text:
Solubility and wetting of substances in water
We have seen that, when we pour salt in water then it gets dissolved, that means it is soluble in water. But when we pour sand into water then it doesn't dissolve in water, that means it is insoluble, but still sand gets wet. But there are certain substances which doesn't get wet by water for example, sulfur particles don't get wet by water but wet in oil, as I was studying about froth floatation method.
My question is that:

What is the difference between solubility and wetting in water ?
What is the reason that the sulfur particle doesn't get wet by water?

Soluble gets dissolved, and wettable just gets wet. It is not going to be any simpler than that. The examples you gave yourself.

But what is the reason that sulphur particle doesn't get wet by water in froath floatation method

The difference is surface tension, and it is, among others, due to differences in polarity.

You're trying to split hairs. For instance there really isn't a good definition of what "soluble" means, so soluble is by definition a fuzzy measurement.  $\ce{SiO2}$ (quartz, sand) does dissolve in water, just not very much.

Polar/hydrophilic soluble substances get dissolved, like table salt or sugar.
Polar/hydrophilic insoluble substances get wet, as they attract water, like sand, or limestone.
Nonpolar/hydrophobic insoluble substances do not get wet, as they repulse water, like wax, teflon or silanized glass.

Nonpolar molecules do not easily dissolve in water and are hydrophobic, they become squeezed together.

What does this squeezed together means?

Water surface tension causes water has more energy if it has larger surface what leads the system to have   as small contact surface water Vs hydrophobic substance via small but real force.

Do all the non polar molecules doesn't get wet in water???

Does any molecule get wet ?

Molecules doesn't get wet

When we see through our naked eyes those salt being added to water then after sometimes salt get vanished which we say salt get dissolved in water

And in molecular level if we see then we say salt molecules get broken down into its constituent ions

And when we see sand being added to water through our naked eyes.... after sometimes when we see sand then we say the sand gets wet by water

But what would be happening in the molecular level...???

The water molecules undergo a sort of attraction between them. This attraction is due to $\ce{H}$ atoms being attracted by the oxygen atoms of neighboring water molecules. At the surface of the water this attraction produces a sort of "skin" due to the attraction of the molecules between them. If a solid is touching this surface, it may produce a similar attraction between its molecules and water. This is the case when the outer part of the solid contains the same $\ce{H}$ and $\ce{O}$ atoms as water. In that case, the attraction between water and these foreign atoms allow water to wet the solid. And even to "climb" a couple of millimeters inside the stuff.
Sulfur has no $\ce{H}$ or $\ce{O}$ atoms. It cannot make bonds with water. So water molecules on the water surface refuse to break their bonds for wetting this foreign solids.

a sort of skin means


text:
Do Neon Lights Work Do To Stokes Shift?
I was wondering whether or not the reason some noble gases, like neon, fluoresce at particular frequencies of the visible spectrum is because of a Stokes effect? I mean, do the neon dimers get stretched out dramatically at a more LUMO state, resonating with high energy electricity due to the high stability of noble gases, and then contract back to a more HOMO state, releasing red light, which has a much lower frequency? Noble gases, like neon, have a high quantum yield, correct?

From wikipedia: "The Ne*2 molecule exists in an excited state in an excimer lamp using a microhollow cathode. This emits strongly in the vacuum ultraviolet between 75 and 90 nm with a peak at 83 nm."

Short answer: no. Please look at the spectra in the second figure in this recent answer   https://chemistry.stackexchange.com/a/126980/79678    It is just a matter of excited atoms and ions emitting their characteristic color photons as they relax back to their ground states. In a simple Ne lamp bulb, Ne atoms are excited and ionized by the electric discharge in the bulb. Same in a He-Ne laser.

Ed has already explained the main ideas. Stokes shift is associated with the concept when you excite a substance with light (or electromagnetic radiation). In neon discharge tubes you are using electrical energy to excite the neon atoms. All the emission you see is from neon atoms. Also recall, that molecular emission has band like structure. Neon light spectrum is nowhere close to this band structure.

Buy a small pocket spectroscope or build your own and see how the spectra look like. The black and white picture has not done justice to the actual extremely beautiful spectrum of Swan bands (observed due to diatomic carbon species).

a more LUMO state, resonating with high energy electricity due to the high stability of noble gases, and then contract back to a more HOMO state, releasing red light, which has a much lower frequency?

This has no scientific meaning, because molecules do not resonate with electricity or electrical energy. I do not know the exact mechanism how electrical energy or even heat is "transformed" into electronic energy, but electrical discharge is quite efficient at exciting atoms or molecules in the gas phase. I am sure it is quite complex, which involves ionization, collisions, recombination etc.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Definition of a cluster
Wikipedia defines a cluster as an ensemble of bound atoms or molecules that is intermediate in size between a molecule and a bulk solid. What is meant by a bulk solid? I can't find a definition anywhere? Further, is this a good enough definition?

https://en.wikipedia.org/wiki/Cluster_chemistry

The properties of a cluster depend on the exact number of constituents, and the geometry of the arrangement.
The properties of a bulk solid otoh are just indifferent to the exact size of the specimen (e.g. crystallite).
(In between you have nanocrystalline substances, where properties change with size, but only gradually.)
I would say that clusters are much like molecular compounds in that they have a distinct structure and absolute composition, however they are held together not by covalent bonds but by polar or dispersive forces (also metallic), and they usually consist of a number of identical particles.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How could I deduce what this organic compound would be?

For compound D (shown in the figure attached above), how can I deduce its structure?
I understand $\ce{CO2}$ has been produced and thus the compound will no longer have the $\ce{-COOH}$ group attached, but what I do not understand is why the double bond will get affected and why a nitrile will be produced?
I am unaware of what type of reaction/decomposition this is for the compound in square brackets (image above) , but I think compound D is supposed to be deduced backwards from its reaction with $\ce{LiAlH4}$. So what I can deduce is that something with $\ce{-CN}$ group in an organic compound can react with $\ce{LiAlH4}$ to produce another type of compound E which would be an amine. On the basis of knowledge I would deduce that compound E would have to be a nitrile.
Also in the image shown below, why is the latter isomer incorrect with that double bond left as it is from the original compounds shown in brackets in the image above?

What do you think the intermediate species is immediately after the loss of CO2?


text:
beer alcohol measurement
I bought an alcohol hydrometer (scale $0-40\%$) to  measure the alcohol in my home made beer, at the end of fermentation and prior to bottling. While it is accurate on pure water (@ $0\%$), home made wine (@ $8\%$), and gin (@ $40\%$), it will not work for beer. I've even tried it on a canned beer of 5% and then it shows a negative figure of -3. I've done all testing @ $\pu{20 ^\circ C}$ and cleared the gas from the liquid. If anybody got any ideas for this misreading, please let me educated.

I'd guess that the beer had bubbles form on the hydrometer.

Is this the kind of hydrometer you are using to measure alcohol content?

I think you are using hydrometer, which has a scale to measure alcohol percentage of the distillate. It should also has a scale to measure specific gravity of your fermentation mixture. I think what you have been doing wrong is you are using the scale to measure alcohol percentage to measure in your fermentation mixture.
This gives you the measure of fermentation mixture’s density as compared to water (if you have used the scale of specific gravity). Sugar increases density, so by measuring the gravity of your beer mixture, you can know how much sugar it contains. By taking one reading at the beginning of fermentation (only contain sugar, yeast, and water) and another at the end of it (any amount of remaining unconverted sugar, yeast, alcohol produced, and water), we can calculate how much sugar has been converted into alcohol, and therefore, the alcohol content of the beer. The density at the beginning should be greater than the final density, because sugar (more dense than water) converted to ethanol (less denser than water). According to Brewer's Friend - ABV-Calculator:
$$ABV(\%) \approx (\text{Initial Gravity} - \text{Final Gravity}) \times 131.25 \tag{1}$$
Thus, if your initial gravity was 1.108, and your final gravity was 1.043, the equation $(1)$ your beer is approximately 8.79% alcohol by volume.
$$ABV(\%) \text{ of your beer} \approx (1.108 - 1.043) \times 131.25 \approx 8.53\%$$
(Note: $ABV$ stands for alcohol by volume).
Few needs to know facts:

Hydrometers are calculated to different temperatures, usually $\pu{60 ^\circ F}$ or $\pu{68 ^\circ F}$. There should be an instruction sheet comes with your hydrometer. Check the temperature of your beer mixture (wort), before you take density reading. Then use the temperature calibration instructions in the chart sheet for how to adjust the gravity if needed (Remember, gravity change with temperature; liquid expands).
As mentioned above, most hydrometers have a few different units of measurements printed on them; you are looking for the gravity reading for this purpose (reading will be a number within $1.000 - 1.160$ or beyond.
Some hydrometers can be used for wine and other liquid, as well (these hydrometers will come with a reading called the Brix Scale, to calculate $ABV$, or you can use Brewer's Friend - Brix Converter).


text:
Molecule with reversible exothermic conformational change?
I proposed some exotic biochemistry on the Worldbuilding stack.
https://worldbuilding.stackexchange.com/questions/166814/digestive-system-of-the-ultimate-omnivore/166829#166829

Your creature oxidizes molecules in its "stomach" which is more of a
  furnace. Anything that can be oxidized is fair game - any reduced
  carbon, or nitrogen, or metal salt. Oxidation produces heat. The
  amount of heat produced is controlled by controlling oxygen ingress to
  the stomach.
Under hot conditions, certain metalloproteins change configuration.
  The circulation brings these proteins into hot conditions near the
  stomach and they capture heat energy with the configuration change.
  When the hot-configured metalloprotein circulates out to the cold
  exterior (possibly radiator plates or fins) it shifts back to the cold
  configuration. This conformational shift is linked to an ATPase and
  generates ATP for the creature to use for its muscles and metabolic
  processes, as Earth life does.

I was challenged to produce a molecule which could capture and release energy in this way. Molecules do this all the time with a phase change - that is a steam engine. But is there any example of a molecule (or a whole type, perhaps) which undergoes a conformational change with heat that it later discharges in a controlled manner, with an exothermic relaxation back to the cold conformation. Not necessarily a biomolecule.

Thing is, they do it all the time! Every single compound with a conformer higher in energy would count.

@Mithoron - that sounds promising.  Can you lay out some examples in an answer?

No, that means, the topic is too broad and question probably on the way to get closed.

@Mithoron - it looks like you are right.  On the WB stack I was challenged that no molecule could do this and here on chemistry stack so many molecules can do it that it is not a question worth answering.  Woe!

:D Well, to be actually able to give off energy surplus to ATPase, the molecule should actually do not lose it - stay in "excited" conformation and only get isomerised by an enzyme and get its energy harvested. Now, that's not really common, and may be just rare enough!

Microscopic reversibility of the endothermic reaction makes it really hard to prevent the reverse happening without the ATP synthesis. What about using a known ion-driven ATP synthase with a molecule that binds ions when cold and releases them when hot to maintain the ion gradient?

@Andrew - I like it.  Can you propose one?


text:
Water becomes cold on mixing energy drink
I noticed that the temperature of water decreases noticeably when I mix an energy drink (Glucon-D) in it. The ingredients mentioned on the pack are majorly glucose (52%) and sucrose (45%). I believe that this is due to the inversion of sucrose being an endothermic reaction
$$\ce{C12H22O11 + H2O + \Delta -> \underset{glucose}{C6H12O6} + \underset{fructose}{C6H12O6}}$$
However, I could not find any sources on the internet to validate this. Please confirm/disprove this hypothesis and also explain the correct reason for this.
EDIT: After everyone's recommendations of sticking a thermometer in my drink, I did so and came up with the following observations:
\begin{array}{|c|c|c c|c|}
\hline
m_\text{drink} & V_\ce{H2O} & T_i & T_f & \Delta T\\
\hline
\pu{14.85 g} & \pu{200 ml} & \pu{20.3°C} & \pu{19.5 °C} & \pu{0.7 °C}\\
\hline
\end{array}
Some elementary thermochemistry gives me the result that $\Delta H_\text{solution}$ for the energy drink is $\approx \pu{39.6 J g-1}$.

Sucrose does not invert upon being dissolved in cold (or room temperature) water. Check out enthalpies of solution.

The dissolution of any substance into water is endothermic, because water has to provide some energy to separate the molecules of the solute to be dissolved This is exactly what you have observed. It is a physical phenomena, and no chemical reaction is involved.

@Maurice Compared to dissolving pure sugar or salt in water, glucose seems to reduce the temperature quite noticeably.

/The dissolution of any substance into water is endothermic/  That is interesting because if you mix borax powder in water, the water becomes warmer.  If you add water to borax powder held in your hand it gets quite hot.  Even more true for calcium chloride.  @Maurice can you write an answer laying out why some solutes dissolving are endothermic and some are exothermic?

@Maurice Willk is right. Here's a kinda related answer I wrote some time ago (for precipitations though) which somewhat goes on the same lines.

@Maurice And you´re really wrong. You cannot dissolve anything without looking closely into the chemistry. Just a few numbers are given here: https://en.wikipedia.org/wiki/Enthalpy_change_of_solution#Dependence_on_the_nature_of_the_solution

@WilliamR.Ebenezer Your answer does provide a good insight, but it deals with electrolytes rather than large organic molecules. Thanks for the help though!

Hydrolysis of sucrose is relatively slow even when boiled and catalyzed by an acid. The idea it is fast even if cold and not catalyzed is wrong.

Dissolving sucrose is endothermic, https://pubs.acs.org/doi/abs/10.1021/ja01164a504, as is dissolving glucose, http://webserver.dmt.upm.es/~isidoro/dat1/Heat%20of%20solution%20data.pdf

From the calculations (and from the enthalpy data posted by Karl, reference here), the following can be calculated:

Enthalpy of solution of glucose in the mixture, given by
$$\Delta H_\text{glucose} = \frac{52}{100} \times 14.85 \times \frac 1 {180} \times 11000 = \pu{471.9 J}$$
Enthalpy of solution of sucrose in the mixture, given by
$$\Delta H_\text{sucrose} = \frac{45}{100} \times 14.85 \times \frac 1 {342} \times 5400 = \pu{105.51 J}$$

By calorimetry, we have
$$Q = mc\Delta T = 200 \times 4.2 \times 0.7 = \pu{588 J}$$
Adding the enthalpies of solution should give us a value close to the heat provided by water, and indeed it does!
$$\Delta H_\text{glucose} + \Delta H_\text{sucrose} = \begin{array}{|c|} \hline \pu{577 J} \approx \pu{588 J}\\ \hline \end{array} $$

Immediately after posting this comment, I am going to upvote both the question and answer because you listened to the suggestions and comments and then DID THE EXPERIMENT! Bravo! Nice work! It would be so nice if everything went like this.

@EdV thank you so much! I really enjoyed doing this experiment: it proved that you don’t need to have access to a lab to enjoy real-world chemistry :)

Well, the solution enthalpy of sugars is positive.
I found these numbers on the internet

$\ce{C12H22O11}$ (sugar(sucrose)) : 5.4 kJ/mol
$\ce{C6H12O6}$ (glucose) : 11 kJ/mol
$\ce{C6H12O6·H2O}$ (glucose monohydrate) : 19 kJ/mol

So if your "energy drink" is a dry powder (and not a readymade drink in an aluminum can), this could explain your observation.
You should however put a thermometer into your experiment, and get us some numbers. As is, my above is just another piece of guesswork.

I checked and Glucon-D is a mix of powdered sugars, flavoring, and so on. The solid mix is added to cold water and stirred until it all dissolves. So I think your answer, minus the next to last sentence, is the correct answer. I am ready to upvote if you concur.

@EdV I have no idea. ;) "Glucon D" is just a powder?

Based on what is on the web, it is just a powder mixture. So I am upvoting you answer. And it would be interesting if the OP had a temperature measurement.

@EdV Hooray for educated guessing. ;))


text:
UV-spectrophotometer analytical method for doxycycline hyclate and tylosin tartrate powder
I am trying to detect tylosin tartrate from the mixture of doxycycline hyclate using $\pu{0.01 M}$ $\ce{HCl}$ as solvent, and waveleghth of $\pu{290 nm}$ for detecting tylosin tartrate in UV-spectrophotometer but the method fails
The aim is  finding an analytical UV-spectro photometer  method to detect tylosin tartrate in mixture of doxcycline hyclate ?


text:
Why are dendrites formed during dilute CuSO4 electrolysis black?
During the weekend, I wanted to conduct a kid's experiment - the copper dendrite formation. I was (partially) successful and now I would like to know what went wrong.
The procedure was taken from the weird science page (unfortunately in polish only).
The procedure was straight forward: Two, thin copper conductors on a microscope slide, connected to DC supply (voltage "over a dozen volts") and few drops of dilute $\ce{CuSO4}$ (according to the procedure, the solution should be transparent or maximum pale blue). It seemed pretty easy and the circuit was built in a few minutes. According to the description (and included pictures), I should see beautiful copper dendrites on the cathode, where $\ce{Cu^2+}$ reduces to $\ce{Cu}$, thanks to electric current (the reaction $\ce{Cu^2+ + 2e- -> Cu}$)…

In fact, the reality was slightly different (see attached figure). Below is my interpretation, but I am aware that it could be wrong:

At the beginning, bubbles on the cathode started to show. I assume it was hydrogen – high voltage (approx. $\pu{12 V}$) caused water electrolysis and $\ce{H2}$ evolution. No gas was evolving on the anode – I presume that the only reaction was oxidation of the copper wire $\ce{Cu-> Cu^2+ + 2e-}$.
A while later green flocs showed on the cathode, progressively moving towards the anode – I presume that $\ce{Cu(OH)2}$ was precipitated, due to increased $\ce{OH-}$ concentration near the cathode, caused by $\ce{H+}$ consumption.
Finally, the dendrites started to show on the cathode – but to my surprise, these were black, not orange like copper.

Why the dendrites are black? It does not look like copper but rather like $\ce{CuO}$. I’ve found the reference stating that $\ce{Cu(OH)2}$ transforms into $\ce{CuO}$ in basic solutions at room temperature, which could be an explanation (Cudennec and Lecerf, The Transformation of $\ce{Cu(OH)2}$ into $\ce{CuO}$, Revisited.). But why are these growing on the cathode instead of pure $\ce{Cu}$? And is it $\ce{CuO}$ or am I wrong?

I believe it is mostly Cu with thin film of oxide. Throw your dendrites in any diluted acid and you'll see the truth right away.

@IvanNeretin You are right. I reversed voltage on the cell and the cathode with dendrites became an anode. After a while, black dendrites changed to orange -  CuO film reduced to pure copper.

To quote a source as you also noted:

Copper hydroxide Cu(OH)2 is metastable. It easily transforms into copper oxide CuO more stable, either in the solid state by a thermal dehydration or at room temperature, in aqueous basic solutions. In the solid state, the transformation is performed at a relatively low temperature, 423 K.

In the current case, a basic solution develops and copper hydroxide breaks down into CuO and water.
However, there may also be heat from applying the overpotential, and this could speculatively be the source of CuO observed.
More precisely related to the cathode, start by noting per this source (Lumen Learning), for example, to quote:

Electrolysis of a solution of sulfuric acid or of a salt, such as NaNO3, results in the decomposition of water at both electrodes. Hydrogen will appear at the cathode and oxygen will appear at the anode.

Assuming on the cathode the formation of an intermediate species (like $\ce{H*}$ mentioned on StackExchange in an advanced discussion of the underlying mechanic on the electrolysis of water) upon self-reaction, to quote:

In an alkaline electrolyte they propose the following mechanism, which only slightly deviates from the above. The intermediate species are the same, that's why they did not need to treat them separately. On the cathode they just summarised as $$\ce{2H2O + 2e- -> 2{}^{-}OH + H2}$$ as the mechanism is bound to be the same.

Next, the action at the cathode of created OH- on Cu++ forms Cu(OH)2 which, in the basic conditions, results in CuO.

I've found two separate sources prooving that indeed CuO can deposit on the cathode if the current density is too high. In such conditions reduction of copper salt produces sponge copper on the cathode which is easily oxidized in presence of air. This fact was even mentioned on this forum: link Therefor I suspect we have two separate reactions - first precipitation of Cu(OH)2 as green flocs due to excess OH- and presence of Cu+2 and second oxidation of porous copper to form black dendrites


text:
How much of Sodium dichloro-s-triazinetrione dihydrate should I use to replace bleach?
I am trying to determine how much sodium dichloro-s-triazinetrione dihydrate powder I should add to water to create a dilution that can be injected into my water supply to react with the Hydrogen Sulfide and remove it.
I have a system in my home that uses sodium hypochlorite (bleach) and water. There is a 30 gallon water barrel that we mix 1 gallon of bleach (5.25% Sodium hypochlorite) with 29 gallons of water. I believe that means the resulting mixture is 0.16666% Sodium hypochlorite. That is injected using a small pump into the water supply where it mixes in a tank and oxidizes the Hydrogen Sulfide.
I want to stop buying bottles of bleach which is expensive and contains additives other than sodium hypochlorite. I would like to switch to sodium dichloro-s-triazinetrione dihydrate which I already have an abundance of in the form of Spa Sanitizer. The Spa Sanitizer label indicates it is 99.99% pure sodium dichloro-s-triazinetrione dihydrate.
So how much of the sodium dichloro-s-triazinetrione dihydrate powder do I need to mix with 29 gallons of water to create a solution similar to the one created by the bleach? (0.16666% sodium hypochlorite)
Or, maybe what I should be asking is what is the amount of free chlorine in the 0.16666% sodium hypochlorite solution and how much sodium dichloro-s-triazinetrione dihydrate powder do I need to mix with 29 gallons of water to achieve the same amount of free chlorine? Since I believe the free chlorine in sodium hypochlorite and sodium dichloro-s-triazinetrione dihydrate are not actually the same.

According to a relevant reference (Ref.1) Following reactions happen when sodium hypochlorite and sodium dichloro-s-triazinetrione dissolve in water, respectively:
$$\ce{NaOCl + H2O -> HOCl + NaOH} \tag {1}$$
$$\ce{NaCl2(NCO)3 + 2 H2O <=> 2 HOCl + NaH2(NCO)3} \tag {1}$$
Hence it is safe to assume that you need only $\pu{0.5 mol}$ of $\ce{NaCl2(NCO)3}$ (sodium dichloro-s-triazinetrione) or in your case $\ce{NaCl2(NCO)3.2H2O}$ (sodium dichloro-s-triazinetrione dihydrate), in place of each $\pu{1.0 mol}$ of $\ce{NaOCl}$ used. Let's see how much of $\ce{NaOCl}$ in $\pu{1.0 US gal}$ of $5.25\%$ bleach solution:
$$5.25\%(w/v) = \\
 \frac{\pu{5.25 g} \;\ce{NaOCl}}{\pu{100 mL} \;\text{solution}}\times \frac{\pu{1.0 mol} \;\ce{NaOCl}}{\pu{74.44 g} \;\ce{NaOCl}}\times \frac{\pu{1000 mL} \;\text{solution}}{\pu{1.0 L} \;\text{solution}}\times \frac{\pu{3.79 L} \;\text{solution}}{\pu{1.00 US gal} \;\text{solution}}\\
 = \pu{2.673 mol/US gal}$$
As states above, you need only half of this amounts in solid $\ce{NaCl2(NCO)3.2H2O}$. You can calculate the amount of $\ce{NaCl2(NCO)3.2H2O}$ in grams as follows:
$$\pu{1 US gal} \: \text{of } 5.25\%(w/v)  = \\
\pu{2.673 mol}  \; \text{of } \ce{NaOCl} \times \frac{\pu{0.5 mol} \; \ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \; \ce{NaOCl}} \times \frac{\pu{255.98 g} \;\ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \;\ce{NaCl2(NCO)3.2H2O}} \\
= \pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}$$
Thus, you can dissolve $\pu{342.1 g}$ of $\ce{NaCl2(NCO)3.2H2O}$ in your 30-gal (US) water barrel to get the same effect of 1-US gal of 5.25% bleach solution. Keep in mind that rate of chlorine release from $\ce{NaCl2(NCO)3.2H2O}$  is slower than that of bleach solution (Ref.1).

Late Edit:
This is about comment made by OP on my post. In that comment, OP states that the cost for dichloro-s-triazinetrione dihydrate (Di-Chlor) and beach are almost same. But I do not agree with that. Just for curiosity, I checked the prices for two chemicals: You can buy $\pu{50 lbs}$ bucket of Di-Chlor granules [online] from Pool Supply World for $\pu{132.99 USD}$; and You can buy $\pu{121 fl \: OZ}$ container of 6% bleach solution from Target for $\pu{4.49 USD}$ (let's assume its about $\pu{1 US\:Gal}$ of 5.25% bleach solution). According to my calculations above, $\pu{50 lbs}$ bucket of Di-Chlor granules equal to:
$$\pu{50.0 lbs} \; \ce{NaCl2(NCO)3.2H2O} \times \frac {\pu{454 g}}{\pu{1.0 lbs}}  \times \frac{\pu{1.0 US\:Gal}  \;\text{5.25% }\ce{NaClO}}{\pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}}\\
 \approx  \pu{66.5 US\:Gal} \text{ of 5.25% }\ce{NaClO} \text{ solution} $$
As a consequence, the cost for $\pu{50 lbs}$ bucket of Di-Chlor granules is $\pu{132.99 USD}$ while the cost for $\pu{66 US\:Gal}$ of 5.25% bleach solution  containers is $\pu{4.49 USD} \times 66  \approx \pu{296.34 USD}$ (more than twice the cost for Di-Chlor granules needed).

References:

Thomas Clasen, Paul Edmondson, "Sodium dichloroisocyanurate (NaDCC) tablets as an alternative to sodium hypochlorite for the routine treatment of drinking water at the household level," International Journal of Hygiene and Environmental Health 2006, 209(2), 173-181 (https://doi.org/10.1016/j.ijheh.2005.11.004).

Your answer makes sense.  I also found this link https://www.oxy.com/OurBusinesses/Chemicals/Products/Documents/isocyanurates/ACL_FAQs.pdf since asking and on page 4 it has conversion chart showing that 1lb of NaCl2(NCO)3⋅2H2O is approximately equal to 1.2 gal of 5.25% bleach.  Your equations roughly support this chart and vise versa.  I appreciate knowing the why behind the chart and you gave me that.  It is in the end worth noting that based on the cost of the shock and the bleach they are about equal for this use.  But i have the spa shock already. Thanks.

$1$ mole of sodium dichloro-s-triazinetrione weighs $220$ g, and is able to liberate $2$ moles of $Cl_2$ in water. If you use the same substance as dehydrate, one mole will weigh $256$ g.
By comparison, $1$ mole of bleach $NaClO$ ($74.5$ g) is able to yield $1$ mole $Cl_2$ in water.
So if you know the mass m of $NaClO$ that you are adding to your water, you may obtain the same effect in solution by using a mass m'  of sodium dichloro-s-triazinetrione dihydrate equal to m' = (m/$74.5$)·($256/2$) g = $1.71$·m.
This means that you must use much more dichloro-s-triazinetrione than bleach.
In an aside I wonder why you would renounce to using bleach. It is cheap and efficient. Apart from $NaClO$ the main "additive" present in bleach is  $NaCl$, or cooking salt, which is not a real nuisance.


text:
Figure out how many deuterium atoms in a 3 microgram ice core sample
I'm in an introductory Chemistry class and ran into a Pearson problem, that no-one in our class including our professor has been able to truly figure out.
I know the solution is $4.69 \times 10^{13}$.
Attached are images.

So far, I have tried two methods. First, I tried to calculate the answer by mass.
1) Calculate mass of deuterium
$3\times10^{-6} \times .0156 \times 10^{-2} = 4.68 \times 10^{-10}$
2) Calculate moles of deuterium
$\frac{4.68 \times 10^{-10}}{2} = 2.34 \times 10^{-10}$
3) Calculate atoms of deuterium
$2.34 \times 10^{-10} \times 6.02214 \times 10^{23} = 1.41 \times 10^{14}$
This answer is 3 times more than the actual solution.
Now, for the real B.S. that is this problem.
Let's try solving this problem by moles/atoms (as in percentage is percent deuterium by atoms)
1) Calculate moles of deuterium
$\frac{3\times10^{-6}}{18} \times 0.0156 \times 10^{-2} = 2.6 \times 10^{-11}$
2) Calculate atoms of deuterium
$2.6 \times 10^{-11} \times 6.02214 \times 10^{23} = 1.57 \times 10^{13}$
This answer is 3 times less than the real answer.
Help. What is 3?

In your second method, you missed a factor of two: you calculated the number of moles of water, but forgot that one mole of water has two moles of H atoms. So moles of deuterium is twice what you got. Then following on, the correct answer is 3.12 times 10 to the thirteenth power and the “book answer” is simply wrong. Sadly, not that uncommon.

What about the case where the molecule is HDO?

How many deuterium atoms (symbol D for deuterium) are present in $\pu {3.00 x 10^{-6} g}$ of water?
Answer:
$$\pu{3.00 x 10^{-6} g} \;\ \ce{H2O} \times \frac{\pu{1 mol} \;\ce{H2O}}{\pu{18.015 g} \;\ce{H2O}}\times \frac{\pu{2 mol} \;\text{H}}{\pu{1 mol} \;\ce{H2O}}\times 0.000156 \times \frac{N_\mathrm A \;\text{D atoms}}{\pu{1 mol} \;\text{D atoms}} = \pu{3.12 x 10^{13} D atoms}$$
where $N_\mathrm A = \pu{6.0221415 x 10^{23}}$ is Avogadro's number and the deuterium fractional isotopic abundance (from the wiki article on "Deuterium") is 1 deuterium atom per 6420 atomic number 1 species, i.e., 1/6420 = 0.000156 = 0.0156%. The units can be mol D per mol H.
Alternate solution: The probability that a water molecule is HOD, rather than HOH, is 2/6420, i.e., 1/3210 = 0.000312 = 0.0312%. This is simply due to ordinary (unsubstituted) water having two hydrogens per molecule, i.e., it is HOH, so it has two independent chances to have H replaced by D. So, on average, there is one HOD molecule out of every 3210 water molecules. Likewise, every 3210 moles of water contains, on average, 1 mole of HOD molecules.
The probability that a water molecule is DOD, rather than HOH, is the square of 1/6420, i.e., $ \pu{2.426 x 10^{-8}}$. This is negligible compared with 1/3210, so it is ignored in what follows. Therefore:
$$\pu{3.00 x 10^{-6} g} \;\ \ce{H2O} \times \frac{\pu{1 mol} \;\ce{H2O}}{\pu{18.015 g} \;\ce{H2O}}\times \frac{\pu{1 mol} \;\ce{HOD}}{\pu{3210 mol} \;\text{HOH}}\times  \frac{N_\mathrm A \;\text{D atoms}}{\pu{1 mol} \;\text{HOD}} = \pu{3.12 x 10^{13} D atoms}$$

This answer is incorrect. The correct answer is showed in the question itself. I'm going to trust my professor on this who showed that the ratio is 3 atoms in H2O to one deuterium atom

My answer is correct.

HDO actually not HDH

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

I go back to your first calculations. $3.00$ microgram water does not contain $3.00$ microgram Hydrogen, as you have calculated. No ! It contains $3.00/18$ micromole of water, and also $2·3.00/18 = 0.333 $ micromole $H$. This is also $0.333$ microgram $H$. I think you can go ahead with this information. Try to calculate the mass of Deuterium in 0.333 microgram $H$.

Can you please elaborate? For example, does 0.0156% represent percentage by atoms or percentage by mass?

In the deuterium article in wiki, it says that the natural abundance of deuterium in the earth’s oceans is one deuterium per 6420 hydrogens. Then 1 divided by 6420 equals 0.000156, which is 0.0156%. Also, remember that answers in books, homeworks, exams, etc., are not always correct!

THANK YOU. This makes sense. I believe Maurice said the same thing but I struggled to understand him. My teacher gave me a similar response just now.  So the factor of 3 comes into play because H2O molecules have 3 atoms and deuterium is just 1? If you make this an answer I will accept.


text:
Energy in chemical reactions
My question concerns how energy is used in chemical reactions. I am working with a reaction between magnesium and hydrochloric acid in a class (although this is not a homework question). What I've found is that the magnesium gives away its valence electrons to either the hydrogen or to the chloride. But in the second case, the chloride ions would first give their valence electrons to the hydrogen, and then they would accept the electrons from a magnesium atom. As I was thinking about which of the cases was more probable, I started thinking about the energy involved. For example, I would think that moving the electrons first from the chloride ions to the hydrogen, and then from the magnesium to the chloride, would use much more energy than the magnesium just giving its electrons to the hydrogen ions.
My question is, as a general rule of thumb, although I wouldn't mind a detailed answer, are reactions that, like this one (as far as I know), have multiple "ways" of reacting going to take the way that uses the minimal amount of energy?  Or are the electrons free to move to whichever atom is closest?

If you dissolve a base metal in an acid, what do you get? (Hint: Chloride ions have an "octet" electron configuration.)

(Can you tell under what context you are "working" with this reaction? It is far easier to give answers if one knows a bit about the background of the audience.)

@Karl I’m a Swedish student so I’m not sure how advanced this is in terms of other courses. We are learning about salts and stuff and this is about a lab we did where we were gonna write a conclusion. Thing is, I’m quite interested in science so that takes me quite some way above what’s needed for this class, I’m just curious :)

@Karl Yeah, you get magnesium chloride :) what I’m trying to get at is if this reaction occurs a certain way, I.e. if the electrons are transferred in a certain way because it would take less energy? :)

@Karl just realized working could be interpreted as a lot of stuff, forgot to mention that it was in a class :P

Magnesium chloride (dissolved) is one thing, and the chloride was there before. What else do you get?

@Karl The total reaction formula would be: Mg + 2HCl —> MgCl2 + H2, so hydrogen gas

"Class" and "student" are equally widely interpretable. ;)

@Karl Problem is I don’t know what it would translate to. I am 15 years old but I’m very interested so I like stuff quite advanced :)

Aha! But that equation is not really what you did, because the whole thing was in aqueous solution. Can you change your equation to reflect that? And then, second step, simplify your equation?

@Karl Um, I can always write (aq) after the hydrochloric acid and the magnesium chloride, but that seems like cheating :) the hydrochloric acid could be written as H+ + Cl- and the magnesium chloride as Mg2+ + 2Cl-

Very good, go on! Write down the equation in full, then strike out what is the same on both sides.

@Karl so, Mg + 2H+ + 2Cl- —> Mg2+ + 2Cl- + H2. So what you mean is that the chloride ions remain unchanged but the magnesium gives away two electrons, supposedly to the hydrogen? :)

That´s it. Just as a side note: You should demand that your teacher thoroughly explains such basic principles. It has been shown again and again that letting students put together things on their own only works if they have already internalised the basics. You and perhaps two others of your class have got it now, the rest has likely given up about chemistry for good, like so many before them.

Magnesium cannot give one or two electrons to the chloride ion. Because Chloride ions have the maximum number of 8 electrons in their external shells. There is no possibility for a new electron to stay on a Chloride ion. And the Chloride ion will be reluctant to accept or to loose one of its electrons, because it has the ideal conformation (8 outer electrons). So there are no "multiple ways" of reacting, as you imagine. The only thing that may happen is for magnesium to loose two electrons per atom, because at the end of this departure, it will have 8 outer electrons, and this seems to be to ultimate goal for all atoms. These electrons will be caught by $H^+$ ions, which are transformed into $H$ atoms and then into $H_2$ molecules.

Yeah, Okay :) what I meant by the chloride accepting the electron is that it would first give its electron to the hydrogen ion, but this reminded me that it might not want to do that because it already has 8 valence electrons, like the octet rule. Right?


text:
Spacing between adjacent planes of a given set in a simple cubic crystal
Today I learned about Miller indices in a cubic crystal, and I learned that adjacent planes $(hkl)$ in a simple cubic crystal are spaced a distance $d_{hkl}$ from each other, with $d_{hkl}$ given by:
$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$$
For example, take the origin any of the below vertices of the cube with z axis directed upwards, the $(0 0 2)$ plane cuts horizontally the middle of the cube, and for this set of planes, $d_{hkl} = \frac{a}{2}$.
May be a silly question, but if we followed the rule of spacing, we would get that the next $(0 0 2)$ plane is the very upper horizontal plane in the cube, which - i guess - is not true since the latter plane has $(0 0 1)$ and not in any case $(0 0 2)$ miller indices. However, if we considered that $(0 0 2)$ planes are only those passing horizontally through the middle of unit cells, then the distance between them will equal $a$ and not $\frac{a}{2}$ (using imagination and not rules).
I must be missing something very trivial here, but it is only today that i started learning these stuff and i am really confused. Thanks for help.

When Miller indices are different only in a common factor, the planes are in the same orientation. The higher order reflections share some planes with the lower order reflections. In the case mentioned, every second plane of the (2 0 0) planes coincides with the (1 0 0) planes, whose spacing is double that of the former.


text:
Can a sulfur atom form six single bonds with six other sulfur atoms?
I found out that diamonds consist entirely of carbons singly bonded to other carbons, and it made me curious as to what other elements could form the same structure. Silicon came to mind, being in the same group as Carbon, but I believe I read about it being too big to form the structure of a diamond.
The next element to come to mind was sulfur because it was the next element that could form solely single bonds, fill its valence shell (by reaching into the d orbital), and have a neutral formal charge.
So can sulfur atoms form 6 single bonds with 6 other sulfur atoms? Can those, in turn, be connected to each other in a three-dimensional shape like a prism? Would having 6 bonds make the molecule stronger than that of a diamond?

Answering the question posed in the title: yes, and under relatively mild conditions.
A hexacoordinated sulfur is present as μ6-sulfido group $(\ce{S^2-})$ linking two $\ce{[Mo3S(S2)3]^4+}$ cores stabilized with three diethyldithiocarbamate $(\ce{dtc^-})$ ligands each into a dimer [1].
The coordination of $\ce{[S7]}$ moiety resembles a heavily distorted trigonal antiprism:

Figure 1. Crystal structure of $\ce{[Mo3S(S2)3(dtc)3]2S}.$
Color code:
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FFFF30}{\Large\bullet}~\ce{S}$;
$\color{#54B5B5}{\Large\bullet}~\ce{Mo}$.
CSD ID: LEBXAR.
Reference

Meienberger, M. D.; Hegetschweiler, K.; Rüegger, H.; Gramlich, V. The Reactivity of Complexes Containing the $\ce{[Mo3(Μ3S)(ΜS2)3]^4+}$ Core. Ligand Substitution, Sulfur Elimination and Sulfide Binding. Inorganica Chim. Acta 1993, 213 (1–2), 157–169. https://doi.org/10.1016/S0020-1693(00)83826-8.

Actually, cyclo-$\ce{S7}$ is an observable molecule with a ring structure. I believe OP was thinking of a octahedral structure for $\ce{S7}$, analogous to the octrahedron of $\ce{SF6}$. Without getting into the debate over d-orbitals$^1$, I would note that this would leave the terminal $\ce{S}$ atoms with seven valence electrons in addition to the steric problems (cf. apparent non-existence of $\ce{SCl6}$).
Update (2023): There is a report, including a crystal structure from X-ray diffraction, of a $\ce{[SCl6]^{-2}}$ species.$^2$
OP also alludes to a 3D polymeric structure ("diamond"), which one could hypothesize like so: a $\ce{S}$ coordinated by six $\ce{S}$ as in the hypothetical octahedron above. Each of the six $\ce{S}$ would participate in two such octahedra each. I think that such a structure may suffer from rather large holes. In addition, the two kinds of $\ce{S}$ atoms are rather different chemically, which I suspect would be energetically worse than the polymeric alternative observed in nature: polymeric 1D chains.

$^1$ which, in my opinion, is settled anyway against their participation.
$^2$
P. Voßnacker, A. Wüst, C. Müller, M. Kleoff, S. Riedel, Angew. Chem. Int. Ed. 2022, 61, e202209684; Angew. Chem. 2022, 134, e202209684.  https://doi.org/10.1002/anie.202209684

A picture of the structure of cyclo-S7 can be found at https://en.wikipedia.org/wiki/Allotropes_of_sulfur#Cyclo-heptasulfur,_cyclo-S7

Under high pressure [a six-coordinate form of elemental sulfur is known 1. As one might expect, a higher pressure is required to achieve this phase with sulfur than with its heavier and more nearly metallic concentration selenium:

S-IV is shown to transform to primitive rhombohedral β-Po phase [see figure below, from the reference] phase at 153(3) GPa, the same transition is found in Se at pressure of around 80 GPa.

Reference

O. Degtyareva, E. Gregoryanz, H. K. Mao & R. J. Hemley (2005). "Crystal structure of sulfur and selenium at pressures up to 160 GPa", High Pressure Research, 25:1, 17-33, DOI: 10.1080/08957950412331331682

Did you have a chance to inspect the crystal structure or do you at least know where it's been deposited? There are several aspects relatively innocent on its own, but when combined would raise a question whether the S-atom coordination can be reliably determined. First, they used Rietveld. Second, the structure of S-IV was incommensurate modulated. Third, "the amplitude of the atomic displacement in S-IV could not be determined due to the lack of the reliable diffraction intensities."

I think the answer would benefit from adding an image of the stacked unit cells (Fig. 11 from the paper could be an inspiration) as well as both structure refinement, displacement, modulation and coordination parameters. I guess we are not getting an ORTEP, but at least some visual aid would add clarity.

@andselisk the figures seem to be behind the paywall. If you gave tge referenced fugure can you please include it?

As a reference, here is Figure 11 from Degtyareva2005: https://i.stack.imgur.com/7gXLs.png

Thanks. I incorporate this.


text:
Is the active mass of every solid and liquid taken to be unity?
I learnt from my textbook and the question/answer - Why is active mass of a pure solid or liquid always taken as unity? that active mass of a solid or liquid is taken to be unity as:
$$[A]=\frac n V=\frac{m}{M\times V}=\frac{\rho}{M}=k$$
where $[A]$ is the active mass, $n$ is the amount of substance, $V$ is the volume, $m$ is the mass, $M$ is the molar mass, $\rho$ is the density and $k$ is a constant.
This is concisely stated in the linked question by:

Molar concentration is directly proportional to density. Since density of solid or liquid always remains constant, the active mass is taken as 1 .

The answer to the linked question explains that, in the course of a reaction mass and volume of solid and liquid substances change but their ratio i.e., density remains constant at a fixed temperature.
As far as I understood, the above explanation holds true for all solids and liquids. So, is the active mass of every solid and liquid taken to be unity? Or are there any exceptions to this fact? If so, how do such substances vary from others for which the active mass is taken to be unity?

This^ is an exception not otherwise. It's a way to simplify things, but a somewhat primitive one. I wager that usually it's either not that useful or even misleading.

Don't ask stuff that was overdone earlier. Probably putting a bounty on earlier question may have been a better idea. https://chemistry.stackexchange.com/questions/59287/why-are-activities-of-solids-and-liquids-taken-to-be-unity

https://chemistry.stackexchange.com/questions/75276/on-the-activity-of-liquids-and-solvents-in-equilibium-constants https://chemistry.stackexchange.com/questions/14058/why-are-solids-and-liquids-not-included-in-the-equilibrium-constant-what-about https://chemistry.stackexchange.com/questions/76307/why-do-solids-and-pure-liquids-have-no-effect-on-yield-and-equilibrium

And there's much more.


text:
How does Gypsum Concrete work?
When mixed with water how does gypsum act as  a binder in creating concrete?

Google is your friend   http://matse1.matse.illinois.edu/concrete/prin.html

Do you mean using Plaster of Paris as the binder to make a concrete ?

The setting of cement involves the reaction of clinker minerals with water, to form a system of interlocking crystals, which lock the materials together.
The setting of cement is primarily due to the reaction of tricalcium aluminate(C3A), which is the most of the four clinker minerals present in concrete. The set regulating action of gypsum is due primarily to its reaction with C3A.
In the absence of gypsum, the reaction of C3A with water is very rapid. The C3A is converted to calcium aluminate hydrate ($\ce{4CaO.Al2O3.19H2O}$) which forms interlynking bridges of solid material between the cement particles. These interlinking bridges reduce the mobility of the cement particles and produce a firm structure, which results in setting of cement mixture. The rapid setting is called a flash set and its irreversible.
The presence of gypsum in cement slows the reaction of the C3A. When mixed with water gypsum dissolves and reacts with C3A to form ettringite ($\ce{3CaO.Al2O3.3CaSO4.32H2O}$). This ettringite is initially formed as very fine-grained crystals, which form a coating on the surface of the C3A particles.

These crytals are too small to bridge the gaps between the particles of cement. The cement mix therefore remains plastic.
After the initial hydration reactions of cement, a dormant period occurs during which the rate of hydration of cement is greatly reduced. During this dormant period, ettrinite crystals continues to grow and eventually are large enough to impede the mobility of cement particles. At this stage setting occurs.


text:
Will the following complex show optical isomerism
So there is this question which asks whether the following complex

$\ce{[Co(NH3)4Br2]+}$

will show optical  isomerism or not.
According to my knowledge it will show optical isomerism in cis form only and not in trans form so overall the compound shouldn’t be optically active.
This what I hav been taught. And moreover shouldn’t it be optically active in every scenario to be called so?
But the ans says that it shows optical isomerism.
I don’t understand why.. is my understanding wrong?

OK, you seem to agree that the trans form does not show optical isomerism. Now what about cis form? Why would it be chiral, when it has an obvious plane of symmetry?

Alright I agree so then is the answer wrong? Cause if I remember correctly this question came in JEE exam (a prestigious exam in India)@IvanNeretin

Yes, the answer is wrong. None of the forms is optically active.

As already commented by Ivan Neretin, the cis form also does not show optical isomerism. Elaborating a bit further on the comments:
Complexes show optical isomerism when they are asymmetric (also known as chiral). They should not have any elements of symmetry, such as planes/axes/centres of symmetry. The trans form has a clearly visible plane of symmetry. It's a bit trickier to find the plane for the cis form. Here's a diagram showing the planes:

(source: Chemistry libretexts)

In the first diagram, the blue plane is not the plane of symmetry. They are rotating the molecules in the left and right panel so that in either case, the two B ligands are in the blue plane. That makes it easier to compare, and to distinguish cis (90 degree angle BMB) vs trans (180 degree angle BMB). It is fine to use this diagram, but you have to specify the location of the mirror planes (there are two for the cis isomer, and five for the trans isomer).


text:
Predicting directions of redox reactions

Hello!
In this textbook example question, how come the chlorine reaction, which has the greatest reducing power will not react? If it has the greatest E of 1.36 V that means it has most reducing stength and can "pull" electrons from both the Br and I reactions. In my mind I believed it would do so, and since the I-reaction has an even lower E than the Br-reaction the extent of oxidation would be greater for I. But in the answer they say the direct opposite, that Cl does not react and Br will only oxidize I.
Why does Cl which is the strongest oxidizing agent not do anything in this case?

You are mixing up $\ce{Cl2}$ and $\ce{Cl-}$. $\ce{Cl2}$ is more reducing than $\ce{Br2}$, but you don't have any $\ce{Cl2}$, you only have $\ce{Cl-}$ from the dissolved $\ce{NaCl}$

Oxidation power  decreases in range $\ce{Cl2}  \gt \ce{Br2} \gt \ce{I2}$,
reduction power decreases in range $\ce{I-}  \gt \ce{Br-} \gt \ce{Cl-}$

Because in the original question there is no elemental chlorine present. It says that you start from a solution of chloride and iodide, so both ions that can be oxidized to the corresponding halogen. And the rest of the answer is already given in your textbook example. Chlorine is a much stronger oxidizing agent. This means that it oxidizes others very well. But what you've got is bromine, and bromine is just between chlorine and iodine, so it can oxidize iodide to iodine but not chloride to chlorine. I think you got confused with the direction and the redox-terms here a bit.

I think I understand what they are saying now thanks to your help. The first thing I needed to understand was that in the solution we have iodide and chloride. Next was that since chlorine has the strongest reducing power it "holds strongest" to its electrons, more so than bromine and iodine which have lower E values. So bromine can't pull away those electrons from chloride. However iodine has a lower E than bromine so bromine can pull away the electrons from iodide, causing formation of iodine. Am i correct?

Basically you are right but elemental chlorine is a strong oxidizing agent, not a reducing agent in your case. It oxidizes others for example bromide to bromine while itself it's being reduced in this process. In cases of halogens I would argument with their oxidation strenght. You could say that bromide is a stronger reducing agent than elemental chlorine, hence it reduces chlorine to chloride.


text:
Setup for Sensing change in concentration of NaCl
I am making a sensor to sense the change in concentration of NaCl. My background is in electronics and chemistry is not my strong area. I got some screen printed electrodes, with Silver (Working Electrode) and Ag/AgCl (Reference Electrode). My initial test is to put the electrodes in separate solutions, with Reference Electrode in a reference solution of known concentration of NaCl and the Working Electrode in the test solution of varying concentration of NaCl. Will this give me a potential difference and will I be able to determine the change in Concentration of NaCl in the working electrode? If not can you make any suggestion on the setup? Thanks.

Thank you. I am making a sweat sensor of some sort. This is the first step to show the basic working principles on the sensing aspect.

This will work with the correct set up. You're constructing a galvanic cell, which will result in a cell voltage. The cell voltage will change when you alter the concentration of NaCl on either side. Important though that your two half cells (each side) are connected both electronically and ionically. Do you have a salt bridge or a membrane between them? https://en.wikipedia.org/wiki/Salt_bridge

Thank you for the info, I do not have a salt bridge. I was not sure what chemical to use for the salt bridge, in my setup. I did a bit of researching on that and KNO3 and KCl look like a strong choice for salt bridge.


text:
Reason for rounded fitting at the top end of Allihn Condenser

I am purchasing an essential oil distillation setup. The Allihn Condenser at the top end, and the attached glassware, are rounded. It is supposedly 24/40. I have never seen this. If one of them broke, I would have a hard time finding a replacement because of the strange shape. Thus, I am wondering if there is a good reason for the shape, or if I should ask the manufacturer to give me a normally-shaped 24/40 connection.
Thank you for sharing your thoughts, in advance.

The metal clip (bottom right of the picture next to the adjustable clamp arm) will hold it in place.

That's a spherical joint. It's standardized laboratory glassware. However, NS 24/40 is not a nominal size for spherical joints. It's probably a KS 35/20 or so.


text:
Is "family" synonymous with "group"?
Is using the term "family" strictly correct when we mean a group of the periodic table or is it just in general use (and should perhaps be discouraged in favour of using "group" exclusively)?  At my highschool, other teachers refer to the transition metals as a family; is this a common practice?

Not a bad question, but you'll get more than a few opinions here. This Wikipedia entry on "Group (periodic table)" will be of use to you.

Family is an informal term. Only a group and a period is an accepted terminology for the periodic table by the International Union of Pure and Applied Chemistry, a widely respected authority on chemical nomenclature and terminology.
The Wiki page of periodic table should be modified because a dubious and layman type reference (www.shmoop.com) for group (= also known as family) is being cited.


text:
Reason for Exothermic reaction?
Why dissolving sodium hydroxide in water is an exothermic process. What causes the evolution of heat when dissolving sodium hydroxide in water?

The enthalpy of solvation of the ions is higher than the lattice energy in the solid, so it is exothermic. See e.g. https://www.chemguide.co.uk/physical/energetics/solution.html

Actually it’s only exothermic at neutral pH because of the exothermic reaction of hydroxide with protons. Once the pH goes up (which takes only a very small amount of hydroxide), the dissolution is endothermic

Let's start with your statement that "dissolving sodium hydroxide in water is an exothermic process." I assume it is based on the reported standard enthalpy of solvation, which is $-44.2$ kJ/mol.
If you go into the lab and try to make a 1 M solution of sodium hydroxide, however, you'll find that the solution gets quite cold. That process is endothermic.
Why the disagreement? The standard enthalpy of solvation is defined for infinite dilution in neutral pH water, so it applies only to the very first tiny amount of $\ce{NaOH}$ that is added to a finite volume of water.
It turns out that the reaction $$\ce{NaOH(s) -> Na+(aq) + HO-(aq)}$$ is actually endothermic. However, the reaction $$\ce{HO-(aq) + H3O+(aq) -> 2H2O}$$ is even more exothermic, so the combined reaction (which is what is occurring at infinite dilution) $$\ce{NaOH(s) + H3O+ -> Na+(aq) + 2 H2O}$$ is exothermic.
At infinite dilution, the added hydroxide ion immediately reacts with hydronium and we observe the net exothermic reaction. But as we add more hyroxide, the hydronium ion concentration is quickly reduced to the point that most of the added hydroxide does not react with hydronium, and we instead observe an endothermic reaction that is just the solvation of the hydroxide and sodium ions without the reaction of hydroxide and hydronium.

This exothermic phenomena is due to the interaction of the ion $OH^-$ with water. The $OH^-$ ion has three free doublets on the Oxygen atoms, which may attract strongly the protons of three neighbouring $H_2O$ molecules. Don't forget that in $H_2O$, the cloud of the only electron of the $H$ atom is nearly entirely placed between the two nuclei $H$ and $O$. So the nucleus of the $H$ atom is practically "nude" in water, I mean not protected by its surrounding electron, as seen from outside. As a consequence, the $H$ nucleus is attracted by any negative charge to be found in the surroundings. And there are plenty available doublets around any $OH^-$ ion in water. Setting up a chemical bond always produces energy, or heat, even if this bond is not an ordinary covalence. When dissolved in water, each $OH^-$ ion may attract $ 3 H_2O$ molecules, which make $3$ new bonds and an ion with the Oxygen atom in the center of a tetrahedron. Three of the four apexes are occupied by a $H$ atom belonging to three different water molecule. The fourth is the $H$ atom coming from the original $OH^-$ ion. There has been long discussions in the literature about the real formula of the ion $OH^-$. It should be written $[OH(H_2O)_3]^-$. But some chemists think that there are more than $ 3 H_2O$ molecules around each $OH^-$ ion. The problem is not definitively solved. Meanwhile, the usual chemist prefers to ignore this difficulty, and to say that the formula of the ion is simply $OH^-$ .

-1 for terrible formatting in spite of my earlier comments.

This answer is incorrect. Simple solvation of sodium cation and hydroxide anion is actually endothermic. The “exothermic” description applies only to infinite dilution in neutral water, where the acid-base reaction of hydroxide with protons occurs


text:
How to test for methane gas purity?
I use methane gas for cooking and there are times in which, despite burners being clean,
the flame doesn't burn that well and/or is quite red.
I suspect that said gas actually has some air added in, either intentionally as a fraud or not,
and would like to test for its purity.
Are there any ways I'd check this?
Thank you

Difficult at home, without a GC MS. And the responsible people at your gas supplier would be criminally liable if they mixed in anything. Unlikely, I´d say.

Pipe line gas , delivered to a home ? Depends where and when to speculate on other stuff may be present. Ethane is probably not unusual . It would increase the heating value so air may be added to keep a constant BTU/ Volume ( usually the basis for billing) , depending where you are /what the rules are.  I think you need to know what may be in it to analyse for it.

And it likely contains a ppm +/- of mercaptan for smell depending where you are.

Btw. gas does contain quite an amount of nitrogen, carbon dioxide, etc., around ten percent. Varied with ethane content, to keep the energy content per volume constant afaik.

During my time of doing Work-Study Chemistry, We have used a bubbler through which a vacuum pump pulled the gas through a solvent, the impurities would then be examined throughout the solvent. Similar in nature to an acetone wash, however, it is done with a molecule in gas format.


text:
Determine H, q, w, and E at 298 K and 1 atm for the complete reaction of 8.170 g of N2O4.?

Determine $H$, $Q$, $W$, and $E$ at $298\ \mathrm K$ and $1\ \mathrm{atm}$ for the complete reaction of $8.170\ \mathrm g$ of $\ce{N2O4}$.
$$\ce{N2O4(g) + 2N2H4(l) -> 3N2(g) + 4H2O(l)}$$
The heat of formation of liquid hydrazine ($\ce{N2H4}$) is $50.63\ \mathrm{kJ/mol}$. Use tabulated data for other heats of formation.
The tolerance on each question is only $0.03\ \mathrm{kJ}$, so express all answers to $0.01\ \mathrm{kJ}$.

Hello, I am having a lot of difficulty trying to do this problem. I found the moles of $\ce{N2O4}$ and got $0.0888$ moles. I got $0.17761$ moles of hydrazine and $0.3552$ moles of water. I found $\Delta H$ by doing $(0.3552)(-2.85.8)-(0.0888)(9.66)+(0.17761)(50.63)$ and got $-93.381$ which is wrong. Also $\Delta H$ and $Q$ should be the same solution right? For $W$ I did $nRT$ $(0.17761)(8.314)(298)= 440.04$.

There is a mistake in the calculation of the heat of reaction. It must be :
$\Delta H = 0.355·(-286)kJ - (0.1776·50.63)kJ - (0.0888·9.7)kJ = - 101.53 kJ -  9.00 kJ -  0.86 kJ = - 111.40 kJ$
Also the work is not  nRT. It is $\Delta n$ RT. And $\Delta n$ is twice the number of moles of $N_2O_4$


text:
What is the @ symbol used for in material science?
I came across the phrase "Silicone Rubber/Paraffin@Silicon Dioxide" also written as $\ce{SR/pa@SiO2}$, utilising the @ symbol.
How can I understand what this means?

The at symbol @ universally means "at the site"  or "at the rate of".
It is routinely used in supramolecular chemistry to denote guest@host relationship because

it follows the original designation of denoting the binding site;
its shape resembles a guest (letter "a") entrapped within a host shell (circle sign "○") providing a visual cue for an intuitive image (e.g. see endohedral fullerene).

The symbol has been introduced by Chai et al. [1] as a part of nomenclature for complex fullerenes:

Ordinarily, a fullerene composed of $n$ carbon atoms with a metal atom, $\ce{M},$ will simply be represented as $\ce{MC},$ regardless of whether
the metal atom is inside or outside the cage.
However, to facilitate discussion of these and more complicated fullerenes with one or more atoms inside, some attached outside and possibly one or more heteroatoms substituting for carbons in the cage network itself, a more explicit symbolism is essential.
We will use a set of parentheses around the symbol $@$ to indicate that the atoms listed within the parentheses are grouped to form a fullerene.
Within this group all atoms listed to the right of the $@$ symbol are
assumed to be part of the cage network, and all to the left are situated somehow inside the cage.
Buckminsterfullerene, under this notation is then $\ce{(@C60)},$ and a $\ce{C60}$-caged metal species is written $\ce{(M@C60)}.$
A more complex example that will be encountered below is $\ce{K2(K@C59B)},$ which denotes a 60-atom fullerene cage with one boron atom substituted for a carbon in the geodesic network, a single potassium trapped inside, and two potassium atoms adhering to the outside.
We have adopted this symbolism since it has the virtue of being
concise, while still containing the critical information.
It is readily printed and transmitted electronically, and it is compatible with ordinary chemical formula conventions.
It also has the virtue of being visually suggestive, and it emphasizes the “superatom” aspects of the fullerenes as new chemical entities.

In the context of the paper by Guo et al. [2] you are referring to, paraffin is encapsulated inside $\ce{SiO2}$ shell (obtained in situ), and these silicon dioxide particles were then embedded in silicone rubber matrix.
Here, in the notation $\ce{SR/Pa@SiO2}$ the forward slash “/” implies composite material (phase separator), and the at sign “@” refers to host-guest interactions as mentioned before.
A  brief infographics to illustrate the concept:

The symbol has also been listed in style guides [3, p. 203]:

3.6.17 At Symbol (@)
The at symbol $(@)$ is most commonly used in email addresses. Its modern use is primarily in accounting (where it means “at the rate of”) and should not be used in scientific writing as a substitute for “at.”
Two legitimate uses of the at symbol are

In chemical formulae, the $@$ is used to denote trapped atoms or molecules. For instance, $\ce{La@C60}$ means lanthanum inside a fullerene cage.
In genetics, an at symbol after a gene symbol indicates that it is part of a gene cluster.

References

Chai, Y.; Guo, T.; Jin, C.; Haufler, R. E.; Chibante, L. P. F.; Fure, J.; Wang, L.; Alford, J. M.; Smalley, R. E. Fullerenes with Metals Inside. J. Phys. Chem. 1991, 95 (20), 7564–7568. DOI: 10.1021/j100173a002.
Guo, Y.; Yang, W.; Jiang, Z.; He, F.; Zhang, K.; He, R.; Wu, J.; Fan, J. Silicone Rubber/Paraffin@silicon Dioxide Form-Stable Phase Change Materials with Thermal Energy Storage and Enhanced Mechanical Property. Solar Energy Materials and Solar Cells 2019, 196, 16–24. DOI: 10.1016/j.solmat.2019.03.034.
The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers, 1st ed.; Rabinowitz, H., Vogel, S., Eds.; Elsevier/Academic Press: Amsterdam; Burlington, MA, 2009. ISBN 978-0-12-373980-3.

Thank you for the second reference. I deleted my incorrect interpretation. Is there a reference for the slash usage?

@M.Farooq I believe the usage of a slash symbol notation in a more or less similar context has already been discussed before, for instance, What is the role of the slash (“/”) in chemical reactions? and Multiple “/” in catalysts

I somehow was convinced we had something on the @ before, but I couldn't find it. However, I found a meta post, and a comment by Loong. This might contain some information worth adding here.

@Martin-マーチン I haven't seen a version of this same question, but there are some prior posts that introduce the @ notation. Examples: https://chemistry.stackexchange.com/questions/63900/how-are-atoms-encapsulated-in-buckminsterfullerene-molecules/63995#63995, https://chemistry.stackexchange.com/questions/24427/whats-the-type-of-bonding-in-lac60/24620#24620,


text:
What is the result of the combustion of the mixture?
I prepared a little experiment about solid rocket propulsion. I have used water ($H_{2}O$), potassium nitrate ($KNO_{3}$) and saccharose ($C_{12}H_{22}O_{11}$).
I made the mixture in a pan by cooking it. I put 100ml of water into the pan. Then I added 50g of potassium nitrate and 100g of saccharose at the same time when the water boiled up. I started to mix it using a spoon and it took like 35-40 minutes to water evoporation. After the cooking there is only a mixture of dough consistency solid left. It had a light-yellow colour.
When the mixture got cold I wanted to test it. I took a little piece of it and then start the fire, it burned like firewoek.
I wonder the equation of the mixture after it mixed. I mean,
$$KNO_{3}(s) + C_{12}H_{22}O_{11}(s) + H_{2}O(l) \rightarrow \dots \text{(Result 1)(s)}$$
And the after burn result of the combustion.
$$\text{(Result 1)(s)} + heat \rightarrow \dots \text{(Result 2)}$$
I hope I gave you enough informations about the mixture and combustion.

If complete combustion occurred, you would be left with potassium oxide, KOH, water, carbon dioxide, and nitrogen oxides. With incomplete combustion, which is likely, you get various pieces of the sugar and leftover potassium nitrate. Probably more than this, actually. Be safe: people have been very badly injured with this homemade rocket mixture!

Thanks @EdV! I really got something like you said. There was a black solid chemical, water, gray gas, and others that I can't see easily. Do you have any idea about the $\text{(Result 1)}$?

Correction: Saccharose is another name for sucrose. Thus, its molecular formula should be $\ce{C12H22O11}$.

@MathewMahindaratne Thanks for the correction. I'm not into these chemistry things, you know :P

My guess is that the light yellow was due to a little caramelization due to the cooking process. But this is just a guess because I never made this stuff.

@EdV Yeah, I agree with your thought. But, I was looking for a formula of the result :P

I don’t think there is one: the longer the cooking time, the more caramel you get. I would guess that caramel is a blanket term to cover a whole range of related species. Maybe someone else here can enlighten us!

@EdV I get what you mean. I'm really grateful to your interpretation. Thanks!


text:
Which has a lower boiling point, Ozone or CO2? Why?
Ozone is a bent molecule, and therefore, Ozone is a polar molecule.
Polar molecules have dipole-dipole intermolecular forces.
CO2 is a linear molecule, and therefore, CO2 is a nonpolar molecule.
Nonpolar molecules have london dispersion intermolecular forces.
Having dipole-dipole intermolecular forces means you have greater intermolecular forces than London dispersion forces, and therefore means that you have a higher boiling point. Therefore, Ozone should have a higher boiling point than CO2.
However, after looking at multiple sources across the internet, Ozone has a lower boiling point than CO2.
Could someone provide an explanation as to why this is true?

CO2 does not melt and hence boil at normal pressure, but sublimes instead. A better comparison would be to find ozone's melting or boiling point at a pressure where CO2 melts or boils. Alternatively consider comparing the triple points.

The primary thing to realize is that, although the ozone molecule has a bent structure, it is a homonuclear molecule.
This means that between the individual oxygens, the bond moments being formed are quite less, as inherently, both the oxygens have the same Pauling electronegativity values. Only a slight difference in the same will occur, due to the asymmetrical bonding scheme in ozone.
It's worth noting that bond moments in carbon dioxide are much more pronounced than those in ozone, even though they end up being canceled vectorially (like you said).
For comparing the boiling points, we need to see the extent of intermolecular attraction as a measure for the same. Due to the difference in electronegativities between $\ce{C}$ and $\ce{O}$, $\ce{CO2}$ ends up having more partial charges on its constituent atoms as compared to ozone. Hence, the boiling point of carbon dioxide is more than ozone


text:
When non-polar particles interact through dispersion forces, how fast does a temporary dipole induce a dipole in the neighboring molecule?
The current question is inspired by this existing question: At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces?. In the comments and answers, it says that there is always a dipole (but the time-average of these dipoles is zero), so it does not make sense to ask how long the dipoles last, or how frequently they appear.
I am wondering how long it takes the neighboring molecule to adjust its electron density to the temporary dipole. Also, I wonder if this is the correct picture: that one molecule has the temporary dipole and the other responds with an induced dipole. Or is it more accurate to say that the neighboring atoms find some common groove of swaying their charge distribution.

Try also Physics SE (if cross posting is fine).

What you describe in the last paragraph is what i would call a resonance condition and is probably what dictates the correlated behavior of the electrons. Speculating further, there is a transition lifetime of the electrons associated with the transition dipole moment that describes the resonance state.

@BuckThorn But this is not a resonance phenomenon, is it? Dispersive forces do not chease to work at low temperatures, when everything is in its ground state.

@Karl I speculate but would not be surprised if it can be described using ideas very similar to resonance, but perhaps I use the word loosely (perhaps I should use the word correlation). Invoking resonance (as correlation) makes sense because electron clouds in the two molecules become correlated in dispersion. As for the low temperature behavior of the dispersion interaction, I don't know the answer. Is there a residual zero K entropy term associated with the dispersion interaction? Does its persistence imply that it is not a "resonance" phenomenon?

@Karl the typical thermodynamics-based derivation of an expression for London dispersion starts from a Boltzmann equilibrium assumption, so I can see why there might be a T dependence and resonance is not considered. However that doesn't seem satisfactory when considering a question involving time-dependent behavior.

@BuckThorn (At zero K, you still have ground state vibration, that´s QM. Not sure what to follows from that.). The electron shell deformations of neighbouring atoms shows a strong correlation at all frequencies below the one Karsten is looking for. That´s the point I made in my answer below. And dielectric interaction is not a resonance phenomenon. What we are looking for is the same effect that microwaves have on permanent dipoles, only for induced dipoles, where it appears even a few orders of magnitude later (at higher freqencies).

@Karl I remember another question comparing resonance spectroscopy versus dielectric spectroscopy was asked here not so long ago and that the answer was that dielectric spectroscopy is not a resonance phenomenon. Still, there is something very intuitive about resonance in this setting (think plasmons). Resonance and correlation seem to me two sides of the same coin.Ah, I see now that it was another answer of yours, which you link to.

The upper limit are typically optical light frequencies. That is where the very last plateau is reached in dielectric spectroscopy. The electron shell becomes purely elastic.
I wrote a bit more about DES in my answer here: Difference between relaxation and resonance leading to an absorption spectral feature?

Karl, I started reading and watching videos about dielectric spectroscopy. Right now I am pondering the limbo dance analogy in this video from Bend research: https://www.youtube.com/watch?v=qkpA0FwfBeI (15 min. in).

@KarstenTheis That limbo dance analogy is ludicrous at best.


text:
How to dissolve salicylic acid into a solution with humic acid
I'm not a chemist but I am trying to make a product to be used as a foliar spray on turfgrass for my own use. The product I'm trying to make is:

12% Humic acid ($\mathrm{pH =}$ 3.5-4.0)
2% Fulvic acid ($\mathrm{pH =}$ 9.6)
6% Kelp extract
0.5% Salicylic acid
79.5% Water (local sourced water $\mathrm{pH }$ is 6.8-7.5)

I'd like to make this in $\pu{19L ( 5 gal.)}$ batches.
The humic, fulvic acids and the kelp extract are all soluble in the solution already, so that's the easy part. What I'm looking for is how to dissolve the salicylic acid so that it can be combined with the other ingredients.
The advice I've been given for dissolving the salicylic acid is to first dissolve in ethanol. Add $\ce{KOH}$ to water until I get a $\mathrm{pH}$ of 12, then add the dissolved salicylic acid to the water.
What I'd like to know is:

Is this the right process?
What quantities of salicylic acid/ethanol/$\ce{KOH}$/water should I be using?
Will adding this mixture to my mixture of humic/fulvic/kelp have any negative effect on any of the ingredients?
Any safety concerns beyond the exothermic reaction of $\ce{KOH}$?

Could you mention the purpose of this exercise? I feel that using KOH/ EtOH will burn your grass.

This will be a concentrate that will be applied at 89ml per 100sq.m. and diluted in water. The KOH/EtOH will be diluted enough to not burn. I am replicating several products together.

I would say stick to natural things for plants. I have heard kelp is good for plants. Even better ask a professional horticulturist or special forum.

The humic acid, fulvic acid and kelp are great for plants. Salicylic acid is great as well, studies have shown it helps with heat stress. I'm looking for a way to put it into solution as it is not very water soluble.

Salicylic acid dissolves @ 2g /L at 20 oC in water. This is a $lot$.  You will have to agitate very well. Secondly, fulvic acid solution, which is at pH 9.6 will easily dissolve it. Read all safety datasheets.

Thank you. If I add it to the fulvic acid solution then add to the water with humic, will the pH change bring it out of solution or will it be ok?


text:
How to determine the relative contribution of resonance structures when different rules give contradictory outcomes?
In order to determine the relative contributions of resonance structures, my textbook gives the following rules (in order):

The more covalent bonds a structure has, the higher it scores.
Structures in which all of the atoms have a complete valence shell of electrons (i.e., the noble gas structure) are especially relevant and make large contributions to the hybrid.
Charge separation decreases the score.
Resonance contributors with negative charge on highly electronegative atoms score higher than ones with negative charge on less or non electronegative atoms. Conversely, resonance contributors with positive charge on highly electronegative atoms score lower than ones with positive charge on non electronegative atoms.

How to determine the relative contribution of resonance structures when different rules give contradictory outcomes?
For example, let's consider the following resonance structures:
$$\ce{\overset{+}{C}H2-\underset{1}{O}-CH3}\longleftrightarrow \ce{CH2=\underset{2}{\overset{+}{O}}-CH3}$$
The following are my conclusions with respect to the above rules:

Structure 2 scores higher than structure 1 as it has more number of covalent bonds.
Structure 2 scores higher than structure 1 as all atoms have noble gas electronic configuration. In structure 1, the leftmost carbon atom has only six valence electrons.
There is no charge separation in either of the two structures and so this rule cannot be used to determine the relative contribution.
Structure 1 scores higher than structure 2 because in the first structure a carbon atom bears a positive charge whereas in the second an oxygen atom carries a positive charge. Since oxygen is more electronegative than carbon or carbon is more electropositive than oxygen, structure 1 scores higher.

Based on conclusions 1 and 2, we could say that structure 2 contributes more. But conclusion 4 contradicts the result given by points 1 and 2. So, how could we determine the relative contributions when the outcomes of the rules contradict each other?
Are some rules (out of the given 4) superior over the other? Or do we determine the overall score by the number of votes on either sides (similar to our community's voting system based on upvotes and downvotes :) ) i.e., structure 2 scores higher since it has two votes (points 1 and 2) supporting it and 1 vote (point 4) against it? If this is the case, what if there are equal number of votes? For example what if points 1 and 2 suggest a structure A scores higher than B whereas points 3 and 4 suggest structure B scores higher than A?
Kindly do not limit your answers to the above two resonating structures. I considered it just to explain the question.
I read the question - How to determine the least stable resonance structures out of a given set? and the answers to it. In fact, that question is based on a problem from the same book I follow. But still, I don't understand how to determine the relative contributions when some rules give contradictory outcomes.

Note: I understood resonance structures are hypothetical, i.e., they don't have an existence in the real world. The molecule is best represented by the hybrid of all the structures. As commented in my previous questions on resonance, I've already read the question/answer - What is resonance, and are resonance structures real? and I feel I don't have any misconceptions on the subject.

Consider that in this case you analyse an ion. This makes consideration about electronegativity to be slippery. The leftmost carbon has only six electrons in the valence shell. Its charged goes along with. Also, in case, the energy of two limiting forms might be little.

@Alchimista: Thanks for the comment. However, rule 4 is applicable only in the case of ions. I've seen the author explaining relative stabilities of ions in terms of whether positive "charge" is on an electronegative or an electropositive atom.

@Alchimista: The linked question answers one of the aspect of the question based on the specific example I considered. However, I don't think it answers all the permutations of rules being satisfied or defied. So my question is not a duplicate of the linked. But it's "related" and thanks for the link. "Note that the accepted answer is right but it contains a wrong incipit." - Yes. I agree; "The fact that resonance [...] by energy, of course." - I understood this already. And finally, I think this is a good reason - "major criteria are covalent bonds number and octets fulfilment".

Your four textbook rules are not supposed to tell you if mesomeric structure the major contributor to hybrid, but, seems to me, are just an simplified way for beginners to "manually" tell which structures actually make sense, i.e. if you can't tell just by looking, then analyse with rules and maybe answer's gonna make some sense.

What you can  tell is that both of these do make sense and are most significant contributors, and that's it. Unless you have some software or find a paper, telling which of these two is the major one is a no go. Unless you had some experience with this and then you wouldn't need to ask.

@Mithoron: Thanks for your comments. So, are these rules meant only for beginners (like me) and are susceptible to errors? If so, could you tell what am I supposed to do when different rules give contradictory conclusions, as mentioned in the question? Is there any priority order for these rules? Actually, the reasons given by the author for these rules was really reasonable and I believed them to be correct. I didn't include them in the question as it would make it longer. It's quite surprising that even something based on good reasons behave in a different manner than intended.

@Alchimista I don't agree with that last part. If they're not actually different, then they can't be ordered by energy. They only appear different because of the limitations of our structural representations.

@Zhe. Which last part? If you refer to the linked answer I agree with you. If you refer to the last part of my comment I can only repeat it. The two limiting forms either drawn or calculated are two different things and they can be ordered by energy, or in other words they enter in the resonance hybrid description of the molecule with different weights. It is right what all this is discussing about. They are different, just they have no individual meaning. This is basic stuff. Of course, in cases they might be different but having more or less the same energy.

@Alchimista Sorry, I should've been more specific. The two resonance structures are only different things because we created the concept of a resonance structure. In reality, they are the same thing and thus cannot have different energies. I think we're ultimately saying the same thing, but my stance on not over-analyzing individual resonance structures sounds more draconian, and that includes not trying to describe which one is more stable.

@Zhe I've suspected that way of thinking and we might have been saying the same. But the things we have sorted to describe the hybrid are nevertheless different. I just came to the point not to confuse the OP, who beside the actual question already had a rather clear picture. With respect, your position sounds draconian. It is not that the two structures are the same, they are different for sure. That both are even not existing at least it is clear for you me OP. I should have been draconian asserting that rules have hierarchy. I didn't because exceptions are existing beyond the current exampl

Resonance structures do not exist; you even state that. They cannot, not even hypothetical, have different energies. Within the concept of Valence bond theory, they are all necessary to describe a molecule. In that way, you can only compare different contributions of those hypothetical structure to the overall wave function. If you cannot agree with this, then your last sentence is wrong as you do have misconceptions on the subject.

Does this answer your question? Rules to identify the most stable resonance structures

The Rules
Actually the rules for determining the relative stabilities (more accurately, contributions) of resonance structures do have their own priorities. Here is one version of them, with descending order of importance listed below (as is indicated by the bold words). The rules can be found in common organic chemistry textbooks (like [1], also supported by [3]).

In any case, the more the octet rule is fulfilled (for appropriate atoms), the better.
Resonance structures with charge separation are usually higher in energy than those in which charges can be neutralized, unless rule 1 is not obeyed.
If charge is separated, then the best structure is consistent with the electronegativity of the atoms (e.g., a negative charge is best on the most electronegative atom).
The more and stronger the bonds, the better. Adjacent atoms should not have the same sign of charges.

Examples
The first rule is undoubtfully the most important. The example of $\ce{CH2^+-O-CH3 <-> CH2=O^+-CH3}$ is already mentioned. Here I would like to provide an example to justify the importance of rule 1.

Consider the Lewis acids $\ce{BX3}$ where $\ce{X}=\ce{F},\ce{Cl},\ce{Br}$. We know that $\ce{BF3}$ is the least Lewis acidic despite the fact that $\ce{F}$ is the most electronegative. This fact can be explained by considering the resonance structures $\ce{X2B-X <-> X2B^-=X+}$. According to rule 1, the second resonance structure is more important, so the electron deficiency of $\ce{B}$ is determined by the orbital overlapping between $\ce{B}$ and $\ce{X}$. Therefore, since $\ce{F}$'s $2\mathrm p$ orbitals interact with $\ce{B}$'s best, $\ce{BF3}$ is the least Lewis acidic.

Another interesting example illustrating the relative importance of rule 3 and 4 comes from Grützmacher's work[2].

The phosphaethynolate anion $\ce{OCP-}$ may attack $\ce{TMSOTf}$ using $\ce{O}$ and $\ce{P}$. At 298K the product is $\ce{TMS-O-C\bond{3}P}$, and at 318K the product is $\ce{TMS-P=C=O}$. To account for this, let's analyze $\ce{OCP-}$'s resonance structures:
  $$\ce{P\bond{3}C-O- <-> O=C=P-}.$$
  The latter has stronger bonds and the former has its negative charge on a more electronegative atom. According to rule 3, the former dominates (actually a ratio of former vs latter 40:52), hence at ambient temperature $\ce{O-}$ attacks more quickly to obtain the kinetic product, while $\ce{TMSPCO}$ is the thermodynamic product.

Theoretical Justifications
You can find theoretical justifications using Hückel Theory as always. I will not show this here because it may be to tedious to do the calculations. However, you can read this article for more information J. Chem. Educ. 2007, 84, 6, 1056.
Ref.
[1]  Grossman, The Art of Writing Reasonable Organic Reaction Mechanisms, Springer, 2002.
[2] Grützmacher et al., Dalton Trans., 2014,43, 5920-5928.
[3] Humbel, J. Chem. Educ., 2007, 84, 6, 1056-1061.

This is mixing two different, though connected, subjects. Any consideration about the stability as per the question has to be carried out in vacuum as well. I appreciate listing of the rules in their order. I should have been more clear in my previous comment. Besides exceptions there is a strict order to follow. It is general enough that I don't even think this is worth a bounty. The only special aspect of the Question is that the two limiting forms are both possible and should both contribute with no too different extent to the description of the real actual molecule. So it is difficult to..

... to do a quantitative assessment using rules not designed to have precision.

It is in principle wrong to say that resonance structure have different energies; in that way most organic textbooks get it wrong. They only exist as a resonance hybrid, also only in the description in terms of Valence bond theory.


text:
Antibacterial properties of Ethanol and Isopropanol
Ethyl alcohol is widely used in beverages, while isopropyl alcohol is commonly sold as an antiseptic. However, some use both ethyl alcohol and isopropyl alcohol as antibacterial protection.
I am curious to know which of the two alcohols will provide better antibacterial protection? Does the chemical structure of the alcohol affect the antibacterial properties?

"Bacteria" is a very diverse species. You cannot generalise this.

This question is really a broad one. For example, see Karl's comment elsewhere. Yet, according to a review on the subject (Ref.1):

Although several alcohols have been shown to be effective antimicrobials, ethyl alcohol (ethanol, alcohol), isopropyl alcohol (isopropanol, propan-2-ol) and n-propanol (in particular in Europe) are the most widely used (Ref.2). Alcohols exhibit rapid broad-spectrum antimicrobial activity against vegetative bacteria (including mycobacteria), viruses, and fungi but are not sporicidal. They are, however, known to inhibit sporulation and spore germination (Ref.3), but this effect is reversible (Ref.4). Because of the lack of sporicidal activity, alcohols are not recommended for sterilization but are widely used for both hard-surface disinfection and skin antisepsis.

Also, Ref.1 instates this interesting fact:

In general, isopropyl alcohol is considered slightly more efficacious against bacteria (Ref.5) and ethyl alcohol is more potent against viruses (Ref.6); however, this is dependent on the concentrations of both the active agent and the test microorganism. For example, isopropyl alcohol has greater lipophilic properties than ethyl alcohol and is less active against hydrophilic viruses (e.g., poliovirus) (Ref.6). Generally, the antimicrobial activity of alcohols is significantly lower at concentrations below 50% and is optimal in the 60 to 90% range.

The the optimal bactericidal nature of 70% ethanol is well documented in Ref. 7:

On the basis of these experimental findings, a routine method of preoperative preparation of the hands has been evolved which, while simple, is a great deal more effective than technics in use at present in many leading hospitals. This procedure has been employed with satisfaction in my operating room for the last four years and, having passed that clinical test, is now recommended to others.
Seventy per cent by weight alcohol is recommended also for use in preparing the field of operation and in disinfecting contaminated hands.
Why exactly 70 per cent should be more toxic for bacteria than any other concentration of alcohol remains an interesting biochemical problem.

Also, the Ref.7 would answer your question about whether the chemical structure of the alcohol affects the antibacterial properties or not, which states:

The activity of the alcohols in inhibiting spore germination increases as their molecular weights and chain lengths increase (Fig. 3). This same relationship is observed for the action of alcohols as bacterial disinfectants. It is of interest that primary alcohols are more effective bacterial disinfectants than are secondary alcohols, and in this study a primary alcohol, n-propanol, was more effective than a secondary alcohol, 2-propanol, in inhibiting the spore germination process.

References:

Gerald McDonnell, A. Denver Russell, “Antiseptics and Disinfectants: Activity, Action, and Resistance,” Clinical Microbiology Reviews 1999, 12(1), 147–179 (DOI: 10.1128/CMR.12.1.147).
S. E. Morton, "Alcohols" in Disinfection, sterilization and preservation; Seymour S Block, Ed.; 3rd ed.; Lea & Febiger: Philadelphia, PA, 1983, pp. 229–235.
Y. Yasuda-Yasuki, S. Namiki-Kanie, Y. Hachisaka, "Inhibition of germination of Bacillus subtilis spores by alcohols," In Spores VII; eds G. Chambliss, J. C. Vary, Eds.; American Society for Microbiology: Washington, DC, 1978, pp. 113–116.
Ralph Trujillo, Nancy Laible, "Reversible inhibition of spore germination by alcohols," Applied & Environmental Microbiology 1970, 20(4), 620–623.
C. E. Coulthard, G. Skyes, "Germicidal effect of alcohol," Pharm. J. 1936, 137, 79–81.
M. Klein, A. Deforest, "Principles of viral inactivation" in Disinfection, sterilization and preservation; Seymour S Block, Ed.; 3rd ed.; Lea & Febiger: Philadelphia, PA, 1983, pp. 422–434.
Philip B. Price, “Ethyl alcohol as a germicide,” Archives of Surgery 1939, 38(3), 528–542 (doi:10.1001/archsurg.1939.01200090135010).

Thanks for asking this question because after reading this source by the CDC (Center for Disease Control) on 'Guideline for Disinfection and Sterilization in Healthcare Facilities' (2008), to quote:

In the healthcare setting, “alcohol” refers to two water-soluble chemical compounds—ethyl alcohol and isopropyl alcohol—that have generally underrated germicidal characteristics 482. FDA has not cleared any liquid chemical sterilant or high-level disinfectant with alcohol as the main active ingredient. These alcohols are rapidly bactericidal rather than bacteriostatic against vegetative forms of bacteria; they also are tuberculocidal, fungicidal, and virucidal but do not destroy bacterial spores. Their cidal activity drops sharply when diluted below 50% concentration, and the optimum bactericidal concentration is 60%–90% solutions in water (volume/volume)..

An important point per above relative to employing alcohol as a disinfectant, in my opinion, is that they are 'generally underrated' with respect to their germicidal abilities. Also, interestingly, ethyl and isopropyl alcohol are each somewhat selective within the broad characterization given above. To quote further:

Ethyl alcohol, at concentrations of 60%–80%, is a potent virucidal agent inactivating all of the lipophilic viruses (e.g., herpes, vaccinia, and influenza virus) and many hydrophilic viruses (e.g., adenovirus, enterovirus, rhinovirus, and rotaviruses but not hepatitis A virus (HAV) 58 or poliovirus) 49. Isopropyl alcohol is not active against the nonlipid enteroviruses but is fully active against the lipid viruses 72. Studies also have demonstrated the ability of ethyl and isopropyl alcohol to inactivate the hepatitis B virus(HBV) 224, 225 and the herpes virus, 490 and ethyl alcohol to inactivate human immunodeficiency virus (HIV) 227, rotavirus, echovirus, and astrovirus 491.
In tests of the effect of ethyl alcohol against M. tuberculosis, 95% ethanol killed the tubercle bacilli in sputum or water suspension within 15 seconds 492. In 1964, Spaulding stated that alcohols were the germicide of choice for tuberculocidal activity,...

So, if one is confronted, for example, by a flu outbreak in your educational institute, perhaps one should lean in the direction of ethyl alcohol as it is effective against the influenza virus. However, always applying both separately remains an albeit inconvenient option, and in either choice, bacterial spores are still a potential issue.


text:
Is cis- or trans-cycloundecene more stable?
While reading about cis/trans isomerism in cycloalkenes, I came across the following statement:
When the number of carbon atoms in a cycloalkane is less than 11, the cis form is more stable than the trans form. When it is greater than 11, the trans form is more stable than the cis form.
An obvious question came to my mind: What if the number of carbon atoms in the ring is "equal" to 11? Which form will be more stable - cis or trans? I think both will be of the same stability, but would like to have some clarification in this regard.
I searched my books and didn't find any specifications on the stability of cis or trans forms, for a ring containing exactly 11 carbon atoms. I read the answers to the question Stability of cis vs trans isomers? but they discuss only the cases "less than 11" and "greater than 11" but not "equal to 11".

https://chemistry.stackexchange.com/questions/30940/relative-stability-of-cis-and-trans-cycloalkenes

There is experimental data for the methoxy derivative: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78289131&Units=SI&Mask=8

I think that means the (Z) form is favored. I would think the situation is similar in the absence of the methoxy group.

Thank you for your answer. Will the relative energy difference between the cis/trans or E/Z forms be less for cycloundecene, compared to other cycloalkenes?

@Magnesium Here is a link to some more data: https://webbook.nist.gov/cgi/cbook.cgi?Source=1980TAS643-646&Units=SI&Mask=8