text:
Long-lived, non-lethal radioisotope for fiction
I'm writing something where the characters are looking for someone among a large set of people. They don't know who this person is and they don't know what they're looking for, so their best bet is to look for anything unusual about this person. I already have a method by using which they'll eventually find him, but I want them to muse over possible, reasonable alternatives. One that comes to mind is looking for any radioactive tracer that might have been injected in the subject. However, for this to make sense within the context of the plot, such tracer should be long-lived enough to be detectable at least for years, and safe enough not to appreciably increase the subject's risk of death over short timespans. (If he's more likely to get cancer at age 90, tough luck, but if the tracer will kill him within five, that's not good.) Any ideas?

I'm not sure. I'm looking for an isotope that exists (if it does) in the real world. I'm afraid I would get more fanciful answers over there, which is not what I'm looking for.

Uranium 238 has a half life of 4.5 billion years. You could look here and find something suitable  https://en.wikipedia.org/wiki/List_of_radioactive_nuclides_by_half-life

@Nicola So the chemical aspect of the question is to suggest "a radioisotope that with a long enough half-life to be detectable in a person after years, from some unknown but supposedly appreciable distance and not so high to kill them by radiation exposure and in quantities similar to radioactive tracers."

The tracer also needs to be something that is absorbed by the body and not excreted

@user79161 something like that. Mostly, my problem is that I don't know if all safe tracers are short-lived, or if there is one that is long-lived. How the characters plan to detect it is their problem, we don't even need to know. I'll look into the list suggested by Waylander, I was looking for something like that but couldn't find it.

The problem is the dichotomy (trichotomy?) between 'long-lived', 'detectable', and 'safe'. To detect it you need enough decays to allow your instrument to respond above the noise level. To be 'safe' you want as few decays that impact biological processes as possible (minimum rem), meaning less energy for your detector. And long-lived implies few decays.

Why not to rather use some kind of RFID tags  ? Increased person radioactivity could trigger alarms at checkpoints. BTW, human body is naturally radioactive due content of $\ce{^{14}C}$ nad $\ce{^{40}}$K.

We're already filled with all sorts of radioisotopes, both those that are naturally-occurring (e.g., $\ce{^{14}C}$), and those from atomic bomb tests. So, as an additional set of requirements, you would need one that is different from those, that has an energy signature that allows it be readily distinguished from those, and whose decay products travel a sufficient distance to allow detection to be be possible.

Good candidates for your research would be the fallout residues of atomic bombs having exploded in the world between $1945$ and $1964$. These bombs were made by USA and the former USSR (Russia today) during the "cold war". Each time one of these superpowers had developed a "more powerful" bomb, the other one managed to create a still more powerful one. And so on for about $20$ years. This stupid course stopped in $1964$, because of the level of the fallout radioactivity measured all over the world. In $1964$, the radioactivity from the fallout had much increased all over the world. The whole planet had been contaminated by the winds to $10$% of the limit value where all food would have to be decontaminated before being eaten. So both Russians and Americans knew about this danger. They managed to meet in Geneva and decided to stop their nuclear rivalry. All that were made for "in the interest of the peace in the world". It was magnificent.
As a result the radioactivity slowly decreased all over the world. At the end of the century it was back again to the level $1945$.
One of the isotopes present in the fallout is the strontium-$90$, who is assimilated like calcium, and milk contains much calcium. So $\ce{^{90}Sr}$ was found in the teeth of the newborns overall. Its half-life is about $28$ y. It was followed in teeth all over the world from $1964$ to today. It was observed that this isotope disappears from the body with much less than $28$ y. Apparently calcium (and strontium) are naturally and continuously renewed in the body.
Another isotope that can be interesting for you is the radium-$226$ who was deposited on the watch dials before $1964$. Like strontium, radium is assimilated by the body like calcium. Workers that were painting the dials with radium used brushes, and have been contaminated by this isotope. So they are radioactive. They emit constantly gamma rays. They can be detected with a Geiger counter. The level of the radioactivity is not dangerous. But they emit rays. Is it what you want ?

The half-life of Sr-90 is 28.79 a.

The OP is requesting something that can be used to identify a particular individual, so you would need a radioisotope not commonly found in the general population.  The fact that fallout residues from atomic bombs are ubiquitous means they would not be good candidates to meet the OP's needs.

@theorist Strontium-90 has disappeared today from the emerged surface of the Earth.  And the OP's story should happen now. So why not Sr-90 ?

My reasoning was that, since that radioisotope was considered ubiquitous in the early 1960's, and since it's only been two $\ce{^{90}Sr}$ half-lives since the above-ground nuclear test ban, the background levels hadn't sufficiently decreased to enable it to be a unique signature for the person being tracked.  I've also read that the only mechanism by which  $\ce{^{90}Sr}$ disappears from the general environment is by nuclear decay. But if that's not the case, and the background levels are instead, say, 100-fold less than they were in the 1960's, then it might work.

@theorist The background is much smaller than 100 times less than the situation in 1965. Because all the strontium-90 has been washed away by the rains all over the planet. These nuclides are now diluted in the oceans.

Do you have a reference for this?  I.e., one that compares the average soil levels of $\ce{^{90}Sr}$ found in surveys done in the early to mid-60's to those found today (and that preferably corrects for systematic differences due to sampling and measurement techniques)?

One possibility is Technetium-99m . According to wikipedia it is the most commonly used medical radioisotope in the world, so get your person involved in an appropriate medical procedure and you should be sorted.
But you have to be a little careful what you are looking for, Technetium-99m itself decays in around 6 hours by emitting a gamma ray. But the result, Technetium-99, has a half life of ~210,000 years. So given that Tc is so rare in the environment any raised levels of it on the body might be enough for the sign you are looking for.

OK, on reading the comments above and the wikipedia article in full I see "In the body, technetium quickly gets converted to the stable TcO4- ion, which is highly water-soluble and quickly excreted" so this probably doesn't tick the box. But I'll leave it in case it is of interest.

Tc-99m is applied as pertechnetate. Tc-99m and Tc-99 are removed from the body faster than by their physical half-life alone. After 7 d, only 0.000000051 % of the initial Tc-99m is left in the body, after 100 d, only 0.23 % of any initial Tc-99 is left in the body. You will probably not be able to detect elevated levels of Tc-99 in the body later  (Tc-99 is hard to detect anyway).


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Which of the following compounds possess more than one percent enol content?

Which of the following compounds possess more than one percent enol content?

I understood that stability of enols are dependent on the hydrogen bonding in the compounds. But here, what is the significance of more than one percent enol content?

I would like to express my disagreement with the assertion that the stability of enols would mainly be dependent on inter- and intra- molecular hydrogen bonding.

@user79161 I mean that h-bonding is one of the factor that could be used for determining the stability of Enols. It need not be the primary factor.

Especially because of the nice representations of the formulae (for easier reading, consider to rotate labels, though [entry i]), a more useful question were the comparison of the entries by line to justify which ones are more likely to enolize, than others -- for example in the line of a to d.  For some, $\log{}K$ may depend on the molecule's structure as well as (if there is) the solvent molecules.  Or -- guessing -- are you literally looking for a tautomer database example?

https://chemistry.stackexchange.com/questions/18904/which-is-the-more-stable-enol-form
This might help you;@Ron's answer was very extensive.
Note that, as Buttonwood pointed out, a fruitful exercise would be to compare likelihood of enolization. To compare the respective percentages would need you to carry out an experiment and interpret that data.

All the compounds that have a relation $\alpha-\beta$ between the carbonyl groups. The hydrogen between two $\ce{CO}$ is very acid because two electrowithdrawing remove electron density from the $\ce{C-O}$ bond. Therefore for this kind of compound the most stable for is the enol. In Organic Chemistry this is very useful in creating new carbon-carbon bonds in mild conditions like in aldoholic reactions, Michael reactions and so on. See the very good book of Marc Loudon, Organic Chemistry.

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Why Inorganic chemistry reaction mechanisms are hard to find?
My textbook Chemistry Part 1,as in the attached photo states that[after the KClO3 reaction]
“This reaction which apparently seems to be of the tenth order is actually a second order reaction. This shows that it takes place in several steps.” If that is so, then as stated above it, all higher order inorganic chemistry reactions must have some sort of mechanism. But it is really hard to find the mechanisms(many a times one cannot find them). Why is that so??.

I think the relative lack of molecular structure plays a role, too.

@Alchimista But even for those whose molecular structures are known the reaction mechanisms are unheard of those as well.

@Rishi well sure see answer. Which is related. We deal with billiard collisions, in a sketchy way.

https://chemistry.stackexchange.com/questions/10959/inorganic-chemistry-reaction-mechanisms https://chemistry.stackexchange.com/questions/126827/why-organic-reaction-mechanisms-so-popular

Does this answer your question? Inorganic chemistry reaction mechanisms

@Nilay Ghosh The answer linked is contracted in organometallic chemistry and coordination chemistry reactions, I mean to ask for the reactions like as the one mentioned in body. Thanks though!

Reactions can be of order $0, 1 , 2$ or $3$ , but never more. No collision may happen between more than three particules simultaneously. If an equation is written with more than $4$ substances on the lefthand side, it means that the corresponding reaction is a sum of several elementary reactions.
Mechanisms in inorganic reactions are hard to establish because they are made of a lot of successive reactions, in cascade. And each step is so quick that it cannot be studied. For example your oxidation of $\ce{Fe^{2+}}$ ion by $\ce{KClO3}$ is probably a series of at least $6$ steps which are all nearly instantaneous : To be able to study such a mechanism, the $6$ intermediate states must be observed for at least some nanoseconds. If the intermediate states are too short living, all we can do then is to make hypothesis without experimental arguments.

Occasionally fractional, mixed, or negative; for once in a while I prefer the nit-pick global order of reaction vs. order with respect to a substance, too.

One might add that 3rd order reactions are extremely rare.

@Karl Not as rare, as one could think. Especially when hydrogen bonds or other strong intermolecular interactions are involved.

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$P 4_{2} 22$ tetragonal group choice of origin for symmetry of special projections
I was looking at the description of the tetragonal group $$P 4_{2} 2 2$$ (No 93 in the International Tables of Crystallography) and there is one aspect that I do not understand. Namely, what is the convention and implications of choosing the origin point? For example, for the two fold rotation along the [110] plane, the origin is choosen to be $$(0, 0, \frac{1}{4})$$ and thus we should have the following transformation:
$$x \rightarrow y, y \rightarrow x, z \rightarrow -z + \frac{1}{2}.$$
Why the origin is not set at $(0, 0, 0)$ and then one would have:
$$x \rightarrow y, y \rightarrow x, z \rightarrow -z?$$

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@Poutnik Thanks! I will have that in mind next time.

You can choose the origin once only, not separately for all the different symmetry elements. So in order to keep one operation simple (like having an axis going through the origin), some other operation will be a bit less simple (like having an axis that does not go through the origin).

@KarstenTheis I think Piotr referred to the origin on the collection of planes of the special projections [110], not the space group origin. That planar origin depends on which exact plane is being considered, hence why it is (x,x,1/4). A different planar origin for these planes may be chosen just as well as p2mm provides multiple options, although none of these options includes (0,0,0).

The space group $P4_222$ has some inherent symmetry on special projections beyond its eight symmetry equivalents. The International Tables for Crystallography, Volume A (DOI: 10.1107/97809553602060000114) notes that planes normal to each of the three projections for tetragonal space groups, namely $[100]$, $[001]$ and $[110]$, having their own symmetry in fact. Here's the relevant section for $P4_222$ from the International Tables:

It looks like you are referring to the planes associated with the projection $[110]$. The planes normal to [110] have symmetry $p2mm$ of their own as well as their own origin at $x,x,\frac{1}{4}$, although only one of these planes has its origin at $0,0,\frac{1}{4}$, and of course none of these planes has its origin at $0,0,0$. Here is how the collection of origins at $x,x,\frac{1}{4}$ looks like:

Lets take an example of one such plane, say $(110)$ normal to the projection $[110]$. Defining that plane is $x+y=1$. Here is how this plane looks like:

Applying the relationship $y=1-x$ to the $P4_222$ symmetry equivalents, the planar symmetry $p2mm$ reveals itself in the following:
$$(x,\bar x,\tfrac{1}{4}), (\bar x,x,\tfrac{1}{4}), (x,x,\tfrac{3}{4}), (\bar x,\bar x,\tfrac{3}{4}),$$ $$(\bar x,\bar x,\tfrac{3}{4}), (x,x,\tfrac{3}{4}), (\bar x,x,\tfrac{1}{4}), (x,\bar x,\tfrac{1}{4})$$
For this one plane $(110)$ the origin at $(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{4})$.
The analysis may be generalized for the rest of the infinite collection of planes normal to the projection $[110]$, each having its origin at $x,x,\tfrac{1}{4}$ as noted above.

As for the origin for the space group $P4_222$ it is at $0,0,0$.

That was the explanation I was looking for. Thanks a lot!

@z1273 I don't understand why you define (x,x,1/4) a collection of origins. This coordinate triplet simply indicates the location and orientation of the symmetry element which corresponds to the symmetry operation. Moreover, I dont' understand how the plane (110) can have the origin at (1/2,1/2,1/4). In your figure the Miller indexing clearly refers to an origin at the intersection of the x,y,z axes, that is at (0,0,0).

@gryphys I added the relevant section of the ITC-A about the Special Projections for P 4sub2 2 2. Each of the planes normal to the projection [110] has the Rectangular p2mm symmetry (Plane group No.6 of the ITC, found on p.97 of 2006 edition) and has its origin of symmetry where that plane crosses the directional vector (x,x,1/4). Although by definition the plane group p2mm has its origin at (0,0), the relevant 2-fold axis in P 4sub2 2 2 crosses the z axis at ¼, not at 0, hence the plane origin at (x,x,1/4).

As stated in the International Tables of Crystallography vol A, as a rule the origin is locatedat a centre of symmetry or at a point of high site symmetry. Anyway, the choice of such a coordinate system is not mandatory since in principle a crystal structure can be referred to any coordinate system. In particular, all centrosymmetric space groups are described with an inversion centre as origin, whereas for noncentrosymmetric space groups, the origin is at a point of highest site symmetry.
In the space group n°93 the origin is at the intersection of 3 different 2-fold rotation axes (site 2a: 0,0,0), not at 0,0,1/4. I think this you're confused on this point, since the origin is always at 0,0,0 by definition.
From the trasnformation that you have assumed (y,x,-z+1/2), I guess you are referring to the operation number (7), that is a 2-fold rotation along the crystallographic [x,x,1/4] axis (not a plane! rotations are along an axis). As stated above, you can choose another coordinate system with a new origin shifted in respect the standard one, but any case the new origin will have again 0,0,0 coordinates in the new coordinate system. If you shift the origin, the coordinates of all the symmetry operations will be obviously redefined in the new coordinate system, since they are not independently distributed in space.

Maybe I was not very precise. I did not mean the general origin, but the origin mentioned in the special symmetry of projections point, where for the projection along [110] it is mentioned that the origin is at [x, x, 1/4]

OK, I was misled by the meaning that you attributed to the term 'origin'. An origin is always a point (with coordinates 0,0,0,), it cannot be an axis (such as [x,x,1/4]). The axis coordinates are defined in respect of the selected origin.

Look at that entry in the table, you will see that they define multiple origins :)

I guess that you mean with 'multiple origin' the origins specified for the symmetry of special projections. Anyway the location of the origin of the plane group is always with respect to the unit cell of the space group (whose origin is at 0,0,0) ;-)


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Does β-hydride elimination in main group metals occur?
As a side reaction of Grignard reagents, the following reaction is introduced on the Internet.
I think this is β-hydride elimination reaction, but magnesium is a main group metal, not a transition metal.
Does β-hydride elimination reaction in main group metals occur?

It rather looks like it.  If anything, magnesium as an alkaline earth metal makes the alpha-carbon more anionic than it would be with a transition metal; thus a more powerful nucleophile for displacing the hydride ion.


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Do you need concentrated sulfuric acid to make Dische diphenylamine reagent and test for DNA?
I'd like to do a DNA quantification experiment for a college project. I don't want to use 16 M sulfuric acid. Is it possible to use more dilute acid and get a reasonably good result?


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How do enzymes affect a reaction's equilibrium?
I am not sure to understand something I read in a educational journal (1)
Introduction
Lets consider the following reversible enzyme-catalysed reversible reaction;
$$E + S \leftrightharpoons ES \leftrightharpoons E + P$$
with forward rate constants $k_1$ and $k_2$ and backward rate constants $k_{-1}$ and $k_{-2}$ respectively.
The equilibrium constant of this reaction can be expressed in terms of reaction quotient at equilibrium, in terms of thermodynamic potential and also in terms of rate constants;
$$K_{eq} = (\frac{[P]}{[S]})_{eq} = e^{-\frac{\Delta_r\mu^\circ}{RT}} = \frac{k_1 \cdot k_2}{k_{-1} \cdot k_{-2}}$$
It is possible to experimentally determine the Michaelis constants of this reaction for S and P by studying it in conditions where $[S]>>[P]$ and $[S]<<[P]$ respectively. The Michaelis constants are in fact only combinations of the rate constants;
$$K_M^S = \frac{k_{-1} + k_2}{k_1}$$
$$K_M^P = \frac{k_{-1} + k_2}{k_{-2}}$$
and the maximum forward and backward rate are respectively
$$v^+_{max} = [E] \cdot k_2$$
$$v^-_{max} = [E] \cdot k_{-1}$$
where $[E]$ is the total enzyme concentration (including $[ES]$). Which leads to the Haldane relationship by rewriting $\frac{k_1 \cdot k_2}{k_{-1} \cdot k_{-2}}$;
$$K_{eq} = \frac{v^+_{max} \cdot K_M^P}{v^-_{max} \cdot K_M^S}$$
Problem
I was thinking that enzymes did not affect the position of the equilibrium of a reaction; that position would always be determined by thermodynamics, and the presence of an enzyme would simply affect the speed at which that equilibrium is reached (by lowering the activation energy). However, I stumbled upon a document (1) suggesting that both forward and backward rate expressions ($v^+ = v^+_{max} \frac{[S]}{K_M^S + [S]}$ and $v^- = v^-_{max} \frac{[P]}{K_M^P + [P]}$ respectively) apply on the whole reaction advancement axis and that the point where $v^+ = v^-$ then determines the apparent equilibrium of the reaction. The point of this document was to show that different values of enzyme parameters could correspond to the same $K_{eq}$, therefore an enzyme can set the apparent equilibrium of a reaction to a value totally different from its thermodynamic equilibrium.
I find that point confusing; if the equilibrium of a reaction changes in the presence of an enzyme, to the point that two different enzymes can change the direction of a given reaction (which is also the subject of another article of the same journal (2)), then the equilibrium constant of that chemical reaction cannot be predicted from thermodynamics (ie based on the chemical potential of its reagents). Then, can it be said that an enzyme modifies the chemical potential change of a reaction?
(1) A. Mellors, "The Haldane relationship, enzymes & equilibrium", Biochemical education 1976 vol 4 no. 4
(2) M. L. Uhr, "The influence of an enzyme on the direction of a reaction", Biochemical education 1979 vol 7 no. 1

Are you sure the equations are correct? I might be getting confused by so many variables, but it seems to me that substituting all the variables in the last equation from the two expressions above it always gives $\mathrm{K_{eq}=1}$, which obviously doesn't make any sense.

@Shoubhik R Maiti: If OP is following Michaelis-Menten kinetics, the last equation should be $\ce{ES* -> E + P}$, which is usually the RDS. It would simplify this a lot.

@Mathew Yes, the last step is generally considered irreversible in Michaelis-Menten. But there are lots of enzymes which are completely reversible, like isomerases. So, the OP's equation isn't impossible.

With respect to the equations you do need to add the $\ce{ES <=> EP}$ step in order to derive the equations as you are doing

Set it up as $\ce{E + S <=>[k_1][k_{-1}] ES <=>[k_2][k_{-2}]  EP <=>[k_3][k_{-3}] E + P}$ with Vmax substrate = $k_2$ and Vmax product = $k_{-2}$ and you should be able to derive the proper Haldane relationship which is the ratio of the respective Vmax/Km's is equal to the Keq, so you can vary Vmax and Km in tandem without affecting Keq. I'll expand on this in my answer if I have time later.

I believe you have misread the first referenced paper. What it says is that the $V_{max}$ and $K_m$ values for the forward and reverse reactions can be varied to optimize a property such as the maximum velocity (represented by $V_{max}$) in one direction, but are always constrained by the fixed value of $K_{eq}$.
The reason that the $K_m$ values (which are pseudo equilibrium constants) can be adjusted is that they represent a reaction in which the enzyme is a reactant. For the substrate, this reaction is approximately $\ce{E + S <=> ES}$.
Likewise, the second paper also describes how the $V_{max}$ for one direction or the other can be maximized, but again subject to the constraint that the respective $K_m$ values must vary such that the overall $K_{eq}$ maintains its fixed value.


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Why are most medicines in salt form?
Is there a reason for why most medicines are in salt form?
Why not have them in conjugate base form?

Salts are generally more water soluble - important for uptake if given orally.

https://chemistry.stackexchange.com/questions/126432/why-are-drugs-mixed-with-their-sodium-salts

In general, salts are more soluble in water than organic acids or organic bases, because they are made of ions. It is a general trend. There are plenty of exceptions. So usually salts are quickly dissolved in the stomach and absorbed in the intestine later on.

thank you for your answer, but how does this work? they dissociate to the ions anyway, so does the conjugate base alone not dissociate @maurice ?

An organic base is a molecule made of covalent bonds. It will not dissociate. A good example is ammonia $\ce{NH3}$ (which is not organic) and amines obtained by replacing one H by $\ce{CH3}$ or $\ce{C2H5}$ or any other organic radical- These molecules are not polar, they cannot dissociate, and are not too much soluble in water. And all they can do is to accept a proton from an acid. In that case it becomes charged and soluble into water.

thank you very much now I get it @maurice

actually I have another question if it’s a hcl salt for example it’s only the h and cl that ionise, the substance still exists in its pure form not an ion... perhaps it’s just a cost thing? @maurice

i think this only applies to sodium salts not hx salts @maurice

@Nickotine I don't understand your question. An $\ce{HCl}$ salt does not contain HCl any more. Let's consider $\ce{NH3}$ or $\ce{OH-}$ which both react with $\ce{HCl}$. A part from the ion $\ce{Cl-}$, the reaction produces $\ce{NH4+}$ or $\ce{H2O}$. which are much weaker acids than $\ce{HCl}$. So where is the problem ?

I’m saying that in the case of Med-HCl, med remains unionised so what’s the point of having it attached to HCl when Med by itself is equivalent and more potent gram for gram as its molar mass is smaller, it remains in covalent form in both cases @maurice

I’m interested in this because how can the drug work while not being an ion? it would just be too unreactive right? does it work in an equilibrium then? @maurice

@ If a drug is a salt, it is part of an equilibrium. And the state of the equilibrium changes. In the stomach, the solutions are acidic. They get neutralized in the intestine.

h and cl dissociate leaviing the covalent form of the drug, this is equivalent to taking the covalent form of the drug by itself you have still not explained this @maurice

@Nickotine. If an insoluble substance $\ce{B}$ is swallowed, it will stay in big grains or blocks on the bottom of the solution. If it is swallowed as an ion $\ce{BH+}$, it will be distributed in all the solution. If it now meets an acid solution, the insoluble $\ce{B}$ will be formed again. But it will make monomolecular particules which stay in suspension in the liquid. They may even form a colloid, which is able to cross the intestinal membrane and pass into the blood. This is the way insoluble substances like cholesterol stay in the blood.

so the reason is because it won’t dissolve? would a work around be to take it as a powder in a capsule? also even thkugh it dissolves how does a covalent molecule react? @maurice

@Nickotine. I don't understand what you mean with "Would a work around be to take it as a powder ?".

you said it will stay in big grains or blocks, so if it was a powder you wouldn't have that problem, and since in both cases the medicine is in its covalent state how does it react? @maurice

The thinnest powders are still made of aggregates containing at least billions of billions of molecules. If the same insoluble substances is made from a solution having reacted with something, the insoluble substances may be present in aggregates containing maybe one hundred or thousand molecules. This is much smaller that the grains of powder as explained above.

and how does a covalent molecule react in the body? since only h and cl ionise @maurice

@Nickotine. Once a covalent molecule is "suspended" like a colloid in the blood, it may go to any part of the body, like cholesterol. The rest is a medical or a biological question. I don't know how it reacts.

Indeed, 50% of the US’s FDA approvals consists of compounds in the salt form, (Paulekuhn et al. 2007) but the choice depends on multiple factors beyond the acid/salt choice as Gupta et al. (2018) noted in their review:

The physicochemical and biological properties of active pharmaceutical ingredients (APIs) are greatly affected by their salt forms. The choice of a particular salt formulation is based on numerous factors such as API chemistry, intended dosage form, pharmacokinetics, and pharmacodynamics. The appropriate salt can improve the overall therapeutic and pharmaceutical effects of an API. However, the incorrect salt form can have the opposite effect, and can be quite detrimental for overall drug development.

References
Paulekuhn, G.S., Dressman, J.B., and Saal, C. (2007) “Trends in active pharmaceutical ingredient salt selection based on analysis of the orange book database.” J. Med. Chem. 50:6665. DOI: 10.1021/jm701032y
Gupta, D., Bhatia, D., Dave, V., Sutariya, V. and Gupta, S.V. (2018) “Salts of Therapeutic Agents: Chemical, Physicochemical, and Biological Considerations.” Molecules 23:1719. DOI: 10.3390/molecules23071719

@z1237 please see my comment to maurice’s answer, I’m trying to understand how it works


text:
Does baking soda permanently eliminate nutritional properties of vitamin C?
(Please forgive me for asking this if it's obvious, but it's not obvious to me.)
EDIT: I am easily confused by chemistry, so please give me a very simple answer, followed by any lengthy further explanation (if you feel it's needed.) Please pretend you are speaking to a 5 year old. Thank you in advance.
Does baking soda (sodium bicarbonate?) permanently eliminate vitamin C's nutritional properties? This question seems to say baking soda would not. Am I correct?

"Denature" isn't the right word, vit. C gets oxidised, by many things, but NaHCO3 isn't one of them.

I don't know what that means in terms of this question, but I'll reword it. Thank you.

Sodium bicarbonate causes mildly alkaline pH, at which L-ascorbic acid ( vitamin C ) oxidizes in solutions by aerial oxygen  faster than alone. Just for illustration, concentrated acid solutions strongly alkalized by sodium hydroxide turns from colourless to thick yellow rather quickly.

So, the charlatans pushing lemon juice and bicarbonate are actually pushing beverages with reduced vit C content....

@Alchimista So, is that a question for Poutnik or a statement/answer for me?

It something funny that the above material makes me realise


text:
What happened to the Na+ ion?

I was in the middle of my acid and base schools questions when I came to a stop at this question. I have posted the answer to tell you what I am confused about. My teacher did not explain this very well, and my friends are not sure either. In the picture I posted, the teacher cancels out the Na from the chemical compound NaC2H2O2. I'm not sure why he did this, and then he proceeded to solve the problem via a separate equation using a hydrolysis reaction with H2O and C2H3O2-. Could someone explain to me why this happens, and would this happen with other acid and bases?

Now that I think about it, does this have something to do with weak acid and bases?

This is what happens to Na+: nothing. That's why it is not there in the equations.

Na+ does not participate in the reaction, i.e. it is a spectator ion. So, you can ignore it for the acid base part of calculation.

The $\ce{Na+}$ ion is a spectator ion in this case. It doesn't participate in the chemical reaction. It is found unchanged on both the reagent and product sides (red top-right and green bottom-right reaction formulas), so it can be canceled out to get the net reaction (red left). From there, he did the usual acid-base calculation to get a formula for $\ce{Kb}$. If you actually know $\ce{Kb}$, you can then find $x$, the $\ce{OH-}$ concentration, and from there, the $\ce{pH}$. Or, if you know the $\ce{pH}$ for this reactant concentration, then you can calculate $\ce{Kb}$.
The reason $\ce{Na+}$ doesn't participate, is as Aditya wrote: it is the conjugate acid of a strong base ($\ce{NaOH}$), so it is a weak acid (weak enough to be ignored). It can't abstract a proton from water, and has no protons of its own to donate. At most, it can combine with $\ce{OH-}$ to form $\ce{NaOH}$, but $\ce{NaOH}$ is very soluble, so it will break back apart right away. In solution, $\ce{NaOH}$ is just separate $\ce{Na+}$ and $\ce{OH-}$ ions.

Sodium ion is the weak conjugate acid of strong base i.e. NaOH. So the show which one ot of the two ions furnished undergoes hydrolysis to a greater extent the Na+(which does not react) is crossed and hydrolysis of stronger base C2H3O2- to produce OH- ions is shown.


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

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

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

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

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

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

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

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

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

References:

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


text:
How many types of 3D packing are possible?
In my book  it's said that there are 3 types of unit cubic cell - primitive , face and body centered . But for 3d packing types they have only mentioned FCC and HCP . I know that these 2 concepts- unit cell and packing types are different but it's the packing that determines the type of unit cell , right ? So are there any packing types like primitive packing and body centered packing ?  .

FCC and HCP are the two types that give 3d close packing i.e. packing atoms as densely as possible. There are many more types of unit cell, it's just that they don't achieve close packing. FCC has face-centred cubic unit cell and HCP has hexagonal unit cell.

https://chemistry.stackexchange.com/questions/35273/why-are-there-only-7-types-of-unit-cells-and-14-types-of-bravais-lattices/35277#35277 May prove useful.

Are there any packing types like primitive packing and body centred packing ?

There are uncountably infinitely many packings.
The fcc structure is the cubic close-packed structure with highest packing density possible in 3 dimensions:
$$ \rho = \frac{\pi}{\sqrt{18}} \approx 74\%$$
This was conjectured by Kepler in 1611 and became a theorem during the last years by the hand of Hales (see more details here). Apart from the fcc lattice, there is another close-packed structure with the same packing density, but different arrangement of the lattice points: the hexagonal close-packing. In both types of packings, each sphere has 12 nearest neighbors and they can be understood as different two-dimensional hexagonal close-packed layers stacking in the form $(X_{1}X_{2}X_{3}\dots)$.

In the above figure, you can see the hcp structure at the left, with a stacking sequence $(AB)_{\infty}$, which is repeated infinite times over space. At the right, we have the fcc close packing with $(ABC)_{\infty}$, where $A,B,C$ means the different ways to put the hexagonal layers on top of the other ones. Well, because we have three different ways to put the layers, there are infinitely many ways to arrange the layers as the number of them is increased. All these packings are usually known as Barlow packings and they have maximum packing density, $\sim74\%$.
Although there are infinite of them, Barlow packings are really rare in nature. If you see the crystal structures in the Periodic Table, most of the elements are either fcc or hcp, but also bcc (a non-close-packed structure). However, some elements like Sm adopt a Barlow structure known as 9R, with a stacking sequence $(ABCBCACAB)_{\infty}$.

Good answer. Another example with infinite variety: amorphous. Who said you need translational symmetry?


text:
Calculating the potential of iron–copper electrochemical cell with Nernst equation

Calculate the ${E_\mathrm{cell}}$ (not ${E^\circ})$ at $\pu{25 °C}.$
$$\ce{Cu(s) | Cu^2+ (\pu{0.10 M}) || Fe^2+ (\pu{0.0030 M}) | Fe(s)}$$
$$
\begin{align}
E^\circ(\ce{Cu^2+/Cu}) &= \pu{0.339 V} \\
E^\circ(\ce{Fe^2+/Fe}) &= \pu{-0.440 V}
\end{align}$$

I found that $\ce{Cu}$ gets oxidized and $\ce{Fe^2+}$ gets reduced and found $E^\circ_\mathrm{cell}$:
$$\ce{Cu(s) + Fe^2+(aq) -> Cu^2+(aq) + Fe(s)}$$
$$
\begin{align}
E^\circ_\mathrm{cell} &= E^\circ(\ce{Cu^2+/Cu}) + E^\circ(\ce{Fe^2+/Fe}) \\
  &= \pu{0.339 V} + (\pu{-0.440 V}) \\
  &= \pu{-0.101 V}
\end{align}
$$
However, I'm not sure if I'm supposed to add them or subtract them because I've seen both done, which is confusing. In what situations do you add, and which situations do you subtract half reaction potential values?
I then plugged this calculated value into the Nernst equation
$$E = E^\circ -\frac{RT}{zF}\ln Q$$
using $z = 2$ and plugged in the correct constants. However, I am not sure if I calculated $Q$ correctly. I did [anode]/[cathode], but is it supposed to be [products]/[reactants]? It is unclear to me because my instructor did the latter, but I keep seeing the former everywhere else because that is how you would calculate it in an equilibrium problem. I know it doesn't make a difference in this specific problem, but I want to know for future problems when concentration of all components are given (such as if they are all aqueous).
From these steps, I calculated that $E_\mathrm{cell} = \pu{-0.164 V}.$ How to approach this question?

When you are using the standard electrode potentials for the two half reactions, check what those half reactions are. For example, the $\mathrm{E_{Cu2+|Cu}}$ represents the half reaction $\ce{Cu^2+ + 2e -> Cu}$. Then you take the total reaction, and write it as a sum of two half reactions. For this you can take the reverse of a half reaction (i.e. the opposite), and then you would have to change the sign of the std. electrode potential as well. At the end, you add the half-reactions and add the electrode potentials at the same time.

@luckyschili Please don't add the $\pu{E}$ values. They should be subtracted.

If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times  \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times  \log0.10} = \pu{0.309 V}$$ $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times  \log0.0030} = \pu{- 0.515 V}$$
So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}=  \pu{0.309 V} - \pu{(-0.515 V)  =  0.824 V}$$

So, the Nernst equation can only be used for half reactions and not the whole cell? If not, I don't understand why what I did was not correct. And Q in the Nernst equation for a half reaction will always be the cation species, regardless of whether or not it is the oxidizing and reducing agent in its respective reaction?

I am resubmtting an answer that was down voted, deleted, hidden and needs 5 votes to be reinstated  but can get none because it is hidden. If the answer needs corrections state them.
The half reaction potentials are written as reductions one of them has to be reversed to an oxidation half reaction and its potential sign changed: Cu++ +2e- = Cu E[O] = +0.339.
Fe++ +2e- = Fe E[O] = -0.440 lets change the iron to oxidation
Cu++ +2e- = Cu E[O] = +0.339.
Fe = Fe++ + 2e- E[O] = +0.440 Add the half reactions the electrons are balanced [if not they must be balanced] and they subtract out. Cu++ + Fe = Cu + Fe++ E[O] = +0.779 Positive voltage means the reaction will proceed as written. Had I changed the other half reaction the voltage would be the same but negative and the reaction would proceed in the opposite direction. The standard concentrations for a solution are 1Molar activity for aqueous solutions for every ion in the equation and the activity of a pure solid is ONE not zero as was stated elsewhere. So to change conditions the complete Nernst equation is used There is a term for each reactant and product. In this case since pure solid metals are used their activities are 1 and it looks like they are omitted. They are not.[were the metals impure the mole fraction would suffice].
Increasing the concentration of the reactants over one increases the potential decreasing the products from one increases the potential. The products are the numerator in Q same as the equilibrium constant.


text:
What did I do wrong in my temperature change calculations?
I am doing an online assignment where I am being asked to find temperature change but there is something wrong in my calculations as I am getting the answer wrong. Can someone help me figure out where I went wrong in my calculations?
Assignment Instructions:
Solve for the unknown variable using the specific heat formula. Input your answer (without units) and using the correct number of sig figs.
Question:
If 31,500J of heat is used to warm 750g of water, what is the temperature change?  The specific heat of water is 4.184J/goC.
My Answer: 10
Since I am being asked to find the change in temperature in this question, I split the original equation up. First, I did (750) * (4.184) and got 3138. Then, I did (31,500) / (3138) and got 10.0382409178. The lowest number of significant figures is two, so I rounded 10.0382409178 to 10. Why is the computer counting my question as wrong?
I used the equation temperature change = (heat) / (mass) * (specific heat). The variables I used are q (heat), m (mass), and c (specific heat).

See my answer below. This is bad software, and our new visitor is a victim of it.

You're not really wrong: the computer program is apparently expecting a float and not an integer, which is really stupid because the question is about temperature change and not the vagaries of significant figures.
Instead of you learning about heat capacity and the relation to temperature change, you're instead learning about significant figures first, and the chemistry/physics part second.
I find this sort of thing patently awful and a frustrating way for you or anyone else to "learn" anything about the principles.
You have to report your answer to 2 significant figures:
750 has 2 (the limiting number); 31500 has 3; and 4.184 has 4.
$$Q = m\times c\times \Delta T$$
or
$$\Delta T = {Q\over m\times c}$$
so
$$\Delta T = {31,500\ {\rm J}\over 750\ {\rm g} \times 4.184\ {\rm J}\cdot {\rm g}^{-1}\cdot {\rm ^{\circ}C}^{-1}}$$
which, to 2 significant figures, is:
$$\Delta T = 10.\ ^{\circ}{\rm C}$$
Notice I have included a decimal point, so the now ridiculous answer that should be accepted is 10. and not 10
Meanwhile, you've lost a lot of time wondering where your calculation went wrong because some programmer doesn't know how to cast. Or maybe you're really supposed to report numbers with a terminal decimal point, and I'll challenge anyone to find any respected publication where that happens:

"The temperature change, in degrees Celsius, is 10.."

Note that 10.0 has three significant figures, so that's also "wrong."
And that, my friend, is why "learning" from a computer is a lazy way out for the "teacher."

Thank you, Todd. I understand where I went wrong now!


text:
Is this shorthand notation correct for the partial derivatives in thermodynamics?
I just have a quick question about partial derivatives/mathematical relations in thermodynamics. Is the following statement correct?
$$\frac{\partial }{\partial v}\left [ \left ( \frac{\partial p}{\partial T} \right )_{v} \right ]_{T}=\left ( \frac{\partial ^2p}{\partial v\partial T} \right )_{v,T}$$

Just realized that this is literally redundant notation. There is no reason to write T and v outside the parentheses.

Are you still seeking an answer? You can either post an answer if you have one or else delete the question if you no longer see value in it.

The independent variables are already given by T and v in
$$\frac{\partial ^2p}{\partial v\partial T}$$
so the notation $$\left ( \frac{\partial ^2p}{\partial v\partial T} \right )_{v,T}$$
is redundant.


text:
Cooling down from −78 °C to −88 °C without cryogenics
Say I have a gas cylinder with nitrous oxide and cool it down to −78 °C with dry ice. Is there an endothermic process/reaction that would extract the remaining heat from $\ce{N2O}$ and bring it down by 10 °C to its boiling point of −88 °C? I know this could be done by using cryogenics like liquid nitrogen, but I want to avoid those.

you could let out a little bit of the gas the pressure drops and it cools down

At -78C the N2O vapor pressure is only 1.8 bar - do you think it would be enough just to let that out?

a dry ice/ether cooling bath will get you below -90C

When dealing with pressurised gases, it is usually better to let the gas flow into a cold place rather than cooling the cylinder. Not least because the gas will still have enough pressure to get out of the cylinder.


text:
Which definition of van der Waals forces is correct?
I've been told by my teacher that van der Waals forces is the formal name for intermolecular forces. But on the internet, many people are saying van der Waals forces is another name for London dispersion forces. If van der Waals forces is another name for dispersion forces, why is it named after another scientist?

I think these terms are often loosely defined. For example in general chemistry, van der Waals would also include dipole-dipole, dipole induced dipole etc. interactions (which are not dispersion), but wouldn't include hydrogen bonds. However, in molecular dynamics, the van der Waals parameters also contain information about hydrogen bonds, but it does not include dipole-dipole etc. forces.

van der Waals forces is not another name for London dispersion forces. In his answer, airhuff has given IUPAC definition for van der Waals forces.
London dispersion forces are the intermolecular forces that occur between atoms and between nonpolar molecules as a result of the motion of electrons in those atoms and molecules. They are induced dipole-induced dipole interaction, better described in following diagram:

These inter-atomic or inter-molecular forces are the weakest of van der Waals intermolecular forces. London dispersion forces got its name due to its discoverer, Fritz London (Ref.1). You can find some details of the distance dependence on London forces here.
References:

Fritz London, "The general theory of molecular forces," Trans. Faraday Soc. 1937, 33, 8b-26 (https://doi.org/10.1039/TF937330008B).


text:
Gibbs Free Energy and Equilibrium
In a problem, it was stated that at 25 degrees Celsius, the K(eq)<1. The question than asked whether T(Δ(S)) was larger than, smaller than, or equal to ΔH for this reaction and the equation ΔG = ΔH - TΔS. Given that the reaction is at equilibrium, I assumed that TΔS had to equal ΔH, since ΔG would equal 0 at equilibrium. However, in the answer the instructor stated that ΔG had to be >0 as K<1. This would make ΔH > TΔS.
I don't understand why this is, as I believe that ΔG = 0 when the reaction is at equilibrium. Am I missing something?

What makes you think this system is at equilibrium?

It is good practice to keep in mind the standard state symbol $^\circ$ as explained in the answer you received. A lot of confusion follows from not differentiating between $\Delta G$ - the difference in free energy between reagents and products in general- and $\Delta G^\circ$ - the difference in free energy between reagents and products when these are in their standard states - and $\Delta G = 0$ - the difference in free energy between reagents and products when these are at equilibrium concentrations.

The free energy difference between reactants and products are given as follows when both are in their STANDARD states:
$$\Delta G^\circ = -RT \ln K_\mathrm{eq}$$
If $K_\mathrm{eq}$ is less than one that means $\Delta G^\circ$ is positive and the reaction as written will go to the left until the equilibrium conditions are satisfied. At this time $\Delta G = 0$ and $\Delta H = T\Delta S$.  You seem to understand this well and are confused by the question. The problem seems to me that the the standard state equation wasn't specified or was misread.
Now the problem:  $K \lt 1$ means $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \lt 0$ and $\Delta H^\circ \gt T\Delta S^\circ$.  I know this seems crazy because we[I] just said that $\Delta H = T\Delta S$ at equilibrium but remember the standard state values correspond to the reactants before they react to the products before they react in the reverse direction. At equilibrium you have both products and reactants present in the equilibrium concentrations NOT in the standard states. It has taken me 50 years to partially understand this so you are well ahead of the game.


text:
Balancing redox reactions
This has gotten me confused since long. Not all equations have clean half- reactions, some cations are in more than one compound while some anions are in some other compounds. Some reactions don't even have any charges and just look like normal equations (if you don't separate ions), should we just balance them like normal reactions then? That takes too much time and we don't even know which side to put $\ce{H2O}$ on.
Also when can you cancel a spectator ions? Beacuse we can't always be sure that after balancing rest of reaction the spectator ions will fit in nicely again without any disturbance to equation.
Can someone give a strategy to balance any and all redox reactions, using this one as an example-
$$\ce{KIO3 + KI + HCl ->  KCl + H2O + I2}$$
(I have seen other answers on stackexchange but none tells a method for all reactions I have doubts in. Some just aren't balanced by the methods I have read).

Do you know how to calculate the $n$ or the $z$ factor of the reaction? And the change in the oxidation states of the reactant species?

Yes I know that. But what after that?

Then there is this shortcut method, of "cross multiplication" of the change in oxidation state. For example, the change in oxidation state of $\ce{I}$ of $\ce{KIO_3}$ is +5, so write 5 in front of $\ce{KI}$ and since there is a change of 1 in oxidation state of $\ce{I}$ in $\ce{KI}$, then write 1 in front of $\ce{KIO_3}$, (which means write nothing). Now, if you see, writing 6 infront of $\ce{KCl}$ and 3 in front of $\ce{I_2}$ does the job.  This method is simply based upon writing the individual half reactions and then multiplying by a factor to cancel the electrons.

1 pack contains n sweets. 1 child requests m sweets. You need m/n * o packs for o children. Many things are trivially easy, if you can forget it is about chemistry.  // Spectator ions and molecules do just fine outside the equation. Involve only what participates.  K+ does not take part in redox reactions in water solutions. There is no KI nor KIO3 nor HCl nor KCl in solutions. There are just hydrated ions.

The usual way of solving such a problem is to write two half-equations, and make the sum of them, after multiplying each one by a factor so as to get the same number of electrons. Each equation is about one of the atoms that changes its oxidation number.
Let's start from the first such atom, iodine in the ion iodate $\ce{IO3-}$. We will build its own half-equation, step by step. It is a long process, made of four steps : balancing the atom that changes its oxidation number, balancing then the oxygen, then the hydrogen, then the charges.
First half-equation.
Here iodine goes from iodate ion $\ce{IO3-}$ whose oxidation number is $+5$ to zero in $\ce{I2}$. Write iodate on the left and iodine on the right, like $$\ce{IO3- + ... -> I2 + ...}$$ Now manage first to get the same number of atoms changing of oxidation number before going further. This gives : $$\ce{2IO3- + .. -> I2 + ...}$$ When it is done, take care of the oxygen atoms, and add enough $\ce{H2O}$ molecules to compensate the excess of oxygen atoms present on one side of the equation. Here there are $2·3$ oxygen atoms on the left. So add $\ce{6H2O}$ on the right. this makes $$\ce{2 IO3- + ... -> I2 + 6 H2O + ..}$$ Now take care of the $\ce{H}$ atoms, and add enough $\ce{H+}$ ions on the side where $\ce{H}$ atoms are missing. Here $\ce{12H}$ are missing at left. This gives : $$\ce{2IO3- + 12 H+ + ... -> I2 + 6H2O + ...}$$ Now all atoms are balanced. The last thing to do is balance the charges. There are none on the righthand side. But on the left there $12+$ and $2-$ This makes $+10$. 10 electrons are needed on the left to get the same total charges equal on both sides of the equation. This gives the final half-equation $$\ce{2IO3- + 12H+ + 10 e- -> I2 + 6 H2O}$$
Second half-equation.
Here it is quicker, because iodine ion $\ce{I-}$ yields $\ce{I2}$ without bringing any hydrogen or oxygen atom. The four steps gives the end half-equation $$\ce{2 I- -> I2 + 2 e-}$$
Adding the two half-equations.
Here the two equations must be added in such a way as to make electrons disappear. So the second half-equation must first be multiplied by $5$ to get the same number of electrons at left and at right. Adding the first half-equation plus $5$ times the second gives $$\ce{2IO3- + 12 H+ + 10 e- + 10 I- -> I2 + 6 H2O + 5 I2 + 10 e-}$$ which can be simplified and gives the final redox equation :
$$\ce{2IO3- + 12 H+  + 10 I- -> 6I2 + 6 H2O }$$
Have you followed my reasoning ?
Last step. Obtaining the non ionic equation
$12$ spectator $\ce{K+}$ cations must still be added on both sides to compensate the $\ce{IO3-}$ and $\ce{I-}$ ions at left. Then $\ce{12 Cl-}$ are to be added to compensate for the $\ce{H+}$ ions at left. This gives $$\ce{2KIO3 + 10 KI + 12 HCl -> 6I2 + 6 H2O + 12 KCl }$$
which can still be divided by $2$ yielding the final equation $$\ce{KIO3 + 5 KI + 6 HCl -> 3I2 + 3 H2O + 6 KCl }$$

(+1). Although this is how I would do it, but I loved the way you have explained it.

This is exactly what I needed and really well explained! Thanks a loads:)


text:
Why does Birch reduction of naphthalene with isoamyl alcohol form tetralin instead of 1,4-dihydrodialin (as formed with C2H5OH)?
I recently came across this question, which required me to know the difference between performing a Birch reduction on naphthalene using ethanol and using isopentanol. Why does using isopentanol lead to that extra hydrogenation?

I suspect that there is little difference between the two primary alcohols, ethanol and isopentanol (isoamyl alcohol) but rather in the conditions of the reaction. This paper might help: https://journals.sagepub.com/doi/pdf/10.3184/030823403322863201

@user55119  you should make this an answer

As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions.
The reaction conditions in the diagram are from this paper.
Naphthalene (1) reversibly adds an electron from sodium under ambient conditions to form the resonance stabilized radical anion 2 that is shown in a canonical form. This species is readily protonated by an alcohol, be it primary or tertiary as in this instance, at the site of highest electron density to afford the resonance stabilized radical 3. The addition of a second electron to 3 in this 2-electron reduction gives rise to the resonance stabilized anion 4, which, in turn, is protonated to form the deconjugated 1,4-dihydronaphthalene 5, the kinetically formed product. Just like cyclohexene is unable to reduce under these reaction conditions, neither can alkene 5.  During the course of this reduction, something else is formed -- namely -- a base, sodium tert-butoxide.
To achieve reduction of the double bond in 5 it is not sufficient to conduct the reaction at reflux but rather to conjugate the double bond with the aromatic ring. This is the role of the base at elevated temperature. Deprotonation of 5 at the benzylic site provides resonance stabilized anion 4 again, which can reversibly protonate to form 1,2-dihydronaphthalene (dialin) 6. This styrene-like double bond has a low reduction potential. Addition of an electron produces the radical anion 7 that protonates to form stabilized benzyl radical 8. Subsequent addition of a second electron affords the benzylic anion 9. This anion is readily protonated by the alcohol to form 1,2,3,4-tetrahydronaphthalene 10 (tetralin).

This was unbelievably helpful, thank you so much. So, just to confirm I got it, the difference doesn't lie in the alcohol used as the proton donor, rather the usage of t-BuOH in the reaction medium.

There is little information to go on in your chart. There is the outside chance that the difference between ethanol and isoamyl alcohol is the boiling point if the reaction was run at reflux. Without proper data, there is no need to speculate further. Glad to be helpful.


text:
Physical Chemistry: Specifically the Dissolution of Carbonates
Please note this question is different to the other that keeps being mentioned. That one is about the concentration of just  $\ce{HCO3-}$ when no $\ce {pH}$ was given.
Whereas, in this one I am trying to find the concentrations of $\ce{H+}$, $\ce{HCO3-}$, $\ce{CO3^{2-}}$, and $\ce{Ca^2+}$, with a given $\ce {pH}$.
Consider the following equations:
\begin{align}
\ce{CO2(g) + H2O(l) &<=> H2CO3(aq)} &\quad K_{\ce{CO2}} &= 10^{-1.47} \tag{1}\\
\ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)} &\quad K_1 &= 10^{-6.35} \tag{2}\\
\ce{HCO3-(aq) &<=> H+(aq) + CO3^{2-}(aq)} &\quad K_2 &= 10^{-10.33} \tag{3}\\
\ce{CaCO3(s) &<=> Ca^{2+}(aq) + CO3^{2-}(aq)} &\quad K_{\text{cal}} &= 10^{-8.48} \tag{4}\\
\end{align}
Atmospheric $\ce{CO2}$ is at $\pu{10^{-3.5} atm}$ and the solution is pure water.
Also, in equation $(4)$ calcite crystals are added in excess ($\ce{CaCO3}$).
So far I have been able to find the equilibrium equations and have got the concentration of $\ce{H2CO3}$ to be $10^{-4.97}$ by multiplying the atmospheric $\ce{CO2}$ by the value of $K_{\ce{CO2}}$.
Having been told the $\mathrm{pH}$ of the solution is 8.26, I then calculated the concentration of $\ce{H+}$ to be $10^{-8.26}$.
Using my values of $\ce{H2CO3}$ and $\ce{H+}$ and substituting them into the equilibrium equation for (2),I have the concentration of $\ce{HCO3-}$ to be $10^{-3.06}$. Is this correct?
Continuing on with the same method I have the following concentrations for $\ce{CO3^2-}$ and $\ce{Ca^2+}$ to be $10^{-5.13}$ and $10^{-3.35}$, respectively. Am I still correct here?
Thank you in advance.

Pure water is in contradiction with ocean mentioned on the title, as sea water contains 1-4 % of sodium chloride. It affects activity coefficients and dissolved calcium and magnesium ions interact with carbonate.

Thank you @Poutnik, I have changed it now.

Better than words and numbers is writing  applicable algebraic equations. And, as I have recently written somewhere, eq. (2) considers as H2CO3(aq) both H2CO3(aq) and CO2(aq).Their equilibrium constant and both kinetic rate constants are known.

@Poutnik , apologies but I do not appear to be able to see this post you mention. Do you mind providing me with the title?

This one  https://chemistry.stackexchange.com/questions/147451/how-to-calculate-molar-concentration-from-equilibrium-equations-with-no-initial

Does this answer your question? How to calculate molar concentration from equilibrium equations with no initial concentration given?

@NisargBhavsar , sorry but it does not. I have updated the question to include why it is different. Thanks,


text:
How do we find the final temperature of a system undergoing an adiabatic change?
When one mole of a monoatomic ideal gas at 'T' (in Kelvin) undergoes adiabatic change under the constant external pressure of 1 atm, changes volume from 1 liter to 2 liters. The final temperature in Kelvin would be (in terms of 'T')?
I've tried using the Adiabatic equation: PV^(gamma) = Const. for gamma = Cp/Cv. And got T(final) = T/(2^(2/3)) . But the answer I've read from some sources gives the answer as T(final) = T - 2/(3 * 0.0821) using the formula Delta(W) = nR(T final - T)/(gamma - 1), I'm not sure which formula to use, because the final temperatures are definitely not equal, but both the formulae aren't wrong either. Please help!

By the way, you may want to look into mathjax formatting. Your equations are hard to read as written.

It is not easy to read your formula. Is it right if I understood it like that ? $\pu{T_f = T - \frac{2}{3·0.0821}}$. This seems an impossible formula because of the unit problem : $0.0821$ is expressed in liter atmosphere par mol and per Kelvin. Well dividing $2$ by this does not give a result in $\pu{K^{-1}}$

The equation you tried to use is for an adiabatic reversible process, and this process is irreversible.
Here are some "leading questions" to help you solve this problem:  For beginners, you know the initial pressure, the initial volume, and the number of moles.  From the ideal gas law, what does this give you for the value of "T", the initial temperature?  For an adiabatic process (Q=0), what does the 1st law of thermodynamics tell you about the relationship between the change in internal energy $\Delta U$ and the work W?  In terms of the number of moles n, the heat capacity at constant volume $C_v$, and the temperature change $\Delta T$, what is the change in internal energy $\Delta U$ for an ideal gas?


text:
Comparing bond lengths in metal carbonyls

Question:

It's solution is given as:

Now I don't understand which d$\pi$ electrons have been considered here. For example, in $\ce{Fe(CO)_5}$, according to this, number of d$\pi$ electrons would be 4. I cannot see how it is 4. Secondly, is it the right method to compare the bond lengths?
I thought of this in terms of electron density, more is the electron density in the complex, stronger would be the metal ligand bond, hence weaker the carbon-oxygen bond, hence more the C-O bond length. But, since the atoms are different, I don't know if this idea would work in this case.

Related: https://chemistry.stackexchange.com/questions/64999/will-the-bond-order-of-co-be-greater-in-its-existence-as-carbon-monoxide-or-in-m. There is a good answer to find bond order.

@MathewMahindaratne Thanks for the link, but unfortunately I don't understand MOT in much detail, just the basics. For example, I am only familiar with $\ce{t_{2g}}$ and $\ce{e_g}$ orbitals. I have now got the part how the electron pairs are 4. It is because since CO is a strong field ligand, the configuration of $\ce{Fe}$ would be $\ce{[Ar] 3d^8}$. But still, my question remains, is this a correct method?

Each $\mathrm{d} \pi$ pair contributed to back bonding with $\ce{C#O}$ carbon so triple bond character get weaken (e.g., $\ce{Fe=C=O}$). That may be the theory behind this calculation. It is a crude method but works.

For the $Fe(CO)_{5}$ part, note that the formula is number of dπ electron $pairs$, hence $0.8$ indicates $4$ pairs of electrons or 8 electrons in total, as seen in the $3d^{8}$ configuration in $Fe$ in the presence of the strong field ligand $CO$. I have the same question in the second part however.

@Metaxylene Doesn't $Fe$ have a $3d^6$ configuration?

@LightYagami could you verify this method of finding the average CO bond length? I haven't seen it anywhere else personally. Is it from a reliable source?

@Ashish The question was asked in FIITJEE AITS, and this is the solution given by them, so I can't say it is from a verified source. Although it is a good method if we are indeed getting correct answers.


text:
Grams of solution of magnesium chloride given molarity of chloride ion
To make 100. mL of a solution that is 0.25 M in chloride ion, how many grams of magnesium chloride would you need to dissolve?
I had a few different approaches to this, but my most immediate thought was to think of each element (monoatomic or polyatomic) as a part of a larger compound. Therefore, molarity of the chloride ion is 0.25, which is 0.66 moles of the entirety of the compound; the rest being magnesium.
This approach led me to 3.52 grams, since I have 0.037 moles (molarity is 0.037 moles/liter). However, the answer is 1.2 grams.
What am I doing wrong?

If you use algebra, you spot any mistake or error much more easily. I am very curious, how you get 0.66 mol. How many moles of chloride and magnesium contains 1 mol of magnesium chloride ?

@Poutnik Yes, I am doing a crash course in chemistry, so I constantly make silly mistakes :D But that’s the fun part. Chloride is just Cl2, so magnesium chloride should contain 2/3 moles of Cl2 and 1/3 mole of magnesium, right?

@Alex No, no, and no. Chloride is not $\ce{Cl2}$. Magnesium chloride doesn't contain any $\ce{Cl2}$. And why do you assume that magnesium chloride contains 1/3 mole magnesium? Do you mean 1 mol $\ce{MgCl2}$ contains 1/3 mol $\ce{Mg}$? No, that would be wrong, too.

It does not make sense to suppose that a solution is a compound. In a solution, the proportion of one compound can change continuously. In a compound the proportion of an atom is fixed and cannot change continuously.
In your problem, the concentration of the $\ce{Cl-}$ ion is $\pu{0.25 M}$. This means that the concentration of Magnesium is half of it, so that $[\ce{MgCl2}]$ = $\pu{0.125 M}$. As  $\ce{MgCl2}$ molar weight is $\pu{95.3 g/mol}$, it means that $1$ liter of $\ce{MgCl2 0.125 M}$ contains $\pu{0.125 · 95.3 g = 11.9 g \ce{MgCl2}}$.
As a consequence, $\pu {100 mL}$ of this solution contains $\ce{1.19 g MgCl2}$

This was very helpful. I think I got confused and read chloride ion as chlorine, which would be Cl2, when we are really referring to the chlorine ION, i.e. Cl-.

The ion $\ce{Cl-}$ is called CHLORIDE, and not chloRINE. The ion CHLORINE does not exist, or it may be the unstable $\ce{Cl+}$.


text:
Find the Linear Packing factor of FCC [-1 -2 1]
I have been trying to figure out this question, and essentially if I assume that the only atoms that the vector touches are the starting and ending atoms, then I get 0.58. I am trying to find a more formal method for determining linear packing factor for any vector direction in a unit cell. Can someone explain how to do this question without the need more inspection?

What is linear packing factor, to begin with?

Linear Packing =  Length occupied by atoms / Magnitude of the direction

Like this? Page 29 + 30 of https://fac.ksu.edu.sa/sites/default/files/chapter3.pdf

Yes, but it is unclear how to figure it out when it isn't an obvious direction as they have presented.

This seems to be relevant to slip planes (https://en.wikipedia.org/wiki/Slip_(materials_science)). Maybe the higher order planes are not relevant for that (you are looking for high-density planes).

I think this would help you.

I watched that, but again, the direction is obvious. Everyone knows that the vector crosses the touching edges of FCC. How do I visualize this in the [-1 -2 1] ?


text:
Is mixture of ammonium hydroxide and ammonium nitrate a buffer solution?
Is a mixture of $\ce{NH4OH}$ and $\ce{NH4NO3}$ a buffer solution?
I have seen some examples with $\ce{NH4OH}$ and $\ce{NH4Cl}$ but never $\ce{NH4NO3}$.

There should be no difference. Both are salts of strong acids, so...

NH4OH does not exist as a molecule in water. There is NH3(aq), reacting with water $\ce{NH3(aq) + H2O <<=> NH4+(aq) + OH-(aq)}$. In extreme conditions, like in interstellar space, it may exist as a donor-acceptor pair NH3.H2O

A buffer is a mixture of a weak acid with the conjugate weak base.

Is a mixture of $\ce{NH4OH}$ and $\ce{NH4NO3}$ a buffer solution? I have seen some examples with $\ce{NH4OH}$ and $\ce{NH4Cl}$ but never $\ce{NH4NO3}$.

Both mixtures contain $\ce{NH4+}$, a weak acid. Where is the conjugate weak base? You can either write $\ce{NH4OH}$ more conventionally (or in a more modern way) as ammonia in aqueous solution, $\ce{NH3(aq)}$, which is a weak base. Or you can recognize $\ce{OH-}$ in ammonium hydroxide as a strong base, which would react with the ammonium to make ammonia. In either way of looking at things, the conjugate base is present as well.
So it is a buffer.


text:
Does glycerin dissolve in fat?
İ couldn't find much info on internet about solubility of glycerol in triglycerides. I suppose it is non soluble because glycerin is polar but fats aren't. But fats have glycerin side in them too, so can they interact?

Glycerol (glycerin) is polar molecule with three hydroxy groups. Fats are tri-esters of glycerol and very long-chained fatty acids, and hence relatively non polar molecules. Thus it is safe to assume that fats might not be dissolved in glycerol and vise versa. This assumption is based on the fact that glycerol is insoluble in biodiesel (Ref.1). An excerpt from Reference 1 states:

Glycerol is essentially insoluble in biodiesel so that almost all glycerol is easily removed by settling or centrifugation. Free glycerol may remain either as suspended droplets [...]

Biodiesel is long-chain fatty acid esters (mostly ethyl, methyl, or propyl esters). The US National Biodiesel Board defines "biodiesel" as a mono-alkyl ester. Thus, if glycerol did not dissolve in long-chain mono-alkyl ester, it is safe to assume that it won't dissolve in long-chain tri-ester such as fats.
OP has also shown interest in the solubility of glycerol in fatty acids, specifically 18-carbon fatty acids (omega 3). That can be done in home easily in smaller scale to find out the solubility. However it is worth noting of this report, which states that:

[..] have studied the purification of glycerol from fatty acids. For the binary systems glycerol-fatty acids, our experimental data show an increasing values of solubility with the number of double-carbon bonds, because of the polarity of this group.

References:

Jon Van Gerpen, Gerhard Knothe, Michael J. Haas, Alfred K. Schultz, Rajiv Banavali, Klaus-Dieter Topp, Mark T. Vandersall, "Chapter 4: Biodiesel  Production," In The Biodiesel Handbook, 2nd Edition;  Gerhard Knothe, Jürgen Krahl, Jon Van Gerpen, Eds.; AOCS Press: Urbana, IL, 2010, pp. 31-96 (ISBN: 978-1-893997-62-2).

What about solubility of glycerol in fatty acids? Specifically 18 carbon fatty acids(omega 3)? İ have a glycerol gelatin surface and i need it to stay inert while i put some fats on it

I'd say they would dissolve each other. I have included anothe paragraph in my answer to give you some help.


text:
Avoiding chemical reactions with carbonated alcoholic drinks
I'm making a cooling beverage dispenser for carbonated drinks. The keg remains room temperature and the beverage rises through a pipe that's coiled up inside a cooling unit filled with water or alcohol.
My first choice would be copper tubing because of its conductive properties, but I remember reading somewhere that it could lead to some kind of chemical reaction that would be poisonous? As I usually prefer not to kill my guests, and avoid it whenever possible, I'd like to find a good alternative that's able to cool the liquid as quickly as possible. I don't actually know what they were referring to or whether or not they were even right, I'd like some input on the uses of copper or aluminium for the piping or whether I should go with stainless steel.
I'd also like to use this for carbonated water, which might be very carbonized and I want to be able to sanitize the system by running a sanitizer like Star San through it.
Suggestions? Thanks!

Stainless steel is the safe choice as it is known to survive harsh cleaning environments (like the inside of a dishwasher). While aluminium should be avoided. Copper might work (is is common in domestic plumbing but that doesn't involve sanitiser). And the heat conductivity won't be a critical factor if you design the system right.

Domestic plumbing altso doesn't involve CO2 and alcohol. Could you elaborate on why aluminium should be avoided?

Copper and many of its compounds are relatively nontoxic when compared to lead. Copper pipes have traditionally been joined using lead-containing solders, which has resulted in poisoning https://www.alcoholproblemsandsolutions.org/moonshine-its-nature-illegality-dangers-and-prevention-important-facts/ . A pure copper cooler might be safe if beverages did not stay in contact with the plumbing for more than a few minutes.


text:
Why does dilithium have a longer bond length than dihydrogen?
I've tried MO theory however they both have a bond order of 1. Which means we can not relate the bond order to the bond length.

My thinking -
I think that the h2 bond length is shorter due to only the 1s electrons participating in the bonding. Where as in dilithum we have the 2s electrons participating only. This makes me think that due to Li having a filled 1s subshell, as the 2 atoms approach the 2s electrons are repelled by the other atoms 1s electrons. Once bonded the filled inner core of 1s electrons will repel the other atoms 2s electrons keeping the atoms at a further distance.
I may be wrong any help would be great!

Perhaps the size of the atoms involved already gives you a hint : the lithium atom is much larger than the hydrogen atom, about 2.5 times.

Also, consider electron shielding, https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/06%3A_The_Periodic_Table/6.17%3A_Electron_Shielding .

Pallas gave a hint. Even more extreme is the comparison between a proton and a lithium ion! Your mention of electron repulsion also hints at possible 2p contributions, which are energetically closer to 2s than to 1s. So the bond length could be longer without being weaker.


text:
Double bonds within cyclopropane ring
With knowledge of ring strain, placing a double bond within a cyclopropane ring would increase the "angle strain" from $109.5 ^\circ$ to $120 ^\circ$ so are cyclopropenes stable? So far, I have only come across the famous aromatic cyclopropene cation but yet to see a neutral cyclopropene ring containing molecule. However, cyclopropanes are quite stable and sometimes occur in nature, so are cyclopropenes well known too?

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

I see, thank you. Is it possible for these rings to exist in naturally occurring molecules like amino acids?

"derivatives are known in some fatty acids.".  From the referenced WP article.

Google sterculic acid.


text:
What does solid matrix mean?
I am reading "Effect of Freeze-Drying on Quality and Grinding
Process of Food Produce": A Review by Timilehin Martins Oyinloye and Won Byong Yoon.
The problem is they are referring to the words "sold matrix" several times. I searched the internet but could not understand what does that mean. Any comments please? Is this term used in some other area not in Chemistry?
Full Text

A glass of orange juice is a liquid matrix. Freeze it solid, and it is a solid matrix. Rocks, like the ones the Perseverance rover will soon be zapping on Mars, are solid matrices. Techniques like LIBS (laser induced breakdown spectroscopy) typically are used with solid matrix samples, e.g., rocks, alloys, etc.

Well, "sold matrix" can be both solid matrix or liquid matrix. Quoted typos have tendency to be perceived as intention, but in food industry context, sold matrix makes sense as well. I suggest reading Wikipedia : Matrix_(chemical_analysis)


text:
Doesn't adding an inert gas at constant volume increase the partial pressures?
When you add an inert gas to a container of constant volume full of gases, you are basically shoving the inert gas into the container and increasing the pressure inside the container, so the partial pressures of all the gases increase don't they?

Ohhhh, I thought because the total pressure increases the partial pressures would increase too. But, doesnt the argon put extra pressure on the N2 and O2?

Why should they ( supposing ideal gas behaviour ) ?
If a volume, containing $\pu{4 mol}$ $\ce{N2}$ and  $\pu{1 mol}$ $\ce{O2}$ has pressure $\pu{5 atm}$, partial pressures are:
$\ce{N2} : \pu{4 atm}$
$\ce{O2} : \pu{1 atm}$
If you add $\pu{5 mol}$ $\ce{Ar}$, the total pressure raises to $\pu{10 atm}$, with partial pressures :
$\ce{Ar} : \pu{5 atm}$
$\ce{N2} : \pu{4 atm}$
$\ce{O2} : \pu{1 atm}$
Note that pressure is macroscopic quantity. Partial pressure of any gas is determined by frequency of collisions with a wall unit area, and by passed impulse by 1 molecule.


text:
Is Km the concentration needed to keep the enzyme running at max speed?
$K_m$ is the Michaelis constant, which is the concentration of substrate needed to achieve a rate of $\frac{V_{max}}{2}$. $V_{max}$ is the maximum number of molecules that can be reacted per second.

My lecturer has said that $K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$.
I do not understand why. To me, I would have assumed that it would need to be at a concentration that would produce $V_{max}$ or something close to it.
I can understand that at a concentration that is too low, then the enzyme will make quick work of the backlog. However, I cannot see why exactly Km should be the point at which this happens. It all feels quite arbitrary to me.
Could someone explain, preferably intuitively, why this is true?

$K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$?

No. $V_{max}$ is never reached, it is the limit of the rate when you increase the substrate concentration indefinitely. Less dramatically said, when the substrate concentration is about ten times higher than $K_m$, the rate will be about 90% of $V_{max}$ (and so on, so hundred times higher - about 99%).
In practice it is often difficult to get to hundred times $K_m$ because the substrate might not be that soluble or there is substrate inhibition at non-physiologically high substrate concentrations.

Could someone explain, preferably intuitively, why this is true?

As I read in a textbook, imagine a bicycle repair shop with 5 people working in it. When the number of bikes brought in doubles from what it is usual, they will repair more bikes. It does not matter, however, if you park a million or a billion bikes in front of their shop, they will be working at capacity in either case.
If you like math, you can also answer your question by looking at the relationship between substrate concentration and rate described by the Michaelis-Menten equation set up by  Leonor Michaelis and Maud Menten:
$$\mathrm{rate} = V_\mathrm{max} \frac{[S]}{K_m + [S]} = V_\mathrm{max} \frac{1}{1 + \frac{K_m}{[S]}}$$
When $[S] \gg K_m$, the fraction approaches 1 and the rate approaches $V_\mathrm{max}$.

My lecturer has said

That is what you heard. If you think the lecturer misspoke or you misheard, it is a good idea to check a textbook or ask here.
The figure
The figure is a conceptual one with two conceptual errors. First, the rate in the figure does exceed $V_\mathrm{max}$.

Secondly, the figure shows a linear increase of the rate for substrate concentrations lower than $K_m$. In the graph showing the real shape of the curve, you can see appreciable curvature at lower substrate concentrations.

For the graph, $V_\mathrm{max}$ was set to 1 and $K_m$ to 0.2. The shape of the curve, however, is the same (if you choose the plotting range in an analogous manner) for any combination of the two parameters.
The problems with the figure are common, as the following gallery shows:

I don't follow, when $[S]<<K_m,; v_0=V_{max}[S]/K_m$

I assume you meant "When [S] >> Km, the fraction approaches 1 and the rate approaches Vmax"?

@porphyrin and Andrew: Thanks for the correction. I was looking for the Latex code for >> and <<  (\gg and \ll ) and did not proof-read.

Still need to change Km to Vmax at the end of the sentence, right?

@Andrew I should drink some coffee before saving half-finished answers... thanks again.


text:
Calculating atomic mass of unknown element in given compound (gas) at STP

A gas $\ce{X2O5}$ has the density of $\pu{5 g L-1}$ at STP. What is the atomic mass of $\ce{X}$ given $A_\mathrm{r}(\ce{O}) = 16?$

Note: The problem is my homework for school and has been translated from another language. It is probably either missing something or it contains wrong data. Just leaving it as a cautionary tale to anyone who might also have it.
I solved, or at least tried to, by expressing the given density $\rho = \pu{5 g L^-1}$ as
$$\rho = \frac{m}{V}.\tag{1}$$
Also,
$$m = Mn\tag{2}$$
$$V = V_\mathrm{m}n,\tag{3}$$
where $n$ is the amount of substance and $V_\mathrm{m} = \pu{22.4 L mol^-1}.$ So:
$$\rho = \frac{m}{V} = \frac{Mn}{V_\mathrm{m}n} = \frac{M}{V_\mathrm{m}}\tag{4}$$
$$\pu{5 g L^-1} = \frac{M}{\pu{22.4 L mol^-1}} \quad\Rightarrow\quad M = \pu{112 g mol^-1}\tag{5}$$
Then obviously by subtracting the atomic mass of oxygen:
$$M_\mathrm{r} = 5A\mathrm{r}(\ce{O}) + 2A_\mathrm{r}(\ce{X}) \quad\Rightarrow\quad A_\mathrm{r}(\ce{X}) = 16\tag{6}$$
But that is not possible, as $16$ is the atomic mass of oxygen and there is no such thing as $\ce{O2O5}$ as far as I know. What am I doing wrong? Or is the problem wrong altogether?

At STP, $V_\mathrm m$ is not $22.4\ \mathrm{l\ mol^{-1}}$; or maybe your school book is forty years old.

@Loong Then what is it? I mean that's what the textbook had and the professor told. Is there a more accurate measurement now?

@Guarav At STP $V_\mathrm{m} = \pu{22.711 L mol^-1}.$ See Wikipedia. Still, that would result in molar mass of about $\pu{16.75 g mol^-1},$ so it would be nice if you could double-check the data; could the density be relative?

@andselisk Hey, thanks! Yeah, I actually researched it, since I didn't know anything about it and I found that since 1982, it's based on 1 bar and the value is $22.7$ approximately. Anyways, I talked with the teacher, and actually, the problem was wrong, posting a question about this might have been dumb, but thanks anyway!

Assuming it is given that $\ce{X2O5}$ behaves like an ideal gas, the best approach to do this problem is manipulating $pv = nRT$ to get the equation, which fits the given data. Suppose $m$ is the mass of the gas, $M$ is the molar mass of the gas, and $\rho$ is the density of the gas. Then $n =  \frac{m}{M}$:
$$ pv = nRT = \frac{m}{M}RT \ \Rightarrow \ M =  \frac{m}{v} \ \cdot \frac{RT}{p} = \rho \frac{RT}{p} \tag1$$
Now, you can apply $\rho = \pu{5 g L-1}$ at STP to the equation $(1)$ to find $M$. Then apply $2M_\ce{X} + 16 \times 5 = M$ to find $M_\ce{X}$, which is the atomic mass of $\ce{X}$.

That seems like a really cool solution. Although unfortunately, we haven't done the ideal gas law yet. So, it really seems like there should be another solution. I'll look into it, if there isn't any other (cause maybe the problem was a bit unclear about something), I'll mark your answer as the accepted one. Thanks!

Actually just to report: Using this equation and entering in the right quantities, it's the same answer I found with the way described above. So, I think it actually has to do with the fact that the problem was wrong. Thanks for your help, though :)

Just FYI: $V_m = \pu{22.7 L mol-1}$ since 1982. Here standard pressure is considered $\pu{1 bar}$ (not $\pu{1 atm}$).

Oh okay, thanks! Guess my textbook is very old. Although I don't know, I'm still in the first grade of high school. Dunno it's weird. Thanks anyways!

@Gaurav Mall: I agree with you about the question been wrong. Ask your teacher whether formula is really $\ce{XO5}$ or not. If it is it make sense, but still $\ce{SO5}$ is not possible.

Yeah, that's what I'm planning to do. Sent him the assignment today in the morning, so probably he'll tell me something. If not, I'll just ask him.

@MathewMahindaratne You are posting a hint, not an answer. Which, by the way, doesn't resolve OP's problem as it leads to the same result OP already obtained on their own, but with added complexity.


text:
How exactly does electrolysis occur if the molar conductivities of cation and anion are different?
I'm having trouble wrapping my head around Kohlrausch law of infinite dilution and molar conductivities in general.
For a salt $\ce{AB}$ the molar conductivity at infinite dilution can be expressed as the sum of molar conductivities at infinite dilution of the respective elements.
$$\Lambda^\circ_\mathrm{m}(\ce{AB}) = \Lambda^\circ_\mathrm{m}(\ce{A^+}) + \Lambda^\circ_\mathrm{m}(\ce{B^-})$$
But I don't get how the conductivities of $\ce{A^+}$ and $\ce{B^-}$ can be different.
I thought it means that the ions take different times to reach the electrodes so there should be, let's say, greater movement of $\ce{A^+}$ towards the electrodes so as time passes there should be more negative charge due to $\ce{B^-}$ accumulating in the solution. That's obviously incorrect.
So what exactly does conductivity signify physically if not different rates of migration?

Charge never (or almost never) accumulates in solution.  It would be extremely unfavorable.  If say A+ had a higher molar conductivity than B-, then A+ ions would carry most of the charge in the bulk solution.  B- ions would still be consumed at the anode.  The difference in conductivities would probably result in different amounts of "concentration polarization", i.e. the thickness of the boundary layer of concentrations very very (100s of nm) near the electrode surfaces would be different.

@CurtF. "A$^+$ ions would carry most of the charge in the bulk solution. B − ions woukd still be consumed at the anode"-Could you elaborate please?

You are completely right that ions migrate at different rates, and the relative rates can be quantified by a property of the electrolyte called the transport number, which is the fraction of the current that can be accounted by one of the ions:
$$t_\circ^+ = \frac{\Lambda_\mathrm{m}^\circ(\ce{A+})}{\Lambda_\mathrm{m}^\circ(\ce{A+})+\Lambda_\mathrm{m}^\circ(\ce{B-})}$$
$$t_\circ^- = \frac{\Lambda_\mathrm{m}^\circ(\ce{B-})}{\Lambda_\mathrm{m}^\circ(\ce{A+})+\Lambda_\mathrm{m}^\circ(\ce{B-})}$$
(here $t_\circ^i$ designates a limiting transport number ie one in the limit of dilute electrolyte solution).
In the early days of the study of electrolytes, Hittorf developed a way of determining the ratio of transport numbers by measuring the concentrations of electrolytes in different parts of an electrolytic cell. Say the cell contains a cathode, an anode, and a bridging chamber. Cations are attracted toward the cathode chamber and depleted at the cathode, and anions are similarly depleted at the anode but also migrate into the anode chamber. The change in the concentration of cations in the cathode chamber is therefore proportional to the current at the electrode minus the fraction of current carried by cations replenishing the cation chamber:
$$\begin{align} \frac{dc_+}{dt} &= \frac{I_+}{z_+F} - \frac{I}{z_+F} \\  &= \frac{(t_+-1)I}{z_+F} \\  &= \frac{-t_-I}{z_+F} \end{align}$$
A similar equation can be written for the anions. The resulting concentrations in the chambers can be related to the migration rates $v_i$ of the ions , which can in turn be related to the transport numbers as
$$\frac{t^+}{t^-} = \frac{|Z^+v^+|}{|Z^+v^-|}$$
Therefore from measurement of electrolyte concentrations in different parts of a cell it is possible to deduce the relative molar conductivities of the components of an electrolyte.
Because the counterions have different mobilities a concentration gradient will develop and this leads to a liquid junction potential. This results in a voltage drop within the cell that counters the EMF, such that the measured cell potential is not the ideal EMF. In Hittorf's experiment this is not a problem. Where it is a problem it might be avoided by using a salt bridge.
For a textbook that provides a reasonably clear introduction to electrochemistry see Atkins Physical Chemistry, from which I derived some of this answer.
Transport numbers are important in practical applications. See for instance Ref. 1 for an interesting thesis that focuses on the determination of transport numbers in cements.
References

J.A. Muka (1992). Transport Number Determination by Analysis of Transport of Aqueous Electrolytes Through Cementitious Materials Using Emf Method. University of Nairobi.