What route to take in converting ethanol to crotonic acid? The question asks to write the theoretical transformation steps to convert ethanol to crotonic acid in considerable yield. One constraint is that you cannot use any external organic compounds but inorganic catalysts and reagents are allowed and aldol condensation is prohibited. I do not know the answer, but I came up with the following steps:-

Treatment with $\ce{HBr}$ to form $\ce{CH3CH2Br}$ via $S_N2$.
Treatment with $\ce{Mg}$ in ether, to form corresponding Grignard reagent.
Oxidation of ethanol by Pyridinium chlorochromate to ethanal, and then treating it with the formed Grignard reaction to form $\ce{CH3CHOHCH2CH3}$.
Acid catalysed dehydration to form $\ce{CH3CH=CHCH3}$.
Using NBS (N-Bromo Succinimide) as an allylic halogenating agent, followed by treatment with aq. $\ce{KOH}$ to give $\ce{CH3CH=CHCH2OH}$.
Oxidising this with PCC followed by Tollen's reagent to selectively oxidise alcohol to acid, to finally yield crotonic acid.

I am unsure if any of these steps are incorrect, i.e if any of these steps will not give appropriate yields of the desired product owing to side reactions. Another problem is that this transformation is too long. Is there any shorter way to carry out this conversion in proper yields?

Allylic bromination and subsequent hydrolysis might be replaced by an allylic oxidation of the butene using selenium dioxide - if that's inorganic enough for you ;)

But I'm wondering whether there's a shorter way. What about the Grignard reaction you described, followed by oxidation (PCC on silica or manganese dioxide) to 2-butenone. Subsequent bromination with bromine in hydrobromic acid should yield an $\alpha,\alpha'$-dibromoketone which, upon treatment with base (Favorskii rearrangement) and, eventually acidic workup should yield crotonic acid.


Is an alcohol-based spray cleaner ok for cleaning an aluminum heat exchanger? Simple alcohols such as ethanol can, apparently, corrode aluminum. However, this corrosion reaction seems to be very slow at room temperature and below. Is an ethanol based, non-foaming spray cleaner, propelled by LPG and with an ethanol concentration of, say, about 20%-30%, some (about 2%-3%) alcohol ethoxylate (this acts as a surfactant) and a slightly basic pH (between 7 and 8), really ok to spray on a cold parallel-flow type heat exchanger without fear of any damage? Or should an alcohol-free cleaner be preferred in any case for that job?

My belief is that the corrosion can only happen when the aluminum object is actually submerged in pure alcohol. That little alcohol conveyed by the spray probably evaporates before having the chance to do any damage, but i'd like to understand the surfactant role better, i.e. if any added surfactant agent can act as a corrosion inhibitor.

When aluminum metal is exposed to the oxygen of the air at room temperture, it forms a protective air-tight layer of aluminum oxide. So as long as nothing is present that will disrupt the oxide layer on aluminium it is OK for aluminium at 20 oC to be in contact with air.

If an aluminium alkoxide is exposed to water then the aluminium alkoxide will react with the water to form aluminium oxide and the alcohol. So in alcohol which contains some water it is not possible for the aluminium to dissolve in the alcohol.

But if the alcohol is anhydrous then it will be able to enter a scratch in the aluminium and start to react with the metal thus causing corrosion. For titanium that anhydrous methanol is very corrosive.

So as long as some water is present and air is present then at room temperture


Why energy of all d orbitals is same? This question first popped up in my head when I learnt that $d_{z^2}$ orbital is used in $sp^3d$ hybridization and $d_{x^2-y^2}$ and  $d_{z^2}$  orbitals are used in $sp^3d^2$ hybridization. If all the $d$ orbitals are said to be degenerated orbitals (i.e. they are said to have equal energy) then why these two orbitals are used in $sp^3d$ and $sp^3d^2$ hybridization?

Some said that $d_{z^2}$  is used because it is closer to the nucleus because of the ring of electron density surrounding the nucleus, but if that is the reason then why do we call the $d$ orbitals as degenerated in the first place when clearly two orbitals are closer to nucleus.

Here's what I want to know:

A) Why $d$ orbitals have same energy when its clearly apparent from their shape that two orbitals have lower energy than other three?

B) Why $d_{z^2}$ orbital is used in $sp^3d$ hybridization and $d_{x^2-y^2}$ and  $d_{z^2}$  orbitals are used in $sp^3d^2$ hybridization?

A. The orbital shapes originate from solutions to the Schrodinger equation for hydrogen like atoms. The derivation of the orbitals come from the section for the solution to Schrodinger equation for Hydrogen-like atoms on Wikipedia: https://en.m.wikipedia.org/wiki/Hydrogen-like_atom

The solutions have the same energy because you solve the equation for specific energies. They really are the same, which can proven by doing an integral across the volume.

The different hybrid orbitals depend on what quantum numbers are possible. This question has been answered here: Why an asymmetric geometry with sp3d and sp3d3 hybridization?


Hydrogen bonding relating to molecular orbital theory In [1, p. 232] here is an explanation of how alcohols hydrogen bond to each other and I do understand how the alcohols form hydrogen bonds to each other:

FIGURE 6.15 Water and alcohols are both Brønsted acids and Brønsted
bases. This figure shows hydrogen bonding between molecules of water $(\ce{R=H})$ and alcohols.

Is it possible to think of hydrogen bonding as a Lewis Acid-Base reaction rather than just a Brønsted Acid-Base reaction, and, by extension, think of this whole scheme as a HOMO-LUMO interaction?
Like, can I think of hydrogen bonding as a HOMO-LUMO interaction in general, arising from the idea that it can be classified as a Lewis Acid-Base reaction?
References

Jones, M.; Gingrich, H. L.; Fleming, S. A. Organic Chemistry, 4th ed.; W.W. Norton: New York, 2010. ISBN 978-0-393-93149-5.

Yes you can think of hydrogen bonding in terms of MO theory. Consider a hydrogen bond between two water molecules as a simple example.

The donor hydrogen donates electron density into the $\pi^*$ orbital of oxygen which is the LUMO of the acceptor oxygen. This thus weakens the donor $\ce{O-H}$ bond, but shortens the distance between the donor hydrogen and acceptor oxygen.

Thus, there are two components to be considered: the pure electrostatic attraction and the stabilization due to covalent interactions.

Because all you're asking about is the MO side of things, as explained above, hydrogen bonding can be viewed as donation of electrons into the $\pi*$ orbital of some acceptor atom. This could indeed be called a Lewis acid-base interaction, but I don't hear this very often.

I'm not sure what your textbook is talking about with hydrogen bonding being a Brønsted-Lowry reaction. Hydrogen bonding is not a reaction (in the traditional sense) so I would disagree with your textbook. The acceptor oxygens are accepting electrons much more than they are accepting the whole proton which would make them Lewis acids. It's certainly possible for water/alcohols to behave as either Lewis acids or bases and Brønsted-Lowry acids or bases, but the former is much more common based on the fact that auto-ionization is quite a small effect but hydrogen-bonding is a very large (and common) one.

Regarding the MO view of things, the exact same logic can be applied in inorganic chemistry when considering carbonyl compounds or dihydrogen/dihydride compounds. In that case the donation of electrons into a $\pi*$ orbital (or $\sigma*$ orbital in the case of dihydrogen compounds) is internal to the molecule however and we call it backbonding.

Here's an MO and orbital diagram showing that interaction.


Which of the following has highest solubility in water? Which of the following has highest solubility in water?

A. $ Ca(OH)_2 $

B. $ Fe(OH)_3 $

C. $ Cr(OH)_3 $

D. $ Zn(OH)_2 $

And also please explain how can we predict their solubilities when we are not provided with their solubility products.

I know that calcium hydroxide is sparingly soluble. So it cannot be the right option. I think Iron hydroxide is also soluble to a good extent. I haven't got any idea about option C. I think D may be the right answer.

You can't predict solubility reasoning, you have to remember solubility rules in your case this one:

  Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and $\ce{Al^3+}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not soluble.

So in this case $\ce{Ca(OH)2}$ even if slightly soluble theoretically should be the right answer.


Thermochemistry: forms of energy & pressure I've been in a few classes now that teach about energy, and I feel they're always so bad. There is always such a rush to start using equations to get problems solved, and I often feel I don't have an intuitive grasp on energy at all. I guess this question is also about pressure.

Questions:

Does pressure decrease with time? If pressure is caused by the force of gas molecules hitting the inside surface of a container, don't they transfer energy and eventually slow down enough to totally reduce pressure?
Along the same lines, energy can be released or absorbed by a reaction as work or heat. If the only change that happens is through work (in something that is totally insulated - where gas is formed and pressure is created), after the pressure is created, doesn't it require continuous energy to maintain pressure against a constant force of 1 atm (or whatever it is)? Heat dissipates to all things, and that makes sense as everything gets colder in time. The same does not seem true for pressure.

Thanks in advance.

Noobsauce

1.Does pressure decrease with time?

The ideal gas law tells us that
$$PV=nRT$$
or
$$\mathrm{P=\frac{[nRT]}{[V]}}$$

If the temperature, number of moles or volume does not change, then the pressure will not change.

  If pressure is caused by the force of gas molecules hitting the inside
  surface of a container, don't they transfer energy and eventually slow
  down enough to totally reduce pressure?

Yes, that's correct.  If energy is transferred out of the system, then the temperature will fall and the pressure will decrease.  But if everything is already at the same temperature, there will be no energy flow in or out and the pressure will remain constant.

  after the pressure is created, doesn't it require continuous energy to
  maintain pressure against a constant force

No, there is no requirement that energy must be supplied to maintain the pressure.  Again, if the gas, the vessel and the surroundings are all at the same temperature the pressure will not change.  However, if the entire system is not at thermal equilibrium, then heat will flow so as to put things at equilibrium and the temperature will change.


Why does 1 cubic meter = 1000000 cubic centimeters? I have the SI chart in front of me. It states that a centi is 10^-2 and since they are already designated cubic meters why does 1 cubic meter equal 1000000 cubic centimeters?

Take a look at the Wikipedia article for SI prefixes, in particular, the following excerpt:

When units occur in exponentiation, for example, in square and cubic forms, the multiplication prefix must be considered part of the unit, and thus included in the exponentiation.

$1\ km^2$ means one square kilometre, or the area of a square of $1000\ m$ by $1000\ m$ and not 1000 square metres.

$2\ Mm^3$ means two cubic megametres, or the volume of two cubes of $1000000\ m$ by $1000000\ m$ by $1000000\ m$ or $2\times10^{18}\ m^3$, and not 2000000 cubic metres ($2\times10^{6}\ m^3$).

Thus, the relationship between the cubic metre and the cubic centimetre is as follows:
$$1\ m^3 = (1\ m)^3 = (100\ cm)^3 = 100\ cm \times 100\ cm \times 100\ cm =  1000000\ cm^3 = 10^6\ cm^3$$
$$1\ cm^3 = (1\ cm)^3 = (0.01\ m)^3 =  0.01\ m \times 0.01\ m \times 0.01\ m = 0.000001\ m^3 = 10^{-6}\ m^3$$
For a real case, converting $243.7\ cm^3$ to $m^3$ step by step:
$$243.7\ cm^3 = 243.7\times 1\ cm^3 = 243.7\times (1\ cm)^3 =243.7\times (0.01\ m)^3 = 243.7\times 0.01\ m \times 0.01\ m \times 0.01\ m = 243.7\times 0.000001\ m^3 = 243.7\times 10^{-6}\ m^3 = 2.437 \times 10^{-4}\ m^3$$
This is simply the way it was defined, and it is the standard everyone adheres to. More examples of SI prefix usage can be found on this page from one of the organizations which created the SI units, the BIPM.


What's the technical term for supercharging an acid with an oxidizer? And does it always work? In a recent Mythbusters episode, where they tested some Breaking Bad chemistry assertions, they debunked the hydrofluoric acid body-dissolving myth referenced elsewhere in Chemistry.SE in the question "Is it actually possible to dispose of a body with hydrofluoric acid?".

The answer is no, not really, and for the same reasons as my answer to the linked question; fluorine's just too electronegative to let the proton go do its acid thing.

To attempt to replicate the results (in this case dissolving out the bottom of the bathtub they were using), they moved to a much stronger acid, sulfuric, and then "supercharged" it with what they simply called their "special sauce" (as they said, they're not in the business of showing people how to dispose of bodies). They gave it away in the end, however, by saying the special sauce was "basically a lot of hydrogen and oxygen", and since it wouldn't be water the obvious candidate is hydrogen peroxide, a strong oxidizer (in high concentrations at least).

This seems to be a well-known trick of household chemistry; add household peroxide to cream of tartar or lemon juice, for instance, and watch those rust stains come right out. It also seems to be the bracingly direct way to form perchlorates in the lab; add hydrogen peroxide to hydrochloric acid, and behold perchloric acid (or a fuming crater, if you weren't managing the exothermic nature of this reaction properly).

So, first question, what's the technical term for this trick? It can't just be "supercharging".

Second and related question, does it work for all acids and oxidizers, or do only certain combinations of acids and oxidizers have this effect?

Lastly, for purely academic purposes, what would the basic reaction have been between sulfuric acid, hydrogen peroxide and organic tissue ($\ce{HxCyOz}$)? The sulfate anion is pretty highly oxygenated as it is, so I'd imagine the combination, unlike with a hydrohalic acid forming an oxoacid, would cause the reaction to be less acid-base, and more rapid oxidation and "combustion" of the organic tissues, leaving behind mainly carbon in water and off-gassing a large amount of steam and sulfur dioxide.

I don't know if it's called anything, but when you use an oxidizer in solution, you usually need a source of hydrogen ions for the oxidized oxygen to be reduced.  In the case of hydrogen peroxide, $\ce{H2O2}$, the oxygen is in the $-1$ state rather than the normal $-2$.  The usual way $\ce{H2O2}$ acts as an oxidizer is to be reduced to water in the presence of 2 hydrogen ions, usually from a strong acid:

$\ce{2H+ + 2e- + H2O2 -> 2H2O}$

Notice that the reaction consumes 2 electrons on the left, which would come from other reactants that have been oxidized (i.e. lost electrons).

The point is that a strong acid by itself has only a corrosive effect, and a highly dehydrating effect in the case of sulfuric acid.  But if you add an oxidizer, then in the presence of a fuel (like an organic compound) plus sufficient energy (say from the heat of reaction), then you are effectively getting a combustion reaction as well.  This is the sort of "combustion" that goes on inside your body to use food as fuel, for example.  I supposed this could be called "supercharged".

EDIT: So to answer your specific questions:

1) No technical term that I know of;

2) Yes, assuming the acid is strong enough to donate protons;

3) In most organic molecules, carbon is more electronegative than hydrogen, so it can be oxidized by oxygen.  And so you get a "combustion" reaction where molecular carbon is oxidized to elemental carbon (i.e. burnt).


Sketching lattice points, primitive axes, primitive cell, and basis of atoms I am currently studying Introduction to Solid State Physics, 8th edition, by Kittel. Chapter 1 provides the following figure and accompanying explanations:

It also says the following:

An ideal crystal is constructed by the infinite repetition of identical groups of atoms. A group is called the basis. The set of mathematical points to which the basis is attached is called the lattice. The lattice in three dimensions may be defined by three translation vectors $\mathbf{a}_1$, $\mathbf{a}_2$, $\mathbf{a}_3$, such that the arrangement of atoms in the crystal looks the same when viewed from the point $\mathbf{r}$ as when viewed from every point $\mathbf{r}^\prime$ translated by an integral multiple of the $\mathbf{a}$'s:
$$\mathbf{r}^\prime = \mathbf{r} + u_1 \mathbf{a}_1 + u_2 \mathbf{a}_2 + u_3 \mathbf{a}_3. \tag{1}$$
Here $u_1, u_2, u_3$ are arbitrary integers The set of points $\mathbf{r}^\prime$ defined by (1) for all $u_1, u_2, u_3$ defines the lattice.
The lattice is said to be primitive if any two points from which the atomic arrangement looks the same always satisfy (1) with a suitable choice of the integers $u_1$.

Here are some lattice points I drew:

It doesn't seem to me that any of these lattice points are primitive; that is, I don't think that any of these lattice points satisfy $\mathbf{r}^\prime = \mathbf{r} + u_1 \mathbf{a}_1 + u_2 \mathbf{a}_2 + u_3\mathbf{a}_3$. The irregular pattern present in this space lattice makes it difficult to see how this could be done.
So, in order to satisfy figure 3c, what are the lattice points of interest?
I would greatly appreciate it if people would please take the time to explain this.
EDIT

I think the cell after EDIT is correct. Here I added a few neighboring cells:

Admittedly, the atom in the right bottom corner is missing. Apparently the author didn't consider it important.


How is the Activation Energy of the Arrhenius equation related to the exponents in rate laws? I know that activation energy for a particular reaction does not depend on the stoichiometry of the reaction.But how can that be justified from Arrhenius Equation ?Doesnt 'k' change with stoichiometry?Or am I going wrong?

Because we know that in all the kinetics equations, when  the reaction stoichiometry changes the rate constant in the equations have to be altered accordingly.

Rate Law

For the reaction:

$\ce{aA + bB -> cP}$

The rate law is as follows:

$R = k[A]^x[B]^y$

where:

  $a$,$b$,and $c$ are stoichiometric constants which have integer values
   $[A]$ is the concentration of species A
   $x$ is the order with respect to species A
   $[B]$ is the concentration of species B
   $y$ is the order with respect to species B
   $k$ is the rate constant.

(1) Neither $x$ nor $y$ depend on $a$, $b$, or $c$.

(2) Neither $x$ nor $y$ is required to be an integer value.

(3) k is measured for a specific temperature. The Arrhenius Equation would predict how k would change with temperature. So k does not depend on any of $x$, $y$, $a$, $b$, or $c$.

Say the reaction was for some sort of gas reaction. I now add a catalyst. I get a new rate:

$R_2 = k_2[A]^{x_2}[B]^{y_2}$

There isn't any way to use the first rate equation to predict what the second rate equation will be. (You could of course try to analyze the function of the catalyst to make predictions on what the best catalyst would be...)


Is the triiodide ion polar? Three professors argue it is non-polar.

My professor argues that it is a monopole, like most ions.

The structure of the triiodide ion places a negative formal charge on the central iodine atom. The molecular geometry is also linear (at least according to VSEPR), and its electronic geometry is trigonal bipyramidal. Given the high amount of symmetry, shouldn't triiodide be non-polar? Sure, it might have a charged central atom, but so does carbon dioxide.

What's the answer?

The structure of the triiodide ion places a negative formal charge on the central iodine atom.

No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).

That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.

The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.


Predicted number of signals in proton NMR spectroscopy Given the molecule,.

I can predict that there will be 4 signals present in the NMR spec. Problem is, the quiz says 5 is the correct answer. My question is, are alkene hydrogen not chemical-shift equivalent?

Since there isn't free rotation around the C=C bond in the alkene, the two hydrogens, Ha and Hb are experiencing different electronic environments. Whenever two hydrogen atoms are experiencing non-identical electronic environments, they are going to possess differing chemical shifts in proton NMR. If you heat this compound, and it doesn't decompose, I would expect the two signals from Ha and Hb to coalesce into one as the hindered rotation about the alkene C=C bond becomes unhindered.

https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_05%3A_Structure_Determination_II/5.2%3A_Chemical_equivalence


Reasonably complete infrared spectral database with data in CSV form? I'm currently developing a neural network that analyses IR spectroscopy signatures. I've found a very complete data set at The Spectral Database of Organic Compounds, but data is only available as images, which are not ideal for training neural nets.

Does anyone know of a reasonably comprehensive open database of spectroscopy data that offers data in CSV/tabulated form?

First off, I would think you could fairly easily design some data extraction tools for the AIST database. After all, the X and Y axes are standardized (i.e., $4000-400 \mathrm{cm}^{-1}$).

Beyond AIST, you can find machine-readable JCAMP files from the NIST WebBook. There are supposedly ~16,000 spectra, although you'd have to script some way to download them one-by-one.

I think a better bet would be to purchase a database:

Bio-Rad sells an IR database. A quick look now indicates that it has ~235,000 spectra and I'm sure it's in an accessible format.
NIST sells a smaller database ~5000 compounds in JCAMP format.

A quick search also turned up:

IR Spectra.com - claiming ~140,000 spectra in various formats.


What is the difference between 'particles', 'molecules' and 'chemical species'? I have read the Chem.SE question about the difference between 'compound', 'mixture', 'element' and 'molecule', but I would like to know the precise definitions for more terms (see below). From the article, these are the definitions:

Elements are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of Isotopes which are physical variations among atoms of the same element).

Molecules are a group of atoms covalently bonded to each other.

Compounds are made up of two or more elements. Compound is often used interchangeably with the word substance.

Mixtures are collectives of different molecules or atoms.

How are compounds substances? Isn't 'substances' a general term for any chemical thing?
These are the terms that I would like definitions for:

chemical species
substance
particle
atom/isotope
salt

I am rather confused about these terms, because my tutor uses particles and molecules (and elements) as if they were the same thing. As for the rest, I want to know how to distinguish between general (such as chemical species, substance) and specific terms (such as particle, atom, isotope, salt).

A chemical species consists of chemically identical atoms, molecules or ions (more generally molecular entities) and is identified as a single species in a spectroscopic experiment. The latter part of the definition is important, because, for example, a compound which is identified as a single chemical species by NMR spectroscopy can still contain different enantiomers (that are separate molecular entities) which cannot be distinguished by an achiral spectroscopic method.

A substance is matter which has a specific and distinguishable chemical composition and specific properties. From this follows that every pure element and chemical compound is a substance. For example, pure $\ce{Fe}$ and $\ce{CH4}$ are substances, while an aqueous solution of $\ce{NaCl}$ and gasoline are not. The latter two are a mixture of two ($\ce{NaCl}$ and $\ce{H2O}$) or more different substances and have therefore properties which depend on the composition.

Particles can be defined as microscopic constituents of matter, like atoms or nuclei. The term can also refer to even smaller constituents like electrons, protons, or neutrons, which are called elementary particles. It is also used in the definition of an atom, the smallest particle which is characteristic for a chemical element. Atoms with the same number of protons $Z$ (and equal number of electrons) represent the same chemical element. Isotopes are atoms of the same chemical element which have different mass numbers, i.e. the same number of protons $Z$, but different numbers of neutrons in the nucleus. In the three isotopes of hydrogen, for example, the nuclei contain different number of neutrons (0, 1 and 2).

A salt is a chemical compound which consists of cations and anions. It should be distinguished from a zwitterionic compound, which contains charged but overall neutral molecules as building blocks. An example for the latter is trimethylamine N-oxide.


Energy a system at room temperature can supply? "This reaction requires XYZ activation energy, therefore it cannot happen at room temperature." is a sentence I find often in organic chemistry textbooks. Activation energies are steadily given in textbooks too. Even though it is a nice piece of info, which gives a feeling of how (not) sluggish a reaction can be, it would be nice to know how much energy can such a system actually supply? Like, I don't know - 30 kJ/mol or 35 kJ/mol.
How much energy do particles in a system at normal conditions (25°c, 1 atm) have? I am note sure if there even is an answer, taking into consideration Maxwell-Boltzmann distribution, and an endless array of reactions... but maybe there is?

Echoing Jon Custer's comment:

  R = 8.314 J/K/mol

Multiply by 298 K, and you have a rough-and-ready rule-of-thumb number:

The "thermal energy" at room temperature is about 2.5 kJ / mol.


Can two lone pairs in different orbitals make a chiral centre? Is it possible that if the two lone pairs in a molecule occupy different orbitals, then could this make the molecule chiral? What I mean is, suppose a carbon atom is attached to two different atoms and has two lone pairs in different orbitals, then could the carbon atom be considered chiral?

Chirality refers only to the positions of the nuclei. So, if we have the situation you described where one Carbon is bonded to two other atoms (e.g. CX$_2$ or CXY) it can never be chiral. (i.e. a three atom molecule can always be superimposed on its mirror image by a single rotation)

Now, why don't we consider the atomic and hybrid orbitals  populations when assigning chirality? This comes down to the fact that the atomic and hybrid orbitals are just a helpful tool for rationalizing/predicting the structure and reactivity, but they are not representations of the actual wavefunction. The true wavefunction is defined by a set of ''molecular orbitals.'' These orbitals always have the same symmetry group as the nuclear coordinates. In this case, there is a reflection through the plane defined by the three atoms (some jargon: this is called the C$_{s}$ symmetry point group). So if the nuclear positions are non-chiral, so is the wavefunction.

(More on chirality and symmetry: http://symmetry.otterbein.edu/tutorial/applications.html)


What non-poisonous chemicals can get rid of SLS effectively? Whenever I brush teeth, there are some kind of bubbles sticking inside the cavity and that's uncomfortable, it should be SLS as found on the internet. Does anyone know how do I get rid of those bubbles things easily?
Eat/Drink sth? Tell me if anything is wrong.

Any help would be appreciated, thanks.

Addition: SLS represents sodium lauryl sulfate, chemical exist in our cleaning product: toothpaste, shampoo,   etc. I don't know what it is for but seems it produce bubbles as long as the product is used.

SLS is highly soluble in water (100g/L)

Therefore, in the specific case of your teeth having lots of bubbles around them after brushing, you should simply gargle your mouth with water to remove the excess SLS.


Why do negative oxidation states not extend to -8? There are oxidation states from -4 to +8

Why do the negative oxidation states not go all the way down to -8? I mean if an atom is hypervalent meaning that it can have more than an octet than that means it can have potentially 8 bonds(counting a double bond as 2 bonds and a triple bond as 3 bonds respectively) to other atoms more electronegative than itself and thus Xenon can have a +8 oxidation state and some metals can as well.

Doesn't that also mean that a hypervalent atom can have 8 bonds to other atoms less electronegative than itself and thus have a -8 oxidation state?

Why do the negative oxidation states not go all the way down to -8?

Let's think about what we would need for this to occur - i.e. an oxidation state of -8.

We would need a central atom that has considerable electronegativity - one that is at least more electronegative than its attached atoms. This already excludes numerous molecules; most molecules' central atoms are, on a relative basis, more electropositive rather than electronegative.

In addition, this atom would need to have one of the following configurations; "allocated" refers to the assignment of oxidation state - since oxidation states are assigned on the premise that all bonding is ionic - the more electronegative element in a bonding configuration is "allocated" all the bonding electrons in the calculation of oxidation states. I.e. in H2O, the oxygen would be allocated both pairs of bonding electrons in the H-O bonds, and oxygen would then "have" 8 electrons, and its oxidation state would be -2.

1 valence electron and be "allocated" 9 electrons

2 valence electrons and be "allocated" 10 electrons

3 valence electrons and be "allocated" 11 electrons,

and on the other extreme:

6 valence electrons and be "allocated" 14 electrons,

7 valence electrons and be "allocated" 15 electrons,

8 valence electrons and be "allocated" 16 electrons.

As we can see, the first set of criteria are basically incompatible. Elements only having 1 or 2 valence electrons are going to be small (on a relative basis). These elements would also much rather be oxidized than reduced. Oxidation allows these elements to easily become isoelectronic with a noble gas.

The other set of criteria for a negative 8 oxidation state is also hard to fulfill. None of the group 6 elements except the possibility of sulfur and elements below it are hypervalent - i.e. they can accommodate more than 8 valence electrons. Even if we consider sulfur as able to accommodate 12 valence electrons (as some depictions of sulfuric acid suggest) we must remember that sulfur is not very electronegative and unlikely to be allocated any electrons.


Rationalising an apparent inconsistency in the determination of reaction spontaneity It is well-known that a reaction is spontaneous if $\Delta G < 0$. This requirement is derived from thermodynamic considerations, as done here. However, there seems to be two different ways of calculating $\Delta G$. As provided by Orthocresol in the answer here, $\Delta G$ is the rate of change of the Gibbs free energy of the reaction mixture with respect to the extent of reaction. This definition based on the rate of change cannot be disputed as a negative gradient of the graph would mean that as the forward reaction proceeds, Gibbs free energy is being minimised.

However, there appears to be another way in which $\Delta G$ is being calculated. In this approach, we take $\Delta G$ as the absolute change in Gibbs free energy going from reactants to products. One way of calculating this would be using $\Delta G = \Delta G_f [Products] - \Delta G_f [Reactants]$. Another would be $\Delta_rG = \Delta_rH - T \Delta_rS$. In this case, we are not calculating the gradient of the $G$ against extent of reaction curve but instead, we are simply calculating the absolute change in $G$ from all reactants to all products.

Clearly, the two approaches, one based on the gradient of the curve and one based on the absolute change of $G$ between reactants and products, are not the same. However, both approaches can be used to calculate $\Delta G$ in the determination of reaction spontaneity. The second approach clearly being more dubious than the first.

My attempt is rationalise this is as follows:

Both approaches are based on the idea of minimisation of the Gibbs free energy. However, they differ in their analysis of the reaction. The first approach of calculating \Delta G$ considers an ensemble of molecules going from reactants to products. It looks at small adjustments to the ratio of reactants to products in the reaction mixture as reaction progresses and subsequent changes to entropy of the mixture (due to mixing of reactants and products)

However, in the second approach, we are trying to look at the energetics in the one-step change from a reactant molecule to a product molecule and we are merely concerned with the question: does $G$ decrease when a reactant molecule changes into a product molecule? Clearly, this process does not occur in small steps but occurs at once, unlike the conversion of a mole of reactants to a mole of products. We make use of the $\Delta G_f$ values for the reactants and products to make an approximation for the conversion of one reactant to one product. This 2nd approach is certainly not exact but it is certainly still chemically correct.

I recommend the following article by Juan Quílez if you have access: DOI: 10.1021/ed100477x. The main point is summarized in their figure:

The lower curve shows your first approach, with $\Delta_r G$ as the slope of the curve. The upper curve (i.e. straight line) shows your second approach, where the slope is non-zero and constant. What the second approach leaves out (and it might be appropriate for certain cases, but usually not) is the entropy of mixing (the $R T \ln(Q)$ term in calculating Gibbs energy for given concentrations).

For some more interpretation of the graph and the different definitions, see Orthocresol's excellent answer here: https://chemistry.stackexchange.com/a/41864/72973


What's the difference between alpha-glucose and beta-glucose? What's the difference between $\alpha$-D-glucose and $\beta$-D-glucose? Are they isomers? Or do they differ in their elemental composition?

$\alpha$-D-glucose and $\beta$-D-glucose are stereoisomers - they differ in the 3-dimensional configuration of atoms/groups at one or more positions.

  $\alpha$-D-glucose

  $\beta$-D-glucose

Note that the structures are almost identical, except that in the $\alpha$ form, the $\ce{OH}$ group on the far right is down, and, in the $\beta$ form, the $\ce{OH}$ group on the far right is up.

More specifically, they are a class of stereoisomer called an anomer. Anomers are capable of interconverting in solution. All cyclic structures of monosaccharides exhibit anomeric $\alpha$ (down) and $\beta$ (up) versions. These differences occur at the anomeric acetal carbon (the only carbon with two $\ce{C-O}$ bonds.

These two forms exist because all monosaccharides also have an open-chain form with one fewer stereocenter. When the chain closes to the cyclic structure, the aldehyde or ketone carbon becomes a stereocenter, and it can do so in either configuration. One configuration is preferred ($\beta$), but both exist.

  Open chain form of glucose:

In the presence of acid or base (although water can fulfill this role if need be), the two anomers interconvert through the open form until dynamic equilibrium is established. The mechanism below starts with $\alpha$ in the upper left and finishes with $\beta$ in the lower right. The open-chain form is in the middle.


What is an imaginary boundary? We know that anything that separates the system with the surroundings is called a boundary.
It is said that a boundary can be either conducting or non-conducting, rigid or nonrigid, and real or imaginary.

What is meant by "imaginary boundary"?

Boundaries are defined with respect to the system in observation.

Imagine you're in a room with the door closed. Say, you are measuring the temperature of the room, i.e. the room is your system. In that case, it's easy to understand that the boundaries are real. In this case, the walls, the ceiling, the closed door comprises the boundary of the system.

Now, say somebody opens the door. Understand that this does not change the system; you are still interested in measuring the temperature of the room only. So, we can say that at the place of the closed door, there is an imaginary boundary. By specifying the boundary, you are now able to define your system more clearly.

This concept of imaginary boundary helps us to clearly define our system.


Unknown Mass in Dalton's Law of Partial Pressures An air with mass of $0.454\ \mathrm{kg}$ and an unknown mass of $\ce{CO2}$ occupy an 85 liters tank at $2068.44\ \mathrm{kPa}$. If the partial pressure of the $\ce{CO2}$ is $344.74\ \mathrm{kPa}$, determine its mass.

Answer is $0.138\ \mathrm{kg}$

Ok lets setup Dalton's Law:

$$\begin{align}
p_\text{total} &= p_\text{air} + p_{\ce{CO2}} \\
2068.44 &= \left(mRT/V\right)_\text{air} + 344.74 \\
m_\text{air} &= 0.454\ \mathrm{kg} \\
R_\text{air} &= 0.287\ \mathrm{kJ/(kg\ K)} \\
T_\text{air}/V_\text{air} &= 18847.73126
\end{align}$$

Ok so I need to get something out of:

$$\begin{align}
T_\text{air}/V_\text{air} &= 18847.73126 \\
V_\text{total} &= V_\text{air} + V_{\ce{CO2}} = 0.085\ \mathrm{m^3}
\end{align}$$

Ok so im basically stuck there. only 2 equations for 3 unknowns.

Any hint? or am I totally approaching this the wrong way?

Assuming that the total pressure $p_\text{total}=2068.44\ \mathrm{kPa}$ is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, the partial pressure of air $p_\text{air}$ can be calculated as follows:

$$\begin{align}
p_\text{total}&=p_\text{air}+p_{\ce{CO2}}\\[6pt]
p_\text{air}&=p_\text{total}-p_{\ce{CO2}}\\[6pt]
&=2068.44\ \mathrm{kPa}-344.74\ \mathrm{kPa}\\[6pt]
&=1723.70\ \mathrm{kPa}
\end{align}$$

Since molar mass $M$ is defined as

$$M=\frac mn$$

where $m$ is mass and $n$ is amount of substance, the ideal gas law

$$p\cdot V = n\cdot R\cdot T$$

where $V$ is volume, $R$ is the molar gas constant, and $T$ is temperature, can be rewritten as

$$\begin{align}
p\cdot V&=\frac{m\cdot R\cdot T}M\\[6pt]
\frac V{R\cdot T}&=\frac m{M\cdot p}
\end{align}$$

This equation applies to air
$$\frac V{R\cdot T}=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}$$
where $m_\text{air}=0.454\ \mathrm{kg}$ is the given mass of air and $M_\text{air}=28.97\ \mathrm{g\ mol^{-1}}$ is the average molar mass of air, as well as to $\ce{CO2}$
$$\frac V{R\cdot T}=\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}$$
where $m_{\ce{CO2}}$ is the unknown mass of $\ce{CO2}$ and $M_{\ce{CO2}}=44.01\ \mathrm{g\ mol^{-1}}$ is the molar mass of $\ce{CO2}$.

Therefore,

$$\begin{align}
\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}&=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
m_{\ce{CO2}}&=\frac {M_{\ce{CO2}}\cdot p_{\ce{CO2}}\cdot m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
&=\frac {44.01\ \mathrm{g\ mol^{-1}}\times344.74\ \mathrm{kPa}\times0.454\ \mathrm{kg}}{28.97\ \mathrm{g\ mol^{-1}}\times1723.70\ \mathrm{kPa}}\\[6pt]
&=0.138\ \mathrm{kg}
\end{align}$$

Note that the given value for the volume $V=85\ \mathrm l$ is not required for the calculation of $m_\text{air}$. Thus, the uncertainty of the volume $V=85\ \mathrm l$ (with only two significant digits) does not affect the uncertainty of the result.


Elimination reaction with bad leaving groups If a halogen is present, as in case of 2-fluorobutane, which is reacted upon by alc. KOH then, shouldn't the product be according to Zaitsev's rule.

Then, why does the removal of hydrogen takes place, does this happens in case of halogens only or in case of all the bad leaving groups?

Alkyl halides with bad-leaving groups (like $\ce{F-}$) having EWG at $\ce{\beta}$-position undergoes these reactions, also known as $\ce{E_{1CB}}$, viz. elimination unimolecular conjugate base.

Example:

aldol condensation

photo-induced decarboxylation

Reagent: bulky or strong bases or both (like $\ce{K+(^tBu)-, LDA, NaOH, KNH2}$, etc) along with some heating.

Mechanism:


Why are some salt solutions coloured? I notice that salt solutions of $\ce{NaCl}$ and $\ce{KCl}$ are colourless while those of $\ce{CuSO4}$ and $\ce{FeSO4}$ are coloured.

I got as far as figuring that it has to do with the transition metal ions, but I can't explain why the salt solution of $\ce{ZnSO4}$ is colourless even though zinc itself is one of the transition metals.

You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes. Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes like $\ce{Cu[(H2O)6]^{2+}}$ (this is what you get when you add water to Copper Sulphate). Thus, metals with no empty $d$ orbitals of comparable energy become colourless.

OK, but where do coordination compounds get their color from? Well, $d$ orbitals look like this:

Even though the last one looks different, these are all identical in many respects, the most important being that they have the same energy. So if an electron jumps from one orbital to another, it on't make any difference. In fact, since this is a quantum mechanical system, it (sort of) doesn't even make sense to talk about electrons jumping from here to there.

All this changes when you have some ligands around. In coordination complexes, the $d$ orbitals are hybridised with the $s$ and $p$ orbitals of the next or previous shell. If we're talking about octahedral complexes, all the orbital lobes have to be aligned along the coordinate axes (since the $p$ orbitals are). This can be said to come from the fact that we're "mixing" these orbitals to get orthogonal hybrid orbitals, but that may not be that good an analogy. Now, the ligands approach along the coordinate axes since their final position will be in hybrid orbitals along the axes. So, they approach the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (as well as the $s$ and $p$ orbitals). The repulsion generated makes these orbitals of higher energy than the other three $d$ orbitals. This is called "d-orbital splitting" {*}

Now, electrons may "jump" between the $d$ orbitals (the energy gap isn't so high that the jump is too improbable). Due to the energy gap, some energy is absorbed/released in the form of a photon and it just happens to fall within the visible spectrum (at least, for most transition metals, it does). This jump is called a $d-d$ transition.

Using this, we can easily show why $\ce{Zn}$ salts are colourless. In $\ce{Zn}$, the $3d$ orbitals are full, so they're useless from the ligand point of view. The $4d$ orbitals are, on the other hand, empty. Ideal for forming a coordination complex with the $4s$ and $4p$, but there won't be any $d-d$ transitions as there are no electrons in $4d$ to jump around. (Remember, this is before the ligand bonds)

*For tetrahedral complexes, the $d$ orbitals take no active part in the bonding, and the ligands approach between the axes. This leads to an opposite gap between the top three(known as $t_{2g}$) and bottom two($e_g$) orbitals, with the $t_{2g}$ orbitals at a higher energy. So we still get $d-d$ transitions and color, though IIRC many tetrahedral compounds do not have a visible color(the energy gap is about half that of the corresponding ocahedral gap--and may not give a photon with a visible frequency)


Reactions of Grignard Reagent: 1,2 vs 1,4 addition So I was reading through Grignard reagent and I came across this answer by jerepierre, which mentions:

  In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone.

Now, I want to know, what is the meaning of 'in general'? Can anyone please provide me some cases where the general rules are not followed.

Also, I came to know that:

give 1,4 addition with $\ce{PhMgBr}$ to give:

(that reference is difficult to provide at the moment).

Can someone please explain why this happened?

This is not a 1,4 addition.  It is a 1,2 addition across a carbon-carbon double bond.  This mode of reaction is unusual for Grignard reagents, but here a highly stabilized "double benzylic" anion is formed and the competitive reaction of adding to the carbonyl group is sterically hindered.


Metallic character of bonds? Why in discussions of percent character of bonds, are only ionic and covalent bondings discussed? Do bonds not have a partial metallic character, and are either metallic and ionic-covalent?

One could think of the Fermi surface and conduction bands as an expression of the degree of metallic bonding, where metals such as aluminum or silver have overlapping empty and filled bands, allowing for electrical conduction, while semiconductors have a small gap between filled and conduction bands, offering more resistance.

As @Mithoron states, metallic bonding is a bulk property. Metals behave differently as nanoparticles -- for example, bulk silver is reflective, but nanometer particles of it are black, as in photographs.


Why isn't this resonance possible in an imidazole ring? My question is why isn't the resonance structure drawn in pencil on the right a valid one? My review book states that after initial protonation on the indicated nitrogen no resonance is possible.

You need to understand the basic reason for the aromatic nature of materials. A molecule is said to be aromatic only if:

A ring in it is planar,
Continuous delocalisation is possible throughout the concerned cyclic ring, and
This delocalisation stabilises the ring.

Now, for continuous delocalisation to be possible, the $\mathrm{\pi}$ orbitals of all the elements in the ring need to be parallel to each other ( They need to overlap sideways for formation of  $\mathrm{\pi}$ bonds).

In your example, consider the carbons in the ring. All of their $\mathrm{\pi}$ orbitals are parallel to each other and PERPENDICULAR to the plane of paper. On the other hand, the imine nitrogen is $\mathrm{sp^2}$ hybrid. The lone pair of electrons is in an  $\mathrm{sp^2}$ hybrid orbital, i.e, IN the plane of the paper. And mind you, it's fixed (can't rotate without distorting the geometrical shape of the ring, that is, which would be too much to expect from the poor guy.)

So the imine Nitrogen cannot give the lone pair anywhere. it can only play around with the  $\mathrm{\pi}$ bond already on it.

Moreover, as soon as the amine Nitrogen takes up a proton, it loses its ability to use the lone pair for delocalisation. The condition for planarity itself is not fulfilled ($\ce{(R)2-N+-(H)2}$ has tetrahedral geometry), so there's no use of discussing aromaticity. Before the proton uptake, the ring was aromatic because the amine Nitrogen was using the lone pair for continuous delocalisation, effectively attaining planar geometry. But now... It can no longer be done.

Conclusion: Resonance is still possible(between all atoms of the ring except the amine Nitrogen), but Continuous Delocalisation is not. So, the system loses its aromaticity.


Titration of reducing sugars on honey So I'm trying to figure out how to titrate the reducing sugars on honey without generating iodine.

The reaction so far I think I could do is do a Fehling reaction with $\ce{Cu}^{2+}$and then back-titrate the remaining with EDTA, but I'm not sure that the EDTA wouldn't solubilize the $\ce{Cu2O}$, or be able to displace the tartrate.

Some sources I've looked say that this can be done as a direct titration using methylene blue as an indicator, but they haven't given a pH range, even if the reaction is fast enough to be done as a direct titration.

Any thoughts?

The usual titration of glucose with Fehling's Reagent is to place the Fehling's Reagent in a flask and titrate the boiling solution with glucose. At the end point, all the $\ce {Cu^{2+}}$ ions are converted to $\ce {Cu^{+}}$; this makes the solution colorless. That is difficult to see since the red copper(I) oxide is floating in the solution.

With methylene blue, after the copper has been reduced, the glucose can then reduce the methylene blue (in a basic solution) to a colorless solution. This makes the end point clearer since the methylene blue is much darker than the $\ce {Cu^{2+}}$ ions. The pH of the solution does not change much in this reaction.


If chemical bonding is quantum mechanic, is saying they bend and vibrate like springs wrong? I don't know much about quantum mechanics, but I do know that it's quite far removed from the easily observable and that anything "quantum" involves things that are discreet. So, how close is the analogy between molecular bending and vibration with that of springs? Is there maybe a better way to think, maybe a better analogy, to describe molecular bonds?

That's a good question, but remember, an analogy is just that: comparing dissimilar things on the basis of structure or function. Though there may not be a physical similarity, the analogy is effective in describing what happens, and can even be used to calculate outcomes, within limits.

In a similar way, the concept of a mass and a spring are used to describe an electrical R-L-C tuned circuit. Both have resonant frequencies and other properties that can be calculated from the physical properties of the components.

One of the difficulties in describing quantum phenomena is that until the physical properties are measured, they may be indeterminate. An electron is not in a specific place, but has a probability of being detected in a specific place. So though the analogy is not physically comparable, it helps understanding of the chemical bond, and can even produce useful results, such as IR spectra, and its use is valid.

This begs the question, though; perhaps there is another analogy that you might find more useful... Any suggestions from others are welcome!


Is there a method for finding the net-ionic equation? I truly don't understand the method for finding the net-ionic equation for a reaction. It is not clear to me how you should decide what the products are, especially when there are multiple possibilities.

Example:

  Excess potassium cyanide solution is added to aluminum bromide solution.

The first thing that came to mind for the molecular equation is:

$\ce{3KCN\+AlBr3 \to 3KBr\+Al(CN)3}$

But apparently the correct molecular equation is:

$\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$

I see no good reason why the one I thought of is incorrect. And to get the net ionic equation, the spectator ions are $\ce{K+}$ and $\ce{Br-}$. But why should I assume the solution is aqueous?

Another example:

  Chlorine gas is bubbled through dilute sodium hydroxide.

Well its easy enough to get the reactants:

$\ce{Cl2\+NaOH \to ?}$

What first comes to mind is $\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$

But the correct molecular equation is $\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$

What should be the thought process for arriving at the correct answer here. I am truly lost.

Another example:

  Lithium oxide reacts with sulfur dioxide.

I recognize this as a synthesis reaction. However how can I predict if the product is lithium sulfite or lithium sulfate?

But apparently the correct molecular equation is $\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$

That's not correct, what happenend to the other 3 potassium atoms?

It could form $\ce{Al2(CN)6}$ but you can't exclude $\ce{Al(CN)3}$ without further information beyond balancing of equations.

There are two main steps to balancing equation:

Balancing of each type of atom
Balancing number of electrons

If you've done both of those correctly, you've done all you can with balancing alone.

  $\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$

Above is clearly wrong because number of oxygen atoms doesn't balance

  $\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$

Above is clearly wrong because number of oxygen atoms doesn't balance


Can glucose and insulin exist in the same container without reacting? Someone asked if glucose and insulin could be mixed in the same container and used as an injection to provide a quick energy boost. I'm not interested about the medical part of this question, but just if glucose and insulin (which is a protein) could exist in the same container without reacting with each other without forming some sort of glycoprotein. Is it possible to make a reasonable prediction what would/not happen in the mixed glucose/insulin solution in the container?

I know that insulin and glucose do not react with in the body, I want to know if in aqueous solution they would react.

Glycoproteins come in O-linked and N-linked forms. For O-linked sugars, a glycosidic bond forms between the sugar and the hydroxyl of a serine or threonine side chain. For N-linked sugars, a glycosidic bond forms between the sugar and the amide of asparagine. The equilibrium for all these reactions lies on the side of hydrolysis in aqueous solution.

  Is it possible to make a reasonable prediction what would/not happen in the mixed glucose/insulin solution in the container?

Yes, in aqueous solution there is no expectation that a glycoprotein forms between insulin and glucose. In vivo, glycoproteins form in a coupled reaction involving the carbohydrate (usually an oligosaccharide rather than a single glucose), the protein and a nucleotide triphosphate.

However, if an individual has high blood sugar levels (i.e. high concentrations of glucose in the blood) for extended periods of time, hemoglobin (specifically, the N-terminal amino group and the sidechain amino group of lysine) will react with the sugar and become glycalated. This is used in the HbA1c diagnostic test for checking if someone has diabetes. After the sugar binds (reversibly), there is a further reaction (Amadori rearrangement) that leads to a long-lasting product.


Is the valence band neutral? While studying about band theory of semiconductors, I observed that when the electrons were excited from the valence band to the conduction band, they left behind holes in the valence band. From my existing knowledge, I believe that the valence electrons alone occupy the valence band which tells me the valence band is negative. For the holes to exist, the valence band has to be neutral. So, why is the valence band neutral?

If my reason for the valence band to be negative was wrong, I would like to know the reason.

Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively.  If you continue this chain extension and wind up with a very large polyene, the energy difference between the all of the bonding (or antibonding) orbitals becomes small and instead of having discreet molecular orbitals, you can imagine them "fuzzing" out into a "band" of molecular orbitals.  All of the bonding orbitals would become what is termed the valence band, while the antibonding orbitals would comprise the conduction band.  Just as ethylene is neutral, so is the large polyene.  If you excite one of ethylene's pi electrons to the antibonding molecular orbital you have created an excited state which is still neutral.  Scientists working with the solid state would use different terms to describe this process.  They would say that the antibonding orbital contains 2 holes and the boding orbital 2 electrons. When the electron is promoted to the excited state a hole is also promoted (holes have lower energy the further they are from the nucleus) from the antibonding orbital to the boding orbital.  Just like the ground state, the excited state molecule is still neutral, the electron and hole are just in different orbitals from where they started.


Understanding and usage of Arrhenius equation A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten degree rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be?

What activation energy would lead to a doubling in the reaction rate
  for a 10 degree increase in temperature, using the Arrhenius equation?

You’re right it is not generally true. Let’s set up an equation to solve the problem.

The Arrhenius equation is
$$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$

In this case

$$2=\frac{A\operatorname{e}^{\frac{-E_\text{a}}{R\cdot(T + 10\ \mathrm{K})}}}{A\operatorname{e}^{\frac{-E_\text{a}}{RT}}}$$

or

$$\ln 2 = \frac{ - E_\text{a}}{R\left( T + 10\ \text{K} \right)} - \frac{ - E_\text{a}}{RT}$$

let’s pick a temperature around room temperature, 300 K is a nice even number, but you can re-solve the problem with any temperature you like.
$$0.69 = \frac{ - E_\text{a}}{R \cdot 310\ \text{K} } - \frac{ - E_\text{a}}{R \cdot 300\ \text{K} }$$
Using $R=8.31\ \mathrm{J/(K\ mol)}$ will produce an answer in J/mol.  Solving we find
$$E_\text{a} = 53\,568.49\ \mathrm{J/mol}$$
or,
$$53.6\ \mathrm{kJ/mol}~~(12.8\ \mathrm{kcal/mol})$$

So around room temperature a reaction that has an activation energy around 54 kJ/mol (or 13 kcal/mol) would show a doubling in rate for a 10 kelvin increase in temperature.


What is the naming convention for Sn and Pb in inter-metallic compounds? How would one go about naming inter-metallic (specifically, transition post-transition) compounds containing tin and lead?

It is usually easy figuring out the naming of such compounds as they follow a simple naming convention, for example ruthenium germanide, niobium aluminiumide and so on.

But what about tin? Would you get ruthenium stannide? Stannumide? Or something completely different? Same goes for lead, but I feel like I've heard the name aurous plumbide referring to $\ce{AuPb2}$ before.

After doing some digging I've found evidence to support that tin based inter-metallic compounds should be named 'stannide' such as in the paper by Fickenscher et al. [1].

Reference

Fickenscher, T.; Rodewald, U. C.; Niehaus, O.; Gerke, B.; Haverkamp, S.; Eckert, H.; Pöttgen, R. The Stannides $\ce{RE3Au6Sn5}$ $(\ce{RE} = \ce{La}, \ce{Ce}, \ce{Pr}, \ce{Nd}, \ce{Sm})$ – Synthesis, Structure, Magnetic Properties and $\ce{^{119}Sn}$ Mössbauer Spectroscopy. Zeitschrift für Naturforschung B 2015, 70 (6), 425–434. DOI: 10.1515/znb-2015-0050.


How to convert 1,4 pentadiene to 1,3 pentadiene? Are there any 'conjugating agents' to accomplish the above transformation? If not, then how to proceed?

My first instinctive answer would be to use a precious metal catalyst. Palladium usually displays this sort of activity. However, this article should cover most methods from basic and metallic to photochemical.
https://www.thieme-connect.de/media/10.1055-b-003-125756/lookinside/10-1055-b-003-125756_chapter012-1.jpg


Electrolysis of aqueous copper (II) nitrate There are two copper blocks sitting in the $\ce{Cu(NO3)2 (aq)}$ solution, a battery is attached onto both of them, providing enough energy to start the reaction.

Since solid pieces of copper are involved, $\ce{Cu}$ must be considered in the reduction potential as well.

However, looking at the half reaction for copper:
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  (E^\circ &= \pu{+0.34 V})
\end{align}

Compared to nitrate:
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) + 3e- &-> NO (g) + 2H2O (l)}&
  (E^\circ &= \pu{+0.96 V})
\end{align}

Since it produces more energy, I believe the second half reaction will occur instead.

I begin by finding all the species/ions I have:

$\ce{Cu^2+}$
$\ce{NO3^-}$
$\ce{H2O}$

Taking a look at the Standard Reduction Potentials at 25°C:

For my SOA (strongest oxidizing agent) half-reaction ($\ce{NO3^-}$):
\begin{align}
  \ce{NO3^- + 4H+ + 3e- &-> NO + 2H2O}&
  (E^\circ = \pu{+0.96 V})
\end{align}

As for my SRA (strongest reducing agent) half-reaction ($\ce{H2O}$):
\begin{align}
  \ce{2H2O &-> O2 + 4H+ + 4e-}&
  (E^\circ &= \pu{-1.23 V})
\end{align}

For the balanced reaction I have:
\begin{align}
  \ce{4NO3- + 4H+ &-> 4NO + 2H2O + 3O2}&
  (E^\circ &= \pu{-0.27 V})
\end{align}

Some observations that can be made are that:

Nitric gas is forming at the anode
Oxygen gas is forming at the cathode
At least $\pu{0.27 V}$ must be put in

My question is:

Did I do all of this correctly? There's no answer key and I'm pretty sure that I might have made a mistake somewhere.
What would happen if you placed a necklace at the cathode?

There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since

$$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$

Your first half-reaction is an oxidation (as written). You have the wrong sign on that $E$ value.
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  E^\circ_\mathrm{ox} &= \pu{-0.34 V}
\end{align}

Your second half-reaction is a reduction (as written)
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) +3e- &-> NO (g) + 2H2O (l)}&
  E^\circ_\mathrm{red} &= \pu{+0.96 V}
\end{align}

The combination of these two half-reactions produces a positive $E^\circ_\mathrm{cell}$ which is a spontaneous reaction. You do not need to use the oxidation of water as your oxidation half-reaction. Copper is a much better reducing agent than water:

$$
\ce{3Cu (s) + 2NO3- (aq) +8H+ (aq) -> 3Cu^2+ (aq) + 3NO (g) +4H2O (l)}\\
E^\circ_\mathrm{cell} = \pu{+0.96 V} +(\pu{-0.34 V}) = \pu{+0.62 V}
$$

How does this change your analysis?


Is the specific heat of salt the same as Na + CL when dissolved? I had this question when my friends were discussing how salt influences boiling time.  When crystal salt is dissolved into water, it turns into a single Na and a single Cl atom.  If knew the total heat capacity of the salt I want to add to my water, could I be confident, that when dissolved, total heat capacity I added to the water is equal to the undissolved salt's heat capacity?

On one hand my intuition tells me that heat capacity should remain the same.  Salt is salt.

On the other, doesn't the salt lose a rotational vibration mode when is dissolves into its two monatomic elements?

I am not trained in chemistry, so sorry if this is a naive question.

Thanks!

Salt does not dissolve to to $\ce{Na}$ and $\ce{Cl}$ atoms, but to hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions, so it's heat capacity is very different to the solid $\ce{NaCl}$.

And yes it looses vibration modes of the crystal lattice, less the rotation, as there is nothing to rotate in the lattice.

There are also vibrations of coordination bonds to water. Dissolved ions also have translational degrees of freedom, what is missing in solids.

Even for the identical substance, its heat capacity in solid and liquid state differ. E.g. liquid water has about 2x bigger heat capacity than ice ( 4.2 versus 2.1 J/g/K ).


What is the relative rate of diffusion of ammonia to hydrogen chloride, both in gaseous states? I encountered this problem during an assignment:

I answered it as C, but the assignment grader marked it as wrong and indicated that A is correct.

I asked another instructor about it and gave me a different answer: D.

I am confused on how to approach this problem given that I have received different answers from what I consider reputable sources. How do I proceed to solve this?

No answer is correct, but "A" is  closest to the correct answer.  Heavier molecules diffuse slower than lighter ones because the same amount of thermal energy has to move more mass.  However, this relation (diffusion rate is inversely proportional to the square root of molecular weight) actually leads to 44 seconds not 64.  The professor who told you "D" probably misunderstood that in the second chamber there is HCl (acidic) instead of ammonia (basic).


Is it possible to measure light absorbance for analysis of lead concentration without photospectrophotmeter I’m designing an experiment which requires I measure final and initial lead concentrations for the determination of aqueous lead uptake by different mediums. Because lead concentrations would be measured on the ppm scale, I’d need to use a spectrophotometer to measure light absorbance by the water samplings. I could then use Beer Lambert’s Law to determine concentration.

With recent protocols surrounding COVID-19, it’s been difficult to contact an institution. I was wondering if there was any way I could measure light absorbance of my samples at home. Is there some other apparatus I could look into using (or perhaps even purchase)?

All the best!

Another possibility is the grease spot photometer (aka Bunsen photometer) German Wiki page. This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light. Also the darker the room, the better.

The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.

If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).

Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48

How to measure photometrically

so schematically we have:

o--U----|----o

+---d---+

with light sources o, cuvette U and the spotted paper |. Measure distance $d$ between equal intensity paper position and the light source with the cuvette.

We have $I_{transmitted} \sim \frac{1}{d^2}$

Next, a calibration series is acquired as usual: blank (which serves as $I_0$) and concentrations over the calibration range. Almost as usual:

$$E = -\lg \frac{I}{I_0} \sim \varepsilon l c$$
$$-\lg \frac{d_0^2}{d^2} = b c$$
$$\lg d = a + b' c$$

$a$ and $b$ are determined by calibration - I use here $a$ instead of $2 \lg d_0$ because several choices are possible and in general the fitted value should be used rather than the theoretical one since it includes some effects in which practice differs from theory (e.g. differences between % transmission in the grease spot from % reflection outside).

Standard addition instead of calibration would be possible as well.


Gaussian Convergence Issue for N2+N PES I am trying to obtain a potential energy surface (PES) for $\ce{N2 + N}$ combination using scan feature of Gaussian. I have tried different combinations of basis sets and methods but unfortunately at some point in the iteration, the convergence criteria is not met and I get an "Aborted Core dump" closure in the log file. The Gaussian code is:

# HF/6-31G test scan

NNN PES

0 2
N
N 1 R1
N 2 R2 1 120.

R1 0.5 80 0.1
R2 0.5 80 0.1

Where am I going wrong. More importantly, I would be really thankful if someone could suggest how to choose a method and basis set for a type of problem. I was thinking if I can get some answer using HF/6-31G, I could use this as an input for a higher order method like cassf. What that be a good idea?

You did not specify at which geometry you have convergence problems, so at first I thought it happens when all three $\ce{N}$ atoms are far away from each other. But then I run the calculation by myself and found out that it is not the case: the SCF convergence failure happens at R1=1.2 and R2=1.1. At this geometry SCF starts to oscillate after 20 iterations back and force as shown in a graph below, so simply increasing the number of SCF iteration would not help.

For that reason I quickly checked few other well known tricks to avoid the problems with SCF convergence (see, for instance, here) and I recommend to use a quadratically convergent SCF (QC-SCF) by adding scf=qc to the route section:

# HF/6-31G test scan scf=qc

Note though that while QC-SCF is more reliable than the default DIIS algorithm it is also slower, so expect an increase in the computation time. For that reason, there is also an interesting option scf=xqc which first tries the default algorithm and switches to a quadratically convergent one only if the default has not converged.

# HF/6-31G test scan scf=xqc

However, switching to a quadratically convergent algorithm did not help, since SCF still fails to converge but at some other geometries. So, I investigated the matter more closely, and in addition to scf=xqc I also recommend to improve both radial and angular flexibility of you basis set by adding diffuse and polarization functions into it

# HF/6-31+G(d,p) test scan scf=xqc

After running the calculation for an hour it reached 2656 points (which is about 40% of all points) without SCF convergence problems, but in case if this variant still fails:

Consider lowering down the convergence criteria by changing scf=xqc to scf=(xqc,conver=n) with n being, say, 6 (it is 8 by default).
If failures happen when at least one $\ce{N}$ is far away from two others, consider lowering down the maximum distance for your PES scan. In my opinion 8.5Å is way to big, 5Å would probably be more than enough.

I finally succeeded with the following route section

# HF/6-31+G(d,p) Scan SCF=(XQC,Conver=6)

Here are the results and below is the plot of the PES. :D


Is it possible to generate electricity by using two inert electrodes in a galvanic cell? The Daniell cell has two active electrodes, meaning that both the copper and the zink electrodes participate in the redox reaction. However, in a lemon battery for example, the copper electrode is inert - it doesn't participate in the redox reaction. When is it possible to only have one active electrode? Is it possible to generate electricity by using two inert electrodes?

A non-chemical way to generate electricity using two inert electrodes in salt water is through magnetohydrodynamics. If the salt water moves past the electrodes in a magnetic field, it generates an electric current without consuming the electrodes or electrolyte.

This could make an interesting "magic trick" or puzzle: use a moving magnetic field, e.g. a magnet attached to a motor, in a box below a beaker filled with salt water, and insert two carbon electrodes connected to an AC current meter. Of course, make sure the motor is very quiet.


Does liquids that emit light when vibrated exist? I am looking for a liquid that illuminates when vibrations are added.

It does not have to be strictly chemical, it might as well be something else as well.

Maybe a crystal in powder form mixed with some kind of oil, or even algae in a thick:ish liquid.

Does such a liquid exist?

Find compounds that emit light under photolysis.

As that may involve radical production pathways, one can also try to apply sonolysis to the same compounds as:

H2O (with dissolved O2, N2,..) + Sonolysis -> *H + *OH

Reference source: 'Free radical formation from sonolysis of water in the presence of different gases'), to quote:

"In this case, O2 reacts with •H to form •OH, indicating that the presence of dissolved O2 in water sonolysis gives additional amount of •OH. "

Also:

H2O2 + Sonolysis -> *OH + *OH

Apparently, with applications, see, 'Bactericidal effect of hydroxyl radicals generated by the sonolysis and photolysis of hydrogen peroxide for endodontic applications'.

where it is likely that one may be able to replicate to varying degrees the action of light treatment.

Here is a source example citing the action of reactive oxygen species (ROS), which can be created by the UV photolysis of, for example, H2O2 (or sonolysis per reference above):

"In biological systems much attention is focused on the two free radicals, superoxide anion and hydroxyl radical....The generation processes of ROS can be monitored using luminescence. Because of the very weak native luminescence of ROS, both luminol and lucigenin have been used in the past to give measurable signals. However, the luminescence of these two substances is very small compared to that of Pholasin®, which allows precise analysis on very small samples."

So, perhaps try performing a sonolysis of, say, luminol or lucigenin, in H2O2 might be interesting.


Mixing hydrochloric acid and sulphuric acid, both concentrated In the chemistry lab, I stupidly decided to mix hydrochloric acid and sulphuric acid, both concentrated, in a random ratio, which was about $\pu{50 ml}$.

When I poured them in one test tube, the result was a vigorous reaction, with the green coloured liquid mixture "jumping" in the test tube and a white gas being released.

Could someone tell me what the resultant chemicals were?

Three things occurred. First, sulfuric acid as an high affinity for water (it is indeed used as a drying agent in some cases. For instance, if you wanted pure and anhydrous bromine, you could bubble wet bromine gas in concentrated sulfuric acid). So, one thing is the "stealing" of water by sulfuric acid, which took the solution one step forward the saturation point (with the release of gas).

The release of gas was aided by the exothermic reaction of sulfuric acid with water: (self) heating of the solution helped releasing the gas.

Edit: the following is an hypothesis, made by assuming that the "extreme" reaction conditions could have aided an unlikely redox reaction, but I am not sure about the oxidation of HCl:

And finally, sulfuric acid is a oxidizing acid: it oxidized part of the hydrochloric acid to chlorine, which made the solution green (which probably was released as a gas, together with HCl).


formation of σ molecular orbital in H2 In the picture, graphs show the constructive interference of two 1s atomic orbitals. My question is, what does the horizontal axis in wavefunction graph represent, does It represent the distance from the nucleus? We see the pick point of wavefunction on H atom, does It mean that highest probability of electron density is on the nucleus, which is impossible.  I am quite confused about the meaning of a wavefunction, can anyone help me?

You are mostly right. The horizontal axis represents the distance from the nucleus. The maximum of the electron density of 1s orbitals is in fact at the nucleus. This is not impossible, remember that electron density can be interpreted as the probability that an appropriate experiment will detect the electron there. In fact, electron density "at" the nucleus is one of the major ways NMR spectroscopy works.


Why are the plastic balls used to reduce evaporation in this reservoir black instead of white? Plastic balls have been added to a reservoir in drought stricken California in order to reduce evaporation.  (Link to news story here.)

Why are the balls colored black instead of white?  I would think that white would decrease heating and therefore evaporation.

Edit:  Apparently the balls are there to block sunlight from converting bromide to carcinogenic bromate.

Youtube video explaining the purpose of the balls

This piece in Bloomberg News is about Sydney Chase, the founder of XavierC, the company which makes the "shade balls" in question (there are two other suppliers, Orange Products, Inc. and Artisan Screen Printing).

From the section of the XavierC site on how they work:

  Two main causes of water evaporation in reservoirs are temperature and surface area. Our hollow plastic conservation balls address these two issues effectively and inexpensively.
  The black conservation balls provide a floating cover on the reservoir taking up approximately 90% of the surface area. Simultaneously the balls prevent UV rays reaching the water and therefore reducing algae.

From the above-mentioned Bloomberg piece:

  The Los Angeles Department of Water and Power has now dumped 96 million balls into local reservoirs to reduce evaporation and block sunlight from encouraging algae growth and toxic chemical reactions. The balls are coated with a chemical that blocks ultraviolet light and helps the spheres last as long as 25 years. Las Virgenes, north of L.A., now uses shade balls, too.

From the Los Angeles Times story on August 12, 2015:

What are they made of?

  The 4-inch-diameter balls are made from high-density polyethylene, which is the same material you would find in a one-gallon milk jug. This plastic is approved to come into contact with drinking water.

Why are they black?

  Carbon black was added to the plastic to stabilize the balls in UV light from the sun.

and, from another LA Times piece about the "shade balls" written in 2008:

  The balls, which cost 40 cents each, are made of polyethylene. The coating contains carbon. Black is the only color strong enough to deflect ultraviolet rays, said Paul Sachdev, president of Orange Products.


Ksp of Manganese Phosphate I can't find the $K_{sp}$ value for Manganese Phosphate ($\ce{Mn_3(PO_4)_2}$) anywhere online. Has it been measured? What is it?

Well, the MSDS entry tells me that the solubility is 50%. Converting, we get 0.5 g per gram of water, or 0.0005 g per liter of water. This is $s=1.4\times 10^{-6}$ moles per liter of water.

$K_{sp}=(3s)^3\times(2s)^2=6.14\times 10^{-27} \rm M^5 $


Is there a sixth type of d-orbitals with index 2z²-x²-y²? There are 5 types of d-orbitals: $\mathrm{d}_{xy}$, $\mathrm{d}_{zx}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{x^2-y^2}$, and $\mathrm{d}_{z^2}$.

I encountered a question which asked about the various d-orbitals, but I came across one that I'd never heard of before: $\mathrm{d}_{2z^2-x^2-y^2}$.
the answer key says that it is a d-orbital.

How is that possible? Is it sixth d-orbital?

There are 5 linearly independent solutions to spherical harmonics for $\ell = 2$.

We generally pick $\mathrm{d}_{z^{2}}$, $\mathrm{d}_{x^{2}-y^{2}}$, $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{xz}$.

Ignoring, $\mathrm{d}_{z^{2}}$, notice that it's really easy to see what the $\mathrm{d}$-orbital looks like figuring out the nodal planes. The lobes of the $\mathrm{d}$-orbitals go in between, and the sign of the function tells you how to switch the sign of the wavefunction. So, in your diagram for (a), (b), and (c), I would take off points for the reason that there is no indication that the diametrically opposite lobes have the same shading.

Here's the question that you want to wrap your head around. Why is there no $\mathrm{d}_{x^{2}-z^{2}}$ and $\mathrm{d}_{y^{2}-z^{2}}$ in analogy to $\mathrm{d}_{x^{2}-y^{2}}$?

The answer is by choice because you can add orbitals together to get new orbitals. The orbitals we write are actually linear combinations of the complex solutions we got from solving the spherical harmonics for the Schrödinger equation. Now, the full name of $\mathrm{d}_{z^{2}}$ is $\mathrm{d}_{2z^{2}-x^{2}-y^{2}}$. This is not hard if you think about the where the node is (a nodal bi-cone) and the signs, the torus is different sign than the lobes. With this and $\mathrm{d}_{x^{2}-y^{2}}$, you can construct the two missing ones (being extremely loose with notation):

$$
-\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y{2}} + \mathrm{d}_{x^{2} - y^{2}})
= -\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2} + x^{2} - y^{2}})
= -\frac{1}{2}\mathrm{d}_{2z^{2} - 2x^{2}}
= \mathrm{d}_{x^{2} - z^{2}}\\
-\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2}} - \mathrm{d}_{x^{2} - y^{2}})
= -\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2} - x^{2} + y^{2}})
= -\frac{1}{2}\mathrm{d}_{2z^{2} - 2y^{2}}
= \mathrm{d}_{y^{2} - z^{2}}
$$

So, all you need is 5.


Formation of Goethite Probably a simple question: When you're mixing potassium hydroxide ($\ce{KOH}$) with iron nitrate ($\ce{(FeNO3)3}$) to form ferrihydrate, does the temperature have to be below $\pu{60 ^\circ C}$?

TL;DR  Yes, the reaction is feasible at $\pu{60 ^\circ C}$. Just extending my comment into an answer.

Goethite and ferrihydrite(not ferrihydrate) are two different things. They are polymorphs of iron oxyhydroxide ($\ce{FeOOH}$). Goethite is $\ce{α-FeO(OH)}$ while ferrihydrite is amorphous/nanocrystalline hydrated iron oxide, officially formulated as $\ce{FeOOH•1.8H2O}$ but also exist in many hydrated states and so formula is theoretically indeterminate as its water content is variable.

Reaction between iron(III) nitrate hydrate in alkaline solution has been studied by Atkinson et al.1 at different concentrations and amounts. One case is: 50g of $\ce{Fe(NO3)3}$ salt dissolved in 825 ml of water is added to $\ce{200 ml 2.5 N}$ potassium hydroxide solution. At pH 12 and maintaining the temperature at $\pu{60 ^\circ C}$, it took 24 hours to complete the reaction. They also noted that in this goethite synthesis, ferrihydrite is the initial precipitate that results from the rapid hydrolysis of Fe(III) solutions, before aging to goethite at higher temperature.
Pure ferrihydrite synthesis is difficult to isolate because it requires careful control of pH and temperature and its composition can changes if any of this parameter changes. Also, ferrihydrite is the only polymorph which is called nanomineral because its crystal size is in the range of 2 nm to 5nm. Nonetheless, its synthesis has also been discussed in the same paper where ferric nitrate is dissolved in potassium hydroxide but the pH is maintained at 7-8. Vigorous stirring and centrifugation yields the ferrihydrite.
An alternative reaction where the pH is not needed to maintain strictly is by adding $\ce{Fe(NO3)3.9H2O}$ and $\ce{NH4HCO3}$ in a 1:3 mole ratio at $\pu{80-100 °C}$ 2 or iron(II) salt or iron(III) salt with silicon or germanium at pH 5-11 3.
Notes and References

Laboratory synthesis of goethite and ferrihydrite of controlled particle sizes by
Milton Villacís-García, Mariana Ugalde-Arzate, Katherine Vaca-Escobar, Mario Villalobos,
Rodolfo Zanella, Nadia Martínez-Villegas, Boletín de la Sociedad Geológica Mexicana
Volumen 67, núm. 3, 2015, p. 433-446 (PDF)
Novel Synthesis and Structural Analysis of Ferrihydrite
Stacey J. Smith, Katharine Page, Hyunjeong Kim, Branton J. Campbell, Juliana Boerio-Goates, and Brian F. Woodfield
Inorganic Chemistry, 2012 51 (11), 6421-6424
DOI: 10.1021/ic300937f
SYNTHESIS OF FERRIHYDRITE AND FEROXYHYTE, NAGANORI YOSHINAGA and NOBORU KANASAKI
Faculty of Agriculture, Ehime University, Matsuyama 790, Japan, Clay Science 9, 43-51 (1993), (PDF)
Simple hydrolysis of iron(III) nitrate yields $\ce{\alpha-Fe2O3}$. Fusion of potassium hydroxide and iron(III) oxide ($\ce{Fe2O3}$) in air produces potassium(VI) ferrate but in laboratory, $\ce{K2FeO4}$ is prepared by oxidizing an alkaline solution of an iron(III) salt with concentrated chlorine bleach.


Why there is no temperature scale tied to normal human body temperature? Fahrenheit:

  The lower defining point, 0 °F, was established as the temperature of
  a solution of brine made from equal parts of ice and salt.

Celsius:

  0 °C was defined as the freezing point of water and 100 °C was
  defined as the boiling point of water

Why hasn't anyone used human body normal temperature as 0? It would be much more logical for humans to base temperature scale on our body, not something that is outside.

Why nobody used human body normal temperature as 0?

That has been done.
See Medical Thermometry and Human Temperature at page 12:

37° Centigrade scale
29.6° Reaumur
98.6° Fahrenheit
77° Walferdin's tetracentigrade
0° of the physiological scale

Then the physiological scale is further discussed starting on page 265, which includes a drawing of a physiological scale thermometer.

physiological 0 is placed in the centre of the drama, whose acts are health, sickness, and death


Non-zero current flow through an electrolytic cell, when an electric potential lesser than that needed for electrolysis is applied? For instance, the potential needed for water splitting is $\pu{1.23V}$. However, there is current flow even at lower potentials (say $\pu{0.5V}$). Clearly, there is no redox reaction happening in the cell as evidenced by lack of bubbles. What is the cause of this current?

As you noticed there is a small amount of current which keeps on flowing when there is no electrolytic decomposition. This "problem" is aptly named as non-Faradaic current. Faradaic current refers to current due to oxidation and reduction. Here is a detailed review: The difference between Faradaic and Nonfaradaic processes in Electrochemistry


Why is no dynamical correlation recovered in the CASSCF method? Configuration interaction is the only certain way to get a calculation to converge toward the exact solution of the many-body Schrödinger equation (within the approximations used for the framework of the method). In such calculations, the wave function is expanded in all possible electron configurations, and will therefore often take too long to be practical. For full CI, all dynamical correlation is recovered.

In truncated CI, part of the dynamical correlation is recovered. The wave function expansion is, for example, limited to include only single and double excitations (CISD).

In a CASSCF calculation, a subset of all orbitals is chosen, and the wave function is expanded in all possible electron configurations within this subset of orbitals. However, here it is said that no dynamical correlation is recovered. This is stated so often I suspect it is common knowledge for computational chemists, but I can provide references if someone want them.

It is my understanding that the CI method recovers dynamical correlation in allowing electrons to be moved into virtual orbitals (and hence "react" to the instantaneous coulombic repulsion from other electrons). In allowing for higher a number of excitations, a larger portion of the dynamical correlation is recovered (CISDT > CISD).

So, why is no dynamical correlation recovered in the CASSCF method? Imagine a good set of orbitals are chosen for the active space, such that no important configurations are left outside of the wave function expansion, surely dynamical correlation is described? What if we extend this to the generalized actice space (GAS) formulation. In developing GAS, the original authors aimed to remove the "deadwood" of configuration state functions; that is, exclude CSF's with near-zero coefficients. So now a larger active space can be used, and hence a larger number of excitations can be used. Why is not more dynamical correlation recovered?

If indeed CASSCF recovers no dynamical correlation, a CASSCF calculation should give the same energy as a HF calculation. Is this the case?

The difference between dynamical and static correlation has always been ill-defined, and there is no clear-cut delineation between them.

Correlation simply means that the probability of finding an electron somewhere at position $a$ and another electron somewhere at position $b$ is not simply the product of the two probabilities.

(In that respect, even Hartree-Fock has some correlation since it forbids two same spin electrons occupying the same state (Fermi correlation).)

Correlation was originally thoguht of as the instantaneous repulsion of electrons. This was dynamic correlation. The origin of the difference between static and dynamical correlation came from studies of bond dissociation in molecules. As you pull a bond apart, the electrons should be further apart, and correlation energy should decrease. However, the opposite was found, and in many cases the correlation energy increases. Therefore, there must be some sort of "other" correlation occurring. This was static correlation.

This is because there are additional (near-)degenerate configurations that contribute strongly to the nature of the wave function during dissociation. You can think of this as a manifestation of entanglement, of which Hartree-Fock, being single determinant, fails miserably.

In short, static correlation came to describe situations where the underlying Hartree-Fock reference was a qualitative and quantitative failure.

Dynamic correlation, on the other hand, was used to describe the cases where the underlying Hartree-Fock reference has the correct qualitative behavior, but was not quantitatively accurate.

CASSCF does a better job at calculating static correlation because most of the important near degenerate configurations are multi-electron excitations within a small valence space. Thus it generally captures the configurations necessary to describe bond breaking, etc. with a well-chosen active space.

Single reference coupled cluster, does better at capturing dynamic correlation because it is usually truncated to few-pair excitations, like the double excitations of CCSD, but cover the whole virtual space.

For high levels of coupled cluster, e.g. CCSDTQ, the distinction between static and dynamic correlation really breaks down, and some high level coupled cluster calculations actually do quite a good job at otherwise "multireference problems." Multireference coupled cluster --- combining the best of CASSCF and CC --- is an active area of research today.


Can chlorine form hydrogen bonds? My question is simple, If chlorine ($\ce{_17Cl^35^.^5}$) and nitrogen ($\ce{_7N^14}$) have similar values of electronegativity, and chlorine has way more lone pairs than nitrogen, what prevents it from forming hydrogen bonds ?

(other than in chloral dihydrate $\ce{CCl3CH(OH)2)}$

Nitrogen is equally electronegative as chlorine but also more compact.  Whether you consider hydrogen bonding as purely electrostatic or having a covalent character (The lone pair on atom X overlaps the molecular orbital in the H-Y bond of an adjacent molecule, thus partially delocalizing the bond), the more compact structure of nitrogen offers an advantage.


Alkaline glycine buffer molarity Having never prepared a buffer solution before, I am hoping for a little help in understanding the process.

This method I plan to use calls for a $\pu{0.1M}$ glycine buffer at pH $10.6$. Some internet searching yielded this recipe (very bottom of the page) for making a glycine-$\ce{NaOH}$ buffer at $10.6$ pH, which I assume is what I'm looking for.

Most of it seems to make sense to me. $\pu{0.2M}$ stock solutions of both glycine and $\ce{NaOH}$ would be prepared as:

0.2M NaOH solution = 0.2 * 39.9971 = ~8g NaOH/l
0.2M glycine solution = 0.2 * 75.07 = ~15g glycine/l
(Molar mass: NaOH = 39.9971; glycine = 75.07 g/mol)

And, to yield a buffer at $10.6$ pH, mix:

25ml glycine stock + 22.75ml NaOH stock + deionized water to 100ml

Where I get lost is figuring out the molar concentration of the resulting solution, since there are two different solutes. Would it be just the total number of moles present of both, something like:

(25ml + 22.75ml) * 0.2M / 100ml = 0.0955M ?

It seems pretty close to the $\pu{0.1M}$ concentration called for, but I can't seem to find any confirmation of this.

As per comments and revisiting the question and my first answer, I am amending my answer to reflect the correct one.

Because you are converting glycine to its sodium salt, the final solution concentration will be 2 times the concentration of glycine.

How many moles of glycine are in $25\,\mathrm{ml}$ of stock solution?
$${0.2\,\mathrm{mol}\,\ce{glycine}\over 1000\,\mathrm{ml}} = {x\,\mathrm{mol}\,\ce{glycine}\over 25\,\mathrm{ml}}\implies x = 0.005\,\mathrm{mol}\,\ce{glycine}$$

You need $0.005\,\mathrm{mol}$ of $\ce{NaOH}$ to convert the glycine to the sodium salt, so:

Total amount of substance in buffer solution:
$$0.005\,\mathrm{mol} + 0.005\,\mathrm{mol} = 0.01\,\mathrm{mol}$$

Total volume of buffer solution:
$$100\,\mathrm{ml} = 0.1\,\mathrm{L}$$

Molarity of buffer solution:
$${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$


How to solvethis problem related to redox titration? A $1.0\ \mathrm{g}$ sample containing $\ce{BaCl2*2H2O}$ was dissolved and an excess of $\ce{K2CrO4}$ solution added. After a suitable period, the $\ce{BaCrO4}$ was filtered, washed and redissolved in $\ce{HCl}$ to convert $\ce{CrO4^2-}$ to $\ce{Cr2O7^2-}$. An excess of $\ce{KI}$ was added, and the liberated iodine was titrated with $84.7\ \mathrm{mL}$ of $0.137\ \mathrm{M}$ sodium thiosulphate. Calculate the percent purity of $\ce{BaCl2*2H2O}$.

I first found the number of mili-equivalents of sodium thiosulphate  and equated that to the number of moles of pure $\ce{BaCl2*2H2O}$ times the n-factor i.e. $2$.

From this equation I get the number of moles of pure $\ce{BaCl2*2H2O}$ and then multiply this by its molecular mass to get the mass of pure $\ce{BaCl2*2H2O}$. But it comes greater than $1\ \mathrm{g}$. How to solve this?

Chromium(VI) is reduced to chromium(III) while oxidizing iodide to iodine. So one equivalent of chromate corresponds to 3 equivalents of iodine ($\ce{I}$, not $\ce{I2}$).

$$\ce{BaCl2 \cdot 2 H2O} \equiv \ce{BaCrO4} ~\mathrm{(chrome(VI))} \equiv \frac{3}{2}~\ce{I2} \equiv 3~\ce{S2O3^{2-}}$$

If you take this into account you should probably end up with 87.5% purity.


why does Zn become oxidised in the first place in operation of voltaic cell? I am so confused that during the operation of a voltaic cell, why does $\ce{Zn}$ gets oxidized? In anode side, there is a zinc metal in $\ce{ZnSO4}$ solution, but why does the $\ce{Zn}$ lost 2 electrons?

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. In anode compartment you have Zn in contact with ZnSO4. In cathode cell you have two other reagents. For example Cu and CuSO4. activity series of the metals suggest that zinc is OK to lose electrons and turn into zinc ion. Copper ions from CuSO4 are strongly willing to turn into reduced copper metal. So zinc gives away electrons to copper ions because Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$.

Why Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$ is a separate question and it belongs to a different field (quantum mechanics/molecular orbital theory rather than electrochemistry).

In short - there is Zn metal that holds electrons weakly and Cu$^{2+}$ that will hold electrons strongly when it forms Cu metal. Hence, electrons flow to Cu$^{2+}$ from Zn.


Acid-molybdate spray of Hanes & Isherwood I am reading a paper where they develop the phosphate spots on a paper chromatogram using what they call " Acid-molybdate spray of Hanes & Isherwood  (1949)".

Any idea what this reagent is?

I tried looking up the original reference but unfortunately I cannot fine a full text version. http://www.ncbi.nlm.nih.gov/pubmed/15398090?dopt=Abstract&holding=npg

Sigma has "Ammonium molybdate tetrahydrate" reagent available which is described as a component of the "Hanes and Isherwood reagent " but not sure what else is supposed to go in there.

http://www.sigmaaldrich.com/catalog/product/fluka/09880?lang=en&region=IN

Any ideas?

I managed to find a re-print of the paper in question– sadly it seems like Nature are mothballing many old papers (dead/missing links).
There are many acidic-cerium/molybdenum type stains that are used nowadays, many of which are commercially available off the shelf, so you might wish to try these before going after something from 50 years ago. (Cf. CAM stain, Hanlessian's stain)
I think this paper may also have pre-dated silica backed TLC plates, which will undoubtedly give you better separation than paper chromatography.

Taken from Nature (Nature Publishing Group)