On the exact definitions of Frenkel and Schottky defects The wiki page on vacancy defects says;

It is also known as a Schottky defect, although in ionic crystals the concepts are not identical.

As far as I'm aware, they're the same. Furthermore, the wiki page for Schottky defects contains the term 'vacancies' while describing the phenomenon, which is linked back to the page for vacancy defects. That seems contradictory.
Are they the same? And is the same true for Frenkel and interstitial defects? Or are the named ones subsets?

Well, you can say that they are kind of similar...the only major difference is that Schottky defects are only for ionic compounds, and the number of missing cations and anions must be equal. Vacancy and interstitial defects can be shown by non-ionic solids. Ionic solids must always maintain electrical neutrality. Rather than simple vacancy or interstitial defects, they show these defects as Frenkel and Schottky defects.
Vacancy Defect: When some of the lattice sites are vacant, the crystal is said to have a vacancy defect. This results in a decrease in the density of the substance. This defect can also develop when a substance is heated.

Schottky Defect: It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal.
Like a simple vacancy defect, the Schottky defect also decreases the density of the substance. The number of such defects in ionic solids is quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. In 1 cm3 there are about 10^22 ions. Thus, there is one Schottky defect per 10^16 ions. The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For example, NaCl, KCl, CsCl and AgBr. It may be noted that AgBr shows both, Frenkel as well as Schottky defects.

Also, with respect to your comment: "I thought Frenkel defects weren't vacancy defects but classified separately under interstitial defects."
Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. Frenkel defect is also called dislocation defect. It does not change the density of the solid. Frenkel defect is shown by ionic substances in which there is a large difference in the size of ions, for example, ZnS, AgCl, AgBr and AgI due to the small size of Zn2+ and Ag+ ions.

Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial defect. This defect increases the density of the substance.

Also, here's why AgBr shows both Schottky and Frenkel defects, even though Frenkel defects are shown by ionic substances in which there is a large difference in the size of ions, and Schottky defects are shown by ionic substances in which the
cation and anion are of almost similar sizes.

Schottky defect in AgBr is exhibited due to precipitation of both cations and anions. In AgBr, Ag+ ion is small in size and when removed from lattice point they can occupy interstitial sites and hence show both Frenkel and Schottky defects.


What is "ölsaurem Alkali" I read a technical German book that refers to "ölsaurem Alkali" which apparently means "oleic acid alkali". How could it be both an acid and an alkali?

They were writing it fancy in old times.
Sodium acetate would have been "essigsaures Natrium", potassium sulfate "schwefelsaures Kalium" and so on.

Therefor, "ölsaures Alkali" is an alkali salt of oleic acid (Ölsäure).


Reaction of aluminum hydroxide and nitric acid I've been focusing a lot on aluminum hydroxide recently, or specifically breaking it down into its base components. I know that when aluminum hydroxide dissolves in nitric acid it forms aluminum nitrate ($\ce{Al(NO3)3}$), which decomposes when it boils into aluminum and nitrogen dioxide. The end goal of this process is to recover the aluminum from the aluminum hydroxide, but that can't be done until the aluminum hydroxide dissolves into the nitric acid. I don't know if aluminum hydroxide readily reacts with nitric acid or not. Can aluminum hydroxide react with nitric acid readily? Or does it need to be catalyzed, or should I use a different reaction to isolate the aluminum?

Aluminum oxide could be the product of fusing Al(NO3)3 at high enough temperature. Aluminum metal would not be a product, in water or in fused salt.

2.Aluminum metal does not readily dissolve in nitric acid because of formation of passive oxide/hydroxide film. The nuclear power industry had need of large amounts of aluminum nitrate, and it was made by dissolving aluminum metal in hydrochloric acid (very fast) and treating the solution with nitric acid, boiling off the HCl.

Aluminum metal is often prepared for welding by removing the oxide film with dilute hydrofluoric acid. The aluminum surface is bare metal for, oh, maybe a microsecond, before oxidizing back to the oxide. But the aluminum oxide film will be very thin for a few hours and easy to weld. The oxide film grows thicker in air as it ages.


Reaction of succinic acid with ammonia When succinic acid is treated with excess $\ce{NH3}$, what will the mechanism be? I feel both $\ce{COOH}$ groups should be converted to $\ce{CONH2}$, but this is not the product.

From Synthesis of Succinimide on Thomas' Chemistry:

  Add 186 g of succinic acid to RBF, place the beaker in an ice bath and start adding 220 ml of 26% ammonia slowly. Assemble short path distillation without a condenser, add some boiling chips. Heat the mixture under a small flame to distil of all the water. When the temperature reaches over 100°C the mixture darkens. Switch the receiver and collect everything that comes off before yellow fumes temp 280-300°C. Cool down to room temperature and place the receiver into a freezer for 24h, recrystallize the crude succinimide in hot ethanol. Then filter the crystals and wash with 40 ml of cold ethanol.

(Note that RBF = round-bottom flask)

The catch is that the amide can attack itself to close the ring, forming succinimide.

There are two rounds of amidification via a nucleophilic addition/elimination mechanism.


Why can't equations of state be replaced by machine learning models? The Peng-Robinson equation of state, for example, has no underlying physical meaning, and is just a model that was fit to data. Computer scientists have developed much, much better models for data than messing around with polynomials and exponentials, like neural networks. Why can't standards organizations take a bunch of data about the chemical, fit a neural network or whatever to it, and publish that as the "EOS", instead of coming up with a meaningless equation that almost certainly wouldn't do as well?

For n data points there exists an n-1 order polynomial that perfectly fits the data.

Therefore there is no basis for "a neural network or whatever" being better.

Furthermore, it simply isn't true that the Peng-Robinson equation "has no underlying physical meaning".

The Peng-Robinson equation (like Van der Waals) recognizes that atoms/molecules occupy space and that attractive forces exist between them.

As Peng and Robinson say in their article:

  Since two-constant equations have their inherent limitations,
  and the equation obtained in this study is no exception,
  the justification for the new equation is the compromise
  of its simplicity and accuracy.

That being said, neural networks have been used to find equations of states.  See for example Equation of state and artificial neural network to predict the thermodynamic properties of pure and mixture of liquid alkali metals

  Generally, ANN is powerful and successful method for complex non-linear systems due to unique advantages such as high speed, simplicity and large capacity which reduce engineering attempt. In recent years, ANN modeling has been successfully used for prediction of thermophysical properties of pure and mixture fluids [24], [25], [26] and [27].


Carbonic acid-hydrogencarbonate buffer system pH calculation? I am learning about the pH of buffer solutions, and am slightly confused by my text book's calculation. For the blood's buffer system of carbonic acid and hydrogencarbonate ions as the conjugate case, the equilibrium is $$\ce{H_2CO_3 <=> H^+ + HCO_3^-}$$

The acid dissociation constant is $4.3 \times 10^{-7}~ \mathrm{mol/dm^{3}}$ and the $pH$ of healthy blood is 7.4.

So $$\ce{\frac{[H_2CO_3]}{[HCO_3^{-}]}=\frac{[H^+]}{K_{a}}}=\frac{1}{10.8}$$

I understand this calculation, however my textbook then concludes that this shows that there are about 20 times more hydrogen carbonate ions than carbonic acid. Should it not be 10 times? Why was this value doubled?

Your calculation is correct. However, there needs to be more bicarbonate in the blood because the majority of all metabolic processes tend to produce more acidic than basic products. There are also other ways to help the body maintain a pH of 7.4, not only the carbonic acid/bicarbonate buffer.

So both you and the book are right. If the calculations are done assuming there is just carbonic acid/bicarbonate acting as a way to maintian the pH in the blood, the ratio should be about 10 to one. Since there is a lot more going on though, the ratio is different than that.


Does clay absorb water because of polarity? From what I have recently learned about ceramics, clay is essentially sheets of silicon and oxygen, bonded by covalent bonds. Between them, water keeps clay from forming a full lattice of $\ce{SiO4^{2-}}$. What I am wondering is, does clay absorb the water because of the water's polarity, i.e. are water molecules bonding with hydrogen bonds to the negatively charged oxygen atoms of the ceramic sheets? Or is water there because pressure from the environment forces it in between the sheets (which I know would then be a physics question)?

I am no expert regarding clay (which is a surprisingly varied field) and will therefore give a more general answer. I suspect that there is also quite complex interaction with ions on the surfaces of the layers and such things.

Materials can be generally categorized on a rather continuous scale, the extremes of which are called hydrophilic and hydrophobic (or lipophobic/lipophilic which are synonyms).

If something is hydrophilic, it "likes" to be wet, i.e. it is energetically favourable for the surface to be wet because this minimized surface energy of the system. The opposite applies to hydrophobic substances.

As you have observed yourself, clay is quite hydrophilic.

One can indeed estimate whether a material is hydrophilic based on its chemistry: water likes to form hydrogen bonds and water complexes (e.g. around Na+). Therefore things are more hydrophilic the more of said things they have. A good example is that sugars and carbohydrates are very soluble in water despite being large, whereas even small alkanes are barely soluble.

The explanation involving air pressure makes no sense, because to move matter there has to be a pressure difference. But if you imagine clay as a porous substance, the pores might as well be filled with air and there is an equilibrium. In fact (as easily observed when burning earthware) clay gets more voluminous with higher water content.


The Workings of Pulse-Height Discrimination Pulse-height discrimination is a common technique to filter out unwanted background signals from the analytical signals. But how exactly does it work?

The analytical signals are not continuously detected, but detected rather in pulses. This is because the detector has a "dead time", and there is a pile-up of ion-pairs (or whatever the principle of detection is) in this dead-time. The intensity of the built up signal - the pulse - is then detected.

Pulse-height discrimination is just that; a discrimination of the detected pulse-heights. However, to be able to discriminate the pulse-heights, surely the system must know what a "desirable" pulse-height is, or, in other words, when a detected pulse deviates from the desirable? How exactly is this done?

In the context of X-ray diffraction, we know the desirable energy of the detected radiation: in theory, this energy should be equal to the K$\alpha_1$ of the X-ray tube's anode material (since diffraction is an elastic interaction, and hence no energy lose occurs). However, the intensity of the signal has nothing to do with the energy. How do we (that is, the ones developing the analytical system) know which pulse-heights are relevant and which are irrelevant to our analysis?

I hope have presented my question clearly, and I appreciate all help. Finding free references online was not very easy, in my experience.

Pule Height Analaysis (PHA) is done on the current pulse from the detector. It only works for detectors which have a current pulse which is proportional to the energy of the x-ray (scintillation counters and gas proportional counters). It does not work for Geiger counters where the current pulse doesn't depend on the x-ray energy.

Think of dropping the current through a resistor. The current pulse then has a voltage. Basically you set a low and a high voltage (a window) for the detector pulses which you will count. Any detector signals within the window are counted, those outside the window (too low or too high) are rejected.

Given that the detector current (or voltage really) should be proportional to the energy of the x-ray detected this improves the S/N ratio. So "low" current pulses don't have enough energy to be a Cu K-alpha x-ray, and high energy pulses have too much energy. High pulses could be for example an x-ray with twice the energy.

It is important to note that the rejected pulses are not "background" counts per se. But there is a twist. The rejected pulses do create detector deadtime. So you have to be careful that the overall count rate of the detector is low enough to still be linear, or do a correction for the total count rate.


Why does solvent leveling occur? Solvent leveling means that strong acids completely dissociate in that solvent so it is impossible to distinguish between two acids that completely dissociate in it. A solvent levels and acid if the $K_\text{a}$ of that acid is greater than 1. However, that’s not completely dissociated. An acid can have a $K_\text{a}$ of 2, 3 or 25 what’s the problem with being able to distinguish between them? In truth, whatever the $K_\text{a}$ it’s not actually COMPLETELY dissociated.

Your final statement:

  in truth, whatever the $K_a$ it's not actually COMPLETELY dissociated

is exactly right.

Strong acid is very often stated to be "completely dissociated", but that's just wrong.

For example, nitric acid has a $K_a$ of 24, so 1M nitric acid about 96% dissociated. If the concentration is higher, it will be less dissociated and if the concentration is lower it will be more dissociated.

I would think of the solvent leveling effect this way:

Suppose you have 1 liter of 0.1M ethanol in water and you add 0.1 moles of sodium amide thinking that it will deprotonate the ethanol.  That's not going to happen because most of the protons that protonate the amide will come from the solvent (water) rather than the ethanol.


Existence of Hyponitrite Molecule Does the hyponitrite molecule exist? I know the hyponitrite ion exists (http://en.wikipedia.org/wiki/Hyponitrite) but what about the molecule?

I have found no reference to the molecular form on Wikipedia. But it does seem that such a molecule could at least come together through the dimerization of nitric oxide.

Yes, see Structural and Stablity of Nitric Oxide Dimer.

Just don't call it "hyponitrite molecule", it might get offended.


Explain the structure of HNC (Hydrogen Isocyanide) What are the different types of bonds present in the structure?

Does Nitrogen donates one electron from its lone pair to carbon and acquires a positive charge?

How do I determine the Oxidation state of Nitrogen and Carbon?

In such type of questions, we can't directly determine Oxidation state by the normal algebra method.

First you have to draw the structure.

1.In a covalent bond, the more electronegative element gets a contribution of -1and the other element gets a contribution of +1 charge(valid for both sigma and pi bonds. Treats double double bond as two single bonds and triple bonds as three singles bonds).

Bonds between same element have contribution 0 to each element.
In a coordinate bond, if the bond is directed from less electronegative element to more electrononegative element, the more elecronegative element gets a contribution of -2 and the less electrononegative element gets a contribution of +2.

4.in case the coordinate bond is directed from more electrononegative to less elecronegative elements, the contribution is 0 to each element.

5.To these 4 steps for each and every bond. The value of Oxidation state of an element is sum of all contributions due to all other elements/bonds.

In your question, due to C-H bond, carbon gets -1 and hydrogen +1. Due to the triple bond between C and N, N gets -3 and C gets +3.

Summing all,

Oxidation state of hydrogen is +1

Oxidation state of carbon is -1+3=+2

And oxidation state of nitrogen is -3


reference states for activities I wanted to ask clarifications about a passage in Atkin's physical chemistry book, chapter 9, in the paragraph 9.2 description of equilibrium.

For studying equilibrium of a reaction, where $\nu_i$ are the stechiometric coefficients it imposes $\sum_i \nu_i \mu_i=0$ with $\mu$ the chemical coefficients. At this point it writes the chemical coefficients as a function of the activities $\mu_i=\mu_i^\circ+RT\ln(a_i)$ and identifies the term $\sum_i \nu_i \mu_i^\circ$ with the standard reaction Gibbs energy. I do not understand this identification. I thought that the standard state for the definition of activities (even dependent on wether the substance is a solute or a solvent) was not the same reference state for computing the reaction Gibbs energy (and does not depend on the role played by the substance in the reaction). I'm surely getting something wrong...

$\mu_i=\mu_i^o+RT ln(a_i)$ is the definition of activity ($a_i$).

Therefore, the standard state for standard chemical potential ($\mu_i^o$) is necessarily part of the definition of activity.

For a pure liquid or solid, the standard state is the pure substance at 1 bar.

For a liquid solvent, the standard state is also the pure liquid at 1 bar.

For a gas, the standard state is the fictitious state of an ideal gas at 1 bar.

For solutes, there are multiple conventions, for example the fictitious state of the solute having a molality of 1 mole/kilogram and exhibiting the behavior it does at infinite dilution.

$\Delta G^o = \sum_i \nu_i \mu_i^o$ by definition.

So there is no one standard state for $\Delta G^o$, because the various substances may have different standard states.


Does Glycerol give silver mirror test with Tollen's Reagent? I did an experiment in my school lab in which we had to identify the given organic compound. In my Lab Manual, it's written:

  Take 2 ml of Tollen's Reagent and add a few drops of organic sample. Heat the contents on boiling water bath for 5 minutes. Silver Mirror is formed. The given organic compound is Glycerol.

As far as I know, the compounds with aldehyde groups in them give silver mirror. Also those which contain $\alpha$,$\beta$-hydroxy ketone are known to react with Tollen's Reagent. I have never heard of Glycerol reacting with it. I searched but couldn't find this thread anywhere. Though I found that on heating Glycerol, it gives us Acrolein which does contain aldehyde group, which I think is the reason but I am not sure.

Your guess of formation of Acrolein is a pretty good one, as indicated in this answer, but a boiling water bath will probably not give you the 160°C temperature you are looking for to convert glycerol to acrolein

This question is most similar to yours and has a pretty good answer by bon. The answer says:

  However, from personal experience, gentle heating of a primary alcohol with Tollens' will cause a small amount of oxidation and result in a fine black precipitate.

It is most likely that Glycerol is undergoing a small amount of oxidation to any of it's oxidation products (most probably glyceraldehyde, since that gives tollen's test) and this oxidized compound is giving a positive tollen's test.


Does fumaric acid form maleic anhydride upon heating? Maleic acid (the cis isomer of the dicarboxylic acid $\ce{C4H4O4}$) readily forms its anhydride on heating. Clayden's organic chemistry textbook mentions this as a point of distinction between it and fumaric acid (the trans isomer), which fails to dehydrate to its anhydride upon heating. However, I have discovered that at temperatures above 200 $^{\circ}\mathrm{C}$, fumaric acid forms maleic anhydride as well.

Which is the correct answer? I don't think it is likely that fumaric acid can form the anhydride - even at higher temperatures - given the trans configuration and restricted rotation.

Fumaric acid does form maleic anhydride due to partial combustion. From wikipedia article of fumaric acid:

  Fumaric acid converts to the irritant maleic anhydride, upon partial
  combustion.

Also, dehydrating fumaric acid with a  dehydrating agent also forms maleic anhydride. This paper describe how to convert $\ce{C^{14}}$ labelled fumaric acid to maleic anhydride.

  Maleic anhydride-2-$\ce{C^{14}}$ is readily obtained from fumaric
  acid-2-$\ce{C^{14}}$ through dehydration using phosphorus pentoxide
  getting 96% yield under controlled condition.


Aqueous copper and how it reacts with NH3 I have learnt that you can test whether an unknown metal is copper by doing a simple test tube reaction. The reaction involves adding aqueous $\ce{NH3}$ to the test tube which contains the copper.

The $\ce{NH3}$ as it is dissolved in water will form $\ce{NH4+}$ ions and $\ce{OH-}$ ions. The $\ce{OH-}$ ions react with the copper forming a precipitate $\ce{([Cu(H2O)4(OH)2])}$ but many other metals form a precipitate with $\ce{NH3}$ as well such as $\ce{Fe^2+}$, $\ce{Fe^3+}$ etc. So to make sure that it is copper present in the test tube you add an excess of $\ce{NH3}$ as this causes a ligand exchange reaction to occur producing $\ce{[Cu(H2O)2(NH3)4]2+}$.

Why doesn't this occur with the other metals? Why doesn't this occur for say $\ce{Fe^2+}$ to form $\ce{[Fe(H2O)2(NH3)4]2+}$ which would also be soluble?

[Fe(H2O)6]2+ is a high-spin d6 complex, while [Fe(H2O)2(NH3)4]2+ is a low-spin d6 complex. The ligand substitution reaction is not favourable in terms of spin, so aqueous Fe2+ does not react with excess ammonia.

However, both [Cu(H2O)6]2+ and [Cu(H2O)2(NH3)4]2+ are d9 complexes. Their spins are the same, and being a stronger field ligand, NH3 increases the crystal field stabilization energy, so ligand substitution is favourable.


Geometrical Isomerism of Metal Complex Does $\ce{Ni((en)2(Cl)2)}$ show cis-trans form ?

And if No

Why not ?

$\ce{en}$ is bidenate so it has to take two neighbour corners from octahedron (let say equatorial position). The second $\ce{en}$ can bind with 1st nitrogen also on equator but then the 2nd has a choice: it can bind in equator or on axial position. So in theory it would be possible to see two isomers. The reality has however more to do with energy. $\ce{en}$ is very good ligand and $\ce{Ni++}$ makes very stable quadratic planar complexes. For that reason both $\ce{en}$ goes into equatorial position and the weaker ligands $\ce{Cl-}$ occupy axial positions.


The difference between elementary and concerted reactions In the IUPAC Gold Book:

elementary reactionA  reaction  for  which  no  reaction  intermediates  have  been  detected  or  need  to  be  postulated  in  order to  describe  the  chemical  reaction  on  a  molecular  scale.  An  elementary  reaction  is  assumed  to  occur in a single step and to pass through a single  transition state.
concerted reaction A  single-step  reaction  through  which  reactants  are  directly  transformed  into  products,  i.e.,  without involvement of any intermediates.

What is the difference between them?

The difference between SN1 and SN2 reaction mechanisms illustrates this quite well.

An SN1 type of reaction can be decomposed into two elementary steps:

formation of the carbocation
substitution of the nucleophile

The rate determining step is the formation of the carbocation, but we do have two distinguishable steps and a reaction intermediate - this is where the other mechanism comes in.

Now, when SN1 is not possible due to - for example - the carbon atom in question being primary, we can have a concerted mechanism - both of the steps mentioned before occur simultaneously, thus avoiding the energetically unstable primary carbocation.

This process can not be broken down into single steps, as it is necessary that the formation and the substitution occur simultaneously. Also, there is no reaction intermediate - hence it is a elementary reaction, where two things happen in a 'concerted' way in a single step.

Also, the experimental rate law will show that the rds is different and the rate depends on both reagents.


Using thermodynamics to predict the acid-base character of fluoride ion in water I was working through the end-of-chapter exercises of the acid-base chapter in Shriver's Inorganic Chemistry when I came across the following problem:

4.7. The effective proton affinity $\ce{A^{'}_{p}}$ of $\ce{F-}$ in water is 1150 kJ $\ce{mol^-}$. Predict whether it will behave as an acid or base in water.

I was initially stumped by this problem since the $\ce{A^{'}_{p}}$ of $\ce{F-}$ is between the $\ce{A^{'}_{p}}$ values of water and hydroxide ion (1130 kJ mol- and 1188 kJ mol-), which tempted me to say $\ce{F-}$ is amphoteric in neutral water even though I know it's basic from general chemistry.
The answer that the book's solution manual provides is the following:

The proton affinities of H2O (l) and OH-(aq) are 1130 kJ mol- and 1188 kJ mol-, respectively. Consequently, the reaction below between $\ce{H3O+}$ and $\ce{F-}$ is exothermic by -20 kJ mol- (1130 kJ mol- minus 1150 kJ mol-):
$\ce{ H3O+ (aq) + F- (aq) -> H2O (l) + HF (aq)}$
The reaction below between $\ce{H2O}$ and $\ce{F-}$ is endothermic by 38 kJ mol- (1188 kJ mol- minus 1150 kJ mol-):
$\ce{ H2O (l) + F- (aq) -> OH- (aq) + HF (aq)}$
Therefore, in neutral water, $\ce{F-}$ will behave as an base.

I understand that the first reaction has an approximate enthalpy change of -20 kJ mol- and is favored because it is the reverse reaction of the following reaction between $\ce{HF}$ and $\ce{H2O}$ and reaction between acids and $\ce{H2O}$ are favorable when the $\ce{A^{'}_{p}}$ of $\ce{H2O}$ is greater than the $\ce{A^{'}_{p}}$ of the conjugate base:

$\ce{ H2O (l) + HF (aq) -> H3O+ (aq) + F- (aq)}$

However, I have a few points I don't understand:

Why does the text consider the second reaction with $\ce{H2O}$? Doesn't $\ce{F-}$ act as a base in both reactions?
If both reactions occur and $\ce{F-}$ acts as a base, then which one is dominant?
If the first reaction is dominant, then how does it proceed given the low concentrations ($\ce{1 x 10^{-7} M}$) of $\ce{H3O+}$ in neutral water?

The textbook walks us through the interpretation of these enthalpies (they also caution us not to use enthalpies as sole criterion for spontaneity of reactions):

So they are asking whether water reacts with the acid HA, and whether water reacts with the conjugate base $\ce{A-}$.

If we try that for the case at hand, we first consider whether HF reacts with water:

$\ce{H2O(l) + HF(aq)->[?] H3O+(aq) + F−(aq)}$

Here, water and fluoride are the bases "fighting for the proton" (reactants: fluoride has the proton; products: water has the proton). Fluoride wins (i.e. HF does not give away the proton) because it has the higher proton affinity, and the reaction does not proceed judged by enthalpy alone.

Second, we consider whether the fluoride ion reacts with water:

$\ce{H2O(l) + F-(aq) ->[?] OH-(aq) + HF(aq)}$

Here, fluoride and hydroxide are the bases "fighting for the proton". Hydroxide wins because it has the higher proton affinity, i.e. this reaction also does not proceed judged by enthalpy alone.

So HF is not a strong acid, and fluoride is not a strong base. Another way of saying this is that hydroxide is a stronger base than fluoride, and hydronium is a stronger acid than HF.

In the textbook walk-through, they conclude that HA is a strong acid and - for a different reaction - $\ce{A-}$ is a strong base. In our case, HF is a weak acid  and fluoride is its conjugate weak base. If you try the same question with HCl, you will see it is a strong acid and chloride is not a strong base. (I don't understand the conclusion of the solution manual - they might have gotten confused because they know fluoride is a weak base, but their calculation together with the textbook argument merely shows that it is not a strong base.)

Now to your questions:

  Why does the text consider the second reaction with H2O? Doesn't F− act as a base in both reactions?

I think they ask "does HF react with water?", and "does fluoride react with water?". I don't know why the answer flips one of the reactions (maybe the solution manual wasn't written by the textbook author, and the original intent of the question got lost).

  If both reactions occur and F− acts as a base, then which one is dominant?

There are three equilibrium reactions, the two shown, and auto-dissociation of water. I don't know the kinetics of the three, but they are all fast and the three equilibrium constants are linked because if you subtract the two reactions you show, you get the auto-dissociation of water. For the equilibrium state, it does not matter which set of two you consider.

  If the first reaction is dominant, then how does it proceed given the low concentrations (1x 10−7 M) of H3O+ in neutral water?

Reaction rate constants vary by many orders of magnitude. The reaction of fluoride and hydronium is nice because the reactants are of opposite charge, so maybe it has a high rate constant. The reaction of fluoride and water is nice because water is the solvent, so it is present at high concentration and already at close proximity. I could not tell you which reaction is faster even if I know the concentrations of all the reactants.

Here is the table of proton affinities as a quick reference:


ELSD detection principle While documenting myself on the detectors used in HPLC, I came across one that I did not know: ELSD.
I understood that the sample with the mobile phase was nebulized at the column outlet but I do not understand what is measured next.
Could you clarify this for me?

ELSD is an acronym for Evaporative Light Scattering Detector. It is used as a quasi-universal detector for HPLC.
The first step is the nebulization of the column effluents with a dry gas as you have mentioned. If the column effluents contain any non-volatile analyte, then those will form minute particles. The mintue particles become suspended in the gas stream and form an aerosol. The gas stream is made to pass through a focused light beam (usually a diode laser) which is perpendicular to the flow direction.
If aersol is present, the light will be scattered, and can be detected/analysed by a detector photodiode which is placed perpendicular to both the light beam and flow direction.
There is a nice picture that I got from this site:

Reference- Analytical Chemistry, G. D. Christian, P. K. Dasgupta, K. A. Schug, 7th ed., Wiley, 2013


What is cheap material that absorbs water in packaging? I am thinking cheap material that can be used between two layers of carton that absorbs water.
I want to use air but I still think that there may be other materials available that

absorb water
can be easily removed when the outer carton layer is damaged

Possible materials

carbon (well absorbing)
air (cheap and easy)
sand (should be bad)

What is cheap material that absorbs water in packaging?

What you are looking for is most probably silica gel.
http://en.wikipedia.org/wiki/Silica_gel

It is a highly water absorbent material with large surface area. It is often used in packaging as well as in laboratory in desiccators.

You can actually buy it packed or in bulk easily (let me not link explicit offers here). You can regenerate used (old) silica by heating / drying / pumping, so you don't need to buy every single time new one if the old gets less effective.


SN1 reactivity order The question is to compare the SN1 reactivity for the following compounds. According to me it should be like R>S>Q>P.

But the correct answer is Q>P>R>S. How can a lone pair donating into a vacant p be more stable than allyl/benzyl carbocation?

These are the four compounds in the order PQRS respectively:

Benzyl and allyl cations are stabilised by the neighbouring $\unicode[Times]{x3C0}$-systems. The electrons in the $\unicode[Times]{x3C0}$-system are actually required for (formally) form double bonds to ensure all-octet structures as well as possible. The cationic structures will contain an electron sextet somewhere (most often drawn on the benzylic carbon for the benzyl cation).

The $\unicode[Times]{x3B1}$-oxygen and $\unicode[Times]{x3B1}$-nitrogen systems on the other hand contain free lone pairs on the heteroatoms. These are not used for double bonds, so they can be shared with the cation to form a new double bond. The resulting mesomeric structure is an all-octet structure which are usually much more stable than non-octet structures (at least for main group elements).

The nitrogen-containing carbocation could even, after the double bond has been formed, loose its proton to create a neutral all-octet structure, the formal $\ce{HCl}$ elimination product.


Why does this reaction produce an alcohol in one case and an alkene in the other? Does it have something to do with the fact that the chlorine atom is bonded with the primary carbon, or with higher temperature in the upper reaction?

Please also see Why substitution and elimination reactions are favored respectively at lower and higher temperature?
The reaction conditions are the clue you are looking for.

The first thing to recognize is that in any situation where there is competition between substitution and elimination, you will always get a mixture of products.  The question is, which one will dominate.

The effect of temperature is subtle, and sometimes misleading. In practice, organic chemists heat many (maybe even most) reactions, in order to get them to occur in a reasonable timeframe.  When we are teaching about the effect of temperature on the competition between substitution and elimination, we often indicate heating ($\Delta$) is present for the elimination and not the substitution.  You should take this to mean that higher temperature favours elimination.

Now, subtleties aside, let's look at why heating favours elimination.

One difference between the two situations is the relative changes in entropy.  We can see that in the case of elimination, the overall entropy of the system is increased after the reaction (there are more different particles) while in the case of substitution, the entropy change is negligible.

We can consider the competition between the two reaction pathways like this:

$\Delta G_{S_N2} = \Delta H_{S_N2} - T\Delta S_{S_N2}$

and

$\Delta G_{E2} = \Delta H_{E2} - T\Delta S_{E2}$

At low temperature, the enthalpy term ($\Delta H$) dominates, while at higher temperature, the entropy term ($\Delta S$) dominates.

Therefore, the increase in temperature favours the reaction with the larger entropy contribution.


What is the use of both mol and mass representing the same thing i.e. amount of something? A mol is a unit but also has a magnitude. How can a unit have a magnitude? And if mole is used to measure an amount of substance than what does mass measure?

Mole itself doesn't have a magnitude, but the value of mole does, in fact it is the magnitude. Any measurement has a magnitude (ie for the mass of 5 kg the magnitude will also be 5).

I think you are confusing it with scalar measurements (ie forces) which also have a direction. For example 5 Newtons 'in that direction'.

Mole is similar to a dozen (just a bit larger). It is used to count how many atoms, molecules, atoms or anything else there are.

If you think about mass as "the amount of stuff there is in an object" you can confuse it with mols

but the actual definition of mass is:

  A property of a physical body and a measure of its resistance to acceleration (a change in its state of motion) when a net force is applied.

which is different to mols (which is used to count objects).


Lowercase "a" in organic chemistry nomenclature? In the IUPAC name for geosmin has a lowercase a mixed into it:
(4S,4aS,8aR)-4,8a-dimethyloctahydronaphthalen-4a(2H)-ol.

Does anyone know what it stands for or can reference me to a source that does?

'4a' and '8a' are locants which refer to a specific atom. They are fundamentally no different from the numbers '1', '2', '3', '4'... which are also locants referring to specific atoms.

The locants in geosmin come from the locants in naphthalene, which are assigned based on the following excerpt from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

  P-25.3.3.1.1 The numbering of peripheral items in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter 'a', 'b', 'c', 'd', etc.


Anthocyanins as pH-indicators: stop the titration at red, purple or blue? If you titrate some hydrogen bromide solution (0.1 M) to some sodium hydroxide in order to determine the exact concentration of hydrogen bromide (0.1 M), and you use anthocyanins as a pH indicator. This anthocyanin will be red if you have a low pH value and purple if it is slightly acidic until it is slightly basic. It will be blue if the solution is basic.

Do you have to titrate until the color of the solution is blue or purple?

Equilibrium from red to purple: $$\ce{H2X+ <=> HX}$$

$K_a = 1.047*10^{-3}$

Equilibrium from purple to blue: $$\ce{HX <=>X^{−}}$$
$K_a = 3.16 * 10^{-8}$

If you titrate some hydrogen bromide solution to some sodium hydroxide [...]

It's probably a bad idea to start with the anthocyane in strongly alkaline solution. Like other pyrylium compounds, flavylium cations do not tolerate (strongly) akaline conditions very well.

The nucleophilic opening of the pyrylium ring to a chalcone is irreversible under the conditions of the titration.


What does "/2" in the structure of calcium benzoate mean? Please see the following figure. This is from the high school chemistry book of my son. He wants to know what does that /2 after $\ce{-COOCa}$ mean.

The book is a non English one. And I have changed the relevant caption to English.

Primary searches over Google have not revealed anything usable. And my major not being Chemistry might have to do something with this.

It means half of calcium.

The organic acid is $\rm{COO}^-$, whereas calcium is $\rm{Ca}^{2+}$. For the structure to be fully correct, there are two benzoic acids per one calcium. Other way round this would be $\rm{(C_6H_5COO)_2Ca}$. Full drawing of this will be way to messy, so it is simpler to put half calcium to one acid.


Tagging functional group Is it a standard practice to name functional group by the greek alphabet of the carbon it is attached to?

Recently read a question in which the order of translation (protein synthesis) was asked and there were a couple of options with the $\ce{NH2}$ was labelled as $\ce{\alpha-NH2}$ and $\ce{COOH}$ as $\ce{\alpha-COOH}$.

I read wiki, it says H atoms are labeled in that way but are functional groups too labeled in the same way?

In the study of amino acids (and peptides, proteins etc), it IS standard practice to label the alpha carbon as such. More carbons leading into the R-group may also be labeled with Greek letters, but this is not as necessary as having a label for the stereogenic center, the alpha carbon. Personally, I've found Greek letters helpful in remembering the structures of the different amino acids, for example I always think of Lysine as "the one with the epsilon-amino side chain". Note that in chemistry in general, whose nomenclature follows IUPAC guidelines, it is NOT "standard" to use Greek lettering, however it is still generally recognized. The practice is probably most prominent in the study of fatty acids (another biochemistry subfield), which results in such Greek -rich categorizations as "omega-3 unsaturated fats".


How to replicate compositions of discontinued paints? I am a house painter, over the last 20 years I've noticed that discontinued lines of paint are much better. These paints were vinyl acrylic. Now the vinyl has been removed. I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula? I think the old ingredient was polyvinyl acetate.

I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula?

No.

Paints are a specific mixture of a number of components designed to fulfill their role well in this specific combination. If you go and add a random additional ingredient, which in itself is not a pure substance but a nondisclosed mixture, you cannot predict what will happen. At best, the additional ingredient does no good and no harm. At worst, you can throw away the entire can of paint.

Paint companies also do not disclose key data about their mixtures for the obvious reason that they don’t want to be copied. So even though it may say X, Y and Z on the back of the can, that doesn’t mean you can infer much about the mixture. Even if they give percentages, you don’t know what all the rest is. Therefore, you cannot duplicate the old formula unless you have access to the full blow of analytical methods and a sample of the old paint.

Finally, I’m not even sure if changing vinyl acrylic on the packaging to just acrylic has any implication on the content of the can. It could just be a new choice of description.


Discrepancy in MO diagrams for Period 2 homonuclear diatomics I've recently noticed that different sources on the internet and in books have different ideas of whcih atomic orbital influences which molecular orbital. In the diagram, where the lines connect the AO to the MO to show their influence, I noticed that in some of the MO diagrams, the $\mathrm{2p}$ orbital does not connect to the $\sigma_1$ orbital (the lowest in energy) and the $\mathrm{2s}$ orbital does not connect to the $\sigma_4$ orbital (the highest in energy). On the other hand, I also see the case where the opposite is true and they DO connect (this is the case for my own chemistry lecturer's notes).

I asked her if she could explain this difference, and she said that the $\sigma_1$ should have some $\mathrm{p}$ character and that the $\sigma_4$ should have some $\mathrm{s}$ character, so they do connect. She didn't however say that the other way was wrong. As a result I'm not sure whether the other way is wrong, or whether it's just like a difference in notation.

I hope someone can help clarify for me. I've attached two photos of the different cases.

Thank you.

$\hspace{15 mm}$

(source)

versus

$\hspace{30 mm}$

As molecular orbital diagrams are qualitative ways to understand bonding, there are different ways in which they may be drawn.

In the simplest possible cases, molecular oxygen (O=O), for instance, it is okay to assume that s-orbitals mix only with s-orbitals (of similar energy), and that the p-orbitals only mix with p-orbitals. This works fine for molecular oxygen, and explains many of its properties: double bond between O and O, paramagnetic etc.

In other cases, molecular nitrogen being the common example, this is not sufficient, and we have to account for orbitals mixing with orbitals of other kinds. This is called s-p mixing.

Firstly, you construct a MO diagram where orbitals of the same kind mix (the same as you would have done for oxygen), you then allow these MO's to interact with other MO's of the same symmetry to produce a new set of mixed orbitals. When the MO's interact in this way, they produce new MO's of higher and lower energy, which accounts for the different order of the MO's between diagrams where s-p mixing has been taken into account.

In general, any homonuclear diatomic molecules below oxygen (lithium, nitrogen, boron etc.) need s-p mixing to be considered.

For more information about s-p mixing (and other kinds of orbital mixing) do a search here on stack exchange. There are plenty of other similar questions with answers to suit.


Is the fact that table salt is composed of Na & Cl a chemical or physical property? So I came across this question asking whether something is physical or chemical.

  Table salt is composed of $\ce{Na}$ & $\ce{Cl}$. Is it a physical property or a
  chemical property?

I really don't understand why the answer says its a chemical property. I asked the others and they did not give me an understandable answer.

Edit: Now as I think about it, I really don't think there is an accurate answer, even if the book said it was chemical. Bad question from the book.

The problem rises when we are mistaken about what a property is.

  The different types of matter can be distinguished through two components: composition and properties. UCDavis

Do a search about chemical property, and you get:

  a property or characteristic of a substance that is observed during a reaction in which the chemical composition or identity of the substance is changed. Dictionary.com

  A chemical property is any of a materials properties that becomes evident during a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity. Simply speaking, chemical properties cannot be determined just by viewing or touching the substance; the substance's internal structure must be affected for its chemical properties to be investigated. However a catalytic property would also be a chemical property. Wikipedia

  Any characteristic that can be determined only by changing a substance's molecular structure.
   Boundless

What's a physical property?

  A physical property is any property that is measurable whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe its transformations or evolutions between its momentary states. Physical properties are often referred to as observables. They are not modal properties. - Physical property in wikipedia.

Basically, a physical property is what that doesn't deal with the composition of matter; but other factors, such as it's physical state. "Physical" isn't at its literal meaning.

Table salt is never pure $\ce{NaCl}$; because simply there's nothing pure in the nature. It's mainly composed of Sodium Chloride crystals. But if, and only if, we consider it to be only made from $\ce{NaCl}$, we are talking about its chemical composition and therefore, it must be a chemical property.

I understand that sometimes there are wrong questions for us to answer, so if composition is to be a property, it must be a chemical one.


Source for literature yields I'm required to compare and justify the yields of my reactions against literature yields, but I cannot find any sources of literature yields for the reactions, or any reactions for that matter. Are there any resources with literature yields available?
The reactions are the formation of tetraphenylporphyrin and then complexing it with Cu and Zn, if relevant.

What databases do you have access to? A Scifinder search would find literature references with no difficulty.
A Google search finds this technical report which has the synthesis of tetraphenylporphyrin and the preparation of Fe(II) and Zn(II) complexes report here.
The Cu(II) complex is prepared in this lab report here. The references in these reports should help further.
The references in this JACS paper here should lead you to other preps


How does the chemical potential allow for open systems to be considered? Gibbs free energy can be defined as: $$dG=VdP-SdT+\sum_i\mu_idn_i$$ where $\mu=(\frac{\partial G}{\partial n})_{P,T}$. Last term, $\sum_i\mu_idn_i$ allows for open systems to be considered (where $dn$ does not equal zero). However, this can also be used for state functions like enthalpy too:
$$dH=TdS+VdP+\sum_i\mu_idn_i$$
But my question is why can this be used for enthalpy, considering that $\mu=(\frac{\partial G}{\partial n})_{P,T}$ (i.e it's a partial derivative of $G$ not $H$). Perhaps it's clearer for me to write the total derivative out like this: $$dH=(\frac{\partial H}{\partial S})_{P,n}dS+(\frac{\partial H}{\partial P})_{S,n}dP+(\frac{\partial G}{\partial n})_{P,T}dn$$ where I have considered a single-component open system (the component may enter or leave the system)

Actually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)

The correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):

$$
\mu_i=(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}
$$

So the total exact differential is expressible as you had originally expected:

$$
dH=TdS+VdP+\sum_i\mu_idn_i=(\frac {\partial H}{\partial S})_{p,n_i}dS+(\frac {\partial H}{\partial p})_{S,n_i}dS+\sum_i(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}dn_i
$$
  The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.

Sorry for any confusion my initial errors may have caused you.  Hope this clears it up.


Why is the lanthanide element Thulium malleable? I was doing some research about Thulium and every website says that it is malleable, but doesn't explain why.

What chemical properties does Thulium possess which makes it malleable?

All the lanthanides are soft metals; there is nothing special about thulium in that regard, AFAIK. For that matter, so are the alkali metals, and gold can be scratched with a fingernail.

Most pure metallic elements are malleable. "This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond." Brittleness (the opposite of malleability and ductility) often occurs because of impurities such as carbon, or a metal alloy which separates on cooling. These impurities "pin" dislocations in the metal, preventing them from moving smoothly. It makes alloys harder, as well as more brittle. The early technologists discovered that the alloy bronze was harder than pure copper.


What makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances? My question is, what makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances?

Charcoal (although activated carbon, a specially processed microporous charcoal product is better) is a material that is mostly carbon. The black allotrope of carbon is graphite, which is made of overlapping planar sheets of graphene. Graphite is very nonpolar (since it has only carbon atoms). Thus, nonpolar substances will adsorb. Most organic substances impurities are nonpolar substances (though not all, for example, sugars). Additionally, the graphene sheets are aromatic. The ring current generates a dipolar magnetic field and a quadrupolar electric field that attract other aromatic substances through $\pi$-stacking. Thus graphite is best at removing aromatic substances, which include all organic dyes. Carbon filtering can also remove contaminants like chlorine, although this is likely a reaction leading to chlorination of the surface.

Silica $(\ce{SiO2})$ and alumina $(\ce{Al2O3})$ are polar materials. They are polar for three reasons. 1) They are composed of polar $\ce{Si-O}$ and $\ce{Al-O}$ covalent bonds. 2) The geometries of these minerals especially around the oxygen atoms are not perfectly in-line with VSEPR predictions, so polarity is not canceled by symmetry (see the silica unit image below).

3) Both materials are hygroscopic and absorb/adsorb water from the atmosphere. The $\ce{Si-O-Si}$ and $\ce{Al-O-Al}$ covalent network on the surface is partially hydrolyzed so that the surfaces of these materials actually present $(\ce{-OH})$ groups, increasing the polarity of the surface. Polar substances are attracted to silica and alumina by dipole-dipole interactions and hydrogen bonding.

$$\ce{-Si-O-Si - + H2O -> -Si-OH + HO-Si -}$$


Is it possible to disable the splash screen of ChemDraw? Is it possible to disable the splash screen of ChemDraw? I can't find a corresponding setting in the preferences dialog. Is there maybe a hidden setting / some configuration file that can be modified?

I assume the question seeks out for a way to use ChemDraw more quickly, for example while being confronted with a larger number of chemical structures already generated, and to be processed further.  Then

If it is to browse through a number of structures aiming "to find the right file", to gain an overview, then Thumbfish (Windows only) may be helpful.  Chemdoodle (Win/Mac/Linux) and MarvinSketch offer in their file-open menu a preview as well.
On the other hand, provided the extensive documentation especially of ChemDraw seconds native, the *.cdxml format, the (batchwise) conversion into other formats is easily done with OpenBabel (supported file formats) and Apple Script ChemBot (application here).


is there any mnemonic for the AXE method? Is there any mnemonic that helps remembering the names of the different geometries (Shapes) that we obtain from applying the AXE method ?

I think you shouldn't abuse with mnemonic. In this case you should start with this rule of VSPR:

each atom in a molecule will be positioned so that there is minimal
repulsion between the valence electrons of that atom.

Now you should begin to train your three dimensional thinking!
This is the method I've adopted to learn AXE, of course everyone has his own but maybe it could help to you to find yours:
Learn to draw the $AX_nE_0$ series:
This for me was the first step: learn to draw the basic geometry 0 lone pair. This is quite intuitive knowing the previous law. Begin with the $A$ then add the $X$s. Note that the linear module could help you to remember it.

Add 1,2,3 lone pairs series
This is the next step. I've brutally modified the wikipedia table to show how you can easily do this step from the Basic geometry 0 lone pair. Simply begin to substitute $X$ with $E$: one $E$ for 1 lone pair, two for 2 lone pair...
Nomenclature
The last step is to add the nomenclature. First of all you should take in account that the lone pair ($E$) don't appear as a part of the final shape from which the nomenclature is derived. If not considering the $E$ group you obtain a planar molecule you have to add the word planar after the name, if you obtain a pyramidal shape you should use pyramidal after the name of the planar pyramid's base.
So imagine that $E$ groups are invisible and try use this method if you  can't find it only with pyramids try to figure out which of the following figures is more close to the actual shape of the molecule.

The first two are platonic solids  (Octhaedron, and Tetrahedron) then there is a T for a T-shaped, and a bent figure. In the original wikipedia table respective molecule configuration is already rotated to the right position, I think you can do it on your own, with your 3D thinking.
For the last, seesaw shaped, there is a powerful "mem":

Here is an old photo of two boys in a $SF_4$ molecule.Good ol' times!


What does C-OXT stand for in PDB files? I've been trying to create 3D structures of unionized tyrosine (and other amino acids, it's just tyrosine is the one I'm up to atm) using the structure of Zwitterions found within PDB files and to do this I need to know what C-OXT (which is found within the carboxylate functional groups) stands for in PDB files as this will help me to decide which oxygen in the carboxylate groups should be represented with the double bond from the associated carbon and which should have the single bond. The other oxygen in the carboxylate group is just called C-O in the PDB file.

The naming conventions for atoms in PDB file require that at the C-terminus of a protein one of the carboxylate oxygens is called O and the other is called OXT.  But since it's a deprotonated carboxylate group they are both chemically equivalent, and you'd expect that bond lengths are equal.


Does water vapor in the air oxidize copper? My friend is installing copper pipes in his house, which is going to be in a humid area, and he says that copper will only be oxidized by the air and the excess water vapor in the humid air won't affect it. However, it is to my understanding that since water vapor is $\ce{H2O}$, which contains oxygen, there will be more oxidation occurring, and thus more corrosion to the pipes because of more oxygen present in the air. Is this correct?

Although $\ce{H_2O}$ contains oxygen, it is there in the most reduced state $\ce{O^{2-}}$ and therefore cannot act as oxidizer. It is the oxygen in the air $\ce{O_{2}}$, which is the main and strongest oxidizing agent.

On the other hand, presence of water as a solvent can speed up some reactions. For a short overview of copper corrosion, see e.g. Why does copper oxidize. As the corrosion products form compact (or even protecting) layer, the corrosion itself is not a problem for copper, as compared to e.g. steel corrosion.


What is actually the difference between valence bond theory and molecular orbital theory? Recently I have read about both of the concepts in my book (Physical Chemistry by Atkins, Paula). It was literally a reading; though I could understand the language and superposition of orbitals, inversion symmetry, and so on, I never could really comprehend how these two theories differ (or agree) with each other. I have googled it but couldn't gain a lucid explanation between the differences in their approaches. So, now my question is:

How do these two theories differ from each other? What are the differences between their approaches?

Also, in VB theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why it is told that "MO theory provides a global, delocalized perspective on chemical bonding"? This portion of the question was split to a separate question: Why is the resonance concept not required in molecular orbital theory?

Edit: After viewing some of the answers, I think IMHO, I had to clear what I was really asking. By saying "difference between MOT & VBT", I am not intending to mean that just write the different predictions the two theories can separately succeed in making. What I really mean is the difference between their approaches & their way of explanation. Of course, the case is not like "Hey, look, one theory is outdated; just follow the other one..."; they are both important and hence do want to know the difference between their pathways of analyzing the cases.

TL;DR

VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears little resemblance to the original atomic orbitals, or to the VB theory description.

Long Version

The key difference between the VB and MO theories lies in their respective assumptions of the underlying mathematical/physical form of the electronic wavefunction.

VB theory holds tightly to the mathematical structure of atomic orbitals (AOs), assuming that the electronic structure of a molecular system is described adequately by retaining the atomic orbitals (possibly hybridized) from each constituent atom, and overlapping them to form the needed description of bonding.  VB calculations are computationally much more complex: the AO basis is generally non-orthogonal, and the linear algebra and other techniques required to cope with this non-orthogonality are much more involved.

More broadly, the need to invoke resonance (superposition of multiple valence-bond structures) and other concepts highlights that, in fact, often this adherence to atomic orbitals and their valence-bond overlaps is not adequate for a proper description, at least not without extensive use of methods analogous to the configuration interaction of MO theory.  VB theory does hold certain advantages, however—for example, I believe the AO basis does sometimes allow for a more direct linkage to fundamental chemical concepts such as hybridization and Lewis electon-pair theory. To note, Sason Shaik is one of the strongest modern proponents of the theory I know of—he describes some of these strengths and their applications in his recent retrospective article (doi:10.1016/j.comptc.2017.02.011).

MO theory, at its core, casts away most assumptions regarding the mathematical structure of the electronic wavefunction, as long as properties like indistinguishability and the Pauli exclusion principle remain satisfied.  Calculations within MO theory can use a wide variety of mathematical structures to try to represent the wavefunction, ranging from gridded numerical quadrature approaches (see here and here, and also orbital-free DFT) to Slater orbital computations to the Gaussian-orbital based calculations used most commonly today.

One potential point of confusion is that MO theory in its most common implementations makes direct use of atomic orbitals in composing the molecular wavefunction (see LCAO theory). The distinction to be made here is that instead of attributing significance to the atomic orbitals as an intrinsically good descriptor of electronic structure, they are used merely because they afford a convenient mathematical form for efficient computation, in trying to construct the MOs.  Particular advantages are obtained from the use of atomic orbitals constructed from Gaussian functions, since the product of two Gaussians is another Gaussian (even in multiple dimensions) and numerical integration of n-dimensional Gaussian functions is straightforward.

It should be noted that, for as much of an improved description it affords, MO theory still is an approximation. For a given system, the electronic wavefunction just IS the highly-abstract mathematical object that satisfies the appropriate Schrodinger equation.  MO theory is used extensively because it provides one of the most computationally tractable paths to getting in the neighborhood of the "real answer."

Practical Aspects of MO theory

The "simplest version" of MO theory for a time-independent system under the Born-Oppenheimer Approximation is the Hartree-Fock method, which provides a remarkably good description considering the significant simplification represented by its mean-field approximation of the electron-electron interactions.  Hartree-Fock is in general not good enough for computations at the level of chemical accuracy (~$1\mathrm{\ kcal/mol}$), however, and numerous methods have been developed to refine it: the most often used techniques are Moller-Plesset perturbation theory, configuration interaction, and coupled cluster.

In-Depth Reading on MO Theory

For a deep dive into the underlying mathematics of MO computations, it's hard to beat the classic Szabo & Ostlund. As a more readable, lighter-weight introduction to MO theory and its implementations, I would recommend Frank Jensen's Introduction to Computational Chemistry.


How do we count charges in this example of a Redox reaction equation? My brother(younger) had a doubt regarding this chemistry book example so I let him here :

Question :

Balance the equation : $\ce{H+ + MnO4- + Fe^2+ -> Fe^3+ + Mn^2+}$

Solution steps(as given by the textbook):

Step 1 :  $$\ce{MnO4-  ->  Mn^2+}$$

Step 2 : $$\ce{MnO4-  -> Mn^2+ + 4 H2O}$$

Step 3 :  $$\ce{MnO4- + 8 H+  -> Mn^2+ + 4 H2O}$$

Step 4 : $$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}\tag{1}$$

Step 5 : $$\ce{Fe^2+ -> Fe^3+ + e-}\tag{2}$$

Step 6 : $5\times(2) + (1)$ and thus we have the balanced equation:

$$\ce{MnO4- + 8 H+ + 5 Fe^2+ -> 5 Fe^3+ + Mn^2+ + 4 H2O}$$

Explanation given in the textbook for step 4:  we equalise the charge by adding 5 electrons on left.

My doubt regarding step 4: how do I know that in step 4 that 5 electrons do actually equalize the charge in the given equation? More accurately how do I know that there is any surplus or deficit of charge in any given equation atleast in this case?

If you look at the charges in step 4, the left side adds up to 7 and the right side adds up 2. The MnO4 on the left has a -1 charge, and the 8 hydrogens add a +8 charge, adding up to a +7 charge on the left overall. The Mn on the right has a +2 charge so that is +2.

To balance out this charge, they must be equal. How can +7 be made equal to +2? By subtracting 5. In this case, adding 5 electrons creates a -5 charge that means both sides of the equation have a +2 charge.
You just need to add up the charges like this for a given equation to balance it.


Entropy of surroundings and heat Apparently the entropy change of a system can be found by

$dS=\frac{\delta q_{rev}}{T}$

I am slightly confused about this because I have seen this being used when the temperature of the system and the surroundings are not the same. However for the heat exchange to be reversible between the system and surroundings, need not the temperature of the system and surroundings be the same?

I have a feeling this might have to do with reversibility with respect to the system and the surroundings. I do not know about this and had not heard of it before I read it in a response to this post:

"So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir)"

Could someone please elaborate on this?

Yes, that is a perfectly acceptable formula for entropy change even outside of thermal equilibrium.

And, as you say, it's related to the irreversibility of heat transfer. The point about ideal thermal reservoirs always exchanging heat reversibly is not necessary to understand this, so let's park it outside for a moment.

Let's imagine a body in an environment. We'll call the temperature of the body $T_b$ and the temperature of the environment $T_e$. If they exchange heat, their respective entropies, $S_b$ and $S_e$, change according to the formula you mentioned:

$dS_b=\cfrac{\delta q}{T_b}$

$dS_e=\cfrac{- \delta q}{T_e}$

Note the sign (the heat flowing into the body is leaving the environment, or vice versa).

Also note, and this is critical, that the magnitude of $\delta q$ is equal in both equations (one receives exactly the same amount of heat as the other loses), but, if $T$ is different, the entropy change won't be the same. The change in entropy (positive or negative) will depend on the temperature of the body or the environment.

Let's have a look at how the total entropy of the system ($S=S_b+S_e$) changes.

If the body and the environment are at thermal equilibrium, in other words, their temperature is the same ($T_b=T_e=T$),

$dS=dS_b+dS_e=\cfrac{\delta q}{T}-\cfrac{\delta q}{T}=0$

In other words, transferring heat between bodies at thermal equilibrium does not change the total entropy of the system.

However, if they are at different temperatures,

$dS=\cfrac{\delta q}{T_1}-\cfrac{\delta q}{T_1}=\delta q \left( \cfrac{1}{T_1}-\cfrac{1}{T_2}\right)$

Now, the sign of the entropy change depends on the relative values of $T_1$ and $T_2$:

If $T_1>T_2 \rightarrow 1/T_1 < 1/T_2 \rightarrow 1/T_1 - 1/T_2 < 0 \rightarrow dS < 0$

If $T_1<T_2 \rightarrow 1/T_1 > 1/T_2 \rightarrow 1/T_1 - 1/T_2 > 0 \rightarrow dS > 0$

Note that I chose the signs of $\delta q$ assuming that the flux of heat went from the environment to the body. So, this means that if the body is hotter than the environment ($T_1 > T_2$), this would imply a reduction in the entropy of the whole system ($dS < 0$) - which the second principle explicitly forbids. And, to the contrary, if the body is colder than the environment, that flux will be spontaneous ($dS > 0$).

So, outside of thermal equilibrium, you still can use that expression to calculate entropy changes - as long as the heat flux goes in the right direction, from hot to cold. It will also apply to heat transfers in equilibrium (and therefore reversible) - although in that case you'll be having some other energy transfer process of equivalent magnitude siphoning that energy off - otherwise the heat would increase the temperature of part of the system, breaking thermal equilibrium (that won't happen).


Grams of bicarbonate vs realeased CO2? So, it's been a good two years since I've done chemistry and I have my first university lab in a few days and I'm sort of freaking out because I can't even remember how to do stoichiometry properly. This a question off of my pre-lab review and I'm having troubles remembering what to do. I believe this is a limiting reagent question.

In Experiment 1, you will place and AlkaSeltzer tablet in each of the dilute $\ce{HCl}$ solutions you prepared. The reaction between $\ce{HCl}$ and $\ce{NaHCO3}$ should take place quickly and release gaseous $\ce{CO2}$. You will then use the mass lost for each flask to determine the amount of $\ce{NaHCO3}$ that reacted.
A student performs this experiment and determines that $\pu{1.60 g}$ of $\ce{CO2}$ were lost. How many grams of $\ce{NaHCO3}$ would have reacted to produce this amount of $\ce{CO2}$?

Firstly, I wrote out part of the equation but I'm also beginning to think it's incorrect too:
$$\ce{NaHCO3(aq) + HCl(aq) <=> CO2(g) + NaCl + H2O(l)}$$
The initial amount for $\ce{NaHCO3}$, I know, is unknown and the final amount for $\ce{CO2}$ is $\pu{1.60 g}$. Other than that though I'm at a loss for what to do next. Please help if you can, thank you!
Edit (see answer): To calculate the amount of moles in $\ce{CO2(g)}$ I used the formula $m/M = n$, which gave me 0.36355 moles for carbon dioxide. Now I use 0.36355 moles for $\ce{NaHCO3(aq)}$ since it's a 1:1 ratio.

Thank you for all the help, I've figured out the answer:

$$\ce{NaHCO3(aq) + HCl(aq) -> NaCl(aq) + CO2(g) + H2O(l)}$$

Since we're given $\pu{1.60 g}$ of $\ce{CO2(g)}$ we can use the mass to calculate the number of moles with the equation:

$$n = \frac{m}{M} = \frac{\pu{1.60 g}}{\pu{44.01 g mol-1}} = \pu{0.036355 mol}$$

Since it's a 1 : 1 ratio between $\ce{NaHCO3(aq)}$ and $\ce{CO2(g)}$, we can directly solve for the mass of $\ce{NaHCO3(aq)}$ using the same formula:

$$n = \frac{m}{M} $$

$$m = n \times M = \pu{0.036355 mol} \times \pu{83.42 g mol-1} = \pu{3.0509116 g}$$

Round to two decimal places: $\pu{3.05 g}$


Are chemical treated as ideal gases when trying to figure out heat of formation using calorimetry? At 36:48 of this video lecture, The professor says that the reaction moving from reactants to products keeping temperature of both same and under standard condition is an isothermal constant volume process.
Some some four minutes later (41:57), he uses the sum of energy of whole cycle property and equates, the $U_{rx}$ to the other definition of other definition of heat change given by calorimeter using the law that sum of internal changes over a loop process is zero.
He gets:
$$ \Delta U_{rx} = - C_{v}^{calo} \Delta T$$
where $C_{v}^{calo}$ is energy change of calorimeter.
Now, what I find weird is that by the definition of the path-3 being isothermal, shouldn't $$ \Delta U_{rx}$$ be zero..?
My resolution for this was that maybe the gas not really ideal and that it may change energy loss by expanding but I'm not sure how I would justify this more authoritatively. So, my question is, when we talk about these chemistry reactions, do we treat gases as non-ideal?

Internal energy of any gas is given by
$$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$
Last three aren't affected by ordinary heating. And $U_\text{intermolecular}$ for ideal gas is zero. That's why for ideal gas $U$ is only function of temperature. But during a chemical reaction (even if it involves an ideal gas)
$U_\text{bonds}$ comes into play. The formation and destruction of bonds results in a decrease of potential energy.
Therefore, for a chemical reaction, even if the temperature remains constant the internal energy changes.


Voltage vs current density - why are both parameters important to control an electrochemical reaction? In many electrochemistry scholarly articles/studies they talk about certain specific reactions occurring at certain voltage potentials, while in other articles they talk about current density as the determiner for which reactions happen.

Why is this? From what I know about electrical engineering you can either set the voltage, and the current will adjust accordingly, or you can set the current and the voltage will adjust to meet the current. You cannot set both of them at the same time. That would defy the laws of physics.

So what do many of these articles mean when they talk about, for instance:

  Reported reactions of this type often occur at potentials smaller than -3 V

How does this relate to the current density that is set by a power supply for instance?

Reaction rate in electrochemistry is limited by diffusion of reactants from solution towards electrode. Once all reactants in near proximity for given type of reactions are exhausted, and if voltage is still applied, other reactions (if voltage is strong enough) may occur. For example, consider positive electrode in $\ce{KCl}$ solution (leaving aside paired negative electrode). It is true, that $\ce{Cl^-}$ anion is oxidized easier then water. However, once all chloride near the electrode is oxidised, the next possible substrate to be oxidised is $\ce{OH^-}$ that is always present is water via reaction $\ce{2H_2O <=> H_3O^+ + OH^-}$. Thus, depending on the current density on the electrode you will have different major products.

In short: resistance of the electrochemical cell is a complex function of temperature, voltage, current and solution composition.

The current density in electrochemistry may be adjusted not only by adjusting voltage applied, but also by adjusting surface area of the electrodes used. Also, it is critical to maintain conditions for effective transportation of reactants towards electrodes.


Formal Charge of Terminal Atoms Would the formal charge of terminal atoms differ from the FC of the non-terminal atoms?

Say you had a sheet of graphite; inside the network covalent solid all the carbons are bound to one another, but the terminal atoms are obviously missing some C-C bonds since they are terminal.

Their formal charges would differ, correct?

What is the consequence of the differing charges?

Yes, the formal charge at such atoms would be different.  Such atoms, typically found at the at the edge of some crystalline or regular solid, are known as "defects" in the solid.  Because they have an incomplete outer shell and lack a stable bonding configuration (bonding vacancies), they usually exist as a radical, anion or cation.  They define an area of extremely high reactivity on the solid's surface and mark the spot where reactions often begin.

Edit: references to charges at defect sites

this link covers inorganic systems, see p.5 for some nice pictures.

this link covers graphene systems, from the "Conclusions" section; "We have performed scanning gate microscopy on graphene and observed such charge inhomogeneity and puddles due to extrinsic doping from possible sources such as metal contacts, graphene edges and surface residues."


How does light emission depend on temperature by fluorescence? I have a glass, which emits green light after charging it with sunlight. I though this is fluorescence, and only photons with shorter wave length (than the green light has) can excite the electrons of this material and can be reemitted as green light. At least I remember on something like that from my studies. Now the interesting part that when I wash it with warm water, it starts to emit light too. How do you explain that?

This phenomenon is thermoluminescence. The incoming photons excite electrons of the material to a higher level, a metastable state, which fall back to a lower state and emit photons. The metastable state may last for only nano- or picoseconds, in which case it is termed fluorescence. If the electrons need a bit of a "nudge" and hang around for a few seconds or hours, it is termed phosphorescence.

In the extreme case of some materials, the metastable state can last for millenia, which can be useful for dating the age of the item. For example, when pottery is fired, atoms in a metastable state relax. From then on, energy accumulates from cosmic rays and radioactive decay of nearby minerals such as thorium or potassium. By heating the ceramic and measuring the amount of light emitted, one can approximate the age of the pottery.

Ceramics need to be heated to a few hundred Celsius, but your glass works in just warm water. There are similar infrared detection cards with phosphors that are charged using visible or UV light, and that emit visible light when triggered by low-energy IR.


Is reduction potential equivalent to the ease of discharge? Loosely speaking, is the reduction potential equivalent to the ease of discharge in the context of electrolysis?

i.e. is it easier to discharge an ion with a higher reduction potential as opposed to one with a lower one?

Easy is a big word.

Lower potential means it requires less energy, but that doesn't say anything about how quick the reaction will proceed.

A low potential reaction may still have slow kinetics and therefore not be "easy".

If you are comparing ions as to which will produce more product as a result of an electrochemical reaction, you need to find something with low potential and fast kinetics.


Energetic Placement of Atomic Orbitals in the HCl Molecular Orbital Diagram How are the 3p orbitals of chlorine lower in energy than the 1s orbital of hydrogen?

I'd say there are two important things to consider here: Firstly, $\ce{Cl}$ is more electronegative than $\ce{H}$. And secondly, as you proceed from the left to the right in any row of the periodic table, the s and p orbitals go down in energy. This is a consequence of the fact that the valence electrons do not screen each other effectively. Thus, the addition of one proton to the nucleus and one electron does not cancel; instead, the valence electrons "feel" an increased nuclear charge.
The electronegativity difference and the reduced screening among the valence electrons seems to be enough to shift the $3\ce{s}$ and even the $3\ce{p}$ orbital of $\ce{Cl}$ below the energy of the $1\ce{s}$ orbital of $\ce{H}$. The experimental state averaged ionization potentials (IP) could serve as a good guide for the placement of starting orbitals in an orbital interaction diagram. For the $\ce{H}$ $1\ce{s}$ orbital you have an IP of $-13.6 \, \mathrm{eV}$. For the $\ce{Cl}$ $3\ce{s}$ and $3\ce{p}$ orbitals you have IP's of $-25.2 \, \mathrm{eV}$ and $-13.7 \, \mathrm{eV}$, respectively.$^{[1]}$ So, this indicates that the $3\ce{p}$ orbital of $\ce{Cl}$ does indeed lie slightly below the $1\ce{s}$ orbital of $\ce{H}$.

Update:

Here is a table of the experimental state averaged ionization potentials, in electron volts, for the main group atoms. However, those $\ce{p}$ atomic orbitals for groups 1 and 2 and for the $\ce{s}$ and $\ce{p}$ orbitals of the sixth row are not experimentally known and hence, calculated values (indicated by an asterisk) have been used.$^{[2,3]}$ The latter include relativistic corrections. The table is taken from the 2nd Edition of the book Orbital Interactions in Chemistry by T. A. Albright, J. K. Burdett, and M.-H. Whangbo.

[1] L. C. Allen, J. Am. Chem. Soc. 1989, 111, 9003 ; L. C. Allen, J. Am. Chem. Soc. 1992, 114, 1510; L. C. Allen, Acc. Chem. Res. 1990, 23, 175.

[2] A. Vela, J. L. Gazquez, J. Phys. Chem. 1988, 92, 5688.

[3] J. P. Desclaux, At. Nucl. Data Tables 1973, 12, 311.


Grignard reagent reaction mechanism: Which bond breaks during the decomposition of the adduct? We just finished our Organic Chem. chapter Aldehydes, Ketones and Carboxylic acids at school, but I still have one lingering doubt.

Considering the reaction between an alkyl magnesium halide and an aldehyde or ketone: (Diagram's from Wikipedia)

Now my teacher tells me that in the adduct, it's the Mg-O bond that gets cleaved during hydrolysis and not the C-O bond, resulting in an alcohol.

He isn't willing to elaborate ("Aaron, it happens... just like that... you shouldn't worry about all the complexities behind everything...") and I can't seem to find any source on the web that deals with the particulars of bond-breakage.

I have doubts regarding that point of his, since the Mg-O bond has significant ionic character (due to the electronegativity difference between the atoms. But in comparison, the C-O bond is more covalent, and therefore must be easier to hydrolyze. Because (the way my high-schooler brain sees it) ionic bonds are established by strong electrostatic forces of attraction between the ions concerned, while this isn't the case for predominantly covalent bonds. So it seems to make sense that the weaker bond is preferably hydrolyzed, hence I'm mooting for the C-O hydrolysis.

Who's right here? Could someone cite any (reliable) sources?

As can be seen above, the end product is the same whether we split the $\ce{Mg-O}$ bond or the $\ce{C-O}$ bond.

The problem is, then, which pathway is more favourable?

This, I have no idea. Maybe both pathways are favoured. It doesn't matter because they give the same products.


Why does milk overflow when boiled? I was wondering that like whenever I boil milk I just have to keep standing in front of it to make sure that I don't waste any milk but then I was wondering like Why does it even overflow ? In case of water it doesn't happen. What makes it to behave like this ?

Milk contains surfactants (primarily lipids), which allow stable bubbles to form. In intense boiling, these bubbles can become mixed with liquid, supporting the liquid above them and forming what I'll call for want of a better term a "wet froth." Imagine a bunch of sponges made out of liquid. They take up more space than they would if you compressed them down and forced out the air. The inclusion of the water vapor in the bubbles lowers the density of the mixture, causing it to expand and overflow the pot.

The state is only stable because the added energy keeps creating new bubbles to replace any that have popped and keeps forcing liquid upward, while it would normally fall.

Water does not experience this mostly because its bubbles are far less stable. In fact, pure water does not form bubbles at all. The bubbles formed in water is typically due to impurities.  If you were to add soap to a boiling pot of water, you would see the exact same effect, if not worse.


Effect of removing the product from equilibrium For the reaction

$\ce{PCl3 + Cl2 <=> PCl5 }$

Upon removing some $\ce{PCl5}$ from the equilibrium, the equilibrium position will shift right and the forward reaction is favored, increasing the yield of $\ce{PCl5}$.

Is the information suggested by the attached graph true? (Do $\ce{PCl3}$ and $\ce{Cl2}$ decrease by the same extent?)

For an arbitrary reaction $\ce{2A + B -> 3C}$, how will the concentrations of each substance change accordingly when say, some $\ce{C}$ is removed from the system? Will $\ce{A}$ and $\ce{B}$ still decrease by the same amount, and why?

First, let’s analyse the situation the graph presents. Obviously the reaction is (just restating for clarity):

$$\ce{PCl3 + Cl2 <=> PCl5}\tag{1}$$

We have already removed some $\ce{PCl5}$ from the mixture and are now looking at the self-reestablishing equilibrium. Obviously, for each molecule of $\ce{PCl5}$ we create we lose one molecule of $\ce{PCl3}$ and one of $\ce{Cl2}$ — this is the information we gain from the reaction coefficients (they are all $1$). This can then be expanded to the amount of substance reacting, namely $$\Delta n(\ce{PCl5}) = -\Delta n (\ce{PCl3}) = -\Delta n(\ce{Cl2})\tag{2}$$

Your graph uses concentration $c$ as $y$ axis. Concentration is defined as:

$$c = \frac nV\tag{3}$$

Since we can assume the volume $V$ to be constant, we see that concentration and amount are simply related by a constant factor; thus any differences in concentration will be analogous to differences in amount. Therefore, the labelled $\Delta x$ in the graph are indeed all identical as the represent the same concentration difference and thus the same difference in amount; and we have shown in equations $(1)$ and $(2)$ that the amounts are identical.

What about a case such as your hypothetical following case:

$$\ce{2A + B <=> 3 C}\tag{4}$$

Well, in general the same idea applies except that you must now take care of the different coefficients. Say we removed some $\ce{C}$ from the mixture. New $\ce{C}$ will be created and the creation of three molecules of $\ce{C}$ will use up two molecules of $\ce{A}$ and one of $\ce{B}$. Thus:

For each mole of $\ce{C}$ created we lose $\frac 13~\mathrm{mol}\ \ce{B}$
For each mole of $\ce{C}$ created we lose $\frac 23~\mathrm{mol}\ \ce{A}$

Therefore the relationship is:

$$\Delta n(\ce{C}) = -3\Delta n(\ce{B}) = -\frac 32\Delta n(\ce{A})\tag{5}$$


Near IR spectroscopy vs. IR spectroscopy I am a backer of this Kickstarter: SCiO Near IR Spectroscopy

Here is a link to their technology: SCiO Technology

Question: How short of full IR spectroscopy do you think this device is? I have asked them whether we could program the SCiO to act as an IR spectrometer, and they have said this would require additional programming (which I am not qualified to do). But I am wondering if there would still be limitations. Conversely, Consumer Physics seems to imply that our taking more observations with the SCiO will make the device smarter and more useful.

This is obviously a question directed at experts in IR spectroscopy, which I am most definitely not.

This is more or less speculation, since they haven't released detailed technical specifications, but given the size of the instrument and the need to avoid moving parts to keep it durable, I would wager that it uses an IR LED for illumination, a diffraction grating for dispersion (although from the picture, it doesn't look like there's room for a conventional reflective diffraction grating setup so they might be doing something else), and probably some kind of silicon CCD or InGaAs diode array for detection. If this is the case, it's likely impossible to convert for use in conventional IR spectroscopy—not as a matter of programming, but simply that the hardware is not equipped to do it. Normal mid-IR spectroscopy (below about $4000\mathrm{\ cm^{-1}}$) involves much longer wavelengths beyond what LEDs can do, which generally means some kind of incandescent lamp, optics that don't involve glass (which blocks mid-IR), and a detector that can deal with such low energy photons (there are a few semiconductor-based sensors that looks promising, but most spectrometers use thermal detectors). There is also the matter of dealing with all the water in the air—many IR spectrometers have gas purge systems in their sample chambers.

There has been a lot of interest in standoff IR spectroscopy for things like explosives detection and advances in mid-IR optics and sensors may make this kind of small instrument possible, but we're not quite there yet. Certainly an instrument designed for NIR is not likely to have any usefulness for mid-IR, without being designed with it in mind.

A related technique that is much better suited to small handheld instruments is raman spectroscopy, since the light involved is actually usually in the NIR. I doubt this thing is sensitive enough to do it well, but it might be possible to add filters to the illumination and measurement ports to give it a shot. It's definitely possible to do raman with a handheld instrument though—I saw one of these at a conference recently. Quite expensive, but very nifty.


Comparing wt% to at% which is normalized to 100% I have measured the element composition of a sample using ICP-OES, and reported my results in wt%. A study which has analyzed the same sample types, but with an electron microprobe, has reported their element concentrations in at%, which in addition are normalized to 100 %. I would like to compare my results to those in the article, but I am not sure how.

I am able to convert the atomic percentages to weight percentages, but I am not sure how to tackle the normalization of the data. When I add the wt% of my sample, I only get to about 50%. This loss is attributed to that I have not analyzed oxygen and carbon, which I suspect would make up a significant portion of the sample weight.

However, the study I want to cite has indeed analyzed both oxygen and carbon, and all the other elements I have measured. I do not suspect that they have a very large portion of at% unaccounted for. Therefore I believe their un-normalized data is quite similar to their normalized data.

So, how should I proceed with comparing my data to theirs? Should I just argue, like I have done here, that I think their data is "comparable enough"?

I would appreciate any insight on the subject!

This question is difficult to answer without knowing more about the data.

First is the issue of "normalization". Do you mean that in the at% study, all measurements added together were only 96% or 90% or 80% or something, and so to get all measurements to add to 100%, the simply divided each measurement by 0.96 (or 0.9 or 0.8)?  This is what I infer from your comments.

Second is the issue of measuring at%.  Are they measuring the raw number of total atoms in their OES experiment? Or are they just comparing their results to some calibration standard for each element measured?  Your comments make me think they are using calibration standards.  From a description of those standards, you should be able to convert the standards to wt%, and thus map from wt% to at%.

Whether you want to map to the "normalized" at% or the unnormalized at% is up to you.


Water softener and RO filter I have a water softener for showers and dishwasher and a reverse osmosis filter for drinking water.  For convenience of the install, the RO filter is using softened water (some calcium carbonate has been exchanged with sodium chloride -- this is my basic understanding; correct me if I'm wrong).  I don't fully understand how an RO filter works except that it is essentially just a molecule-sized physical filter.

My question is, Does the RO filter care whether the supply water is salty or hard?  Is it more effective at filtering one or the other?  Is one going to influence the life of the RO filter more than the other.

My apologies if this is not a "true" chemistry question, but I'm looking for a bit more of an thorough explanation than I got at diy.stackexchange.

You are not wrong in your understanding of a water softener, nor are you wrong in your understanding of a reverse osmosis filtration system.

  Does the RO filter care whether the supply water is salty or hard?

I would like to make the point that your RO filter is incapable of caring about the hardness of the water. This is more about carefully choosing your words more than anything else. I was also tripped up by this when I first began chemistry, but it was something I learned to do and I challenge you to do the same.

  Is it more effective at filtering one or the other?

Yes, but only slightly. Here you can find a list of the typical rejection characteristics of RO filters. The lower bound for the percent amount of multivalent ions ($\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$) filtered is higher than the lower bound for the percent amount of monovalent ions ($\ce{Na+}$ and $\ce{K+}$) filtered, but the upper bounds are very close in value.

Overall you could say that RO systems are more efficient at filtering hard water, but the difference is not very large. Most systems will likely be performing considerably above the lower bounds of the typical rejection characteristics anyway, so this should not be a deciding factor in deciding to feed your RO hard or soft water.

  Is one going to influence the life of the RO filter more than the other?

Yes! In an ideal RO filtration system, all of the filtered materials would simply drain away, and pure water would flow to the storage tank, but nothing is ideal. If you feed hard water into your RO system, there will be scaling on the semi-permeable membrane that will eventually reduce its filtered water output. This can cause lack of flow will create a potentially damaging buildup of pressure elsewhere in the system. I would therefore recommend that you keep your setup the way it is!


What is exactly molar conductivity? I know molar conductivity is conductivity of solutions by molarity,  but then I see on next page of my book it defines molar conductivity $\Lambda$ as follows:
$$\Lambda = \frac{\kappa a}{l},$$
where $\kappa$ is conductivity, $l$ is the distance between electrodes and $a$ is the area of cross section. Where does this formula come from?

Say conductivity of the electrolyte is represented by $\kappa$, molar conductivity by $G$, resistance by $R$ and resistivity of the electrolyte by $\rho$
I'm assuming you know the definitions to the above since you've asked this question.
Now, resistance $R$ = $\frac{\rho\space l}{A}$ where $l$ is the length of the conductor and $A$ is the area of cross section.
For an electrolyte, $l$ can be taken equal to the distance between the electrodes and $A$ can be taken as the surface area of the electrodes.
Since conductivity is the reciprocal of resistivity ($\kappa$ = $\frac{1}{\rho}$),
$R$ = $\frac{\space l}{\kappa\space A}$
Reciprocating both sides,
$\frac{1}{R}$ = $\frac{\kappa\space A}{l}$
Since molar conductitivy is defined as the conductance of the entire solution having 1 mole of electrolyte dissolved in it and conductance is $\frac{1}{R}$,
$G$ = $\frac{\kappa\space A}{l}$
where $G$ is molar conductivity


Why is 1-Ethyl-1-methylcyclohexane achiral? I was wondering why the 1-Ethyl-1-methylcyclohexane molecule is achiral.
Thank you in advance.

The functional groups are on the same carbon, so either configuration can be rotated into the other. If the functional groups were on different carbons, then depending on their relative locations the molecule could be chiral.


Growing crystals that are in a magnetic field or subjected to an electric field I know different crystals resonant at different frequencies but is it possible to grow crystal at a certain frequency.  An example would be growing crystals using Epsom salt or Alum in a magnetic field or when it's subjected to an electric field while two electrodes are submerged into the saturated solution.

The reason I ask is I know they can "program" magnetic fields one company doing this is called Correlated Magnetics

A video that explains "programming" magnets

I would like to do something similar but with crystals.  I do know there are ways to test the resonance of crystal oscillators Video showing how to find resonance frequency of crystal oscillator

How to test Rochelle salt piezoelectric crystals voltage oscilloscope

Now I'm just trying to see and find out if I can "program" crystals at various frequencies by subjecting them to magnetic fields or electric fields.  Has anyone tried this or know of research being done?

The main influence on resonant frequency of a crystal is mechanical: the mass of the crystal and the restoring force of elasticity cause the crystal to vibrate at a specific frequency when disturbed. To fine-tune the crystal, bits may be chemically or abrasively removed or the mass may be increased slightly by rubbing with graphite.

That said, crystal orientation does influence resonant frequency because the modulus of elasticity of a crystal such as quartz is anisotropic, i.e. it varies significantly in which plane it is measured. Perhaps applying a strong magnetic or electrostatic field would change the resonant frequency -- but you would need to isolate the effect on the crystal from that of its mounting, particularly if there are metallic contacts.


Does sulfuric acid formed in a previous step dehydrate a secondary alcohol? In the following reaction won't the sulphuric acid dehydrate the secondary alcohol after it's formed?

It could, but the addition of water would push equilibrium toward the hydrated product, and the $\ce{H2SO4}$ would likely transfer into the water layer that you made by adding $\ce{H2O}$.

Plus, you would only make stoichiometric sulfuric acid. Depending on the amount of water, the pH of the solution formed could be fairly neutral. Also, usually, the acid (despite being a catalyst) is used in large excess for dehydration reactions.

Therefore, it is possible (given the right conditions) but unlikely.


In a single cell battery, is voltage exactly equal to cell potential? I've been researching electrochemical cells. In chemistry-focused sources, you usually see the potential difference between the electrodes referred to as cell potential, but in engineering/industrial sources(where you talk about "batteries") you usually see plain "voltage." Are the two always the same, or are there some situations where they're different(excluding multiple cells)? I want to know whether you can use the cell potential formula $$E^o_{cell} = E^o_{cathode} - E^o_{anode}$$ to calculate an accurate standard conditions voltage for a battery. If not, what gets in the way?

It is not uncommon for two identical phenomena to be described by different terminology in different contexts or scientific disciplines. In this case, voltage and electric potential difference refer to the same thing. According to this Wikipedia article:

  Voltage, electric potential difference, electric pressure or electric tension (formally denoted ∆V or ∆U, but more often simply as V or U, for instance in the context of Ohm's or Kirchhoff's circuit laws) is the difference in electric potential energy between two points per unit electric charge. The voltage between two points is equal to the work done per unit of charge against a static electric field to move the test charge between two points. This is measured in units of volts (a joule per coulomb).