How can oxygen atoms suddenly appear on the product side of an equation? The question states the following  equation:
 $$ \ce{C(s) + 2H2(g) -> CO2(g) + 2H2O}$$
 And then asks you to calculate the enthalpy change for it, given enthalpies of combustion.

The stated values are $C(s)$ -393,  $\ce{H2}(g)$ -286,  and $\ce{CH4}(g)$ -890

I'm curious as to why oxygen atoms have suddenly appeared on the products side. The only thing I have been able to think is that it refers to the enthalpy of formation of methane, and that the answer would be -393 + (2*-286) + 890. But it still doesn't make sense to me considering the equation provided. I understand the products are the products expected from combustion, but no oxygen has been added on the reactants side.

They probably want you to calculate the enthalpy change required to convert the reactants into the products. Another way of looking at it is the enthaply difference between the products and reactants.

Remember, $\ce{O2(g)}$ has 0 enthalpy of formation in standard state. (so do $\ce{C(s)}$ and $\ce{H2(g)}$, but we don't need that)

So, adding $\ce{2O2}$ to the reactants side won't change $\Delta H^o=H^o_{products}-H^o_{reactants}$. So calculating $\Delta H^o$ for the given "reaction" is the same as calculating it for $$\ce{C(s) + 2H2(g) +2O2 -> CO2(g) + 2H2O}$$. And $\Delta S^o$ for this reaction can be calculated very easily using enthalpies of combustion. You don't even need to use the given value for methane.

So, while the question is wrong in calling it a reaction, it is still asking us to do something reasonable -- that is, calculate $$\Delta H^o=H^o_{\ce{CO2(g) + 2H2O}} - H^o_{\ce{C(s) + 2H2(g)}}$$. Since we can subtract $H^o_{\ce{2O2(g)}}=0$ easily, the problem can be solved by taking a combustion path.


Generating geometries for substituted molecules I would like to automate some of my calculations, and I am looking for a tool to generate reasonable good initial geometries.

The details: I have already optimized molecular geometries, and I would like to perform calculations of several different substituted version of the molecule. Because of the size of the system, it would make sense to keep the already optimized geometry and exchange e.g. selected -H atoms to some R1, R2 groups. Is there any library that can manipulate 3D geometries this way?

I am aware I can do similar things with much graphical software (GausView etc).  However, the point would be automation: a command line tool or a library or something similar that I can call from Python scripts, instead of me clicking things on a screen one by one.

I am also aware that I can generate 3D geometry from 2D or such. However, I would like to keep the already optimized main part of the molecule (it can be pretty big), for more efficient calculations, and insert new groups. I appreaciate any help, thanks.

For most of the geometry creation/manipulation I do, I use the Atomic Simulation Environment (ASE) - all in python.

It's pretty straightforward to use and has a gui so you can make sure you're doing the right thing as you go along. In particular, I'd recommend playing with it in a jupyter notebook.

Checkout https://wiki.fysik.dtu.dk/ase/tutorials/manipulating_atoms.html for some examples.

The ASE also has a dataset of existing molecules that you can import and add to your structure in various places.


Historic Importance of Tin(Sn) How important is tin for a bronze age society when there's many copper alloys including arsenical bronze. If you're out of tin whats the next best copper alloy(s) for making shortswords and axe blades and is it such a downgrade. This is a chemistry and history question.

Some tools were made of "pure" copper. But tin was the best and only available addition to copper for strength. At the time tin was the only type of bronze. Today a variety of copper alloys are called "bronze"  that do not contain tin. Lead could possible have been available depending on time and place but it does not improve strength of copper and could actually weaken it. I see in alloy listings there are silicon bronzes ( today) that contain 1 to 2 % . Silicon was available, but it takes modern technology to alloy it into copper. Silver and gold could be added to copper ( they are all mutually soluble ) but make little increase in strength. There is a copper alloy today that contains about a quarter % of each phosphorus and arsenic ( C14200) , it is a wrought product so not available in the bronze age. The Copper Development Association  presumable has some information on the internet.


ortho/para-Selectivity in the Reimer-Tiemann Reaction I can't figure out why the dichlorocarbene molecule produced in the Reimer-Tiemann reaction gets attached to the ortho position of phenol.
The ortho position is sterically slightly more hindered due to the oxygen, so why dosen't the carbene attach to the para position?

The Reimer–Tiemann reaction will give a mixture of products unless the formation of one product is highly disfavoured. The Kürti-Csakó notes in its introductory paragraph on the Reimer–Tiemann:[1]

the regioselectivity is not high, but ortho-formyl products tend to predominate;

The statement already hints at a product mixture being obtained. To further show that both ortho and para attacks are possible, the 1960 review published by Wynberg contains the following scheme as its very first scheme in the introduction:[2]

So while both products are obtained, a back of the envelope calculation already shows that the distribution is not a perfect $2:1$ ratio. This can apparently be explaned by the effect of positive counterions, as Hine and van der Veen report:[3]

The ratio of ortho to para product was found to be 2.21 [under high base concentrations], showing that the tendency towards o-substitution is indeed increased under conditions where ion-pair formation is encouraged. One factor that would certainly be expected to be present and that would tend to favor o-substitution is an electrostatic effect. When a dichloromethylene molecule attacks the o-position of a sodium phenoxide ion-pair to yield the probable initial product, there is less separation of unlike charges than when the analogous para product is formed.

So while the steric hindrance of a single oxygen atom is not much more than that of a hydrogen atom and thus does not play a huge role, the presence of positive counterions – although increasing steric congestion – may serve to enhance the formation of the ortho-product through favourable electronic effects.

References:
[1]: L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 378.
[2]: H. Wynberg, Chem. Rev. 1960, 60, 169–184. DOI: 10.1021/cr60204a003.
[3]: J. Hine, J. M. van der Veen, J. Am. Chem. Soc. 1959, 81, 6446–6449. DOI: 10.1021/ja01533a028.


How to derive partial gas equation? If the following gasses are to mix, what is the partial pressure of neon?

The correct answer is 0.1. I cannot figure out how my professor derived this equation. From the ideal gas law, all I can arrive to is this:

$P_x = \frac{P_tV_t}{V_x}$

This gives me an answer of 4.4, which is obviously incorrect.

I've determined the equation he used is as follows:

$P_x = \frac{P_xV_x}{V_t}$

which gives the correct answer of 0.1. How did he arrive to this answer?

The total pressure of the system is:

$\frac{1.0\times0.5+1.0\times0.4+2.0\times0.2}{1.0+1.0+2.0}=1.3/4=0.325$

i.e. just the sum of the $PV$s divided by the sum of the $V$s.  Then $0.325$ is the sum of the partial pressures by the Law of Partial Pressures.  So this at least suggest that the individual partial pressures are near $0.1$.

Can you figure out how to parcel out the individual pressures from here?


Charge consistency in fragment qm/mm methods I'm calculating the multipole moments on several fragments of a molecule. What are some procedures to ensure that the charges etc are representative of the complete molecule? How do I ensure I'm not getting an overall charge in a neutral molecule?

If you're doing any sort of computational run (either quantum chemistry, solid state, or molecular dynamics), you have to set the charge on the overall system.

You said in your comments that you're doing a distributed multipole analysis across multiple fragments of a protein.

In this case, one of the constraints on the calculation is exactly that the overall charge must match that specified. The program won't arbitrarily change the charge to be an anion or a cation if you specified a neutral system.

Now, that's not to say that the individual fragments won't have fractional or integer charges. Depending on the method, it's entirely reasonable that there's some amount of partial charge transfer between fragments.

It's just that there's absolutely no way that the sum of the charges will be different than the constraint.


Concentration of excess reactant If increasing/decreasing the concentration of an excess reactant influenced the percentage yield of a product experimentally, how could this be explained?

I know that in theory, the excess shouldn't affect the yield- since it's the limiting that decides how much is produced. But is there any possible explanation for this result?

A very typical situation in organic chemistry is the following:

$\ce{A -> undisired products}$
$\ce{A + B -> target product}$

Here it is easy to see that increasing the concentration of $\ce{B}$, we will get a higher yield of the target product.
The other reaction can be of many kinds.
$\ce{A}$ can decompose in the reaction medium, can react with itself, with the solvent, with byproducts form the desired reaction, etc.


Newman Projection with t-butyl group? So again in Chapter 4 David Klein book, we have this organic molecule with a t-butyl group (just as an aside, this is t-butyl ethane, right?

Now how is it that the correct Newman projection must be the following?

What I glean from the Lewis structure is that because there are no wedges or dashed lines, the methyl groups ought to be in the same plane. How can there be a methyl group jutting out upwards and to the right, and another methyl group jutting out upwards and to the left in the t-butyl group and not be illustrated as such with dashed lines and wedges? Thanks in advance!

Lewis structures and skeletal structures show which atoms are connected to which other atoms, and the type of bonds (single, double, triple). They do not show the shape (geometry) of the molecule.

For example, the two skeletal structures below refer to the same molecule, butane:

When there is ambiguity (i.e. multiple molecules with the same skeletal structure), you can add information about the stereochemistry by adding dashes bonds and wedges. In your case, the molecule is achiral, so that does not apply.

To figure out the geometry, remember that carbon atoms with four single bonds have a tetrahedral geometry. You get other geometries when there is a double bond (trigonal planar) or a triple bond (linear) or other, more unusual features associated with a carbon atom (lone pair, non-zero formal charge, radicals, two double bonds).


Ambiguity in synthesis (crystallization) A paper says the following,

  Dichloroglyoxime was prepared by chlorination of glyoxime in 10 % HCl at 0 deg C and crystallized from toluene.

When they say "crystallized from toluene" do they unambiguously mean recrystallized from boiling toluene at standard pressure? Thermally these compounds are unstable and instead of a white nice product I've ended up with a purple one much to my dismay and I'm not sure what actually causes the purple.

I've had problems with this synthesis because the $^1$H NMR always shows more than one oxime environment.

No.  This could also mean applying a vacuum to the toluene-dichloroglyoxime solution, and letting the solvent evaporate-away at room temperature.


How does different diffraction grating effect resolution and sharpness of lines in a line spectrum? In my lab notebook it asks us to think about the spectroscope further than the reading goes on.

It asks if we use a diffraction grating that contains 600 lines/mm in the spectroscope what do we expect would happen to the lines of the line spectrum in terms of sharpness and resolution.

Is the answer (assuming 600 is more than normal) that the lines would be both more sharper and have a higher resolution? My first thought was that the lines would be skinnier but more clear.

Sharpness is colloquial for how narrow a peak is. Resolution is how well we can see two features as they move closer together. You might do well to re-read your book to learn these definitions better. As already noted, skinnier and more clear have no meaning in spectroscopy.

A search on terms such as diffraction grating equation will give you further reading to support or invalidate the guess (hypothesis) that you propose.


19F NMR of hexafluorobromine(VII) cation $\ce{[BrF6]+}$ is an octahedral species, and should have only one fluorine environment due to the symmetry. $\ce{Br}$ has nuclear spin $I=3/2$ and therefore is unlikely to cause splitting due to rapid interconversion between its spin up and spin down states averaging to a zero magnetic effect. Therefore I would expect one singlet in the $\ce{^19F}$ NMR spectrum.

However... the $\ce{^19F}$ NMR is actually two $1:1:1:1$ quartets (at low temperatures). Can someone help me find the flaws in my reasoning?

$\ce{BrF6^+}$ does have an octahedral structure consequently all of the fluorines are equivalent.  As you point out, bromine is a spin = 3/2 nucleus, therefore, it is also a quadrupolar nucleus (all nuclei with spin > 1/2 are also quadrupolar). A spin 3/2 nucleus that is coupled to another nucleus (fluorine in this case) will split the nmr signal for the other nucleus into a 1:1:1:1 quartet.  In the case of bromine we don't normally see this coupling because bromine typically undergoes rapid quadrupolar relaxation (some quadrupolar nuclei do, some don't) which washes the coupling out (e.g. rapid quadrupolar relaxation effectively decouples the nuclei, that's why the protons in $\ce{CH3Br}$ are a singlet, for example).  However, when the quadrupolar nucleus exists in a highly symmetric environment (octahedral, tetrahedral geometry) the electric filed gradient at the nucleus is near zero and quadrupolar relaxation becomes slow (on the nmr timescale) and coupling can often be observed in such cases.

Finally, bromine occurs naturally as a mixture of two isotopes, $\ce{^{79}Br}$ (50.7%) and $\ce{^{81}Br}$ (49.3%).  Both isotopes are quadrupolar with spin = 3/2.  Because of the slight difference in electronegativity of the two bromine isotopes, they will produce a slight difference in chemical shift for the fluorines bonded to them. Hence, this mixture of $\ce{^{79}BrF6^+}$ and $\ce{^{81}BrF6^+}$ will produce two 1:1:1:1 quartets in the $\ce{^{19}F}$ nmr spectrum.


Which mutated enzymes act as better drugs than their non-mutated form? Is there an enzyme which is used as a drug (let it be called version $A$) and a mutated version of it ($M$), where the activity of $M$ was found to be higher than $A$?

Any references on such a case would be greatly acknowledged.

There are mutants of clotting factor VIII that have longer half lives and better activity than natural Factor VIII. Factor VIII is one of the missing proteins commonly involved in hemophilia. Hemophiliacs usually use injections of Factor VIII to prevent bleeding. The mutant form should last longer in the blood, with better activity, resulting in fewer injections.

source: https://encrypted.google.com/patents/US7211559

Since this is a 2007 patent, I don't know if this product is actually being used to treat patients, drug approval takes a long time.


In a redox reaction, why does more than one oxidation state of an element form? Consider the reaction 100ml 10M $\ce{NaOH_{(aq)}} +$ 100ml 10M $\ce{NaHSO3_{(aq)} +}$ 100mL 0.01M $\ce{KMnO4_{(aq)}}$

In the picture posted here, the left columns give information on the Molarity of species present before 100mL 0.01 M $\ce{KMnO4}$ is added. The right column is after $\ce{KMnO4}$ is added.

In the resulting solution, $\ce{Mn}$ is oxidized to various states: $\ce{Mn^2+ Mn^4+ Mn^6+ Mn^7+}$ (where $\ce{Mn^7+}$ is the original form).

I would like to know, why do different oxidation states of $\ce{Mn}$ appear? How can we predict the concentration of oxidation states?

They all appear because they are all finite amount of energy apart, but note that some concentrations are many orders of magnitude greater than others, and the concentration of $\ce{MnO4-}$ is so low that the table shows a zero.

Predicting the concentration requires being able to read tables of potentials. In standard tables like http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm  you will find equations such as $$\ce{MnO2{}+4H+ +4e-  \rightleftharpoons Mn + 2H2O}\quad (1.23\mathrm{V})$$ and $$\ce{Mn^{2+}{}+2e^-  \rightleftharpoons Mn}\quad (-1.185\mathrm{V})$$ (the numbers in parentheses are the standard reduction potentials, $E^\circ$, always measured in volts).

With this and the Nernst equation you can relate the concentrations of any set of species that form a balanced reaction.  In a complicated mix like this you don't know a priori the concentration of any species, only the certain sums (for instance, for all sulfur-containing ions) but you still can, perhaps iteratively, solve the corresponding set of algebraic equations and find the individual concentrations.  All the equilibrium equations have to be satisfied simultaneously; if they aren't the system is not in equilibrium.

(In this answer I've used the word "concentration" but the equations written with concentrations are only approximately satisfied; to account for second-order effects the notion of "activity" was introduced, and while it is an essentially empirical notion, it's still useful if the dependence of activity on concentration for a particular problem has been tabulated.)


Can superoxide ion act as a ligand in a similar fashion as peroxide? There is a question in my chemistry book:

  Find oxidation state of platinum ion in the complex $\ce{[Pt(O2)(en)2Br]+}$.

My books says that there is peroxide, but I want to ask why it is peroxide? Can it be superoxide?

My answer was that oxidation state of platinum is $+3$, but the answer in the book given is $+4$ with explanation that $\ce{O2}$ here is peroxide. How can I get to know from the formula that $\ce{O2}$ is peroxide and not superoxide?

Here the oxygen "molecule" in its complex with hemoglobin is rendered as a superoxide ligand.  But remember, electrons are shared between the metal and the ligand; so there may be some latitude in how you can apportion them to determine "oxidation states".


Yield of alkyl chloride in the Hunsdiecker reaction Apparently the yield of alkyl chlorides is low as compared to that of alkyl bromides when preparing alkyl halides using the Hunsdiecker reaction.
Why does this occur? The only factor I can see is the lower stability of the free radical of chlorine as compared to that of bromine, as the reaction proceeds by the free-radical mechanism.
Is that what causes the lower yield?

The Hunsdiecker reaction is a halogenation-decarboxylation reaction of silver salts of carboxylic acids.

To understand what causes the yield to change, first we have to consider the mechanism of the reaction.

As seen from the mechanism the homolytic cleavage of the $\ce{O-X}$ bond is the slowest step in the reaction and thus is the rate determining step. So as you predicted it all comes upon the stability of halogen radical. It's commonly known that the bromine radical is fairly stable than the chlorine radical and thus the Hunsdiecker reaction gives better yield with bromine rather than with chlorine.


Determining aromaticity of compounds Which of the following planar structures would be aromatic?

Could someone help me determining the aromaticity of the compounds?

My Attempt:

(i) Not an aromatic compound since it only has 4 pi electrons that does not satisfy Hückel's (4n+2) rule.

(ii) Not aromatic because it only has 4 pi electrons, however I'm not sure if the oxygen contributes a p-orbital, but it shouldn't matter since 5 pi electrons still does not satisfy Hückel's rule.

(iii) I'm very confused because I'm not sure if the ketones are considered "part of the ring" therefore do they also contribute p orbitals? However if we ignore the ketones, then I believe the Nitrogens contribute 2 e- since they are sp2 hybridized? Therefore the 2 N atoms and 2 C atoms contribute a total of 4-not aromatic either?

(i) that's correct

(ii) is aromatic, it fit's the 4n+2 rule with n=1.  Here is a drawing of furan showing 2 lone pair electrons in a p-orbital that completes the cyclic aromatic ring. Furan's other lone pair of electrons is perpendicular to the pi system and does not play a role in furan's aromaticity.

(image source)

(iii) Edit to incorporate @jerepierre's comment:

Just like phenol can exist in equilibrium with its non-aromatic tautomeric keto form

so can derivatives of pyrimidinedione (or uracil).  However, unlike phenol were the equilibrium lies far to the aromatic phenol side, these compounds usually (it depends on the substituents) exist in the keto form.  The strength of the keto and amide bonds outweigh the weaker aromaticity of the pyrimidine skeleton and shift the equilibrium to the keto side.

(image source)

The question remains, is the keto form aromatic?  As jerepierre points out, you can draw the following resonance structure for the keto form.  In this double-dipolar resonance structure there are 6 pi electrons 2 in the double bond and 2 [a lone pair] on each nitrogen) suggesting that this resonance structure would be aromatic.  Since this resonance structure contributes to the overall description of the molecule, it would be fair to say that this molecule has some aromatic character; certainly not as much as benzene or furan - but some.


Acidic KMnO4 oxidizes H2S to S. Why doesn't the oxidation proceed further to sulfur oxide, sulfur dioxide or sulfur trioxide? I have some idea about the question. I think it has to do with oxidation/reduction potentials but I am not able to find any resource for the exact values or how to make sense of them in this context. Any reference would be appreciated.

Yes, the reaction can further proceed to sulfur dioxide and sulfur trioxide, but only appears for certain, per my research, to occur in alkaline conditions.
Here is a supporting opinion to this effect, which cites the corresponding reaction in alkaline conditions as:

$\ce{3 S(2-) + 8 KMnO4 → 3 SO4(2-) + 8 MnO2 }$

that implies the further conversion of $\ce{S}$ to $\ce{SO2}$ to $\ce{SO3}$, which in alkaline conditions is represented by sulfate formation.
Note, the cited source notes only in acidic conditions, to quote:

Neutral-Acid pH:
$\ce{3 H2S + 2 KMnO4 → 3S + 2 MnO2 }$

Wikipedia on KMnO4 supports this pH effect, to quote:

The Mn-containing products from redox reactions depend on the pH. Acidic solutions of permanganate are reduced to the faintly pink manganese(II) ion (Mn2+) and water. In neutral solution, permanganate is only reduced by three electrons to give manganese dioxide (MnO2), wherein manganese is in a +4 oxidation state.

Interestingly, the corresponding reaction replacing KMnO4 with either H2O2 or HOCl, can results in $\ce{SO2}$, ..., in acidic conditions.


Solubility of Potassium Dichromate in Water According to Wikipedia, the solubility of Potassium Dichromate increases very significantly with the temperature.

At 20°C, 13g/100ml, and at 100°C, 104g/100ml.

Therefore, it should be possible to dissolve about 520g of potassium dichromate in 500 ml of boiling water, but only about 65g in the same water, but at room temperature.

Today, I acquired about one kg of technical potassium dichromate and tried recrystallizing it, by weighing 507g of raw dichromate and dissolving it in about 500ml of water, room temperature. I expected it to no dissolve fully until the water was heated to boiling.

However, as soon as the water got slightly warm (45°C-50°C) everything dissolved. I added more raw material, and it kept dissolving until in the end everything (~970g) was added and dissolved before reaching boiling temperature.

Now, it is very slowly cooling down, and at about 35°C I seeded some small crystals and they're growing.

What went wrong in my calculations of the solubility? Anything I should've considered and missed?

Thanks.

PS.: Things I've already considered and ruled out:
Crystallization water: As far as I reseached, potassium dichromate doesn't have crystallization water, unlike, for example, copper (II) sulphate.
High humidity: My raw material was a free flowing powder, < 1mm particle size, uniformly deep orange colored. Had I high humidity, it'd clumping and not free flowing.
Contamination with potassium chromate, sodium chromate: Any of those wouldn't be soluble enough to explain what I observed.

What happened was very simple.

I've been duped.

The seller sold me (intentionally or not) sodium dichromate and not potassium dichromate, or at least very contaminated with sodium dichromate.

According to online source (https://en.wikipedia.org/wiki/Solubility_table#S), the solubility of sodium dichromate in water is about 198g/100ml of water at 30°C, and a very high 415g/100ml at 100°C.

That explains clearly why everything dissolved promptly while the water was only slightly warm (40°C? Didn't come to mind put a thermometer in since I was expecting it to dissolve only near 100°C), since 500ml of solution would be able to dissolve 970g (my raw material) at only 30°C.

Also, when I crystallized some of it overnight on a cold (~15°C?) water bath, the solubility would drop to about 175g/100ml, causing the precipitation of about 100g of salt.

Finally, to settle matters about the presence of sodium dichromate, I dropped some saturated potassium chloride on a small sample of the solution in a test tube, and observed the instant formation of an orange precipitate.

The same occurred when I picked some of the crystals I separated before and made a small amount of saturated solution - therefore the crystals were made of sodium chromate and not dichromate - since they have much higher solubility they shouldn't've precipitated at all before potassium dichromate.

Also, I picked some potassium dichromate of known procedence and made some saturated solution and added a few drops of a saturated solution of potassium chloride, and nothing happened.

In the end, everything said, for what I intend to make, sodium dichromate or potassium dichromate are interchangeable after adjusting quantities, but I'll have to execute extra steps to purify it from this solution.

As an piece of advice (for me and everyone else), I'd test any new potassium/sodium dichromate from unknown sources before doing anything with it, simply by making some saturated solution and adding saturated potassium chloride.

I'll have to devise a procedure for separating one from another given unknown proportions of either.

Thanks for everyone time.


Calculating base dissociation constant given the molarity and pH At $\pu{25°C}$, a $\pu{0.100 M}$ aqueous solution of $\ce{C_6H_5NH_2}$ has $\mathrm{pH}=8.80$. Find the $K_{\mathrm{b}}$.

I figured I would have to use an ICE table, but I am not too sure how to set it up.

I can find the equation, which is $$\ce{C6H5NH2 -> C6H5NH3^+ +OH-}$$ but I don't think it helps. I only have an elementary knowledge of acid/bases, so I am hoping for a simple answer.

$\ce{C6H5NH2}$ (aniline) is a weak base, and therefore, in its aqueous solution, it takes a proton from water and forms $\ce{C6H5NH3^+}$ and $\ce{OH-}$, which are correct as you have written.

Suppose the extent of protonation is $\alpha$. After taking up the proton, $[\ce{C6H5NH2}] = c (1 - \alpha)$ and $[\ce{C6H5NH3^+}]= c \alpha = [\ce{OH-}]$.

So, we can write the following, from the above mentioned reaction:

$$
K_{\mathrm{b}} = \frac{[\ce{C6H5NH3^+}][\ce{OH-}]}{[\ce{C6H5NH2}]} = \frac{c^{2}\alpha^{2}}{c(1 - \alpha)} \approx c \alpha^{2}
$$

As the degree of protonation is very less than 1 ($\alpha << 1$), we get $\alpha = \sqrt{K_{\mathrm{b}} / c}$. Now,

\begin{align*}
\mathrm{p}\ce{OH} &= -\log[\ce{OH-}] \\
& = -\log(c\alpha)=-\log(c\sqrt{K_{\mathrm{b}}/ c})=-\log(\sqrt{cK_{\mathrm{b}}}) \\
&= \frac{\mathrm{p}K_{\mathrm{b}}}{2} - \frac{\log c}{2}
\end{align*}

(as $-\log K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{b}}$).

In the question, $\mathrm{pH} = 8.8$, so, $\mathrm{pOH} = 14 - 8.8 = 5.2$, and $c = 0.1$. Putting these values in the equation we will get

$$
\mathrm{p}K_{\mathrm{b}} = 2.(5.2 + \frac{\log(0.1)}{2}) = 9.4 \implies K_{\mathrm{b}} = 10^{-9.4} = 3.98 \times 10^{-10}
$$

Thus we get the $K_{\mathrm{b}}$ of aniline as $3.98 \times 10^{-10}$.

Note that, if you have any aqueous solution consisting of only weak base, you can apply the direct formula as $\mathrm{pOH} = \frac{\mathrm{p}K_{\mathrm{b}}-\log c}{2}$, and easily get the $K_{\mathrm{b}}$.


Iron Out and CLR mixed - toxic gas? Someone I know mixed the liquid forms of Iron Out and CLR (Calcium, Lime, & Rust Remover) together to clean out a plastic bathtub.

I have the SDS sheets for each:

https://images-na.ssl-images-amazon.com/images/I/61Dkhh0JXRL.pdf - CLR
https://summitbrands.com/wp-content/uploads/2018/03/Iron-OUT-liquid-English-GHS-Rev-01-29-18.pdf - Iron Out

The bathroom smells terrible, has produced a cough in the person using it, and now my eye burns a little from very brief contact. I was wondering what kind of gas (if any) was created when these two chemical cleaners were mixed?

Most likely, sulfur dioxide. $\ce{SO2}$. was produced. The ingredients sodium hydrosulfite, $\ce{Na2S2O4}$, and sodium metabisulfite, $\ce{Na2S2O5}$, when mixed with acid, release $\ce{SO2}$, which oxidizes to $\ce{SO3}$ eventually.

$\ce{SO2}$ is quite irritating, but not immediately damaging in low doses, <100 ppm, though it can certainly induce a life-threatening asthma attack. "Sulfur dioxide is mildly toxic and can be hazardous in high concentrations. Long-term exposure to low concentrations is also problematic." $\ce{SO3}$ is even more dangerous, since it forms sulfuric acid when it meets water, e.g. in tears in the eye.

Air the room *thoroughly. There is not much else to do: theoretically, ammonia could be used to form a salt, removing the gas from air, but there is no way you could balance the reaction, leaving you with an excess of one irritant or another.


Can sodium carbonate-sodium bicarbonate buffer be used for EDTA titration of calcium? I am planning to measure general/total hardness (calcium and magnesium) of aquarium water using EDTA titration with Eriochrome Black T indicator. From what I understand, the unknown solution must be kept basic, at pH 10, for the EDTA to work in chelating the calcium ions off of the dye indicator complex. The procedures I can find online all use ammonia-ammonium chloride buffer. I would prefer using another buffer if possible as I find ammoniacal solutions difficult to handle and relatively hard to source.

Can I use a sodium carbonate-sodium bicarbonate buffer? Will it interfere with the titrant: will the EDTA form chelates with sodium ions?

From a similar titration procedure for exclusively determining calcium ion concentration using EDTA but with a Patton-Reeder indicator (calconcarboxylic acid), pH was raised to 12.5 using sodium hydroxide to precipitate magnesium off the solution. I suppose sodium will not interfere then?

Thank you!

Sodium hydroxide $(\ce{KOH})$ works as $\mathrm{pH}$ modifier (measure the $\mathrm{pH}$ by adding drop by drop of hydroxide until $\mathrm{pH}$ is approx. $10$ to $12).$

Carbonate ion will be combined with calcium(2+) cation to form calcium carbonate. You can measure the calcium content by precipitating the calcium using soda ash light solution. Good luck.


Pubchem, InChI, SMILES, and uniqueness PubChem compound 6140 is L-phenylalanine in its neutral (not zwitterionic) form.  According this PubChem, this molecule has the following SMILES and InChI indentifiers:

SMILES: C1=CC=C(C=C1)CC(C(=O)O)N
InChI: InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/t8-/m0/s1

PubChem compound 6925665 is the zwitterionic form of L-phenylalanine (protonated amine and deprotonated carboxylate).  Pubchem has decided that this species should be called "(2S)-2-azaniumyl-3-phenylpropanoate".  The SMILES and InChI identifiers are:

SMILES: C1=CC=C(C=C1)CC(C(=O)[O-])[NH3+]
Inchi: InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/t8-/m0/s1

Confusion.  My confusion is why these different compounds have different PubChem entries (CIDs), and different SMILES identifiers...but the same InChI structure.  The different SMILES identifiers each appear to reflect the respective structures displayed by PubChem, but the single InChI identifier given for both compounds seems to reflect only the neutral form.  I even put the InChI into rdkit and converted it back to InChI.  The result was the same and rdkit interpreted this InChI as the neutral (not zwitterionic) species.  What is the reason for this discrepancy between InChI duplicity yet structural uniqueness?

Here are some possibilities:

PubChem is in error.  They should change the InChI for the zwitterionic compound.  (If so , to what?)
PubChem is right but is using definitions of compound and CID that are different than mine.
I have some kind of fundamental misunderstanding of the purpose of InChI, which I had thought would uniquely specify a molecular structure.  But InChI is designed to handle ambiguities like zwitterion vs. neutral.
rdkit interprets InChIs weird.
Something else.

Unfortunately Pubchem is right, the two structures have the same InChI string and key, since the protonation state is the same in the zwitterion and the neutral form. So the reason for the discrepancy is by design.
I also always thought, InChI was designed for distinguishing between these conformations, but it turns out to just be one of the limitations of the system. The issue is addressed in section 13.2 of the technical FAQ of the InChI trust:

  The different protonation states of the same compound will have InChIKeys differing only by the protonation indicator (unless both states have a number of inserted/removed protons greater than 12; in this case the protonation flag will also be the same, ‘A’).
  This is exemplified below by standard InChIKeys as well as standard InChI strings for neutral, zwitterionic, anionic and cationic states of glycine (note that neutral and zwitterionic states do not differ in the total number of protons so they have the same standard InChI/InChIKey):


'Hard' soap vs. 'Soft' soap: Why do they work this way? In school, we did an experiment using fat and $\ce{KOH}$ to create 'soft' soap.

Afterwards, we added $\ce{NaOH}$ to change it into 'hard' soap.

I read on multiple places on the internet that the difference between using Potassium and Sodium is what determines if it becomes a 'hard' soap or 'soft' soap.

Am I correct in the notion that the $\ce{K+}$, resp. $\ce{Na+}$ ions connect to the $\ce{O-}$ ends of the Fatty Acid?

Is it possible to skip the first step and create 'hard' soap by using $\ce{NaOH}$ right away?

Why does 'Soft' soap dissolve so much easier than 'Hard' soap? Or in other words: Why is it much easier for the $\ce{K+}$ to fall off than for the $\ce{Na+}$ ?

Basically, yes, by common terms, soft soap has $\ce{K+}$ and hard soap has $\ce{Na+}$.
They are both alkali metals, but $\ce{K+}$ is below $\ce{Na+}$ on the periodic table so maybe we should consider periodicity trends. As you go down a table the atoms become more reactive, so $\ce{K+}$ is more likely to react with polar parts of water than $\ce{Na+}$. Here are solubilities for both (though there are other values that would tell this story, perhaps better than solubility of the product):
[Solubility for $\ce{KOH}$] in $\ce{H2O}$ at 25C: 121 g / 100 mL
[Solubility for $\ce{NaOH}$] in $\ce{H2O}$ at 25C: 100 g / 100 mL
Regarding your question of going straight into hard soap, based on this link it seems doable; he just used $\ce{NaOH}$ (lye) and oils with no mention of $\ce{KOH}$ (potash?) to speak of.
And yes, the positive ions connect to the $\ce{O-}$ of the fatty acid as seen in this diagram.

(img src: http://jan.ucc.nau.edu/~doetqp-p/courses/env440/env440_2/lectures/lec19/lec19.html)


Permanent cloudy colloid with water I need to make a cloudy white liquid where the particles won't eventually settle out. It will be exhibiting the Tyndall effect for a very long time.

I'd use milk, but I don't want it to become rancid. I've tried white inks and fabric softener, but both eventually tend toward the bottom half of the container. I think it's a matter of particle size.

What substance would hit the sweet spot - small enough to stay dispersed, large enough to scatter light, and not reactive?

Starch in water might work, perhaps with table salt or other bacterial/fungal growth inhibitor. Follow directions in most recipes: mix a bit of starch in cold water and then add gradually to hot water to make a more stable "soup".

BTW, blasting a colloid with ultrasound (e.g. from an ultrasonic humidifier) homogenizes it and may delay separation.


Octahedral Crystal Field Splitting Orbital Degeneracy How are the $\mathrm{e_g}$ orbitals degenerate with each other?

Note: This isn't a homework question. After the semester ended (I don't go to MIT), I ended up on MIT open course-ware to watch some videos about areas of chemistry I haven't covered yet or haven't covered well. I am asking the question here because I have no other avenue in which to ask questions to other people. I'm just trying to use my time during the pandemic to build my knowledge of chemistry. I don't have a lot of understanding about this, I'm just looking for some help

It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$  are degenerate.  If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the  $\mathrm d_{yz},$ $\mathrm d_{zx}$ lowered.


How do salt additives and impurities affect a critical micelle concentration? So I understand that a CMC typically decreases as more salt additives are added. What I am not clear on is how exactly.

The head groups are charged in this example (one anionic and one cationic) so I am not sure how the salts would impact this? There is a counter-ion present as well.

Thank you so much for your response, its really helpful. I'm just learning about CMC and micelle so please forgive my silly questions. When you lower a CMC, are the micelles themselves now smaller or bigger? And when the literature describes micelles as smaller, does this mean the circumference is smaller, or the number of detergent molecules is smaller.

Would the presence of salt additives and impurities be of greater concern in measuring CMC by absorbance, fluorescence or conductivity???

In a solution with two dissociated ionic detergent molecules of equal charge (say two anionic detergent molecules $\ce{A^-}$), their interaction will be dominated by two effects:

Electrostatic headgroup repulsion
Hydrophobic interactions

The first contribution, electrostatic repulsion, will oppose the second and discourage association of the detergent molecules.

If you instead mix two oppositely charged detergents, $\ce{A^-}$ and $\ce{B^+}$, the electrostatic headgroup interaction will now encourage association. All else (e.g. tail length) being equal, this means mixed micelles will have a lower critical micelle concentration (CMC) than pure detergent at the same total concentration.

Adding salt (say $\ce{NaCl}$) to a charged detergent will also lower the CMC, for essentially the same reason as for mixed micelles (electrostatic repulsion is reduced).

Now comes the tricky part. If you add salt to a mixed detergent system, you set up possible competition between electrostatic attraction of salt and headgroups, with electrostatic attraction of oppositely charged headgroups. Individually, either attractive interaction has a tendency to lower the CMC. The question is what effect they will have together. My educated guess is that the effect of salt on mixed micelle systems will be subtle and complicated. The CMC will be further reduced, but interactions between oppositely charged detergent molecules will predominate so the effect of added salt will be minor.

  When you lower a CMC, are the micelles themselves now smaller or bigger? And when the literature describes micelles as smaller, does this mean the circumference is smaller, or the number of detergent molecules is smaller?

I don't think there is a universal answer to the first question.  I will have to wave my hands a little, and say it depends on the detergent, solvent, and temperature. However if you look at a table of aggregation numbers and CMCs for different detergents, you'll usually see that the ones with a lower CMC (for instance, nonionic detergents) also have larger aggregation numbers.

The second question is easier to answer than the first.  The size of a micelle can be measured by a number of different techniques. The aggregation number is often derived by making assumptions about packing in the micelle and starts from estimates of the sizes of the micelle and detergent molecule. So it's pretty safe to generalize that both the dimensions of the micelle (the effective diameter say) and aggregation number are reduced in smaller micelles.


What is the quickest way to dissolve steel? What is the quickest way to dissolve a large chunk of steel? I mean a very quick reaction, assuming that the interaction occurs in a sealed or semi-sealed environment? Reaction times of the order of a few minutes versus tens of minutes, hours or days are the objective.

I'm curious as to how quickly acids like $\ce{HNO3}$ and $\ce{H2SO4}$ can react with steel containers.

The fastest way is dissolving the stainless steel electrochemically using $\ce{H2SO4}$ at elevated temperature where the steel block is attached as an anode. Higher current densities lead to higher removal rate.

Alternatively a $\ce{FeCl3}$ with minimal content of $\ce{HCl}$ (up to 3%) heated at 45°C or higher.


How can I measure a droplets contact angle without a setup? I have to use a phone camera to replicate the Sessile drop analyzer, but there doesn't seem to be a standard to different parameters, such as how far away should my camera be from the droplet or how much liquid should I use for my droplets? Is it possible to get accurate results this way, and how should I go about it?

In 2010, Borguet et al. published in Journal of Chemical Education, DOI: 10.1021/ed100468u an intentionally simplified setup which may be useful as a reference.  A Sony Cybershot camera was placed in front of a sample scrutinized under diffuse illumination:

The additional lens shown is focussed on the drop (both for water and $n$-hexadecane, the authors recommend a volume of $\pu{5 uL}$; focal length of the length used equals $\pu{50 mm}$):.

With photos recorded like this

the contact angle determination was performed with the contact angle plugin of imageJ.  As Fiji, it is open source, free ware, and java based/OS portable.  To quote from their publication:

  For each easurement, the user must choose two points to manually
  define the baseline and three points along the drop profile (see
  the supporting information that features a set of instructions
  supplying all the details on the procedure used to make the
  measurement). The program then fits the profile of the drop and
  calculates the contact angle using the sphere approximation
  or the ellipse approximation. In our study, the ellipse approximation gave consistent results for contact angles $> 40^\circ{}$. For
  drops with contact angles $< 40^\circ{}$, the sphere approximation was
  used.

In comparison with reference data, the authors judge contact angles recorded as well as the subsequently calculated surface tensions as good enough for instruction / lab classes and research lab.


For covalent compounds, what is the microscopic difference between a solid and a liquid? I've got a question regarding the microscopic difference between liquids and solids. I know that there exists a precise phase transition between solids and liquids, so some discrete parameter must change in the transition. I'm letting the molecules under examination be covalent, so we can't assume that a crystal lattice is formed in the solid case.

(1) Are different forces dominant in between solid and liquid molecules? By "forces" I mean ionic bonds, dipole-dipole interactions, London forces, etcetera.

(2) What is the difference in the geometric structure of the molecules' interactions? Solids are fixed and gases are mobile, but what precisely are liquids? Solid structures with enough degrees of freedom to pivot fluidly? Do liquids generally have the same structure (simple cubic, hexagonal, etcetera) as their solid counterparts when they do interact?

I guess my primary confusion is that the distance between molecules in a material (solid or liquid) should intuitively be a continuous parameter dependent on the density of particles and their temperature, not a discrete parameter taking the values present in the "solid" and "liquid" states. Why the discrete nature?

Let's break this down, and keeping it to simple covalent compounds...

  Are different forces dominant in between solid and liquid molecules? By "forces" I mean ionic bonds, dipole-dipole interactions, London forces, etcetera.

All those forces (if applicable) will still exert their expected effect and there is no 'dominant force' which increases in strength when temperature rises. Just that when we raise the temperature, the molecule may have enough kinetic energy to overcome London Dispersion but not ionic forces, etc.

  What is the difference in the geometric structure of the molecules' interactions? Solids are fixed and gases are mobile, but what precisely are liquids? Solid structures with enough degrees of freedom to pivot fluidly? Do liquids generally have the same structure (simple cubic, hexagonal, etcetera) as their solid counterparts when they do interact?

So let's take a look at a typical phase transition diagram, that of water:

At the flat stages, water is changing discreetly and exists either as solid (rigid lattice) or liquid (completely formless and structure-less). And if you look at ice melting, that is exactly what you see and experience.

But the source of this question is probably the fact that you've also seen certain solids soften under heat without melting completely. These covalent materials are called thermoplastic and have a property called the "Glass Transition Temperature". The Stiffness-vs Temperature graph looks like this:

In this state, the melting solid is able to overcome the forces that hold it to a lattice, but is not able to overcome the forces that hold the molecules together and so the molecules cannot 'flow'. The reason this exists for some compounds and not others is largely a property of how their molecules hold each other together.


Preparation of Barium Hexaaluminate using co-precipitation reaction mechanism I am preparing Barium Hexaaluminate (BHA) using $\ce{Ba(NO3)2, Al(NO3)3}$, and $\ce{(NH4)2CO3}$. For $\pu{1 mol}$ of each $\ce{Ba(NO3)2}$ and $\ce{Al(NO3)4}$, I am using $\pu{1.5 mol}$ of $\ce{(NH4)2CO3}$.

Following are my questions:

What temperature to select for reaction, according to some research papers we should use $\pu{60 ^\circ C}$. But the problem with keeping reaction temperature at $\pu{60 ^\circ C}$ will cause $\ce{(NH4)2CO3}$ to decompose into $\ce{NH3}$ and $\ce{CO2}$ and it will not take place in double displacement reaction.
Can I replace $\ce{(NH4)2CO3}$ with some other carbonates?

@Maurice, I always thought that barium nitrate and aluminium nitrate will react with ammonium carbonate and barium carbonate and aluminium hydroxide will precipitate down.

$$
\begin{align}
\ce{2 Al(NO3)3 (aq) + 3 (NH4)2CO3 (aq) &-> Al2(CO3)3 (s) + 6 NH4NO3 (aq)}\tag{R1}\\
\ce{Ba(NO3)2 (aq) + (NH4)2CO3 (aq) &-> BaCO3 (s) + 2 NH4NO3 (aq)}\tag{R2}
\end{align}
$$

(As aluminium carbonate is not stable and readily decomposes to form aluminium hydroxide.)

After that, if we keep the ratio of barium and aluminium $1:12$ and raise the temperature above $\pu{1200 °C}$ by calcinating the dried precipitate, barium hexaaluminate forms.

Also, can you clarify what $\ce{BaAl6(OH)20}$ is, because according to my understanding the general formula of barium hexaluminate is $\ce{BaO.6(Al2O3)}.$

Reference

Li, J. Q.; Wang, R. K.; Chen, C. Y. Preparation of Barium Aluminate with $\ce{BaCO3}$ and $\ce{Al(OH)3}$. AMR 2015, 1096, 156–160. DOI: 10.4028/www.scientific.net/AMR.1096.156.


What exactly happened when I added KOH to hydrogen peroxide? I recently added a potassium hydroxide, $\ce{KOH}$, pellet to a small droplet of hydrogen peroxide (30 %), $\ce{H2O2}$. You could clearly see the potassium hydroxide dissolving in the droplet when instantaneously the solutions began to fizz strongly (boil and/or release oxygen gas, I'm not sure) and release a lot of heat.

I'm not sure what happened, but I've got a few ideas of what might have happened:

There is some kind of reaction between the $\ce{KOH}$ and $\ce{H2O2}$.
The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of  150.2 °C in which it also triggered the decomposition of $\ce{H2O2}$ into water and oxygen gas.
On Wikipedia I read:

  The rate of decomposition increases with rising temperature, concentration and pH.

So when $\ce{KOH}$ is dissolved, the solution has a high pH and this might trigger the decomposition of hydrogen peroxide

I did notice after a pH test that the plastic lid on which I had condensed some of the evaporated water had a very high pH, so this would most likely contain some dissolved potassium hydroxide.

In case a reaction happened, what could that specific reaction be?

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:

$\ce{H2O2 + OH- <=> HO2- + H2O}$

$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

  It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.


Is the -OCH3 group bulky? Our teacher told us that $\ce{OCH3}$ is a less bulky group when compared to other bulky groups like $\ce{CH3}$, $\ce{NO2}$ etc. Is this true? If yes, why?

In organic chemistry bulkiness isn't just a matter of geometry; one can find the volume of the group using vdW radii, but it's a crude approximation.
Quantitatively steric bulk relies on energy measurements and can be described with a cyclohexane A-value — energy required for a group to switch between axial and equatorial positions in cyclohexane.
From the table of A-values methoxy group is indeed less bulk in comparison to nitro and methyl groups:

$$
\begin{array}{lc}
\hline
\text{Substituent} & \text{A-Value}/\pu{kcal mol-1} \\
\hline
\ce{-OCH3} & 0.6 \\
\ce{-NO2} & 1.1  \\
\ce{-CH3} & 1.7 \\
\hline
\end{array}
$$


Polyurethane epoxy resin chemistry advice please? My friend can buy polyurethane epoxy from his workplace for 3 dollars a litre, except that he doesn't know precisely the data sheet for it. His factory gets it in 20 ton deliveries.

He says that the mix has to be around 99% Polyurethane, 0.2% cobalt, and 0.8% of a toxic transparent catalyst liquid.

It's a generic industrial chemistry used everywhere, and can be mixed roughly by eye to achieve different drying speeds.

Do you know what the name of this resin system is called and where i can find precise information on it? What is the catalyst?

The resin is the same variety commonly used to produce fiberglass baths, boats and many other uses.

The resin is unsaturated polyester, using methyl ethyl ketone peroxide (MEKP) as catalyst, cobalt solution for example cobalt octoate (Coct) as an accelerator.

It isn't often sold to consumers because it is too explosive and corrosive, the MEKP reacts explosively with raw cobalt, and it's dangerous to the eyes, it can cause blindness in 15 seconds if splashed in the eye.

The MEKP doseage is 1-2% per volume of polyester, depending on desired drying time. The cobalt is mixed first and dyes the polyester a whiskey color, you can dose it by color, the darker whisky color the faster it will dry.

Industry standards are 1% and 6% cobalt solution, 1% is dosed at 0.5% to 2%, however there is no way to tell without the product data sheet if the cobalt solution 1%, 6%, 10%.

safety:

MEKP and cobalt mix = flames
MEKP in eyes = blindness
wear goggles and rinse with water and go to hospital if you get some MEKP eye contact.
wear protective clothing and avoid skin contact.

"Cobalt octoate is the most reactive and most commonly used paint drier. Cobalt octoate catalyses the oxygen uptake and accelerates peroxide formation causing a surface film to form rapidly and hence known as oxidation catalyst. It basically a metal caroxylate used to accelerate the drying process like linseed oil. It accelerates the decomposition of the organic peroxide initiators (often called catalysts) and promotes the cross-linking of polyester resins.
"

Note that baths are finished with a gloss product called gel coat.


Vacuum good inside a box? I have no clue whatsoever about Chemistry on a academic point of view.

We wish to transport some goods.
Let's say we vacuum some of those goods, by doing like so:

We place those goods on a box, we close the box, and then we vacuum seal the box (not the goods).

Can we consider the goods to be vacuum protected?

Does that mean, that no hight heat will affect those goods?

Thank you in advance, and sorry for not using specific chemical terminology.

Let's imagine that your transport box has a decently closing lid. It you place the herbs in the box, close the lid , stuff the box into a vacuum sealer bag, remove the air from the bag until it is shrunk around the box and finally seal it, you may consider it sealed.

This means:

According to "household standards", it is air-tight to the outside. No air from the environment will enter the bag, unless it is damaged.
There is however still air in the box. Whatever degradation, loss of flavour, etc. may happen to the herbs during transport still can, since there is oxygen.

However, if you vaccum-seal the box, or replace the air in the box with nitrogen, you will minimize (or fully prevent) degradation of the herbs.


Is the number of tetrahedral voids always equal to double the number of octahedral voids in any crystal? I came accross the question whether the number of tetrahedal voids is always equal to twice the number of octahedral voids (in the case where both the voids are present). The key was given as the statement was true. Could anyone justify this?
I know that it is true for ccp and hcp but can this be generalized for all types of crystals?

Note that, in case of tetrahedral voids, each sphere is in contact with three tetrahedral sites above, and three tetrahedral sites below, hence, there are two tetrahedral sites associated with each sphere. Thus,

Number of tetrahedral voids $(N_{th}) = 2 \times$ (Rank of unit cell)

Or alternatively, if you divide a FCC lattice into 8 octants, each octant will have a tetrahedral site at it's body centre.

For octahedral sites, each sphere is associated with one octahedral sites, thus,

Number of octahedral voids $(N_o) =$ Rank of unit cell

For FCC, besides the body center, there is one site at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit cell, and the three belonging to two adjacent unit cells.

For clarity see:

For HCP,


How can I prepare Sn(NO3)2? What is the method to be followed to prepare $\ce{Sn(NO3)2}$?

I tried to prepare $\ce{Sn(NO3)2}$ from reaction below:

$$\ce{SnCl2 + 2HNO3-> Sn(NO3)2 + 2HCl}$$

But $\ce{SnCl2}$ is not soluble. We are getting a turbid solution.

Tin nitrate can be prepared by mixing tin and nitric acid at various proportions.

$$\ce{5Sn + 12HNO3 -> 5Sn(NO3)2 + N2 + 6H2O}$$

  Tin react with nitric acid to produce nitrate tin(II), nitrogen and
  water. Nitric acid - diluted solution. This reaction takes place
  slowly. Impurities: nitrogen(II) oxide (chemiday 1)

$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$

  Tin react with nitric acid to produce nitrate tin(II), ammonium
  nitrate and water. Nitric acid - very dilute solution. This reaction
  takes place slowly.(chemiday 2)

$$\ce{3Sn + 8HNO3 -> 3Sn(NO3)2 + 2NO + 4H2O}$$

  Tin react with nitric acid to produce nitrate tin (II), oxide nitrogen
  (II) and water. Nitric acid - diluted solution.(chemiday 3)

A brief information on tin nitrate can be found here*

  Stannous Nitrate is obtained in white leaflets having the
  composition $\ce{Sn(NO3)2.20H2O}$ by cooling to -20° C. a solution of
  stannous oxide in nitric acid of density 1.2. It is also formed by the
  action of highly diluted nitric acid on the metal, some of the acid
  thus suffering maximum reduction to ammonia:

  $$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$

  According to C. H. H. Walker, tin dissolves in nitric acid forming
  stannous and stannic nitrates, the relative proportion of these two
  salts produced depending on the temperature and the strength of the
  acid; moreover, the yellowish white precipitate which separates when
  the somewhat concentrated acid is employed is said to be a hydrated,
  ill-defined, stannic nitrate.

*http://tin.atomistry.com/stannous_nitrate.html


What is the industrial yield of paracetamol? Anyone know how I can find out the typical yield of paracetamol (acetaminophen) from the acetylation of 4-aminophenol?

Please provide the source if you happen to know!

For a variety of explanations of industrial techniques as applied to acetylation,  including acetaminophen synthesis,  try here
https://www.google.com/patents/US20120065423
The examples are right at the bottom and are only scaled down lab versions of the industrial technique. Over 90% yield should be easily achievable if acetic anhydride is used (and is achieved in the industrial synthesis of ibuprofen).  The patent examples give yields for the use of acetic anhydride and acetic acid, as well as the technology employed e.g. continuous stirred tank reactor (cstr)  vs plug flow reactor (pfr or tubular reactor)  vs reactive distillation.  Enjoy.


Why does the Grignard reagent of CH3I can only be prepared using diethyl ether as solvent but there is no reaction when using THF? Does anyone know why Grignard reagent of CH3I can only be prepared using diethyl ether as a solvent and yields no reaction when developed in THF? Theoretically, I is the most reactive halogen for Grignard reactions, so using THF or diethyl ether should make no difference.

MeMgI is usually prepared in ether, but it can also be prepared in THF. However, magnesium iodide is sparingly soluble in THF, so RMgI reagent in THF should not be very concentrated, so that no MgI2 precipitate forms.


Preparation of p-chlorobenzoic acid from p-chloronitrobenzene I was reading my book where I spotted the preparation of p-chlorobenzoic acid from p-chloronitrobenzene.

The preparation method there was given as such:

$\ce{Sn/HCl}$
$\ce{NaNO2 + HCl}$ ($\pu{0-5 ^\circ C}$)
$\ce{CuCN}$
$\ce{H2O/H+}$

Now, the thing is that, I recall the first two steps from the process of forming diazonium salts but I have never seen the last two steps give the $\ce{-COOH}$ group.

Also I recall that $\ce{CuCN}$ gives us the nucleophile $\ce{CN-}$ where the negative charge is on the nitrogen atom and also that $\ce{H2O/H+}$ is used for hydration but I don't see how these two could help us form p-chloronitrobenzene from Benzenediazonium chloride.

This is simply the summery of all comments involved:

$$\ce{Cl-C6H4-NO2 ->[Sn/HCl] Cl-C6H4-NH3+Cl-} \tag1$$

$$\ce{Cl-C6H4-NH3+Cl- ->[\text{a. } NaNO2/HCl \ (\pu{0-5 ^\circ C})][\text{b. } CuCN] Cl-C6H4-CN} \\ \tag2$$

$$\ce{Cl-C6H4-CN ->[H+/H2O][\Delta] Cl-C6H4-COOH} \\ \tag3$$

Note: The reaction $(1)$ is a typical reduction reaction, which reduce nitro group to amino group. The reaction $(2)$ is called Sandmeyer Reaction. Finally, The reaction $(3)$ is a hydrolysis condition to convert cyano group to carboxylic acid group. That hydrolysis can be done under basic conditions as well (then, final product is carboxalate).

Also note that even though it is not necessary, I put this answer up just to avoid people asking same kind of questions again and again. When they have a visual answer they might not do so.


Difference between aldehydes and ketones Why are ketones and aldehydes two different categories? To me it seems like aldehydes should be a subset of ketones. Since the general formula for ketones is CnH2n+1COCmH2m+1 and for aldehydes it is CnH2n+1COH which is just the formula for ketones but with m=0.

I am not looking for an opinion, but rather an explanation behind the reasons to have two different groups. Are there any inherent differences between them in how they react with other compounds or how they are produced?

Ketones are to aldehydes as ethers are to alcohols. You could extend this to water. Is water formalcohol? Once you begin to explore the periodic table deeply enough, you will see that there is no distinctions really exist. Only trends. The same is true of organic structure. Organic chemistry is a continuum and the groups we create are constructs -- but they are not arbitrary. Especially in organic chemistry, we group things into these categories because they help us to make assumptions based on consistent trends in reactivity. Ketones and aldehydes have significantly different chemistry. There is only a minor difference between the chemistry of acetone and butanone, but there is a pretty big difference between the chemistry of acetone and acetaldehyde.

To make a long story short, aldehydes are categorized differently because their chemistry varies significantly enough that different sorts of reagents are needed for the same types of reactions.

Aldehydes are much more electrophilic. This is because they don't have the electron donating effect of an alkyl group. Aldehydes have much more acidic alpha-protons. Aldehydes are also more easily oxidized. Aldehydes pose problems different problems in synthesis. For Their non-acidic proton, for example can be hard to remove for substitution.

An interesting example of a useful tool for dealing with aldehyde protons is dithiane chemistry. You can use a dithiane protecting group to make the aldehyde proton acidic, substitute it, then put the oxygen back on in a deprotection step.


Why is UV-Vis Spectroscopy not like IR Spectroscopy UV/Vis Spectroscopy is where electromagnetic radiation in the UV to Visible range is shone at a sample to be analysed. The absorbance of each wavelength is recorded and then plotted (very similar to Infrared spectroscopy). That's why it surprised me so much.

This graph is so smooth and with only 1 peak. This is completely different to the IR graph (below):

Why is it that UV-Vis spectroscopy only produces 1 peak? Is it something to do with orbitals overlapping and hence only producing one major peak?

I'm collecting ideas from the excellent comments and adding some of my own:

The typical UV/Vis spectrum has a narrow range

On the low energy side, it ends where the visible spectrum ends, at about 800 nm. On the high energy side, it ends where water and quartz start absorbing too much, at about 220 nm. Compared to an IR spectrum, it has a narrow range.

Most functional groups have no electronic transitions in that range

As far as organic compounds are concerned, the most common functional group to absorb in the UV is the benzene ring, and extended conjugated systems (e.g. beta carotene) will have electronic transitions in the visible. On the other hand, most organic functional group will show vibrational transitions in the range of a typical IR spectrum.

Peak widths

The spectrum of an electronic transition is broad because it is coupled with vibrational and rotational transitions. An example is the electronic spectrum of iodine:

Source: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/13%3A_Molecular_Spectroscopy/13-06._Electronic_Spectra_Contain_Electronic%2C_Vibrational%2C_and_Rotational_Information

In solution, the bands are not resolved, so the spectrum appears as a single broad peak.


Is it possible to make a disaccharide from two fructose monomers? If so, what is its name? Maltose is a disaccharide made from two molecules of glucose; and sucrose (table sugar) is made from glucose and fructose. Is it possible to combine two fructose molecules to make a disaccharide?

If so, what is that called? Or if not, what prevents the glycosidic bond from forming?

in agave plants the typical sugars are fructans (fructose-polimers) like inulin, levan, agavin - with sometimes really complex structures comparable to starch. As starch breaks down into glucose, maltose, etc. you have also a need for break-down process e.g. in the tequila/mezcal industry which is usually a thermic hidrolisis (no enzymes like in beer). This hidrolisis is never complete so you have dimers and trimers (or higher) fructose-sugars, which are even metabolizable by yeast-strains. They are referred to as nistose or kestose (https://pubchem.ncbi.nlm.nih.gov/compound/1-Kestose#section=Top).
I hope this answer helps you to finally resolve your question - check out also user55119 coment.


Why does Cu(OH)2, which is generally insoluble due to low Ksp become soluble in excess of a base, such as NaOH? Where does the "excess" criteria come into the picture? Also, can the equilibrium
$$\ce{Cu^2+ + 6 H2O <=> [Cu(H2O)6]^2+}$$

participate in the solubility process in any case? So there are 2 questions: one regarding the solubility of $\ce{Cu(OH)2}$ and other regarding the equilibrium mentioned above?

Also, can the equilibrium $$\ce{Cu^{2+} + 6H2O <=> [Cu(H2O)6]^{2+}}$$ participate in the solubility process in any case?

This equilibrium does contribute to the solubility of copper(II), but probably not in the way it appears you are envisioning it to.  When copper(II) is fully dissolved in moderately acidic solution, as long as the anion of the acid is a weak ligand for copper (e.g., $\ce{ClO4-}$ or $\ce{NO3-}$), the copper ions are dissolved as the aquo complex, $\ce{[Cu(H2O)6]^{2+}}$.

As the $\mathrm{pH}$ is adjusted upward with, e.g., $\ce{NaOH}$, though, other reactions come into play (see below) and the hexaaquo species contributes negligibly to the overall solubility behavior.

  Where does the "excess" criteria come into the picture?

To answer this question, one really must first understand what's actually going on when solvated metal ions undergo acid/base reactions.  As I indicated above, in acidic solution copper(II) exists as an aquo complex, typically written as the hexaaquo species, $\ce{[Cu(H2O)6]^{2+}}$.  As the $\mathrm{pH}$ rises, the following series of acid-base ("hydrolysis") reactions occurs$^\ddagger$:

$$
\ce{[Cu(H2O)6]^2+ + OH- <=> [Cu(H2O)5(OH)]+ + H2O}\tag{1}
$$

$$
\ce{[Cu(H2O)5(OH)]+ + OH- <=> [Cu(H2O)4(OH)2]^0 + H2O}\tag{2}
$$

$$
\ce{[Cu(H2O)4(OH)2]^0 + OH- <=> [Cu(H2O)3(OH)3]- + H2O}\tag{3}
$$

$$
\ce{[Cu(H2O)3(OH)3]- + OH- <=> [Cu(H2O)2(OH)4]^2- + H2O}\tag{4}
$$

(For simplicity, I have omitted the $\ce{M2L2}$ hydrolysis complex reported in Vuceta$^\ddagger$, and have written the above in terms of addition of alkali so as to highlight the similarities among the series.)

So, to some extent, the copper can be treated much like a polyprotic acid: as hydroxide ion is added, the copper is able to "absorb" it, forming water/hydroxide complexes with increasing numbers of hydroxides attached.

Per the low $K_\mathrm{sp}$ of $\ce{Cu(OH)2}$, however, the following (fictitious) solubility relation must also be considered:

$$
\ce{ [Cu(H2O)4(OH)2]^0 (aq) <=> Cu(OH)2 (s) + 4H2O} \tag{5}
$$

Eq. $(5)$ is a re-interpretation of the typical solubility expression:

$$
\ce{[Cu^2+][OH^{-}]^2} \leq K_\mathrm{sp} \tag{6}
$$

Qualitatively, Eq. $(5)$/Eq. $(6)$ indicates that the copper existing as $\ce{[Cu(H2O)4(OH)2]^0}$ is limited to a certain concentration in solution, depending on the conditions of the system.

Thus, as base is added to an acidic solution of copper and the equilibria are swung to a range that favors $\ce{[Cu(H2O)4(OH)2]^0}$/$\ce{Cu(OH)2}$, much of the copper precipitates out.  But, as more base is added, eventually there's enough hydroxide present that the equilibria can start to shift toward $\ce{[Cu(H2O)3(OH)3]-}$ and $\ce{[Cu(H2O)2(OH)4]^2-}$, and the copper will start to redissolve.

This, finally, leads back to question: what does "excess" mean in this context?

To help answer this question, I input the stability constants from Vuceta$^\ddagger$ into HySS, a program that is designed to simulate these sorts of equilibria and that is able to handle precipitates like $\ce{Cu(OH)2 (s)}$.  The HySS file used to generate the below figures can be downloaded from Google Drive here.  The below figure shows the simulated speciation diagram for $10~\mu\mathrm M$ total copper over a $\mathrm{pH}$ range of $4$ to $14$ (all ligating water molecules are omitted from the labels; click the image for a larger version):

As expected, the copper is fully dissolved at an acidic $\mathrm{pH}$ of $4$, but starts to precipitate at around $\mathrm{pH}~7$.  Coincidentally, the concentration of the dissolved $\ce{[Cu(H2O)4(OH)2]^0}$ is predicted to be equal to the "concentration" of the precipitated $\ce{Cu(OH)2 (s)}$ (mass of solid divided by the liquid volume) over much of the insoluble range. Starting at about $\mathrm{pH}~11.5$, the base is in sufficient excess (~$316$-fold) over the copper to begin to re-dissolve it.  Only around $\mathrm{pH}~13$ or so does all of the solid finally re-dissolve, at a ~$10,000$-fold excess of base.

At the higher copper concentration of $10~\mathrm{mM}$, re-dissolution has hardly begun to happen at all by $\mathrm{pH~14}$ (click image for larger version):

This is perfectly sensible, as the base is only in excess of the copper by ~$100$-fold at that $\mathrm{pH}$.  In order to observe re-dissolution at this copper concentration, one would presumably have to use extraordinarily concentrated alkali in order to achieve a sufficient excess of base.

$^\ddagger$ Vuceta & Morgan, Limnology and Oceanography 22(4): 742, 1977, doi:10.4319/lo.1977.22.4.0742


Calculate the second ionization energy for the Mg from the following data? here is a problem :

What I did " as shown in the picture below "

I rewrote the information given in terms of reactants and products ( equations )// in Part 1

Then I used the Hess law to obtain the desired result //in part 2

in part 3 // I added and cancelled the terms that appear in both sides

but I left with $O_{(g)}$ !! and the enthalpy is 1192 which is inconsistent with the correct value 1442 !!

What's wrong?

Thanks for showing your work.  We're given that
$$\ce{\Delta H_{f} (MgO (s))~ =~ -601$.$8 ~kJ/mol}$$
So any balanced equation we write that has MgO as the product will have this same heat of formation.  An equation that produces MgO and allows us to solve the problem is
$$\ce{Mg^{+2}(g) + O^{-2}(g) -> MgO (g)}$$
The heat of formation of $\ce{O^{-2}}$ is given by
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2 the ~O2~ bond~ energy~ (only~ half~ because~ we~ only~ need~ 1~ oxygen~ atom) + the ~first ~ionization~ potential~ for~oxygen + the~second~ionization~ potential~ for~ oxygen}$$
or
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2(498.7)~-~141+780~ = ~$888.4$~ kJ/mol}$$
Similarly the heat of formation of $\ce{Mg^{+2}}$ is given by
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~ \Delta H_{sublimation}Mg(s) + the ~first ~ionization~ potential~ for~Mg + the~second~ionization~ potential~ for~ Mg}$$
or
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~130 + $738.1$ + IP_2(Mg)}$$
Before solving for $\ce{IP_2(Mg)}$ there is one more adjustment to make.  Everything we've done so far has been for the gas phase except for our starting heat of formation for $\ce{MgO}$ which was for the solid, so let's adjust this heat of formation to its gas phase value.
$$\ce{\Delta H_{f} (MgO (g))~ =~ \Delta H_{f} (MgO (s)) + lattice~ energy~ MgO }$$
or
$$\ce{\Delta H_{f} (MgO (g))~ =~ $-601.8$+ 3800 ~=~$3198.2$~ kJ/mol}$$
Now, inserting all of these values into our equation we get,
$$\ce{130 + $738.1$ + IP_2(Mg) + $888.4$~=~$3198.2$}$$
or
$$\ce{IP_2(Mg)~=~$1441.7$~kJ/mol}$$


Concentration of a substance in a solution Will the Concentration of any dissolvable substance be uniformly distributed in any type of liquid?

I am willing to read more about this topic , so if you have good suggestions for helpful books , please share it with us

Using a FULLY MIXED sugar solution (in a closed container to prevent evaporation) as an example, if you could measure out successive 1-ml samples with perfect accuracy, and could count the sugar molecules in each sample with perfect accuracy, you would expect to find that the number of sugar molecules varies and follows a normal distribution with a mean value corresponding to the "nominal" concentration of the solution.

You can observe this phenomenon experimentally in the case of standard solutions containing a given DNA sequence. When you create a series of increasingly diluted samples and then analyze each dilution multiple times using a PCR assay designed to detect that sequence (modern PCR assays can literally detect one molecule of DNA!), you begin to observe the expected proportion of negative sample results as you approach a dilution factor corresponding to one copy of DNA per sample. That is because when you repeatedly sample your solution containing a nominal concentration of one DNA copy per sample, some physical samples will have zero copies, some will have one (highest probability), some will have two, etc. with decreasing probability as you go up.


Reaction of 2,4-DNPH with aromatic aldehydes Does 2,4-dinitrophenylhydrazine react with aromatic aldehydes, e.g. benzaldehyde (phenylmethanal), in a condensation reaction similar to 2,4-DNPH's reaction with aldehydes and ketones ?

2,4-DNPH is a derivatization reagent for the identification of ketones and aldehydes.
This was extensively used in organic analysis before instrumental methods were readily available.
You determined the boiling point of the problem, which in the case of benzaldehyde is $179\ ^{\circ}\mathrm{C}$ , then you made a solid derivative and determined it's melting point.
In the case of benzaldehyde and 2,4-DNPH that would be $237\ ^{\circ}\mathrm{C}$ .
Then you go to the tables, cross check both values and had benzaldehyde identified.

For aldehydes and ketones there are different options: oxime, semicarbazone, phenylhydrazone, p-nitrophenylhydrazone and 2,4-dinitrophenylhydrazone.
The latter one was usualy preferred because they have higher melting points and so it was easier to get a solid even if not totally pure.


Chemical equilibrium constant We define chemical equilibrium as the point in a reaction where the rates of the forward and reverse reactions become equal. We also write

$$K_c = \frac{k_f}{k_r}$$

If at equilibrium the rate of the forward reaction and the rate of the reverse reaction become equal then their rate constants should also be same. If rate constants $k_f$ and $k_r$  become the same, then $K_c$ should always be equal to 1. However, the value of $K_c$ is not always 1. Why is it so?

For an example reaction $\ce{A + B <=> C + D}$

the forward rate is given by

$$r_\mathrm{f} = k_\mathrm{f}[\ce{A}][\ce{B}]$$

and the backward rate is given by

$$r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$

For the rates to be equal one simply needs

$$k_\mathrm{f}[\ce{A}][\ce{B}] = r_\mathrm{f} = r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$

Note that this does not imply that the two rate constants are the same

$$k_\mathrm{f} \neq k_\mathrm{b} \quad \text{(in general)}$$


Ca energy release when gaining an electron Why does $\ce{Ca}$ have a negative electron affinity, i.e. energy is released when it gets an electron (ca. $\pu{-2 kJ/mol}$), when $\ce{Be}$ and $\ce{Mg}$ have positive values?

I know that the electron will be settled in the $\ce{3d}$ orbital which has a lower principal quantum number than $\ce{4s}$, but still, why would energy be released?

Two sign conventions: (1) the more common one states that a positive
  electron affinity value represents energy release when an electron is
  added to an atom; (2) the other states that a negative electron
  affinity represents a release of energy.

http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf

  Recommended values for these electron affinities, in the units
  commonly employed in introductory texts and with the sign convention
  used here, are 2.37, 5.03, and 13.95 kJ/mol for Ca, Sr, and Ba,
  respectively. The endothermic electron affinities often quoted for Be
  and Mg are also too large and should be reported simply as ">0". An
  argument for a return to the original sign convention for the electron
  affinity is presented in this paper.

http://pubs.acs.org/doi/abs/10.1021/ed074p123

  OP: I know that the electron will be settled in the 3d orbital

Actually, as explained in the above article, the electron occupies 4p.

  OP: but still, why would energy be released?

Very little energy is released and the lifetime of Ca- is on the microsecond scale in practical experiments.  The reference Contributions to the electron affinity of calcium and scandium may be the best source for theoretical models that do and don't predict a slight energy release.  According to Atomic negative ions: structure, dynamics and collisions, core electrons need to be considered in addition to the valence electrons in calculating the stability of Ca-.  Early calculations that considered only valence electrons predicted that Ca- was unstable.  Configuration-interaction study of differential correlation energies in Ca-, Ca, and Ca+ presents detailed calculations of energy levels.


How hydrate isomerism results difference in colours? The standard example is $\ce{CrCl3.6H2O}$, which can have three distinctly different
crystalline compounds now known as

$\ce{[Cr(H2O)6]Cl3}$ (violet),

$\ce{[CrCl(H2O)5]Cl .H2O}$
(blue-green), and

$\ce{[CrCl2(H2O)4]Cl.2H2O}$ (dark green).

How does this difference of ligands in octahedral system affect the colour of compound? I think it may be due to the change of splitting energy but if that so then how does it actually work?

Your idea appears to be correct.

Initially, the structure of $\ce{CrCl3.6H2O}$ corresponds to the dark green compound $\ce{[CrCl3(H2O)3].3H2O}$. This complex shows hydrate isomerism; i.e. the chlorido ligands are step by step replaced by water molecules:

$$\ce{$\underset{\text{dark green}}{\ce{[CrCl3(H2O)3].3H2O}}$ <=> $\underset{\text{dark green}}{\ce{[CrCl2(H2O)4]Cl.2H2O}}$ <=> $\underset{\text{blue green}}{\ce{[CrCl(H2O)5]Cl2.H2O}}$ <=> $\underset{\text{violet}}{\ce{[Cr(H2O)6]Cl3}}$}$$

Thus, the actual complexes are:

$\ce{[CrCl3(H2O)3]}$ (dark green)
$\ce{[CrCl2(H2O)4]+}$ (dark green)
$\ce{[CrCl(H2O)5]^2+}$ (blue green)
$\ce{[Cr(H2O)6]^3+}$ (violet)

Each of these complexes has octahedral geometry. Cr(III) has a $\mathrm{d^3}$ electronic configuration. In the ground state, three electrons occupy the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$). The two $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2{-}y^2}$ and $\mathrm{d}_{z^2}$) are empty.

Three spin allowed d–d transitions are expected for Cr(III) complexes. The corresponding absorption maxima of $\ce{[Cr(H2O)6]^3+}$ are:

$$\begin{alignat}{2}
&\mathrm{^4A_{2g}\to{^4T_{2g}}}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(F)}}\quad&&\tilde\nu_2=24\,500\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(P)}}\quad&&\tilde\nu_3=38\,600\ \mathrm{cm^{-1}}
\end{alignat}$$

The first transition $\mathrm{^4A_{2g}\to{^4T_{2g}}}$ corresponds to promoting one electron from the $\mathrm{t_{2g}}$ level to the $\mathrm{e_g}$ level. Thus, it directly gives the value of $10\ \mathrm{Dq}$ as $17\,400\ \mathrm{cm^{-1}}$.

The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the low-field ligand $\ce{Cl-}$, the value of $10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are lowered:

$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\\
&\vdots \\
&\ce{[CrCl6]^3-}\quad&&\tilde\nu_1=13\,200\ \mathrm{cm^{-1}}
\end{alignat}$$

By way of comparison, if the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{CN-}$, the wavenumber of the corresponding absorption maximum is increased ($\tilde\nu_1=26\,700\ \mathrm{cm^{-1}}$ for $\ce{[CrCN6]^3-}$).

The wavenumbers can be converted to the wavelength of the absorbed light:

$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\quad\lambda=575\ \mathrm{nm}\ \text{(yellow)}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\quad\lambda=610\ \mathrm{nm}\ \text{(orange)}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\quad\lambda=640\ \mathrm{nm}\ \text{(red)}\\
\end{alignat}$$

Since the colour of the absorbed light in this series changes from yellow to orange to red, the corresponding colour of the remaining visible light changes from violet to blue green to green.


Can you turn lead dioxide into lead? I've read that you can reduce lead oxide to lead by heating it with carbon. Does this also work with lead dioxide?
I took apart an old lead-acid car battery and realised then, that there are almost equal parts of lead and lead dioxide.
Already having saved and distilled the acid, I started to think there was a lot more effort involved giving all the safety steps and time than buying the items separately.

Decomposition of lead(IV) oxide at different temperatures forms different types of oxides. At around $\pu{600 ^\circ C}$, lead(II) oxide forms. Wikipedia adopts the scheme of lead(IV) oxide decomposition in air summarized by White and Roy [1]:

$$\ce{PbO2 ->[\pu{293 ^\circ C}] Pb12O19 ->[\pu{351 ^\circ C}] Pb12O17 ->[\pu{375 ^\circ C}] Pb3O4 ->[\pu{605 ^\circ C}] PbO}$$

You can then reduce lead(II) oxide to lead using carbon or carbon monoxide at around $\pu{1200 ^\circ C}$. The effect of dissolved inorganic carbon is discussed by Wang et al. [2]
$$
\begin{align}
\ce{PbO + CO &->[\pu{1200 ^\circ C}] Pb + CO2}\\
\ce{2 PbO + C &->[\Delta] 2 Pb + CO2}
\end{align}
$$
Note that lead(II) oxide can also decompose to release toxic fumes of lead which can be harmful and also doesn't serves your purpose of getting solid lead.
References

White, W. B.; Roy, R. Phase Relations in the System Lead-Oxygen. J American Ceramic Society 1964, 47 (5), 242–249. DOI: 10.1111/j.1151-2916.1964.tb14404.x.
Wang, Y.; Xie, Y.; Li, W.; Wang, Z.; Giammar, D. E. Formation of Lead(IV) Oxides from Lead(II) Compounds. Environ. Sci. Technol. 2010, 44 (23), 8950–8956. DOI: 10.1021/es102318z.


Kinetic Isotopic Effect in EAS It is observed that rate of reaction of two isotopologues of benzene, $\ce{C6D6}$ and $\ce{C6H6}$, is the same for electrophilic aromatic reactions, as the deprotonation is not the rate limiting step.  Instead the carbocation is usually limiting.

Do iodination and sulphonation somehow do show different rates with these isotopologues?  Is $\ce{C6H6}$ the faster of the two, due to stronger C-D bond (smaller size of D)? If this is correct what is the reason?

Does the protonation becomes the rate limiting step? if yes, then why in only with iodination and sulphonation?

Update : Found potential energy graphs for suphonation and nitration and it does seem like protonation either becomes the rate determining step or just smaller than carbocation hill, but why in sulphonation, the deprotonation curve was high is still a mystery.

In general, deprotonation of a $\ce{C-D}$ bond would require more energy than that for $\ce{C-H}$ bond because the zero point vibrational energy of the bond is lowered. By referring to our harmonic oscillator zero point energy equation,
$E_0 = \frac{1}{2}h \nu$
where frequency is,
$\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}$
and the reduced mass ($\mu$) being,
$\mu = \frac{m_1m_2}{m_1 + m_2}$
we see that as the mass for either atom increases, the zero point energy decreases, therefore, the bond is relatively more stable than its lighter isotopologue. However, swapping out isotopes in a molecule does not change the potential energy surface for the reaction [1]
1.) Sulphonation
Changes in the rate determining step may arise from the type of sulfuric acid used. Since fuming sulfuric acid has a higher concentration of $\ce{SO3}$ (the electrophile) then more dilute forms of $\ce{H2SO4}$, the rate limiting step is no longer forming the bond between benzene and $\ce{SO3}$, rather, it is re-forming the aromatic ring through deprotonation. Ergo, if you did a KIE study with different concentrations of sulfuric acid, you may see that only fuming $\ce{H2SO4}$ shows significant values of $\frac{K_\ce{H}}{K_\ce{D}}$.
2.) Iodination
This article [2] studied the iodination of benzene with the source of iodine being generated through the "oxidation of 0.5 g of $\ce{I2}$ in acetonitrile-tetraethylammonium perchlorate to 2.4 faradays/mol."

The article reported that the secondary kinetic isotope effect for $k_1$ and $k_{-1}$ is negligible, however, $\frac{K_\ce{H}}{K_\ce{D}} = 2.25$ for $k_2$.
Here is a thesis I found that discusses the iodination of aromatic rings [3]
The thesis, in short, suggests that whether the deprotonation is rate limiting has to do with the concentration of the reactants, along with the transition state of the molecule. The transition state is affected by the strength of the base, since a higher $\mathrm{p}K_\mathrm{b}$ may give a transition state where the hydrogen is partially bound to both the ring and the base, making this step in the reaction faster (therefore, not rate limiting). However, it did only investigate substituted aromatic rings, where substituents can play a significant role in the transition state, making the reaction with benzene a little less comparable.
3.) Nitration
This article reports no KIE observed in nitration of benzene [4].


Acetaldoxime Synthesis: Hydroxylamine vs Hydroxylamine HCl Wikipedia says acetaldoxime can be prepared by reacting Acetaldehyde & Hydroxylamine in the presence of NaOH. https://en.wikipedia.org/wiki/Acetaldoxime

But the reaction shown uses Hydroxylamine Hydrochloride instead of Hydroxylamine.

Does it not matter? Do both work? Would Hydroxylamine Sulfate work as well?

Hydroxylamine is a low-boiling liquid readily oxidized to explosive nitrogen-oxygen compounds.  It is stored as the hydroxylammonium hydrochloride salt for convenience and safety, since the salts do not oxidize as easily.  Additionally, hydroxylamine itself may explode when heated.

The sodium hydroxide is to render the freebase which then may participate in the reaction.  In principle, the hydroxylammonium bisulphate would work as well.

It is somewhat important that an exact stoichiometry be used for base to hydroxylamine to prevent aldol side-reactions if relevant (which for acetaldehyde, they are).  Additionally, this reaction is reversible in the presence of acid and water, so all the acid must be consumed, since water is a byproduct of the condensation.  Given the explosive nature of hydroxylamine, azeotropic distillation via Dean-Stark trap is not viable for dehydration.  Chemical dehydration may be effective.


Defining an acid-base reaction from the perspective of the Arrhenius theory of acids and bases $$\ce{NH3 + HCl -> NH4Cl}$$

The above equation is the reaction of ammonia with hydrochloric acid (all substances are aqueous). I understand that ammonia counts as a base because it can react with water to give hydroxide ions, and $\ce{HCl}$ is an acid. By logic, this should be an 'acid-base' reaction and this link
says that it is indeed an acid-base reaction but how is this so if no water is produced?

I am learning acid-base theory from the beginning and can understand why this is an acid base reaction from the perspective of the Brønsted-Lowry theory and the Lewis theory, but I am now attempting to understand the Arrhenius theory, so please explain to me, in terms of the Arrhenius theory, why this is an acid-base reaction and a 'neutralistaion' if there is no water produced and no hydroxide ions which take place in the reaction what so ever?

I understand that ammonia counts as a base because it can react with water to give hydroxide ions, and HCl is an acid.

This is where you went wrong.
The Arrhenius theory is the oldest of the common acid-base theories. According to Arrhenius:

An acid is a compound that will dissociate in aquaeous solution forming $\ce{H+}$ ions.
A base is a compound that will dissociate in aquaeous solution forming $\ce{OH-}$ ions.

As per the Arrhenius theory, ammonia is not a base. Also, contrary to all later theories (Brønsted-Lowry, Lewis and Pearson), Arrhenius did not account for any acid-base reactions; instead, he intended to define only those substances whose solutions reacted acidicly or basicly. It is therefore meaningless to define an acid-base reaction according to the Arrhenius theory.
What Arrhenius did account for was neutralisation reactions. Indeed, if an Arrhenius acid and an Arrhenius base are mixed, the form water and a salt, which can be easily shown in the equation below:
$$\ce{(H+ + A^-) + (B+ + OH^-) -> HOH + A^+B^-}$$
But it was not until Brønsted’s and Lowry’s theory that it was realised how a reaction should always be acid-base and not one or the other.


A glass of water with ice-cubes in it. Where's the water the coldest; at the top or bottom? Suppose that I fill a glass with ice water.  As the ice melts, it cools the water around it.  Given that cold water is denser than hot water, I would presume that the cold water would sink to the bottom … but it would warm as it sinks, reducing the density.  Meanwhile, the ice is still melting and giving off its cold to the surrounding water.

So is drinking ice water with a straw going to get you cooler or warmer water than drinking from the lip of the glass?

Interesting question!

A few things first:

  As the ice melts, it cools the water around it.

Technically, the ice cube melts because the water cools down. This may sound ridiculous at first, but you must consider the fact that the ice melts because it has drawn "heat" (energy) from its surroundings. The "surroundings" being the air and water that surround it (but the water's more important since it's a better conductor of thermal energy).

  Given that cold water is denser than hot water, I would presume that the cold water would sink to the bottom...but it would warm as it sinks, reducing the density.

You're right, cold water is denser than hot water. It is helpful to note that it shouldn't be too cold though. As the temperature of water drops to 4 °C, the density of water gradually increases. However, as the temperature drops below 4 °C the density of water actually begins to decrease and water in this range easily "floats" over water in the room temperature range.

  Meanwhile, the ice is still melting and giving off its cold to the surrounding water.

Ice isn't giving off its "cold", rather, it takes in the water's "heat" (thermal energy).

Back to your question.

As Max mentions in his answer, you have done a particularly good job of indicating what physical parameters we're dealing with; the really important ones being the temperature of ice, temperature of water (at the time you put the ice in) and the quantity of ice used (at least with respect to the water).

But assuming you're drinking water (originally at room temperature) out of a 250 ml styrofoam or plastic cup, and you used two (normal-sized) ice cubes and that you began drinking the water a minute after you plonk in the ice cubes, the water should be colder at the top than at the bottom.

Consider minute, imaginary layers/regions/packets of water in the cup (thinking about this in terms of water "packets" rather than water molecules is easier to comprehend). Also, think of the cup as having three (crudely demarcated) regions: Top, middle, and bottom.

Packets of water immediately adjacent to the ice cubes are in thermal equilibrium with the outermost regions of the ice. However, these packets soon gain some thermal energy from other water packets that are adjacent to them. So as these packets slowly rise in temperature, from zero degrees to past 4 °C, they sink and new packets occupy locations adjacent to the ice. The cycle repeats for as long as the ice is there.

Now, as those packets of ice sink, they gain more thermal energy from the packets of water they come in contact with on their way down. This, coupled with the viscous effects of water results in the mild "warming up" of the sinking packets.

Now since they warm up a bit, they tend to rise back up. Back at the top, they're get cooled and sink again. This process repeats for as long as the ice remains in the water.

Take a step back, and you'll see that the middle of the cup ought to be cold, the bottom of the cup ought to be colder, and the top of the cup is the coldest.

So even if the ice cubes aren't actually touching your lips, you'll find sipping the water at the top to be colder than sucking out water from the bottom through a straw.


Why n-alkanes can be found in organic solvent extract of rubber? Sorry for the confusing title, I didn't know how to describe this better. So, if we douse rubber, say, a bicycle tire or a certain type of rubber glove, with organic solvent (n-heptane), we can find n-alkanes being dissolved in the solvent. Why is this? I'm talking especially about gas chromatography to analyse the paraffinic hydrocarbons here.

I guess this is quite a trivial question but I can't figure it out. As far as I know, natural rubber consists mainly of isoprene polymer chains. But isoprene has a double bond and in GC chromatogram we can see n-alkanes being the largest peaks. And not only that, we can also find straight chain hydrocarbons in synthetic rubber (like NBR or SBR) solvent extracts. SBR has a benzene ring in its structure but still n-alkanes are the dominant finding in its solvent extract chromatogram.

This is a major subject in my bachelor's thesis but I'm an analytical chemistry major and struggling a bit with materials science in this depth. Any advice and ideas would be greatly appreciated.

To make my point clearer I present these two chromatograms:

(GC-FID run of pure n-heptane)

(GC-FID run of n-heptane extract of bicycle tire)

This question still haunts me. I found this. Could that be a partial explanation for the aliphatic hydrocarbon finding?

Although natural rubber and synthetic rubber display a poor chemical resistance to alkanes among others, they do not  dissolve in alkanes. Your GC/MS traces (I believe that so because of alkane/alkene identification) have confirmed that since they have shown n-heptane as the main peak (I believe everything else are trace amounts exhibiting below your set-up threshold).

There is a reason for your finding of straight chain hydrocarbons in synthetic rubber such as SBR in your solvent extracts. For example, SBR is the shorten form for synthetic material, styrene butadiene rubber, which is made by anionic copolymerization of styrene & butadiene. For this anionic copolymerization, frequently used initiator is a solution of lithium alkyl compound in small paraffins, for instance, n-butyllithium in hexanes. The rubber may still have few molecules of these small paraffins, which may diffuse into the n-alkane you use to extract, and hence, would show up in your GC/MS chromatogram.