Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently? Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently?

What could happen if I would strongly boil it.
Would the OH- react into the carboxyl groups to form a diol?

What could happen if I would strongly boil it

Mechanistically, nothing different will occur.  The same reaction that occurs at lower temperature will still occur at higher temperature.  The only difference will be the rate of the reaction; the reaction will go faster as you increase the reaction temperature.  In a classroom setting the procedure might involve simmering rather than boiling because

The reaction is usually followed by making uv-vis absorbance measurements every 15, or so, minutes.  Depending upon the reaction temperature and the concentration of base, the reaction might proceed too fast at boiling to allow for convenient measurements (e.g. the reaction might be complete in 10 minutes).
Safety reasons, if something goes wrong (the flask cracks) a warm basic solution won't burn the student as bad as a boiling basic solution would.


Symmetry and stability of a molecule (Jahn-Teller Distortion) I've taught in my class that if a molecule posses high degree of symmetry it's stability will be more,in other words symmetry and stability are directly proportional. Is this statement correct? And if this is correct than why Jahn teller distortion stabilizes a molecule when it's symmetry gets low eventually because of the distortion.

The symmetry and stability are not directly related per se; is benzene more stable that naphthalene for example? Benzene is more 'symmetric' than the naphthalene but both are stable molecules; $D_{6h}$ vs $D_{2h}$ point groups.

Jahn  and Teller discovered that in a non-linear molecule in an orbitally degenerate electronic state, say $e_g$ or $t_{2g}$ in an octahedral complex, that there is an interaction between the electronic orbital motion and nuclear vibrational motion, which causes the minimum energy position  to differ from the symmetric structure. This motion lifts the orbital degeneracy and lowers the energy. Thus the lower symmetry state has lower energy.

An example might be the lengthening of a pair of axial bonds in a octahedral complex, $\ce{[CoF6]}^{3-}$ for example, however, the J_T theorem does not give any indication of how big the distortion might be, i.e. it does not predict the geometrical nature of the distortion. There is also a restriction which is that if the degenerate structure has a centre of symmetry so does the distorted structure.

note: The electronic spin of the molecule (singlet, triplet etc.) is not important it is the orbital degeneracy only that matters. The Jahn-Teller effect is a breakdown of the Born-Oppenheimer approximation which is based on separating nuclear and electronic motion.


IUPAC nomenclature of aspartame I'd like some help in constructing the IUPAC name of aspartame:

From what I understand there are carboxylic acid, amine, amide, ester, and phenyl groups - but I'm having trouble figuring out the order in which to name them.

Aspartame is (a methyl ester of) a dipeptide. Therefore, you could use the nomenclature of amino acids to find a descriptive name for aspartame.

However, since you are asking about the correct order of the individual groups, apparently, you want to generate the fully systematic name according to nomenclature for organic chemistry.

Since there are various groups, you have to identify the principal characteristic group in accordance with the seniority of classes and suffixes first. In this case, the characteristic group with the highest seniority is the carboxylic acid group.

The principal characteristic group is chosen for citation at the end of a name by means of a suffix or a class name (or implied by a trivial name). Since there is only one carboxylic acid group in aspartame, the principal characteristic group also decides the choice of a principal chain. Thus, the parent structure of aspartame is butanoic acid.

The remaining substituents are expressed as prefixes in the name in alphabetical order. In this case, the remaining substituents are one amino group, one oxo group, and the complex substituent group [(2S)-1-methoxy-1-oxo-3-phenylpropan-2-yl]amino.

Therefore, the complete systematic name of aspartame is (3S)-3-amino-4-{[(2S)-1-methoxy-1-oxo-3-phenylpropan-2-yl]amino}-4-oxobutanoic acid.


Butane Burning Color Why does Butane burn blue? From my (very) basic understanding of chemistry, carbon and hydrogen do not burn blue, their radiation spectra are different-so why should the combination burn blue?
Please elaborate as much as possible, about bonds, orbitals, electrons, and the like!

The blue flame from butane (or many other hydrocarbons) is due to the complete combustion of the gas.  From the Wikipedia article Flame color it mentions,

  In the most common type of flame, hydrocarbon flames, the most important factor determining color is oxygen supply and the extent of fuel-oxygen pre-mixing, which determines the rate of combustion and thus the temperature and reaction paths, thereby producing different color hues.

Specifically, making the link to the common tool in the laboratory, a Bunsen Burner, which typically has two 'types of flame (as seen below):

Image Source - which states that:

  The Bunsen Burners we use at school use a mixture of alkane gases like propane and butane

The yellow flame is from an incomplete combustion, producing $\ce{CO}$ and the colour is due to (according to the Wikipedia article):

  incandescence of very fine soot particles that are produced in the flame.

With more oxygen, a complete combustion of butane is possible (without the soot):

$$\ce{2C4H10 +13O2 -> 8CO2 + 10H2O}$$

Which provides

  enough energy to excite and ionize gas molecules in the flame.

and the blue flame colour is due to the

  emission of excited molecular radicals in the flame, which emit most of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.


Why is the deprotonated form of dopamine more reactive than the protonated/neutral form of dopamine? See this (from the Wikipedia article on dopamine):

  Dopamine, like most amines, is an organic base. At neutral or acidic
  pH levels it is generally protonated. The protonated form is highly
  water-soluble and relatively stable, though it is capable of oxidizing
  if exposed to oxygen or other oxidants. At basic pH levels, dopamine
  becomes deprotonated. In this free base form it is less soluble and
  also highly reactive and easily oxidized

I'm very curious about this since I'm curious about dopamine auto-oxidation, as it is often a cause of neurodegeneration observed in those who take amphetamines (and seems more likely in the cytosol, than the vesicles it would otherwise be in). Though I'm also curious how much auto-oxidation would happen for extracellular dopamine as well.

In general the free amine will react as a nucleophile i.e. by "donating" an available lone pair of electrons. In its protonated form, this lone pair is "blocked" and thus it has to lose a "blocking" proton to react (I use the words in "quotes" more freely - although I'm a chemist and not biochemist but hopefully it will help).

However re oxidation of dopamine, the reactive site is not the nitrogen (at least not the primary one - but will come back later) but the catechol part. Because of the electron donating effect of the 2 neighboring OH's, the aromatic ring is electron rich and more prone to oxidation to give o-quinone products. This oxidation reaction depends to an extend to the pH but mainly to the presence of (obviously enough) reactive oxygen species (ROS) and also enzymes (oxidases) or even Fe ions that can catalyse this oxidation.

Coming back to the nitrogen, after oxidation occurs with creation of reactive sites on the quinone part, it can act now as a nucleophile and attack, creating indolines that will further oxidize to the more stable but toxic indole products.


What are some good examples of rate equations for a math class? I'm a mathematician who's currently teaching a course on differential equations.  Though I don't know much about chemistry, I like to include examples from chemistry in my course, and I prefer for the details to be accurate.  Here is a typical exam problem:

  When a container of gaseous nitrogen dioxide is heated above 150 °C, the gas begins to decompose into oxygen and nitric oxide:
  $$\ce{2NO2 -> O2 + 2NO}$$
  The rate of this reaction is determined by the equation
  $$\frac{\mathrm d\left[\ce{NO2}\right]}{\mathrm dt} = -k\left[\ce{NO}_2\right]^2$$
  where $k$ is a constant.

  (a) Find the general solution to the above equation.

  (b) A large container holds 50.0 moles of $\ce{NO2}$ at a constant temperature of 600 °C.  After one hour, only 34.3 moles remain.  How much $\ce{NO2}$ will there be after another hour?

So my questions are:

Is the science in this problem reasonably accurate?  Is there anything you would change?  (I looked up a suitable value of the rate constant $k$ to make sure that the time in part b was reasonable.)
What are some other examples of reactions that are governed by simple rate laws?  Ideally, I'd like to have several examples each of reactions governed by the equations
$$
\frac{\mathrm dy}{\mathrm dt} = -ky,\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^2,\qquad\text{and}\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^3.
$$
(Is $y^3$ really possible?  Are non-integer powers of $y$ possible?)

How thoughtful of you to include chemistry in your differential equations course! We appreciate your effort, especially going the extra mile to make it realistic.

I hope you've seen the rate equation page on Wikipedia as it contains a good deal of the mathematics in several examples of reactions. You should also be interested in the reaction order page (you can find examples here of reactions with fraction or negative order with respect to some reagents).

Let me do a quick recap on basic chemical kinetics theory. Consider the general reaction equation:

$$\ce{a\ A + b\ B + c\ C + d\ D + ... \longrightarrow w\ W + x\ X + y\ Y + z\ Z + ...}$$

, where uppercase letters indicate different molecules in a gas or solution, and lowercase letters indicate the stoichiometric coefficients (negative for the reactants, positive for the products). Naively, we can assume the reaction happens because, while the molecules are jostling about, at some point all of the reactants ($a$ molecules of $\ce{A}$, $b$ molecules of $\ce{B}$, $c$ molecules of $\ce{C}$, etc.) will bump into each other all at the same time and undergo reaction. If you think of collisions as independent events with a probability of happening proportional to the concentration of each species, then it isn't too hard to understand that the frequency with which reactions happens is proportional to $[A]^a[B]^b[C]^c[D]^d...$, and so the rate equation for the reaction would be:

$$r=k[A]^a[B]^b[C]^c[D]^d...$$

, where r is the reaction rate and k is a proportionality constant. The consumption or production rate for any species $\Gamma$ with stoichiometric coefficient $\gamma$ is trivially related to the reaction rate by:

$$\frac{d\Gamma}{dt}=\gamma r$$

However, experimentally things aren't so simple (fortunately for nature, unfortunately for our minds). A reaction that requires the simultaneous collision of many molecules would be highly unlikely, meaning that it would happen very, very slowly. Even so, experimentally we observe many reactions that involve a large amount of molecules. For example, the combustion of cyclohexane in air is formally given by the equation:

$$\ce{C6H12 + 9 O2 → 6CO2 + 6 H2O}$$

If the reaction really required ten molecules to bump into eachother at the same time with sufficient energy and in the right geometry, then this reaction probably could not happen in air, as oxygen would be too rarefied to compensate the extremely low proportionality constant k for a single-step reaction. In actuality, a gaseous mix of cyclohexane and air can react so fast as to cause an explosion, converting all the reactants into products in a miniscule fraction of a second. Clearly something is wrong.

It turns out the assumption that reactions occur in one step is incorrect in general. Usually, there is more than one step involved, and indeed there generally is more than one path from reactants to products. The speed at which a reaction occurs is really described by considering all paths at once, with all their steps, and adding their contributions (like a start and end point in a complex tree diagram). In general this is pretty difficult and unnecessarily complex. It is often a reasonable approximation to model the reaction by selecting only the fastest route from reactants to products, and the reaction rate is bottlenecked by the slowest step of the fastest route, which becomes responsible for determining the rate equation.

Now, since the reaction rate depends on selecting the fastest route, you shouldn't expect any of the steps in it to contain some very huge bottleneck, such as requiring six molecules to react simultaneously; there would likely be a slightly different route, perhaps with more steps, but in which each step doesn't involve so many molecules interacting at once, and hence being a faster route. Because of this, it turns out that most of the fastest reaction pathways involve slowest steps that very rarely depend on more than two molecules at once. Most reactions which can be modeled simply like this therefore have rate equations of the type $r=k[A]$, $r=k[A]^2$, $r=k[A][B]$ or even $r=k$. Few reactions rates are of the third order, of the type  $r=k[A]^3$,  $r=k[A]^2[B]$,  $r=k[A][B][C]$ or similar. I don't know of any reaction that goes as $r=k[A]^3$ specifically. Fourth order reaction rates are so rare that they are the focus of research when found. I don't expect anyone to know a fifth order reaction.

At the end of the day, kinetic theory is hard enough that reaction kinetics end up being determined simply by parameter fitting from experiments. Even if the underlying reaction mechanism is not completely understood, if a fit of the type $r=k[A]^{3/2}[B]^{-1}$ happens to be good, then so be it. Reactions which have been studied in depth are significantly more complex, such as the misleading simplicity of the reaction between hydrogen and bromine.

Edit: I don't know why I didn't search for a fifth order reaction rate! Turns out some exist, such as this one. The reaction rate is of the type $r=k[A][B]^4$, so you could have an example where $\frac{d[B]}{dt}=k[B]^4$.


Non-toxic organic bases? I am trying to find out if there are any non-toxic organic bases that could be given orally on a non-empty stomach to treat acidosis. From my research, histidine and diluted choline hydroxide were the best candidates. Some phosphazenes, including polyphosphazenes, could make the cut, but I don't know which ones exactly.

Wikipedia [1] lists the following compounds as organic bases: pyridine, alkanamines (such as methylamine), imidazole, benzimidazole, histidine, guanidine, phosphazene bases, hydroxides of quaternary ammonium cations or some other organic cations.

The LDLO of guanidine given orally to rabbits is 500 mg/kg [2]. The LD50 of imidazole (rabbits, oral) is 950 mg/kg. Pyridine seems to be in the same range [3]. Benzimidazoles are also toxic.

Histidine, which is considered the most toxic amino acid, can be therapeutic in doses of up to 4.5 g/d, but is toxic in the range 24–64 [4].

The last ones from this list are choline hydroxide and phosphazenes. The UTL of choline is 3.5 g/d; in the 10-16 g/d range it causes fishy smell so it probably overloads us with TMA, which is a carcinogen. Polyphosphazenes are being studied for drug delivery; apparently their metabolism generates low-toxicity products such as urea and phosphates. [5].

Naturally occuring amino acid with basic sidechain - L-Arginine:

  Oral supplementation with L-arginine at doses up to 15 grams daily are generally well tolerated.

source here


Which dihydroindenyl cation is the most stabilised? I am facing difficulty in the following problem.

The carbocation which is most stabilised among the following is …

In my view c) should be most stabilised because even if one hydrogen is abstracted it will help in restoring the aromatic character. However my book gives the answer as a).

The carbocation a) is mostly stabilised because it has maximum resonance structures which leads to maximum delocalisation of the charge.

None of the others has so many resonance structures.Therefore a) is most stable.


What additive can I add to Jelly Wax/Gel Wax to increase its cooling time/setting time? Jelly Wax. Jelly wax, also commonly known as gel wax, is a clear, dimensionally stable, jelly-like material made from specially processed mineral oils. It can be used to create a variety of special effects and is also suitable for making decorative container candles. It is normally heated until it liquifies and then poured into a mould. It sets quickly back into a jelly.

I want to use it to pour into a transparent PVC pipe for a sculpture. Unfortunately I have found that it cools rather quickly for it to be useful for this purpose.

Anyone knows how I can increase the cooling time, using an additive maybe?

At your own risk, you can try using plasticizers for waxes. Usually, these plasticizers are wax esters of palmitic acid (well-known example is spermaceti, which can be replaced with jojoba oil, or you can buy/obtain those esters). But it's hard to predict exact amount of plasticizer needed. Add too much - and it will be creamy I-will-never-cool substance. I hope we have chemical engineers here who know more about wax plasticizers and can add details to my superficial statement.


How to calculate the equilibrium pressure when ammonia and hydrogen gasses react in a closed system? There is a container that has a volume of $5~\mathrm{L}$ on one side and $3~\mathrm{L}$ on the other and is separated in the middle. There are $2~\mathrm{mol}$ of $\ce{N2}$ on one side at $0~^\circ\mathrm{C}$ and $4~\mathrm{mol}$ of $\ce{H2}$ on the other at $0~^\circ\mathrm{C}$. When the valve is turned and the gases react, what will the total pressure be?

The balanced reaction is $$\ce{N2 + 3H2 <=> 2NH3}$$
My thinking was that the reaction occurs at STP so the pressure would stay at 1  atm?
Or that you could use
\begin{align}
pV &= nRT\\
p(8~\mathrm{L}) &=6(.082)(273.15)\\
p &= 0.06
\end{align}
The $8~\mathrm{L}$ comes from adding the two volumes, and the 6 from adding the number moles together.

Note: Not all the reactants will react to form the products, but rather exist in equilibrium. However since the question doesn't mention anything about the equilibrium constant, I will assume that the reaction goes to completion as otherwise it will be impossible to do this question.

If you are aware of the various gas laws, doing this question is actually really simple and doesn't involved many calculations.

Step 1: Finding Initial Pressure

Firstly, the question doesn't specify what the initial pressure of the container is. Therefore you can't just assume that it is $1\ \mathrm{atm}$. So the first step would be finding the initial pressure: $$p = \frac{nRT}{V} = \frac{6\ \mathrm{mol}\times 8.3125\ \mathrm{L\ kPa\ mol^{-1}\ K^{-1}}\times273\ \mathrm{K}}{8\ \mathrm{L}} = 1702.4\ \mathrm{kPa} = 16.8\ \mathrm{atm}$$

Step 2: Find Moles of Products

Looking at the stoichiometry of the chemical equation, $4$ moles of gas react to form $2$ moles of gas. Therefore the ratio is $2:1$. In this case, we originally had $6$ moles of gas. Therefore at the end of the reaction, there should be $3$ moles of gas.

Step 3: Find Final Pressure

Now we know that the total number of moles of gas has halved (from $6$ moles to $3$ moles). One of the gas laws state that the number of moles of gas is directly proportional to the pressure of the system. This is clearly obviously by looking at the ideal gas equation. Therefore, by halving the original amount of moles of gas should result in halving of the original pressure of the system.

Therefore the final pressure will be: $8.4\ \mathrm{atm}$


Exact meaning of "degree of crosslinking" in polymer chemistry In the ion-exchange chromatography literature, a term "degree of crosslinking" appears frequently. Ion exchange resins are typically made from divinylbenzene and styrene. One of the stated characteristics is the "degree of crosslink" which is often stated as 5% to 55% crosslink with DVB.
The term degree of crosslink does not appear in the IUPAC's glossary; yet it is a very well established terminology. What is the physical significance of, say, 55% crosslink when the manufacturers quote this number? Is it simply the mole percent of DVB present in the polymer?
I am looking for its exact definition from a reliable source but it seems to be elusive.

Normally, in practice the amount fraction of DVB in the feed is used as a nominal number for the degree of crosslinking. IUPAC provides a definition in a somewhat unexpected place [1, p. 2352]:

Crosslinking: Property of a solid support prepared from polymeric materials with interconnected strands. Often results from the inclusion of multifunctional monomers in the polymerization reaction, e.g. divinylbenzene in polystyrene production. In such cases, the degree of crosslinking is often quoted as the proportion of the multifunctional monomer in the reaction mixture. The extent of crosslinking is important for physical properties of the solid support, such as the propensity to swell in different solvents [18].

Following reference [18], one can find further clarification [2, p. 2278]:

Care is therefore required in defining or interpreting the DVB content of crosslinked polymers since this might be quoted as a percentage of technical DVB used to make the
polymer, or the figure can be adjusted to reflect only the content of actual DVB isomers present. Since for convenience the nominal crosslink ratio or degree of crosslinking of a polymer network is often quoted as the mol% of crosslinker used to prepare the network, defining the actual percentage of DVB isomers employed is more informative since this equates with the nominal crosslink ratio (Fig. 4). Bearing in mind that there is no unambiguous method, and certainly no simple and rapid method, for determining the real crosslink ratio in a polymer network, the nominal figure based on the actual DVB feed is a
very useful parameter.

Fig. 4 Polymerisation of styrene and divinylbenzene to form an infinite polymer network.

Further, the paper lists several reasons as to why the real level of crosslinking is always accompanied with a level of uncertainty:

The effective crosslink ratio significantly decreases with greater quantity of DVB being fed due to growing number of unreacted vinyl groups.
Entanglement crosslinking occurs at higher rates of polymerization in slowly-agitated systems and results in additional mobile crosslinking.

Fig. 5 Permanent entanglement crosslink

Additional crosslinks can also be created as the result of intramolecular side-reactions.

These factors are tricky to estimate quantitatively; a method allowing to do so would be a single pulse excitation (SPE) 13C solid state NMR.
References

Maclean, D.; Baldwin, J. J.; Ivanov, V. T.; Kato, Y.; Shaw, A.; Schneider, P.; Gordon, E. M. Glossary of Terms Used in Combinatorial Chemistry. Pure and Applied Chemistry 1999, 71 (12), 2349–2365. DOI: 10.1351/pac199971122349.
Sherrington, D. C. Preparation, Structure and Morphology of Polymer Supports. Chem. Commun. 1998, No. 21, 2275–2286. DOI: 10.1039/a803757d.


Comparing pKa of hydrocarbons We have to compare their acidic strength, and we know acidic strength is directly proportional to stability of the anion.
So the first compound is more acidic due to its being aromatic after deprotonation. The second most acidic is the fifth compound, because of $\mathrm{\% s}$ character of $\ce{C-H}$ bond. And the least acidic is sixth compound because its anion is antiaromatic.
But how would we compare the second, third and fourth compounds?

Source: MS Chouhan; Advanced Problems in Organic Chemistry 11th Edition; Chapter 1 General Organic Chemistry problem 27

Nice work! You clearly understand how factors like aromaticity/antiaromaticity and % s-character in a $\ce{C-H}$ bond can affect acidity.

You can use Coulson's Theorem to estimate the % s-character in the other molecules. In its general form Coulson's Theorem describes the relationship between the angle $\ce{X-Y-Z}$ in a molecule and the hybridization of the $\ce{Y-X}$ and $\ce{Y-Z}$ bonds according to the following equation:

$$\mathrm{1+\lambda_{XY}\lambda_{YZ}cos(\Theta_{XYZ})=0}$$

(where $\mathrm{\lambda}$ represents the square root of the hybridization index of a bond, e.g. $\mathrm{\lambda=\sqrt{3}}$ and the hybridization index $\mathrm{=3}$ for an $\mathrm{sp^3}$ hybridized bond)

In molecule III where the two $\ce{C-H}$ bonds are equivalent, Coulson's Theorem simplifies to

$$\mathrm{1+\lambda^2cos(117)=0}$$

and we find that $\mathrm{\lambda^2 = 2.2}$, or the $\ce{C-H}$ bond in III is $\mathrm{sp^{2.2}}$ hybridized.

As the hybridization index gets smaller, there is more s-character in the $\ce{C-H}$ bond which results in a more stable carbanion and consequently increased acidity. This is consistent with our thinking that, in general, acidity follows the order $\mathrm{sp > sp^2 > sp^3}$; or, the larger the $\ce{X-C-H}$ bond angle, the more acidic the $\ce{C-H}$ bond.


Cathodic protection of tubes in Shell & Tube Heat Exchanger Our $\ce{Cl2}$ liquifier is of the following type:

It is entirely made of mild steel (shell as well as tubes plus the connecting pipelines). At shell side there is Freon (R-12) with its refrigeration cycle (not shown here); at the tube side there is dried & compressed $\ce{Cl2}$ gas from compressor section (~2bar max). One must realise that drying of $\ce{Cl2}$ is done physically via passing it through packed glass wool columns & then washing it in $\ce{H2SO4}$ scrubbers & then again through glass wool columns...Compression is done via $\ce{H2SO4}$ liquid ring compressors.

Problem we face is excess corrosion of tubes which thus lose their thickness & may in time compromise...For instance every 2-3 weeks we open up this liquifier & wash the tubes (via high pressure demi/RO water) & we found quiet an amount of dark-green/brown sludge indicating ineffectiveness of our drying section. Lab test are yet to be done so that I can clearly say what this sludge is made of. From the WIKI what I understand the compounds of such colour can either be oxides or chlorides of Iron because sulphates of iron are yellow in colour...

In a nutshell what I can do is increase the number of drying towers or increase the number of glass wool columns, but the most active measure I am thinking of cathodic protection via impressed currents. Have any user here applied impressed currents to a S&T heat exchanger & how can it be done? I would love any literature on it.

As a side note our liquifier can be electrically isolated using gaskets & insulated teflon bolts on any flange fitting.

Cathodic protection requires an electrolyte to work. You could just pass current through the tubes but how is it going to return to the power supply? If there is no electrolyte to conduct ions it cannot provide any protection to your system. Have you considered replacing your heat exchanger with glass or glass-coated metal? If the capital costs of this are prohibitive, I would think your best bet would be to try drying the gas some more, as you mentioned. Do you happen to know the water content of the gas going into the exchanger now?


Why is the pH level of carbonation loss decreasing then increases afterwards? We have done an experiment of the titration of unsealed sprite soda to find the amount of carbonic acid being loss on various duration (0 to 1 hour), due to decrease pressure.
The calculation was to determine the pH of carbonic acid overtime, thus the number of concentration of flat and unsealed soda are subtracted and is converted to pH level using -log[concentration of flat soda - unsealed soda].This was done multiple times on multiple duration until 1 hour. Now the results has demonstrated that carbonic acid pH level was decreasing in the first 30 minutes but then afterwards it was increasing. This contradicts on what my hypothesis is as it states that due to pressure being decreased (by unsealing the soda) the pH will increase turning more basic. Is this due to the factors of pressure and temperature? Can you please describe this thoroughly? Thank you for your time on reading this.

At the beginning, there was titrated mostly just the citric acid and the equilibrium amount of carbonic acid.

After that, carbon dioxide slowly converted to carbonic acid and $\mathrm{pH}$ has been decreasing.

Finally, the long term process took dominance as dissolved carbon dioxide together with carbonic acid were  leaving solution as gaseous carbon dioxide and $\mathrm{pH}$ have been raising.


What reaction takes place when potassium iodide is added to a tetraammine copper(II) complex? I am given a blue solid W and it contains a tetraamine copper(II) complex and a $\ce{SO_4^{2-}}$. I added dilute sulfuric acid until the solution is pale blue; then added potassium iodide solution. A white precipitate forms ($\ce{CuI}$) in a brown solution ($\ce{I_2}$). However, I am confused about what reaction is actually going on.

Copper(II) is a hard Lewis acid that would be octahedrally coordinated but for Jahn-Teller distortion.  Instead, it is square-planar coordinated with two distant weakly bound apical ligands.  The lovely deep blue color, hard Lewis base ammonia versus hard Lewis acid water coordination, illustrates the nephelauxetic effect.  What would obtain with triethylamine coordination of copper?  Cu(II) is substitution labile.

Jahn-Teller effect
Nephelauxetic effect
Kinetics (and thermodynamics) of substitution reactions

Adding acid protonates the ammonia, its lone pair now in a dative bond with a proton and unavailable for coordination.  Tetraamminecopper(II) (note the spelling of "ammine" when coordinated) becomes tetraaquocopper(II) with fast kinetics.

Aqueous Cu(II) is a modest oxidizing agent; iodide is a modest reducing agent.  Cu(I) is a soft Lewis acid, iodide is a soft Lewis base, $\ce{CuI}$ is very insoluble (look up its $\ce{K_{sp}}$ -- see here.)

Iodine reacts with excess iodide to give triiodide anion.  However, isolate the "$\ce{CuI}$," wash, and titrate with thiosulfate.  $\ce{CuI}$ strongly adsorbs elemental iodine.  It is also modesty light sensitive.  Cyanide, being a pseudohalide, follows the same path to afford exceptionally insoluble CuCN plus cyanogen.

Remember who earned the "A," or do your chair parade for better references, then understanding for the next question.


why we can use the equipartition theorem for translational motion of molecules at room temperature and above because quantization is unimportant From the book Chemical Principles The Quest for Insight, 5th Edition by Peter Atkins, Loretta Jones

The equipartition theorem is a result from classical mechanics; so we can use it for
translational and rotational motion of molecules at room temperature and above, where
quantization is unimportant.

I know quantization describes that photons, separate "packets,"are what energy is thought to happen in a subatomic level. However, I don't how quantization is related here: because quantization is unimportant, we can use it for translational and rotational motion of molecules at room temperature and above? and also why quantization is unimportant here? I believe translational and rotational motion of molecules happen in a subatomic level. Moreover, why can't we use the equipartition theorem for vibrational motion?

What they mean is that for translational motion the quanta are so small that using classical mechanics produces no error, i.e $k_BT$ is far far bigger than the energy gaps between translational quanta, ($k_B$ is the Boltzmann constant) the same may be true of rotational motion where quanta may be only be separated by a few wavenumbers vs $k_BT\approx 210$ wavenumbers at room temperature. It is generally true for rotational motion but not all e.g. H$_2$. Vibrations have quanta of at least hundreds of wavenumbers  and $v=1 , v=2$ etc. are hardly populated at room temperature.
We can only use the equipartition theorem if $k_BT \gg \Delta E$ where $\Delta E$ s the gap between energy levels, i.e the fact that there are discrete levels is no longer important. This can occur if the energy gaps are v small or the temperature very high or both.
(Technically the 'equipartition theorem' of classical statistical mechanics states that the mean (average) value of each independent quadratic term in the energy is equal to $k_BT/2$)


How can I derive the preferred oxidation states of a transition metal from its electronic configuration? For example $\ce{Fe}\ (Z=26)$ Short formula: $\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^6\ 4s^2}$. If we make the electronic formula we get 4 single electrons ($s=1/2$), which means that the valence in normal state is 4, but in the periodic table it is $2/3/6$. How can I find its valence?

Firstly you need to appreciate that the 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the oposite). This means that the two 4s electrons are lost first giving a valency of two (which is common to all the first row transition elements). The next point is that one of the 3d orbitals is fully ocupied so one of these two electrons will be preferentialy released due to the repulsion of its pair. This results in a valency of three. These are by far the most common valencies of iron. In fact compounds of 4, 5, and 6 oxidation states (valencies) are known but are very unstable, so of little importance.


How do we know that water has [H+] = 10^-7? In the derivation of $K_\mathrm{w}$, the ionic product of water, it is stated that

$$[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$$

and substituting $[\ce{H+}] = [\ce{OH-}] = 10^{-7}$ we can find that $K_\mathrm{w} = 10^{-14}$.

How do we know, though, that $[\ce{H+}] = 10^{-7}$? And why is $[\ce{H+}] = [\ce{OH-}]$?

Why is $[\ce{H+}] = [\ce{OH-}]$?

Because the water is electrically neutral.

In general, for all solutions, the sum of the concentration of each ion, multiplied by the ion's respective charge, must be zero.

  How do we know $[\ce{H+}] = 10^{-7}$?

This is found by experimentally by determining $K_\mathrm{w}$ and substituting, or by measuring the pH.

For details see: Bandura, A. V.; Lvov, S. N. The Ionization Constant of Water over Wide Ranges of Temperature and Density. J. Phys. Chem. Ref. Data 2006, 35 (1), 15–30. DOI: 10.1063/1.1928231.


Is there any difference between a completely filled orbital and an half-filled one? Is there any reduction in size of the orbital for a half-filled orbital? Is the probability at any point of finding an electron doubled if there are two electrons instead of one? Is there any difference at all? I'm trying to understand the quantum mechanical model. Thank you in advance!

While what ssavec says about not taking orbitals away from a structure is true, it's possible to make some generalisations, particularly for lone atoms where the symmetry constraints make the orbital divisions more clear.

When you generate orbitals for an atom, you get the set of s, p, d, f (and so on) orbitals separated out, and you can generally say that for the same principal quantum number (usually denoted n), a doubly-filled orbital will produce more diffuse orbitals than a singly-filled one. This is an obvious consequence of Coulombic repulsion. But, again, all the other orbitals you generate for the altered wave-function will be affected by the change in occupancy and overall charge.

It would be possible to construct a somewhat artificial pair of calculations in which you kept fixed all but one of the occupied orbitals, and then optimised the wavefunction using one electron in one calculation and two in the other, and I would expect you to find that in the calculation with two electrons, the varied orbital would be more diffuse.

Because the two orbitals would be different in spatial form, you wouldn't expect the value at a given point to be a multiple of the other.


Which ethanol derivatives display the gauche effect? How many of the following ethanol derivatives of the form $\ce{GCH2CH2OH}$ have a more stable gauche conformer compared to their anti conformer?

  $G =  \ce{F}, \ce{Cl}, \ce{OH}, \ce{NH2}, \ce{OMe}, \ce{NHMe}, \ce{NMe2}, \ce{NO2}$

I was sure that $\ce{F}$, $\ce{OH}$, $\ce{NH2}$ and $\ce{NO2}$ display the gauche effect. I'm not so sure regarding all the other groups: all of them seem bulky and won't form hydrogen bonds with the $\ce{OH}$ group on $\ce{C}_1$. How many of the other substituents are capable of displaying this effect? Also, is there a quantitative way to determine if the gauche conformer would be more stable than the anti conformer?

As per this source, there is evidence for the gauche effect for most of these groups on basis of intramolecular hydrogen bonding:

I tried looking for a similar evidence for 2-nitroethanol, and found this paper on the same:

  Figure 4. OCCN gauche structures with an intramolecular H-bond for 2-aminoethanol (7) and 2-nitroethanol (9); Conformations 8 and 10 indicate disrupted H-bonds after rotations by approximately 120º about the O-C axes

So it seems like all the groups shown here can show the gauche effect on the grounds of intramolecular hydrogen bonding,and so your answer should be 8


Why do we neglect the protons from water's autoprotolysis? Why don't we need to consider the $\ce{H+}$ ions of water when finding the $\ce{pH}$ of $0.01~\mathrm{M}$ of $\ce{HCl}$ solution?

The $\ce{pH}$ value of something is depending on the molarity of $\ce{H+}$ ions. But the teachers says the calculation is just $-\log(0.01)$, don't we need to consider the $\ce{H+}$ ions of water?

You could consider them, but what would be the result?

$0.01 M$ of $\ce{HCl}$ with $1\times10^{-7} M$ of $\ce{H^+}$ from water gives you $0.0100001 M$ of $H^+$. Find the pH of this concentration of $\ce{H^+}$ and compare with the pH you would obtain if you used $0.01 M$ directly.

You shouldn't see that the resulting pH differ significantly.


Equilibrium Constant in terms of Molar Concentration of Gases versus Equilibrium Constant Suppose the gases involved in a reaction behave as real gases with
significant intermolecular attractions.

I just learned that $$K_c(RT)^{\Delta n_r} = K$$

Would the value of Kc calculated
from the equation above be too high, too low, or the same as the actual value?

I thought that it would be the same, but I don't think that's right because deviation from ideal behavior should deviate from equilibrium.

I believe that the real $K_c$ would be higher. This is because in ideal gases london dispersion forces, hydrogen bonding, etc. is ignored. These would increase the number of collisions between particles since they would be attracted to each other more.
$$\ce k=zp*e^{-E_a/RT}$$
where k= rate constant, z= frequency of collisions, p= fractions of collisions with proper orientation to cause a reaction, $E_a$= activation energy, T= temperature. So Since z would increase, the K would increase. Hence the $K_c$ would increase.


What is the hybridisation in HCN molecule? According to VSEPR theory hybridisation of the central atom should be sp. However, my teacher says that there exists a near-100% s-character in the carbon orbital that bonds with hydrogen due to some other reasons. I couldn't get why that would be the case. One can think of it as the nitrogen that polarizes the carbon atom, but shouldn't it lower the s-character of the C-H bond? What am I missing?
Is it based on quantum calculations (which are beyond my scope) , as a commentator points out?

$\ce{HCN}$ and $\ce{HC#CH}$ are linear, triple bonded, with a $π$ system consisting of two perpendicular $π$ bonds. They would be symmetrical in $\ce{HC#CH}$, and slightly distorted in $\ce{HCN}$, and they leave two orbitals for the sigma system.
In $\ce{HCN}$, we hybridize/combine the two remaining orbitals on the carbon atom to form two bonding orbitals, one to the hydrogen, another to the atom on the other side of the carbon (a $\ce{C}$ or an $\ce{N}$). The natural first combination is a $50$–$50$ split to form two $\mathrm{sp}$ orbitals, one directed to the $\ce{H}$ and the other directed toward the $\ce{N}$.
That's enough most of the time, but if you get picky, you could point out that the electronegativities of $\ce{H}$ and $\ce{N}$ are quite different ($\ce{H}$ $2.1$, $\ce{C}$ $2.5$, $\ce{N}$ $3.0$), so the nitrogen will be pulling on its $\mathrm{sp}$ bond more than hydrogen pulls on its $\ce{sp}$ bond, so the $50$-$50$ split readjusts to maybe $70$-$30$ (did your teacher say $~100\%$-$0\%$?), meaning that the hydrogen gets less of the carbon orbital (i.e., less $\mathrm p$-character, more $\mathrm s$-character from the carbon orbitals), while the nitrogen gets more of carbon's $\mathrm p$-orbital.
If this is so, the $\ce{H}$ atom in $\ce{HCN}$ should be more easily removed than a H atom in $\ce{HC#CH}$. This appears to be correct: the $pK_\mathrm a$ of acetylene is $24$ (which is considered to be quite acidic for a hydrocarbon), whereas the $pK_\mathrm a$ of hydrocyanic acid is $9.21$, much more acidic (although a weak acid by any other measure).
Your teacher pointed out something interesting, but the exact split ratio remains to be calculated.


Corrosive/ Acid cabinet gets wet from time to time I have a metallic acid cabinet in a chemistry-based lab which I use store hydrochloric acid, sulfuric acid and ammonia. I also store nitric acid together inside but they are physically separated from the rest by an additional PP plastic case inserted into the acid cabinet.

The issue is that from time to time, I realized that the doors and secondary trays in my acid cabinet will get wet, especially the area near the sulfuric acid, and white powdery crystals sometimes form on the walls of my HCl bottles. I have ensured that I tightly sealed each of the bottles. Can anyone help me out? Is this dangerous? I'm trying to seek a solution before anything serious that might happen.

Also I have a few bottles of unopened, still sealed sulfuric acid which have turned black but the acid is still clear and usable when poured out, is it still alright to use for experiments? Will the results be affected?

The white powdery crystals are probably $\ce{NH4Cl}$, that is formed thanks to leakage form amonia. You can try to measure the pH of the wet layer, should be acidic because of the leakgae from other acids bottles that condensate on inner walls. Most experiment results should be ok, but if you are working in an analytical lab, the results can be shifted.


Why does the 31P{1H} NMR spectrum of cis-[Mo(CO)2(dppe)2] show two signals? I've tried searching for literature references that explain this, finding only a single reference which refers to the two signals as originating from phosphine environments cis/trans to carbonyls, however I'm having trouble rationalising why this occurs.

In this complex there are two different 31P environments which are not related by symmetry:

The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted.

As extra proof, consider that the green P is cis to both carbonyl ligands whereas the purple P is cis to one and trans to the other.

They therefore have different chemical shifts and in the spectrum you would expect to see two different peaks. Presumably they would show coupling to each other, so I would expect two triplets, if we ignore satellites from all other nuclei (e.g. 13C).


What is happening when KCl burns in a bunsen flame? I did the experiment to qualitatively observe the different colours given off by different salts when placed in a flame.
But what is actually happening to the salt ?
Take $\ce {KCl}$ for example.
Does the $\ce{ KCl}$ react with $\ce {O2}$ ?
What does the reaction equation look like ?
Other salts include $\ce {LiCl}, \ce{BaCl2}$, $\ce{SrCl2}$, $\ce{CuCl2}$ and $\ce {NaCl}$.

I wanted to elaborate a little bit more so that the OP does not develop misconceptions. It is the potassium atoms not potassium ions that color the flame purple. While I was writing the answer, another poster emphasized the same point.
Caveat: In ordinary Bunsen burner type flames, emission from calcium, Sr, and Ba is from molecular compounds not atoms. The story of atomic emission is only true for alkali metals. The flame temperature is quite low in Bunsen burners. You need special flames to "see" the atomic emission of Ca, Ba, Sr.
Assume the processes when you introduce KCl in the flame:
a) The salt has to melt KCl (l)
b) It has to vaporize KCl (gas)
c) It has to decompose into constituents. Flames have free electrons, so potassium ions are converted to potassium atoms.
d) From the thermal energy of the flame, potassium atoms are excited. Once they de-excite, you see the violet/lilac color. It is an ultrafast process!
e) The flame temperature is so "low" on the scale of universal temperatures (stars, Sun) is that only 1-2% atoms get excited.
Now you may ask, what about the chlorine atoms, well, they emit in the deep ultraviolet and we cannot see them. The flame temperature is so low that it cannot excite chlorine atoms.
Keep in mind it is not a single wavelength. If you were lucky enough, you could use a pocket spectroscope and see red and violet lines (plus a persistent yellow line of sodium contamination).
Actually the strongest lines for potassium are 764 and 769 nm which are deep red. Many humans cannot see this deep red. So your flame would rather be deep red. Potassium atoms also emit less strongly at the other extreme end of the spectrum, corresponding to deep purple lines around 404 nm. The resulting "mix" of wavelengths appears to our eyes as lilac color!


Sign of work done by the system in first law of thermodynamics equation In the equation, $\Delta E = q - W$, according to my book, $W$ is positive when it is done by the system and negative when it is done on the system.

But in this problem:

  $\pu{3600 kJ}$ heat is entered in a gas cylinder, calculate the value of $∆E$ if the volume is not constant and work of $\pu{-800 kJ}$ was performed by the gas.

the sign is different. How should I make the equation?

I always got lost in remembering the rules for the signs, so I just ignore them and think about the two parts from the standpoint of the system (cylinder):

If heat enters the system, then it gets hotter so that contribution is positive. This part's fairly intuitive.

If I let the cylinder expand, it's relaxing to a less compressed state and so loses energy, which means the work contribution is negative. In order to compress the cylinder I have to push on it, adding some of my own energy to the system, so the work contribution is positive.

In your case, the change in energy from the heating (positive), and loss of energy through expansion (negative) is:
$$ \Delta E = 3600kJ - 800kJ $$

Edit: The problem itself is ambiguous regarding whether it's expansion of contraction. My answer assumes expansion since that's what would naturally happen upon heating a gas cylinder. Gaurang's answer is correct if you read it as contraction.


How does optimizing the molecular orbital coefficients in CASSCF improve its multi-reference capabilties? The complete active space self-consistent field (CASSCF) method is a special kind of multi-configurational SCF procedure, attempting to combine Hartree-Fock (HF) and configuration interaction (CI). In HF, the molecular orbital (MO) expansion coefficients are varied during the procedure, but the wave function contains just one Slater determinant. In CI, the multielectron determinant expansion coeffcients are varied, but the MO's used to build the excited Slater determinants are taken from HF, and are not varied during the energy-minimization.
During CASSCF, on the other hand, all MO expansion coefficients are optimized during the energy-minimization procedure, in addition to the multielectron expansion coefficients.

CI is able to recover the dynamical correlation of the system (configuration with large weight, many configurations with small weight), whereas (properly selected) CASSCF recovers the static correlation of the system (a few configurations with about the same weight, many configurations with small weight). By having a wave function that have some configurations with about the same weight, means that the wave function has some multi-reference character. I think that it is the MO optimization that leads to this multi-reference capabilities of CASSCF, but I'm not sure I understand why.

If can be shown mathematically, instead of just by concepts, that would be great.

I think you are maybe confusing how dynamical and static correlations are treated with different method. Also CASSCF by itself is not a multi-reference method.
CI in general is able to describe both dynamical and static correlation (FCI does at least). What is treating dynamical correlations (but not static) is the truncation scheme using different degrees of excitation, or example CISD, or in a similar way CCSD. Those are called single-reference methods, because they generate configurations based a single reference wave function.
To deal with static correlations you can for example use the Active Space approach (CASCI), or select configurations by hand (since you might not need to many here).
The term multi-reference refers to doing both, dynamic and static correlation, by generating excitations starting from multiple configurations. One first does a calculation for static correlation (usually MCSCF or CASSCF). Then a second calculation adds dynamical correlations by using the configurations from the first calculation (all or only the most important ones) as multiple reference points (e.g. MRCI-SD or MRCC-SD).
Optimizing the MO coefficients in a MCSCF calculation (e.g. CASSCF) is about optimizing the one-electron basis for your specific CI problem. Since you include more than just one configuration, the HF orbitals are not optimal anymore. In principle you could combine MCSCF with CISD as well, but usually CISD generates to many configurations making such an approach unfeasible.
TL;DR
The optimization of the orbitals in MCSCF does not directly improve the multi-reference capabilities. But it improves the description of static correlation, so you get a better starting point for your multi-reference calculation.
Maybe this explains better the need to optimize both sets of coefficients:

[...] The presence of several important configurations poses a difficult challenge for ab initio electronic-structure theory. The single-configuration Hartree-Fock approximation, by its very construction, is incapable of representing systems dominated by several configurations. By the same argument, methods designed to improve on the Hartree-Fock description by taking into account the effects of dynamical correlation, such as the coupled-cluster and Moller-Plesset methods, are also not suitable for such systems. Furthermore, to carry out a CI calculation 'on top of the Hartree-Fock calculation'is problematic as well since the Hartree-Fock model is inappropriate as an orbital generator: the orbitals generated self-consistently in the field of a single electronic configuration may have little or no relevance to a mutliconfigurational system.
An obvious solution to the multiconfigurational problem is to carry out a CI calculation where the orbitals are variationally optimized simultaneously with the coefficients of the electronic configurations, thereby ensuring that the orbitals employed in the wave function are optimal for the problem at hand and do not introduce a bias (towards a particular configuration) in the calculations. This approach is referred to as the multiconfigurational self-consistent field (MCSCF) method. [...] it may be used either as a wave function in its own right (for a qualitative description of the electronic system) or as an orbital generator for more elaborate treatments of the electronic structure.

Taken from the introduction to the MCSCF chapter in the book T. Helgaker, P. Jorgensen, and J. Olsen. Molecular electronic-structure theory. New York: Wiley, 2000.


Why search new elements? I've read this today. From the little I know, they discover new elements by bombarding some atoms in particle accelerators (I guess) and these new elements last a few miliseconds. So what's the insight they reveal to us? Why are these discoveries useful?

There are multiple reasons:

To improve our experimental prowess: There's no single catch-all process to create higher elements. While the basic process is to smash stuff together, once can't get much more specific than that. For each new element, the techniques used differ. And, as with any experiment, the complications which arise differ. Solving these lead to a deeper understanding of how to conduct experiments. Note that there have been gaps of years between the synthesis of various transuranium elements. This is not because scientists find it a useless task; it is because designing an experiment to synthesize these elements is hard.
To test stability: We have various nuclear models that can predict the stability of the elements. Creating new atoms helps validate these theories.
Islands of stability: Some higher elements are considered islands of stability in that they are more stable than their neighbors. If we can improve our experimental techniques to the extent that these can be generated, we may find some use for them. Of course, these may or may not be useful — these still are short-lived, just not extremely short-lived
Because we can!


What is the true caloric impact of Alcohol? I've only had 1 year of basic Chemistry at school, so I am waay too uneducated to answer this question myself:
What is the true caloric impact of ingested alcohol (ethanol) in the human body?

Here are a few things I presume to be fact:

When testing, how many calories a substance contains, it is done by setting that substance on fire, then measuring the missing oxygen, maybe the resulting CO2 and the heat.
Alcohol is poisonous to the human body, and cannot be used directly as an energy source
the human body has to expend energy to turn alcohol (and starch ect) into some usable form. (by oxygenating or hydrogenation or something alike)
every energetic substance we consume will be turned into adenosine triphosphate (ATP), which is the only 'energy carrier' our body uses when powering muscles

Conclusions I draw:

When we test things for calories, we get a good estimate of how much 'energy' it would serve us
yet the testing is just a guideline, as human digestion and internal biochemical processes are a bit more convoluted than the simple 'set it on fire' combustion
When we have two different substances with the same caloric value, consuming them might yield different energetic value to the body

So, how to formulate this mathematically:

tcv is the tested caloric value
ecv is the effective caloric value for the human body
tcv >= ecv
for those sugars, that can be used directly (e.g. converted into ATP): tcv = ecv
for most things, where internal biochems have to turn it into usable form : tcv > ecv
alcohol: tcv > ecv

So how energetically effective is alcohol to the human body, compared to sugars?

on a pure chemical level: ecv/tcv=? 50%? 70%?
considering neutralizing / storing / discarding unwanted byproducts

An interesting study, 'Can calories from ethanol contribute to body weight preservation by malnourished rats?', to quote:

  In terms of caloric content, ethanol is a unique drug whose oxidation by alcohol dehydrogenase (ADH) produces 7.1 kcal per gram of ethanol oxidized. However, it is not known to what extent these calories are converted into body weight. There is no controversy about the malnutrition produced by chronic ethanol intake in alcoholics. In addition to the known toxic effects on absorption and synthesis of essential nutrients (1), some important metabolic changes occur. Pirola and Lieber (2) reported that activation of the microsomal ethanol oxidizing system (MEOS) in hepatocytes of alcoholics may consume one third of the caloric content of ethanol and impair ATP synthesis. Lands and Zakhari (3) suggested the existence of a futile cycle in ethanol metabolism which would account for a loss of six ATPs per gram of ethanol oxidized. They proposed that ethanol may be oxidized to acetaldehyde, consuming three ATPs, and acetaldehyde may be again reduced to ethanol, consuming three extra ATPs.
  However, there is some controversy about what happens to ethanol calories when consumption is moderate and not related to the metabolic changes of the liver that characterize chronic ethanol intake. It has been suggested that calories derived from ethanol may function as do calories derived from carbohydrates, producing a thermogenesis value of about 10% (4,5). This thermogenic value may be higher than that of fatty food (3%) and lower than that of proteins (about 20%) (6).

  The contribution of calories provided by ethanol to body weight gain has also been studied but there seems to be no consensus about the data from animal and human studies.

  Moderate ethanol consumption seems to favor an increase in fat storage and could result in weight gain, especially in overweight individuals (7). In an epidemiological study involving 89,538 women and 48,493 men, Colditz et al. (8) reported a strong negative association between alcohol intake and body weight index for women and the absence of this association for men.

  Animal studies are consistent in reporting a decrease in the body weight of rats receiving ethanol solutions as the only source of liquids. Concentrations of ethanol as low as 5% (v/v), which are similar to the ethanol content of a Brazilian beer, or as high as 40% (v/v), solution similar to spirit drinks, are related to decreased body weight gain (9). Similar results have been reported for 20% (v/v) ethanol solution (10).

  Different results have been obtained for malnourished animals. Da-Silva et al. (11), studying rats which had been treated with ethanol for 90 days, reported a significant weight gain by malnourished rats (50% food restriction) drinking a 20% (v/v) ethanol solution when compared to malnourished rats drinking water. A more recent study (12) reported improvement in somatic and motor development and a decrease in the mortality rate of the offspring of malnourished rats drinking low doses of ethanol (5%, v/v). These data suggest that malnourished rats can benefit from ethanol calories.

  In summary, in spite of the large number of studies on the effects of ethanol in well-nourished animals and humans, there is still controversy about how well ethanol-derived calories can be utilized. Fewer studies are available about special physiological conditions such as malnutrition. Over the last few years, scientific research has mainly focused on obesity, an increasing problem in developed countries, which led us to the false belief that malnutrition was no longer a problem worth investigating. However, there are still 800 million malnourished people in the world (13). The decreasing interest of the scientific community in problems related to malnutrition has left many questions without an answer. Ethanol consumption and its consequences on the malnourished organism are among them.

  In view of the importance of malnutrition in Brazil - 22% of the population or 40 million people are malnourished (14), as well as ethanol consumption and alcoholism - 11% of Brazilian population are alcoholics (15), the aim of the present study was to assess the use of ethanol calories in a dose/effect model by evaluating body weight before and after the installation of malnutrition.

My take, ethanol is apparently somewhat unique and there remains some controversy about how well ethanol-derived calories can be utilized. Study classification factors include well-nourished vs. malnourished, low doses (5%) vs high doses (20%), producing significant variations in effects.


How to switch Mercury-vapor lamp between UVA and UVC bands? I have Mercury-vapor lamp emitting UV light. I am looking for the nice solution to cut off spectral lines for UVA or UVC, to make it one day to generate ozone and another day to use it for disinfection as UVC band lamp. I mean using precoated tubes and put mercury vapor lamp inside or using coated glass as a shield.

Related: What property allows ozone to absorb uv light

TL;DR - You need two different lamps, one that has a phosphor to absorb the 254 nm light, and one that doesn't. Alternatively, you can use one lamp, and use a filter to select the wavelengths.

Mercury vapour lamps emit a significant amount of both UV-C light (around 254 nm) and UV-A light (around 365 nm).  This includes fluorescent bulbs used for lighting a room. The ultraviolet emission is a fundamental property of the element, not some kind of engineering accident. Light at these wavelengths is harmful, so you don't want it flying all over the place. The solution? Coat the inside of the lamp with a substance that absorbs the UV light. In some cases (but not all), the glass of the lightbulb itself can absorb the UV light.

Except now you're turning lots of luminous energy into heat, which is wasteful. A smarter solution is to coat the lamp with a phosphor, which absorbs the UV-C and UV-A light and reemits the energy at longer wavelengths. Depending on the phosphor, the reemitted light will have a different spectrum.

If you look at a 254 nm UV-C bulb and a 365 nm UV-A bulb (while they're turned off..), you'll notice that the 254 nm lamp is completely see-through, whereas the 365 nm lamp often has a white powdery coating the inside. This powder is a phosphor, which is chosen to absorb the 254 nm UV-C light, while letting the 365 nm UV-A light through (as well as other lower energy visible wavelengths). Some 365 nm UV-A lamps also have a coating to absorb the any visible light generated, such that relativity pure invisible 365 nm light shines through. This is popularly called a "black light".

Note that this does NOT imply every mercury lamp with a phosphor coating blocks the UV-C light it generates.


stability of a negative charge in halogens and that of orbitals when we talk about stability of halogen anions we say,
$\ce{I-} \gt \ce{Br-} \gt \ce{Cl-} \gt \ce{F-}$, and the reason is that "the negative charge (the electrons) have more space to move about in the larger halogen". But when we talk about stability of a negative charge on an orbital, why do we say that the negative charge on the $sp$ hybridized orbital (which is the smaller one) is more stable than $sp^2$ and then $sp^3$ as the least stable where as the  $sp^3$ orbital is the biggest orbital among all of the three and can allow the electrons to spread out more.

When we talk about hybrid orbitals and stability of charge on them, the electronegativity factor plays a more prominent role. The sp hybridised orbital is more electronegative than the others, owing to Bent’s rule which states - Greater the s character of a hybrid orbital, greater is its electronegativity.
Therefore, species having the negative charge on an sp hybridised atom tend to be less basic.
(You might be interested in knowing that, an sp hybridised carbon atom is actually more electronegative than a nitrogen atom.)


Does an acetal need to have one hydrogen attached to its carbon? I was asked to identify the functional groups in this molecule above.

So far I have found an amide ($\ce{-NHAc}$), an ether ($\ce{-OBn}$), and a conjugated diene, but I was also told that there is an acetal ($\ce{RHC(OR')2}$).

However, in the only part of the molecule where I can see a carbon atom attached to two other oxygen atoms, the carbon atom is not attached to a hydrogen atom.

Where is the acetal group in this molecule?

According to the IUPAC definition, an acetal is any compound with a formula $\ce{R2C(OR')2}$ (R' ≠ H). In the past, "acetal" referred only to compounds of this type, which were derived from aldehydes (one R = H), however, it now broadly encompasses all compounds of this type, including those derived from ketones, which were formerly only referred to as ketals. Interestingly, there was a period of time in which use of the term "ketal" was discouraged (IUPAC says 'abandoned', but many chemists kept using it anyway), but it has now been re-embraced as a term, for the subset of acetals that are derived from ketones.

One last note - the molecule shown in this question is a "spiroacetal" - a type of acetal where the acetal carbon is the only atom joining two rings.


How to calculate the vapor pressure of water? $48~\mathrm{l}$ of dry $\ce{N2}$ passes through $36~\mathrm{g}$ $\ce{H2O}$ at $300~\mathrm{K}$, and this results in a water loss of $1.20~\mathrm{g}$. The vapor pressure of water is?

The solution says, that water loss is observed because of escape of water molecules with $\ce{N2}$ gas. Then it says that water vapours occupy the volume of $\ce{N2}$, i.e  $48~\mathrm{L}$, this part I am not able to understand. Why would water occupy the same volume as that of $\ce{N2}$, it isn't mentioned that these two are in some container of volume of $48~\mathrm{L}$. What is the logic behind this? Then the solution proceeds, by applying $pV=\frac{m\mathcal{R}T}{M}$.

It doesn't matter if there is a container.  I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.

The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption.

But adding 1.2 grams of water would increase the total volume by about 1 L.  I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%.


Molecules made of molecules? When a molecule of water ($\ce{H2O}$) and a molecule of hydrochloric acid ($\ce{HCl}$) react together, would they become a molecule made of molecules? What's the name for such thing? How would it be represented - just as $\ce{H2OHCl}$?

The key to your question lies in the word "bound". Without mentioning the state of your species we can say in general, species which are joined by covalent bonds are single molecules, there are countless examples of such reactions in organic chemistry where two small molecules condense into a single molecule.

Species with intramolecular forces (hydrogen bonds, van der waals etc) can be called bound or perhaps "clusters", which is quite a broad term. Real bonding however is not so easily defined as the molecular orbitals may  generate a wide spectrum of charge distributions etc. Quantum mechanics gives a much clearer description of the bonding of a particular molecule or extended system.

The terminology you write your $\ce{H2OHCl}$ is similar to how we write water of crystallisation in a solid. If this is where your idea came from, it should be remarked that this is a distinct concept.

To conclude; species that are covalently bound (that could have strong polarisation even to bordering ionic bonding) are molecules. Intermolecular forces are precisely that, between distinct molecules. We should remark that we can also have extended solids with covalent bonding (e.g. group 14). In addition some regard metals as molecules too, molecules whose molecular orbitals are distributed over the entire structure etc (research tight binding model for more information). But I feel that wasn't the focus of your question?


Small Propellanes Stability Why is [1.1.1] propellane more stable than other small members in the propellane family even though it seems to be under the greatest strain?  What makes a small propellanes stable, or in general, any propellane stable?

Fulfilling the request from @Mockingbird, there is a quote from (1, p. 90)

  Surprisingly, [1.1.1]propellane is somewhat more stable to thermal
  decomposition than the next larger propellane, [2.1.1]propellane,
  indicating a reversal in the trend of increased reactivity with
  increased strain. To understand this observation, it is important to
  recognized that the energy of both the reactant and intermediate
  influence the rate of unimolecular reactions that lead to
  decomposition. In the case of propellanes, homolytic rupture of the
  central bond is expected to be the initial step in decomposition. This
  bond rupture is very endothermic for [1.1.1]propellane. Because
  relatively less strain is released in the case of [1.1.1]propellane
  than in the [2.1.1]- and [2.2.1]-homologs, [1.1.1]propellane is
  kinetically most stable.(2)

  Another manifestation of the relatively small release of strain
  associated with breaking the central bond comes from MP4/6-31G∗
  calculations on the energy of the reverse ring closure.(3)

  The thermal decomposition of [1.1.1]propellane has been studied both
  experimentally and by computation(4)

The book chapter further discusses reactivity of [1.1.1]propellane, if OP is really interested in this area, I would suggest to grab the book in the library or focus on the papers (2-4) the book references to.

(1) Carey, F. A.; Sundberg, R. J. Advanced organic chemistry. Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
(2) Wiberg, K. B. Angew. Chem. Int. Ed. Engl. 1986, 25 (4), 312–322. DOI: 10.1002/anie.198603121
(3) Adcock, W.; Binmore, G. T.; Krstic, A. R.; Walton, J. C.; Wilkie, J. J. Am. Chem. Soc. 1995, 117 (10), 2758–2766. DOI: 10.1021/ja00115a011
(4) Jarosch, O.; Walsh, R.; Szeimies, G. J. Am. Chem. Soc. 2000, 122 (35), 8490–8494. DOI: 10.1021/ja994043v


What's the minimum amount of energy needed to remove CO2 from the atmosphere? I've been learning about CDR (Carbon Dioxide Removal) techniques to combat climate change. These all seem to involve extracting $\ce{CO2}$ from the atmosphere and converting to a mineral, which is then buried.
I am not a chemist, but something seems wrong about this to me.  Wouldn't the decrease in entropy need to balanced by an increase of entropy somewhere else, e.g., by using up a low entropy source of energy?
Is it possible to come up with a theoretical lower bound for the amount of energy needed to extract $\ce{CO2}$ as a proportion of the energy obtained by burning the fuel that released it in the first place?
If this is more than 100% then CDR schemes cannot possibly be an alternative to decarbonising our energy sector asap.  Even if it's less than 100% the calculation is still useful as it allows us to compare the true cost of fossil fuels with renewables, e.g., a value of 33% would indicate that the cost of fossil fuels is 50% higher than advertised (assuming we remove the $\ce{CO2}$, otherwise it's infinitely more expensive!)
(Edit)
Assume the original CO2 came from coal at 24MJ/Kg.
I’m not going to specify the mineral to which this is converted because I’m looking for a theoretical lower bound across all (scalable) technologies.

The OP is specifically interested in the efficiency of removing $\ce{CO_2}$ from ambient air, rather than from point sources like power plants.  The former is obviously less efficient because the $\ce{CO_2}$ is far less concentrated.
According to p. 50 of the the SAPEA* Evidence Review Report No. 2
(Schlögl, Robert, et al. Novel carbon capture and utilisation technologies: research and climate aspects. 2018.  Available at : https://www.sapea.info/wp-content/uploads/CCU-report-proof3-for-23-May.pdf),
"the thermodynamic minimum energy required to extract CO2 from ambient air is about 250 kWh/(ton $\ce{CO_2}$)."  This equals 0.99 MJ/(kg $\ce{CO_2}$).
The OP asks that we assume the $\ce{CO_2}$ came from a coal-fired plant that produces 24 MJ/(kg $\ce{CO_2}$).
Hence (assuming SAPEA's number, and the OP's, are correct), at at the theoretical maximum thermodynamic efficiency, the energetic cost of capturing $\ce{CO_2}$ from ambient air is 0.99/24 x 100 = 4% of the cost of the energy generated in producing that $\ce{CO_2}$ from a coal-fired plant.
Of course, this is not a lifecycle analysis.  One needs not only to determine the actual thermodynamic efficiency of the process, one also needs to account for the energetic costs of putting the infrastructure in place for the carbon capture, and for maintaining that infrastructure.
*For more information on SAPEA, see:
https://en.wikipedia.org/wiki/All_European_Academies


What is the meaning of the slope in a graded dose-response curve? A graded dose-response relationship is shown below with the drug concentration on the $\log$ scale on the $x$ axis and the response on the $y$ axis :

According to this website:

Slope:  a useful parameter - when it is shallow, it suggests a greater chance of overlap between desired effects and side-effects, whereas a slope which is too steep suggests that it will be difficult to achieve precise control of the effects.

I want to understand how does the slope as $\frac {Response} {concentration}$ have the meaning mentioned above?

Since the drug concentration axis is logarithmic, even a small change in the nominal drug concentration can have a large effect if the slope is steep, so getting the correct dosage could be problematic. Accordingly, this requires care in drug concentration preparation and, in particular, it might make it harder to avoid overdoses.
For a shallow slope, a relatively large change in drug concentration may be needed in order to achieve the desired result. But side effects of the drug do not necessarily follow the same dose response relationship. So increasing dosage by a factor of 10, for example, might pose a much higher than factor of 10 risk of side effects.


Can fluorine ever have a positive oxidation state? I know fluorine is the most electronegative element but can we humans ever synthesize fluorine in a positive oxidation state like +1?

As noted in the referenced question, fluorine is not in the +1 oxidation state in hypofluorous acid (which, incidentally, is the only hypohalous acid that has been isolated) nor in any other compound where its bonds give it a complete octet.  Even where the fluorine has a positive formal charge, attaching it to less electronegative atoms will lead to an oxidation state of -1.
However, it is possible for fluorine to show a zero oxidation state and for a fluorine atom to apparently show a positive charge.  As one might expect, this occurs in one of the most unusual compounds in all of chemistry.
Striking gold -- and a lucky 7
That compound is "gold heptafluoride", $\ce{AuF7}$, whose unusual structure is reflected in the IUPAC name: instead of gold(VII) fluoride, it is officially called difluorinegold(V) fluoride.  It is more accurately formulated $\ce{AuF5\cdot F2}$.
Himmel and Riedel [1] give calculations showing that unlike other heptafluorides, not all seven fluorine atoms are bonded to the central (gold) atom.  Rather, the gold is bonded octahedrally to five individual fluorine atoms and a difluorine ($\ce{F2}$) ligand, the latter attached end on and rotated so that in the most stable conformation (according to most of the several optimization schemes used), the remote fluorine atom is eclipsed with one of the other ligands as if attracted to it (a, below, picture from [1]).  The staggered arrangement one might expect (b) is instead a transition state in this fluxional molecule.  If we assume that the individual fluorine ligands are negatively charged as usual, this interaction indicates that at least the remote fluorine atom, labeled $\text{F}_b$, is positively charged.

And indeed, Table 2 of the paper shows the following charge distribution:

Thus the relatively electronegative gold atom, it's effective elecyltronegativity further enhanced by the other fluorine ligands, actually imparts positive charge to both fluorine atoms in the molecular fluorine ligand, as if there were a more electronegative moiety even than atomic fluorine.  Note the significantly greater portion of positive charge on the remote fluorine, the one which wheels around to attract one of the fluoride ligands in the eclipsed conformation.
To sum up, difluorine gold (V) fluoride has:

A difluorine ligand, apparently the first known case of thus

A gold-fluorine moiety, either $\ce{AuF5}$ or $\ce{AuF6}$ depending on how one views the structure, that is effectively more electronegative than fluorine itself

A resulting positive charge on some fluorine atoms.

Reference
1.
Daniel Himmel and Sebastian Riedel, "After 20 Years, Theoretical Evidence That 'AuF7' Is Actually AuF5‚F2", Inorganic Chemistry 46 (13), 5338-5342 (2007).


The history of ethylene According to the history section in Ethylene, it is written that:

the name ethylene was used in this sense as early as 1852

However, the earliest mention of it I could find was in 1859 (Wurtz, "Sur l'oxyde d'ethylene"; Comptus Rendus 48, p.101). I also found mentions of "ethylene" in 1862 (Wurtz, "[...]On oxide of ethylene, considered as a link between organic and mineral chemistry[...]"), and in 1878 (Frankland and Thorne, "[...]On the Luminosity of Benzol when burnt with non-luminous Combustible Gases[...]"). Can anyone find any earlier publications that mention "ethylene"?

According to the Oxford English Dictionary entry on "ethylene", the first English usage was in 1849 in the Chemical Gazette (alternate name Journal of Practical Chemistry), published in London. The specific citation is Vol. 7 p. 116 (Mar. 15, 1849). The sentence in which it is used is in an article by "A. Wurtz" titled "On a Series of Organic Alkalies Homologous with Ammonia.", and the relevant sentence reads "By adding to ammonia the elements of ethylene, $\ce{C4H4}$, ethylic ammonia, $\ce{C4H7N}$, is obtained."
The full text of the journal is available digitally via Google Books.


Nomenclature of aryl halides How do you know when to name a benzene ring substituted by a simple haloalkane as a derivative of benzene or as a derivative of the alkyl halide.

For example, is it 1-chloro-4-isopropylbenzene or 2-(4-chlorophenyl)propane?

In the usual substitutive nomenclature, simple halogen compounds are always expressed by prefixes (‘bromo’, ‘chloro’, etc.). Hence, the senior parent structure of such compounds is not determined by a principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the benzene ring is selected as the senior parent stucture for the preferred IUPAC name of the compound given in the question, which leads to ‘1-chloro-4-isopropylbenzene’. However, according to current recommendations, the preferred prefix for the isopropyl substituent is ‘propan-2-yl’; the prefix ‘isopropyl’ is retained for use in general nomenclature. Thus, the complete preferred IUPAC name is ‘1-chloro-4-(propan-2-yl)benzene’.


How would you measure the rate of this SN1 reaction? I am performing the following $\mathrm{S_N1}$ reaction:

$$\ce{t-BuOH + HCl -> t-BuCl + H2O}$$

I would also like to measure the rate of this reaction, but I cannot seem to find a way. I initially envisaged measuring the rate by using an indicator to tell when the conc. $\ce{HCl}$ had been used up.

However, I do not believe the reaction has gone to completion, thus why, when using my method of using an indicator, the indicator never changed colour, due to leftover acid - also Clayden's Organic Chemistry [1, p. 348] cites a 90% yield, presumably meaning not all of the acid reacts.

The reaction should take 20 minutes [1] at room temperature and so I don't think I can measure the rate by looking measuring a change in temperature. I do not know how else I could measure the rate.

Clayden, J.; Greeves, N.; Warren, S. G. Organic Chemistry, 2nd ed.; Oxford University Press: Oxford; New York, 2012. ISBN 978-0-19-927029-3.

If you want to measure it with an electronic device I would recommend some spectroscopic technique such as IR, Raman, 1H-NMR etc. or using a digital pH meter. For the spectroscopic techniques you should probably cool down the solution to minimize error during the measurement or alkalify it to avoid further reaction.
However if you want to avoid these methods and you have enough soultion-volume you could try titrating it with NaOH solution and a proper indicator.


Will lye damage vinyl pool liners? I have a nutty neighbor that said someone told them to put liquid lye into their above ground pool.  I said it would eat away the lining, their answer was to only use a little bit.

Lye (Sodium Hydroxide) is sometimes added to pool water to bring about increase in pH, without making significant increases to total alkalinity. It will bring about quite rapid changes in pH. I assume the intention is to bring the overall pH back within a sensible range of 7.4-7.6.

Adding high concentrations of sodium hydroxide (liquid lye is about 50% in water) might degrade the vinyl over long periods, but so too will high levels of acid, bleach, exposure to sun, and just about every other chemical used in pools. Vinyl liner can be damaged by low pH (below 7.0), but there will be other symptoms of high pH (above 7.8) (cloudiness, calcification, precipitation) long before damage to the vinyl liner kicks in, I think.

I think the nutty neighbour may be not too far from the mark to say "only use a little bit". All things in moderation.


Is there any difference between isotonic and isosmotic solution? Is there any difference between isotonic and isosmotic solution? Can we say isotonic solution do not show osmosis because they have same osmotic pressure?

Yes, there is a difference. Isotonicity implies a biological compatibility, whereas isoosmoticity implies similarity of chemical and/or physical composition. Solutions that are isoosmotic to biological fluids/blood are not necessarily isotonic as tonicity refers to a given cell membrane [1, p. 230]:

  The 0.9% sodium chloride solution is said to be iso-osmotic with physiological fluids. In medicine, the term isotonic, meaning equal tone, is commonly used interchangeably with iso-osmotic. However, terms such as isotonic and tonicity should be used only with reference to a physiological fluid. Iso-osmotic actually is a physical term that compares the osmotic pressure (or another colligative property, such as freezing-point depression) of two liquids, neither of which may be a physiological fluid, or which may be a physiological fluid only under certain circumstances. For example, a solution of boric acid that is iso-osmotic with both blood and lacrimal fluid is isotonic only with the lacrimal fluid. This solution causes hemolysis of red blood cells because molecules of boric acid pass freely through the erythrocyte membrane regardless of concentration. Thus, isotonicity infers a sense of physiological compatibility where iso-osmoticity need not. As another example, a chemically defined elemental diet or enteral nutritional fluid can be iso-osmotic with the contents of the GI tract, but would not be considered a physiological fluid, or suitable for parenteral use.

References

Gennaro, A. R. Remington: The Science And Practice Of Pharmacy, 21st ed.; Lippincott Williams & Wilkins: Philadelphia, 2005. ISBN 978-0-7817-6378-3.


Amount of Phenolphthalein required We just had chemistry practical few days a go. The topic was strong acid-base titration, in which we had to find the normality and concentration of $\ce{HCl}$.

My teacher instructed me to add 2 drops of Phenolphthalein  in 10 ml $\ce{HCl}$, and then add $\ce{NaOH}$ from buret and take a reading.

We needed to repeat the process 4 times and then find the constant reading, but in that I never used to take 2 drops of Phenolphthalein, instead I used to take 5-6 drops and then also used to get almost same reading as of pilot reading.

So my question is whether it is compulsory to take 2 drops of Phenolphthalein or it was just a instruction so students don't waste it?

In this same instruction given. (Procedure point 2)

Using a little extra phenolphthalein won't matter. You just need enough so that the color is visible once the pH becomes basic enough. If you were titrating a large volume, you might need more than a couple of drops, but 2 drops is enough to be sure that the color will be visible in a typical lab-scale titration vessel. One drop would probably even work. The reason they say "two drops" and not "at least two drops" is that they don't want you to be wasteful.


Converting a CIF file to a .com or .gjf file I would like to perform DFT calculations and extract atom coordinates for calculations in Gaussian.
I am using Mercury and have been able to isolate the atoms that I am interested in simulating. However, when I save the CIF files and open them again i see something very strange.
Atoms that have fractional occupancy from the XRD are shown to be fully occupied, this gives me nonphysical molecules that contain to many atoms. How can I get rid of these fractional occupied atom positions?
I am attaching an image to illustrate what I mean.

When i retrieve the data from CCDC's webpage, the fractional occupied positions are marked with crosses, is there a way to clean the file?:

In general, OpenBabel may convert some of the information of a .cif into a Gaussian .gjf.  If your are fine with using the CLI, the command would follow the syntax pattern of
obabel -icif input.cif -ogjf -O output.gjf
The parameters -icif and -ogjf hereby instruct OpenBabel about the expected input / output format of the files, while -O actually triggers the program to write the output into a permanent record.  Further information may be obtained here.
Depending on the computation intended, the crosses Mercury displays may be very valuable information.  It is Mercury's way to tell the user that the structure is disordered, i.e. there is more than one orientation of a molecule in the unit cell.  For example, that a conformational flexible $\ce{-CH3}$ or $\ce{-CF3}$ group has been refined into more than one position (think staggered / eclipsed conformation of butane, there are more than one energetic minima, and one is the absolute one.)  But this orientational disorder may extend over larger portions and complete molecules, too (e.g., small solvent molecules).  It would be better if you could display more of the structure, and tell a bit in addition about the intended computation.
Do you know to read a .cif file about site occupancy factors?  E.g., for molecules with an up / down or ($\uparrow$ / $\downarrow$) orientational disorder and only two realizations, these often are refined to add up to $p(\uparrow{}) + p(\downarrow) = 1.00$.


Why are phosphate-phosphate bonds weaker than phosphate-water bonds? According to the Wikipedia entry on ATP,

  ATP is an unstable molecule in unbuffered water, in which it hydrolyses to ADP and phosphate. This is because the strength of the bonds between the phosphate groups in ATP is less than the strength of the hydrogen bonds (hydration bonds), between its products (ADP + phosphate), and water

Why are the bonds between a water and a phosphate stronger than the bonds between two phosphates?

I'm studying first year chemistry, so an explanation that appealed to basic principles would be best.

(I've thought about this for a bit, and looked at the molecular diagrams in hope of seeing the answer. I'd include my thoughts on the matter in this question, however those thoughts amount to a half dozen inconclusive lines of speculation, so I opted to leave them out. Saying that to say, I have put effort into trying to figure out the answer, but I haven't gotten anywhere.)

ATP is an unstable molecule in unbuffered water, in which it hydrolyses to ADP and phosphate. This is because the strength of the bonds between the phosphate groups in ATP is less than the strength of the hydrogen bonds (hydration bonds), between its products (ADP + phosphate), and water acid anhydrides are generally prone to hydrolysis into two acid molecules.

Ignore that description Wikipedia just gave you, it is just blatantly bad from every perspective.

Phosphorous(V) is generally a strong electrophile and will be readily attacked nucleophilicly by water molecules — think of an $\mathrm{S_N2}$ mechanism. On the other side, $\ce{ROPO3^2-}$ is a good leaving group, too.

While the exact details may or may not be like I presented here, all acid anhydrides can easily hydrolysed similarly, be it disulphonic acid ($\ce{H2S2O7}$), organic acid anhydrides ($\ce{R-CO-O-CO-R}$) etc. Technically, the $\ce{P-O-P}$ bond in ATP/ADP is no different from the bonds in $\ce{P4O10}$ which we all know to be very readily hydrolysed.


Can there be two IUPAC names of a compound? What is the difference between propan-2-amine and 1-methylethanamine?

Generally, since many compounds can have two or more names in accordance with several methods recommended by IUPAC, a compound may be named correctly in more ways than one. Therefore, current IUPAC recommendations include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names.

In particular for primary amines, the rules given in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) are as follows.

  P-62.2.1.2 Primary amines, $\ce{R-NH2}$, are systematically named in the following ways:

  (1) by adding the suffix ‘amine’ to the name of the parent hydride;

  (2) by adding the name of the substituent group $\ce{R-{}}$ to the parent hydride ‘azane’;

  (3) by adding the name of the substituent group $\ce{R-{}}$ to the term ‘amine’ used as a preselected parent hydride name for NH3; this method is used only with monoamines.

  (…)

  Method (1) leads to preferred IUPAC names.

Since, according to the current recommendations, the substituent group $\ce{(CH3)2CH-{}}$ can be named ‘propan-2-yl’, ‘isopropyl’, or ‘1-methylethyl’ (where ‘propan-2-yl’ is the preferred prefix), the compound given in the question can have the following seven names in accordance with methods recommended by IUPAC. The PIN is ‘propan-2-amine’.

Method (1):
propan-2-amine (PIN)

Method (2):
propan-2-ylazane (propan-2-yl is the preferred prefix)
(1-methylethyl)azane
isopropylazane

Method (3):
propan-2-ylamine (propan-2-yl is the preferred prefix)
1-methylethylamine
isopropylamine


Balancing reactions algebraic method limitations How do I balance this reaction using the algebraic method?

Seems that the equation can't be solved.

How do I balance such reaction and does the algebraic method has limitations and wont work for balancing some reactions?

$$\ce{XeF4 + H2O -> XeO3 + Xe + O2 + HF}$$

$$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF}$$

$$4 a = f$$

$$2 b = f$$

$$a =  c +  d$$

$$ b = 3 c + 2 e $$

As per the OP's request, here is a solution. The equation is:

$$\ce{a XeF4 + b H2O  -> c XeO3 + d Xe  + e O2  + f HF  \tag 1}$$

So atom balance yields:

$$4a=1f     \tag 2$$
$$2b=1f     \tag 3$$
$$1a=1c + 1d     \tag 4$$
$$1b = 3c + 2e     \tag 5$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:

Assume $a = 1$, which is the minimum non-zero value it might have. Then, from equation $(4)$, $1 = a = c + d$, which is impossible because either $c$ or $d$ would have to be zero.
Assume $a = 2$. Then $b = 4$ because equations $(2)$ and $(3)$ show that $b = 2a$. But then equation $(5)$ shows that $4 = b = 3c + 2e$. Then if $c = 1, e = 0.5$, which is impossible. Obviously, $c$ cannot equal $2$, because that would give a negative value for $e$.
Assume $a = 3$. Then $b = 6$, because $b = 2a$, as above. Then equation $(5)$ shows that $6 = b = 3c + 2e$. Then if $c = 1, e = 1.5$, which is impossible. If $c = 2$, then $e = 0$, which is impossible. As before, $c$ cannot be greater than $2$, because that would give a negative value for $e$.
Assume $a = 4$. Then $b = 8$ and $f = 2b = 16$. Then equation $(5)$ shows that $8 = b = 3c + 2e$. If $c = 1$, then $2e = 5$, so $e = 2.5$, which is impossible. If $c = 2$, then $2e = 2$, so $e = 1$. Then equation $(4)$ shows that $4 = 2 + d$, so $d = 2$. This is a solution. Lastly, $c$ cannot equal $3$, because that would give a negative value for $e$.

Almost final result: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence

$$\ce{4 XeF4 + 8 H2O  -> 2 XeO3 + 2 Xe  + 1 O2  + 16 HF  \tag 6}$$

For a = 5, the solution is a = 5, b = 10, c = 2, d = 3, e = 2 and f = 20. This is the same as Oscar Lanzi's example solution. For a = 6, the solution is a = 6, b = 12, c = 2, d = 4, e = 3 and f = 24.

For a = 7, a double solution occurs: a = 7, b = 14, c = 2, d = 5, e = 4 and f = 28 and also a = 7, b = 14, c = 4, d = 3, e = 1 and f = 28.

There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is 11a.

Bonus: Mithoron provided a link to a problem Balancing disproportionation reactions answered by Ivan Neretin years ago. The equation that was to be balanced was:

$$\ce{a XeF2 + b H2O  -> c Xe + d XeO3  + e O2  + f HF  \tag 7}$$

So atom balance yields:

$$1a=1c + 1d     \tag 8$$
$$2a=1f     \tag 9$$
$$2b=1f     \tag {10}$$
$$1b = 3d + 2e     \tag {11}$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I first note that equations $(9)$ and $(10)$ show that $a = b$. Then equation $(11)$ shows that $b$ must at least equal $5$. Otherwise either $d = 0$ or $e = 0$, which is impossible.

Now I proceed by assuming $a = b = 5$. Then equation $(8)$ gives rise to the following four possibilities: $c = 1$ and $d = 4$; $c = 2$ and $d = 3$; $c = 3$ and $d = 2$; $c = 4$ and $d = 1$. If $c = 1$ and $d = 4$, then equation $(11)$ yields $5 = 12 + 2e$, which is impossible. Likewise for $c = 2$ and $d = 3$ and also for $c = 3$ and $d = 2$. But if $c = 4$ and $d = 1$, then equation $(11)$ yields $5 = 3 + 2e$, so $e = 1$. This is a solution: $a = 5, b = 5, c = 4, d = 1, e = 1,$ and $f = 10$. Hence the balanced equation is:

$$\ce{5 XeF2 + 5 H2O  -> 4 Xe + 1 XeO3  + 1 O2  + 10 HF   }$$

However, as Ivan Neretin beautifully pointed out in his answer, there is also the simpler redox reaction given by

$$\ce{2 XeF2 + 2 H2O  -> 2 Xe   + 1 O2  + 4 HF   }$$

So any linear combination of these, with positive integer coefficients, is also a valid balanced equation. For example, just add the last two equations together together:

$$\ce{7 XeF2 + 7 H2O  -> 6 Xe + 1 XeO3  + 2 O2  + 14 HF   }$$


Conformations of butane Would these be valid conformations of butane? I tried matching these up with the group but my drawings all seem rotated versus the book.
I'm fairly sure my anti conformer is correct, along with my gauche conformer. I'm not too sure about the two eclipsed conformers; from what I understand, there are two eclipsed conformations - one with strong torsional strain resulting from the methyl group interacting with the other methyl group, and the other resulting from the methyl groups interacting with the hydrogens, right?
Also, can we consider the dihedral angle to be the angle between the methyl groups?

Another problem is that what I've drawn doesn't seem to match with the book's energy diagram too. I'd expect the anti-conformer to be the lowest in energy, but I'd also expect a difference in E for the two eclipsed conformations. One should have more torsional strain from placing the two methyl groups next to each other and the other, less, because the methyl groups are eclipsing the comparatively small hydrogen.

Some years ago I did a (relaxed) rotational scan of n-butane for the purpose of showing that fatty acids are all-trans. That is also why I found the graph in Awesome's answer some kind of disturbing - it should form a sinus like wave. I include this graph here:

Edited out the error noticed by ron. Thanks again.


Lifetime measurement: double peak in instrument response function I am measuring flourescence lifetimes with a picosecond pulse laser, using the LifeSpec II with an MCP detector. Instead of a single peak rapidly decaying, as would be expected, my IRF contains two peaks. I am suspecting some sort of reflection causing this. Does anyone get what I mean here? Could add a graph soon if that might help.

EDIT: I have realized this might not be an error, but I would still like to know why.

I am suspecting some sort of reflection causing this. Does anyone get what I mean here?

The time between the two peaks looks like 0.4 nanoseconds.

This time (0.4 ns) should be compared to the dimensions of the vessel (chamber, cuvette, I don't know the right word) and the speed of light in the medium being observed (water, air, diamond).

So for example in vacuum or gas the light would travel 0.4ns x (0.3m/ns) = 0.12m.

In water where the index of refraction is 1.33 more like 0.09m.