text:
How to make pyramid-shaped NaCl salt crystals
I'm attempting to make sea salt pyramids like the one below.

I understand this needs to be evaporated at a reasonable speed to stop the crystals forming a cube, as the when formed quickly, the edges grow faster than the faces, creating the inverted pyramid shape.
When evaporating the saturated water at around 75 degrees celsius, I start with what look like great pyramids forming (like the below image). However, the crystals then begin to join together forming a large flat 'crust', with the individual small pyramids losing their unique shape.
I've attempted reducing the temperature of evaporation to a point where this mass of crust does not form, however at that point, the crystals which now form are cubic instead of pyramids.
Any help would be great!

When you say you get an unwanted crust, is this in the solid surface of the container you're heating the salt solution in, or is it on the surface of the liquid exposed to the atmosphere? If it is the latter, a completely 100% wild guess is that you may get something different by evaporating the water using a thin layer of, say, olive oil on top, to reduce excessively fast evaporation of the upper layer of the brine.

Hi Nicolau, thanks for your response.  The salt is forming on the liquid surface which is exposed to the atmosphere.
I've considered adding something like olive oil as an experiment. However, as the idea is to produce a food grade product, I'm not wanting to add anything with may leave residue on the salt.
As shown in the video, https://www.youtube.com/watch?v=gCXTLRxqVmo to achieve the result I'm after, the salt should form and then drift to the bottom.  However, once the individual crystals fuse into one layer, this no longer happens.

Nice video, though unsurprisingly lacking details about the process. Adding olive oil certainly wouldn't affect the food grade status of the salt. Furthermore, it may be that a single drop of oil already has an effect, because over water oil will spread as much as it can, going all the way to a layer a single molecule thick. You may think such a thin layer would be inconsequential, but check this video out. Absolutely nothing useful may happen, but I figure it's simple enough to try in a small scale.

Wow, interesting video, I've read about this with sailing before but didn't understand how it would make a difference.  Will give this a go and let you know the result.

I gave this a go with the olive oil at first a couple of drops, then a bit more.  At around 75 degrees celsius, it did seem to calm the movement of the water a bit and had the effect of isolating some crystals as the oil created a barrier.  However, once the crystals met they would fuse and crust over like previously, though maybe a little slower.  Appreciate the idea.

Ah sorry to waste your time then. Hopefully you'll find the trick!

I too am working on this. There's lots of factors which can effect it in regards to the brine's makeup and the amount of time it takes to bring it to 75C-80C. We should connect to see what we've learned!

TL;DR: Use 10-20% urea solution for octahedral crystals and a water-ethanol mix for pyramidal crystals.

There is no universal method to predict and to control crystal habits. Very often the experimental results are explained with molecular dynamics; there are models (Wulff condition, Donnay–Harker correlation, Hartman–Perdok theory, Hartman–Bennema attachment energy) that can only be used to represent vapor-grown crystals. Crystal morphology is vastly dependent on many factors:

phase it's growing from (gas, liquid, melt);
growing rate (temperature);
concentration of the mother solution;
ionic strength of the solution;
presence of impurities (habit modifiers);
solvent etc.

Rock salt can take one of the three morphologies dictated by its FCC lattice with a strong tendency to prefer cubic crystals:

Figure 1. Morphologies of $\ce{NaCl}$ crystal denoted with Miller planes: cube $\{100\}$; octahedron $\{111\}$; rhomb-dodecahedron $\{110\}$.
There are numerous literature sources providing experimental information on crystal growth. For sodium chloride there is a recent review by Aquilano et al. [1], not only providing a brief description of common habit modifiers, but also giving an insight on to why one crystal direction is preferred:

The morphology of natural halite is largely dominated by the cube $\{100\}$; the $\{110\}$ and $\{111\}$ forms are seldom present. When crystals grow by evaporation from
  aqueous solution the perfection of the cube faces is usually lost; owing to the high values of the supersaturation, hopper shaped $\{100\}$ faces appear along with dendritic
  branches developing in the $❬111❭$ directions. This is the main reason of the halite tendency to cake. Hence, the best way to face this problem is to search for additives
  able to favor the appearance of the $\{111\}$ form, thus avoiding the morphological instability such as the “hazardous” dendritic growth.

Here is a brief summary for the non-standard crystal shapes of rock salt and corresponding modifiers:
Octahedra. Formamide; urea; sodium tetraphosphate $\ce{Na6P4O13}$; cadmium chloride $\ce{CdCl2}$; polymer additive containing an amide functional group.
Dendrites and needles. Hexacyanoferrate(II) $\ce{Fe(CN)6^4-}$; polyvinyl alcohol.
Pyramides. Water-ethanol solution; growth from a supersaturated solution [2].
Presence of heavy metals ($\ce{Pb^2+}$, $\ce{Cd^2+}$) generally promotes formation of large single crystals instead of many small crystals. For more practical details, photos of crystals and manuals you can refer to Crystal Growing Wiki (originally in Russian, but now most articles are translated to English).
References

Aquilano, D.; Otálora, F.; Pastero, L.; García-Ruiz, J. M. Three Study Cases of Growth Morphology in Minerals: Halite, Calcite and Gypsum. Progress in Crystal Growth and Characterization of Materials 2016, 62 (2), 227–251. https://doi.org/10.1016/j.pcrysgrow.2016.04.012.
Desarnaud, J.; Derluyn, H.; Carmeliet, J.; Bonn, D.; Shahidzadeh, N. Hopper Growth of Salt Crystals. J Phys Chem Lett 2018, 9 (11), 2961–2966. https://doi.org/10.1021/acs.jpclett.8b01082.


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What type of gloves should be used while diluting sulfuric and hydrochloric acids?
What type of gloves should we use on hands while diluting sulfuric and hydrochloric acid?
Can we use these gloves in chemistry lab to handle other acids?

And why do you think normal ones won't do?

@Mithoron define normal -- for some that would still mean latex

There are numerous sources online comparing polymer materials resistivity in hazardous environments, but most of them agree that in your case ordinary nitrile gloves would protect in case of incidental contact, and butyl rubber/neoprene/PVC gloves are more suitable if a prolonged exposure is expected.
Nitrile gloves are disposable and cheap, and always should be changed when contaminated with chemicals as they only offer basic protection.

source: Stack Exchange

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The major product of 1,4-pentadiene NBS bromination
Considering penta-1,4-diene, if allylic substitution is done with NBS, which product will be major, 3-bromo or 1-bromo?
In the first case I get stable free radical , in the second case I get conjugated diene.
Should I consider stability of free radical first as it is formed first , or the stability of the product which is formed in the end?

Continuous conjugation is better than cross conjugation. Therefore, the radical in which continuous conjugation occurs should form the major product. 1 bromo product should be the major.

Does this mean that here the terminal free radical is more stable due to continuous conjugation than the free radical with cross conjugation?

Or is it that the end product is more stable?

What should be seen first? stability of free radical or stability of the end product.

Is your question, does 3-bromination occur faster than the double bonds move into conjugation to give the radical in the 1- position?

The Main reason for looking at the stability of the intermediate is Hammond's Postulate. Now the intermediate has the electron density spread all over it's body. SO it's reasonable to look at the activation energy of the transition state itself.
When the Br radical attacks at the end position, you have a much sterically favored transition state and the other two bonds are in conjugation and hence we would expect the activation energy would be much lower for this and not for the other transition state where the Br attacks at $3^{rd}$ position.
Hence we can conclude that conjugated diene is most probable product !.
Please correct me if I am wrong.

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Typography of multi-letter variables
Chemical elements are typeset with upright font ($\ce{H2O}$ for water) while variables are typeset with an italic font ($V$ for volume).
But is there a convention what to do with multi-letter variables like IC50, EC50 (or others without subscripts)? Is there any official convention for this?
Special cases may be pH and pOH where on could read p as an operator and H and OH as chemical abbreviations, so I would always write them upright. But I am not sure about the others.

In the International System of Quantities, i.e. the system of quantities that corresponds to the International System of Units (SI), as described in ISO 80000, the symbols for quantities are generally single letters from the Latin or Greek alphabet, sometimes with subscripts or other modifying signs. However, symbols for so-called characteristic numbers are written with two letters from the Latin alphabet, the initial of which is always capital. For example:

Reynolds number $Re$
$$Re=\frac{\rho vl}\eta=\frac{vl}\nu$$
Nusselt number $Nu$
$$Nu=\frac{Kl}\lambda$$
Prandtl number $Pr$
$$Pr=\frac{\eta c_p}\lambda=\frac\nu a$$

(Note that such two-letter symbols should be separated from other symbols if they occur as factors in a product.)
Quantity symbols are always written in italic type.
Various other widespread multi-letter symbols are not quantity symbols in accordance with the above-mentioned systems; they are actually just abbreviations. For example in toxicology, LD50 is used as abbreviation for “median lethal dose” (or “lethal dose, 50 %“). Such abbreviations are typeset normally, i.e. generally upright.
Also note that abbreviations for quantities shall not be used in equations.

Do you have an authoritative argument for "abbreviations shall not be used in equations"? In real life, they often are, e.g. https://en.wikipedia.org/wiki/IC50

@mhchem I believe they are referring to a paragraph in section 5.1 of the SI Brochure: "It is not permissible to use abbreviations for unit symbols or unit names, such as sec (for either s or second), sq. mm (for either mm² or square millimetre), cc (for either cm³ or cubic centimetre), or mps (for either m/s or metre per second)."

In my chem engineering education, I've always seen dimensionless numbers written in upright characters: Re, Pr, Nu, etc. That might be an engineering convention, in contradiction to that of chemistry, though.

Two-letter physical quantities
In the IUPAP Red Book, two-letter physical quantities are written in sloping/slanted Roman.[1, pp 44] There is quite the extensive list[1, pp 45–46]. (ISO instead recommends usual italic, more on that below.[1, p 44] Note also that I am currently having trouble generating sloping Roman here, or at least fail to spot the difference. There is supposed to be a subtle variation! See, e.g., the answer by Gonzalo Medina to the question 'What is the difference between italics and slanted text?'.)

a. Dimensionless constants of matter
Prandtl number; nombre de Prandtl: $\nu/a\ \ \ \ Pr$
Schmidt number; nombre de Schmidt: $\nu/D\ \ \ \ Sc$
[---]
b. Momentum transport
Reynolds number; nombre de Reynolds: $vl/\nu\ \ \ \ Re$
[---]

The IUPAC Green Book often uses the IUPAP Red Book as a primary reference.[1, preface] Therefore, a similar table is available there.[2, pp 82] These quantities have in common the fact they are all dimensionless, i.e. have the dimension $1$. The sloping Roman style is used to distinguish the two-letter physical quantity from a product of two independent quantities carrying one-letter italic symbols.[1, pp 44]

The ISO recommendation is that two-letter dimensionless parameters be printed in sloping type in the same way as single-letter quantities.  When such a symbols is a factor in a product it should be separated from other symbols by a thin space, a multiplicaton sign or brackets. This disagrees with some journals that set two-letter symbols in roman type to distiguish them from ordinary products. In this report sloping roman is used to distiguish a two-letter symbol from the product of two italic single-letter symbols.

The symbol $\ce{pH}$
The official exception is, as you mentioned, $\ce{pH}$ and similar symbols.[2, p 70] Note, however, that the letter '$\ce{p}$' here cannot directly designate the operator $-\lg()$ since an operator cannot act on something that isn't a mathematical function, and $\ce{H}$ is not a function. I am aware that this is splitting hairs, but thought was deserving of mention since this has the notation tag. It is specifically pointed out in the IUPAC Green Book as well.[2, p 75]
Furthermore, there are a lot of misconceptions surrounding the origin of the letter $\ce{p}$ in $\ce{pH}$. It most likely stems from a variable designation for an electrochemical cell in an experimental setup, not 'power' or anything similar usually quoted. We will proabably never know for sure, however.
For further information, read Nørby, J. G. 'The Origin and the Meaning of the Little p in pH'. Trends in Biochemical Sciences, 2000, 25 (1), 36–37. DOI: 10.1016/S0968-0004(99)01517-0.
The specific examples given in the OP
See andelisk's answer and Loong's answer for more references. I couldn't find $\ce{IC50}$, $\ce{EC50}$ specifically but $\ce{LD50}$ and $\ce{ID50}$ are given in ACS's Style Guide.[3, p 189]

$\ce{ID50}$ dose that is infective in 50% of test subjects (also ID50)
[---]
$\ce{LD50}$ dose that is lethal to 50% of test subjects (also LD50)

[1] E. Richard Cohen, Pierre Giacomo. (1987). Symbols, Units, Nomenclature and Fundamental Constants in Physics. ('IUPAP Red Book'). International Union of Pure and Applied Physics. Commission C2 – Sunamco. (2010 reprint)
[2] E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. (2007). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. (2008 IUPAC & RSC reprint). ISBN: 978-0-85404-433-7.
[3] Anne M. Coghill, Lorrin R. Garson. (2006). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online).

Interesting that your source recommends sloping type to distinguish symbols from italic symbols. I am not sure whether all readers would recognise the difference. Usually sloping type is just considered a poor substitute if italic type is not available. And often the word “sloping” is just used to explain “italic”, although these are actually two different things.

@Loong Agreed. Not normally one to share pictures, yet in this case I feel something like this is pertinent. Do you happen to know if sloping Roman vs italic side-by-side comparison is possible in Mathjax?

You write "The sloping Roman style is used to distinguish a product two letter italic symbols.". I do not read this in the original. When they write "printed in sloping type in the same way as single-letter quantities", they do not make any distinction between slanted and italic, but they mean "in the same way", identical. Don't they?

@mhchem Sorry, not sure I fully follow your question; if I misunderstood your comment, please clarify. According to the IUPAP Red Book, ISO recommends writing two-letter quantities in italic in exactly the same way as single-letter quantities. This is in contrast to some journals which use upright style for two-letter quantities. Both of these are in contrast to IUPAP's style used in the Red Book which is sloping ~ slanted Roman (not italic Roman). See, e.g., this answer for more clarification.

I re-read the IUPAP Red Book a couple more times. It is really confused in itself. pp 2 and 67: "italic" and "slanted" are synonyms; p. 47 unclear; p. 44: "d in sloping type in the same way as single-letter quantities" = synonym, BUT three sentences later, they make a distinction: " In this report sloping roman is used to distiguish a two-letter symbol from the product of two italic single-letter symbols."

@mhchem I think it might be because, in some fonts, the two options match. IUPAP doesn't necessarily specify any one font that is correct to use, this includes Roman. They just mention that (for the font they've chosen in their document) they don't match, hence giving them an opportunity to use upright, italic and slated / sloping / oblique Roman in different context. This is only my opinion, sadly. I agree that they should have been more clear on the matter.

According to IUPAC Glossary of Terms Used in Toxicology [1, pp. 1211, 1245] these and similar notations are just ordinary upright characters with corresponding indices:

effective concentration (EC)
Concentration of a substance that causes a defined magnitude of response in a given system.
  Note: EC50 is the median concentration that causes 50 % of maximal response.
[...]
inhibitory concentration (IC)
Concentration of a substance that causes a defined inhibition of a given system.
  Note: IC50 is the median concentration that causes 50 % inhibition.

Reference

Duffus, J. H.; Nordberg, M.; Templeton, D. M. Pure and Applied Chemistry 2007, 79 (7). DOI: 10.1351/pac200779071153.

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Why these chemical reactions do not occur?
I have this list of reactions which don't occur:
$$\ce{MgCl2(aq)} + \ce{K2SO4(aq)}$$
$$\ce{Fe(NO3)2(aq)} + \ce{SnCl4(aq)}$$
$$\ce{Cu(NO3)2(aq)} + \ce{FeBr3(aq)}$$
$$\ce{KCl(aq)} + \ce{CuSO4(aq)}$$
I would like to know why and how can I recognise when a reaction occurs or it does not.
My thought: I think the products are salts, which they remain dissolved in water. In this case reactions would not happen because we are considering an aqueous solution. Am I right?

I think you have the write idea. The reactions you are looking at in this case are double replacement reactions. The starting reactants exist in solution as their component ions so unless a particular pair of ions is insoluble when they are mixed, they will just remain ions.


text:
Enthalpy vs. energy change in a gas reaction with fewer moles of gas produced

If $\Delta H$ is the change in enthalpy, $\Delta U$ the change in internal energy accompanying a gaseous reaction, then $\Delta H < \Delta U$, only if the number of moles of products is less than the number of moles of reactants.

Could you explain the above statement? Moreover, I have read that enthalpy and internal energy are extensive properties. How will they be affected because of the change in the number of moles?
I believe the changes are related according to
$$
\Delta H = \Delta U + P\Delta V.
$$

If a chemical reaction occurs in a closed system at constant temperature (i.e., the initial and final temperatures are the same), then, from the ideal gas law, $$\Delta (PV)=RT\Delta n $$So, $$\Delta H=\Delta U+RT\Delta n$$


text:
How is using Jones reagent different than Potassium dichromate in Sulfuric acid for the oxidation of alcohols?
I was taught at an institution I study in that using $\ce{K2Cr2O7}$ in the presence of $\operatorname{dil}.\ce{H2SO4}$ will oxidise $1^{\circ}$ alcohols to aldehydes, and $2^{\circ}$ alcohols to ketones. This is supported in here too.
At the same time, I came across Jones reagent ($\ce{Na2Cr2O7}$ in the presence of $\operatorname{dil}.\ce{H2SO4}$) on Khan Academy. But this says that it is an extremely strong oxidising agent and can oxidise Aldehydes to Carboxylic Acids too (thus giving acids on the oxidation of $1^{\circ}$ alcohols).
I do not understand how the above two reagents have different oxidising abilities. Is my premise itself wrong or is there any reason for it to have different oxidising abilities?

Jones Reagent usually oxidises primary alcohols to acids via the aldehyde. Only aldehydes which do not form hydrates (eg benzylic or allylic) are stable to Jones conditions and are not oxidised further. The reagents you refer to are all the same regardless of how they ate prepared.

@Waylander So the sources that say in case of Potassium dichromate with Sulfuric acid oxidises primary alcohols to aldehydes are partially correct, and those aldehydes will be oxidised to acids? Please add your comment as an answer.

Sodium or potassium dichromate or chromium trioxide in dilute aq. sulfuric acid all give rise to Jones reagent. This is used, though less than it used to be due to toxicity and cost issues, to oxidize secondary alcohols to ketones and primary alcohols to carboxylic acids via the aldehyde hydrate. In the case where the aldehyde does NOT form a hydrate with water (e.g. benzylic and allylic aldehydes), then that aldehyde will be stable to the conditions and be the major reaction product. Aliphatic aldehydes will generally give carboxylic acids.
Read more here.


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What is the reason behind the 2% increase in silver metal composition just after melting the raw silver alloy (55% silver)?
I determined the silver composition from an alloy obtained from a jeweler to be 55% and wanted to extract silver out of it. However, upon melting around 5 kgs of alloy, the silver composition went up to 57%. I noticed similar increase for another 2 times (different alloys of silver, i.e. different samples)
Could it be something related to oxidation/reduction of silver or any metal resulting in the increase in composition while in the furnace?
What are some of the analytical tests out there I could use to accurately determine the elemental composition and the compounds present in the original alloy vs the one just after melting it?

How have you been determining silver content before and after?

I had them determined by a jeweler. I guess he used acid-base titration to determine silver. Almost all of my jeweler friends get the composition determined from them.

I find it hard to tell what exactly may cause this without knowing the approximate composition of the alloy and how silver has been quantitatively analyzed. It might be degasification of some impurities, or melt fractionation in the crucible, or something entirely different.

How then would I be able to determine the alloy composition? Scanning electron microscope comes to mind. I am not sure about how would an inorganic chemist go to determine the alloy composition without completely knowing what all is inside it.

XRF, assuming the sample is homogeneous.

Does this technique also give out the composition of metal oxides?

No, it gives you the amount of elements only.

Gravimetric analysis might still work at jewelry level. I do not not how jeweler work but refinement in the crucible as suggested is a possibility. Link on silver in alloy are easily found one is http://strippolichemistry.weebly.com/uploads/9/7/8/2/9782140/analysis_of_silver_alloy.doc

I believe most jewelers in my town quantify silver using both gravimetric analysis followed by the Volhard method which involves back titration of silver nitrate.

Zinc is used in some silver alloys; Zn would be easy to lose into dross or vaporization.

I am so sorry I posted the question incorrectly. Actually, I noticed a 'decrease' of 2% and NOT the 'increase.' The composition of Ag went down to 55% from 57% Do you have any insight as to why that might happen?


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What does this three-number isotope notation mean?
What does the superscript $3+$ mean?
$$_{21}^{45}\mathrm{Sc}^{3+}$$
I understand that $45$ is the atomic mass number and $21$ is the number of protons. Is the $3+$ to indicate that it is a particular isotope, or does it have some other meaning?

The superscript '$3+$' denotes a normalised electric charge. Charges are quantised, that is they take on only certain discrete values.

Neutrons do not have a net charge ($q = 0 $).
Electrons and protons have the same absolute charge[footnote]. They have opposite, arbitrary signs. By historical convention, protons are assigned positive charges.

Hence,
$$\text{electric charge} = \text{no. of protons} - \text{no. of electrons.}$$
For shorthand purposes, we usually normalise the electric charge by dividing the total electric charge of our agreed system with the elementary charge $e$.[footnote] This provides us an opportunity to operate with integer values in chemistry; there are fractional charges for quarks which were discovered later but these particles cannot be observed separately by current understanding.
Note also that we write '$3+$' instead of '$+3$' since we are, again by convention, counting charges (that is $+ + + $), not denoting the mathematical normalised charge itself.

As of 2011, this is verified up to essentially $1$ part in $10^{21}$. The absolute charge of an electron (or charge of a proton) is called the elementary charge $e$, and
$$e=1.6021766208(98) \cdot 10^{-19}\ \mathrm{C}$$
where the parentheses designate a standard uncertainty (assuming a normal distribution, $\approx 0.68$ probability, coverage factor $k =1$).
ref: Mohr, P. J.; Newell, D. B.; Taylor, B. N. 'CODATA Recommended Values of the Fundamental Physical Constants: 2014'. Reviews of Modern Physics, 2016, 88 (3). DOI: 10.1103/RevModPhys.88.035009, freely available on arXiv here, also archived link.

Disclaimer: originally, I claimed that proton–electron charge symmetry is verified up to $1$ part in $10^{26}$. This was wrong, and based on a naive approach. Again, see How accurately is it known that protons have the same charge as electrons? for a correction.

+1 For the OP's benefit, perhaps it'd help to point out that what he's dealing with is called an "ion" ;-)


text:
Scent neutralizer
I'm an avid outdoorsman and outdoors photographer. I use a lot of scent neutralizers to help me stay hidden. Buying the store brands get expensive so I found a recipe to make your own. the recipe is 1.5 quart 3% hydrogen peroxide, 2 quarts distilled water, .5 cup baking soda, 1 ounce unscented laundry detergent.  The recipe does work  but goes bad fairly quickly, and can't use the brown bottles of hydrogen peroxide, it has additives and stabilizers that have an odor.
Is there anything that I can use as a stabilizer to it have a longer shelf life that doesn't have an odor and safe to the skin? I know the store brands have an additive to give it a longer shelf life but they don't say what they use.

Keep your 1% peroxide and the rest separated until use.


text:
Why does adipic acid give a cyclic ketone on heating while glutaric acid and succinic acid give a cyclic anhydride?
Why does adipic acid give cyclic ketone on heating while glutaric and succinic acid give cyclic anhydride?

It's different reaction, with Ba(OH)2. I guess you should find more precise source of information.

Seven-rings are notoriously unstable.

https://en.wikipedia.org/wiki/Ketonic_decarboxylation and http://www.orgsyn.org/demo.aspx?prep=CV1P0192

The calcium salt of adipic acid also works. Essentially a Dieckmann cyclization with decarboxylation. Been there. Done it. No metal salt needed with succinic and glutaric acid.

@RitwikDas Can you specify the source? Thanks!

According to Blanc Rule for Dibasic Carboxylic acids, A cyclic Ketone is formed unless a five or six membered cyclic acid anhydride can be formed. This rule is valid for 1,4; 1,5; 1,6. dicarboxylic acids

Because cyclic anhydride formed will be a 7 Member ring compound which is less stable than 5 membered cyclic ketone

Can you please cite some sources? Also, note that the rings are substituted. Btw, welcome to Chem.SE!

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text:
Substitute for alkali-iodide-azide reagent for BOD test of water samples
To measure the biological oxygen demand (BOD) of water samples, I am following a procedure that uses manganese sulphate, $\ce{H2SO4}$, and titration with standard sodium thiosulfate solution. Alkali-iodide-azide reagent is an important part of the procedure, but my school lab does not have this reagent. Is there a substitute for it?

There are two types of oxygen demand of water: easily oxidated substances and  oxidated with difficulties. Thus potassium permanganate, $\ce{KMnO4}$ (permanganate index) are used for first group or potassium bichromate $\ce{K2Cr2O7}$ (Chemical oxygen demand)  for both groups. Both $\ce{KMnO4}$ and $\ce{K2Cr2O7}$ are used in acidic environment.
Alternatives are iodates and cerium(IV) compounds.
I recommend using $\ce{KMnO4}$, but it is prohibited as precursor in some countries. also it is not primary standard, thus standardization by titration of solution of primary standard like oxalic acid needed.
$\ce{K2Cr2O7}$ is primary standard, but it is highly toxic, carcinogen. Thus be very careful with that chemical.
You can try iodates for determination but they are not very common at school. Some bleaching agents contain iodates, but I can not guarantee their usability for analysis.


text:
How to separate platinum from steel?
I'm currently extracting some platinum from an electrode. One part seems stuck in some way (possibly welded) to a piece of steel. I'm guessing it's steel because it's magnetic. I did a heat test on the piece. Results:

Left side is platinum, right side suspected steel, and the platinum is not oxidized, as expected. What would be the best/easiest way to

determine if this is actually steel/iron;
dissolve or otherwise separate it?

My resources are limited, I've got a propane torch and household items, and can of course order chemicals if needed, but no acetylene torch or anything.

Maybe $\ce{HCl}$, which will attack steel and not platinum?

Cut off as much of the "steel" as possible (hacksaw or angle grinder will work - use gloves & eye protection) then immerse the stub in 33% hydrochloric acid until the "steel" remnants are fully dissolved. Neutralise the acid with bicarbonate before disposal.


text:
Viscosity of liquid interrupts accuracy of micropipette?
Does viscosity of liquid interrupt accuracy of micropipette?
I would like to dilute 85% $\ce{H3PO4}$ acid to $\pu{1.194 mol/dm3}$ so I had to pipette 366 microliter into 4.634 mL of water. Then, I adjusted the pH to 11 by dropping 25% $\ce{NH3}$ solution. I did 4 times but final volume was not equal (about 50, 50,100, 200 mL). I think the problem is viscosity of liquid when I pipetted concentration acid.

Why error at microliter scale should give difference as high as about 50 or 100 ml? In addition for non routine multi samples work microsyringes are preferable.

In general, using a micropipette to work with highly corrosive substances is usually discouraged. Aside from this, I don't really see the problem in higher viscosity. Acid shouldn't be stuck in the tip, and, if you took the probe slowly, the proper amount of acid can be transferred with no losses. I also agree with @Alchimista that you bring tremendous error if you dilute little quantities in a large portion of water. This works for homeopathy, but not really for chemistry.

Also, pipettes tend to be designed to leave a little drop in the end because people have had this problem before.

I think you try to use glass pipette instead. It may decrease viscosity effect.

What exactly is this "viscosity effect" you are talking about which can be reduced by switching to a glass pipette?


text:
An interesting oxonium ion
On p. 429, when introducing the idea of bromonium ions, Clayden et. al.[1] mention that:

You can think of the bromonium ion as a carbocation that has been stabilised by interaction with a nearby bromine atom. You have seen a similar effect - this 'oxonium ion' was an intermediate, for example, in the $\ce {S_N1}$ substitution of MOM chloride on p. 338 of Chapter 15.

This carbocation stabilisation described in the text above is illustrated in the image below. The case described seems to be that of an adjacent carbon with a positive charge. However, I was wondering if a carbon one atom away with a positive charge could be stabilised in the same manner as well. This would be more similar to the bromonium ion. I have illustrated what I am referring to in the image below.
I would like to ask if this "interesting oxonium ion", as an oxygen analogue of the bromonium ion, does exist in reactions, such as $\ce {S_N1}$ substitutions.

Reference:

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012.

Could you please be a bit clearer? I didn't understand what you are asking.

@EashaanGodbole I have edited to clarify my question

there were NMR studies dating back at least to 1966 http://pubs.acs.org/doi/abs/10.1021/ja00965a030?journalCode=jacsat

These should form in SN-NGP reactions too

That's quite common type of oxonium cation, there are more interesting ones. Also title is rather ambiguous now.


text:
Which one of displayed formula below does not represent an isomer of heptane?
All of these appeared to be correct. I’ve already counted their hydrogen and carbon number. Please help!
[1]: https://i.stack.imgur.com/tnQvO.jpg

(D) is n-heptane only. So, it should be the answer.

If you were ever asked to count the total number of structural isomers of heptane, you will definitely count n-heptane as one of them. However, this question has explicitly drawn what "heptane" looks like (actually it's n-heptane), and then asks you to identify which of the following is NOT an isomer of "heptane". Clearly, option D is not an isomer of  "heptane", rather "heptane" itself. Hence, D.

@GaurangTandon Is it because D’s vertical chain isn’t less than D’s horizontal chain.

@user28603 It is a single unbroken carbon chain of seven atoms...

@GaurangTandon So you're saying that the structure drawn in D is the principle chain of Heptane. The structure that is above the question isn't the principle chain of Heptane. Am I right?

@user28603 Both the structures (one drawn above the question and the option D) are same (and yes, both are unbranched, n-heptane). But, option D is drawn in a way to make it look different from the other one.  See carefully.

@user28603 You should try to use ball and stick models to visualize better... https://www.amazon.com/Molecular-Model-Organic-Inorganic-Chemistry/dp/B016XJN2G0

The structure in the question and option (D) are the same thing. Note that a branching as apparently seems is not a real branch on the principle chain that would give an isomer. Basically (D) is same as that given in the question.

But a, b, c, d’s amounts of hydrocarbons are the same right?

@user28603 I didn't get it.What do you mean by amount of hydrocarbons?

Amount of Hydrogen and Carbon.

@user28603 Yes. The amount is same as structural formula is similar.

The option d should be correct since the name of the compound is the same as the given in the question...
They are homomers


text:
Origin of the names “thionyl” and “sulfuryl”
Compounds in question:

Thionyl chloride - $\ce{SOCl2}$
Sulfuroyl dichloride/Sulfuryl chloride - $\ce{SO2Cl2}$

I understand that the "thio" prefix is often used when sulphur replaces oxygen in its compounds, like "thioalcohol" or "thiosulphate". I also note that sulphur and sulfur for the same element $\ce{S}$.
However, even after countless attempts, I manage to swap the two compounds with each other, and have not been able to accurately identify their formulae when their names are given (and vice-versa). So, if someone asked me the name of $\ce{SO2Cl2}$, I would respond with thionyl chloride or sulfuryl chloride, and never be sure which one is which.
The names of other inorganic compounds are generally logical and memorable. The use of prefixes like meta, ortho, per, etc. is logical as well. However, these two sulphur compounds are very confusing. This makes me ask the question:

What is the rule/logic/reason behind the naming of these two inorganic compounds?

I hope if I know the reason I'll be better able to recall their formulae from their names and vice-versa.

I guess these compounds were named by the wish of the person who discovered or studied them, and then the names were accepted by other chemists. Take for example, aniline, phenol, anisole, benzene, ammonia etc.

Perhaps you can use IUPAC names, but I don't know if those names are known universally.

The general convention, as it seems, is that $\ce{thionyl->O=S< , sulfuryl->O2S<, and sulfonyl->O2S<}$. Sulfonyl is used in organic chemistry and sulfuryl in inorganic chemistry.

@ShoubhikRajMaiti "Take for example, aniline, phenol, anisole, benzene, ammonia etc." at least these names do not clash with each other as the above sulphur compounds' names do.

@ShoubhikRajMaiti " IUPAC names" Sulfuroyl dichloride is the IUPAC name...

@ShoubhikRajMaiti " general convention, as it seems" yes, thanks! perhaps you're right on that.

I don't understand your points.  It might be difficult to memorize them, But they are different compounds with different names under all circumstances

@Alchimista As I said in my original post, inorganic nomenclature is generally systematic. Yes, different compounds may have different names. But the prefixes themselves follow a certain rule or logic. As an example the "thio" prefix occurs when sulphur replaces oxygen in its compound. But the compounds in the above question are very closely related in their structure and their names are also almost similar. However, their does not seem to be a reason as to why one compound got the "thionyl" prefix, and the other got the "sulfuryl" prefix, besides, what do these prefixes stand for in general?

@Alchimista Ok, I know the thionyl group has a wikipedia entry and so does sulfuryl (and what groups these prefixes stand for), but that page does not do a good job of explaining why one group is called thionyl and the other is called sulfuryl? Are the two names just randomly assigned to these two groups? There must have been some structural feature/differences that made one group get the name thionyl, and the other got sulfuryl. "thio" suggests sulfur, but "sulfuryl" also suggests sulfur. It seemed very hard for me to relate their names to their structure, hence I asked this question.

This is right what I do not get. Thionyl and sulfuryl are different they must get different names.  Perhaps compare the alternative names sulphinyl and sulphonyl for  the two and you see the rationale behind.

@Alchimista Yes, they are different compounds and yes they have and should have different names. However, I feel their names should have some sort of underlying rule as to which group got what name. Consider the series of oxoacids pf phosphorus (list), they too have structures similar to each other, one or two of them even have peroxide linkages. They are all different compounds. Yet, their names are closely tied to their structure and oxidation states, unlike the two sulphur groups we are discussing above.

Thio- , as  azo- when N is involved, etc is replacement nomenclature . It is reserved to compound in which one original atom dictating a name of a compound or of a class of compound is formally replaced. What is curious is that the simpler IUPAC recommended names are never (perhaps rarely) used.  At least in organic chemistry. Just to be clear, this is confusing me as well it confuses you.

@Alchimista I honestly don't see how your new comment connects/agrees/disagrees with my previous comment? I mean are we both agreed on the fact that the names "sulfuryl" and "thionyl" are rather arbitrary seeming and deserve some systematic analysis?

Or compare sulforous and sulfuryl than. Obviously you cannot compare name belonging to different systems or asystematically given.

Different nomenclatures by IUPAC
For the convenience of the reader, here are the names of the compounds as mentioned in the IUPAC's 2005 Red Book[1, p 129]:

$\ce{SO2Cl2 = [SCl2O2]}$ – sulfuryl dichloride or sulfur dichloride dioxide[a], sulfuryl dichloride[b], dichloridodioxidosulfur[c];
$\ce{SOCl2 = [SCl2O]}$ – thionyl dichloride or sulfur dichloride oxide[a], sulfurous dichloride[b], dichloridooxidosulfur[c]

[a] – acceptable common name, [b] – functional replacement name, [c] – systematic (additive) name
The names which will be pertinent to this discussion are marked with italics, they should otherwise still be upright.
Also note that $\ce{H2SO4 = [SO2(OH)2]}$ is sulfuric acid[a], dihydroxidodioxidosulfur[c]. The molecular entity $\ce{H2SO3 = [SO(OH)2]}$ is sulfurous acid[a], dihydroxidooxidosulfur[c].
TL; DR: The general scheme of things is acid $\to$ radical, and the suffix used in this purpose is -yl. The difference comes from overlap as "properly" both radicals would have the same result: sulfuryl.
Etymology and history
Historically, the terms 'sulfureous' (sulphureous) and 'sulfurous' (sulphurous) date back to the 16th century. Lavosier was first to introduce the adjectives 'sulfuric' and 'sulfurous' to name acids containing sulfur in 1787[2]:

SULPHURIC, SULPHUROUS
The adjectives sulphureous and sulphurous meaning "pertaining to, or of the nature of, sulphur", date from the 16th century. The well-known oxy-acid was then known as the oil of vitriol (q.v.). In 1787, in the Méthode de Nomenclature Chimique, Lavosier Proposed the name sulphuric acid (Fr. sulfurique) for oil of vitriol and sulphurous acid (Fr. sulfureux) for the related acid containing proportionately less oxygen.[2, p 208]

The suffix -yl designates (hypothetically) as having derived from something.[2]

-YL
The suffix -yl, adapted from Fr. -yle, is derived from Gk. hylē which, in addition to meaning "wood", also means "the stuff or raw material (of any kind) from which a thing is made.[2, p 231]
[---]
(2) In the name of a radical formed by the removal of a hydroxyl group $\ce{-OH}$ from an acid.[2, p 232]
[---]
(b) Inorganic acids. A number of oxygenated inorganic radicals may be regarded as derived from acids (or from hypothetical acids) by the removal of one or more $\ce{-OH}$ groups. Such a radical is normally named by replacing the suffix -ic of the name ofthe acid by -yl. The more important of these radicals are sulphuryl $\ce{SO2\!\!:}$ (cf. sulphuric acid $\ce{SO2(OH)2}$) ... nitrosyl $\ce{NO}-$ (cf. nitrous acid $\ce{NO$.$\!OH}$)[2, p 232] [---]

As we can see, the -yl name does not care about the oxidation state of the acid. For sulfur, we see that both sulfuric + -yl and + sulfurous + -yl would give the same result: sulfuryl. One of them had to be different. For this purpose, the Greek language was used for the radical of sulfurous acid, and a Latin (possibly Sanskrit[2, p 20]) form remained for the radical of sulfuric acid.[2] This is an historic accident, it could have easily been the other way round.

THIONYL
The name given by Schiff in 1857 to the radical $\ce{SO\!\!:}$ as in thionyl chloride $\ce{SOCl2}$. It was formed from Gk. theion, sulphur, + -yl (q.v.). The radical may be regarded as the characteristic group of sulphurous acid $\ce{SO(OH2)2}$; compare sulphuryl $\ce{SO2\!\!:}$, the characteristic group of sulfuric acid.
(It is fortunate that the L. and Gk. names for sulphur (sulfur and theion) are quite different so that two different names could be formed for the two radicals.)[2, p 214]
SULPHURYL
The radical $\ce{SO2\!\!:}$, the characteristic group of sulphuric acid $\ce{SO2(OH)2}$ (see -yl), e.g. sulphuryl chloride $\ce{SO2Cl2}$ (1867)[2, p 208]

For the functional replacement name, $\ce{SOCl2}$ is named sulfurous dichloride to tie the radical more directly to its acid as well, sulfurous acid. Historically, as noted, the approach here was to use different languages of origin for the main element. This was possible because the words were very different.
If you wish to operate with common names, there is always some historic memorisation. Since this is a binary system, you could remember one and automatically derive the other. But as stated, this will not work for other similar cases: nitrosyl is for nitrous acid radical, not nitric acid.

References and bibliography
[1] Connelly, N. G.; Damhus, T.; Hartshorn, R. M.; Hutton, A. T. Nomenclature of Inorganic Chemistry; The Royal Society of Chemistry, 2005. ('IUPAC Red Book')
[2] Flood, W. E. The Origins of Chemical Names, 1st Edition.; Oldbourne, 1963.


text:
Do covalent salts produce sodium carbonate extract?
In case of qualitative analysis of anions, sodium carbonate extract is prepared for insoluble salts.
It is done by heating (for 10 minutes) an aqueous solution of a mixture of the solid salt and sodium carbonate in 1:4 ratio (by mass). As far as I understand, the principle behind preparing sodium carbonate extract is to prepare a clear solution containing the anion by precipitating the cation as insoluble oxide or carbonate.
But is this process applicable for salts that do not ionize at all in water, or covalent salts like $\ce{SnCl2}$, $\ce{SbCl3}$ etc.? I have never tried this myself.

You mean precipitate in reaction with sodium carbonate? "Extract" isn't correct, so edit it.

Do you mean by "extract" that you mix the finely ground solid analyte with six times or so the amount of sodium carbonate and then melt the whole thing over a bunsen burner? (That's called "Aufschluß" in German, i don't know an English term. Transfers insoluble sulfates into carbonates.)

@Karl , No, the solid analyte is mixed with 4 times the amount of sodium carbonate and dissolved in water. Then the solution is boiled for 10 minutes.

@Mithoron , See http://jscienceclass.blogspot.in/2011/05/preparation-of-sodium-carbonate-extract.html .

@Karl , All chemistry laboratory manuals in my state recommends this process. Otherwise I would not have written this.

Sorry, I remember now, after getting out old lecture notes. ;-) That's indeed the standard method ("Sodaauszug"=soda extract) for qualitative analysis of anions. https://de.wikipedia.org/wiki/Sodaauszug

Those chlorides are well soluble even in water, so that should work.

For BaSO4 and SrSO4, it doesn't work at all, their solubility is too low. You have to do what I proposed above, do it with soda in the melt, not in solution.

Soda extract is done to remove multivalent cations from your sample solution, by precipitating them as (mixed) carbonates and hydroxides.
Antimony or tin chlorides are well soluble even in water, and even if the solids have a strong covalent character, they are still ionic in solution.
The sulfates of barium, strontium and calcium are however very badly soluble, so it doesn't work for those. You can however do it in the melt, instead of solution. After cooling and powdering that melt, you add water. What dissolves are sodium sulfate and soda. Add $\ce{BaCl2}$ to detect the sulfate. The residue are the alkaline earth metal carbonates, which dissolve in $\ce{HCl}$.
(Please check your textbook/lab manual for the actual procedure. I have Jander, Blasius, Einführung in das anorganisch-chemische Praktikum. Perhaps someone else can recommend a good English book.)

By melt do you mean molten sodium carbonate?

Yes. In a sodium carbonate melt, the sulfates dissolve.

But I am confused by that. The wikipedia page shows that sodium carbonate does not melt but decomposes at about 800 degree celsius. Is it possible to heat the mixture to that high a temperature? Because glass should start to melt at that temperature.

The melting point of mixtures is depressed, works perfectly in a borosilicate test tube on a bunsen burner.

Ok, I understand now.


text:
What is the reaction involved in a childrens paper Christmas tree crystal growth?
For Christmas I bought my children a "Magic Growing Christmas Tree Crystal Paper Tree Christmas Decoration Science Toy". It looks like this:

It came with a bag of liquid which you add to the tray at the bottom and then 'evaporates' up the tree.
Presumably the liquid has an ionic solution which then crystalises as the solution rises up the tree and the liquid evaporates.
From my limited knowledge of playing with growing crystals as a boy - the two ionic solutions that came to mind were:

table salt - sodium chloride - white crystals
sodium sulphate - blue crystals

The shape of the crystals was fine fingers - not like either of those solutions - which tend towards large crystals. In addition the colour came from the tint on the paper - so it must have been white originally.
My question is: What is the reaction involved in a childrens paper Christmas tree crystal growth?

The crystals might look different as they are growing from the porous paper. In diy recipes table salt is used. Dyes are likely those for food.

Thanks @Alchimista - can you expand that into an answer?

What does that tree look like before the experiment? Picture?

@Karl image with before and after, timelapsed video, DYI video

Tnx. No chemistry, just dye from the paper tree and a salt solution from which the water evaporates.

@Karl - quibble - there are more ingredients involved: https://chemistry.stackexchange.com/questions/89333/what-is-the-modern-hardware-supermarket-term-for-laundry-bluing


text:
Use of repeated numerical prefixes for substituents on methane
Take trichloromethane for example. Is its preferred IUPAC name:

1,1,1-trichloromethane

or simply

trichloromethane

And which is that rule in IUPAC that governs this naming? Thank you!

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013,

P-14.3.4.2 The locant '1' is omitted in substituted mononuclear parent hydrides.

For the compound $\ce{CHCl3}$ (chloroform), the parent hydride is methane, which is mononuclear. Hence the locant '1' should be omitted in the preferred IUPAC name (PIN); it is simply trichloromethane.


text:
How to acurately measure pH in heterogeneous solution?
Following my experiment, I adjust pH of $\ce{Ca3PO4}$ from acid to base (about 11)with ammonia solution by using pH meter for precipitating $\ce{Ca3PO4}$. I waited and found pH wasn't constant. In addition, I tried to wash a pH probe and be placed in the solution again but pH was changed from last value even if I didn't add more acid or base.

Please explain the procedure better. Of course the pH is not constant while you are still precipitating. What do you mean by "you washed a probe and placed in solution"? I assume by "a probe" you mean "sample"? Or do you mean the pH probe?

Please don't just accept the first answer that turns up.

@Karl It’s a pH probe. I think solid which is on a pH probe affect measured pH so I cleaned a pH probe.

The article ("the") makes a huge difference to the meaning in English. It's NOT just decoration.

Why do you call it a "heterogeneous" solution?

The $\mathrm{pK}_{a3}$ of phosphoric acid is 12.319 according to Wikipedia. So at a pH of 11 most of the phosphate species would be in the form of $\ce{HPO4^{2-}}$. Thus as $\ce{Ca3(PO4)2}$ precipitates the solution will become more acidic. So an unanswered question here is how fast would equilibrium be established in the solution. Normally once precipitation starts, then equilibrium is established fairly quickly.
Thus it seems likely that equilibrium would depend a lot on how well the solution was stirred, thus swirling the ppt in the solution rather than having it sit at the bottom...
Also the solution is basic enough that $\ce{CO2}$ from the atmosphere will be forming carbonate species which would also slowly make the solution more acidic. (not sure how much drift in pH that you are observing...).

Makes a huge difference for CO2 how closely you a observing the experiment, closely as in cm of distance from your nose to the test tube. ;-)

How to measure pH ? Use NMR spectroscopy, and in particular 31 P NMR, after making a calibration line using solutions of phosphoric acid. One possibility for having a reference is to add in the NMR tube a capillary tube containing D2O for locking as well as medronic acid.

Possible, if you have an NMR spectrometer at your disposal. :-/


text:
Is toulene be a part of the systematic IUPAC name?
Consider the following compound:

The name engraved below it is what ACD ChemSketch generated for me. However, my book mentions that "3-chlorotoluene" is the iupac name for this compound.
This is another compound:

My book mentions "2,4,6-trichlorotoluene" is the IUPAC name, while the above is the name generated by ChemSketch.
Now, I am confused as to which name to agree with.

Which name is correct?
  Which IUPAC rule says that toluene (and other related siblings like phenol) should be/should not be a part of the IUPAC name?

NOTE: I know that even common names can be considered as PINs, but in my post I am dealing with systematic IUPAC name only.

Phenol is different, though... 4-chlorophenol is OK.

@orthocresol Is it preferred IUPAC or systematic IUPAC?

Yes, 4-chlorophenol is a PIN (P-63.1.1.1), whereas 4-chlorotoluene is not (explained in duplicate question). The reason is because phenol is designated as a preferred name which is substitutable at any position, whereas toluene is designated as a preferred name which cannot be substituted.

@orthocresol Ok. And are non-toluene names, like the one generated by ACD above, always valid and are correct IUPAC names?


text:
Do we prefer the substituent or the longest chain in IUPAC nomenclature?
I've seen too many related questions already, but mine is a simple and slightly different query

Take this compound as an example:

The name written below it is generated by ChemSketch. I agree with this generated name, because according to me, "bromo" is not a functional group, but rather a substituent only. Hence, it is not necessary to include it in the parent chain. Thus, the above name has preferred the longest chain (length) over substituent inclusion.
However, my book gives the answer as 1-bromo-2-ethyl-2-methylbutane, preferring substituent inclusion over longest chain (length).
My query thus is:

In any hydrocarbon where the substituent does not lie on the longest chain, do we prefer the longer chain, or sacrifice it for the inclusion of the substituent?

The related question and answer that were already linked in this question also cover this case. If the other answer is too long and confusing, we can limit it to the rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) that are relevant in this case:

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)

P-45.0 INTRODUCTION
  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. A parent structure is defined (P-15.1) as a parent hydride, for example, benzene, a functionalized parent hydride, for example cyclohexanol, or a functional parent compound, for example, acetic acid. Application of the criteria in this section will generate the preferred IUPAC name.
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

Therefore, for a correct name, only the longest chain is relevant for the choice of a principal chain in this case. Only if there is still a choice, substituent groups that are expressed as prefixes (e.g. ‘bromo-’) are taken into account for the preferred name according to P-45.2.1 and P-45.2.2.
Thus, the correct name for the compound that is given in the question is indeed 3-(bromomethyl)-3-methylpentane and not 1-bromo-2-ethyl-2-methylbutane since the pentane chain has the greater number of skeletal atoms.

Thanks! But the "related question" also had a functional group (hydroxy) in the compound. I believed that would skew the assignment of locants to favor minimum locant for hydroxy group. So, I didn't delve much deep into it and instead posed a new question. Though you're right my question is a subcase of that question. Thanks again!


text:
Effect of increase in volume of container on concentration of chlorine at equilibrium conditions

How the concentration of the chlorine would be affected if the volume of the container is increased?
  $$\ce{4 HCl + O2 <=> 2 Cl2 + 2 H2O + $Q$}$$

My initial thought was that the total and partial pressures were decreased on increasing the volume of the container. But I did not account for the fact that the $\ce{HCl}$ did not play a role in this pressure.
So, according to my previous process, the reaction should proceed in the backward direction, concentration of $\ce{Cl2}$ should decrease. On the other hand, if we exclude the $\ce{HCl}$ factor, the reaction should proceed in forward direction increasing the concentration if $\ce{Cl2}$.
Please help me understand what will be the actual effect.

According Le Chatelier's Principle, when a change in concentration or pressure is applied to an equilibrium, the equilibrium changes to counteract the change.
In your equilibrium, the volume of the container is increased, causing a decrease in pressure. To counteract this, the side of the equilibrium with the highest number of moles is favoured. I.o.w. the equilibrium will move somewhat to the left ($\ce{4HCl + O2}$).


text:
Assigning locants when both unsaturation and substituents are present in a hydrocarbon
For the following compound:

I understand that the longest chain is of five carbon atoms. However, I am having trouble assigning the locants here.
Going bottom to top we get the locant set as: 223.
Going top to bottom we get the locant set as: 244.
Obviously, there are no functional groups in this compound so the first point of difference rule should govern the numbering of locants. Hence, according to me, bottom-to-top locant assignment is correct here.
Unfortunately, ChemSketch uses top-to-bottom assignment :(

What rule am I missing here?

I figure out the unsaturation is being given preference here, but I do not know any rule regarding it.

PS:
My IUPAC name: 2-bromo-2-methylpent-3-ene.
ChemSketch's name: 4-bromo-4-methylpent-2-ene

Double bond is prioritized over halogens. Besides, ChemSketch does generate IUPAC name by default, and "My IUPAC name" is a somewhat confusing term. It can be "My little pony", but "My IUPAC name" sounds weird to me.

@andselisk "My IUPAC name is a somewhat confusing term" yeah obviously it is! I realized it now. Though given the context I hope it can be safely be assumed to mean that "The IUPAC name I expected it to be" without meaning that "My assigned name is Gaurang Tandon; My IUPAC name is 2-bromo-2-methylpent-3-ene" xD Anyway, regarding the first part, "Double bond is prioritized over halogens" I didn't know that. So unsaturation does not follow the first point of difference rule?

It is correct that the principal characteristic group (the characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name) has a higher priority for numbering than any other substituents (expressed as prefixes). It is also correct that the compound that is given in the question does not have any principal characteristic group.
However, you are missing the priority for unsaturation (i.e. multiple bonds), which is between the priority of the principal characteristic group. The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Since the compound that is given in the question does not have any principal characteristic group, the first criterion (greater number of suffixes) given in P-14.4 (c) is not relevant in this case.
According to (e), the correct name is (2E)-4-bromo-4-methylpent-2-ene and not (3E)-2-bromo-2-methylpent-3-ene since the locant ‘2’ for the double bond is lower than ‘3’.


text:
Are there caffeine analogs or substituents, and what are they?
I have heard of designer chemists that would take well known drugs, change a hydrogen to a methyl group, and then see what the effect would be in people.
So my question is, what happens in caffeine when these kinds of substitutions are done? What happens if you take the methyl group coming out of one of the nitrogens and make it an Ethyl, or if you change a methyl into a hydrogen?

Did you bother to google this a bit? Wikipedia perhaps? https://en.wikipedia.org/wiki/Theobromine

@Karl, how was I to know that Theobromine was the analog? Wikipedia does not have a search-by-structure feature.

Really. The link in the first line in the wikipedia article about coffeine: https://en.wikipedia.org/wiki/Xanthine

Yes there actually a few analogs of  caffeine (methylxanthines).
Other than caffeine, the naturally occurring methylxanthines are  theophylline,
and theobromine:

Caffeine is a widely used CNS stimulant. Theophylline
has some medical use as a CNS stimulant, but its CNS stimulant
properties are encountered more often as sometimes
severe, and potentially life-threatening, side effects
of its use in bronchial asthma therapy. Theobromine has
very little CNS activity (probably because of poor physicochemical
properties for distribution to the CNS).

Thanks! This is precisely the info that I was looking for.


text:
How to dissolve white dextrin in water?
Dextrin is labelled as highly soluble in water. I have white dextrin (analytical from Sigma Aldrich). When I mix it in water, no matter how long I stir and how little dextrin I put, I get a suspension. The dextrin does not dissolve.
I am not a "real" chemist. If I search on Google with keyword such as dextrin water solution, nothing pops up, except that dextrin is highly soluble in water.
I must miss something obvious. I just don't know what. If someone could point me in the right direction...

Try heating the water.
Sugar is also very water soluble, but just dumping sugar into cold water is going to take forever to dissolve. You'll age before you get a clear solution. So, add heat and stir it, and it should dissolve.
PS: Some dextrins are only partially solvable. If sigma aldrich says that your particular dextrin is fully solvable, then it should be. Check the label.

Rule of thumb: the darker the color of dextrin is, the better it's soluble in water. White dextrin is practically always only partially soluble in water, whereas yellow one is often fully miscible.
If the main goal is to get a homogeneous solution, use yellow dextrin. Yellow dextrin (the "burnt" one) is obtained by heating dry starch powder till the color changes to yellowish-brown. You can pretty much do the same by putting some potato starch on a sheet of paper lying in a baking tray at $\pu{120 .. 180 ^\circ C}$ in oven for a couple of hours.

Thanks a lot will try that this week as we just need a good dextrin solution.


text:
Is matter really divided in two categories?
Page 1 of almost every Textbook start with the ordinary divison of matter in 2 categories: Substances and Mixtures.
Substances, or Pure Substances sometimes, are defined

as a form of matter that has constant chemical composition and
characteristic properties.It cannot be separated into components
by physical separation method.

This  is according to Wikipedia, at least.
Mixtures always citing wikipedia, instead are

materials made up of two or more different substances which are mixed
but are not combined chemically....
Mixtures are one product of a mechanical blending or mixing chemical
substances such as elements and compounds, without chemical bonding or
other chemical change, so that each ingredient substance retains its
own chemical properties and makeup.

Is this separation so radical?
Is there something in between? What about alloys?

What do you think, can one physically separate an alloy?

related https://chemistry.stackexchange.com/questions/2879/what-is-the-definition-of-of-compound-mixture-element-and-molecule https://chemistry.stackexchange.com/questions/2870/what-is-a-pure-substance

It can be "divided" only in people's heads.

Please explain downvotes so that I can try to edit my question., or at least understand what's wrong with it. It would be helpful

I don't exactly get what you wanna know. Such categorisations are more or less arbitrary. And your thought about alloys is maybe better then one can think, as there's kinda spectrum between intermetallic compounds and alloys.

Thank you for your explanation. I'm a beginner I just wanted some clarification or insights on the topic.  I would have loved maybe a list of all the "in-between elements" or "border-line" cases, with some explanation :)

Lists are off-topic - too broad. My thought can be summarised that it is sometimes more or less arbitrary if something is considered chemical compound - "single substance" or mixture of compounds/elements.

Matter....
Matter is generally categorized as either mixtures or pure substances.  Pure substances are elements or compounds.  Mixtures are either heterogeneous or homogeneous.  For heterogeneous mixtures one should be easily able to discern the different components, while if the mixture is homogeneous it appears uniform.  Solutions and alloys are homogeneous mixtures.
A concept map which I give to my students: https://c7chemistry.wikispaces.com/file/view/CW%20matter%20concept%20map%20new%20rev%20091211.pdf/300804074/CW%20matter%20concept%20map%20new%20rev%20091211.pdf


text:
What does “to the left” mean?

However, the concentration of hydronium and hydroxide ions will be very, very small; in fact, the equilibrium that's established in solution lies so far to the left, that only 18 in 1010 molecules of water will undergo autoionization.

What does "to the left" mean in the extract above? Does it mean the left side of the chemical equation?

Yes, "to the left" refers to the left side of an equilibrium expression.
Traditionally, the autoionization of water is written as
$$\ce{2H2O <=> H3O+ + OH-}$$
When we talk about equilibrium lying "to the left", it means that the educt/reactant is favored, i.e. more $\ce{H2O}$ than $\ce{H3O+}$ or $\ce{OH-}$.
Conversely, an equilibrium that lies "to the right" is one that favors products. In this case, it would favor the production of $\ce{H3O+}$ and $\ce{OH-}$ over $\ce{H2O}$,


text:
Why exactly does precipitation occur?
In a solution, we have ions floating around but when we have a precipitate, they're arranged as they would be in a solid. This conversion should cause a decrement in the entropy of the system and cause the reaction to be non-spontaneous, but that is not the case. Data of several $K_\mathrm{sp}$'s suggests that the reactions are very spontaneous, with $\log{K_\mathrm{eq}}$ around $30-40$ for some salts. Why does this happen?
One possible explanation I could think of was that all precipitation processes are enthalpy driven instead of entropy driven but in those cases enthalpy should either be really high and all solubilities must increase upon increase in temperature. I haven't studied these enough to know if that is the case or not, so help from someone more experienced would be appreciated.

Two words: Crystallisation enthalpy. You are basically asking why isn't everything completely soluble in water

To some extent, most if not all of the massive equilibrium constants are misleading, because they only happen in compounds that dissociate into multiple ions. A brief look at a solubility product table suggests that even the most insoluble ionic compounds have a solubility of ~0.1 nM, which is admittedly quite small, but less impressive than may be inferred from just looking at a $log\ (K_{sp}^{-1})$ of  >30.

Related

Short story
Maybe it will help to think first about a ridiculous case. Say you have a large hydrophilic solid, and a small drop of water is added on top. We would  not expect the entire solid to disappear into the liquid. Yes, there will be some adsorption and solubility here, yet most of the solid will remain undisturbed. In some sense, the solid 'does not fit' into the liquid droplet over a certain limit.
Similar things are true,  again in a certain interpretation, for more realistic situations. What does this 'fitting' mean exactly? It definitely does not mean there is not enough physical space per se. Instead as you increase the concentration of your solute, the distances between solute entities starts to decrease on average. Hence, at a certain radii the intermolecular forces become strong enough to pull the solutes together. There is often some energy barrier to this approaching, as colloid chemists will tell you, but if the end result is some energy minimum, the process will occur at some rate.
If the ratio of the amount of solvent to the amount of solute increases, that is we add more solvent, we should expect the solubility to increase. This is indeed what we observe. In summary, the amount of 'space' with respect to solubility is a function of the strength of intermolecular forces and the amount of solvent, among other things. Here you might also discover yourself that smaller particles dissolve more (not only faster!) than bigger particles.
You may compare precipitation to condensation of water in air. There is plenty of actual physical space to go around but once a certain limit is crossed (the dew point) the water will become liquid. Nitrogen and oxygen will not condense at these temperatures since their intermolecular forces are much weaker.[a]
An equally short story
Note first that for $\Delta_\pu{r} G \leq 0$ ($\Delta G$ was poor notation on my part) to be the criterion (necessary and sufficient) of spontaneity, some five conditions must be met: constant temperature, no pressure change, only expansion work is done, and the application of the thermodynamic limit. Fifthly, only one chemical reaction is assumed to take place.[b] Provided these prerequisites stand, we may look at the terms in
$$\Delta_\pu{r} H-T\Delta_\pu{r} S\le0.$$
(We should not really compare entropy directly to enthalpy because they have different dimensions.) Jumping in quickly, we might say immediately that if the solute can wriggle around more in a solution than in a solid, then certainly $\Delta_\pu{r} S>0$ for the transition $\text{solid}\to\text{solute}$. The only thing standing in our way is that damned enthalpy! Sure enough, that may be the case for some solutions but it is not the whole picture.
Since the solute entities have a solvent layer in solution, there is an entropy effect for the solvent as well. If the solvent is tightly bound to the solute, entropy decreases for this solvent entity. There is also an enthalpy contribution from solvation, not only from the act of breaking or forming bonds in a solid. I would not really dare to generalise which is the bigger contributor 'for most cases'. In general,
\begin{align}
\Delta_\pu{r} H_{\pu{solid to ideal gas}} & + \Delta_\pu{r} H_{\pu{breaking bonds in solvent}} + \Delta_\pu{r} H_{\pu{solute forming bonds with solvent}}\ + \\ &+ \Delta_\pu{r} H_{\pu{residual}} - T\cdot(\text{corresponding ‘entropy’ terms}) \le 0
\end{align}
where 'residual' signals the  solute-solute interactions in the solution.
A sentence or two on $K_{\pu{sp}}$ (with concentrations)
From the IUPAC Gold Book,

solubility
The analytical composition of a saturated solution, expressed in terms of the proportion of a designated solute in a designated solvent, is the solubility of that solute. The solubility may be expressed as a concentration, molality, mole fraction, mole ratio, etc.

As pointed out by Nicolau Saker Neto in a comment, we should not overemphasise the role of $K_{\pu{sp}}$ in determining solubility. Again there are cases where it is a good indicator, and situations where it fails poorly. There are assumptions which must be satisfied:

there actully exists an equilibrium $$\ce{AB(sln) <=> A+(sln) + B+(sln),}$$ that is not all of the solute "molecules" are dissociated, not really the case for strong electrolytes in a dilute solution,
that the simpification of $a \approx c$ is good enough ($a$ is activity, $c$ is some type of concentration),
provided that the above holds, the concentration of $\ce{AB(aq)}$ should be constant wrt additon of another electrolyte (in some calculations).

If these do not hold, $K_{\pu{sp}}$ (with concentrations alone) is a poor approximation for true solubility. See Denbigh, K. G. 'The Theory of Solubility Product'. J. Chem. Educ., 1941, 18 (3), 126. DOI: 10.1021/ed018p126.

[a] I have largely ignored the discussion of surface energies and surface tension, and how the interaction is often more favourable in the bulk than on the surface, and the fact that $r^3$ grows faster than $r^2$, except for small $r$ when it does not, and hence we need some seeds or nuclei to start the process, and supersaturation, and how particle sizes change etc. The example of smaller particles dissolving more is, of course, a surface tension effect.
[b] In the thermodynamic limit, the general condition of spontaneity comes from Clausius inequality. This is written as
$$\small{T\pu{d}S \ge \delta q.}$$
With the first law of thermodynamics, this is equivalent to
$$\small{T\pu{d}S\ge\pu{d}U-\delta w}$$
or
$$\small{-(\pu{d}U +p\pu{d}V - T\pu{d}S)\ge-\delta w(\text {other})}.$$
With constant temperature and pressure,
$$\small{\pu{d}(U +pV - TS)\le\delta w(\text {other})}.$$
Finally, if only expansion work is done,
$$\small{\Delta_\pu{r} G\le 0}$$
where $\small{\Delta_\pu{r} \equiv \partial / \partial \xi}$; $\small{\xi}$ is the extent of a reaction. The last result also assumes only one reaction takes place, otherwise it is no longer a necessary–sufficient condition for spontaneity.

The state symbol of AB should be (s), not (aq)? (aq) suggests that the solid is already dissolved into ions.

@TanYongBoon I actually did mean $\ce{(aq)}$. The '$\ce{(aq)}$' signifies 'in aqueous solution' instead of in the bulk solid. If you like, they can be intimate ion pairs (i.e., no solvent entities in between), but I don't really say whether the bond itself is mainly ionic or not. I should change it to '$\ce{sln}$' to be more general, though.

Ah, thanks for that explanation with physical space, I didn't think of it that way. And coming to the thermodynamic explanation, I can see that there are a lot of enthalpies at work here, so basically this is an enthalpy driven process, right? So basically almost all solubilities should decrease with increment in temperature?

@Shikhar_Mohan For conserving visual space, I didn't write out the entropy contribution but that's what I meant by 'corresponding entropy terms'. Solubilities need not be monotonic functions (e.g., solubility curves for salts with crystal water in water); furthermore, you will find plenty of examples of solubility decreasing with an increase in temperature (many non-polar gases in water in some temperature range). Note also that enthalpy and entropy are functions of temperature, therefore the equation needs to change to take this into account.

In short, I'd stick with my statement: I would not really dare to generalise which is the bigger contributor 'for most cases'. For some common inorganic salts (without crystal water) in aqueous solutions in low enough temperatures (but enough above freezing) solubility might indeed increase with an increase in temperature but there are still exceptions.

@Linear Christmas Any source you could point me to, to learn more about the entropy contribution in such cases? I am new to this nuanced take on these processes.

Although you ask specifically about forming precipitates your question is also about the solubility of one substance in  another. The simplest answer as to why precipitates form is because the free energy $G$ is minimised when some solid, its solute and solvent exist in equilibrium compared to the situation of only solute in a solvent or only solid and solvent. To estimate how quickly the precipitate forms one must appeal to chemical kinetics not thermodynamics.
If the solute is A and solvent B, when a solution forms the molecules will mix freely and the number of ways they they are distributed with respect to one another is vastly larger, compared to the unmixed state, and thus the entropy is greater.  However, molecules have attractive intermolecular forces and these are of the form AA, BB and AB. (We need not consider the exact form these take unless we wish to calculate their exact values.)
In an isolated system suppose that the attractive force between solute and solvent (AB) is of a larger magnitude than of A with themselves or B with themselves, then complete mixing would imply that the potential energy of the whole system will have to increase and this can only happen, as the system is isolated, by withdrawing (thermal) energy from the motion of the molecules. This means reducing the number of translational, vibrational and rotational quanta in  the molecules. This implies a decrease in the entropy as the number of ways energy can be distributed among the molecules is reduced. Thus it may be the case that the overall entropy maximum occurs when A and B are not completely mixed, i.e. some precipitate is formed or two liquid phases form if the temperature is high enough.
To discuss the same system in equilibrium at constant pressure and temperature then the Gibbs free energy is important $G=H-TS$ and clearly a decrease in $G$ is favoured by a decrease in $H$ and an increase in $S$. We are now considering the system in the presence of a heat bath which keeps the temperature constant and so prevents a decrease in thermal energy. The entropy is now mostly unchanged and the intermolecular forces now greatly influence the enthalpy and consequently the solubility of A in B.
If $E_{AA}$ is the increase in potential energy between two A molecules and $E_{BB}$ for B and $E_{AB}$ between A and B then the increase  in potential energy by transferring one molecule of A from the pure liquid A to pure B and vice versa is $\approx  2(E_{AB}-(E_{AA}+A_{BB})/2)\, )$, which means that the interaction between A and B is to be compared to the average of the energies between AA and BB. If $E_{AB}$ is a smaller negative number than the mean of $E_{AA}$ plus $E_{BB}$ then there is an increase of enthalpy on mixing, and this favours increasing $G$ and works in the opposite direction as any increase in $S$. This situation thus favours forming a precipitate, but, of course, there will always be some molecules of the  solid, (as solute) in the solvent even if the amount is nanomolar.
This is another way of saying that the solid will dissolve in the solvent until the chemical potential of the solution is equal to the chemical potential of the pure solid. If $x$ is the mole fraction of the solid (substance A the solute in solution) then for an ideal solution
$$\mu_{(s)}^*=\mu_{(l)}=\mu_{(l)}^* + RT\ln(x)$$
and rearranging gives
$$ \ln(x)=\frac{\mu^*_{(s)}- \mu^*_({l)}}{RT}=\frac{-\Delta G_m(T)}{RT}$$
where $\Delta G_m(T) $ is the free energy of melting at temperature $T$. The Gibbs free energy of melting at the melting point is zero thus although it is a bit of a mathematical 'trick' we can add zero the the equation in the form $\Delta G_m(T_m)/RT_m$ to give
$$\ln(x)= \frac{-\Delta G_m(T)}{RT}+ \frac{\Delta G_m(T_m)}{RT_m}$$
Using $\Delta G=\Delta H - T\Delta S$ produces
$$\ln(x)= \frac{\Delta H_m}{R}\left(\frac{1}{T_m}-\frac{1}{T}  \right)$$
This equation describes the ideal solubility of solids in liquids and was first derived 1893, a surprisingly long time ago. In older literature $\Delta H_m$ is called the Latent heat of melting). This equation can also be derived without the 'trick' used above using $\displaystyle d(\mu_{(s)}/T)=d(\mu_{(l)}/T)$.
For molecules such as naphthalene in benzene, carbon tetrachloride this equation provides a fairly good description of the solubility vs temperature. A more accurate expression could be realised by considereing the temperature dependence of $\Delta H_m$. This equation is entirely analogous to that for the depression of the meltig point and a plot of temperature vs mole fraction can illustrate both depression of the melting-point (the freezing-point curve) and the solubility curve as shown in the rough sketch.

The curve M-B is the freezing-point curve, $T_m$ the normal melting point of the solvent and $x$ in this case refers to the solvent. The solubility curve BA with $x$ that of the solute gives the temperature dependence of the solubility.
More detailed analyses are to be found in the textbooks by Denbigh 'Chemical Equilibrium' and  Murrell and Jenkins 'Properties of liquids and solutions'

Solubility is the concentration at which the chemical potentials of the solute in itself and in the solution are equal.
The chemical potential of a substance is the change in Gibbs free energy upon adding a molecule to the system and is determined by the interactions between the molecule and the rest of the system. This is where we get the "like-dissolves-like" rule of thumb. The chemical potential of a substance in a system that doesn't already contain it is very low as the system can "tolerate" its presence without much entropic penalty. As the concentration increases, so does the chemical potential and at some concentration it will become equal to that of the bulk state of the pure solute. In the case of miscibility, this equality never occurs.
In the case of inducing precipitation, you are driving the chemical potential of the solute in the solution higher so the solute molecules attract each other more strongly than the solvent molecules. This requires an activation energy to overcome entropy and large barriers lead to supersaturation.


text:
Why do t-butyl bromide and silver(I) nitrite form the nitroalkane, not the nitrite ester?
We know that reaction of $\ce{AgNO2}$ with alkyl halides is SN1 type reaction and follows HSAB principle. So in the reaction of $\ce{(CH3)3C Br}$ we would expect $\ce{AgBr}$ to form and also a carbocation $\ce{(CH3)3C+}$.
According to HSAB, since the electrophile is hard, we expect the hard nucleophile to attack. So attack is expected from oxygen, and expected product is $\ce{(CH3)3C-ONO}$
But my book states that reaction product is  $\ce{(CH3)3C-NO2}$. Why is this product formed?

Even though it is widely taught, HSAB is quite inappropriate in this context. I don't have the time now to explain that thoroughly, but the key reference is https://doi.org/10.1002/anie.201007100. See also Jan's answer about cyanide alkylation. I also somewhat question the claim that the nitroalkane RNO2 is formed. According to https://doi.org/10.1021/ja01626a028 the nitrite ester RONO is obtained in ~65% yield and the nitroalkane  RNO2 cannot be observed.


text:
How is chemical and dynamic equilibrium related?
I have been told that dynamic equilibrium is attained when the reactants and products of a system has achieved the same concentration or when their rates of formation are the same. And I also know that even though nothing seems to go on in a dynamic equilibrium there's still so much activity.
Now assuming I've got that right. I want to know how is chemical equilibrium any different from what I mentioned abt dynamic equilibrium.
Also can chemical and dynamic equilibrium coexist??

Duplicate of https://chemistry.stackexchange.com/questions/88509/what-is-the-difference-between-a-chemical-and-dynamic-equilibrium-and-pls-expla ?

@Mith Well, it's a dupe as far as the title goes ;-) Then again, as pointed out in the comments there, the title and the actual question aren't matching. So I suppose this post is safe O:)

@paracetamol sigh

Background: It is generally the case that there is an energy barrier between reactants and products, sometimes called an activation energy, and this is smaller for reactants than products so that a reaction proceeds towards products.
Suppose now that there is a reaction such that $\ce{ A \to B}$ and initially only A is present so that the rate of forward reaction is large, i.e. lots of A and almost no B make the forwards reaction faster than the reverse reaction. As soon as some B is present it can react back to make A, but $\ce{ B \to A}$ will be slow compared to $\ce{ A \to B}$ because there is less B (smaller concentration) and the activation barrier is larger so the 'rate constant' is smaller.
Eventually the amount of A is reduced and B increased until both rates are the same and equilibrium is achieved. Note that the concentrations need not be the same, in fact they usually are not. Also note that molecules of A continuously convert to B; $\ce{ A \to B}$ and similarly B convert to A; $\ce{ B \to A}$ which is to say that the normal state of affairs is the 'dynamic equilibrium' and simply using the word 'equilibrium' implicitly assumes that this is the case.


text:
Carbon tetraradical
I was just reading through Reactions: The private life of atoms by Peter Atkins and I noticed that in Chapter 3, the chapter on the combustion reaction, the author writes:

As we watch we see $\ce {CH_4}$ being whittled down to naked $\ce {C}$ as its $\ce{H}$ atoms are stripped away by radical attack.

"Naked $\ce {C}$" here seems to refer to a carbon atom, with four unpaired electrons. This seems rather surprising to me as I was thinking that such a reactive and unstable species could not possibly form. Thus, I would like to ask if a "naked $\ce {C}$" can possibly exist, though it may have a really short lifetime. Additionally, I would like to ask if this "naked $\ce {C}$" is considered a tetraradical or a biradical because in the ground state of carbon, there are only two unpaired electrons.

https://chemistry.stackexchange.com/questions/16477/is-a-single-carbon-atom-stable

Nice question.  Never saw such a species even in combustion mechanisms. Surely flame chemistry might be different . It is matter of T how many H and electrons are stripped.

Atomic carbon isn't unstable, but reactive. Book you're citing may simplify things for popular audience.

It exhibits very interesting reactivity. You can check it in "Reactive Intermediate Chemistry" textbook by Robert A. Moss, Matthew S. Platz, Maitland Jones, Jr.

In its ground state, naked carbon is triplet $^3P$, with two metastable singlet states $^1D$ and $^1S$ ($^1D$ being the one that participates in most reactions) while the tetraradical is the least stable one:

It can be synthesized by several different methods. Graphite vaporization by an electric arc or laser can be employed. Photolysis of suitable precursors can also produce atomic carbon:

Finely designed diazo compounds could be good precursors for atomic carbon, e.g. the diazo compound in the image below produces on heating a carbene, which immediately collapses to form benzene and atomic carbon. The problem with this approach is that the products can react further.

Another approach is to use diazotetrazole, but the drawback here is that diazonoium chloride obtained after the first step is extremely explosive.

I would also like to comment about the reactivity of atomic carbon since it is rather interesting. This reactive species is known to insert in C-H bonds and it can also abstract carbonyl oxygens to form carbenes:

If it is reacted with water, carbohydrates form:

When reacted with alkenes, cumulenes are usually obtained:

They insert into C-Br and C-Cl bonds, but abstract fluorine from compounds containing C-F bonds.
More about this rather interesting chemistry can be read in  "Reactive Intermediate Chemistry" by Robert A. Moss, Matthew S. Platz and Maitland Jones, Jr.

So would its reactivity be that of a biradical or tetradical? From the reactions it undergoes, it seems to be more of a biradical. But could u elaborate on this?

Sorry, I am also not very knowledgeable regarding the triplet and singlet states notation. Could you elaborate on those as well? If you could just explain those a bit more, I would be more than happy to award you the 50 points bounty. Thanks.

I have added a scheme to clarify the states for you. Assigning the notations should be a separate question, but here is a tutorial for you: https://youtu.be/Rerp-G5fdjg.

The $^{5}S^{\circ}$ state, which lies just above the two singlet excited states referenced elsewhere here, could be considered a "tetraradical":
https://physics.nist.gov/cgi-bin/ASD/energy1.pl
The quintet state seems to be of little importance, however.


text:
Can a structure like this exist?
I was studying carbon compounds and this popped up in my mind:

You can try googling "[3.3.3.3]fenestrane" or "pyramidane" for some info

Wikipedia says this: Pyramidane ([3.3.3.3]fenestrane) is the smallest possible fenestrane, and has never been synthesised

fenetre is french for window. nice name for this compound!

If that matter window is  fenestram in latin. I never understood why there are continuing references to French words. Surely french chemists did a lot. But the assonance to a french word will be likely found for the same word in all neo latin languages. Just to say that the history of a named compound might or not have to do with France or french chemists.

@samjoe FYI fenestra is window in Latin.

hold your horses @alchimista! I was just sharing my (little?) knowledge. I am not even french btw :)

@samjoe nothing specific to your comment! I read often such comparison to french here and elsewhere.  I wanted to say not to assume chemical-historic motivations behind.  :)

Boron clusters of square antiprism geometry capped with carbon atoms (axial positions) is the closest thing that comes to mind (e.g. in your structure 4 vertices of the square are B atoms). But as of now, there is no known crystal structures of the composition you've drawn.


text:
Arrhenius Equation?
I'm a bit confused. So we have an Arrhenius equation, which gives the dependence of the rate constant of a chemical reaction on the absolute temperature. What will happen to the value of the rate constant when the temperature (T) is 0, when the activationn energy is 0, and when the temperature (T) is extremely high? Thank you in advance!

Leaving the maths aside we can think of what is actually happening. The high temperature case is the easiest to explain since here the reactant molecules have a huge amount of energy and so the activation energy is irrelevant and the reaction proceeds with a rate constant independent of temperature and this is the 'pre-exponential' term A.
The low temperature and low activation cases are trickier and is only really answered by noting the the Arrhenius equation is only an empirical equation and is not applicable when the activation energy is very small or the temperature very low.
When the activation energy is low the reaction proceeds with a measured rate constant determined by how quickly the reactant species can diffuse together, since when they meet reaction is effectively instantaneous and faster than diffusion. The reaction is called 'diffusion limited'. If the reactant species are held apart as may be done in some electron transfer reactions then from the Marcus theory of electron transfer reactions the rate constant has a maximum value when the activation energy is zero, of about $10^{13}\mathrm{s}^{-1}$ and this is the 'pre exponential' value.
As we approach zero K the reactants will become solidified so that different conditions apply here as species cannot diffuse together on any sensible time scale. The Arrhenius equation does not apply here.
If you study Statistical Mechanics and Quantum Theory then it is possible to derive 'Arrhenius type' equations from a sound basis.

What you are asking is simple maths:
$$
k = A \cdot \exp\left(-\frac{E_a}{RT}\right)
$$
As can be seen this equation is not defined for $T=\pu{0 K}$. However if one inspects the limit of the function then it is obvious that $k$ will converge towards $k=0$. Therefore one can say that at zero kelvin no reaction will take place.
For very high temperatures one can see that $k$ will converge towards $k=A$.
If the activation energy is zero then the rate constant will be $k=A$ for all temperatures (except $T=\pu{0 K}$).

Suggestion: use $\pu{0 K}$ $\pu{0 K}$. Apart from that, $0/0$ is undefined, so I don't see how one can "do the math".

Oh yeah true... Somehow had infinity/zero in mind. What do you mean by "use $\pu{0 K}$"? Where should I use it?

When you’re typesetting quantities such as temperature. Andselisk’s edit already did it for you, though,


text:
If we start heating a substance, does it melt when we break its intermolecular bonds?
I was reading the 2 following posts:

Do substances melt when we break their intermolecular bonds?
Is melting/boiling point of ionically bonded substance higher than of covalently bound?

Altough they both contain wonderful answers, I did not understand if melting can be caused just by the break of the intermolecular weak bonds, while we're heating the substance.

I have a sugar crystal and smash it into grains. Obviously I broke intermolecular bonds. Did the sugar melt?

@MaxW Thank you for your insight. I was assuming the substance to be heated up. I edited to make it clearer.

Also important to ask is when you break the bonds, do they reform?

This question is a bit unclear. Could you please elaborate?

I'm somewhat befuddled as to the confusion. I think it is perhaps the problem is in the phrase "intermolecular weak bonds."
If you look at a diamond crystal there aren't any diamond "molecules." Rather the whole crystal is one big "molecule."
Likewise we write the chemical formula for sodium chloride as $\ce{NaCl}$, but in solid and liquid sodium chloride there are no sodium chloride molecules. True sodium chloride molecules can only exist in the gaseous state.
If you look at $\ce{H2O}$ the hydrogen atoms are extremely labile. So the two hydrogen atoms and the one oxygen atom that were a molecule when the ice froze might not be part of the same molecule when the ice melts.
So to fit the overall criteria of "intermolecular weak bonds," we need a molecule which has stronger intramolecular bonds than intermolecular bonds. Say methane for example. So when methane freezes the same molecule that freezes in the solid will be the same molecule that is freed when the solid methane melts.
So in the case of methane adding energy in the form of of heat to solid methane will eventually reach the melting point of methane. At that point the weaker intermolecular bonds of the crystal break and the bulk solid turns into liquid methane.
So the overall notion is that the solid form exists as colder  temperatures than the liquid form and that the liquid form exists at cold temperatures that the gaseous form. So generally we take solid and heat it. When (IF!) it turns to a liquid then the temperature of the transition is the melting point. Likewise if the liquid can be turned into a vapor then the temperature of the transition is the boiling point.  Thus solid methane and sodium chloride will both "melt," but for different reasons.

Yeah It was the "intermolecular weak bonds" phrase that created confusion, I did mean just intermolecular bonds. Btw thank you very much for your answer, I did not understand if you consider Diamond to have intermolecular bonds. And I'm little bit confused, because your answer seems to suggest that not every susbstance melt by breaking "intermolecular bonds", but the answer below seems to express a different view, in some way.  I would love some clarification.

@GabrieleScarlatti - The point is that a diamond crystal doesn't have discrete molecules in its crystal structure. So with no discrete molecules there are not any intermolecular bonds which need breaking. Same thing for NaCl.

When you heat a substance to the melting point, all permanent intermolecular (or inter-ionic) bonds break. That's why you call it melting point. You cannot just randomly break bonds, except by heating to the melting point, so it's pointless to ask what happens if you break intermolecular bonds.
Of course you can break bonds at a lower temperature (by radiation, aggressive chemicals, etc), but then you destroy your sample, not melt it.

Thank you for your interesting answer, btw where did I ask "What happens if we break intermolecular bonds?".