How to determine the correct InChI for a certain compound? Let's say I have a compound L-xylulose 1-phosphate and I want to know its correct InChI, how to do this? The reason I ask is because when I go to different databases I get different results:

ChEBI:

$\ce{C5H11O8P}$

InChI=1S/C5H11O8P/c6-1-3(7)5(9)4(8)2-13-14(10,11)12/h3,5-7,9H,1-2H2,(H2,10,11,12)/t3-,5+/m0/s1

NIKKAJI; link taken from KEGG:

$\ce{C5H11O8P}$

InChI=1S/C5H11O8P/c6-3-1-12-5(8,4(3)7)2-13-14(9,10)11/h3-4,6-8H,1-2H2,(H2,9,10,11)/t3-,4+,5?/m0/s1

ModelSeed:

$\ce{C5H10O8P}$

InChI=1S/C5H11O8P/c6-3-1-12-5(8,4(3)7)2-13-14(9,10)11/h3-4,6-8H,1-2H2,(H2,9,10,11)/t3-,4+,5?/m0/s1

So, ChEBI and NIKKAJI do show the same chemical formula but different InChI expressions, while NIKKAJI and ModelSeed differ regarding their chemical formula but have the same InChI (which seems wrong, as ModelSeed's InChI suggests that it also als 11 Hs). But this would still not explain the differences for the first two...

That all looks quite fishy to me; how can I now decide which of the databases show me the correct information? Is it possible that one gets the same InChI for compounds that differ in their chemical formula (I thought this wouldn't be possible)?

One of the goals of the InChI project was to ensure uniqueness: [1]

Strict uniqueness of identifier
The same label always means the same substance, and the same substance always receives the same label (under the same labelling conditions). This is achieved through a well-defined procedure of obtaining canonical numbering of atoms.

Whilst this is often the case for InChI strings, there are some (complex) examples where the above goal has not been met, for example in the natural product Spongistatin (below) where two isomers incidentally have the same InChI key: [2]

InChI=1S/C63H95ClO21/c1-33(19-42(67)18-17-35(3)64)20-53-55(72)57-39(7)58(79-53)59(73)63(75)31-51(70)37(5)52(85-63)16-14-12-13-15-44-22-43(68)27-61(81-44)29-47(76-11)23-45(82-61)25-50(69)38(6)56(78-41(9)66)36(4)34(2)21-49-28-60(10,74)32-62(84-49)30-48(77-40(8)65)24-46(83-62)26-54(71)80-57/h13,15,17-18,36-39,42-49,51-53,55-59,67-68,70,72-75H,1-3,12,14,16,19-32H2,4-11H3/b15-13-,18-17+/t36-,37+,38+,39+,42+,43+,44+,45-,46+,47+,48-,49+,51-,52-,53-,55+,56+,57-,58+,59+,60+,61+,62+,63-/m0/s1
Although InChI can sometimes break (as per above) most of the issues with InChI strings are with implementation (how the structure is parsed to the thing generating the string in the first place). Several (common) features are at present un-supported by the InChI implementation:

Polymers
Complex organometallics
Markush structures
Mixtures
Conformers
Excited state and spin isomers
Local stereochemistry/chirality
Topological isomers
Cluster molecules
Polymorphs
Unspecific isotopic enrichment
Reactions

Generation of InChI strings
InChI is, by nature, an algorithm designed to be ran by a computer. Whilst strings can be parsed by humans (with ennough effort), the complexity of the strings is such that it is challenging to ensure they're correct. The InChI FAQ specifically deals with this:

You should not do so (though you of course can). This may give apparently reasonable answers but it is error-prone and may break relations in the InChI.

The most recent implementation of InChI is provided by the InChI trust, open source and free of charge.[3]
As alluded to above, a common source of InChI errors is the way in which the structure of interest is passed to the algorithm rather than with a fundamental flaw with the process used to generate it. Sh*t in, sh*t out, so to speak.
To give a concrete example, consider the heterocyclic system below:

Clearly, the two tautomeric forms cannot be distinguished chemically, but depending on how the InChI string is generated, they may end up having the same or different strings. In this case, we need to specify to the InChI algorithm whether we want to fix the hydrogens (to show a single tautomer, each of which would have a unique InChI string), or not fix them (such that both tautomers have the same InChI string)
Validation
The real question you're possibly interested in isn't How to determine the correct InChI for a certain compound?, but rather How to validate an InChI string for a certain compound?.
Given the complexity of the InChI strings, this is a challenging thing, and to my knowledge there is no tool which allows a string to be provided and says whether it is valid or not (similarly you can't provide a proposed IUPAC name to any tool which will say whether it is the IUPAC preferred name).
One thing you can do is use the InChI string to generate a structure (ChemDraw does this). Using your strings from the question, it's evident that they actually refer to different forms of the molecule (cyclic vs acyclic, just imagine the terminal primary alcohol attacking the ketone).

Chemically, it may be that the cyclic and acyclic forms are in equilibrium- in this case there is no way for InChI to represent the mixture (and hence which structure used to generate the string is ambiguous). It could also be that they are separable chemically and not interconverting, in which case quite rightly they should have different InChI strings.

[1]: Journal of Cheminformatics, 2015, 7, 23
[2]: http://www-jmg.ch.cam.ac.uk/data/inchi/ Accessed 3-Sept-2017
[3]: http://www.inchi-trust.org/about-the-inchi-standard/
Accessed 3-Sept-2017


Effect of pressure on gas solubility in liquids By Henry's Law ($K=\frac{x}{p}$), we know that increasing the partial pressure of a gas also increases its solubility on liquids.
Now a lot of sources (this, for instance) illustrate this by presenting "the bends", a condition that divers may experience when ascending too quickly due to the rapid decrease of the N$_2$ solubility on the blood.
So, my question is: why is the N$_2$ solubility greater deep on the ocean than on the surface?
Henry's Law says nothing about the total pressure, only about the partial pressure of the gas in question. When deep in the ocean, the pressure rises thanks to the water above, not due to a increase in N$_2$ partial pressure.
Bonus: does this video illustrate a greater CO$_2$ solubility in coke because of the greater total pressure? I feel like this is the same situation. Why would the CO$_2$ partial pressure be greater?

The pressure underwater increases by roughly 1 bar every 10 meters, that is, ambient pressure is doubled every 10 meters compared to standard atmospheric pressure.
When you breathe air underwater using a SCUBA tank, your lungs expand against the surrounding ambient pressure in order to increase their volume. As the lungs expand they fill with gas which props up the lungs (balances pressure within the lungs with the surrounding pressure). The SCUBA device adjusts the pressure of the delivered air so that it balances the ambient pressure. The partial pressure of the component gases in the gas mixture inhaled from the SCUBA tank is not the same at the water surface and when a diver is submerged. That means the pressure of the air in the diver's lungs is not ~1 bar, it can be far greater, within limitations imposed by the potential toxicity of high blood concentrations of oxygen and nitrogen.


I know PV=nRT but that's total what about partial? What is the formula for finding the partial pressure ?
I know PV=nRT but that's total what about partial ?

To find partial pressures of a mixture of non-reactive gases, you can use Dalton's Law of partial pressures.

If you know the total pressure of the mixture and the number of moles of each gas, you can find the pressure of each gas by multiplying its mole fraction with the total pressure.


What are solvents that can dissolve sebum that is safe for topical use? I do not know much about these properties and I need some feedback regarding these.

Sebum is a semi-fluid that in humans contains a mixture of triglyerides, cholesterol, ester waxes, and squalene. 57.5% of sebum is triglycerides and fatty acids. The next most abundant molecules are ester waxes, which comprise 26% of sebum. Squalene makes up 12%, and cholesterol is the least abundant lipid in sebum.

Dissolving sebum (the oily substance that protects and waterproofs our skin) sounds like a neat chemical way of removing it - a way to remove skin oiliness. But the solvents that dissolve sebum in a test tube are likely to be too aggressive for extensive use, or to have an oily nature in themselves.

Of the safe and common solvents that might be useful, isopropanol (91% if available, or 70%) would be somewhat invigorating - if not used in excess. It will almost certainly dry the skin, perhaps too much - so that you resort to applying a skin lotion to provide some oiliness.

An alternate way of removing sebum from skin would be to emulsify it with a detergent. Sodium stearate is a common, safe detergent (actually, a soap) which is somewhat soluble in water and available commercially (Ivory Soap is 99.44% pure!). Applied to a washcloth, it will remove much of surface sebum. Rinse, lather, repeat.

Some chemical and cosmetic improvements have been made to other products: added colors, fragrances and acids to reduce the pH from neutral to slightly acidic, for a gentler action on skin; added water for semi-liquid soap (Soft Soap).


Why does the Rosenmund reduction stop at the aldehyde stage? In the Rosenmund reduction, an acyl chloride can be reduced to an aldehyde by hydrogenation over a palladium catalyst poisoned with barium sulfate.

Why isn't the aldehyde product further reduced to an alcohol?

In general, the Rosenmund reduction isn't actually used very often, mainly since the catalyst is also very good at reducing other functional groups which may be present in the molecule (the catalyst is similar to that used for a Lindlar reduction).

To quote from Comprehensive Organic Name Reactions and Reagents:

  [...] the Rosenmund reduction is an unreliable and irreproducible method for organic synthesis, and it is often in competition with other side reactions. For example, alcohols are often the by-products of the Rosenmund reduction.

How the reaction works/why we get an aldehyde

As with many reactions involving catalyst systems in which palladium has been absorbed onto something (the classic one being palladium on carbon/charcoal), the mechanism of the Rosenmund reduction isn't well defined, and indeed there are likely to be different mechanisms operating depending on the quality/specification of the catalyst used.

One possible pathway has been proposed, and is given below. The critical thing here is that the acyl chloride is first activated by oxidative addition into the acyl chloride – this activation allows the reduction to take place.

The product of the reaction is often therefore the aldehyde, which should be kinetically inert to further reduction (this has been discussed on chemistry.SE previously, and can be found here), though the alcohol is often obtained in small amounts.


Thionyl Chloride and enols, carboxylic acids, alcohols \begin{align}
\ce{Propanol + SOCl2 &-> Expected~ alkyl~ halide~ product}\\
\ce{Propanoic~ acid + SOCl2 &-> Expected ~acid ~chloride ~product}\\
\ce{Acetone ~enol + SOCl2 &-> doesn't~ work ~in~ the ~expected ~manner}\\
\end{align}

My prof says the $\ce{SOCl2}$ only works on $\ce{sp^3}$ substrates.

But ... both propanoic acid and acetone enol involve an $\ce{sp^2}$ carbon bound to an alcohol group ... why does $\ce{SOCl2}$ work in one case but not in the other?

I think it instead has to do with the nucleophilicity of the electrons involved with oxygen ... the mechanism starts with electrons associated with oxygen attacking thionyl chloride. A regular alcohol has an oxygen with lone pairs that aren't fairly "free," so to speak. An carboxylic acid has resonance donation enhancing the nucleophilicity of one oxygen's lone pairs. In acetone, however, there is no resonance donation ....... (inductive donation, sure).

So does the $\ce{sp^2}$ vs. $\ce{sp^3}$ explanation hold water?

Obviously it's not true that thionyl chloride only works with sp3 substrates. As you say, a carboxylic acid, an sp2 hybridized substrate, can be chlorinated with thionyl chloride.

Ultimately, the difference is that the carbon of an enol is not readily attacked by a nucleophile, while the carbons in the other substrates can be. Each of the reactions go through the same first step(s): addition of the hydroxyl to sulfur with chloride leaving sulfur (likely through addition/elimination, not shown below). However the next steps are different depending on the functional group of the substrate. For an alcohol, chloride attacks the activated alcohol (looks very much like a sulfonate) in an SN2 reaction to give the alkyl chloride. For a carboxylic acid, chloride attacks the carbonyl carbon, and after addition/elimination (not shown in detail), the acid chloride is produced. Starting from an enol, reaction with thionyl chloride gives an enol chloro sulfate. Unlike the sp2-carboxylic acid case, the pi bond doesn't have an effective electron sink (such as oxygen), which means that the reaction would have to be a direct SN2 displacement. As I'm sure you learned, SN2 at sp2 carbons is very slow.

A SciFinder search shows that other reagents, notably phosphorus pentachloride can accomplish the transformation.


difference between solving and dissolving something What is the difference between the terms 'solve' and 'dissolve'? Up until now I have used the two terms interchangeably, not having any sound scientific definitions for each. I have thought them both to mean to make a solid into a solution through the addition of a solvent (not sure if only for solids into solutions; is the mixing of two fluids also dissolving/solving because $\ce{CO2 (g)}$ can dissolve into water?).

I have not been able to find any scientific definitions on the internet that differentiate the two. it would also be helpful if you could clearly define any other terms related to the mixing of substances without a chemical reaction occurring (eg. immiscible, which I understand to be the state of two FLUIDS being unable to homogeneously mix)

Also, if both substances are liquids, which one is the solvent and which one the solute?

According to the Merriam Webster dictionary, dissolve originates from the Middle English word solve, meaning to loosen or dissolve. There is no official modern English definition for solve other than 1) to find a solution, explanation, or answer or 2)  to pay (as a debt) in full.

The IUPAC Gold Book actually gives no definition for either solve or dissolve, though it defines solvation:

  Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.

When two miscible liquids are combined, they are said to form a solution.


E2 reaction — which one is faster? Will an E2 reaction that has 2 potential products go faster than one that has 3 potential products? The specific question involves both molecules below reacting with methanol. I figured it's an E2 reaction (I hope I'm right) and the question is, which one will go faster?

For the E1 pathway, molecule A will react faster since the intermediate is more stable than from molecule B. According to this web site, the major product for A is 1,2-dimethylcyclohexene; for B, it is 1-methylcyclohexene.

For the E2 pathway, both reactions will have nearly the same rate. Depending on the orientation of the α-methyl group (cis or trans to Br), the major product is the same for cis, but for trans it will be 2,3-dimethylcyclohexene for A and 2-methylcyclohexene for B.


what wavelength should I use to measure the transmittance of iodine colouring? I am doing an experiment on how pH buffers may affect amylase's break down of starch into glucose in a starch solution. I am using iodine reagent to identify how much starch is left after amylase is added. I am wondering, to which wavelength should I set the spectrophotometer to in order to measure the absorbance and percent transmittance values of the solution after iodine solution is added to measure the amount of starch?

You know that in first approximation the absorption recorded is a product of the optical path length $d$, the extinction coefficient $\varepsilon$, and the analyte concentration $c$ as $$\mathrm{abs} = c \cdot \varepsilon \cdot d$$

Now, depending on your specific task, consider the three variables to your advantage.  Because for the wavelength eventually elected to record your data, you want to stay in a linear relationship between analyte concentration and detected absorption as the read-out.

The pathlength $d$ is something typically fixed to $1\,\pu{cm}$, but be aware that there are optical cells offering a shorter path length (e.g., $5\,\pu{mm}$, $1\,\pu{mm}$ as here or even $0.1\,\pu{mm}$) or longer paths (e.g., $50\,\pu{mm}$ (example), or $100\,\pu{mm}$.  It depends on the design of the spectrometer if mounts are interchangeable, or not, to benefit from an intended adjustment to your needs here.
What is the range of concentrations $c$ to record?  As the extinction coefficient depends on the wavelength of recording, it may be beneficial to record at a wavelength maximizing $\varepsilon$ of your amylose iodine complex, provided you want to detect low concentrations of this specis.  (Of couse it shouldn't be in the range other molecules of your solutions absorb intensely.)

But: if you want to determine solutions with a high concentration of your analyte, than it makes sense to record the absorption at a wavelength where $\varepsilon$ of your analyte just is high (not maximal), while «all the rest» of the sample solution does not contribute significantly to the absorption.  This approach will guard you to encounter absorptions too high to be sensible for detectors in the spectrometer.  Look up the manual of the spectrometer or ask colleagues already using the instrument; for some, an absorption greater than $2.0$ equates to «skip it, there is no light on the detector at all».

The iodine-amylose complex is known to absorb intensely.  This equally may justify to record $\mathrm{abs}$ in function of analyte concentration $c$ a little «off» from $\varepsilon_\mathrm{max}$.  By recording the spectra of your blank, and of a typical solution with your complex to quantify, you may justify how many nanometers above or below $\lambda$ yielding $\varepsilon_\mathrm{max}$ of your analyte is fine.  If time and resouces permit, you may record at multiple wavelengths, too; eventually, the readings should lead to the same concentrations per sample.


Why are some chemicals packaged in incompatable containers? I've noticed that some organic chemicals are sold in HDPE containers despite compatibility charts indicating that such containers are not appropriate acetone.

For example, this acetone listing from Walmart above uses an HDPE container, but when incompatible solvents are packaged in HDPE containers the containers can experience a relaxation in the polymer stress from blow molding causing the bottle to contract as shown below.

How can they use incompatible containers and still safely package solvent products?

The shrinkage is called panelling and bottles that are otherwise incompatible are made compatible by a process called fluorination. In fluorination the HDPE is exposed to fluoride gas with precise control in time, temperature and pressure. This exposure converts the outer layers of polyethylene to a highly fluorinated polymer similar to teflon.

$$\ce{-(C2H2)_n-~+ 4 F2~~ \longrightarrow ~~-(C2F4)_n-~ + 4HF}$$

This fluoropolymer coating greatly reduces diffusion of the solvent into the plastic and thus the solvent is unable to soften the plastic, preventing paneling.


Accurate method to calculate the pH of a salt from a weak acid and weak base Can someone help me with this?

The problem is as follows:

  Find the $\mathrm{pH}$ of $\pu{1M}$ solution of $\ce{NH4CN}$. Their respective
  hydrolysis constants are:

  $K_h = 5.6\times 10^{-10}$

  $K_h = 2.5 \times 10^{-5}$

The alternatives given are as follows:

$\begin{array}{ll}
1.&2.8\\
2.&3.5\\
3.&6.4\\
4.&11.7\\
5.&12.2\\
\end{array}$

What I've attempted to do was to establish the following reactions:

$\begin{array}{lllll}
\ce{NH4+ + H2O <=> NH3 + H3O+} &k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\ce{CN- + H2O <=> HCN + OH-} &k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{a}}\\
\ce{H3O+ + OH- <=> 2H2O}&k_{h}=\frac{1}{K_\mathrm{w}}\\
\end{array}$

This meant that:

$K=\frac{K_\mathrm{w}}{K_\mathrm{a}\times K_\mathrm{b}}$

For the given information it can be calculated:

$K=\frac{10^{-14}}{\frac{10^{-14}}{5.6\times 10^{-10}}\times\frac{10^{-14}}{2.5 \times 10^{-5}}}=1.4$

All of this is translated into:

$\ce{NH4+ + CN- <=> NH3 + HCN}$

From this equation it can be established the equilibrium equation:

$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$

Returning to the previous equation I'm getting:

$\begin{array}{cccccc}
&\ce{NH4+ + CN-  & <=> &  NH3 + HCN }\\
i&1&1 & & &\\
c&-x&-x &  &+x &+x\\ \hline
e&(1-x)&(1-x) &  &x &x\\
\end{array}$

Using this relation with what was established earlier it becomes into:

$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$

$1.4=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$

$1.4=\frac{x^2}{(1-x)^{2}}$

Solving this for $x$ results into:

$x=0.54196$

and

$x=6.45804$

The thing here comes on which to choose for the concentration. I decided that the first seems more logical as it will not produce negative concentration in the denominator.

Therefore the concentration for $x=0.54196$

This means:

$\ce{[NH3]=[HCN]}=0.54196$

$\ce{[CN-]=[NH4+]}=1-0.54196= 0.45804$

Then to get the $\mathrm{pH}$ can be established from the hydrolyzation of $\ce{HCN}$:

$\ce{HCN + H2O <=> CN- + H3O+}$

Then:

$K=\frac{10^{-14}}{2.5\times 10^{-5}}$

$K=\frac{\ce{[CN-][H3O+]}}{\ce{[HCN]}}$

$4.0\times 10^{-10}=\frac{\ce{[H3O+]}(0.45804)}{(0.54196)}$

$[\ce{H3O+}]=4.73286\times 10^{-10}$

$\mathrm{pH}=-\log(4.73286\times 10^{-10})=9.3248$

Which seems kind of logical that the $\mathrm{pH}$ is within the range of an alkaline solution as $K_\mathrm{b}>K_\mathrm{a}$.

But since this option does not appear within the alternatives. I'm not convinced whether my solution is right or not.

I've found this equation which relates the $\mathrm{pH}$ with the $\mathrm{p}K_\mathrm{w}$ and $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ and the concentration:

It states:

$\mathrm{pH}=\frac{1}{2}[\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} + \log c]$

where $c = \text{concentration}$ of the salt.

Inserting the given values in the above equation yields:

$\mathrm{pH}=(0.5)\left[-\log(10^{-14})+\log\left(\frac{10^{-14}}{5.6\times 10^{-10}}\right)-\log\left(\frac{10^{-14}}{2.5\times 10^{-5}}\right)+\log(1)\right]$

$\mathrm{pH}=9.3248$

which also checks with what was obtained previously.

But again none of these seem to check with any of the alternatives given. Could it be that my procedure was wrong?

The thing which it bothers me here is if this could be accepted as a general procedure to calculate the $\mathrm{pH}$ of a salt which has a weak acid and weak base ions. Can somebody help me here? And more importantly, how to prove that equation for $\mathrm{pH}$? I'm not sure if the hydrolyzation constants given in the problem are right or accurate. Does somebody has access to cross reference and check if those values are correct?

Can somebody help me with that part too?

Ignoring activities, activity coefficients, ionic strength effects, and so on, I get a pH of around 9.3. Given the 1 M concentration, I would not bet a lot of money on this, though. My solution, assuming a more typical solution concentration, is given in the two figures below. Sorry these are figures rather than proper formatting: I have been away for 3 months and am beyond rusty at this point. If the mods wish to delete this answer, they should feel totally free to do so.

There are obviously other ways to do this problem and what I have posted is simply an approximation: any pH problem solution, of this type, that does not explicitly involve the formal concentration of the salt, i.e., "F", always gives me pause. But maybe this helps a bit.


What is the difference between the equilibrium position and the equilibrium constant? I used to think that the equilibrium constant was a quantitative measure of the equilibrium. However, changing concentration and pressure in a system at equilibrium that shifts the equilibrium doesn't affect the equilibrium constant, which is solely temperature dependent. This means that my definition was wrong. What is the real relationship between the two, or what do they actually mean? (I haven't been able to find any suitable definition for "equilibrium position," and the only definition I know how "equilibrium constant" is the value of the reaction quotient when a system has reached equilibrium).

Here's a more simplistic response.

Equilibrium position: a particular set of concentrations of reactant and product species.
Equilibrium constant: a number that describes how far "forward" equilibrium lies for a particular system (as described by a particular equation).

Example: for the system $\ce{A <=> B}$, the mass action expression is $\ce{\frac{[B]}{[A]}}$.  Let's say $K = 200$.
This means, at equilibrium, $\ce{[B]}$ will be 200 times greater than $\ce{[A]}$. That's what the equilibrium constant tells you - equilibrium for this system lies to the right, and the K value is greater than 1.

There are infinitely many different equilibrium positions that satisfy K.  We could have $\ce{[B]} = 20~\mathrm{M}$ and $\ce{[A]} = 0.1~\mathrm{M}$.  Or we could have $\ce{[B]} = 1~\mathrm{M}$ and $\ce{[A]} = 0.005~\mathrm{M}$.  Those are two different equilibrium positions that are both at equilibrium.  In both cases, $Q = K$ and the system is at equilibrium.

Some notes:

If we describe the exact same system with a different equation, $\ce{B <=> A}$, then the mass action expression is $\ce{\frac{[A]}{[B]}}$. We would find that $K = 0.005$ (that is, $1/200$) and we would say that equilibrium lies to the left.  Of course, the molecules don't behave any differently, but the equation we use to describe the system is different, and that difference is reflected in a different $K$.
If we have a system with more than one product or reactant or with stoichiometric coefficients other than 1, then it's a bit more complicated than just saying something like "product concentrations are 200x greater than reactant concentrations". Instead, equilibrium may be achieved at many different equilibrium positions. You have to look at the mass action expression to interpret the meaning of $K$.
The equilibrium constant $K$ is just one way of expressing how far "forward" a particular reaction will proceed.  Gibbs free energy change ($\Delta G$) is another way. And, for redox systems, cell potential is yet another way.


How is the acetoxy group ortho/para-directing in electrophilic aromatic substitution? Why is the $\ce{-OAc}$ group o,p-directing in electrophilic aromatic substitution?

How can I compare the activating effect of $\ce{-OAc}$ with $\ce{-OH}$ and $\ce{O-}$? Is it to do with the number of lone pairs on oxygen?

Phenoxide is more reactive than phenol towards electrophilic substitution. However, the number of lone pairs isn't the factor that controls it, because it is not possible for both lone pairs to be simultaneously in conjugation with the benzene ring. If one lone pair is in conjugation with the benzene ring, the other lone pairs will not be oriented correctly to overlap with the pi system of the ring. So, you can have one, two, or three lone pairs - it does not matter because any lone pairs beyond the first are useless.

Only the availability of the lone pairs matters. That has to be discussed with inductive or resonance effects. With an $\ce{-OH}$ group you have a hydrogen atom fighting for some of the electrons (inductive effect), and with a $\ce{-O^-}$ group there is nothing else.

Moving on to the acetoxy group $\ce{-OC(O)CH3}$, the lone pair on oxygen is delocalised into the electron-withdrawing carbonyl group, via resonance. You can think of it as the $\ce{C=O}$ and the benzene ring both fighting over the oxygen lone pair, which makes it less available for donation to the pi system. Consequently, the acetoxy group is still o/p-directing (because of the presence of at least one lone pair), but is not as strongly activating as the hydroxy group (because the lone pair is less available).


What is the effect of pressure upon the melting and boiling points of a substance? I want to know about the inter-relation between the effect of pressure upon the melting and boiling points of a substance. Can anyone help?

When we apply pressure on a substance, its melting point increases or decreases depending upon the change of volume which occurs during the phase change of that substance. To be specific, if the volume of the substance's liquid phase is greater than the volume of the solid phase, its melting point will increase upon the increase of volume. Again, if the volume of the substance's liquid phase is less than the volume of the solid phase, its melting point will decrease upon the increase of volume.
The boiling point of liquids always increases when pressure is applied on that liquid. This is because the molecules of the liquid will require comparatively more energy to turn into gaseous state when pressure is applied on that substance.


How can I figure out if my store bought chlorine is the percentage it says it is? Chlorine is an important part of maintaining any pool. I've read horror stories of people buying chlorine from pool stores or big box stores and the chlorine being ineffective.

If stored long enough Chlorine can degrade, lose it's strength. If put in the sun i've read that it can speed up that process.

Like most pool owners, I have a pool test kit that tests how much chlorine is in a pool water sample. I specifically own a Taylor K-2006c Kit which test for Free Chlorine using the FAS-DPD Method.

At most pool stores they supply 12.5% Sodium Hypochlorite and 87.5% Other Ingredients.

With my given test kit would it be possible to accurately test whether or not it's actually 12.5% Sodium Hypochlorite or not? If so, how would I go about doing so?

If not; What do you recommend using in order to test the chlorine.

Bleach solutions can be analyzed by addition of iodide and back titration with thiosulfite - somewhat difficult for non-laboratory conditions. Kits are available, but expensive (~$250).

A less precise (but probably adequate) method involves measuring the heat evolved as bleach oxidizes acetone. https://www.flinnsci.com/api/library/Download/650c3e8ed0574ed18c7b37b8832cccae It seems that you might be able to modify this process to do a rough quantification (+/- 10%) quickly. It was also discussed briefly on Quora: https://www.quora.com/When-happens-when-acetone-and-bleach-are-mixed

Just realize that if you drop solid hypochlorite into pure acetone, it could explode, catch fire, and make you wish you had never tried it. Creep up on this reaction: add only small amounts of acetone (1-2 grams) to a dilute hypochlorite solution (~5% ?) until you get the feel of how fast and how hot this reaction will go.

Eye protection is mandatory. Also have some water available (like a hose). Just be safe by expecting danger.


Optical rotation in inversion of sucrose In reading about the inversion of sucrose I came across the equation for the reaction rate constant below:

$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(0)}\right] \tag{5}$$

What is the logic for difference of the rotations' sign for the logarithmic term? i.e conceptuatlly why is $\alpha(0)$ the minuend in the numerator and the subtrahend in the denominator?

I have seen other quantities like volume or pressure represented in this form but I don't understand the reason why reacting sugar would be presented this way.

Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.

The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.


Volatile nature and odour Will it be correct to make a general statement that all compounds having odour are volatile?

My reasoning: if a compound has odour then to smell that it must evaporate ( ie. some molecules must diffuse in the surrounding) . So I conclude that for smelling the compound must be volatile.

Am I right?

It's true that all odors come from compounds with at least some volatility.

However, some of the smells we associate with certain materials (like some metals) aren't the metal themselves but products of reactions caused by those materials, so you have to be careful about that.


Atomic shielding 2s to 1s How much is the effecting shielding of the 1s electrons from the 2s electrons? . Because the 2s subshell overlaps with the 1s subshell this will cause the 1s electrons to 'feel' less charge than the nucleus charge.

However due to the fact that the 2s subshell is much bigger than the 1s subshell the effect of shielding will be small correct?

The effect of "reverse shielding", as I call it, is on the order of one volt of potential difference in light elements where the $n=2$ electrons are the valence electrons.

Let's look at lithium.  From this data page the effective nuclear charge on the $1s$ electrons in the neutral atom is 2.691.  If we assume that the effective nuclear charge is related to the electronic energy level by

Energy$=(-13.6 eV)(Z_{eff}^2)$

then we infer that the orbital energy level is -98.5 eV for each electron.

Now what is that energy level if we take away the 2s electron? For this we need the electronic energy level of the $1s$ orbital in the lithium ion, which is obtained by averaging the second and third ionization energies of lithium given here.  That leads to -99.1 eV per electron.  The 2s electron has effectively increased the energy level of each 1s electron by 0.6 eV.

A similar calculation for beryllium gives a difference of about 1.1 eV per $1s$ electron (-184.7 eV from the effective charge on $1s$ in neutral Be vs -185.8 eV  from the third and fourth ionization energies after forming the di-cation).  The second 2s electron has basically doubled the shielding.

The electron volt differences determined above have some impact on chemical reactions, whose energies are on the order of one electron volt per electron transferred in a redox reaction.


Why is a squared cosine window function commonly used for 2D spectra? In nearly all default parameter sets for 2D NMR experiments on Bruker spectrometers, the window function is set to a pure squared cosine function for both dimensions (QSINE with SSB = 2). In contrast to that, 1D spectra are often processed with the efp command, which uses an exponential window function.

Why is the squared cosine window function the default? Which advantages and disadvantages does it have compared to other common window functions?

The difference in window function is a result of the parameter being optimized for the respective experiment.  Exponential multiplication of the Free-Induction Decay (FID) weights the beginning of the FID where the amplitude is at it's greatest.  Weighting the beginning of the FID results increases the signal-to-noise ratio, since the noise should have the least impact with the strong signal.

For 2D NMR, we are usually more interested in resolution to aid in differentiating the various peaks and cross-peaks.  We can always take more scans to try and increase the signal strength, so resolution enhancement becomes the key objective.  A pure sine function offers increased resolution at the loss of signal enhancement.  The shifted sine function (or pure cosine function with SSB=2), provides a good mixture of the two objectives.  The enhancement doesn't weight the early parts of the FID as heavily as exponential weighting, but more than a sine function.

The use of SSB=2 for a cosine function is used most often in 2D experiments where we need a usually more significant amount of signal enhancement, such as TOCSY, NOESY etc.  sine enhancement is used most often for COSY, to reduce filtering out peaks from overlap.

--UPDATE--

To answer some further questions posted in the comments.  Using a gaussian weighting for 2D-NMR data is not as preferable as cosine or sine weighting.  The gaussian weighting has a much more narrow distribution, and rapidly approaches zero.  This can prematurely truncate the FID (by essentially weighting the later portion as zero), which introduces artificats into the resulting spectrum known as "truncation wiggles".

Squaring the cosine (which remember is just the sine function shifted) is done both to slow the increase and decrease of the weighting function over the course of the FID.  At the beginning of the FID, this allows for a greater amount of sensitivity to the early parts, while sacrificing a small amount of resolution.  Additionally, the slow decay also reduces the truncation artifacts discussed in the first question of the update.


Solubility of Alkali Metal Chlorides Why does Potassium Chloride less soluble than Sodium Chloride in water, isn't it supposed to be other way round? Since Potassium Chloride would be more ionic than Sodium Chloride?

When dissolving, the ionic crystal’s lattice enthalpy must be broken while at the same time the hydration shell is generated around the cations and anions. The latter process is typically exergonic while the former is strongly endergonic.

How well a salt dissolves depends not only on the strength of solvation but also on the strength of the crystal lattice. If the lattice is strong, the solubility will be lower because there is an increased tendency to reform the crystal lattice.


Reference states for molecular orbital energies in DFT calculations When doing a population analysis using density functional theory (specifically as implemented in Gaussian), what are the orbital energies (i.e. alpha and beta eigenvalues) with respect to? If I have two different but closely related compounds, can I directly compare their HOMO (or LUMO) energies? It should be noted that I am not referring to comparing their HOMO-LUMO gaps, but rather the energy of a single molecular orbital.

In molecular electronic structure, electronic/orbital energies are measured with respect to an energy zero of an isolated electron in free space.

Hence a lower LUMO energy does correspond to an orbital that is more apt to accept an electron/a more negative electron affinity compared to another molecule.

Correspondingly a higher HOMO energy corresponds to an orbital that is more liable to give up an electron/a less positive ionisation energy compared to another molecule.

This is the basis of Koopmans' theorem in Hartree–Fock theory and it's equivalent in DFT.


Is reduction potential and stability related? Suppose the half reactions:

$$\begin{split}\ce{Na^+    +  e^-   &->  Na} ~~~~~~~~~~~~&\text{ $E^o = -2.71$}\\
\ce{Cu^{2+} + 2 e^-  &->  Cu} &\text{ $E^o = +0.34$}\end{split}$$

Can I say that $\ce{Na+}$ is more stable than $\ce{Na}$, so $\ce{Na+}$ has negative potential to reduce? If so then can I also say that $\ce{Cu^2+}$ is less stable than $\ce{Cu}$ so $\ce{Cu^2+}$ has a positive potential to reduce?

Remember that reduction potentials do not have any meaning by themselves. They are measured against a hydrogen cell to which arbitrarily we assign the value $E^0=0$. You can compare between different potentials. For example, in your case, you could say that that $Cu^{2+}$ has a greater tendency to be reduced than $Na^+$  because its reduction potential is more positive. However, I do not think the potential itself tells you anything with regard to the stability (and, in any case, what do you mean by that) of the species involved in a particular half reaction.


What do ε₀ and V mean in a Pourbaix diagram? From Wikipedia:

What does this graph mean?
What is $\varepsilon$?  It looks like $\varepsilon_0$, like the vacuum permittivity, but what does this have to do with the polymorph and/or isomers of ferrite or hematite?
And what is the V or 'five' in square brackets after the $\varepsilon_0$?

This is a Pourbaix Diagram (I assume this is where you got this figure). While it's not a common notation as far as I'm aware, $\epsilon_0$ is being used to represent the reduction potential in volts (V). As mentioned at the top of the Wikipedia page, this is basically a phase diagram, that tells you the ranges of voltage/pH for which a particular species of (in this case) iron is most thermodynamically favorable. The dividing lines coincide with an even mixture of the surrounding species.


How to calculate the number of atoms of each element in a urea? I have a stoichiometry question which I answered incorrectly and I am not sure where I went wrong. The question is as follows:

  Urea, $\ce{(NH2)2CO}$, is used for fertilizer and many other things. Calculate the number of $\ce{N}$, $\ce{C}$, $\ce{O}$, and $\ce{H}$ atoms in $1.68 \times 10^4~\mathrm{g}$ of urea.

First, I found the total atomic weights of each element in this compound to find the percent composition:

N: 28.02 amu
H: 4.032 amu
C: 12.01 amu
O: 16 amu
Total atomic weight of urea: 60.062

Then, I found the percent composition of urea:

N: 46.65%
H: 6.71%
C: 20.00%
O: 26.64%

Then, I converted $1.68 \times 10^4~\mathrm{g}$ into moles by dividing this amount in grams by 60.062, and found that it is approximately 279.71 moles.

After that, I took percentages of this to calculate the number of moles of each element in the sample of urea (for example, 46.65% of 279.71 to find the number of moles of nitrogen).

Finally, I multiplied each of these numbers in moles by Avogadro's number ($6.022\times10^{23}$) to obtain the number of atoms. However, when I checked my answers in my textbook, I was wrong by a significant amount.

\begin{array}{lrr}
       & \text{My answer}    & \text{Correct answer}\\\hline
\ce{N}:& 7.83 \times 10^{25} & 3.37 \times 10^{37}\\
\ce{H}:& 1.13 \times 10^{25} & 6.74 \times 10^{26}\\
\ce{C}:& 3.37 \times 10^{25} & 1.69 \times 10^{26}\\
\ce{O}:& 4.49 \times 10^{25} & 6.74 \times 10^{26}\\\hline
\end{array}

Do you know where I am going wrong?

You were correct in converting the grams of urea to mol of urea, but the next step would be to convert the mol of urea to molecules of urea using avogadro's number (which converts mol to molecules or atoms).  Then you want to multiply the molecules of urea by the number of atoms of the element in question that are present in Urea.  So for example there are 2 atoms of N per molecule of Urea, therefore multiplying the molecules of urea you calculated earlier by 2 will yield the total number on nitrogen atoms present.  As was mentioned in the comments, the value for N atoms present given by your book was incorrect and should have been 3.37 × 10^26 atoms of Nitrogen.  See if you can calculate the correct number of each of the remaining elements, as the answers given were correct.


Is it safe to drink reboiled water? There have been some claims saying that drinking reboiled water is dangerous because it produces arsenic, nitrates, etc. Is this true?

Reboiling water will do precisely three things:

Some volatile contaminants in the water that survived the first boiling might be driven out by the second boiling.
Some additional evaporation will occur, causing anything dissolved in the water (salts, contaminants) to be slightly more concentrated.
Some additional material from the surface of the vessel you're boiling the water in might dissolve into the water.

It's virtually impossible to imagine any scenario in which this would matter, unless you were taking a huge amount of barely drinkable water and boiling away most of it or using a vessel whose lining could produce toxins in your water. Needless to say, you really shouldn't be drinking barely drinkable water at all, nor should you be boiling water even once in a vessel whose lining can produce toxins. If you're doing these kinds of things, the additional risk from reboiling would be miniscule.

It will definitely not summon aresenic or lead from nowhere.


Electrolysis of water: Which equations to use? (IB Chem) There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells?

$$
\begin{array}{cc}
\hline
\ce{\text{Oxidized species} <=> \text{Reduced species}} & E^⦵(\pu{V}) \\
\hline
  \begin{align}
    \ce{H2O(l) + e- &<=> 0.5 H2(g) + OH-(aq)} \\
    \ce{H+(aq) + e- &<=> 0.5 H2(g)} \\
    \ce{0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq)} \\
    \ce{0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)}
  \end{align}
&
  \begin{array}{r}
   -0.83 \\
    0.00 \\
   +0.40 \\
   +1.23
  \end{array}
\\
\hline
\end{array}
$$

(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)

For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs.

As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$\begin{align}
\ce{O2(g) + 4H+(aq) + 4e-  &<=> 2 H2O(l)}\\
\ce{2H+(aq) + 2e- &<=> H2(g)}
\end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $\ce{OH}$- occurs.

$$\begin{align}
\ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\\
\ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
\end{align}$$


How do I find molar mass using gas laws? A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound?

When I worked the problem I got $\pu{29.9 g/mol}$ is that right?

A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound?

The ideal gas law states $$pV = nRT\tag{1}\label{id-gas}$$ where $p$ is defined as the pressure, any units, $V$ is the volume in liters, $n$ is the amount of substance of gas, $R$ is the universal gas constant, and $T$ is temperature in kelvin.

The information given to us is the density of $\pu{0.977 g/L}$, a pressure of $\pu{710.0 torr}$, and temperature of $\pu{100 ^\circ C}$.

Knowing that amount of substance, $$n = \frac{M}{m},$$ where $M$ is molecular mass in $\pu{g/mol}$, and $m$ is the mass of gas in gram, we can see the unit gram cancels out and we are left with the unit mole.

However, we are given a density, $$d = \frac{m}{V} = \pu{0.977 g/L}.$$

Plugging into $\eqref{id-gas}$ the molecular mass and mass for $n$ we get:

$$ pV = \frac{M}{m}RT .$$

Isolating $m$ in density, $m = d\cdot V$, and isolating $m$ in $n = m/M\Leftrightarrow m = n\cdot M$, we can combine those two equations for $m$:

$$n\cdot M = d\cdot V$$

Isolate $n$, and plug back into our ideal gas equation $\eqref{id-gas}$:
\begin{align}
 n &= \frac{dV}{M}&
 \implies&&
 pV &= \frac{dV}{M}RT
\end{align}

Volume cancels out, and we isolate for $M$:
$$ M = d \cdot \frac{RT}{p}. $$

Plug in our known values, knowing that $\pu{1 torr} = \pu{1 mm Hg}$ and $\pu{760 mm Hg} = \pu{1 atm}$:

$$ M = \pu{0.977 g//L} \cdot \frac{
  \pu{0.082057 L * atm // mol * K}\cdot\pu{373.15K}}{
  \pu{710 torr}\cdot \pu{mmHg//torr}\cdot\pu{{1atm}//760mmHg}}
     = \pu{32.0 g//mol}$$

And our answer has three significant figures.


Moving water vapour between flasks I have a flask of silica gel connected to a flask that contains a bit of water, so when I heat the silica gel the water absorbed gets evaporated and I want it to move to the other flask, but it always condenses on the roof of the first flask, so is there any way to force the vapour to move flasks?

To prevent the condensation, you need to create the conditions under which the vapor does not become saturated. The way to do that is to heat up the whole transfer line (the roof of the first flask and all the tubes that follow) - the higher the temperature, the higher the saturated vapor pressure. Heated transfer lines are often employed in commercial equipment when there's need to get condensible gases from one setup (e.g. thermogravimetric) to the other (e.g. mass spectrometer).


calculating pH - is my answer correct? I'am working on this exercice, Is my answer correct? and can you please help me to solve the second part?

exercice:

1/Calculate the ionization coefficient and pH (with two methods) of a 0.1M ammonia solution of pkb = 4.7

2/ to one liter of the previous solution, an equal volume of water is added. Calculate the new ionization coefficient and deduce the pH of this new solution

1/ For the first part my answer is :

$\ce{NH3  +  H2O = NH4+   +  OH^-}\tag{1}$

$Ka = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3+]}}\tag{2}$

The coefficient of ionization $\alpha$

$\alpha = \sqrt{Ka/C°}\tag{3}$

$Pkb = \log{kb}\tag{4}$

$Kb= 10-pkb\tag{5}$

$\alpha =0,014\tag{6}$

Two methods:

First one:

$pH= \frac{1}{2} (pka + 14 + \log{C°})\tag{7} $

$pka + pkb=14\tag{8}$

$pH = 11,15\tag{9}$

second method:

$14 + \log (\alpha C°) = 11,15\tag{10}$

for the second part, i am lost

Thank you

The general method to solve problems of this type can be found in the answer here, How to set up equation for buffer reaction?. In your case you need to modify this equation. To follow the text below you need to look at the main result in the linked answer (rather than repeat all this here). The derivation is straightforward although the equation is a bit complicated.

The base constant $K_\mathrm{B}$ is given but as $K_\mathrm{w}=K_\mathrm{A}K_\mathrm{B}$ we can start with  main eqn in linked page and convert to $k_\mathrm{B}$ later on. In this case as ammonia is a base (with concentration $c_b= 0.1$ M) we may suppose that any acid concentration $c_\mathrm{a} \approx 0$ then,

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} + \mathrm{[H^+]} - \mathrm{[OH^-])} }{(- \mathrm{[H^+]} + \mathrm{[OH^-]} )}  $$

Additionally we can reasonably assume that $\mathrm{[OH^-] \gg [H^+]}$ and so

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} - \mathrm{[OH^-])} }{ \mathrm{[OH^-]} }  $$

This can be solved by substituting for $K_\mathrm{A}=K_\mathrm{w}/K_\mathrm{B}$  and for $\ce{[H^+]}$ as $K_\mathrm{w}=\ce{[H^+][OH^-]}$ and solve the quadratic equation.  But as $c_\mathrm{b} \gg \ce{[OH^-]}$ we can simplify by letting $c_\mathrm{b} - \ce{[OH^-]} \to c_\mathrm{b}$ then the equation becomes

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{c_\mathrm{b}  }{ \mathrm{[OH^-]} }  $$

We can now get

$$\frac{K_\mathrm{w}}{K_\mathrm{B}}= \frac{K_\mathrm{w}c_\mathrm{b}} { \mathrm{[OH^-]^2}}   $$

from which

$$\ce{[OH]^- } =\sqrt{K_\mathrm{b}c_\mathrm{b}}$$

and so you can obtain the pH.

When you dilute, the concentration of base $c_\mathrm{b}$ is halved.


How to predict disproportionation products from Frost diagram? I read that concave points in a Frost diagram disproportionate. But what are the products of the disproportionation?

For example, in the Frost diagram for manganese at pH = 0, the species $\ce{HMnO4-}$ is concave with respect to $\ce{MnO4-}$ and $\ce{H3MnO4}$. But at the same time, it is also concave with respect to $\ce{MnO4-}$ and $\ce{MnO2}$. How do I know which set of products $\ce{HMnO4-}$ disproportionates to in acidic solution at pH = 0?

Interesting question. First of all understand what is on the x-axis and on the y-axis. You y-axis is Gibbs free energy/Faraday constant. Choose three consecutive oxidation states, the middle oxidation state, if it lies above the two, is likely to disproportionate.Read more here on Frost and Latimer digram

In your case look at the example of $Mn^{3+}$, if you connect $Mn^{2+}$ and $MnO_2$, $Mn^{3+}$ lies above the two, so 3+ oxidation of Mn is unstable with to disproportionation.

In short, just by using Frost diagram you cannot predict disproportionation of Mn (VII) because if it were to disproportionate, you will need Mn(VIII), which is not shown, so Frost diagram analysis only works for three points and you cannot say anything about the boundary values.


How to calculate the energy to dissociate a bond into neutral atoms? I am self studying chemistry through MiT ocw 5.111 . On practice exam 2 problem 2 there is a question which states the following

    element           ionization energy          electron   affinity
    Potassium(K)      418 kJ/mol                 48 kJ/mol
    Fluorine(F)       1680 kJ/mol                328 kJ/mol
    Chlorine(Cl)      1255 kJ/mol                349 kJ/mol

  (a) (12 points)   The ionic bond  length for KF is 0.217 nm. Calculate
  the energy (in units  of kJ/mol required  to  dissociate  a single
  molecule of KF into the neutral atoms K and F, using information
  provided above. For this calculation, assume that the potassium and
  fluorine ions are point   charges.

I proceeded to calculate the bond dissociation energy using the following formula $U(r)=\frac{z_1z_2e^2}{4\pi\varepsilon_0r}$  from which I obtained a dissociation energy of $-640$ kJ/mol. Then left with two ions F$^{-1}$ and K$^{+1}$.
 I went on to determine their ionization energies and electron affinities respectively.

To make F$^{-1}\to$ F + $e^{-}$ an electron must be lost thus warranting an ionization energy of $-1680$ kJ/mol and. To make K$^{+1}$+ $e^{-} \to$ K and electron must be gained thus referring to the election affinity of K of $18$ kJ/mol. When equated as (Energy Required - Energy Released) I arrive at

 (Dissociation Energy + Ionization Energy) - Electron Affinity = (-640 kJ/ mol - 1680 kJ/mol) + 48 kJ/mol = -2272 kJ/mol required

However I see in the answer key this is incorrect could someone guide me as to where I am going wrong here thank you.

Your $\pu{640 kJ mol^-1}$ value is correct, so the next step is to neutralize the ions. First add an electron to a potassium ion to get a potassium atom. This releases $\pu{418 kJ mol^-1}$ because it is the reverse of the potassium atom ionization. Then take away an electron from fluoride ion to get a fluorine atom. This requires an input of $\pu{328 kJ mol^-1}$ because it is the reverse of the fluorine electron affinity. So the result is $\pu{550 kJ mol^-1}$, i.e., $\pu{640 kJ mol^-1}$ - $\pu{418 kJ mol^-1}$ + $\pu{328 kJ mol^-1}$.

This figure makes it clear:

Figure copyright information: D.W. Oxtoby, H.P. Gillis, L.J. Butler, Principles of Modern Chemistry, 8th Ed., Cengage Learning, Section 3.8, p. 87, © 2016 Cengage Learning.


Do most elements generally exist in a planet during early formation, or is it a small subset? I'm interested in whether there are a small subset of elements that are by far the most common in a planet's early stages of formation and a wider range are formed slowly over time through chemical reaction, or if a very wide range of the periodic table can be initially found from the start.
Obviously every planet it unique, but I'm asking about whether there's a common scenario / elemental ratio present in the hot cataclysmic early stages of a planet's formation, or rather it's pretty much diverse from the start. Do most elements generally exist in a planet during early formation, or is it a small subset?

Elements are not formed through chemical reactions or in planets (except through nuclear decay) . They are formed in stars through nuclear reactions.
Your heavier elements can't even be formed in your typical star. It takes a supernova.
So pretty much everything the planet is made of is there from the start.


Why is bread warm at room temperature? If I buy a loaf of freshly-made bread and leave it out at room temperature for a day or two, when I reach into the bag the air inside is much warmer than the rest of the room. Why?

After reading some more stuff I actually think that crystallization of starch might be the answer to this question. There is a lot of literature about this and it might very well be that it can be sufficiently exothermic to warm the small volume of air in a bag.


Extent of dehydration by concentrated sulphuric acid I know that conc $\ce{H2SO4}$ is a dehydrating agent. And the most easiest mechanism for example would be this:

But I want to know how to find the extent of dehydration of it.
I came across this question:

And the answer is:

 d

My thinking:
(a) $\ce{O}$ is EN and will pull $\ce{-OH}$ towards itself inductively, thus decreasing the electron density on $\ce{-OH}$, and decreasing its power to pull $\ce{H}$ from concentrated $\ce{H2SO4}$.
Similarly for (b) and (c) and I think due to inductivity (c) will have more chances than (b).
As for the (d) I think $\ce{-CH3}$ group is an electron pusher so it should increase the electron density, and hence reducing the effect on $\ce{-OH}$ group .
1) But my doubt is: is this the only reason for the extent of dehydration by conc $\ce{H2SO4}$ on (d) to be more than (c)?
2) I don't think there would be much difference if it tautomerises to enol form.

The most stable Alkene is the most highly substituted Alkene. (d) gives a tri-substituted alkene, all others give the less disubstituted alkenes. Most stable Alkene (d) therefore more of it is formed! (Edited thanks to Bon for highlighting my error)

Also a factor to keep in mind is conjugation. (a), (b) and (d) will have conjugated double-bonds, while (c) will not. This conjugation also adds stability to the system. But the major stability contributor is the higher substituted alkene in (d).


What is the oxidation state of Boron in Ammonia Borane? So I was searching about the difference between redox reactions and Lewis Acid-Base reactions and came across classical example of latter,

$$\ce{NH3 + BH3  \rightarrow NH3BH3}$$
Now in this example what is the oxidation state of borane? Is it $+4$, if yes then can this example be considered a type of redox reaction also, since boron went from $+3$ oxidation state in the products to $+4$ oxidation state?

As per this source, the idea and definition of oxidation state is based on the following principle:

  The oxidation number of an atom in a molecule is based
  on a formalism that forces a covalent compound to possess
  complete ionic character and may be defined as the charge
  that an atom would have if all bonds to it are broken such
  that the ligands retain a closed-shell configuration ;
  an exception, however, refers to homonuclear bonds, in which
  case the bond is broken homolytically and a single electron
  is transferred to each atom.

  The oxidation number may
  thus simply be expressed as
  Oxidation number = charge on compound - charge on ligands

So,evidently, for the presented molecule $\ce{NH3BH3}$, we see that $\ce{NH3}$ is a ligand for the $\ce{BH3}$ moiety. So,you can heterolytically cleave the dative bond between $\ce{N}$ and $\ce{B}$ towards $\ce{N}$(as per the electronegativity trends),leaving no charge on the $\ce{B}$ as of now,as after this action,it's valence shell contains 3 electrons.Now,for finding oxidation state on $\ce{B}$,an interesting thing crops up due to the presence of the three $\ce{B-H}$ bonds.
The article further enumerates:

  In many cases, the charges assigned to simple monoatomic
  ligands do not vary from compound to compound, as illus-
  trated by $\ce{F-}$, $\ce{Cl-}$, and $\ce{O^2-}$. However, a notable exception is
  provided by hydrogen for which both $\ce{H+}$ and $\ce{H-}$ have per-
  missible closed-shell configurations ($\ce{1s^0}$ and $\ce{1s^2}$
  , respectively).
  In this case, the charge assigned to hydrogen is determined by
  the relative electronegativity of the atom to which it is attached.

So again, due to sligthly higher electronegativity value of $\ce{H}$ than $\ce{B}$,the $\ce{H}$ atom becomes the ligand for the $\ce{B-H}$ bond. Hence,all the $\ce{B-H}$ heterolytically cleave towards $\ce{H}$,each cleavage leading to a +1 charge on $\ce{B}$ and -1 on $\ce{H}$. In totality,$\ce{B}$ ends up with +3 oxidation state, as it has lost all three electrons from it's valence shell.

Note: I highly recommend everybody to read that paper cited above. It's really insightful into the difference between valence,oxidation number and coordination number,which are often used interchangeably


Why is the principal energy of an electron lower for excited electrons in a higher energy state? Several places state the 'principal energy of an electron' can be calculated as such:

$$E = \frac{2π^2mZ^2e^4}{n^2h^2}$$

Another equation I found was:

$$E = -\frac{E_0}{n^2},$$

where $$E_0 = \pu{13.6 eV}~(\pu{1 eV} = \pu{1.602e-19 J})$$

As seen in these equations, the greater the principal number ($n$) of the electron, the lower the principal energy $E$ of this electron.

However, the principal number $n$ is associated with higher energy states. The farther from the electron, the higher the energy state of this electron.

I seem to be fumbling the concept of 'principal energy of an electron.' What is the difference between the 'energy state' associated with the principal number, and the 'principal energy' of an electron? What exactly do the 'principal energy' equations mean? I read somewhere that it would be the energy it would take to "unbind" or ionize the electron, which would make sense, but I have not seen 'principal energy' explained as the ionization energy of an electron anywhere else.

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~\mathrm{eV}
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~\mathrm{eV}
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $n\rightarrow\infty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.


What is the equation for solving the concentration of the unknown from a Single-point Standard Addition Calibration for Potentiometry? I know how to use the standard addition equation for a single-point analysis from an equation I got from the Analytical Chemistry book "Quantitative Chemical Analysis" 9ed by D. Harris.

However, solving for a potentiometry problem, the equation doesn't seem to be effective anymore. I keep on getting negative value for the initial concentration of the unknown sample.

The problem goes like this:

The Na+ concentration of a solution was determined by measurement with a sodium ion-selective electrode. The electrode system developed a potential of -0.2462 V when immersed in 10.0 mL of the solution of unknown concentration. After addition of 1.00 mL of 2.00 x 10-2 M NaCl, the potential changed to -0.1994 V. Calculate the Na+ concentration of the original solution.

Is there a specific equation I can use so I can correctly solve for the concentration of the unknwon solution?

The little problem with the standard addition in potentiometry is, that the response to the addition is not linear, but logarithmic.

If we consider the simplified Nernst equation in context of parameters $A, B$,

and if we consider $c, c_\mathrm{0}$ as unknown concentration and  concentration increment, respectively,

we can follow this derivation:

$$\begin{align}
E_\mathrm{1} &= A + B \cdot \log c\\
E_\mathrm{2} &= A + B \cdot \log ( c + c_\mathrm{0} )\\
E_\mathrm{2}- E_\mathrm{1} &= B . \log( 1 + \frac {c_\mathrm{0}}{c} )\\
10^{\frac{E_\mathrm{2}- E_\mathrm{1}}{B}} &= 1 + \frac {c_\mathrm{0}}{c}\\
c &= \frac {c_\mathrm{0}}{10^{\frac{E_\mathrm{2}- E_\mathrm{1}}{B}} -1}
\end{align}$$

where $B = \frac{RT}{nF} \cdot \ln(10)$


Reaction between zinc and sulfur Would the reaction between zinc and sulfur be

$$\ce{Zn_{(s)} + S_{(s)} -> ZnS_{(s)}}$$
or
$$\ce{8 Zn_{(s)} + S_8\ _{(s)} -> 8ZnS_{(s)}}$$

I know that $\ce{S}$ and $\ce{S8}$ are allotropes of sulfur, but which one should I use for a reaction with sulfur at room temperature?

Either reaction is valid for representing the reaction between zinc and sulfur. The important thing is that both equations stipulate the same stoichiometry (one atom of zinc for one atom of sulfur). The mathematics of chemical equations work just the same whether you add structural information about the reactants or not; the reaction equation is just shorthand for a system of mathematical equations involving the amounts and types of atoms.

In truth, neither equation is a correct description of the reaction at the microscopic scale (though I again restate that's not really the point of a reaction equation). Free zinc atoms do not interact with free sulfur atoms or $\ce{S8}$ rings. Rather, after a little bit of initial heating to get the reaction going *, liquid sulfur (a very complex mixture of sulfur chains ranging from $\ce{S6}$ to $\ce{S_{n}}$, where n is as large as 100,000) is in contact with a surface of zinc (or possibly the molten metal), and electrons are exchanged by some mechanism which probably hasn't even been elucidated, as electron transfer reactions are very quick and hard to monitor.

* You mention considering the reaction at room temperature. Even though thermodynamically, the reaction between zinc and sulfur is spontaneous at room temperature, it is kinetically very slow because both reactants are solid. Even if you mix them very very well, the reaction is diffusion-limited in the solid mixture; the atoms can't "reach" each other very well. Heating a little bit creates a small amount of liquid sulfur, which can then mix better with the zinc.


Lucas test in the presence of EWG like -CN Does the given compound give Lucas test?

My teacher says due to the presence of a strong electron withdrawing group the carbocation formed is unstable.
But I say if it undergoes $\mathrm{S_N2}$ mechanism then an electron withdrawing group should increase its reactivity. Why then does this compound not give the test?

Lucas reagent follows SN1 pathway. $\ce{Zn}$ is a lewis acid that tends to bond with $\ce{-OH}$ and form a partial positive charge leading to the formation of a carbocation.
According to Hammond Postulate, the stability of the transition state should allow successive formation of product or intermediate. The stability of the positive charge induced on the carbon if stabilised in the transition state should therefore stabilise the carbocation. The rate determining step is the formation of the carbocation.
Since tertiary carbocations are the most stable the carbocation is formed quite easily and the product formation happens quickly.
Similarly, the primary and secondary alcohols, need to form a primary and secondary carbocation, which is not stable. Hence the reaction is slow. Zinc forms a weaker bond with $\ce{-OH}$ and induces a partial charge on them, thus not forming the whole carbocation.
So I think that in such cases where the carbocation is not stable, the reaction is slow, and may not have a a very well defined SN1 or SN2 classification i.e. truly borderline reaction may not occur, due to accompanying facets like solvation and ion pairs.


Outermost electronic configuration of f block elements Why is it $(n-2)\mathrm{f}^{1-14}(n-1)\mathrm{d}^{0-1}n\mathrm{s}^2$.

The only exception I thought to this formula was Thorium $\ce{5f^{0} 6d^{2} 7s^{2}}$.

So I feel it should be $(n-2)\mathrm{f}^{1-14}(n-1)\mathrm{d}^{0-2}n\mathrm{s}^2$.

As you said, if you look at the electron configurations of Lanthanides and Actinides, Thorium seems to be the only exception from this empirical rule. You could expand the rule to also include the case of Thorium but I would say it is more useful to have the rule describe the majority of cases and remember $\ce{Th}$ as an exception.


Relation between ionisation enthalpy and nature of bonds in boron My notes given by institute states that:

  Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form $+3$ ions and forces it to form only covalent compounds.

How is the covalent nature of bond of boron related to the sum of its first three ionization enthalpies?

EDIT :- I am directly or indirectly asking to prove fajan's rule for boron.

At first, let me give a picture of what ‘relatively high’ actually means. I present the sum of the first three ionisation enthalpies of some elements typically found as metal(III) ions:

Boron $6887.4~\mathrm{\frac{kJ}{mol}}$
Aluminium $5139~\mathrm{\frac{kJ}{mol}}$
Scandium $4256.7~\mathrm{\frac{kJ}{mol}}$
Iron $5281.4~\mathrm{\frac{kJ}{mol}}$
Cobalt $5640.4~\mathrm{\frac{kJ}{mol}}$
Ytterbium $4195.2~\mathrm{\frac{kJ}{mol}}$

We see that boron’s ionisation energy sum is $1.2~\mathrm{\frac{MJ}{mol}}$ larger than that of the next largest in my sample. Also keep in mind that boron is rather small per se so a $\ce{B^3+}$ ion must be even smaller.

But what does that actually mean? To generate an ionic compound from its elements, one usually considers an overall formation enthalpy composed of bringing both elements into the gas phase, turning them into ions and letting the crystal lattice form — this last step liberates most energy and is usually the driving force for salt creation. In boron’s case, the cations of such a salt would be $41~\mathrm{pm}$ — exceptionally small; way smaller than $\ce{Al^3+}$ ($68~\mathrm{pm}$) which is already tiny and smaller than $\ce{Be^2+}$ whose $59~\mathrm{pm}$ are typically cited as the smallest ionic radius known ($\ce{H+}$ doesn’t count).[1]

The structure of $\ce{BeF2}$ can already be considered as the greatest possible compromise in keeping the cations and — more importantly — the anions apart from themselves; hence its spacious structure. These problems would be multiplied if an even smaller cation were used here: It would be extremely hard to generate a crystal lattice that would allow for sufficient separation of the anions from each other since the cation is so tiny (and we need lots of anions: three negative charges per boron!).

So all things considered, ionic boron salts would have a very unstable crystal structure and would hence be hard to generate, especially since the lattice enthalpy cannot counteract the large ionisation enthalpy required for boron adequately. Luckily, covalent compounds are always a possibility and boron’s atomic orbitals are well shaped and well placed in energy to allow covalent bonding.

Note that boron’s electronegativity of $2.0$ on the Pauling scale also shows how badly boron would form ionic bonds: The tiny sizes of ions would draw electrons in in an even stronger way, potentially reducing boron before it would reach the ionic state.

The Polish chemist Kazimierz Fajans simplified this in his set of Fajans’ rules:

A compound will be covalent, if:

there is a high positive charge;
the anion is large; and/or
the cation is small.

A compound will be ionic, if:

there is a low positive charge;
the anion is small; and/or
the cation is large.

Note that in ambiguous cases such as $\ce{AlF3}$, (small cation, high positive charge but small anion) the rules fail to predict a mainly ionic compound, but correctly make us assume a rather high covalent character of said ionic bonds.

[1] Even beryllium forms rather covalent compounds most of the time. Due to the electronegativity difference of $2.5$ on the Pauling scale, we may just be able to consider solid $\ce{BeF2}$ as an ionic compound. All of these are borderline cases in one way or the other, though.


How can the equilibrium shift, while Kc remains constant? Consider the following reversible reaction.
$$\ce{Cr2O7^2-(aq) + H2O(l) <=> 2 CrO4^2-(aq) + 2 H+(aq)}$$
What will happen to the position of equilibrium and the value of $K_c$ when more $\ce{H+}$ ions are added at constant temperature?
$$
\begin{array}{lcc}
\hline
  & \text{Position of equilibrium} & \text{Value of}~K_c \\
\hline
\text{A.} & \text{shifts to the left} & \text{decreases} \\
\text{B.} & \text{shifts to the right} & \text{increases} \\
\text{C.} & \text{shifts to the right} & \text{does not change} \\
\text{D.} & \text{shifts to the left} & \text{does not change} \\
\hline
\end{array}
$$

The answer given is D.
I understand that when more $\ce{H+}$ ions are added, the reaction shifts in reverse. So I've understood that this is known as having the equilibrium shift to the left, toward the reactants.
What confuses me is why $K_c$ doesn't change. If the equilibrium shifts to the left, I suppose that must mean more reactants are formed in relation to the products, so $K_c$ would decrease.
But the answer key says that $K_c$ remains unchanged. This is coherent with what I've been taught, but I don't understand how the equilibrium can change yet have $K_c$ remain constant.
I'm just afraid I've got some of the theory mixed up, so I wanted to get some clarity. Could someone explain this please?

$K_c$ is related to the ratio of reactants to products at equilibrium.

If the reaction is currently at equilibrium, and you add more products then the reaction is now out of equilibrium and the reverse reaction will happen until it is back in equilibrium.

I don't like the wording of "equilibrium shifts to the left" myself, I would say that reverse reactions occurs to restore equilibrium. But $K_c$ doesn't change based on reactants/products concentrations since its the ratio at equilibrium. $K_c$ will normally depend on temperature though.

Based on the options you have, $K_c$ not changing and the reverse reaction occurring, the closest answer would be D. (If you take equilibrium shifting to the left to mean the reaction goes in that direction.)


Why do chiral biological molecules only exist as one enantiomer? Does it have any advantage? Why is it that chiral biological molecules are enantiomerically pure? The other enantiomer would have the same reactivity, and the only difference is their angle of rotation of plane polarized light. Why, then, is one enantiomer preferred over the other?

Is it that the enantiomer found in our bodies has some advantage over the other enantiomer, or is it random and luck that the structure we see naturally was chosen?

Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too?

Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too?

Many molecules exist in both forms in nature. One fun example are the enantiomeric terpenoids R-(–)-carvone and S-(+)-carvone. The R-form smells like spearmint while the S-form smells like caraway. The difference in smell shows that properties other than the optical activity are different for two enantiomers.

  Why, then, is one enantiomer preferred over the other?

Nucleic acids

The_Vinz stated correctly in the comments that for replicating structures like RNA, choice of the enatiomer was random; once established, one chiral form prevailed. DNA building blocks have the same chirality as those of RNA because they are made by the same biochemical pathway.

Amino acids

One amino acid, glycine, is not chiral. Many amino acids are found in both forms. L-amino acids are used to make proteins, but D-amino acids are made in bacteria and used in the context of cell walls and natural antibiotics.

Proteins

Proteins made by ribosomes (from amino acids attached to tRNA by tRNA-synthetases) use L-amino acids exclusively. The tRNA-synthetases are highly specific (including stereo specific), and they don't link tRNA to D-amino acids. It helps that there are very little D-amino acids made in a typical cell. Why one form was chosen over the other is probably luck again. Why all amino acids have the same chirality at the alpha carbon is more intriguing. Some are made from the same precursor, so that will contribute. Right-handed alpha helices require that the amino acids in them be L-amino acids. If proteins had a mixture of L- and D-amino acids (e.g. all alanines are D-alanines but all aspartates are L-aspartate), alpha helices would be more constrained in the possible sequences, and some of them would be left-handed.

All other molecules

Most steps in the synthesis of biomolecules are catalyzed by enzymes. Enzymes, as chiral catalysts that have lots of interactions with reactants, are often highly stereospecific. So the presence or absence of enzymes catalyzing certain reactions largely determines which products are made, and there is no additional cost of making a chiral product from non-chiral precursors (very different from a typical lab synthesis).


Presumptive test for modafinil? Given a sample that is probably modafinil plus common fillers, but could be fake, what reagent tests would best determine whether it contains modafinil? Assume a known-good sample for comparison.

(Modafinil is a sulfur-containing atypical stimulant which has become common in some circles, but which, due to off-label use and the high price of the US name brand, is mostly obtained by mail order from slightly-sketchy websites in India and the UK. There are rumors that some fraction of the modafinil imported this way is fake, but this is only rumor because the effects vary to begin with, and most people don't seem willing to pay for laboratories to do proper testing.)

Complex structure

Modafinil is a pretty complex molecule and judging by its structure (see below) I think it will be extremely difficult to find a single reagent test that will identify it.

Post-consumption identification

Interestingly it is much easier to find after consumption, either by the strong sulfur smell of your urine or through chromatography techniques (also here).

'Sketchy' tests

There are apparently some simple tests you can try like dissolving it in hot water and vinegar (which should happen without formation of gas), but these tests are far from fool-proof and might even cause you to think something is a fake while it is not.

Combination of physical/chemical tests as a 'solution'

The best thing to do if you really want 'kitchen-level' tests is to test for a whole range of chemical/physical properties. For example this reference has some information on the pH, water solubility and melting point which are fairly easy to test. Passing the tests will certainly not guarantee anything, but at least it will take away a significant part of the doubts that the product is fake.

Something to consider when designing these tests is that there will always be fillers (e.g. lactose monohydrate, microcrystalline cellulose, pregelatinized starch, croscarmellose sodium, povidone, and magnesium stearate) in the sample so you have to find ways to correct for that.


At what composition is Maximum deviation from Raoult's law observed? Raoult's law states that in an ideal binary solution of two volatile components, the partial vapour pressure of each component is proportional to its mole fraction in the solution.

A plot of total pressure to mole fraction of one of the components gives:

If A and B form a non-ideal solution with positive deviation, then the following graph is observed:

At what composition of the two components is the deviation (difference in total pressure) maximum?
My guess would be that maximum deviation is observed when mole fraction of both the components are equal.
My reasoning is that the deviation is observed because of the fact that interactions between A-A and B-B particles are not equal to the A-B particle interactions. When the quantity of both the components are equal, there is a maximum amount of A-B interactions. Hence, the deviation must be maximum.
Is this reasoning correct?

There's no definitive answer to your question. You might want to look up azeotropic mixtures.
Essentially, to make an azeotrope, you need a set of liquids with really high deviation from "ideal" behaviour, and at the composition where the deviation is maximum, the slope is zero. At that composition, there isn't much variation in vapor pressure as you vary the mole fractions, and so both components of the liquid boil equally, as stated here:

An azeotrope or a constant boiling point mixture is a mixture of two or more liquids whose proportions cannot be altered or changed by simple distillation.Because their composition is unchanged by distillation, azeotropes are also called (especially in older texts) constant boiling point mixtures.

Now, if you look at a few examples of azeotropes, it becomes clear that the maximum deviation is not achieved at the same composition ratio:

Sulfuric acid (98.3%) / water, boils at 338 °C
||Nitric acid (68%) / water, boils at 120.2 °C at 1 atm||
Perchloric acid (71.6%) / water, boils at 203 °C||
Hydrofluoric acid (35.6%) / water, boils at 111.35 °C

References: Azeotropes, wikipedia


Can there be more solvent in a solute? Something I know that leads to this question:

  Water dissolved in ethanol.

It might sounds weird to say water as a solute. Textbooks always says that solvent is the one present in larger quantity. This leads to a contradiction with the saying that water is an universal solvent.

There can be water present in smaller quantity than ethanol in a water-ethanol solution. But which is the solute and which is the solvent?

How to identify the solute and the solvent in a solution besides locating the word "aqueous"?

The textbook statement you gave ("solvent is always the one present in larger quantity") is simply a convention that makes it easier for us to talk to each other about mixtures of substances.  We have other words for other types of mixtures.  For example, in your ethanol-water example, you are correct in stating that it doesn't make much sense to talk about solute and solvent.  So, we call these two liquids "miscible".  They mix together in all proportions.  A good demonstration that captures the essence of different types of mixtures is the phase diagram of phenol-water:

Here we have a feature that is called a "miscibility gap".  There is a region, defined by the rainbow shape in the diagram above in which phenol and water separate into two distinct phases.  If we had pure water and added just a little bit of phenol to it at room temperature, we would find that it would dissolve into the water up to about 7 or 8 wt%. So, phenol would be the solute and water the solvent.  Adding more than this amount pushes the mixture into the miscibility gap so that now the phenol and water would mostly separate into two separate phases.  Let's say we add enough phenol to get us to 40 wt%.  If we heat the two phase mixture up to 67 C then it will suddenly become miscible.  Now only one phase exists and we can add as much phenol as we want to it without having to worry about two phases forming again.  There is no solute or solvent here.  Just a mixture of two liquids that exist in a single phase.


Which single organic molecule has strong absorbance throughout the visible range? Is there a single organic molecule that has strong absorbance throughout the visible range? For example a black dye that is based on a single molecule?

Quite a few dyestuffs will have strong absorption across the visible range of wavelengths, though few will have an entirely even absorbance. This won't matter much is they are concentrated enough and absorb most of the light at each wavelength.

Acid Black 1, for example, shows strong absorbance from the UV to just under 700 nm, which is most of the visible. Absorbance is stronger in the red but overall absorbance is large enough that it is sometimes used in black inks. See this paper for a sample absorbance spectrum:

There are quite a few other dyes with strong absorbance at most visible wavelengths and whether they appear black depends on the concentration.


Le Chatelier's Principle and Rate constants Consider the reaction: $$\ce{H2O2(g) <=> 2OH^.(g)}\label{a}\tag{1}$$ to be at at equilibrium ($\ce{OH^.}$ is a radical). As the pressure is increased, the reaction moves towards forming $\ce{H2O2}$ as this direction reduces the amount of substance in the system.

However, it is well-established that the forward rate constant of $\ref{a}$ increases with increase in pressure, meaning forming more $\ce{OH}$. This seems to be in conflict to what the Le Chatelier's principle suggests. Could someone help me get my head around this?

It is true that the forward reaction rate increases as you increase pressure. However, the reverse reaction rate will increase as well, and it will increase more than the forward reaction rate.

The key here is that the system is in equilibrium; the forward and backward reaction rates are equal. Altering the pressure affects both reactions.


Is a weak deactivator considered a stronger activator than a strong deactivator for the purposes of directing effects A nitro group is a strong deactivator whereas a halogen is a weak deactivator. The directing effects of these groups oppose each other: Nitro is meta directing whereas halogens are ortho/para directing. It is said that the strongest activator determines the directing effects in a disubstituted benzene ring. In a ring with no activators, such as one containing a halogen and a nitro group, is the weakest deactivator considered the "strongest activator", so that in the example I have just given the halogen would control the directing effects and said ring would be ortho/para directed?

I think you are reaching the limitations of directing group rules. These rules are not perfect and can often break.

In this example, chloride and nitro meta to each other, the group should add meta to the nitro group because the nitro group has a much more significant effect on the electronics of the molecule.


Adsorption and distribution chromatography difference? Well, yes, that is the question. What is the difference between the absorption chromatography and distribution chromatography?

Adsorption Chromatography = Utilizes a mobile liquid or gaseous phase that is adsorbed onto the surface of a stationary solid phase. The equilibriation between the mobile and stationary phase accounts for the separation of different solutes.

Distribution Chromatography (aka Liquid-Liquid extraction) = Due to its liquid stationary phase, the ingredients are only separated by their respective polarity, i.e. according to their distribution coefficient KD. Distribution chromatography is also ideal for the concentration of trace quantities. This is more closely related to Flash, Column, and Gas Chromatography.


When I place a straw in carbontated drink and blow, more gas bubbles rise to the surface? When I place a straw in carbontated drink and blow (making bubbling noises), more gas bubbles rise to the surface. The gas bubbles which rise to the surface are from the carbon dioxide gas dissolved in the drink, I'm not talking about the air bubbles I made with the straw?

Why does this happen?

When you blow into a solution, your warm breath (~98.6°F) warms the solution up. As a solution warms up, less $\ce{CO2}$ is soluble.  Therefore, blow into a solution, warm it up, less $\ce{CO2}$ is soluble, $\ce{CO2}$ escapes (bubbles out) from the warmer solution.

(image source)


Carbocation stability + determining position of attack on a carbocation in SN1 reaction It's implied that the right structure is the more stable carbocation because it undergoes the second half of an SN1 reaction, but I'm not entirely sure why:

I'm also not sure why the $\ce{MeOH}$ attacks the carbon next to the carbocation and not the oxygen itself, where it has the positive charge (doesn't the nucleophile attack the carbocation, or whichever atom bears the positive charge?)

It is important to note that the two structures shown aren't actually different. That's what resonance implies that in reality, it exists in between both of the as a 'hybrid'.
So it is not right to state that the nucleophile attacks on the left structure 'or' the right.

Coming to the carbon, (again, it's not right to take the structures to be different)

"I just don't know why the methanol attacks the $\ce{sp2}$ carbon in the right structure instead of the carbon with the $+1$ formal charge on the left structure"

It's actually the same carbon! and the hybridisation is still $\ce{sp2}$ no matter what, as the carbocation is also a planar species($\ce{sp2}$). And this is where the attack occurs.

"I'm also not sure why the MeOH attacks the carbon next to the carbocation and not the oxygen itself, where it has the positive charge"

'Carbon next to the carbocation' ? I believe you are confusing the +1 formal charge on oxygen to be a carbocation which is not true( Carbocation has the positive charge exclusively on Carbon) .
The attack is ocuring at the carbocation.

Our nucleophile here, is $\ce{OH-}$ while oxygen only has a +1 formal charge and also, the positive charge is more localised on the carbon as Zhe mentioned.
This is because oxygen being a more electronegative species, doesn't really 'like' the positive formal charge on it.
But still,

"doesn't the nucleophile attack the carbocation, or whichever atom bears the positive charge?"

Well, a nucleophile is a species that donates electrons at electron deficient sites. But oxygen still has it's lone pair! and the $\ce{OH-}$ will face repulsions from them if it tried to attack here.
The real electron deficient species here is only the carbocation and thus it is the preferred site for attack by the nucleophile.


Rate of reaction for hundred reactants I'd like to programm a toy chemical reaction program. Rough simplifications that deviate from industry and academia are totally acceptable. Not very educationaly accurate, but fun.

Are there methods to calculate rate of reaction for dozens and hundreds of reactants in a closed system?
Or I can blatantly use original r = k(T)[A][B] for every pair of reactants while selectively ignoring presence of other molecules?

Or should I use something else to evaluate what will happen with hundred of reactants after two seconds?

Sample gas composition below:

The questions has at least two aspects to address.

The first thing is, there is no guarantee that the reaction is first-order with respect to each of the reactants in first place. In other words, without further investigation you can not just take $r=k[\ce{A}][\ce{B}]$ because $r=k[\ce{A}][\ce{B}]^2$ or $r=[\ce{A}]^2[\ce{B}]$ or something more exotic may happen, depending on the nature of the reaction.

OK. So let's say for simplicity we take $r=k[\ce{A}][\ce{B}]$ when the system has only $\ce{A}$ and $\ce{B}$, as we don't need to be quite educationally correct. Then yes, after mixing many reactants, the equation still holds for each pair of the reactants, provided they are elementary reactions. It will be a completely different story if, for example, neither of two of the reactions are elementary and they share the same reaction intermediates.

I also have a reminder that when stepping from the expression of the rate to the expression of concentration with respect to time, you have to be careful and do the integration properly instead of just naively extrapolating the concentration expression when only two reactants exists. Let's say that we don't consider chemical equilibrium for the moment (i.e. consider all the reverse reactions to be prohibitively difficult to happen), then as far as I can see, the $r-t$ curve will not be smooth throughout the time and will have some turning point every time a reactant is fully consumed.

Side note: I suggest that you do your formatting properly in your program for the superscripts and subscripts.


Naming amitriptyline by IUPAC Can someone guide me through step by step how amitriptyline is named by IUPAC?
I know the name per se, but it is very confusing to me, and I'd like someone to explain it to me. The systematic name is

  3-(10,11-Dihydro-5H-dibenzo[a,d]cycloheptene-5-ylidene)-N,N-dimethylpropan-1-amine.

Note: As stated in a comment to the question by Loong, the preferred IUPAC name according to the 2013 nomenclature recommendations is:

  3-(10,11-dihydro-5H-dibenzo[a,d][7]annulen-5-ylidene)-N,N-dimethylpropan-1-amine

The only change this makes to the analysis it that we can ignore the ‘incorrect’ cycloheptene, because [7]annulen is automatically a cycloheptatriene-type molecule.

First, let’s separate the name into its fragments. We have:

3-something-propane-1-amine
N,N-dimethyl
(10,11-dihydro-something)
(something-5H-dibenzo[a,d]cycloheptene-something)
(something-5-ylidene)

I’m going to start with item 4, since that is the main bicycle. Cycloheptene sounds like this molecule:

But the [a,d] and the dibenzo is telling us that we have a fused polycyclic unsaturated hydrocarbon, so we will need to add two more double bonds. (Skipping a picture of this step.) Dibenzo means that there are two fused benzene rings, and the letters in square brackets symbolise the attachment points. a stands for 1,2; d for 4,5 (analogously for other letters). With that, we get a tricyclic compound with many double bonds. However, it is still ambiguous how exactly to place the double bonds, we need the information 5H — it tells us that the ‘additional’ hydrogen must be at the 5-position. Beware: The 5-position does not reference the initial numbering we used for fusing the rings; the atoms are renumbered according to the rules for aromatic polycycles: Starting at the top right and going clockwise, ignoring all bridgehead atoms. So we get 5H-dibenzo[a,d]cycloheptene which looks like this:

Next up, item 3: 10,11-dihydro. This means that the double bond in the seven-membered ring is hydrated, i.e. there is a single bond instead.

And finally, item 5: 5-ylidene. We are attaching something to carbon 5 via a double bond. We’ll get back to that.

Item 1: 3-something-propane-1-amine. That’s nice and simple, we’ll first consider propane-1-amine:

N,N-dimethyl (item 2) tells us that there are two methyl groups attached to the nitrogen, giving us the following:

And now we need to add the entire tricyclo-blurb to the propane’s 3-position using a double bond:

And voilà, there is amitriptylene!


Recovering iodine from activated carbon I have a sample of an iodine solution that has been dyed brown to mask the amount of iodine within.

I have added activated carbon to 100 mL of the sample and mixed for 3 hours to remove the iodine from the solution but now I'm having trouble recovering the iodine back into solution.

Based on Recovery of Iodine with Activated Carbon from Dilute Aqueous Solutions,Petrochemical Center of Excellence, Department of Chemical Engineering, Amirkabir University of Technology, No.424, Hafez Ave., Tehran, Iran , you should get good recovery of your iodine by heating in a solution of NaOH. According to the paper:

  The optimal desorption conditions obtained by the software are as
  follows: temperature of NaOH solution: $\mathrm{50^oC}$, concentration of NaOH
  solution: 0.5 % and agitation speed: 300rpm. Desorption experiments
  under these optimal conditions showed 98.9% desorption. The Taguchi
  design software results also showed that in the desorption section,
  the contributions of each factor on desorption percentage are as
  follows: concentration of NaOH solution (55%), temperature of NaOH
  solution (39.4%) and agitation speed (4.6%).
  [...]
  the necessary time for equilibrium desorption is obtained as 6h.

The paper does not discuss the algorithms used to calculate the relative importance of the different parameters, but obviously temperature and NaOH concentration are the important factors to getting such good recovery.


Two monoprotic acids vs. one diprotic acid My intuition says that the titration curve for titrating a $1:1$ mixture of the  monoprotic acids $\ce{HA_1}$ and $\ce{HA_2}$ (having ionization constants $K_a = x$ and $K_a = y$ respectively) is identical to the titration curve obtained by titrating a diprotic acid $\ce{H2X}$, with successive ionization constants $Ka_1 = x$ and $Ka_2 = y$.

I feel that this should be correct because the "concentration" and "strength" of the acids are the same, but I am not sure how to prove if it is correct.

Your intuition is right, because a potentiometric sensor (a $\mathrm{pH}$ probe in your case) does not know the source of hydrogren ions, it only senses the presence of hydrogen ions. The same is true for any other detection mode. All we care is $\ce{H+ + OH- -> H2O}$.

It can come from two monoprotic acids or it can come from a diprotic acid. In order to make your wish true, $\mathrm{p}K_\mathrm{a}$ must be less than $\mathrm{p}K_\mathrm{b}$. They cannot vary randomly and they cannot be equal.

However intuition is not enough to prove a point. You have to pick numbers and simulate. Here is useful online tool to calculate titration curve and see how well they match.

pH Titration Curve Generator

Choose to add $\ce{HA}$ or $\ce{H2A}$ and $\ce{NaOH}$ as a base.


How does the use of inert gas in fluorination of saturated aliphatic hydrocarbons make it less violent? While studying alkanes, I found this written in my book:

Fluoridation can be achieved without violence when alkane is treated with fluorine diluted with an inert gas, or by the action of inorganic fluorides like $\ce{AsF3}$, $\ce{SbF3}$, $\ce{AgF}$, $\ce{HgF2}$, $\ce{Hg2F2}$, etc., or by use of bromo or iodo derivatives.
The picture is given below for clarification:

Inert gases are chemically inactive, except xenon and radon, which are radioactive and may form some compounds.
How do the non-radioactive inert gases help in reducing the violence of fluorination? If I put a certain volume of fluorine and an alkane in a jar for reaction, then even if inert gases are present, they won't react. What is their role?

Let's say the gas mixture has something like 10:1 molecules ratio of inert gas : fluorine. The alkanes in the gas would have 10 times more chance of colliding with an inert gas molecule as compared to a fluorine molecule. This reduces the rate at which alkanes and fluorine react together, making the reaction a lot safer (fluorination is highly exothermic to the point where it can be explosive).

The inert gases are basically a necessity, without which it's impossible to do this reaction safely and responsibly.