text:
Details of Boltzmann distribution derivation
I have 2 questions regarding the derivation of the formula which calculates the probability of molecules having a particular amount of kinetic energy $E_x$ in a system of $N$ molecules. It states that:
$$p(E_x) = \frac{\exp(-E_x/kT)}{\sum_{i=0}^\infty \exp(-E_i/kT)}$$
This formula is initially derived from finding the maximum number of permutations $\Omega$ with $N$ molecules distributed over $n$ energy compartments within a system according to:
$$\Omega = \frac{N!}{n_1!n_2!n_3!...n_i!}$$
Question 1. It is explained that one must find the maximum number of permutations by differentiating this permutation function and solving for $0$. However, I can already see from the formula that the maximum number of permutations is achieved if the denominator is equal to 1 (it can't be less), which means that there should be 1 molecule in each energy compartment $n_i$. Why isn’t it possible to reason the maximum that way? I'm aware this would need $N$ number of energy compartments resulting in some having very improbable high energy levels, but can't the difference in energy levels between the compartments be very small?
Question 2.  Suppose the molecules only possess translational kinetic energy. I understand that translational kinetic energy is continuous: it does not have discrete energy levels like rotational or vibrational kinetic energy. Since it is continuous, there are no energy compartments with exact energy values. The number of possible energy compartments within such a system should therefore be infinite and one could only derive probability densities. How is it possible nonetheless to calculate a limited number of permutations using a limited number of energy compartments? Does each energy compartment cover a certain range of energy levels?

My stat mech is rusty so I'll avoid commenting on the first question, but will just note that as long as the particle is confined, its translational energy is quantised as well (think particle in a box); just that the gaps between the energy levels are so small such that behaviour is almost classical

What you say is a good idea, but is not quite correct because we must maximize the distribute subject to two constraints. I will reiterate the derivation following along with McQuarrie's Statistical Mechanics.
Constraints in the Ensemble:
The constraints I mention are actually the properties that determine the ensemble we are working with:
$$
\sum_ja_j=A
$$
$$
\sum_j\epsilon_ja_j=E
$$
The first constraint says that sum over all states of the population numbers for each state $j$, given by $a_j$, must equal the total number of members in the system $A$. This is the number of molecules if you like.
The second constraint says that the energy of each state, $\epsilon_j$, multiplied by the population of that state, $a_j$, must equal the total energy, $E$, of the system when the sum is taken over all $j$. These constraints are so intuitive physically that it is easy to forget they even have to be enforced.
By the way, we are not actually constraining the total energy. We only constrain that whatever the energy is, it can be found from the second constraint. In fact, in McQuarrie, this derivation is undertaken for the canonical (NVT) ensemble.
Method of the Most Probable Distribution:
Now, as you say, the distribution which we primarily observe will be the one which maximizes the number of ways to arrange the members of the ensemble. Notice I say primarily observe. This is discussed below. This number of ways is given by the equation you write down,
$$
W(\mathbf{a})=\frac{A!}{\prod_ka_k!}
$$
There are many possible distributions that satisfy the constraints we have given. Given this indeterminate (but probably infinite) number of ensembles which satisfy our constraints, the probability of being found in state $j$ is given by,
$$
P_j=\frac{\bar{a}_j}{A}=\frac{1}{A}\frac{\sum_{\mathbf{a}}a_j(\mathbf{a})W(\mathbf{a})}{\sum_{\mathbf{a}}W(\mathbf{a})}
$$
Notice that we are now summing over the vector of all the quantum states for each ensemble, $\mathbf{a}$, because we want to find $\bar{a}_j$, the average population of quantum state $j$ in each different ensemble satisfying our constraints.
This is clearly not very useful to us because there are so many possible ensembles we will never be able to do something with them. We can, however, take advantage of the form of $W(\mathbf{a})$. Namely, by just playing around with the numbers, you will find that the multinomial coefficients are very strongly peaked at their maximum when all of the $a_j$'s are large. Fortunately, this is exactly the regime we live in because we are thinking about something on the order of $10^{23}$ particles, so the population of states will be very large.
This means we can make an approximation, which is formally exact in the limit as $A\rightarrow\infty$, that $\bar{a}_j/A=a_j^*/A$ where $a_j^*$ is the $a_j$ in the distribution which maximizes $W(\mathbf{a})$. So, the fact we are going to maximize this distribution comes about through an approximation, albeit one which becomes exact in the limit near where we are working.
So, we can simplify our equation for the probabilities to,
$$
P_j=\frac{1}{A}\frac{a_j^*W(\mathbf{a^*})}{W(\mathbf{a^*})}=\frac{a^*_j}{A}
$$
Method of Lagrange Multiplies:
We want to find $a_j^*$ which means we must perform a maximization subject to the two constraints we listed at the beginning. Maximizing something subject to constraints does not seem to be easy, but there is a very nice method for doing it called the method of Lagrange multipliers. Essentially, we introduce some undetermined coefficients to the maximization equation so that the constraints are included, but in reality we are just adding zero to the equation.
What I mean by this is that I can rewrite the constraints as follows:
$$
\alpha\left(\sum_ja_j-A\right)=0
$$
$$
\beta\left(\sum_j\epsilon_ja_j-E\right)=0
$$
where $\alpha$ and $\beta$ are the so-called Lagrange multipliers. Then, we can simply maximize our distribution as normal:
$$
\frac{\partial}{\partial a_k}\left(\ln W(\mathbf{a^*})-\alpha\left(\sum_ja_j-A\right)-\beta\left(\sum_j\epsilon_ja_j-E\right)\right)=0
$$
It really isn't obvious that this should maximize the system while satisfying the constraints, so you will have to read elsewhere why this works, but the fact is that it does work.
Notice, I have actually played a trick. Instead of maximizing $W(\mathbf{a^*})$, I am maximizing $\ln W(\mathbf{a^*})$. This is because $W(\mathbf{a^*})$ has factorials in it, so by taking the logarithm we can employ Stirling's approximation. This is also more or less exact in the limit of a large number of particles, which we always satisfy. The reason we can do this is because the logarithm of a function will have a maximum at the same coordinate as the function itself and the logarithm is smooth such that will not somehow create another maximum or something which would spoil the trick.
After taking the derivative, we lose the constant and every term in the sum where $k\ne j$. So, we have,
$$
-\ln a_j^*-\alpha-1-\beta E_j=0
$$
Or, after rearranging and solving for $a_j^*$:
$$
a_j^*=e^{-(\alpha+1)}e^{-\beta E_j}
$$
Putting it All Together:
So, all that remains is to solve for the undetermined mutlipliers and plug it all back into our probability statement, $P_j=a_J^*/A$.
This can be done very simply (and we actually don't care about the mutliplier $\alpha$). If we take the sum over all $j$ of the equation we wrote for $a_j^*$, we realize that the left side is equal to $A$ due to our first constraint and we can rearrange to write $\alpha$ in terms of $\beta$:
$$
e^{\alpha+1}=\frac{1}{A}\sum_je^{-\beta E_j}
$$
Then, we can find the final probabilities from,
$$
P_j=\frac{a_j^*}{A}=\frac{e^{-(\alpha+1)}e^{-\beta E_j}}{A}=\frac{A}{\sum_je^{-\beta E_j}}\frac{e^{-\beta E_j}}{A}
$$
We finally arrive at the final probability distribution which defines Boltzmann statistics:
$$
P_j=\frac{e^{-\beta E_j}}{\sum_je^{-\beta E_j}}
$$
Thus, the only one of these Lagrange multipliers we care about is $\beta$. We have not actually found what this multipliers is yet, but I will stop here as it should be no surprise that the multiplier is given by $\beta=1/k_bT$.

So, I believe the above answers your question 1. That is, you have to worry about the constraints associated with you ensemble so what you mention does not quite work.
As to your second question, it is not quite true that translational energy is continuous. Pretty much everyone pretends it is, but kinetic energy is quantized. It is, however, very rare that the boundaries a particle is confined to are sufficiently small (and impermeable) that quantization of the kinetic energy is noticeable. Also, because there are many of these states which are closely spaced, one must reach very low temperatures for the quantization to be important.
What I have described above is the origin of Boltzmann statistics (not Maxwell–Boltzmann, which is often confused with it). This is a common confusion because of the famous Maxwell–Boltzmann distribution of velocities. I only bring this up because the Maxwell–Boltzmann distribution of velocities makes exactly the approximation which you describe by pretending the kinetic energies are continuous.
That is, if we simply plug the kinetic energy of a particle $E=1/2mv_x^2$ into our final result (and note the deonominator is the partition function), we get,
$$
P_{v_x}=\frac{e^{-\frac{mv_x^2}{2kT}}}{Q}
$$
Then, the partition function, $Q$, is just a normalization constant. Because the states of different kinetic energies are so closely spaced, we can elevate $P_{v_x}$ to a function, $P(v_x)$, and integrate over all velocities to determine the probability of finding a particle with a velocity between $v_x$ and $v_x+dv_x$. I will leave this as an exercise. Notice this is just an integral over a gaussian function.

Thank you so much for your detailed explanation. Is there a way to prove or reason that having 1 molecule in each energy compartment violates your mentioned constraints? Perhaps it's because there are discrete energy levels which forces such a large amount of compartments to exceed the maximum amount of allowed total energy?

Ah. I think I understand now. What you have said assumes that each compartment gives an equal contribution to the energy which is not in fact the case. If one distributes the particles equally, there are only N (or N! if the particles are distinguishable) ways to do this. In reality, however, the bins have different energies associated with them so one has to find the balance between putting particles in lower energy bins and still allowing for many possible rearrangements. Also, you seem to say the number of states and particles are the same which need not be the case.

Yes, this is indeed what I meant; each bin containing one molecule with a different energy. I thought this would also give the maximum  number of permutations (denominator is at its minimum at 1) while not violating your mentioned constraints. But by the way you're explaining it, an amount of N! possible ways is actually NOT the maximum amount? Perhaps I'm misunderstanding the way of how the permutations are meant to be imagined then.

I believe what you have written does satisfy the constraints but we have a parameter to worry about which is the temperature of the system. As it stands, you have not said anything about the probability of occupying a particular state given a value of the temperature. Once you take this into consideration, you get the result I have given above which is that lower temperatures make low energy states more likely. The Boltzmann distribution adopts the configuration you suggest as $T/rightarrow/infty$ and all states become equally likely to occupy.

Sorry, I'm still not sure I understand how it is initally decided that it is impossible for each energy bin to contain only one molecule at a low temperature, which sounds possible to me if the energy bins don't differ very much from each other (very narrow energy range). Does there exist an amount of permutations higher than N!?

(1) You have to maximize $\Omega$ subject to the constraints that the total number of particles and the total energy is constant, which is more complicated than just taking the derivative and setting it equal to zero.  You can do it using Lagrange multipliers or (which is more typical in an undergraduate course) move to the canonical ensemble where the requirement of constant energy can be relaxed.
(2) Translational kinetic energy is only continuous in the limit of a particle that has amplitude over the entire universe.  For any finite-sized container, the energy eigenstates are quantized.  Mind you, they are very closely spaced.  Nevertheless, you'll find that the translational partition function contains the volume of the container, and in the limit of an infinite container goes to infinity.  Fortunately, Z itself is only of theoretical interest, any practical calculation requires a derivative of Z, and these are all nice and finite.

Thanks for the explanation. Regarding 1), I can't see how having 1 molecule in each compartment would violate the contraints. Is there a reason or mathematical proof for this? Or is N! not the maximum amount of possible permutations?

That depends on the number of compartments, and on the energy of a particle in each.  In other words, your approach will work for one specific state, but is not generalizable to all similar states in the microcanonical ensemble.  To get the general answer, you need to either use the method of Lagrange multipliers or shift to the canonical ensemble where the problem is easier.

Ok, I thought the whole purpose was to find one specific state (not several) that can have the highest amount of permutations regarding which molecule can be in which energy bin, while not violating the mentioned constraints.

I think I now know what my misunderstanding was.
What I did not realise is that, to have a certain microstate, the system must have a certain total energy U and thus a certain temperature. Therefore, the maximum number of possible ways $Ω$ is a function of the temperature. In other words, each temperature has its own maximum number of possible ways $Ω$.
If I solve the equation for 1 particle per energylevel, then I basically have already decided which microstate the system is in and which temperature it has. I have not given the formula freedom to deploy itself as a function of temperature.
Please correct me if my reasoning is wrong.


text:
Is there a concise notation for a solution of one substance in another?
I'm aware of phase notation of $\ce{(aq)}$ for aqueous solutions. Is there something similar for solutions of chemical X dissolved in chemical Y?

Welcome to Chem.SE! I edited your post a bit for clarity -- if you don't like my changes, please feel free to roll them back (click either "edit", or the "edited [time]" above my name).  I feel like I've come across something like this at some point, but I don't remember anything specific. Hopefully someone'll know.  Please take a look at the [tour] and the [help] for more info about the site -- enjoy!

X / Y will do$!$

Does this answer your question? Phase abbreviations for non-aqueous solutions

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text:
difference between elimination reactions and oxidation reactions
I know that oxidation reactions involve the loss of hydrogen. But is the mechanism the same

That rather depends on what oxidation or reduction conditions you are considering

sorry, I'm not sure what you mean

Whether the oxidation of an alochol or the reduction of a carbonyl group follows a certain mechanism depends on the reagents/conditions.
For example:
The sodium borohydride reduction of a ketone follows a nucleophilic addition mechanism:
$$\ce{R2C=O ->[1)\ \ce{NaBH4},\ \ce{CH3OH}][2)\ \ce{H3O+}] R2CHOH}$$

However, some metal catalyzed hydrogenation reactions are considered to be (nearly) concerted additions. For example, here is the transition state of the key step of the mechanism of the Noyori reduction.
$$\ce{R2C=O ->[\ce{H2}][\ce{RuCl2en2/BINAP}] R2CHOH}$$

Many oxidations of alcohols follow something of an elimination mechanism. For example, the key step of the Swern oxidation looks like an elimination.
$$\ce{R2C=O ->[1)\ \ce{DMSO, (COCl)2}][2) \ce{Et3N}] R2CHOH}$$

There is a second important mechanism: hydride transfer oxidation. This is the prevailing mechanism biochemically, where the alochol is used as a hydride source to simultaneously reduce another compounds.
$$\ce{R2C=O ->[\ce{NAD+}][\ce{base}] R2CHOH}$$

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text:
Is Cracking of Every HydroCarbon Family possible? (Alkane ,Alkenes,Arenes)
I know Cracking is usually used to crack alkanes to get smaller hydrocarbons (as they are easier to ignite) and alkenes(as they have more industrial uses)...but can we crack arenes and alkenes as they are also included in hydrocarbons definition.
But I really doubt about cracking of benzene as lot of energy is required to destroy the delocalised ring. But given infinite energy that is surely possible.... but the product won't be stable... :/

Hint: All those .... make a text hard to read.

Simple arenes (benzene, toluene, xylenes) are a typical product of a steamcracker. https://en.wikipedia.org/wiki/Cracking_(chemistry)#Thermal_methods
Catalytic hydrocracking otoh. (with added hydrogen at high pressure) leads to an increase in the aliphatic fraction and lowers the aromatic content. https://en.wikipedia.org/wiki/Cracking_(chemistry)#Hydrocracking
in itself, benzene is quite stable. It's a questions if it has sth. to react with.


text:
Chemistry - Solutions
Please give me a few hints as to how I can approach this problem
A mixture of two miscible volatile ideal liquids P and Q (obeying Raoult’s law) is kept in a vessel (molar ratio of P and Q in the mixture is m). At a suitable temperature T, the vapour above the liquid is condensed in another vessel. The liquid obtained on condensation is allowed to evaporate and establish equilibrium with its vapour. The vapour is then condensed in another vessel. The process of such evaporation and condensation is repeated for n times. If the ratio of the vapour pressure of pure P to that of pure Q is p, the molar ratio of P and Q in the condensed liquid obtained after nth cycle (for finite n > 1) is..?
In each condensation-evaporation cycle, shouldn't the ratios of moles be the same?

 Hint:
The vapor pressure of a particular component in a solution is:
(Vapor pressure of that pure substance)*(Its mole fraction in solution)
Use this to find the ratio of the vapor pressure of P and Q above the initial solution. This ratio is equal to the ratio of their moles in the liquid which is obtained by condensing. Repeat n times.


text:
Relationship between Hartree-Fock and Hückel coefficients
I am new in computational chemistry. I'm studying the Hartree–Fock–Roothaan (HFR) method and have some of questions about the HFR basis function coefficients.

Do we have the condition on HFR coefficients like in the Hückel method?

In the Hückel method, we have a normalization condition on the coefficients $\{c_i\}$:
$$c_1^2 + c_2^2 + \dots + c_i^2 = 1 \tag{1}\label{1}$$
In the HF literature I didn't see that condition, so my guess is that we don't have it.

Do we have similar to Hückel method condition on HFR coefficients of bf?

I'm wondering because going deeper in understanding HF I find out that we have certain conditions that looks like $\eqref{1}$. For example, a normalization condition for the Gaussian primitives that are not normalized:
$$
S = \left< \Phi_{\textrm{1s}}^{\text{STO-3G}} | \Phi_{\textrm{1s}}^{\text{STO-3G}} \right> = 1
$$
$$
\Phi_{\textrm{1s}}^{\text{STO-3G}} = N_1C_1e^{-a_1r^2} + N_2C_2e^{-a_3r^2} + N_3C_3e^{-a_3r^2}
$$
This is not the same as $\eqref{1}$ but we can represent it as
$$f_1(c_1^2) + f_2(c_2^2) + \dots + f_i(c_i^2) = 1 \tag{2}\label{2}$$
Because HF is more complex than Hückel, $\eqref{2}$ can make a sense. So can I think about $\eqref{2}$ as complex version of $\eqref{1}$? I was hoping to find $\eqref{2}$ or similar equation in HF literature intro part, but those sources that I have do not mention $\eqref{2}$ in intro part. Maybe my attention was scattered or it is just a mess in my head, anyway I will be grateful for any help.

What are the conditions imposed on the coefficients of the basis functions that use in the calculation of the HFR equations?

For example, the electronic charge distribution is
$$
\sum_{mn} P_{mn}S_{mn} = N =\sum_{mn} 2\sum_{a}^{N/2} c_{ma}c_{na}^*\int{f_m(r)f_n^*(r)\mathop{dr}},
$$
where
$$
S_{mn} =\int{f_m(r)f_n^*(r)\mathop{dr}} = \delta_{mn},
$$
and N is the total number of electrons. Can we consider this as a condition?

I would highly recommend getting Szabo and Ostlund's Modern Quantum Chemistry if you are trying to learn about Hartree Fock and the Roothaan-Hall method. @AlexMal

Let start by definition the Hartree-Fock-Roothaan equation:
$$\sum_\nu F_{\mu\nu}C_{\nu i}=\varepsilon_i\sum_\nu S_{\mu\nu}C_{\nu i}$$
Where the Fock matrix elements are giving as:
$$F_{\mu \nu} = \left< \phi_\mu \left| \hat{f} \right|\phi_\nu\right>$$
And the overlap matrix is given as:
$$S_{\mu \nu} = \left<  \left. \phi_\mu \right|\phi_\nu\right>$$
Here $\phi_\nu$ is our atomic orbitals (i.e. what you get from your basis set).
When we do a Hartree-Fock calculation we want to find a set of orthonormal orbitals, also known as molecular orbitals. The molecular orbitals are constructed as:
$$\psi_i=\sum_{\mu=1}^KC_{\nu i}\phi_\nu$$
For the molecular orbitals to be orthonormal we can write the following condition:
$$\left< \left. \psi_i \right|\psi_j\right>=\delta_{ij}$$
This means that the orbitals have zero overlap with one-another and an overlap of $1$ with them self. If we insert the expansion of the molecular orbitals in atomic orbitals in the above equation we get:
$$\sum_{\mu=1}^K\sum_{\nu=1}^LC_{\nu i}^\dagger C_{\mu j}\left< \left.\phi_\nu \right|\phi_\mu\right>=\delta_{ij}$$
I.e. we can see that when we do a Hartree-Fock calculation we have the constraint on our molecular orbital coefficients that have to transform our atomic orbital overlap to a matrix with ones on the diagonal and zero elsewhere. This can also be written more compact in matrix form:
$$\mathbb{C^\dagger SC=1}$$
Which is a condition equivalent to the one you know from Hückel theory. At last let me make a breif comment about basis set, since you also talked about those in your question. In general we use contracted Gaussian type basis sets, just like STO3G, that you mentioned. Our atomic orbitals can thus be written as:
$$\phi_\nu = \sum_{k=1}^M N_k d_k P(r)\cdot e^{-\alpha_k\cdot r^2}$$
Here $P(r)$ is a function related to the angular moment quantum number.The way the contraction coefficients $d_k$ are choose implies that we normalize our Gaussian functions, i.e:
$$N_k^2 \int_{-\infty}^\infty \left( P(r)\cdot e^{-\alpha_k\cdot r^2} \right)^2 dr = 1$$
We can now write $\left< \left.\phi_\nu \right|\phi_\nu\right> = 1$ as:
$$\overline{N}\sum_{k=1}^M N_k d_k\sum_{l=1}^T N_l d_l\left< \left.P(r)\cdot e^{-\alpha_k\cdot r^2} \right| P(r)\cdot e^{-\alpha_l\cdot r^2}\right> = 1$$
Where $\overline{N}$ is the normalization we need. As we can see there is no easy condition we can put on the contraction coefficients, in order to avoid evaluation $\overline{N}$. It can be shown (I will not show it here) that $\overline{N}$ only depends on the total angular moment and on the Gaussian exponents, thus it is possible to choose the contraction coefficients in such a way, that we do not need to evaluate $\overline{N}$.

Thank you very much. I appreciate this. But one thing is confuse me. The coefficient C^2 should be less than 1 to hold the equality. But this resource http://phys.ubbcluj.ro/~vchis/cursuri/cspm/part3.pdf  (page 17) gives for some coefficients value more than one.  What is my problem? Maybe they are not normalized?

No, the pdf is right. I just made a mistake. The final condition is the CSC = 1. The other part was a brain fart from my side.


text:
How would you write the empirical formula of an ion?
How would you write the empirical formula of e.g. X2Y42-?

Divide the charge too and write $XY_2^-$?

This seems like a really good question--I've never thought about it before. I recommend against closure. Since in the experimental analyses that involve the use of empirical formulas it is impossible to have large quantities of isolated ions, practically speaking it's not of very much use to define an empirical formula for an ion.

I think we should keep the question, if you allow me to digress into neutral things. Keep in mind that solid CrO3 forms Cr3O9 in the gas phase.


text:
Would it be possible to destroy gold?
I'm a writer. I have a scenario in which a sizable amount of gold needs to be rendered unusable, preferably completely destroyed. I know an acid like aqua regia is able to dissolve gold, but would there be a way to make it impossible (or at least not worth the cost) to chemically salvage the gold from the acid, or another way to permanently destroy gold?

Remember the plot of Goldfinger?

Short of nuclear reactions, you can't permanently destroy a chemical element. Then again, dissolving it and pouring the solution into the sea might be a viable alternative.

@ivan neretin nobody has mentioned dissolving it in the ocean in an answer yet, you should

@arp I did. ;-)

Just so we know what the wager is, how much gold exactly are we talking? `Cause I will definitely try to blow up that volcano if I can ruin the world's gold market for years with it.

If cost is no object, then fling into the sun. Crash it into Jupiter, it won't be usable or retrievable.

Another place to ask these sort of questions is https://worldbuilding.stackexchange.com/

@IvanNeretin More simple way: it could be also town in the toilet.

Not a chemical means, but you could pulverize the gold into dust by mechanical means and then release it from an airplane or balloon at high altitude.  Good luck recovering more than a tiny fraction of it.

Marianas Trench?

Oddly no one has asked you to specify the amount.  "Sizable" is vague, and the amount makes a huge difference to what is practical.

Other than a nuclear reactor, the only chance is to dump it into a volcano. Having a much higher density than magma, it will just fall through until it hits earth's mantle. Then it's really gone. OK, dissolved and diluted in the Atlantic is also quite safe.
P.S. Warning: If the bullion gets stuck in the volcano, which later explodes Krakatau-style, your gold brick might land on someone's front lawn. Making a hole five meters deep, but it may be salvaged easily.

They way of the magma in the volcanos are probably not linear, so it won't sink until the core of the Earth, but probably even this will be enough.

@peterh You're probably right, but proving that will require some serious spelunking. ;-)

Can you please remove the line about the Earth's core because that's complete rubbish. There's a whole 2800 km of solid mantle for it to get through before it reaches the core, even aside from the fact that it would melt in many magmas, especially if it sinks a bit first.

@bon You're right, good point, but a several kg gold brick will also fall through the mantle within a few thousand years I wager. "Solid" is not so much an elastic but more a pliable argument. ;-)

No the mantle really is very solid on short timescales. It has a viscosity of $\mathrm{\sim 10^{20} - 10^{21} ~Pa~s}$. Also, it won't even reach the mantle if you dump it in a volcano. It might get a hundred metres down if you're very lucky before it melts.

@bon So what if it melts? It'll keep falling. I doubt gold is soluble in molten rock. $10^{20}$ pascalseconds is a word. Didn't know what the actual number was. That will take a few million years to get through.

Why wouldn't gold be 'soluble' in molten rock. There will be convection in the magma and it will get mixed to some extent...

"Next to the nuclear reactor, the only other chance is to dump it into a volcano. " Are there any volcanoes next to a nuclear reactor? That seems quite unsafe.

"Next to means "Next best thibg to" here, not "adjacent to".

With full understanding of the intended meaning...pretty sure some of the nuclear reactors in japan are physically quite close to volcanoes :P

Does it really matter how deep it sinks or how well it mixes? The gripping hand is, your gold is now under a lot of red-hot lava, making recovery infeasible over the next several human lifetimes.

@Karl no, your gold will not even get to the mantle. It will not even penetrate the crust. You’re welcome to ask a question in earth science SE on why that is the case.

So in other words, Frodo had it right.

As other answers have pointed out, you can't 'destroy' gold chemically.
Putting it in a nuclear reactor for an extended period of time will functionally convert most of it to an isotope of mercury. If you really want to get rid of it, and you have lots of funding, you could launch it on a rocket into the sun (or another star, or better yet a black hole).
If you just want it to be very difficult to recover, you have lots of other options depending on the desired difficulty:

Dissolving it in aqua regia is a decent start, as you can then dump the result into almost any large body of water, which would make it extremely difficult to recover.
Dropping it into an active but not erupting volcano would be decent too, it will melt reasonably quickly (gold has a somewhat low melting point), and then diffuse throughout the magma.
Somewhat lower tech, but if the setting is before submarines became common, drop it in the ocean over a subduction zone.  It will sink very quickly, and slowly be pulled into the mantle by the subduction zone, so if you drop it close enough, it will end up being unrecoverable by the time people can go that deep.
This won't really get rid of it, but gold dissolves readily in mercury at room temperature to form an amalgam.  The resulting amalgum can actually be pretty easily separated by just letting the mercury evaporate, but it is a pretty neat way to get rid of the gold temporarily.

Thank you for the reply! I think the option of aqua regia dissolution, then dumping it into a large body of water would be a somewhat "realistic" (atleast in the realistic writing sense) solution.

I'm not sure about the black hole idea. Because of time dilation, an object sent towards a black hole will never cross the event horizon from our perspective, leaving open the possibility to catch up to it and retrieve it before it crosses over. Another issue is that the nearest black holes are at least a few thousand light years away. Presumably the OP wants the gold destroyed in a faster time scale than this. +1 for some good suggestions though.

@JBentley Sorry for that the Community user rejected your edit suggestion, because it conflicted with my ongoing one. But I dug after in the logs and I did everything as you suggested. Tyvm!

@JBently Time dilation affects everything moving inwards though, so it's still functionally unrecoverable because the vessel recovering it will also be delayed.

"you could launch it on a rocket into the sun" Because that means killing 30km/s of Earth's orbital velocity, it's easier to instead use it to your advantage and launch it away from the Sun with only about 16 km/s. While it's possible to catch it, it's all but impossible to bring it back.

@Schwern  You could just as easily set up a hyperbolic orbit around the earth to slingshot it into the sun.  In fact, if you do it right, you only need enough fuel to get the payload into the right position such that the earth passing by pulls it into the appropriate orbit.  This would of course take a long time to achieve, but is still doable.

Not destroying gold, but still giving a good appearance, is this time-honored story of how two golden Nobel prizes were "hidden" from the Nazis by dissolving the gold into aqua regia.  The aqua regia, of couse, had only a limited shelf life; but the soluble gold species it produced could be retained indefinitely.  After the war the gold was reduced, recovered, and recast into the Nobel prizes.

From the point of the view of the OP, would simply dissolving it in this fashion, and then dumping the dissolved gold into a suitable water supply suffice? Or even pouring it onto the ground to soak into the soil? Will the gold chloride remain in solution in these circumstances?

Pouring into water is better.  Pouring into the ground could leave the gold concentrated enough to recover.  In the actual case it was just left in the spent aqua regia solution(s), and the Nazis never caught on.

@OscarLanzi If you pour it into one spot on the ground, certainly. If you slowly dribbled it across a large area (say, at a drop or two a second out of a car while speeding down the highway), then it would be much harder to recover.

If you don't use any nuclear technology, you can't destroy gold.
Everything that you can do to the gold, can be undone.1
What you can do: you can make its reconstruction economically infeasible.
However, gold is

Valuable
Very stable (chemically) and non-reactive.

Because of these, gold can be easily extracted from practically any gold compound.
What could work: Maybe mixing it into high-reactive nuclear waste, for example into freshly depleted nuclear fuel rods, and then turning off their cooling. (Depleted fuel cells should be cooled for a while after they are removed from the reactor, to avoid them melting because their own radiation.) The result will be that they melt, and the resulting corium will mix with the gold therein.
1 As @IvanNeretin mentioned in a comment: dissolving the gold in an acid, and then pouring this into the ocean (or throwing in the toilet) would work.

I was just about to recommend "keep it in a nuclear reactor until its nicely activated".

@Karl Wow, how much simpler :-) Note, gold has only a single stable isotope, gold-197. Gold-197 with a neutron can make either platinum-197 (due to neutron-proton reaction), which decays back to gold-197 with beta decay. Or it can make Gold-198, which decays quickly to mercury-198, which is stable.

OK, Pt-197 has a half-life time of 20h, so that will only make the gold unuseable for a few months. As our present hates nuclear fuel reprocessing for some reason, your idea is safer. ;-)

Gold can be practically extracted from most compounds when the concentration is sufficient.  The key to destroying gold's practical value is to "dilute" it either chemically or physically.  A nice variation (which, according to The Disappearing Spoon, has accidentally occurred in reality) is to hide gold in materials that will be used to construct something else valuable before it is discovered.

@supercat: Now I'm curious about the story.

@R..: I recommend the book, but this particular story, in brief, was about a community that discovered that many of the stones that had been used as building material would, if heated, release gold.  So gold which could have been easily and cheaply extracted if people knew about it before using the stones to build things ended up being built into structures.

Gold is extremely difficult to get rid of, either chemically or physically.
Gold is produced when a massive star goes supernova. Every single atom of gold in the entire universe is the result of the death of a star. So, a lot of energy went into the actual creation of the gold to begin with. More energy will be needed to destroy it's very atomic structure.
Ways to get rid ot it:

Grind it to dust and disperse it;
Dissolve it in aqua regia (as mentioned in various other answers);
Melt it in a kiln and mix it with something else (metals or ceramics);
Others suggested to dump it in a volcano or something like that to make it disappear into the planet's interior
Reacting it to a halogen to make AuCl3, AuBr3 or AuI. Any aurohalide can of course be reduced back to it's constituent parts.

Are you sure there are no gold atoms that have been artificially created? What about gold in nuclear reactors? Do you consider gold that has been converted to an unstable isotope of platinum and then decayed back to gold created by the death of a star?

The exception proves the rule ;-)

Gold has one principal isotope (gold-197) and all other isotopes are unstable and decay mostly by beta emission into isotopes of mercury. Given that its cross section is 98.7, it will act as a neutron poison in reactors transmuting with a 2.7 day half life into mercury-198, which is stable. Given the cross section and short half life, gold was considered a salting agent in radiological weapons.  It has been reported that the plutonium pits of US nuclear weapons are all gold-plated to prevent corrosion with the air. The only way to get lighter isotopes would be through spallation.
As you know you cannot destroy an element.  Aqua regia could dissolve it but not destroy it.  As a matter of fact, George de Hevesy dissolved Max von Laue and James Franck's gold Nobel Prizes medals in aqua regia during the Nazi occupation of Holland. The stupid Nazis were oblivious to the solution sitting in a flask on a shelf at the Niels Bohr Institute. After the war, de Hevesy precipitated the gold and the Nobel society minted them new medals from the comingled gold of their original medals.

The medals were in Kopenhagen, in Nils Bohr's Institute (which today carries his name). Hevesy worked there, and dissolved them when Denmark was occupied in 1940.

In William Gibson's novel Spook Country, a large quantity of cash is exposed to a radioactive source making it radioactive and permanently unusable. The same thing could be done to "ruin" gold, by contaminating it with a radioactive substance.

Gold has only a single stable isotope, ${}^{197}Au$. For neutron radiation, it can become either ${}^{197}Pt$, which decays quickly back to ${}^{197}Au$, or ${}^{198}Au$, which decays quickly to ${}^{198}Hg$, which is stable.

I remember reading many years ago that you could transmute mercury into gold by a suitable nuclear reaction, but it was actually cheaper to transmute gold into a very pure mercury isotope using a different nuclear reaction.

It was already mentioned, with much more detail...

Use something that can shot waves of protons with great dynamic to make protons and neutrons separate in gold atoms so it not gonna be gold again! (but if you can decomposite it so you can reform it in some way)
( Just like you use your hand's force to cut a cheese off)
Like an explosion of a neutron star.
Or in an imaginary scenario when you use a rocket to carry all the gold amount and let it detonate itself with a nuclear bomb so there will only be gold dust in the vacuum.


text:
Identifying Donation of an Electron Pair
Which one of the following reactions does not involve the donation of an electron pair?
A) $\ce{H+ + CH3NH2 -> CH3NH3+}$
B) $\ce{AlCl3 + Cl- -> AlCl4-}$
C) $\ce{CH3Cl + CN- -> CH3CN + Cl-}$
D) $\ce{Cl2 + 2I- -> 2Cl- + I2}$
The answer is D, but I don't know why.
The equation could be split into two half equations:
$\ce{Cl2 +2e- -> 2Cl-}$
$\ce{2I- -> I2 + 2e-}$
But surely this shows the iodide ion is donating a pair of electrons to the chlorine atom?
I'm sure really what approach I should take for this question.

D donates 2 electrons, not an "electron pair." There is a subtle but important difference here because D is a reaction that takes place via the transfer of single electrons (though in the end, 2 are transferred in total).
So, A, B, and C are examples of Lewis acid-base neutralizations, but D is not. D is classical redox reaction.

How could one differentiate between the two (electron donor and electron pair donor) ?

Mechanistic studies, or in most cases, pattern recognition.


text:
What forms the emulsion in chocolate?
Some days ago, before buying a chocolate I read the ingredients and saw that it contains a natural emulsifier. A Google search didn't reveal anything about of what the emulsion formed is, do you have any idea?
The general composition of chocolate:

The main components in milk chocolate is cocoa butter, sugar, and milk
  powder.

Then, a concrete example:

Hershey’s chocolate is made of sugar, chocolate, cocoa butter, cocoa
  processed with alkali, milk fat, lactose (milk), soy lecithin, PGPR
  (emulsifier), vanillin, artificial flavor, and milk.

This is taken from The Chemistry of Chocolate on ChemistryIsLife.com
I continue wondering what the emulsion is formed by.

Liquid chocolate is emulsion, afair lecithin emulsified.

@Mithoron yes, I'm asking why chocolate is an emulsion...

Only molten. And what else could be pretty homogeneous liquid containing fat and water?

There isn't enough information in the question to provide a definitive answer. Do you have an ingredients list?

Chocolate consists of tiny clumps of cocoa solids, sugar, and milk solids dispersed in cocoa butter.  Because the solids aren't soluble in the butter, they tend to clump up, which reduces the quality of the texture of the chocolate, it feels less "smooth."  When you add an emulsifier, the hydrophilic end tends to stick to the relatively more polar surface of the solids (the proteins in the cocoa solids, for example, or the carbohydrates) and then the hydrophobic end interacts favorably with the cocoa butter.  The coating of emulsifier tends to help the clumps stay small and well dispersed in the fat.  That makes it easier and faster to process the chocolate, and it also lets the chocolate last longer before the fat and chocolate solids start to phase separate on a macroscopic scale.


text:
Methionine/Cysteine under redox control
Methionine and cysteine are the principal sulfur-containing amino acids, and
oxidation of cysteine and methionine residues had been well studied. How come NOT all proteins carry Methionine and/or Cysteine are under Redox Control?


text:
Preventing silver tarnish with copper
I am examining methods for storing silver bullion (both .999 fine and 90% silver) in such a way that it does not tarnish and want you to evaluate a suggested solution.  I know there are commercial products intended to prevent or slow tarnish, such as cloth bags that are infused with silver particles.  However, they aren't really suitable for bullion in my opinion.  A proper solution should be

Very cheap
Work for a lot of silver, in varying sizes and shapes
Work for a long time

The suggestion in question is to store the silver in a somewhat air-tight container (like a ziploc bag or rigid plastic container) in which you also store some clean copper (perhaps bare electrical wire).  In this situation, it has been suggested that atmospheric sulfur that gets in will react preferentially with the copper, leaving the silver pretty free of tarnish.  From time to time the copper could be cleaned or replaced.
Unfortunately, the source who presented this solution did not seem very authoritative to me. Can you analyze the solution and give me your opinion?  Please assume that tarnish is unacceptable, even though the silver is being kept as bullion and there are methods for removing tarnish.
Bonus question 1: Are there other atmospheric chemicals aside from sulfur that attack silver over time that one should worry about?
Bonus question 2: It as been suggested by some that the low density polyethylene in plastic bags will off-gas over time and cause additional tarnish or some other chemical reaction.  I can't imagine this is sulfur but perhaps there is a chemical released that affects copper?

Give this a read when you can. Also, you might find this table useful in that regard O:)

Thanks!  Am I to infer that in order for a copper object to best protect silver, it must be in physical (well, electrical) contact?  I have only heard of this kind of thing in a liquid solution.  Just looking to understand better.

The classic way to protect silver is probably to coat it with cellulose nitrate varnish.

@farnsy The way I see it, yeah. Galvanization works on the same principle. Even if we don't expect trash cans, lamp posts, nails, etc to spend time submerged in water, we still galvanize them. This is because corrosion is an electrochemical surface process; presence of moisture in air is conducive to this. Of course, if you can keep your container totally dry (which isn't practically feasible), then you needn't worry about corrosion O:)

It seems like aluminum (or zinc) is the usual choice for cathodic protection, not copper.  But I'm not used to seeing aluminum get corroded by atmospheric sulfur.  I'm sorry to be ignorant (high school chemistry only), but is the silver-sulfur reaction a good candidate for cathodic protection?  If so, what's the right metal to use (should I be thinking of using aluminum foil instead of bits of copper)?

Aluminium is good, Zinc would do it


text:
Why does the Leidenfrost effect (seemingly) not apply to the case of molten NaCl when it is poured into water?
On pouring molten sodium chloride into water you can see that when molten $\ce{NaCl}$ (table salt, i.e. sodium chloride) is poured into water ($\ce{H2O}$), the Leidenfrost effect appears to be nonexistent, whereas it clearly manifests itself in instances where other molten salts such as sodium tetraborate (1) (borax) and sodium carbonate (2) are poured into water.

Why does the Leidenfrost effect in molten NaCl last much shorter than that of the other salts, appearing to be nonexistent?
Input from paracetamol:

"The Leidenfrost effect (physical phenomenon) does show up. See the "cushion" of water vapour around the blob of molten salt in the video? Though short-lived, it's pretty apparent. Now why molten table salt would react with water faster than molten borax or soda (as implied in your post) is an interesting question."

The Leidenfrost effect (physical phenomenon) does show up. See the "cushion" of water vapour around the blob of molten salt in the video? Though short-lived, it's pretty apparent. Now why molten table salt would react with water faster than molten borax or soda (as implied in your post) is an interesting question. My fingers are crossed for the answers! O:)

Related: https://chemistry.stackexchange.com/questions/53840/why-exactly-does-molten-nacl-explode-when-it-is-poured-into-water

It could be that the Leidenfrost effect actually applies just fine.
The violent reaction may instead be a coulombic explosion, not simply a classical heat transfer.

In the video, TheBackyardScientist offered an explanation for the huge explosive effect when 50 grams of molten NaCl is poured into water. He said that some water penetrated the blob of molten NaCl and blew it apart from the inside. A Leidenfrost effect on the outside of the blob is visible, but not the major effect.
The borax blob did not explode, so it must not have been penetrated by a significant amount of water. Borax is a tetrahydrate; when it melts and then solidifies, it goes thru a glassy stage which might be expected to be somewhat cohesive, perhaps resembling Silly Putty. In addition, it loses a lot of water on melting: an initial amount of 50 grams melts to only 26 grams. (Boric acid solidifies similarly.)
NaCl does not exhibit a glassy stage on cooling (note the crystalline surface on the patty from solidified NaCl). Perhaps 26 grams of NaCl would also not explode but merely show the expected Leidenfrost effect. As a 50-gram blob of NaCl cools, solid (strong) regions form over weak molten interior; the weak part of the blob splits open, allowing water into an interior hot enough to boil the water quickly and blast the blob skins apart. The hardened skins of the blob serve as a partial container - like a pipe bomb.
50 grams of washing soda (sodium carbonate decahydrate) will melt down to 18 grams; again, perhaps this is too little to give an explosion.
The Leidenfrost effect should be expected for small blobs of anything. The explosive effect might be expected only for larger quantities, but including even materials cooling thru a glassy stage, if the amount were large enough.

Thanks for the answer, I'll do a bit more research with your answer in mind.

+1 for the mass analysis and mentioning the possibly glassy state of borax. The rest is a little bit speculative, imho.

The Leidenfrost effect breaks down when stable film boiling stops and transition boiling starts. The temperature when this occurs is called the Leidenfrost point and its value is about $\pu{100°C}$ above the boiling point of water. This means that when the surface temperature of the salt drop is below $\approx \pu{200°C}$ the Leidenfrost effect breaks down.
Now consider liquid sodium chloride at its melting point temperature of $\pu{801°C}$. The thermal energy it contains compared to $\pu{100°C}$ is $\pu{28 kJ mol^-1}$ heat of fusion and $\pu{26 kJ mol^-1}$ from heating the solid from $\pu{100°C}$ to $\pu{801°C}$.
From this we can calculate that the sodium chloride must have released about
$$\frac{601°C}{701°C} \cdot \pu{26 kJ mol^-1} + \pu{28 kJ mol^-1} = \pu{50.3 kJ mol^-1}$$
or 93% of its effective thermal energy before the Leidenfrost effect breaks down.
Given this value it does not seem plausible to me that the vapor explosion observed when pouring molten sodium chloride into water is related to a global breakdown of the Leidenfrost effect.
Instead it seems to be more plausible that the high temperature difference between the water and the molten sodium chloride allows for sufficient heat transfer, even in the film boiling regime, if the drop of liquid sodium chloride is fragmented into small droplets by dynamic processes when the hot liquid enters water. This assumption is backed by the observation of Matsumura and Naria that the occurrence of vapor explosions when dropping liquid tin into water depends (besides other factors) on the dropping distance.
Parameters that might affect the fragmentation process might include

temperature of the hot liquid
temperature of the cold liquid
dropping distance
thermal conductivity
viscosity
surface tension
density
mass of the drop
shape of the drop
energy content

and other factors.
Viscosity probably explains why the glassy sodium borate melt does not show a vapor explosion.

I want to acknowledge that I think the answer by James Gaidis is close and identifies key details but makes a wrong analysis. The Idea that Lidenforsting water could penetrate a small crack is an inconsistent explanation.  If you watch the video, there is a clear "detonation" of some sort; An instantaneous point in time where the hot sodium chloride violently expands. The final size of the blob might explain the difference in results, but I am unconvinced.

The first thing to note is that sodium chloride will always solidify as a crystalline material even when rapidly cooled, but borax and boric acid will not. This means that on cooling sodium chloride undergoes a first-order phase transition releasing energy from the latent heat of fusion and experiencing discontinuous viscosity that borax and boric acid do not.
The second thing to realize is that sodium chloride shrinks upon solidification from about $\pu{1.54g {cm}^{-3}}$ as a liquid to $\pu{2.17g {cm}^{-3}}$ as a solid.  This creates a phenomenon similar to producing Prince Rupert's drops that creates a very tempered glass.

With these concepts in mind, we can now know that upon pouring sodium chloride into water it will cool rapidly starting at the leading edge entering the water first. This rapid cooling causes the sodium to solidify on the outside at net shape and will not shrink appreciably further since it is in a crystalline, not a glassy state. As the inside cools and solidifies it shrinks and pulls the outer hardened shell of the salt inward causing it to be loaded under a great deal of pressure similar to the creation of a Prince Rupert's drop.
Salt does not have the covalent bonding needed to tolerate this tensile force well and eventually the inner part will fracture off the outer part. This fracture causes other fractures and the outer layer to decompress and spring outward similar to when a Prince Rupert's drop is scratched or fractured (usually at the tail). As the outer layer of salt springs outward, it further fractures and the pieces come closer to the water which rapidly increases the rate of vaporization causing the observed explosion.

Because, physically speaking, the salt water should never come in contact with the water, thanks to something called the Leidenfrost effect. This effect occurs when a liquid comes into contact with a substance significantly hotter than its boiling point and forms an insulating layer of vapour

@PeterJohnmeyer-Here if you notice my friend,the salt water comes in contact with the water as soon as you pour molten $\color{aqua}{NaCl}$ into it


text:
Phantom Material: why formaldehyde doesn't rise my jelly melting point?
Just to be clear, I'm not a chemical but only a electronic engineer which want create a "phantom material" to reproduce the electrical characteristic of epidermis for the analysis of electrodes impedance.
I found this paper which explain very well the process to create the phantom. As described, the use of formaldehyde is very useful to rise up the melting point of jelly to menage the final product even at room temperature. The quantity I need to fill my container is 285 ml so I scaled the doses in Table 3 at pg. 31 in the paper. I got

285 ml distilled water
8.55 g of gelatine foil (in the paper is used the powder but I found only the foils. The product says 12, or 20 g, foils for 1 litre of water but I rised this dose to get a more solid jelly)
0.96 ml of formaldehyde (37%)
0 g of NaCl (I need the lower conductive type)
0 g of ethyl 4-hydroxybenzoate (I haven't found this product)

I dissolved the gelatine foils in hot distilled water, then added the formaldehyde and, after the temperature of mixture has dropped, putted all in the fridge.
Now the problem is that the my phantom melts even at room temperature. It seems the formaldehyde doesn't his job.
Did I do something wrong?


text:
What causes energy peaks during a molecular dynamics energy minimization?
I'm minimizing the energy of a protein with NAMD using the fixedatom feature. I put my protein in a water box, and I allow all atoms other than the protein backbone to be unconstrained. After running the minimization, I used VMD's NAMD Plot for NAMD's log file and the TOTAL3 total energy option, which resulted in some energy peaks that weren't present until this step:

Are those peaks normal to have? If not, what causes them?
Here is the NAMD log file. The NAMD configuration file used for the minimization is:
#############################################################
## JOB DESCRIPTION                                         ##
#############################################################

# Minimization and Equilibration of
# Human Pi GST in a Water Box
# N, V, E simulation

#############################################################
## ADJUSTABLE PARAMETERS                                   ##
#############################################################

structure          ionized.psf
coordinates        ionized_h_free_min.pdb

set temperature    310

firsttimestep      0

# output
set output              Cry1_side_free
outputname              $output
dcdfile                 ${output}.dcd
xstFile                 ${output}.xst
dcdfreq                 50
xstFreq                 50

binaryoutput            no
binaryrestart           no
outputEnergies          100
restartfreq             1000

#############################################################
## SIMULATION PARAMETERS                                   ##
#############################################################

# Input
paraTypeCharmm      on
parameters          /usr/local/lib/vmd/plugins/noarch/tcl/readcharmmpar1.3/par_all27_prot_lipid_na.inp
temperature         $temperature

# Force-Field Parameters
exclude             scaled1-4
1-4scaling          1.0
cutoff              12.
switching           on
switchdist          10.
pairlistdist        14.0

# Integrator Parameters
timestep            1.0  ;# 2fs/step
#rigidBonds          all  ;# needed for 2fs steps
nonbondedFreq       1
fullElectFrequency  2
stepspercycle       10

# Constant Temperature Control
langevin            on    ;# do langevin dynamics
langevinDamping     5     ;# damping coefficient (gamma) of 5/ps
langevinTemp        $temperature
langevinHydrogen    off    ;# don't couple langevin bath to hydrogens

# Periodic Boundary Conditions
cellBasisVector1     -75.1629982   0.0   0.0
cellBasisVector2     0.0    -78.31999779 0.0
cellBasisVector3     0.0    0.0  -98.59599972
cellOrigin          -12.303732872009277 19.206361770629883 -36.690704345703125

wrapAll             on

# PME (for full-system periodic electrostatics)
PME                 yes
PMEGridSizeX        80
PMEGridSizeY        80
PMEGridSizeZ        100
dcdUnitCell     yes
# mobile atom selection:
# protein
fixedAtoms              on
fixedAtomsFile          fixedatoms_backbone.pdb
fixedAtomsCol           B

#############################################################
## EXECUTION SCRIPT                                        ##
#############################################################

# Minimization
minimize            10000
reinitvels          $temperature

The downvote(not by me)  was given because Chemistry SE does not generally support software queries. However, a new materials (and molecular) modeling stack exchange just launched in a private beta to address questions like this. For example see https://materials.meta.stackexchange.com/questions/38/are-support-like-queries-for-software-packages-on-topic

At the top of your config file, you imply that you want to perform an NVE calculation, but you’re actually doing NVT, because you have the Langevin thermostat turned on. Langevin works by giving the system a random “kick”, which is probably spiking the energy.

Thank you for your answer. The configuration file was from one of the NAMD tutorials that they supplied on the website. I changed that file but I didn't remove commentations. That is why unrelated comments are in that file.

@OnurOzcan That doesn’t change the fact that using a thermostat will occasionally cause the energy to jump due to the bath coupling.


text:
How to balance redox reactions possessing oxidation number change in multiple elements?

Hello, I am bit confused about balancing these type of redox reactions where multiple elements are involved in oxidation and reduction reactions.
If I try to use half reaction ion exchange method and write reduction and oxidation half separately and then try to neutralise the electrons then I am not able to go to RIGHT solution of this particular reaction.
Is there any simple way?

Although, i got right answer but not by typical process,  i got it by combining ion electron method by elemental level balancing with hit and trial approach...  I want to know, if there is any predefined way to do so?

If what you show in the picture is supposed to be the final balanced equation, it isn't correct. If that's not the final answer, perhaps you should show what you have as the correctly balanced equation.

That isn't a final balanced equation, it is my effort which is wrong.

Here is an easy way to balance this.
Balance in order: (1) metals; (2)ions; (3) non metals; (4) oxygen
FeS2 + O2 ==> Fe2O3 + SO2
Balance metals (Fe)
2FeS2 + O2 ==> Fe2O3 + SO2
There are no ions, so balance non metals next (S)
2FeS2 + O2 ==> Fe2O3 + 4SO2
Finally balance O
2FeS2 + 5.5 O2 ==> Fe2O3 + 4SO2 (and multiply through by 2 to get)
4FeS2 + 11O2 ==> 2Fe2O3 + 8SO2

Write the components in their dissociated (ionic) forms, such as Fe2+ + S22-
Then write sub reactions, or more properly actual redox reactions. Here 1 for Fe(II) oxidation to Fe(III) by oxygen, and disulfide ion oxidation to sulfur dioxide by again oxygen.
Solve these two reactions separately then combine reactants and products in the last step.
That is the simple and safe way.

Tried,  but didn't got it balanced & right :(

"Aquatic" is the wrong word here. Other than that, the method should work.

It says "disulfide ion oxidation to sulfate", but I don't see any sulfate anion.  How can this method work?

Take the sulphur dioxide instead of sulfate. I just had a mistake there. And I also inclined to take this reaction in solution, I am working on them for a long time, but it does not have to be so. Rather than aquatic, dissociated forms are more appropriate I guess.


text:
How to develop a Reactive Force Field (ReaxFF)? (Molecular simulation)
Currently, I am working on CO2 capture on Hydrotalcites as solid adsorbents. I want to run molecular simulations to analyse the behaviour of the material under different temperatures, pressures and CO2 concentrations.
In order to do so, I am working in ADF, with the ReaxFF method. Nevertheless, there are no force fields generated specially for this kind of material. That is why I am building one by comparing data from existing force fields and selecting the data from the most similar to my project. However, I am not sure this approach is the best. I have read the ReaxFF manual (https://www.scm.com/wp-content/uploads/ReaxFF_training.pdf) but could not make a lot of sense of it. The ReaxFF manual of van Duin (https://www.scm.com/wp-content/uploads/ReaxFF-users-manual-2002.pdf) explains better but fails to say how to manipulate the data.
Does anyone know how to develop a Force Field input file for ReaxFF?

What level are you at, this to my eyes looks like either a Post Doc, PhD student, MSc student or Undergrad asking for help on how to do their project. The more junior you are the more likely someone will feel the urge to help you

If you're using ADF, they have very comprehensive support. I'd contact them first - it's a large part of the license cost for that software.

While I don't have any direct experience with ReaxFF, my understanding is that the van Duin group are one of the few groups that can reliably develop new parameters. You may also want to contact them if ADF can't help.

Check out SCM website: they have regular workshops on this topic all around the world. If you cannot find one in your area and cannot travel for one, they have a list of previous ones with some of the slides, exercises uploaded to the site. Also, just as @GeoffHutchison recommended, contact them directly, they can help you. Mind, developing a new forcefield is a long and bumpy road.

I dislike that this is being marked as too broad. A good answer would have a (potentially simplistic) worked example. I heard through the grapevine that they do this in order to force a collaboration with them that results in paper co-authorship. That is "closed source" science, and it would be great to have an answer here as an open resource.

I am in my first PhD year.

I am in my first PhD year.  The links I shared are found on their support website. I have also read some of the articles of existing FF in order to compare methodologies, but they do not give much away. I have contacted the ADF mailing list, but they have not answered yet. Thank you for your suggestions.


text:
Does the concentration affect the voltage in electrochemistry?
For my electrochemistry lab I'm using $\pu{10 mL}$ of $\ce{ZnSO4}$ and $\pu{10 mL}$ of $\ce{CuSO4}$ solutions connected with a salt bridge. I connect $\ce{Zn}$ metal and $\ce{Cu}$ metal to a voltmeter and dip those metals in the correct beakers. The voltage I get is $\pu{1.044 V}.$
However when I add $\pu{10 mL}$ of water to both beakers, I still receive the same result. Does diluting two solutions affect the voltage?

Do you have reason not to trust your experimental results?

What result were you expecting? What does the Nernst equation say?

Cell potential definitely depends on the concentration of the two solutions. This dependence of the cell potential on the concentration can be described by the Nernst equation.
For the reaction you are suggesting the half cells are:
$$
\begin{align}
\ce{Cu^2+ + 2 e- &-> Cu} &\quad E_1^\circ &= \pu{0.34 V}\tag{R1}\\
\ce{Zn^2+ + 2 e- &-> Zn} &\quad E_2^\circ &= \pu{-0.76 V}\tag{R2}\\
\end{align}
$$
Since zinc is more reactive than copper, or since $\ce{Zn^2+}$ reduction is less thermodynamically favored than the reduction of $\ce{Cu^2+}$, then the half-equation of the reduction of $\ce{Zn^2+}$ must be reversed and so does the sign of its potential.
Combining the half equations gives:
$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}\tag{R3}$$
$E_\mathrm{cell}^\circ = \pu{1.10 V}$ at $\pu{25 °C}$ and $\pu{1 atm}$ and $\pu{1 M}$ of solutions.
From the Nernst equation:
$$E = E_\mathrm{cell}^\circ  - \frac{0.0591}{n}\ln Q\tag{1}$$
where $n$ is number of electrons and $Q$ is reaction quotient which is the concentrations of product solutions raised to the power of their coefficients over concentrations of reactant solutions raised to the power of their coefficients:
$$Q = \frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\tag{2}$$
If you halve both the concentrations of $\ce{Zn^2+}$ and $\ce{Cu^2+}$, then $Q$ will remain the same and so will the the cell potential.


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Hybridization molecule
Why is SO32- sp3 hybridized even though it forms a double bond with the oxygen (in resonance)?
I thought double bonds needed a p orbital so I would think it would be sp2d so it can still have an unhybridized p orbital for the double bond. Thanks!

This is a point of continuing drama.  Theoretical calculations as long ago as the 60s (cf. J. Am. Chem. Soc., 1994, 116 (10), pp 4414–4426) strongly suggest there is no d orbital participation in hybridized orbitals for Period 3 "hypervalent" compounds like the sulfite anion.  In fact, these calculations appear to suggest the observed bond shortening is probably explained quite simply by a significant ionic component to the bond, which means the best Lewis structures may well be the simple octet-rule following one, with 4 sp3 bonds around the central S, occupied by 3 bonding pairs (giving 3 sigma bonds) and 1 lone pair. No pi bond at all. This gives you higher formal charges than the resonance structure with a single double bond -- but that's what's meant by saying the calculations suggest the best VB description simply has more ionic content to the bonding between the S and O atoms (represented by the higher formal charges), rather than any d mixing.

Not 4 "sp3 bonds"... Hybridisation should not be applied to bonds and only to orbitals. Be careful of your terminology.

Would be good if u could provide additional insights from such papers, especially those that do calculations on sulfite in particular. Perhaps, show charge distribution for the structure.


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What effect does an increase in concentration of copper sulphate solution have on the rate at which copper is deposited at the cathode?
I have been studying electrolysis in school and was wondering whether a higher amount of copper sulfate solution would produce a greater mass of copper in an electrolytic cell with two copper electrodes in a given amount of time or not.

No, the amount of copper deposited is dependent on the current passed (unless you actually run out of copper in solution)


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H-NMR Equivalent Protons and Coupling of Glycerol
I've been searching for explanations to understand why the labeled protons on carbons 1 and 3 are not equivalent in the below figure.

Another figure that shows the same results:

SDBS confirms the same, although the assignments of the signals show that all 3 hydroxyl Hs are equivalent, which I don't understand.

I would think that all 4 hydrogens on C1 and C2 would be in identical environments but it seems this is incorrect. In fact, the figures show that the two hydrogens on C1 (and C2) are not equivalent to each other. Why is it that the hydrogens on the terminal carbons are not equivalent, but rather each hydrogen on one terminal carbon has an equivalent hydrogen partner on the other?
Also, what is the explanation for why SDBS shows all 3 hydroxyl protons as equivalent? Why would the -OH on C2 be equivalent to the ones on the terminal carbons?
Edit: As for my first question, I think I found the correct explanation. I believe it's because the hydrogens pairs are either diastereotopic or enantiotopic.

They're diastereotopic as you suspect!

The couplings shown at the top are wrong. $\ce{H_c}$ to $\ce{H_a}$ is not $\pu{12 Hz},$ as it's way too big and is in fact the geminal coupling between $\ce{H_a}$ and $\ce{H_b}.$ The actual couplings are $11.7,$ $6.3$ and $\pu{4.2 Hz}$ for $\ce{H_a}/\ce{H_b},$ and $\ce{H_c}$ to $\ce{H_a}$ or $\ce{H_b},$ respectively. The hydrogens on the $\ce{OH}$ will not show up in exchanging solvents like $\ce{D2O}/\ce{H2O},$ but can in $\ce{CDCl3}.$

See a newer answer to this question: https://chemistry.stackexchange.com/questions/144088/diastereotopic-methylene-protons-in-achiral-molecules/144096#144096

This might be confusing since both ends of the molecule are the same. The difference is that the alcohol points to only one side of the molecule.

So, while the hydrogen atoms on carbons 1 and 3 on paper have the same environment, they are not chemically equivalent when looking at the whole molecule.
Concerning your second question: The chemical shift of protons on heteroatoms such as Oxygen, Sulfur, or Nitrogen is highly dependent on the recording conditions (concentration, temperature, solvent) of the NMR. Frequently, the signals are broadened and show no coupling (broad singlett). The reason for this is the rapid intermolecular proton exchange. If there is a rapid exchange between two chemically non-equivalent groups within a molecule (like in this one), only a single, averaged signal is observed.


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Combustion/dispersment of mercury (Hg)
recently tuning a motorcycle using a manometer that contained liquid mercury.  Small amount probably a little more than a thermometer but not much more. Unfortunately the meter was upended and the mercury was sucked through the 4 vacumn ports on the motorcycle cylinders and cycled through the running engines combustion chambers and exhausted out of the exhaust system assumably in vapor form? The mercury did not go through the carburetor but was sucked into the cylinders through vacuum ports just after them but just before the intake valve. So it was directly sucked into the 4 cylinders.   Only a minute trace of mercury was found in one or two of the rubber hoses on the tool afterward. My concern short term was the obvious vapors from the accident but the garage area was mostly open with front and rear overheads open to the outdoors being a garage of course.
This brings me to the questions. After much research I found that Hg burns at 350f I believe but I can’t find any real information that pertains to a situation as this. Would the mercury burn completely into a vapor in the combustion chamber of an engine or would the extremely dense nature and slow evaporation rate effect the time frame that it takes as mixed with a fuel in this case to completely burn?  Do you think the mercury was completely burned and vaporized in the time it was in the combustion chambers of the motorcycle or did millions of tiny uncombusted liquid mercury droplets get spewed all over the rear area of the garage and are simply sitting around evaporating and presenting a dangerous situation?  Any insight on this would be extremely helpful. I have searched for signs of any physical mercury but have found none. I am concerned that if the liquid mercury wasn’t vaporized and exited as tiny liquid mercury droplets into the garage that it may be still there of course.  I spend many hours in this garage woodworking and rebuilding motorcycles etc with the doors closed. In the summer of course I open the doors but in the evenings and most of the week the garage is closed up. In the summer like now it gets very hot inside which I’m sure helps evaporate the mercury if there is any leftover but I am still concerned about how much may or may not have been burned. Thanks in advance, J

Mercury doesn't really "burn" as fuel would. The wording is confusing - burning (as in a reaction) or evaporation (just physical change)?

What can we say? Some is probably still in the motorcycle and some was probably expelled in the exhaust. You'd need very sophisticated equipment to detect the traces now.

“Burning” was in regard to the fact that the engine was running and the combustion process was actively happening.  Was just interested in knowing if there was a chance that it would combust and be turned to a vapor that would be evacuated by the normal process of the room being open to the outdoors etc. or if it just passed through in liquid form unchanged by the combustion process and then mostly just scattered around by being expelled through the exhaust.

I understand the principle that some may be both in the engine and some expelled. I was just looking for an educated answer about the combustion properties (if any) and the “odds” of a percentage being combusted into vapor. Thanks for the info.

Knowing it’s effect on aluminum I am hopeful that very little if any remains, seeing as how the entire engine block and many other parts are solid aluminum. Not a great mix unfortunately. The engine has been fun a few times and ridden roughly 100 miles so it appears to be unaffected as yet.

Mercury won't "combust" but under the conditions you describe it will vaporise. Whether this vapour is a problem inside your garage depends on several things. If it doesn't condense (and you might not notice if it does as the droplets will be very small) the vapour level will depend on the airflow through your garage. So leave the door open a day or so and it will all be gone. But if there are traces of liquid (even unseen) vapour will accumulate any time the air circulation isn't good. It might be worth a professional check of the vapour level.

Can you identify the manufacturer or other info on the manometer? I tried to find a mercury manometer on the internet to estimate how much mercury went into the engine, but could not find any. I can visualize an amount from about 2 to 10 mL, or 25 to 125 grams. Maybe I'm way off.

Does your motorcycle have a muffler or baffle?

The unit was made by Race Pro. I have requested info but haven’t received any yet. They sell a refill it also doesn’t specify a measurement.

Yes the motorcycle has a muffler system on it which includes the factory installed baffle system. Which I assume has collected a bit of the mercury as well. I have ridden the bike several times and have probably over 100 miles on it since the event so I’m sure the heat and flow of the hot exhaust gases has at least minimized if not evacuated the remaining bit in there?

Here you go:
https://link.springer.com/article/10.1007%2FBF00342256
Their estimate is about 50% of the Hg emerges as elemental, the rest as various Hg(II) compounds.  That's for pretty high temperature flue gases, and it's unclear to me that your exhaust stream contains the assorted trace gases in a coal plant's exhaust with which the Hg reacts.  Nevertheless, I'd say 50% is a reasonable WAG for how much of your Hg exited as elemental.


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What is the structure of 92 m/z fragment for aspirin
Aspirin has the following structure:

What would the fragment structure be for m/z = 92?
Even though $\ce{C6H4O}$ is a molecular structure with mass 92, I don't think this will be formed since two bonds needs to be broken. Also the 92 peak is a major peak and the probability of two bonds breaking to lead to major peak is unlikely. The mass spectrum is as follows

C6H4O has appropriate mass not C4 BTW there's no problem at all with losing two groups, there are further losses too, and peak 43 is even higher.

@Mithoron I have edited the question to change the C4H4O to $\ce{C6H4O}$

This spectrum is recorded using electron impact. That means that the molecule is positively charged by losing one electron.
Since that is a very unstable structure, the molecule starts fragmenting losing neutral fragments.
More processes occur, but these are the ones that we see in the spectrum.
The peaks are the positively charged fragments.
A fragment with m/z of 92, 93 or 94 is typical for phenolic compounds.
The aromatic $\ce{C-O}$ bond is very strong and therefore first all the rest is lost.
In this case the fragmentations are:

Loss of the acetyl group: $180-42=138$
Loss of water from the acid group: $138-18=120$
Loss of $\ce{CO}$ from the acid group: $120-28=92$
Loss of $\ce{CO}$ from the phenolic group: $92-28=64$

These are all typical fragmentations.
The height of the peaks indicate the relative stability of the positively charged fragments.

Just a comment on the first paragraph: it is generally not true to state that "Since that is a very unstable structure". In general, radical cations are minimums on the potential energy surface and some energy must be brought in to allow fragmentation. The point is that electron ionization deposits an extremely variable amount of energy in the system. The sequential fragmentation in the case of aspirin is in itself a nice illustration of this spread in internal energy: some ions are formed with enough energy to break 4 bonds, whereas others remain unfragmented as the molecular ion m/z 180.

Hi there! I've been looking at this answer and you stated a fragment with m/z of 92, 93 or 94 is typical of phenolic compounds alongside loss of CO from the phenolic group: 92-28 = 64 but I can't find any source that confirms such an observation. Where can I find a source that backs up this observation?


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How much heat is created by absorbent-type sunscreens?
I have a friend who claims her dermatologist told her than sunscreen was what was burning her skin. (I think it's much more likely that her rash/burn was either a sunburn, a skin reaction to some other product, or a "sun poisoning" type rash from using a sunscreen that doesn't block UVA -- my wife gets that.)
Said doctor told her it's because sunscreen converts the light energy to heat, and that heat was burning her skin. I know this is technically the way those organic sunscreen ingredients work, but can anyone quantify how much heat would actually be produced by this process?
Some web sites claim that some users feel hot when they're using absorbent-type sunscreens -- and again I find that hard to believe. Make me want to slather a few hot dogs with different sunscreens (and a control with none) and leave them in the sun and stick them with my meat thermometer to see if there's any discernable difference.

Bwahahaha :D I suggest not to go to this doc unless to laugh him out.

Well, the sunscreen is obviously not creating any additional heat.  All the light that falls on skin is either reflected/scattered immediately or absorbed and turned to heat.  Unless the sunscreen reduces the albedo of the skin -- that is, unless it reduces the amount of light that gets reflected or scattered -- then it can't possibly increase the amount of heat energy that is generated by the absorption of light.
What it does do is cause the absorption of UV light to occur right there in the uppermost layers of the skin, where the sunscreen is absorbed, instead of further down in the living dermis, where it can do damage.  So I suppose the amount of heating in the upper (dead) layers of the epidermis goes up, with a compensating decrease in the amount of heating in the lower (live) layers.
To roughly quantify the heat, we could take the solar irradiance at sea level, which obviously varies by latitude, of about 1 kW/m^2, and then multiply by the reflectance of human skin, which varies considerably by individual and race, but let's say about 50%.  That means about 500 W/m^2 = 50 mW/cm^2 of light energy is being turned into heat for your average white person on a clear day.   If 1/4 of one's total skin area (about 2 m^2) is exposed, that's about 250W of energy you need to get rid of.  Ordinary resting metabolism is about 80 W, so you're asking the body to get rid of 3x more heat than it normally does.  Hence you start to sweat, become attracted to the idea of going into the ocean, et cetera.  If we want to quantify the sweating necessary, we divide by the latent heat of vaporization for water (2260 kJ/kg) and find about 100 mg/s of sweat need to be produced and evaporate to get rid of 250W.


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Bredt's rule applied to fused bicyclic systems
Ultimate question: Is Bredt's rule equally valid for fused bicyclic system (no bridgehead alkenes without a ring 8 carbons or more), but for other reasons (historical?) it is only defined as applying to bridged systems? Or am I missing something else fundamental about bridged vs fused bicyclic systems?
--Background--
Everywhere I can look (online, reference texts), Bredt's rule is specifically in reference to bridged bicyclic systems. However, shouldn't the rationale behind Bredt's rule -- that by having a bicyclic alkene with the double bond at the bridgehead means it has to have a trans alkene and therefore needs 8 carbons or more -- also apply to a fused bicyclic alkene as well?
The only example my book gives students is a bicyclo[4.4.0]dec-1-ene. It says "well this doesn't need Bredt's rule, because it's not a bridged system, but it's still stable because the largest ring has 10 carbons to accommodate the trans bond". That answer is unsatisfying to me, because the basis for them recognizing that the molecule has a trans alkene in it, comes from the discussion about Bredt's rule in the first place.

It never gives an example of something like bicyclo[3.1.0]hex-1-ene (above)...but I assume based upon a trans-alkene with less than 8 carbons this should also not be stable? Would this just be an application of the idea that trans-alkenes need to be in rings 8 or larger to be stable (without being part of Bredt's rule specifically)?

Here's a way to look at it - bicyclo[3.1.0]hex-1-ene is a bridged compound, the bridge just has "0" units as the name indicates.

@ron That's fair, and sort of how I planned on presenting this particular topic (if I couldn't find a more complete answer). However, it still leaves me wanting, because Bredt's rule is presented as specifically for bridged (as in non-zero carbon) bicyclic. So...I think my original question still remains :)


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Carbon chains isomers, why is 1-methyl-hexane not a thing?
As title reads, why shouldn't 1-methyl-hexane work but 2-methyl-hexane and 3-methyl-hexane does?

You need to always number your longest chain first. If you draw 1-methylhexane, you'll see the longest chain isn't actually 6 carbons anymore, but 7. Then all the atoms are part of the longest chain, and the name simply becomes heptane.


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origin of use of differential equations for modeling chemical reactions?
what are some of the original examples of uses of differential equations for modeling and analyzing chemical reactions, particularly those relevant to biochemistry, involving proteins and enzymes? Michaelis and Menten's work in 1910s is one example.
What are the earlier examples or antecedents to this work? (references to good historical accounts of this modeling endeavor would also be helpful).

This might be better suited for History of Science/Mathematics

The inversion of sucrose by strong acids by Wilhelmy in 1850 is considered to be the first  kinetic study . Daniel Bernoulli studied the spread if disease (reported in 1766) using differential equations effectively the same as the rate equations we use today.


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Heterocyclic Compounds - Glucose
Why cyclic glucose is not a heterocyclic compounds if it consist Oxygen as a Heteroatom ?

It is. However sugar chemistry is most often treated as a niche of its own.

Waylander is right, many sugars are heterocyclic compounds. While sugars such as glucose can exist in an open chain form, many of them are normally found in the form of six or five membered rings which consist of carbons and an oxygen.
Thus glucose fits the very definition of a heterocylic compound. It is important to keep in mind that any cylic (ring) compound where more than one element type is found in the atoms which make up the ring is a heterocylic compound.
For example S4N4, pyridine, ethylene oxide, glucose, 9-BBN and tetrahydrofuran are all heterocycles. They have very different chemistry to each other but they are all heterocylic.


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How does chloroform react with cyclopentadiene in base?

Is my reaction mechanism feasible?

Isn't it abnormal Riemann tiemann reaction??

The worry I had is that the normal way to make dichlorocarbene from chloroform is to treat it with a strong base, this will convert the cyclopentadiene into the cyclopentadienyl anion. This anion is an aromatic species which is unlikely to be able to react with the carbene.
But in one paper (Grignon-Dubois, M.; Dunogues, J.; Ahra, M. - Recueil des Travaux Chimiques des Pays-Bas, 1988, vol. 107, # 3, p. 216 - 225) it has been reported that chloroform, base and cyclopentadiene will react together to form the first intermidate.
I think that the reaction step showing the formation of the cyclopropane ring is perfectly reasonable. We need to consider how we would form the carbene. I know that it is possible to form carbene by a photochemical reaction on ketene which forms the carbene and carbon monoxide. I know from personal experience that chloroketene reacts in a [2+2] reaction with cyclopentadiene to form a cyclobutanone. So that route might not be a good idea.

The abstract of this paper http://pubs.rsc.org/en/content/articlelanding/1971/c2/c29710000461#!divAbstract  records that under very similar conditions the product of cyclopentadiene and dichlorocarbene is 6-Cl-fulvene. It is behind a paywall so I cannot read the full paper.

@NuclearChemist In your first paragraph, last line you state that, "* This anion is an aromatic species which is unlikely to be able to react with the carbene.*" Why?, In reimer teimann reaction it is completely fine for the aromaticity to break momentarily?


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Raoult's law & Dalton's law of partial pressures
I am confused between Raoult's law and Dalton's law of partial pressure as they state completely different things about same thing...
Raoult's law claims $p = P \cdot x$
where $p$ is the partial pressure of a gas, $P$ is the pressure of the same gas at the same temperature and volume and $x$ is the mole fraction.
Dalton's law says $p = p_\text{total} \cdot x$
where $p_\text{total}$ is the total pressure of the mixture.

I think you are forgetting that Rault's law is used for ideal Liquid mixtures while Dalton's law is for non-reacting ideal gases.
They look similar due to basic assumption of ideal interactions.
In Dalton's law partial pressures add up (PT=PA+PB), but in Rault's law since liquid has to evaporate from a surface which is approximately shared based on mole fraction their vapour pressure do not simply add up and is averaged based on mole fraction PT=xAPA+xBPB.

You are using the symbol x to describe two different things.  In Raoult's law, $p = Px$, where x is the mole fraction of the substance in the liquid phase.  In Dalton's law, $p =P_{tot}y$, where y is the mole fraction of the same substance in the vapor phase.  So, combining these, you have $P_{tot}y=Px$, which is really Raoult's law.

In Raoult's Law, $p$ is the partial pressure of the vapor in equilibrium with some liquid which has vapor pressure of $P$. $x$ is the molar fraction of this liquid (it can be mixed with other liquid or solid.

Raoult's law states that the partial vapor pressure of a substance is equal to the total vapor pressure of the solvent multiplied by the mole fraction of the substance. This also takes into account things like ion dissociation in a solution [Van't Hoff factor].
Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure of these gases is equal to the sum of the partial pressures of the individual gases (where the partial pressure of a given gas is calculated by multiplying the mole fraction of that gas by the total pressure).
Therefore, Raoult’s Law deals with vapor pressures while Dalton’s Law deals with non-reacting gases. However, the sum of pressures given by Raoult’s Law will results in the total amount of pressure of the [vapor] solution (which follows Dalton’s Law). So Dalton’s Law still holds in that the sum of vapor pressures is the total vapor pressure.


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Calculating PH of solution after adding strong acid (equilibrium)
Consider the classic example when talking about acids and bases:
Say you have 1M of $\ce{HCL}$ and you throw it into water, calculate the resulting pH (= concentration of $\ce{H3O+}$).
The way this problem is done: just take $-log(1) = 0 \rightarrow $ PH of resulting solution is $0$.
I don't understand why this is a valid approach. If you throw a concentration of a strong acid in water, wouldn't $\ce{H2O}$ first establish a new equilibrium with (the new total of) $\ce{H3O+}$ and $\ce{HO-}$ according to the autoionization reaction of water?
$$Kw=[\ce{H3O+}][\ce{HO−}]$$
It seems to me like we're mixing up the reaction quotiënt and the equilibrium constant. If this is right, why can we do this?
I know that the autoionization doesn't happen very much but we still use the expression for $Kw$ (wich stays constant) to calculate the concentration of $\ce{HO-}$.
I would think the way to solve this problem is to wait till the new equilibrium has been reached (new concentration of $\ce{H3O+}$) and then we can calculate the pH of the solution.
If the reaction quotiënt and equilibrium constant are negligibly close to eachother, could someone show the proces of throwing the acid in the water in the style of the second answer to the following question please?
How can the equilibrium shift, while Kc remains constant?

In order for you to completely understand this, you first have to write the mass balance, charge balance and equilibrium constants. This manner of solving $pH$ always works.
Mass balance:
$$[HCl]_0=[HCl]+[Cl^-]=[Cl^-]$$ Since $HCl$ is a strong acid, there are no free molecules of $HCl$ after equilibrium has been reached, just chloride ions and hydronium ions. Here, the index $0$ means before equilibrium has been reached.
Charge balance:
$$[H_3O^+]=[OH^-]+[Cl^-]$$
Equilibrium constants:
$$K_w=[H_3O^+][OH^-]$$
$$K_a=[H_3O^+][Cl^-]\rightarrow\infty$$
I.e., as $HCl$ is a strong acid, it's acid constant tends to infinity for our purposes.
Solving this system of equations we get:
$$1=[Cl^-]$$
$$[H_3O^+]=\frac{10^{-14}}{[H_3O^+]}+1$$
$$[H_3O^+]^2-[H_3O^+]-10^{-14}=0$$
You might want to solve this second degree equation, find the concentration of hydronium ions and find the exact $pH$. But you will notice that it is very close to $0$. That is because the term $\frac{10^{-14}}{[H_3O^+]}$ in the last equation, which means the concentration of hydroxyl ions on equilibrium is very small compared to the others. So you can approximate it to $0$ when comparing it to the others and get:
$$[H_3O^+]=1$$

How is [H3O+]=[OH−]+[Cl−] charge balanced? Isn't there 1+ on the left and 2- on the right? Just asking!

The charge neutrality in a solution is a constraint. The sum of positive charges must be equal to the sum of negative charges, since it was obtained from mixing neutral species.


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What is the difference between "adding a liquid" and "diluting to a liquid"?

Place $\pu{20.0 g}$ $\ce{NaOH(s)}$ in a flask and dilute to $\pu{1.00 L}$ with water.
Place $\pu{20.0 g}$ $\ce{NaOH(s)}$ in a flask and add $\pu{1.00 L}$ of water.

How exactly do these two statements differ? Wouldn't adding $\pu{1.00 L}$ of water to $\pu{20.0 g}$ of $\ce{NaOH}$ dilute it anyways?

There is no difference between adding and diluting. There is, however, a difference between "adding 1 L" and "adding 0.5 L".

The gist here is that adding a volume $x$ of solid and a volume $y$ of liquid does not result in a solution of volume $y$ nor $x+y$. The volume of the solution is unpredictable.

Place $\pu{20.0 g}$ $\ce{NaOH (s)}$ in a flask and dilute to $\pu{1.00 L}$ with water.

Chemists do this frequently. Typically by using a volumetric flask to be the container. Since the molecular mass of $\ce{NaOH}$ is $\pu{40.00 g/mol}$, this will yield a $0.500$ molar solution, $c(\ce{NaOH}) = \pu{0.500 mol/L}$.

Place $\pu{20.0 g}$ $\ce{NaOH (s)}$ in a flask and add $\pu{1.00 L}$ of water.

In this case the volume of the solution is unpredictable with only the information given. However the volume will be a little more than $1.00$ liter.
In chemistry there are also molal solutions. The units are moles per kilogram of solvent. In this case there are $\pu{0.02 kg}$ of $\ce{NaOH}$ ($\pu{0.500 mol}$) per $1.00$ kilogram solution, so the molality of the $\ce{NaOH}$ is $b(\ce{NaOH}) = \pu{0.500 mol/kg}$.

"Dilute to" means add solvent until the designated volume is reached. That is, the final volume is known and definite.
"Add x liters of water" means to add the designated volume of water with no pre-defined final volume.  The final volume will be whatever it happens to be when the solute and solvent are mixed.

If you are diluting to 1.00 L with water, you are adding water until the solution equals 1.00 L. You will be adding less than 1.00 L of water to bring the volume up to 1.00 L. Therefore the final volume will be 1.00 L of solution.
However, if you are adding 1.00 L of water then the final volume will be more than 1.00 L. That final volume will depend on the density of the solution that results, and how much material that was already present.


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Slaked lime: Why does calcium hydroxide have to be used in conjunction with sodium hydroxide? Why can't it be used alone?
To absorb carbon dioxide. I haven't been able to find an answer to the above anywhere.

The overall reaction for a solid soda lime scrubber according to the Wikipedia page here 1 is:
$$\ce{CO2 + Ca(OH)2 -> CaCO3 + H2O + {heat (in the presence of water)}}$$
Each mole of $\ce{CO2}$ (44 g) reacting with calcium hydroxide produces one mole of water (18 g).
The reaction can be considered as a strong-base-catalysed, water-facilitated reaction.
The reaction mechanism of carbon dioxide with soda lime can be decomposed in three elementary steps:

$\ce{CO2 (g) -> CO2 (aq)}$
($\ce{CO2}$ dissolves in water - slow and rate-determining)
$\ce{CO2 (aq) + NaOH -> NaHCO3}$
(bicarbonate formation at high pH)
$\ce{NaHCO3 + Ca(OH)2 -> CaCO3 + H2O + NaOH}$
(($\ce{NaOH}$ recycled to step 2) - hence a catalyst)

This sequence of reactions explains the catalytic role played by sodium hydroxide in the system and why soda lime is faster in chemical reactivity than calcium hydroxide alone. It reacts much more quickly and so contributes to a faster elimination of the $\ce{CO2}$ from the rebreathing circuit. The formation of water and the moisture from the respiration also act as a solvent for the reaction. Reactions in aqueous phase are generally faster than between a dry gas and a dry solid.

But if the first step, the dissolving of CO2 is the rate determining step then the efficiency of steps 2 and 3 shouldn't make any noticeable difference?

I'm assuming that you are using a solution to trap the $\ce{CO2}$. It would seem that you're probably filtering out the $\ce{CaCO3}$.
The gist is that $\ce{CaOH2}$ has a low solubility in water (1.73 g/L at 20 °C). So adding $\ce{NaOH}$ increases the efficiency of trapping the $\ce{CO2}$ in bubbles of air as $\ce{CO3^{2-}}$ in the solution.

Bad assumption. I think the OP is asking about solid soda lime scrubbers