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Generating PDB file from electron density without force field minimization
Is there a step-by-step tutorial on how to generate a PDB file from an electron density file (*.ccp4)?
For my current project, I am trying to take some statistics of some PDB files. However, I do not think that the empirical energy function (MM) implemented in x-plorer could accurately reproduce experimental or QM data. As a result, I would like to use x-plorer to generate the raw PDB file myself (without minimization), but since I have never done this before, I was wondering if somebody could provide me some guidance on how this could be done.

Is it a PDB (Protein Data Bank) file with macromolecular structure?

@marcin, yes it is

"the raw PDB file (without minimization)" - this is not clear to me

all PDB structures undergo force field minimization after the x-ray diffraction data is obtained. This is done for refinement purposes. However, force fields are far from perfect, and I was interested in the structures from the raw diffraction data.

No, they usually don't. Perhaps you think about restraints that are used during the refinement.

@marcin I don't agree. Crystal structure refinement uses a priori information, which is derived from the CHARMM force field within the XPLORER program. The restraints which you talk about are the crystal structure diffraction data. The refinement is a combination of the two processes, force field and diffraction data.

Saying that restraints are the diffraction data makes no sense. Restraints are the a priori information.You may google "Engh and Huber refinement restraints" to read about the most popular restraints. I'm not familiar x-plorer. The vast majority of structures in the PDB is refined with Phenix.refine, Refmac or Buster.

@marcin It looks like my terminology was swapped. It seems like restraints are the force field data. In any case, it is well known that x-ray crystal structures undergo MM minimization (just look at the paper you suggested I google), so there is no point arguing about this.

The process of making structural models of macromolecules (the content of PDB files) is called model building, rather than generating, because it normally involves tedious manual building of the model in a graphical program such as Coot. A few programs can do automatic model building, but then the structure needs to be finished manually. Maybe this will change in the future.
All these programs have rather steep learning curve, so it's probably not what you are looking for.
The next problem is that you plan to start from an electron density map.
Assuming that the experimental method was crystallography, not Cryo-EM, the electron density map is derived from both reflection data and the model (and at the same time the model is built into the map, iteratively). So for the purpose of re-building or validation people generally use reflection data, not the map.


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Weak acid and weak base salt hydrolysis
While solving for weak acid and weak base salt hydrolysis, why do we take the degree of hydrolysis for both the anion and cation to be the same?
If we assume they are the same, when solving for $K_\mathrm{a}$ and $K_\mathrm{b}$ for the cation and anion using simultaneous equilibrium, I always get them to be the same, resulting in the solution always being neutral. I feel that both should have different degrees of hydrolysis.

I'm struggling to understand what you mean. Can you include a simple example calculation that gives an erroneously neutral solution?

I currently can only find a definition of "degree of hydrolysis" $h$ in terms of protein hydrolysis, where it refers to the number of cleaved bonds. In a more general setting, for a generic reaction $$\ce{AB <=> A+ + B-},$$ one could define this as $$h = \frac{n(\ce{A+})}{n(\ce{AB})} = \frac{n(\ce{B-})}{n(\ce{AB})},$$ because it is obvious that the solution has to be neutral. I am really unsure what you're getting at, please add more context.

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How do the pKa and pH indicate to the proper environment of protonation?
I have been reading an abstract of a paper for a couple of minutes (reference below) but I cannot understand a piece of it. The abstract:

The ionization state and $\mathrm pK_\mathrm a$ of the inhibitor 6,8-dimethyl-N5-deazapterin bound to the recombinant human dihydrofolate reductase (rhH2folate reductase) complex with NADPH was determined by a spectrofluorimetric method. The excitation spectra for bound ligand as a function of pH from 6.1 to 9.7 indicated it was the same cationic form as for unbound ligand, which is protonated on N3. However, the lower limit for the $\mathrm pK_\mathrm a$ of the bound form was determined to be 9.1, a value about pH 2.5 higher than that for free ligand, indicating that ligand bound to the enzyme is protonated at neutral pH. The excitation spectra for bound ligand as a function of pH were generated by computer simulation by employing corrections for the pH dependence of the proportion of bound ligand (variable $K_\mathrm d$; ligand-dissociation constant) and taking account of the different $\mathrm pK_\mathrm a$ values for bound and unbound forms. A plot of $K_\mathrm d$ values against pH showed a bell-shaped curve indicating that 6,8-dimethyl-N5-deazapterin bound to rhH2folate reductase. NADPH to form a ternary complex of ionised enzyme with protonated ligand and/or protonated enzyme with unprotonated ligand; the spectrofluorimetric results are consistent with the first alternative.

I made the not-understandable part (for me) in bold. Could some one help me to understand how do the previous sentences help to indicate to this sentence? Maybe I am missing something in the previous sentences.

Jeong, S.-S.; Gready, J. E. Ionization state and pka of pterin-analogue ligands bound to dihydrofolate reductase. Eur. J. Biochem. 1994, 221 (3), 1055–1062. DOI: 10.1111/j.1432-1033.1994.tb18824.x.

The sentence immediately before the bolded one indicates that the $\mathrm pK_\mathrm a$ value of the ligand bound to the enzyme was determined to be $9.1$. It also indicates that the $\mathrm pK_\mathrm a$ value o the free ligand is 2.5 logarithmic units less, i.e. $\mathrm pK_\mathrm a(\ce{Lig}) \approx 6.6$.
The human cell is well-buffered to ensure that the pH stays within a very narrow range. There are enough buffering compounds around, that we never need to use any Henderson-Hasselbalch equation or other difficult calculation methods to determine the protonation state of a given compound. We can simply examine its $\mathrm pK_\mathrm a$ value; if that is lower than the cell’s pH value, the compound is deprotonated. In all other cases, it is not.
The $\mathrm pK_\mathrm a$ value of the free ligand indicates it should be deprotonated in the cell. However, upon binding to the protein with the $\mathrm pK_\mathrm a$ value rising by $2.5$  logarithmic units, the ligand suddenly becomes protonated. This is what the bolded sentence implies.

Aah ok, thank you a lot! so protonated here means simply, not ionized, right?

@MohammedNoureldin Protonated means that the most acidic proton has not been lost. Whether this corresponds to a neutral structure (e.g. $\ce{HCN}$) or to a cationic one (e.g. $\ce{NH4+}$) I cannot say without checking the structure.

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Physical Explanation for variation of Gibbs Free Energy with Temperature and Pressure
Reading Atkins Elements of Physical Chemistry right now, and the book says:
"At constant pressure and entropy, an increase in temperature ($\Delta T\gt0$) results in a decrease in $G_\mathrm m$ (molar Gibbs energy) ($\Delta G_\mathrm m\lt0$)."
"At constant molar volume and temperature, an increase in pressure ($\Delta p >0$) results in an increase in molar Gibbs energy ($\Delta G_\mathrm m\gt0$)."
They have equations to describe this but I'm not really interested in the mathematical derivations (those are understandable) but rather in a logical, physical, and qualitative explanation of why these relations occur.
Given that $\Delta G$ indicates the maximum non-expansion work derivable from a system, what does this mean for the above relations? Does increasing the pressure of the system require work input ($\Delta G$ is positive) while increasing temperature result in the system doing work/an output of work ($\Delta G$ is negative)?
So, given a system (like a gas or something), why does $G$ change the way it does in response to $T$ and $p$ changes, and what does this physically mean for the system?
I'm a high school student (went through AP Chem) who's just trying to delve deeper into chemistry, so the simplest and most "layman's terms" explanation would probably help.

Probably the easiest way to picture it is that, since the Gibbs function is indeed a measure of the potential of the system to transfer energy in certain ways, these transformations affect the energy that can be transferred.
For a certain system at constant pressure and entropy (which therefore cannot absorb thermal energy from its surroundings), an increase in temperature is necessarily linked to an internal transformation - an internal work - that reduces the potential energy available to exert work on its surroundings; that increase in temperature is "using up" part of the potential reversible work as measured by the Gibbs function.
On the other hand, a system at constant volume and temperature which increases its pressure (and which, given these constrains, is therefore receiving some form of energy input from the outside; note that there isn't a pressure-volume tradeoff) will increase the potential reversible work it can exert on its surroundings, as measured by the Gibbs function.
It is important to note the implications of the constant conditions that each of those two cases set, because these constant conditions are telling you which forms of energy transfer are permitted or prohibited in each case.

While the stated problem seeks a non-mathematical answer, it is a good idea to review the equation that is at the source of the problem. For a a homogeneous phase at constant composition we can write:
$$\mathrm dG = V\mathrm dP - S\mathrm dT \tag1$$
This equation describes how the free energy for the system depends on T and P.
A possible source of confusion with the way this problem is formulated is that it compares G at different T or P. Another answer does a good job of explaining how G might be affected by changes in T or P. Still, it remains that $\Delta G$ represents the max amount of of non-$PV$ work that can be performed at constant $T$ and $P$. In that sense, comparing $\Delta G$ computed between two different temperatures or pressures should not be interpreted as the maximum possible non-expansion work that can be performed. It is important to keep this in mind. In chemistry we are most often interested in the effect of changes in $T$ or $P$ on $\Delta G $ for some process such as a phase change or chemical reaction which is otherwise carried out at constant $T$ and $P$. At the risk of redundancy I restate: we are interested in $\Delta G $ specifically because it is intended for the analysis of processes at constant $T$ and $P$.
For such a process from one state to another it is proabably clearer to write
$$\mathrm d \Delta G = \Delta V \mathrm dP - \Delta S \mathrm dT \tag2$$
where we are comparing initial and final states during the process ($\Delta G = G_{final} - G_{initial}$).
Once that is clear, the question remains: why does a decrease in $P$ at constant volumes and temperature, or an increase in $T$ at constant entropies and pressure, result in an increase in $\Delta G$? (note my use of the plural since we are referring to initial and final states of $V$ and $S$)
The short answer is that it doesn't have to!
Or rather, the change in $\Delta G$ depends on whether the difference between the initial and final entropies and volumes is positive or negative. For instance, if the difference in volume is positive - that is, the volume increases when the transformation is carried out - then an increase in pressure will result in a less stable (higher free energy) final state relative to the initial state. In fact, increasing the pressure will require more work to be performed on the system for the process to be completed. Such a response is implicitly described by Le Chatelier's principle.

Total entropy and Gibbs energy:
An alternative view on the Gibbs energy may help to get some insight:
$$\Delta S_{sys} + \Delta S_{surr} = \Delta S_{tot} = -\frac{\Delta G}{T}$$
The second law of thermodynamics tells us that a process is spontaneous only if the total entropy $S_{tot}$ does not decrease. The sign of $\Delta G$ tells us whether $S_{tot}$ increases or decreases. Note that this relation is only valid at constant pressure, and thus Gibbs energy is only really useful for evaluating the feasibility of a process at constant pressure. Let's have a look how $S_{tot}$ changes in the first case.
Increasing temperature at constant pressure and entropy:
It is important to note that the "at constant entropy" refers to $S_{sys}$ only.
How do we increase temperature at constant pressure and entropy? We can for instance have an exothermic reaction. One part of the generated heat is used to increase the temperature of the system itself. This would inevitably increase $S_{sys}$, but we can compensate it by transferring the rest of the generated heat to the surroundings. Hence, we increase $S_{surr}$ and in return increases $S_{tot}$:
$$\Delta S_{tot} = \Delta S_{surr} = \int_{T_i}^{T_f}\frac{S_{sys}}{T} dT$$
Increasing pressure at constant volume and temperature:
As mentioned in the beginning, Gibbs energy is not really useful for processes that are not at constant pressure. We can still rationalize the increase of Gibbs energy with increase of pressure by comparing two constant pressure experiments at different pressures, but same initial volume and temperature. Let's look at the following reaction:
$$\ce{A_2 <=> 2 A}\quad\text{and}\ \Delta G_1 = 2G_A-G_{A_2}$$
If we now increase the pressure at constant volume and temperature, the Gibbs energy of each component increases by $V\Delta p$ and we end up with:
$$\Delta G_2 = \Delta G_1 + V\Delta p$$
Hence, our reaction is less feavored / more disfavored at higher pressure. What energy does $V\Delta p$ correspond to? Our reaction doubles the number of gas molecules. Increasing the number of gas molecules becomes more expensive at higher pressure and $V\Delta p$ is the additional energy we have to pay. It's basically just the manifestation of the Le Chatelier's principle.

For the last case (V constant) $q_V=\Delta U = \Delta H -V\Delta p = -T \Delta S_{surr}$

@BuckThorn Yeah, you are absolutely right. My second explanation was garbage. I found a better example to rationalize the increase of Gibbs energy with pressure.

"The second law of thermodynamics tells us that a process is spontaneous only if the total entropy  does not decrease." That doesn't quite capture it. The 2nd law states that  does not decrease for any process* and that, for spontaneous processes,  must increase. [*And, for completeness, I need to add: more specifically, for any process involving a sufficiently large number of particles for the 2nd law to apply.]

"The sign of Δ tells us whether  increases or decreases. Note that this relation is only valid at constant pressure..."  For this relationship to hold, you also need to specify constant $T$, and no non-$pV$-work.  See my answer here:  https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/124470#124470

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NaN3 quantity for the generation of N2

You wish to design an airbag for a car. The gas that is released must be at a pressure higher than that of air – $\pu{110 kPa}$ at a temperature of $\pu{22 ^\circ C}$. The bag that you develop can expand to a volume of $\pu{45.5 L}$. What quantity of sodium azide $\ce{NaN3}$ should be used to generate the required mas of $\ce{N2}$ gas?

I've been given this question on my revision sheet. So far I've figured out using Boyle's law that I need 2 mol of $\ce{N2}$ gas. However, I am unsure of how to transfer this over to the $\ce{NaN3}$ and use the ratio or what not to find the number of moles there.

Use $\ce{2NaN3 -> 2Na + 3N2}$ and stoichiometry to reach the final answer.

$$\ce{2NaN3 -> 2Na + 3N2}$$
As you already found the amount of $\ce{N2}$ required, I'll just give you a hint on how to get the amount of sodium azide required to produce that much nitrogen.
We know that 2 moles of sodium azide gives us 3 moles of nitrogen. So $x$ moles of sodium azide gives us $2$ moles of nitrogen. Solve for $x$. That's all!


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Rationalising the trend of standard electrode potentials between the +II and +III oxidation state of first-row transition metals
Trends in standard electrode potentials of d block elements:

An examination ofthe $E^0 (\ce{M^3+}/\ce{M^2+})$ values (Table 8.2) shows the varying trends. The low value for Sc reflects the stability of $\ce{Sc^3+}$ which has a noble gas configuration. The highest value for Zn is due to the removal of an electron from the stable $\mathrm d^{10}$ configuration of $\ce{Zn^2+}$. The comparatively high value for Mn shows that $\ce{Mn^2+}\ (\mathrm d^5)$ is particularly stable, whereas comparatively low value for Fe shows the extra stability of $\ce{Fe^3+}\ (\mathrm d^5)$.The comparatively low value for V is related to the stability of $\ce{V^2+}$ half-filled $\mathrm{t_{2g}}$ level, Unit 9).

It says that low value of Sc is due to stability of $\ce{Sc^3+}$. At the same time, low value of V is due to stability of $\ce{V^2+}$.
Don't these statements contradict each other? If not, could someone please explain this paragraph in an easier way?
I m not able to understand it.

https://chemistry.stackexchange.com/a/66172/9961

Yes, those statements contradict each other, so you might be better off ditching this book. My take on the matter is already linked in the comment above.

It is a standard ncert book

Here low value of 'V' is low value in magnitude(as it is negative). Some chemistry text don't bother to write clearly in these type of cases.


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Isn't the dual behaviour of subatomic particles contradictory?
I recently came across Molecular Orbital Theory. In this, if two electrons are bonding, their waves merge and they form a proper bond. But this is directly opposite to the particle theory, according to which they should repel each other as they have like charges.

Take a nice simple system like a hydrogen molecule. It's true that the two electrons repel each other, but at the same time they are attracted by the positive charge of the two protons.
If we start with an $\ce{H2+}$ ion, with just a single electron, it should be obvious that the electron is tightly bound because it feels an attractive force from both the protons.
Now imagine adding the second electron to form a neutral hydrogen molecule. The second electron will be repelled by the first electron that is already present but it will be attracted by both the protons, and the attraction from the two protons wins. The binding energy of the second electron in a hydrogen molecule is about 15.4 eV, so it is even more strongly bound than the electron in a hydrogen atom.
Note that this is not related to the wave/particle duality. Wave particle duality is in any case a misguided concept since particles are quantum fields not waves or particles – they simply behave approximately like waves in some circumstances and approximately like particles in others. See this post on the Physics Stack Exchange for more on this.

For some quantitative analysis, see this question & my answer.

I would add the influence of the Pauli exclusion principle.


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Handling reagent bottles
I have done an experiment to prepare a standard solution. To keep this standard solution , I'm going to put it into a reagent bottle .
If this reagent bottle is clean but not dry , what must I do ?
my attempt is to use small amount of ethanol to absorb the water and throw away the ethanol and allow the rest to evaporate away. Is this the correct method if I were to only use ethanol ? If not what must I do ?

Normally in a lab you will use acetone after ethanol.  Then you dry the bottle in hoven . At home, using only ethanol and a hairdryer can be an alternative.
However, especially without water sensitive materials, and perhaps having a standard solution in water, the main point is to "normalise" the bottle.
In other words you wash your bottle with a little amount of standards solution that you pour away.  And repeat this step at least 3 times.
As such, whatever is residual in your bottle, will already be standard solution.
This applies if the bottle is already reasonably clean, as in your case.


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Why does old gasoline seem to have more "kick"?
From my personal experience, old gasoline (leftover in a container from the previous year) has more "kick" (power) when I actually use it. Higher idle RPM, higher throttle sensitivity.
It says here that it's not the gasoline itself that burns, but the vapour. And old gasoline is bad because it has vaporised over time. Thus, old gasoline must be weaker.
However, I read somewhere that gasoline vapour has nothing to do with octane rate, so there's no difference in octane between the gas left out in the open for a while and the gas sealed in a container.
I'm confused. Does old gasoline seem to have more "kick" because it needs more compression to combust? Or what is it?

Gasoline on storage leads to gum separation. Maybe the heavier hydrocarbons separate?

Similar as in the answer by @AJKOER I think that gasoline - a blend of hydrocarbons and other compounds (e.g., MTBE ether) - may alter over time.  Without experimental evidence, assuming it is stored only for about a year in a dry and cool place in the (metal) jerrycan, I speculate its compounds with a lower boiling point evaporate easier than its compounds with a higher boiling point across the cap of the jerrycan.
As stated correctly in the question, it is not the liquid gasoline driving the engine.  It is the compressed aerosol of the gasoline ignited by the spark plug.  With said relative enrichment of the compounds of higher boiling point in comparison to those of lower boiling point, I speculate a higher compression ratio may be attained without premature (auto)ignition of the fuel.  And because of the higher compression ratio accessible, the combustion may offer more mechanical work (or, as mentioned in the question, offer «more kick»).

My candidate to explain the increased performance is the removal of water as part of the corrosion reaction occurring in metal storage containers. To quote a reference:

Since oxygen has different solubilities (higher for the organic and lower for the aqueous phase), anode or cathode regions are formed on the steel or metal surface. This creates favorable conditions for electrochemical corrosion in a given environment.

As such, I would argue that not only does oxygen content impact corrosion rate, but potentially more important, is that the removal of water over time (from corrosion chemistry) also increases O2 solubility (per above). This could be especially favorable when the fuel is vaporized and mixed with air in the ignition chamber.
[EDIT] Further comments, same source:

Oxidation of biodiesel leads to the formation of various products such as peroxides and hydroperoxides. During degradation, these products transform into shorter-chain compounds such as low-molecular weight acids, aldehydes, ketones, and alcohols [9].

Per another source:

In the winter, companies produce a gasoline containing lighter hydrocarbons, making the liquid more volatile and therefore easier to ignite.

This source also notes with more volatile fuels, evaporation can leave behind a heavier fuel mix.
Corrosion induced oxidation of biodiesel, in particular, appears to not only impact potential oxygen content of the fuel, but may add peroxides as well. The creation of lower molecular weight organics should also increase flammability.
An interesting point once presented in Wikipedia on Methanol, to quote:
"As a fuel for mud racers, methanol mixed with gasoline and nitrous oxide produces more power than gasoline and nitrous oxide alone."
Which, in the current context, supports my opinion that fuel mix chemistry (whether corrosion based, or otherwise, involving O2 or N2O) CAN IMPACT PERFORMANCE.
So, more oxygen content, less heavy fuel mix, more complete ignition, may explain some of the so-called 'kick' that has been referenced. On the other hand, significant evaporation is likely detrimental.
[EDIT] Yet another path as a consequence of corrosion chemistry, to quote from Wikipedia on Tetraethyllead:

At the temperatures found in internal combustion engines, (CH3CH2)4Pb decomposes completely into lead and lead oxides as well as combustible, short-lived ethyl radicals. Lead and lead oxide scavenge radical intermediates in combustion reactions. Engine knock is caused by a cool flame, an oscillating low-temperature combustion reaction that occurs before the proper, hot ignition. Lead quenches the pyrolysed radicals and thus kills the radical chain reaction that would sustain a cool flame, preventing it from disturbing the smooth ignition of the hot flame front.

Also on the performance effect of Tetraethyllead (TEL):

Antiknock agents allow the use of higher compression ratios for greater efficiency[17] and peak power.[18] Adding varying amounts of additives such as low percentage TEL or high percentage ethanol to gasoline allowed easy inexpensive control of octane ratings... Aviation spirits with TEL used in WWII reached 150 octane to enable supercharged engines such as the Rolls-Royce Merlin and Griffon to reach high horsepower ratings at altitude.[19] In military aviation, TEL manipulation allowed a range of different fuels to be tailored for particular flight conditions.

So, the corrosion process on even minor alloy metals, found in the metal storage container, may have also introduced an organometal compound with similar properties to TEL in its ability to scavenge radical intermediates.

I'm not sure if I could be served leaded gas at stations in NJ.  (Because I never needed it, I didn't asked the gents yet.)  Even if available (i.e., leaded aviation fuel probably still is in use), lead would damage catalytic converters installed in cars driven by an Otto engine; and I assume a large population of these cars built since the 1990s have them.  Thus I think nowadays (2020s), the section about TEL becomes less and less important to consider.

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What are the ways to make radioactive water drinkable again?
Let's say survivors in the area contaminated by radioactive fallout need to somehow process the water from rivers, rain or pipes. How could they go about it? Is it even possible to 'remove radioactivity' from the water?

The water itself isn't radioactive, rather particles dissolved in it, because water is a quite good solvent. Therefore, filtration (e.g. with activated carbon) or precipitation can remove such particles. Also a ion exchanger can remove radioactive species. But this can take quite long, especially if the needed amount is large.

Thermonuclear explosions produce a large amount of tritium. Can't filter that out, because it's likely bound in water.

I'll point out that all surface water has tritium, even before mankind split the atom.

@Karl OP was not specific about the nature of the radioactive contamination, but the amount of tritium produced by nuclear weapons are generally not a serious threat due to there extreme dilution and short lifetime in body. You also can effectively cure it by drinking normal water.

@Karl - what mechanism produces ‘large’ amounts of tritium? Particularly compared with all the fission byproducts?

@Jon Custer: Compared to the total natural amount of $\pu{3.5 kg}$ tritium in the atmosphere, the amount produced by nuclear explosions or in nuclear rectors is large (CANDU reactors e.g. produce $\pu{1 kg}$ tritium per $\pu{5 GWa}$).

@aventurin - a stated mechanism would allow rough quantification. An unboosted primary will make essentially no tritium.

The following link shows tritium measured in Ottawa (Canada). It has a large spike in 1964 due to nuclear testing (note that the scale is logarithmic) http://snobear.colorado.edu/Daniel/isotopes/tritium.html

@JonCuster A thermonuclear bomb breeds tritium out of lithium (Li+n->T+He), which then gets fused with deuterium (D+T->He+n). Naturally some of the T escapes the fusion.

@Karl - I don't dispute the 'some' - I dispute that it is a 'large amount'. Beyond the tritium either from boost or the secondary, that is about it. The cross section for D(n,g)T is quite small (heck, the 3He(n,p)T reaction has a higher cross section).

@JonCuster The main "explosive" is lithiumdeuteride, and Li gets split into T + He-4 (plus one additional neutron from Li-7). D + Li + n -> D + T + He -> 2 He + n. There is necessarily one equivalent of tritium in the reaction, isn't there? Natural water does not get activated strongly, of course.

@Karl - the main explosive in the secondary is the U or Pu - the D-T reaction is there to kick-start it with a pulse of neutrons. The energy from D-T fusion is a small percentage of the total output.

@JonCuster Either you or I have some wrong idea (or different nomenclature) about thermonuclear bombs. My understanding is primary=Pu fission, secondary=LiD (10-30 times as "large"), optional tertiary=U-238. But OK, de:wp also says that tritium is used up almost completely. Not sure how, do they put in some extra deuterium?


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Fajans' Rules Versus Lattice Enthalpy
In analyzing the solubility among $\ce{CaF2, CaCl2, CaBr2}$, I found that $\ce{CaF2}$ is insoluble due to high ionization enthalpy.
But comparing among $\ce{AgF, AgCl, AgBr }$, it was said  that $\ce{AgF}$ is more soluble because of Fajans' rules (consequence of polarization). $\ce{AgCl, AgBr}$ have high polarizing effect and so they have high covalent characteristics.
Now question is why can't I use the Fajans' rules at $\ce{CaF2}$? Where can I use Fajan's Rule and Where should I use Lattice enthalpy rule?
This problem is also present at comparing the melting point of $\ce{NaCl, KCl}$. Fajans' rules cannot be applied here also.

The thing you have to understand is that the enthalpy of dissolution is actually the sum of lattice enthalpy and hydration enthalpy. So being a thermodynamic concept, we will have to turn to Gibbs free energy i.e. ΔG= ΔH-TΔS for completely understanding the solubility trends. For the process to be spontaneous, ΔG(dissolution)<0 . Now, entropy will generally increase while dissolving, so we can safely assume ΔS>0 for now. So obviously, the factor of -TΔS will be negative for T>0, so this contributes to the solvation of the said compound.(From now on, ΔH will be referring to enthalpy of dissolution)
Now, it all comes down to the ΔH part : it's magnitude and sign. If ΔH is positive and initially greater than -TΔS, then ΔG will be positive, and as temperature is increased, the -TΔS factor will ultimately overcome the positive ΔH and then, ΔG will become negative. And indeed, this is why you will find that $\ce{PbCl2}$ is insoluble in cold but quite soluble in hot water. Similarly, you can consider the implications of variations in the magnitude and sign of ΔH on solubility (in a polar solvent) yourself.
Now, coming to your specific problem, you see,  you first of all have to consider the configurations of $\ce{Ag+}$ and $\ce{Ca^2+}$
$\ce{ Ag^+ -  [Kr] 4d10}$ and
$\ce{ Ca^2+ - [Ar]}$
So you see, the calcium ion has 8 electrons in it's outermost shell ( 2 in 3s and 6 in 3p) while the silver ion has 18 electrons in it's outermost shell( 4s2 4p6 4d10). Now, due to the ineffective shielding by the d-electrons, effective nuclear charge on the valence shell and hence it's polarising power increases tremendously. Here, the calcium ion possesses an "inert gas configuration" while the silver cation has a kind of  "pseudo noble gas configuration"  (as it is not exactly a noble gas configuration, but still the cation is quite stable). On moving further down the periodic table,you get cations like $\ce{Tl^+, Bi^3+}$ etc. which have neither 8 nor 18 valence electrons (hence they can be called "non-inert gas configurations" ) and owing to the even poorer shielding by f and d electrons, there polarising power is even higher than the previous two configurations. So, the general order of polarising power of cations can be : Non-inert> Pseudo-inert> Inert gas configuration .( Note: All this can also be derived quantitavely by actually calculating effective nuclear charge on valence electrons by Slater's rule)
So, between $\ce{Ag^+}$ and $\ce{Ca^2+}$, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as compared to $\ce{Ca^2+}$ . As to your question about when to apply which concept: You are basically required to compare the sum of the hydration enthalpy and lattice enthalpy. If the lattice is very strong(example: LiF due to compatibility of ions and less interatomic distance) then the hydration enthalpy can be ignored, while for something like $\ce{LiNO3}$ the lattice is weak and hydration enthalpy is highly negative, so lattice enthalpy can be ignored for the final ΔH. You can use any concept as per your convenience, be it Fajan's rule, or extent of hydration or lattice strength; they will genrally complement each other provided you apply them correctly
For alkali metal chlorides like KCl and NaCl, the cations have only got a unit positive charge (which is not much to cause a lot of polarization) and more importantly, these cations belong to the inert gas configuration type, so here lattice energy will be the dominant factor

It might be a good idea to brush up your usage of MathJax a bit. I'm quite certain you mean $\ce{Ca^2+}$ ($\ce{Ca^2+}$), not $\ce{Ca2+}$ ($\ce{Ca2+}$). If you want to know more, please have a look here and here. Please don't use any edit statements, if someone really needs to know what has changed and how, the edit history is available. Otherwise these are just superfluous, distracting noise for the (first time) reader.

Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number larger than two, the bond is ionic.
\begin{array}{lr}
\text{Compound}  & \Delta(\text{EN})\\\hline
\ce{AgF}   &  2.03\\
\ce{AgCl}  &  1.23\\
\ce{AgBr}  &  1.03\\
\ce{CaF2}  &  2.96\\
\ce{CaCl2} &  2.16\\
\ce{CaBr2} &  1.96\\
\ce{KCl}   &  2.34\\
\ce{NaCl}  &  2.18\\
\end{array}
For $\ce{Ag}$-stuff and $\ce{KCl}$ and $\ce{NaCl}$ it's obvious: most ionic is least stable. But why do we have $\ce{Ca}$-stuff, ignoring this rule? The problem is ... of rivalry kind.
The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For $\ce{Ag}$, $\ce{K}$, $\ce{Na}$ (monovalent atoms) it's true. For bivalent atoms, such as $\ce{Ca}$, delocalization comes into play: those two little anions, fighting for electron pair.
Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak ($\ce{Ca}$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as $\ce{F}$) would stretch the rubber harder than not very strong (such as $\ce{Br}$), delocalizing electrons and making $\ce{CaF2}$ more stable.
I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made.

[edited]Picture source: http://www.chemhume.co.uk/ASCHEM/Unit%201/Ch3Chemstr/Images%203/electronegativity_values.jpg

'Bond strength depends on electronegativity of bonded atoms.' This is an assessment so fuzzy it says everything and nothing at all. The difference of electronegativities is also a quite poor indicator of whether a bond is ionic or not. It is also not a sharp line, no matter what arbitrary number you are assigning. AgBr is a salt with almost exclusively ionic bonds and the difference is only about 1.0. HF on the other hand is a covalent compound and the difference is about 1.8.

'The weakness of ionic bond is it localization of electrons - they are comparatively well-localized.' That is just plain wrong. Ionic bonds are not localised at all, they have no direction as covalent bonds would have. The dominant force it electrostatic attraction here. And after that how well the ions fit into a crystal lattice, which defines the lattice energy and ultimately how well the crystal can be dissolved. The whole fighting story is very much misleading, because there is nothing like a rubber involved.

Please also add the sources and citations to images you have not compiled yourself.

"Ionic bonds are not localised at all", but electron pairs, which are "bonds", are well-localized, and I've written about electrons (as you can see from citation you've used).
The whole story is metaphoric, and I've mentioned that. You can't fight for probabilistic wave with your fellow, at least, not as for physical object. Rubber can be stretched, and it's easy to "physically" imagine delocalization via stretching (the easiest way I know, at least).  Thank you for the tip about lattice energy - I will review this topic, +1.


text:
Carbon in Glass - XPS Analysis
I was reading a paper and found this elemental XPS analysis of an unmodified microscope glass slide surface.
C (1s): 17.7%
Si (2p): 28.47%
O (1s): 53.83%
It seems strange for a piece of normal glass to have almost 20% of Carbon, since most of it is silicon dioxide and other metal oxides in smaller amounts.
I am not used to XPS analysis so there might be something I am overlooking (maybe adsorbed CO2 on the surface ?), can someone tell me if this is a normal result for an XPS analysis of glass, and if it is, where does the carbon come from?

Was the slide washed or cleaned before analysis? Almost every surface gets a few to several at% C on the surface just by air exposure (hydrocarbons etc). XPS is hugely surface sensitive after all, so contamination can have a big effect on the data. A "dirty" slide could certainly approach 20 at% carbon.

Well, XPS analysis should detect atoms as light as Carbon, but not atoms as light as Hydrogen.
Far be it from me to claim any expertise on microscope sliders, but I have used one or two in my career, and I am hazarding a guess. It is likely a pure glass ($\ce{SiO2}$) sheet laminated with some acrylic glass. The hydrogen in the acrylic polymer are too light to appear on the XPS.
There is a slight possibility, although I consider it very unlikely, that it is glass with a substrate of moissanite on top, which would in theory make it scratch proof, due to the hardness of $\ce{SiC}$.
If you would include whether the %'s are weight percent or atomic percent, it would be easier. If it is atomic, then the case for $\ce{SiC}$ is more likely, since the ratio ($\ce{Si:O} > \ce{1:2}$) and the extra Si would be in SiC. If it is weight percent then there is some math to be done. I still think it is just glass and acrylic glass.

It is in atomic ratio... I think it's possible that the glass is laminated with some sort of C-containing layer, although no information is given in the paper.

Do you think it could be surface contamination of a pure silicate-based glass (i.e no acrylic)?

This carbon they have reported (considering the amount present and without knowing the sample preparation) is likely adventitious carbon.
This carbon is on every surface and so perhaps the glass was not cleaned well before the measurement. Even then it is likely some carbon will still be present.
Adventitious carbon (binding energy 284.8 or 285.0 eV) is often used as a charge reference in XPS binding energy analysis.

My question is did they crush the material and kept as a powder or did they keep it a slab of glass?
If they have done XPS on the powder then they just reported the atomic percentage coming from the carbon tape which is not part of the materials.
It is simple to get the new atomic percentages after you don't count in the carbon contribution.
If they have used the glass slab as a whole then there is a possibility of contamination from the carbon tape.


text:
Nomenclature explanation for a bicyclic molecule consisting of two fused pyrimidines
From Sigma Aldrich, the following molecule is named 1,3,4,6,7,8-hexahydro-2H-pyrimido[1,2-a]pyrimidine

I would like to understand how it came to be named this.
My main issue is reconciling the bridge nomenclature [1,2-a] with the hydrogen nomenclature 1,3,4,6,7,8-hexahydro-2H.
I came up with a system that works, but I feel it's likely incorrect.

What I did
First, it seemed apparent I must use two different numbering systems. For specifying the bond fusing the two pyrimidines, I have the following counting:

This would explain where the [1,2-a] came from.
However, starting with 1 on the bridging nitrogen fails when numbering for the hydrogens. The only way I can get the hydrogens right is to change the numbering to:

† (Reasoning at the bottom of post).

Questions
I have two problems understanding this.

It seems pretty non-intuitive to require two separate counting systems. Is that a mistake?

Even if I accept two separate counting systems, the molecule name designates the hydrogen numbers strangely. Based on how I counted the hydrogens, I would expect the name to be 1,2,3,4,6,7,8-heptahydro-pyrimido[1,2-a]pyrimidine. Instead, the 2 position is strangely labeled separately as 2H. That's so strange; it must indicate a mistake from me, right?

Thank you for advising.

†  I based this on examples from page 8 of a springer document:

where it's shown the numbering always starts in the peri position, but it does stop to number the bridge if a heteroatom is there.

side note
Interestingly, there's another page on sigma aldrich which names this moleculue as 1,5,7-triazabicyclo[4.4.0]dec-5-ene. This name I can understand, although it does seem to contradict the numbering always beginning at the peri position (it begins with the bridging nitrogen here, presumably to minimize the double bond number to 5).
I would assume this name applies to the standalone molecule, and that as a ligand or coordinating compound it uses the pyrimido pyrimidine formulation from above, correct?

Edit - IUPAC citations
So I managed to find a way to peek directly into the IUPAC book (2013 edition) without having a physical copy via previews on google books. If this isn't allowed feel free to edit it out of my post.
The relevant section begins on page 615. You can manually flip to this page by clicking in the top right. I was able to scroll through the entire relevant chapter this way.
There's a brief distilled version on the same topic from rsc; relevant pages are 36(d) and 48.
At the end of the day, I didn't really understand despite having the IUPAC in front of me. It seems very nuanced, and with no way to check my answer, I don't know if the sigma aldrich source is up to date with the current standard or if I'm just not getting it.
A couple things I suspect I might have learned:

In heterocycles deviating from the trivial name (such as pyrimidine, pyrrole, pyran, furan etc), a functional group is demarcated with the H nomenclature. In this context, H actually refers to a locant (ring carbon) in a valence that deviates from the trivial structure. This explains the 2H by the carbonyl in the bottom molecule below (the top molecule follows rule 2 discussed next).

However, a greater priority rule also exists. Namely, when a double bond in the trivial structure becomes saturated, the H will be placed here instead. If next to a functional group (such as a carbonyl group), the H will be placed alpha to the carbonyl and not on the carbonyl locant. This can be seen in the first example of napthalen-1(2H)-one, as well as a couple others below.

I believe this rule accounts for the 2H in my molecule above
Honestly, I found this rule really tricky, and I'm sure I still don't get it. My impression is it's important to draw the trivial structure first to note exactly where the double bonds are. Even though it's a resonance, the precise location of the double bonds matters in the nomenclature, at least that's how I understood the explanation here for 1,3,4,5-tetrahydronaphthalene-4a(2H)-carboxylic acid.

Incidentally, anthracene always follows a numbering pattern with 9/10 in the center aromatic cycle. I found 2 other molecules that also followed such a fixed numbering pattern even though it deviates from IUPAC standards for every other heterocycle (I believe one was phenanthrine but I'm not sure). The pyrazine on the left is an example of an exception to case 1.

I found a couple examples somewhat reminiscent of my own:

The closest example I could find was a fused dipyran:

What we see is an example of case 2 above. Pyran should have two double bonds in each ring. Each pyran ring is marked with an H in the lowest numbering possible. After that, "hydro" is used for the remaining positions. It seems the H is only needed once, as if it's to announce something has deviated from the trivial structure, and after that normal "hydro" is used.
Why my ring, which is a fused pyrimidine, only uses H for one ring and not the other remains a mystery to me.
In general, the exact details, from numbering schemes to H assignments, are a mystery to me, like the 5,8 carbonyls not being demarcated with an H in 1H-cyclopenta[b]naphthalene-1,5,8-trione.

or 6H instead of 7H in

But, this is as far as I got. Maybe all these examples will help someone else better elucidate the pattern going on here.

The numbering is put in place after fusion of  the rings took already place.... look for fused system and perhaps H-S nomenclature.  I should go through the rules myself, so comment only.

Okay. I can do the numbering after fusion. Is my numbering for hydrogens incorrect? I am referring to my 2nd question.

Don't know at the moment I would give a locator 1.... I can't look for it . Mention me if you can't find the rules as I have hard notes on  them somewhere. I might try to find them

I have looked at the rules just to get as far as I did. While I don't have the IUPAC book, the acdlabs website labels it rule B-3.4, and the springer document I linked discusses the same concepts. It's probably some amalgamation of multiple rules that fit together in a way I can't see, or the molecule is just misnamed. Don't worry if you can't find your notes; you'll probably just spend hours fooling around with it like I did. Maybe someone else already understands it.

Concerning your doubt about the two names , assuming both are right, they are two accepted names for the same molecules.  There are many cases in which it happens ( at least at the time of my organic II  course iupac was packed with exceptions and things like this ). About final numbering I really don't know.

I'm voting to close this question as off-topic because despite possibly being an interesting question about complex nomenclature, it currently reads like a stream of consciousness with no clear single question, using large blocks of text copied directly from the IUPAC Blue Book.

I think your edit could be the base of an answer.


text:
Can I approximate the CO concentration of a combustion product, if I know the CO2 and O2 concentration?
Assume I have the product (exhaust gas) from a combusted air/petrol mix. The mixing ratio was richer than stoichiometric (i.e., air:fuel < 14.67 in this case), so there was no excess of oxygen.
I know the $\ce{CO2}$ and $\ce{O2}$ concentration of the combustion product, say 11% and 2% respectively. I also know that the combustion product only consists of $\ce{CO2}$, $\ce{H2O}$, $\ce{CO}$ and leftover $\ce{O2}$. (All fuel combusts to $\ce{H2O}$ and either $\ce{CO2}$ or $\ce{CO}$.)
Is it possible to approximate the $\ce{CO}$ concentration of the combustion product with that information?

No, that wouldn't work since there are too many unknowns. Combustion is a very complex process. In most chemistry problems "complete" combustion is typically assumed.

You’re ignoring uncombusted and partially combusted hydrocarbons in the equation.

If you knew the H2O concentration you could estimate CO from an oxygen balance.  If you knew the uncombusted fuel concentration you could estimate CO from a carbon balance.

@MaxW What unknowns would have to be known then? I know that real combustions are more complex, but i'm looking for an approximation, so the calculation can be kept simple.

@Jan I know, I said that in my question; that the combustion only results in CO2, H2O and CO. I assume al HC to be combusted to either CO or CO2 to keep things simple.

@CurtF. If I don't know the H2O concentration, and I assume no uncombusted HC. Is an oxygen balance also calculable with that? If you know the CO2 concentration that would be achieved by a completely burned stoich mixture,(around 16% i believe) would that make an oxygen balance calculable?

I guess if you are willing to assume that all fuel carbon that was not converted to CO2 was converted to CO, you could estimate CO.

@CurtF. I can assume that, because in reality, only a very small part of HC will remain. If I ignore that, it won't be of significant influence on my outcome; the accuracy of my sensors is of way more influence. Could you help me approximate the CO?


text:
Quantitative determination of chemical composition of a simple unknown mixture

Unknown mixture A [$\ce{H2SO4}$ and $\ce{CuSO4}$]
Unknown mixture B [$\ce{H2SO4}$ and $\ce{HNO3}$] (ranged between $\pu{0.1 - 0.3 M}$ for the solution)

(E.g. if you are using Unknown mixture B [$\ce{H2SO4}$ and $\ce{HNO3}$], concentration of $\ce{H2SO4}$ will be $\pu{0.1 – 0.3 mol/dm3}$, and concentration of $\ce{HNO3}$ will be $\pu{0.1 – 0.3 mol/dm3}$.)

Sodium hydroxide [$\ce{NaOH}$] solution (approximately 2.0 M 1.0M)
Potassium hydrogen phthalate [$\ce{C8H5KO4}$] powder
Barium chloride [$\ce{BaCl2}$] solution ($\pu{0.3 M}$)
Copper sulfate pentahydrate [$\ce{CuSO4.5H2O}$] powder

You may choose to use spectrophotometric, titrimetric, or gravimetric methods to execute your analysis. There are no major restrictions apart from non-wasteful use of chemical reagents.

How do I approach this?
I tend to use the gravimetric method to perform crystallization of anhydrous copper sulfate so I can measure the mass of $\ce{CuSO4}$ in Mixture A, but for this experiment I will need $\ce{(NH4)2SO4}$, which not provided.
Using spectrophotometry, I will be able to find the concentration of $\ce{CuSO4}$ in the mixture as copper sulfate pentahydrate ($\ce{CuSO4.5H2O}$) powder is given, and I can make a blank out of the chemical is it true?

Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?


text:
Chemistry underlying blacks spots in banana bread
I've noticed that when I bake banana bread with certain brands of self-raising flour, the finished product looks more or less like any other cake.

However, when I use flour with sodium bicarbonate as the leavening agent, black spots appear peppered throughout the cooked cake.

What is the chemistry behind this phenomenon?

Can those be banana seeds? :)

Totally. It's just that they're only visible when I used sodium bicarbonate as the raising agent.

Bananas are seedless (at least the ones we eat today).

@RaulLuciano Sometimes there are rudimentary/unformed seeds which look similar to poppy seed, at least in those imported in Europe from Africa, and they look pretty much like those stains on the OP's photo. But yes, you are right, it's more like an exception these days.

Yes, but judging from the photo above (I am not sure whether this is the picture of the actual banana bread or an image from google) those are definitely not only those "rudimentary seeds" we see in bananas, there's just too much of them. That's why I thought of the two reactions mentioned in my answer.

Both images were sourced from Google. I've replaced the one with the 'dots' with an image of my own banana bread.

There are, at least, two reactions taking place and causing the brown colour of your bread. One is the oxidation of monophenols by an enzyme called PPO and the other is Maillard reaction, which is the same reaction that makes your steak looks deliciously brown.
Reaction 1. Oxidation of monophenols.
Bananas have a lot of polyphenol oxidase (also known as phenolase or PPO), an enzyme that catalyzes the oxidation of polyphenols naturally present in the banana fruit. The product is a colored ortho-quinone, which is brown or dark reddish. When you smash bananas to make a banana bread you end up releasing a lot of PPO which will catalyze the polymerization of the phenols and, therefore, make your bred looks darker then others. Below is a representation of the oxidation of monophenols into the colored ortho-quinone.

Reaction 2. Maillard reaction
The Maillard reaction is one of the most important reactions in food chemistry. Its products can add sensory features to many preparations, from steaks to coffee. Basically, sugars from the ingredients react with an aminoacid's amine group to form a glycosylamine. This compound is then isomerized to a ketosamine, which breaks into smaller substances. These substances are the Maillard products, and one class is called Melanoidins, responsible for the dark spots in your bread.
The reason why it is formed when you add sodium bicarbonate to the mixture is because alkaline media speed up the fragmentation of the ketosamine, yielding more Maillard products. 2
I have included an infographic from CompoundChem with other information and the Maillard reaction in chemical notation.

2Owusu-Apenten R., Introduction to Food Chemistry, CRC Press, 2004

I'm having hard times trying to understand how this answers the question. There is only a general explanation of why bred is brown(ish), but I don't see anything explaining the nature of black spots inside, presumably caused by sodium bicarbonate.

@andselisk I am also not 100% sure about the black spots, I have searched everywhere and could find no mention about this effect. However, I believe that small parts of the banana fruit which were not 'dissolved' in the pastry, thus remaining as small banana chunks, would be the source of brown spots... I think this might be the answer because when I bake a banana bread with some banana chunks in it the bananas turn to an intense brown after cooking. I am not sure though if this is best explanation possible although I believe these two reactions are responsible for it.

In banana cells they carry amyloplasts. Amyloplasts contain starches and when they are exposed to certain chemicals the starches turn black.

It’s because bananas are acidic and when it reacts with baking soda it’s creates the black specks in the banana bread

If you’re saying that there’s an acid-base reaction, most salts (and certainly nearly all sodium salts) are colourless/white, so we’re gonna need a bit more explanation on why that makes black spots.


text:
What is wrong with my enthalpy calculations?
I need to calculate the enthalpy of combustion for butane using bond formation energies. Thus I first balance the equation:
$$\ce{C4H10 + 6.5 O2 -> 4 CO2 + 5 H2O}$$
Then, I calculate the total bond energy difference:
$$-3 \cdot 346.0 - 10 \cdot 414.2 - 6.5 \cdot 479.9 + 4 \cdot 2 \cdot 803.3 + 5 \cdot 2 \cdot 464.4 = 2771.05~[\pu{kJ mol-1}]$$
This is different than the correct value of $\pu{2877.5 kJ mol-1}$. Why?

To me the equation is conceptually correct and I guess the deviation in numbers lies in different sources of information for average bond energies. Note that C-H BEs even in butane only deviate a lot in different sources (from 400.4 to 421.3 kJ/mol), averaging these values bring more error to the table. I'd say you did it correctly and call it a day; if someone will argue, point them to Luo's Comprehensive handbook of chemical bond energies.

Thank you. That's what I'll do. I see you corrected the markup too, could you point me to a good guide for LaTeX for chemistry?

Sure, here is the mini-guide for chemistry and a comprehensive tutorial for math. Welcome aboard:)

I assume you did not mix energies and enthalpies? You found tabulated values for bond enthalpies?

The thing is, the values you used are possibly the mean bond enthalpy of those bonds and not the specific bond enthalpy of the bonds in butane. What is the difference? The chemical environment of a bond will affect its enthalpy. As the mean bond enthalpy uses enthalpy of a bond in various compounds it is never the same as the actual bond enthalpy in a given compound.


text:
The Molar mass of the hydrocarbon is 42.0gmol^-1. Determine its molecular formula

As you can see, I have answered both the part (a) & (b), but I am unsure how to answer part (c). I am 100% sure I answered part (a) right; not quite positive about part (b) though. I'm lost on part (c).

I don't understand what you did in (b). What I'd do is write an equation for mass percentage of hydrogen in unknown hydrocarbon $\ce{C_xH_y}$ (you can also do it for carbon with the equal result) and find $x : y$ ratio:
$$\frac{y}{12x + y} = 0.143 \implies y = 2.00 x$$
Knowing that the empirical formula is $\ce{C_nH_{2n}}$, we can solve (c):
$$12n + 1 \cdot 2n = 42 \implies n = 3$$
The molecular formula of hydrocarbon is $\ce{C3H6}$.

You don't even need the empirical formula. You can't get anything else that makes sense.

@Zhe True; my intention was to try to show a common approach for solving these types of questions.


text:
How does Ozone disinfect water?
How does ozone remove the impurities in water? I know that there is a process which some European countries use water purification systems based on ozone. It would be beneficial if you could provide with some chemical equations to describe the process of how impurities are removed in water.

Ozone is a powerful oxidizer (redox potential of 2.07) and will kill bacteria and destroy viruses (not sure if I can say it kills viruses) by transferring oxygen atoms to them. It also decomposes in water forming OH radicals, which quickly oxidize organic and inorganic compounds. Ozone can also react by ozonolysis, breaking compounds which have double or triple bonds into smaller molecules, some of which will act as oxidizers themselves.
The chemical equation (mechanism) for ozonolysis is as follows:


text:
Making 250mL of 0.1M NaOH
I measured out approximately 1g of NaOH pellets . I then transfer it into a volumetric flask and transfer some amount of distilled water into the flask and dissolve the pellets before filling the volumetric flash to the mark (250ml).
Why i can't make sodium hydroxide by transferring a fixed volume (250ml) into the flask ?

I don’t get what you’re asking, could you clarify?

@Jan I'm trying to ask if it is possible to prepare a standard NaOH solution by dissolving an accurate amount of NaOH solid in a known volume of water and i think it's not possible , but I'm not sure why

Oh, so using a random beaker to mix a fixed weight of sodium hydroxide with a fixed volume of water, both determined separately?

@Jan both using a volumetric flask . Can I get the same result of concentration ? 1. If I add a small volume of water and mix it before adding it up to 250ml after dissolving . 2. If I add exact 250ml and mix it .

If you add a fixed volume of water to a volumetric flask (e.g. $\pu{250ml}$) and then attempt to add sodium hydroxide to dissolve, the resulting volume will no longer be $\pu{250ml}$. Depending on the salt you are dissolving, it can be larger or smaller but it will never be identical. Thus, your resulting solution will not have the molarity you think it has.
The recommended procedure is to add a solid, add enough water to dissolve it (but well less than the volume of the volumetric flask!) and then slowly add water until the line is reached. This means you know exactly how much solute you have added (in grams and thus moles) and also how much the volume of the solution is.


text:
What is the Chemical Compound that has a NFPA 704 (Fire Diamond) of 2-2-1-ox?
I recently saw this fire diamond configuration of:
2 - Flammability;
2 - Health;
1 - Reactivity;
OX - Extra Information
I have been attempting to find a database of terms, but after turning up with nothing, and seeing a similar answer about fire diamonds answered here, I posted this question.
I would be thankful if someone happened to know what compound that this fire diamond is labeled to.

Unless it's something incredibly common, I feel like there are too many possibilities. There are only so many possible combinations for the safety diamond, and well over an Avogadro's number of possible compounds that could exist.

From what I found NIOSH lists 6 compounds with 2-2-1 pattern, but none of them are labeled with oxidizing properties:

ethylhexaldehyde: NFPA 704 2-2-1;
benzyl chloride: NFPA 704 2-2-1;
2-methyl-1-butanol: NFPA 704 2-2-1;
n-octyl mercaptan: NFPA 704 2-2-1;
vinyl toluene: NFPA 704 2-2-1;
methyl chloroacetate: NFPA 704 2-2-1.

On the other hand, there are oxidizers with slightly different NFPA diamond which much as well be the candidates:

ammonium dichromate NFPA 704 2-1-1-OX;
amyl nitrate NFPA 704 2-2-0-OX.

The values might deviate a little in different sources and they also depend on conditioning.


text:
How to identify if a chemical equation has a 'fractional order'?
Consider the following reaction:
$$\ce{CHCl3 + Cl2 -> CCl4 + HCl}$$
The rate law expression is:
$$\text{Rate} = [\ce{CHCl3}]\cdot[\ce{Cl2}]^\frac{1}{2}$$
with an order of $1.5$. But 'how can we determine' this fact?
And why is it $1.5$ and not $2$?

It is not possible to determine the rate law of a reaction based on solely the balanced reaction. We must determine the rate law by experiment or by an established mechanism for the reaction.
What is meant by a mechanism is a series of elementary steps, a set of reactions whose linear combination yields the balanced reaction. Each elementary step represents a microscopic collision that happens during the process of the reaction, and these elementary steps obey the law of mass action, which allows us to write the rate law for each step in the reaction.
We then make a few approximations, perhaps guided by experimental observation, to obtain the rate law for the reaction. For example, we may conclude that one elementary step is rate-determining, so that the rate law for the reaction is governed by the rate law for that step; we may make a quasi-equilibrium approximation, which means that the species involved in an elementary step are equilibrated; or we may make a pseudo-steady-state assumption, which states that the net rate of production of an intermediate, presumed to be short-lived, is zero.
How do we know that a proposed mechanism is correct? Well, we don't. All we can do is prove that a mechanism is incorrect, by establishing that it posits the wrong rate law as found from experiment; we cannot distinguish between different mechanisms that result in the correct rate law. Nevertheless, we can often obtain much indirect evidence to support a given rate law, and we can frequently make good guesses based on patterns from experimental observations.
For example, while I wouldn't be able to accurately guess the rate law for the reaction you've provided, I can at least say that it is likely to be of half-integer order in chlorine, because such reactions often follow a radical mechanism, which often lead to non-integral powers in rate laws.

In special cases we can get direct information even about transition state.


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How to deal with exact numbers on error propagation?
In Analytical Chemistry, 7th edition by Christian et al., in its Example 3.10 (screenshot below; see blue circle for the main matter), when it computed for the standard deviation after getting the average of the three measurements, the authors further divided the standard deviation by 3 to get its absolute uncertainty.

However in its Example 3.12 (screenshot below; see blue circle for the main matter) it did not divide further the calculated standard deviation by 100 to get the absolute uncertainty, even though it's essentially the same operation as in Example 3.10. In addition, you can see in the price calculation near the end of the solution that he multiplied it too to the standard deviation to get the absolute uncertainty.

Of the two, which is the correct action for the error when dividing or multiplying with an exact number? If they're both wrong, or if they're both right and I'm just misinterpreting the concept, what is, then, the proper way to deal with this situation?
Follow-up question:
On the other hand, what should I do with the error when adding/subtracting a measured number from an exact number or vice-versa?

Note that the first instance is a case of averaging (i.e. separate measurements), whereas the second case is making the percentage numbers  that go into the calculation look nicer. This amounts to a (albeit exact) unit conversion, which probably should be treated differently.

1/2 Actually, I think that in the case of the percent, the 100 really needs to be distributed to the absolute uncertainty. In the example, the (36.28 +- 0.04%)/100 is actually the conversion step of the percent to the actual factor. If the conversion factor, 100, was not applied to the absolute uncertainty, it would look like that the author was multiplying the mass to (0.3628 +- 0.04). By making the factor like this it would seem that the uncertain digit is actually the hundredths place, instead of the ten-thousandths place (0.3628 +- 0.0004) --

2/2 which would occur if we divided the absolute uncertainty too with 100.

A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.


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Will squeezing lemon juice on top of foods with ferrocyanide additive release hydrogen cyanide gas?
For example, will squeezing lemon juice on top of table salt with sodium ferrocyanide ($\ce{Na4[Fe(CN)6]}$) additive (E 535) release hydrogen cyanide gas?
Examples of ferrocyanide additives:

sodium ferrocyanide (E 535)
potassium ferrocyanide (E 536)
calcium ferrocyanide (E 538)

You can see this video at 2:02 for reference but seriously don't try this.

@NilayGhosh your video is irrelevant. This question is about hexacyanoferrate not sodium cyanide.

No. The hexacyanidoferrate(II) complex is a very inert complex that does not undergo ligand exchange easily. This is somewhat exemplified by the harsh conditions alchemists used to synthesise hexacyanidoferrate(II), which included high temperatures and strong bases.
Furthermore, whatever amount of lemon juice you add the resulting acidity will be much less than in your stomach. If you swallow a cyanide source such as potassium cyanide, this will get protonated by your stomach acid to liberate hydrogen cyanide and potentially kill you. Food additives — especially approved ones — undergo various testing procedures to make sure that does not happen.

Yes, it has been demonstrated in this video.

@NilayGhosh what you reference is totally misleading and irrelevant. Cody is playing with sodium cyanide. This question is about hexacyanoferrate salts which are totally different, behave totally differently and are far safer. It is easy to get HCN from sodium cyanide and almost impossible from hexacyanoferrate, which is why it is safe as a food additive.

Hexacyanoferrate(II) maybe?

Is the complex as stable with Fe(III) as with Fe(II)?

@OscarLanzi Hexacyanidoferrate(II) is actually more stable than hexacyanidoferrate(III) (which is an oxidising agent) due to the $\mathrm{(t_{2g})^6}$ state.

The iron(II) complex is more stable due to the fact that it is a low spin d6 compound with a high ligand field. It so happens that Prussian blue is a cyanide complex of iron which is a cation exchanger. Some forms of prussian blue can be used to remove cesium from people and animals. The normal dose for a Cs137 contaminated adult is 3 grams per day. But in extreme cases the dose can be increased to 30 grams a day. It is interesting that prussian blue varries a lot. I knew a man who used to work on the decontamination of farmyard animals he said he never had two batches which were the same.


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Electronegativity trends: Is it more important to prioritize families or periods?
I know that an atom that an atom is more electronegative as it moves up a group or right on a period (in terms of the periodic table).
So if you have two atom that is diagonally across from each other, which one is more electronegative? Do you prioritize groups or periods when determining electronegativity?
For example: Which is more electronegative - Sulfur, or Bromine?
Finally, does this priority also work for first ionization and atomic radius trends. (in other words, if you prioritize groups when determining electronegativity, do you also prioritize groups when determining the other two periodic table trends?)
My guess is that they would be equal, but I may be wrong.
P.S. I would be most likely to give best answer to the person who answers all my questions as a numbered list.

The easiest thing to do is google for a table of electronegativities. As far as I know, there's no simple rule. Period seems more important, but after the d-elements the group is more important because of the higher charge density of additional protons. More exceptions: noble gases who have a full octet where additional electrons add into a new shell, and transition metals with varying d-electron configurations. Imo best to just reference a table.

There's no hard and fast rule.  In fact, there's actually no rule per se that states that electronegativity does increase along the periods and decreases down the groups.  It's just a general trend, which can be and is broken on occasion.  For instance, Zinc, to the right of Copper, has an electronegativity of 1.66, while copper is 1.75, which "breaks" the rule that electronegativity increases along the periods.
As far as which one takes precedence, it's a case by case scenario.  In some cases, down one and right one would be very close, like you guessed, as with Scandium (1.20) and Zirconium (1.22).  However, if you go over to the non-metals, some of them increase radically in a down-one-right-one scenario (particularly from 3rd period to 4th period, such as Aluminum [1.47] to Germanium [2.02], Phosphorus [2.06] to Selenium [2.48], etc.), while some of them decrease radically (particularly from 2nd to 3rd period and 4th to 5th, such as Boron to Silicon, Oxygen to Chlorine, Arsenic to Tellurium, etc.)
Bottom line:  There's no rule.  You just have to either learn them by heart or check a periodic table such as this one:


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Do dyed-alcohol thermometers tolerate cold/hot stems?
How do the red organic solvent thermometers respond to having the tip submerged while the stem is at a different temperature? I understand mercury thermometers could easily handle this, and bimetallic strip probe thermometers fail utterly.
To put it simply, if I'm heating just 2-3 cm of liquid, can an organic solvent thermometer still give accurate readings?

I removed the non-related question. If it remains, this question would have to be closed as too broad. You are welcome to ask a separate question, though! (Assuming a duplicate does not already exist)

@Jan Thanks, I'll search and may open another thread afterwards.

The very shortest answer is no. Since asking this question, I've learned that thermometers have an insertion depth they are calibrated to, presumably assuming the rest of the thermometer is somewhere between room temperature and the bath temperature. Some thermometers are "total immersion", meaning the conductivity of the fluid doesn't matter if used as intended, because it's expected to conduct through the sides as well.
Given what I've learned, the first half of the answer to this question is that if a thermometer is calibrate to be immersed 2 cm, then immersing it deeper will give a wrong reading.
This article explains further: https://www.coleparmer.com/tech-article/glass-thermometer-selection-guide. In particular, correction factors are given for when a thermometer is not inserted at its calibrated depth:

Emergent Stem Corrections. Avoid inaccurate readings when a total
  immersion thermometer cannot be properly immersed. Determine the
  approximate stem correction with these formulas:
For mercury Fahrenheit thermometers:
°Correction = 0.00009°F x n x (T-t)
For mercury Celsius thermometers:
°Correction = 0.00016°C x n x (T-t)
For spirit-filled Fahrenheit thermometers:
°Correction = 0.0006°F x n x (T-t)
For spirit-filled Celsius thermometers:
°Correction = 0.001°C x n x (T-t)

where T is the bath temperature (the temperature indicated on the
  thermometer), t is the average temperature of the emergent stem, and n
  is the number of emergent thermometer degrees. Hold an auxiliary
  thermometer next to emergent stem to determine t.

While these corrections aren't meant to be applied backwards (for a partial immersion thermometer that's immersed too deep), the numbers speak for themselves: a colored organic thermometer is not nearly as accurate when it's either immersed too far or too shallowly. I plugged in some numbers for a very nearly worst case scenario (cold room with a strong draft where the thermometer is calibrated to be fully inserted but is actually only inserted up to the 0⁰C mark):
T=100⁰C
t=20⁰C
n=100
mercury °C correction = 0.00016°C x n x (T-t) = 1.28⁰C
spirit °C correction = 0.001°C x n x (T-t) = 8⁰C

The final answer is that no, dyed-spirit thermometers do not tolerate being at the wrong insertion depth, at least not if your experiment's parameters are too far off from what the thermometer manufacturer intended.

That is typically the way these thermometers are used in the lab. They are submerged a few centimetres into the liquid — ideally making sure that the entire lower ‘bulb’ is submerged, but if there is not enough liquid there is not enough liquid — and this reading is taken to be accurate.
If it is a low-temperature thermometer, you can confirm the reading’s accuracy by testing a few known low temperatures such as dry ice/acetone. This, of course, also works for high-temperature thermometers, e.g. by using boiling dichloromethane, but should be done in proper glassware with a reflux condenser for safety reasons.


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Regioselectivity in the aldolcondensation towards (Z)-Jasmon
In the synthesis of (Z)-Jasmone via aldol-condensation as shown below,
how can the observed regioselectivity be explained?
After all, it is also possible to form the side product shown below. I thought that at least with LDA, the sideproduct should be favored, while I assume that for milder bases such as KOH, the thermodynamic enolate is formed, finally leading to the desired (Z)-Jasmone. Are there any other reasons for the regioselectivity?

The entire reaction (not just enolate formation) is under thermodynamic control, so you get the tetrasubstituted alkene preferentially over the trisubstituted alkene.

@orthocresol OK, I see that the the tetrasubstituted ketone is thermodynamically favored, but isn't the last step, i.e. the formation of the unsaturated ketone, effectively irreversible?

The reverse of an E1cb step is conjugate addition of hydroxide to an enone, which isn’t unrealistic.

Not only ‘is not unrealistic’ but also rather likely, depending on the conditions involved. I’m too lazy to write up an answer now though … D=


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What primary standard acid can be used to standardise NaOH solutions by titration?
I've done an experiment to standardise $\ce{NaOH}$ using oxalic acid.
What other primary standard acid can be used to standardise $\ce{NaOH}$?
Can $\ce{HCl, H2SO4, HNO3}$ be used?
I read up and found that:
$\ce{HCl}$ and $\ce{H2SO4}$ are easily diluted, however, the concentration of the "concentrated" solution is not accurately known.
$\ce{HNO3}$ contains a little nitrous acid, $\ce{HNO2}$, which has a destructive action on many acid-base indicators.
Therefore I only found potassium hydrogen phthalate. Is there other primary standard acid that can standardise $\ce{NaOH}$?

What actually does having a "destructive action on many acid-base indicators" actually mean? And are there concrete examples of it? Thanks!

Oxalic acid dihydrate and KHP are the most comfortable acids for standardization, as their content per a unit of measurement (in this case it's obviously grams, as they are solid) hardly changes. When you have liquid acids, besides that what you've already mentioned, they can absorb water and other compounds from air. As they are liquid, people usually measure amount of those compounds by volume, and volume is a variable which depends on temperature, pressure.

I don't think any analytical chemist would use oxalic acid dihydrate over KHP.

@MaxW True, but oxalic acid has benefits for the amateur chemist in terms of price and availability and it is definitely used as a primary standard.  It may not be the professional's choice, but there's nothing wrong with using it.  It has all of a primary standard's properties, after all.


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Why H3O+ exists and H3O- doesn't?
I really don't know if it is even logical to ask that $\ce{H3O-}$ doesn't exist but someone asked me and I wasn't clear. Hope that some one will make me clear.

Well, it seems a valid question, while not particularly good. Your H3O- is H2O + H- and these react violently.

Someone should answer this with the MO scheme of water …

To me, this sounds like $\ce{H2}$ and $\ce{OH-}$. Both particles are fine, but have absolutely no intention to stay together.

It does exist. It has been observed in the gas phase.

@Raditz_35 What kind of gas phase? Very low pressure and temperature, like in space? There many thoroughly unstable particles can exist.

@Karl Maybe you should write an answer. What you call "unstable" is not 100% correct or well defined I think, but it leads to the answer of this question I feel like.

I'm not a big fan of MOs when this can be explained using just AOs. For sure, MO theory gives a much better and profound explanation, and I'm surely waiting for someone to do that, but that's what I would answer if the kids asked me this.
Oxygen has 6 electrons on its outer shell, which has $\mathrm{2s}$ and $\mathrm{2p}$ orbitals. The arrangement of these electrons is such that $\mathrm{2s}$ sublevel is completed, and $\mathrm{2p}$ sublevel contains 4 electrons. Following Hund's rule,  two electrons occupy one of three $\mathrm{2p}$ orbitals and other two come by one to each of $\mathrm{2p}$ orbitals left.

When water forms, oxygen shares two non-bonded electrons with hydrogen atoms, forming a covalent bond with them.
$\ce{H3O+}$ forms via donor-acceptor bonding, when oxygen supplies its lone pair, which it has two, to form a bond with $\ce{H+}$ ion.
To form $\ce{H3O-}$, you'll need an interaction between water molecule and $\ce{H-}$ ion. Even if we ignore the fact that hydride is a powerful base ($\mathrm pK_\mathrm a$ is around $42$) and will react violently with water, it would be impossible. In this theoretical case neither of the species has an orbital suitable to accept a lone pair from another species. Each has a complete outer shell.
Edit: it really exists though. Wow.
Sources:

Probing the transition state via photoelectron and photodetachment spectroscopy of H3O−
Calculation of the photodetachment spectrum for H3O−

@Raditz_35 Yeap, never checked this. Edited my answer. Thanks.

Thanks! I think the important part is why it is irrelevant for everyday chemistry = the reason for the question.

Please take your time and check out the mhchem package that this site has loaded for MathJax. Using the \ce command automatically gives the correct output (especially non-italicised element symbols!) for a number of intuitive inputs. See this post on meta for more information.


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Why is decomposition of ammonia gas on quartz a 1st order reaction?
In general, decomposition of a gas (like ammonia) on a metal surface (like platinum) is a zero order reaction.
This is because the gas is adsorbed on the metal surface due to pressure and the initial concentration is immaterial.
But, why and HOW does it behave differently with quartz?

The metal surface has catalytic centres, quartz obviously hasn't?

Firstly a correction and general clarification: your first general assertion is not true. The metal surface acts as a heterogenous catalysts (supported by paper linked later), and the kinetics can be sufficiently modeled by the Langmuir-Hinshelwood model for a monomolecular reaction in which the rate determining step is the surface reaction
$$ \ce{A_{ad} -> B_{ad}} $$
so
$$ r = \frac{\mathrm{d}[B]}{\mathrm{dt}} =k_r [A_{ad}] = k_r \theta_A $$
$\theta_A$ is the surface coverage of A. Applying the steady state approximation to $\theta_A$ over the adsoprtion, desorption and reaction steps and using the Langmuir adsorption model gives
$$ \theta_A = \frac{k_{ad}\:p_A}{k_{ad}\:p_A+k_{de}+k_{r}}$$
so the rate is given by
$$r = k_r \; \frac{\bar{k} \: p_A}{1+\bar{k} \: p_A}$$
where $\bar{k}$ is a complex rate constant. For low partial pressures of A, the reaction thus approximately follows first order kinetics and zeroth order for high $p_A$. Here is a nice paper verifying this for several metals.
Now lets get to the quartz. The authors of the paper also mention a possible bimolecular reaction mechanism, of which the key step is the formation of hydrazin from two adsorbed ammonia molecules. They disregard this mechanism due to experimental evidence, but lets apply some metal organic chemistry here.
$\ce{NH3}$ is a great $\sigma$ donor, so a $\sigma$ acidic metal center will happily bond. A simplistic view might be that the nonbonding lone pair is being donated to the metal - taking a look at a MO diagram of $\ce{NH3}$ you will however find that the HOMO (from which we donate) is slightly bonding.

Source
Therefore, this particular $\sigma$ donation will decrease N-H bond strength, which supports the monomolecular reaction mechanism for such a surface (this does put us somewhere in the grey zone between heterogenous and homogenous catalysis).
$\ce{SiO2}$ is not a substance that comes to mind when we are looking for a $\sigma$ acid. So maybe, under these conditions the bimolecular mechanism becomes the preferred pathway. Bimolecular reaction kinetics of this sort can be modeled by Langmuir-Hinshelwood (both reacting species adsorbed) or Eley Rideal (one adsorbed, one in gas phase). Langmuir-Hinshelwood would give us the boundary cases of second order rate at low $p_A$ and zeroth order at high $p_A$, and I could find no experimental results supporting that anywhere. An Eley-Rideal model would give us a rate law of the form
$$ r = k_r \; \frac{\bar{k} \: p_A^2}{(1+\bar{k} \: p_A)}$$
which would give us first order behaviour for high partial pressures of ammonia! Here is a paper finding experimental data supporting this kinetic model. Possibly the reaction product in the rate-determining step is not hydrazin, but an intermediate hydrazinium which then decomposes.
It thus seems plausible that ammmonia decomposition on a quartz surface follows a bimolecular Eley-Rideal mechanism, compared to the monomolecular mechanism on a metal surface - as a result of the different interactions between surface and adsorbent.

Thanks for the answer here on an old question!

Made a couple of clarifying/correction edits -- please feel free to revert if you desire, though I would recommend not reverting back to the 'RDS' acronym.  Separately, is there an ammonia MO diagram on the web somewhere that would help to illustrate the slight bonding character of the HOMO?

Included one now. Thanks for any recommendations, my idea was to use my summer break to improve my writing so any help is appreciated!


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What types of energies can be determined by molecular dynamics in gas phase or explicit solvents?
For molecular dynamics in the gas phase I've seen that electronic energy can be determined and also decomposed in kinetic and potential energy. I've also seen that we can include ZPE. But can we determine (changes in) $\Delta H $, $\Delta S $ and $\Delta G$?
And how about explicit solvent systems? In particular I'm interested if it's possible to determine the free energy of activation and the entropy of activation for reactions in explicit solvent systems. Is it possible to determine such energies for the whole system or even for only the reacting molecules (if that makes sense to do)?
To add: I'm talking about using QM (DFT) for the molecular dynamics. Not molecular mechanics.

First, whether you use DFT or molecular mechanics to get your energies and forces doesn't really matter. If you have equations of motion, forces, and energies then you have the dynamics necessary to sample an ensemble and from there you should be able to get any physical observable you desire.
It is often very difficult to determine free energy changes using straight up molecular dynamics. The reason for this is that even very fast reactions are rare events when compared to typical molecular motions such as vibrations and rotations. For this reason, many methods have been developed to determine free energies without having to run simulations long enough to get good sampling of reactions. Usually, this is not possible anyways.
First, you may wish to read about umbrella sampling. Umbrella sampling, as a general idea, does not sample a trajectory on the potential energy surface determined the potential energy of the system (your output from DFT if you like), but rather, a harmonic potential energy is added to the system as well. This has the effect of lifting the molecule out of the potential well it may be stuck in so that sampling of rare events like reactions becomes much more common.
It is not obvious at all that any physically meaningful quantities should be able to be derived from this new system where the potential energy is constantly being modified. Yet, surprisingly and rather beautifully, it is possible. You should probably see ref. [1] and the citations therein.
The second way to do this I will mention is so-called meta-dynamics. The conceptual idea here is to first define the so-called "collective variables" of the system. These are literally any variables which can be used to describe the properties of a system. Typically one would choose some set of internal coordinates, particularly for a gas-phase system as you describe. However, my understanding is that it is also possible to choose various other coordinates such as intermolecular distances. One could imagine this being quite useful if you are interested in something like radial distribution functions in some hard-to-sample region.
Using these collective variables, again, a biasing potential is applied to the system, but this potential is said to be "history-dependent" in the sense that the form of the biasing potential might change based on what parts of the free energy landscape have already been seen. This differs from umbrella sampling where the potential is always harmonic and the rate at which the potential is biased and how extreme the bias is, is chosen by the user. You should see ref. [2] and references therein for more details on metadynamics.
So, yes, it is possible and even fairly routine to calculate free energies, enthalpies, entropies and pretty much anything else from molecular dynamics simulations. However, just because it is routing does not mean these methods are perfect. It can be very hard to measure error with these sorts of methods because one does not generally know the correct answer particularly for dynamical processes. That is, if your aim is to study details of dynamics which can't be observed experimentally, then it is very hard to know if the dynamics under your biased potential mean much beyond the fact they should give you pretty good ensemble averages. Also, there is the inherent subjectivity in metadynamics involved in choosing collective variables which has been known to cause problems. How many variables are enough and which ones?
Hopefully that is about what you were looking for.
I don't address the differences between explicit solvent and gas-phase much because these methods will work in any context (even implicit solvent) for which forces and energies can be supplied.

[1]: Kästner, J. (2011). Umbrella sampling. Wiley Interdisciplinary Reviews: Computational Molecular Science, 1(6), 932-942.
[2]: Barducci, A., Bonomi, M., & Parrinello, M. (2011). Metadynamics. Wiley Interdisciplinary Reviews: Computational Molecular Science, 1(5), 826-843.

Thanks for your answer.If I understand it correct the first problem is to actually get constructive trajectories but once we got them we can calculate pretty much anything we want? or do we need a distribution between constructive and non constructive trajectories? The reactions I'm looking at are extremely fast (no enthalpy barrier, fully entropy controlled) and time in the transition state zone should be quite close to that of a C-C vibration. There are examples using normal mode sampling at the transition state and propagate it forward and in reverse to get mainly constructive trajectories.

Once you have a trajectory, you can calculate anything at all that statistical mechanics allows you to calculate, which is basically anything. The difficulty comes from getting enough sampling to actually have reliable averages. As for your specific case, it's hard to know what would work without more information (do you have an advisor?). I think that either need described could be used for this because your problem will be getting sufficient sampling of such a quick event, so you have to force the system to stay near the transition state.

thanks that's what I wanted to hear. I just wanted to get sure about some of the absolute basics before I approach the people who have the knowledge to run such calculations. They have done it before for similar systems and got several hundred trajectories but they mostly looked at hydrogen bonds during the reaction and not thermodynamic values.


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How can you determine whether a solution is dilute or concentrated given its concentration?
How can you determine whether a solution is dilute or concentrated given its concentration? I don't whether this is a stupid question or not but I couldn't find any answer.

I wouldn't say it's stupid. But stop and consider if I point to a random object of known dimensions and ask if it is big or small.

Unfortunately it is near impossible in chemistry to give absolute rules of thumb. Rather a rules of thumb is just a "rule" that typically works, but for which there are always exceptions. So for aqueous solutions I'd say that anything less than 0.1 molar would be "dilute." The key notion here being that the various equilibria are dependent on activities not concentrations. But for 0.1 molar solutions or less, the activity should be well approximated by the molar concentration.

https://chemistry.stackexchange.com/questions/54049/under-which-criteria-hno3-is-a-concentrated-or-a-diluted-acid

There are definite answers for saturated or not (or even supersaturated), which have definite limits, but concentrated is a matter of opinion, except at the extrema.

There is no hard and fast definition of "concentrated" or dilute.  For example, concentrated HCl generally refers to a 12 mol/L solution of HCl.  Concentrated sulfuric acid is 18 mol/L.  When we refer to a "concentrated" solution, we are generally referring to the most concentrated solution that you can prepare, or that you can buy from a supplied.
https://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/reagent-concentrations.html


text:
When an iodide/iodate solution is acidified with H2SO4 instead of thiosulfate, why should it be titrated immediately?
My class is currently performing a reaction to determine the oxidizing power (i.e. the amount of hypochlorite) of household bleach. The general process is as follows:
Standardize thiosulfate to be used as a titrant of $\ce{I2}$ solution.

An excess of I- ions are allowed to react with a known amount of $\ce{KIO3}$ in acidic solution.
The iodine formed from this reaction is titrated with thiosulfate using starch as an indicator of triiodide anion.
The relation between moles $\ce{KIO3}$ and thiosulfate is used to determine molarity of thiosulfate.

Analysis of oxidizing capacity of liquid bleach

Excess iodide ion solution added to bleach.
Iodide ions are oxidized to I2 after this solution is acidified simultaneously with the reduction of hypochlorite.
The iodine that is formed is then titrated with the above standardized thiosulfate solution.
The indicator (starch) is not added until the dark-brownish color of the iodine has changed to pale yellow. When it is added, the solution color changes to blue black.
The end point of titration is indicated when the solution becomes colorless.
(Note that if the starch is added too soon in the titration, the formation of the blue-black complex is not easily reversed, making the end point very slow and difficult to detect.

Pertinent reactions are given below
Reactions Pertaining to Titration
$\ce{HOCl + 2I- + H+ -> I2 + Cl- + H2O}$
$\ce{2S2O3^--  + I2 -> 2I- + S4O6^--}$
$\ce{HOCl + 2S2O3^-- + H+ -> Cl- + S4O6^-- + H2O}$ [the addition of the above two reactions]
$\ce{I3- + starch <-> starch*I3-(complex)}$  [blue-black in solution]
Reactions pertaining to the standardization of thiosulfate
$\ce{IO3- + 5I- +6H+ -> 3I2 + 3H2O}$
$\ce{3I2 + S2O3^-- -> 6I- + 3S4O6^--}$
The question
We were told that, during the standardization process, if the iodate-iodide solution is acidified with $\ce{H2SO4}$, it should be titrated immediately. My question is why? Using thiosulfate, this is not the case, correct? Why does this change when sulfuric acid is used in its stead?

After acidification with $\ce{H2SO4}$, starch starts to hydrolyze to its component sugars. Lugol's iodine does not react with those sugars to form the blue complex.

Our amylose is gone, today,
Its trace we'll see no more,
For what we thought was $\ce{H2O}$
Was $\ce{H2SO4}$.


text:
Why are lithium (Z)-enolates more stereoselective than (E)-enolates in aldol reactions?
In aldol reactions, it is apparently a general rule of thumb that (Z)-enolates give higher stereoselectivity (for the syn aldol product) than (E)-enolates do for the anti product. Quoting Carey & Sundberg, Advanced Organic Chemistry: Part B (5th ed.), p 69:

From these and many related examples the following generalizations can be made about kinetic stereoselection in aldol additions of lithium enolates.
  [...]
  (3) Z-Enolates are more stereoselective than E-enolates.

and the example given (taken from a Clayton Heathcock paper, J. Org. Chem. 1980, 45 (6), 1066–1081) is

Evans also mentions this in his paper comparing the Cornforth vs polar Felkin–Anh models (Angew. Chem. Int. Ed. 2003, 42 (15), 1761–1765):

While Z and E lithium enolates are also formed with good configurational purity, the characteristic decline in syn/anti selectivity is observed, particularly with the E enolate.

Why is this the case? Or, why is $\Delta \Delta G^\ddagger$ for the (Z)-enolate larger (corresponding to greater selectivity) than for the (E)-enolate? I don't see anything that stands out when looking at the six-membered Zimmerman–Traxler transition states - within this model, the difference $\Delta \Delta G^\ddagger$ for both enolate geometries is simply the cost of putting a phenyl group axial instead of equatorial.

I can't get my thoughts into coherent sentences right now, but Heathcock (J. Org. Chem 1980, 45, 1066), and the nice (old) review In Comprehensive Organic Synthesis (Vol II, 190) give an okay explanation. tl;dr is that the aldol is kinetic but ΔG is so small that retro-aldol easily occurs to give  a thermodynamic component. Keep in mind that Li-O bond long so TS energies essentially the same.   Dave Evans' talks about this in his lectures and essentially says its a lie if the right enolate (boron not lithium) is chosen (its maybe more surprising that any Li aldol is selective).

All the boron aldols I performed on ketones during my PhD thesis were fully stereoselective. I either only got a single product (I wanted anti) or I got two different anti-diastereomers stemming from the Felkin and the anti-Felkin attack. This supports what NotEvans has written about NotNotEvans.


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Why does sodium bicarbonate taste salty?
I just made myself sodium bicarbonate solution to treat heartburn. I never did this before, I rarely suffer from this condition.
What surprised me is how salty did the solution taste. I thought there is no similarity between table salt and baking soda. Yet they taste almost the same it seems.
Why is that?

Table salt is $\ce{NaCl}$ and sodium bicarbonate is $\ce{NaHCO3}$ so then what gives the "salty" taste?

@Hexacoordinate-C So it's something with sodium?

It should be related don't you think?

Related

Due to the fact that the link from @Hexacoordinate-C basically answers the question, I am voting the question to be closed

Hmm, this:  https://chemistry.stackexchange.com/a/19662/9961 is better answer, but feel both questions aren't good duplicates.