg-shaped atoms to grasp it, can you repeat?

Important that we agree. It was to say that the current molecule cannot be different on the two sides of the plane....It will take me start to atom and so on. ...I wanted to  make a (needed , in my opinion) clarification :)


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Is it theoretically possible that some STABLE isotopes from periods 8-10 exist on some distant planet somewhere in the universe?
By stable I mean a half-life of thousands of years at the very least.
We may never have enough resources to synthesize stable isotopes from those periods, but is it theoretically possible that somewhere in the universe there has been a stellar nucleosynthesis with enough energy to naturally produce them, so that they may exist on some distant planet somewhere in the universe?
If so, we can only imagine to ourselves what sorts of materials may be abundant on those planets...

Neutron stars may contain some of these, but things are getting more and more weird the deeper they are in there.

Neutron stars do not contain what we would recognize as atoms in their bulk.  You would have to find them in the thin outer crust and even that would be hard to extract from their powerful gravitational fields.

@OscarLanzi Well, "extraction" from there is indeed rather out of question ;)

There are no reliable models for nuclei that can be used to make reasonable predictions.
The island of stability is really not well defined in theory and it is disputed if such an island even exists.  "Stability" for physicists doing this work means quite a long way short of thousands of years - thousands of seconds would be considered pretty stable by these standards.
Heavy elements are generally less abundant, so don't expect them to be abundant anywhere even if they exist.  These super heavy elements are being manufactured, and not discovered in nature.

This is not answering my question. I did not ask for predictions or observations. I was asking if it's theoretically possible, or otherwise our current models suggest that heavier elements are necessarily less stable, i.e. we know for sure there can't be a stable isotope with an atomic number larger than 82.

We don't know that for sure.  You're essentially asking me if we can completely rule out something I've told you we can't model properly, so the answer to such a question is always going to be "yes, it's possible", simply because we cannot disprove it.  As I tried to explain it seems unlikely.  Even if we had models we had great confidence in, it is the nature of science that the word "impossible" is rarely used in such a context


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How to identify Enantiomers
AP Biology Question
So I know that enantiomers are like mirror images of each other, but not identical (superimposable) upon one another. On Wikipedia, I read that it was like our hands, similar but not identical (you can't rotate one hand into another. Makes sense. But I'm not sure how this would help me identify the enantiomers. All the molecules look identical.
The answer is (D).
What is different about D?

Curious as to why this post was downvoted without explanation. I believe I provided proper context and any attempts, because this problem is just identifying a property.

Most likely it was downvoted because it's a homework like question with very little effort shown towards solving it. You haven't demonstrated drawing any mirror images or superimposing them.

i've done literally all i can. i can't post my drawings here because i don't have a camera. all the molecules are literally identical no matter how hard i try to differentiate them from one another. @Zhe do you know any way in which D is different?

In your defence I’ll admit that these are exceedingly terrible representations because the way the central tetrahedron is drawn makes everything look like the same old methane. But if you look closely notice that not every corner contains a hydrogen. Imagine if the corners were coloured differently for each of the groups (methyl, $\ce{COOH}$ or $\ce{NH2}$) and then try again.

The key point is that for four of the five pairs, the structure on the left can be rotated to make a structure exactly like the one on the right. For example, the left structure in (A), if rotated 120° about the C-CO$_2$H bond, is equivalent to the structure on the right.
For two "enantiomeric" structures, it is not possible to do this. It might be helpful to draw every possible rotation of the structures in (D), to see that it is impossible to produce the structure on the right by rotating the one on the left.

i never noticed the clockwise/counerclcokwise aspect ahhh THANKS

If you see duplicated groups attached to the C (e.g. 2 H's or 2 CH3's), then you can rule them out. For molecule to be chiral it should have a chiral center (C connected to 4 different groups). This helps you easily identify that the only possible answer is D.
Then you could check if molecules are mirror images of each other. If not, you need to rotate them to be so. If it's impossible - they are not enantiomers.
And finally when you have mirror images you can try to rotate one of the molecule to superimpose them. It still may be possible if molecule has symmetry.


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Amount of substance expressed in yoctomole units
I stumbled upon a couple of papers where amounts were expressed in yoctomoles ([ymol], $\pu{10^{-24} mol}$) and I find it somewhat bothersome as 1 ymol would correspond to about 60% of atom/molecule/ion etc., which doesn't make much sense to me:
$$N_\mathrm{A} \times \pu{10^{-24} mol} = \pu{6.022 \times 10^{23} mol-1} \times \pu{10^{-24} mol} \approx \pu{0.6}\,\text{(elementary entities)}$$
Simply put, I'd say that using zeptomoles ([zmol], $\pu{10^{-21} mol}$) is already a stretch as one of the ideas behind a mole concept is to allow to handle large numbers of entities more conveniently, and at this scale there is no reason not to use direct quantities of molecules/atoms/ions.
Should the yoctomoles units really be used, and if so, then how to rationalize fractional elementary entities?

I don't follow your conversion, but agree that a yoctomole would correspond to 0.6 molecules. Using the unit seems fine to me.

When it comes to homeopathic dilutions even this unit seems a little bit coarse-grained. Oscillococcinum e.g. has one part duck offal in $10^{400}$ parts water.

Well, the closer you get to atomic detection limits, the less sense bulk units make, that is correct.
However, coming from lab scale units it is much easier to scale down directly with SI prefixes — the exact reason what they were designed for. Therefore, it makes sense that these papers might say ‘Our predecessors could only detect $\pu{10zmol}$, we have lowered the detection limit by a factor of $100$ to $\pu{0.1zmol}= \pu{100ymol}$.’ And that is basically what they say. I note that both abstracts immediately place a molecule count behind the amount in ymol, so they are well aware of the scale of their unit with respect to molecules.
Once you get to the full molecular scale, you will be dealing with a type of quantisation this way or that; whether you are reporting energy, charge, mass or amount. In none of the other cases is anything special described for atomic scales so why should we make an exception for moles? Just make sure that the value itself corresponds to a whole number of molecules within the boundaries of rounding errors.

An addition to the self-contained Jan's answer: an existence of yoctomole can be justified by using it as the systematic unit for the absolute detection limit of $N_\mathrm{A}^{-1}\approx\pu{1.66 ymol}$ below which a molecule seize to exist.
The first instrumental method to allow for such high sensitivity was, as it appears, capillary electrophoresis with laser-induced fluorescence detection which approached yoctomolar limit of detection during 1980s (and the yocto- prefix officially adopted in 1991).
I cannot help not to quote a part of Castagnola's brief paper [1] mainly because it shares concerns raised in OP and because of the illustration (addressing aventurin's comment on homeopathy):

While we can count a dozen eggs one egg at a time, the magnitude of Avogadro’s number normally rules out direct counting of objects. … As far as egg counting goes, it is better and simpler to say, ‘Please, give me four eggs’, than to ask for ‘one third of a dozen eggs’. It should surely therefore be better to discuss sensitivity levels in term of hundreds of molecules than in terms of a ‘zeptomole sensitivity level’. … Moreover, it is obvious that the manipulation of a twentieth of a dozen inevitably leads to broken eggs and, therefore, researchers can only introduce the yoctomole limit if they are
  planning to expand their culinary skill towards the preparation of a sub-molecular omelette!

References

Castagnola, M. Sensitive to the Yoctomole Limit. Trends in Biochemical Sciences 1998, 23 (8), 283. DOI: 10.1016/S0968-0004(98)01251-1.

It isn't just yoctomoles.  Synthesis attempts for super heavy elements involve cross-section for nuclear reactions involving small fractions of the cross-sectional area associated with one proton or neutron.  Compare the cross-secions in this passage from Wikipedia with roughly 0.01 barn, the cross-section typically rendered for a single nucleon:

In August–October 2011, a different team at the GSI using the TASCA facility tried a new, even more asymmetrical reaction: …
$$ \ce{^{249}_{98}Cf + ^{50}_{22}Ti → ^{299}_{120}Ubn^* → no atoms}$$
Because of its asymmetry, … the reaction between $\ce{^{249}Cf}$ and $\ce{^{50}Ti}$ was predicted to be the most favorable practical reaction for synthesizing unbinilium, although it is also somewhat cold, and is further away from the neutron shell closure at $N = 184$ than any of the other three reactions attempted. No unbinilium atoms were identified, implying a limiting cross section of 200 fb. … Jens Volker Kratz predicted the actual maximum cross section for producing unbinilium by any of the four reactions $\ce{^{238}U + ^{64}Ni}$, $\ce{^{244}Pu + ^{58}Fe}$, $\ce{^{248}Cm + ^{54}Cr}$, or $\ce{^{249}Cf + ^{50}Ti}$ to be around 0.1 fb; … in comparison, the world record for the smallest cross section of a successful reaction was 30 fb for the reaction $\ce{^{209}Bi(^{70}Zn,n)^{278}Nh}$, … and Kratz predicted a maximum cross section of 20 fb for producing ununennium. … If these predictions are accurate, then synthesizing ununennium would be at the limits of current technology, and synthesizing unbinilium would require new methods.

The sub-particle cross-section comes from incorporating the low reaction probability into the calculation.


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What are introductory level books on chemical engineering?
I'm currently enrolled in a chemical engineering course in my college, and in my first year, I'd like to get a nice introductory book on that field of study.
I'd prefer that the book covers several topics, but less in-depth analysis of each.
Unfortunately the big-list in the Resources for learning Chemistry post does not yet cover this topic yet. With respect to the meta discussion the answer to this question will be amended to that list (and this question can be closed as duplicate after the fact).

As standard texts---though not quite introductory---one might consider Bird, Stewart, and Lightfoot's Transport Phenomena, Koretsky's Engineering and Chemical Thermodynamics, and Fogler's Elements of Chemical Reaction Engineering. I do not enjoy the chemical engineering curriculum---I find it quite boring and restrictive, but maybe you'll have better luck.

I can recommend to check out this list as well: https://accessengineeringlibrary.com/subject/chemical_engineering. Also, note that chemical engineering requires basic knowledge of computer algebra and CAD-systems, as well as some programming knowledge.

I'm studying chemical engineering for more than two years and what I can tell you is that there is not really a lot of "good" books you may read as an introduction.
You have to understand that Chemical Engineering contains both "Chemical" and "Engineering". The first one refers to chemistry whereas the second refers to the way to make something. A chemical engineer who works at Metalor (which makes precious metal bars amidst other things) really has not the same problems as another one working at Air Liquide (which produce liquid gas).

Now if you are looking for what we call the basics, any chemical engineer must be good at, fluid mechanics, thermodynamics and kinetics, mathematics and informatics are a good start.

Fluid mechanics because in any plant, whatever you produce, you need to be able to size pipes, reactors, etc. It's also a must if you encounter a two-phase flow;
Thermodynamics because when you are sizing somethings you must ensure safety and security so you need to know what properties the molecules involved in your plant have. Kinetics because this is important to know the rate of a reaction and a lof other stuff related to catalysts etc.;
Mathematics because you can't do the first three without it;
Informatics because this is pretty hard to perform all the correlations and models by hand when sizing equipment and knowing how to use the common software (Aspen, Matlab) plus having some skills in coding a few lines in either Fortran/C/C++/Python is always a must. Especially if your company does not have the budget to pay the licences and Excel is limited in many ways;
Finally, mathematics and informatics are a strong requirement to design process and safety control loops efficiently and test them with data from the process before implementation.

If you speak French, the book "Le génie chimique à l'usage des chimistes" by Joseph Lieto gives a pretty good overview about Chemical Engineering. How to make a balance (of mass or energy or momentum), basic operation units (distillation and crystallisation), mass and heat transfer, thermodynamics, kinetics and process control.

The best reference I have for chemical engineering is the McGraw-Hill Chemical Engineering Series which contains more than you should know as a student. After you will need to get strong knowledge in many areas of chemistry (electrochemistry is sometimes a big plus) because as an engineer and a chemist especially if you are looking for sizing operation, you need to think about details (Production practices, HSE, Quality), which are not in the books, unfortunately.

Could you do me the favour and adding a small paragraph instead of the place holder in the resources post?

@Martin-マーチン I don't understand what do you mean :/

I have updated the section in the resources post please have a look and extend it as necessary.

@Martin-マーチン well you did it pretty good I have nothing to add for the moment. I'll look at it again if I have some more ideas.


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NBO natural bond orbital
Is it always reliable the bonding description of a molecular system you obtain using the NBO method? In which cases do you consider that it is necessary to have special care when analyzing the NBO results?

Welcome to Chemistry! Take the tour to get familiar with this site. Right now, your question is very broad. In general, localization-based methods have a tough time with delocalized systems. It would be better if you gave more concrete examples of systems you're trying to study.


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Theoretical comparison of dipole moments of N,N-dimethyl-4-nitroaniline and N,N,2,3,5,6-hexamethyl-4-nitroaniline
These two compounds are given in my workbook, and I am supposed to compare their net dipole:

The solution given in the workbook is that:

Molecule A (N,N-dimethyl-4-nitroaniline) has a "net shifting of electrons" in the downward direction - due to $+M$ effect of $\ce{NMe2}$ and $-M$ effect of $\ce{-NO2}$;
Molecule B (N,N,2,3,5,6-hexamethyl-4-nitroaniline) - due to steric hindrance - does not undergo resonance to an appreciable extent, hence, there is negligible "net shifting of electrons" in any direction. The lone pair of $\ce{NMe2}$ almost remains in its place.

Hence, molecule A has a higher dipole than molecule B. I am content with this explanation.
However, my question is:

Is this explanation actually correct, or is it just a theoretical mask to explain an experimental observation (i.e. they just made up a random theory on their own to explain an observation)?

Are such types of explanations common, universally accepted, etc?

What is "actually correct"? We can know for sure what are the dipole moments, but not why. To me, the explanation sounds good.

@IvanNeretin i wonder if this explanation is a "standard" one, universally accepted for such compounds?

Yes, it is a fairly common explanation, but only if the groups are large enough to actually cause steric repulsions (in this case, two methyl groups are just fine, because of the flatness of the molecule).

Try googling “steric inhibition of resonance”

@orthocresol I know (and have solved several workbook questions based on) that topic, unless there was a specific link which you wanted to point me to?

Oh, no, nothing specific. This is just the same concept, applied in a different context, that's all...

The workbook's explanation could have been more specific. If you build a 3D model, you may see that the orbital interaction between the lone pair on the $\ce{NMe_2}$ and the $\pi$-system on the ring is somewhat hindered by bad overlap because the steric interaction of the methyl groups (on the ring and the $\ce{N}$) prevents perfect alignment. Same for the nitro substituent. A higher interaction would lead to more charge shift.


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Where to find good worksheets for balancing chemical equations
I am preparing for the Chemistry SAT in a couple of weeks. I am looking for worksheets like this that will give me practice balancing complex chemical equations with lots of compounds. Is there any website where I can find such questions in bulk? (And also, check my answers?) Thanks.

Textbooks and Khan Academy....


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How do ions bind with complex ions
I understand how ligands bond to a complex metal ion. e.g $\ce{[Mn(CN)6]^3-}$ but how would 3 potassium ions $\ce{K3}$ bond with the complex ion to give $$\ce{K3[Mn(CN)6]}$$
I hope my question isn't two ambiguous.

Like what besides "ionic bond" would you like to hear?

I didn't think it was just that simple. I suppose what kind of confused me is that I though all the Mn bonds would be used up and what is there for the $K$ to bond to but now I realise it bonds to the valence electrons.

Well, there's probably some weak coordination  with lone pairs of N atoms, but that's details.

It is $\ce{K+ + [Mn(CN)6]^{3-}}$. A simple ionic bond.

It is important to understand the bonding situation in complexes which can be difficult and, yes, complex. You probably should spend a fair amount of time discussing the σ and π interactions between manganese and the cyanido ligands including pointing out the high covalency of the $\ce{Mn-C}$ bonds causing a larger orbital difference and thereby a greater field split.
You may now also expect some complex interactions of the potassium cations with this complex. However, there is none. In a crystal structure, you would basically assume very footballs (the anionic complex) and much smaller golf balls (the potassium ions) that form an ionic crystal much like lithium iodide has two ions that vastly differ in size. As soon as you dissolve the solid in water, these ionic interactions break up completely to give way to dipole-ion interactions as if you had dissolved a simple ionic compound. Thus, while the solid state is adequately described as $\ce{K3[Mn(CN)6]}$, the dissolved state is better written as $\ce{3K+ + [Mn(CN)6]^3-}$.
Yes, it was indeed that simple.


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Can you decay to half an atom?
In school we are learning about half-life, and I was wondering if it is possible to decay to half an atom?

No, much like it is impossible to die to half a man.

This is a weird question. When the atom decays, there's still an atom left. In fact, there may be 2.

@IvanNeretin Well, you could get cut in half; not that unsimilar to fission ;)

If I cut a piece of string in half, and discard one of the pieces, am I left with half a piece of string?

To clear up what are becoming confusing comments.
An atom is a nucleus of protons and neutrons surrounded by electrons.
An atom can decay by fission to make two or more atoms with a smaller number of protons and neutrons in each new atom.  The electrons get distributed between them.  Now in physics we'd normally describe this as a nuclear decay, rather than thinking about electrons around the nucleus which, to a certain extent, are irrelevant to the underlying process.
Now depending on how you want to think about this you're either increasing the number of atoms by splitting one into two, or you're creating two (or more) smaller sized atoms which you might loosely describe as halves of the original.
Fission products (the results of fission) do not generally contain exactly half the number of protons and neutrons and electrons in the original atom, even when those numbers are even, so we can't say they're generally exact halves.
Fission can also release neutrons - so a nucleus looses e.g. one neutron but still has the same number of protons left.  Even when the split is closer to 50-50, the process can still release some neutrons, and in fact that's exactly what is used in nuclear weapons and reactors.
Another typical decay type is by alpha particle emission.  An alpha particle is, in fact, a Helium nucleus, so again that's a split to two different nuclei.

The photodisintegration of $\ce{^{9}_{4}Be}$ might come close to what you are looking for.
When high energy photons excite a beryllium-9 nucleus it may decay into a neutron and two helium-4 nuclei ($\alpha$ particles). Neglecting the neutron, the latter are roughly half of the original nucleus.
$$\ce{^{9}_{4}Be + \gamma -> 2 ^{4}_{2}He} + n$$

No. When an atom decays, it splits into two smaller atoms, and its protons are divided between those two new atoms.  The smallest number of protons you can split off an atom is one, and a single proton by itself is still a hydrogen atom.
Note that you can split a proton into sub-atomic particles, but this would not be called "decay" anymore.

A proton splitting spontaneously into other particles would be a proton decay.  No one has ever observed a proton decay and it's an open question in physics how long the proton half life is or even if it does decay.  Anyone seeing a proton decay, report it at once. :-)


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Treating a sample with water, how to deduce present compounds given precipitate and solution parameters?
A sample consists of one or more of the following salts: $\ce{Cu(CH3CO2)2}$ (acetate), $\ce{BaCl2}$, $\ce{(NH4)2SO4}$, $\ce{SrCO3}$, $\ce{KNO3}$. The sample, when treated with water, gives a white precipitate that is filtered off and a blue solution. The precipitate is a white solid that is completely insoluble, both in water and in $\pu{6M HCl}$ solution.
Which compounds MUST be present? Explain why.
But $\ce{Cl}$ is soluble with all the given metals! Another question gave the parameter that the precipitate is insoluble in water and $\ce{HI}$, but that was easy because there was a salt with Hg given, so it was obvious that $\ce{HgI}$ must be the precipitate. Please help. Thank you very much.

$\ce{BaSO4}$, obviously.

Well, what else could it be? SrCO3 is soluble in acids, and the rest are just soluble in water.

But Cl is soluble with Ba

True, but SO4 is not.

You correctly point out that Cl "is soluble with all the given metals". This statement is a little unclear, though, which might be why you're confused. What this means, exactly, is that no precipitate that incorporates Cl can form. For example, the solid $\ce{KCl}$ cannot precipitate, because the combination of $\ce{K}$ and $\ce{Cl}$ is soluble.
This fact alone doesn't forbid precipitates of other species. For example, the solid $\ce{BaSO4}$ is insoluble in water. We know that $\ce{BaCl2}$ alone would be soluble in water, but the barium atoms in $\ce{BaSO4}$ don't care that there are $\ce{Cl-}$ ions in the solution; all that matters is that the combination of $\ce{Ba}$ and $\ce{SO4}$ prevents the two from separating. As a consequence, when $\ce{Ba^2+}$ and $\ce{SO4^2-}$ meet in solution, they merrily form the insoluble product $\ce{BaSO4}$, regardless of the other ions present in the solution.


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Do auto-catalytic aromatic compounds capable of forming self-replicating molecules exist?
I want to know are there any aromatic compounds able to catalyze their own synthesis? Do aromatic compounds with molecules capable of self-replication exist?
P.S.I'm looking specifically for heterocyclic aromatic compounds (preferably containing nitrogen in the ring) capable of self-replication which could self-assemble into sheets of planar structures capable of self-replication. Is such thing possible and are such structures known in modern organic chemistry and/or nanotechnology?

According to the wikipedia page on autocatalysis, the bromination of acetophenone is autocatalytic. Do you mean a reaction where neither of the reactants is aromatic, but the product is?

Are you asking about autocatalysis or about self-replication which are two related but distinct topics?

I'm asking for an aromatic compound able to catalyze its own synthesis and in the process create a molecule capable of self-replication. Does such a compound exist?

You ask for way too much. We only know one example of self-replicating system (AKA life), and even that we hadn't made ourselves.

May be the one system I'm looking for here and trying to describe now is the simplest version of that self-replicating system you mentioned @IvanNeretin . If that is the case, then, we already know enough about that system! The only question which remains is can we reproduce this evolution in the lab or are we not that far yet? And if we aren't that far how could we get there?

No, we don't know enough. And no, when it comes to reproducing this or any other evolution, we aren't there yet, and that by a wide margin.

@IvanNeretin Bzzt! Wrong! Actually, quite a few man-made self-replicating systems (using the chemical definition of self-replication) are known. I can throw you the name Rebek Jr if you want to look up some papers. Unfortunately, this question is far too broad and does not ask for self-replicating molecules so I can’t really write an answer about it here

I recall seeing a system that consisted of two planar components (and I am fairly certain those were aromatic) that could couple to form the product. The self-catalysis consisted of a product molecule acting as a template for two component molecules to absorb to, which was claimed to have increased the reaction rate.
Edit: I have found the system I was thinking about. The Asker may decide for themselves if this fits the bill. The example is given in K. Ariga, T. Kunitake: "Supramolecular Chemistry - Fundamentals and Applications", Springer, 2006, p. 31. doi: 10.1007/b137036. As of this writing, the page in question is available through Google Books and the reference list for the section (2.10) is on the publisher's page. I was not able to definitively select the correct reference and I do not have access to them, anyway. Note that Ariga & Kunitake seem to have up their mind on what to call it:

In this system, [the product] acts as a template for self-replication. It may be surprising that a simple host-guest system like this mimics the fundamental activity associated with life, self-replication.

I’m pretty sure I know which system you are talking about but unfortunately it doesn’t fit OP’s context.

@Jan In any event, this may be a starting point. If you have the exact reference, please edit or answer, I could not get it.

Comments are not for extended discussion; this conversation has been moved to chat.

@TAR86 Unfortunately, I am not able to find the reference you are referring to. I did find the book on Google Books but the page in question was not part of the preview.

You are asking for a very specific, very advanced case namely a compound that:

self-replicates on a per-molecule basis
associates into sheets; and
these sheets again self-replicate

To the best of my knowledge, this is much more advanced than the self-replicating systems chemists have discovered so the answer to your full question is no.
In fact, we can tick off either of bullet points 1 and 2. A number of self-replicating systems have been discovered and published; a full overview is out of scope of this question. Also, self-association is known for many systems. However, there are no systems that perform both and to the best of my knowledge there are also no macro-aggregate sheets that self-replicate as per your outline.


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Error and effect - Dumas method
The Dumas method is the easiest approach to measure the molarity of Gases. However, this method assumes that all the vapor (i.e. from a vessel) is condensed in, for example, an ice bath.
What are the errors from the Dumas method? I believe that Henri-Victor Regnault stated this error but I cannot find any good explanation.


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How to convert the factors of radiative forcing of methane compared to carbon dioxide from ratio by weight to ratio by molecule?
The intergovernmental panel on climate change's fifth assessment report (IPCC AR 5) lists the radiative forcing (RF) of methane in chapter 8 as being $28\times$ and $84\times$ as potent as $\ce{CO2}$ by weight (or tonne of methane for tonne of $\ce{CO2}$) for the global warming potential integrated over $20$ and $100$ years, $\mathrm{GWP}_{20}$ and $\mathrm{GWP}_{100}$ respectively. In order to calculate the ratios of the gases RF potential molecule for molecule in the 20 and 100 year time spans, I'd need to apply the ratio of atomic weights for these molecules. But do I need to consider the average concentration in the atmosphere of $\ce{CO2}$ when multiplying the RF ratios.
Adding to the complexity of the problem might be the fact that at various levels in the atmosphere methane may have a different lifespan, concentration and RF, i.e. in the troposphere methane has a role as a precursor to tropospheric ozone, which takes the overall RF ratio of atmospheric methane to $105\times$ $\ce{CO2}$ (Shindell et al, IPCC AR5, NASA). Or can I ignore this complexity and just derive an indicative molecule for molecule ratio of RF potency for $\ce{CH4}$ to $\ce{CO2}$ by dividing each side of the equation by the molecular weight?

My attempt at answering:

Molecular weight of methane is $\pu{16.04 g/mol}$ and $\ce{CO2}$ is $\pu{44.01 g/mol}$.
Assuming it’s safe not to consider atmospheric concentrations in this,
$\pu{1 kg}$ of of methane has $44.01/16.04 = 2.744$ fewer molecules than $\ce{1 kg}$ of $\ce{CO2}$
If methane is $105\times$ as potent as $\ce{CO2}$ by weight, it is $105 \times 2.744 = 288\times$ as potent as $\ce{CO2}$ molecule for molecule


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Is there a connection between UV-degradation and the photoelectric effect
Recently I've been learning about the photoelectric effect in my physics classes.
I only have a rudimentary understanding of chemistry, but I've been wondering what the connection is between the photoelectric effect and UV degredation.
For the sake of simplicity let's focus on the degredation of polypropylene.
My hypothesis is, that the UV light frees some electrons from the material. These will either weaken or break the bond between two molecules or atoms in the molecule itself.
I would ultimately except the molecule chains to be broken into smaller pieces due to this.
Is that a reasonable assumption or does the process work completely different?

Yes and no.
In a metal, the photoelectric effect is when a UV photon has enough energy to push an electron from it's current energy level to at least the $n=\infty$ level. Extra energy goes into electron kinetic energy.
For UV degradation, we are promoting elections to higher energy orbitals in an analogous way. This produces an energized state that is highly reactive and may undergo radical reactions that ultimately lead to the formation of different species.
The difference is that we assume the electron is essentially removed from the atom in the photoelectric effect. Elections are weakly enough bound in a metal for a UV photon to achieve this. But in general, for UV degradation, the UV photon can't actually remove the electron from the molecule. But the photon can energize electrons that lead to degradation reactions.

Zhe is correct . However would like to point out that PE can be measured in many materials. Things of the such are even more current nowadays that organic materials are used or explored for photonic and electronic applications. A big however is that if limit yourself to e.g. degradation of polymers in sunlight,  than the second part of Zhe answer applies

Thanks for the interesting answer. I usually hear/read about UV-degradation and not say IR-degradation or anything alike - am I correct in the assumption that only for light with a short wavelength the photons carry enough energy to promote electrons to higher energy orbitals?


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Elements with no solid state
Helium superfluid-3 has got at around 2K a special liquid state with 0 viscosity with interesting properties. I am really interested in what would happen at 0K. With no kinetic energy in the particles of the compound, would it be solid? Do elements with no solid state at all exist?

Helium won't really turn solid at atmospheric pressure, but at higher pressures it can. Google image search "helium phase diagram" for an illustration

I'm not sure "at 0K" is well defined enough (since 0K is not reachable) to answer the first part of your question.

In quantum mechanics there is no such thing as zero kinetic energy.  The Heisenberg Uncertainty Principle does not allow it.  In helium the base, minimum level of kinetic energy allowed by the HUP is enough to never have a solid unless it's compressed.

In fact superfluid helium below critical temperature can not be solidified only lowering temperature. Below critical temperature if pressure is increased 25 atm or more then it solidifies.

https://chemistry.stackexchange.com/questions/60860/why-is-absolute-zero-unattainable

https://chemistry.stackexchange.com/questions/31870/is-helium-4-a-liquid-at-zero-temperature-and-zero-pressure

First of all, we cannot achieve temperatures of 0 kelvin in experiments, that is forbidden by quantum thermodynamics.
Currently, experiments can go down to seriously low temperatures of nanokelvins (source: a friend is working in that field, but spontaneously the only reference I can give  you is Wikipedia.)
At those temperatures the phases of 'solid' or 'liquid' don't really exist anymore. Rather collective quantum effects play a role, that make new phases appear, namely Bose-Einstein condensates and Fermi gases. What phase appears depends on the particles spin, which allows or disallows them to occupy the lowermost energy level.
Those exotic phases of matter count however still as 'condensed matter', which all elements at cold temperatures will have. The classical solid phase, which you're asking about, can be circumvented this way, but not at all pressures, as others correctly pointed out.