Correct Rate Equation for a Reversible Reaction I'm helping my son with Matlab as he is having trouble learning differential equations as they apply to chemical kinetics.   His task is to solve for the equilibrium concentrations for a reaction
$$\ce{A + B <=>[$k_{\mathrm{f}}$][$k_{\mathrm{b}}$] C}.$$
We are given the forward and backward rates and the initial concentrations of the reactants and products.   I helped him to set up the required equations and ran them through the ode integrator in R and Matlab but the results don't look right.   I'm pretty sure the integration process is working so I think maybe the actual formulae are wrong:
\begin{align}
\frac{\mathrm{d[A]}}{\mathrm{d}t} &= {k}_\mathrm{b}[C]-{k}_\mathrm{f}[A][B]\\
\frac{\mathrm{d[B]}}{\mathrm{d}t} &= {k}_\mathrm{b}[C]-{k}_\mathrm{f}[A][B]\\
\frac{\mathrm{d[C]}}{\mathrm{d}t} &= {k}_\mathrm{f}[A][B] -{k}_\mathrm{b}[C]
\end{align}
When I run this through the integrator with $k_\mathrm{f}=3$, $k_\mathrm{b}=1.5$,
$[A]=\pu{1M}$, $[B]=\pu{1M}$, and $[C]=0$,
the concentrations all reach equilibrium at the same level of $\pu{0.5M}$.  Is that correct?  I would have expected them to be different with the forward rate being bigger.
Really sorry for the basic question but as I said, I'm a programmer not a chemist.  I got the formulae from this How to set up an equation to solve a rate law computationally? but assumed that I didn't need to square any of the concentrations since the reaction is $\ce{1+1<=>1}$.

If all you want to find is the final concentration of all the species at equilibrium than you don't need to solve these differential equations.
To solve this types of questions easily in chemistry we define a quantity $K_{eq}$. Where $K_{eq}=\frac{K_f}{K_b}$ and $K_{eq}=\frac{\text{product of concentrations of products at equilibrium to the power of their stoichiometric coefficient}}{\text{product of concentrations of reactants at equilibrium to the power of their stoichiometric coefficient}}$.
So for your case we can set the equations as such:
$$
\begin{align}
\begin{array}{cccccc}
& \ce{A} & \ce{+} & \ce{B} & \ce{<=>} & \ce{C}\\
\text{Initial} & \pu{1 M} & & \pu{1 M} & & \pu{0 M}\\
\text{Final} & (1-x)\pu{M} & & (1-x)\pu{M} & & x\pu{M}
\end{array}
\end{align}
$$
$$
K_{eq}=\frac{K_f}{K_b}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\
2=\frac{x}{(1-x)^2}\\
2x^2-5x+2=0\\
(2x^2-4x)+(-x+2)=0\\
(2x-1)(x-2)=0\\
x=\{\pu{0.5M}, \pu{2M}\}\\
$$
But as $[\ce{B}]\leq \pu{1M}$, $x$ has to be equal to $\pu{0.5 M}$.
Which gives, $[\ce{A}]=[\ce{B}]=[\ce{C}]=\pu{0.5 M}$.
This also shows that your solving of differential equations was also correct!


Ionic compound having highest solubility in water Which ionic compound has highest solubility in water? I can find CsBr having highest solubility with 1230 g/L at 25 °C.

Note: compounds like ethanol are soluble to any extent in water, but they are covalent, not ionic.

Caesium salts are unapologetically ionic, and they typically have quite high mass solubilities in many solvents, including water.

Assuming organic ions are allowed, caesium acetate ($\ce{H3CCO2^-Cs+}$) in particular has a remarkably high solubility of 9451 g/kg water at −2.5 °C, increasing to 13 455 g/L water at 88.5 °C.

Caesium formate ($\ce{HCO2^-Cs+}$) is also quite soluble, with a solubility of 4880 g/kg water at 20 °C, resulting in 2.56 L of solution with a density of 2.297 g/mL (reference, .docx file). However, its solubility increases much faster with increasing temperature (J. Chem. Soc., Trans., 1922, 121, 1837-1843). At 100 °C, it reaches an outstanding value of 20 071 g/kg water! That's 11.6 molal, or roughly 20-25 mol/L assuming the density doesn't change too much. This saturated solution is 67.7% cesium formate by number of moles, which means more than two caesium ions and two formate ions per molecule of water.

I believe I have read somewhere that caesium formate is the record holder for highest mass solubility in water (evidently only at high temperatures). If this is not true, then I can scarcely believe it will be topped by much.

Tangentially, Ivan mentions Clerici's solution, which is actually a mixture of thallium(I) formate and thallium(I) malonate. The mixture doesn't count (though the individual components are quite soluble themselves), but it interesting to analyse. Apparently 300 g of each compound will dissolve in 40 g of water without saturating it room temperature (ref), giving a lower bound to their combined solubility of 15 000 g/kg water. This value rises with heating, and is the only way I can see to beat the mass solubility of caesium formate.

For further entertainment, I recommend these two solubility tables with a large number of entries (1, 2). The second one can be ordered by solubility at different temperatures. It's interesting to see the variety of cations and anions which can be combined to display extreme mass solubility.

Edit: If room-temperature ionic liquids are allowed, then it is quite likely some of them are miscible with water in any proportion, which is effectively "infinite solubility in water". Something as simple as ethylammonium nitrate ($\ce{C2H5NH3+NO3-}$) may suffice.


Why will a strong base neutralize as much acid as a weak base? I don't understand how the volume of 0.200 M NaOH needed to neutralize 250.0 mL of 0.010 M HCl (0.0125 mL) is the same volume of 0.200 M NH3 that would be needed to neutralize 250.0 mL of 0.010 M HCl.

Well it's because the strength of an acid or base is a measure of it's dissociation in water. But for the the reaction of an acid with a base the equilibrium lies completely to the right as the products (salt and water) are much more stable than the acid and base.

You're trying to compare the dissociation of a species in water with the equilibrium constant of an actual reaction. I understand why, but think of them as two different situations.


How do I prepare citric acid solution (from anhydrous powder) of accurate concentration, given that the solution volume expands when water is added? I attempted to prepare saturated citric acid solution from anhydrous powder. According to online data, the solubility limit was 59.2 w/w at 20 °C, i.e. 59.2 g citric acid in 100 mL water. To ensure full saturation (as in the case of preparing brine solution), I ended up weighing 67 g of the powder and tried adding 100 mL of water to it.
However, when I had added about 67 mL of water, the 100 mL Schott bottle was already full, and all of the powder dissolved (estimated 5.22 M solution). Since there is an apparent volume expansion, how then should this solution (or a citric acid solution of any other desired concentration) be prepared?

The "expanding solution" problem is irrelevant if you use standard techniques
First, you need to realise that a standard solution is a known amount of solid in a known volume of solution not a known amount of solid and a known amount of the solvent. You don't need to know how much solvent you use. So you need to target getting a result that occupies, say, 100mL not a solution made from 100mL of water.
There is an easy way to achieve this. Use a volumetric flask with an accurate volume marking. Add the solid to the flask. Fill the flask slowly (while shaking enough to ensure the solid dissolves before the flask is full). Keep adding small increments of the solvent until the total volume reaches the graduation. You now have an exact volume of a solution containing an exact amount of the solid. At no point do you need to worry about any volume expansion or contraction.


Electron configuration of non-hybridized carbon I'm currently learning about orbital hybridization in carbon.
I see that carbon has an electron configuration of 1s2 2s2 2p2.
However, there are actually three p-orbitals in the second shell, namely px, py and pz, which could each hold two electrons. However, there are only two electrons to be distributed on these three orbitals. Which orbitals do they occupy? Since their energy levels are identical, is it random?
I also learned that the three p orbitals can be combined with the 2s orbital (sp3 hybridization) to form four new hybrid orbitals. It then makes sense that the four electrons available each occupy one of the four new orbitals (so that each of them holds one electron), because this way, they can have the greatest spatial distance to each other (I assume they "want" this as they repel each other).
However, how are the second shell electrons distributed in a non-hybridized carbon atom? Does such an atom even naturally occur or does it always hybridize?

Update: I found this question and this answer which both state that non-hybridized carbon has the configuration 1s2 2s2 2px1 2py1.
Why is this the case, and not, for example, 1s2 2s2 2py1 2pz1?

The answer comes in several layers, as Martin has alluded to. I'll try and give a short summary: each point goes slightly deeper than the previous one.

For an isolated atom, the labelling of the orbitals is arbitrary. That is to say, the $p_x$, $p_y$, and $p_z$ orbitals are all interchangeable. More formally, this reflects the isotropy of the system, which means that all directions are equal. There is nothing that can physically differentiate the $x$-axis from the $y$-axis or the $z$-axis.

You can still arbitrarily assign $x$-, $y$-, and $z$-labels to three perpendicular axes. If you kept these coordinate axes consistent between different atoms, such that every $p_x$ orbital pointed in the same direction, you should not expect to find in all molecules that the $p_x$ and $p_y$ orbitals are always occupied. Instead, on average, you would have one-third of them with the configuration $(p_x)^1(p_y)^1$, one-third with $p_xp_z$, and one-third with $p_yp_z$. Again, this symmetry stems from the fact that there is no physical factor that favours any of the three coordinate axes over each other.

The previous point assumes that you have made a measurement of the electrons, such that you can definitively say that they are in a specific set of orbitals.* But prior to this measurement, there is no reason that they should already be in that state. It's better to posit that an isolated carbon atom is in a superposition of those three states, i.e. $p_xp_y$, $p_xp_z$, and $p_yp_z$. My old question: What constraints are imposed on a wavefunction by the symmetry of the system? is somewhat related, except that it deals with boron (which has the same issue).

Having gone through this, there is still the question of why we write $p_xp_y$ when it's not really accurate. The answer is that it's far easier than dealing with a superposition of states (in (3)), and since space is isotropic, $xy$ is the same as $xz$ and $yz$ as long as you relabel the axes, which means that (2) is more of a technicality rather than a real "gotcha". Also, it's perhaps not worth overthinking this: once you form a bond, all of this ceases to have any meaning. In a molecule, electrons don't remember "where" they come from, so it becomes pointless to quibble over whether it was in the $p_x$ or $p_y$ orbital.

* You can't measure individual electrons in an atom, you can only measure their collective properties. Also, I'm skeptical about whether you can measure a state of $p_xp_z$ given that $p_x$ and $p_z$ are eigenstates of non-commuting operators, but let's leave this aside for now (or assume, perhaps, that we're dealing with $p_{-1}$, $p_0$, and $p_{+1}$, which are all eigenstates of $\hat{L}_z$).


Coupled and Coupling Reactions Is Coupling Reactions, Heck Reaction for example, different from Coupled Reactions like the production of Glucose 6-phosphate in living organism?

A simplistic answer is as follows (though, please do refer to the links given by Greg in a comment to the question):

According to the University of Wisconsin page Coupling Reactions, coupling reactions are

  a reactant-favored reaction is linked to a product-favored reaction so that both reactions yield products.

A main point in the related University of Wisconsin website is that if the reaction is reactant-favoured ($\Delta{G} > 0$), then it is not going to happen, so it is coupled with a product-favoured reaction ($\Delta{G} < 0$), leading to coupled reactions are

  When chemical reactions are coupled, the free energy released by the exergonic reaction is used to drive the endergonic one toward the formation of products.

where the overall reation is product-favoured.

(The website uses an analogy of a coupled train).

A great explanation is in this YouTube clip from Dr. John Pollard, of the University of Arizona.


How to measure ppm of copper What is the best way to measure ppm of copper in colloidal solution.  My TDS meter only shows at 2ppm which cannot be correct.

The absolute approach:

Note: colloidal copper is small flecks of copper mixed with purified water. Typically colloidal copper is sold in dilute quanitities (like 100 ppm).

Measure and record the density of the colloidal copper solution (in grams per liter). Measure and record the weight of an empty flask. Put a sample of the colloidal copper solution in the flask. Measure and record the total weight of the sample (and flask). Subtract the weight of the flask from the total weight of the sample and record the total weight of the sample (only). Boil off the aqueous portion of the sample. Weigh and record the total weight the container of dried copper. Subtract the weight of the flask from the total weight of the container of dried copper and record the weight the dried copper.

Use the following conversions to determine the ppm of copper: 1 ppm = 1 mg per liter; and sample weight/density = liters of sample.

Example:

sample density = 1000g/L
(total) sample weight = 250g
(dry) copper weight = 25mg

250g/(1000g/L) = 0.25L

25mg/0.25L = 100 ppm.


What trends exist in the Periodic Table? I wish to know the trends in the periodic table and why that trend exists, for the following parameters:

Atomic size
Ionization energy
Metallic character
Positive ion size
Negative ion size
Electronegativity

My progress (if my reasoning is wrong for any of these please correct me):

Atomic size - I memorized from my class lecture that the greatest atomic size is towards the bottom-left of the periodic table. I think the reason for this is because as you go down the groups (IA, IIA, etc.) you add more electrons. Adding more electrons means the atom will logically grow bigger. I am not sure why the answer is not bottom-right, though.
Ionization energy - This is the energy required to remove an electron from an atom. If there are more electrons then the electron will be further from the nucleus and by Coulomb's law, since the distance increases, the force attraction that keeps the electron in the nucleus's orbit is weak. Thus, the ionization energy is the highest toward the top-left.
I have no clue what to think for metallic character.
Positive ion size I know the answer is that it increases towards the bottom left, but I don't know why.
Negative ion size I know the answer is that it increases towards the bottom left, but I don't know why.
Electronegativity I am pretty sure is highest at F which is 4.0 and is lowest towards the bottom left. So the trend increases towards the top-right.

You are correct on atomic size being due to number of electrons
and their shells and the reason why atomic size decreases from
left to right is due to the number of the protons being greater
on the right than on the left of the periodic table with same number of shells due to shell theory. (Reference).
The concept you are forgetting is
Valence Electrons.
Atoms are stabilized most often when there outer most shell is
filled with 8 valence electrons (except for hydrogen and helium with
0 or 2). The elements on the left side of the periodic table lose
electrons easily since they have a very large atomic radius allowing
for lower ionization energy. on the right side of the periodic
table(nitrogen, oxygen , and halogens) have a smaller atomic radius
creating a stronger pull making it harder to lose an electron thus
they will steal an electron to become stable.
The best way to summarize this is to quote Chemistry.about.com

  "There are trends in
  metallic character as you move across and down the periodic table.
  Metallic character decreases as you move across a period in the
  periodic table from left to right. This occurs as atoms more readily
  accept electrons to fill a valence shell than lose them to remove
  the unfilled shell. Metallic character increases as you move down an
  element group in the periodic table. This is because electrons
  become easier to lose as the atomic radius increases, where there is
  less attraction between the nucleus and the valence electrons
  because of the increased distance between them."
  (Reference)
  Positive ion size decreases due to the lack of electrons which allows for the protons to pull stronger on fewer electrons.
  The inverse applies for negative ion size. Due to the fact that there are more electrons, the size will increase since they will occupy more space.
  Electronegativity is the pull a elements has on electrons. This determines how it bonds with other elements i.e.(whether it shares or looses electrons). If a element has over a 1.7 Electronegativity difference then these the element with the smaller electronegativity will give away several electrons 1-4. Elements on the left side of the periodic table have the smallest Electronegativity due to their atomic size and inability to gain enough electrons to be stable ions. Elements from left to right in a period show increasing elctronegativity, and electrons form up to down in a group will have a lower electronegativity. Also fluorine has a 3.98 elctronegativity. (Reference).


How to calculate the Gibbs free energy from the equilibrium constant for the decomposition of sodium bicarbonate? Solid $\ce{NaHCO3}$ is heated to $90~^\circ\mathrm{C}$. At equilibrium the total pressure of the gases produced is $0.545~\mathrm{atm}$. Calculate $\Delta G^\circ$ at $90~^\circ\mathrm{C}$ for the reaction.

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

Using the $\Delta G^\circ = -\mathcal{R}T\cdot\ln(K)$ formula, I did the work below:

$$\Delta G^\circ = -(8.31)(273.15 + 90)\cdot\ln(0.545)$$

This gave me $1.83~\mathrm{kJ/mol}$ and therefore $3.66~\mathrm{kJ}$ for the entire reaction, $2~\mathrm{mol}$. Unfortunately, the answer is marked as $7.85~\mathrm{kJ}$, suggesting that $K=1$, why is this?

Shouldn't $K=(0.545)/1$ because of the gases present?

For your reaction:

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

$$K=P_{\ce{CO2}}\cdot P_{\ce{H2O}}$$

However, $K\ne0.545\text{ atm}$ because the sum of the pressures is $0.545\text{ atm}$ and the equilibrium constant is the product of the pressures.

$$P_{\ce{CO2}}+P_{\ce{H2O}}=0.545\text{ atm}$$

Since you started with no gas, and the stoichiometric coefficients of both gases are equal, then

$$P_{\ce{CO2}}=P_{\ce{H2O}}=\dfrac{0.545\text{ atm}}{2}$$

  What is the value of $K$ now?


Viscosity of two different solutions Suppose that I have 10 kg of plastic/polymer that is non solvable in water. I then divide the plastic/polymer into two equal parts.  Let us name the parts A and B in order to identify them.  I then proceed to make beads from each part. I make $n_A$ beads of radius $r_A$ from part A. I also make $n_B$ beads of radius $r_B$ from part B. Here $r_A>r_B$ and consequently $n_B>n_A$. Suppose that I put the beads from part A into 10 liters of water and I do the same for the beads from part B.  Which one is more viscous, the "solution" from part A or from part B? Note: here $n_A,n_B$ stand for some arbitrary number.

Disclaimer: I will be talking about the shear viscosity. I think that most of what I write is also applicable to extensional viscosity, although with different numerical constants, but I am not fully sure so I will limit myself to shear viscosity.

Most theoretical models predict that the shear viscosity of both solutions will be equal. The reason is that these models describe the viscosity of the suspension $\mu$ as the viscosity of the 'empty' liquid $\mu_0$ $+$ a correction term that depends on the volume concentration of solids, $\phi$.

The most famous one is the Einstein viscosity equation that simply reads:
$$\frac{\mu}{\mu_0}=1+2.5\phi $$

This model works for low solid concentrations when inter-particle interactions can be neglected. Later Batchelor showed that including long-range hydrodynamic interactions an additional correction of $+6.2\phi^2$ should be included, but still only a dependence on the volume concentration is there.

A nice summary of more models can be found in this paper: K.D Danov, J. Colloid Interface Sci., 235, 144–149 (2001). That paper also includes a discussion that shows that the factor 2.5 should be dependent on the mobility of the particle. This would mean that it will be higher for smaller particles, so in your case that the solution with part B will be somewhat more viscosity.

Another overview of models, including models which have an increasing viscosity with increasing particle size can be found here. So that would point to part A giving the more viscous solution.

So to summarize: to first order (i.e. low concentrations) the two viscosities will be the same. At higher orders you have competing effects of the particle radius that should increase the viscosity and the particle number that should also increase it. Which effect is dominant is probably hard to tell theoretically.


Strip away a small amount of aluminium with household chemicals I'm trying to turn a 6″ diameter mirror (I have spares) into a two-way mirror. It shouldn't be too hard to remove the backing, but then I need to remove most, but not all of the Aluminium plating.

My current plan is to remove the backing with paint-stripper, then soak the mirror in citric acid. However, I'm not sure how long it needs to be left in the citric acid, or even if that will work at all.

Are there any household, or easily acquired, chemicals I can use that will react with Aluminium in a controlled fashion?

Any help or pointers in the right direction would be appreciated.

Citric acid will not etch pure aluminum on any reasonable timescale, at room temperature, though commonly-used aluminum alloys, high temperatures, or the presence of salt can change that. However, it may be better to try some other chemicals: here is a decent resource for finding metal etchant solutions. The two that are probably accessible with household chemicals are hydrochloric acid (can be found at hardware stores with pool supplies) and sodium hydroxide (often found at grocery stores as lye). Neither is super dangerous, but you'll want to take precautions to avoid getting them in your eyes, on your skin, or on your clothes and HCl produces hydrogen gas as it dissolves aluminum, so don't put big pieces of aluminum in concentrated HCl unless you have adequate ventilation.

It will probably take some experimentation to get the concentration right: you want an etch rate fast enough that you're not there for hours etching, but slow enough that you can control it. The big problem with wet etching aluminum is that most etchants take a long time to get through the surface layer of aluminum oxide (and the time it takes is not really reproduceable due to differences in thickness and such) so when you put a piece in the etchant, nothing may happen for several minutes, then you look away for a few seconds and turn back to find all the aluminum gone. This can be avoided by adding HF, which happily removes the oxide, but is very nasty stuff that is rather dangerous to work with.

It might be possible to do this by picking a low concentration etchant and checking the piece frequently, but you may run into issues if the surface is dirty (I recommend washing the surface with acetone, isopropanol, and deionized water in sequence) or depending on how the aluminum was deposited. Some deposition techniques produce large variations in film thickness across the surface and you might run into the situation where the aluminum will be completely etched away in one place but is still totally reflective in another.


Software for technical calculation of chemical reactors I'm not a chemical engineer and don't have good background in it.
Just my father asked me to do something for him, as I'm a computer engineering graduate.

This is a software for calculation of technical parameters of reactor. I haven't seen it. I've just printed output of it. Fields are like following:

5  REM**************
10 REM***********
20 REM  Name etc.
~~~~~~~~~~~~
35 REM ********** PRODUCTIVITY********
~~~~~~~~~~~~
155 REM ***** INPUTS*******
~~~~~~~~~~~~~~~~~~~~~~~~
220 DTB
240  DTS
~~~~~~~~~~~~~~~~~~~~
555 REM *********** CALCULATIONS****************
~~~~~~~~~~~~~~~~~~~~~~~~~
~
~
~
********RESULTS*************
DTO=75
G=543.2
Q=XXXX
XXXXXX

Can anyone help me to find what is the software for calculation. I'd be very grateful if you can help.

Years ago a colleague of mine mentioned ASPEN as a typical software (bundle) used to model chemical plants.  Despite this and this lack of examples how the print-out will look like -- and I did not work with it, either.


How does hydraulic equilibrium work on the particulate level? I recently learned that osmosis occurs when there is a semipermeable membrane between pure water and a solution of water. The pure water then fills up the solution's side until, as my textbook put it,

"The pressure exerted by this column of solution counterbalances the pressure exerted by the water moving through the membrane from the pure water side."

My question is how does water just naturally want to move from one gradient to the other? Doesn't water move through the membrane both ways? Also, why does water "want" to go to the more concentrated side when the particles move at random?
In other words, why doesn't the flow of water molecules depend on the number of water molecules in the membrane instead of the concentration gradient?

The membrane has plenty of holes that are a little bit bigger than small molecules like water. Bigger molecules or bigger ions like usual solutes cannot go through these holes. So water has the tendency to cross the membrane both ways. When a water molecule hits a hole in the membrane from the pure water side, it can cross it. Nothing prevents it. When a solute molecule hits a hole of the same membrane, it blocks the passage for a couple of microseconds. No water can go through this hole from the solution side of the membrane. So on average, more water can cross the holes of the membrane in the direction water to solution, then backwards.
Is it what you wanted to know ?


What's the difference between pipettes calibrated "TD", "TC", "Blow-out" and related terms? I've never blown-out the last drop of any glass pipette I've ever used during my undergraduate studies. Recently I re-read some notes I had lying around in order to prepare a class I had to give to a group of first year undergraduates and found out that there is a class of glass pipettes which need to be "blown-out" in order to deliver the nominal volume correctly.

I researched a little bit and found what seems to me to be completely inconsistent terminology and conventions to refer to this class of pipettes.

In https://en.wikipedia.org/wiki/Graduated_pipette it is clearly stated that there are "two kinds" of graduated pipettes. "serologic/blow-out" and "Mohr/drain-out". It is also stated that serologic pipettes have calibration marks all the way to the tip, whereas Mohr pipettes do not. The same page states that pipettes are either calibrated "to deliver" (TD) or "to contain" (TC). It states that "TD" is a far more common calibration than "TC" and explains that pipettes calibrated TC need to be blown-out in order to deliver the expected volume correctly, one would assume from this that TD is a synonym fro "blow-out". Judging from this there would seem to be two kinds of pipettes, "serologic/blow-out/TD" and "Mohr/drain-out/TC". This really contradicts my experience, I've been using pipettes with marks going all the way to the tip and never been told to blow out the last drop, however, there is always the chance I've been doing it wrong this whole time.

Digging a little bit deeper I found that this document http://www.brand.de/fileadmin/user/pdf/Information_Vol/Brochuere_Volumenmessung_EN.pdf (link downloads pdf) that wikipedia lists as a source doesn't mention "blow-out" pipettes at all, but does mention pipettes calibrated "TC". In page 13 there are some pictures and descriptions of pipettes resembling what one would think are "Mohr" and "Serologic" types, and both are said to be calibrated "to deliver" actually, reading this text one would think that ALL graduated and volumetric pipettes are calibrated to deliver (however, this document is from a lab equipment manufacturer, perhaps they only manufacture TD pipettes...?).

Digging yet deeper http://faculty.clinton.edu/faculty/Mike.Lawliss/My_webpage/Sci110/Other%20documents/Pipetting%20A%20Practical%20Guide_Guzman2001.pdf is another article cited as a source by wikipedia. Here the author pretty much identifies serologic pipettes with blow-out ones, but also states that blow-out pipettes "usually" have double rings (non-colored) near the top. There is, however, a drawing of a "serologic" pipette WITHOUT rings near the end of the article, close to an "Ostwald-Folin" pipette WITH rings which the author specifically identifies as blow-out. The definition of the term "serologic" pipette is, according to the author, "a type of measuring pipette used to measure variable volumes" which in my view would make ALL graduated pipettes serologic.

To add to all this confusion Sigma-Aldrich seems to manufacture "TD, blow-out, serologic pippetes" http://www.sigmaaldrich.com/catalog/product/sigma/z325422?lang=es&region=AR and also "TD/TC volumetric pippetes", which makes me lose all hope of making sense of any of the terms listed in the question at all.

In short, is there a STANDARD way in which these terms (TD, TC, IN, EX, Blow-out, Serologic, etc) are used? Is there a standard definition regulated by some sort of organization, or at least generally agreed upon? If you think that there is, could you please provide trustworthy sources? Thanks a lot.

This probably seems like a silly question (it is =D) but it kind of bugs me not to know something so basic as "what are the different kinds of graduated pipettes?"... It seemed like something that would be quite easy to settle when I started looking into it... guess I was wrong.

Since you are explicitly asking for standards, you might want to start with ISO 4787 Laboratory glassware – Volumetric instruments – Methods for testing of capacity and for use. This international standard provides methods for testing as well as for use of various volumetric instruments made from glass. Further requirements for individual volumetric instruments can be found in

ISO 648 Laboratory glassware – Single-volume pipettes
ISO 835 Laboratory glassware – Graduated pipettes
ISO 385 Laboratory glassware – Burettes
ISO 1042 Laboratory glassware – One-mark volumetric flasks
ISO 4788:2005 Laboratory glassware – Graduated measuring cylinders

Most commercially available laboratory glassware is in accordance with these standards. However, some nonconforming volumetric instruments are also available, for example in non-standard sizes.
Due to retention of liquid on the inner surface of the volumetric instrument, the volume of liquid delivered is not identical with the volume of liquid contained by the volumetric instrument. Volumetric instruments are either adjusted ‘to contain’ (i.e. the capacity printed on the instrument corresponds to the contained quantity of liquid) or ‘to deliver’ (i.e. the capacity printed on the instrument corresponds to the delivered quantity of liquid; the wetting residue remaining in the instrument has already been taken into account in the calibration). According to international standards, the marking ‘Ex’ is used to indicate that the volumetric instrument has been adjusted to deliver; the marking ‘In’ is used to indicate that the volumetric instrument has been adjusted to contain. Most typical graduated pipettes, bulb pipettes, and burettes are usually adjusted to deliver, whereas typical graduated cylinders, volumetric flasks, and capillary pipettes (i.e. very small measuring pipettes with a capacity of up to 0.2 ml) are usually adjusted to contain.
ISO 648 and ISO 835 specify three classes of accuracy for pipettes:

Class A (higher grade, without waiting time)
Class AS (higher grade, with a specified waiting time)
Class B (lower grade, without waiting time)

‘Waiting time’ is the period to be observed in order to ensure complete delivery after apparent completion of the liquid delivery of the pipette (i.e. after the meniscus appeared to come to rest) and before the final reading of the delivered volume is taken (i.e. before the tip of the jet is removed from the inner surface of the receiving vessel).
Furthermore, four types of graduated measuring pipettes are standardized in ISO 835:

Type 1: Partial delivery (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is represented by the lowest graduation line)
Type 2: Total delivery (adjusted to deliver from any graduation line down to the jet; nominal capacity is represented by the highest graduation line)
Type 3: Total and partial delivery (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is obtained by delivery down to the jet)
Type 4: Blow-out (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is obtained by delivery down to the jet where the last drop of liquid in the jet is expelled by blowing. Blow-out pipettes are only adjusted to accuracy Class B.)

Typical graduated measuring pipettes with a capacity from 0.5 ml to 25 ml are commercially available in accordance with ISO 835 as Type 1, 2, or 3 and in Class A, AS, or B. Bulb pipettes with a capacity from 0.5 ml to 100 ml are available in accordance with ISO 648 in Class A, AS, or B. They are all adjusted to deliver (‘Ex’); i.e. the residual liquid still left in the tip has already been taken into account during calibration and must not be expelled into the vessel by blowing out.
Very small measuring pipettes with a capacity of up to 0.2 ml (capillary pipettes) are designed as Type 4 (blow-out) in Class B; however, most available capillary pipettes are actually adjusted to contain (‘In’), which is strictly not conforming with ISO 835. Though, the formal difference between blow-out pipettes adjusted to deliver and pipettes adjusted to contain may be small for total delivery since the last drop of retained liquid in the jet can be expelled by blowing in both cases. Nevertheless, when the greatest possible accuracy is desired, it is not sufficient to simply empty pipettes that are adjusted to contain by blowing out the liquid with a pipetting aid; such pipettes should also be rinsed two to three times with the diluting medium.
The general procedure for the use of pipettes that are adjusted to deliver (‘Ex’) is described in ISO 4787 as follows. However, when the greatest possible accuracy is desired, pipettes should be used as closely as possible to the manner in which they have been calibrated (which is also specified in ISO 4787, ISO 648, and ISO 648).

10.5.1 Pipettes adjusted to deliver (see ISO 648 and ISO 835)
After rinsing with the liquid or reagent to be used, fill the pipette by suction to a few millimetres above the selected graduation line. Remove any liquid remaining on the outside of the jet. The final setting of the meniscus shall then be made by dispensing the surplus liquid through the jet. Remove any drops of liquid adhering to the jet by bringing an inclined ground glass vessel into contact with the tip of the jet. Delivery shall then be made with the tip of the jet in contact with the inner surface of the inclined receiving vessel.
If the setting after delivery is done at a lower graduation line, the liquid flow has to be nearly stopped a few millimetres above the graduation line. After observing a waiting time, if specified, complete the final setting quickly.
A waiting time, if specified, shall be observed before making the final setting for delivery of a given volume.

The corresponding procedure for the use of pipettes that are adjusted to contain (‘In’) is described in ISO 4787 as follows. Note that, when the greatest possible accuracy is desired, it is not sufficient to simply empty the pipette by blowing out the liquid with a pipetting aid; the pipette should also be rinsed two to three times with the diluting medium.

10.5.2 Pipettes adjusted to contain
Rinse the pipette with the reagent to be used to a few millimetres below the desired graduation line. Fill the pipette by suction to as close as possible above the selected graduation line. Remove any liquid remaining on the outside of the jet. Make the final setting of the meniscus to the line by withdrawing the surplus liquid by means of filter paper. For the discharge, rinse the pipette several times with the diluting medium.


What is the chemical structure of ceramic nonstick coating on pans? There are lots of pans nowadays that have a ceramic nonstick coating instead of a PTFE coating. However I can not find anything online describing what kind of ceramic such nonstick coatings are. Does anyone have an idea what these nonstick ceramics are on a chemical level?

There is a near continuum (and thus a potential infinitude) of relative concentrations of the different components of ceramics and their means of fabrication, but US Patent US7879449 B2 (PDF) gives a thorough description of their own non-stick ceramic invention as well as descriptions and references to others.  Below are several excerpts from the cited patent:

The novel coating includes a $\ce{SiO2}$ ceramic matrix that includes the condensation product of hydrolyzed alkoxysilane and a colloidal silica sol. A series of new abrasion resistant and high hardness non-stick coating materials have been developed with ceramic matrix and organic compounds by sol-gel process.
This non-stick material has two main component of silicon dioxide. One is silica sol which is pure $\ce{SiO2}$, another is an organoalkoxysilane which is an organic-inorganic hybrid material. Both materials can make strong chemical bond from chemical reaction, condensation. $\ce{(OR')_{4-x}}$ groups are hydrolyzed with water and then make condensation with silica sol and each other. The ceramic non-stick hybrid coating is applied by a sol-gel process. Two kinds of non-stick organic compounds make a chemical bond with silica ceramic network or matrix and it can exist in the ceramic structure in a stable state.
The main matrix is composed of a network of silica formed by the chemical reaction or condensation of colloidal silica sol (water base) and an organoalkoxysilane such as MTMS (methyltrimethoxysilane). The colloidal silica sol is a very stable state because of the electrostatic repulsion of silica particle surface.
The organoalkoxysilane preferably has the general formula $\ce{R_xSi(OR')_{4-x}}$. Wherein $\ce{R}$ is one or more organic radicals chosen independently from among methyl, ethyl, and propyl, $\ce{R'}$ is a low molecular weight alkyl radical, and x is at least one and less than four. Commonly x is one, so that the organoalkoxysilane has three hydrolysable sites. This increases cohesion and bonding of the components and also provides a thickened layer and increases the links with the inorganic silica surface.
The surface of colloidal silica is coated by organoalkoxysilanes, in other words, hydrolyzed organoalkoxysilane make a thin layer or coating on the silica surface as a result of the condensation reaction with the hydroxyl groups of the silica surface.

Again, this is just an overview of a single process. The patent is well written and diagrammed and I suggest reading in detail to best answer your question (Images > Full Pages).


Conversion process of lanolin to 7-Dehydrocholestrol What is the Chemical process to convert lanolin to 7-Dehydrocholestrol?
I don't know about the chemcial reaction, if I get the process I can try in lab scale.

The conversion of lanolin to 7-dehydrocholestrol is an important procedure in the production of vitamin D.
Lanolin is converted to cholesterol by a method which comprises of following steps:

dissociating cholesterol from lanolin by transesterification
reaction;

performing short-path distillation to obtain cholesterol crude
product

purifying by silica gel column chromatography

crystallizing to obtain refined cholesterol with purity higher than
95% and yield higher than 70%.

The extracted cholesterol is converted to 7-dehydrocholesterol and then treated with UV light to produce vitamin D3. The method has such advantages as high product consumption, little pollution, simple process and easy operation, thus is suitable for mass production.
References

https://patents.google.com/patent/CN1958596A/en
https://patents.google.com/patent/CN1263769C/en
https://www.sigmaaldrich.com/technical-documents/articles/biofiles/nutrition-research.html


Determine molecular weight of biomass based on ultimate analysis The ultimate analysis (by weight percent) for beech wood is:
$$\ce{C}=49.05\,\%,\; \ce{H}=5.83\,\%,\; \ce{O}=45\,\%,\; \ce{N}=0.12\,\%$$
Using the ultimate analysis values, the molar mass of each element, and assuming 100 grams of wood (therefore C = 49.05 g, H = 5.83 g, etc.), a representative formula for the beech wood is:
$$
\ce{C} = 49.05/12.01 = 4.0841\; \ce{mol} \\
\ce{H} = 5.83/1.01 = 5.77\; \ce{mol} \\
\ce{O} = 45/16 = 2.81\; \ce{mol} \\
\ce{N} = 0.12/14.01 = 0.0086\; \ce{mol} \\
\ce{C}_{\,4.0841\,}\ce{H}_{\,5.77\,}\ce{O}_{\,2.81\,}\ce{N}_{\,0.0086\,}
$$

What is the molecular weight of the beech wood based on the representative formula?

As a general rule, it is not a good idea to treat complex materials like wood as something you can calculate a mole of. You can of course analyze the material and get elemental rates, but that does not mean they are bonded in such way that this rate forms a minimum fraction which represents the whole thing.

When you calculate the number of moles of each atom in 100g, what you have is an estimation of the expected quantities in that sample. Undoubtedly, if you took different samples of the wood you would find different values, specially if sampling from different parts of the material.

As an exaggerated example, think about the elemental constitution of the human body, which is mainly 65% Oxygen, 18% Carbon, 10% Hydrogen and 3% Nitrogen in mass (source). With that in mind you can calculate a mole ratio of these elements and create a "representative formula". Obviously, though, there's no such thing as the human molecule, and you can't say that someone's body is X or Y moles of human.

Yet, like I originally answered, if you define what the representative formula is, you can calculate it's molar mass.

$$\ce {M}=4.0841*12.0107 + 5.77*1.00794+2.81*15.9994+0.0086*14.0067 = 99.947 \ \ce{g/mol}$$

Knowing ${\ce M}$, the number of moles is ${\ce 700,000/99.947 \approx 7004}$, and the concentration of biomass is $7004\ \ce{mol/m^3}$


Is (2R,3S)-3,4-dimethylpentan-2-ol feasible to be dehydrated with POCl3 and pyridine? I was given a problem of reaction between (2R,3S)-3,4-dimethylpentan-2-ol and $\ce{POCl3}$ and pyridine. I know that $\ce{POCl3}$ will be attacked with hydroxyl group and the hydrogen on hydroxyl will be cleaved with pyridine so the hydroxyl group becomes good leaving group.

The reaction mechanism proceeds on E2 mechanism no matter whether the alcohol is primary, secondary, or tertiary. The problem is E2 mechanism can proceed if the stereochemistry of antiperiplanarity is fulfilled. When I drawn the stereochemistry of the alcohol, it turned that methyl on alpha-carbon and beta-carbon made the beta-hydrogen on these carbons cis to the hydroxyl group (the methyl on alpha carbon is wedge so the bonded hydrogen and hydroxyl group are dash), thus the reaction can not proceed.

Is that correct? If it is correct, will the reaction mechanism proceed on other mechanism?

I am uncertain whether or not you are assuming that the E2 product of a "cis elimination" is the (E)-alkene. Alternatively, you were given the geometry of the alkene and were asked to provide a mechanism for its formation. In either case, let's go through the steps.
You are correct that the alcohol 1 reacts with phosphorus oxychloride to form derivative 2a in addition to pyridine hydrochloride. Recall that there is free rotation about the C2-C3 bond in 2a, which is shown as its Newman projection in 2b in the anti-periplanar conformation ready for E2-elimination. Clearly, this elimination leads to (Z)-3,4-dimethylpent-2-ene. (Z)-4. If indeed it is the stereoisomer (E)-4 that is the target, then one can invoke SN2 displacement on 2a by chloride to invert the configuation at C2 leading to chloride 3a. Now E2-elimination of 3b leads to (E)-4.


Why do NaHCO₃ and KHCO₃ have different crystal structures? Why do $\ce{NaHCO3}$ and $\ce{KHCO3}$ have different crystal structures?

We had this in a test today and this got me stumped...

My questions are:

Do $\ce{NaHCO3}$ and $\ce{KHCO3}$ possess crystal structures?
If they do, why should they have different crystal structures, because they are in the same group and must possess almost similar physical properties.

The reason why the crystal structure of $\ce{NaHCO3}$ is face centered and the crystal structure of $\ce{KHCO3}$ is body centered is due to the fact that the cation/anion ratio is different, meaning that certain types of hole are now too small for the cation to fit into them and as a result it must go into a different hole.


What is the difference between molar conductance and molar conductivity? I'm learning (or at least trying to learn) about electrochemistry, but a major obstacle to that, is that different books I refer use different terms for the same symbols. So in a last ditch attempt to clear stuff up, I've resorted to Chem.SE.

So here's what I intend to do; I'll list out everything I think I've understood, as well as pose a couple of questions regarding some of them. I'd really appreciate it if someone would take the time to go through what I've listed out, checking them for errors and then clearing those queries which I've got. So here I go

  Symbols used:

  Resistance ($R$) , Resistivity or specific resistance ($\rho$), Conductance ($C$), Conductivity or specific conductance ($\kappa$), Area of cross-section of the electrode ($A$), distance between the electrodes ($L$),

and the REALLY confusing bit, Molar conductance according to some books and Molar conductivity according to others and one book uses both terms, both represented by $\mathrm{Λ_{m}}$

Now,
$$
R = \rho \frac{L}{A}\\
R = \frac1C\\
$$
Therefore
$$\frac1\rho = \frac1R\cdot\frac{L}A = C\cdot\frac{L}A = \kappa$$

Now if I've got this right, then,

Conductance is the degree to which the solution conducts electricity.
Conductivity is the conductance per unit volume of the solution; it may also be considered as the concentration of ions per unit volume of solution.
Molar Conductivity is the conductance of the entire solution having 1 mole of electrolyte dissolved in it.

Q1. So what's Molar Conductance?

Q2. Is there a difference between Molar Conductivity and Molar Conductance?

Also, according to Ostwald's Dilution Law, greater the dilution, greater the dissociation of the electrolyte in solution.

Regarding dilution of an electrolyte solution, this is what I've understood

As dilution increases, Conductivity (ion concentration per unit volume) DECREASES.
As dilution increases, Molar conductivity (Conductance of 1 mole of electrolyte in the total solution) should INCREASE in accordance with Ostwald's Law

Q3. How does dilution affect Molar Conductance?

Q4. How is Conductance affected upon dilution?

I suppose if the above statements are proof-read and the queries answered, I might get fairly good idea about this....

Also if you feel there is are any additional points worth mentioning, by all means go ahead and put it in the answer.

And finally, if anyone could recommend a decent site that deals with the above-mentioned terms and concepts in a fairly lucid manner, it'd be appreciated.

I can understand your frustration. The use of terminology is often inconsistent and confused (much to my chagrin). I think you've got the general idea, the conductance ($G$) can be defined as follows:

$$G = \frac{1}{R}$$

i.e. the ease with which a current can flow. As you said,  $$R = \rho \frac{l}{A}$$

one can now identify, $$G = \kappa\frac{A}{l}$$ where the conductivity $$\kappa = \frac{1}{\rho}$$

Molar "any quantity" always has the dimensions (it is helpful to think in terms of dimensions) $\text{"quantity" } \mathrm{mol^{-1}}$

so, it follows molar conductivity $$ \Lambda_m = \frac{\kappa}{c}$$ where $c$ is the molar concentration. It is useful to define molar conductivity because, as you already know, conductivity changes with concentration.

Now, to address the effect of change of concentration on molar conductivity, we need to consider the case of weak and strong electrolytes separately.

For a strong electrolyte, we can assume ~ 100% disassociation into constituent ions. A typical example is an $\ce{MX}$ salt like $\ce{KCl}$

$$\ce{MX} \rightleftharpoons \ce{M^+} + \ce{X^-}$$ The equilibrium constant for this reaction is
$$ K = \frac{[\ce{M^+}][\ce{X^-}]}{[\ce{MX}]}$$ and thus, with decreasing molar concentration of the electrolyte, equilibrium shifts towards the disassociated ions.

At sufficiently low concentrations, the following relations are obeyed:

$$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt  {c}}$$

where $ \Lambda _{m}^{0}$  is known as the limiting molar conductivity, $K$ is an empirical constant and $c$ is the electrolyte concentration (Limiting here means "at the limit of the infinite dilution").

In effect, the observed conductivity of a strong electrolyte becomes directly proportional to concentration, at sufficiently low concentrations

As the concentration is increased however, the conductivity no longer rises in proportion.

Moreover, the conductivity of a solution of a salt is equal to the sum of conductivity contributions from the cation and anion.

$$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$
where:
$ \nu _{+}$ and  $\nu _{-}$ are the number of moles of cations and anions, respectively, which are created from the dissociation of $\pu{1 mol}$ of the dissolved electrolyte, and $\lambda _{+}^{0}$, $\lambda _{-}^{0}$ are the limiting molar conductivity of each individual ion.

The situation becomes slightly more complex for weak electrolytes, which never fully disassociate into their constituent ions. We no longer have a  limit of dilution below which the relationship between conductivity and concentration becomes linear.
We always have a mixture of ions and complete molecules in equilibrium. Hence, the solution becomes ever more fully dissociated at weaker concentrations.

For low concentrations of "well behaved" weak electrolytes, the degree of dissociation of the weak electrolyte becomes proportional to the inverse square root of the concentration.

A typical example would be a monoprotic weak acid like acetic acid (again, from your graph):

$$\ce{AB} \rightleftharpoons \ce{A^+} + \ce{B^-} $$

Let $\alpha$ be the fraction of dissociated electrolyte, then $ \alpha c_0$ is the concentration of each ionic species. And $(1 - \alpha)$, and $(1 - \alpha)c_0 $ gives the fraction, and concentration of undissociated electrolyte. The dissociation constant is:

$$K = \frac{\alpha^2 c_0}{1-\alpha}$$

for weak electrolytes, $\alpha$ is tiny, so the denominator is nearly equal to one so, $$ K \approxeq \alpha^2 c_0$$ and $$ \alpha = \sqrt{\frac{K}{c_0}}$$ (like I said earlier)

for conductivities, one can now write the following relation
$$\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^0} + \frac{\Lambda_mc}{K (\Lambda_m^0)^2}$$ This fits the curve seen in your graph.

Caveat, all these arguments hold for dilute solutions. Things get out of hand at high concentrations, and one has to account for some additional phenomenon (for example, acetic acid will form hydrogen bonded dimers).

Anyway, long story short, conductivity increases with increasing dilution (though differently for strong and weak electrolytes). From the definitions I outlined at the very start, guessing how conductance changes is trivial.


Why is the melting point of tert-butyl alcohol 140 °C higher than that of sec-butyl alcohol? This is one of the most drastic differences in a physical property I've ever seen for two such similar molecules, and in a simplistic sense anyway the difference lies in the opposite direction from what one might expect.  tert-butyl alcohol is completely sterically hindered from participating in H-O hydrogen bonding so it would seem to be limited to van-der-Waals interactions. sec-butyl alcohol interactions should be dominated by van-der-Waals forces, but should also have some ability to participate in hydrogen bonding.  Furthermore, sec-buty alcohol has a slightly larger van-der-Waals surface.

These are such common reagents and the unusually high melting point of tert-butyl alcohol is so well known that there is a surprising dearth of information on theoretical or experimental evidence available (in my searching anyway) explaining the reason for this $\pu{140^oC}$ disparity in melting points.

Does anyone know of a concise, coherent explanation for this observation?

I am not aware of a solid scientific publication on this matter, but the following points should elucidate what is happening:

The high melting point sort of proves that there is strong hydrogen bonding in tBuOH. I also don't see how the tBu-group could be so bulky that each OH group cannot form two hydrogen bonds.
n-, sec- and iso-butanol all have pairs of methyl- and/or OH-groups that can be in gauche or anti position with respect to each other. The  energy difference is in the range of 4 kJ/mol, which is not much higher than the thermal energy $k_BT$ at 300K. Those alcohols exists more or less as a mixture of conformational isomers, which precludes crystallisation. In tBuOH, there is only one conformation.
Rotation of the methyl groups has a low excitation barrier, but it can also rotate in the solid phase (there is some tunneling involved), so this does not preclude crystallisation.

(The energy of ~4kJ/mol is from n-butane, but I expect no large difference. The gauche conformation of the OH group is probably even lower, energetically.)


Can pressure remain constant in isothermal expansion? An ideal gas undergoes isothermal expansion at constant pressure. During the process:

  enthalpy increases but entropy decreases.
  enthalpy remains constant but entropy increases.
  enthalpy decreases but entropy increases.
  Both enthalpy and entropy remain constant.

I applied this formula.
$\Delta H = \Delta U + p\Delta V$ (pressure is constant)
$\Delta U = 0$ (since, process is isothermal)
$\Delta V >0$ (gas is expanding)
So $\Delta H > 0$
So enthalpy must be increasing $\left(\Delta S =\frac{\Delta Q_{\text{rev}}}{T}\right)$

Since heat is supplied to expand gas $\Delta S$ must be positive. And entropy must increase. But none of the option is matching with my answer.

Also I think question is wrong since $PV= nRT$ and, at constant temperature (isothermal) and constant pressure, volume can not change.

$$\Delta H = \Delta U + \Delta PV$$
Here you were wrong you wrote that
$$\Delta H = \Delta U + P\Delta V,$$ thus the product of
 $PV$  remains constant and the $\Delta H$ is zero, so that the enthalpy is constant. And you are wrote that entropy will increase, hence the answer is B.


How does an electrolyte conduct electricity at low potential? My understanding is that electrolytes conduct electricity by:

1) movement of ions, the negative ions migrate to the positive electrode and vice versa. This is however not enough, if these ions just remain ions, the migration and separation of charges will create a potential difference in the solution counteracting the applied potential and at that moment migration of ions will stop. This means no more current will flow.

So therefore:

2) For current to keep flowing the ions must undergo electrolysis at the electrodes to form their neutral forms. In this way their is no build-up of counterpotential, ions keep flowing and the electrons exchanged in the electrolysis can flow from one electrode to the other. (This argument is often given as the reason electrolytes are conductive when I was doing an internet search, also on this site)

Now if we have say a NaCl solution, this would mean that for electrolysis Na+ needs to be converted in to Na, and Cl- into Cl2-gas. The potential needed for the electrolysis of NaCl as calculated from my tables is more than 4V.

So does this mean that if the potential difference is less than 4 V a salt solution does not conduct electricity, ie has very low or zero conductance? If not, how exactly is the current created?

I've read that conductivity is measured in a way that electrolysis is avoided (alternating voltage, wikipedia), so I guess then only the migration of ions is accounted for, but the very definition of conductivity is the ability of a substance to conduct electricity. According to my understanding to maintain a current this requires electrolysis.

Under low potential differences, electrolytes conducts electricity in a similar way that they does at high potentials differences. I think, after reading the question and the comments, that the key point to understand this is: electric potential difference between which spatial points?

The conductivity of a solution is measure of its ability to conduct electricity when the solution is under a difference of electric potential. That is: the solution must be under the influence of an electric field (remember that $\boldsymbol E = - \nabla \boldsymbol \phi$) for having a non zero flux of charges.

You have to look the potential differences inside the solution and not at the electrodes.

You are right in that in order to maintain for long time a continuous current  electrolysis is required (at least for simple experimental devices). If there is not electrolysis you won't see a current. However, this is not because the solution (lets say the bulk, that is where the solution as such is (homogeneous)) does not have the ability of conduct electricity, but because there would not be a net external electric field at all in that region.

In short, the conductivity of the solution is not dependent of your ability for creating an external electric field on it.

EDIT

  But then again, what if the potential difference passes a treshold and
  at a certain point allows electrolysis, wouldn't the conductivity
  increase very steeply at that point?

If the potential difference between electrodes is low enough to ignore electrolysis, then there will be a charge accumulation in their surroundings that generates a contrary effect in the potential of the solution "far" from electrodes. So, the mean electric field in the bulk will be null.

According to Ohm's law:

$$ \boldsymbol J = \sigma \, \boldsymbol E$$

For the reasons stated above: $\boldsymbol E = \boldsymbol 0$, and so $\boldsymbol J = \boldsymbol 0$. What is the value of $\sigma$ for the solution? As any value satisfies the equation: We can say nothing from Ohm's law
! (technically, it can be computed even if no electric field is present, but it is not important here).

If the potential allows electrolysis, then a current appears so in principle you can measure the conductivity of the solution using Ohm's law. Notice that it is not an step function the current vs. potential. Due to this, there is not a truly step, but it seems because the growing is almost exponential, so we do not notice the increment till we notice a big increment. You can check out the Butler-Volmer equation for more details.

Once the current due to migration (allowed by electrolysis) is measurable, you could measure the solution conductivity. BUT it is a very bad idea, as normally the solutes concentrations are very small and the process of electrolysis change them! (and of course, many times (but not always) we want to know the conductivity to estimate concentrations)


Why is DNA negatively charged and what makes it so? What part in the strand contributes to the overall non neutral charge?
DNA is not isolated in the body, so what keeps it stable while being charged?
Why is it important for DNA to be charged?

As cibrail said, DNA is a polymer of nucleotides. They join themselves through phosphodiester bonds (a specific kind of covalent bond) that can grow to as much as millions of nucleotides.

What part in the strand contributes to the overall non neutral charge?

The reason why DNA is negatively charged is the phosphate group that makes up every nucleotide (pentose + nitrogenous base + phosphate). When forming part of the phosphodiester bond, they retain 1 of 2 negative charges (the other being lost to form the other ester bond to a new pentose, that's why the bond is called "phospho-di-ester").

Lehninger 5th edition. Part of figure 25-5

DNA is not isolated in the body, so what keeps it stable while being charged?

The main molecules in charge of that are proteins called histones. They have alkali features and positive charge, and so, they neutralize these negative charges. On the other hand, that's not the main point of histones. Histones play a really important key role in the regulation and compaction of DNA within the nucleus of the cell, and are important targets of diverse molecules that want to alter DNA structure and function:

Lehninger 5th edition.Figure 24-26

Why is it important for DNA to be charged?

It has great importance because cell machinery is adapted to that property. If DNA wasn't negatively charged, the structural fundamentals of interactions between DNA and transcription factors would be completely different.
Thanks to the fact that it is negatively charged, it becomes spontaneously pretty straight when denaturalised, because negative charges repel themselves. It forms very wide curves.
And of course, histones wouldn't be able to join DNA and exert their key regulating functions.


Ring opening, nucleophilic substitution Are the mechanism and the product for this given reaction right or is there anything wrong with it?
1,2 - Epoxybutane has originally the 2R configuration and the product got an R configuration as well.
I was not sure about, where the nucleophile attacks. Under acidic conditions steric hindrance will be ignored, right? In this case the hydrogen atom will be added to the less substituted carbon atom and the ring will be opened.

The free electron pairs of the oxirane may act as a Lewis base and coordinate with a Lewis acid like $\ce{Li+}$.  This form of activation facilitates the attack of the alanate $\ce{[AlH4]-}$ as a $\sigma$-nucleophile at the less substituted side of the epoxide.

Once opened, the nucleophilic oxygen attacks at the electron deficient centre of $\ce{AlH3}$ (Lewis acid).  But the $\ce{Al-O}$ bond hydrolyzes upon the workup, yielding the alcohol and the hydroxides of lithium and aluminum.


Why is the iron(II) on the heme molecule depicted as having 6 electrons in its 5 d non- and anti-bonding orbitals? Approximating the $\ce{Fe}$ complex as a square base pyramid ($D_\mathrm{4h}$ symmetry) with 5 $\ce{N}$ ligands, My electron count of the system gives me 16 electrons – 10 from the $2\times 5$ $\ce{N}$ electron pairs and 6 from the iron(II). Filling the molecular orbitals of the valence shells, I ended up with $\ce{4 e-}$ in the d non-bonding orbitals – for there to be 6 in the d orbitals, I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).

Looking at depictions of the orbital diagram however, such as this one:

and this one:

I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it. Can anyone help me make sense of this situation?

If I'm being a little unclear about my approximation, just look at the $\ce{Fe}$ molecular orbitals on the lefthand side of the images and subtract 2 electrons

I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it.

Iron is in the 8th group (or group VIIIb in older terminology). Thus, in its ground state it has eight electrons; two of which are in the 4s orbital. However, practically always once you oxidise a transition metal and enter coordination chemistry, the orbital energies shift relatively to each other and it becomes more stable to have all the energies in the d-orbitals. Thus, the typical configuration for iron(II) is not $[\ce{Ar}]\,\mathrm{3d^4\,4s^2}$ but $[\ce{Ar}]\,\mathrm{3d^6\,4s^0}$.

  I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).

I think, you’re not aware of the looseness of the 18-electron-rule guideline. Most of the time it does nothing except for stating a maximum that can occur. However, there are still many complexes that violate it; e.g. anything octahedral with zinc(II) (22 electrons), anything with copper(II) (21), octahedral nickel(II) (20) etc. It is much more valuable when going to neutral carbonyl complexes; e.g. iron typically forms pentacarbonyliron $\ce{[Fe(CO)5]}$ due to the 18 electron rule. But in the context of standard cationic complexes. you can safely ignore it.


Preparation of oxalic acid from acetylene I would like to, theoretically, prepare oxalic acid from acetylene.

I think that using ozonolysis would be a good idea to do so. Upon ozonolysis of acetylene, we would get:

$$\ce{O=CH_2-H_2C=O}$$

On oxidizing this using hydrogen peroxide, would I obtain the desired product?

Ozonolysis of acetylene won't produce oxalic acid.  The triple bond holding the two acetylenic carbons together will be attacked as shown below.  The reaction is complex and depending upon reaction conditions, different products can result (ref_1, ref_2).

Oxalic acid can be produced from acetylene by reacting acetylene with alkaline $\ce{KMnO4}$ (ref_3).

$\ce{H-C#C-H ->C[KMnO4/OH^-]\ HOOC-COOH}$

EDIT: Response to OPs comment
Here is a rough idea of what is going on mechanistically:

1) the permanganate forms a cyclic intermediate

2) that produces a diol

3) since a double bond is still present the reaction can be repeated to yield 2 gem-diols that can equilibrate to an alpha-diketone

4) now the standard alkaline permanganate oxidation of a carbonyl occurs

5) hydroxide can add to a carbonyl and

6) the manganese ion complexes with the carbonyl oxygen

7) the alpha-proton donates its electrons to reform the carbonyl which is now part of a carboxylic acid

8) repeat steps 5-7 on the other carbonyl and you have oxalic acid


Strength of a nucleophile A nucleophile should get better at its job if it can donate a pair of electrons more easily. We say that a more electronegative atom should be less nucleophilic , generally. I understand this is because the more EN an atom gets, the more strongly it pulls electrons towards itself. But why does the fact that such an atom would ,generally, have more electron density not override this?

Electronegativity and nucleophilicity are different concepts. Electronegativity refers to how strongly an atom pulls bonding electrons towards itself. Nucleophilicity refers to ability of atom to donate a lone pair of electrons. Different factors affect nucleophilicity, such as orbital size, charge of ion, size of molecule, hybridisation etc etc.


Why is buffer capacity at a maximum when the ratio of its components is 1? A buffer consists of a weak acid and its salt or weak base and its salt.

When the ratio of weak acid and its salt in a buffer (or the ration of weak base and its salt) is equal to 1, we say that the buffer capacity is maximum. How does this tell us that buffer capacity is maximum?

With a buffer, you're trying to maximize your ability to neutralize acid or base. The log dependence of the pH on the ratio of interest in the Henderson-Hasselbalch equation is why the buffer works well:

$$\mathrm{pH} = \mathrm{pK}_{a} + \log\frac{\ce{[A-]}}{\ce{[HA]}}$$

Now, your ability to neutralize acid is determined by the amount of the conjugate base $\ce{A-}$. Your ability to neutralize base is determined by the amount of the acid $\ce{HA}$. When the ratio between these is 1, you have equal ability to neutralize either. If you're skewed too far one way or the other, then you will not have much of the respective species around to neutralize added acid or base.


Calculating the E of a galvanic cell So, I have $\ce{Cr2O7^2-(aq) + 14 H+(aq) + 6 I-(aq) -> 2 Cr^3+(aq) + 3 I2(s) + 7 H2O(l)}$

and I know the concentrations to be like so:

$\ce{[Cr2O7^2-]}= 1.6\ \mathrm M$

$\ce{[H+]}= 2\ \mathrm M$

$\ce{[I-]}= 1.6\ \mathrm M$

$\ce{[Cr^3+]}= \pu{0.05271 M}$

And I want to calculate the $E(25\ \mathrm{^{\circ}C})$ of this galvanic process.

I know that I have to use $E=E^{\circ}-\frac{RT}{nF}\ln Q $, but I have a few questions:

How can I find $E^{\circ}$?
I know that $n$ represents the amount of electrons moles moving per mole of transformed material, but how can I find it when I have only the complete reaction like here?
With $Q=\ce{\frac{[Cr^{3+}]^2}{[H^+]^14[I^-]^6[Cr_2O_7^{2-}]}}$
I get $Q=\pu{6.32*10^{-9}V}$, which is very very small - does this seem to be the correct $Q$ for $\ln Q$?

Standard electrode potential can only be measured experimentally by comparing the half cells, in your case $\ce{Cr^{VI}/Cr^{III}}$ and $\ce{I2/I-}$ with a standard hydrogen electrode or a calomel electrode. The net standard electrode potential will be obtained by subtracting those of cathode-anode. You can find the standard electrode potential of most half cells in your textbook or online.
See in your reaction, two moles of $\ce{Cr^{VI}}$ turns to $\ce{Cr^{III}}$. So $n = 2 \times (6 - 3) = 6$. You can try out the same thing with the other half cell. The answer will be the same.


Can someone clear this confusion I have about the first law of thermodynamics In our school we were taught that the first law of thermodynamics represented mathematically is $\Delta U=Q+W$. Also we were taught that for an isothermal process $\Delta U=0$ and therefore $Q=-W$, but we were taught that for an isothermal free expansion of a gas against vacuum $W=0$ and therefore $\Delta U=Q$ but shouldn't $\Delta U$ be zero here as well, I mean after all the process is still isothermal. Can somebody pls help with regards to this, also pls try to explain in a simple manner (High School Level). Thanks for the help

Everything you've written is correct (for a free expansion, $\Delta U = q = w = 0$), but it applies only if the system is an ideal gas, or has ideal-gas-like properties, namely that there are no attractive or repulsive interactions between the particles.
It also requires that the system is closed (no matter can flow into or out of the system). Internal energy is an extensive property, so its value is proportional to the amount of substance in the system. If matter is added to the system, internal energy can increase even if no there is no flow of work or heat.
Your teacher probably had both of these restrictions in mind.
Thus, if you expand an ideal gas isothermally into a vacuum, the system doesn't do any work, so $w=0$. And there is thus no need to flow heat into the system to compensate for the work to maintain the temperature, nor is there any need to flow heat into the system to provide energy to pull the particles apart (no interaction between the particles), so $q=0$.
If you want a molecular explanation, imagine an expansion against a constant external pressure.  As the piston retreats, the particles hitting the piston move more slowly after they reflect off of it (this is because they are colliding with a retreating object rather than a stationary object), which has the effect of cooling the gas (the temperature is a measure of how fast the particles are moving).  Thus, for an isothermal expansion in which work is done against some external pressure, to maintain the temperature, heat must flow into the system to compensate for this.
But for a free expansion, the gas isn't expanding against a retreating piston. So particles are not robbed of their kinetic energy and thus, in a free isothermal expansion (of an ideal gas), no compensating heat flow is required to maintain the temperature.


How could scientists produce accurate chemical structures before they could capture images of molecules? How were scientists able to produce these accurate structural diagrams long before scientists were able to capture images of the actual molecules?

Reference: Scientists capture first images of molecules before and after reaction

They ran chemical reactions on the molecule and identified the fragments.  Then like a jigsaw puzzle, using additional information derived from other chemical reactions, they pieced the molecule back together.  Some Nobel prizes were awarded for such structure determinations.


Which theta should be taken from x-ray diffraction pattern for calculating Scherrer's equation? We use Scherrer's equation to calculate crystalline size of a particle:

$$d = \frac{Bλ}{β\cos{θ}}$$

But which theta should we take from the XRD pattern for this calculation?

For example, the following plot shows the diffraction pattern of nano (above) and bulk (below) silicon and it has two diffraction peaks at $2θ \approx 69^\circ$ and $2θ \approx 76^\circ$. Which theta should we take for calculating crystalline size of nano silicon and why?

Scherrer’s formula is a simple equation for the simple case of a single peak broadened only due to crystalline particle size. Application of the formula to other, more complex cases, gives only an estimate. Quoting Wikipedia:

  It is important to realize that the Scherrer formula provides a lower bound on the particle size. The reason for this is that a variety of factors can contribute to the width of a diffraction peak; besides crystallite size, the most important of these are usually inhomogeneous strain and instrumental effects. If all of these other contributions to the peak width were zero, then the peak width would be determined solely by the crystallite size and the Scherrer formula would apply. If the other contributions to the width are non-zero, then the crystallite size can be larger than that predicted by the Scherrer formula, with the "extra" peak width coming from the other factors.

Scherrer formula gives an estimate, an order of magnitude. Here, calculate the particle sizes for the two peaks, and they should be close. The differences are covered by the large uncertainty of your resulting value.


Exceptions to the Madelung rule for electronic configurations The electronic configuration of platinum is

$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$

and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or $\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$.

I see this question has already been answered, but here is what Wikipedia says:

  Other exceptions to Madelung's rule:
  For example, in the transition metals, the 4s orbital is of a higher energy than the $\ce{3d}$ orbitals; and in the lanthanides, the $\ce{6s}$ is higher than the $\ce{4f}$ and $\ce{5d}$.

I don't understand the reason behind this.

Shouldn't it be that the $\ce{4s}$ orbital is of a lower energy than the $\ce{3d}$ orbitals (that's why it is filled first) and in the lanthanides, the $\ce{6s}$ is lower than the $\ce{4f}$ and $\ce{5d}$?

It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?

There is a list of exceptions arising from electronic configurations of lanthanides and transition series.

The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?

So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

A lot of questions in here.

It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?

In the Schrödinger Equation, $\hat H\Psi = E\Psi$, the Hamilton operator $\hat H$ is generally considered to consist of a kinetic term $\hat T$ and a potential term $\hat V$. In general, $\hat T$ is defined via an electron’s impulse $\vec p$ rather than its velocity $\vec v$, meaning that we end up with a term $$\hat T = -\frac{\hbar^2}{2m}\nabla^2$$
Within this operator, the kinetic energy of an electron is considered and since $\hat T$ forms part of $\hat H$, we use an electron’s kinetic energy in the original Schrödinger equation and thus it is considered in the calculations.
Using relativistic quantum mechanics will complicate the calculations but the general idea will remain the same.

The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?

That’s … overly simplistic. In reality, it is a series of considerations we could make at the drawing board than when calculated leads to one electronic configuration being preferred over another. However, this question already covers that part.

Other exceptions to Madelung's rule:
For example, in the transition metals, the 4s orbital is of a higher energy than the 3d orbitals; and in the lanthanides, the 6s is higher than the 4f and 5d.

I don't understand the reason behind this.
Shouldn't it be that the 4s orbital is of a lower energy than the 3d orbitals (that's why it is filled first) and in the lanthanides, the 6s is lower than the 4f and 5d?

While your sentence is correct according to the aufbau principle or Madelung’s rule, this is no longer true when it gets to everything that’s not a neutral atom. Indeed, as soon as you turn scandium into a scandium(I) ion, the 4s orbital becomes less favourable than the 3d orbitals because the latter are still ‘closer to the nucleus’ and experience more stabilisation from an overall net positive charge. In general, once you go from neutral atoms to cations, the aufbau principle loses its relevance and a more ‘logical’ 3d-4s-4p order is favourable.

So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

Lanthanides tend to be rather subtle in their electronic effects if I recall my lecture correctly; as far as I am aware there is no explicit rule of any kind for them. But as for memorising: yes, you would have to do that.


Do crown ethers have to have two carbons between the oxygens? If so, why? I just began learning about ethers and I have noticed that pictures of crown ethers, no matter what size they are, always have two carbon atoms left and right of each oxygen atom before you reach another oxygen atom in the ring. It's almost like it is a poly-ethyl ether. So my question is can you have crown ethers with more than 2 carbon atoms between the oxygen atoms?

To be a crown ether, the compound in question must be an ether, i.e. must contain oxygen atoms which are bonded to two carbon atoms neither of which is bonded to another oxygen atom. Thus, if you had just one carbon between two subsequent oxygens you would no longer be dealing with a crown ether; if you wanted, you could call it a crown acetal. But it would have vastly different properties; most notably it would break down in aquaeous solution under acid catalysis.

To the best of my knowledge, there is no formal restriction other than that given in the first paragraph for crown ethers. 20-crown-5, which would have three $\ce{CH2}$ groups between each ether oxygen, would be a valid crown ether in my humble opinion. However, most cations that are coplexed by oxygen prefer five-membered rings for various reasons, resulting in many complexing agents — $\ce{en, dmg, m2g, phen}$ to name just a few examples — having exactly two other atoms between the ligating atoms. Thus, practically all crown ethers one hears of have two carbons between the oxygens.


Cutting a Solid with a sharp edged tool The third point of difference suggests that amorphous solids can't be cut into two parts of regular surfaces
But glass being an amorphous solid is cut into regular pieces often.
What am I getting wrong?

Your textbook is correct. Glass surface may appear smooth, but in reality it is really rough and irregular if viewed under a microscope. Here you will find some pictures. Have you ever seen an insect walking on a window? It is possible because of those irregularities on the glass surface. Insect legs have tiny hooks which graple to all these lumps, cracks and holes on the glass.


What is the IUPAC name of borax? Please don't say sodium tetraborate decahydrate since that is not the real composition of borax.

Borax contains tetranuclear coordination complexes $\ce{[B4O5(OH)4]^-^2}$. I'm asking about the name of this complex in borax.

The name suggested here (via the question comments), the anion-corresponding part, ‘bicyclo[3.3.1]tetraboroxane-3,7-diolate’ ($\ce{[B4O5(O)2]^2-}$) obviously does not match with the tetraborate formula $\ce{[B4O5(OH)4]^2-}$ present in borax, whose structure, confirmed by X-ray crystallography, including hydrogens[1], is

However, the ‘bicyclo[3.3.1]tetraboroxane’ part is correct, for the parent hydride:

This type of names is known e.g. for the ‘cyclotriboroxane’ name for boroxine, or ‘cyclotetrasiloxane’ (in octamethylcyclotetrasiloxane).
So the name would be rather 1,3,5,7-tetrahydroxybicyclo[3.3.1]tetraboroxane, or bicyclo[3.3.1]tetraboroxane-1,3,5,7-tetrol, however it still does not explicitly tell that it's an anion.
Interestingly, this boroxane, siloxane type names (interchanging element/oxygen pattern) is rather marginally mentioned in the inorganic nomenclature, notably more detailed in the organic nomenclature.[2]
Inorganic additive nomenclature is able to constructs names for similar skeletons,[3] but somewhat differently than organic nomenclature.
E.g. bicyclo[x.y.z]Nane (where N is the numerical prefix root, with the exception for 1..4, ‘meth’..‘but’, for the organic names) would be named: Ncarby-[0a.z1,b]dicycle, where a = x+y+2 and b = x+2. If the structure is an ion, it's ‘…cyclate’ followed by the charge in parentheses. One advantage is that the first numeral is the size of the (largest) cycle, not just a branch.
Some examples from the book:

1,7-diazyundecasulfy-[012.11,7]dicycle
(IR-1.5.3.4 Example 6)

cyclo-triphosporic acid
2,4,6-trihydroxido-2,4,6-trioxido-1,3,5-trioxy-2,4,6-triphosphy-[6]cycle
(IR-8.3 Example 5)

tetraaluminy-1-lithy-[05.01,301,402,5]tetracyclate(1−)
(IR-7.4.3 Example 11)

(The leading zero in the first numeral in the square brackets is intentional for the cycles, however in the latter example, it's not used, despite IR-7.4.2 Nodal descriptor, example 3, that states that the nodal descriptor for hexacycle is ‘06’.)
With this nomenclature we can get very descriptive (but almost unusable) names
1,3,5,7-tetrahydroxido-1,3,5,7-tetrabory-2,4,6,8,9-pentaoxy-[08.11,5]dicyclate(2−),
1,3,5,7-tetrahydroxidotetrabory-2,4,6,8,9-pentaoxy-[08.11,5]dicyclate(2−), or
1,3,5,7-tetrahydroxido-1,3,5,7-tetraborypentaoxy-[08.11,5]dicyclate(2−).
References:

Levy H.A., Lisensky G.C. Crystal structures of sodium sulfate decahydrate (Glauber's salt) and sodium tetraborate decahydrate (borax). Redetermination by neutron diffraction Acta Crystallographica Section B 34 (1978): p. 3507
IUPAC Blue Book 2013, P-52.1.6.2
IUPAC Red Book 2005, IR-7.4 Additive nomenclature: Inorganic chains and rings


Why is the tetraamminecopper(II) ion square planar? The tetraamminecopper(II) ion is square planar in shape. But why? How can I rationalise this?

Can I rationalise this without the concept of hybridization? And even if I take hybridization into account, which hybridization takes place here and why?

Well it's not really square planar, it's just a streched octahedron. For copper(II) you end up with a $d^9$-system, a $d^8$ system might be able to do a real square plane but that one electron in a $d^9$ would cost too much energy if it was in a true square plane. The effect is called Jahn-Teller effect. You can simply imagine the $z$-axis for example of the octahedron to be elongated. If you did this till it breaks, you end up with the square base of an octahedron. And this is also what happens in a real square planar complex. All $d$-orbitals with a $z$-component are lowered in energy so they become occupied by electrons and, due to repulsion with the ligand's electrons, ligands will not come from the $z$-axis anymore. On the other hand if you lower the $z$-part in energy due to lever principle you will have to increase the energy of other orbitals as well.

You can see this quite beautifully demonstrated here:

Let's imagine the right case with a square plane and a strong ligand so you start filling orbitals with two electrons before continuing to fill the next higher level or orbitals. Then you could add 8 electrons in total and leave the $d_{x^2 - y^2}$ empty. In total this gives you more energy than in the case of an octahedron. If you look to the left at the octahedron, you can see that for 8 electrons you would have to fill the $e_g$ as well and in total you'd end up with less energy which is why a plane geometry is favored by low spin $d^8$ systems.

Now copper(II) with its $d^9$ is in between that, a plane would be favored of course but it would require a lot of energy to put that extra electron into the $d_{x^2 - y^2}$. So you go in between, a really stretched octahedron, this is the second picture. Here the $d_{x^2 - y^2}$ is not as high as in the other case but the other orbitals you fill are already lower than in a symmetric octahedron so you end up with more energy already. And this is why copper(II) tends to do that. On the other hand this also changes the $d-d$ gap and explains that blue color of copper(II) compounds.

The other two ligands which are on the streched positions can be water, it could be a water and a sulfate, like in $\ce{CuSO4 * 5 H2O}$, but there is still something usually.


Can thiols be created using a Grignard reagent? I'm trying to find information on the synthesis of thiols via Grignard reagents. While looking at the synthesis routes on Wikipedia, it said that a Grignard reagent will react with sulphur to form thiolates which will hydrolyse easily to the thiol; however, I'm struggling to find any references for this. Has anyone ever heard of this before or maybe tried it themselves?

I've done this with aromatic Grignards, but I have no reference and no database access any longer.

I think when I did it, I was relying on an older colleague's instructions. It was pretty much generate your Grignard from ArBr + Mg in THF, cool in ice bath, stir vigorously and add elemental sulfur. You may need to introduce a reductive step as you can generate polysulfides.

Grignards will also add to disulfides so you could react with dibenzyl disulfide then reduce off the benzyl group.

This reference is the nearest that Google has managed to find Can. J. Chem. 70 (1992) 30


Something I don't understand about batteries someone explained how a zinc copper battery works but there's already something I don't understand in the beginning.

Here is the reaction this person said happens in the battery

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

I understand how $\ce{Zn}$ transforms in $\ce{Zn^2+}$ (because it gives electrons to the wire, the $\ce{Zn^2+}$ then goes in the electrolyte) But I don't understand where the $\ce{Cu^2+}$ comes from. I understand that the $\ce{Cu^2+}$ gets electrons from the wire and turns into $Cu$ but the copper electrode that is in the battery is made out of $\ce{Cu}$ not $\ce{Cu^2+}$ so where does it come from?

It comes from the first answer here: https://physics.stackexchange.com/questions/247725/why-electrons-flow-through-a-wire-connected-to-a-battery

We manually put Cu ions on the electrolyte, it then reacts with the excess electrons on the electrode and makes Cu atoms.


What is a functional group? I know that a functional group gives definitive property to an organic compund.
But my text book claims carboxylic acid is a functional group but isnt carboxyl the functional group with a formula -COOH , is the text book wrong or am i missing something?
And similarly isnt hydroxyl the functional group of alcohols rather than alcohol being the functional group.
Could you please clear this out ?

Your books statement that "carboxylic acid is a functional group" doesn't seem quite right to me. I think it's better to say that a carboxylic acid is an organic molecule containing a carboxyl functional group.

Similarly, I wouldn't say that an alcohol is a functional group, but that an organic molecule containing a hydroxyl group is an alcohol.

Of course you can have single large molecules with multiple functional groups that could be called, at once, for example, an alcohol, a ketone and a carboxylic acid, because it contains a hydroxyl functionality, a carbonyl functionality and a carboxyl functionality.

I hope this addresses what you were asking, don't hesitate to ask if I missed something.


What's the difference between Zinc Ascorbate and Zinc Gluconate? I've been looking at a dental product for pets, called MAXI/GUARD Oral Cleansing Gel and it states it uses zinc ascorbate/taurine, however in the ingredients I found elsewhere it instead lists zinc gluconate.

Active Ingredient(s): Deionized water, zinc gluconate, ascorbic acid (vitamin C), methylcellulose, taurine, methylparaben, propylparaben, F.D.&C. blue no. 1. May also contain zinc sulfate.

What is the difference between the two and any idea why they are listed as such?

As I understand your question, “the two” refers to Zinc Gluconate and Zinc Ascorbate.
Zinc gluconate and zinc ascorbate are distinctly different compounds.  They are both adducts of Zn with one Zn and two bidentate oxygen-based ligands of organic acids.  Some basic information from PubChem for both zinc gluconate and zinc ascorbate is given below.
Zinc gluconate: $\ce{C12H22O14Zn}$
Zinc gluconate [Gluconic Acid Zinc (II) Salt] is a zinc salt of gluconic acid comprised of two gluconic acid molecules for each zinc cation ($\ce{Zn^2+}$). Zinc gluconate is a generally recognized as safe (GRAS) substance by FDA.

Of interest, as related to periodontic issues note that a phase 4 clinical trial was completed 2014-12-16 “The effect of Zinc-D-gluconate on bleeding propensity of the periodontium in patients with chronic parodontitis - a randomised, placebo-controlled, double-blind interventional study”.
Zinc ascorbate: $\ce{C12H14O12Zn}$
Zinc ascorbate is a zinc salt of ascorbic acid[Vitamin C] comprised of two ascorbic acid molecules for each zinc cation ($\ce{Zn^2+}$). Zinc ascorbate goes by a variety of different names, L-Ascorbic acid, zinc salt (2:1), L-Ascorbic acid, zinc salt, (-)-Ascorbic acid, zinc salt to name a few.

A ResearchGate reference discusses zinc-gluconate, taurine in reducing halitosis and gingivitis.  A portion of the abstract follows:

Objectives: to evaluate the efficacy in reducing halitosis and gingivitis insurgence in fixed orthodontic patients of a
zinc-gluconate, taurine and cetylpyridinium-chloride (CPC) oral-rinse,
sole or combined with a toothpaste with the same active principles.

References:

https://pubchem.ncbi.nlm.nih.gov/compound/443445
https://pubchem.ncbi.nlm.nih.gov/compound/Zinc-ascorbate#section=Structures
Dalessandri, Michela & Dalessandri, Domenico. (2011). Halitosis And Gingivitis Reduction Using A Zinc-gluconate-taurine-cetylpyridinium-chloride Oral-rinse And Toothpaste. (link)


Why does ammonia accept a hydrogen to form ammonium when its octet is already complete? Why does ammonia accept a hydrogen to form ammonium when its octet is already complete as $\ce{NH3}$ and it should be, seemingly, stable as $\ce{NH3}$?

"Hydrogen" means different things at different times. It can mean $\ce{H+}$, $\ce{H.}$, or $\ce{H2}$. (Not to mention deuterium and tritium!)

A bare proton ($\ce{H+}$) does not bring an electron to the filled octet on nitrogen, but attaches to the lone pair. The positive charge is now shared over 5 atoms, and the energy of formation of a new bond stabilizes the ammonium cation. The octet on nitrogen is still only an octet.

Interestingly, a hydrogen atom ($\ce{H.}$) can also attach to an ammonia molecule to form ammonium radical. Electrolysis of ammonium salts with a mercury cathode  yields ammonium amalgam, which rapidly decomposes to $\ce{NH3}$ plus $\ce{H2}$. In this case, the extra electron in $\ce{NH4.}$ is also shared among the 5 atoms, and is more stable than when it is attached to just one proton. But not that much more stable, and the bond between two hydrogens is even stronger.


Bolt head corrosion in soup. Toxic? Zinc? Cadmium? This may not be quite the right forum for this question, but please help me if you can.

I recently repaired the lid of my grandma's crockpot using a bolt I found in a miscellaneous fastener box in the garage. I've been using the crockpot quite a lot since then, and I just noticed that the bolt head is corroding with white gunk. I also noticed a white film floating on the middle of the soup I left cooking all night, obviously from some of the corrosion dripping onto the soup. My grandma and I had both tasted the soup before I noticed the corrosion and film. I've been doing a little research to try and identify the bolt's coating by its appearance, and I am concerned that it might be zinc or cadmium, which are apparently highly toxic. [EDIT: apparently zinc is not highly toxic, so I guess my main concern is: is this cadmium? Also, the bolt may well be over 50 years old, as it's from my grandparents' garage]

I'm hoping someone here who's more familiar with these materials can tell me:

What metal this corrosion looks like
Whether I should be concerned about toxicity considering the quantities present

I'm including a picture of another bolt that matches the one in question, with yellow zinc plated wood screws and some U.S. currency for visual reference.

I will of course be replacing the bolt with something explicitly food-safe, but having already consumed some of the soup, I want to know if we should be high-tailing it to the emergency room, or if in fact it's safe to finish the soup (it's good soup!).

You can test for cadmium at home, but the ~US$30 price for the test kit (with 1,5-diphenylcarbazone) is probably more than the soup is worth, and the test might not be sensitive enough to detect sub-acute toxic amounts of Cd.

Considering the highly toxic nature of the metal, you might check with your family physician if you and anyone else sampling the soup should have a blood or urine test.

As has been mentioned, the small amount of zinc that might have been absorbed had the bolt been galvanized (unlikely, anyway) is not significant. The US NIH recommends ~10 mg of Zn in the diet per day as a nutrient.


Molecular orbitals of CH2 an SLACs I was wondering about molecular orbitals of methylene different states 1B1 3B1 1A1
and how are they connected to symmetry point group lower case 1a1, etc orbitals what those numbers and alphabets mean?

It would be great if someone can derive those orbitals from symmetry and show me which states 1B1, 3B1 and 1A1 are referred to and what does up script, subscripts and alphabets stand for and connect them to lower case 1a1 ,etc.

The first question to always ask when you are trying to build an MO diagram is what point group does my molecule fall into?  In this case, the $\ce{CH2}$ molecule has C2V symmetry (like the water example shown above).  Each of the atomic orbitals (AO's) on carbon (left) and the combination of $1s$ orbitals on the hydrogen atoms (right) has a symmetry label that corresponds to one of the irreducible representations of the point group.

You can determine by symmetry label for the AO's by operating on the orbital with each of the symmetry operators in the C2V group (E, C2, σ, σ').  For example, the 2py orbital is in the plane of the nuclei and so it is symmetric with respect to E and σ' but anti-symmetric with respect to C2 and σ and has b2 symmetry from the C2V character table.

For the hydrogens, you have to consider the combination of both 1s orbitals at the same time.  The a1 combination is when the phase of the wavefunction is the same and the b2 combination is when the phase is opposite (i.e. H1 is positive and H2 is negative or vice versa).

AO's that have the same symmetry can combine to form MO's with that symmetry label.  In the above example, the 2s and 2pz orbitals on oxygen have a1 symmetry, and they combine with the a1 combination of hydrogen 1s orbitals to form three MO's with a1 symmetry.  The 2py (b2 symmetry) orbital combines with the b2 combination of the hydrogen atoms to form two MO's. Finally, the 2px orbital remains unchanged because there are no other SLAC's with b1 symmetry.

The MO's are qualitatively ordered based on the number of nodes that are found when you combine the SLAC's.  The numbers in front of the MO's differentiate the orbitals with the same symmetry.


How to predict a COSY NMR spectra for an AMX spin system? Does anyone know how to draw (predict) a hypothetical COSY NMR spectra for AMX spin system (the spin of A M X nuclei is 1/2)?

An AMX spin system represents three nuclei coupled to each other, with their signals separated by a large chemical shifts compared to the coupling between them (such that $\Delta \delta >> 3J$). This system is said to be first-order, and the A, M and X signals are each a doublet of doublets, and the couplings can be extracted by inspection.

The simple COSY spectrum of this will show the three core signals along the diagonal, with cross peaks appearing for all signals ie each signal will have two cross peaks. Below is a (very basic) schematic for an AMX COSY.

The appearance of the cross peaks will depend on the type of COSY experiment you are running, such as a COSY-45 or COSY-dqf etc, but this should give you the general picture for what you are asking.