Is there a systematic approach to determing structural formula from NMR spectra? I have not been able to find a methodologically consistent approach. If NMR spectroscopy is a rational analysis, it must not be beyond logical description in principle.

Here is what I have surmised so far:

Calculate degrees of unsaturation (I understand this)
Determine functional groups from IR spectra (I understand this)
Figure out chemical shift of each peak (I usually understand this)
Calculate integration of each peak (I usually understand this)
Determine number of neighboring protons (I usually understand this)
"Put the molecule together" (Variable results)

I "usually" understand steps 3-5, but sometimes I can't honestly figure out how many peaks I am looking at. How can I reliably tell the difference between lines within a peak due to coupling and distinct peaks when they are sometimes so close together?

Step 6 is difficult. Sometimes I can deduce an answer, and sometimes I am wondering how to tell the differences between hundreds of isomers. How can I systematically determine the right answer (or even be able to recognize IF there is single valid answer)?

Of course there is a systematic approach to solving structure. The key is that it is an iterative process, and it is not uncommon to have to repeat the cycle several times. For complicated molecular structures (for example, alkaloids with C35+ frameworks) you might get 20 or more possible candidate structures that you have to work through very closely.

Of course, the method you use will depend on the class of compound, but for unknown organic molecules, I would start with a molecular mass and NMR data from 1H, COSY, HSQC, HMBC. Subsequent iterations or refinements might require NOESY, TOCSY and if there is enough material maybe even an ADEQUATE.

The general approach that I teach is as follows. The first step is to identify the molecular formula from mass spec data and calculate the double bond equivalence. Subsequently:

1. Environments

From 1H and 13C data, determine number of carbon environments, and account for all carbon atoms, as these will represent the individual jigsaw pieces
Determine proton environments attached to 13C - tally up CH3, CH2, CH groups.
Tally up integrals, identify symmetry and equivalence.
This should account for almost all of your C and H atoms

2. Chemical shifts

Establish functional group types based on chemical shifts (aromatic, alkene etc).
Look for possibility of heteroatom inclusion that ties in with the molecular formula

3. Couplings and spin systems

From obvious 1D coupling and COSY data, start assembling spin system fragments
From HMBC start assembling C–C connectivities
Consider drawing out a correlation map to show connectivities, especially for complicated systems
You should be reasonably confident about the spin system fragments leading into the next stage

4. Structure hypothesis

Assemble spin system fragments into allowable possible structures, and test the structure against the data - that is rationalise the chemical shifts and couplings for all signals to see that they fit the model. If not, refine and try again.

A number of structure elucidation software packages exist, and they essentially base their method on building up a molecular correlation map, and then doing a chemical shift prediction for the proposed structure to give an overall confidence score. The sort of molecular map you might get looks something like the picture below. On the left are all the explained correlations, and on the right the incorrect correlations. What is important is look closely and the incorrect correlations, and rationalise them. In this case, all of the incorrect correlations would be allowable, and therefore we would have good confidence in this structure (which is understandable, as we know exactly what it is!). If we couldn't rationalise some of the correlations, we need to either reconsider our molecular structure, or re-analyse our data.


Why would there be differences in UV/Vis spectrum? $\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{MeCN}$ and $\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{H2O}$ have different UV/visible spectra in the visible region. How would they differ, for example would one of them have more peaks? Also how would the molar extinciton coefficients differ and what do these numbers tell you?

$\ce{H2O}$ does not absorb the radiation almost at all, so it will not interfere in the absorbance of the analyte we are analysing, thats why we use water almost always as a solvent in instrumental methods of analysis.Therefore $\ce{MeCN}$ wil absorb some of the radiation so then of course the absorbance of the analyte will also change ( decrease ) and the spectrum will be different ( the peaks of the analyte will be lower ).


Distinction between whole molecules and functional groups in IR spectra Are infrared spectra specific to compounds or to functional groups?

For example, which of the molecules below best fits the IR spectrum shown?

I'd have to say #2, which is a ketone based off of looking at a similar graph of a linear chain ketone. Is there a better way to approach IR graphs than learning them for functional groups?

Your question contains several points to address here.   And following Martin's suggestion, it is time to wrap up the points already mentioned in comments into a conclusive answer, too.

"IR spectrum graphs"-- there is nothing such.  The graph (or, the figure) represents the IR spectrum.  (And, for future reference, should there be need to deploy the plural form, it is "spectra".)
"Are IR spectra specific to compounds or to functional groups?"  If you record IR spectra in the typical interval of 4000 to 600 (or, depending on the window materials built in, about 500) 1/cm, then your IR spectrum will contain both: absorption bands that are characteristic for functional groups, like the classic ca. 1600 1/cm for carbonyls, or the ca. 2270 1/cm for the nitrile; AND between about 1400 to 500 1/cm what is called "fingerprint region" which contains absorption pattern that are characterisitic for the very molecule.
Attribution which absorption band in the fingerprint region belongs to what structural feature may be complicated enough that looking up a reference cataloge (by Sadtler, Aldrich, or other corporations) does not help, yet thankfully, modern computer chemical methods often allow a prediction of what vibrations may occur and where these should be observable in the IR spectrum. To render this fingerprint region more legible, often the $\tilde\nu$-scale is depicted with changes the scaling, like at 2200 and 1100 1/cm.
Which of the four compounds is more likely to be the one for the spectrum depicted?
 Well at first sight, 3 (1,4-dioxane) differs by the absence of a carbonyl group, typically yielding an intense absorption band around 1600 1/cm.  However, such an absorption band is recorded -- so it is either acetophenone (1), cyclohexanone (2), or $\delta$-valerolactone (4).
 There is not so much absorption in the region of 500--600 1/cm that were typical for aromatic rings, rendering 1 unlikely.
 Now, facing the distinction between either cyclohexanon 2, or the ester 4, it is indeed helpful that these compounds are literature known, already characterized by IR spectroscopy, and -- thankfully -- their data are made public; for example by AIST.  And as it turns out, it is likely cyclohexanone 2 which's spectrum was depicted here.

As a comment to the spectra by AIST:  They are searchable by chemical name, formula, or CAS number.  In the case of IR spectra, a little table ($\tilde\nu$), observed remnant transmittance) and a small image of the molecular structure is provided, too.

The excluded acetophenone 1:

The likely cyclohexanone 2:

The early exluded 1,4-dioxane 3:

The eventually excluded cyclic ester (lactone) 4:

So in turn, deciphering IR spectra is both learning / training to spot features (here: absorption bands), and comparison with reference data (here: spectra cataloges).  And typically supported by other means of analysis or characterization, like melting points / refractive indices compared with tables, or further NMR spectroscopy, for example.


Finding moles of water evolved How many moles of water are released on gentle heating of 1 mole of $\ce{Al2Cl6•12H2O}$?

Since the heating is gentle, the water molecules that are not present in inner orbital complex of compound and not directly attached to aluminium will be evolved, but the structure of $\ce{Al2Cl6}$ is a bridged tetrahedron. Then again, water is a better ligand than chlorine, so overall I cannot decide what the structure of the compound should be.

Anhydrous aluminium chloride ($\ce{AlCl3}$) exists; however, it cannot be obtained by heating of the hexahydrate ($\ce{AlCl3.6H2O}$).

$$\ce{AlCl3.6H2O \rlap{\ \ \not}->[\quad] AlCl3 + 6 H2O}$$

The crystalline hexahydrate $\ce{AlCl3.6H2O}$ actually is the complex salt $\ce{[Al(H2O)6]^3+ 3 Cl-}$.

The acidic hexaaquaaluminium(3+) ion is susceptible to hydrolysis.

$$\ce{[Al(H2O)6]^3+ <=> Al(OH)3 + 3 H3O+}$$

Therefore, the hexahydrate $\ce{AlCl3.6H2O}$ hydrolyses upon gentle heating and releases three equivalents of water and hydrogen chloride.

$$\ce{AlCl3.6H2O -> Al(OH)3 + 3 HCl + 3 H2O}$$


Why do SN1 and SN2 reactions not occur at sp2 centres? I was told in my organic chemistry course that $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ reactions did not occur at $\text{sp}^2$ centres. When I asked why, I was not given a satisfactory explanation. For $\text{S}_\text{N}2$ it was suggested that the reaction could not proceed with inversion of configuration, as this would disrupt the orbital overlap causing the $\pi$ bond. I couldn't find any explanation in Clayden's Organic Chemistry (2nd ed.), or even a mention of it.

Could anyone please explain why? I have attached some curly arrow diagrams that seem reasonable enough to me, although I realise that being able to draw a reaction mechanism does not mean it is valid.

Sometimes, especially in introductory courses the instructor will try to keep things "focused" in order to promote learning.  Still, it's unfortunate that the instructor couldn't respond in a more positive and stimulating way to your question.

These reactions do occur at $\ce{sp^2}$ hybridized carbon atoms, they are often just energetically more costly, and therefore somewhat less common. Consider when a nucleophile reacts with a carbonyl compound, the nucleophile attacks the carbonyl carbon atom in an $\ce{S_{N}2}$ manner.  The electrons in the C-O $\pi$–bond can be considered as the leaving group and a tetrahedral intermediate is formed with a negative charge on oxygen. It is harder to do this with a carbon-carbon double bond (energetically more costly) because you would wind up with a negative charge on carbon (instead of oxygen), which is energetically less desirable (because of the relative electronegativities of carbon and oxygen).

If you look at the Michael addition reaction, the 1,4-addition of a nucleophile to the carbon-carbon double bond in an $\ce{\alpha-\beta}$ unsaturated carbonyl system, this could be viewed as an $\ce{S_{N}2}$ attack on a carbon-carbon double bond, but again, it is favored (lower in energy) because you create an intermediate with a negative charge on oxygen.

$\ce{S_{N}1}$ reactions at $\ce{sp^2}$ carbon are well documented.  Solvolysis of vinyl halides in very acidic media is an example.  The resultant vinylic carbocations are actually stable enough to be observed using nmr spectroscopy.  The picture below helps explain why this reaction is so much more difficult (energetically more costly) than the more common solvolysis of an alkyl halide.  In the solvolysis of the alkyl halide we produce a traditional carbocation with an empty p orbital.  In the solvolysis of the vinyl halide we produce a carbocation with the positive charge residing in an $\ce{sp^2}$ orbital. Placing positive charge in an $\ce{sp^2}$ orbital is a higher energy situation compared to placing it in a p orbital (electrons prefer to be in orbitals with higher s density, it stabilizes them because the more s character in an orbital the lower its energy; conversely, in the absence of electrons, an orbital prefers to have high p character and mix the remaining s character into other bonding orbitals that do contain electrons in order to lower their energy).


Why does coumarin predominantly exist in keto form? By applying the logic that phenol exists in enol form rather than keto form as it attains aromatic character, why is it not the case in coumarin?

The answer is as simple as it will be unhelpful to your original question: There is no tautomer possible.

Coumarin does not have a keto group, and there are no enolisable protons either. Coumarin consists of a benzene ring, connected to a heterocycle with the functional group lactone. The molecule is flat and the conjugation extends throughout it.
What you have drawn are two resonance structures. Both of them are valid configurations (there are more), but neither actually exist; see What is resonance, and are resonance structures real? In both of your drawn structures the nuclei positions (in the clamped nuclei approximation) do not move.

The concept which applies here is completely different from keto-enol-tautomrie, where there are two different, observable molecules.

The question which you are probably asking is: Which of the drawn structures is the major contributor to the (so-called) resonance hybrid.

I have run a quick calculation of the DF-B97D3(BJ)/def2-SVP//GFN2-xTB level of theory, which suggests that the conjugation is skewed towards the lactone group. More investigations would be necessary, but the π orbitals should give you a general idea.


The anode and cathode when corrosion happens Let's say $\ce{Fe}$ reacts with $\ce{Cu^{2+}}$ ions. $\ce{Fe}$ would oxidize and therefore give electrons to $\ce{Cu^{2+}}$so that:

$$\ce{Fe-> Fe^{2+}  +2e-}$$
$$\ce{Cu^{2+} +2e^- ->Cu}$$
The overall reaction:

$$\ce{Cu^{2+} +Fe ->Cu +Fe^{2+}}$$

Now this is an example of corrosion, right? And when this type of corrosion happens, the anode is the electrode where oxidation happens and the cathode is the electrode where reduction happens, right?
Therefore iron ($\ce{Fe}$) is the anode where oxidation happens and Copper 2+ ions ($\ce{Cu^{2+}}$) is the cathode where the reduction happens.

My question is, is this all correct? Have I successfully described corrosion (of this type) and given an example of it? Or is there something wrong? I'm 15 and soon I'm having a test in chemistry, and because of my age there's no point in making it "super-advanced", but if there's something I missed or did wrong please point it out, I would be grateful for that!

You are right! What you have written is an example of Galvanic corrosion (Galvanic corrosion occurs when two different metals have physical or electrical contact with each other and are immersed in a common electrolyte).
You have  successfully described Galvanic corrosion and given an example of it


Reaction Rate Limiting Reaction Why exactly does the slowest reaction limit the instantaneous rate of a reaction even if the faster processes come after? Theoretically, if the products of the slow reaction are required for the fast reaction, the fast reaction limits the rate because even though it is fast, it will not finish in exact synchronization with the end of the slow reaction. There will be a lag between when the net reaction is done and when the slow reaction is done.

It is the measured overall reaction rate that counts and not which reaction finishes the sequence.
If you measure the rate of a reaction that consists of two elemental reactions with rate constants $k_1$ and $k_2$ where $k_1 \gg k_2$ you will get an overall rate constant $k_{\mathrm{tot}} \approx k_2$.
This is so, because if the faster reaction takes place after the slower reaction, the faster reaction uses up its reactants (which are the products of the slower reaction) faster than the slower reaction produces its products, so that the faster reaction will consume every new product molecule produced by the slower reaction nearly instantly compared to the time it takes the slower reaction to produce it.
The time it takes the faster reaction to consume the newly formed molecule is insignificant.


Is there such thing as too much fermentation When you make a wine the more you age it the better the taste. But is there such a thing as too much fermentation where it makes the wine taste worse? If not is there a time where fermentation doesn't really affect the taste anymore?

Continued fermentation of wine produces vinegar, which is not necessarily a bad thing. However, further fermentation may encourage growth of "vinegar eels" (harmless worms, actually, but not very appetizing).

Vinegar itself can be metabolized by microorganisms. According to the Compendium of the Microbiological Spoilage of Foods and Beverages, "several molds, yeast and bacteria are capable of spoiling vinegar... [such as the appropriately named] Lactobacillus acetotolerans."


What is an 'acidic' oxide? My book says that:

Oxides in higher oxidation states of elements of group 14 are generally more acidic than those in lower oxidation states.The dioxides $\ce{CO2}$, $\ce{SiO2}$ and $\ce{GeO2}$ are acidic whereas $\ce{SnO2}$ and $\ce{PbO2}$ are amphoteric in nature. Among monoxides, $\ce{CO}$ is neutral, $\ce{GeO}$ is distinctly acidic whereas $\ce{SnO}$ and $\ce{PbO}$ are amphoteric.

First and foremost:
How can an oxide be acidic? Is it because of their reactions with other compounds? I know about Arrhenius, Lewis and Bronsted acids and bases but still can't seem to connect!

Secondly, why does the book say "Oxides in higher oxidation states of elements of group 14 are generally more acidic than those in lower oxidation states"?  Is there a logical reason behind this?

Is the trend mentioned in the second question followed by all groups of the periodic table and not just for group 14?
By 'all groups', I am referring to s and p block elements. I don't know much about the chemistry of the d and f block elements - I will study them in a few weeks time, but I reckon they should also follow this trend.

Oxides are acidic or basic based on their reaction with a base or an acid.
Here, in group 14 elements, $\ce{CO2}$ is acidic ie; it reacts with a base. When the oxides are amphoteric, they react with both acids and bases.
Oxides of higher oxidation states are more acidic because of higher positive charge. On going down the group the ability to attract electron decreases and electropositive character increases. Therefore an element becomes less acidic and more basic down the group.
I did not understand your last question, but each period has a trend for their acidic and basic nature of oxides.


Methanol-water mix freezing points - clarify contradictory online data > 60% w/w? Extending an exploration of windshield washer-fluid choices to freezing points of methanol-water mixes, all web-sources located agreed to within a couple of degrees from 0-60% w/w, yet above this proportion (and below 100%), repeatedly came across conflicting data (which fell into three groups). (Yes, the disagreements are all outside the range of winter driving on this planet - except perhaps for around Vostok Station...)

w/w
various graphs by eye
various tables
engineering toolbox table
all˚C

60%
- 71
- 71
- 71
˚C

70%
-102
-137
- 82
˚C

80%
-116
-143
- 87
˚C

90%
-108
-146
- 90
˚C

100%
- 96
- 98
- 98
˚C

examples of sources for above data:
http://www.methanol.org/wp-content/uploads/2016/06/FreezingPointsMethanol-WaterSolutions.pdf    graph
https://patents.google.com/patent/WO2011153147A1/en table (~ 1/3 way into the document)
https://www.engineeringtoolbox.com/methanol-water-d_987.html        table
Thanks for any comment.

From Solid-Liquid Phase Diagram of the System Methanol-Water by G. A. Miller and D. K. Carpenter in J. Chem. Eng. Data 1964, 9, 3, 371–373 (July 1, 1964) [https://doi.org/10.1021/je60022a017]

"A solution of methanol and water that is methanol-rich tends to form
a glass when cooled below the melting point. The liquid is very
viscous at such temperatures (below -100° C.) because of the presence
of hydrogen-bonded polymers. Because it is difficult to induce
crystallization, the melting point is determined from a warming curve.
Much of the wide divergence of melting point data for the composition
range of 55 to 95 mole per cent* methanol results from the use of
cooling curves."

* by a back-of-a-napkin calculation (assuming 32.042 g/mol MeOH and 18.015 g/mol H2O) the mole percentage mentioned is approximately 68% to 97% methanol (w/w).
Further - successful - online research was inspired by a comment; it appears the failure to find information with common search terms was due to a language difference. It seems specialists (presumably physical chemists) use "solid-liquid equilibrium" or "solid-liquid phase diagram" in documents which may completely exclude the term "freezing-point" - such as the publication quoted above which answers the original question.


How to determine the percentage of iron(II) and iron(III) in a solution? Outline a plan of an experiment to determine the percentage of iron
  present as iron(III) in a solution containing $\ce{Fe^3+(aq)}$ and $\ce{Fe^2+(aq)}$
  ions. You are provided with zinc, a standard solution of potassium
  dichromate(VI) and dilute sulfuric acid. How you would use the zinc and how you would calculate the
  final answer.

So I know that the following equations will be involved

$\ce{2Fe^3+ + Zn -> 2Fe^2+ + Zn^2+}$
$\ce{6 Fe^2+ + Cr2O7^2- + 14 H+ → 6 Fe^3+ + 2 Cr^3+ + 7 H2O}$

I am very unsure on the method to use, I know that I can titrate a sample of the solution against $\ce{K2Cr2O7}$ to eliminate $\ce{Fe2+}$ ions, but I am clueless on how to proceed after.

You’re already very close. You have realised that:

You can titrate the $\ce{Fe^{+II}}$ content with a potassium dichromate standard solution; and
You can turn $\ce{Fe^{+III}}$ into $\ce{Fe^{+II}}$ by adding metallic zinc.

Note that in the first bullet point of mine, your second reaction, you are creating $\ce{Fe^{+III}}$, hence iron(III) that was present before you started won’t be modified. Similarly for the other reaction and iron(II). That should lead us to following methodology, hidden in a spoiler tag:

   1. Titrate the iron(II) content in an aliquot of your solution. That gives you the content of iron(II) and only iron(II).

 2. Take another aliquot and reduce all iron(III) to iron(II). That will give you the total iron content.

 3. Substract the value from 1. from the value from 2. to get the content of iron(III).


Calculation of the enthalpy change of reactions First, I’m a high school student and I wonder why every textbook just tells you how to set up those equations and cancel them to get the entropy change but never tells you why you can cancel the same species on both sides.

Second, for example,

$\ce{2B + 3H2O -> B2H6 + 3/2O2} \tag{  $\Delta H= 762~\mathrm{kJ}$}$

$\ce{H2 + 1/2O2 -> H2O} \tag{  $\Delta H=-286~\mathrm{kJ}$}$

Here the textbook tells us we should add up two reactions in order to get what we want (I didn’t post the whole question, just go with this)

Now, why do we have to add the two whole reactions up instead of just change the $\ce{H2O}$ on the left side into $\ce{H2 + 1/2O2}$?  What I understand about this $\ce{H2 + 1/2 O2 ->H2O}$ is if we reverse it, we can say if $\ce{H2O}$ takes in $286~\mathrm{kJ}$ of energy, it will turn into $\ce{H2}$ and $\ce{1/2 O2}$. So why we can’t just add $286~\mathrm{kJ}$ to the equation? Why we still need to add the whole thing $\ce{H2 + 1/2 O2 -> H2O}$ to the other equation?

You always need to add the equations since the break of bonds or the formation of bonds leads to an exchange in energy.
Here is a simple case

$$\begin{alignat}{2}
\ce{2 B + 3 H2O &-> B2H6 + 3/2 O2}\quad&&(\Delta H = 762\ \mathrm{kJ})\\
\ce{H2 + 1/2 O2 &-> H2O}\quad&&(\Delta H = −286\ \mathrm{kJ}) \times 3\\
\hline
\ce{3 H2 + $\boldsymbol{\ce{3/2 O2}}$ &-> $\boldsymbol{\ce{3H2O}}$}\quad&&(\Delta H = −858\ \mathrm{kJ})\\
\ce{2 B + $\boldsymbol{\ce{3 H2O}}$ &-> B2H6 + $\boldsymbol{\ce{3/2 O2}}$}\quad&&(\Delta H = 762\ \mathrm{kJ})
\end{alignat}$$

Notice the ones bold get canceled since the product is the reactant in the other equation.

$\Delta H_\mathrm{r}$ = sum of the $\Delta H$s so it’s equal to $-96\ \mathrm{kJ}$.

And in the second equation you were talking about one $\ce{H2O}$ molecule however in the first you are talking about three.


Do all double displacement reactions lead to precipitates? I know that most of the double displacement reactions form some kind of precipitates as their products. But do all of the double displacement reactions form precipitates? I have searched on this site as well as on Google, I've found answers, but they are not clarifying my doubts. Maybe I need a simpler explanation.

So, if you answer is yes, they all lead to precipitation, please justify.

If it's no, not all of them give precipitates, please justify with an example.

No, recall the definition of a precipitate "insoluble ionic compound". Take this reaction into consideration $$\ce{ NaOH~(s) + HCl~(l) <=> H2O~(l) + NaCl~(aq)}$$ none of the products are insoluble. You'll learn more when you go over gas evolution reactions.


Oxide and chloride of a metal I found this problem:

  The oxide of a metal contains 60.0% metal and the chloride of the same metal contains 25.26% metal (in mass percents). What are the possible formulas for the oxide and the chloride?

I looked up the solution and it states that the results are $\ce{MgO}$, $\ce{MgCl2}$, $\ce{TiO2}$ and $\ce{TiCl4}$, but I don't understand how they got it. How should I approach this kind of problem?

Because an oxygen ion is $\ce{O^2-}$ and a chlorine ion $\ce{Cl^-}$, you know that the salts have to be of the form $\ce{M_xO_y}$ and $\ce{M_xCl_{2y}}$.

Let $o$ be the atomic mass of oxygen, $c$ of chlorine and $m$ of the unknown metal M. The problem states that

$$0.6(xm+yo)=xm$$

which implies

$$0.6yo = 0.4xm \implies m = \frac{3y}{2x}o$$

If you recall some formulas for salts, you know that $x$ can only be 1 or 2 (e.g. $\ce{M_5O_3}$ makes no sense). This means that $m$ is a multiple of $\frac34o$ which is (approximately) 12. So we only need to check the elements with an atomic weight which is a multiple of 12:

12: Carbon (C) isn't a metal
24: Magnesium (Mg) is, and you can check the oxide/chloride satisfy the equations
36: n/a
48: Titanium (Ti) checks out
60: n/a
72: n/a
84: Krypton (Kr) isn't a metal

and so on. Even if the atomic weight corresponds to a metal, you can verify the value of $y$ doesn't match with the oxide that's formed.


Which gas among radon and hydrogen chloride has higher ideal behavior? Which one among the following shows maximum ideal behaviour?

$\ce{NH3}$
$\ce{HCl}$
$\ce{Rn}$
$\ce{H2O}$

Ammonia and water have hydrogen bonding so they're ruled out. That leaves radon and $\ce{HCl}$. Radon is a noble gas which means inertness but it's too big in size. That could mean greater van der Waal forces. And $\ce{HCl}$ on the other hand has covalent bonding.
I want to confirm which one of them is more ideal.

Some molecules are highly polar and thus are subject to electrostatic (dipole-dipole) interactions, which are typically far stronger than the interactions of induced dipoles (commonly referred to as van der Waals forces). See where this gets you.


Does "soluble in alcohol" imply ethanol? Sources like Wikipedia and SolubilityOfThings say that copper (II) acetate is soluble in alcohol.

Does that mean any alcohol, or does "alcohol" in this context mean ethanol?

The term "alcohol", without modification (i.e. "propyl" alcohol, etc.), in the context of something like a solubility table, almost always refers to ethanol (as it does in your sources for copper (II) acetate solubility).

Note, for example, that both pubchem and Wikipedia list alcohol as a synonym for ethanol. This is not the case for other alcohols like isopropanol.


Assigning E–Z configuration for alkenes with cyclic substituent groups I was taught that according to CIP rules, we must check the priority order for the substituents. But how can we do the same for molecules with cyclic groups as substituents?
How do we compare the substituents for the following molecule?

How to determine E–Z configuration for such molecules?

The compound is a tetra-substituted alkene with four different groups. Thus, it could be a (E)- or (Z)-ethene. To find the priority of the groups, you must follow CIP rules. The four groups are cyclobut-1-enyl, cyclobut-2-enyl, 2-methylcyclobutyl, and 1-methylcyclobutyl. Since it is an alkene (planner sp2–sp2 bond), we don't have to consider all four groups at once. What you have to do is consider two groups at once on each side of the double bond (say they are a and b):

On side a the first carbon of each cyclobutyl group (cyclobut-1-enyl and 1-methylcyclobutyl) has equal priority ({C,C,C} and {C,C,C}) as indicated in the image. However, the second carbon of cyclobut-1-enyl group has the priority {C,C,H}, while the second carbon of 1-methylcyclobutyl group has the lower priority {C,H,H} compared to that in cyclobut-1-enyl group. Therefore, between these two groups, cyclobut-1-enyl group has the higher priority.
On side b the first two carbons of each cyclobutyl group (cyclobut-2-enyl and 2-methylcyclobutyl) have equal priority: First carbons of both rings have {C,C,H} and the second carbons of both rings have {C,C,H} as well (see the image). Yet, the third carbon of cyclobut-2-enyl group has the priority {C,C,H} while the that of 2-methylcyclobutyl group has the lower priority {C,H,H} compared to cyclobut-1-enyl group. Therefore, between these two groups, the cyclobut-2-enyl group has a higher priority.
Now, you see higher priority groups of a and b sides are oriented on the same side of the double bond. Therefore, the compound is (Z)-isomer. The name is (Z)-1-(cyclobut-1-enyl)-2-(cyclobut-2-enyl)-1-(1-methylcyclobutyl)-2-(2-methylcyclobutyl)ethene.
Also, note that this molecule also has three chiral centers as indicated in the image. However, their stereochemistry is not given, so I neglect them.


How to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution? Carbon dioxide reacts with calcium hydroxide as shown by the equation:
  $$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$
  A solution containing $\pu{0.1 mol}$ of calcium hydroxide reacts with
  excess carbon dioxide at $\pu{298 K}$ and $\pu{100 kPa}$. Which statement about the reaction is correct?
  (a) It consumes $\pu{4.4 L}$ of carbon dioxide gas
  (b) It produces $\pu{24.5 L}$ of liquid water
  (c) It consumes $\pu{0.5 mol}$ of carbon dioxide
  (d) It produces $\pu{10.0 g}$ of solid calcium carbonate

I have assumed that if there are $\pu{0.1 mol}$ of calcium hydroxide would there also be $\pu{0.1 mol}$ of carbon dioxide (since they both have a stoichiometric coefficient of 1?), please correct me if I am wrong.

Under this assumption, I have calculated the volume of carbon dioxide gas using the equation:
$$n=\frac{V}{V_\mathrm{m}},$$
where $n = \pu{0.1 mol}$ and $V_\mathrm{m} = \pu{24.79 L/mol}$, as this is the volume  one mole of particles will be contained in at a temperature of $\pu{298 K}$ and a pressure of $\pu{100 kPa}$.

Solving for $V$, I got:
$$
  V(\ce{CO2}) = n \cdot V_\mathrm{m}
  = \pu{0.1 mol} \cdot \pu{24.79 L//mol}
  = \pu{2.5 L} \text{(1 dp)}$$

However, from the statements above, $\pu{2.5 L}$ of carbon dioxide gas is not an option. I am not sure whether I have calculated the volume wrong or if I am supposed to use another approach for this question.

(a) It consumes $\pu{4.4 L}$ of carbon dioxide gas

(wrong)

You've correctly calculated the volume of carbon dioxide $V(\ce{CO2})$ via the molar volume of ideal gas $V_\mathrm{m} = \pu{24.790 L mol-3}$ at SATP (Standard Ambient Temperature and Pressure):

$$V(\ce{CO2}) = nV_\mathrm{m} = \pu{0.1 mol}\cdot\pu{24.790 L mol-3} \approx \pu{2.5 L}$$

I personally wouldn't recommend to remember molar volumes at STP, NTP, SATP etc. as the numbers are close, but still different, and it brings up a lot of confusion.
Instead, only remember the ideal gas constant $R$, and remember it well as it's being used ubiquitously.
Pretending you forgot the molar volume for these conditions, you can use ideal gas law directly:

$$V(\ce{CO2}) = \frac{nRT}{p} = \frac{\pu{0.1 mol}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}}{\pu{1e5 Pa}} \approx\pu{2.5 L}$$

Here, keep in mind that $\pu{1 Pa} = \pu{1 J m-3}$ and $\pu{1 m3} = \pu{1e3 L}$.

  (b) It produces $\pu{24.5 L}$ of liquid water

(wrong)

Liquid water and SATP both imply that the density of water $ρ(\ce{H2O})\approx\pu{1 g mL-1}$, and its volume can be found via molecular mass $M(\ce{H2O})$:

$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{ρ(\ce{H2O})} = \frac{nM(\ce{H2O})}{ρ(\ce{H2O})} = \frac{\pu{0.1 mol}\cdot\pu{18.02 g mol-1}}{\pu{1 g mL-1}} \approx \pu{1.8 mL}$$

  (c) It consumes $\pu{0.5 mol}$ of carbon dioxide

(wrong)

The simplest out of four.
As you've already ruled out, there are equimolar amounts of both reactants and products according to the stoichiometry, so $n(\ce{CO2}) = n(\ce{Ca(OH)2} = \pu{0.1 mol})$:

$$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$

  (d) It produces $\pu{10.0 g}$ of solid calcium carbonate

(correct)

Mass of calcium carbonate $m(\ce{CaCO3})$ can be found trivially from its molar mass $M(\ce{CaCO3})$:

$$m(\ce{CaCO3}) = nM(\ce{CaCO3}) = \pu{0.1 mol}\cdot\pu{100.09 g mol-1} \approx \pu{10.0 g}$$


How does DeepHF (CRISPR sgRNA design tool) compute $\Delta G_{\text{binding}}$? I am looking at the code of DeepHF that computes $\Delta G_{\text{binding}}$, and I see that it breaks a sequence into overlapping 2-mers and compute a weighted sum according to a dictionary with 16 key values:
def dG_binding(seq):
    seq = seq.lower()
    dG = {'aa': -0.2, 'tt': -1, 'at': -0.9,
          'ta': -0.6, 'ca': -1.6, 'tg': -0.9,
          'ct': -1.8, 'ag': -0.9, 'ga': -1.5,
          'tc': -1.3, 'gt': -2.1, 'ac': -1.1,
          'cg': -1.7, 'gc': -2.7, 'gg': -2.1, 'cc': -2.9}

    seq = seq.replace( 'u', 't' )
    binding_dG = 0
    dGi = 3.1
    for i in range( 0, len( seq ) - 1 ):
        key = seq[i:i + 2]
        binding_dG += dG[key]
    binding_dG += dGi
    return binding_dG

Where do the values of each 2-mer come from? I couldn't find where these values in dG stem from.

The simplest model for estimating binding energies (or melting temperatures) is to consider each base pair individually, without regard to sequence context. This is the basis of the 4 deg / 2 deg rule of thumb for DNA duplex. You can refine this model by considering the effect of ionic strength. More sophisticated models go beyond just considering the composition to include sequence information.
The nearest-neighbor model for binding is the simplest of these more sophisticated models, and it looks like that's what is used here. For DNA:DNA duplexes, you can find  parameters on Wikipedia. The DNA:DNA model has less parameters because of symmetry lacking for the RNA:DNA hybrids (i.e. AG/CT is the same as TC/GA for the former, but AG/CU is different from TC/GA).
The code does not indicate the units, which might be kJ/mol or kcal/mol. The parameters are usually derived from experimental data and some fitting algorithm, see e.g. https://pubs.acs.org/doi/abs/10.1021/bi00035a029
Missing in this implementation is the contribution of the first and the last base pair.


Choice of Lewis Acid for EAS reaction Why is AlCl3 or FeCl3 might be preferred for chlorination reactions and AlBr3 or FeBr3 for bromination reactions? Why not use AlCl3 for bromination reactions?

Source: https://www.masterorganicchemistry.com/2011/07/22/reagent-friday-aluminum-chloride-alcl3/

Honestly speaking I think there won't be a difference as in clayden they have used $\ce{AlCl3}$ for bromination

If I had to give a reason it would be that when the $\ce{ AlCl3-Br2}$ complex is made there might be a teeny tiny chance that instead of bromine breaking away the chlorine  breaks away.

My basis of saying that the probability of chlorine breaking away is low is because there's a +tive charge on $\ce{Br}$ in the complex.Chlorine on the other hand is free from any such effects. There are no factors supporting the breakage of the $\ce{Al-Cl}$ bond. So the chances of $\ce{Br-Br}$ bond breaking are higher than the $\ce{Al-Cl}$ bond.

The same reasoning can be applied for $\ce{AlBr3}$ with $\ce{Cl2}$

That's why in Master Organic they've written preferred not necessary.


why do we call lone *pairs*? According to VSEPR theory, molecules adjust their shape to minimize the effect of repulsion.

Suppose I take the structure of $\ce{ClF3}$. It has three bond pairs and two lone pairs.
So to minimize the repulsion it gets T-shape (the two lone pairs are at corners of a trigonal bipyramidal shape).

But my doubt is why do we call two lone electrons a lone pair? I mean, if we keep a lone pair at some point in a molecule, how can two lone electrons be so near? Why don't we take electrons individually but always lone pairs?

We call them "pairs" because there are two electrons in each grouping. Why are there two instead of three or just one?

Spin and orbital energies.

One electron up and one electron down, in order to have all electrons in the lowest energy orbital available. Having them separate is a vastly higher energy state for the particles as a whole.

If you have not yet studied orbital theory - your question will be answered in much more depth when you do.


How to convert g/L to mol/L I know I should divide $\pu{g L-1}$ by the molar mass of the substance, but I don't seem to find the specific answer on Google. So just to be sure:
If I have $10^{-5}~\pu{g L-1}~\ce{Cu^2+}$ solution, do I have $1.57 \times 10^{-7}~\pu{mol L-1}$?
I think I'm missing something.

Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$
So $$c = \frac{m}{VM}$$

Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).


Raoult's law & Dalton's law of partial pressures I am confused between Raoult's law and Dalton's law of partial pressure as they state completely different things about same thing...

Raoult's law claims $p = P \cdot x$
where $p$ is the partial pressure of a gas, $P$ is the pressure of the same gas at the same temperature and volume and $x$ is the mole fraction.

Dalton's law says $p = p_\text{total} \cdot x$
where $p_\text{total}$ is the total pressure of the mixture.

In Raoult's Law, $p$ is the partial pressure of the vapor in equilibrium with some liquid which has vapor pressure of $P$. $x$ is the molar fraction of this liquid (it can be mixed with other liquid or solid.


Precipitation of Cu2+ and Pb2+ ions Hi chemistry enthusiasts!
I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge although the standard may be varied across the world. And I'm not quite sure what I did wrong. Could you please have a look?

  There is an aqueous solution containing Cu2+ and Pb2+ ions. The most
  suitable reagent to precipitate one of the two ions from the solution
  is (1) nitric acid (2) sodium carbonate (3) sulfuric acid (4) hydrogen
  sulfide

I picked answer (4) hydrogen sulfide because I think that it will react with copper (II) ions and will precipitate out. However, the correct answer is (3) sulfuric acid. How is this answer correct and what did I do wrong for hydrogen sulfide. Your advice will be much appreciated!

If you read the question carefully, they wish to precipitate only one ion, either copper or lead. The facts that must be in your mind are:

(i) Lead sulfate is insoluble in water but copper sulfate is.
(ii) Both lead sulfide and copper sulfide are water insoluble.


Tungsten 4+ ion magnetism? I just had a quiz and struggled on a particular question:

  Tungsten forms a $+4$ ion. It is observed to be diamagnetic. The electron configuration for this ion may not be what you would expect. Write the electron configuration for tungsten(IV) that is diamagnetic.

So if tungsten has a configuration of $\ce{[Xe] 6s^1 4f^{14} 5d^5}$ and it loses 4 electrons wouldn't it be $\ce{[Xe] 6s^0 4f^1^4 5d^2}$? And this would be paramagnetic since there are 2 unpaired electrons in the 5d subshell. How could it be diamagnetic?

The electron configuration for Tungsten is [Xe] 6s2 5d4 4f14. If it were to lose its electrons, and the answer is that it is diamagnetic, then it would lose its 4 d-electrons.

References

https://www.quora.com/What-are-the-differences-between-paramagnetic-and-diamagnetic-elements
https://www.tabulka.cz/english/elements/configuration.asp?id=74
https://socratic.org/questions/what-are-columbic-and-exchange-energy-and-how-it-it-determined


Oxidation number approximately I argued with my lecturer when he is saying that during the calculation of oxidation number, and your answer give you like 3.46 you can approximate it to whole number, I know tell him NO that if the answer is in decimal that means your calculations is wrong, he now tell me to go and research on my thought.
So please can anyone help me out

For complex or non stoichiometric ( many metal oxides ) compounds, it makes sense to calculate the average oxidation number that can be a rational number.
This can be useful if we are to enumerate chemical reactions with non-stoichiometric metal oxides, or if just a summary elemental composition is known, like hydrocarbon mixtures.
Oxidation numbers of particular atoms are integer numbers, with exceptions when   extra electrons ( or theirs deficit ) are shared among different number of atoms. The particular example of this are superoxides $\ce{M^{I}O2}$.


Zinc Picolinate - Positive, negative or neutral charge? I'm going to be using a finings agent with a positive charge to remove negative trub from a solution (beer).

I would like to keep some chelated zinc in the solution - ideally zinc picolinate - so wondering what its charge is? I'm hoping it's neutral...

Depending on the answer, are there any zinc chelates that carry a neutral charge?

It's all so zinc can remain in the beer after fining to reduce the estrogenic effects of hops.

Zinc picolinate with two ligands per metal ion ($\ce{Zn(C6H4NO2)2}$) is a neutral chelate complex. Deprotonated picolinic acid has one negative charge at the carboxylate group ($\ce{NC5H4-COO^-}$). Two picolinate ligands thus compensate the two positive charges of $\ce{Zn^2+}$, forming a neutral complex in which they act as bidentate ligands. The nitrogen of the pyridine ring and the carboxylate coordinate the Zn center ion.

Image source: http://www.lookchem.com/Zinc-picolinate/


How does KMnO₄ affect electrolysis? I'm just blindly experimenting. I tried to dissolve some potassium permanganate in water before performing electrolysis.

I can't see much difference. How is $\ce{KMnO4}$ dissolved in water? How does it react during electrolysis?

I added salt and the electrolysis was so violent that the copper started to burn (probably reacting with something).

A lot of questions have been posed on electrolysis and electrochemistry on this forum recently, with many recurring themes. I will explain electrolysis in depth in order for a true understanding to come out of this; if you're in a hurry then skip to the last three paragraphs.

One of the most important things about learning electrochemistry is the abstract property of voltage. Voltage can be considered a driving force, which denotes which direction charges are going to flow in a system. A high positive voltage means that charges most definitely wish to flow in that direction, and a high negative voltage means that the charges wish to flow strongly in the opposite direction. If we really want charges to flow in the direction of negative voltage, we must supply energy to the system in order for that to happen—in other words, we're supplying energy to move a charged particle against an electric field (work must therefore be done to do this, as with mass in a gravitational field).

And so it is with $\ce{H2O}$! The reaction corresponding to our notion of electrolysis is:

$$\ce{2H2O -> O2 + 2H2}$$

At least two of the intermediate species in electrolysis are going to be $\ce{OH-}$ and $\ce{H+}$, and so the energy requirement for this process takes into account the energy required to move an electron from $\ce{OH-}$ to the anode, and also to move an electron from the cathode to $\ce{H+}$.The voltage requirement for this process is  $\pu{-1.23V}$. In reality the voltage requirement is closer to  $\pu{-2V}$ taking into account overpotential (the electrochemical equivalent to activation energy).

So this negative voltage tells us that this process will not happen spontaneously, and we will need to supply a voltage of  $\pu{2V}$ in order for this to happen. So far so good: we never observe electrolysis in a standard beaker of water just sitting in the lab!

Your question asked what happened to the $\ce{KMnO4}$ during electrolysis. $\ce{Mn}$ is quite happy in a $\ce{Mn^7+}$ and $\ce{Mn^4+}$ oxidation state. Therefore in solution, the following reaction can happen:

$$\ce{Mn^7+ + 3e- -> Mn^4+}$$

Note that the above is a simplification, $\ce{Mn^7+}$ and $\ce{Mn^4+}$ don't occur simply as isolated ions in solution. In our case, these oxidation states of manganese are tied up in molecules:

$$\ce{MnO4^- -> MnO2}$$

But the above isn't balanced, so we ask ourselves what other likely molecules are going to be produced in order to balance our oxygens. It's common for these redox systems to produce $\ce{H2O}$, so we add $\ce{H2O}$ on the right to balance the oxygens, and add $\ce{H+}$ on the left so we have enough hydrogens. Thus:

$$\ce{MnO4^- + 4 H+ -> MnO2 + 2 H2O}$$

The voltage requirement for this reaction is $\pu{+1.68 V}$, so this means that the electron transfer happens spontaneously. In other words, this reaction is more likely to occur than the electrolysis of water at a less than neutral pH. If you start your electrolysis of water using distilled water (neutral pH implied) then the reduction of $\ce{Mn^7+}$ doesn't happen initially. It's only when the electrolysis starts and $\ce{H+}$ is generated that the reduction of $\ce{Mn^7+}$ can happen; this is then favoured over electrolysis on thermodynamic grounds.

As a last point, it's unlikely your copper electrode was burning, but these can get very hot if the voltage is high enough and hence melt. Turn the voltage down as only $\pu{2V}$ is needed for water electrolysis.

Experimental hint: try using $\ce{NaOH}$ and $\ce{KMnO4}$ in combination and see if there is any change in your observations. Then try increasing the amount of $\ce{NaOH}$ relative to $\ce{KMnO4}$ to see if you observe any changes which would suggest my theory. Note that other manganese reactions (pdf via The Internet Archive) can still happen at lower voltages, which does complicate things.

To answer your question directly, nothing additional is observed using $\ce{KMnO4}$ as it reacts instead of behaving as a charge carrier for electrolysis. A quality of a good electrolyte is that it does not interfere with your reaction of interest.


How to know if computational calculations are correct I've recently been modelling some transition states of a reaction I've been doing (read: struggling to do).

Whilst I have a lot of technical support in doing this (both step by step help and also academics who research in the field of computational chemistry), one thing that bugs me is the difficulty in knowing whether the calculation is meaningful.

In synthetic chemistry, I can do an experiment, then take an NMR/crystal structure, and be confident that i've got what I was intending to do. Whilst different bases may promote different reactions, its always possible to work out whats happened.

In computational chemistry, I do a calculation and get an answer, however depending on the methods used the answer might change.

This can be minimal: isomer A is favoured over isomer B, but the extent changes depending on the basis set/experiment. Or can be significant: isomer A and B change their order.

I was wondering if there is any way of telling whether a calculation is relevant. Obviously experiment can do this, but the reason I'm doing the calculation to begin with is to try and rationalise the difficulties in reaction .

Is there a concrete way to quantify the 'meaningfulness' of a computational result, or is it always a case of just fitting the computational data to experiment and then working backwards.

The only guaranteed way to achieve absolute correctness is to use asymptotically better methods and see the asymptotic behavior yourself. This, however, ofter requires prohibitively complex approaches.

Outside of that, one needs an experiment to check your data against. A good way (with limits) is to perform vibration frequency calculations and see if that fit the experiment. (making correction of unharmonic behavior). Another way is to run through similar well-studied cases and fit computational results. Most probably, a systematic error would be found, that can be easily eliminated. Luckily, most probably you are not the first to work with your particular kind of systems, so using google scholar it is usually possible to find a review or an establishing work that would list what work and what does not. Most practical methods have known 'glitches', and the ways to deal with them were considered. In particular, DFT cannot catch dispersion interactions, requiring empirical correction schemes.


Does increasing the number of interferometers increase signal in FT-IR? I'm trying to understand why most modern FT-IRs have three interferometers. Does this improve the signal in someway or does it have another effect?

I use FTIR in Physics, pretty much top-of the-range commercial instruments (actually more than just IR - ours go down to the UV and would do THz if we had the detectors; this isn't uncommon so I'll mainly refer to FT rather than FTIR).  There's definitely only one interferometer in each of our systems (and the ones used by our friends in Chemistry).  In astronomy FT spectroscopy has been used on the ground, but the best IR work these days is done from space using dispersive instruments (e.g. Spitzer's IRS).
The only common commercial use of three series-connected spectral devices (trying to coin a deliberately vague term) is dispersive instruments, which use three diffraction gratings for high resolution spectra and, for Raman spectroscopy, laser rejection with low stray light. A good benchtop FT instrument (with one interferometer) can give just as good resolution, though slowly and without the dynamic range. I mention this because if you were giving a lecture introducing a range of systems and techniques, and skipped a slide or two, you could jump from FTIR to triple dispersive systems and cause a lot of confusion.


How do I find the final concentration after an additional dilution? I took poor notes and didn't record what the final concentration of an amylase solution I  made last year. I recorded that I took $\rm 0.2\ g/10\ ml$ of the amylase solution, and then took $\rm 1\ ml$  of that and diluted to $35 ml$.

How would I solve for the final concentration?

Would I use :

$$c_1\ V_1 = c_2\ V_2$$

$$\rm(0.2\ g/10ml) \cdot (1\ ml) = (\mathit{c_2}) * ( \mathrm{35\ ml})$$

$$c_2\rm = \frac{0.02\ g}{35\ ml} $$

Therefore, I would take $\rm 0.02 g$ of amalyase powder and add up to $\rm 35\ ml$  with distilled $\ce{H2O}$ to achieve the final concentration that I would have previously had after dilution?

Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c_2 $.

Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.

e.g.
$$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$
$$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$
$$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$

The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).

The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).

Then I rearranged this to express the mass of solute needed to obtain this concentration.


pKa explanation through empirical formula? Suppose two compounds with, say, $\mathrm{p}K_{\mathrm{a}1}=2.8$ and $\mathrm{p}K_{\mathrm{a}2}=3.2$, and formulae $\ce{C_{$a$}H_{$b$}O_{$c$}}=\ce{C_{$p$}O_{$q$}(OH)_{$r$}}$ and $\ce{C_{$d$}H_{$e$}O_{$f$}}=\ce{C_{$u$}O_{$v$}(OH)_{$w$}}$, respectively, with given $abcdef$, $pqruvw$. Can we explain the above $\mathrm{p}K_\mathrm{a}$ with (ONLY) the knowledge of the given numbers of atoms of $\ce{C}$, $\ce{H}$ and $\ce{O}$ using the theory of Usanovich, and no knowledge of the internal chemical structure of the chemical bonds inside the two molecules?

Can we explain the above pka with (ONLY) the knowledge of the given
  numbers of atoms of C, H and O using the theory of Usanovich

No, nobody's theory (including Usanovich's) can explain $\ce{pK_{a}}$ differences based on molecular formula coefficients.  This is best illustrated by the observation that even compounds with the same coefficients (same values for "abcdef") can have different $\ce{pK_{a}}$ values.  For example, $\ce{C3H6O}$

\begin{aligned}
\ce{&acetone &&pK_{a}=20}\\
\ce{&allyl~ alcohol &&pK_{a}=$15.5$}\\
\end{aligned}


Determine if stuff is Potassium Alum How can one determine (at home) if the product shown below is Potassium Alum? I held a piece of it in a gas flame (assuming that the flame would become violet if potassium were present), but there was no change in the color of the flame. [Got product for use as astringent and antiseptic after-shave -- and people say (do not know why) that for such use, it has to be potassium-(not something else)-alum.]

Update: As mentioned in the comments to the chosen answer (by @James Gaidis), the product is ammonia alum. Just to be double sure, I repeated the tests with McCormick's Alum powder whose ingredient is potassium alum. (1) The picture of the flame test is below.

(2) With the McCormick Alum, the test involving sodium bicarbonate resulted in sound and definitely no ammonia gas.  (3) The solution of Fatkari had a salty taste that became sour in a second or so; the K-alum was tasteless and became sour in a second or so.

From Wikipedia:

An alum is a type of chemical compound, usually a hydrated double sulfate salt of aluminum with the general formula $\ce{XAl(SO4)2 · 12 H2O},$ where $\ce{X}$ is a monovalent cation such as potassium or ammonium.

There are also chromium alums and ferric alums, but the color in your sample does not seem to suggest either of them.
The alums melt at relatively low temperatures, because they already have a lot of water in the crystal structure, so ammonium alum and potassium alum can not be differentiated on the basis of melting points. Solubilities are fairly close, too:
$$
\begin{array}{llc}
\hline
\text{Compound} & \text{mp}/\pu{°C} & \text{Solubility}/\% \\
\hline
\ce{(NH4)Al(SO4)2 · 12 H2O} & 93 & 15 \\
\ce{KAl(SO4)2 · 12 H2O} & 92.5 & 11 \\
\hline
\end{array}
$$
If there is no violet color to a flame, there may be no potassium, but the test can be sensitive and may be indeterminate. Eliminate the possibility that the alum is an ammonium salt by dissolving some in water and adding a strong alkali, then sniffing it gently. If you smell ammonia, you have an ammonia alum; it will still work as a flocculant for muddy water, but is different chemically. A strong alkali readily available at home, well, at the grocery store anyway, is sodium carbonate (washing soda).


NaOH solution not stable $\ce{NaOH}$ solution can react with $\ce{CO2}$ to form sodium carbonate.
Is there any other reason why $\ce{NaOH}$ solutions for accurate titrations cannot be stored for long (or the titre has to be updated every time). Wouldn't filling the vessel with argon fix the problem.

Your questions has two a two part answer.

(1) How would you get the $\ce{NaOH}$ out of a sealed bottle?

(2) A concentrated stock solution can be made for $\ce{NaOH}$. In the top of the bottle you use a drying tube stuffed with solid $\ce{NaOH}$ to remove $\ce{CO2}$ from air. On the bottom you have an outlet from which you pour the $\ce{NaOH}$. The gist is that $\ce{Na2CO3}$ is the least soluble carbonate. So most of the $\ce{CO3^2-}$ in the original solution participates. So if you let the solution sit for a couple of days it will be very stable.

The point with the stock solution as outlined is that $\ce{Na2CO3}$ "sticks" to inside of the container and doesn't flow out with solution. (You don't shake bottle up like you do with tomato juice that you drink to resuspend the solids.)

So for (1) it is high near impossible to make a $\ce{NaOH}$ solution without some $\ce{CO3^2-}$ in it. The $\ce{NaOH}$ itself has $\ce{CO3^2-}$ unless it is specially made, sealed in argon and so on.

So your stock solution is concentrated. I boil some distilled water to get rid of $\ce{CO2}$ and seal it till it cools. I then take $10~\mathrm{ml}$ of my stock $\ce{NaOH}$ solution and dilute to $250~\mathrm{ml}$ (or whatever ratios) to get the desired concentration of $\ce{NaOH}$. (That I can check with KHP titration.)


How can you estimate how much of a substance will dissolve in a given lipid? How can you estimate how much of a substance with a known partition coefficient and other physical/chemical data regarding the substance will dissolve in a given lipid? In particular I would like to know the answer to this question in the context of the blood-brain barrier (BBB).

The major issue with the blood-brain barrier is that aside from small hydrophobic/lipophilic compounds (including $\ce{O_{2}}$ and $\ce{CO_{2}}$), almost everything else is prevented from passing due to cellular tight junctions and will only (potentially) pass through via active transport.

However, for small lipophilic molecules passing through a phospholipid bilayer (such as on a cell surface in the blood-brain barrier), this post provides information on equations: http://www.ncbi.nlm.nih.gov/books/NBK21626/

Where:

$\frac{dn}{dt}$ is diffusion rate (mol/s)

$K$ is the partition coefficient (the equilibrium constant for its partition between oil and water) - e.g. for Urea, K = 0.0002 while for diethylurea which is more hydrophobic, K = 0.01

$C^{m}$ is the concentration in the hydrophobic part of the phosopholipid bilayer (mol/L)

$C^{aq}$ is the concentration in aqueous solution (mol/L)

$P$ is permeability coefficient

$D$ is diffusion coefficient of the substance within the phospholipid membrane

$A$ is the area of the membrane (in $m^{2}$ I think?)

$x$ is the membrane thickness (in m I think?)

$C_{1}^{aq}$ and $C_{2}^{aq}$ are the aqueous concentrations on either side of the layer (one will be the serum or CSF concentration, the other will be the intracellular concentration)

$K = \frac{C^{m}}{C^{aq}}$
$\frac{dn}{dt}=PA(C_{1}^{aq}-C_{2}^{aq})$ and $P = \frac{KD}{x}$
$\frac{dn}{dt}=A\frac{KD}{x}(C_{1}^{aq}-C_{2}^{aq})$

You would have a similar equation to 2. on the other side of the cells lining the blood-brain barrier.

It seems from my reading that $P$ tends to be better known than $D$, which is actually usually calculated using equation 2. from $P$, $K$ and $x$.

Hope that helps!


Why does Gibbs free energy correspond solely to non-expansion work? First of all: I know this question has been asked before (Why does the Gibbs free energy only correspond to non-expansion work?), and although I understand the mathematical derivation given there, I struggle with the concept.

If both $\Delta G$ is negative and the change in enthalpy is negative (exothermic), I don't see why we cannot use that heat to do mechanical work? So how is it that it only cooresponds to non-mechanical work?

For a closed system in contact with a constant temperature reservoir held at the initial system temperature T, and in contact with a surroundings at an external pressure held at the initial system pressure P, and experiencing a change between two thermodynamic equilibrium states, the change in internal energy is given by:  $$\Delta U=Q-W$$If we express the work W as the sum of expansion work and non-expansion work (the latter, say, obtained by generating electricity or turning a shaft), then $$W=P\Delta V+W_{NE}$$And, because there is a single constant temperature bath comprising the surroundings, the heat added to the system is related to the change in entropy between the initial and final thermodynamic equilibrium states of the system by:
$$\Delta S=\frac{Q}{T}+S_{gen}$$where $S_{gen}$ is the entropy generated as a result of any irreversibility's that may be present.  So, if we combine these equations, we obtain:$$W_{NE}=-\Delta G-TS_{gen}$$where $\Delta G=\Delta U-T\Delta S+P\Delta V$.  So, for any process that goes between the same two thermodynamic equilibrium  states, the maximum non-expansion work that the system can do on its surroundings is $-\Delta G$ (since $S_{gen}$ can only be positive).  The maximum corresponds to a reversible path between the two end states.


Soluble compounds to absorb radio-frequency electromagnetic radiation Are there any soluble compounds capable of absorbing radio-frequency or long microwave  radio-frequency electromagnetic radiation? I have noted that metals are often used to absorb these signals, but I cannot find any absorption spectra for radio/long microwave wavelengths.

Of course! Salt water is effective, because it is an ionic solution. Submarines and SCUBA divers cannot use radio-frequency (RF) communication over any distance because of that issue. Very high power, very-low-frequency (VLF) land-based transmitters can penetrate to perhaps 20 meters depth to subs, but salt water is essentially opaque to RF and microwaves.


Is there any chemical preservative better than sodium benzoate? I'm developing a fruit juice for humans, I just want a second opinion about allowed food preservatives. Does anyone happen to know if there are any chemical preservatives that can extend the shelf life of fruit juice better than sodium benzoate?

I just want a second opinion about allowed food preservatives.

Allowed is crucial here and it might strongly depend of where the product is to be sold. You might want to have a look at national legislations.

For Germany, the Zusatzstoff-Zulassungsverordnung lists allowed preservatives and their maximum concentrations in different products.
For Canada, Health Canada provides a List of Permitted Preservatives.


Calculate pressure equilibrium constant from concentration equilibrium constant $$\ce{N2(g) + 3H2 (g) -> 2NH3(g)}$$
  Calculate $K_\mathrm p$ from $K_\mathrm c$, given that $K_\mathrm c = \pu{4.2 M^-2}$
  and that there is 0.2 moles of $\ce{N2}$, 0.6 moles of $\ce{H2}$ and 0.425 moles of $\ce{NH3}$. Temperature = $\pu{600K}$

My attempt at the question:

$K_\mathrm p= K_\mathrm c(RT)^{\Delta n}$

$K_\mathrm p= 4.2(8.314\times600)^{-2}$

$K_\mathrm p= \pu{1.69\times10^{-7} Pa^-2}$

However, in our answer sheet the value of $K_\mathrm c$ is divided by $10^6$ and the answer is $\pu{1.69x10^-13 Pa^-2}$. Could someone please explain why $K_\mathrm c$ is divded by $10^6$?

If you intend to get an answer in Pascal, which is an SI unit, you need to have your original values in SI units as well.

Note that:

$$K_\mathrm c=\pu{4.2M^-2}=4.2~\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}=4.2~\left(\frac{\text{moles}}{\pu{10^-3m^3}}\right)^{-2}$$

The conversion factor for litre (non-SI) to metre cubed (SI unit) is $10^{-3}$. Since you have $K_\mathrm c$ in units of $\pu{M^-2}$, the final answer would be $10^{-6}$ of the answer you had arrived at.

Elaboration:

Let's do dimensional analysis on the RHS factor-by-factor.

$K_\mathrm c$ has the units $\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}$ as shown above.
$RT$ has the units = $\pu{Jmol^-1T^-1}\cdot\pu{T}=\pu{J\cdot mol^-1}$
$(RT)^{-2}$ has the units = $\pu{J^-2\cdot mol^2}$.
Since one joule $=\pu{Pa\cdot m^3}$, hence $(RT)^{-2}$ has the units = $\pu{Pa^-2\cdot m^-6\cdot mol^2}$
Multiply $K_\mathrm c$ and $(RT)^{-2}$ we get
$$\text{Units of }K_\mathrm p=\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}\cdot\pu{Pa^-2\cdot m^-6\cdot mol^2}=\pu{Pa^-2\cdot m^-6\cdot litre^2}$$

Thus your final answer is not in the units of $\pu{Pa^-2}$ but rather $\pu{Pa^-2\cdot m^-6\cdot litre^2}$. Can you now see why we need the conversion factor?


Is there a means to measure pH without indicators? I have created a solution that should contain hydroxide ions. Therefore, the pH > 7, but is there any way to roughly estimate the pH in absence of any indicator? Maybe using current, or checking whether it dissolves certain chemicals?

Please mind that I am limited in equipment and so complex science is not feasible for me.

Have you tried with red or black tea? It turns lighter in acidic media and turns dark in basic media. I tried once adding the pulp juice of one lemon to a cup of tea and it lighted up (mainly for acetic acid), then I added NaOH-based salt and turned very black and opaque.

Maybe that suits for your solution.


Can protons be swapped in a chemical reaction? Can the protons of two elements be swapped in a chemical reaction?

Roughly speaking, no.   This would amount to nuclear fission/fusion, and is generally not considered "chemistry" in any colloquial sense of the term, but rather physics.

In general, chemical reactions consider nuclei to be tightly bound by the strong force, and immutable over the course of the reaction.

Disclaimer: there is an ill-defined border between physics and chemistry, and so one could argue that I am wrong about this.  Indeed, alchemists had/have been trying to e.g. convert lead to gold for centuries, and this is related to the reaction that you propose.


What metal is under the stainless layer when it wears of? I have this belt for several years now and the belt buckle is naturally subject to wear and tear.

On the most exposed spots the outer layer is now completely worn of and I wonder

Is this an example of passivation of the inner material?
What is the outer layer made of? Stainless steel?
What is the metal under the outer layer? It looks (and smells) like copper but why would they make it out of copper? Copper is – as far as I know – more expensive (by volume comparison, because the belt buckle can't be made smaller) than stainless steel?
Wouldn't it be cheaper to make it only from stainless steel (assuming the outer layer is from stainless steel)? It would also be better from health perspective because I suspect the inner layer causes skin irritation.

I intentionally scrubbed the bottom edge of the buckle before taking the photo to reveal the real color of the metal, on the top edge it is presumably covered by its natural oxide layer (copper has it green, right? But it's just a thin layer so it can look different I suppose).

I am kind of ashamed answers to these questions aren't already known to me because this effect happens on a lot of daily used metal objects...

1 =no , 2= no, 3 Likely brass , 4= no. The buckle is likely a brass although the bottom edge where you polished it is copper color . That could result if you used some cleaning agent. It could be made from copper but modern mass production would use brass. It was plated with nickle and / or chromium. Nickel plating is usually put down first , under chromium plating.


Removing unreacted Mg metal in a Grignard reaction A Grignard reagant was prepared from an alkyl halide and magnesium chips, then reacted with a ketone. The resulting solution has been stirring for a few days at low temperature (0-5°C). It's now time to transfer the reaction solution slowly to a vessel containing an acid quench. My problem is the unreacted magnesium chips (smaller than when we started, of course) have settled, and are plugging the tubing and stopper valve.

What solvents could I use to flush the magnesium through, breaking it up, and possibly dissolve it? This can't be added to the reaction solution (the Drain-O method), but only through the tubing itself. Obviously water is a bad idea. Any insights?

Pure water would certainly be a bad idea, but a 1% solution of 1 M hydrochloric acid in ether would probably work without a vigorous reaction.


Cp vs Cv for liquid water @ < 4 °C I'm looking at a textbook question (Engel & Reid, Thermodynamics, Statistical Thermodynamics & Kinetics, 4 ed, Q3.1) which states the following:

  The heat capacity $C_{\mathrm m,p}$ is less than $C_{\mathrm m,V}$ for $\ce{H2O(l)}$ near $\pu{4 °C}$. Explain this result.

Their answer is that water contracts as you heat it in this regime, thus (at constant $p$) the surroundings are doing work on the system, hence $C_{\mathrm m,p}\lt C_{\mathrm m,V}$.

But the textbook also contains the following equation (which is correct, and whose only restrictions are that the system is of fixed phase and composition, and $\text{đ}w = –pdV$, i.e., pV-work only):

$C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \beta^2/\kappa$

where $\beta$ is the isobaric thermal expansivity, and $\kappa$ is the isothermal compressibility.

Based on this, for a substance of fixed phase and composition, $C_{\mathrm m,p} \ge C_{\mathrm m,V}$, always, because, while $\beta$ can be $\le0$* (water near $\pu{4 °C}$ being a notable example), $\beta^2$, as well as $T, V_\mathrm m$ and  $\kappa$, are always positive. [*Edited from  $\lt 0$ to $\le 0$ based on Night Writer's answer.]

So is the statement in the textbook's question just a flat-out mistake, or am I missing something here?

It appears their error is in assuming the difference between Cp and Cv is due to pV expansion alone, when in fact there are two terms:

$$C_p - C_V = p \left(\partial V\over\partial T\right)_p+ \left(\partial U\over\partial V\right)_T \left(\partial V\over\partial T\right)_p,$$

where the first term on the RHS is the PV work per unit change in T, while the second term is the change in internal energy with respect to volume (which results from changing the intermolecular distance between interacting particles) times the rate at which the volume changes with temperature.

It would be nice to have direct reference-quality experimental values for $C_{\mathrm m,p}$ and $C_{\mathrm m,V}$ for $\ce{H2O(l)}$ near $\pu{4 °C}$, but I've not been able to locate them, nor do I expect to: because of the difficulty of accurately measuring $C_{\mathrm m,V}$ for a liquid, measurements are typically made at constant pressure, giving $C_{\mathrm m,p}$, and then $C_{\mathrm m,V}$ is calculated using the above equation.

Repeating your line of reasoning, if $C_{\mathrm m,p}<C_{\mathrm m,V}$ you would conclude based on the expression you provide ($C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \alpha^2/\kappa$) that $\kappa<0$ which in turn would mean that

$$\left(\frac{\partial V_m}{\partial P}\right)_T > 0,\label{eqn:1}\tag{1}$$

which is not physically reasonable except presumably in some very unusual circumstances. Therefore, either the question statement contains a fallacy, or the above analysis contains a fallacy, i.e. $\eqref{eqn:1}$ is reasonable. My bet is on the first scenario, as you suggest.

What seems to happen is that near $\pu{4 °C}$

$$C_{\mathrm m,p} = C_{\mathrm m,V}\label{eqn:2}\tag{2}$$

because the change in the density or molar volume with $T$ (and therefore also the thermal expansion coefficient $\beta$ - usually designated with the symbol $\alpha$) becomes zero, that is there is a (well known) minimum in the dependence of the density or molar volume on $T$ at constant $P$, illustrated in the following figure (source: wikipedia)

In agreement with this, if you inspect available heat capacity data you will see that the isochoric and isobaric heat capacities converge as the temperature of liquid water nears $\pu{0 °C}$.

The International Association for the Properties of Water and Steam has released a document (IAPWS R7-97(2012)) containing expressions for the heat capacities of water parameterized based on experimental data (see Table 3).


Nitrogen triodide energetics Ammonia has a heat of formation of $-46 \:\mathrm{kJ/mol}$ (which means that it's formation from the elements releases heat). When mixed with iodine at room temperature, it makes nitrogen triodide (heat of formation +146 complexed with ammonia, +287 pure). With the slightest touch it decomposes explosively back into nitrogen and iodine. Thus the net reaction for both steps seems to be ammonia decomposing into its elements.

Iodine seems to be catalyzing its decomposition. However, such a highly endothermic reaction seems unlikely (evaporating water is only $40.65 \:\mathrm{kJ/mol}$). Also, ammonia's formation is spontaneous ($-26 \:\mathrm{kJ/mol}$) despite the entropy cost with $1 \:\mathrm{atm}$ reactants. This shouldn't go backwards.

So where do we get the energy to make this unstable compound? Iodine is a weak halogen, it barely reacts with hydrogen. Could oxygen drag this reaction forward by stealing hydrogen from ammonia?

I'm not sure what your question is. The reaction from NI3 is largely exothermic (Wikipedia gives –290 kJ/mol as the reaction enthalpy), which is consistent with its explosiveness. The reaction from NI3 · NH3 is also exothermic: you quote the heat of formation of the NI3 · NH3 as being +146 kJ/mol, and the heat of formation of product NH4I is –202 kJ/mol. Both are largely favored entropically because they release gas.


Aromaticity of Caffeine In the Caffeine molecule, is aromaticity confined only to the 5 membered ring? The entire molecule is planar and the 6 membered ring also seems to have 6π electrons.
The entire molecule has 10π electrons.(inside)

No. The 6 membered ring is also a part of its aromaticity.
This can be better understood by considering the Zwitterion of Caffeine.

The entire molecule is planar, conjugated and has 10π electrons(4n+2)
Reference:Slides by Nevina Keskineva


Conductivity and Molar Conductivity of Equimolar Solutions of LiCl and KCl If one took equimolar solutions of $\ce{Li}$ and $\ce{K}$, which would have higher conductivity? Is it the same for molar conductivity?

Say, $\ce{LiCl}$ and $\ce{KCl}$, of a litre each. Also, as Molar Conductivity = Conductivity / Molarity, are molar conductivity and conductivity directly proportional? Why or why not?

Since Li is highly solvated due to its small size, it's mobility is reduced and hence conductivity decreased.


How are lowercase m and uppercase M used in general chemistry courses? Where do lowercase m and uppercase M make appearances in general chemistry, and what do they signify?

Lowercase m

$m$ is the official symbol for the quantity mass.
m is an SI-prefix that stands for milli ($\pu{e-3}$), for example in mg (milligram).
$m$ is sometimes uses as the symbol for molality. This is not IUPAC recommended (they recommend $b$) and leads to the confusing definition $$m_\text{solute} = n_\text{solute} / m_\text{solvent}$$ where the first $m$ is for molality and the second one for mass.

Uppercase M

$M$ is used as symbol for the quantity molar mass (dimensions mass per amount of substance).
M is an SI-prefix that stands for mega ($\pu{e6}$), for example in MW (megawatts).
M is often used as an abbreviation for mol/L, and mM for mmol/L.
$M$ is sometimes used as a symbol for the quantity molarity (dimensions are amount of substance per volume, official IUPAC name is amount of substance concentration, and official IUPAC symbol is $c$). Textbooks that follow this convention sometimes use MM or ℳ as an alternate symbol for molar mass to avoid name conflicts. Combining the quantity $M$ with the unit M gives silly statements such as: $$M = 5~\text{M}$$

Other useful information

Make sure your lowercase m and uppercase M look different when you write or type, otherwise you are confusing your readers. If you type, quantities should be italicized and units should not. That also helps to distinguish the different meanings of m.
You can't cancel m with m if they don't have the same meaning. It is fine to cancel the milli in mg/mL to obtain g/L, but not cancel $m$/mg to obtain 1/g.


Is azirene an antiaromatic compound I believe that a compound which is planar and having parallel p orbital with $4n\,\pi$ electrons is antiaromatic, but azirene has a lone pair in a $\text{sp}^3$ hybrid orbital.

Shouldn't it be non-aromatic rather than antiaromatic?

Wikipedia, in a short article, indicates that there are actually two types of azirenes:

1H-azirenes, where there is one ligand on each ring atom as shown in the question.  Since all the ring atoms have ligands the lone pair contributed by nitrogen enters the conjugated pi system, which then has four conjugated electrons and is antiaromatic.  Such a molecule is thereby unstable and prone to rearrange into the alternative structure that follows.
2H-azirenes have two ligands on one carbon, saturating that atom so we no longer have a conjugated ring.  The other ring carbon has one ligand and is still unsaturated.
The nitrogen atom is left without a ligand, and its lone pair now replaces the missing ligand in an outward pointing in-plane orbital.  The only pi bonding is an isolated bond between the nitrogen and the remaining unsaturated carbon.  These compounds are non- (not anti-)aromatic and may be isolated.


What is the word for how polyprotic an acid is? What is the word to describe the number of acidic hydrogens an acid has? I am currently thinking something like "proticity", but I am not sure  The use case world be for a table of things like this:

Acid    | Proticity
--------------------
CH₃COOH |  1
H₂SO₄   |  2
H₃BO₃   |  3

Or in a set of statements like this:

  The volume of gas produced by reaction A was 4.6 L. The proticity of Acid B was 3. The tare mass of object C was 4.76 g. The density of substance D was 7.8 g/mL.

Is there a specific word used to refer to this quantity? What word should I use to refer to it?

The term that I learned was (ironically) "basicity."

I'm not sure if that's the standard term, or even if there is a standard term, but this refers to how many bases you need to neutralize the acid completely.

EDIT: Thanks for the correction from LDC3: the "basicity" of an acid refers to how many bases you've already added to it. Thus, $\ce{NaH2PO4}$ would be "sodium phosphate, monobasic," $\ce{Na2HPO4}$ would be "sodium phosphate, dibasic," and so on.

Again, not sure if this is standard terminology.


What makes the enthalpy of hydration of calcium chloride dihydrate unique? I have recently done an experiment in an attempt to find the relationship between the water ratio in a hydrate and the enthalpy of hydration, and my results showed a correlation, but an anomaly for calcium chloride. My results were:
Sodium carbonate monohydrate: 10.03 kJ
Calcium chloride dihydrate: -19.51 kJ
Sodium acetate trihydrate: 28.98 kJ
Sodium thiosulfate pentahydrate: 47.37 kJ
Sodium sulfate decahydrate: 66.32 kJ

I have tried researching about what makes calcium chloride dihydrate unique, but I can't seem to find anything. And this is not a random error, as each of these salt compounds had 3 trials each, with consistent results.

The heats of hydration of your salts, plus some others, can be obtained by subtracting the heat of formation of the anhydrous from the heat of formation of the hydrated salt, using the numerical values from the Handbook of Chemistry and Physics, 43rd Ed., 1961, pages 1807 - 1831. The so obtained values are rather different from yours, but the tendency is the same. Here are the obtained molar heats of hydration, divided by the number of water molecules. They are strangely similar.
$\ce{Na2CO3·H2O....... 295.3 kJ (mol H2O)^{-1}}$
$\ce{Na2CO3·10 H2O..... 295.3 kJ (mol H2O)^{-1}}$
$\ce{NaCH3COO·3H2O ... 297.6 kJ (mol H2O)^{-1}}$
$\ce{Na2SO4·10H2O ..... 294.3 kJ (mol H2O)^{-1}}$
$\ce{Na2S·4.5H2O ...... 305.0 kJ (mol H2O)^{-1}}$
$\ce{Na2S·5H2O....... 304.6 kJ (mol H2O)}^{-1}$
$\ce{Na2S·9H2O ....... 301.1 kJ (mol H2O)^{-1}}$
$\ce{Na2S2O3·5H2O..... 297.2 kJ (mol H2O)^{-1}}$
$\ce{Ca(NO3)2·4H2O.... 297.7 kJ (mol H2O)^{-1}}$
$\ce{Ca(CH3COO)2·H2O . 291.2 kJ (mole H2O)^{-1}}$
$\ce{CaCl2·2H2O ...... 179.6 kJ (mol H2O)^{-1}}$  ........ (!)
$\ce{CaCl2·6 H2O ......301.8 kJ (mol H2O)^{-1}}$
It is immediately clear that the heat of hydration of one molecule $\ce{H2O}$ in these compounds is $\ce{300 ± 10 kJ (mol H2O)^{-1}}$. But there is one odd exception : $\ce{CaCl2·2H2O}$. Has anybody an explanation ?
I would have been happy to show my previous numerical values in a table, or an array with lines and several columns, showing the different heats of formation. But I have not been able to write it correctly. There was always a message saying : \hline incorrect. Can somebody tell me how to proceed ? Thank you in advance.


A sample must not contain more than 2% contaminants to be pure. Are there instances where 2.12% is ok? I do not have any background in chemistry, I am asking this on behalf of someone else, twice translated, so I really hope it makes sense.

"A maximum of 2%" seems quite a clear specification, but since  practice involves rounding of measurements and significant decimals, that might not be the case.

It may seem silly but which of the following options are valid work protocol?

at most 2% means less than 2.5% after rounding
at most 2.0% means less than 2.05% and option 1 still holds
at most 2% is the same as at most 2.0% and means at most 2.000...%

Off topic: An optimal answer would be in the context of the EuAB (Europäisches Arzneibuch).

It depends mainly on the accuracy of your measuring instrument in practice and the accuracy required by your work.

For example:

If you want to make a solution with 2g solute in it, and to measure amount of solute you have an measuring instrument that have least count of 0.1g, then 2.05g is acceptable as the instrument will show 2.0g only but if the least count of intrument is 0.01g then 2.05g is not acceptable if you want 2g.

In real life the least count varies for different situations.

For example if you are manager of a water bottle manufacturing unit you won't need much accuracy to fortify the water with minerals, but if you are carrying out a titration in lab to find strength of some oxidising agent you need very high accuracy that is why burette are made such accurate.


Why does Coca-Cola Zero Sugar Contain Sodium Despite no Sodium Ingredients? According to the Coca-Cola website and every label I read, Coca-Cola Zero Sugar is a mixture of organic compounds:

  CARBONATED WATER (OC(OH)₂), CARAMEL COLOR (Various sugars), PHOSPHORIC ACID (H₃PO₄) , ASPARTAME (C₁₄H₁₈N₂O₅), POTASSIUM BENZOATE (C₇H₅KO₂), NATURAL FLAVORS, POTASSIUM CITRATE (C₆H₅K₃O₇), ACESULFAME POTASSIUM (C₄H₄KNO₄S), CAFFEINE (C₈H₁₀N₄O₂).

The label also says 40 mg sodium. There's a perceptible salty taste in the beverage, which suggests to me that something activates the taste receptors for saltiness, but a potassium cation can do that just fine.

There is no sodium in any of those molecules. That means "Natural Flavors" must contain the sodium... but in what form? The essential oils in things like OpenCola seem to all be hydrocarbons without sodium. And if it's plain old NaCl table salt, why wouldn't they just list it as "salt"?

Where's the sodium in Coke coming from?

EDIT: The discussion has centered around the sodium content of "still" drinking water; what about the process of carbonating the water? Some processes add a sodium carbonate to the solution; could that be the source, contained in the ingredients list as "carbonated water"?

Two sources: water and natural flavors

According to the Wisconsin department of healt natural sodium levels in water is $\pu{40-60 mg/L}$ meaning a $\pu{20 fl oz (\! 591 mL)}$ Coke has about $\pu{25 mg}$ of sodium from water. Thus the other $\pu{50 mg}$ of sodium must come from natural flavors. This makes sense since Coca-Cola uses plant products like coca leaves which would naturally add sodium to the carbonated sugar water.


How can I measure carcinogenic risks of an M-ENDO-Agar LES substrate after heat destruction? I'm about to conduct a study of possible toxic traces from the substance M-ENDO-Agar LES in substrate after destruction with heat (I think it is 160℃) for 180 minutes. The substance is carcinogenic.

I want to find a way to measure if the substrate is still toxic in any way after the heat treatment. How could I conduct such a study?

Most assays of carcinogenicity are designed to measure DNA damage. The 2 most common of those assays are the Ames test and the Comet Assay.

The Ames test uses a special set of strains of bacteria that have known phenotypes. You expose the bacteria to the suspected carcinogen and then look for changes in phenotype. Different strains will be susceptible to different types of mutations, deletions, insertions, frameshifts, etc.

The comet assay uses mammalian cell lines. You have to suspend the cells in an agar and break them up. This will create little pockets of cell debris in the agar. You fluorescently stain the DNA and run an electric current through it. If the DNA is damaged and broken up, it will migrate through the gel, creating tails, hence the "comet". If the DNA is intact, it will not migrate.

I've never done either assay, just heard about them in class. Wikipedia has more detail than I do.

http://en.wikipedia.org/wiki/Ames_test
http://en.wikipedia.org/wiki/Comet_assay


Water-stable organoaluminum compounds in which aluminum is part of the aromatic ring The more common organoaluminum compounds, such as trialkylaluminum, decompose on contact with oxygen-containing molecules (water, ethanol etc). However, my guess is that organoaluminum compounds with an aluminum atom being part of the aromatic system might be more stable towards oxygen compounds, since the high affinity for oxygen of aluminum could be (at least partially) offset by the high energy barrier to destroying the aromatic ring.

My questions are- is my guess correct, and is there any specific aromatic organoaluminum compound that is not destroyed (or acts reversibly at the worst) with water, ethanol, THF (etc)?

You're probably not going to get what you want putting aluminum into an aromatic ring.  Aluminum and carbon have relatively poor pi overlap, so your ring will not have the full delocalization necessary for string aromatic stabilization.  And even strong aromatic stabilization may not be all that strong compared with aluminum preferring to bond with oxygen or nitrogen rather than carbon.

Instead of trying to design a compound that somehow tames the reactivity of aluminum-carbon bonds, choose the right polar solvent.  Protic solvents are bad.  According to Wikipedia, organoaluminum compounds "readily form adducts with bases such as pyridine, THF and tertiary amines. These adducts are tetrahedral at Al."  The named examples are aprotic and coordinate through nitrogen or oxygen.


Speed up polymerisation of formaldehyde Is there a way to speed up the formation of para-formaldehyde in a formalin solution?

According to this patent, alkaline catalyst of composition sodium and potassium hydroxides, carbonates, formates, and acetates can be used to speed up formation of paraformaldehyde from formalin solution (aq. formaldehyde).

  This invention relates to a new and improved process for the
  production of formaldehyde polymers. More particularly it relates to a
  process whereby high strength aqueous formaldehyde solutions can be
  concentrated to paraformaldehyde at a greatly accelerated rate.

Moreover, the patent describes the process of this reaction in details. Have a good read.


Coconut aldehyde mechanism How would you draw the mechanism going from non-3-enoic acid to coconut aldehyde (5-pentyloxolan-2-one)? I know I added sulfuric acid and sodium carbonate, which I was confused how they worked into the mechanism. I am certain there is some type of internal reaction to create the five membered ring.

Those are the only chemicals that could, in my opinion, be involved. The others are not needed in the mechanism. Pretty much confused on the first step of what happens when sulfuric acid is added. What does it protonate? How does this lead to an internal reaction and where does that reaction take place? Does the C-C double bond have anything to do with it?

Would the lone pairs on the oxygen readily take the proton? But how would that affect the overall state of the compound in order to form that ring. I'm not seeing how that could be right.

I have the impression that you a very close to find the answer yourself!

I will not give a full answer here, but some hints.

  [...] I know I added Sulfuric acid and sodium carbonate [...]

Let's assume that you did not add them simultaneously. Then, it seems likely that sulfuric acid does the magic and sodium carbonate is added to neutralize the reaction mixture after your product has been formed.

  I am certain there is some type of internal reaction to create the five membered ring.

Good point! Let's draw the starting material and the product.

3-nonenoic acid is a carboxylic acid with a non-conjugated $\ce{C=C}$ double bond. With other words, there are two functional groups.

Now think in the sulfuric acid! It's a strong acid while its anion is a rather weak nucleophile.

"Coconut aldehyde" is  not an aldehyde at all, but a furanone, a five-membered lactone. Think intramolecular ester.

EDIT

Is it conceivable that the $\ce{C=C}$ double bond is protonated?
If so, what could possibly add to the resulting cation?


Chemical synthesis via 3D printing? I happened to stumble upon this Nature Chemistry article today. It claims that

  [...] use a 3D printer to initiate chemical reactions by printing the reagents directly into a 3D reactionware matrix, and so put reactionware design, construction and operation under digital control.

Unfortunately, I cannot obtain the full article to read, but even the abstract does not seem plausible in my opinion. My questions are...

How can one "print reagents"? What is meant by this?
Most synthetic reactions require hours of refluxing and whatnot, not to mention isolation/purification afterwards. How can all of these techniques be performed with a printer?

Perhaps I'm not "getting it", but the abstract of the article isn't providing me with the explanations and theory I'm looking for. How can something like this be possible?

References

Symes, Mark D., Philip J. Kitson, Jun Yan, Craig J. Richmond, Geoffrey J. T. Cooper, Richard W. Bowman, Turlif Vilbrandt, and Leroy Cronin. “Integrated 3D-printed Reactionware for Chemical Synthesis and Analysis.” Nat Chem 4, no. 5 (May 2012): 349–354.

The authors use a 3D printer to generate the labware, and to place the reagents in its inlets.

So we're not printing standard ink on paper but using a the print head to exactly place very amounts of monomers at certain positions. These monomers cure (polymerize) upon heat or irradiation to a solid resin. By this, small reactors of arbitrary design (think in 3-neck reactors) with a volume of ~ 10 mL may be printed layer by layer.

Apparently, it is even possible to create reactors with windows to allow for later reaction monitoring or print and assemble electrodes for spectroelectrochemistry.

  How can one "print reagents"? What is meant by this?

This is all about precise positioning of the printhead! Thus same printhead used for the ink can be used to fill the inlets (think in dropping funnels of normal labware) of the reactor with componends whose reaction you want to monitor. Instead of dropping, material transfer is achieved by applying a vacuum to the reactor through a self-healing hole.

  Most synthetic reactions require hours of refluxing and whatnot [...]

Well, a lot of reactions do proceed quite well at room temperature, think in

Diels-Alder reactions of furanes and maleimides
electrochemically triggered reactions
photoinduced electron transfer reactions

just to mention a few.

  [...] not to mention isolation/purification afterwards. How can all of these techniques be performed with a printer?

Oh, you got that one slightly wrong. The "printer" is just used to build the reaction sphere (the flask, if you like it) and to place the reagents in the right inlet of your vessel. The rest is "just chemistry" ;)

I hope i made myself clear enough - at least after the n-th edit ;)

I think the concept is pretty cool.


Which species gives electron in the following reaction? In a given reaction like this:
$$\ce{FeO4^2-  + 3e^-  + 8H+ <=> Fe^3+ + 4H2O}$$

From where does the electron come  in the acidic medium, as there is no other species to get oxidised in the medium? Also if this were in basic medium then why there is no oxidation of any other species in the medium?

What you have here is a reduction half-equation as the $\ce{Fe(IV)}$ is accepting electrons to get reduced to $\ce{Fe(III)}$. As the name suggests, this only shows half of the actual reaction as if something is getting reduced, something else must be getting oxidised. So what you need is the oxidation half-equation which shows the species that is getting oxidised and donating electrons. When you add these two half equations the electrons will cancel each other out and you will get your equation for the reaction.

For example, one possible oxidation half-equation is the following:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
Now if you times the co-efficients of the reduction half-equation by 2 and times the co-efficients of the oxidation half-equation by 3 and then add them together you will get the following equation:
$$\ce{2FeO4^{2-} 3Mg + 16H+ -> 2Fe^{3+} 3Mg^{2+} + 8H2O}$$
As you can see all the electrons balance out and that is your complete equation for the reaction.

Sorry, I don't think I really understand your second question about basic medium. Could you please expand on it to clarify what you mean.


Explanation for increasing wave number trends for hydrocarbons At my university, we had an organic chemistry exam question that stated,"The trend in the values of the ir absorbtions for the c-h bonds of methane ($\ce{CH4}$: $2995\,\mathrm{cm}^{-1}$), (ethylene: $3080\,\mathrm{cm}^{-1}$), and (acetylene: $3305\,\mathrm{cm}^{-1}$) can be explained by...

The answer was " by the increasing bond strength of the carbon-carbon single, double, and triple bonds. I thought the correct answer was  "increasing s character of the carbon atoms". Could I get an explanation as to why the first response is valid?

Thanks!

As orthocresol suggested in his comment, you are correct. First of all, methane does not contain a $\ce{C-C}$ bond, so that answer could not explain the wave number form methane. In addition, you can think of bonds as tiny harmonic oscillators. Specifically, they behave like a mass on a spring, absorbing frequencies that match the compound's frequency of oscillation. Remember that for a mass on a spring, $\omega = \sqrt{\frac{k}{m}}$, where k is the spring's stiffness (related to the strength of the bond) and m is its mass. This is why stronger bonds (such as a $\ce{C=C}$ bond over a $\ce{C-C}$ bond) have a higher frequency, and therefore wavenumber (these two quantities are directly proportional). Likewise, the $sp$ orbitals in acetylene are stiffer than the $sp^2$ orbitals in ethylene, which are in turn stiffer than the $sp^3$ orbitals in methane. This explains why methane has a lower wave number than ethylene, which in turn has a lower wave number than acetylene.


thinking about osmotic pressure when liquid is replaced by gas An option in my test says:

  "The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature."

I'm not able to think how should I relate between a solution and a gaseous mixture. Am I missing something?

In both the case of the osmotic pressure of a dilute solution and the case of the pressure exerted by an ideal gas, the solute or gas may be described as composed of non-interacting (ideal) particles, and the mathematical expressions (equations of state) describing the two situations are very similar (in one case $p=cRT$, in the other $\pi = cRT$)$^\dagger$. However it might be less confusing if equivalent to state that in both cases the equations describe similar relationships between the work required to change the volume of the system and the accompanying change in the concentration of gas or solute. In both cases work can be done by the system through an expansion, but in one case the expansion results from pressure exerted by the gas, while in the other it results from pressure exerted by the solvent. In the case of osmotic pressure, since the chemical potential of the solvent is coupled to that of the solute (as described by the Gibbs-Duhem relation) it is possible to relate the osmotic pressure to the solute concentration  (in the limit of an ideal solution as described by Henry's Law).

$^\dagger$As I commented, the equations are analogous but in my opinion, "The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature." is a misstatement. The entire solution, if evaporated, would not exert the same pressure. It is more subtle than that. It is the solute that would exert an equivalent pressure if a gas.


Reaction of unsaturated lactone and alcohol Image source
I found this equation to turn this compound into an ketoester but I am unsure of the reaction mechanism.
I postulate that the alcohol attacks the carboxyl carbon, the O becomes formally negative. It reforms the double bond and the leaving is the O in the ring which takes the electrons in the bond.
Here my thought differs from the image (gotten from one of the United States Patent Office Patents). I thought the end product would be an ester OHCCCCOOR. Alcohol then alkene and at the end an ester but the picture shows a ketone and no alcohol.

You are correct that the alcohol atacks the carbonyl then the ring oxygen leaves in a manner analagous to ester transesterification. The intermediate generated is an enol which picks up a proton to for the ketone.

image from here: 1