text:
Symbolic form of equilibrium constant for ammonium carbamate dissociation

$$\ce{NH2COONH4(s) <=> 2 NH3(g) + CO2(g)} \qquad K = \left(\frac{p(\ce{NH3})}{p^\circ}\right)^2\times\frac{p(\ce{CO2})}{p^\circ}$$
Assuming that you start this experiment with pure solid $\ce{NH2COONH4}$ in an evacuated container and allow it to dissociate until equilibrium is reached, show that the equilibrium constant can be rewritten as
$$K = \frac{4}{27}\left(\frac{p_\mathrm{tot}}{p^\circ}\right)^3.$$

Let $n_0$ be the initial amount of ammonium carbamate and $\alpha$ be the factor describing the extent of dissociation:
$$
\begin{array}{cccccccc}
\ce{&NH2COONH4(s) &<=> &2 NH3(g) &+ &CO2(g)} &\quad &\text{Total}\\
& n_0 && 0 && 0 && n_0\\
& (1 - \alpha)n_0 && 2\alpha n_0 && \alpha n_0 && (1 + 2\alpha)n_0\\
\end{array}
$$
We can express mole fraction $x_i$ and partial pressures $p_i$ for $i$th component as follows:
$$x(\ce{NH2COONH4}) = \frac{n(\ce{NH2COONH4})}{n_\mathrm{tot}} = \frac{(1 - \alpha)n_0}{(1 + 2\alpha)n_0} = \frac{1 - \alpha}{1 + 2\alpha}\tag{1}$$
$$x(\ce{NH3}) =  \frac{n(\ce{NH3})}{n_\mathrm{tot}} = \frac{2\alpha}{1 + 2\alpha}\tag{2}$$
$$x(\ce{NH3}) =  \frac{n(\ce{CO2})}{n_\mathrm{tot}} = \frac{\alpha}{1 + 2\alpha}\tag{3}$$
$$p(\ce{NH3}) = x(\ce{NH3})\cdot p_\mathrm{tot} = \frac{2\alpha p_\mathrm{tot}}{1 + 2\alpha}\tag{4}$$
$$p(\ce{CO2}) = x(\ce{CO2})\cdot p_\mathrm{tot} = \frac{\alpha p_\mathrm{tot}}{1 + 2\alpha}\tag{5}$$
Finally, plugging in the partial pressures into the given expression for the equilibrium constant:
$$
\begin{align}
K &= \left(\frac{p(\ce{NH3})}{p^\circ}\right)^2 \times \frac{p(\ce{CO2})}{p^\circ} \\
  &= \left(\frac{2\alpha p_\mathrm{tot}}{(1 + 2\alpha)p^\circ}\right)^2 \times \frac{\alpha p_\mathrm{tot}}{(1 + 2\alpha)p^\circ} \\
  &= \frac{4\alpha^3p_\mathrm{tot}^3}{(1 + 2\alpha)^3(p^\circ)^3} \tag{6}
\end{align}
$$
This leads me to the right answer if I make an assumption that all of the ammonium carbamate dissociates. The question doesn’t say I could do this, hence why I’m not sure how to proceed. Any suggestions are greatly appreciated for deriving this.

Twice as many moles of $\ce{NH3}$ are formed as $\ce{CO2},$ so the partial pressure of $\ce{NH3}$ is $\frac{2}{3}p_\mathrm{tot}$ and the partial pressure of $\ce{CO2}$ is $\frac{1}{3}p_\mathrm{tot}$, where $p_\mathrm{tot}$ is the total pressure of the gases formed. Also, the carbamate is a solid species, so it is not involved in the equilibrium constant.

Let:
$C$ represent $\ce{NH3}$
$D$ represent $\ce{CO2}$
The expression of $K$ given can be set in terms of $K_P$:
$$K=P_C^2\;P_D\;\left(\frac{1}{P^o}\right)^{\Delta n}$$
$$K=K_P\;\left(\frac{1}{P^o}\right)^{\Delta n}$$
$K_P$ can be set in terms of $K_X$:
$$K_P=K_X\;\left(P\right)^{\Delta n}$$
Substituting above:
$$K=K_X\;\left(\frac{P}{P^o}\right)^{\Delta n}$$
For this reaction, $\Delta n=3$, so we have:
$$K=K_X\;\left(\frac{P}{P^o}\right)^3$$
Now, all we need to do is evaluate $K_X$.
Since total pressure $P$ only depends on products $C$ and $D$, and considering the fact that $C$ and $D$ are in a $2:1$ molar ratio:
$$X_C=\frac{c}{\Delta n}=\frac{2}{3}$$
$$X_D=\frac{d}{\Delta n}=\frac{1}{3}$$
We can now evaluate $K_X$:
$$K_X=X_C^2\;X_D=\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)=\frac{4}{27}$$
Finally, substituting this value for $K_X$ above, we get:
$$K=\frac{4}{27}\;\left(\frac{P}{P^o}\right)^3$$


text:
How does Combinatorial Chemistry differ from Rational Drug Design?
I stumbled on Vertex Pharmaceuticals - Wikipedia.

It was one of the first biotech firms to use an explicit strategy of rational drug design rather than combinatorial chemistry.

Then I looked up Combinatorial Chemistry. What is Combinatorial Chemistry

     Combinatorial chemistry is a laboratory technique in which millions of molecular constructions can be synthesized and tested for biological activity.  It has generated massive numbers of targeted molecules for testing and the developing techniques of high throughput screening has automated the screening process so larger numbers of biological assays can be done.  All this together has reduced the discovery-to-market time from what used to be 10-14 years to 5-8 years.

But this doesn't explain how Combinatorial Chemistry differs from Rational Drug Design? Isn't Combinatorial Chemistry rational too? Mustn't chemists pick reactants rationally? Even if she can't predict the products, she won't picked her reactants randomly.

From a medical point of view the problem in drug discovery is to find a compound that causes a desired outcome, not one that interacts with a specific target. For instance, if you want to reduce mortality due to infection with Covid, you have many strategies that may be pursued at the molecular level, from vaccination to interference with viral propagation to reduction of symptoms or boosting of the immune response.
The difference between a combinatorial strategy and a rational strategy is to a first approximation simple: in a combinatorial strategy compounds are tested for activity without immediate concern for how they bring about an outcome. There is no "a priori" information to guide compound selection other than the common properties of known drugs (as encoded eg by Lipinski's rules) and the ability to procure or synthesize compounds. Therefore the "rational" approach in this case is to create as large a library and to screen as many compounds as is reasonable given cost and time.
Rational drug design takes into consideration a molecular target (such as a protein) and perhaps the mechanism of action of a known lead compound and designs future potential drug compounds around that target and/or lead. That narrows the set of compounds under consideration reducing time and cost.
In general, the strategies of combinatorial and rational design are not mutually exclusive. It is typical to start a rational strategy after early leads have been identified some other way, say through combinatorial chemistry. It is also common to design a set of compounds during a stage in rational drug design using a combinatorial strategy applied to develop next generation leads from a starting compound.

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text:
Is Combinatorial Chemistry related to combinatorics in math?
If the answer is no, then I don't understand the adjective "Combinatorial". Barring decompositions, aren't the five main types of chemical reactions combinatorial? You must combine at least two reactants for a reaction!

Combinatorial chemistry usually refers to changing the choice of A and B, and when the reaction products differ according to the type of reaction, also the type of reaction.

Yes, the term "combinatorial" chemistry is related to mathematical combinatorics. Imagine that we have three positions (numbered 1, 2, 3) on a molecule, and at each position, we can place any one of 5 functional groups (A, B, C, D, E). We can synthesize all possible molecules in parallel by making all possible derivatives at position 1 (5 total), then reacting all of those molecules to make all possible derivatives at position 2 (5 for each of the 5 intermediates) and then again at position 3 (again, 5 new molecules for each product of step 2). In this example, we would produce 5x5x5=125 combinations as our final output.

In addition to the examples given, e.g., on Wikipedia's entry, combinatorial chemistry is not constrained to small organic and larger biomolecules.  Folks in catalysis screen by means of combinatorial chemistry libraries of e.g., mixed metal oxides, too.

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text:
Derivation and visualization of cell potential dependency on temperature and entropy
Working from the thermodynamic identity
$$\Delta{S} = -\left(\frac{\partial{\Delta{G}}}{\partial{T}}\right)_p,$$
is it allowed to integrate both sides to obtain an expression derived from the Nernst equation
$$E_\mathrm{cell} = \frac{T\,\Delta S}{zF}?$$
Can one plot the results and obtain $\Delta S$ from the gradient, assuming $\Delta S$ is constant for the reaction and $E_\mathrm{cell}$ values at some temperatures are given?

You need to decide how to integrate both sides of $\displaystyle \Delta S=\left( \frac{\partial E}{\partial T}\right)_pzF$.

What you are suggesting in your equation is that $\Delta G_m=-T\Delta S_m$ but you know that in fact $\Delta G_m=\Delta H_m-T\Delta S_m$. Your math is incorrect because you are missing a constant of integration.
The expression $\Delta G_m=\Delta H_m-T\Delta S_m$ can be rewritten using the Nernst equation as
$$-nFE = \Delta H_m -nFT\left(\frac{\partial E}{\partial T} \right)_p$$
by using the definition you provide for $\Delta S_m$ as the temperature-derivative of $\Delta G_m$, and the Nernst equation $\Delta G_m = -nFE$. In fact
$$\Delta S_m = nF\left(\frac{\partial E}{\partial T} \right)_p$$
Therefore if you plot $nFE$ as a function of T the slope of the curve at a given temperature will provide the value of $\Delta S_m$ at that T.


text:
Why is there no salt bridge for hydrogen–silver bromide cell?
What is the reason there is no salt bridge for this cell?
$$\ce{Pt(s) | H2(g) | HBr(aq) | AgBr(s) | Ag(s)}$$

The function of a salt bridge or a diaphragm is to avoid mixing of 2 different electrolytes or electrolysis products from the respective half-cells, while allowing ion electro-migration between them.
This cell does not need a salt bridge as it uses the only electrolyte $\ce{HBr(aq)}$ and the is no need to separate the $\mathrm{AgBr}$ electrode from gaseous hydrogen.


text:
Can cis-trans isomers interconvert?
Can cis-trans isomers interconvert via tunneling ?

Cis trans tunnelling in a search renders quite some papers. Samples were specially prepared, tunnelling involves hydrogen/deuterium positional exchange rather than a "rotation".

I have just searched for "geometrical isomers tunnelling". Found things involving D as a label so I imagine it is easy for hydrogen atoms to tunnel.

geometric isomerism [obsolete] Obsolete synonym for cis-trans isomerism. (Usage strongly discouraged). I emphasise: strongly. Otherwise the question is good.

@MilanPaul that was a pictorial way to say that the tunneling involves somehow one nucleus tunneling to the other side (but I am not even sure such a decomposition makes sense, I just want to suggest that a long alkyl chain won't tunnel. Or, it makes sense but it is implicit in the concept of tunnelling).

If you have, say, HDC=CHD then if both H and D on the same carbon could exchange via tunnelling a close to 1:1 cis/trans mixture would exist: it might be possible  to measure this ratio precisely as one species may be slightly more stable than the other. Do you know if this type experiment has been done, I would be surprised it it hadn't.

@porphyrin If we started with pure cis/trans and the barrier is high, the probability will be very low right? So how can we expect close to 1:1 mixture ?

The barrier is often too high for thermal interchange (absorbing a photon will work though) but protons and deuterons can tunnel so the height of the barrier is less important. I'm assuming that H/D have similar tunnelling rates when saying 1:1 but in fact its more complicated and is $\sim e^{-c\sqrt{m}}$ where $c$ is the area of potential energy barrier above any given energy $E$.

This conversion is possible photochemically, after absorption of a sufficiently energetic UV photon.

Dear <@porphyrin> Does e stand for electric charge or Euler's number?  M is mass?  And c is area of potential energy, not potential itself?

In my comment $e$ is the exponential i.e.  $\exp(-c\sqrt{m})$

Electron tunneling
Tunneling refers to elementary particles moving "through" a potential barrier in the absence of sufficient energy to go "over the barrier". This is roughly equivalent to saying that a particle can occupy two locations and traverse between them even though all paths between those locations have a zero probability of the particle being there.
Some orbitals such as the p-orbital describe a probability density with areas of non-zero probability ("lobes") separated by space with zero probability ("nodes").
When electrons are transferred from one electron carrier to another, the transfer can occur through bonds or through space; the through-space transfer is either based on orbital overlap or on tunneling.
Proton tunneling
Proton tunneling is much slower because the proton is about 2000-times heavier than the electron. Tunneling has been used to explain reactions involving protons that are faster than expected from inspecting the potential energy landscape. This proton tunneling also explains proton transfer in hydrogen bonds.
For formic acid and acetic acid, proton tunneling explains cis-trans isomerism at ultra-low temperatures. Acetic acid, at room temperature, has a cis and a trans conformation.

There is a rotational barrier (which one may explain via resonance involving the carbonyl double bond) which is lower than thermal energy at room temperature [Ref 1]. For related systems like the peptide bond, cis and trans forms are isomers because the rotation barrier is higher.
At ultra-low temperature in a nobel gas matrix, the trans form is dominant and does not convert into the cis form unless you excited the molecule at the right wavelength. This means that the cis and the trans form are isomers with individual properties. Under these conditions, it is possible to enrich the cis form, and to observe the conversion to the trans form in the dark [Ref 2]. The temperature-dependence of the kinetics (and kinetic isotope effects) suggest that this cis-to-trans reaction occurs via tunneling.
Hydride tunneling in enzymes
Many redox reactions involve hydride transfer, e.g. from alcohol to NAD+ in the alcohol dehydrogenase catalyzed reaction. For enzymes, it is typically not possible to measure reactions at ultra-low temperatures; instead, a key experiment is to study the primary and secondary kinetic isotope effect [Ref 3]. From studies of multiple enzymes that catalyze hydride transfer it is evident that tunneling does occur, and the extent of tunneling is modulated by mutations in the active site. The review makes the argument that tunneling is easier to study for hydrogen (i.e. proton, hydrogen radical and hydride) transfer than for electron transfer because in the former case, isotopes (with a different deBroglie wavelength) are available as a probe. While the examples in the review are all redox reactions rather than E/Z isomerization reactions, one could envision the latter based on hydride tunneling as well.
"Double" tunneling
If there are two atoms connected to a double bond (such as in the molecule $\ce{H2C=CDH}$ mentioned in the comments), isomerization through tunneling would be much harder because the hydrogen and the deuterium have to somehow switch position. To avoid a clash, they would have to tunnel simultaneously, and could not both take a path along the saddle point of the potential energy landscape. Therefor, a cis-trans isomerization of $\ce{H2C=CDH}$ seems much less feasible, and there does not seem to be any experimental evidence for it.
References

Barriers to rotation adjacent to double bonds. 3. The carbon-oxygen barrier in formic acid, methyl formate, acetic acid, and methyl acetate. The origin of ester and amide resonance. Kenneth B. Wiberg and Keith E. Laidig. Journal of the American Chemical Society 1987 109 (20), 5935-5943. DOI: 10.1021/ja00254a006
Rotational isomerism of acetic acid isolated in rare-gas matrices: Effect of medium and isotopic substitution on IR-induced isomerization quantum yield and cis→trans tunneling rate. E. M. S. Macoas, L. Khriachtchev, M. Pettersson, R. Fausto, M. Rasanen
J. Chem. Phys. 121, 1331 (2004); DOI:10.1063/1.1760733
Tunneling and Dynamics in Enzymatic Hydride Transfer.
Zachary D. Nagel and Judith P. Klinman, Chem. Rev. 2006, 106, 3095-3118, DOI:10.1021/cr050301x

Tunneling between atoms may be difficult according to the other answer, but in some environments a scrambling mechanism may work.  In the presence of superacid, a bridgehead hydrogen in bicyclic compounds can be removed as a hydride ion, forming a carbocation.
Upon re-attachment the hydride ion may return to either of two positions mixing the cis and trans isomers, and eventually equilibrating them.  This process is documented for two such hydrocarbons by Olah et al. [1].

From Ref. [1]
Reference

George A. Olah, G. K. Surya Prakash, Jean Sommer, Arpad Molnar, Superacid Chemistry (John Wiley & Sons, Mar 26, 2009), p. 533.

Do the authors claim this is due to tunneling? The question asks about tunneling in the main body but not in the title. I'm not sure a hydride ion (i.e. a multiple particle system with electrons attached to protons) can tunnel.

This is entirely non-QM in nature.  I specifically point out that this is an alternative to tunneling, which appears not to work for atoms based on another answer.  The hydride ion scrambles (by switching between substrare and graphene layer) rather than tunneling.

@karsten theis I’m not familiar with this specific example, but there are many other examples of claims of hydride tunneling. See for example Judith klinman’s work on alcohol dehydrogenase reactions

@Andrew Thanks for the pointer, I included a reference to and discussion of  DOI:10.1021/cr050301x in my answer.

Have a look at the literature report by Helen Leung et al. (2013):

We synthesized cis-1,2-difluoroethylene from the dechlorination of 1,2-dichloro-1,2-difluoroethane. The products contain a cis/trans mixture of 1,2-difluoroethylene (...).

Here's the experimental evidence of tunneling using microwave spectroscopy (Leung, 2013, fig. 4a):

Edit
I think we should strive to use E/Z instead of cis/trans.

Reference
Leung, H.O., Marshall, M.W., Mueller, J.L., and Amberger, B.K. (2013) "The molecular structure of and interconversion tunneling in the argon-cis-1,2-difluoroethylene complex" J. Chem. Phys. 139:134303. DOI: 10.1063/1.4823494

We get a mixture of isomers, but do they interconvert?  The mixed product could come from mixed confirmations of the reactant.


text:
Why does having equivalent resonance structures give more stability (Related)?
This question is based on @Heisenberg's Question with the same title.
I recently offered a bounty for a more detailed answer for the same question.
These are the things I would like to be in the answer:

Can molecules with 'n' canonical structures be taken as an n-state  system with coupling(At least as an approximation) ?

Suppose if it is possible.
Consider a molecule with 2 canonical structures represented by $|A\rangle$ and $|B\rangle$
Without the coupling let the energies of the two states be $E_1$ and $E_2$ respectively.
So the Hamiltonian is represented by:
$H=\pmatrix{E_1& 0\\0&E_2}$

Now with coupling the Hamiltonian in the most general case will be:
$H=\pmatrix{H_{11}& H_{12}\\H_{12}^*&H_{22}}$
The energy eigen-value with the lower energy in this case is:
$E_1 = \frac{H_{11}+H_{12}}{2} - \frac{\sqrt{(H_{11}-H_{22})^2 + 4H_{12}H_{12}^*}}{2}$
So my doubt is if the 2 states are symmetrical then $H_{11}$ and $H_{22}$ are equal right?
Then the term inside the square-root $(H_{11}-H_{22})$ becomes zero, and it seems to me that when this term becomes zero, the energy splitting becomes the least.
And yet it is resonance with equivalent canonical structures that are more stable. Why?

Are molecules with equivalent resonance structures actually more stable? Please include references or experimental evidence if the it is true.

If molecules with equivalent resonance structures are in fact more stable than others, then does it mean that resonance in molecules cannot be approximated as a discrete n-state system even for ones with 2 canonical structures (Since there might be a contradiction with arguments in 2)) Can resonance be understood only using MOT or Huckel Theory ?

Note: Please add your answer in Heisenberg's question so that I can award you the bounty.
You might understand why I'm having these doubts and the doubts itself if you read these chapter of the Feynman lectures on Physics Chapter 8 , Chapter 9 , Chapter 10

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

2) is answered here


text:
Writing a chemical equation with a reactant of unspecified chain length
Would appreciate it if I got some general tips and hints and not the solution cause I'm trying to learn from this question :)
The question is as follows:

FA1 is a solution containing $\pu{5.00 g dm-3}$ of a dibasic acid, $\ce{HOOC-(CH_2)_n -COOH}$.
FA2 is a solution containing $\pu{5.00 g dm-3}$ of $\ce{NaOH}$.
$\pu{25.0 cm3}$ of the acid solution FA1 needed $\pu{17.0 cm3}$ of the sodium hydroxide solution, FA2, for complete neutralization.

Determine the value of $n$ in the formula of the acid.

The issue I have is in starting the question. I genuinely have no clue how to start a chemical equation with something like $\ce{HOOC-(CH2)n -COOH}$. Tips and hints would be greatly appreciated thanks.

Start by writing the equation of the reaction
from the volume added to achieve neutralization, you can write a relation between the quantities of matter so the unknown will be the molar mass of FA1
you will be able to determine FA1 since you know its structure

Tips and hints would be greatly appreciated.

My tips and hints for you (I apologize for if my tips and hints direct to me treating OP as a novice):

Your dibasic acid is $\ce{HO2C-(CH2)n-CO2H}$. Thus, $\ce{A^2-}$ in the equation $(2)$ of Karsten Theis's answer is $\ce{^-OOC-(CH2)n-COO^-}$.
Molar mass of this dibasic acid is: $2 \times (12+32+1) + 14n = 90 + 14n$
Find the concentration of FA2 solution in $\pu{mol dm-3}$ (in other words, in $\pu{M}$. Hence, you can find the the amount of $\ce{NaOH}$ (in $\pu{mol}$) in FA2 solution ($\pu{17.0 cm3}$).
Determine the amount of dibasic acid (in $m_\ce{H2A} \ \pu{ mol}$) in FA1 solution ($\pu{25.0 cm3}$) using the equation $(2)$ of Karsten Theis's answer.
Now, you can find $n$ using following equation:
$$m_\ce{H2A} \ \pu{mol} \times \pu{(90 + 14n) g mol-1} = \pu{5.00 g dm-3} \times \pu{0.025 dm-3}$$

Since you have already calculated $m_\ce{H2A}$ in step 4, you can calculate the $n$ $(\pu{1.0 dm-3} = \pu{1000 cm-3})$.
Note that $n$ is an integer so final answer should be rounded to closest integer.

I genuinely have no clue how to start a chemical equation with something like $\ce{HOOC(CH2)_nCOOH}$. Tips and hints would be greatly appreciated thanks.

The only thing that an acid does in an acid/base reaction is to lose hydrogen ions. You can write the neutralization of a monoprotic acid with $\ce{NaOH}$ like this:
$$\ce{HA(aq) + NaOH(aq) -> A-(aq) + Na+(aq) + H2O(l)}\tag{1}$$
This is already balanced. For a diprotic acid, the unbalanced equation would be
$$\ce{H2A(aq) + NaOH(aq) -> A^2-(aq) + Na+(aq) + H2O(aq)}\tag{2}$$
It is possible to balance this even though we did not specify what $\ce{A}$ is.
$\ce{HOOC(CH2)_nCOOH}$ is a diprotic acid as well, so its neutralization is described by a similar chemical equation. The bothersome $\ce{-(CH2)_n -}$ group prevents you from counting atoms on both sides of the equation; however, as it does not partake in the reaction, you can ignore it - it will appear on the reactant and product side in the same form.

Tips and hints would be greatly appreciated thanks.

You got some tips and hints in the comments how to proceed once you figured out the stoichiometry. Another way of stating the steps is:

Determine the mass of FA1
Determine the amount of FA1 (how many moles)
Determine the molar mass of FA1
Figure out what $n$ is

Thanks! This was helpful. Am I right to think that in this case A is OOC(CH2)COOH?

almost ! Here you assay a diacid $ \ce {H_2A} $ -> the group A would therefore be $ \ce {OOC-(CH_2)_n-COO} $

Ahhhh. So the salt that will be formed is Na2(OOC−(CH2)−COO) am I correct?

Yes, if you want to write it as salt. I chose to write it in the dissociated form in equation (2)


text:
How to carry out cyclization of 1‐(2‐methylbutyl)‐2‐nitrocyclohexane?
How the cyclization can take place as there is no other functional group than $\ce{NO2}?$

One way that I can think of is photochemical bromination which occurs selectively at tertiary center which then can be promoted to give cation in polar medium. Nitro group on being reduced to amino group will attack on carbocation, thus closing the ring.
The problem here is that there are three tertiary centers and radical close to functional group will be more stable than the one at the far tertiary center on side chain.
How would you handle this conversion?

You might want to read up on the Hofmann-Loeffler-Freytag reaction. https://en.wikipedia.org/wiki/Hofmann%E2%80%93L%C3%B6ffler_reaction

The Hofmann-Loeffler-Freytag reaction as suggested by @user55119 in the comments above is the way to go.
First you need to reduce your nitro group. The reduction of alkyl nitro compounds is less simple than aromatic nitro groups and lower yielding in general. I would suggest the use of Al/Hg amalgam.
Acetylate the resulting amine.
N-chlorinate the amide - trichloroisocyanuric acid is good for this conditions
Irradiate the chlorinated amide to form the [5] ring similar to this example Baldwin gelsimicine synthesis
Finally hydrolyse off the acetyl group with cHCl.


text:
Does partially/completely removing solid shift the equilibrium?
For the reaction: $$\ce{P4(s) + 5O2(g) <=> P4O10(s)}$$
Let's say the reaction is at equilibrium, and SOME $\ce{P4O10(s)}$ is removed from the reaction. Will that affect equilibrium? My assumption is no, since $\ce{P4O10(s)}$ is not included in the equilibrium expression.
However, for me, it gets confusing when let's say $\ce{P4O10(s)}$ is COMPLETELY removed. In this case, I would assume that the reaction shifts to the products (since no products are present). I'm not sure if this is right.
Help is appreciated. Thank you!

There isn't much of an equilibrium here; the reaction is shifted pretty much all the way to the right anyway. That being said, your reasoning is correct.

@IvanNeretin That makes sense. Thank you. I'm just confused because I thought solids/pure liquids do not effect the equilibrium?

If there's no equilibrium then there's no effect on equilibrium. Find a valid example.

Hmm but in my example, there was an equilibrium. It's just that I'm saying what happens in  P4O10 is completely removed?

I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate.
If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer dissolving because there aren't any. Less obviously, there is no crystallization process anymore (there are no crystals that solvated ions could attach to). So this is not a dynamic equilibrium, i.e. the reaction or process no longer is observed at all.
This is different from a system out of equilibrium where one direction of the process is faster than the other direction. In that case, the system will approach equilibrium, and the process (in either direction) will continue to happen when equilibrium is reached (which is why it is called a dynamic equilibrium).
You might ask how we can ever get solid sodium chloride again. If the solution is sufficiently supersaturated, nuclei will form (in solution, or on the container or on some impurity), and then crystal growth will be observed again.

What do you mean by "it is not longer an equilibrium?" If it isn't, wouldn't a change follow?

@BuckThorn Sorry, sloppy language. Either we have a reaction or process where both forward and reverse reactions are happening. This can either be at equilibrium or approaching equilibrium. Or, like in this case, neither forward nor reverse reactions are happening.

Ok, I see your point. @theorist made the additional good point that you have to keep things constant, e.g. pressure, to speak of equilibtrium when you remove the salt.

Nothing would happen. The reaction would not shift to produce more product.
Let's suppose you allow this reaction to equilibrate in a sealed container.   Equilibrate means you've reached the equilibrium value for $p_{\text{O}_2}$ at that temperature.  Now further suppose you were able to remove all the product without changing the free volume (say you slightly decreased the volume in the container by an amount equal to the volume of the solid product you removed).  The system is still at equilibrium ($p_{\text{O}_2}$ is unchanged).  Thus nothing will happen—no new product would be formed.
If you're not convinced, imagine that you had some fluctuation in the equilibrium such that a bit of product was produced.  In that case, $p_{\text{O}_2}$ would increase above its equilibrium value, and thus the product would need to disappear (the reaction would need to shift back to the left) for the system to go back to equilibrium.

I agree with "nothing would happen" but think it is important to say that something was happening before (dynamic equilibrium), and now the underlying molecular processes (forward and reverse reaction) are no longer in place.

@KarstenTheis  That's a v. good point—that while nothing changes macroscopically, there are changes microscopically. But while you wouldn't have fwd and rev rxns as they would normally occur in an equil sys, I wouldn't go so far as to say that the fwd and rev rxns no longer take place, since there would still be fwd and rev rxns due to fluctuations about equil (as describred in my last paragraph).  So this raises a very interesting ques: How would the nature and magnitude of the fluctuations about equil differ between the original equil sys (where product was present) and this "one-sided" sys?

I would agree with your last paragraph for a homogeneous equilibrium. If the reaction involves forming a solid (as in a precipitation reaction) and the solution is not saturated, no solid is formed. On the microscopic level, there might be some ion pairs formed, or even some transient nuclei but I don't think they rise to the concept of solid.

It might help to think of the system as containing three phases: $\ce{P4(s)}$, $\ce{O2(g)}$ and  $\ce{P4O10(s)}$. As long as the pressure, temperature and compositions of each phase remain constant (and P,T and chemical potentials across phases are in equilibrium i.e. equal), we can alter their extents arbitrarily without altering the condition for equilibrium to be sustained.
An example is freezing of water. When water is cooled to $\pu{0^\circ C}$ at 1 atm pressure it can remain at equilibrium without forming ice (it will not freeze), provided no energy is removed (as heat) from the liquid phase. Otherwise ice will begin to form. If you remove an ice cube from a glass of water at  $\pu{0^\circ C}$ more ice does not spontaneously form, provided T and p are not altered (say by providing a means for the water in the glass to lose heat to the surroundings, which would of course describe a nonequilibrium situation).
This is admittedly somewhat strange and not the end of the story because we have considered the system to be an abstract body in which the interfaces between phases are not important. In fact they can be, and we might describe a system in which substituting the interface between two phases with another type of interface might cause a disequilibrium.


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Chemical Volcano- Experiment
In a science exhibition in 2011 , I and my  friend made a volcano that used vinegar and baking soda for the 'effects'. There was another group that made volcano too and they had brought their own supplies. But before the exhibition started, the members of other group came to us and asked us if we could share our baking soda because their baking soda wasn't 'working'. When I asked them why it wasn't working they told me that there wasn't 'salt' in it. (We were only 8-9 years old). When I tried using it it was forming bubbles but very slowly. Eventually our supplies ran out before the exhibition was over. One of my friends suggested that we should use chalk powder instead of baking soda(I don't know how he figured it out at age 9). So we ground up chalks and tried using it. In this case , it was just like the baking soda that didn't work , it was producing bubbles ,but very slowly.
What could be the issue in the two cases? Was it a matter of purity?
We had used synthetic vinegar used for food purposes(5-10%)(There was no issue with the vinegar because it worked well with our own baking soda). The chalks were ordinary chalks used in classrooms. Regarding the other group's Baking soda I do not have any other information cause it happened 10 years earlier.(And I'm no longer in contact with them)
Was reaction less fast in case 2(Chalk powder) because of the low solubility product of Calcium carbonate? But Solubility product is a thermodynamic factor and not a kinetic factor right ?

It must just have been a question of dimension of the grains. It is well known that when powders reacts, the reaction rate is proportional to the contact surface, So the contact surface is biggest when the grains are thinnest. If the baking soda had been ground with care, it must react quicker with vinegar.

It always works properly with real baking soda. No salt is needed. Calcium carbonate has poor solubility in water, e.g., white cliffs of Dover, etc., so you would need to grind it to fine powder to get high surface area contact with the vinegar. Even then, the bubbles impede contact between the vinegar and powdered chalk. Sodium bicarbonate has reasonable water solubility, and vinegar is just a dilute aqueous solution of acetic acid. So mixing them always works well. Not much more to say: use inexpensive fresh baking soda.

Baking soda is sodium bicarbonate NaHCO$_3$. It is stable in air indefinitely. Baking powder is a mix of sodium bicarbonate and a powdered acid, typically an aluminum or calcium hydrogen phosphate, which, in the absence of moisture, is stable for a long time (maybe a year? maybe more). After exposure to moisture, it will release CO$_2$ and become inactivated.
The third item, "chalk", has many definitions. Blackboard chalk is not just a stick of CaCO$_3$. A stick of CaCO$_3$ could be marble, or limestone - oh - no, you would like it to be nice and powdery but sticking together. Yep, it can be done: glue. Nah, too scratchy. Plaster of Paris! Mix with water, add some CaCO$_3$ chalk, maybe some white clay... You get the idea. Blackboard chalk is not just CaCO$_3$. Not even that, but CaCO$_3$ reacts very fast with hydrochloric acid (but that's too hazardous for grade school experiments), and its reaction with vinegar (5% CH$_3$CO$_2$H) is slower initially, and slows down markedly as the reaction progresses because the acetate ion produced inhibits the ionization of the remaining (weak acid) CH$_3$CO$_2$H.
The key lesson is that demonstrations are very temperamental, and depend on doing the same thing with the same materials many times, to see how the "same" experiment can vary so much. You may even begin to doubt that any experiment can ever be duplicated - but it can, if you keep everything constant.


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Why is it that protons and electrons undergo the same amount of deflection in an electric field if they have the same energy?
The question I have is with respect to this diagram. Which depicts a stream of protons and electrons entering a proton field with equal energy.

Why is it that in the case where the energy of protons and electrons are equal, their angle of deflection is exactly the same? However when speed is kept the same, the deflection of electrons is greater. Why is this so?

a better quality image would be appreciated and appreciated! In an electrostatic field, the force undergone is worth $\vec{F} = q \vec{E} $ therefore the force depends only on the charge of the particle

I’m voting to close this question because it's not about chemistry.

@Mithoron it's about mass spectrometry, isn't it?

@LevelRiverSt: It's a question about basic electromagnetism, which'd commonly be accepted as a straightforward Physics topic.

@Nat it had not escaped my notice that was about basic electromagnetism. I was pointing out that OP had presumably posted it due to its application in mass spectrometry, an important technique in analytical chemistry. If so, I think it has a place here.

Because they have equal (but opposite) charges. That is all.

Now that's a mildly non-trivial observation. Why would they be equal, really?
Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the acceleration $a={qE\over m}$. By the time the particle reaches the right side of the picture, it gets deflected by $y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$.
That pretty much sums it up. $E$ and $L$ are the same for everyone, $q$ is the same in absolute value for electron and proton, so the only difference is $mv^2$. Now, considering your two situations:

Same speed: same $v$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
Same energy: same $mv^2$, same deflection (just to the opposite sides).

Things get hairy in relativity, but that's another story.
So it goes.


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How can you increase the pH of an acidic solution by 1?
In order to increase the pH of a solution of some acid $\ce{HA}$ by 1, how many times must it be diluted by? The pH log scale suggests that the required dilution factor is 10. Is that generally valid for strong and/or weak acids, or do we need to take into account the dissociation constant for weak acids?

You can't tell without more data.

Since you have not given any values some of the important values need to be considered as variables,
Also assuming its a monobasic acid $\ce{HA}$
$$\ce{HA_{aq} <=> H^+_{aq} + A^-_{aq}}$$
The important values are the

Acid dissociation constent $K_\mathrm{a}$
Initial concentration $c$

Writing the equation for the acid dissociation constant
$$K_\mathrm{a} = \ce{\frac{[H^{+}][A^{-}]}{[HA]}}$$
Initially a $x$ amount of acid gets dissociated equation simplifies to
$$K_\mathrm{a} = \frac{x^{2}}{(c-x)}$$
This is a $ax^2+bx+c=0$ type equation which can be solved easily simplified,
$$x^2 + K_\mathrm{a}x - K_\mathrm{a}c = 0$$
Take the accepted value for $x$ then (Not thinking that you will get complex number for real $K_\mathrm{a}$ and $c$ values),

Optionally you can consider that $c \gg x$ so $c - x \approx c$ simplified $x = \sqrt{K_\mathrm{a}c}$

Since $x$ is found now, we now know the initial $\mathrm{pH}$ value or $\mathrm{pH}_1$
$$\mathrm{pH}=-\log(\ce{[H^+]})$$
so,
$$\mathrm{pH}_1=-\log([x])$$
Final $\mathrm{pH}$ value $\mathrm{pH}_2$ goes up by 1 so,
$$\mathrm{pH}_2=-\log([x]) + 1$$
So the new dissociated $x_2$ value,
$$x_2 = 10^{-(1-\log([x]))}$$
Since $\ce{[H^+]=[A^-]}$ new equation,
$$K_\mathrm{a} = \frac{x_{2}^{2}}{(c_{2}-x_{2})} = \frac{10^{-2(1-\log(x))}}{(c_{2}-10^{-(1-\log(x))})}$$
Now you can solve this equation and get $c_2$ new concentration now the answer is almost complete just need to get the dilution factor
$D = \dfrac{c}{c_{2}}$
$D$ is how many times you need to dilute
But in the non monobasic acids this is a bit more complex since it has more than one dissociation constant $K_{\mathrm{a}_{1}}$ $K_{\mathrm{a}_{2}}$ $K_{\mathrm{a}_{3}}$ etc. need to check if the ratios of the dissociation constants if so that they are effective in the $\mathrm{pH}$ calculation.

I assume the task author goes for the simplified formula, based on $c_0 \gg [\ce{H+}] \gg [\ce{OH-}]$

Hmm yeah I have now added that in a block just now

why does 2=10^−(1−([]))  and not x2=10^-log[x]+1

Its just $\ce{pH = -\log{([H^{+}])}}$ substitute the new $\ce{pH_2}$ to the equation and try to get the new $\ce{H^+}$ concentration you made a mathematical mistake watch out there is a minus sign there


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Why does whey protein powder make porridge much less viscous?
When I mix whey protein powder into warm porridge (that I cooked in the microwave), the thick porridge becomes much less thick.
I searched this, and I found several forum threads discussing this too so it seems to be a common phenomenon. Happens when cooking porridge on the hob too, not just from mixing protein in after cooking.
I make the porridge through simply heating porridge oats + cow's milk. It doesn't matter if the oats are steel-cut or rolled, part of a fast-cook microwave pack, or completely natural.
How come the porridge becomes more liquidy when whey is stirred into it?

Seasoned Advice SE often plays with disparate metallurgy and rheology stuff.

I suspect that the thickness of the porridge is in fact, due to it being a non-Newtonian, shear thickening fluid. When the protein powder is mixed, perhaps the mixture becomes less non-Newtonian. That's just my guess though.

@ShoubhikRMaiti according to this podcast I just found, porridge is shear-thinning

I found that whey proteins are used as thickeners, at least in yogurt.

@Alchimista that's why this behaviour in porridge is very surprising

@theonlygusti you have a complicated matrix, so one has to identify and study really in details. What is thickening a fluid yogurt can reduce the viscosity of a thick porridge. In the latter case it suffices, but I am inventing, that the proteins in the powder are smaller than those of milk. Voilà, now the ibterchain interactions are between long and shorter chains, disrupting gels or anyway acting as "lubricants". It is just a possibile mechanism. Or there can be more cysteine, etc.

Does it contain lecithin as an emulsifier? https://www.renegadedad.net/why-lecithin-is-used-in-whey-protein-powder Also, if hydrolyzed, that also could include chemicals breaking down the starch.


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Number of π electrons in all trans-2,4,6-octatrienoic acid

Sketch the Lewis structure of all trans-2,4,6-octatrienoic acid.
(a) Consider the delocalized π electrons. Employ the particle-in-a-box model. Drawing on your knowledge about the occupation of orbitals in atoms what is the highest occupied state? Sketch the occupation of the energy levels with electrons using ↑ and ↓.

I wondered if the electron lone pairs on the oxygen atoms also counted as π electrons since resonance structures exist where these lone pairs can be delocalised over the molecule.

If so, how would this correspond to using them to find the length of the conjugated system because wouldn’t that give extra length where there was no double bond?

No, discard them. Also, are you sure that such resonance structures exist?

Which oxygen do you mean? The carbonyl oxygen (=O) lone pairs are orthogonal and cannot participate in the resonance. The carboxyl oxygen (-OH) can participate in resonance to some extent, but since the question asks about particle in a box model, you can't really consider that.

In UV-Vis spectroscopy, these two electron pairs on carbonyl-O and alcohol-O are named n electron pairs because these are not shared with the extended $\pi$ electron system.


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Addition of 1 equivalent bromine to 1,3 pentadiene at temperature below 0 degree Celsius gives which compound?
As far as I know reactivity of alkenes towards addition with halogens is directly proportional to their nucleophilicity, so bromine in this particular case should attack the double bond at Carbon-3 but in my textbook answer Bromine has been made to react with double bond at carbon-1

I am confused. What are you trying to ask?

What is the solvent? At which pH and at which temperature is this?

Temperature below 0 degrees celsius and ph wasn't mentioned in the question


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Is 2 ethyl 3 methyl oxirane optically active?
According to me this molecule does not have any any plane of symmetry nor centre of symmetry, but in my textbook it is given that this is optically inactive.

Then your textbook is wrong. Or maybe they mean a racemic mixture.

Please provide a reference for your textbook and some context. Also, the name in title doesn't match the structure (which is 2-ethyl-3-methyloxirane).

I am sorry, I have edited it. Thanks!
It was a question in the book "Advanced Problems in Organic Chemistry for JEE"
By Himanshu Pandey

You may draw the structure in a chemistry-aware search engine like Chemspider.  With title structure as sketched, omitting stereochemical information like here you see there is discern about the stereogenic centres to be either in RR, RS, or SS configuration.  One points to this noting chiral resolution.

Both oxirane carbon atoms are optically active, because both contain 4 different substituents. The left one has a methyl group, a hydrogen atom, the other oxirane carbon atom and the oxygen atom from the oxirane ring. The other oxirane carbon has the following substituents: a hydrogen, an ethyl group, the oxirane oxigen and the other oxirane carbon. I assume the author of the textbook made the mistake to count the oxirane ring as one functional group, and overlooked that this one functional group functions as two substituents.

Maybe the author of the book forgot the non visible H atoms.


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How to determine the number of π-electrons in a conjugated system from UV-Vis data?
Is there a way to determine the number of π-electrons in a conjugated system from UV-Vis data? I am modelling the electrons as particles in an infinite box of length $L$ equal to the length of the conjugated system. I am unsure how to interpret the spectra to obtain the number of π-electrons, including the systems with more than one peak.

I don't know if you can determine the number of pi-electrons from UV-vis, because UV-vis is basically the HOMO-LUMO transition. In some cases, there are additional bands, but they might come from sigma or non-bonding electrons being excited. So, it's hard to say what the total number of pi electron is, even if the peak comes from pi->pi* excitation. Aside from that, particle in a box is very inaccurate quantitatively, so I don't know what you are trying to achieve with that.

Yes and no. I might come back to this but it depends on the level of prediction you have in mind and on the classes of compounds. In all cases a good amount of knowledge is required as starting point. The peak to look at would be that associated with the pi-pi* transition. You can right "when the particle in a box is somehow a decent model".

I think the point of the exercise I’m doing is to show how inaccurate this is. How would I go about finding the number of pi electrons? I just don’t see it with this theory

Not really. A uv-vis spectrum normally contains peaks due to vibrational transitions on top of an electronic transition, but your particle in a box calculation only produces electronic transitions and in any case is a very crude approximation. If you can see two or more  electronic transitions and work out the transition energy that would be better. So you can only have an inspired guess at best, not a proper prediction.


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Haworth structures of sucrose and lactose
The haworth structures of lactose and sucrose  is

My question is regarding the glyosidic linkage, in lactose it is represented as a zig zag and in sucrose it is a "V". What does that mean. Also, in sucrose why is the the glyosidic linkage displaced down w.r.t the compound

The link in lactose structure is beta (up position)-1(1-position around the ring)-4(glucose's 4-position around its ring, in which the OH is down for glucose). The link in sucrose is alpha (down)-1(glucose's 1-position)-2(fructose's 2-position which is at the down position because fructose is "facing" the glucose unit and it is upside down).

Haworth structures are cumbersome to draw in this regard. The bridging oxygens are attached directly to the rings and not to CH2 groups as the "zig-zag" and "V" might imply.


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How does the force constant change when a heavier atom is used in vibrational spectroscopy?

(iii) The fundamental band in the IR spectrum of $\ce{^1H^{127}I}$ is found at $\pu{2308 cm^-1}.$ Calculate the force constant for the bond. Assume the molecule can be treated as a harmonic oscillator.
(iv) Calculate how the position of the fundamental band would be expected to differ for molecules of $\ce{^2H^{127}I}.$

I have found the force constant in (iii), but I am struggling with (iv). I don’t know how to calculate the difference. Any insight would be greatly appreciated.

You can calculate the ratio of the two frequencies as you know all the masses. Usually homework questions such as this get closed unless you explain what you have done so far and where you are stuck. As a start I suggest that you look a the vibrational spectroscopy part of your textbook where you are bound to find the equations you need.

In the harmonic oscillator approximation, the wavenumber of vibration for a diatomic molecule is
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
Here, $k$ is the force constant of your bond, and $\mu$ is the reduced mass for the two atoms.
$$\mu=\frac{m_1m_2}{m_1+m_2}$$
When you change isotopes, only $\mu$ changes. The force constant depends on the strength of the bond, which depends on the chemical nature of the elements, and can be assumed to stay constant regardless of changing isotopes.
You have to calculate the $\mu$ for the second problem, and use that with the $k$ you got from the first problem to calculate the wavenumber of the fundamental band.

Can the factor $k$ be obtained by some calculation ?

@Maurice Well, usually it's the other way round. You can measure the wavenumber by spectroscopy, and then you can use it to calculate the force constant (k). You can however, use computational chemistry, which will directly calculate the force constants of the bond.

As suggested by @porpyrin, in the second calculation you're assuming the force constant $k$ is constant in both molecules so a ratio can be taken. Let $\tilde{\nu_H}$ and $\tilde{\nu_D}$ be the hydrogen and deuterated versions respectively,
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
$$\frac{\tilde{\nu_H}}{\tilde{\nu_D}} = \frac{\sqrt{k/\mu_H}}{\sqrt{k/\mu_D}} = \sqrt{\frac{\mu_D}{\mu_H}}$$
Note: the bond strength changes as the zero point energy of the bond changes upon isotopic substitution hence this is only an approximation.


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Why does π-backbonding in a metal-carbonyl complex lead to a decrease in C-O bond strength?
I understand that π-backbonding in this complex involves the σ-donation of carbonyl to an empty d orbital of the metal, and the π-donation of the d electrons of the metal to the π* MO of the carbonyl. But why exactly does this lead to the weakening of the C-O bond (relative to free C-O)?

Do you know what a $\pi^*$ orbital is?

In simple words, the pi* orbital is antibonding in nature. So, putting electron density there will weaken the C-O pi bond. The CO bond contains both sigma and pi components. The pi component will be weakened.


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What are the limits of size differences in a host–guest complex?
We learn in my chemistry course that there must be a snug fit for a host-guest complex to form. So, something like benzene (0.6 nm in diameter) can form a host-guest complex with a β-cyclodextrin host (0.7 nm cavity).
Other than this vaguely-worded ‘snug-fit’ rule, we are not given a lot of information of what molecules would fit.
Would something like methane (0.38 nm diameter, according to Wikipedia) form a stable host–guest complex with β-cyclodextrin, or would it be too small?
I know that methyl red fits in a β-cyclodextrin cavity, but what about something slightly larger like anthracene?
These questions are a small part of something that contributes a small amount towards my overall course grade. Please, if you are able to, give me some kind of guidance so that I can work out answers on my own. I've tried to look independently online but I come up with nothing.

Why do you think anthracene is larger then methyl red? Would orange be bigger then banana if they had same volume?

The size of the cavity is one of the many factors collaborating to host-guest chemistry with cyclodextrin (CD). For example, depending on the size and the properties of the guest molecule, CD may form complexes with either the entire molecule or only a part of the molecule (The diameter and the volume of the cavity of β-CD are around $\pu{7.8 Å}$ ($\pu{0.78 nm}$) and $\pu{262 Å^3}$, respectively, and β-CD complexes well with aromatic rings; Ref.1). It is accepted in general (Ref.2) that the binding forces involved in complex (host-guest) formation are:

Van der Waals type interactions (or hydrophobic interactions) between the hydrophobic units of the guest molecules and the CD cavity.
Hydrogen bond between the polar functional groups of the guest molecules and the hydroxyl groups of the CD rims.
Release of high energy water molecules from the cavity during the complex formation process.
Release of strain energy into the ring structure system of the CD.

Note that the role of hydrogen bonding is not universal since stable complexes are formed with hosts and hydrophobic guests such as benzene derivatives (Ref.3), which do not form hydrogen bonds.

Also, regardless of which types of stabilizing forces are involved, the most important factors in determining the stability of the inclusion of the complexes are: (i) The geometric capability (which is the OP's concern); (ii) the polarity of the guest molecule; (iii) medium; and (iv) the temperature. The reference 2 states that:

Geometric, rather than the chemical factors, are critical in determining the type of guest molecules that can penetrate into the cavity. If the guest is too small, it passes easily through the cavity and the bond will be weak or will not occur. The formation of complexes with the molecules significantly larger than the cavity is also possible, but only some limited groups or the side chains penetrate into the cavity.

It is also important to note that cavity size of CD seems to depend on the method used by the researcher (Ref.4). For instance, Ref.2 reported that inner diameters of $\pu{5.7 Å}$ for $\alpha$-cyclodextrin, $\pu{7.8 Å}$ for $\beta$-cyclodextrin, and $\pu{9.5 Å}$ for $\gamma$-cyclodextrin, but did not verify each was the upper or the lower or middle diameter. Meantime, Ref.3 reported that top and bottom inner-diameters of $5.3$ and $\pu{4.7 Å}$ for $\alpha$-cyclodextrin, $6.5$ and $\pu{6.0 Å}$ for $\beta$-cyclodextrin, and $8.3$ and $\pu{7.5 Å}$ for $\gamma$-cyclodextrin. Therefore, it is hard to predict which molecule would bind to appropriate CD without knowing exact parameters of their cavity. Regardless of the reported cavity sizes, Ref.3 shows that the grafting of $\beta$-cyclodextrin to silica gel conveniently removed water soluble organic molecules with different sizes such as p-nitrophenol, p-nitroaniline, m-nitrophenol, p-chlorophenol, and phenol.
It is also evident that cyclodextrin-catalyzed organic synthesis can be achieved using $\beta$-cyclodextrin as the catalyst. In these reactions, you may find part of starting materials with different sizes making stable host-guest complexes with $\beta$-CD (Ref.5).
On the other hand, an identical molecule can be bound to all three CDs ($\alpha$-, $\beta$-, and $\gamma$-cyclodextrin) grafted to a biomolecule with different impact (Ref.5). For example, $\mathrm{^1H \ NMR}$ studies suggested that methyl-$\beta$-CD forms inclusion complexes with the lipid guest molecules such as cholesterol, linoleic acid, and $\beta$-sitosterol within the CD cavity regardless of their sizes (Ref.6):

Note: Although it is not relevant to the question in hand, it is worth noting that  the chiral nature of the cavity in these molecules makes them useful for enantiomeric separation applications (Ref.7).

References:

Vijaykumar Parmar, Gayatri Patel, Nedal Y. Abu-Thabit, “Chapter 20:  Responsive cyclodextrins as polymeric carriers for drug delivery applications,” In Stimuli Responsive Polymeric Nanocarriers for Drug Delivery Applications, Volume 1: Types and Triggers (Woodhead Publishing Series in Biometerials); Abdel Salam Hamdy Makhlouf, Nedal Y. Abu-Thabit, Eds.; Woodhead Publishing, an imprint of Elsevier: Duxford, United Kingdom, 2018, pp. 555-580 (ISBN: 978-0-08-101977-9).
Julia Martin, Enrique Jacobo Díaz-Montaña, Agustín G. Asuero, “Chapter 1:  Cyclodextrins: Past and Present,” In Cyclodextrin: A Versatile Ingredient; Poonam Arora, Neelima Dhingra, Eds.; IntechOpen: London, United Kingdom, 2018, pp. 3-44 (ISBN: 978-1-78923-068-0).
A. Bibby, L. Mercier, "Adsorption and separation of water-soluble aromatic molecules by cyclodextrin-functionalized mesoporous silica," Green Chem. 2003, 5, 15-19 (DOI: 10.1039/B209251B).
Avilasha A. Sandilya, Upendra Natarajan, M. Hamsa Priya, "Molecular View into the Cyclodextrin Cavity: Structure and Hydration," ACS Omega 2020, 5(40), 25655–25667 (DOI: https://doi.org/10.1021/acsomega.0c02760).
Chang Cai Bai, Bing Ren Tian, Tian Zhao, Qing Huang, Zhi Zhong Wang, "Cyclodextrin-Catalyzed Organic Synthesis: Reactions, Mechanisms, and Applications," Molecules 2017, 22(9), 1475 (17 pages) (DOI: https://doi.org/10.3390/molecules22091475).
Lajos Szente, Ashutosh Singhal, Andras Domokos, Byeongwoon Song, "Cyclodextrins: Assessing the Impact of Cavity Size, Occupancy, and Substitutions on Cytotoxicity and Cholesterol Homeostasis," Molecules 2018, 23(5), 1228 (15 pages) (DOI:  https://doi.org/10.3390/molecules23051228).
Mikhail V. Rekharsky, Yoshihisa Inoue, "Complexation Thermodynamics of Cyclodextrins," Chem. Rev. 1998, 98(5), 1875-1918 (DOI: https://doi.org/10.1021/cr970015o).


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Comparing the zero point vibrational energies of DCl and HCl

(a) Compare the zero point energy (i.e. $E_0)$ in joules for $\ce{H^{35}Cl}$ and $\ce{D^{35}Cl}$ (assuming both are anharmonic oscillators).
(b) Explain the effect this will this have on their chemistry (you may wish to use a diagram to help here).

I have calculated a value for the vibrational frequency and have a value of the anharmonicity constant for $\ce{H^{35}Cl}$ (but not the anharmonicity constant for $\ce{D^{35}Cl}).$ I am unsure on how to proceed with these as all I know is an equation linking vibrational energy to the vibrational frequency and anharmonicity constant. I would be so grateful for any insight into this.

Hint: look at the values of H$_2$ and D$_2$ in the X state and see if you can find the mass scaling of $\omega_e$ and $\omega_e x_e$.

Marina, I have edited the answer a little bit, the previous answer was slightly wrong btw.

I don't know which equations you are using, but this is the one I was taught:
$$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$
Here, $\tilde{\nu}$ is the wavenumber of the vibration; $v$ is the quantum number that represents the vibrational levels $(v= 0, 1, 2,\ldots);$ $\chi$ is the anharmonicity constant:
$$\chi=\frac{\tilde{\nu}}{4\,D_\mathrm{e}},\tag{2}$$
where $D_\mathrm{e}$ is the depth of the lowest point of the potential well (not the depth of the $v = 0$ vibrational level!).
The zero-point energy is the energy when $v=0$ i.e. the molecule is in the lowest vibrational state. So you have to substitute that in the energy equation.
Now, we can assume that the well depth $(D_\mathrm{e})$ of the $\ce{H-Cl}$ and $\ce{D-Cl}$ bonds are same.
The only difference between $\ce{H-Cl}$ and $\ce{D-Cl}$ would be in the value of the vibrational wavenumber $(\tilde{\nu}),$ which you can get from the vibrational frequencies that you already have, as you wrote in the question. Once you substitute that in the equation, you will get your answer.

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


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Should Birge–Sponer plots always have a negative gradient?
I have a set of data I would like to use the Birge–Sponer extrapolation with, but the data is the wavenumbers of the bands in the vibrational-electronic spectrum of a diatomic molecule for $v' = 0, 1, 2, 3, 4, 5, 6.$
I thought the $y$ axis should have the wavenumber values given as they are (in $\pu{cm^-1}),$ but the values given increase with $v'$ giving a positive gradient of the graph.
Every plot I have seen has had a negative gradient. Have I interpreted what should go on the $y$-axis wrongly?

Yes, they should have a negative gradient. The energy gap between each level decreases as the vibrational quantum number increases. Maybe your data is for the energy difference between $v=0$ and $v$ as opposed to the difference between the levels?

Yes! I apologise if this seems a basic question. I am finding this field quite tricky- how should I think about converting this to the energy gap between the levels?

Don't apologies, we're all here to learn!!! To get the energy gaps between each of the levels you will need to subtract the values to leave you the difference. I would recommended drawing a diagram. If you need more help I can write a longer answer explaining how to get each value.

Let me know when you've got your full plot all looking good :)

Yay! I think I have it. It certainly looks nice and gives sensible values. Thank you again for your help, I wouldn't have understood this properly at all without you.

Let us continue this discussion in chat.


text:
Getting an “index” of 7 for a set of powder diffraction data
I tried to index powder diffractogram from the following problem:

A metal (of density $\pu{0.862 g cm^-3})$ gives the following powder X-ray diffraction data $(\lambda = \pu{1.5418 Å}).$
$$2\theta \qquad 23.6^\circ \quad 33.6^\circ \quad 41.5^\circ \quad 48.3^\circ \quad 54.4^\circ \quad 60.1^\circ \quad 65.5^\circ $$

I calculated $\displaystyle d = \frac{\lambda}{2\sin\theta}$ and compiled a table:
$$
\begin{array}{cclccc}
\hline
2\theta/^\circ & d/\pu{Å} & \frac{1}{d^2}/\pu{Å^-2} & \frac{1}{d^2}/Z & h^2 + k^2 + l^2 & hkl \\
\hline
23.6 & 3.769754932 & 0.07036776606~(= Z) & 1 & 1 & 100 \\
33.6 & 2.667180594 & 0.1405708123 & 1.997659159 & 2 & 110 \\
41.5 & 2.175894723 & 0.2112147493 & 3.001583838 & 3 & 111 \\
48.3 & 1.884256436 & 0.2816566342 & 4.002637145 & 4 & 200 \\
54.4 & 1.686509508 & 0.3515785836 & 4.996301621 & 5 & 210 \\
60.1 & 1.539473673 & 0.4219446333 & 5.996277229 & 6 & 211 \\
65.5 & 1.425021033 & 0.4924446812 & 6.998157094 & 7 & \color{red}{?} \\
\hline
\end{array}
$$
I know a value of $h^2 + k^2 + l^2 = 7$ is not allowed, but I cannot understand where I am going wrong with the indexing here. Any insight would be greatly appreciated as I am wondering if this is made up data that has a typo?

Assuming metal = element (which may be wrong), and this table here, did you give potassium a trial (COD, a = 5.247 A), bcc?  And, what is the source of this table you cite?

I'm not sure there is any problem as the unit cell parameters were assumed to be identical, such as in the case of a cubic space group. How about in the case of a tetragonal space group?

@z1273 Three nitpicks: #1 a space group (typically one of the 230 standard settings, like $P2_1$, or $P2_1/c$) may belong to a crystal class (like for the two mentioned, both monoclinic). #2 The tetragonal class is of lower symmetry than the cubic one. It is possible to describe anything in monoclinic $P1$, the lowest symmetry.  But even then, one strives to identify evidence for higher symmetry in the data (some computer programs proceed this way early / lately, e.g. Platon's ADDSYM).

@z1273 #3:  It is possible the experimental data leading to the above mentioned COD .cif entry differ to the (maybe constructed) diffractogram of the above question differ.  If however the above mentioned .cif is feed, e.g. into CCDC's Mercury, after adjusting for this approx Cu K$\alpha$ radiation, then the position of the predicted diffraction peaks along $2\theta$ are very close to the ones predicted.  (Only, that the diffraction indices don't match the ones in the OP ...)  By this & the proximity of the density stated in the OP and in the .cif (calculated Rx density of 0.859), K is likely.

@Buttonwood These are many interesting bits of information, but I'm not sure of their relevancy to the question at hand. I am also leery of coincidence, such as the stated density matching K's figure. It seems to me a≠c is a simpler explanation.

@z1273 True, #1 and #2 are not required here, address a meta level.  And on occasion, with the information given, there may be more than one solution.

TL;DR: The data from the assignment is correct.
You confused yourself by assigning $h_i^2 + k_i^2 + l_i^2$ to the ratio of $\sin^2\theta_i$ values, and keep in mind $(100)$ might be absent.

First, let's get bold and assume that we are indexing a cubic system.
This is usually the first step in analytical indexing, plus it's an extremely common system for metals.
Combining Wulff–Bragg's condition
$$2d_i\sin\theta_i = n\lambda\tag{1}$$
with the relationship between the quadratic form of the relation between interplanar spacing $d_{hkl}$ and the cell parameter $a$
$$\frac{1}{d_i^2} = \frac{h_i^2 + k_i^2 + l_i^2}{a^2},\tag{2}$$
the following relation between the scattering angles and Miller indices is evident:
$$\sin^2\theta_i = \frac{\lambda^2}{4a^2}\left(h_i^2 + k_i^2 + l_i^2\right)\tag{3}$$
Since $h_i, k_i, l_i \in \mathbb{Z}$ and $\lambda^2/(4a^2)$ is a constant, normalized series of $\sin^2\theta_i$ ratios corresponds to the $h_i^2 + k_i^2 + l_i^2$ sums.
Your mistake was that you took normalized $d_i^2/Z$ values (by the way, $Z$ is the number of formula units in crystallography) as $h_i^2 + k_i^2 + l_i^2.$
$$
\begin{array}{cccccccc}
\hline
i & 2\theta/^\circ & d_i/Å & \sin^2\theta_i & \displaystyle\frac{\sin^2\theta_i}{\sin^2\theta_1} & h_i^2 + k_i^2 + l_i^2 & hkl & a_i/Å \\
\hline
1 & 23.6 & 3.77 & 0.042 & 1 & 2 & 110 & 5.33 \\
2 & 33.6 & 2.67 & 0.084 & 2 & 4 & 200 & 5.33 \\
3 & 41.5 & 2.18 & 0.126 & 3 & 6 & 211 & 5.33 \\
4 & 48.3 & 1.88 & 0.167 & 4 & 8 & 220 & 5.33 \\
5 & 54.4 & 1.69 & 0.209 & 5 & 10 & 310 & 5.33 \\
6 & 60.1 & 1.54 & 0.251 & 6 & 12 & 222 & 5.33 \\
7 & 65.5 & 1.43 & 0.293 & \color{red}{7} & 14 & 321 & 5.33 \\
\hline
\end{array}
$$
As you can see, we have a number seven again which is … good news!
Not only our initial suggestion to go with the cubic system was right (integer relations), but we also stumbled upon $\sin^2\theta_7/\sin^2\theta_1 = 7$ thanks to the author of the assignment who was kind enough to give us at least seven peaks.
As you have rightfully suggested, it is impossible to find the sum of the squares of three integers for the number seven, but we don't have to — these are ratios, and can be multiplied by two.
At this point I pre-calculated cell parameter
$$a_i = d_i\sqrt{h_i^2 + k_i^2 + l_i^2}\tag{4}$$
to find the average from each signal, but thanks to the rounding error that wasn't necessary, and $a = \pu{5.33 Å}.$
Multiplying by two is equivalent to assuming that the first reflection in the diffractogram is obtained as a result of x-ray scattering from the atomic plane with indices $(110).$
At this point any number can be represented as the sum of the squares of three integers and we populate the table to the end with Miller indices.
Now thanks to that seven we also know what out of three cubic systems we have exactly: body-centered cubic (bcc) with $Z = 1 + 8 \cdot 1/8 = 2:$

This means we can check ourselves and deduce what metal has been analyzed by finding its molar mass $M$:
$$
\begin{align}
M &= \frac{\rho a^3N_\mathrm{A}}{Z} \\
  &= \frac{\left(\pu{0.862 g cm^-3}\right)\left(\pu{5.33E-8 cm}\right)^3\left(\pu{6.02E23 mol^-1}\right)}{2} \\
  &\approx \pu{39.3 g mol^-1},\tag{5}
\end{align}
$$
which leads us to believe the metal we analyzed was potassium with the density $\pu{0.89 g cm^-3}$ as reported by Wikipedia, which in my opinion is close enough to the value of $\pu{0.862 g cm^-3}$ from the assignment.
Final note: keep track of significant figures.
You are not allowed to present values with ten sig figs calculated from the scattering angles given with three sig figs accuracy.
“Indexing is more an art than a science” —A. Le Bail [1], aka the quote that crystallographers love to insert into presentations and emails for no reason:)
Reference

Bail, A. L. Monte Carlo Indexing with McMaille. Powder Diffraction 2004, 19 (3), 249–254. DOI: 10.1154/1.1763152.


text:
Why can you sanitize paper money notes with a Clothes Iron?
I have some Canadian and British bank notes. The first issue is that I don't know if they're paper or plastic.
Second, I don't understand this answer. Why's it safe  to iron paper bills? Won't the heat or pressure ruin the paper bill?

You can iron paper bills, like the American ones, on rather high heat. Just the same iron you use for your clothing, at the hottest setting, but no steam. It will kill all viruses.
Plastic bills can be hand washed, with the same soap you would use on your hand, you can dry them the same way you dry your hands, with a clean cloth or even the hot air dryer you use for your hands. Or you can let them air dry.
These bills can withstand some heat but I have heard not enough to iron them clean.

Ever hear of a book titled “Fahrenheit 451”? Paper money in the USA has to meet stringent requirements, by design, so a quick ironing, without steam, will not damage it. Of course, this is a testable matter: iron a one dollar bill. As for polymer currencies, I would not do that: too risky. But, again, this is testable.

@EdV Even this got an entry on wikipedia...

For the same reason that a laser printer doesn't ruin its printouts - paper can withstand the heat and pressure. Viruses can't

ps; the corona situation has people doing some crazy things.. but i suppose if one is sticking rolled up paper notes up one's nose for some reason, one would want them to be virus free, I'm sure!

In short: This is possible because «paper money» actually is not referring to the paper like paper from the office supply for (your) desktop printer or xerox, but to cotton paper.
Because of the cotton, if folded and forgotten in a pocket of your jeans close to $\pu{40 ^\circ{}C}$, they may physically survive a pass in the washing machine with a mild non-bleaching detergent (including the spinning).  They obviously suffer from the dye if e.g., the jeans blead, or mechanically if you try to unfold them after if passing a dryer.  Of course different to the ones made of polymers as (but not limited to) in Australia.

Among other things, USA paper money has to withstand submersion in 5% NaOH solution for 24 hours.

While I don't let my cloths this long in the tide (TM), it is good to know.

Honestly, "may physically survive a pass in the washing machine" seems like an understatement.  Dollar bills go through the wash just fine.


text:
Why does methane have 4-different bond dissociation energies?
I will elaborate on my question here.
In my Lecture Notes it is mentioned that for methane, there are 4 different $\text{C-H}$ bond energies. I plotted a quick graph of the bond energies to analyse the trend.

Here the vertical axis represents the bond dissociation energy while the horizontal axis represents the bond dissociation number. Looking at the graph, there is no clear trend so I decided to turn my attention to the structure.
$\text{CH$_4$}$ has the following structure

It is from here that I am unable to proceed. Here are my questions.

Am I barking up the wrong tree by looking at the structure and how do I account for the changing bond energies
Is there a VSEPR structure for molecules like $\text{CH$_3$}$ $\text{CH$_2$}$ $\text{CH}$? If so how do I figure it out.

All bonds in methane are not broken together. Say one C-H bond is broken by heteroolysis, it is no longer methane. It becomes a methyl cation, where C-H bond enthalpy is different. Likewise, the pKa values of the two hydrogen atoms in sulphuric acid are not same.

But why? It is still a bond between C-H @AdityaRoychowdhury

@SanjayAdhith because molecules are more complicated than localized bonds between parts of them. And even if everithing would be fully localized, which is quite true for molecules as CH4, CH3 ions or radical, and so on, the potential felt by electrons isn't / can't be the same.

VSEPR is consistent with the geometries of the various radicals, but the energetics are more difficult to unravel.
The following article gives this a shot:
https://www.sciencedirect.com/science/article/abs/pii/S2210271X14001285

Also, it looks like the 2nd and 3rd bond dissociation energy values are switched (see for instance the article referenced in the answer).

See Average Bond Energies: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/atomization.html

The dissociation energies are different because bonds don't all break at once
Most chemical processes happen step by step. Even in simple molecules like methane where there are only 4 bonds, bond dissociation almost always occurs stepwise.
The implication of this is that you would not expect each step to involve the same energy (and this applies whatever sort of bond-breaking occurs; some might create a $\ce{H+}$ and a $\ce{CH3-}$ others a $\ce{H.}$ and a $\ce{CH3.}$ ).
The issue is, in general, that the second bond-breaking step is happening to a different molecule to an isolated methane. Whatever the second step is, the starting point is not a complete methane. And, since the structure of a molecule affects the properties of all the bonds in it (though the effect might be small in a large molecule where the bonds are far apart), the second dissociation will, in general, be different from the first (and the bond energy will vary as a consequence).
While, hypothetically, a single concerted breaking of all 4 bonds could happen, this is likely to be vanishing rare or extraordinarily hard to achieve.  If you could somehow do that reaction, then you would measure a different result. But that isn't what chemists are usually talking about when measuring or reporting bond strengths so any such result would be a rare oddity unrelated to most achievable results.

Ah okay that makes sense. Am I right to say that the MO will change every time a bond is broken?

@SanjayAdhith yes. The bonding structure of a molecule changes when atoms are removed.

Bond energies are important, but you have to start somewhere. Actually, I'm surprised that the bond energies for the four dissociations are so similar. But that's a good start.
Now let's get into a bit more interesting compound: ethane. Very interesting!

The table is only part of the information on ethane given in an open access article (Ref 1). Note that the first C-H bond dissociation requires 422 kJ/mol, while the second requires 150 to 465 kJ/mol, depending on which bond you are breaking, because of what product you are making. Each line in the table is worth several paragraphs of explanation.
I admire your desire to tie up your knowledge neatly here - get a lot of string, because it's a very big package!
Ref 1. https://pubs.acs.org/doi/pdf/10.1021/acs.jpca.5b01346   Branko Ruscic, J. Phys. Chem. A 2015, 119, 28, 7810–7837

This is a well put together answer but I am afraid to say that I don't think that it answers OP's question.

@Nisarg: Thanks, you're right. I was answering the question so it satisfied me. In particular, I left out VSEPR calculations because the methane data really just say that the bonding between C and H is pretty uninteresting as far as CH4 goes.  CH4 is too simple. BUT, bond energies help us to understand more complex molecules, and the bigger picture involves looking not just at the original molecule, but also at the final product.


text:
Why is the disordered CuAu structure more stable at high temperatures?
I know that the disordered CuAu structure is based around FCC whereas the ordered CuAu structure is tetragonal. Why would the FCC structure be more stable at high temperatures?
For context, I was asked to explain this with the aid of a graph. I don't really know where to start

Pretty much all disordered things are more stable at high temperatures. Think of liquids vs solids, to begin with.

That makes sense. But why would a closely packed structure ever be more disordered than one which isn't?

Because every single atomic position in that close-packed structure has some freedom: it may be Cu, or it may be Au. That's a lot of disorder.

Fabulous , I understand this. Thank you!

I'm not a metallurgist or even not a good physical chemist. However, I'd like to explain what's happening in $\ce{AuCu}$ alloy with temperature and any expert can intervene.
According to Ref.1, $\ce{Cu}$ and $\ce{Au}$ are both univalent group IB metals with the atomic size difference ~12%. The $\ce{AuCu}$ alloy forms the tetragonally distorted fcc lattice where alternate (00h) planes contain either $\ce{Cu}$ or $\ce{Au}$ atoms and cause a contraction in c-direction. Resulting tetragonal face-centered structure has c/a ratio of 0.92 $(\frac{c}{a} = \frac{\pu{367 pm}}{\pu{396 pm}} = 0.93)$. In the temperature range ~$\pu{380 ^\circ C}$ to $\pu{410 ^\circ C}$, the superlattice $\ce{CuAu}$-II is formed, which consist of $\ce{CuAu}$-bct unit cells with the antiphase domains along the b-direction (Ref.2). There is a lattice shift of $\frac12(a+c)$ at each five unit-cell length. The superlattice $\ce{CuAu}$-II is described as orthorhombic cell with 10 cells along one of a direction, $oI40$.

Suggested reading: Ref.3 and 4.

References:

Valentina F. Degtyareva, Nataliya S. Afonikova, "Simple Metal and Binary Alloy Phases Based on the fcc Structure: Electronic Origin of Distortions, Superlattices and Vacancies," Crystals 2017, 7(2), 34 (13 pages) (DOI: https://doi.org/10.3390/cryst7020034).
V. Ozoliņš, C. Wolverton, Alex Zunger, "Cu-Au, Ag-Au, Cu-Ag, and Ni-Au intermetallics: First-principles study of temperature-composition phase diagrams and structures," Phys. Rev. B 1998, 57(11), 6427 (DOI: https://doi.org/10.1103/PhysRevB.57.6427).
M. Sanati, L. G. Wang, Alex Zunger, "Adaptive Crystal Structures: $\ce{CuAu}$ and $\ce{NiPt}$," Phys. Rev. Lett. 2003, 90(4), 045502 (DOI: https://doi.org/10.1103/PhysRevLett.90.045502).
S. -H. Wei, A. A. Mbaye, L. G. Ferreira, Alex Zunger, "First-principles calculations of the phase diagrams of noble metals: Cu-Au, Cu-Ag, and Ag-Au," Phys. Rev. B 1987, 36(8), 4163 (DOI: https://doi.org/10.1103/PhysRevB.36.4163).


text:
Ought I add salt to water bottles, if I'm putting the bottles in front of a fan to improvise an air conditioner?

Is this advice correct? Are they alluding to Freezing Point Depression? Adding salt to tap water lowers the Crystalloid's freezing point.

Pour 3 tbsp (51 g) of salt into each of your 3 plastic water bottles. Use disposable plastic bottles for the easiest set-up and clean up. Pour 3 tablespoons (51 g) of table salt per bottle. Put the caps back on and shake the bottles to thoroughly mix the salt.

I live in a studio flat. I know of the Law of Thermodynamics, also here  – but my plan is different. Where I live, night time is way cooler than day time. I'll freeze the bottles over night, because even if my freezer heats up more than usual, my night room temperature will still be cool enough. During day time when it's hot, I'll place the frozen bottles  in front of my fan to air condition my studio flat. Does this work? I read Quora.

It makes sense physically, but the question is if you would have enough bottles for that for sufficient effect.
Twice water  heat capacity than ice about compensate lower amount of water, so it can be considered about the same.
The major effect is increased heat transfer keeping it colder for most of the time, as melting, absorbing the most of heat, would occur at lower temperature. Also, liquid provides better heat transfer due convection.
The strongest effect would be using as much salt for solution to still get frozen in the freezer.
About 330 g / 1 L water makes solution with freezing point near $\pu{-20^{\circ}C}$.
50 g / 1.5L PET would be like $\pu{-2^{\circ}C}$.
If we take the heat for ice melting as 100%, water warming up by $\pu{12^{\circ}C}$ takes  15% and ice warming up by $\pu{16^{\circ}C}$ takes just 10%. Shifting melting point by $\pu{2^{\circ}C}$ causes increasing total water heat capacity by 1.25%.


text:
What is left over when water and sodium hypochorate decompose, i.e. household bleach?
I am having some confusion in understanding the decompostion of household bleach. One website says bleach decomposes to  salt and sodium chlorate, simple enough.
Orther wedsites say bleach decomposes to salt and nothing else. I called Clorox customer service and their vote is it converts to salt only. I have 3 votes for salt only and 2 votes for salt and sodium chlorate thus far.

@Nilay Ghosh.  Yes my motivation for my question was a safely concern as you guessed. Thank you for the link. It is very helpful

It depends on temperature. At higher temperatures, hypochlorite has tendency to disproportionate: $$\ce{3 ClO- -> 2 Cl- + ClO3-}$$ Otherwise it mostly releases atomic oxygen : $$\ce{ClO- -> Cl- + O}$$
In reality, it would be combination of both in various ratio, depending on relative reaction rates,  all ending as mixture of sodium hydroxide, carbonate, chloride and chlorate.


text:
Finding the concentration from a dilution
This must be a very basic question but I have gotten confused over it.
How would I find the concentration of each component of the reaction for each experiment from this data? I am guessing it can’t be the concentrations given, is that for some stock solution?

Solving one out of the four would be good enough to understand the concept behind this. Therefore, I shall explain the first experiment and the concentrations of the components in the solution produced.
In experiment 1 all four are added in equal quantities. Finding the total volume, we get the total volume
$$V_\text{tot} = \pu{4 ml}.$$
Assuming the amount of $i$th solute to be denoted by $n_i,$ its molarity $c_i$ can be found as
$$c_i = \frac{n_i}{V_\mathrm{tot}}.$$
Now, we add $\pu{1 ml}$ of $\pu{2.0 M}$ acetone. This is equivalent to saying we add $1/1000$ of $\pu{2 mol}$ of acetone. Therefore we add $\pu{2 mmol}$ (millimoles) of acetone.
This means that we have $\pu{2 mmol}$ of acetone in $\pu{4 mL}$ of water. Using the formula for molarity we get that acetone is in $\pu{0.5 M}$ in solution.
Hope you can follow on for the other parts as well, the concept remains the same.


text:
How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas?
How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas? What are the conditions of that happening? The related underlying chemical reactions are hydrogen and butane combustion in oxygen, right? What is the reaction mechanism? We should avoid using any electrical appliances, electric light, electric door bells and smartphone , just anything that could produce sparks and turn off any flames, right? So I should turn off the electrical appliances immediately after noticing a leakage or should I just keep them on?

Aside of open flame and operating electronic devices, another danger hard to eliminate is static electricity, mostly due synthetic material of clothes, or rubber soles of shoes.
The safest and the most available prevention choice at emergency scenario is intense ventilation. All others are rather hope it would be enough, as it is hard to eliminate or not to forget all possible ignition sources.
Be aware that switching devices off may be more dangerous than keeping them on, as interrupting high voltage  AC circuit usually creates sparks, unless this is addressed by parallel capacitors. This is typical for devices transforming AC to  low voltage, like most low power electronics.

Coal gas is of little use these days, with major fuel components  carbon monoxide and hydrogen.
Much more common is natural gas, based on methane, or LPG, based on butane (typically 80%) and propane.
The oxidant is in all cases aerial oxygen.

the chemical element(s)/compound(s) (reactants) involving are hydrogen and oxygen in the case of coal gas and alkane and oxygen in the case of LPG gas, right?

See the answer update.


text:
Are the orders of reactants with respect to a reaction different for different starting concentrations?

If the reaction is repeated with $\pu{2 M}$ ethyl iodide the pyridine concentration decreases as shown below. Give the rate law of the reaction in terms of pyridine and ethyl iodide. Explain your reasoning.
$$
\begin{array}{l|rrrrr}
\hline
t/\pu{min} & 0 & 20 & 40 & 80 & 120\\
\hline
[\ce{C5H5N}]/\pu{M} & 0.1 & 0.0746 & 0.0559 & 0.0312 & 0.0175\\
\hline
\end{array}
$$

I have found the rate constant and found that the orders of ethyl iodide and pyridine are both $1$ with respect to the reaction when their initial concentrations were both $\pu{0.1 M}.$ I would have thought this would be the case for any set of initial concentrations.
I would like to confirm this. In that case, I am thinking the only thing different would be the rate constant. But how would I go about finding that?
Could I plot the data, interpret the half life and go from there? Or is my approach totally wrong?

the order of a reaction is independent of starting concentration, if it were so then order would have kept changing at every moment sinc each moment can be considered as a starting point.

When the concentration of one reactant is much higher than the other, then it can appear to be zero order in the high concentration reactant because the percent change in that concentration is so small that it has a negligible effect on the overall rate. But it is still first-order in that reactant even if it's not observable under those conditions

This makes sense, thank you both. The rate constant would be a different value though wouldn’t it?

Disregard that. I read it over and realise that wouldn’t make sense

A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of  reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$
It seems that this reaction is a second order reaction since the rate order appears to be:$$\text{Rate} = k[A][B] $$
As @Andrew astutely pointed out in the comments, a reaction can be of a lower order if one of the reactant's concentration is much higher than the other's.
Let's suppose, that one reactant is in great excess (say $10 \text{M of B vs } 0.01\text{M A}$)
So what we're observing now is that the reactant B is barely consumed, yet the reaction terminates since reactant A, the limiting reagent is exhausted.
In other words, the rate equation looks something like this: $$\text{Rate}=k'[A]$$, where $k^{'}=k[B]_o$ .Now, the reaction appears to be a first order reaction. This is known as a pseudo first-order reaction, since the rate of the reaction ultimately depends only on one reactant.
An example that I found here:
$$ \rm{CH_3Br + OH^- \rightarrow CH_3OH + Br^-}$$

[...]"Imagine we had an initial concentration of CH3Br of 100 μM and and an initial concentration of OH- of 10 mM.   Now even if all of the CH3Br has reacted the concentration of OH- will be essentially unchanged.  Therefore during the course of the reaction, the concentration of OH- will be essentially constant.  This makes the reaction "like a first order reaction", thus the name pseudo-first order."

Effectively, the rate now is: $$\ce{rate= k'[CH_3Br]}$$
This can also be mathematically represented in the integrated rate law:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} = kt \quad \ce{where [OH-]\neq[CH3Br]}$$
Now, since $\ce{[OH-]_0 \gg [CH3Br]_0}$, we can write the equation as:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} \approx \dfrac{1}{\ce{[OH-]}}\ln \dfrac{\ce{[CH3Br]_0}}{\ce{[CH3Br]}}=kt $$
Remember, however that the rate and order of a reaction usually depends upon its mechanism.
For further reading, check here and here as well.


text:
Using the steady state approximation in the decomposition of nitramide to determine rate law

$\ce{O2NNH2 (aq) ->[\mathit{k}_1] O2NNH- (aq) + H+ (aq)}$
$\ce{O2NNH- (aq) + H+ (aq) ->[\mathit{k}_{-1}]  O2NNH2 (aq)}$
$\ce{O2NNH- (aq) ->[\mathit{k}_2] N2O(g) + OH- (aq)}$
$\ce{H+ (aq) + OH- (aq) ->[\mathit{k}_3] H2O(l)}$
We assumed the first two steps were fast and gave a rapid pre-equilibrium, while the third step that generated $\ce{N2O}$ was slow. The last step is very fast.
(i) For this question, use the steady state approximation for [$\ce{O2NNH-}$], assuming it is a reactive intermediate, to derive an expression for the rate law.

 The solution given is
 $$\frac{\mathrm d [\ce{N2O}]}{\mathrm d t} = k_\mathrm{obs}\frac{\ce{[O2NNH2]}}{\ce{[H+]}}$$

My attempt at the solution is as follows:
$$\frac{\mathrm d[\ce{N2O}]}{\mathrm dt} = k_2\ce{[O2NNH-]}$$
Using the steady state approximation,
\begin{align}
\frac{\mathrm d[\ce{O2NNH-}]}{\mathrm dt} = k_1[\ce{O2NNH2}] - &k_{-1}[\ce{O2NNH-}][\ce{H+}] - k_2[\ce{O2NNH-}] = 0 \\
\ce{[O2NNH-]} &= \frac{k_1[\ce{O2NNH2}]}{(k_{-1}\ce{[H+]} + k_2)} \\
\frac{\mathrm d[\ce{N2O}]}{\mathrm dt} &= \frac{k_2k_1[\ce{O2NNH2}]}{(k_{-1}\ce{[H+]} + k_2)}
\end{align}
I have done this so far, but this doesn’t seem right to me. I’ve been looking over at it for seemingly ages but I can’t see where I am going wrong or if I have misunderstood something about the steady state approximation. Any insight would be greatly appreciated.

https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
This should help for future reference.. I'll fix the text up this time. Take care that you format the text so it is easier for people with text readers.

You must next assume $k_2$ is small wrt $k_{-1}[H^+]$  to get the answer given.

How have you come up with your equation for the production of N2O, the one just under "Using the steady state approximation". The last term, −k2[O2NNH−], is telling me that the rate of production of N2O decreases if there is more O2NNH−, do you think this can be right? Also apply the steady state approximation to the fast equation - the whole point is that the quick ones reach steady state compared to the slow one.

That should be for the rate of [O2NNH-], apologies someone edited this and I don’t know how to change it

@MarinaCalder - ah, OK, that makes sense now. I've submitted an edit correcting, hopefully it will get accepted soon. In that case the earlier comment by porphyrin has the answer for you


text:
Why is Ka of weak acid increasing while being diluted?
I am calculating $K_\mathrm{a}$ and percent dissociation of acetic acid being diluted to $\pu{1.00 \times 10^{-1} M}$, $\pu{1.00 \times 10^{-2} M}$, and $\pu{1.00 \times 10^{-4} M}$.
The $\mathrm{pH}$ goes up as the weak acid is diluted, which is to be expected.
The % dissociation is inversely related to the initial concentration of acetic acid.
However, I fail to understand why the $K_\mathrm{a}$ value increases (slightly) as the solution gets more dilute. Isn't $K_\mathrm{a}$ supposed to be a constant (acid dissociation constant) therefore stay constant as long as the temperature stays constant? Isn't according to Le Chatelier's principle, $K_\mathrm{a}$ is a function of temperature?

The simplest solution is that your calculation is wrong or not accurate enough. As you say, it's a constant.

The scale at which you are working, the standard formulas won't work. You will have to solve a quadratic equation to get the results. The standard formulas are just an approximation to make our lives easy.

Have you involved ion activity coefficients ? How exactly have you calculated Ka ? Have you applied any implicit or explict simplifying assumptions ?

E.g. the famous formula pH=0.5(pKa - log c) assumes c >> [H+] >> [OH-].

Can you please edit the question to include the readings you got during the experiment. That might help us understand where the problem is, approximations or just experimental error.

$K_a ≠ x^2/0.100$, rather $K_a ≈ x^2/0.100$. This expression assumes the condition mentioned by @Poutnik. Divide by $0.100-x$ instead. Even after that the results might me screwed because when $x<10^{-6}$,  even self ionization of water causes significant change in $\mathrm{pH}$.

You should rather involve in the question the used algebraic equations explicitly.   A claim not making assumptions may be an assumption itself, not necesserily correct. By other words, if you do not say what you did, we cannot say what you did wrong.

Try to make it complete. And be aware of x^2/(c-x)  not x^2/c-x .Consider also Activity coefficients. Another case is oH meter calibration. How did you do it for low pH values ?

Remember that with increasing ionic strength, activity coeefficients decrease, according to Debye-Hueckel equation. So while activity based Ka is constant, concentration based Ka grows with the acid and therefore ion concentration.


text:
Flavonoid Xanthine Oxydase Inhibitors vs Hydroxychloroquine
Regarding COVID-19 and hydroxychloroquine (HCQ), HCQ is known to increase pH in lysosomes by 1-2 orders of magnitude, and also results in lethal intracellular heme overload in parasites responsible for hemolytic diseases.  Question is, if circulating levels of xanthine oxidase (XO) are also known to be increased in hemolytic diseases, if one considers the natural flavonoids Kaempferol, quercetin, or luteolin, which are known xanthine oxidase (XO) inhibitors, antivirals, and anti-oxidants, what would be different about pharmacology of natural XO inhibitors and HCQ?
Is HCQ an XO inhibitor because it results in heme overload in parasites?  What about anti-oxidant properties of HCQ in terms of radical scavenging?  There is no information supporting the notion that HCQ is an antiviral.
Sticking with hemolytic diseases, for which the goal is overload of intracellular heme in parasites, what would happen if a human took high levels of an XO inhibitor that wasn't HCQ during malaria?
XO inhibitors usually result in an overall decrease in uric acid, essentially rendering them as a natural supplement for gout -- i.e., replacement to allopurinol, which has horrendous side effects.

HCQ does show antiviral properties in vitro, but, in vivo (in humans), when taken after an initial dose of 800mg for 4 days at 600 mg/day, it was not found to prevent COVID-19 in high-risk exposure settings.  Thus, it didn't stop people from getting SARS-CoV-2 infection followed by onset of symptoms/severity.
I suspect this is exactly what would happen with flavonoids, many of which have shown antiviral properties in vitro (laboratory settings).  The information on flavonoids and SARS-CoV (2003) is very sparse, but several flavonoids did inhibit SARS-CoV in vitro (see last paragraph below).  I do know there was a clinical trial approved in 2020 by NIH in which the investigators received an IND (From FDA)  to use diosmin as a supplement to treat patients with COVID-19 (i.e. hospitalized).  However, I don't know if was completed.  Diosmin is available OTC as a supplement, and used via prescription under the brand names Dafflon® and Vasculera® for chronic venous insufficiency (CVI), which is caused by venous hypertension (VH). VH, in turn, aggravates metabolic imbalances leading to a self perpetuating cycle of further metabolic changes, including venous acidosis. These changes promote further inflammation in vascular tissue leading to edema, skin damage and possible ulceration and deep vein thrombosis (DVT). Varicose veins and hemorrhoids are also part of the spectrum of CVI disorders.
Interestingly, it took 16 years before a group published the most comprehensive paper on in vitro inhibition if SARS-CoV (2003) by many flavonoids that were considered.


text:
How to convert cell diagram to a chemical equation?
How would one convert on the right side of the cell diagram:
$$\ce{Zn(s) | Zn^2+(aq) || MnO4-(aq), Mn^2+(aq), H+(aq) | Pt(s)}$$
$$
\begin{align}
\ce{Zn(s) &-> Zn^2+(aq)} \tag{ox}\\
\ce{MnO4-(aq) &-> Mn^2+(aq)}\tag{red}
\end{align}
$$
Obviously, one would then balance and combine the half-equations. I just need to know if the unbalanced half-equations are correct.

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.
See also upright vs italic and Math SE MathJax tutorial.

The reduction «or something like that» does not make sense.  Most often classes introducing redox chemistry  refer to the reduction of permanganate $\ce{MnO^-_4}$ into $\ce{Mn^{2+}}$ if the reaction is conducted in acidic media.  Reductions in neuter, and basic media yield other products.

Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced.
The correct half cell reactions would be:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\
$$
And so the overall balanced reaction would come out to be:
$$\ce{5Zn + 2MnO^-4 + 16H+ -> 5Zn^{2+} + 2Mn^{2+} + 8H2O}$$

It was typed wrong on the tutorial sheet and I did not spot the error before submitting the question. The corrected cell diagram makes far more sense; before it was conflicting against cell diagram convention.

To clarify: The H+ on the RHS of the cell diagram simply means the reaction is being performed under acidic conditions?

@jseissmann Yes, it shows that the medium is acidic.

@jseissmann For a reaction equation to be balanced, the number of atoms on the right, and on the left hand side must be ... equal. Simultaneously, the sum of the charges on the left must equate the sum of the charges on the right hand side of the arrow, too. The reduction, as written by you: $\ce{ MnO^-4 -> Mn^{2+} }$. Left hand side: four atoms of oxygen, right hand side: none; thus not balanced.

@NisargBhavsar This is true, I forgot to set the @ sign; the comment was to the OP.

@Buttonwood I am aware they are unbalanced - I simply wanted to check that my translation of the cell diagram to the unbalance half-equations was correct as the H+ on the RHS threw me. I can balance half-equations just fine and therefore didn't need full answers on that part