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Are nucleophilic substitutions reversible?
When $\ce{OH-}$ interacts with a halogenoalkane, for instance $\ce{CH3CH2Cl}$, it acts as a nucleophile and undergoes $\mathrm{S_N2}$ mechanism, which results in the following reaction:
$$\ce{CH3CH2Cl + OH- -> CH3CH2OH + Cl-}$$
Since $\ce{Cl-}$ has a free pair of electrons, can it act as a nucleophile that will reverse the previous reaction?
$$\ce{CH3CH2OH + Cl- -> CH3CH2Cl + OH-}$$
Since an $\ce{OH}$ group is more electronegative than $\ce{Cl}$ group, the carbon atom to which it attaches would have a greater partial positive charge ($\delta+$), does this mean it's more prone to nucleophilic substitution?

Can such nucleophilic substitution occur on halogenoalkanes with fluorine groups? My textbook says it occurs very slowly, does this mean it occurs?

A nucleophilic substitution will occur only when the nucleophile approaches the substrate and also attack at the carbon to displace the leaving group. More partial positive charge means that the chloride might solvate the substrate slightly better as compared to OH- in the alkyl halide case, but due to higher electronegativity and charge density of the former,hydroxide will be the species which "actually attacks" better

Fluorine is a very bad leaving group,so you can't really use simple SN2 on fluoroalkanes for getting an appreciable yield

In general, yes, $\mathrm{S_N2}$ reactions are reversible. But not in this particular case.
Rates of $\mathrm{S_N2}$ reactions depend on several factors: the nucleophile, the leaving group, the alkyl group undergoing substitution, and so on. In the context of your question, the leaving group ability is quite possibly the most important. Iodide, $\ce{I-}$, is an excellent leaving group. Fluorine, $\ce{F-}$, is an extremely poor leaving group. Again, there are several reasons which dictate leaving group ability, but at the simplest level, it has to do with the stability of the negative charge on the leaving group. This broadly correlates with basicity: iodide is hardly basic at all (its conjugate acid, $\ce{HI}$, is a very strong acid) and is quite happy to exist as $\ce{I-}$. Conversely, fluoride is reasonably basic (its conjugate acid, $\ce{HF}$, is a weak acid) and is not all that happy existing as $\ce{F-}$.
Using this logic you will find that hydroxide ion $\ce{OH-}$ is a very poor leaving group. Consequently, it's almost impossible to get ethanol (or other alcohols) to directly react in an $\ce{S_N2}$ fashion. So in the particular case you have raised, no, the reaction is effectively irreversible.*

The way around the leaving group problem is to convert the hydroxyl group to something that leaves better. Because alcohols are common, lots of people have devoted effort into making sophisticated ways of doing this. But by far the simplest way is to simply protonate it:
$$\ce{ROH + H+ <=> ROH2+}$$
The leaving group now is not hydroxide $\ce{OH-}$; it's now water $\ce{H2O}$, which is a much, much better leaving group. Consequently, if you treat alcohols with $\ce{HCl}$, you have a much better chance of getting an alkyl chloride, as opposed to if you were to just add chloride ion (in the form of e.g. $\ce{NaCl}$). The $\ce{H+}$ protonates the alcohol, and the $\ce{Cl-}$ does the actual substitution.

For an example of a reversible $\mathrm{S_N2}$ reaction, the best example is one where the nucleophile is the same as the leaving group. For example, consider a chiral alkyl iodide $\ce{R1R_2R_3CI}$. Let's say we have an enantiomerically pure sample of this alkyl iodide. Since $\mathrm{S_N2}$ reactions occur with inversion, if we treat it with sodium iodide, can we effect an $\mathrm{S_N2}$ reaction which will convert it entirely to the opposite enantiomer ?
The answer is no, and that's precisely because the reverse reaction will also occur. Both the forward and reverse reactions have the same nucleophile and the same leaving group, so in this case there aren't any considerations which would make one less effective than the other.

* "effectively" because technically speaking every reaction is reversible.

Hoe is every reaction reversible? Are reversible reactions ideal by definition and thus not possible to realise in real life? Like so for reversible processes

no, the usage of reversible here is not the same as the idea of a reversible process. Reversible reactions simply refer to those which can happen in either the forward or backward direction, which in principle, is true of every chemical reaction (https://en.wikipedia.org/wiki/Microscopic_reversibility). Sometimes the rate of one far outstrips the other and the position of equilibrium lies very far to the left or to the right. Colloquially we call those "irreversible". Technically speaking they are still reversible in the original sense, though.


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What is the effect of polar protic solvent on nucleophilicty?
Question

Attempt:
I could figure out the answer correctly using simple rules governing nucleophilicity . But my doubt is how does polar protic solvent affect nucleophicity. And what would be the answer in case of polar aprotic solvent?. Please help.

Protic solvents affect nucleophilicity because they form a shell around the nucleophile via their hydrogen which has a positive partial charge and therefore interacts with it, thus the nucleophilicity will decrease. This effect is stronger for smaller nucleophiles as they can be "trapped" more easily. So taking the classical nucleophiles you obtain this order:
$$\ce{I^-} > \ce{Br^-} > \ce{Cl^-} > \ce{F^-}$$
Aprotic solvents affect nucleophilicity in a more complex way. You can sometimes read that the previous order of nucleophile strength is inversed and small nucleophiles are favored:
$$\ce{F^-} > \ce{Cl^-} > \ce{Br^-} > \ce{I^-}$$
But be careful, this is not always true. There is more consideration to take.

A "soft" anion is highly polarizable and has more electron density to give, making it a good nucleophile compared to "hard" ones which are tightly holding their electrons.
This order is not respected by some polar aprotic solvents, acetone is the only one coming in my mind.

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What would the following NGP reaction yield?
Question

Attempt
I believe that in first reaction there would be simple Sn2. In the second reaction there Should be NGP. Hence i think the products should be diastereomers. But answer is (A)they are enatiomers. How?. Please help.

Reaction-1 is a simple example of an $\mathrm{S_N2}$ reaction, where the acetate group ($\ce{AcO-}$) substitutes the tosyl group ($\ce{OTs}$) forming the product (1) as follows:

But for Reaction-2, the lone pairs present on the $\ce{NMe2}$ provide anchimeric assistance in the removal of the tosyl group, leading to the formation of a bicyclic intermediate (both rings are not coplanar though). The acetate anion can now open the smaller ring in two ways and hence, lead to two different products labelled (1) and (2) as follows:

Finally,to deduce the enantiomeric relation between (1) and (2), the following transformations can be done:

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Rubbing alcohol gets warm after diluting with water
I've just diluted a 95% rubbing alcohol bottle with water to bring it down to 70%. However, the bottle got noticeably warmer after I did so. I've searched online and I couldn't find any exothermic reaction that happens when alcohol is mixed with water. I did use mineral water by mistake to do so, so my best guess is that alcohol reacted with... one of the salts in the water... maybe?! (I have no idea)
Why did this happen?

Why the downvote? It is a good question

Ethanol does the same. It is the consequence of deviation from the Raoult law.
The other consequences are existence of azeotropes ( affects distillation ) and mixing volume deviations. If you mix 1 L of ethanol and 1 L of water, it warms up. When it cools down back, its total volume is not 2 L, but about 1.9 L.
It is closely related to total energy of inter-molecular bonds in pure solvents and in their mixture.


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Which one is the correct molecule of the given properties of IR Spectroscopy?

$\ce{CH3CH=CHCO2CH3}$
$\ce{CH3COOCH=CHCH3}$

$\pu{2950 cm-1}, \ \pu{2870 cm-1}, \ \pu{1740 cm-1}$ are the observed peaks.
No peaks observed $\pu{1700-1500 cm-1}$.
The formula of the molecule is $\ce{C5H8O2}$.
My pre-answer is the first option. Because $\pu{1500-1700 cm-1}$ includes unsaturated carbons.

Does not correspond to the literature spectrum for 1  https://www.chemicalbook.com/SpectrumEN_623-43-8_IR1.htm

So, the first one is wrong. And, the answer is 2. Nothing else?@Waylander

I'm not a fan of IR as a diagnostic for structural determination, and have not used it much for the last 30 years. NMR is much stronger.

Thanks for your comments. My professor is awaiting replies from me.@Waylander

1740 is a bit low for the carbonyl of a vinyl ester according to my old copy of Williams & Fleming but it could be that signal


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XRD: Can stacking faults cause pseudo-centring?
Consider a cP lattice with a = 10 Å and an octahedron residing on each of the unit cell's vertices as shown below:

The maximum symmetry will be Pm-3 m.
Now, let's consider there are stacking defects along half of the diagonal of the unit cell's faces. Let's look at displacement by a translation of [0 0.5 0.5]:

If we look onto the bc plane, this will look like an A-centred cell:

If the stacking fault energy is low, random displacement along [0.5 0.5 0], [0.5 0 0.5], and [0 0.5 0.5] is to be expected. Will the situation described cause the powder X-ray diffraction pattern to resemble space group symmetry Fm-3 m, e.g. by integral extinctions of reflections where h + k, h + l or k + l is odd? From my perspective, the average structure the X-ray beam will 'see' is the following:

The outcome depends on the typical length between stacking faults and the coherence length of the X-ray beam.
If the length between stacking faults is large compared to the coherence length, you will get a phenomenon similar to that of twinning (unit cells in same orientation) or of broadened diffraction spots similar to the model of high mosaicity (unit cells in slightly different orientations).
If the length between stacking faults is small compared to the coherence length, you will get a phenomenon similar to partial occupancy of sites in an otherwise perfect crystal (i.e. static disorder - same diffraction pattern with different intensities compared to absence of partially occupied sites).
I found a statement about coherence lengths of synchroton radiation (https://www.researchgate.net/post/What_are_typical_requirements_on_coherence_lengths_for_coherence_experiments_using_3rd_gen_Synchrotron_Light_Sources):

[Keoki Seu, March 21, 2012] I work at a coherent scattering beamline at the Advanced Light Source. It is an undulator source optimized for coherent flux. At 500 eV, which is usually where we work, coherence lengths are on the order of 6 um vertical and 3-4 um horizontal. We'd love to have larger coherence lengths so we can coherently illuminate larger sample areas, but for the soft x-ray, the values we see here are typical values.

The question was about power patterns rather than single crystal diffraction.

[OP] Will the situation described cause the powder X-ray reflection pattern to resemble space group symmetry Fm-3 m, e.g. by integral extinctions of reflections where h + k, h + l or k + l is odd?

I don't know what the typical dimensions of the solids in the powder would be, and I don't know what the coherence length of the Xray radiation used in the experiments are, but if you consider all of these factors, you should be able to say in which regime (i.e. twinning-like or static disorder-like) the experiment at hand is in.

Thanks. I'm refering to measurements acquired using in-house instrumentation. I will try to rephrase your answer: We are trying to discriminate between (i) addition of scattered intensities (twins) and (ii) addition of amplitudes of scattered waves, thus allowing for interference (disorder). Due to added intensities, twins should never cause reflections to become extinct. The coherence length of the beam indicates the length regime in which interference can be expected. After reviewing the pXRDs, at least one very broad reflection of the P-centred phase seems to be present in every case.


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How to calculate bicarbonate and carbonate from total alkalinity
It's been a long time since I did my chemistry classes and I'm currently trying to analyze groundwater samples for hydrogeology purposes. I would like to evaluate carbonate and bicarbonate concentration from groundwater samples, but I only have values of total alkalinity as $\ce{CaCO3}$, $\mathrm{pH}$, and temperature. Is it possible? Some of the $\mathrm{pH}$ values are above 8.3.
I know that:

If $\mathrm{pH}$ is above 8.3: $[\mathrm{alk}_{tot}]=[\ce{HCO3-}]+2[\ce{CO3^2-}]+[\ce{OH-}]-[\ce{H+}]$
If $\mathrm{pH}$ is under 8.3: $[\mathrm{alk}_{tot}]=[\ce{HCO3-}]+[\ce{OH-}]-[\ce{H+}]$

With the $\mathrm{pH}$, I can find calculate $[\ce{OH-}]$ and $[\ce{H+}]$. But how can I calculate $[\ce{HCO3-}]$ and $[\ce{CO3^2-}]$?
Thank you!

I remember getting 2 values, for titration to  phenolphthaleinum ( if alkalic enough ) and methyl orange titration ends. The first was took for carbonates only and MO for carbonate + bicarbonate weighed sum. But it is my memory for chemical high school, focused on analytical chemistry in 1980-84 and subsequest undergrad lectures and labs.

Remember that Henderson-Hasselbalch provides the equilibrium ratio of concentrations at a given pH.

If I understood your question correctly, you have solutions where you know there is a given amount of calcium carbonate dissolved, and would like to know the distribution of this carbonate between all the species present. This proportion is commonly refered as the alpha($\alpha$) for a given species, that varies from 0 to 1(0% - 100%). From your question, I can make some assumptions:

As a groundwater sample, any solids dissolved are very diluted, so we don't need to worry about ionic strength;
pH is not fixed;
Temperature is not fixed, but I will assume its close to room temperature;
As other components are not mentioned, I will assume all carbonate comes from calcium carbonate.

Carbonic acid, $\ce{H2CO3}$, has two ionizable hydrogens, so it may assume three forms: The free acid itself, bicarbonate ion, $\ce{HCO3-}$(first-stage ionized form) and carbonate ion $\ce{CO3^2+}$(second-stage ionized form). The respective proportions in comparison with the total concentration of calcium carbonate dissolved are $\alpha0$, $\alpha1$ and $\alpha2$. They must sum to 1(100%), as in chemical reactions matter is neither created or destroyed, only changing between forms. When the calcium carbonate dissolves, a equilibrium is established between its three forms, expressed by the respective equilibrium equations:
First stage:
$$\ce{H2O + H2CO3 <=> H3O+ + HCO3-}$$
$$K1 = \frac{\ce{[H3O+][HCO3-]}}{\ce{[H2CO3]}} \approx 4.47*10^-7 $$
Second stage:
$$\ce{H2O + HCO3- <=> H3O+ + CO3^2-}$$
$$K2 = \frac{\ce{[H3O+][CO3^2-]}}{\ce{[HCO3-]}} \approx 4.69*10^-11 $$
You can also write a equation for the overrall reaction, by sum of each stage (and multiplication of the respective equilibrium constants):
$$\ce{2H2O + H2CO3 <=> 2H3O+ + CO3^2-}$$
$$K1K2 = \frac{\ce{[H3O+]^2[CO3^2-]}}{\ce{[H2CO3]}}$$
Analysing our system, to give a full treatment, if we know the solution pH, we can calculate $\ce{[H3O+]}$. So we are left with three unknown variables, $\ce{[H2CO3]}$, $\ce{[HCO3-]}$ and $\ce{[CO3^2+]}$. But so far we have only two independent mathematical equations, for K1 and K2 (the overrall equation does't count as independent, as it's only the merging together of the other two). To solve it, we need at least one more independent equation, to match the number of unknows. What we need is the equation for the material balance of the system. As we assumed all carbonate came from calcium carbonate, we can write:
$$Cs = \ce{[CaCO3]} = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$
Where Cs here stands for the known concentration of the salt, calcium carbonate. Now we can start replacing values taken from the equilibrium expressions into the material balance, isolating each unknow. For the bicarbonate, for example:
$$Cs = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$
$$Cs = \ce{\frac{[HCO3-][H3O+]}{K1} + [HCO3-] + \frac{K2[HCO3-]}{[H3O+]}}$$
$$Cs = \ce{\frac{[HCO3-][H3O+]^2 + K1[HCO3-][H3O+] + K1K2[HCO3-]}{K1[H3O+]}}$$
$$\frac{\ce{[HCO3-]}}{Cs} = \ce{\frac{K1[H3O+]}{[H3O+]^2 + K1[H3O+] + K1K2}} = \alpha1$$
So we got the expression for $\alpha1$, that has a curious structure: a fraction, where the denominator is a polynomial of degree 2, and the numerator its middle term. The same procedure can be repeated to find the expressions for the alphas  of the other dissolved species. For sake of brevity, I won't do it, but the final result will be:
$$\alpha0 = \frac{\ce{[H2CO3]}}{Cs} = \ce{\frac{[H3O+]^2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$
$$\alpha2 = \frac{\ce{[CO3^2-]}}{Cs} = \ce{\frac{K1K2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$
Note that a interesting pattern emerges. The expressions for the remaining two species have the same structure, just changing the term that goes in the numerator.
With the expressions for all species, it's helpful to use a spreadsheet to automate the calculations for a entire range of pH values, to grasp in a visual way what happens with carbonates as pH changes. I did just that, look at the results (here the spreadsheet, to whomever wants to download and play with it):

We see that in lower pH the predominant form for carbonate is the free carbonic acid. A bit over 6 bicarbonate ion takes over, and reigns up to pH a bit over 10, from where fully ionized carbonate ion takes over. For a given pH, the concentration of each species can be computed multiplying the respective $\alpha$ by the concentration of total calcium carbonate originally present.
The plot that looks like a "XX" also allows us to see a interesting property of carbonates. Nowhere in the plot you will find a pH value where we have the three species all in significant amounts. In the lower pH region you can find both bicarbonate and carbonic acid. But carbonate only shows up when carbonic acid goes away. It's like the unconfortable situation where you have two close friends who both hate each other. In a given moment I can see you in a room talking with either friend, but I will never see you three in the same room, or both friends of yours. The dividing line is close to the pH 8.6 you mentioned in your question. However, that sad situation has a upside. It makes the problem easier to calculate.
The full treatment I gave to this problem was indeed overkill. If I have three species, but only two show up together at any given time, I can "forget" I'm dealing with a diprotic acid. I need only to see the dividing line I've found, around pH 8.6. If I'm above it, free carbonic acid concentration is zero, and I have to deal only with the pair bicarbonate/carbonate, pretending the bicarbonate anion is just a monoprotic acid. From the equilibrium, we have:
$$\ce{[H3O+]} = \frac{\ce{K2[HCO3-]}}{\ce{[CO3^2-]}}$$
Or in logarithimic form:
$$pH = pK2 + log(\frac{\ce{[HCO3-]}}{[CO3^2-]})$$
This is the old Henderson–Hasselbalch equation you surely heard about before. As we know the pH and K2, we can calculate the ratio between carbonate and bicarbonate.
In the other side, if I'm below my dividing line near 8.6, carbonate ion concentration is zero, now I have to deal only with the pair carbonic acid/bicarbonate, pretending carbonic acid is just other monoprotic acid. From the equilibrium, we have:
$$\ce{[H3O+]} = \frac{\ce{K1[H2CO3]}}{\ce{[HCO3-]}}$$
Or in logarithimic form:
$$pH = pK1 + log(\frac{\ce{[H2CO3]}}{[HCO3-]})$$
As we know the pH and K1, we can calculate the ratio between carbonic acid and bicarbonate.
If you want to study in depth such calculations, I recommend this book: Butler, James N. Ionic Equilibrium: Solubility and PH Calculations. John Wiley & Sons, 1998.

Thank you so much! What if the temperature is lower than or higher than room temperature? Does it change the "K" values?

Yes, they do. If a exact result is desired, it's necessary to account for that, and use the constants corrected for the actual temperature. But unless the difference in temperature is big, the error will be probably acceptable. Look this question: https://chemistry.stackexchange.com/questions/9108/why-does-the-equilibrium-constant-depend-on-the-temperature-but-not-on-pressure


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Under what circumstances would azeotrope exist in non ideal solutions? (when would the dew point curve and the bubble point curve meet?)
[solved. For anyone who's interested in this topic, please view MaxW's comment under his/her answer. A thousand thanks to Poutnik and MaxW.]
I looked up several books on thermodynamics, and they all states that for non ideal solutions, the dew point curve and the boiling point curve sometimes meet at some point. Azeotropes forms there.
What I couldn't figure out was, under what circumstances would the two curves meet? Do they always meet if both of them have a local minimum? What is the logic under this phenomenon?

Edited: I corrected the wrong statement. The books just mentioned the meeting of the curve, but they didn't say "always".

Binary Mixture Which is an Ideal Liquid
Well let's start with what an ideal liquid binary mixture would be. An ideal liquid binary mixture would obey Raoult's law as shown in the figure below from the Wikipedia article.

A ideal liquid is a special cases of Raoult's law where the mole fraction of both phases form a straight line from the boiling point of A to the boiling point of B. Thus for an ideal liquid the dew point curve and the boiling point curve have exactly the same mole fraction at any temperature. Thus, by definition, an ideal liquid can not have an azeotrope.
Binary Mixture Which is an Non-Ideal Liquid
Now in the following figure there is a region bounded by drawing a horizontal line from A across the diagram, and a horizontal line from B across the diagram. Consider this the bounded region.

So the correct statement is that if the dew point curve goes outside the bounded region, then there must be an azeotrope.
The statement:

For non ideal solutions, the dew point curve and the boiling point curve must meet at some point.

is just wrong. A lot of ideal behavior can occur with the dew point curve and the boiling point curve in the bounded region.
In chemistry there is usually some weird outlier for any general statement. I can't think of a binary system which has an azetrope where the dew point curve and the boiling point curve stay the bounded region but I'd bet that there is at least one known.

Edit - I realize now that if the dew point curve crosses the ideal line then there must be an azeotrope at the crossing point.

Why must azeotrope exist if the dew point curve goes outside the bounded region?(I think this equals to "if the dew point curve goes outside the bounded region, then it must meet the bubble point curve". I hope this equivalence isn't wrong.)

@Wang - yes that statement would be equivalent to mine.

So what is the reason behind this? sorry for keep asking but I really want to know. Have been searching for this all day.

@Wang - My calculus is too rusty to work it out, but I'm sure that it is the calculus rather than the chemistry. If the dew curve goes outside the bounded region, then it must form a maxima or a minima. Remember if the vapor phase is changing composition, so must the liquid phase.

The math seems to have been worked out in "New Trends in Statistical Physics" by Alfredo Maacias and Leonardo Dagdug, Chapter 6 - Thermodynamic Properties and Model for Vapor-Liquid Azeotropic Binary mixtures

Raoult's law doesn't say that the vapor and liquid have identical compositions. It sounds like you are suggesting that.

@BuckThorn - If I qualify the statement to say that the mole fractions in the vapor and liquid are the same would that assuage your objection?

@MaxW At the azeotrope they have identical compositions (in terms of mole fraction) of course, but this is not generally true.

@BuckThorn - OK, I broke the post into sections and explicitly stated that an ideal liquid is a special case of Raoult's Law. // An ideal liquid has no azeotrope of course. // Think that the changes made the article better.

Azeotropes do not exist for all non-ideal solvent solutions. All solvent solutions are more or less non-ideal, but not all solvent combinations form azeotropes.
E.g. ethanol and isopropanol form azeotropes with water, but methanol does not.
Azeotropes exist if there is large enough deviation from the Raoult law. It means if the azeotrope point of the diagram is pushed enough away from the endpoint connections.
For near ideal solutions, like n-pentane + n-hexane. there is no bottleneck and onle one convex/concave chart segment.
Back to the meeting of boiling and condensation curves, they meet at the pure solvent points and eventually at the azeotrope point, if it exists.

Existance of such minimums ( or maximums ) ob the vapour diagram means there is an azeotrope. An azeotrope behaves like if it were the 3rd substance, with  the same composition of liquid and vapour above it, so the curves meet there.
Minimums meet if and only if liquid and vapour compositions are the same. That happens if and only if either the liquid is pure solvent either it has the composition of the azeotrope ( if it exists ).
You can create simulations yourself by defining nonideal vapour pressure equations as $p_i(T)<>p_{i,0}(T) \cdot x_i$. It cannot be exactly calculated, as neither the mixture of pure solvent and azeotropic mixture has the ideas behaviour.

Is there any picture with minimum in dew point curve and bubble point curve, and those minimum does not meet?
Also I can't find out if the Txy diagram of ethanol-water solution is calculable by some equation, or is it pure experiment results?

So how does the picture change, when the deviation from Raoult law is small to large enough to form an azeotrope? I mean why would the minimum always meet when they exist?

@Wang Non-ideal vapor-liquid equilibrium curves can be generated using Wilson's Equation of State and Antoine's Equation.

According to statistical mechanics and algebraic analysis (10.26434/chemrxiv-2021-xsj92-v2), the bounded region is related to $|ln(p_A^*/p_B^*)|$ $k_B T$. Only if the intermolecular interaction between A and B components is stronger than this amount, the binary mixtures can form azeotropes.


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Probability density of a particle in a 2D box with given coordinates

Wavefunction of particle in a $2$D box with sides $L$ in the $x$ and $y$-axis:
  $$φ = \sqrt{\frac {4}{L^2}} \sin\frac{n_1πx}{L}\sin\frac{n_2πy}{L}$$
For the different states ($n_1=1$, $n_2=2$), ($n_1=2$, $n_2=2$) and ($n_1=1$, $n_2=3$), which has the highest probability density at the coordinate ($0.25L$, $0.25L$)

I know that the probability density is the integral of $φ^2$ but I don't know the interval for the integral. I would assume that it is $0<x<L$ but when I do this I get something completely different from the given answers. The given answers are:

For ($n_1=2$, $n_2=2$) $=>$ the value of the wavefunction is $\frac{4}{L^2}$
For ($n_1=1$, $n_2=2$) $=>$ the value of the wavefunction is $\frac{2}{L^2}$
For ($n_1=1$, $n_2=3$) $=>$ the value of the wavefunction is $\frac{1}{L^2}$
The dimension of the probability density must be $\frac{1}{length^2}$, so ($n_1=2$, $n_2=2$) is the correct answer.

I don't understand how they have gotten this and why ($n_1=2$, $n_2=2$) is the correct answer. Could someone please explain this to me. Thank you!

You are close. The probability of being in some area is the integral of $|\phi|^2$ over that area. The probability density at a point is just the value of $|\phi|^2$ at that point. In your case, you want the probability density at the point (0.25L,0.25L), so just plug this into $|\phi|^2$ with each choice of $n_1$ and $n_2$, no integration involved.

@Tyberius Oh, thank you very much! But why would (n$_1=2$, n$_2=2$) be the correct answer when $\frac{1}{length^2}$ corresponds to (n$_1=1$, n$_2=3$)?

They phrase it confusingly here, but that is just the units of the probability density, which are the same for any choice of the quantum numbers. It does help to check that you got a reasonable answer, as it must have these units, but to choose between the options given you need to find the one with the largest value.

@Tyberius Okay, thank you very much for the help! :)


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Best way to store gasoline for a generator
For my lawn power-tools, my normal regime is to store 2-galleons of 85-octane gas, purchased in the spring, and treated with Sta-Bil. This lasts me through the season. Now, however, I'm looking into getting a generator. This gas may not be used within the season, so longer-term storage considerations come into play (though, some rotation scheme inclusive of my power-tool usage may make sense). So what combinations of octanes, ethonal/non-ethanol containing gas, and sta-bil should I consider and why? And what form of rotation - should I put ever everything in a larger tank and mix it with "fresher" gas. Keep things separate in smaller containers? Seal some up with a treatment of some kind and not touch it? And again, why?

How much gas does the generator use per day? How many days supply for generator? Where do you store the gas?

What is that "sta-bil" supposed to do? If you keep the fuel in a closed cannister, and outside of the sunlight (because of the ethanol), it will stay fine forever. Temperature swings in a ventilated tank make water condense into the fuel, but that´s already the worst that could happen, imo.

"What is that "sta-bil" supposed to do" - I don't know, that is why I asked on the chemistry forum ;)

"Where do you store the gas?" - backyard shed.

" How many days supply for generator? " - I'd like to keep enough for a week of continuous daytime-use. I'm still looking into different models, so I don't yet know consumption rates.

I have a small generator that burns a gallon in about 15 hours Storage is about 6 gal in 1 and 2 gal containers. I use the same gas in mower and other small engines, so I rotate the inventory . My combined use is about 4 gal per year so the average storage life is roughly 16 months , this has been no problem. Another option is to use old gasoline in a car in order to refresh the stock. I store it in a garden shed as I don't want that volume in a garage.


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heat capacity in kelvin vs celsius
I understand
Fact 1) Average kinetic energy of an atom in a noble gas is proportional to the temperature in Kelvin.
Fact 2) Average kinetic energy of an atom is not linearly proportional to temperature in Celsius or Fahrenheit.
Fact 3) The heat capacity equation
delta q = mc delta T
shows increase in energy is linearly proportional to temperature.
However, Textbooks frequently use this equation in degrees Celsius.  How is this not a contradiction of Fact 2?
Thanks

You wrote it yourself: delta. Doesn't matter if the difference is in degrees Celsius or in kelvins. It's the same number.

$\mathrm{T(K)=T , ^\circ C + 273.14}$ so, if $\mathrm{K , \alpha , T(K)}$. Then what about $\mathrm{T , ^\circ C}$?

If we for sake of simplicity assume the heat capacity is temperature independent, then the thermal content, expressed as functions of $K$, resp. $^{\circ}C$, are 2 parallel lines:
$$Q = C \cdot T $$
$$Q = C \cdot (t + 273.15) = C \cdot t + C \cdot 273.15)$$
Heat capacity C=dQ/dT, resp. C=dQ/dt is their slope. Slopes of parallel lines are equal.
But the kinetic energy $E$ of particles, proportional to thermal energy $Q$ ( as  generally, not all thermal energy is kinetic* ) is proportional to T, but not t.
$$\frac{E}{T} = \frac{a \cdot Q}{T} = a \cdot C$$
$$\frac{E}{t} = \frac{a \cdot Q}{t} = a \cdot C \cdot \frac{t + 273.15}{t}= a \cdot C \cdot (1 + \frac{273.15}{t})$$

*) I missed the noble gas mentioning.  Well, for noble gases, the amounts of their thermal and kinetic energy is almost identical.
The very subtle difference is

Their slight non-ideality, caused by non-zero intermolecular interaction of their atoms, like collision potential energy and van der Waals forces.
Electron excitation and ionization at high and very high temperatures.

when you say that thermal energy has forms other than kinetic energy, are you referring to the interconversion between kinetic and electrostatic potential energy during collisions of the noble gas atoms?

See the answer update.


text:
Percent degree of dissociation of gas by pressure
I stumbled across the following interesting problem:

Assume that you have started to live on a new planet where standard pressure condition is $\pu{2 bar}$, standard concentration is $\pu{1 M},$ and all types of gases behave as an ideal gas. On this planet, you are asked to determine equilibrium conditions for the reaction below:
$$\ce{XY4(g) <=> X(s) + 2 Y2(g)}$$
$K^\circ = \pu{1.1455E-9}$ (at $\pu{298 K}$).
Calculate the percent degree of dissociation for $\ce{XY4}$ at $\pu{298 K}$ where total pressure is $\pu{0.2 bar}.$

Since the equilibrium constant is very small, I approximated that the pressure of $\ce{Y2}$ at equilibrium is insignificant. Therefore I got:
$$p(\ce{Y2}) + p(\ce{XY4}) = \pu{0.2 bar}$$
$$p(\ce{XY4}) = \pu{0.2 bar}$$
I don't know how to proceed from here though and how to calculate the degree of dissociation.

I think I see how to solve the problem, do you have an answer?

Unfortunately, no. Could you share any tips?

First what does the circle on the equilibrium constant $K$ mean?

Not sure, but I assumed it doesn't make a difference. The equilibrium constant given is at 298 K by the way.

The circle means standard conditions. // I didn't read the problem carefully enough. I don't see how standard conditions could be (1) pressure of 2 bar, (2)  concentration of 1 molar and (3) gases behave as ideal gases (about 11 liters per mole at 2 bar and 298 K) . // I was thinking that the problem needed you to convert $K_c$ to $K_p$ at 298 K.

I still don't understand the degree of dissociation: with what should I compare the pressure of $\ce{XY4}$?

Yes, $\ce{(XY4_\mathrm{inital}/XY4_\mathrm{final})\times100%}$

Isn't 0,2 bar the final pressure of $\ce{XY4}$? If so what is the initial pressure? I'm a bit confused.

You can't use approximations as you did because the values you care about are the values you approximated out!

If $\alpha$ is the degree of dissociation, at equilibrium $1-\alpha$ of XY is present and $\alpha$ of X and $2\alpha$ of Y making a total of $1+2\alpha$. The partial pressure of XY is $\displaystyle p_{XY}=\frac{1-\alpha}{1+2\alpha}\frac{P}{P^\mathrm{o}}$. The $P^\mathrm{o}$ is the standard pressure used to make the equilibrium constant dimensionless. In our world this is 1 bar, in your new world it is 2 bar, thus $P/P^\mathrm{o}=0.1$. If you calculate the partial pressures for the other species in terms of $\alpha$ and form the equilibrium constant then $\displaystyle K=\frac{4\alpha^3}{(1+2\alpha)^2(1-\alpha)}\left(\frac{P}{P^\mathrm{o}}\right)^2$.(please check).  As $K$ is small we will assume that $\alpha$ is also then $\displaystyle K=4\alpha^3\left(\frac{P}{P^\mathrm{o}}\right)^2$ which produces $\alpha \approx 3.10^{-3}$.

In my solution, I approximated the pressure of $\ce{XY4}$ to be 0,2 bar and calculated the pressure of $\ce{Y2}$ from K. How should I take the standard pressure of 2 bar into account in my solution? Should the pressure of $\ce{XY4}$ be divided by 2?

All the pressures used in calculating $K$ have to be dimensionless which means dividing by the standard.

This results in 0.0107% as the degree of dissociation (I divided the pressure of $\ce{Y2}$ by the pressure of $\ce{XY4}$, which is 0.1 bar). So is there an error in my solution?


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How is a reaction with more products than reactants spontaneous?
If I have a reaction that has $\Delta G^\circ = 0,$ where there are more products than reactants (for example, $\ce{A <=> 2B + 2C}),$ then if I change the concentration of all the reactants and products (i.e. to 20 mM), then shouldn't the reaction favor the reactants, since you have more total product than reactants (in my example, you would have 40 mM of both $\ce{B}$ and $\ce{C}$ and then just 20 mM of $\ce{A}),$ and therefore produce a positive $\Delta G$ value? I get that if you use the equation $\Delta G = \Delta G^\circ + RT\ln Q,$ you would get a negative $\Delta G$ for my example, showing that it is a spontaneous reaction.
However, can anyone explain this problem from a conceptual perspective, since it would seem that if you have more products than reactants with $Q,$ that would drive the reaction to favor the reactants, according to Le Chatelier's principle?

Hint:  Write the expression for Q for this reaction, then substitute in 20mM for the concentrations of A, B, and C.   Then solve for $\Delta G$.  You'll see that it's positive.

You are using Le Chatelier's principle the wrong way. It doesnt state that if product is more than the reaction moves to reactant side. It states that if the reaction has attained equilibrium then change in concentrations or conditions will change the reaction direction accordingly.

Note that whether the forward or reverse reaction is favorable depends entirely on the relative magnitudes of Q and K (the reaction quotient versus the equilibrium constant):
$$\begin{align} Q&>K \rightarrow \text{reactants favored}  \\ K&>Q \rightarrow \text{products favored}  \end{align}$$
Next, consider that when you declare a value for $\Delta G^\circ$ you can translate that into a value for K, using the following equation:
$$\Delta G^\circ = -RT\log K$$
For instance, if $\Delta G^\circ = 0$ then $K=1$.
For the example here, this means that
$$\begin{align} Q&>1 \rightarrow \text{reactants favored}  \\ 1&>Q \rightarrow \text{products favored}  \end{align}$$
Next, consider the standard state of all substances. For solutions this is typically $\pu{c^\circ = 1 M}$. Standard states are important because the reaction quotients and equilibrium constants, when described in terms of concentrations, contain ratios of the concentrations in solution to the standard concentrations, for instance
$$Q= \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}$$
If all substances are present at the standard concentration, $Q=1$. In the present example $K=1$ so the equilibrium state consists of all substances being at standard concentrations ($c_i=\pu{1 M}$).
Finally, consider the following way to relate Q, K and $\Delta G$:
$$\Delta G = RT\log \left( \frac{Q}{K}\right)$$
If $K=1$ as in the present example then
$$\Delta G = RT\log Q $$
This just reinforces what was stated earlier for the special example presented here: when $Q<1$ the free energy difference is negative and products will be favored, and vice-versa when $Q>1$.
Inserting the definition of the reaction quotient
$$\begin{align} \Delta G &= RT\log \left( \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}\right) \\ &= RT\log \left( \frac{c_B^2c_C^2}{c_Ac^{\circ\,3}}\right) \end{align}$$
If we assume all substances are present at the same concentration c, then
$$  \Delta G= 3RT\log \left( \frac{c}{c^{\circ}}\right) $$
Clearly if $c<\pu{1 M}$ then $\Delta G<0$ and products are favored.

One of my professors in a graduate Reaction Engineering course gave us a conceptual justification for a similar question.
Consider two reactions,

$\ce{A + B <=> C + D}$
$\ce{A + B <=> 2C + 2D}$

For reaction 1, it requires 2 mols in the case of both the forward (1 mol A, 1 mol B) and reverse cases (1 mol C, 1 mol D) to drive the reaction forward. Conversely in reaction 2, while it requires 2 mols for the forward reaction, it requires 4 mols for the reverse.
When one thinks of a reaction, the fundamental mechanism will involve the assumption of some level of simultaneous collision or interaction (for example in the liquid of gaseous phases). From a probability perspective, it is significantly more likely for 2 molecules to collide perfectly to initiate a reaction, than it is for 4 molecules to do the same.
This is usually an assumption that one makes when modeling a reaction from first principles. For example, for the same reaction $\ce{A + B <=> 2C + 2D}$, we could reasonably assume that this reaction system would likely favor the forward over the reverse reaction, i.e: $\ce{A + B -> 2C + 2D}$-or that $K_{forward}$ >> $K_{reverse}$, just based on the intuition provided in the previous paragraph.
This is not a one-size fits all explanation; chemical reaction systems are very complicated, but I hope this conceptual perspective helps.


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Reaction between anisole and concentrated HI in excess
Question:

My Attempt:  In most of the ethers reaction of HI in excess yields alkane iodide products. But in case of anisole two scenarios are possible, I believe:

The HI breaks bond between methyl and oxygen leading to formation of phenol and methyliodide.
The phenol formed further reacts to form phenyliodide.

But would HI be strong enough to break partial double bond in phenol?

Try drawing a possible mechanism: what would the first step be?

Protonation of O ?

Correct, and then where is the easier site for I- to attack? Remember how big iodide is relative to carbon

It would be easier to attack methyl group. But as we normally see in case of reaction of ethers with alkane iodides, would the product be phenol or can HI attack attack it too? Had it been dilute i wouldn't have thought that, but conc makes me think.

The correct answer here should be C. $\ce{I-}$ is an excellent nucleophile, it would attack the $\ce{C-O}$ bond in the methly group because phenoxide is a good leaving group as it is resonance stabilized and quite polarizable. So, the products are phenol and methyliodide. As for the excess of conc. $\ce{HI}$, it is just there so that some other nucleophile like $\ce{OH-}$ doesn't grab the limelight. I don't think it is energetically favourable to remove hydroxide from phenol to make it iodobenzene.


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Effect of EWG on groups undergoing decarboxylation
I have been reading Peter's Sykes and there I came across certain compounds which show decarboxylation.
My doubt is that how the acidity of the groups attached to the carboxyl groups will help in decarboxylation.
For example we have one EWG and one EDG then which will show faster decarboxylation.
I am quite aware of the mechanism but somehow I can't relate here.

EWG stabilise anions. The key step of decarboxylation is the formation of an anion from the loss of CO2.

Yes but here we are proceeding via 6 membered cyclic TS not by anion formation

A diagram would assist

What I mean to say is the decarboxylation mechanism at higher temperature where we assume co2 leaves

Anyone can help?


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Why are Tertiary Radicals more stable than Tertiary Carbocations?

I think tertiary carbocation must be more stable due to Hyperconjugation and +I effect....

That wouldn't be homolysis, to begin with.

The comparison is not useful. Carbocations and radicals are produced in different ways, react with different types of partners, and cannot interconvert.

It's about homolysis, not heterolysis. If you start wit a neutral molecule, product has to be  radicals.

The magnitude of the effects needs to be considered. To form a radical from isobutane, you break a C-H bond. To form t-butyl radical requires about 93 kcal/mol (390 kJ/mol) https://en.wikipedia.org/wiki/Carbon–hydrogen_bond. To form a cation, you break the C-H bond AND ionize the radical, which costs you another 6.7 +/- eV (154 kcal/mol or 646 kJ/mol http://www.colby.edu/chemistry/PChem/Hartree.html) https://webbook.nist.gov/cgi/cbook.cgi?ID=C1605738&Mask=20
So, while the t-butyl carbocation is indeed stabilized by the indictive effect, and hyperconjugation, the energy level at which this is occurring is much higher than the energy level of the radical.
Hyperconjugation is a qualitative concept, with a possible energy of about 5 kcal/mol.  Like a fudge factor, which explains all possible effects if you need it and apply it correctly, but in this comparison, its effect is negligible compared to bond-breaking and ionization.  https://en.wikipedia.org/wiki/Hyperconjugation


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What does the phase diagram mean by "liquid" at NTP? (The liquid and vapor coexist even when the situation is not on the curve)
[Solved. For anyone who is also confused, see the comments under Poutnik's answer. As Poutnik reminds me, remember there is no air involved. I successfully picture that with the piston experiment in Karl's answer.]
When I read the phase diagram, I think it suggests that "at certain temperature and pressure, the phase of water would be (solid, liquid, gas) or two/three of them coexisting."
But when I look up NTP on the diagram, it says "liquid", while the saturation pressure of water at 25°C is 0.03atm. So the situation can't exist since the vapor is going to condense until the pressure drops to 0.03atm.
Then what actually is the phase diagram suggesting by "liquid"?
What am I missing here?
Edited: Poutnik reminds me that the pressure is the total pressure, not the vapor pressure. Then my question should be "If the vapor and liquid coexist not only at the curve, then what is the meaning of 'liquid' on the diagram?"

Then under a total pressure of 1atm, I think liquid water and vapor still coexist when there's 0.03atm vapor in the gas? What is the meaning of "liquid"? I'm confused because in the textbook it says water and vapor coexist at the curve, but I think they also coexist elsewhere like the NTP.

This diagram is for one compound, just one compound and nothing else. There are no other compounds. There is nothing else. There is no air, in particular. Total pressure is the same as vapor pressure. That's what it says.

Imagine filling the e.g. water into a cylinder with a piston on top. No air inside.
Now the experiment is to control the temperature inside the cylinder, and move the piston to get the wanted pressure inside. If the pressure is above the vapour pressure, your cylinder is completely filled with liquid water, if it is below, all the water will be evaporated.
Of course you have to pull out the piston very far to make room for the complete phase transition.
While you still have two phases in the cylinder, the equillibrium T/p value pair is always on one of the lines in your phase diagram.

So if I keep the temperature at 25°C, press the piston so hard that the pressure reaches the ice area in this picture Poutnik posted: https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase_diagram_of_water.svg , I will get ice and no gas, no liquid?

@Wang Of course!

@Wang Yup.  At about 10,000 atmospheres you get Ice VI.  At 21,000 atmosphere Ice VII, then on up through Ice X and Ice XI as you run the pressure even higher.  You better have a very strong piston!  Note that at sufficient pressure you can make a baking dish out of ice, but anything for the stove would be problematic.

The phase equilibrium diagram is for a system consisting of just water.
The diagram pressure is the total system pressure.
It may be equal to the saturated vapor pressure, if no external pressure is imposed on the system.
Or, it can be ( much ) higher, if some external pressure higher than the saturated vapour pressure exists.  In such a case all vapor is at equilibrium transformed to liquid or solid phase.
Really high pressures can be seen in the water phase diagram for very high pressures, displaying rare ice phases.

So, can I interpret the water phase diagram at NTP as "at NTP, liquid water can exist if the vapor pressure is saturated(0.03atm) " ? I'm saying this because at a total pressure of 1 atm, liquid water can't exist if the vapor pressure is under 0.03atm (right?), which I think contradicts the diagram.
Also, does the more detailed phase diagram you posted here gives the information of how the high pressure ice would exist? By how I mean like the vapor pressure of the ice.

To be clear, I'm asking all these questions because I can't find out how all the phases would coexist ( or not) if I'm looking at a phase diagram I'm not familiar with. I wonder what information phase diagram gives, and what it doesn't.

No. at 1 atm And 0 deg C, all vapour condensates. But essential part is "water only system".

Withing areas, oy 1 phase exists. At curves 2, at the triple point all 3.

If only one phase exists within areas, how does it not contradict the fact that at 25°C / 1 atm, liquid water coexists with water vapor?

Does have it vapour pressure 1 atm ? Remember, water only system. What happens to vapour if it has pressure 1 ATM at 25 Deg C?

I'm confused whether the pressure should be vapor pressure of water, or the total pressure.

I think in a water only system, we have a 1 atm/ 25°C water vapor. Then it would form liquid water right? And the pressure would drop to 0.03atm on the diagram ( on the curve ), and reach equilibrium.
Is my statement right?

No. If pressure is higher than saturated vapour pressure at given temperature, vapour cannot exist in equilibrium.

@Wang If you apply a pressure of more than 0.03atm on water at 25°C, the gas phase condenses completely.

I guess he is confused, as he may subconsciously still count with air pressure. There is none.

I think I understand now. Yes I was subconsciously still counting air pressure! Big mistake here. Thanks for this clarification!


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Why Calcium Oxalate is insoluble in acetic acid?

Calcium oxalate is insoluble in acetic acid, but not it's carbonate, oxide and hydroxide.

From internet, it is even considered as a test to distinguish calcium oxalate from it's carbonate and oxide. The reason from the resources which i had searched, only gave factual reasons such as

High Lattice Energy and
Low solvation energy of $Ca(C_2O_4)$

But intuitively the "short sized" compounds like $CaO$ and $Ca(OH)_2$ must have better packing and more lattice energy, then why much bigger calcium oxalate have higher lattice energy?
Also, i am clueless about how to compare solvation energy of different compounds. I think it has to do with polarity of the solute and the solvent. Any help will be appreciated.
Edit:
I had omitted the relative basicity of oxide, hydroxide and carbonate of calcium but i am more interested in knowing how to compare the lattice energy and solubility of the given compounds.

You have omitted to consider acid-basis relations, particularly the relative acidity/alkalinity of respective cunjugated acid/base pairs.

Yes, but what may be the reason for other facts, i am more interested in knowing how to compare properties like solubility and lattice energy

Such things are more related to solubility  product. For solubility itself, you have to involved the side acid-base equilibrium reactions. The play essential role.

Calcium oxide, hydroxide and carbonate react with acetic acid to form calcium acetate. Calcium oxalate does not react.
Edit: In real life, you'll of course will end up with a little oxalic acid and calcium acetate, as calcium oxalate is very slightly soluble and then you'll mix all available ions.

If you take a solution of calcium acetate and add oxalic acid, you will precipitate calcium oxalate, leaving acetic acid in solution. That's because oxalic acid is a stronger acid than acetic, as Poutnik suggested.

"That's because oxalic acid is a stronger acid than acetic". Probably doesn't work like this. Hydrochloric acid is stronger that oxalic acid, but calcium oxalate won't dissolve in it.
It's rather the solubility product of calcium oxalate that's driving the precipitation.

Reverse the comparison: Since oxalic acid is stronger than acetic acid, acetic acid is weaker than oxalic acid, and hence won't protonate oxalate (at least, not very much). Since acetic acid won't protonate oxalate, it can't bring calcium oxalate into solution. You are correct that the solubility product drives the precipitation, but to undrive it requires protonation of oxalate anion. CRC Handbook says calcium oxalate is soluble in acids; stronger acids will do better because a strong acid could protonate oxalate and drive down the concentration of oxalate anion as CaC2O4 dissolves.

You're right, James. There is the equilibrium between free oxalate and oxalic acid which will draw the calcium oxalate into solution at low pH. My bad.

James may I know edition of CRC Handbook says "calcium oxalate is soluble in acids; stronger acids will do better because a strong acid could protonate oxalate and drive down the concentration of oxalate anion as CaC2O4 dissolves". Thanks

@ yeni Sulastri: CRC 62nd edition, 1981-2. Should also be in earlier and later editions.


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1,2 Aryl and alkyl shift
What is the criteria in selecting which of the available groups adjacent to a carbocation is the most favourable to be shifted?
I have two compounds: (I'm unable to post the structures, so I have to write it here)
1) A methyl and an ethyl available for shift to a carbocation
2) A nitrobenzyl and anisole as the substituents for aryl shift.
Our teacher said that in 1)Ethyl shifts on the grounds of stability of the resultant carbocation and
In 2) anisole shifts because it has "more electron donating tendency than nitrobenzene"
The two answers seem contradictory to me. Is it correct?

Carbocations are stabilised by electron donation. I fail to see any contradiction in the two statements

The first reason was stability of the resulting carbocation and the second, the stability of the carbocation before the shift

https://chemistry.stackexchange.com/questions/31599/migratory-preference-in-a-rearrangement-involving-carbocations


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How do CO2 Fire extinguishers procure their CO2
How do CO2 Fire extinguisher manufacturers get such a large amount of Carbon Dioxide? Do they get it through a chemical reaction or burning hydrocarbons?

They just buy it from companies producing CO2.
Actual sources may be sequestration from combustion (e.g. power plants) or fermentation (e.g. ethanol production) processes.


text:
Are inner transition metals also part of the transition group?
My book doesn't say anything about this and leaves it ambiguous. The periodic table they gave colored the inner transition metals as "transition metals". So, are inner transition metals also part of the transition group?

Why wouldn’t they be?

Of possible interest: a huge number of different periodic tables to examine! Go here and follow the link in the comment by @Buttonwood.

Well to categorize them, you need to first define what transition elements are. One definition is etymological, that, 'the chemical properties of transition elements are transitional between those of s and p-block elements'.
A more refined IUPAC definition is that 'transition elements have an incomplete d-subshell, either in the neutral or ionic state'. The f-block elements/inner transition elements meet this criterion, so yes, they are transition metals. You can read more about it here.
The reason why they are in separate blocks is due to the reason that they have successively varying f-subshells and frankly, to accommodate the periodic table on one regular-sized page.


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Microscopic View of Adding Water to A Basic Solution
I confronted a problem, stating that we have a solution of potassium hydroxide. By adding extra water, when the volume of solution reaches 630 mL, the pH decreases by 0.5 units. How much extra water have we added to the solution?
And I know this is a simple problem, but I wondered, when we add water to the strong basic solution, the volume increases, which means a decrease in $[OH^-]$. and a decrease in $[OH^-]$ means an increase in $[H^+]$, and therefore a decrease in pH.
Everything sounds good in a macroscopic view, but what really happens to $[H^+]$ when we add more water? Why doesn't it decrease just like $[OH^-]$ when we increase the volume of soultion? What happens to $[H^+][OH^-] = constant$ from a microscopic view?

The volume unit is liter or cubic meters, but not millimeters.. A volume of 630 mm has no meaning. Anyway ad data is missing : the initial volume.

You are right. I will edit it. Thanks. But the initial volume is not given.

OK, If the pH decreased by $0.5$, it means that the concentration changes by $\ce{10^{-0.5} = 0.3162}$. So the initial volume was $\ce{630·0.3162 = 199.2 = 200}$ mL . So you can subtract the initial volume (200 mL) from the final volume, to get the amount of extra water,

And if you add water to an solution of a strong acid, then the pH does rise, which means that $\ce{[H+]{\mathrm{final}} < [H+]{\mathrm{initial}}}$. Of course to maintain $K_w$ then $\ce{[OH-]{\mathrm{final}} > [OH-]{\mathrm{initial}}}$.

@MaxW Here, it is adding water to a strong base. But your argument works by simply looking at the pOH instead of the pH.

@Maurice The question does not have a single answer. Let's say that the pH was 7.5 initially. To get it to pH = 7.0, you would have to add a lot more water.

@KarstenTheis - I was reading the OP's last paragraph to be asking about an acid solution.

@MaxW I don’t think that’s what the OP meant. I think H+ refers to the minor species in the  basic solution

Microscopic view often means looking at the kinetics (thermodynamics is easier for looking at bulk where concentration, temperature, and equilibrium make sense).
In the initial solution, the auto-dissociation of water is at equilibrium:
$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$
Forward and reverse reactions occur at the same rate, with a zero net reaction.
If you add water, the concentration of hydrogen ions and hydroxide ions both decrease by the same factor, while the concentration of water hardly changes. As a consequence, the rate of the forward reaction stays the same while the reverse reactions slows down.
Now, we have a net forward reaction making additional hydrogen ions and hydroxide ions. The hydroxide ion is the major species, so it's concentration does not change much (it stays lower than before the dilution). The hydrogen ion is a minor species. It was present at very low concentration (potassium hydroxide solutions are basic...). The net forward reaction increases its concentration by a lot (0.5 pH units is a factor three, approximately), to compensate for the decrease caused by dilution, and beyond, giving a concentration that is higher than the initial concentration.
Once you solve the quantitative question, you can make a table listing the effects of dilution and of the reaction attaining equilibrium again to get a better picture.


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Can Ferrofluids vaporise to gaseous state and Condense back to liquid state and do it many times over?
Here is the Link for Ferrofluids for basic idea. After going through this Wiki link, probably many people will come to a conclusion like when Ferrofluid boils, its carrier liquid vaporises and nano particles gets settle down. Therefore Ferrofluid looses its properties(destroyed). I come to this conclusion but not 100% sure. There may be some wild science going over because "nano" particles and its size. So I decided to ask here and get some help from some bright minds across the world.

No ! A ferrofluid is made of tiny magnetic particles dispersed in a liquid. If you try to heat it up, the liquid will boil first. The magnetic particles will melt then boil at a temperature impossible to reach in an ordinary lab.

Also, even the liquid portion alone would likely decompose below its B.P.

Part of the ferrofluid could theoretically be vaporized. As ferrofluid is made from tiny magnetic particles(eg. iron powder coated with a surfactant to inhibit clumping) suspended in a fluid. The fluid can be organic solvents, water and most commonly ink. Some of these fluids can evaporate and recondense. However, as a whole ferrofluid, cannot be vaporized. You can imagine ferrofluid as a cup of mixed sand and water that wouldn't settle(suspended).


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Risks of burning melamine formaldehyde adhesives
Background- I make biochar for gardening, and small scale carbon capture and storage. Most of my feedstock is clean, unpainted, untreated, end of life timber, but there is a small proportion of engineered particle board- MDF, chipboard, plywood and composite blocks containing, I believe, melamine formaldehyde binding resin. No halogens from PVC, no metals from CCA treated timber or old lead paint.  I want to establish what pollutants may be produced- I would in a worst case scenario be burning less than 50kg of the suspect material per hour.
All the charring happens in TLUDs- Top Lit Up Draft burners where all the pyrolysis gases are ducted through the hottest zone of the burn- round about 800-900C. I'm currently working to improve residence times and temperature gradients- carbon particulates are a known issue.
Does any of the above sound like it'll produce anything nasty from the carbon, hydrogen, oxygen and nitrogen going into the mix? PAHs? Anything else above and beyond nitrogen oxides and carbon particulates from clean wood?
To put it another way, do I err in favour of creating a pollution factory, or in favour of condemning a potential feedstock to be a disposal nuisance?
Thanks for any replies- I'd like to lay this to rest.

I am by no means an expert in the pyrolysis of different woods by any means. But after spending some time reading through some papers for you, I found one which burns MDF in a slightly different burner to you (around 450-600C) and the ash produced contained formaldehyde at 1681 mg/kg.
doi:10.1016/j.enconman.2014.12.032 - incase you want a read yourself.
This document goes through different methods in producing biochars and says that providing the temperature of pyrolysis is above 500C, it shouldn't contain any PAH's or Dioxins within the biochar and if it does, it'll be several times lower than required clean up set by the Model Toxic Control Act.
Everything else I have read seems to not have specific analysis on what exactly is left within the biochar. There are plenty of organics released as gases during the process specifically above 500C which you could argue should no longer be in the biochar. Have a look around at papers, I know it's not a definitive answer but it might be a start for you.


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Equilibrium cell potential question
In an attempt to wrap my head around the basics of electrochemistry I'm working my way through Wesley R. Browne's 'Electrochemistry' primer.
With regards to the equilibrium potential of an electrochemical cell he states: "The equilibrium potential is defined as the potential at which the forward and reverse electron transfer rates are equal and the net current flow is zero. ... When the electrode potential is more positive or more negative than the equilibrium electrode potential, the forward and reverse rates are no longer equal and a net current will flow."
My understanding is that the only equilibrium potential possible for a two-electrode electrochemical cell is 0V, as anything above or below 0V will result in the electrons 'wanting' to go to one electrode or the other = a net current.
However, in a three-electrode setup (working electrode, reference electrode and counter electrode) the potential of the working electrode is stated relative to the reference electrode (eg, 'vs Ag/AgCl'). In this setup, would the equilibrium potential of the working electrode be 0V relative to the counter electrode (as it's the counter electrode with which the vast majority of electrons are exchanged) but therefore not 0V relative to the reference electrode and thus you would be correct in stating that the equilibrium potential for working electrode X is (for example) +0.21V vs an Ag/AgCl reference electrode?
Any and all help appreciated!

Maybe this will help a bit, but potentiostats always seem a bit strange!

One thing that might be useful to remember is that the current depends on the voltage and the resistance. If you place a reference voltage electrode behind a large resistance then it will not draw significant current. That's in part what the link Ed V posts explains.

You must have two half-cells to get a voltage. So in a three-electrode setup there could be three different equilibria (O volts). Number electrodes 1, 2, and 3. So 1-2, 1-3, and 2-3.

Thank you for the responses all. @MaxW yes there would be - so in that case the voltage could be 0V between the WE and the CE (they're at their equilibrium potential) whilst simultaneously the voltage could be (for example) +0.21V between the WE and RE.

If there is current flowing through the solution impedance between the WE and CE, then there is a non-zero potential difference.


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Comparison between groups with -I effects
So I have a few questions in my book that require me to compare between two different groups on the basis of their -I effects. I have a few questions on this that have been bugging me for quite a while. I am preparing for an exam, and please not we will not be given any energy data or charts so we wont be able to reference them while giving the exam, so all I can do I understand.

What happens if I have to compare between a compound that has a single group with a stronger -I effect (like $\ce{NO2}$) and a compound that has more than one group with a comparatively weaker -I effect (like $\ce{Cl}$)? In which case is the -I effect stronger, and how to understand that?
What is the actual order of -I effect? Wikipedia and my books sometimes are conflicting in their orders. For example, Wikipedia states it as:

$\ce{–NH3+} > \ce{–NO2} > \ce{–SO2R} > \ce{–CN} > \ce{–SO3H} > \ce{–CHO} > \ce{–CO} > \ce{–COOH} > \ce{–COCl} > \ce{-CONH2} >\ce{-F} > \ce{–Cl} > \ce{–Br} > \ce{–I} > \ce{–OR} > \ce{-OH} > \ce{-NR2} > \ce{–NH2} > \ce{–C6H5} > \ce{–CH=CH2} > \ce{–H}$
One of my books states this as:
$\ce{-NMe3+}>\ce{-NH3+}>\ce{-NO2}>\ce{-CN}>\ce{-F}>\ce{-Cl}>\ce{-Br}>\ce{-I}>\ce{-OAr}>\ce{-COOR}>\ce{-OR}>\ce{-OH}>\ce{-C6H5}>\ce{-CH=CH2}>\ce{-H}$
And another states this as:
$\ce{-NF3+}>\ce{-NH3+}>\ce{-CF3}>\ce{-NO2}>\ce{-CN}>\ce{-SOO-OH}>\ce{-COOH}>\ce{-COCl}>\ce{-COH}>\ce{-COR}>\ce{-F}>\ce{-Cl}>\ce{-Br}>\ce{-I}>\ce{-OH}>\ce{-OC6H5}>\ce{-OR}>\ce{-NH2}>\ce{-CH=CH2}>\ce{-C6H5}>\ce{-H}$
Clearly some do not contain some compounds while in some cases the orders are a tiny bit switched up. Which one should I use? I know that some of this is really close comparisons but a definitive order if possible would reall help me out. Thanks.

Basically, the inductive effect is determined quantitatively by Heemat's equation. I haven't study it yet. I also have the same doubt. My teacher told me that those groups which have conflicting orders between them have approximately same result from Heemat's equation. I think in examination, examiner ask about only general groups for which order is correctly determined. I myself had never encountered any inductive effect question for which the order of groups was conflicting. Same answer for the first part.

@Manu I have encountered such questions. I wouldn't have asked this question if I hadn't.

Oh sorry. But I think studying Heemat's equation will help.

@Manu Thanks, that's a lead atleast. :)


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Different direction of dipole moment in physics and chemistry
In chemistry, we learn that dipole moment tells us about measure of polarity in a molecule. In books, it is written that "conventionlly the direction of dipole moment is from electropositive to electronegative element or less electronegative to more electronegative element"
My question is that in physics we are taught that the direction of dipole moment is from negative charge to positive charge as convention, so what is the basis behind the conventions of dipole moment in the two subjects which leads to different signs of dipole moment?

I understand the first answer but I have one doubt that if we apply electric fields on molecules and do analysis as a chemist's point of view, then the analysis which we have done in electromagnetism course in physics can't be directly applied in chemistry, we have to change our results according to the direction of dipole moment which we used in chemistry.

LOL - Physicists got the charge of the electron wrong. Why blame that on chemists?

Worse - it was Benjamin Franklin :)


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Acidity of naphthol vs hydroxynaphthoquinone
I have the following two compounds whose acidity I need to compare. I tried drawing out the resonating structures but I seem to not be able to draw final conclusion.

I know that the conjugate base of Q has more resonance structures, but R is still more acidic according to the book, with its resonance looking like this:

Why such resonance makes it more acidic?

What even is a structure that isn't "valid", something tells me I'm better of not knowing, but not being able to tell the acidity may be your last problem... BTW You should write structures for resulting anion, rather then the compound itself.

@Mithoron I do know how to get at them. I just feel like that the conjugate base of Q has more resonating structures but R's conjugate base has fewer. BUT the book states R to be more acidic than Q. This is my EXACT problem

You should have started with this. That's true and R is more acidic, just like acetic acid is. Now, edit the question.

@Mithoron Is that due to Equivalent resonating structures?

Mhm, you're on right track. You got your hint, now improve the question, or it may get closed.

@Mithoron I still cant understand why R is said to have equivalent resonance structures. Could you lease explain with a diagram?

Let us continue this discussion in chat.

For naphtol both the acidic and the basic forms have many resonance structures along the two fused benzene cycles. However, in the case of 2-hydroxy-1,4-naphthoquinone, the basic conjugated form creates novel resonance structures that engage the double bond of the quinone cycle as seen in the written resonance structure in the question. The novel resonance structures of the basic forms will stabilize the basic form.
Therefore, the hydroxyl group in 2-hydroxy-1,4-naphthoquinone has higher tendency to lose the hydrogen ion than the hydroxyl group in naphtol. This explains the higher acidity of 2-hydroxy-1,4-naphthoquinone in comparison with naphtol.


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Non-existence of time operator in quantum mechanics
In quantum mechanics, there is no operator for time instead of this we  we have $\psi(t)$ which is related to $\psi(0)$.
In a book, I read that if there is an operator for time, then it will not be Hermitian".
I am not able to understand how exactly why? What is the reasoning behind this statement.

Hm, time is not an observable like energy or impulse.

Time is not just a physical value like any other. What would you want this operator to return, had it existed? Time in milliseconds since the Big Bang?

No, I mean time elapsed.

If you go to Physics and search for "time operator" you will find plenty of answers...

Define "Hermitian".

I think part of the problem here is that relativistic quantum mechanics may be necessary to more adequately describe the role of time. Definitely should take it up with the physics folks.

On the not Hermitian point.
Let $\hat{F}$ be a non-stationary observable (but does not depend explicitly on time $t$, hence $\partial \hat{F}/\partial t = 0)$. in the Heisenberg picture it satisfies the Heisenberg equation of motion for $\hat{F}$, namely
$$i \frac{d \hat{F}}{dt} = [\hat{F},\hat{H}] \tag{1}$$.
Now assume that whenever $d \hat{F}/dt \neq 0$ then it has an inverse
$$(d \hat{F}/dt)^{-1} = i([\hat{F},\hat{H}])^{-1}\tag{2}$$
Now, if
\begin{align}
[(d \hat{F}/dt)^{-1}, \hat{H}] &= 0 \\
\implies [([\hat{F},\hat{H}])^{-1},\hat{H}] &=0.
\end{align}
Now introduce a Time operator, $\hat{T}$ as
\begin{align}
\hat{T} &= \frac{1}{2}\left(\hat{F}\left(\frac{d \hat{F}}{dt}\right)^{-1} + \left(\frac{d \hat{F}}{dt}\right)^{-1} \hat{F}\right) \\
&=\frac{1}{2}\left(\hat{F} ([\hat{F},\hat{H}])^{-1}+([\hat{F},\hat{H}])^{-1}\hat{F}\right)\tag{3}
\end{align}
Now using equations $(1)$ and $(2)$ into $(3)$ one finds that
$$[\hat{H}, \hat{T}] = i \tag{4}$$
That is, $\hat{T}$ and $\hat{H}$ form a canonically conjugate pair. Owing to the fact that the Hamiltonian spectrum is bounded from below, the time operator $\hat{T}$ is not self adjoint, in other words, not Hermitian.


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Orbital Approximation
We determined the hydrogen wavefunctions exactly, and the orbital approximation says that we approximate the orbitals of multi-electron atoms with hydrogen wavefunctions. How do we know that this is an okay approximation to make? How much error is there in assuming this?

no we don't approximate multi-electron atoms with hydrogen wavefunctions, we model them with gaussian basis functions that somewhat resemble hydrogen atoms. In the limit of infinite basis functions the approximation for the true shape of the molecular orbital is exact.

We know it is kind of OK because it kind of works out. Until it doesn't, of course. Which causes the many questions here about the transition metals and $d$-orbitals and why they aren't all behaving just like a hydrogen atom says they should.


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HNMR spectroscopy mono-carboxylated acid
During lab, I have done following reaction:

I did HNMR of the product, to be certain if the product really was formed.
The HNMR gave following results:

And looked like this:

So my question is: has the reaction succesfully taken place?
I know that the peaks between 6.5 and 8 ppm represent protons, attached to an aromatic molecule. But there are only four peaks between 6.5 and 8 ppm, instead of 5 (the amount that I expected). At 3.89 ppm, there is a singlet with 5 protons present, which is weird because I expected only 1 singlet with 3 protons. The 7 protons at 1.53 ppm I don't understand at all.
Please help me out.
Thanks.

Do you have the NMR spectrum of the starting material?

No, I unfortunately do not .

I think your questions should rather be "what has happened instead of the reaction intended?"

How did you work up this reaction? Have you got a fifth proton under the CDCl3 signal at 7.25?


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Attack of HClO4 to dipropylcycloprop-2-en-1-one
What are the products when $\ce{HClO4}$ attacks this compound?

I expect $\ce{HClO4}$ to protonate the ketone, forming an (aromatic?) carbocation. I don't know where to go from here.
Answer suggests formation of butanoic acid and production of $\ce{CO2}$. Unsure what really happens here. All hints appreciated.
This question appeared in the JEE 1997.

That is one very strained system, so once it protonates it is going to rearrange

@Waylander thanks for your insight - it would be amazing if you could elaborate yourself further; what rearrangement do you see happening? Do you see ring cleavage?

I wonder where the extra oxygens come from if you are getting butanoic acid and CO2? Presumably the perchloric acid is an aq solution.

See this prior thread on StackExchange at https://chemistry.stackexchange.com/questions/31670/reaction-of-perchloric-acid-with-cyclopropenol

AJKOER, I've already seen that question as well as the accepted answer - but I'm unable to see how anything from there could possibly help me with this question.... :(
@Waylander If we assume that the acid is an aqueous solution, what do you feel may happen? I'm unable to see what potential difference it could make...

Unfortunately this was exam for high school students required to answer in a limited time. May be this is of the fragmentation reactions grob's fragmentation reaction refer to this in Wikipedia                                     https://en.m.wikipedia.org/wiki/Grob_fragmentation#Reaction_mechanism

Assuming that your end products are correct I think this reaction would proceed in a similar manner to the way it will proceed if you use hot $\ce{KMnO4}$ as a reagent.
The reason I think that is because $\ce{HClO4}$ is practically similar to $\ce{KMnO4}$ structure and is an even stronger oxidizing agent ($\ce{Cl^{+7}}$ is much more unstable (and hence much more willing to oxidize) than $\ce{Mn^{+7}}$ ion) which should be able to imitate the behavior of hot $\ce{KMnO4}$ reagent (presuming that the greater oxidative strength compensates the level of reactivity  $\ce{KMnO4}$ shows at a higher temperature)
If the similarity is established then the reaction would proceed in this manner giving the products matching your answer:-

regarding the formation of aromatic cyclopropenium ion, I think cleavage of alkene relieves the angle strain which provides a higher stability than aromaticity.
N.B. :- Here I have presumed that $\ce{HClO4}$ can behave similar to hot $\ce{KMnO4}$ however I was unable to find a source for that.

Here is the source https://chemistry.stackexchange.com/a/131374/106370

@Rishi it is for $\ce{HIO4}$ and it doesn't suggest that $\ce{HIO4}$ can do oxidative cleavage of alkenes.


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How is it that the Inner Periodic Table Sections (have 15 elements each) versus their f-block (with only 14 element’s electrons each)?
Given:
1.  the “Lanthanide and Actinide” Inner Periodic Table Sections (each have 15 elements) versus the f-block Orbitals (which only have a capacity for 14 out of the 15 element’s electrons each) - leaving 1 element’s electrons each in the Inner Periodic Table Sections out in the cold.
2.  the f-block Orbital Structure is different from say the d-block, and not interchangeable.
How do we complete the correct electron configuration notation without substituting the d-block which has its own different unique Orbital Structure for its element’s electrons?

This is a little confusing to follow could you elaborate?

at the bottom of every Periodic Table (https://applets.kcvs.ca/IPTEI/IPTEI.html) you will find the section for the “Lanthanide and Actinide” Inner Periodic Table Sections - bearing  the 1st element name for each of the 2 Inner rows of 15 elements each - which are squeezed into rows 6 & 7, column 3.

If you check out YouTube's "Electron Configurations: Writing Condensed and Expanded Versions" URL "https://www.youtube.com/watch?v=5omUQRmTCiU&list=PLVKTjV4NoC88AHv6Q9063cpqS9QgCjRZB&index=3", you will find the Electron Configuration information where f-block has 14 spaces for 1 electron each. Each f-block is short by 1 electron each when compared with the number of elements in the Inner Table. This is why 5d1 is often utilized when writing out the Electron Configuration for Lanthanide for example.

I hope that the above will be helpful to you.

Unless I'm mistaken the Lanthanide and Actinides have 14 elements each, which match with the amount of electrons needed to fill the 4f block. You might be confused because they range from Lanthanum ([Xe]5d1 6s2) to Lutetium ([Xe] 4f14 5d1 6s2) on a peroidic table. The inclusion of Lutetium is because thats where the block would normally be replaced with an asterisk out of the table and so the element has to be shown somewhere.
Good example of a periodic table which shows this, Lutetium (atomic number 71) would be normally placed next to Barium (atomic number 56).

https://applets.kcvs.ca/IPTEI/IPTEI.html

1.the above URL will clearly show 15 elements for each of the Inner Periodic Table

2.please note that your Noble Shorthand "Lanthanum ([Xe]5d1 6s2) to Lutetium ([Xe] 4f14 5d1 6s2) " shows that you used up all of the 14 places for 4f14 + plus an additional 5d1 for the shortfall of 1.

By the way lanthanum-57 comes after barium-56.


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Why is the net dipole moment of carbon dioxide equals zero
Why is the net dipole moment of carbon dioxide equal to zero? And what is the partial charge of carbon in co2? Is it +2 because it forms two polar bonds or if they cancel the effect of each other due to their linear stereo structure then is it +1? What is the relation between being linear and having +1 partial charge? (my book says it have +1 and I couldn't understand why as I believe it is +2 because it forms 2 polar bonds so it has +1 from each oxygen atom)

it's zero because of symmetry, the dipole moments of the C-O bonds point in equal and opposite directions. The partial charge of carbon is not +1, most likely the book says its partial positive, i.e. not an integer. The oxygens are partial negative because they are more electronegative, and therefore there are two dipole moments pointing from the carbon to each oxygen.

Then carbon must have +2 partial charges

sure if you want to think about it that way, it is twice as more partially positive than the oxygen will be negative. it would not make sense to have two partial positive signs since they don't represent integers and it's not always the case that the partial charge signs are exact multiples of one another.

What is the exact reason that compounds such as carbon dioxide cannot interact with other compounds through dipole-dipole interactions?

Does this answer your question? Why does a tetrahedral molecule like methane have a dipole moment of zero?

In my book there is a question that says "what is the partial charge on carbon in co2 is is it +1 , +2 , or 0 " and the answer is +1 and I don't get why it is +1 although it forms two polar bonds even if they cancel the effect of each other

Get a better book, such question doesn't make sense. Partial charges aren't integers at all.

This question was discussed to exhaustion here: https://chemistry.stackexchange.com/q/130801/89908


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A basic physical chemistry problem
I am a high school student and have just begin my classes(online). I just started physical chemistry but I know about moles, percentage composition, empirical formula. I haven't been able to solve this question and due to my inability to approach my teachers(most of the time they are busy recording lectures), I post my question here.
In particular I am confused by that silver chloride line.

The idea here would be that any chlorine in the sample reacts to form silver chloride. Use stoichiometry to back calculate how much chlorine was in the sample.

I  have not done that much, please post an elaborate answer

You haven't provided enough details to effectively help you. Where are you stuck? Can you show some work? Do you have a book where you can look at examples of determining empirical formula?

A basic tenet of Stack Exchange is to have the questions and answers succinct that others can search for and answer questions. However your title is generic and your question is an image which is not searchable.

The example problem is not exactly the same as the problem you have. You need to use some stoichiometry to back compute the mass of the individual elements first. From there, you can try to determine the empirical formula, similar to what is done in the example.

@JordanLawson, There is a very famous reaction in which chloride ions are quantitatively precipitated by silver nitrate. 1 mol silver nitrate reacts with 1 mol chloride ion to form 1 mol silver chloride. This hint should help you to proceed further.

You can use each of the emperical formulas given and produce a balanced complete combustion question for each one as well as working out the moles of each compound given. You already know how many moles of CO2 and water were produced because you were given weights. From there you can see which compound it is most likely to be because the moles of that compound should produce the moles of the products when the stoichiometry is taken into account.
As for the silver chloride: (Correction) You cannot directly react AgNO3 with organic solvents but when you convert the compound into a sodium salt and dissociate the chloride ions, the chloride ions will then react with the AgNO3 in a 1:1 molar ratio. So working out the moles of AgCl produced and comparing with the moles potentially produced from each compound will help you determine if there is one or two Cl atoms in the compound you're looking for.

Could you please make a correction? AgNO3 does not react with an organic compound which contains chlorine directly like chloroform or dichloromethane. AgNO3 is not even soluble in organic solvents or pure compounds. You have to heat an organic compound with sodium and then you make a solution of it. That solution is reacted with AgNO3.


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Is this a counterexample to ΔU = 0 J for isothermal processes?
In class, I have learned that for isothermal processes the following is true at all times.
$$\Delta U=q+w=0\text{ J}$$
But suppose I lift an object against the force of gravity. Then I (the surroundings) have done work on the object (the system) without exchanging heat (neglecting friction).
Now, I claim that because some nonzero work has been done on the system, and no heat has been exchanged, the change in internal energy is nonzero:
$$\Delta U=w\neq0\text{ J}$$
Are my assumptions about internal energy incorrect or is there something totally wrong about my approach?

For isothermal process $\Delta U =0$ is true only for ideal gases. Isn't mechanical work considered for gases only, not sure?

If I had assumed that initially, I think this question would still valid because I could be lifting the gas, too. Edit: I see your edit now, I am not sure about it either.

Adiabatic, isothermal expansion of ideal gases must be free expansion...

Sure, but even if it doesn't expand, work can be done on it as I'm showing in my question

This was cross-posted to [physics.se] at https://physics.stackexchange.com/questions/545690. @awalllllll, which site would you like your question to appear on?

The situation you cite calls for the use of the more general form of the first law of thermodynamics (which I assume your textbook provides), given by $$\Delta U+\Delta (KE)+\Delta (PE)=q+w$$where KE is the organized kinetic energy of the system and PE is the potential energy.  Applying this to the problem at hand gives $$\Delta (PE)=w$$and the change in internal energy is zero.  So the thing that was wrong with your approach was not using the proper form of the first law.

Is $\Delta U=q+w=0\text{ J}$ valid only for gaseous system?

@Zenix No.  Even a gaseous system can experience a change in kinetic energy and/or gravitational potential energy.

I have always seen $\Delta U=q+w=0\text{ J}$... ;(

So here U is internal energy whereas KE and PE are the energy of the system with respect to an external reference frame (ie the PE and KE of the internal reference frame)?

@zenix Get yourself a decent textbook.

@BuckThorn Yes.  U includes the random kinetic energy of the molecules and the potential energy of the molecular interactions (the latter for the case of a non-ideal gas).


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Can other group IV elements form graphene analogs with interesting properties?
Graphene has gathered quite a bit of press due to its unique physical properties which hold enormous promise in a wide range of applications from semiconductors to high strength materials. Given that carbon is similar to other Group IV element chemically, I wonder if there are graphene analogs of other Group IV elements such as silicon, tin or lead and what fascinating properties and useful applications they may hold.

https://chemistry.stackexchange.com/questions/123524/are-all-stable-xenes-graphene-like-2d-honeycomb-sheets-buckled

Be careful with tagging, the ones you used, were all wrong. That led to zero links in "related" column.

https://chemistry.stackexchange.com/questions/34971/why-do-we-not-see-silicon-in-a-structure-like-graphite


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Effect of pressure on transition temperature
I was wondering if someone can explain why the answer to the following question is volume instead of the others.

If pressure has no effect on the transition temperature between two
crystalline forms of matter the two forms have the same molar
(a) volume
(b) energy
(c) enthalpy
(d) entropy
(e) heat capacity

I know the answer is volume but I'm not sure which equation that comes from. I was thinking maybe the law of reduced properties but at this point I'm just guessing based on the answer and I know that that particular equation is for gases. Does anyone know if there's an analog for crystalline forms of matter or if there's a more intuitive explanation? Thanks!

The mathematical answer is based on Clapeyron's equation (written in what follows in form inverted relative to the more usual):
$$\frac{dT}{dp}= \frac{\Delta V}{\Delta S}$$
which must hold on any point on the phase coexistence line.
If $\frac{dT}{dp}=0$ then it follows that $\Delta V=0$.

You can understand this using Le Chatelier's principle. In transitioning from a smaller-volume form to a larger-volume form, the substance takes up more space, and in doing so has to push against the external pressure to make more room for itself. The greater the external pressure, the more unfavorable this transition becomes.
Conversely, if there is no difference in volume, the two forms are equally favorable with respect to external pressure. Hence changes in pressure have no effect on the transition.


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How to calculate the concentration of reagents and products from Kc?

In the synthesis of ammonia at equilibrium constant is $K_c = 1.2$ according to the reaction:
  $$\ce{N2 (g) + 3H2 (g) <=> 2NH3 (g)}$$
  If an equilibrium is proposed based on initial concentrations of $[\ce{H2}] = \pu{ 0.76 M}$, $[\ce{N2}] = \pu{ 0.60 M}]$, $[\ce{NH3}] = \pu{ 0.48 M}$, with $x = 0.014$ as the change in the concentration of $\ce{N2}$, what are the equilibrium concentrations?

We'll guide you to solve such a problem but not just do it for you. // If the initial concentration of $\ce{N2}$ is 0.60 M, and there is a change of 0.014, what is the final concentration of $\ce{N2}$?

The final concentration of N2 could be 0.60M-0.014M=0.586

If that is correct what should I do next?

What else could he final concentration of $\ce{N2}$ be?

To me it isn't clear if the change is positive or negative. It seems the problem statement is deliberately ambiguous.

Independently if the change is positive or negative, because it is not clear for me too, could I use the final concentration of N2 and initial concentration of the other elements to calculate something? How would be the process of calculating the concentration of the other elements?

You're on the right idea! Let's start with the final concentration of $\ce{N2}$ being 0.586. What must the concentrations of $\ce{H2}$ and $\ce{NH3}$  be? // Then consider the positive change. What must the concentrations of $\ce{H2}$ and $\ce{NH3}$  be?

Mmm I am not understanding it at all. The negative change is N2=0.586M and the positive change is N2=0.614. Then with these changes, does kc=1.2 change too?

How does the change of N2 affect the other elements?

No $K_c$ must be a constant (for the given temperature...). // Back to the concentrations for a moment. If there is less nitrogen when the equilibrium is established, where did it go? // LOL - Scotty on the Starship Enterprise didn't just beam the nitrogen out!

HINT the course is chemistry, not Transwarp Wormhole Fluid Dynamics. ;-)

If there is less nitrogen in the equilibruim, is it now represented in an increase of NH3?

Yes, if the nitrogen decreases then the ammonia increases. What else must decrease? Use the chemical equation!!

One molecule of nitrogen reacts with 3 molecules of hydrogen and the reaction yields two molecules of ammonia. // A mole is just a big counting unit like a dozen or a million. So one mole of nitrogen reacts with 3 moles of hydrogen and the reaction yields two moles of ammonia. // So how do all the concentrations change?

Well for example if we have 0.586 M and the volume is despicable we have 0.586 moles of N2 too. Then if we have one mole of N2 for every three moles of H2 we will have to multiply 0.586 * 3 to get the moles of H2. Then the same with NH2. And then the concentrations are going to be the same as the moles we found because the volume is despicable. Is this correct?

The hydrogen change is not 0.586 * 3.

The given chemical reaction is:
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$
So:

One molecule of nitrogen reacts with 3 molecules of hydrogen and the reaction yields two molecules of ammonia.
A mole is just a big counting unit like a dozen or a million. So one mole of nitrogen reacts with 3 moles of hydrogen and the reaction yields two moles of ammonia.

The problem states "with x = 0.014 as the change in the concentration of $\ce{N2}$". It isn't clear if the change is positive or negative. It seems the problem statement is deliberately ambiguous.
So using the stoichiometry from the chemical reaction, if nitrogen changes by $-x$, then hydrogen changes by $-3x$, and the nitrogen changes by $+2x$. Conversely if nitrogen changes by $+x$, then hydrogen changes by $+3x$, and the nitrogen changes by $-2x$.
Now for the equilibrium:
$$K_\mathrm{c} = 1.2 = \dfrac{\ce{[NH3]^2}}{\ce{[N2][H2]^3}}$$
So we can calculate $K_\mathrm{c}$ for the various compositions of the gases and see if any is equal to 1.2.
\begin{array}{|c|c|c|c|} \hline
& \pu{\ce{[N2]}} & \pu{\ce{[H2]}} & \pu{\ce{[NH3]}} & \pu{calc}\ K_\mathrm{c} \\ \hline
\pu{initial} & 0.600 & 0.760 & 0.48 & 0.875\\ \hline
\pu{\Delta \ce{N2 = +0.014}} & 0.614 & 0.802  & 0.452 & 0.645\\ \hline
\pu{\Delta \ce{N2 = -0.014}} & 0.586 & 0.718 & 0.508 & 1.19 \\ \hline
\end{array}
So the composition of the gases must be equal to the last line in the table.

Interestingly the question poses a reaction in the gas phase, but then uses molarity. I don't think it is asked very well.

@Martin-マーチン - I agree. I also don't like problem specifying 2 digit concentrations when to follow the change need three digits are needed. So 0.60 molar for $\ce{N2}$ magically must turn into 0.600 molar.


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Reaction of copper oxide with acid
Copper metal is less electropositive than hydrogen and thus less reactive. It is unable to displace hydrogen ions from a solution of sulfuric(IV) acid.
Why then would copper oxide or copper carbonate react with sulfuric acid?
$$CuO(s)+H_2SO_4(aq)\longrightarrow CuSO_4 (aq)+ H_2O(l)$$
$$CuCO_3(s)+H_2SO_4(aq)\longrightarrow CuSO_4(aq) + H_2O(l)+CO_2(g)$$
Is it because copper ions are just spectator ions, where the real reaction happens between hydrogen ions and oxygen ions?
$$Cu^\mathrm{2+} O^\mathrm{2-} + 2H^+ +SO_4^\mathrm{2-}(aq)\longrightarrow Cu^\mathrm{2+} + SO_4^\mathrm{2-} + H_2O$$
Therefore the ionic equation is only:
$$2H^+ + O^\mathrm{2-} \longrightarrow H_2O$$
Therefore it does not matter whether copper is less electropositive than hydrogen, since the only reaction that happens is between hydrogen and oxygen?

This analysis seems vague to me there are many considerations such as lattice energy, hydration energy we cannot simply write the compound as it has split into ions

If it was true all oxides would be solube in water and there would be no classification of an oxide such as acidic or basic

In any case, regardless of quantification of CuO solubility in water or acides, it is not a redox reaction, but acido-basic reaction. In final state, $\ce{Cu^2+}$ goes to solution, and $\ce{O^2-}$  as a very strong base reacts with $\ce{H+}$.

I don't understand quite the premise of your question. The net reaction says that there is no redox couple involving copper, which is what you expected. The reactions that you've written down are all consistent with that observation. What's the problem?

Is it because copper ions are just spectator ions, where the real reaction happens between hydrogen ions and oxygen ions?
...
Therefore it does not matter whether copper is less electropositive than hydrogen, since the only reaction that happens is between hydrogen and oxygen?

Absolutely. Copper does not change its oxidation state.
Edit: To ???, please justify your downvotes. The question is about reactivity which may happen at solid-liquid-interfaces, too, not about solubility.


text:
Comparing between SN2 and E2
Question
I am confused between the questions B. And C. The earlier is SN2 and later is E2. But what makes the difference?

Look at the transition state in SN2. How likely (energy barrier?) is an inversion in C)?

It is pretty clear that you can't perform any kind of elimination in B due to valency constraints, it has to be SN2. For C,you can look at the chair form and decide

B seems more likely to be $\mathrm{S_N1}$, in C won't we require high temperature for $\mathrm{E_2}$ to occur?

@Zenix Oops yes, I made a slip up,it should be SN1 for C

@YusufHasan Bro, shouldn't SN1 be for B?

With methoxide and benzyl chloride, the substitution is more likely to be bimolecular. Regular solvolysis of benzyl halides is on the border between the two mechanisms.

@Zhe Won't argue with aikido player ;) how did you then decide between two, when it's on borderline?

I didn't. Solvolysis would be in methanol, without methoxide. Since methoxide is better at SN2, I think we're breaking the tie.

For B you get a benzylic cation in a protic solvent. Surely SN1 would be favoured.

The most important factor in these reactions are nucleophiles. Here, the use of strong base (Sodium methoxide) in both of the reactions favors the bi-molecular mechanisms ($\ce{S_N2}$ and $\ce{E2}$).
Now, the second factor would be degree of halogen-linked carbon. Lower degree carbons ($\ce{\text{deg.} < 2}$) favor $\ce{S_N2}$, while higher degree favors $\ce{E2, E1, S_N1}$. So, $\ce{S_N2}$ is favored in first and $\ce{E2}$ in second.
Temperature conditions and nature of solvent are least important factors, and should be looked up when above two factors are similar.