Where can I find ranges of isotope abundances? I'd like to know how much stable isotopes like $\ce{^{13}C}$, $\ce{^2H}$, $\ce{^{17}O}$ and $\ce{^{18}O}$ deviate from their natural abundances in the earth's crust among all possible compounds we may come across, especially organic molecules. I'd like to know a rough minimum and maximum percentage of such isotopes that a given compound is likely to contain, and if possible how likely it is to contain such a percentage. It might be really good to have a histogram showing how common compounds with different percentages of such isotopes are. Does anyone know where I could find such data?

I am aware of this range diagram about $\ce{^{81}Br}$ that appeared in the recurrent (open access) IUPAC recommendations of atomic weights in Pure and Applied Chemistry:

(source: doi 10.1351/PAC-REP-13-03-02)

which cites data of a publication in Environmental Chemistry extending a little to $\ce{^{37}Cl}$ (doi 10.1071/EN10090, open access), too.

Even if the elder (2000) IUPAC compilation in the same journal includes a graphical representation of the range in occurrence of $\ce{^2H}$ as known at this time:

(source: doi 10.1351/pac200375060683, open access)

it may be worth to read more in detail, as -- in contrast to the 2011 edition (32 pages) -- the 2000 edition (120 pages) includes a dedicated section "Isotope fractionation from natural processes" and "Element-by-element review of the standard atomic weights" for a detailed view.

Otherwise, if you were "just" interested in an overall overview of isotopes, their masses and their percentage of occurrence, NIST has a searchable dedicated database here.


Does variations in air pressure affect the melting point of PLA and ABS while 3D Printing? Tight temperature controls are needed to ensure that the filament used in fabrication does not begin to "ooze". In places where the air pressure rises and drops during the course of the day there may be more or less ooze taking place. Controlling ooze is paramount to obtain better print quality.

Question: Given Boyle's law, shouldn't the melting point of PLA and other plastics used in the Fused Filament Fabrication 3D printing methods change based on the ambient air pressure of the machine's location? Should air pressure be taken into account when calculating how much heat to apply to the print nozzles? Does humidity affect PLA's melting point?

Boyle's law refers to properties of an idealized gas, where the product of pressure and volume remains constant.  Taking into account your description, as well from filament-based 3D printers seen so far, my answer is no, Boyle's law does not apply here.

If running by extrusion, polylactic acid (PLA) is heated to overpass melt, however this liquid is not heated to become a gas. Temperature control clearly is of importance, because the viscosity of the polymer depends on it.  And viscosity co-influences the precision of details you may print, actually down to the scale of less as a tenth of a millimetre.  Without experience to run one 3D print in at sea level, and the same again during an intercontinental flight, I assume it is more likely, machine control takes into account the pressure difference between ambient air and polymer melt, and not (yet) the (in comparison) lesser variations of air pressure.

The chemical structure of PLA shown in Wikipedia's article shows a relatively low number of functional groups susceptible to water, relatively low to the inner backbone of the polymer (in the structural scheme, this is written in the square brackets).  The article refers to typical molecular weights of 130 kDa when PLA is produced.  These two let me say, humidity does not play an important role.


References for Transition State Theory and Quantum Transtion State Theory Having asked recently this, I started to look for good references and notes on Transition State Theory and, also, modern versions of it including things like tunneling (which isn't taken into account in Eyring's equation).

Unfortunately, I haven't been able to find any satisfactory reference. I have downloaded a bunch of lecture notes and gained conceptual understanding of TST, but nowhere do I find a explanation with all the mathematical details.

What could you recommend for a thorough and detailed exposition of the TST?

Also, I would like to find stuff about how the theory has been improved over the years (the theory is mainly classical). I guess there are more accurate quantum mechanical versions of the theory.

Thanks a lot.

There are many sources that you can check out, in particular the reviews of Truhlar:

Current Status of Transition-State Theory by Truhlar et al.

This article was used in my undergraduate course of chemical kinetics, so I think that it could be a good source. Also we used:

Variational transition-state theory by Truhlar and Garrett

Maybe they are a bit outdated, but they are good as start points. The abstract of this newer article make it seems to be useful for you

Transition state theory: Variational formulation, dynamical corrections, and error estimates by Vanden-Eijnden and Tal

To deeply understand the theory statistical physics is a must. A good source (that also deals with TST) is the Ralph H. Fowler's book.


How does the silver nucleate on glass in Tollen's reagent mirroring? I want to know the exact mechanism for how silver would start to grow during a mirroring with Tollen's reagent. I've made two pictures to illustrate how I see the nucleation/growth process of silver on silica:

As you can see, I didn't draw any chemical bonds, because I think that the silver adherence is due to Van Der Waals forces. Am I correct? If that is the case, then I believe a place with high permanent polarisation would be better suited for the nucleation of silver.
Now, let's follow the logic. We have a greater electronegativity difference between $\ce{O}$ and $\ce{Si}$ ($3.44 - 1.90 = 1.54$) than between $\ce{O}$ and $\ce{H}$ ($3.44 - 2.2 = 1.24$).
For the upper oxygen, assuming an angle of 109 degrees, you get a total dipole moment of: $2(1.54)\cos\left(\frac{109}{2}\right) = 1.79$
For the lower oxygen with the bonded hydrogen, assuming an angle of 120 degrees , you would get a total dipole moment of: $1.54 + 1.24 = 2.78$
(please tell me if I'm wrong to assume a straight line!)
This would mean that you have a greater accumulation of electrons at the bottom oxygen, which would favor Van Der Waals attraction (Debye interaction, I guess, since you have a permanent dipole and an induced dipole). This, in turn, would mean that it would be a better nucleation site.
If you make a surface treatment using silane groups, I believe you promote the presence of $\ce{OH}$ groups at your silica surface, though I'm unsure about this. I do know that it makes better silver mirroring though. I've made another drawing:

Following my logic, this one is wrong in many way. First of all, I drew angles on the $\ce{Si-O-H}$ bonds, which I didn't draw on my first pictures. Secondly, the silver is not even near the oxygen atoms, which I believe are the nucleation points.
I'm confused, could you please point out to me where I'm right or wrong in my logic/knowledge?
Mainly:

Is there a bond forming between the silica surface and the silver? Or is there only Van Der Waals (Debye) attraction?
Is my depiction of nucleation & growth correct (for the first two pictures)?
Is my logic regarding the electronegativity correct?
Is my logic of " higher total electronegativity difference, hence better nucleation site" correct? (assuming I'm right about the Van Der Waals Forces)
Is my depiction of silanisation correct?
Am I right to say that the higher abundance of surface $\ce{OH}$ groups by silanisation implies better nucleation?
Could I do the same when using a strong acid on my glass surface?

"If you make a surface treatment using silane groups, I think that you promote the presence of OH groups at your silica surface". This is NOT correct. You siliconize glass to cap the free SiOH groups

2 silica_Si-OH + n SiCl2Me2 + HO-Si_silica -> silica_Si-[O(SiMe2)]$_n$-O-Si_silica. As a result you decrease the number of active OH groups and make it highly non-polar.

Before siliconization silica is surrounded by a hydration sphere. Ag$^+$ can bind it well, but we need to precipitate neutral Ag. In case of non-siliconized glass we have two effects:

(1)  It is easier for silver to grow on the surface that already has a silver nano-crystal
(2) Before treatment glass have inhomogeneities, so nucliation in some places is heavily favored.

Both effects are negated by siliconization.

Note that I didn't account for surface charges which change as a result of siliconization and might play a role.


Wurtz Reaction with acidic alkyl halides My chemistry book says we cannot use alkyl halides containing acidic hydrogen for Wurtz reaction. Why is this so?

The Wurtz reaction involves the reaction of alkyl halide with sodium in dry ether. In this reaction, two molecules of alkyl halide react to form a alkyl-alkyl chain and sodium halide is also produced -

$$\ce{R-X + 2Na + X-R ->R-R + 2NaX} \tag{$\ce{X}$ = halogen}$$

Now, sodium is a strongly electropositive metal, and is above hydrogen in the electrochemical series. Thus it replaces acidic hydrogen wherever it occurs. For example,

$$\ce{CH3COOH + Na ->CH3COO- Na+ + 1/2 H2}$$

So, sodium reacts with the acidic hydrogen present in the starting alkyl halide.

Also, acid-base reactions are, in general, faster than a substitution reaction. If there was no acidic hydrogen, alkyl sodium intermediate would have formed and an SN2 reaction would have taken place.

However, in this case, if any alkyl sodium intermediate is formed, it would accept the acidic hydrogen and convert into an alkane, stopping the reaction.

$$\ce{R- Na+ + H-M -> R-H + NaM}$$

(R = alkyl group, M represents the functional group or chain attached to the acidic hydrogen)


Will fire corrode stainless steel? I am considering building a stove out of stainless steel. I want to us stainless steel because the stove will have a double outer jacket to pass water through and heat it.

Will a wood fire sitting directly on stainless steel corrode it? Would a higher grade of stainless steel, say 316 instead of 304, prevent this?

Should the stainless steel be protected in some way, perhaps by a bed of fire bricks (at least) under the fire, to prolong its life? Or will stainless steel stand up to exposure to fire without corroding for a long time?

Here's a device that's been around for at least several decades, as far as I know:

(source: teploterm.ru)
I've been using something similar for around 15 years. The idea is you put firewood into the oven you can see at the bottom, and the chimney goes through a 90 liter water tank. The bottom is something like .5 - 1 cm thick cast iron, the rest is made from stainless steel sheet metal.
There are variants to the construction, but I think here are the two main points, coming back to your question:

This thing needs replacement around once a decade, being in use around 500-1000 hours a year.

If the stainless steel is thick, then rusting gradually becomes irrelevant. Does make the thing heavy, though.

Yes, there is corrosion, especially around the fiery hot parts.

Also, note that the plumbing within this thing is kind of special, as it needs to refill cold water while simultaneously mixing your hot and cold water. I remember that it was very damaging to run it empty.
And yeah, this is more engineery than chemicky...


Stern-Volmer plot and calculation of rate constant for quenching I did an experiment irradiating benzophenone in UV light and worked out the rate constant for the reaction. Then I repeated the experiment using various concentrations of a quencher and worked out their respective rate constants. Now I need to calculate the rate constant for quenching $k_\mathrm{q}$ using a Stern-Volmer plot. How would I do this?

The equation I have used before for the Stern-Volmer relationship is:

$$\frac{1}{I_\mathrm{f}} = \frac{1}{I_\mathrm{abs}}\left(1 + \frac{k_\mathrm{q}[\ce{Q}]}{k_\mathrm{lum}}\right)$$

where the x-axis is $[\ce{Q}]$, the concentration of the quencher, and the y-axis is $1/I_\mathrm{f}$, and to work out $k_\mathrm{q}$ you calculate the gradient/y-intercept = $k_\mathrm{q}/k_\mathrm{lum}$.

I think I would plot a graph of concentration of quencher $[\ce{Q}]$ on the $x$-axis against 1/rate constant on the y axis. But how would I work out $k_\mathrm{q}$?

Benzophenone is a notorious case where prompt fluorescence, phosphorescence and delayed fluorescence can mix. Have a look at Modern Molecular Photochemistry by N.J. Turro for some general information and Prompt and delayed fluorescence from benzophenone by
R.E. Brown and L.A. Singer for the original reference.

\[
\begin{align*}
\ce{BP &->[h\nu]\ ^{1}BP^{\ast}}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(prompt\ fluorescence)}\\
\ce{^{1}BP^{\ast} &->[ISC]\ ^{3}BP^{\ast} &(intersystem\ crossing)}\\
\ce{^{3}BP^{\ast} &-> BP + h\nu''\quad &(phosphorscence)}\\
\ce{^{3}BP^{\ast} &->\ ^{1}BP^{\ast} &(repopulation\ of\ S_1)}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(delayed\ fluorescence)}\\
\end{align*}
\]

Let's ignore all this (for a moment).

Fluorescence quenching, i.e the observed decrease of fluorescence intensity in the presence of a quencher, may occcur through two different pathways:

Static fluorescence quenching

In this case, formation of a ground state complex between the fluorophore $\ce{F}$ a quencher $\ce{Q}$ decreases the concentration of $\ce{F}$, hence the concentration of the excited state $\ce{^1F^{\ast}}$ and the fluorescence intensity. This has no impact on the fluorescence lifetimes! These are the same in the absence ($\tau_0$) and presence ($\tau$) of a quencher.

\[
\frac{\tau_0}{\tau} = 1
\]

Dynamic fluorescence quenching

Here, quenching occurs from the excited state $\ce{^1F^{\ast}}$. With other words, the presence of a quencher introduces an additional, non-radiative pathway for the deactivation of the excited singlet state. Consequently, the fluorescence lifetime $\tau$ is shortened!

Let's try to derive the Stern-Volmer equation and see if that leads to some insights:

(You probably know all of this already, but I'm just in the mood...)

In an experiment on dynamic fluorescence quenching, the following processes and their rate constants have to be considered:

\[
\begin{align*}
\ce{F + h\nu &->[k_{abs}]\ ^1F^{\ast}\quad &\mathrm{(excitation)} } \\
\ce{^1F^{\ast} &->[k_{ic}]\ F + heat\quad &(internal\ conversion)}\\
\ce{^1F^{\ast} &->[k_{f}]\ F + h\nu'\quad &(fluorescence)}\\
\ce{^1F^{\ast} + Q &->[k_{q}]\ F + Q'\quad &(quenching)}\\
\end{align*}
\]

The fluorescence quantum yield $\Phi$ is given as the quotient of emitted ($I_{em}$) and absorbed ($I_{abs}$) photons:

\[
\Phi = \frac{I_{em}}{I_{abs}}
\]

The number of emitted photons depends on the concentration of the fluorophore in its excited singlet state $\ce{^1F^{\ast}}$ and the rate constant $k_{f}$:

\[
I_{em} = k_{f}\cdot[^1F^{\ast}]
\]

The number of absorbed photons depends on the concentration of the fluorophore $\ce{F}$ and the rate constant $k_{abs}$:

\[
I_{abs} = k_{abs}\cdot[F]
\]

The rate of absorption processes is difficult to measure, they're bloody fast. Fortunately, we don't have to! Under steady-state conditions, population of the excited singlet state and its deactivation balances out.

Consequently, we can express $I_{abs}$ as:

\[
I_{abs} = k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]
\]

The expression for the fluorescence quantum yield thus is:

\[
\begin{align*}
\Phi &= \frac{k_{f}\cdot[^1F^{\ast}]}{k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]}\\
&= \color{\red}{\frac{k_{f}}{k_{f} + k_{ic} +k_{q}\cdot[Q]}}
\end{align*}
\]

The latter is pretty convenient since we don't have to think about the pesky $[^1F^{\ast}]$ anymore! YAY!

If we do perform a fluorescence measurement in the absence of the quencher,  $[\ce{Q}] = 0$, $\Phi_0$, the fluorescence quantum yield under this condition is:

\[
\Phi_0= \frac{k_{f}}{k_{f} + k_{ic}}
\]

The effect of a quencher $\ce{Q}$ on the fluorescence is expressed as:

\[
\begin{align*}
\color{red}{\frac{\Phi_0}{\Phi}} &= \frac{k_{f}}{k_{f} + k_{ic}} \cdot \frac{k_{f} + k_{ic} +k_{q}\cdot[Q]}{k_{f}}\\
&= \color{\red}{1+ \frac{k_{q}\cdot[Q]}{k_{f} + k_{ic}}}
\end{align*}
\]

Too bad that we don't know all these rate constants and using just a steady-state spectrometer, we won't be able to measure them! And for a lab chemist, there's too much stuff in the equation anyway! Bummer!

But we can argue about what influences the fluorescence lifetimes, which, btw, can be measured!

Lifetimes are pretty much the inverse of the (sum of the) rate constants.

Without the quencher, the fluorescence lifetime $\tau_0$ is:

\[
\tau_0 = \frac{1}{k_{f} + k_{ic}}
\]

In the presence of a quencher, the fluorescence lifetime $\tau$ is:

\[
\tau = \frac{1}{k_{q}\cdot[Q] + k_{f} + k_{ic}}
\]

If we plug that in again and rearrange a bit, we get:

\[
\begin{align*}
\frac{\Phi_0}{\Phi} &= 1 + \tau_0\cdot k_q\cdot[\ce{Q}]\\
&= 1+ {K_{SV}}\cdot[\ce{Q}]
\end{align*}
\]

We name $K_{SV}$ the Stern-Volmer constant!

Summary

Plotting $\frac{\Phi_0}{\Phi}$ or the quotient of the areas of the fluorescence spectra vs the concentration of the quencher should give a straight line, that cuts the $y$ axis at 1.
The slope of the curve is the Stern-Volmer constant $K_{SV}$.
The real rate constant $k_q$ for the reaction of the excited state $\ce{^1F^{\ast}}$ with the quencher can only be obtained from $K_{SV}$ if the fluorescence lifetime $\tau_0$ has been obtained from a time-resolved fluorescence measurement!


Proper notation of doubly charged ions I have always notated doubly charged ions with a superscript $++,$ such as $\ce{He^{++}}.$  For more highly-charged ions I would denote them with the number of charges, such as $\ce{Pt^{4+}}.$

Recently I have been challenged that doubly charged ions should be similarly written as e.g. $\ce{He^{2+}}.$  My internal consideration of the matter is that the fully consistent application of such a convention would necessitate writing $\ce{He^{1+}},$ which wastes space.  I have a vague recollection from my student days of being told that we use a superscript $++$ because it is the same character count as $2+,$ while avoiding the logical conclusion that one should use superscript $1+,$ and thus the logic of the naming convention was to preclude excess typesetting.  However, my Google skills have failed to turn up documentation of such a rationale $−−$ or any rationale, for that matter!

Can anyone point me to some documentation that might describe/explain the notation convention?

TL;DR: $\ce{He^2+}$ is the only preferred notation.
Notations $\ce{He^{++}},$ $\ce{He^{+2}}$ or $\ce{He^{1+}}$ are obsolete and should be avoided.

From IUPAC “Green Book” [1, p. 49], section 2.10.1 Other symbols and conventions in chemistry, subsection (i) The symbols for the chemical elements:

  The ionic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.

  $$
\begin{array}{lll}
\textit{Examples} & \ce{Na+} & \text{a sodium positive ion (cation)} \\
  & \ce{^{79}Br−} & \text{a bromine-79 negative ion (anion, bromide ion)} \\
  & \ce{Al^3+}~\text{or}~\ce{Al^{+3}} & \text{aluminium triply positive ion} \\
  & \ce{3 S^2-}~\text{or}~\ce{3 S^{-2}} & \text{three sulfur doubly negative ions (sulﬁde ions)}
\end{array}
$$

  $\ce{Al^3+}$ is commonly used in chemistry and recommended by [74]. The forms $\ce{Al^{+3}}$ and $\ce{S^{−2}},$ although widely used, are obsolete [74], as well as the old notation $\ce{Al^{+++}},$ $\ce{S^=},$ and $\ce{S^{−−}}.$

For the historical perspective, see the corresponding article by Jensen [2]:

  In contrast, the German chemist, Walther Nernst, in his equally inﬂuential 1893 textbook of theoretical chemistry, chose to place an appropriate number of superscripted $+$ or $−$ signs directly above the ion’s atomic symbol (4), a practice which was soon modified by placing them instead to the immediate right of the symbol, as in $\ce{Ba^{++}}$ and $\ce{PO4^{---}}$(5).

  The IUPAC guide to Quantities, Units and Symbols claims that yet a third “algebraic” method of indicating ionic charges was also used in the past in which the charge preceded the numerical value, as in $\ce{Ba^{+2}}$ and
  $\ce{PO4^{-3}},$ even though this particular sequence of symbols was originally intended to represent the inherent sign of a number or exponent and not the number of signs (6). However, inspection of nearly three dozen general, inorganic, and analytical textbooks, spanning the period 1909–1975, revealed that the vast majority employed the modified Nernst notation, with a smaller number — mostly of European or Russian origin — using the Ostwald notation instead. Rather surprisingly, very few examples of texts using the algebraic notation could be found, all of them post-1970 (7).

  Since at least the 1950s IUPAC has ruled that ionic charges or “charge numbers,” as they are now officially called, should be written instead with the number preceding the charge sign, as in $\ce{Ba^2+}$ and $\ce{PO4^3-}$
  (6, 8, 9). There are several reasons for this decision. It is more concise than the typographically inelegant Nernst approach and more physically meaningful than the Ostwald notation. Unlike the algebraic notation, it
  avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7. (PDF)
Jensen, W. B. The Proper Writing of Ionic Charges. J. Chem. Educ. 2012, 89 (8), 1084–1085. DOI: 10.1021/ed2001335.


Can Rydberg constant be in joules? In my textbook (Chemistry Part - I for Class XI published by NCERT), there is an equation for the energy of an electron in an energy state: $$E_n = -R_\mathrm H\left(\frac{1}{n^2}\right)$$ and there is a paragraph below it with the following text:

  where $R_\mathrm H$ is called Rydberg constant and its value is $2.18\times10^{-18}\ \text{J}$.

There is another section with the expression for the wavenumber ($\overline{\nu}$):
$$\overline{\nu}=109\,677 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\ \text{cm}^{-1}$$
with a paragraph with the following text:

  The value $109\,677 \space\text{cm}^{-1}$ is called the Rydberg constant for hydrogen.

I checked online and found that in most (all) websites (incl. Wikipedia), the value of Rydberg constant is $109\,677 \space\text{cm}^{-1}$. But when I searched for its value in joules, I found this website with the value of Rydberg constant $= 2.18\times10^{-18}\ \text{J}$.

How can Rydberg constant be written in joules?

Authors may be sloppy about notation in this matter. I recommend considering $R_\ce{H} \approx \pu{10973 cm-1}$ and $Ry \approx \pu{2.18e-18 J}$, noting $Ry = hc \cdot R_\ce{H}$. Units of wavenumbers $(\pu{cm-1})$ and energy are often considered interchangeable in practice because they are proportional to each other by the constant value $hc$.

In my notes, I would always be sure to write $R_\ce{H}$ or $Ry$ to explicitly remind myself "which" Rydberg constant I was using (in fact I merged the R and y into a single symbol because I didn't like the suggestion of multiplication.)

Note also that there is a unit of energy known as a Rydberg, with $\pu{1 Ry} = Ry = hc \cdot R_\ce{H}$.


Can you concentrate alcohol by evaporating the aqueous solution at room temp and collecting on a cold surface? I have a bowl containing cheap wine and an immersible fish tank heater. The heater warms the wine to 30 deg C. The bowl is located in a large closed bucket, where it sits on a support. The bucket is sited out of doors in winter when the air temperature is Ca 1 deg C. Evaporation occurs and the vapour condenses on the inner surface of the bucket. Droplets of condensate run down and form a pool of liquid on the floor of the bucket.

Question: what can be predicted about the composition of the condensate?

Question #2 What effect is achieved by raising or lowering the temperature of the wine on the composition of the condensate. In plain language what temperature of the wine will give me the highest concentration of alcohol in the condensate? How can the strength of the alcohol be measured?

Question #3 What is the effect of different air temperatures outside of the bucket (on the composition of the condensate)?

Question #4 Why is this method not commonly used to concentrate alcohol?

Yes you can do that, this technique is called 'distillation' (it requires a minimum difference of 10 degrees celsius to seperate two liquids with different boiling points, for efficiency), I don't know what temeperature the fish tank heater runs at, but you should know that ethanol i,e. what you are trying to separate evaporates at 37 degrees celsius same as normal body temperature, and the temperature of the cooling plate should be < 27 degrees for efficiency

Edit -
yes, exactly the boiling point of ethanol is 78.3 degrees but ethanol is highly volatile and evaporates very fast at 37 degrees (without taking much water vapour with it) as far as my observation is concerned(with room temperature around, 1 atm pressure). the values I mentioned are the optimal conditions with experimental background. your goal as I can percieve is to avoid as much water as possible, if you take the condenser temperature to below 10 degrees the water vapour present in air will also start to condense at a good rate


Why assume a neutral amino acid is given for acid-base reaction? The questions is:

  $1.812\ \mathrm{g}$ of a crystallized α-amino acid ($\mathrm{p}K_\mathrm{a1} = 2.4, \mathrm{p}K_\mathrm{a2} = 9.7$) has a $\mathrm{pH}$ of $10.4$ when dissolved in
  $100\ \mathrm{mL}$ of $0.1~\mathrm{M}\ \ce{NaOH}$. Calculate the molecular mass of this amino acid.

As usual we can obtain the ratio of the conjugate base and acid around $\mathrm{p}K_\mathrm{a2}$ by using the Henderson-Hasselbalch equation. However, once we get the ratio, the solutions key says that we know $c(\text{Base}) = 0.1$ because all of the $\ce{OH-}$ reacted with our acid. While I totally agree that all the $\ce{OH-}$ reacted with the acid, since it is a strong base, how do we know there was no $c(\text{Base})$ before the addition of $\ce{NaOH}$? Thus making the concentration $c_0(\text{Base}) + 0.1$.
Is this because it came from a solid crystal where all of the amino acid was in the acid form?

There are a few observations to make.

The amino acid only has two $\mathrm{p}K_\mathrm{a}$ values. Thus, it cannot have a side chain that is in any way acidic or basic. We only have three different possibilities: $\ce{H2AA+, HAA, AA-}$.
It was crystalline when added. Therefore, it must have been in its neutral $\ce{HAA}$ state. Otherwise, it would have been given as ‘the $\ce{HCl}$ salt …’ or ‘the $\ce{NaOH}$ salt …’

If we had any excess base in there, it would have to have been given in the question as the sodium hydroxide crystal adduct or something.


What is the oxidation number of the oxygen in BaO2 According to the book for a binary compound, first we assign the element with greater electronegativity its oxidation number (oxygen always -2 except in peroxides). So, we have
-2 x2 =-2. So, each oxygen would have an oxidation number of -2. The book says the answer is -1.

The only thing I can think of, is since barium is an alkaline metal, it has an oxidation number of +2, so each oxygen would be -1. But, it contradicts itself because in the rules. It says for a binary compound, "the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element."

"oxygen always -2 except in peroxides"

This exception is in effect here, because $\ce{BaO2}$ is a peroxide consisting of $\ce{Ba^{2+}}$ and $\ce{O2^{2-}}$ ions. The oxidation number of Ba is +II, and the oxidation number of each of the oxygens in the peroxide anion is -I. This fits with the charge of the peroxide anion ($2 \times -1 = -2$), and as $\ce{BaO2}$ is a neutral compound, the sum of all oxidation numbers is 0. Moreover, there are even more exceptions to the rule of thumb cited above, for example, the superoxide radical anion $\ce{O2-}$ with a fractional oxidation numer of $-\frac{1}{2}$, or the dioxygenyl cation $\ce{O2+}$ with a formal oxidation number of $+\frac{1}{2}$ for each oxygen.

In contrast to peroxides, "simple ionic compounds" of oxygen are oxides, which contain the anion $\ce{O^{2-}}$. The charge of the oxide anion is $2-$, so the oxidation number of oxygen in those compounds is -II.


Is the Arrhenius Equation only valid for reactants in the gas state? I couldn't find a source that explicitly say this, but given the gas constant R is used, is the Arrhenius equation only valid when all reactants are gases? Do they have to be ideal gases?

If the above is true, is there a similar equation for reactions where at least one reactant is in liquid/solid form?

As Ben Norris explained, $R$ can be just more convenient so that one can deal with quantities per mole.

I.e. one could write either $e^{(-E_a/RT)}$ or $e^{(-E_a/K_BT)}$, where in the first instance one would use $E_a$ in $\ce{J~mol^{-1}}$, in the second instance just in $\ce{J}$. The term in the exponent should of course be dimensionless.

Since the numbers in $\ce{kJ~mol^{-1}}$ are much more familiar for chemists, one would often use the first one to avoid having to use very small numbers.

The crux is that the Boltzmann factor in the Arrhenius expression is related to the chance for particles having enough energy at a given temperature, $T$, to surmount the activation barrier.


Are all esterification reactions reversible? What is their general mechanism? For example the reaction of ethanoic acid with ethanol is reversible, however is this true for all esters? I'm not quite whether it is as easy for a reversible reaction to occur both ways. Is there a limit to the complexity an ester can be? What would be the general mechanism for these reactions? I assume there is no limit on the complexity (carbon chain length) of esters, other than the frequency of longer-chain carboxylic acids and alcohols needed to form them. Is this true?

I'll start this answer off with saying that both of these answers were the first result when searching for Google.

Yes, The esterification reaction is both slow and reversible

and here is the mechanism from an image search


Which has a higher lattice energy CaO or NaCl? I know that as charge increase the lattice energy increases as well. However when the size increases, the lattice energy decreases. Can you please explain this to me? I think it is CaO but i'm not sure. Thanks!

Lattice enthalpy of an ionic solid is the minimum thermal energy required to completely convert one mole of that ionic solid to isolated gaseous ions. Here "isolated ions" in the sense that the distance between the separated ions is sufficient to reduce the electrostatic interaction between them to zero practically.

It is also the energy released for the reverse process of  settling of isolated  ions into the solid lattice.

Greater the charge densities of ions higher the electrostatic interaction and so higher the lattice energy by magnitude. A smaller anion/cation with more charges will have higher charge density. Here calcium ion (114pm)is  not only smaller than sodium ion(116pm) but also having twice the charge as that of sodium ion.
Similarly oxide ion(126pm) is much smaller than chloride ion(167pm). Former is doubly charged where as latter has only single charge. For both these reasons CaO has a much high lattice enthalpy (3414kJ/mol) compared that of NaCl (787kJ/mol)


Is there an electrolyte that will halt the production of Hydrogen in water electrolysis? Care must be taken in choosing an electrolyte, since an anion from the electrolyte is in competition with the hydroxide ions to give up an electron. An electrolyte anion with less standard electrode potential than hydroxide will be oxidized instead of the hydroxide, and no oxygen gas will be produced. A cation with a greater standard electrode potential than a hydrogen ion will be reduced in its stead, and no hydrogen gas will be produced.

From what I understand, that bit states that certain electrolytes will stop the production of hydrogen gas and only oxygen gas and whatever gas from the electrolyte will be produced.

So I have two questions:

What are some electrolytes that will stop the production of hydrogen gas in water electrolysis?
What kind of water will be left behind if distilled water and an aforementioned electrolyte runs through electrolysis until all of the oxygen has been exhausted from the solution?

Please correct me if I'm misunderstanding anything.

The reduction of hydrogen is what the reduction potential of every other reduction potential is based off of, so:

$$\ce{2 H+_{(aq)} + 2e^{–} ->H2_{(g)}}~~~~~\Delta E^{\mathrm{o}}=0\,\mathrm{V}$$

So if a species has a has a reduction potential $>0\,\mathrm{V}$, then it will be be reduced before the naturally occurring $\ce{H+}$ ions are (these exist in solution due to the auto-protolysis of water, $\ce{H2O -> H+ + OH-}$, thereby not producing any $\ce{H2_{(g)}}$.

If your goal is to produce $\ce{O2_{(g)}}$ from the oxidation of $\ce{OH^{-}_{(aq)}}$, the reaction for this is:

$$\ce{4OH^{−}_{(aq)} -> O2_{(g)} + 2 H2O_{(l)} + 4 e^{−}}~~~~~\Delta E^{\mathrm{o}}=-0.401\,\mathrm{V}$$

Similarly, if a species has a has a oxidation potential $>-.401\,\mathrm{V}$, then it will be be oxidized before the  $\ce{OH-}$ ions, thereby not producing any $\ce{O2_{(g)}}$.

Wikipedia has a rather comprehensive standard reduction potential table, so use that to compare the reduction potential of ions. Choose your salts wisely!

To answer your other question, if this reaction continued indefinitely, a very acidic solution would result, as for every $\ce{OH-}$ that is oxidized, a $\ce{H+}$ is left in solution. Once all of the other cation from the salt is reduced, however, the $\ce{H+}$ would then start being reduced, producing $\ce{H2_{(g)}}$.


What exactly is "B-strain"? On factors that influence the stability of carbocations, I came across this peculiar one:

  Bulky groups attached to the positively charged $C$-atom (of the carbocation), stabilize it by B-strain, which prevents the C-atom for returning to the $sp^{3}$ state.

   Source: Pearson's Organic Chemistry (7e), Morrison and Boyd

The issue? I've never heard of "B-strain" before. I mean, it's possible that bulky groups stabilize the positive $C$-atom via inductive (and other) effects...but what's with "B-strain"?

I ran a cursory Google search, but I couldn't find any reliable Chemistry-literature that deals with this. In fact, it raised more questions than answers, most notably, the presence of another effect called the "F-strain".

So I must ask:

What is the "B-strain", and how does it contribute to stabilization of carbocations?

Additionally, is there such a thing as the "F-strain"? If so, what's it all about?

B-strain

Both B- and F-strains are the terms I found were originally used to describe steric hindrance in complexes [1, pp. 66-67]:

  The concept that steric hindrance in a monodentate ligand can affect the affinity of that ligand for a metal was first introduced by Brown and Bartholomay (1944) who suggested that co-ordination of a tertiary amine with bulky substituents would create strain within the amine. This strain, termed B-strain on account of it occurring behind the ligand atom with respect to the metal, would arise because although in the unco-ordinated amine any steric interaction between the bulky substituents could be relieved to some extent by an increase in the $\ce{C-N-C}$ angles.

Interestingly, further it is pointed out that at the same time there is no other solid evidences that B strain exists. Although modern organic chemistry textbooks are widely using this term for tertiary compounds.

From [2, p. 417]:

  Steric effects of another kind become important in highly branched substrates,
  and ionization can be facilitated by relief of steric crowding in going from the tetrahedral ground state to the TS for ionization. The relative hydrolysis rates in 80% aqueous acetone of $t$-butyl $p$-nitrobenzoate and 2,3,3-trimethyl-2-butyl $p$-nitrobenzoate are 1:4.4.

  This effect has been called B-strain (back strain), and in this example only a modest rate enhancement is observed. As the size of the groups is increased, the effect on rate becomes larger. When all three of the groups in the above example are t-butyl, the solvolysis occurs 13,500 times faster than in t-butyl p-nitrobenzoate.

According to [3, p. 350]:

  Not all steric effects decrease reaction rates. In the hydrolysis of $\ce{RCl}$ by an $\mathrm{S_N1}$ mechanism ..., the ﬁrst step, which is rate determining, involves ionization of the alkyl chloride to a carbocation:

  The central carbon in the alkyl chloride is $\mathrm{sp^3}$ hybridized, with angles of $\sim 109.5^\circ$, but when it is converted to the carbocation, the hybridization becomes $\mathrm{sp^2}$ and the preferred angle is $120^\circ$. If the halide is tertiary and the three alkyl groups are large enough, they will
  be pushed together by the enforced tetrahedral angle, resulting in strain. This type of strain is called B strain (for back strain), and can be relieved by ionization to the carbocation.

  [...]

  Substitution of ethyl groups for the methyl groups of tert-butyl chloride
  does not cause B strain; the increase in rate is relatively small, and the rate smoothly rises with the increasing number of ethyl groups. The increase is caused by normal ﬁeld and resonance (hyperconjugation) effects. Substitution by one isopropyl group is not greatly different. But with the second isopropyl group the crowding is now great enough to cause B strain, and the rate is increased 10-fold. Substitution of a third isopropyl group increases the rate still more.

Further in [3, p. 419]:

  For some tertiary substrates, the rate of $S_N1$ reactions is greatly increased by the relief of B strain in the formation of the carbocation. Except where B strain is involved, $\beta$ branching has little effect on the $S_N1$ mechanism, except that carbocations with $\beta$ branching undergo rearrangements readily.

There is also a Baeyer strain aka angle strain most noticeable for three-membered rings ($60^\circ$ is far from expected tetrahedral angles), which can also probably be abbreviated alike. [3, p. 193]

It also seems like earlier sources used B-strain as a reference to H. C. Brown [4, p. 291]:

  Hammond has proposed a steric "B-strain" (see H. C. Brown, J. Chem. Soc. (London) p. 1248 (1956)) argument to explain the greater stability of primary over secondary or tertiary carbanions. From the viewpoint of hybridization, structure A with $90^\circ$ geometry is preferred in carbanions. But this conformation brings the substituents too close together, and a compromise structure with larger bond angles results. Methide ion, $\ce{^-CH3}$, with the smallest substiuent groups, should be the most stable simple carbanion, since it can more nearly adopt the small angled conformation A preferred on electronic grounds. The more alkyl groups are added, the less able sterically are
  the anions to adopt the favorable conformations with smaller angles, and
  the less stable and more reactive the intermediates become.

F-strain

[1, p. 68] illustrates the origins of the term F-strain:

  The boron trifluoride and boron trichloride complexes of diethyl ether are less stable than those of dimethyl ether, and the same order of stability is observed for the complexes of diethyl and dimethyl sulphides. As expected, the steric interaction is decreased as the distance between the metal and ligand atom is increased. Thus it is decreased when the metal atom is changed from boron to aluminium, and when the ligand atom is changed from oxygen to sulphur. It is largely eliminated, of course, if the substituents on the ligand atom are held back away from the metal, and thus very stable complexes are formed by the tertiary amine quinuclidine, while complexes of tetrahydrofuran and tetrahydropyran (and also the analogous sulphur heterocycles) are far more stable than those of their acyclic analogues.

  Strain which arises as the result of this type of steric interaction has been termed F-strain by H. C. Brown -- who has carried out extensive investigations in this area -- on the grounds that the interactions responsible for the strain occurs mainly in front of the ligand atom with respect to the metal. For a critical but now rather dated review of H. C. Brown’s work, see Hammond (1956).

References

Houghton, R. P. Metal Complexes-Organic Chemistry, 1st ed.; Cambridge University Press: Cambridge; New York, 1979.
Carey, F. A.; Sundberg, R. J. Advanced organic chemistry Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013.
Hartley, H.; Karabatsos, G. J. Advances in alicyclic chemistry; 1966; Vol. 1.


Sensitive test for lead in water? Some of my friends are in Flint and I'm horrified by their situation. I would like to learn how to test for lead independently.

After all if you can't trust the state sending them water to test is not sufficient. The testing procedures I have read about give too many type II errors. A procedure with a bias toward type I errors would be better since then I could pay a professional lab to see if it was real.

For tests that only detect high levels of lead, could they be made more sensitive by evaporating the water then testing a larger sample?

I used to work in an environmental water testing laboratory.  The regulatory standard method for testing for lead in water uses an Atomic Absorption (AA) spectrophotometer.

It seems that the issue with Flint's water appears to be that the new source of water is corrosive, the water has not been treated for corrosion, and pipes in Flint are commonly made out of lead.  So if your pipes at home are made of lead and are corroding, you may have a problem that the lab at the water treatment plant was not able to test for.

If you are curious about checking your water at home, you can obtain a water testing kit from a local hardware store.  These kits come with test strips that allow you to check your water for things like nitrates, pH, alkalinity, hardness etc.  Make sure your kit allows you to test for lead.


Are all strong acids the same strength? Since the six strong acids dissociate completely into their ions, and the reaction goes to completion, does this mean that acids such as $\ce{HI}$ and $\ce{HCl}$ have the same acidic strength?

I know that 0.10 M $\ce{HI_{(aq)}}$ and 0.10 M $\ce{HCl_{(aq)}}$ both contain 0.10 M $\ce{H3O+_{(aq)}}$, which leads me to believe they are equally strong acids. Is this correct?

There's something called the "solvent leveling effect". In short, there's a lowest pKa in a particular solvent, based on the basicity of the conjugate base.

In water, you're limited by $\ce{OH-}$. So given the same concentration of $\ce{HI}$ and $\ce{HCl}$ in water, they will indeed have the same pH.

In other solvents, you may have a difference between acids, depending on the pKa.


Number of resonating structures (hypothetical) To my knowledge, resonance happens where more than one lewis stucture can represent the compound with no shift in sigma bonds

so, suppose in a simple conjugated system even like 1-3 pentadiene , two forms are well known, i want to know if a third form where each atom has completely filled or completely empty p orbitals should also be taken into consideration, as technically, there is nothing wrong with it, I accept that it's "contribution" will be negligible, but it should be counted in all POSSIBLE resonating structures

https://commons.wikimedia.org/wiki/File:Naphthalene_resonance_structure.svg

similar argument here, a lot more can be created if we allow charge separation of pi bond

EDIT: The question is whether there is a convention of some sort which tells us which ones to count which ones not (if we cant tell the magnitude of difference in stability of the structure)

Resonances structures are a model of the quantum mechanical orbital mixing.  To my mind they are a complete farce.  They are also quite useful

A model is a simplification of a problem, usually based on ignoring negligible factors.  So this question really distills down to what is negligible and what isn't?

That depends on roughly two factors:

The conditions of reaction.

Are the conditions strongly ionic or acidic?  Then charge separation may become relevant.

The rigor of the problem.

Is this for a high school course? A Ph.D. thesis? A bar-room chat with your mates?

Unfortunately, like most of chemistry, there is not an absolute answer here.


What is the temperature of evaporating water (not boiling)? The vapor pressure of water at $\pu{70 ^\circ C}$ is $\pu{233.7torr}$.
If I were to put a temperature probe directly above the liquid water, what would it read?

In other words, does a water molecule need to have $\frac32kT$ (where $T=\pu{373 K}=\pu{100 ^\circ C}$) to escape the liquid?

No, the average thermal energy doesn't need to be the one corresponding to 100C.
Solo water molecules can exist in vacuum or in a gas phase with whatever kinetic energy. Liquid water condensation starts when the concentration of the water molecules is enough high (there is a tipping point), i.e. when the partial pressure of water reaches the vapor pressure for the given temperature.

From the classical thermodynamic point of view (similar argument to Hexacoordinate-C's comment), there is a chemical potential of water in the gas phase which is dependent on the partial pressure of water, and there is a chemical potential of condensed water for a given temperature. When the partial pressure is low, the gas phase chemical potential is lower than it would in the condensed phase -> it is stable in gas phase. When the partial pressure reaches the vapor pressure, the chemical potential is equal to the gas phase and condensed phase.


Why must there be degeneracy in the orbitals to have a Jahn-Teller effect? Why is it the case that for Jahn-Teller effects to occur in transition metals there must be degeneracy in either the $\mathrm{t_{2g}}$ or $\mathrm{e_g}$ orbitals?

In the abstract to the original Jahn-Teller paper it is stated:[1]

  We shall show that stability and degeneracy are not possible simultaneously unless the molecule is a linear one.

Thus the Jahn-Teller effect is specifically about the removal of a degeneracy, and that that will always lower the total energy of the system. Thus you can not have a Jahn-Teller effect without a degeneracy in the reference, fully symmetric state.

Reference:
[1]: H. A. Jahn, E. Teller, Proc. Royal Soc. A 1937, 161, 220–235. DOI: 10.1098/rspa.1937.0142.


Why does lead nitrate powder on heating starts turning yellow from the top rather than from the bottom? Today while in school, I observed the thermal decomposition of lead nitrate. The peculiar thing I noticed was that the lead nitrate powder in the test tube starts turning yellow (due to the formation of lead oxide) from the top and then towards the bottom. This indicates that the certain decomposition temperature is first reached at the top and later at the bottom.

I tried to apply a logic that as soon as some heat was provided to the bottom of the powder in the test tube, it was quicky transferred to the powder present above it and so on till the top. At this point, since the heat had nowhere to go (since air is a bad conductor is heat) it got accumulated at the top and hence it reached a higher temperature.

However I'm unsure of my logic because for the top powder to reach the specified temperature, the lower powder must first reach either the same or a higher temperature (otherwise the powder present above it would not get heated).

Lead oxide is a peculiar type of salt. It's brown when it's hot and turns yellow when it's cool. The lower most part remaining white could be due to two possible reasons:

The top-powder becoming yellow indicates the region is more cooler than the salt with is present at the lower part of the test tube. The lower most part is in direct contact with the flame and the  upper part is receiving heat due to conduction and losing heat to the atmosphere; many factors could be at play here.
Another possible deduction could be the paucity of atmospheric oxygen at the bottom of the test tube than at the top and hence the rate of conversion to lead oxide to the top would be faster than at the bottom.

In case you haven't done the experiment and want to see what the OP is trying to convey, here's a youtube video that demonstrates the whole lab-based thermal decomposition of lead nitrate (it starts at 1.32  incase the timed setting of the link fails):
http://www.youtube.com/watch?v=csEnz9ZsAFM&t=1m32s


Trying to understand why Sn2 is not possible in this reaction I am trying to understand why the red Br doesn't react? Is it because there can't be a flip of groups? Or steric hindrance?

Thank you!

Backside attack on the red bromide is not possible. The rest of the norbornyl skeleton completely blocks the trajectory required for $\mathrm{S}_{N}2$ reaction.


In what approximations, if any, can we assume a constant temperature in a Bomb calorimeter? In most, if not all, general chemistry books, you will find that in from constant volume calorimeters, $\Delta H = \Delta U + RT \Delta n_\mathrm{g}$, which is, of course derived from $H = U + PV$.

My question is, given that the measured quantity is the change of temperature of the calorimeter, how can we assume that $T$ is constant to get to $\Delta H = \Delta U + RT\Delta n_\mathrm{g}$?

Edit:

As said in the comments, I found this link where it says that this approximation is valid for a small $\Delta T$ value. But if we have the values, why not just use them to get a more accurate number? Furthermore, for example, let's assume an experiment we measure a $\pu{10 kJ mol-1}$ heat from a $T_\mathrm{i} = \pu{298 K}$ to $T_\mathrm{f} = \pu{303 K}$, and a $n_\mathrm{i} = 0$, $n_\mathrm{f} = 2$. Then the approximation gives:

$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$

$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(2)(\pu{303 K}) - (0)(\pu 298 K)\right] = \pu{-4.96 kJ mol-1}$$

And the error is, indeed, small ($\pu{80 J}$) (side question, isn't $\Delta n$ supposed to be in moles? my dimensional analysis doesn't match if it does). However, if the reaction was from $n_\mathrm{i} = 2$ to $n_\mathrm{f} = 4$, and we keep the other values, we get:

$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$

$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(4)(\pu{303 K}) - (2)(\pu 298 K)\right] = \pu{-4.88 kJ mol-1}$$

And the error is, yes, still small but it doubled (of course). So I'm guessing the $\Delta T \approx 0$ to be valid only for small values of $\Delta T$, but also dependent on the actual values for $n_\mathrm{i}$ and $n_\mathrm{f}$. Is there a rule I'm missing?

You have more or less answered your own question. Enthalpy is a function of state, so you can write an expression of the form $\Delta {H} = {H_f}-{H_i}$, and we know that this depends only on the initial and final states, and not the path taken between them.

You also have the relation, $H = U + PV$, using this and the previous equation, and the fact that $U$ is also a state function (so we can write $\Delta {U} = {U_f}-{U_i}$)

$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i \tag{1}$

One comment, in the link that you posted equation (6) namely,

$\Delta {H} = \Delta {U} + \Delta(PV) = \Delta {U} + P\Delta V + V\Delta P $ seems absurd, for finite, non-infinitesimal differences.

For infinitesimal changes, i.e if we write the differential

$\mathrm{d}H = \mathrm{d}U + \mathrm{d}(PV) = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P $ this holds true; (chain rule from mathematics)

Anyway, for solids in liquids, the $PV$ term can be neglected on account of their small molar volume. I guess this would be one approximation/assumption you are interested in.

Second, assumption/approximation would be that a gas behaves ideally, and thus follows the equation of state: $PV = nRT$

Plugging it into (1);

$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i = \Delta {U} + R(n_fT_f-n_iT_i) $

This expression is true for finite differences. Now, the finally approximation is indeed $T_f \approx T_i \equiv T$, i.e $\delta T$ is "sufficiently" small, then indeed we can write

$\Delta {H} = \Delta {U}+ RT\Delta n $

Note however, that we don't necessarily require $\Delta T \to 0$ in a strict, mathematical sense. What can be considered "sufficiently" small depends on your system, your problem, and the kind of solution you need.
What I mean to say is in some cases/applications $-4.96\ \mathrm{kJ}$ is just as good as $-4.88\ \mathrm{kJ}$

There are many other simplifying assumptions in calorimetry when it is taught such as heat-capacities are independent of temperature for example; this only, really true over narrow ranges of temperature and in general, heat capacity is in fact a function of temperature.


Why is the carboxyl proton in salicylic acid more acidic than the phenol proton? I know I'm just wrong about this but I'd appreciate some help seeing why.  It seems to me like the phenol proton in salicylic acid should be more acidic than the carboxyl proton.  When I try to draw resonance structures for the conjugate base, it looks to me like the charge gets distributed over two oxygen atoms and the whole ring (if anyone can tell me what software makes the nice pictures in some answers here I'd take it):

Then when I try to draw the conjugate base when the carboxyl group is deprotonated, as far as I can tell only the oxygens take the charge, and trying to distribute over the ring results in invalid resonance structures because you aren't supposed to add formal charges to already charged species.

What am I doing wrong?

More resonance structures does not necessarily mean more stabilization.

Phenolate has a negative charge at an oxygen atom which cannot be shared without "disturbing" the aromaticity of the ring. This is in part because in the resonance structures drawn, there are negative charged carbons within the aromatic ring, which either implies that:

The ring is still aromatic, so there must be a free electron pair in a p orbital (very high in energy).

Or:

The free electron pair is in an sp³ orbital, which means the ring is no longer aromatic.

In the case of the carboxylic acid, the charge is distributed between two atoms more electronegative than carbon and the ring retains its aromaticity.


Finding percentage of the gas in a binary gaseous mixture of the given density At STP, the density of a gas in a vessel is $0.9002$. If the gas is a mixture of argon and helium, what percentage of the gas is argon?

I am stuck on this. From what I can gather, the only influencing characters would be moles and grams. This idea is based of off of STP and given constants.

I have a couple equations written down but I can’t seem to the flow going.

R, T, P are all known along with acccording densities. I also want to say $22.4$ is also a known at $V_\mathrm{tot}$ and therefore $n_\mathrm{tot} = 1$.

$$n(\ce{Ar}) + n(\ce{He}) = 1$$

$$\frac{n(\ce{Ar})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{Ar}) + \frac{n(\ce{He})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{He}) = d_\mathrm{tot}$$

$$d = PM/RT$$

Well, I can see all the relations you require knowledge of in the question itself!

Calculate the effective molecular mass from

$$M = \frac{dRT}{P}$$

$R$ is known, $d$ given. $P$ and $T$ are available from the fact that it is at STP.

$M$ comes out to be ${20.176 u}$.

Effective molar mass is easily calculated below:

$$M_\mathrm{eq} = M_1x_1 + M_2x_2$$

where $x_i$ is the mole fraction of each gas. You can replace $x_2$ by $1-x_1$.

Plugging in the values and solving for $x_1$, argon mole fraction comes out to be $0.4493$.

I believe you can now calculate the demanded percentage, be it by mass or by moles.


Reducible representation of planar molecule N2H2 with bond lengths as a basis I'm studying about molecular symmetry and its representations. Today I got a little confused about the planar $\ce{N2H2}$ molecule.

It looks like this

The basis in the example was chosen as $\Delta r_1, \Delta r_2$, i.e. lengths of N-H bonds. Then, it was stated, that the molecule belongs to $\mathrm C_{2h}$ point group, which I agree with.

The problem is, the example stated, that the reducible representation is

\begin{array}{|l|l|l|l|}
\hline
{} & \mathrm E & \mathrm C_2 & i & \sigma_h\\ \hline
\Gamma & 1+1=2 & 0+0=0 & 0+0=0 & 1+1=2\\
\hline
\end{array}

And here I don't understand, what's going on. I understand, that $\mathrm E$ and $\sigma_h$ are 2, as both $\Delta r_1$ and $\Delta r_2$ stay the same after those operations, i.e. bond length doesn't change.

But, bond lengths don't change after $\mathrm C_2$ and $i$ either, so why are there zeros in the table?

I'd understand it if we used vectors of Cartesian coordinates on every atom as a basis, but in this case, I'm quite lost.

Could you, please, explain it to me?

The example is taken from this presentation (slide 18).

Instead of using the $\Delta r$ as in your figure redraw it with a vector  shown as an arrow parallel to each NH bond and label them 1 and 2. You know from the point group that the NH bonds are the same length and at the same angle so these vectors are similar but point in different directions.

Operate on the whole diagram including arrows (with labels) according to the operations in the point group. If, after an operation the figure is indistinguishable from your starting point, then count $1$ for each vector otherwise count $0$.

In your figure the point in the point group is at the centre of the NN bond. The $C_2$ axis points out of the page centred at this point. The $\sigma_h$ plane is the plane of the figure.

The  identity, which is an 'I exist' or 'do nothing' operation count 2. The $C_2$ operation is rotation by 180 degrees and so interchanges the vectors 1 and 2 so is not indistinguishable  and so each vector counts 0. Similarly for the inversion $i$ through the centre. Reflection in the mirror plane (the plane of the image) leaves the molecule indistinguishable so counts 2. Thus you have the reducible representation you give in your question.


Irreducible representations and system states connection Let's say I have the molecule $\ce{N2+}$. Its symmetry point group is clearly $D_{\infty\mathrm h}$. But I'm confused by its irreducible representations. I know there are 8 of them, but as I understand it, they're all representations of the same group, just with different bases.

But today I found out, that they're not completely "equal", i.e. that they mark different states of the system ($\mathrm{A_g}$ is the ground state, $\mathrm{B_{1u}}$ is one of the excited states etc.).

What I'm completely missing is - how are irreducible representations linked with the system states?

Please, try some "low-level" explanation, I'm a beginner both in quantum chemistry and group theory.

The symmetry of a state is the MO (molecular orbital) product of its occupied molecular orbitals. Let's take the example of water molecule. We can see the symmetry label of each of the MO in the diagram. Each MO is assigned an irreducible symmetry label (associated with the point group of molecule) which can be deduced from the symmetry of MOs in space (or from the LCAOs). The wavefunction associated with the MOs also have the same symmetry species. Now, this arrangement of electrons represents the ground electronic state of water, so the symmetry label of the ground state of water can be obtained as

$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2=(A_1)^2\bigotimes (B_2)^2\bigotimes (A_1)^2\bigotimes (B_1)^2 =A_1$

Follow the same procedure to obtain ground state of $N_2^+$. Let me know if you have any questions.


Need some help proving the following statements in Thermodyamics? So in my second year thermodynamics class we are currently talking about entropy and deriving its definition from the 1st law.

$$\mathrm{d}q_{rev} = dU - dw_{rev}$$
$$= C_{V}(T)dT + PdV$$
$$= C_{V}(T)dT + \frac{nRT}{V}dV$$

And then it just says since the mixed second partial derivatives aren't equal, i.e.

$$(\frac{\delta Cv(T)}{\delta V})_T \neq (\frac{\delta(\frac{nRT}{V})}{\delta T})_V$$ dq is not an exact differential (no proof of this is given). It then stipulates by dividing both sides of the equation by T

$$\frac{\mathrm{d}q_{rev}}{T} = \frac{C_{V}(T)}{T}dT + \frac{nR}{V}dV$$
suddenly it becomes an exact differential because

$$(\frac{\delta Cv(T)/T}{\delta V})_T = (\frac{\delta(\frac{nR}{V})}{\delta T})_V$$
Can anybody give me a proof of this with calculus because my textbook gives 0 proof whatsoever and it irks me because I want to understand the theory underlying this. Thanks!

These statements can be proved in general i.e. irrespective of whether you take your system as ideal gas or not. As you have used in the question $PV = nRT$, you inherently assuming that your system of thermodynamic interest is an ideal gas. So, let's first prove this for an ideal gas and then do the general proof.

$\textbf{Case I : Ideal Gas}$

For an ideal gas, $C_V(T)$ is independent of volume $V$, which means that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. (We will prove the general relation in the next section, and this statement will be obvious from that proof. ), But on the other hand $\big(\frac{\partial (nR T/V)}{\partial T}\big)_V = \frac{nR}{V}$  by simple partial derivative. So, thus you get the proof that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T \neq  \big(\frac{\partial (nR T/V)}{\partial T}\big)_V$. Thus $\delta q_{rev}$ is not an exact differential.

On the other hand if you divide both sides by $T$, you still have, $\big(\frac{\partial (C_V(T)/T)}{\partial V}\big)_T = \frac{1}{T}\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. And also on the other hand you have, $\big(\frac{\partial (nR /V)}{\partial T}\big)_V = -\frac{nR}{V^2}\big(\frac{\partial V}{\partial T}\big)_V = 0$ (obviously, as we are taking partial derivative of a constant term. Thus, by the equality of mixed partial derivatives of $\frac{\delta q_{rev}}{T}$ it becomes exact differential.

$\textbf{Case II : General System}$

For general system we have, $\delta q_{rev} = dU +PdV$. But remember we can't write, $dU = C_V(T) dT $ and $PV =nRT$. We have to write $dU = C_V(T)dT + \big(\frac{\partial U}{\partial V}\big)_T \  dV$ .

So, now we have $\delta q_{rev} = C_V(T)dT + \Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big)dV $.

Also, from the definition of Helmholtz free energy ($A = U - TS $ and $dA = -SdT -PdV$) and by using Maxwell's relation it can be shown that $\Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big) = T \big(\frac{\partial P}{\partial T}\big)_V$

Now, it can be shown properly that
$$\left(\frac{\partial C_V(T)}{\partial V}\right)_T = \frac{\partial^2 U}{\partial T \partial V} = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \qquad(= 0\text{ if } PV=nRT)\tag{i} $$
$$\left(\frac{\partial (\frac{C_V(T)}{T})}{\partial V}\right)_T = \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V} =  \left(\frac{\partial^2 P}{\partial T^2}\right)_V \tag{ii}$$
and for the right hand side
$$\frac{\partial}{\partial T}\Bigg(T\Big(\frac{\partial P}{\partial T}\Big)_V\Bigg)_V = T\Big(\frac{\partial^2 P}{\partial T^2}\Big)_V + \left(\frac{\partial P}{\partial T}\right)_V\tag{iii}$$
and if you divide the right side by $T$ and then check the partial derivative criterion, you will ultimately get $\Big(\frac{\partial^2 P}{\partial
 T^2}\Big)_V$ on the r.h.s also (exactly same as $(ii)$ ) , But note that, $(i) \neq (iii) $. Thus $\frac{\delta q_{rev}}{T}$ will become exact differntial but $\delta q_{rev}$  will remain inexact.


Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine Unsymmetrical dimethylhydrazine ($\ce{H2NN(CH3)2}$) has a melting point of −57 °C, but its isomer, 1,2-dimethylhydrazine ($\ce{(CH3NH)2}$) has a melting point of −9 °C. That looks to me like a significant difference, so there must be some mechanism which makes these two compounds extremely different in this respect. However, I'm unable to figure out why that difference exists; could you please explain that?

These are the structures of the compounds (from wikimedia commons):

Unsymmetrical dimethylhydrazine:

1,2-dimethylhydrazine:

Here're some of the arguments which I considered:

The packing in crystal structures could be 'different' in a significant way. This could potentially be supported by the fact that the difference between their boiling points is much lower: unsymmetrical dimethylhydrazine has a boiling point of 64 °C while 1,2-dimethylhydrazine's is 87 °C. However, this is an extremely vague argument; it could be strengthened by some details regarding how the different packings contribute to the melting points.
Hydrogen bonding should be a pretty significant factor to consider while analyzing these compounds. I suspected that steric hindrance may allow only one $\ce{N}$ atom for hydrogen bonding in unsymmetrical dimethylhydrazine. However, this would suggest that 1,2-dimethylhydrazine should have a lower melting point, which is clearly not the case.

We had a discussion about these two compounds' melting points in the Physics SE chat a while back, so I thought I'd pop over here and ask.

According to the Carnelley’s Rule:

  That of two or more isomeric compounds, those whose
  atoms are the more symmetrically and the more compactly
  arranged melt higher than those in which the atomic
  arrangement is asymmetrical or in the form of long chains

If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.

For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:

$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$

where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.

References and further reading:

Effect of molecular symmetry on melting temperature and solubility

Melting Point and Molecular Symmetry

Carnelley’s Rule and the Prediction of Melting Point


Why do we sometimes write reactants above the reaction arrow? We write:

\[\ce{Alcohol ->[\ce{KMnO4}] Acid}\]

The alcohol forms a carboxylic acid, i.e. gains an $\ce{O}$ from $\ce{KMnO4}$. This means $\ce{KMnO4}$ is a reactant! Then why is it written above the arrow?

What else do we write above the arrow?

Things written on arrows are chemicals that are part of the reaction in some way, but not one of the reactants or products. In this case, the $\ce{KMnO4}$ is your oxidizing agent, which can be swapped out for any other strong oxidizer, and is not an inherent part of the reaction.

Other things you could see on the arrow would be $\ce{H2O}$ or Diethyl Ether, both of which are solvents. Other things include $h\nu$ and $\Delta T$, both of which are conditions under which the reaction takes place. Lastly the most common thing you'll see are catalysts such as $\ce{Pt}$. As you  can see, all of these things are necessary to the reaction, but aren't really what we care about in the reaction/can be substituted for other, similar chemicals.


Using the mass of calcium carbonate, what volume of carbon dioxide could theoretically be obtained if the reaction progressed to completion? Using the mass of calcium carbonate $m(\ce{CaCO3}) = \pu{0.500 g}$, what volume of carbon dioxide $V(\ce{CO2})$ (in $\pu{mL})$ could theoretically be obtained if the reaction progressed to completion? Assume that the pressure was $\pu{101.3 kPa}$ and the temperature was $\pu{22.5 ^\circ C}$ and that the gas was behaving ideally $(R = \pu{8.314 kPa L mol-1 K-1})$.

I know that I will work with
$$pV = nRT$$
and to bring the $n$ (amount of substance), I have to divide the mass by the molar mass and I assumed that $\ce{CO2}$ mass will be $\pu{0.5 g}$ as well because $\ce{CaCO3}$ will give $\ce{CaO + CO_2}$ so the ratio is $1:1$ so $\ce{CO2}$ mass is $\pu{0.5 g}$, but my problem is that when I bring the amount of substance it should be for $\ce{CO_2}$ or $\ce{CaCO3}$? Because it will differ, $\frac{0.5}{44.01}$ or $\frac{0.5}{100.09}$, that's what I am stuck on.

You are confused about chemical equations. The stoichiometric coefficient (the numbers before a compound) do not indicate the mass of the compound, they indicate the amount of substance. For example
$$\ce{CaCO3 -> CaO + CO2}.$$
This equation means that $\pu{1 mol}$ of calcium carbonate will decompose to give $\pu{1 mol}$ of $\ce{CaO}$ and $\pu{1 mol}$ of $\ce{CO2}$.
It does not mean that if 1 gram calcium carbonate were to decompose, 1 gram of calcium oxide and 1 gram of carbon dioxide will be obtained.
In your question the mass of $\ce{CaCO3}$ is given. From it you can calculate the moles of $\ce{CaCO3}$ involved. Same number of moles of $\ce{CO2}$ will be obtained and you can put that in the ideal gas equation.
Just so you know,
$$n_{\ce{CO_2}} = n_{\ce{CaCO_3}} = \frac{\pu{0.5 g}}{\pu{100.09 g/mol}}.$$
I recommend you study the chapter about the mole concept again.
As a side note: You mentioned in your question that the ratio of products is $1 : 1$. That's actually the ratio of the amounts of substances, not the mass ratio.


Calculate concentration gradients in FPLC Our FPLC works with two solutions:

A: 20 mM Tris.HCl, 0.2 M NaCl, pH 7.8
B: 20 mM Tris.HCl, 1.6 M NaCl, pH 7.8

I'm interested in sample volume from 102 mL to 117 mL. The FPLC will have a linear concentration gradient: At 24 mL it will be 75% A and at 144 mL it will be 5% A (the rest is filled up with B).
For visualization, I made a graph (y = -7/1200*x + 0.89):

Now, I have a sample with the whole output between the two indicated points (102 mL and 117 mL, so I have 15 mL in my tube) and would like to calculate the NaCl concentration.

Percentage of both A and B would change 70% from 24 to 144mL, with A from 75% to 5% and B from 25% up to 95%. The change rate would be 70% during the 120mL period or $0.70/120$ per mL. At any given point $x$ mL, between 24 and 144mL, concentration of $\ce{NaCl}$ in $\ce{mol/L}$ is:  $$\ce{[NaCl]} = (0.75 - 0.7\times (x-24)/120)\times0.2 + (0.25 + 0.7\times (x-24)/120)\times1.6$$

For a 15mL solution between 102 and 117mL, the concentration should be the average of the two points when x = 102, and 117mL.


What does x mean in (1-x)A – xB composite materials, molar ratio, weight ratio or volume fraction? I've been collecting data about ceramic materials and often I get to see composite materials are described as $(1-x)\ce{A} - x\ce{B}$. I am confused what this $x$ means. And I haven't seen any of the papers mention what $x$ is unless %wt is given specifically.

For an example in $(1-x)\ce{MgO} - x\ce{LiF}$ ceramic material, does $x$ mean molar ratio or weight ratio? For me it doesn't make sense to think of $x$ as a volume fraction as most of the times powders of these materials are taken for sintering.

If nothing is mentioned, is it all right to assume $x$ as the molar ratio?
Can we assume volume fraction is equal to molar ratio even in solids or liquids?

Here, $x$ is a variable $(x\in[0;1])$ denoting mole fraction of a given constituent in a phase of variable composition (non-stoichiometric compound).
This is an algebraic way of denoting a given compound from the phase diagram for a solid solution: $x$ corresponds to its abscissa axis.

And no, volume fraction is not equal to mole fraction and is overall a very inconvenient variable to use, so it's usually being avoided.
For example, volume fraction would deviate for the different morphologies and allotropes of the same solid, which is not useful for the description of chemical composition, not to mention it's also temperature-dependent.


Degredation of Anusol and Sunscreen Hopefully this is not TMI. I mistakenly used some expired Anusol recently, and noticed that the product is 12 years expired. One of the listed ingredients is Zinc Oxide, which is also used in sunscreen. This led me to ask myself "why do consumer drug products expire?", which led to the question, "How does Zinc Oxide, one of the active ingredients of sunscreen and Anusol, degrade over time?"

Existence in an oxidative atmosphere means many commercial products may be, and often are, exposed to oxygen.  Many things, and almost all drugs, react with oxygen.  The products of these reactions (oxidations, normally) are different molecules that may be less effective than the original drug.  As such, the efficacy of a drug that is beyond its expiration date is presumably lower, meaning dosing instructions become invalid.  Similarly, water content in the atmosphere may alter reactivity of drugs.

In general, studies to determine expiration dates take large amounts of time, and are hence expensive.  The standard for such an expiration would normally be a remaining purity level of the drug in question.  For example a company might study their drug for a year and note it is still 99% unchanged.  This means it meets a pretty good standard after one year.  From a marketing perspective they just label the drug as expiring after one year.  This allows them to be out of the legal liability because they are disclosing actual data indicating the efficacy of the drug over a period of time, and if a consumer takes the drug outside the expiration date, they are using the drug counter to instructions.  It also tends to generate extra sales for people who keep old drug stocks around, and must buy new ones to guarantee efficacy.  The actual drug may be still therapeutically useful even if "expired".

Specifically related to zinc oxide, I see no pathway for further degradation.  Wikipedia reports Anusol actually contains zinc sulphate (monohydrate), which I expect to be similarly robust.  However, the other but the qualities of the product ( presumably an ointment or lotion ) may have degraded such that its application may be less effective.  Additionally the Anusol plus contains a drug called Pramocaine Hydrochloride.  This compound is likely to oxidize overtime.


Drug design methods - docking software, binding affinity Lately I've seen videos and some articles explaining drug design. I would also like to specifically cite this guy, who isn't the most credible guy in the planet, but some of the stuff in the video makes sense to some degree, and he just confused me further.

Now I work on a completely different field on my day job, but I would like to ask a few questions, out of curiosity, because the techniques described in the video really piqued my interest.

Is it the case that when designing drugs (i.e. these proteins) we attempt to optimise binding affinity (maximise?)
How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute docking score? (can you recommend further reading for me in this regard)
Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?
Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?
How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?

To clarify, I'm not looking for detailed answers, since I am completely off the domain, I just want to have a basic idea for these questions, and perhaps recommendations on further reading from you guys.

Thank you, and stay strong!

I mainly work with simulation (MD), so take these answers with a grain of salt.

  Is it the case that when designing drugs (i.e. these proteins) we attempt to optimize binding affinity (maximize?)

Computational docking usually employes two methods: A force field based estimation of the free energy of binding, and exploration of the conformational space for the ligand.

After a docking run you will end up with set of different ligands, a set of conformations (or poses) for each ligand, and a corresponding score value for each conformation. The score is usually based on the binding constant and Gibbs free energy (which is based on various numbers of physicochemical parameters, and may or may not include parameters like desolvation and entropic effects). Different software use different scoring functions and methods to estimate the free energy.

  How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute docking score? (can you recommend further reading for me in this regard).

As far as I can tell, it is hard or impossible to make general statements about which piece of software is the best, since it depends a lot on what you are trying to use it for. Some of the limitations of the AutoDock suite (implicit hydrogens, spherically symmetric H-bond potentials etc.) are given in this publication.

  Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?

Not necessarily. A docking run gives you an (rough) estimation of the ligand affinity for the receptor you chose. It might be that the ligand also binds strongly to a lot of other proteins, making it less effective in an in vivo setting.

  Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?

You can, but not necessarily with the AutoDock suite. Large compounds have too many degrees of freedom, and the software is not capable of properly exploring the conformational space. Usually these large compounds are split in to smaller fragments which are then docked individually.

  How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?

Again, depends on the software. You listed some of the features of AutoDock, and you can find the details of the receptor preparation scheme from their website. It is definitely not accurate in every case. As far as I know, most docking runs are performed on protein structures obtained with X-ray methods, and there is no guarantee that the conformation of the protein in crystal is a good representation of the conformation in the solvent phase.


What is this wave function in Chemical bonding? Was studying chemical bonding(Valence bond theory) but there came this topic called wave equation but  have not studied quantum mechanics,so finding hard to understand these terms.Please help!!!

$\Psi$ is the wavefunction describing an electron's motion.  $\Psi^2$ is the probability density function for the electron. $\Psi^2$ is usually interpreted as the probability of finding the electron at a given set of coordinates.

In the case that $\Psi$ is a complex function (in that it contains imaginary numbers and not that it is complicated, which it usually is), $\Psi^2$ is defined as the product of $\Psi$ with its complex conjugate: $\Psi \Psi^*=|\Psi|^2$ so that the probability density function contains only real numbers.

Consider a simple case of $\Psi=a+bi$. The complex conjugate would be $\Psi^*=a-bi$.

$$\Psi \Psi^* = (a+bi)(a-bi)=a^2+b^2$$

Often the probability density function is normalized so that its integral over all space is 1 (i.e. there is 100% chance of finding the electron in all of space). In the equation below, $A$ is the normalization constant.

$$1=A \int_o^{2\pi} \int_0^{2 \pi} \int_o^\infty \Psi \Psi^* dr d\theta \phi$$


Electrochemical potentials I was carrying out an experiment into the effect of temperature changes on the standard potential of a cell.

$$\ce{Zn(s) +2Fe^{3+}->Zn^{2+} +2Fe^{2+}}$$
I used equi-molar concentrations of $\ce{Fe(CN)_6^{4-}}$and $\ce{Fe(CN)_6^{3-}}$. Thus the Nernst equation

$$\mathrm{E^{\theta}_{cell}=E_{cell}-{\frac{RT}{nF}}*ln1}$$ reduces to $$\mathrm{E^{\theta}_{cell}=E_{cell}}$$

We plotted our $\mathrm{E^{\theta}_{cell}}$ values versus T and noticed that our $\mathrm{E^{\theta}_{cell}}$ values were actually increasing with temperature.

I would have expected that these values go down as the concentration of the reactant falls.

I asked my demonstrator and sourced information from different web resources, but can't access a definitive answer.

Any hints even?

A few thoughts:

For starters, $\ce{Zn^{2+}}$ is also part of the reaction quotient, so the log term in the Nernst equation will not necessarily be zero if the hexacyanoferrate concentrations are equal, depending on the $\ce{Zn^{2+}}$ concentration:

$$E = E°+\frac{RT}{zF}\ln\frac{[\ce{Zn^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$
Depending on whether the reaction quotient is greater than or less than 1, the potential can have either a positive or negative temperature dependance.

Second, the Nernst equation uses activities not concentrations so depending on the ionic strength and other factors, the activities of the ions in solution may not be what you think. This deviation is greater for multiply charged ions and may be significant for concentrations over 1 mM.

Finally, since the hexacyanoferrate species are both soluble, I'm assuming you're using some kind of reference electrode for the second electrode. This is also temperature dependant in the same way as the rest of the cell.


Is it correct to consider dissolving HCl gas in water as oxidation –reduction process? References define oxidation –reduction process as change in oxidation number ,Accordingly it is not appropriate to describe the mentioned  process as oxidation –reduction because there is  no change in oxidation number of atoms and it is not a chemical reaction .

But the teacher consider it as oxidation –reduction reaction because HCl gas ionize in water .
Iam confused , so any idea will greatly help.

Given the dissolution of $\ce{HCl}$ gas in water:

$$\ce{HCl(g) + H2O(l) <-->H3O+(aq) +Cl-(aq)}$$

The oxidation state of all hydrogen atoms is +1 and that of all chlorine atoms is -1. Thus, nothing has been oxidized or reduced, and no redox reaction has taken place.


How to calculate the electromotive force of a silver chloride/ silver bromide cell? For the galvanic cell: $\ce{Ag|AgCl(s)|KCl(0.2M)||KBr(0.001M)|AgBr(s)|Ag}$, calculate the electromotive force (EMF) generated.
  $K_\mathrm{sp}(\ce{AgCl}) = 2.8\times10^{-10}$;
  $K_\mathrm{sp}(\ce{AgBr}) = 3.3\times10^{-13}$

I have tried this forming the cell reaction which is coming out to be
$$\ce{AgCl + Br- <=> AgBr + Cl-}$$
then using the $K_\mathrm{sp}$ of both salts I found the equilibrium constant which should be
$$K_\mathrm{eq} = \sqrt{\frac{K_\mathrm{sp}(\ce{AgCl})}{K_\mathrm{sp}(\ce{AgBr})}}$$
then I put it in the formula
$$E_\mathrm{cell} = -0.059 \log(K_\mathrm{eq})$$ but my answer is coming wrong.


What does the "D" group represent on a chiral center? I am trying to figure out the configuration for this chiral carbon to answer a homework question, however they arbitrarily include a "D" group as shown in the image. My guess is that it is some higher priority group than the carbon, and lower priority than the halide, making the configuration (S). Is this correct? What does the D group represent?

I assume it's deuterium, the isotope of hydrogen that has a single neutron as well as a proton in the nucleus. As such, it has the same atomic number as hydrogen, which means you break the tie on atomic mass --  deuterium is higher priority than hydrogen but lower than everything else.


Biomolecular energy functions and entropy From what I've been reading, it seems that entropy (rather than enthalpy) is the biggest driver of protein folding (especially the burying of hydrophobic residues). However, popular energy functions like AMBER, Charmm, etc. don't model entropy. They are however able to identify the native fold within a few angstroms (i.e. the minimum energy structure per the force fields is not more than a few angstroms away from the native structure) (http://onlinelibrary.wiley.com/wol1/doi/10.1002/jcc.20720/full).

If entropic changes are the primary driver of folding, then how is it that energy functions that don't even model entropy are still able to get close in terms of identifying the native structure (in particular, the burying of hydrophobic residues as well)? Is the entropic information incorporated in the solvent models?

I am sure that a more definitive answer could be provided by someone more familiar with the field, but here are a couple of observations.

First, I think it is a bit of an exaggeration to say that protein folding is almost entirely the result of entropy because the protein will have larger entropy in the unfolded state than in the folded state. The main entropic increase on protein folding will be the entropy of the solvent (and the system solvent + protein is what matters thermodynamically). This is the so-called hydrophobic effect which is what you're referring to by saying folding buries the hydrophobic residues.

Notice, however, that an energy function which contains solvent-solvent, solvent-residue, and residue-residue interactions implicitly contains this entropic effect. This comes by maximizing solvent-solvent and solvent-residue interactions. In order for this to happen, the hydrophobic residues will naturally become buried because this will increase the average number of hydrogen bonds between solvent molecules (I'm assuming the solvent is water) and strengthen the solvent-residue interactions at the exterior of the protein.

Second, the reason an energy-only model of protein folding might get "lucky" by ignoring entropy is that even if the net enthalpy change for protein folding is roughly zero, this does not mean that energy did not play a significant role in the folding process. People often talk about how proteins cannot simply be randomly walking the conformation space because it would take an endless amount of time to find the energy minimum by doing this. This is exactly why intermolecular interactions between residues get discussed so frequently.

So, an energy-only approach will very likely identify the residue-residue interactions which are important structural features of the final folded state. It is entirely possible that there are steps which are energetically destabilizing, but when these steps take place (if they are to take place in the real world), there must be a very constrained conformation space for the step to take place by randomly sampling states. Nonetheless, each step in the protein folding process must hit an energy minimum at some point, so just searching the energetics is a perfectly reasonable first approach. Furthermore, when this approach fails, it tells you that whatever is going on must be quite sensitive to the entropy of the system, so it is not like you are flying blind.

Third, it could easily be the case that energy-only approaches sometimes arrive at the correct structures completely accidentally. That is, there is more than one path from the unfolded state (which unfolded state?) to the folded state. The "real" path, if there is a single path that proteins take, might be quite different from the one that a model like this will take. In fact, if the protein is being sort of randomly perturbed and energy minima are being sought out, the pathway is likely complete nonsense, but this doesn't really matter as long the as protein is folding. It would be interesting to see how models like this do with proteins that have very energetically similar but distinct folded states. If what I'm saying in this part is true, one would expect the model randomly ends up in either state when you run the simulation. This would indicate that in the real world, there is a kinetic effect which forces the folding to follow a certain path and the model does not capture this.


Formation of silica from TEOS Usually tetraethyl orthosilicate/tetraethoxy silane (TEOS) is mixed with both ethanol and water to produce silica in a sol-gel process. I am curious what happens if one adds TEOS to water at a high pH, without any alcohol. Would TEOS simply not mix and not react with water? Or it would hydrolyze too rapidly and not condense at all?

It depends. if your pH is really high, silicate $\ce{SiO3^2-}$ or $\ce{SiO4^4-}$ ions will form and will become a stable dissolved solute in the solution. if it is moderately high the rate of hydrolysis will increase greatly and the silicon will crash out in the form of orthosilicate hydrates [$\ce{Si(OH)4}$], which may agglomerate to form silica particles. These results are a bit of a spectrum and there is no pH limit where the reaction gives a binary change in result, as it is also sensitive to extrinsic properties of the solution such as the amount of base.


Does bleach turn a substance white or colorless? From what I understand, bleach works to weaken the ability of a substance to absorb light. Thus this seems to be "whitening" i.e. reflecting all light. In contrast, colorlessness (according to Wikipedia) implies that light can pass through the object without scattering.

So is 'colorless' a misnomer when used to describe bleach?

(Also, to double check when you bleach something, is it still there? Only that it is now white?)

First, let's think about what makes a substance coloured: a substance will appear to be coloured if it absorbs light in the visible spectrum. Beta carotene:

absorbs blue/green light (400–500 nm), so it looks red-orange when white light is shined on it as the blue/green light is subtracted from the white and what remains is what reaches our eyes. The details aren't dreadfully important, but you'll notice that the molecule has a lot of conjugated double bonds. This is what allows the molecule to absorb visible light and this conjugation is what produces the colour in many things that might stain clothing or the like. The size of this conjugated system affects how it absorbs light. In general, the more conjugation, the more absorption occurs and at longer wavelengths (towards red).

Substances used as bleaches tend to be oxidizing agents. Again, the details aren't too important, but one thing oxidation can do to these conjugated molecules is break them apart or convert double bonds to other functionalities. By disrupting the conjugated system, one changes the absorption properties of the molecule. Breaking the molecule apart or removing double bonds shrinks the extent of conjugation, reducing absorption and shifting it to shorter wavelengths. If the wavelength is short enough, the absorption is the in the UV range, which is invisible to the human eye, making it colourless.

Now, to your questions:

  From what I understand, bleach works to weaken the ability of a substance to absorb light.

This is as described above.

  Thus this seems to be "whitening" i.e. reflecting all light.

Reflective/scattering properties are distinct from absorption. e.g. if I add bleach to an aqueous solution of a dye, it may go from a coloured clear solution to a colourless clear solution, but if I bleach a shirt stained with a dye, the shirt doesn't become transparent—the reflective properties are determined by the fabric, not the dye. Bleach simply changes the structure and thus, absorption of the dye, it doesn't render them reflective.

  So is 'colorless' a misnomer when used to describe bleach?

Colourless is the correct term to describe the effects of bleach—it makes coloured substances colourless.

  Also, to double check when you bleach something, is it still there? Only that it is now white??

The coloured molecules undergo a chemical change so they aren't the same substances with which you started. Whether the products of the reaction are still there depends on what they are. It may very well be that the oxidation products of a dye are more soluble and easier to remove when you wash a piece of clothing. The bleaching itself implies a chemical reaction that alters the absorption of coloured substances, not their removal.


Why is polyethylene not called polymethylene? I understand that the monomer that polymerises to give polyethylene is an ethylene, but after polymerisation, the smallest repeating monomer in polyethylene seems to be a methyl and not an ethyl. So why don’t we call it polymethylene instead?

According to the Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008, i.e. the ‘Purple Book’), three different types of names can be used for polymers:

traditional names,
structure-based names, and
source-based names.

When traditional names fit into the general pattern of systematic nomenclature, they are retained, in this case: polyethylene (PE)

The systematic name for a polymer requires the naming of a preferred constitutional repeating unit (CRU). If necessary, this basic name is then modified by prefixes. The systematic structure-based name for polyethylene according to the Purple Book is indeed poly(methylene).

Polymers can also be named as being derived from a monomer (or precursors), which is named according to IUPAC rules. Such names are referred to as source-based names. The systematic source-based name for polyethylene according to the Purple Book is polyethene.


Remove labels from jars? How can I remove labels from jars?

Sometimes I like to recycle jars for my own use, but normally I want to remove the label first. I find this extremely difficult to do. With ordinary stickers, like the stickers found as price tags on books, acetone works very reliably.

However, the labels on food jars are stuck on with a much more tenacious glue that does not respond to vegetable oil, acetone, methyl ethyl ketone, hot vinegar or ethanol. What is this stuff and how can I dissolve it?

Right now the only way I have found is to laboriously scrape them off with razor blades, but it is extremely time consuming to do this.

(Note that the tough labels I suspect are "hot melt" adhesives used to affix polypropylene labels. For example, traditional hot melt compositions are ethylene/vinyl acetate copolymer and ethylene/(meth)acrylate copolymers, but obviously there are others. I am hoping somebody with expertise can identify the range of typical compositions and corresponding solvents.)

I think this question is impossible to answer. There are dozens of chemical types for label adhesives. Most are thermoplastic, meaning they don't form new covalent bonds with the surface. But some are thermosets, meaning they react and cross-link.

In general, it is quite difficult to dissolve a crosslinked polymer since the dissolution process requires destruction (scission) of covalent bonds. The ways to do that are obvious: heat, oxidation, reduction, acids, or bases.

One thing that needs to be said here is consideration of both the opportunity cost as well as the environmental cost of your method generally makes all but the easiest methods inadvisable. When you say the adhesive "does not respond" to X, do you soak the label in X? (I typically use a slightly larger container (say part of a PET bottle) and a shallow layer of the solvent and allow it to soak overnight.

It's always a good idea, in my opinion, to start with water with a drop or two of surfactant in it. If hot water doesn't work, I move on to 70% IsoPrOH and give that a few hrs. Next, I use an alkaline (basic) detergent solution and give that another overnight soak. If it's worth my time, I end with an acetone soak (I have acetone which I reuse for this purpose in its own container.) One of these works most of the time. When none of these work, I toss (recycle) the jar. You may or may not know that not all polymers are soluble. So, your assumption that there is a solvent for the label you want to remove isn't necessarily true.

The most aggressive solvent I am aware of is mixtures of methylene chloride, dodecylbenzene sulfonic acid, and N-methyl pyrrolidone. That was used to de-pot Soviet microelectronic back in the Cold War days to reverse-engineer them. Anyway, glass is mostly Si-O-Si bonds and alkali bases should loosen any organic's adhesion to it.

Be sure to remove any surface layer on the label - abrasion is generally quickest. Oh, speaking of dangerous chemicals (methylene chloride is a carcinogen, and ddbsa is corrosive to the skin), I've also used aqueous HF to clean glass. The HF actually removes some of the silica, but if done carefully enough doesn't visibly damage the glass surface (never would try this with glass jars; why bother? but with expensive glass flasks, and other apparatuses, it was worth the hassle. The disposal of the waste solvent was a pain, though.)


Why is the 8-electron rule more important than the 2- or 18-electron rule Why is the fulfilled electronic configuration of only $p$ orbital  is stable. I mean why $II-B $ group with fulfilled $d$ orbital,$II-A$ group with fulfilled $s$ orbital...are not stable. Why makes $p$ orbital so special?

By the time the $p$ orbital is filled it is still the valence shell, you can see this by filling order e.g. $3s^2$ $3p^6$ $4s^2$ $3d^{10}$ ; You can see there isn't really an '18 electron rule' because there can never be 18 electrons in the valence shell and since chemistry is under most circumstances about the interaction of valence electrons then you can see that the $d$ shell electrons rarely have much to do with reactions because they are always 'one level down' from the valence shell.

Your question is a little unclear so I can't really say much more but I think you seem to be trying to ask why Neon or an analogous noble gas is so stable whilst Magnesium with a full $s$ orbital is not. Think about the difference in energy levels between going from one shell to another and you'll realise that an $\ce{Mg^2+}$ ion and Neon are very similar in terms of their stability because they are both in a low energy state.