What is "mechanistic duality"? I am doing a general overview of different kinds of dualities present in science. I am quite well versed with the dualities in physics and mathematics. I was searching for dualities in chemistry, biology and other sciences. In that process, I encountered "mechanistic duality" in chemistry. For instance, see [1] and [2].

But I am not able to find a good introduction or description about "mechanisitic duality". It would also be nice if someone can briefly explain what this duality mean. Does someone have clue about its history (who coined it first and for what purpose)? Also, please point me to relevant book/paper which discuss this.

Denmark, S. E.; Su, X.; Nishigaichi, Y. The Chemistry of Trichlorosilyl Enolates. 6. Mechanistic Duality in the Lewis Base-Catalyzed Aldol Addition Reaction. J. Am. Chem. Soc. 1998, 120 (49), 12990–12991. DOI: 10.1021/ja982993v.
Ross, S. P.; Baire, B.; Hoye, T. R. Mechanistic Duality in Tertiary Amine Additions to Thermally Generated Hexadehydro-Diels–Alder Benzynes. Org. Lett. 2017, 19 (20), 5705–5708. DOI: 10.1021/acs.orglett.7b02888.

This just means that the same type of reaction may proceed by two different mechanisms depending on the exact chemical being used. It's not a particularly widely used term (it is not a "standard" phrase such as wave-particle duality) and there are many other ways of saying the same thing ("divergence", "pathway", etc.).

As an extremely simple example, alkyl bromides can react with nucleophiles via either an SN1 or SN2 pathway, depending on the degree of substitution of the carbon bonded to bromine. This is an example of how a different substrate may follow a different mechanism, although the end result is the same (replacement of Br with the nucleophile).

The papers you cited are just slightly more sophisticated examples of the same concept.


Hydrogen bonding in concave rings Can hydrogen bonding occur if both O and H are in a ring having 6 members  however the ring is concave .
Example
3 hydroxycyclohexanone

It doesn't work.  The ring geometry holds the carbonyl oxygen and hydroxyl hydrogen too far apart.  If you have a molecular model kit, construct the molecule with it and see what happens to the oxygen and hydrogen atoms.

Hydrogen bonding between different molecules is not constrained by the ring geometry, and thus works perfectly well.


How to convert isopropanol to acetone and hydrogen From Wikipedia:

  Isopropyl alcohol can be oxidized to acetone, which is the corresponding ketone. This can be achieved using oxidizing agents such as chromic acid, or  by dehydrogenation of isopropyl alcohol over a heated copper catalyst: $$\ce{C3H8O -> C3H6O + H2}$$

However, how exactly do I do this? Is it possible to do this by simply mixing isopropanol with an oxidizer (e.g. sodium chlorate)? Also, if the copper catalyst is actually required, what temperature is needed to drive the reaction? Neither the Wikipedia article nor the cited source provide much information on this. I want acetone as opposed to isopropanol, as a fuel, because its flash point is lower and it is much easier to ignite.

This is just an experiment, and if this method of producing acetone is impractical, I will just make/obtain it in other ways.

There are several ways to convert isopropanol to acetone.

Use of $\ce{Cr2O3}$ as catalyst[1]

  The conversion of isopropyl alcohol to acetone has been studied over $\ce{Cr2O3}$ powder at 473 K in a flowing microcatalytic reactor. Appreciable steady-state catalytic activity was achieved only when oxygen was present in the reactant gas flow. The effect of mild pretreatment of $\ce{Cr2O3}$ in $\ce{O2–N2}$ at 473 K on the initial activity, and the observed dependence of the process on $\ce{O2}$ pressure has been related to the oxidation–reduction cycle of the surface chromium species. $\ce{Cr^{(3+n)+}}$ on the surface of $\ce{Cr2O3}$ have been considered to be the active species, where n= 1–3.

2 . Catalytic dehydrogenation of isopropanol

a) Source 1.

  Catalytic dehydrogenation of isopropanol can be chosen as an
  alternative    synthetic route when high-purity acetone is required,
  such as in biomedical applications. Turton et.al. mentions that a
  single pass conversion of 85-92% with respect to isopropanol, with
  reactor conditions of 2 bar and 350° C, is generally achieved. A
  molten salt stream will be used a heat source for the endothermic
  reaction:

  $$\ce{(CH3)2CHOH -> (CH3)3CO + H2}$$

  The acetone produced in the reactor passes into a phase separator and
  then into a separation system that includes one stripping and two
  distillation columns. A recycle stream takes a mixture of unreacted
  isopropyl alcohol and water, with a trace amount of acetone, back into
  a mixer that feeds the reaction system. Using the catalyst which will
  be employed throughout this analysis, the reaction is first order with
  respect to the concentration of isopropanol and has an Arrhenius
  dependence on temperature with E=72.38 MJ/kmol and k=351,000 cubic m
  gas/cubic m reactor sec.

b) Source 2

  At the beginning of the process, feed including i-propyl alcohol and
  water, and recycle stream are mixed in feed drum. From here, this
  mixture is send to vaporizer to changestream’s phase as vapour. After
  vaporizer, mixture is heated to reaction temperature in theheater.
  Reactor used is a tubular flow reactor. Acetone, hydrogen gas (H2) are
  produced andwater and i-propyl-alcohol are discharged. The mixture
  with acetone, hydrogen, water, i- propyl-alcohol are sent to cooler
  and then to condenser. After condenser the mixture is sent toflash
  unit. Hydrogen, acetone, i-propyl-alcohol and water are obtained as
  top product. This top product is sent to scrubber to remove hydrogen.
  The bottom product of flash unit which isformed by acetone, water,
  i-propyl-alcohol are mixed with the bottom product of scrubber  before
  acetone column. In acetone column, acetone is obtained from top
  product with 99 wt%. i-propyl alcohol and water and also 0.1% of
  acetone is sent to i-propyl-alcohol column from bottom product. The
  top product of this column is sent to feed drum and bottom product is
  thrown away as waste water.

Using chromic acid [2]

  Oxidation at 25' to 40' is accordingly specified, requiring the
  mixture to be chilled in ice after each addition of oxidant. In 1926
  the writer worked out a student experiment which involves the same
  reaction, the oxidation being effected by gradual addition of chromic
  acid solution to the boiling solution of isopropyl alcohol in water.
  The reaction is carried out in a flask surmounted by a short
  fractionating column, through which the acetone is distilled during
  and after the oxidation. The crude acetone (collected below 65'), and
  several subsequent fractions, are redistilled through the column,
  permitting the easy isolation of a high yield (70% to 80% or more) of
  acetone boiling 57' to 60' and over 99% pure, and also the isolation
  of any unchanged isopropyl alcohol (76' to 82').

References:

Ilyas, M.; Shah, S.; Nigar, R.; Khan, H. Conversion of isopropyl alcohol to acetone catalysed by Cr2O3 at 473 K: role of molecular oxygen. Faraday Trans. 1994, 90 (16), 2413. DOI: 10.1039/FT9949002413.
Wagner, E. C. Oxidation of isopropyl alcohol to acetone. A student experiment. J. Chem. Educ. 1934, 11 (5), 309. DOI: 10.1021/ed011p309.


Confusion on ion product of water The ion product of water is defined as $$ K_w = [\ce{H+}][\ce{OH-}] $$ and in pure water $ [\ce{H+}] = [\ce{OH-}]= 1.0 \times 10^{-7}$ and $ K_w = 1.0 \times 10^{-14}$. It is said that if we dissolve some acidic substance in water then $[\ce{H+}]$ will increase and $[\ce{OH-}]$ will decrease and the value of $K_w$ remains constant. I understand that $[\ce{H+}]$ increases because acids donate $\ce{H+}$ (according to Brønsted definition of acids) but how $[\ce{OH-}]$ decreases. I mean a water molecule i.e. $\ce{H2O}$ will certainly break into $\ce{H+}$ and $\ce{OH-}$ and to this $\ce{H+}$ our acid has added it's part and hence it's concentration has increased but why the value of $[\ce{OH-}]$ gets lowered from $1.0 \times 10^{-14}$ , if water molecule were to ionize then whenever $\ce{H+}$ gets formed simultaneously we would get $\ce{OH-}$.

I can cite a common example problem

  The concentration of $\ce{OH-}$ ions in a certain household ammonia cleansing solution is $0.0025$. Calculate the concentration of $\ce{H+}$ ions.

We can solve this problem by using the equation $$ K_w = 1.0 \times 10^{-14}$$
$$ [\ce{H+}] [\ce{OH-}] = 1.0 \times 10^{-14}$$
and if put the value of $[\ce{OH-}]$ in the above equation and solve $[\ce{H+}]$, then we would get $ 4.0 \times 10^{-12}$. Here we observe that the value of $[\ce{H+}]$ has decreased from the it's original value in pure water, well this is understandable because ammonia being a base would consume $\ce{H+}$ but how the value of $\ce{OH-}$ has increased.

I want to know that how the chemical reaction can cause the increase or decrease of $\ce{OH-}$ when each molecule of $\ce{H2O}$ always going to yield one $\ce{H+}$ and one $\ce{OH-}$ always.

Thank you. Any help will be much appreciated.

It is not a chemical problem, but a trivial mathematical problem.

If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.

The key part is to understand that  chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same rate. Like

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

where the rate of ion creation equals the rate of ion recombination.

If there is an excess or deficit  of either of  $\ce{H+}$   or $\ce{OH-}$ ions, the rate of their recombination changes, while the rate of their creation remains the same. As consequence, the product of their concentrations converges quickly towards $K_\mathrm{w}$ to be in the equilibrium again.

The ammonia reacts :

$$\ce{NH3 + H+ <=>> NH4+}$$

That creates deficit of $\ce{H+}$. As consequence, dissociation

$$\ce{H2O -> H+ + OH-}$$

Is faster then recombination

$$\ce{ H+ + OH- -> H2O}$$

An alternative reaction mechanism is
ammonia reacting with water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

what directly produces the excess of $\ce{OH-}$ that recombines with the most of the present $\ce{H+}$.

The production of $\ce{OH-}$ and elimination of $\ce{H+}$ continues, until their concentrations satisfy both the basicity constant of $\ce{NH3}$ ( or equivalently acidity constant of $\ce{NH4+}$ ) and the ion product of water  $K_\mathrm{w}$

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

$$K_\mathrm{b, \ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}=\frac{[\ce{NH4+}]K_\mathrm{w}}{[\ce{NH3}] [\ce{H+}] }=\frac{K_\mathrm{w}}{K_\mathrm{a, \ce{NH4+}}}$$


Why are there only 7 periods in the periodic table? Is seven a hard theoretical limit (there can’t be an eighth period) or is it there just because we haven’t discovered/synthesized anything beyond that?

There is no theoretical limit to the number of rows, but...

On one hand, as Tom Lehrer states, there may be many others but they haven't been discovard.
On the other hand, after all s, p, d, f orbitals are filled in the seventh period, there might be a new row, or there might be an extension of the seventh row, as a new type of orbital (g? Anyone know it there's a name?). Each orbital has double an odd number (double because the Pauli Exclusion Principle allows pairs of electrons of opposite spins), so the next orbital would have 18 electrons. Is it possible? Inner electrons would be moving at very high speeds, and they'd also shield outer electrons from attraction by the positive nucleus.
On the gripping hand, transuranics may last a few seconds... or a few attoseconds, so getting spectra might be problematic.


Trouble with Converting from Structural Formulas into Newman Projections I have the following question from my Organic Chemistry textbook:

I'm asked to convert the following line diagram into a Newman Projection.

The arrow pointing to the carbon signifies "looking down the carbon".

My logic was the following:

At the carbon that is being pointed at by the arrow, there are 3 bonds to other carbons, all of which are in the same plane due to the absence of any 'dashed lines' or 'solid wedges' that would indicate otherwise.

I then concluded that the geometry of that carbon is trigonal planar and that it has a lone pair, since all the bonds to other carbons are in the same plane.

However, drawing a Newman Projection for an atom with trigonal planar geometry proves to be impossible.

I'm not sure how to go about converting this line diagram into a Newman projection.

How can I do so?

Thank you.

this is how you can convert the str.s into each other. It should have had some wedge/dash or sawhorse information... for the conversion.


Why do *tert*-butyl ethers in diluted sulfuric acid undergo elimation instead of solvolysis? My organic textbook states the pathway for propyl tert-butyl ether in dilute sulfuric acid will follow E1 with a tertiary carbocation intermediate. It mentions that Sn1 will only occur with good nucleophiles.

It does not explain why however. Why is this any different than a tertiary haloalkane in water? I expect that the alcohol is a good leaving group, similar to water.

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 7th edition; W. H. Freeman: New York, 2014, p350

It isn't necessarily the same situation as the haloalkane in water. In the sulfuric acid case, there is a large amount of acid present, and as you rightly say the alcohol and water are similar leaving groups, so the SN1 product is easily decomposed back to the tertiary carbocation. As a result, the mechanism tends towards the thermodynamic product, which apparently is the alkene (probably for entropic reasons.

Compare this with the alkyl halide in water. Here, the halide is a much better leaving group than water. Additionally, there is a low concentration of acid in solution (at most 1 equivalent, from the generation of HX). This means that loss of water from the SN1 product is much less favourable.

That said, I'm sure you'd get a mixture of E1 and SN1 products from the haloalkane reaction you describe, I merely wanted to point out that the ether and haloalkane cases aren't as similar as you may have presumed.


Any side effects to using calcium chloride as a desiccant? Based on cost, availability, and moisture absorption capacity, calcium chloride seems to be too good to be true. Are there any side effects to using calcium chloride as a desiccant?

I'm interested in effects that are directly attributed to the desiccant itself, not so interested in the effects of maintaining an extremely low moisture level in an enclosed environment.

For example, if I were to put a container of it in my vehicle, am I exposing the inside of the vehicle to corrosive fumes? Is there a danger of the calcium-chlorine bond breaking and releasing chlorine gas? I realize that the upholstery may dry up and crack if the moisture level is too low, but that can be managed. Rust build-up in the vehicle frame, not so much.

The only issue with $\ce{CaCl2}$ I've encountered is that eventually it pulls in so much water that it dissolves.

To keep dust down and to prevent the $\ce{CaCl2}$ solution from sloshing around, place it in a plastic container with a tightly-fitting sponge atop.

$\ce{CaCl2}$ is relatively non-toxic (it's used to make pickles), is inexpensive and is readily available, in cold climates, for melting snow and ice.


The relationship between energy and air pressure? When boiling water at room temperature, what is the relationship energy has with the air pressure that is making the water boil cold instead of hot?

Boiling occurs when the vapor pressure of a liquid reaches ambient pressure.  In the example you asked about, water can be made to boil by removing the air in the chamber in which the vessel holding the water is kept.  When the air pressure gets low enough, there is not enough pressure to keep bubbles of water created by the water's vapor pressure from expanding.  We call this boiling.  At atmospheric pressure the water still exerts vapor pressure but the air keeps the bubbles from expanding.  The only way for water to reach the vapor phase here is for it to evaporate from the surface so we see no bubbles.


What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d? What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?

It keeps mentioning that my regio-selectivity is incorrect. And that I should consider the charge distributions of the diene and dienophile to figure it out correctly, but I honestly have no idea.

These resonance structures and the polarization shown within explain the regioselectivity (1,4 isomer instead of 1,3) in the product.


How to I neutralize the smell of (VERY) spoiled milk Let's say I have a cup full of milk that had been at room temperature for months (apparently my son was running an experiment...). The smell started seeping out of the makeshift lid, and it's horrible. I have no background in chemistry but I imagine there might be a way to neutralize the smell before getting rid of the contents.
I'm worried if I just flush it down the pipes it might linger and make the whole house smell. Also, ideally I would like to salvage the cup because it's a very nice handmade glassware (he was running a fancy experiment).
Again, this has been there for months so I hope there isn't any dangerous bacteria still around, and any advice to get rid of the smell and clean the thing would be enormously appreciated.

The disgusting smell is probably butyric acid. Try adding a base: For example, washing soda (sodium carbonate), or perhaps drain cleaner (sodium hydroxide, be careful with this one, this is very corrosive to people/eyes), or even ammonia. This should turn the butyric acid into butyrate ion, which should be much less disagreeable. Then flush down the drain.


Why do lattice structures form into powders? For example, when you react magnesium with oxygen over heat, it forms a lattice of ions. Why does a strong lattice of ions like MgO turn out to be a powder, rather than a solid block?

(Sorry if the answer is obvious; I'm only just being introduced to this)

This is simply because the combustion of magnesium doesn't occur orderly enough to create one big solid block of magnesium oxide.

When magnesium reacts with air, the way of formation of the oxide lies at the mercy of the chaotic motion of oxygen due to it's entropy in the air. Often, the molecules react simultaneously at various locations in the air around the part of the ribbon undergoing combustion, creating several micro-crystals, which fall down due to gravity on reaching a certain weight. These micro-crystals appear as powder in the macroscopic scale.


Balancing using oxidation numbers For the reaction

KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O

I have Mn going under oxidation from +4 to +2
I have Cl going under reduction from -1 to 0

Then after using my teachers method o multiplying the oxidation and reduction atoms I have

4KCl + 1MnO2 + H2SO4 -> K2SO4 + 1MnSO4 + 2Cl2 + H2O

However I'm not sure I'm on the right track, but if I am, what are the next steps to completely balance the equation?

It is really easy to solve by the half reaction method, here is my answer:

Half reaction method

Global eq: KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O

Cathode: 2Cl-   --> Cl2 + 2e-

Anode: MnO2 + 4H+ + 2e- --> Mn+2 + 2H2O

So the final balanced eq is this one:

2KCl (ac)  + MnO2 (s) + 2H2SO4 (ac) --> K2SO4 (ac) + MnSO4 (ac) + Cl2 (g)  + 2H2O

Variations method

Without half-reaction method will be like this:
You have one Mn+4 reduced to Mn+2
That is 2e x 1 atom (ion indeed) so this is a net variation of 2
On the other hand you have Cl- wich oxidizes to form a Cl2 molecule so this is
1e x 2 atoms wich also is a net variation of 2
Now you use these numbers as the coefficient of the products that contain these atoms and you'll have

KCl (ac)  + MnO2 (s) + H2SO4 (ac) --> K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + H2O

and considering this as valid just left to adjust the rest of the elements
and finally you'll have:
4KCl (ac)  + 2MnO2 (s) + 4H2SO4 (ac) --> 2K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + 4H2O
Wich is also a valid adjust

I always recommend using half-reaction method or undetermined coefficient method instead. Good luck!!


What are the units of P and V when R is expressed in terms of J/(mol K)? I have learnt the ideal gas equation $PV=nRT$. However, I have some doubts regarding the units of the quantities in this equation.

When we take the gas constant $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$, what should be the units of $P$ and $V$?

Would you agree that pressure is expressed as a force acting on a surface? If so, let's juggle with units and start with the pascal.
$1\ \mathrm{Pa} = 1\ \mathrm{Nm^{-2}}$
will certainly be familiar to you ;)

Would you further agree that we can express the joule as $1\ \mathrm J = 1\ \mathrm{Nm}$?


What is the result of this elimination reaction? $$\ce{(R)-2-Bromooctane + alcoholic~KOH -> E2~reaction}$$

My answer is that this reaction will have a minor and a major product. In relation to the regiochemistry, the major product is the one which follows the Zaitsev rule, but this major product will still have two forms the trans- and the cis-, of course the major is the trans-. so there will be 3 different products.

Was I correct?

I am not an expert but I think you are correct. The alkylhalide is secondary, the nucleophile is a strong base and the solvent is polar protic. That leads to the conclusion that the most likely reaction to happen is E2. According to Zaitsev rule the major product should be the most substituted alkene because its transition state is lower in energy. That means that the main product would be 2-octene. In reference to the cis and trans structure, the trans one would be more stable. Taking all of the above under consideration, the main product of the reaction would be trans-2-octene and the minor products: cis-2-octene and 1-octene.


Does the concentration of water accounts for determining the rate of a reaction? I found this problem in my book and I'm not sure how to assess this question.

It is as follows:

  Regarding the relationship between the rates of the chemical species
  in the following reaction. Which of the following alternatives is the
  correct one?

$\ce{2H2 + O2 -> 2 H2O}$

$\begin{array}{ll}
1.&v_{H_2}=2v_{H_{2}O}\,, v_{O_2}=2v_{H_{2}}\\
2.&v_{O_2}=2v_{H_{2}O}\,, v_{H_2}=2v_{O_{2}}\\
3.&v_{H_2}=2v_{O_{2}}\,, v_{O_2}=v_{H_{2}O}\\
4.&v_{H_2}=v_{H_{2}O}\,, v_{O_2}=\frac{1}{2}v_{H_{2}}\\
5.&v_{H_2}=v_{H_{2}O}\,, v_{O_2}=v_{H_{2}}\\
\end{array}$

In the original source (which doesn't list a specific authorship) doesn't indicate the state of matter for each species. So I'm assuming perhaps that the reagents are gases and the products is liquid?.

Anyways what I've attempted to do is to establish the rate law as follows using the law of mass action:

$\ce{2H2 + O2 -> 2 H2O}$

For Hydrogen gas:

$v_{H_{2}}=-k[H_{2}]^2[O_{2}]$

For Oxygen gas:

$v_{O_{2}}=-k[H_{2}]^2[O_{2}]$

From this it can be established that:

$v_{H_{2}}=v_{O_{2}}$

For water (since it is on the products it must be established as if it were a reagent):

$v_{H_{2}O}=k[H_{2}O]^2$

And that's where I'm stuck.

Supposedly the answer is

$v_{H_2}=2v_{O_{2}}\,, v_{O_2}=v_{H_{2}O}$

But I don't see any way of how to get there. Can somebody point me exactly what did I missed?.

Rate law vs. definition of rate

For this problem, you don't need the rate law. And, just for the record, you can't get the rate law from the stoichiometry - it depends on the mechanism and has to be determined empirically.

Definition of rate of reaction

The rate of reaction is based on the change of concentrations. For a homogeneous reaction (i.e. all species in the same gas phase, or all species part of the same solution), the definition is the following:

$$\mathrm{rate} = \frac{d[\ce{X}]}{\nu_X \cdot dt}$$

where $\nu_X$ is the stoichiometric coefficient of the species $\ce{X}$. This does not apply to heterogeneous reactions. For one, if a species is a pure solid or liquid, its concentration will not change. Also, the stoichiometry of a reaction refers to amount of substance - only if all species are in the same solution (or the same gas phase) does this translate into easy relationships of concentration changes (or partial pressure changes).

Definition of "rates of the chemical species"

This is not a technical term I am familiar with. In order to be able to answer the question with answer 3., you would have to define it as:

$$v_\ce{X} = \frac{\mid dn_\ce{X} \mid}{dt}$$

Once you define it such, it does not matter what the physical state of reactants and products is, it does not matter whether a species is a reactant or a product, the only thing you need to related the different rates are the stoichiometric coefficients.

So for example, the ratio of the amount of hydrogen and oxygen used is equal to the ratio of stoichiometric coefficients:

$$\frac{\Delta n_\ce{H2}}{\Delta n_\ce{O2}} = \frac{\nu_\ce{H2}}{\nu_\ce{O2}} = \frac{2}{1}$$


Electronic oxygen concentration sensor I bought an electronic meter which measures the concentration of oxygen in air. But I think the sensor has failed because it is too old, as it had a crusty substance coming from it.but I can no longer find spare sensors for it.

How might this sensor have worked? I presume the property of some substance was measured; this property changes with the concentration of oxygen.

Perhaps one of these:

Its resistance changed as air was passed through some sort of chemical, and the resistance measured to determine the oxygen concentration.
Or perhaps its opacity to light measured with an LED and an appropriate optic sensor.
Or the volume of the substance changed, or some other property, which ultimately varies the capacitance.

What substance might this have been? I want to make my own device - I don't intend to fix this one, so I don't need to identify the substance used in this particular sensor.

I cannot say as to your specific sensor, but most oxygen sensors are electrochemical sensors built around concentration cells.

Take a look at the following diagram of a simple zirconia-based oxygen probe from Wikipedia. This type of sensor is very like the one in automobiles.

In the middle of the sensor is a gas permeable zirconia $(\ce{ZrO2})$ membrane between two gas permeable electrodes. The membrane and electrodes form the electrochemical cell. In many cases, an electrochemical cell will use a reference electrode with a standard reaction going on for comparison. In this sensor, the standard reaction is the same one that is going on at the other electrode. The difference is that at the standard electrode, the partial pressure ("concentration") of oxygen is known and at the other electrode it is unknown.

  But wait, how does that work?

Let's take a generic electrochemical process, where $\ce{M}$ in the following equation represents something generic that we can oxidize:

$$\ce{M(s) + O2(g) -> MO2(s)}$$

This cell has a standard potential $E^\circ$. The actual measured potential, however, will depend on the partial pressure of oxygen $P_{\ce{O2}}$ because the standard potential is defined at $P_{\ce{O2}} = 1\ \text{atm}$. The relationship that gives us the actual electrode potential at some other oxygen level is the Nernst equation;

$$E=E^\circ - \frac{RT}{nF}\ln{Q}$$

Where:

$R$ is the ideal gas constant
$T$ is the temperature
$n$ is the number of electrons transferred (4 in my example)
$F$ is the Faraday constant
$Q$ is the reaction quotient, in this case $Q=\frac{1}{P_{\ce{O2}}}$

If $P_{\ce{O2}}$ is large, then $E$ is small and vice versa.

Now, we have two electrodes, our unknown and our reference, each with their own potential:

$$E_{ref}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},ref}}}$$
$$E_{unk}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},unk}}}$$

The voltmeter in the device measures the difference between these two potentials and does some maths to calculate back $P_{\ce{O2},unk}$ and then convert it to concentration or percent. Note that it does not even matter what $\ce{M}$ and $\ce{MO2}$ were, since they are not in the equation and even $E^\circ$ cancels out.

$$E_{unk}-E_{ref} = \frac{RT}{nF}\ln{\frac{P_{\ce{O2},ref}}{P_{\ce{O2},unk}}}$$

  What substance might this have been?

If your sensor was this kind of sensor, the crusty mess is probably zirconia.


Will monosodium dihydrogen phosphate react as base or acid in water? Will $\ce{NaH2PO4}$ react as as an acid or a base if you make it react with water?

I got stuck here. I can figure out the reaction:

$$\ce{NaH2PO4 + H2O -> Na^+ + H2PO4^- + H2O}$$

But when I look $\ce{H2PO4^{-}}$ up, it says that it can both be an acid and a base.
Any ideas?

Monosodium dihydrogen phosphate, $\ce{H2PO4^-}$, is an amphoteric species and it will act as both an acid and a base.
These are the related chemical equations:

\begin{align}
  \ce{H2PO4- + H2O &<=> HPO4^2- + H3O+}  & \mathrm{p}K_\mathrm{a} &= 7.21\\
  \ce{H2PO4- + H2O &<=> H3PO4 + OH-}     & \mathrm{p}K_\mathrm{b} &= 11.88
\end{align}

As an example, $\pu{0.1 M}$ solution of such a salt will have a pH of about $4.7$.
In other words, the acidic behaviour dominates over the alkaline one. This is due to the value of $\mathrm{p}K_\mathrm{a}=7.21$ (related to the acidic behaviour) compared to the $\mathrm{p}K_\mathrm{b}=11.88$ (related to the alkaline behaviour).


Why can't hot water break disulfide bonds when the energy storage in hot water exceeds the bond energy? I teach Biochemistry and know that heating water can break weak bonds like hydrogen bond or ionic interaction, but not disulfide bond (without reducing agent). The bond energy of hydrogen bond in water is about 12–30 kJ/mol, while the disulfide bond is 251 kJ/mol. I initially thought the heated water would have the energy much lower than 251 kJ/mol, but higher than 30 kJ/mol. But when I looked at the energy storage in 1 L of water at 90 degrees Celsius from here, it was calculated to 294 kJ. Now I wonder whether I can compare this energy to the bond energy. What do I miss here?

Chemical reactions do not simply occur because there is enough energy around to make something happen. They also require a mechanism that enables something to occur. It isn't enough that the product will release energy on some reaction happening.

Some reactions will happen spontaneously with enough input of thermal energy to start them off (burning coal, for example). Others, like breaking disulphide bonds) will not. Even when there is a plausible mechanism for a reaction, you have to add enough energy to make the reaction happen which may be much more energy than you expect. Most reactions have an "activation energy" to get past the barriers specific mechanisms holding back the reaction.


Reaction between copper(II) iodate and iodide I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $\ce{Cu(II)}$ makes things a bit more complicated since it can also be reduced to give $\ce{CuI}$.

To determine the solubility product of copper(II) iodate, $\ce{Cu(IO3)2}$, by iodometric titration in an acidic solution ($\pu{25 ^\circ C}$), $\pu{30.00 cm3}$ of a $0.100$ molar sodium thiosulfate solution are needed to titrate $\pu{20.00 cm3}$ of a saturated aqueous solution of $\ce{Cu(IO3)2}$.
1.1 Write the sequence of balanced equations for the above described reactions. [...]

After realising that $\ce {Cu^2+}$ is reduced as well, I wrote the following overall equation :
$$\ce{2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O} \tag{1}$$
which is based on
$$\begin{align}
\ce{2Cu^2+ + 4I^- &-> 2CuI + I2} \tag{2} \label{eq:cuii} \\
\ce{2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O} \tag{3} \label{eq:iodate}
\end{align}$$
However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $(\ref{eq:iodate})$ by $2$ and the overall equation given is:
$$\ce{2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O} \tag{4}$$
Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?

In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.


Does nitrogen inversion affect the basicity of amines? If I were to compare the basic strength of 1-azabicyclo[2.2.1]heptane and triethylamine:

Can I say that 1-azabicyclo[2.2.1]heptane is more basic than triethylamine because the lone pair of electrons is less available in the latter due to rapid nitrogen inversion? Nitrogen inversion is not possible in the bicyclic amine.

There is not much difference in the inductive effect. In both cases, the nitrogen atoms are $\mathrm{sp^3}$ hybridized.

Background

Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons.  The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$) than the lone pair in pyridine (the lone pair is in an $\ce{sp^2}$ orbital), which is less acidic then the lone pair in a nitrile (the lone pair is in an $\ce{sp}$ orbital).

(image source)

The rate of nitrogen inversion also correlates with, among other things, hybridzation. But in cases where these "other things", such as ring constraints, limit nitrogen inversion, then the rate of nitrogen inversion may not correlate with hybridization. So the rate of nitrogen inversion does not always correlate with hybridization and is therefore not a good indicator of lone pair "availability".

Answer

Looking at your two compounds, a reasonable first guess might be that the nitrogens in these compounds are both roughly $\ce{sp^3}$ hybridized.  Therefore, even though one compound can undergo nitrogen inversion and the other can't, their hybridzations are similar and we would expect similar basicities.

We could refine this prediction by noting that in the bicyclic compound, due to geometric constraints, the carbon-nitrogen bonds around nitrogen are slightly pulled back a bit from the tetrahedral angle.  This would increase the p-character in these bonds (as a bond angle moves from 109° to 90° the p-character increases with 90° being pure p).  As a result, the p-character in the lone pair orbital decreases and the s-character increases.  As mentioned above, more s-character in the lone pair orbital leads to lower basicity.

Therefore we might expect both compounds to have similar basicity with the bicyclic compound perhaps being a bit less basic.

This analysis is corroborated by the $\mathrm pK_\mathrm a$ values of the conjugate acids (a more positive number means that the amine is more basic). According to J. Org. Chem. 1987, 52 (10), 2091–2094, quinuclidine has a $\mathrm pK_\mathrm a$ of 10.90, and 1-azabicyclo[2.2.1]heptane a $\mathrm pK_\mathrm a$ of 10.53. The decrease in basicity can be attributed to the increased strain and greater s-character of the lone pair. Triethylamine itself has a $\mathrm pK_\mathrm a$ of 10.75 (Evans table), making it marginally more basic than 1-azabicyclo[2.2.1]heptane.


Why can an s orbital only hold two electrons? According to Pauli’s exclusion principle, an $s$ orbital contains at most two electrons with the opposite spin (up and down). Why can't an $s$ orbital contain a third electron whose state is the linear combination of spin up and down?

Electrons are magnets, they have magnetic fields. Those fields have only two possible orientations, and a single orbital can only be occupied by two electrons if those orientations are mutually opposed.

OK, but why? Why are electrons magnets? Why are there only two possible orientations for the fields? Why can the two electrons in the orbital only exist with opposite directions for the magnetic moments?

Well, there are answers to those questions, but they don't make sense a lot of sense. The magnetic moment of electrons is due to a property called "spin"; the spin value of an electron is 1/2, so it can adopt one of two spin quantum states, +1/2 and -1/2, which correspond to the "up" and "down" orientations of the magnetic moment. Since it has a half-integer spin, an electron is a member of a class of sub-atomic particles called "fermions" which obey rules called the Pauli Exclusion Principle and Fermi-Dirac statistics - one key result of these rules is that no two identical fermions can simultaneously occupy the same quantum state: you cannot have two electrons in the same orbital if they have the same spin orientation, so if one is +1/2, the other must be -1/2, and no more can be added because a third would have to adopt the same quantum state as the one of the first two.

OK, but why? Why does an electron have a spin value of 1/2? Why can't two fermions adopt the same quantum state (when integer spin particles, called bosons, can do so)? Why does the property of "spin" even exist? You need quantum mechanics to even begin to answer those questions -- again, that's more math than anybody is prepared to show you for at least five years -- and even then, the "why" of the Pauli Exclusion Principle is an empirical observation (it has no theoretical basis, it's just an after-the-fact thing that happens) without relativity, but there is a firm theoretical rationale and a reason "why" if one revisits the postulates of quantum mechanics and takes into general relativity into account. But relativistic quantum field theory is more math than any mortal should be forced to endure.

OK, but why? Why is quantum mechanics how the universe behaves? Why is general relativity the way the universe behaves?

You can play the "why" game for a very long time. None of your high school teachers will be able to give you more than one or two repetitions. University professors will be able to go a lot farther. Eventually, you get to questions to which the only possible answer is "nobody knows.... yet". And those are the questions that scientists -- including those university professors -- spend their day trying to answer.

Which is terrific fun. So keep asking "why", even to the answers your teachers give you. And if they can't answer, go ask someone else, or read about it, or go to university. And if they don't know, be a scientist, and figure it out yourself, and tell the rest of us.


Organisation in Periodic table Why do we find big spaces in the periods of the Periodic Table between $\ce{H}$ and $\ce{He}$, $\ce{Be}$ and $\ce{B}$, and $\ce{Mg}$ and $\ce{Al}$?

What is the logic of such organization of the periods?

From a historical perspective, this was done to account for the commonalities between how the various elements behaved. In other words, the Alkali metals all exhibited a strong reaction with $H_2O$, the noble gasses all exhibited inertness in reactions, and so on.

This is explained in modern day science via the quantum description of electronic structure which indicates that elements within the same group tend to have the same number of valence electrons in the valence shell. Simply put, the reason why there are such gaps between $H$ and $He$; $Be$ and $B$, and otherwise is because there are simply different quantum arrangements of the electrons which are readily accessible to the element. That is, the ground state of $He$ has both accessible $l=0$ states occupied by electrons, which constitutes the valence shell and so is in an energetically favorable state to remain inert - becoming of the noble gasses. Hence, it’s moved far to the right from $H$ to the 8th group; consistent with the historical structure of the table.

That’s a crude description, but about what’s happening here. I agree however that the chemists may feel more at home discussing the periodic table. I’ll stick to the chart of nuclides.

Also, the Wikipedia article is fairly informative on this topic:
http://en.wikipedia.org/wiki/Periodic_table


Relationship between rate equation and equilibrium constant I have read mannaia's answer on the question of 'rate order and confusion', as well as Nicolau Saker Neto's answer to a related question. They have both been very helpful.

As I understand, the equilibrium is actually derived from the rate constants of a reaction. At dynamic equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction, hence in a hypothetical reaction

$$
\ce{pA + qB <=> rC + sD}
$$

with the rate equations $Rate_{fwd} = k_{fwd}[A]^{a}[B]^{b}$ and $Rate_{bck} = k_{bck}[C]^c[D]^d$, with $a, b, c,$ and $d$ not related to $p, q, r$ and $s$.

The equilibrium is derived as such:
\begin{align}
  R_{fwd} &= R_{bck}\\
  k_{fwd}[A]^{a}[B]^{b}&=k_{bck}[C]^c[D]^d\\
  \frac{k_{fwd}}{k_{bck}} &= \frac{[C]^c[D]^d}{[A]^a[B]^b}\\
  \therefore k&=\frac {k_{fwd}} {k_{bck}}\\
\end{align}

Based on the derivation, and the fact that the stoichiometric coefficients of the reactants in the rate-determining step (RDS) of the reaction is equal to the order of the reactants in the rate equation, is it correct to state that the reaction equations given as the equilibrium equation is actually the rate-determining step of two reactions?

At equilibrium it does not matter if the reaction has multiple steps; your equilibrium constant only has to involve the initial reactants and final products.

For instance, in an S$_\mathrm{N}$1 reaction, obviously the RDS is the departure of the leaving group, but you don't have to (and usually are not in a position to) know the concentration of the intermediate carbocation in order to calculate the equilibrium of the overall reaction.  You can simply write an  equilibrium equation for the initial reactants and final products: $$K=\frac{[\ce{LG}][\ce{product}]}{[\ce{nucleophile}][\ce{substrate}]}.$$

Of course finding $K$ may be not be easy, even if you have access to good thermodynamic data for the reactants and products, because $\Delta G_0$ (and therefore $K$) depends on solvation, temperature and other conditions. But given identical conditions, $K$  is constant over a wide range of concentrations, so you might be able to find it in the literature or measure it with a single experiment.

All this assumes that the RDS is fast enough that you're able to reach  equilibrium before the solvent evaporates, or your reactants and products decompose via other pathways, or you die of boredom!

Note that you can also write an equilibrium equation for the RDS alone:
$$
K'=\frac{[\ce{LG}][\ce{carbocation}]}{[\ce{substrate}]}
$$
But since you seldom know the concentration of the carbocation, this equation isn't as useful in practice (it is of course extremely useful for mechanistic studies).

Edit: This page has a more complete and very nice answer!


How to store small amounts of gas I need to store small amounts of gas. I already saw devices that look similar to the image below, but I don't know how they are called. Does someone here know how these devices are called? I am also open for alternative ideas. I mainly need some kind of storing devices that works with low pressure and that is capable of pushing the gas back if pressure is removed. A ballon for example wouldnt work since it requires to much force to get the gas inside. It would be awesome if someone here know more and can help me with this.

A rubber balloon does require some force to get filled. A mylar balloon (typically filled with helium) can be purchased empty (or can be emptied) and requires almost no force to fill; then can be squashed manually to eject the gas. They are available in many sizes.

Rubber balloons also tend to go flat or leak, so their storage utility, especially for low m.w. gases, is short-term (a day or so). Aluminized mylar balloons can hold helium or hydrogen for long periods (weeks and even a month or two). For short-term, you could also use a plain polyethylene sandwich bag - after all, many objects are shipped with bags of air in the boxes to help protect them from damage. Use a needle to insert the gas and retrieve it; tape the area around the needle insertion - like an intravenous operation! Or, you can tape a plastic or glass tube into the bag.

Longer storage would require a gas sampling or collection tube, a glass container with a valve at one or both ends. The internet has many references to these.


How to prepare and store anisaldehyde tlc staining solution? What will be the color of this solution? What is its shelf life? I prepared  anisaldehyde charring solution,  and refrigerated it.  Its color was yellow.
However when I am using it for staining tlc plate,  it is giving  dark pink color on tlc plate surface and unclear blue stains...  Also not all the stains are visible.  If I am not wrong it should have white background  with dark brown spots appearing clearly.

What mistake I am doing here possibly?

p-Anisaldehyde (PAA) stain

This posting from www.ochemonline.com (which is available under a CC-By-SA-3.0 licence) provides the following information about the stain and its behavior. The information below is consistent with my personal experience using PAA stains.

Color: Range of colors on a light pink background (my experience was mostly blue spots)
Medium shelf life

The post on stains goes on to say:

  This stain is light and oxidation sensitive and will gradually turn pink/orange. The stain should be kept in an aluminum-covered jar while in use, and the excess should be kept cold and in the dark. Once the stain turns dark red, it should be discarded and made fresh again.

Now you have your answer - the stain should turn pink over time, it should produce (mainly) blue spots on a pink background, and you have a visual indicator of its shelf-life. Once it is dark red, it is done. To keep the stain around longer, wrap the jar in foil.

Brown spots on white background

The iodine stain is the only stain I know that does this for most compounds.

Not all the spots are visible

PAA is principally an electrophile and stains nucleophilic compounds more strongly than other types of compounds.

Personal recommendation

I grew frustrated with the PAA stain because it degraded quickly and switched to the vanillin stain. The recipe is in the link posted, and reproduced below scaled down. Make small batches to avoid waste. You want to use the stain faster than it decomposes.

  Dissolve 3 g of vanillin in 50 mL of ethanol. Slowly add 0.5 mL of concentrated sulfuric acid.

This stain is very sensitive to functional groups and produces a wider range of functional group specific colors than PAA. Vanillin (the structure of vanillin is simiular to PAA, but with an additional OH on the ring) is both an electrophile and a nucleophile and so reacts with a wider range of compounds.

Ochemonline indicates that it has the same "medium shelf-life" as PAA, but in my experience, PAA decomposed more quickly. I left my vanillin stain in an amber glass jar without foil wrapping in my hood at room temperature, and since I was doing lots of TLC every day, I used it faster than it degraded. PAA stain degraded much more quickly for me under these conditions, and I was throwing half of it out despite the number of TLC plates I was doing. If you are not doing TLC every day, then wrap in foil and refrigerate.

  Best reason to use vanillin: The awesome smell!

Poor spot quality

This main not be a problem with your stain.

Big streaky spots are indicative of too high concentration in your solutions.
Alternatively, you may need a more optimal solvent. For example, when doing amines on silica, you may want to add 1% triethylamine to your TLC solvent since silica is acidic. Amines and other basic compounds will streak less in these conditions.
Big streaky spots may also mean poorly resolved ocmpounds
Faint spots may mean poor compatibility with stain or that the concentration was too low.
If spots run together, perhaps your lanes are too close, the bottom of the TLC plate is not level, or your plate touched the side of the developing chamber.


Conjugate acids and bases Consider the reaction

$\ce{NH3 + H2O -> NH4+ + OH-}$

Here, $\ce{NH3}$ is a weak base, and thus $\ce{H2O}$ becomes a weak acid. Thus, shouldn't its conjugate be a strong acid? However, $\ce{NH4+}$ is a weak acid, instead of the strong conjugate acid it should be. Why is this?

Consider the $pK_b$ of the base $\ce{NH3}$ and the $pK_a$ of the acid $\ce{NH4+}$.

$pK_b$ = $4.8$

$pK_a$ = $9.2$

$\ce{NH3}$ is a stronger base than what $\ce{NH4+}$ is as an acid, being $4.8$ smaller than $9.2$: let's compare $pH$ for equal concentrations, say $0.1 M$.

For $\ce{NH3}$, you get $11.1$, while for $\ce{NH4+}$, you get $5.1$.

$11.1$ is about 100 times more "distant" from neutral $pH$ than $5.1$, if we move from the $pH$ scale to the concentration scale. That's what also stems from the comparison of the $pK$ values.

To simplify the problem, take the $pK_a$ for the acid/base couple, e.g. $\ce{NH4+/NH3}$, that is equal to $9.2$. If that value is larger than $7$, then, within that couple, the base will be stronger than its conjugated acid. The other way round, if $pK_a$ is lower than $7$.


In which cases do you say "oxide" when naming a molecule? Based on the names of following molecules I got quite confused about the correct nomenclature:

$\ce{NaHCO3}$
$\ce{K2SiO3}$
$\ce{PbCrO4}$

In none of the above molecule names you ever say "oxide" although there is a oxygen atom. Why? What is the applied rule for the nomenclature? In which cases do you actually say the word "oxide" in the name of the molecule?

The naming nomenclature for inorganic molecules is confusing, but if you are studying chemistry then you just have to grit your teeth and learn.

First there are household names, common names and chemical names. Vinegar, acetic acid, and ethanoic acid are talking about the same molecule. We say water, but aqueous solutions. $\ce{SiO2}$ is silicon dioxide to a chemist, but quartz to a geologist. The list could go on and on...

In general an oxide is compound which has just some element, X, and oxygen in some combination - ${\text{X}_m\text{O}_n}$

Some anion combinations get a name.

$\ce{CO3^{2-}}$ is carbonate
$\ce{SiO3^{2-}}$ is metasilicate
$\ce{CrO4^{2-}}$ is chromate


Order of −I effect among ethenyl and 1-propenyl groups We know that both ethenyl $\ce{-CH=CH}_2$ and 1-propenyl $\ce{-CH=CH-CH3}$ are electron-withdrawing groups when attached to hydrocarbon chain. Which one has greater electron withdrawing power?
The $\ce{-CH3}$ is electron-donating, hence allyl should have lesser electron withdrawing effect. Is this correct?

I would say yes, the -CH3 group adds a little +I effect, making the -CH=CH-CH3 a little less electron withdrawing compared to the vinyl.


Why are (3-methylbut-2-en-2-yl)benzene and (3-methylbut-1-en-2-yl)benzene not stereoisomeric? My answer key says neither product exhibits stereoisomerism, but both products are substituted. Why is it that I cannot assign E/Z configuration? I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.

I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.

Well, you cannot assign E/Z when it doesn't make any sense.

tl;dr: When on at least one side of double bond two identical groups or atoms (e.g., Hydrogen), there's only one way to arrange substituents near double bond.

E/Z is used to explain how substituents on different sides of double bond located, which excludes 1,1-substituted ethylenes (which is the case with minor product on your figure).

When on one side of double bond is two identical groups, e.g., methyl (which is the case with major product) there's no E/Z isomerism too, because there's only one way to arrange substituents. You can actually think of first case as of the second, minor product has 2 identical atoms at one side of double bond.

Hope I haven't made more confusion than there was.

Also, I would suggest to play with some ball and stick models (They could be at the uni library), I think that building couple of examples and trying to rearrange them to stereoisomeric form will help with understanding stereochemistry in general.


How to calculate Absolute error under square root sign? I was calculating standard deviation of a data set. Could someone help me find the square root of 15.13 with an absolute error of 0.005? Is there a rule for calculating absolute errors under square root sign?

A slightly different approach to @Zhe's answer is to use the 'propagation of errors' formula. Your formula is $y=\sqrt{x}$  with $x=15.13$ and an error $\sigma_x=0.005$ then the formula becomes $\displaystyle \sigma_y=\frac{\mathrm dy}{\mathrm dx}\sigma_x = \frac{\sigma_x}{2\sqrt{x}}= 6.4\cdot 10^{-4}$.

[if you have a formula in more than one variable , say $x$ and $z$ then $y=f(x,z)$ and the equation is more complicated but still quite easy if done in steps and is $\displaystyle \sigma_y^2=\left(\frac{\mathrm dy}{\mathrm dx} \right)^2_z\sigma^2_x+ \left(\frac{\mathrm dy}{\mathrm dz} \right)^2_x\sigma^2_z$. Notice that the square of the final error is calculated in this case.]


Is a strontium-fluorine battery the highest voltage battery using pure elements? Strontium has a very low standard electrode potential and fluorine has a very high one.

$$ E_{\ce{F2 + 2e^- -> 2F^-}} = 2.87\mathrm{V} $$
$$ E_{\ce{Sr -> Sr^+ + e^-}} = -4.10\mathrm{V} $$

In theory a strontium-fluorine battery would have a voltage of $6.97\mathrm{V}$ although there are many practical reasons such as danger and rarity of materials for such batteries not to be made.

Is a strontium-fluorine battery the theoretically highest voltage chemical battery using pure elements or is it possible to obtain a higher one?

The half-potential you've given for strontium is only for the first ionisation. The half-potential you'd actually get is -2.899V for the stable dication to give a cell potential of 5.769V. From the CRC Handbook[1], lithium has the lowest element->stable ion potential of -3.0401V, which is why it's common in batteries.

[1]: Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.


how to destroy excess iodomethane in reaction mixture? I have used iodomethane to make a phenolic $\ce{OH}$ group to $\ce{OMe}$ group in Aq+Methanol medium. I have added water to precipitate the product of mine and will evaporate the solvents but I cant use rotavap until I am sure that any residual iodomethane is not contaminating the condenser of rotavap. How shall I destroy excess iodomethane? will aq. solution of $\ce{KOH}$ work?

Yes, you can destroy small amounts of methyl iodide by hydrolysis with an aqueous solution of sodium hydroxide. You can just add some sodium hydroxide solution to your reaction mixture, if your molecule is stable to basic conditions.

The reaction is: $\ce{NaOH + MeI -> MeOH + NaI}$

From: "A Comprehensive Guide to the Hazardous Properties of Chemical Substances" by Pradyot Patnaik, 2007


I am trying to find the solubility of benzoic acid in different solvents to prepare a liquid-liquid extraction I need to prepare a liquid-liquid extraction using the partition coefficient (K) of benzoic acid in different solvents and water:

K ~ solubility of benzoic acid in solvent/solubility of benzoic acid in water

Several solvents are offered to me (e.g. ethyl acetate, chloroform,...), and I am trying to find the solubility of benzoic acid in each of those solvents. The problem is that I can't find all of them (e.g. benzoic acid in ethyl acetate). I can't seem to find a database of such solubilities. I would greatly appreciate if someone could please re-direct me to such database. Conversely, can solubility be predicted efficiently?

Thank you very much for your time!!

Funny you're looking at solubility of benzoic acid. A few years ago, there was an Open Notebook Science challenge to measure solubility:

  The first round of this challenge calls upon people with access to materials and equipment to measure the solubility of compounds (aldehydes, amines and carboxylic acids are a priority) in organic solvents and report their findings using Open Notebook Science

So the raw data is available through Google Sheets here

There is a general web service for predicting solubility.

For example, you can predict the solubility of benzoic acid using Abraham solvation parameters

There used to be a search mechanism, but I'll have to ask the authors - it's been a few years.


How to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation? I am struggling with what appears to be an extremely easy pH problem that uses the Henderson-Hasselbalch equation. The problem and answer provided by the book is given below. I understand where the book is getting the answer but it appears that they are overlooking something.

From what I understand, the Henderson-Hasselbalch equation allows you to calculate the pH of a solution based on the concentrations of acid and conjugate base at equilibrium. In the problem, you start off with a solution of just the conjugate base. Then you add pure acid to the solution to get a desired pH. However, the book seems to assume that the acid you add completely contributes to the concentration of undissociated acid at equilibrium [HA]. How can this be the case? Some of the acid you add will surely dissociate. Anyways, here's what the book did...

Boric acid and sodium borate forms an acidic buffer (A mixture of weak acid and salt of weak acid and strong base). The acid is weak hence the concentration of basic ion from acid is much less compared to the concentration of basic ion obtained from the salt which can ionize completely. Yes. Some of the acid dissociates to form hydronium ion and a basic ion. That's how a buffer works.

Consider a weak acid HA and the salt BA.
Its easy to show that the concentration of $H^+$ due to ionisation of HA is $$[H^+]=k_a\frac{[HA]}{[A^-]}$$
It is reasonable to assume that the concentration of $A^-$ from BA is much larger than that from HA due to complete ionisation of BA. Hence $[A^-]=[BA]$=[salt]
So, $$[H^+]=k_a\frac{[acid]}{[salt]}$$
Taking logarithm, you get the desired equation.


How effective is sulfur in detecting mercury vapors I've read that powdered elemental sulfur will react to the presence of mercury vapor in the air creating red mercury sulfide, and that an open container of sulfur can be used as a means to verify leaks in a container of mercury that's being kept in a storage cabinet. But is the sulfur reaction/color change sensitive enough relative to the permissible levels of mercury vapor that might exist?

The vapour pressure of mercury is appreciable, but the evaporation kinetics are slow.  In a well-ventilated lab there shouldn't be concern about mercury poisoning from broken thermometers, Schlenk lines &c.

That said, there is Alfred Stock's famous review (Angewandte Chemie, 1926) on mercury poisoning.  He suffered from a chronic case of mercury poisoning, which IIRC really took off after the war, when they had to save on ventilation.  In old laboratory buildings the quantities found under the wood flooring during renovation are at times staggering.  Our safety guys liked to show pictures of puddles.

To detect mercury there are commercially available detectors, usually build around the resistance of a vacuum-deposited gold film (a few 100 atoms thick).  A low-pressure mercury lamp might also work.


Suppose a heated 5.00 g sample of sodium bicarbonate lost a mass of 1.00 g, what percent of impurity is in the sample? The 5.00 g is a mixture of sodium bicarbonate and an unknown substance, making it impure.

Balanced decomposition equation:

$$\ce{2NaHCO3(s) -> NaCO3(s) + H2O(g) + CO2(g)}$$

There are 0.06 mol of $\ce{NaHCO3}$ in a pure 5.00 g sample of sodium bicarbonate.
Theoretically, 0.0298 moles of $\ce{H2O}$ and $\ce{CO2}$ would be lost durin the decomposition of 5.00 g of sodium bicarbonate.
0.0298 mol converted to grams is:
0.536 g of $\ce{H2O}$
1.31 g of $\ce{CO2}$

First there is a big gotcha. We'll let $x$ be the grams of sodium bicarbonate and $y$ be the grams of the impurity. Since we were given no information about the impurity we must assume that the impurity neither loses mass nor gains mass when heated with the sodium carbonate. So: $$x + y = 5.00$$

Well the balanced equation is really:

$\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$

so 2 moles of $\ce{NaHCO3}$ yields 1 mole of $\ce{Na2CO3}$. Thus the mass conversion factor is: $$\dfrac{0.5*\text{MW(}\ce{Na2CO3}\text{)}}{\text{MW(}\ce{NaHCO3}\text{)}} = \dfrac{0.5*106.0}{84.0}= 0.6310$$

So we now also have: $$0.6310x + y = 4.00$$

Subtracting the second equation from the first we get: $$0.3690x=1.000$$ so $$x=2.71g$$ and $$\text{Purity} = \dfrac{2.71}{5} = 54.2\%$$ $$\text{Impurity} = 100\% - 54.2\% = 45.8\%$$


Decrease in temperature of an aqueous salt solution decreases conductivity Why does the conductivity of a water solution drop as the temperature decreases?

How are these two related?

Decrease in temperature has two effects, both attributing to lower electrolytic conductivity:

decreases the mobility of the charge carriers (e.g. $\ce{H3O+}$ and $\ce{OH-}$ for pure water);
suppresses auto-ionization of water (higher $\mathrm{p}K_\mathrm{w}$), reducing the total number of charge carriers.


Are there any datasets containing molecules with more than 38 heavy atoms? I have been testing a machine learning approach for molecular energy prediction. The current dataset that I have is QM9, which is consist of molecules with up to 9 heavy atoms.

I was wondering if anyone knows of available datasets that contain molecules with large numbers of heavy atoms. I will be testing ZINC, which has up to 38 atoms. Anyone knows of a larger dataset available?

This sounds like you were exploring work at least related to the work by the Lilienfeld group equally hosting a dedicated site here about data sets already used in their earlier and ongoing exploration of chemical space, programs used to work with the data, and publications.

To go considerably higher in molecule count than QM9, you could either go for

GDB-11 about small organic molecules up to 11 atoms of C, N, O and F which «contains 26.4 million molecules (110.9 million stereoisomers), including three- and four-membered rings and triple bonds», described in J. Chem. Inf. Model. 2007, 47, 342-353 (doi.org/10.1021/ci600423u), or
GDB-13, about «small organic molecules up to 13 atoms of C, N, O, S and Cl following simple chemical stability and synthetic feasibility rules. With 977 468 314 structures, GDB-13 is the largest publicly available small organic molecule database to date».  This one was described in J. Am. Chem. Soc. 2009, 131, 8732-8733 (doi.org/10.1021/ja902302h)

Convienently, you can download both -- including sub-sets like «containing only carbon and nitrogen», or «chlorine and sulfur», or «fragrance like» in case you don't want to fetch 2GB of already compressed data -- from the Reymond group. To quote: «All the molecules are stored in dearomatized, canonized SMILES format.»

The even larger GDB-17 («of up to 17 atoms of C, N, O, S, and halogens» with an universe of 166 billion entries, described in J. Chem. Inf. Model. 2012, 52, 2864-2875, [doi.org/10.1021/ci300415d, open access]) is accessible to the public on this site as a 50 million random subset only, partly because the gzipped archive is about 400GByte.  Among the publications citing this work is for example the Lilienfeld group again for machine learning (J. Chem. Phys. 143, 084111 (2015), doi.org/10.1063/1.4928757).

Initially, I misinterpreted the question but think the answer may be more rounded by addition of the following complementary publication:  «Chemical diversity in molecular orbital energy predictions with kernel ridge regression» (J. Chem. Phys. 150, 204121 (2019), doi.org/10.1063/1.5086105, preprint available here).  Aiming for a machine-learning analysis, the authors first compared QM9, 44k conformers of proteinogenic amino acids (AA), and a 64k set of organic molecules extracted from the CCDC potentially suitable for organic electronics (OE) for the content of atoms per molecule and found the following distribution:

To shed some light on them:

QM9 represents 133,814 small organic molecules with up to 9 heavy atoms (C, N, O and F)
AA is about «44,004 isolated and cation-coordinated conformers of 20 proteinogenic amino acids and their amino-methylated and acetylated (capped)
dipeptides. The molecular structures are made of up to 39 atoms including H, C, N, O, S, Ca, Sr, Cd, Ba, Hg and Pb.»
OE is about «64,710 large organic molecules with up to 174 atoms extracted from organic crystals in the Cambridge Structural Database (CSD). [...] The OE dataset is not yet publicly available. OE offers the largest chemical diversity among the sets in this work both in terms of size as well as number of different elements (Fig. 2). It contains the 16 different element types H, Li, B, C, N, O,  F, Si, P, S, Cl, As, Se, Br, Te and I.»

(The mentioned restriction sharing the original data relates to the user agreement with the CCDC.)

Further DFT-based property computations with these OE extracted molecular geometries lead to an ensemble of equilibrium molecular structures, and these derived geometries are accessible within a public Jupyter notebook.  Shared with the public here, the deposit comes with a guiding tutorial.ipynb, including an example how to retrieve these optimized geometries and display them with Jmol.


What are some easily-undertaken reactions that will precipitate or restore the iron from rust, in a non-lab context? Unusually, this question arises in a DIY context rather than a chemistry one.

I'm doing some work in my garden with reinforced concrete (a retaining wall against a  high bank of soil). As some will know, it usually doesn't matter if the steel used in reinforced concrete  gets rusty, because it will still have its inner bulk of steel for strength, still be gripped when concrete pours around it, and won't corrode further provided it is deep enough inside the body of concrete.

There's one exception. If the steel is used to join two sections of concrete below ground level that are poured separately, without a "water stopper", water will inevitably enter the point where the two sections of concrete meet, and in favourable conditions can slowly corrode the steel there. It's a low risk, and a long term issue, but can happen.

That's the situation I'm in, and I didn't realise the issue early enough to take precautions. Having already poured one section with its steel last year, I'm trying to retrospectively prevent possible issues before pouring the second part.

To counter any corrosion issues, I plan to use a zinc based galvanising coat to retrospectively protect the steel in the region of the join between the two lots of concrete. But there's a problem. Galvanisation relies on electrical conductivity to create the sacrificial aspect, and rust doesn't conduct. Physically stripping rust back to steel is also hard on these bars, as they are designed with an irregular surface, to enhance grip.

So I'm wondering if there is a solution via chemistry - some easily accessible way to revert a rust coat back to metallic form, or remove rust to expose metal. Most rust removal products as far as I suspect, attempt to passivate or stabilise rust, not actually return a surface to metallic conductive form.

Is there anything I can easily do to return the rust layer to an electrically conductive/metallic layer? If not, any ideas how else I might chemically reduce the scope for future corrosion over the years?

Update: noting for clarity, the join is below groundwater level, with some degree of hydrostatic pressure behind it (it's a 1m retaining wall where a soil slope in the garden was cut away, and I can't add drainage below the joint because the bottom slab is already set and the joint will be 30cm below ground level even on the lower side, hence below groundwater level). Also the concrete mix is self-consolidating/self-compacting (SCC) so air bubbles as mentioned in one answer are not an issue - both mixes are luckily, specifically designed to preclude them, and the formulation sets pre-compacted without vibrating.

If you have a plastic bowl or container large enough to accommodate your item, fill it with water with with some salt (e.g., kitchen salt).  Then you need Aluminum foil lightly wrapping your rusty piece of iron which altogether are immersed in this said solution (luke warm is enough).  One important detail is that the foil is in mechanical contact to the iron, while the solution is able to be in contact with the foil and the cast iron simultaneously.

Over the course of the next hours, Aluminum will be "eaten away" -- or, more chemically speaking, Aluminum will be oxidized, and the rusty iron will be reduced.  To facilitate the process, it does not harm to scratch the Aluminum surface with a brusch on the surface exposed toward your iron pieces.

To direct the work of this oxy-reduction reaction toward your rusty iron pieces, it is important that the container for this setup itself is not made of metal.  Depending on the size of the item "to recover" things like a glass / china bowl, or a plastic bucket are suitable.  After half a day or so, remove your item, rinse it with water and a brush; dry it and protect it against new corrosion.  As already mentioned by you, this may be done by galvanizing, or by galvanic anodes.

An alternative to using a salt solution is the addition of vinegar to the water.  The key point of both (salt or vinegar) is that they increase the electric conductivity of the solution, hence accelerating the process.  If using vinegar, however, do not forget to protect your hands with gloves and eyes (goggle) against the acid.


Reaction of cyclohexene oxide with LDA The question given above appeared in the 2016 Indian National Chemistry Olympiad (INChO). What I think is that the epoxide ring is being opened by a base-catalysed mechanism ($\mathrm{S_N2}$). I proposed the following few steps for this:

The diisopropylamide ion, which is a very strong base, would not be a good leaving group. To obtain the desired products, how could the mechanism then proceed? And if not, then what would be the real mechanism? I did not find any reliable information on the Internet regarding ring opening using LDA.

Because of the steric hindrance of the two iPr groups, LDA is a very poor nucleophile. It is, as you note, a very strong base. The two products observed both arise from deprotonation of cyclohexene oxide.
Deprotonation of one of the methylenes alpha to the epoxide ring opens it to give the allylic alcohol.
Removal of one of the protons on the epoxide ring carbons opens the ring to the enol, which on quench gives cyclohexanone.


How long does a linear hydrocarbon chain have to be before we call it polyethylene? Might be a really silly question here but here goes. Polyethylene is a polymer with a repeating unit of just 2 carbon atoms attacked to 2 hydrogens each which connect to other repeat units.

This structure is just an aliphatic hydrocarbon like butane or hexane or decane etc. My question is, at what carbon chain length does a hydrocarbon cease to be a normal alkane but instead becomes a polymer?

Actually, polyethylene is still an alkane as it fits IUPAC definition perfectly fine [1, p. 1313]:

  alkanes:
  Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms.

Smaller alkane molecules you are referring to, are oligomers. IUPAC defines the prefix oligo- as follows [1, p. 1353]:

  oligo:
  A prefix meaning 'a few', and used for compounds with a number of repeating units intermediate between those in monomers and those in high polymers. The limits are not precisely defined, and in practice vary with the type of structure being considered, but are generally from 3 to 10.

At the same time removing numerical restraints when referring to oligomer molecules [2, p. 2289]:

  1.2 oligomer molecule
  A molecule of intermediate relative molecular mass, the structure of which essentially comprises a small plurality of units derived, actually or conceptually, from molecules of lower relative molecular mass.

Obviously, the boundary between oligomers and polymers is blurred. Usually for the emergence of "polymer" sufficient degree of polymerization $N$ starts with $10^2 - 10^4$ for synthetic polymers and about $100 - 200$ for globular proteins.

A significant feature of any polymer as compared to its molecular weight counterpart is that when one or several units are added or removed into the polymer molecule, the properties of the polymer material remain almost unchanged. Properties of oligomers substantially depend on molecular weight.

As a typical example, melting points of low molecular weight alkanes gradually increases with polymerization degree (Source):

reaching a plateau of approx. $\pu{400 K}$ for linear low density polyethylene.

References

Moss, G. P.; Smith, P. A. S.; Tavernier, D. Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995). Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375. https://doi.org/10.1351/pac199567081307.
Jenkins, A. D.; Kratochvíl, P.; Stepto, R. F. T.; Suter, U. W. Glossary of Basic Terms in Polymer Science (IUPAC Recommendations 1996). Pure and Applied Chemistry 1996, 68 (12), 2287–2311. https://doi.org/10.1351/pac199668122287.


What is the meaning when it is said "Temperature of an Oxyacetylene Torch is 3500 degrees Celsius"? I don't think flame is anything but the product at that temperature but I am not sure. What exactly is flame? And I think the correct statement should be that the exothermic reaction here produces heat at such level that the temperature of the product gas reaches 3500 degrees.
What will be the exact idea behind the statement?

Oxyacetylene is a mixture of acetylene (ethyne) and oxygen (not air). The mentioned temperature of about $T=3500\ \mathrm{^\circ C}$ is the adiabatic flame temperature for the combustion reaction of acetylene and oxygen.

An adiabatic process is one that occurs without transfer of heat. The adiabatic flame temperature is a theoretical temperature, which would be achieved in a combustion system in which there are no heat losses (e.g. no radiation losses from the flame). Note that this cannot be achieved in a real flame; therefore, adiabatic flame temperatures are calculated values.

The heat $Q$ released from the combustion reaction of acetylene and oxygen is the heat of combustion. If all of the energy released by this chemical reaction is used to increase the temperature $T$ of the products ($\ce{CO2}$ and $\ce{H2O}$) with no heat losses, the resulting adiabatic flame temperature can be calculated from the heat capacity $C$ of the products.

Hence, the adiabatic flame temperature represents the maximum possible theoretical temperature for a particular fuel/oxidant combustion.


Compare stability of transition metal cations in aqueous solution I have to compare the stability of the following transition metal cations: $\ce{Co^3+, Fe^3+, Cr^3+, Sc^3+}$ in aqueous solution. (Source: Joint Entrance Exam (JEE) 2013 Mains India)

The first thing I looked at were the ions' electronic configurations:

$$
\begin{align}
\ce{Co^3+}:&\ce{[Ar] 3d^6}&\\
\ce{Fe^3+}:&\ce{[Ar] 3d^5}\\
\ce{Cr^3+}:&\ce{[Ar] 3d^3}\\
\ce{Sc^3+}:&\ce{[Ar] 3d^0}\\
\end{align}
$$

It suggests that $\ce{Fe^3+}$ should be the most stable, but the answer key given for increasing stability instead is $\ce{Co^3+ < Fe^3+ < Cr^3+ < Sc^3+}$. This seems to suggest that lower number of d-electrons leads to more stability, but I am not sure why that is true.

I also thought of the fact that more water of hydration would probably lead to higher stability, and that depends on charge/radius ratio. However, that would suggest $\ce{Sc^3+}$ to be the least stable, but that isn't the case here.

So, what factor is at play instead? And why are the stabilizing factors of half-filled d-subshell and greater water of hydration ignored?

Update: As pointed out in the comments below, to narrow down the cases, we may look at the enthalpies of formation, Gibbs free energy of formation, or the standard electrode potential - whichever one of these is the most applicable to ions in aqueous solution - to compare their relative stability.

It is written in NCERT Chemistry class 12 "In a medium like water, d3 is more stable than d5 (see CFSE).

CFSE stands for Crystal Field Stabilisation Energy. Basically in water there are h2o molecules which complexate the ions. Assuming octahedral geometry the d orbitals split into d3 (lower in energy compared to barycentre) and d2 (higher in energy compared to barycentre). If you only fill the d3 then you would actually have a more stable configuration as compared to d5 situation where the e.g. levels would also be occupied.

This is a more prominent factor than half filled orbital stability. In $\ce{Co^3+}$ there are no stabilizations involved and hence it is least stable.

$\ce{Sc^3+}$ has a noble gas configuration and this dominates all effects in terms of stabilizing the ion.


The dissolution of calcium carbonates in water Kindly help out. It is generally know that calcium carbonates has poor solubility in distilled water. Hence its ability to form slurry becomes difficult. Studies have shown that eggshells contain basically calcium carbonates.  I conducted an experiment in modification of eggshell powders with anionic surfactant in order to improve its solubility in water.  I carried out solubility and dissolution test on the modified eggshell powder. This was compared against the raw eggshell in order to confirm its solubility rate.  ICP was used to calculate the calcium  content in the solvent filtrate.  The reusults is showing that the calcium content in filtrate was higher than that of the modified eggshell.  Please who can assist in explaining this phenomenon. Thanks.

Regarding the slurry statement, you can only make slurries with insoluble compounds or they would dissolve and your slurry would go away leaving you with a clear solution!

Your attempt to alter the eggshell composition by extracting with the surfactant should have had little effect.  You didn't give any quantitative results from your ICP analysis, but I think one or both of two things happened:
1) There was residual Ca from the surfactant remaining on your eggshell
 and/or
2) You may have extracted the few percent of protein that makes up eggshells, leaving a more "concentrated" $\ce{CaCO3}$ residue as compared to the untreated shells.


Significance of single point energy when calculating interaction energies I am currently investigating about the interaction behavior of a few atoms in certain conditions.

Is it possible to use the concept of single point energy to represent the atomic interaction energies or I have to go other way around?
What is the basic difference between potential energy and single point energy?

Single point energy arises in the framework of the Born–Oppenheimer approximation and corresponds to just one point on the potential energy surface. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-called clamped nuclei approximation.

Mathematically, if you develop the Born–Oppenheimer approximation step-by-step you can easily see that single point energy it is the sum of the electronic energy and nuclear repulsion potential energy,
$$
U = E_{\mathrm{e}} + V_{\mathrm{nn}} \, ,
$$
where the electronic energy $E_{\mathrm{e}}$ is the solution of the electronic Schrödinger equation,
$$
    \hat{H}_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) = E_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) \, .
$$
The fact that at this point we use the symbol $U$ which (alongside with $V$) is usually used for potential energy to mean the single point energy is justified a little later. Namely, when we introduce the Born-Oppenheimer approximation which give rise to the nuclear Schrödinger equation,
$$
    \Big( \hat{T}_{\mathrm{n}} + U(\vec{r}_{\mathrm{n}}) \Big) \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) = E \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) \, ,
$$
it is easy to recognize that the values of the single point energy $U$ for all possible nuclear configurations define the potential energy for nuclear motion. So, it is in this sense that the single point energy is related to the potential energy.

Update: it became clear that OP misunderstood the notion of the single point energy $U$. Indeed, once we do few single point calculations for different nuclear configurations, the resulting $U(\vec{r}_{\mathrm{n}})$ is the potential energy for nuclear motion. However, it is not the interaction energy between some fragments, though, the interaction energy contributes to it. So if one wants to obtain the interaction energy one has to decompose the potential energy $U(\vec{r}_{\mathrm{n}})$ into its parts.

There different ways to perform the energy decomposition, just to name a few without any particular order:

SAPT (Symmetry-Adapted Perturbation Theory) a separate program (few of them, to be more precise) which can be interfaced with different quantum chemistry codes.
NEDA (Natural Energy Decomposition Analysis) which is available as a part of NBO package.
Morokuma decomposition already available in some quantum chemistry codes (GAMESS-US, for instance).
LMO-EDA (Localised Molecular Orbital Energy Decomposition Analysis) also available in some quantum chemistry codes (GAMESS-US, for instance).


Why is sea salt coarse? What makes sea salt coarser than other salt?
I have tried checking online but no one seems to discuss this. It seems that the grain of sea salt is larger, but I don't understand since the salt is dissolved in the sea water so how does it become coarse.

The method of production of sea salt usually involves bulk evaporation of seawater. This allows the salt to crystallise and due to it's bigger size it is sold without much processing. On the other hand, the table salt has much smaller size and is processed a lot more than we could assume before it is sold.
This could be one reason why sea salt has coarse texture.
See also: Maxine Builder and Maxine Builder. What's the Difference Between Kosher Salt and Sea Salt? (via the Internet Archive)
I was interested to answer this question because I have always thought that both the salts are just the same. The tiny size of salt grains just seems less coarse but it essentially is having the same texture is what I would believe.
But then I looked at sugar as well. The tiny grains of sugar or a little bigger sugar diamonds seem smooth compared to the rough textured bigger blocks of sugar. And so I was convinced that it was due to it being processed. But then think of powdered sugar? That appears to be smooth.


Why exactly are standard potentials additive? I don't really study chemistry so while my question may be very obvious, its not obvious to me. If we take an electrochemical reaction like

$$\ce{2Fe^2+ + Au^3+ -> 2Fe^3+ + Au+}$$

we can find its standard potential and standard Gibb's energy by summing half reactions whose are known, and can be consulted in tables.

It's intuitive for me that the standard Gibb's energy will be given by a sum of the corresponding reactions, and that I can multiply a reaction by minus one, or whatever, and use that to construct my complete reaction.

The reason I see Gibbs energy as intuitive is because you're essentially just taking

$$\sum_{\mathrm{products}} \mu_{\mathrm{products}} - \sum_{\mathrm{reactants}} \mu_{\mathrm{reactants}}$$

and this will prove the additive property of "summing up reactions".

But I don't see any reason for additivity being true when considering standard potentials. I don't even see why inverting the reaction should invert my potential if my half cell is in equilibrium and is a kind of "whole" that is compared to the standard hydrogen electrode and should therefore have the same potential no matter how I write it.

How am I screwing up here? Why can we add up potentials?

Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look at this from the perspective of Hess's Law.

How so?

Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new free energy will equal the sum of the individual ones.

This is a critical issue when trying to figure out the potential of a new reaction that is not the sum of an oxidation and a reduction.

For instance, take a look at these reactions:
$$ \ce{Fe^3+ + e- -> Fe^2+} \qquad {E^\circ}_1 = 0.771\, \text{V}_{SHE}  $$
$$ \ce{Fe^2+ + 2e- -> Fe_{(s)}} \qquad {E^\circ}_2 = -0.447 \, \text{V}_{SHE}  $$
$$ \ce{Fe^3+ + 3e- -> Fe_{(s)}} \qquad {E^\circ}_3 =\, ? \, \text{V}_{SHE}  $$

It is tempting to simply add these potentials and get the third value, but we would be incurring on a major error. The proper way to proceed would be to calculate each reaction free energy and then figure out the standard potential.

$$ \ce{Fe^3+ + e- -> Fe^2+} \qquad {\Delta G^\circ}_1 = -(1)\mathrm{F}{E^\circ}_1  $$
$$ \ce{Fe^2+ + 2e- -> Fe_{(s)}} \qquad {\Delta G^\circ}_2 = -(2)\mathrm{F}{E^\circ}_2  $$
$$ \ce{Fe^3+ + 3e- -> Fe_{(s)}} \qquad {\Delta G^\circ}_3 = -(3)\mathrm{F}{E^\circ}_3  $$

Since we know that ${\Delta G^\circ}_3 = {\Delta G^\circ}_1 + {\Delta G^\circ}_2$, we have:
$${E^\circ}_3=\frac{{E^\circ}_1 +2{E^\circ}_2}{3} = -0.041\, \text{V}_{SHE}  $$

Why don't we go through all this way when working with cell potentials? Because you always work your equations so that the number of electrons transferred on each of them has the same value. That way, the sum of the potentials is an indirect way of summing the free energies, which are now normalized by a common value.

This trick is evidently very useful, but sometimes it may be very misleading.


Who developed the concept of ICE tables? Many general chemistry courses (at least in the US) or online tutorials solve problems of chemical equilibrium by using ICE (sometimes RICE) tables, which track how the moles of reactants and products change going from an initial state to equilibrium.

I'm curious when this technique became so widespread and who developed it? My understanding is that ICE tables are a somewhat modern pedagogical tool, but I have yet to find a source for the idea and when it was popularized.

As expected the concept of equilibrium calculations is more than a century old idea. The algebra has not changed. Check internet archives for classical general chemistry textbooks to see how equilibrium calculations were done in 60s-70s. If you are specifically looking for the representation of "initial, change, equilibrium" tables, this is also four decades old. Books from the early 80s used this approach. See for example "Chemistry with inorganic qualitative analysis" by Thenard (1980). These days everyone likes to coin fancy acronyms and words. Gold colloids became gold nanoparticles, and "initial, change, equilibrium" became ICE!


Why do crown ether look the way they do? Like, the structure of lines of the rings of crown ether looks very different from regular cycloalkane rings. For example, the oxygen in crown ether is pulled more inside the ring. The line between carbon atoms consecutively bonded look almost horizontal. Why?

Crown ethers don't look like that except in structure drawings

You confuse two completely different things: the drawings we use to describe chemical structures and what molecules actually look like in real chemical environments. We draw crown ethers like this because the drawing accurately describes the connectivity in the structure. In reality crown ethers are flexible and floppy and certainly not flat. we don't add that extra complexity to 2D structure drawing as they get very messy very quickly and obscure the essential features that describe the connectivity in a simple way.

When they are in solution with appropriate ions (for 18-crown-6, potassium ions would be a good example) they form complexes driven by the attraction between the oxygens and the charged ions. These complexes have 3-dimensional structures (with K+ the complex is approximately octahedral as potassium fits neatly into the cavity created by the oxygens when the molecule folds up). It looks a little like this as a 3D structure:

Wikipedia has some good descriptions of this and further pictures.

Crown ethers are not rigid molecules and the actual shapes they take up in 3D space demand a lot on what else is in solution with them. Other, similar, complexing agents are far more rigid. Calixarenes, for example, have a 3D shape like a chalice which is even more misrepresented by a 2D drawing.


1,2,3-Trimethylcyclopropane stereoisomers 1,2,3-Trimethylcyclopropane has two stereoisomers. When one of the methyl groups is above the plane and other two are below the plane, then there are two chiral centres as well as a plane of symmetry. By definition this compound should have a meso isomer. However I read that this compound has zero meso isomers since it does not have any optically active stereoisomers.
So what is the exact definition of meso?

The way that a meso compound is often described in introductory organic chemistry courses is a compound that has stereocenters but is not chiral (due to a symmetry element). This description works for most cases, such as those in the previous questions (see this or this). By this description, the stereoisomers of 1,2,3-trimethylcyclopropane are meso.

However, the IUPAC Gold Book definition of a meso compound is:

  A term for the achiral member(s) of a set of diastereoisomers which
  also includes one or more chiral members.

As is pointed out in the question, none of the diastereoisomers of 1,2,3-trimethylcyclopropane are chiral, so by this definition, none of these compounds are meso.

The Gold Book also has a description of the meso prefix:

  A prefix to indicate the presence of a symmetry element of the second
  kind (see chirality) in a chemical species which is usually one of a
  set of diastereoisomers that also includes a chiral member.

The usage of meso in this way, to make a particular call out for a member in a stereoisomeric set of compounds, fails for 1,2,3-trimethylcyclopropane. Because both of the stereoisomers are achiral, neither can unambiguously be called "meso-1,2,3-trimethylcyclopropane." Since that's the case, neither are meso.


Why does oxidation of the electrode take place in a solution? What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?

So probably if you place it in distilled water it will not ionize, right?

Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).

So why do I use electrode potentials if I have lets say a bowl made purely of $\ce{Zn}$ atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

Sorry to be so catty but the overall line of questioning indicates a profound lack of understanding. I can't explain all about of redox reactions in a few paragraphs to let me try to answer some of the questions in a general way and hope that those answers will provide some focus.

  What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?

Every redox reaction has both a reduction reaction and an oxidation reaction.

I'm going to assume two electrodes in the solution. Thus in a redox reaction one electrode will have an oxidation reaction and the other will have a reduction reaction. The two electrodes will be connected via a wire through which electrons flow. Without the wire no reaction will occur.

  So probably if you place it in distilled water it will not ionize, right?

There are two aspects to the answer to this question.

First distilled water doesn't conduct well. So even with our two electrodes in the distilled water we wouldn't see much current flow through the wire connecting the two electrodes. In this case it isn't the wire limiting current flow but the counter flow of ions through the distilled water.

Second chemists chose "electrodes" that will be inert in the solution being used. For example if a sodium metal "electrode" was dipped into water then a violent reaction would occur spontaneously on the contact surface between the electrode and the water. The products would be hydrogen gas and sodium hydroxide. So a sodium metal electrode in an aqueous solution just won't work.

  Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).

An overall redox reaction can be written as two half-cell reactions. The gist is that we can use electrons to balance the formula. This makes it easy to balance an overall redox reaction since we just have to balance the electrons.

So for an oxidation half-cell for a zinc electrode we have:
$\ce{Zn -> Zn^{2+} + 2e^-}$

For a reduction half-cell for a copper electrode we have
$\ce{Cu^{2+} + 2e^- -> Cu}$

Since we have two electrons in both half cells we can just add the equations as is to get:

$\ce{Cu^{2+} + Zn -> Cu + Zn^{2+}}$

Now there is a quirk in this particular reaction. If the copper ions in solution, which are to be reduced, come in contact with the Zn electrode then the copper ions would plate on the Zn electrode. So a membrane of some sort must be used as a barrier which keeps the copper ions away from the Zn electrode but still allows anions to flow through the barrier. Of course if we used ordinary glass as the barrier then neither anions nor cations could flow through the barrier and the battery formed by the two half-cells wouldn't work.

  So why do I use electrode potentials if I have lets say a bowl made purely of Zn atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

Well a bowl can't be pure Zn and have Cu too.

So let me go to a hypothetical situation. I haven't checked Cu/Zn alloys (which would be possible but complicates the general idea here). So let's just assume a Cu/Zn alloy. When metals or alloys become solid they tend to form grains which are interlocked together by their 3D surfaces rather than one homogeneous material. The grains will vary in chemical composition. So the different grains have different electromotive potentials because of their compositional differences.  Thus corrosion will occur. Some grains will dissolve and other grains will plate the dissolved ions back to metallic atoms.

Now let me clarify that corrosion is a very complicated subject for which I have made a gross generality. The actual corrosion reactions will depend on the metals in contact, and the exact composition of the solution with which they are in contact.


Horizontal symmetry plane of cis-dichloroosmiumtetracarbonyl Consider the molecule $\ce{Os(CO)4Cl2}$ with a drawing given below via Chemtube3d:

I thought in this molecule, if I drew a horizontal plane, then a $σ_\mathrm{h}$ would be present i.e. A ‘horizontal’ mirror plane is perpendicular to the principle axes seeing as the mirror would, in my view, cut the atoms in half so the top would be the reflection of the bottom.

However, my professor said in a lecture that this molecule shown here does not have $σ_\mathrm{h}$ and I'm slightly puzzled as to why.

Why does this molecule not have $σ_\mathrm{h}$ plane?

Your compound has only one proper axis of rotation and it is $C_2$ as shown in the diagram. Therefore, it is the principle axis of rotation. By definition, $\sigma_\mathrm{h}$ is a plane of symmetry perpendicular to the principle axis of rotation. However, the plain you were talking about is paralleled to the principle axis of rotation. Thus, it is not a $\sigma_\mathrm{h}$ plane. Nonetheless, it is a $\sigma_\mathrm{v}$ plane, by definition.


Is aluminium hydroxide a strong base? Is $\ce{Al(OH)3}$ in a very dilute solution deemed as a strong base? Are there any actual experimental data for it?

No, it is not a strong base.

It is a weak acid with pKa of 5

See DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION at the page numbered 143 (page 12 of the file).