Hybridization of Cu2+ in tetraamminecopper(II) sulfate complex $\ce{Cu^2+} $ has 9 electrons and the $\mathrm d$-orbital is almost completely filled (except for 1 electron vacancy), eventually forming $\mathrm{sp^3}$ hybrid orbitals, where each $\ce{NH3}$ donates a lone pair of electrons.
But it has been found to have a square planar geometry, hence completely contradicting the above reasoning. So, please explain what is the hybridization of the $\ce{Cu^2+}$ in the given complex?

I feel a bit like a broken record. Do not use hybridisation to describe coordination compounds; it is not helpful.

The cation you called tetraamminecopper(II) is more accurately described as tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$, because two aqua ligands are there, just further away. The four ammine ligands donate two electrons each, as do both the aqua ligands, totalling 21 electrons overall. Due to the strongly Jahn-Teller distorted geometry and other reasons, the bonding interactions are overall stronger than antibonding ones making the complex favourable.

The interaction between the metal and ligand involves $\mathrm{e_g, a_{1g}}$ and $\mathrm{t_{2u}}$ orbitals (when viewed as an undistorted octahedral complex). That would include two d-orbitals, three p and one s. If your life depended on it and you are the last person on Earth, you may attempt to extract a hybridisation from that. But don’t do it.


Concentration of NaOH to dissolve glass I am doing a physics experiment in which it is required to etch the outermost glass layer from a coverslip (regular lab apparatus). What concentration of NaOH should be dissolved in water, and how long should the coverslip be placed in the base?
Any idea or information that can help me figure out myself is welcome.

The thickness of the attacked layer (or the corresponding loss of weight per surface area) due to alkali attack on glass mainly depends on the following parameters:

type of glass (most laboratory glass is borosilicate glass 3.3)
concentration of the alkaline solution (or the corresponding pH)
temperature
exposure time

Note that the density of typical borosilicate glass is 2.23 g/cm³ (at 25 °C). Thus, an attacked thickness of 1.00 µm corresponds to a loss of 22.3 mg/100 cm².

The following figures are taken from product information for Schott DURAN® glass:

Alkali attack on borosilicate glass as a function of pH, exposure time: 3 h

Alkali attack on borosilicate glass as a function of temperature, calculated from the weight losses, c(NaOH) = 1 mol/l, exposure time: 1 h

Alkali attack on borosilicate glass as a function of time

For example, in order to remove a layer of about 1 µm, you can expose the glass for a period of approximately 40 min to an alkaline solution with c(NaOH) = 1 mol/l at 100 °C.


What's the densest alkali metal? This was asked in a school exam of mine, GCSE curriculum, exactly as it was worded on the title. I found this question a bit vague, and answered it francium, however of course francium is just an estimation. I couldn't be sure what the answer to the question would be, so can anyone give me some insight?

It's hard to tell what is the answer you were supposed to provide, but yes, generally speaking, you are correct.
According to CRC handbook of chemistry and physics [1, p. 4-14] (emphasis mine):

Francium, the heaviest known member of the alkali metal series, occurs as a result of an alpha disintegration of actinium. […]
Because all known isotopes of francium are highly unstable, knowledge of the chemical properties of this element comes from radiochemical techniques. No weighable quantity
of the element has been prepared or isolated.

As the result, the majority of physical properties of francium, including density, were obtained by extrapolation and not by direct measurements.
According to Bonchev and Kamenska [2], $ρ_\mathrm{calcd.}(\ce{Fr}) \approx \pu{2.9 g cm-3}$, which is in a good agreement with the trend for the measured densities across the group.
For comparison, the following data (NTP conditions) is collected from the table PHYSICAL CONSTANTS OF INORGANIC COMPOUNDS [1, p. 4-43]:
$$
\begin{array}{cl}
\hline
\text{Element} & ρ/\pu{g cm-3} \\
\hline
\ce{Li} & 0.534 \\
\ce{Na} & 0.97 \\
\ce{K} & 0.89 \\
\ce{Rb} & 1.53 \\
\ce{Cs} & 1.873 \\
\hline
\end{array}
$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Bonchev, D.; Kamenska, V. Predicting the Properties of the 113-120 Transactinide Elements. J. Phys. Chem. 1981, 85 (9), 1177–1186. https://doi.org/10.1021/j150609a021.


Sigma and pi symbols in molecular orbital theory In molecular orbitals diagram, sigma ($\sigma$) and pi ($\pi$) symbols are used. What do these symbols really mean?

These symbols are for bonding orbitals. A covalent bond is formed when two orbitals in different atoms hybridize. $\sigma$ and $\pi$ refer to two different configurations in which this can happen.

In a $\sigma$ bond, two $s$ orbitals or two $p$ orbitals oriented towards each other join into one orbital, with the electron density concentrated between the two atoms.

In a $\pi$ bond, two $p$ orbitals oriented perpendicular to the bond axis merge, and, in the resulting orbital, electron density is concentrated above and below the bond axis.


Enzyme Specificity vs Selectivity Background:
I can see a difference in stereoselective and stereospecific reactions in organic chemistry. However, both terms are used interchangeably as synonyms in a biochemistry class.

Question:
What differs selectivity from specificity in enzymatic reactions. What is meant when we talk about enzyme specificity and enzyme selectivity?

Specificity is the term used to define the selectivity of enzymes
for their substrates. The selective qualities of an enzyme are collectively recognized as its specificity.

Other texts have synonymised enzyme selectivity with substrate specificity:

The non-covalent forces through which substrates and other molecules bind to enzymes are similar in character to the forces
that dictate the conformations of the proteins themselves : Both involve van der Waals, electrostatic, hydrogen
bonding, and hydrophobic interactions.
Molecules that differ in shape or functional group distribution
from the substrate cannot productively bind to the
enzyme; that is, they cannot form enzyme–substrate complexes
that lead to the formation of products.

The extraordinary ability of an enzyme to catalyse only one particular reaction is a quality known as specificity.

Specificity means an enzyme acts only on a specific substance,
its substrate, invariably transforming it into a specific product. That is, an enzyme
binds only certain compounds, and then, only a specific reaction ensues.
Some enzymes show absolute specificity, catalyzing the transformation of only one
specific substrate to yield a unique product. Other enzymes carry out a particular
reaction but act on a class of compounds. For example, hexokinase (ATP-hexose-6-
phosphotransferase) will carry out the ATP-dependent phosphorylation of a number
of hexoses at the 6-position, including glucose.

  Pioneering enzyme specificity studies at the turn of the 20th century
  by the great organic chemist Emil Fischer led to the notion of an
  enzyme resembling a “lock” and its particular substrate the “key.”
  The specificity of an enzyme (the lock) for its substrate (the key) arises from their geometrically complementary shapes. This was later
  modified to a more accurate induced fit hypothesis given the high
  flexibility, conformationally dynamic properties of enzyme molecules.

References

Voet and Voet Biochemistry 4th ed: Introduction to Enzymes
Biochemistry Grisham & Garret How Can Enzymes Be So Specific?


Cr(II) and Mn(III) - their oxidizing and reducing properties? My textbook states that $\ce{Cr^2+}$ is a reducing agent while $\ce{Mn^3+}$ is an oxidizing agent in spite of both having $\ce{d^4}$ configuration. The explanation states that when $\ce{Cr^2+}$ gets oxidized to $\ce{Cr^3+}$, it attains a stable half-filled $\mathrm{t_{2g}}$ orbitals but when $\ce{Mn^3+}$ gets reduced to $\ce{Mn^2+}$, it attains a stable half-filled $\mathrm{d}$ orbitals.

First, is it a reasonable explanation? If this is the case, why can't $\ce{Cr^2+}$ get reduced to $\ce{Cr+}$ to attain a stable half-filled $\mathrm{d}$ orbitals and $\ce{Mn^3+}$ get oxidized to $\ce{Mn^4+}$ to attain a stable half-filled $\mathrm{t_{2g}}$ orbitals, i.e. in-short is the vice-versa true?

Related question with same answer but in a different context of the 4f block: Why don't we see these lanthanide species?

You have a misconception regarding the stability of oxidation states. The factors you have listed are honestly not very important in determining the stability of a certain oxidation state. They will tip a delicate balance in favour of one oxidation state or another, but they are hardly the sole determining factors.

If you are going from $\ce{M+}$ to $\ce{M^2+}$, you need to consider two main things.

You need to ionise the second electron from $\ce{M+}$, denoted $I_2$. That represents an input of energy.
$\ce{M^2+}$ is smaller and more highly charged than $\ce{M+}$, so you can gain some energy back on the basis that the $\ce{M^2+}$ ion is more strongly solvated or forms stronger ionic/covalent bonds with anions.

Loosely speaking, if the energy you get out of (2) is enough to compensate for the energy you lose in (1), then $\ce{M+}$ will act as a reducing agent. And the converse is true as well. It is for this reason that one does not observe Mg(I) compounds (apart from some esoteric organometallics). The $I_2$ of Mg is so small that it can be recouped extremely easily. Likewise, the $I_2$ of K is so large that there's no way any stronger bonding can compensate for that.

For the cases you have described:

$\ce{Cr^2+}$ simply does not act as an oxidising agent because $I_2$ of Cr is rather small. The same is true of the whole 3d block. How many 3d compounds do you know of with the metal in a +1 oxidation state? Without resorting to organometallic compounds, all I can suggest is copper(I). That's precisely because their $I_2$ values are small enough.
Similarly, $\ce{Mn^3+}$ does not act as a reducing agent because $I_4$ is way too large. Remember that successive ionisation is extremely difficult.

What your textbook should have written is:

The $I_3$ of Mn is anomalously low because you are "losing the stability of a half-filled 3d subshell".
The hydration enthalpy of $\ce{Cr^3+}$ is anomalously large (for a 3+ ion) because there is a large ligand-field stabilisation energy associated with the $(t_{2g})^3$ configuration. (This is factor number 2, not factor number 1).

These factors are small compared to the raw size of the ionisation energies. Yes, the $d^3$ ion $\ce{Mn^4+}$ also enjoys a large LFSE and its hydration enthalpy would be larger than expected. However, that 4th ionisation energy is just too big and that simply cannot be compensated for by the comparatively tiny increase in LFSE.


What is the perfect definition for chirality? Why is chirality defined differently for organic and inorganic compounds?

Why are inorganic compounds deemed to be optically active if they have more than one of the same ligands attached to the central metal atom, but glycine is optically inactive for having two $\ce{H}$ atoms attached to the carbon atom?

Finally, what should be the criterion for chirality (optical activity) and the perfect definition for it?

Could anyone explain how $\ce{$cis$-[PtCl2(en)2]^2+}$ is optically active, but the $trans-$isomer is not?

The correct definition of chirality is given in the IUPAC gold book as follows:

  chirality
  The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

From this we can deduce, that all organic compounds, which have tetrahedral carbons with four different ligands, are chiral. This is a rule of thumb and an easy way to recognise chirality.
Glycine is optically inactive as it has an internal mirror plane:

Often we teach this as a rule or a definition, because it is so easy to comprehend and to see.

In inorganic chemistry, we usually deal with more complex structures, where it is not necessary, that all ligands have to be different. One example for this is the cation in tris(ethylenediamine)cobalt(III) chloride:

Another interesting example is (Λ/Δ)-cis-dichlorobis(ethylenediamine)cobalt(III) chloride.


Is a buffer solution really inactive beyond a range of pKa±1? I have learnt from several books that the range of a buffer solution is always from $\ce{pH = pKa - 1}$ to $\ce{pH = pKa + 1}$. But is there any solid reason to this and any exception?

The 10:1 ratio is really just something convenient for the chemists.  The buffer still works outside this range.  True, its effectiveness against base addition decreases with the amount of available acid at high pH, and vice versa for acid addition at low pH.  But if enough capacity remains, you can still live with the buffer even outside the 10:1 range.

Er, actually you do.  Our main blood buffer, based on dissolved carbon dioxide as the acid and bicarbonate as the base, has about 20 times as much bicarbonate as carbon dioxide (https://en.m.wikipedia.org/wiki/Bicarbonate_buffer_system).  Factors other than what a laboratory chemist calls "optimal" should be given their due.  Metabolic processes commonly produce (weak) acids, and the solubility of carbon dioxide is more limited than that of bicarbonate salts.  Both of these factors make a base-rich buffer desirable.


Is it possible to curve fit GC, Mass-spectrometry and NMR peaks? I am trying to measure the amount of compounds by the peak area and I want to increase accuracy to 0.1% by curve fitting. Can I do it and what functions should I use?

You should not assume any given peak shape as in practice this may not be what it should be from theory.

You have a set of data points so once you have determined the baseline a numerical integration should work well. There are many of these in the literature. When fitting use the chi squared statistic as a measure of fit, the Pearson R is just useless and should never be used in fitting. (This is because a poor fit has, say, an R = .99 and a good fit .999 so its not anywhere near discriminating enough). You should also look at the shape of the residuals, autocorrelation function, runs tests etc to make sure you have a good fit. A one parameter good/bad test is often not good enough.

If you have overlapping peaks then you will have to assume some shape simply to use non-linear least squares to estimate one peak from another. This will lead to some more uncertainty but depending on what you are trying to determine this may not matter.

In gc data the peak shape is not necessarily gaussian, often it is lopsided so a numerical method is better here.

It may not be necessary to fit your data to try to get such high precision, the data simply may not allow it due to instrumental factors. But what is important is results not data, i.e. its what the data tells you that is important; this may not need such high precision as you seek. I'm absolutely not being patronising, but I would go as far as to say that if you need 0.1% accuracy you should perhaps think of another way of trying to learn what you want to know.


How to calculate the pH of a redox reaction between nickel(IV) oxide and silver? For the following reaction:
\begin{gather}
\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2H2O(l) + 2Ag+(aq)}\\
E^\circ = \pu{2.48 V}
\end{gather}
Calculate the $\mathrm{pH}$ of the solution if $E = \pu{2.23 V}$ and $[\ce{Ag+}] = [\ce{Ni^2+}] = \pu{0.023 mol/l}$.

I know that there is the Nernst equation:
$$E = E^\circ-\left(0.0592/n\right)\log Q$$
Where $E^\circ$ is the standard cell potential, $E$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations,
$$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$
I know that for this problem, $n(\ce{e-}) = 4\ \mathrm{mol}$, and that
$$Q = \frac{[0.023]^3}{[\ce{H+}]^4}.$$
So:
\begin{align}
  2.23
  &= 2.48 - \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{1}\\
  0.25
  &= \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{2}\\
  1.00
  &= \frac{0.0592}{1}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{3}\\
  1.00
  &= 0.0592\log\left(\frac{0.023^3}{[H^+]^4}\right)
  \tag{4}\\
  \frac{1.00}{0.0592}
  &= \log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{5}\\
  \frac{1.00}{0.0592}
  &= \log(0.023^3)-\log\left([\ce{H+}]^4\right)
  \tag{6}\\
  \frac{1.00}{0.0592} - \log(0.023^3)
  &= -\log\left([\ce{H+}]^4\right)
  \tag{7}\\
  \frac{1.00}{0.0592} - \log(0.023^3)
  &= -4\log\left([\ce{H+}]\right)
  \tag{8}\\
  \frac{\frac{1.00}{0.0592} - \log(0.023^3)}{4}
  &= -\log\left([\ce{H+}]\right)
  \tag{9}\\
  \mathrm{pH} &= -\log\left([\ce{H+}]\right)
  \tag{10}\\
  \frac{\frac{1.00}{0.0592} - \log\left(0.023^3\right)}{4}
  &= 5.45
  \tag{11}
\end{align}
Therefore, $$\mathrm{pH} = 5.45\tag{12}.$$
Yet, the online assignment here says my answer is wrong.

It even gives me practice versions for other variations of this problem, and yet I still always get a wrong answer using the same methods.  Would someone be so kind as to point out any errors I may have made while calculating this answer?

For the following reaction:
\begin{gather}
\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\\
E^\circ = \pu{2.48 V}
\end{gather}
Calculate the $\mathrm{pH}$ of the solution if $E = \pu{2.23 V}$ and $[\ce{Ag^+}] = [\ce{Ni^2+}] = \pu{0.023 M}$.

The important thing to understand that was not immediately obvious to me and a lot of other people was the fact that $n = 2$, and this changes how you calculate your answer.
$\ce{Ni}$ does not change oxidation state in its transition from $\ce{NiO2}$ to $\ce{Ni^2+}$, as its oxidation state in $\ce{NiO2}$ is $+2$.  The 4 $\ce{H^+}$ ions also do not change oxidation state in their transition from $\ce{H^+}$ to $\ce{H2O}$, oxidation state in $\ce{H^+}$ is $+1$ and oxidation state in $\ce{H2O}$ for $\ce{H}$ is $+1$ for 2 $\ce{H}$ atoms.  That leaves us with the only $\ce{e^-}$ transfer occurring to $\ce{Ag}$.  There are 2 moles of $\ce{Ag}$ atoms going to oxidation state $+1$.  Therefore, 2 moles of $\ce{e^-}$ are transferred.  Hence, $n=2$
The Nernst Equation:
$$E = E^\circ  -  (0.0592/n)\log Q$$
Where $E˚$ is the standard cell potential, $E$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$
For this problem, $n(\ce{e-}) = \pu{2 mol}$, and that
$$Q = \frac{[0.023]^3}{[\ce{H^+}]^4}.$$
Following the above calculation with this difference leads to $\mathrm{pH} = 3.34$, which is the correct solution.
\begin{align}
\tag1
2.23 &= 2.48
- \frac{0.0592}{2}\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag2
0.25 &= \frac{0.0592}{2} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag3
0.50 &= \frac{0.0592}{1} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag4
0.50 &= 0.0592\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag5
\frac{0.50}{0.0592} &= \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag6
\frac{0.50}{0.0592}
&= \log\left(0.023^3\right) -\log\left([\ce{H^+}]^4\right)\\
\tag7
\frac{0.50}{0.0592} - \log\left(0.023^3\right)
&= -\log\left([\ce{H^+}]^4\right)\\
\tag8
\frac{0.50}{0.0592} - \log\left(0.023^3\right)
&= -4\log\left([\ce{H^+}]\right)\\
\tag9
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4}
&= -\log\left([\ce{H^+}]\right)\\
\tag{10}
\mathrm{pH} &= -\log\left([\ce{H^+}]\right)\\
\tag{11}
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} &= 3.34
\end{align}


Nickel salts from hypochlorites (JUST A HEADS UP I’m 15 years old and new to chemistry) So I was recently looking around the house looking for products that I can extract chemicals from and I found a bottle of cleaning Vinegar ( 5% acetic acid), so I decided to make some acetate compounds, but when I first started I thought it would be a simple 1 step synthesis. I was wrong. I soon realized you must have hydrogen peroxide and acetic acid to make zink and copper Acetate.

So I tried making nickel acetate, with no luck. So then I tried to make nickel chloride knowing this new hydrogen peroxide step. I took a pure nickel sheet and put it in a beaker then added calcium hypochlorite and hydrogen peroxide. And it worked! The solution turned light green over a few hours, and it had a gas being produced. And also a black Precipitate which was most likely nickel oxide (I filtered the precipitate off before taking the picture)

But just to check myself I tried to find this process online, to no avail. So now I’m wondering if I did something wrong and got the wrong product or no one does it through this process. If you can check my process to make sure it’s possible and I didn’t waste my time I would Appreciate it. Thanks in advance.

Your result is surprising, for the following reasons : 1) Nickel compounds and ions are mainly green. And you say your solution is blue. 2) When treated by hypochlorite ions, Nickel will produce a black precipitate of $\ce{Ni_2O_3}$, which is not what you got . 3) You obtain a solution without any precipitate, although the hypochlorite solution is always basic, which must produce a Nickel Hydroxyde precipitate. 4) The only possibility to have a blue solution with Nickel, is adding ammonia $\ce{NH_3}$, which produces a blue Nickel-Ammonia complex in solution. So my question : Are you sure that you have taken pure Nickel in the beginning ? Other question : With this method you will obtain a lot of calcium chloride with a small amount of nickel (or copper ?) compound.


Can someone explain my professor's answers for these alkene additions? (a) Why would cyclohexane expand into cycloheptane? Cyclohexane is stable.

(b) Why would cyclopentane not ring expand into cyclohexane? In the process a tertiary carbocation would be formed just like with the methyl shift, but you would also get rid of ring strain.

(c) Why is there only one product? Wouldn't the double bond have hydrogen addition on both carbon sides leading to a potential cyclobutane to cyclohexane expansion?

(e) Once again like in (c), why is there only one product shown?

(f) Wouldn't there be two products once more?

POTENTIAL EXPLANATION: By examining more mechanisms in class, I have noted that potentially the methyl/alkyl shifts could be occurring in concert with the hydrobromination so perhaps that is the reason why.

In example A, you are right that the cyclohexane ring is more stable than the cycloheptane ring. There are two choices for rearrangement in this case:

1) Ring expansion to form the cycloheptane ring

2) 1,2-hydride shift to yield the tertiary alkyl bromide

The second product will be more stable and thus be the major product

In example B, you are right, a ring expansion is possible, so there should be the ring expansion product too. You will have a mixture of products in this case (1,2-alkyl shift, ring expansion, and no carbocation rearrangement)

In examples C, E and F, it is possible for the H-A to add across the double bond either way, so again, a mixture of products would be obtained

A lot of times, only one product is shown for addition reactions such as hydrohalogenation; however, it is important to remember that there are often multiple products possible.

Chemists very rarely utilize reactions that involve a carbocation rearrangement because a mixture of products is obtained, which results in very poor yields


Synthesis for turning synephrenine into N-methylphenethylamine I am trying to devise a synthesis to change synephrine (4-[1-hydroxy-2-(methylamino)ethyl]phenol) into N-methylphenethylamine.

I was figuring that I could remove the hydroxyl group attached to the ethyl part using thionyl chloride to convert the hydroxyl into a chloride and then remove the superior leaving group using $\ce{LiAlH4}$ as a reagent and $\ce{AlCl3}$ as a catalyst, but thionyl chloride affects amine groups and I couldn’t find any information about how to protect the amine group when performing this reaction to reduce the hydroxyl.

I also researched how to remove the hydroxyl from the benzene ring but that involved using zinc and distilling the benzene which wouldn’t work for this because the boiling point of the substance is around 341 °C for synephrine.

Any insight would be really helpful.

Suggested route - step 1 acetylate the amine with 1 eq of acetyl chloride in pyridine.

step 2 hydrogenation of the benzylic alcohol over Pd/C in the presence of acid or use TFA/triethylsilane

step 3 form the triflate of the phenol

step 4 catalytic transfer hydrogenation of the triflate, conditions here

Step 5 acid hydrolysis of the N-acetyl group with cHCl to give product

Thanks to @PCK for the suggestion of forming and reducing the triflate


Constructing a grenade (for a novel I'm writing.) Ok, before any eyebrows get raised, this is for a novel that I'm writing and inspired by this answer from worldbuilding and this reference answer. I just want to make sure the device I'm trying to use is as close to accurate as can be.

So, in the novel, I've devised a compartmentalized glass urn. The urn contains three compartments, one will house a mixture of a finely powdered combination of charcoal, sulfur, and saltpeter. The second compartment would contain slightly acidic water and the third one would contain limestone. What I'm expecting is that when the urn is dropped, the shattering of the glass would cause the water to react with lime resulting in an exothermic reaction which would help to ignite the gunpowder mixture. Is this idea liable to work or is too quixotic and fanciful to consider ? If so, what adjustments can I make to it so that it works? Do keep in mind that the sources have to be natural as it is an apocalyptic world.

Also, can I just call it a mixture of a finely powdered combination of charcoal, sulfur, and saltpeter or does it need to be prepared in a special way or maybe there is a special term for it?

Apologies if this is not the right place to ask the question but I could not think of any other apt place.

In short-no your technique will not work.  Reacting limestone with acid does not release enough energy to reach an appropriate temperature to ignite gunpowder and the acid would hinder combustion.

Fear not though, spontaneous ignition may be achieved by mixing glycerol and potassium permanganate which works well and is a classic example.  If you would like to be more out of the ordinary, zinc powder and ammonium nitrate with a minor amount of copper or ferric chloride and water will also create a spontaneous ignition for your gunpowder.


What exactly is Q in the Nerst equation for a ¨lemon battery¨ of Cu and Zn in varying concentrations of acid? I want to vary $\ce{[H+]}$ and see its affect on the voltage produced by a Zn-Cu in acid "lemon battery".  My question is about the expression for $Q$ in the Nernst equation for this cell.  I think $Q$ will have three species but I'm unclear about their values.

$\ce{[H+]}$ in mol/L, which is easy to quantify as its my independent variable.
$\ce{[Zn2+]}$ in mol/L which would be initially $0$, making Q undefined. Is this right?
The partial pressure of $\ce{H2}$, is this right, and in what units?
Can Q have partial pressures and molarities?

The 'Q' in Nerst equation is the same Q which you might've used while dealing with chemical equilibria, otherwise known as the Reaction Quotient. The expression is same as the equilibrium constant (K) and yes, they both include partial pressures and concentrations.

The units of partial pressure is the same as the unit of pressure (atm or equivalent), which can pinned down by knowing what partial pressure is in the first place. If I give you a container to begin with, filled with hydrogen and neon, and ask you about the partial pressure of hydrogen, I'd expect that you'd say:

  "Had the container been filled with only  hydrogen and then the pressure exerted by hydrogen would've been measured, then I'd say that partial pressure of hydrogen (in the first case) would have been equal to this measured amount."

Now, to write down Q we need to have the equilibrium reactions with us:

At the cathode (copper in this case), hydrogen ions are forming molecular oxygen due to their reduction, so the situation appears to be like:
$$ \ce{2H+ \text{(aq.)} + 2e- -> H2 (g)}$$
At the anode, zinc is getting oxidized:
$$ \ce{Zn (s) -> Zn^2+ \text{(aq.)} + 2e-}$$

Recall that solid species are omitted from the equilibrium expression, so Q of the reaction would be written as:

$$\displaystyle\pu{\frac{P_{H_2} \cdot [Zn^{2+}]}{[H^+]^2}}$$


Why does the addition of oxygen render materials colourless? In our book, it is mentioned that $\ce{H2O2}$ (hydrogen peroxide) is used as a bleaching agent because of its oxidizing nature:
\begin{align}
  \ce{H2O2 &-> H2O + [O]}\\
 \ce{\text{coloured material} + [O] &-> \text{colourless material}}
\end{align}

Why does the second equation happen? In other words, what role does oxygen play in the absorption of light here?

Many organic dyes (arguably, all of them) produce color due to electron transitions in conjugated $\pi$ systems. Strong oxidants (not just the peroxide, mind you, but also hypochlorite bleach and others) snap the double bonds and thus destroy these conjugated systems. So the color vanishes.

This is not applicable to inorganic dyes, which usually would not get bleached by peroxide. On the contrary, some transition metals ($\ce{Ti^4+,\;Cr^6+}$, and some others) which have little or no color by themselves would produce intensely colored complexes when hydrogen peroxide is added. These reactions are sometimes used in analytic chemistry to discover their presence.


What chemical substance is obtained by dissolving C60 in olive oil? What chemical substance/component would be obtained of dissolving $\ce{C60}$ in olive oil? Would buckminsterfullerene (bucky-ball) change its spherical fullerene structure (soccer ball-like) into something else ($\ce{C60OO}$) or will it stay the same?

See:

Fullerene - Safety and toxicity at Wikipedia
Fathi Moussa's experiment study on rats in 2012 which doubled their lifespan
The color of the Carbon-60 in olive oil
Buckminsterfullerene-olive oil production process

There is no chemical reaction between fullerene and olive oil, that is to say that the two molecules do not break or make chemical bonds to produce a new chemical species.  However, fullerene is soluble in olive oil and so there are typical solvent-solute interactions.  A homogenous solution of fullerene in olive oil produces a light purple color typical of many fullerene solutions.

Since Moussa's 2012 experiment where he fed fullerene in olive oil to rats and found that their life span roughly doubled, many companies have been marketing a dark-red liquid comprised of olive oil and fullerene.  It is likely that in many of these highly colored liquids, the fullerene is not completely dissolved and light scattering by the fine particulates are producing the observed color.


correct psilocybin molecular structure I'm writing a research paper on psilocybin.  I need to know the correct chemical structure.  When I look online it it shows different ones.  There is one with two OH molecules attached to the phosphorous and there is one with one OH and one O- attached to the phosphorous.  There is also one that I found with a O+ attached to the phosphorous.  Are they all they same or are they interchangeable?  Are there different ones?  I can't find a clear answer.  Any help is greatly appreciated.

Clearly, it cannot be an $\ce{O+}$, as that would require a triple bond to the oxygen, making the phosphorus in the middle quite unhappy.

Of the other two alternatives, both are correct -- however, it matters what environment you're considering the molecule. In neutral or slightly basic conditions, it will be an $\ce{O-},$ while in slightly acidic conditions, it will be an $\ce{OH}.$

Let me know if you need any clarification on why this is the case -- I have no idea what your level of knowledge is.


How to select for 1-4 carbon sugar bond in Chemical Glycosylation? I am interested in Chemical Glycosylation, specifically the production of cellulose by synthetic means. However, it is difficult to ensure that the glycosidic bond is formed between the 1 and 4 carbons.

Are there any techniques that can select for this reaction?

I am looking for good references on this subject.

You have to protect the OH groups you don't want to react with protecting groups.

In this case you could install a benzylidene acetale between the 4 and 6 OH, then protect all the other OH and then do selective reductive opening of the benzylidene so your benzyl group will be on position 6 and you got a free OH.
That said: it's quite expensive, labor intensive and has very bad atom economy. And you might run into problems with deprotection once you got polymer structures, so that's not a feasible option for synthesis of cellulose.

I don't think it's possible to optimize a synthesis without protecting groups, so the best way would be doing it enzymatically, which could be very expensive too.


Relative abundance of the isotopes of iridium There are two stable isotopes of iridium, $\ce{^191Ir}$ and $\ce{^193Ir}$. If you randomly pick an iridium atom from a large collection of iridium atoms, which isotope are you more likely to select?

An urn contains nothing but

63 white balls and
37 black balls.

You grab into the urn, pick one item, look at it and put it back.

What is the probability that you drew:

a ball
a white ball
a black ball
a red ball?


Difference in peaks in EDS and XPS spectra I have been reading some papers on XPS and EDS and I saw that there's a difference between the width of the peaks (of the same material) in XPS and EDS. Is the reason for this difference because of the better energy resolution of XPS (spectral resolution)? EDS uses the signal from emitted X-rays, whereas XPS uses the emitted electron. There's a limit to what how much energy of the X-rays can be detected (about 2 eV) and electrons are easier in that respect and allow for measurement of lower eV.
Or is there another reason for the difference in the peaks?

The two techniques are used for different purposes.
EDS - Is a "bulk" analysis technique.

Since the exciting electron is of high energy it penetrates relatively deeply into the sample.

The emitted x-rays have a characteristic energy which is unaffected by the chemical bonding. (There is some effect due to bonding of course, but the difference in energy is infinitesimally small and hence insignificant.)

High energy x-rays can escape through the sample to a "great" depth.

XPS is a surface analysis technique.

Exciting electrons of low energy and only penetrate the surface layers of the sample.

The bonding energies of the valence electrons are low in energy.

The emitted electrons have a low energy (few eV) and hence don't penetrate out of the sample except for the surface layers.


How would one determine an element simply by looking at its binding energy? I am self studying MIT OCW chemistry 5.111 2014, one of the lecture questions states the following:

  Consider a neutral atom with 8 distinct electron binding energies: −14 eV, −28 eV, −94 eV, −218 eV, −293 eV, −1730 eV, −1921 eV, and −14326 eV.
  (a) Name all of the possible ground state atoms that could have these binding energies (without looking up any values).

It seems like the values are quite arbitrary without having a reference place to begin from i.e. electron configuration, quantum numbers, period and so on.

I thought maybe to use equation

$$E_{nl} = -\frac{Z_\mathrm{eff}^2R_\ce{H}}{n^2}$$

as it has an energy parameter, but it appears I have one equation and two unknowns.

Any ideas on how to go about this?

There are a few clues on how to solve this question:

you are given binding energies. The energies in question are just enough to liberate electrons from their bound state.
The equation you've written $$E_{nl} = -\frac{Z_\mathrm{eff}^2R_\ce{H}}{n^2}$$ describes the binding energy. The important part is that it can be assumed to be only a function of n and l, that is, of the subshell to which an electron belongs ($Z_\mathrm{eff}$ is a function of n and l quantum numbers. Unfortunately this is probably a somewhat complicated function but that is not very important just now)

Finally consider the Aufbau principle, outlined in this libretext:

  In essence, you can think of electrons filling in "n+l " order. If you have an equal "n + l"  value for two orbitals, then electrons will occupy the orbital with the lower "n" value.

which is followed by a nice illustration (shown below). The number of distinct binding energies equals the number of subshells. The energy for emission from each subshell decreases in the same order in which the subshells are filled. All electrons in a subshell share a common binding energy. Following the Aufbau pattern you can identify the highest-energy occupied (valence) subshell, which corresponds to the smallest binding energy. I count 8 subshells and assign the minimum $E_{nl}$ to the 4p subshell (the electron configuration is $\ce{[Ar] 3d^{10} 4s^2  4p^?}$). This means there are six possible atoms that match the number of binding energies, namely Ga through Kr.

If we cheat and check a table of ionization energies we see that the lowest binding energy (-14 eV) matches that of Kr (other values in the table don't match those in the problem since the table lists the ionization energy for multiple electron emissions whereas the problem lists energies for all possible single electron emissions $\ce{A->A+ + e-}$, starting from the neutral atom, an emission from one subshell corresponding to one energy). You should note the pattern of decreasing binding energies (magnitudes) starting from the 1s electrons (−14326 eV) up to the valence shell (-14 eV). The following diagram adapted from the libretext quoted above illustrates how this works:


CuCl2 dissolved in water We dissolve $10\ \mathrm{g}$ of $\ce{CuCl2}$ in $0.1\ \mathrm{L}$ of water.
Given that:

$M(\ce{Cl})=35.5\ \mathrm{g/mol}$
$M(\ce{Cu})=63.5\ \mathrm{g/mol}$

How can one calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$?

First of all I calculated the the quantity of matter of $\ce{CuCl2}$:

$n(\ce{CuCl2})=0.074\ \mathrm{mol}$

The the molar concentration:

$c=\frac{0.074\ \mathrm{mol}}{0.1\ \mathrm{L}}=0.74\ \mathrm{mol/L}$

Then I don’t know how can I calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$ because I don’t know the quantity of matter of them!

But here im blocked!

When a salt is fully dissociated, than the atoms in a solution exist separately:
$$\ce{CuCl2<=>Cu^{2+}}+2\ce{Cl-}$$
Amount of substance is 0.074 mol. The molecule is made of 1 atom of Cu and 2 atoms of Cl.

$$ n=\frac{m}{M}$$
$$ n \cdot M =m$$
Amount of copper in the mixture then is: $$0.074~\mathrm{mol} \cdot 63.5~\mathrm{g/mol} = 4.7~\mathrm{g} $$ and the rest is formed by chlorine. $$ 10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$


How is phosphorus-32 produced? I'm doing an assignment where I have to research a radioisotope of my choice. I decided to do phosphorus-32 and my assignment requires me to explain how the isotope is made. There is very minimal information on the production of phosphorus-32 so I'm in a little trouble.

A brief explanation would be much appreciated!

The most important nuclear reactions for the generation of anthropogenic P-32 are
Cl-35 (n, α) P-32,
S-32 (n, p) P-32, and
P-31 (n, γ) P-32.
Since especially chloride, sulfate, and phosphate can occur as impurities in water, typical P-32 activity concentrations in the reactor coolant of light water reactors are in the range of several kBq per litre. (Sources for impurities in the reactor water of BWR’s can be leaking condensers; in the primary coolant of PWR’s, phosphorous compounds in the added boric acid. Sulfate might also be released from cation exchange resins.)

In the laboratory, P-32 can be produced (even as an experiment with students) using the S-32 (n, p) P-32 reaction with fast neutrons. A sample of pressed elemental sulfur is placed close to a neutron source for several days. After that, the sulfur is washed with boiling water. The conversion of the α-sulfur allotrope to β-sulfur at 95.3 °C helps to release the phosphorous compounds from the solid sulfur matrix to the water. Thus, the resulting P-32 solution can be made carrier-free.


Electron density definition of single, double, triple bonds? I was reading this question and the answer by user EricBrown to it, and this got me thinking about covalent chemical bonds.

The way I was always taught is that a single bond contains 2 paired electrons, a double bond 4 etc (The Lewis dots drawings always show this nicely). In his answer Eric mentions that the whole definition of single, double etc bonds is ambiguous because there haven't been measurements of the electron density. This notion strikes me as odd, because a simple googling gives me articles like this: How similar is a molecule to another? An electron density measure of similarity between two molecular structures Int. J. Quantum Chem. 1980, 17 (6), 1185–1189. But it did make me think about the way the different types of bonds are defined.

I read up a bit on the Theory of atoms in molecules and indeed this theory only specifies 1 bond type: the line with maximum electron density between two nuclei. According to the theory this line is unique for a given pair of nuclei in a given molecule. I could imagine that a molecule which has a triple bond (6 bonding electrons in the 'classical' theoretical sense) would have a relatively high electron density along this maximum line as compared to e.g. a single bond.

My question is: if we would be able to measure the electron density in a molecule (are we?) couldn't we use this to define a relation between the value of the maximum electron density and the notion of a single, double or triple bond? Or perhaps use the spread in electron density between the nuclei for this?

Below are plots of the electron density for staggered ethane, ethene, and ethyne, computed from electronic structure theory.  For ethene, which is planar, imagine that the hydogens are jutting forward and backward into the paper, so that the contour plots should contain any putative $\pi$ bonds. It is the orthogonal plane that still contains the two carbon atoms.

A couple of points that I wish to make:

You are correct (in this case for sure) that there is some association between stronger bonds and higher electron density at the C-C bond critical point $(\star)$.  This was noted by Bader and co-workers.
There is no evidence (to my eyes) that there are any side-wise $p$ orbital interactions forming $\pi$ bonds. These look like single $\sigma$ bonds!

My question is, where do we draw the dividing line in the correlation between electron density at the bond critical point, and what constitutes single, double, and triple bonds?

Responding to @michaelm: I might have expected this, if for example, ethene were to have a "double bond" between the carbons:


Choosing a balance for home chemistry A good balance is an important piece of equipment for chemistry. For home chemistry, the price of the balance is unfortunately limiting ..

How should one go about picking a balance (for home chemistry; general guidelines are welcome too)? Specifically, I'm thinking whether to buy a model with 0.01g precision (up to 200g), or a one with 0.001g precision (up to 20g) (the latter is a so called "diamond balance" but probably does the job). Which one would serve me better?

Home chemistry can be difficult sometimes due to the expense of the essential working materials and instruments required. As a mathematician all I need is a sharp pen (which is sometimes still quite difficult to obtain!) and blank paper. I can give you the following general (personal) advice regarding purchasing chemistry lab equipment:

Determine your budget and the needs for the experiments you want to perform. Do you really need a scale that has to weight with an accuracy of 0.001g or is 0.01g sufficient (or even 0.1g)? Ask yourself what you REALLY need for an experiment BEFORE searching for instruments. If you do it in the converse order you often buy something way more fancier (and pricier) than what you actually needed in the first place.
Look at specialized chemistry equipment supply websites to orientate on what is available and in what price range certain products are (think of scales, chemicals, glassware, etc.)
A lot of stuff can be bought incredibly cheap on websites where you not expect (at least I did not) people to sell lab supplies/instruments. Think of ebay.com and amazon.com as examples. They sometimes list items at a fifth of the usual price that would have to be paid at a specialized store.
Take a look at local non-specialized stores like the construction market (you know, the place where they sell all kinds of construction material). They sometimes hide hidden (cheap) gems! As an example I was able to obtain a vacuum source (membrane pump) for < 100\$, which would have cost me easily > 400\$ at a chemical supply website.

About your particular question: I really recommend a 0.01g precision 200g (or 600g) scale. It brings the benefit of having relatively accurate weighting for even small samples, while still being able to weight bulk mass. Furthermore, they are most common in the lab/home environment and hence relatively cheap in comparison with professional analytical scales. If you plan to do a lot of microscale experiments (yielding a product < 200 mg) the analytical scale (precision of 0.001g) may suit your needs better.

I wish you much success, think and buy wisely!


Why does the lone pair of phosphine not contribute to its dipole moment? Unlike ammonia, the lone pair of phosphorous does not add to the dipole moment of phosphine. My textbook says this is due to the presence of lone pair of phosphorous in $\ce{PH3}$ in an s-orbital. I couldn't understand my book's explanation. How can the orbital of lone pair determine its contribution to the dipole moment?

Let's see if we can use the following picture to help clarify things.

Let's look at the molecule $\ce{H3X-Z}$.  Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$.  Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment.  Are you with me so far?

If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased.  If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$.  These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture.  The orbital and the electrons in it still "point" towards where $\ce{Z}$ used to be.  The orbital and the electrons in it still have directionality.  They still contribute to a dipole moment.  Can you see this?

Let's continue assuming you can.  Let's repeat all that we've done above with one change.  Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized.  Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital.  An $\ce{s}$ orbital is spherically symmetric.  It has no directionality, it doesn't "point" in any specific direction because it is spherically symmetric.  So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.


What's the difference between aqueous NaOH and dilute NaOH? Dilute $\ce{NaOH}$ can be prepared by adding a suitable amount of concentrated $\ce{NaOH}$.

What's the difference between "dilute $\ce{NaOH}$" and "aqueous $\ce{NaOH}$"?

Dilute simply means that the concentration of the solute of interest, in this case $\ce{NaOH}$, is low relative to some reference.  One could imagine a non-aqueous, dilute solution of $\ce{NaOH}$ in, say, methanol.

An aqueous solution is one where the solvent is water, regardless of the concentration of the dissolved solute(s).

So, for example, the following combinations of the terms are possible:

Concentrated aqueous solution - High $\ce{NaOH}$ concentration in a water solvent
Dilute aqueous solution - Low $\ce{NaOH}$ concentration in a water solvent
Concentrated methanolic solution - High $\ce{NaOH}$ concentration in a methanol solvent
Dilute methanolic solution - Low $\ce{NaOH}$ concentration in a methanol solvent


Why doesn't Oxygen form coordinate covalent bonds? We know that an Oxygen atom has two lone pairs. Why doesn't oxygen then form coordinate covalent bonds using those since it has nothing to lose?

Oxygen very much does form bonds in which both electrons come from the oxygen atom.  Examples include:

1.
The $\ce{H3O^+}$ ion at the center of the solvated proton in aqueous acids, also available as salts of some of the strongest acids such as $\ce{(H3O)(ClO4)}$

2.
Carbon monoxide, with its triple rather than double bond.

3.
Ozone, in which the oxygen atoms form a bent chain and one of the three bonds must have both electrons coming from the middle atom.  The resulting charge separation makes ozone one of the few one-element substances with a nonzero dipole moment.


Finding new pH after adding potassium hydroxide to citrate buffer Calculate the $\mathrm{pH}$ change of citrate buffer containing $\pu{10.26 mmol}$ of citric acid and $\pu{6.98 mmol}$ citrate caused by adding $\pu{4.72 mmol}$ of potassium hydroxide. $\mathrm{p}K_\mathrm{a} = 3.06.$

I'm trying to calculate it like initial $\mathrm{pH}$

$$\mathrm{pH_0} = 3.06 - \lg\frac{\pu{10.26 mmol}}{\pu{6.98 mmol}} = 2.89,$$

then if we add $\pu{4.72 mmol}$ $\ce{KOH},$ it's $1:3$ to citric acid. Final $\mathrm{pH}$ will be

$$\mathrm{pH_1} = 3.06 - \lg\frac{\pu{10.26 mmol} - \pu{1.57 mmol}}{\pu{6.98 mmol} + \pu{1.57 mmol}} = 3.05,$$

but this answer is incorrect. Where is the mistake?

$\mathrm{p}K_\mathrm{a} = 3.06.$

Citric acid is a triprotic acid:

$$\ce{H3C6H5O7 <=>[K_1] H2C6H5O7- <=>[K_2] HC6H5O7^2- <=>[K_3] C6H5O7^3-}$$

The pKa given is for the second step. Citric acid is the fully protonated species, and citrate the fully deprotonated one. When you add the two, they neutralize to give the intermediate species.

  I'm trying to calculate it like initial $\mathrm{pH}$

The easiest way is to calculate the amount of protons available (3 x 10.26 mmol = 30.78 mmol) and the total amount of organic species (17.24 mmol). There would be no citrate (17.24 mmol of protons to protonate all with one proton, 13.54 mmol protons left), and no citric acid. 17.24 mmol - 13.54 mmol = 3.70 mmol would be monoprotonated and 13.54 mmol would be diprotonated. You can plug those numbers in (assuming that your volume is not crazy high or low) to get the initial pH.

  then if we add $\pu{4.72 mmol}$ $\ce{KOH},$ it's $1:3$ to citric acid.

No, because citrate is not a major species before or after adding the KOH. Instead, you would deprotonate 4.72 mmol of the diprotonated species to get 4.72 mmol more of the monoprotonated species.

$$\mathrm{pH_1} = 3.06 - \lg\frac{\pu{13.54 mmol} - \pu{4.72 mmol}}{\pu{3.70 mmol} + \pu{4.72 mmol}} = 3.04$$

  but this answer is incorrect. Where is the mistake?

I think both answers are wrong, for the reasons given above.


Why would decreasing ionic strength of buffer disrupt hydrogen bonds I'm reading about cDNA synthesis in my textbook and it mentions that:

  The captured mRNAs are subsequently released and collected, for example, by decreasing the ionic strength of the buffer so that the hydrogen bonds will be disrupted

So my question is why would this happen? What does the ionic strength of the buffer have to do with the hydrogen bonds the mRNA has formed?

At high ionic strength, the poly-A tails of mRNA bind oligo-dT that is covalently attached to beads in a column (source). The binding interaction is based on base stacking and hydrogen bonding. When you decrease the ionic strength, the repulsion between the (negatively charged) phosphate groups becomes stronger (is shielded less by counter ions), lowering the binding strength to the point that the mRNA elutes.

The length of the oligo-dT on the beads is chosen such that the mRNA binds sufficiently strong at high ionic strength, but is released at low ionic strength. The appropriate length of oligo-dT is probably on the order of 8-10 nucleotides.

  decreasing the ionic strength of the buffer so that the hydrogen bonds will be disrupted

This statement does not capture the situation accurately. What is really being disrupted is the overall binding of mRNA to the beads. The hydrogen bonds, as short range interactions, should not be directly affected by ionic strength. The stacking interactions might weaken with decreasing ionic strength. Long range ionic interactions, however, are significantly weakened by increasing ionic strength because counter-ions bound to the phosphate backbone will strongly shield the ionic interactions (water also shields it compared to vacuum or non-polar solvents). In this case, the ionic interactions are repulsive, so binding will get weaker at low ionic strength because that is when the repulsive interactions are strongest.


Oxygen in water: How is it single if it's always double? This is simple question, I think. At college (I'm an Intro Chem. student), the professor explained that oxygen is always found in molecules of 2 oxygen atoms combined together - $\ce{O2}$. On the other hand, molecule of the water is $\ce{H2O}$. So I don't get, how is it that O is single in this molecule, if it's always found coupled?

I think you are confusing something here: Molecular oxygen is most commonly found as the gas we know, $\ce{O2}$ and ozone, $\ce{O3}$. However, this doesn't limit it to only that. Oxygen can appear in a wide variety of (bio-) chemicals and there, it only needs two bonds to other atoms (or a double bond to one atom).

Good examples for this are:

The aforementioned water $\ce{H2O}$ where the oxygen is bonded to two hydrogen atoms.
One of my most loved molecules: Ethanol. Here, the oxygen is bonded to a hydrogen and a carbon atom.
In an ester, which usually are quite fragrant compounds, there actually are two oxygen atoms. But they are not connected to each other. Instead, one has a double bond and the other two bonds, each of them to carbon respectively.

I hope this has made it clear that oxygen doesn't have to bond to itself (but it can, see peroxides), and that oxygen doesn't have to come in pair of twos for the molecule to not fall apart.


Why does glass shatter when heat is applied to it? When I poured hot tea into a cup, the cup shattered after about three seconds. The cup was made of glass. It was pretty thin. How did the glass and the tea react in terms of chemicals. This is a bit odd because wasn't heat applied to sand in order to make glass, shouldn't the glass break up into the reactants that made it?

This is not a chemical process.

Objects typically expand when heated. Think in railway profiles in the summer, think in loosening a stuck nut by heating it with a blowtorch.

However, when the material is under stress and a bad heat conductor, unevenly heating will result in cracking.

While it is a pity for the cup, the breaking of glassware to thousands of rounded (!) pieces upon uneven or punctual (mechanical) impact may be exactly what you want under different circumstances. Think in tempered (toughed) glass used in windows of railroad passenger cars.

On a side note, there's the Bologna Bottle. In the making, these bottles are slowly cooled on the outside - but very rapidly on the inside, which maximizes the stress.

As a consequence, you might use the bottle as a hammer to drive a nail into something, but beware if you drop something in the bottle. Here's a video to demonstrate the effect. (Be warned that there's a lot of screaming kids in it.)


Who Was the First Chemist? I know chemistry has been around for a long time, but I was curious who the first chemist in recorded history. I am aware that Antoine Lavoisier was one of the first modern chemistry and is considered to be the "Father of Modern Chemistry", but was he is the first chemist?

That depends how you define "chemist."

Today we think of chemists as working at the molecular level. However Lavoisier didn't have any concrete evidence of an atom's existence. He worked with Sulfates (SO4-2), metal oxides, and other molecular groups in an empirical an algebraic fashion.

Throughout much of the 1800s there was enormous debate about the very existence of atoms. Much of the problem arose from confusion over molecular weights and stoichiometry. Physicists were the most stubborn opponents of atoms, which we still can't fully explain even today. Eventually Advodagro's number and accepted stoichiometry combined to form irrefutable evidence of molecules. Many scientists were completely convinced that atoms existed long before they were proven though. Dalton was the first to introduce stoichiometry so he would be a good candidate as well.


Is it possible to prevent sacrificial anode contaminate the soil? I have a mild steel pipe (black pipe, NOT zinc plated) water well. My question: is it possible to somehow put a sacrificial anode in an electrolyte contained in a bucket instead of burying it in the ground, which allows zinc to eventually enter and contaminate ground water?

I wonder if a sacrificial anode near the surface could prevent corrosion at the deepest part of your pipe, where the water and wet soil are. Usually, sacrificial anodes are placed fairly close to the item being protected.

It would be worth testing some of the water (that comes up thru the pipe) for pH and corrosivity. If the water is alkaline (hard water), it may not be very corrosive. The oxygen level far below ground would be low, perhaps not enough for rapid corrosion.

Maybe a sacrificial anode is unnecessary.


Does the first step in electrophilic aromatic substitution follow equilibrium? If so, can equilibrium constant be calculated? Using chlorine (a Lewis base) and ferric chloride (a Lewis base) as an example, the first step I saw in the textbook looks like this:

The picture above depicts the Step 1 in the EAS mechanism using chlorine as an example.

I noticed that this step has two reversible arrows. I think it means chlorine, ferric chloride, the molecular complex and the ion pair will all be present in a system and their concentration values follow chemical equilibrium.

From high school knowledge (I am a freshman), we can calculate equilibrium constant using concentration. Can we still do it in this case? If so, how?

Let me put it this way: Every compound exist in a 3N dimensional configuration space (may be simple XYZ coordinates of the system, may be internal coordinates, based on your choice), and based on the configuration, its potential energy is defined. The potential energy surface (PES) can be calculated quantum mechanically or using some force field classically. The probability of the system to be found in the given set of coordinates is given by the distribution function defined according to statistical mechanics. For a canonical ensemble, the formula is given as (Z is the partition function):

$$P(\vec{r}_{3N})= \frac{1}{Z}e^{-E(\vec{r}_{3N})/k_BT}$$

Now imagine this, a reaction between compounds containing M and N atoms is actually an exploration of the potential energy surface in 3(M+N) dimensions. Reactants, product and intermediates are given by local minima in the PES while transition state is given by maxima along one dimension and minima along all other dimensions. In other words, for a stable state (reactant, product, intermediate, etc.) all the frequencies (square root of the second derivate of the PES) is real, while for transition state one of the frequencies is imaginary.

Now coming to your question: You will require to calculate the energetics of each of the step of reaction considering the presence of solvent (this is probably happening in presence of solvent), plug in your values to the appropriate statistical ensemble (if it is inside solvent without release of gaseous products, canonical ensemble should be good enough) probability function and you will get an idea about the percentage of intermediates you get.

In most high school textbooks, the reaction marked with double arrows simple means the reactant, product (and in your case the intermediate) is not too much different in terms of potential energy. For a single headed arrow, it means that the difference between potential energies of the minimas (reactant and product) is significant and most of the compounds (in an ensemble) will be trapped in the lower minima.

In case you want to read more about it, I suggest the following books:

Statistical Mechanics by McQuarrie
Physical Chemistry by Atkins
Physical Chemistry by Levine

For more details about reaction dynamics, you may refer to:

Molecular Reaction Dynamics by R. D. Levine

Anyway, please reply in comments if you have more doubts, will be happy to help.

Greetings from India. जय हिन्द.  Jai Hind.


Calculate the moles of acid and conjugate base needed Question

  Using the following salts listed below, calculate the moles of acid and conjugate base needed to make $V = \pu{100 mL}$ of $c = \pu{0.50 M}$, $pH \in 6.5, 6.7, 6.8$ phosphate buffer solutions. ($\mathrm pK_\mathrm a(\ce{H2PO4}) = 6.64$)

  Molas masses of salts:

  $\ce{KH2PO4} = 136.09 ~\pu{g/mol}$
  $\ce{NaH2PO\cdot H2O} = \pu{137.99 g/mol}$
  $\ce{Na2HPO4 \cdot 7H2O} = \pu{268.09 g/mol}$
  $\ce{K2HPO4} = \pu{174.2 g/mol}$

I tried using the Henderson-Hasselbalch equation but I am stuck at the point where I have found the base-acid ratio and I need to find the moles of the acid and the moles of the base.

$$
\mathrm pH = \mathrm pK_\mathrm a + \log\left(\frac{[\text{Conjugate base}]}{[\text{Acid}]}\right)\\
n = \pu{100 ml} \times \pu{0.5 M} =\pu{ 0.05 mol}
$$

For pH 6.5:
$$
6.5 = 6.64 + \log\frac{[B]}{[A]}\\
\frac{[B]}{[A]} = 0.72
$$
Here I got the base-acid ratio as 0.72 but what should I do next? I tried looking up for solutions and a person suggests using $A + B = 0.05$ But how do I know what A and B are just by knowing the ratio is 0.72?

Since it is a buffer solution, the volume is constant ($V =\pu{0.1 L}$), hence
$$
\frac{c(b)}{c(a)} = \frac{n(b)}{n(a)}\overset{calc.}{=}0.72
$$
 and with
$$
n_{\text{tot}}=n(b)+n(a) \overset{\text{def}.}{=}\pu{0.05 mol}\\
n(a)=\frac1{1.72}\cdot\pu{0.05 mol}\\
n(b)=\frac{0.72}{1.72}\cdot\pu{0.05 mol}
$$
Then $$m=M\cdot n$$ and you should be fine.


Hydrogen halide acidity Why is it that
$$\mathrm{p}K_\mathrm{a}(\ce{HF}) < \mathrm{p}K_\mathrm{a}(\ce{HCl}) < \mathrm{p}K_\mathrm{a}(\ce{HBr}) < \mathrm{p}K_\mathrm{a}(\ce{HI}),$$
even though the electronegativity decreases down the group? The more electronegative the atom accompanying hydrogen, the lower the energy of the σ* bond. The lower the energy of the σ* bond, the easier it is for nucleophiles to attach to it (i.e. hydrogen is more easily removed from the compound, thus making it more acidic).
Obviously, something is wrong with this reasoning. I could guess it is because hydrogen binds to different $\mathrm s$ orbitals with the different halogens. For $\ce{F},$ $\ce{Cl},$ $\ce{Br},$ $\ce{I}$ these are $\mathrm{2sp^3},$ $\mathrm{3sp^3},$ $\mathrm{4sp^3},$ $\mathrm{5sp^3},$ respectively.

Yes, you are on the right track, let's look at the situation in more detail.  In the haloacid equilibrium $$\ce{HX <=> H+ + X-}$$ anything that stabilizes HX will push the equilibrium to the left and make the $\mathrm{p}K_\mathrm{a}$ more positive. Anything that stabilizes the proton and the halogen anion will push the equilibrium to the right and make the $\mathrm{p}K_\mathrm{a}$ less positive. Here's a table that summarizes some of the haloacid data:

$$\begin{array}{cccc}
\text{X} & \text{H-X bond strength / kJ mol}^{-1} & \text{Electronegativity of X} & \mathrm{p}K_\mathrm{a}(\ce{HX}) \\
\hline
\ce{F} & 565 & 4.0 & 3.1 \\
\ce{Cl} & 427 & 3.0 & -7.0 \\
\ce{Br} & 363 & 2.8 & -9.0 \\
\ce{I} & 295 & 2.5 & -11.0
\end{array}$$

There are 3 main factors that influence this equilibrium: bond strength, electronegativity and polarizabilty.

In the "Bond Strength" column we see that the $\ce{HX}$ bond strength decreases as we move down the column. This is because, as you noted, overlap between the hydrogen $\mathrm{1s}$ orbital and the halogen orbital is most effective with the fluorine $\mathrm{2sp^3}$ orbital (in fact, overlap is so good in the fluorine case that the bond has a significant covalent nature) and the effectiveness of the overlap decreases as we move down the column. Therefore, $\ce{HX}$ bond strengths should make the $\mathrm{p}K_\mathrm{a}$ decrease (less positive, more negative) as we move down the halogen column.

Your analysis of electronegativity is correct. The more electronegative the halogen, the more stable the halide anion. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to increase (more positive) as we move down the halogen column.

The larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to decrease (more negative) as we move down the halogen column.

So we have two factors, bond strength and polarizability, that work together to push the $\mathrm{p}K_\mathrm{a}$ lower as we move down the halogen column; and one factor, electronegativity, that works in the opposite direction. Based on the observed $\mathrm{p}K_\mathrm{a}$'s, the first two factors win out over the electronegativity.

BTW, I like your $\sigma$, $\sigma^*$ reasoning. Here's how I think of it in those terms.  As you bring two atomic orbitals closer together, they will split and form two new molecular orbitals. The degree of splitting is dependent upon the overlap of the two atomic orbitals.  Overlap is a measure of electron density between the atoms.  An electronegative atom will share its electron less readily decreasing the splitting.  If there is a strong bond, then the electrons are well shared and the splitting will be larger.


Evaporation of a mixture of salts I was wondering what happens when you boil dry a solution of several salts for example a solution of KCl and NaNO3. Since the ions dissociate I was wondering if the ions would combine randomly upon being boiled out of solution making the original salts as well as KNO3 and NaCl? Does this happen? If not why?

Also in a related question when you mix two solutions to form a precipitate in a double displacement reaction will the insoluble salt always form or are there situations where an insoluble salt can form but doesn't?

Thanks in advance.

As the solvent evaporates the least soluble salts precipitate in succession.  This is used during seawater and salt lake evaporations to isolate values beyond NaCl. Overall, you will get a mixture of combinations of cations and anions.

Supersaturation is not uncommon.  Metastable solutions usually require nucleation sites to get over their initial activation energy for solids' crystallization.  High melting points (strong lattice forces) are difficult to supersaturate.  Benzil in xylenes remarkably supersaturates, and is stable for a week or more in silanized glassware and cooled very slowly within styrofoam and rags.


Hosoya Z Index and Correlation with Boiling Point I have recently read about the Z topological index found by Hosoya. The papers/sources The Z index is the sum of the coefficients of the matching polynomials (no. of ways to choose an independent edge from the set of edges in the graph - if we let each carbon atom be a vertex and each bond between a pair of carbon atoms is an edge.). The paper states that in particular for alkanes or saturated hydrocarbons, there is a correlation found between the Z index (which measures the extent of branching - in what way??) and the boiling point (bp) of these alkanes.

I know that the bp of alkanes increases with its size (due to strong force of attractions). However, I am confused about how the Z index measures the extent of branching, and any underlying reasons that the Z index will correlate with the bp and other physiochemical properties.

The papers I have read so far fail to enlighten me on these questions above. Please help.
Thank you.

  Some ideas I have so far:
  Since Z index measures branching (or at least correlates to branching since more compacted graph will have a smaller sum of matching coefficient?), and branching correlates to the bp of alkanes (due to less surface area sometimes if you have more branching), perhaps thats why the Z index correlates to bp...

Refer to Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?

For a given hydrocarbon C$_n$H$_{n+2}$ more branched hydrocarbons have lower boiling point. This is because they have less surface area to have Van der Waals interactions to entangle with each other (ropes have more chances to entangle with each other than balls). For example C(CH3)4 (n-pentane) is a gas, but all other pentanes are liquids.

Their method is qualitative only. For instance, o-p-nitrophenol can be steam distilled and has much higher vapor pressure at a given temperature than p-nitrophenol, but the table would predict that they have the same properties. The difference is due to intramolecular h-bond present in o-, but ansent in p- isomer.


Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure? What happens during this reductive amination?

I got the first product; the lithium aluminium hydride will reduce the nitrile to an amine. But what about the reductive animation step? As far as I can see the C=O will be protonated by the strong tosic acid, and then there will be a competition between the methanol and the internal amine to attack C=O. Actually, the amine is able to attack the C=O without the C=O being protonated, but still, this intramolecular seem too favorable given that we are forming a 4-membered ring. So will the methanol outcompete the amine? In that case what's up with the reductive animation step?

This is what I think happens after the second step ... I am going through the intramolecular route ... since that's the only way we get an iminium ion, on which the sodium cyanoborohydride can act.

Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.

But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:

The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.

In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):

But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.


Are there metals that react with sulfuric acid but not hydrochloric (or vice versa)? In school practicals introducing the reactions of metals with acids, it seems to be standard (as in this one from the RSC) to use both hydrochloric and sulfuric acid, but with metals which react to both or neither (e.g. magnesium, zinc and copper). It would be nice to be able to make the point that different things react with different acids, but I'm at a loss for what might fit the bill. Is there anything?

OK, let's clean things up.

Reaction of an acid with a metal is generally a redox reaction. The oxidizing agent are protons generated in the process of dissociation of the acid (let's forget about anions for a sec). The property that decides whether a metallic element can be oxidized by protons is its standard reduction potential. In general, any metal with a negative reduction potential will react with any Brønsted acid (one that produces protons), no matter whether it is a weak (e.g., acetic) or strong acid (e.g., hydrochloric). And if a metal has a positive reduction potential, it won't react with acids. Therefore, if we do not consider other factors, there are no examples to make the point that different acids react differently with a given metal, in fact they all should behave in the same way.

But what makes Chemistry fun are those 'other factors'. The anions produced in the dissociation process of an acid can also act as an oxidizer. For example, nitrate ions produced from nitric acid will oxidize copper, so the metal will dissolve in nitric acid. Sulfate ions are also oxidizing agents, but much weaker than nitrate ions, so copper won't dissolve in diluted sulfuric acid, but a hot and concentrated sulfuric acid will do the trick. And hydrochloric acid in the absence of some other oxidants will not attack copper at all. Another factor to consider is the passivation of a metal. A thin, invisible layer of an oxide or salt can protect a metal from an acid attack. Aluminum, which is a very reactive metal, will not dissolve in oxidizing acids (e.g., nitric), because it will develop a protective layer of aluminum oxide on its surface. Non-oxidizing acids, on the other hand, will easily do the job. Metal salts that are insoluble in water can also create a passivation layer.

Regarding the original question, there are some examples for sulfuric and hydrochloric acid, where they will behave differently due to the formation of insoluble salts. One example is thallium - it dissolves in sulfuric acid but not in hydrochloric acid. An opposite is barium - it dissolves in hydrochloric acid but not in sulfuric acid.


How are these three things equal? In this answer, I don't understand how this step worked.

Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$
\begin{align}
\Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\
\Delta G &= \Delta G^\circ + RT\ln Q
\end{align}

How does $P_i=\dfrac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$? And how does that equal the quotient $Q$?
If this is wrong, then please provide me the correct way to proceed from there.

You have, for each compound, $\mu_i = \mu^\circ_i + RT\ln(P_i/P^\circ)$.
So, replacing each potential in $\Delta G$ by its expression, assuming $P^\circ = 1$ bar, we get :
$$\Delta G = c(\mu^\circ_C + RT\ln(P_C)) + d(\mu^\circ_D + RT\ln(P_D)) - a((\mu^\circ_A + RT\ln(P_A)) - b(\mu^\circ_B + RT\ln(P_B))$$
Then we develop and group standard potentials and logs:
$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT(c\ln(P_C) + d\ln(P_D) - a\ln(P_A)-b\ln(P_B))$$
Then, using calculation rules on logarithms:
$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\left(\ln({P_C}^c)+\ln({P_D}^d)-\ln({P_A}^a)-\ln({P_B}^b)\right)$$
And finally, since $-\ln(x) = \ln(1/x)$ and $\ln(x)+\ln(y)=\ln(xy)$,
$$\boxed{\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\ln\left(\frac{{P_C}^c{P_D}^d}{{P_A}^a{P_B}^b}\right)}$$
Then, try expressing the quotient of the reaction in terms of partial pressures, assuming standard pressure is 1 bar and you'll find out it gives the inside of the log!


What does non-integer bond order mean? I am studying molecular bond theory, and I see that sometimes this happens.
Does this have some significance to the molecule?

It is one of the man-made tools to conceptualize the nature of a bond between two atoms, to know if there is a single / double / triple / quadruple *) bond between them, and to account for both electrons involved in bonding / antibonding molecular orbitals.  Because the nature of electrons is neither "true and only" particle, nor wave alone, hence sometimes escaping our day-to-day experience with macroscopic objects.

In addition, rarely you look at the bond order of two atoms in a molecule without keeping in mind the situation with other atoms present in the same molecule, too.

*)  Sorry, I forgot at least the three-centre-two electron bonds in diborane...


Reversible and irreversible cyclic processes The answer in Arihant's book is given as (b), but wouldn't (c) be correct as well? This is a question of IIT JAM Chemistry.

  A system undergoes two cyclic processes 1 and 2. Process 1 is reversible and Process 2 is irreversible. The correct statement relating to the two processes is
  (a) $\delta S=0$ for Process 1, while $\Delta S\neq0$ for Process 2
  (b) $Q_\text{cyclic}=0$ for Process 1 and $Q_\text{cyclic}\neq0$ for Process 2
  (c) More heat can be converted to work in Process 1 than in Process 2
  (d) More work can be converted to heat in Process 1 than in Process 2

Answer b is incorrect for the reversible Carnot cycle.  The book has given the wrong answer. Answer c is correct if only one cycle of each process is carried out.  You can obviously convert more heat to work if you run process 2 through the same cycle several times, which also counts as a cyclic process.


Definition of relative atomic mass I found out there are two definitions of relative atomic mass $A_\mathrm r$.
First definition is $A_\mathrm r$ is the mass of 1 atom of an element relative to 1/12 the mass of carbon-12 atom. It can be found in any chemistry school textbook.
Second definition is what I just found and dont understand. It states $A_\mathrm r$ is the mass of 1 mole of atoms relative to 1 mole of carbon-12 atoms.
Can you kindly explain second definition with examples?

These are all mathematically identical. Here’s why:

$$\frac{\text{mass of one atom}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}}\\=\frac{\text{mass of one atom}\times N_\mathrm{A}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}\times N_\mathrm{A}}\\=\frac{\text{mass of one mole of substance}}{\frac1{12}\text{ mass of one mole of }\ce{^12C}}$$


Why catalyst does not alter the entropy of a reaction? In my textbook it is written that "Catalyst does not alter the entropy of a reaction". But why? There is literally no explanation given in my textbook as to why this is true. Could you please explain why this is true? I mean if a catalyst speeds up or slows down a reaction so shouldn't it affect the entropy of a reaction? I am really confused. Please explain. Would greatly appreciate it!

TL;DR Catalysts by definition do not alter the magnitude of changes in thermodynamic properties that accompany conversion of reactants to products, they only alter the conversion (reaction) rates.
The reason a catalyst does not alter the entropy of a reaction is that this is one part of the definition of a catalyst! Ideally none of the changes in thermodynamic functions accompanying a reaction are altered by the catalyst. However, that is a circular and not very helpful argument. The fundamental reason is that catalysts are not consumed in chemical reactions. Since the products and reagents remain the same, it follows from Hess's law that catalysts do not contribute to thermodynamic changes associated with the reaction.  Only the rate with which reactants and products interconvert (kinetics) changes.
Catalysts alter the reaction rate by altering the rate of encounter of the reagents or products or by assisting in formation of new lower energy (more stable) reaction intermediates. A catalyst may be as simple as a material that increases the proximity of reagents by increasing their local concentration, without otherwise altering their reactivity (bond enthalpies). That can be the case in either enzymes or solid supports. However catalysts often also alter reactivity by weakening specific bonds. An example of a catalyst that provides both functions is platinum.


Freundlich Equation's dimensions So, we have the Freundlich's Isotherm equation as $\frac{x}{m} = kp^{\frac{1}n}$, where the LHS measures the extent of adsorption with the corresponding pressure $p$ in the RHS with the two constants, $k$ and $n > 1$. But, I got a little confused with the dimensions on both the sides. Since $\frac{x}{m}$ is dimensionless, so should be the RHS. Therefore, I reasoned that the constant $k$ having units should cancel out the units of $p^{\frac{1}n}$. But yet I'm confused, and I don't think its right. What is going on?

Technically speaking, one typically wouldn't write the left hand side unitless, as it is the ratio of the mass of two different components (usually mg of some chemical per gram of the surface).
But, otherwise your intuition is correct and $k$ just needs units that will cancel with $p^{\frac{1}{n}}$ to give the units on the left hand side. So $k$ could have units of $\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{bar^{\frac{-1}{n}}}\right)$.
We could also reformulate Freundlich's equation in terms of concentration (since we are dealing with an isotherm) as: $$q=KC^{\frac{1}{n}}$$ where $q=m/x$, $C$ is concentration of the chemical and $K$ is a new constant with units of $\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{{\dfrac{L}{\text{mg chemical}}}}\right)^{\frac{1}{n}}$.
Source


Acrylic container and electrolysis Normally when you do electrolysis, you would use a glass container. Would it be wise to use a acrylic container? I have one spare but not sure if this would affect the electrolysis process. Is there any reaction between the acrylic container in the electrolysis process?

Poly(methyl methacrylate), PMMA withstands acids and bases of low concentration. So it should be safe for electrolysis of water.


Removal of tetrabutylammonium from an organic reaction I need to synthesize some organic compound for my Ph.D. research. Everything went OK during the synthetic route but got messed up at the very end. In my final step (once the final compound has been obtained) I remove a silyl protecting (TBS) group using TBAF. The problem comes when I try to purify the reaction. No matter how carefully I run my flash column (FCC, flash column chromatography), I always obtain my compound with traces of TBA.

I've tried other deprotection procedures (mainly acidic ones) but they hydrolyze my final compounds since they are sensitive to acidic conditions.
Does anyone know of an alternative purification technique (or reaction) that would remove the TBA?

I know it is a relative old issue but I had the same problem with nucleosides during my PhD. I tried a wide range of reagents but all of them lead to the decomposition of compounds or no reaction occurred. The only one that worked was TBAF 1 M in THF. I could not extract them because they are too polar. FCC did not work at all.

Solution (at least it worked for me): after FCC, I dissolved my product with TBAF in water and pass it through a Supelco SPE DSC-C8 3 mL, 500 mg for up to 10 mg of my compound. Use only water and do it slowly. I always get the desired compound from tube one to 6-7 and no TBAF is present. It is useful if you can use a LC-MS to analyze each tube. I tried other brands and no one worked except this one. I don't work for them, just want to share my solution.


How to extract force when the center of mass of molecules is put in the box in GROMACS? To extract the force as in the title I used the gmx trjconv command like below.

gmx trjconv -f alad.trr -s alad.tpr -n alad.ndx -pbc mol -novel -force -o alad.gro

However, the force was not extracted in alad.gro file.
Of course, I got the coordinates and the force at equal intervals when getting the alad.trr file.

Another way, I used gmx traj command to get force.

gmx traj -f alad.trr -s alad.tpr -n alad.ndx -pbc -mol -ox coordi.xvg -of force.xvg

However, the force obtained from force.xvg didn't look like the value I wanted.

So, how do I get the force of molecules put in a single periodic box?

I solved the problem. As gmx trjconv -pbc mol option only make molecules translate, the cooridinates obtained from alad.gro can correspond to forces obtained from gmx traj -f alad.trr -s alad.tpr -n alad.ndx -of force-dump.xvg.


Equilibrium constants for lewis acids such as Ca2+ I need to calculate the theoretical $\mathrm{pH}$ in a $50~\mathrm{mL}$ water solution where $6.4\times10^{-4}~\mathrm{g}$ of $\ce{Ca^{2+}}$ is present. The reaction I base this on is the following

$$
\ce{Ca^{2+}_{(aq)} + 2H2O_{(l)} \rightleftarrows Ca(OH)2_{(s)} + 2H^+_{(aq)}}
$$

$$
K = \frac{[\ce{H+}]^2}{[\ce{Ca^{2+}}]} \Rightarrow [\ce{H+}] = \sqrt{K \cdot [\ce{Ca^{2+}}]}
$$

Is there a database of standard equilibrium constants for such Lewis acid reactions?

I appreciate any help!

You're confusing terms here. $\ce{K_{a}}$ is the (Brønsted) acid dissociation constant. This suggests that $\ce{K_a}$ refers to the degree that a Brønsted acid dissociates into its conjugate base and hydrogen protons.

$\ce{K_{a}}$ is inapplicable to Lewis acids because Lewis acids are not proton donors.

Brønsted acids are Lewis acids in that Brønsted acids such as the hydrogen proton can accept a pair of electrons. But the converse is not true.

Lewis acids are electron acceptors. How do we measure the strength of Lewis acids? There is no universally accepted measure as there is for Brønsted acids. We could look at the enthalpy of reaction (Lewis adduct formation) between a Lewis acid and a Lewis base. This is a common method.


Chemical Kinetics / Rate Question -- Error in Textbook Question/Answer? I am reading Kaplan's General Chemistry MCAT Review 2018-2019, and think that one of its chemical kinetic questions is wrong. (Or perhaps I am simply missing something). I have attached an image of the question, which is easier than typing out as it includes the standard table of concentrations at different trials.

The answer reads: "In the first two trials, the concentration of $\ce{XH4}$ is held constant while the concentration of $\ce{O2}$ is multiplied by 4, and the rate of the reaction also increases by a factor of approximately 4..." It ultimately comes up with a rate of $k\ce{[XH4]^2[O2]}$

Am I completely missing something? My first question is, is the book's answer right? If it is not, is the problem solvable (i.e. does the table make any sense at all)? I've tried to come up with rates using the table and encountered the following issues. Please let me know if I am right in approaching the question in this way:

If you try to find the rate with respect to $\ce{XH4}$ using trials 1 and 3, I believe you get an order of 3. However, if you try to find the same rate using trials 1 and 4, you get 4. This isn't normal, right? Maybe an example of a mixed order reaction? Or just a mess up?
How do we interpret the data from trials 3 and 4? The concentration of neither reactant changes and yet the rate doubled. If we ever see something like this in a table, does it suggest that the reaction is 0 order? Or can we not make that leap? And is it even possible to have this data alongside the other data in the table?
It seems to me that, based on Trials 1 and 2, the rate is 0 order with respect to $\ce{[O2]}$. Is that a correct interpretation?

Thank you so much!

The exercise is absurd. None of the proposed answers can be right, as for any answer A,B,C,D the rate is a function of the concentrations [XH4],[O2] but the table suggests that the rate is constant over time, e.g. 2.0$~$M/min for entry 1. Thus:

$$
R = 2.0~\text{M/min} = \text{constant} \cdot [\mathrm{XH_4}]^0 \cdot [\mathrm{O_2}]^0
$$
and is therefore a zeroth order kinetics.

I assume that the initial rate $R_0$ or a rate at some point $R_{t>0}$ is meant, and the question just heavily confuses things. So from here on, let's assume that $R$ means some rate at some point in time.

But even then, it is not possible to answer the question. This is because entries 3 and 4 have the same domain ([XH4] and [O2]) but a different codomain (different rates R). This means that there is no function mapping from the concentration of the reactants to the rate of the reaction.

However, there obviously is such a functional relation for all answers A,B,C,D.

Therefore, the question is absurd. It is probably meant to be that [O2] is 2.0 in entry 4.


What are the equations giving the molar flow of different molecules linked by a reaction? Let consider a steady state CSTR in which occurs the following reaction:

$$\ce{aA + bB \leftrightarrow cC}$$

Notations:

$F_i$ is the molar flow of $i$, $r$ is the rate of reaction, $V$ is the volume, $X$ is the conversion at the end of the reactor, $\nu_i$ is the stoichiometric coefficient of $i$ and $\xi$ is the extent of reaction.

Mass balance:

$$\ce{IN + PROD - CONS = OUT + VAR}$$

Therefore,

$$
\begin{cases}
F_{A,in} + 0 - arV = F_{A,out} \\
F_{B,in} + 0 - brV = F_{B,out} \\
F_{C,in} + crV - 0 = F_{C,out} \\
\end{cases}
$$

As $$\xi = -rV = -X\frac{F_{i,in}}{\nu_i}$$ we have for $\ce{A}$:

$$\begin{alignat}{}
F_{A,out} &= F_{A,in} - arV \\
& = F_{A,in} - a(-\xi) \\
& = F_{A,in} - a\left(X\frac{F_{A,in}}{a}\right) \\
& = \left(1 - X \right)F_{A,in}
\end{alignat}$$

For $B$ with can find as well:

$$F_{B,out} = \left(1 - X \right)F_{B,in} = \frac{b}{a}\left(1 - X \right)F_{A,in}$$

My problem with C:

Now if I follow my calculus for $\ce{C}$, here is what I get:

$$\begin{alignat}{}
F_{C,out} &= F_{C,in} + crV \\
& = F_{C,in} + c(-\xi) \\
& = F_{C,in} + c\left(X\frac{F_{C,in}}{c}\right) \\
& = \left(1 + X \right)F_{C,in}
\end{alignat}$$

Which states that if there is no $C$ at the beginning, there is no $C$ produced at all! Which is wrong. Can you tell me where I am wrong?

NB: I have no clue, which tags fit best for this question.

I can't tell you precisely where your reasoning starts to be wrong, but here is another way to think about the problem.

Notice that for a specific reaction as you gave here, the term $rV$ is always the same independently of the molecule you choose to determine it. It does vary for sure over time though (don't read I said it was a constant).

As you wrote:

$$\begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases}$$

Thus, $$rV = \frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{B,in} - F_{B,out}}{b} = \frac{F_{C,in} - F_{C,out}}{-c}$$

You can write what happens for $C$ in terms of $A$: $$\frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{C,in} - F_{C,out}}{-c} \Leftrightarrow F_{C,out} = F_{C,in} + \frac{c}{a} \left(F_{A,in} - F_{A,out}\right)$$

Which boils down to, $$F_{C,out} = F_{C,in} + \frac{c}{a} X F_{A,in}$$

And if you write equations in function of a particular molecule, $A$, as you did here, make sure that the conversion you wrote $X$, is to be more precise, the conversion of $A$ at the exit of the reactor. Otherwise, it doesn't make sense.

EDIT:

Note as well that your equation for $B$ is wrong as well, using the same reasoning as before you should find: $$F_{B,out} = F_{B,in} - \frac{b}{a} X F_{A,in}$$ which make sense because the flow of $B$ must be less at the end of the reactor if this reaction really occurs.


Why does 3-hydroxy-butan-2-one give positive Tollens' test? Why does 3-hydroxy-butan-2-one give a positive Tollens' test despite the absence of an aldehyde group? What is the mechanism?

From the Wikipedia entry on Tollens 1

  Tollens' reagent is a chemical reagent used to determine the presence of an aldehyde, aromatic aldehyde and alpha-hydroxy ketone functional groups

The substrate you have drawn is an alpha-hydroxy ketone.


Why are NMR typically taken for a power of 2 scans? Why do we typically take $n^2$ scans, even in the basic NMR experiment? I've heard various explanations, including that it helps the fast Fourier transform, and would like to know the real reason.

If you mean why do we process raw NMR data using $n^2$ data points, then @Klaus's answer is correct. Although, it is actually usually possible to record an arbitrary number of time domain data points; it is just the processing algorithm is limited to how many data points it can handle. You will either lose the extra time domain data (if using less frequency domain data points) or add zero filling (if using more frequency domain points).

If you mean the number of scans as the number of transients added together to give the final raw FID, then this comes about from the requirement for phase cycling a pulse sequence in order to remove unwanted NMR signals that may arise from

imperfections in hardware phase (giving rise to quad images)
coherent noise
generation of undesired magnetization arising in multi-pulse experiments

Here, though, the number of scans is not limited to $n^2$. Historically, the number of scans for a 1D spectrum would typically have been confined to a multiple of 8, as a phase cycle requirement. As you start to need to run greater numbers of scans due to sensitivity issues, then you would need to consider the accumulation of noise as well as signal, as @ron has alluded to (doubling the number of scans only give a $\sqrt{2}$ improvement in S/N; you need to run 4 times as many scans to double S/N). It is through this combined relationship that the number of scans, then, adheres to a $n^2$ increase. If 8 scans (required by phase cycling) didn't give sufficient S/N, then we need to double or quadruple the number of scans to see a real improvement in S/N. And so of for ns=16 or 32 or 64.
For more complicated experiments, phase cycling may have required multiples of 16 or more. I remember running a dqf-COSY experiment which had a phase cycle of 24, so the number of scans had to be a multiple of 24. Yes, these experiments took many, many hours to run.

However, (there's always a however) in modern spectrometers (anything newer than about 20 years old), the phase cycling restrictions are largely removed through improvements in hardware design and implementation of gradient pulses. So now we see for most basic 1D experiments, such as a 1H, there is a minimum number of scans of 1. We run many high quality, high resolution, quantitative analyses of samples using long acquisition, single scan experiments.
For multi-pulse experiments, there is still a requirement to phase cycle, but even for many 2D and 3D experiments, this is limited to multiples of 1, 2 or 4. Very few experiments these days require large numbers of scans to account for phase cycling.


How does heat energy interact between two bodies of different heat capacities in order to attain thermal equilibrium? Suppose there are two objects $A$ & $B$ of same mass; the former has specific heat capacity $C_A$ and the later has $C_B$ such that $C_A\gt C_B.$ Let $A$ be at temperature $T$ while $B$ be at $T+\Delta T.$ So, in order to attain thermal equilibrium, they have to attain the same temperature $T+ \frac{\Delta T}{2}.$ That means there would be a change of $\frac{\Delta T}{2}$ in both bodies. For that, $A$ must gain heat energy $Q_A= m\;C_A\cdot \frac{\Delta T}{2}$ while $B$ has to loss heat energy $Q_B=m\;C_B\cdot \frac{\Delta T}{2}.$ Now, $Q_A-Q_B\gt 0$ since $C_A\gt C_B.$

Then in order to attain the equilibrium temperature, $A$ has to get $Q_A$ energy which it receives partly from $B$ viz. $Q_B$ but where does the additional heat energy $Q_A-Q_B$ come from? It can't come from $B$ as that would deviate it from achieving the thermal temperature; $B$ can lose only $Q_B$ in order to attain $T+\frac{\Delta T}{2}.$ But where does $A$ get the extra energy $Q_A- Q_B$ from?

I'm confused as I thought so-far that when two bodies achieve thermal equilibrium, they do it by exchanging heat energy between themselves - the cooler body receives heat energy fro the hotter body till the equilibrium is reached.

But here, it seems that $A$ in order to achieve thermal equilibrium, needs energy $Q_A$ which cannot be provided by $B$ since the energy $Q_B$ it releases  is less than $Q_A.$ Where would then the extra energy $Q_A-Q_B$ come from?

Let's say, somehow $A$ gained that energy from the surroundings; however what would happen if they were isolated from the surroundings?

I don't know where I'm mistaking. Is it true that the hotter body itself provide the required energy to the cooler body to attain thermal equilibrium?

Please help.

Say the initial temperature of $A$ is $T_1$ and the initial temperature of $B$ is $T_2$. To determine the temperature ($T_f$) of $A$ and $B$ at equilibrium, we start with the equation $q_A=-q_B$, where $q_A$ and $q_B$ are the heat gained by $A$ and $B$ respectively as they reach equilibrium. We have $q_A=mc_A\Delta T_A=mc_A(T_f-T_1)$ and $q_B=mc_B\Delta T_B=mc_B(T_f-T_2)$. Thus the equation becomes $$mc_A(T_f-T_1)=-mc_B(T_f-T_2)$$ From here, we can solve for $T_f$. We can also note that if $c_A=c_B$, $T_f=\frac{T_1+T_2}{2}$ , which in your example is $T+\frac{\Delta T}{2}$. Otherwise, $T_f$ will not be the midpoint between $T_1$ and $T_2$.