Does dinitratomethane exist? In 2016 synthesis and structure of tetranitratoethane $\ce{C2H2N4O12}$ were reported [1]. Currently (as of 2017-Q4) this is the only dinitrato-carbon compound with the well-established crystal structure and composition:

This article also mentions dinitratomethane (reference number updated from 6 to 2):

While nitrate esters such as nitroglycerin (NG) and pentaerythritoltetranitrate (PETN) (Figure 2A and B) are well known,  only very few examples of germinal C-nitrato compounds are known. The simplest one, dinitratomethane (Figure 2C) is a liquid which is obtained from nitration of 1,3,5-trioxane in a $\ce{HNO3/H2SO4}$ mixture [2].

Figure 2. Structural formula of nitroglycerin (A), PETN (B), and dinitratomethane (C).

However, I wasn't able neither to get the full-text of original paper from 1938, nor any more recent proofs of existence for dinitratomethane. Google Scholar returns only a couple of entries one of which is [1], and another one is a PhD thesis. There are also zero Ngrams for "dinitratomethane", which is unusual. ChemSpider also lists this compound as nitrooxymethyl nitrate (CAS 38483-28-2).
Can anyone approve or disapprove the existence of dinitratomethane $\ce{CH2N2O6}$ (feel free to utilize computational chemistry too)?
References

Fischer, D.; Klapötke, T. M.; Stierstorfer, J. Chem. Comm. 2016, 52 (5), 916–918. DOI: 10.1039/C5CC09010E.
G. Travagli, Gazz. Chim. Ital., 1938, 68, 718.

After some more research I'm pretty much convinced in its existence. As Mithoron noticed in the comments, the name dinitratomethane is somewhat confusing, and is indeed not preferred.

I suppose the most accurate IUPAC-compatible one would be (nitrooxy)methyl nitrate. It also turned out that in the literature (mostly on regulated and hazardous chemicals) this compound is generally known as methylene glycol dinitrate.

The second volume of a classic handbook "Chemistry and Technology of Explosives" [1, p. 155] gives the most extensive description I discovered (reference numbers are updated):

  Methylene glycol dinitrate, or "nitromethylene glycol", is an oily liquid having a boiling point of $\pu{75-77 ^\circ C}$ at $\pu{20 mm Hg}$.

  It is a strong explosive, capable of dissolving collodion cotton. However it is not stable, being readily hydrolysed and therefore it is of no practical significance.

  A method of obtaining this explosive, by the nitration of moist
  paraformaldehyde was patented by Travagli and Torboli [2]. They utilised the well-known property of formaldehyde of reacting like methylene glycol in the presence of water.

  According to Travagli [3,4] methylene glycol dinitrate is obtained as follows:
  paraformaldehyde is dissolved in sulphuric acid and added drop-wise at $\pu{3-5 ^\circ C}$ into a mixture of nitric and sulphuric acid. The resulting oil is separated and washed with water to give a final yield of $35\%$.

The recent handbook of liquid explosives from 2015 [5, p. 227] contains virtually the same information without any updates, which indicates there simply hasn't been much of an interest in this compound due to its instability and lack of possible applications.

References

Tadeusz Urbanski. Chemistry and Technology of Explosives, 1st ed., Pergamon Press GmbH, 1965; Vol. 2.
G. O. Travagli and Torboli, Ital. Pat. 338080 1935.
G. O. Travagli, Gazz. Chim. Ital. 1938, 68, 718.
G. O. Travagli, Chimica ed Ind. 1938, 20, 389.
Jiping, L. Liquid explosives; Springer Berlin Heidelberg: New York, NY, 2015. ISBN 978-3-662-45847-1.


Why Cl⁻ can't act as bidentate ligand? $\ce{Cl-}$ is a monodentate ligand. It has four lone pairs then why it can't donate two lone pairs to central atom and act as a bidentate ligand?

A bidentate ligand is one that forms bonds to the metal center through two different atoms, not through two lone pairs. While chloride does have more than one pair of valence electrons, it only has one atom. Thus it cannot be bidentate.

If the chloride shares a second pair of electrons with the metal center, there would still only be one bond site between the ligand and the metal:

$$\ce{[M+]-Cl <-> [M]=Cl+}$$

A bidentate ligand needs two lone pairs on two separate atoms. Consider ethylenediamine (en), one of the classic bidentate ligands: $\ce{H2NCH2CH2NH2}$. Each nitrogen atom has a lone pair that can be used to form a bond to the metal center, creating two bonds from two different sites on the ligand. See the structure of the tris(ethylenediamine)cobalt(III) ion from tris(ethylenediamine)cobalt(III) chloride $(\ce{Co(en)3Cl3})$.

Update:
As Nicolau points out in a comment, chloride can bridge two metal centers. This is not the denticity phenomenon, which is one ligand with two bonding sites to the same metal. This is one ligand with bonds to two metal centers. See the structure of niobium pentachloride (really $\ce{Nb2Cl10}$ or better $\ce{(NbCl5)2}$.


How is the mol highway used in this problem? How many moles of propane gas, $\ce{C3H8}$, are contained in 11 grams of the gas at standard conditions? Also, can you please explain step-by-step (and include sigfigs?)

I think I might know how to do this, but I'm not sure if this process is correct.

I found online that $\ce{C3H8}$ has a mass of 44.11 grams per mol. So do I just use a proportion to find the number of moles this 11 grams of propane has?

You are on the right track. Every element has a molar mass that is defined under standard conditions, which is in this case the natural abundance of an element taken into account. Pressure, volume and temperature will not change the mass of a compound. Since every element has a specific molar mass, every compound does too. The molar mass is a measurement of how much mass does one mole of a substance have.

The molar mass is usually referred to with $M$, while the mass is referred to as $m$. The amount of substance is $n$. This gives you the following relationship: $$M=\frac{m}{n}$$

Since you have given $m(\ce{C3H8})=11~\mathrm{g}$ and you already looked up $M(\ce{C3H8})=44.1~\mathrm{g\,mol^{-1}}$, you can use this formula to determine $n(\ce{C3H8})$.

In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.


Determining number of unique hydrogens in lactupicrinal 1H NMR I'm having trouble understanding which hydrogens are magnetically distinct in the following molecule (lactupicrinal):

I've counted all the hydrogens (there are 20) and I know that hydrogens in a methyl group are all chemically the same (I think), but I don't know about the hydrogens in the ethyl group (double bonded on the lower right side). Am I right in that all other hydrogens outside of those groups are distinct, or are the mirrored hydrogens on the benzene ring also equivalent? Please help, I'm completely at a loss and I've been trying desperately to understand hydrogen equivalence for weeks.

Your question doesn't sound like you care about exact splitting, so I'm going to ignore magnetic equivalence/inequivalence. If you do care, you need to take that into account.

For the purposes of this discussion, equivalent means chemically equivalent.

You also need to keep in mind that NMR is achiral. In other words, opposite enantiomers will produce the same spectrum.

The standard procedure is to replace the atom of interest with something that is distinguishable from the regular atom. I usually replace an $\ce{H}$ atom with an $\ce{H}*$, but anything will do, provided it's still a hydrogen but different from the others.

With this replacement you've created a new molecule.

For two specific hydrogen atoms, you can perform the replacement operation to generate two molecules.

Now, compare these two molecules.

If they are different molecules (e.g., geometric isomers, (E)-/(Z)- isomers, cis-/trans- isomers), then the original hydrogen atoms are heterotopic and should have different chemical shifts. Their environments are different.
If the molecules are diastereomers, then the two hydrogens are diastereotopic. Since we can distinguish diastereomers by achiral methods, so can NMR. That means that the two original hydrogen atoms will have different shifts.
If the molecules are enantiomers, then the two hydrogens are enantiotopic. They are different, but because NMR can't distinguish between enantiomers, it can't tell that these two hydrogen atoms are different. They will have the same shift.
If the two molecules are identical, then the two hydrogens are homotopic. Nothing can distinguish them. They are chemically equivalent, and by NMR, they will have the same chemical shift.

Try applying this test to the hydrogen atoms on the ethyl group and the hydrogen atoms on the ring.


What is a "tri-salt polymer" as described by this product that eliminates and prevents mold? This answer mentions an anti-mold product called Concrobium Mold Control. The FAQ linked on their website says:

Q: I see Sodium Carbonate listed on the label. Is Concrobium Mold Control just Sodium Carbonate (washing soda) and water?
A: No. Sodium Carbonate is the listed “active ingredient” on the Concrobium Mold Control label, but in fact there are two other ingredients in the solution which, when combined with the sodium carbonate and the water, result in this patented tri-salt polymer that eliminates and prevents mold. (Sodium Carbonate and water alone are not effective against mold.) The solution contains no bleach, ammonia or VOCs. Our Material Safety Data Sheet is posted on our website.

The MSDS lists two particular ingredients:
Chemical Name            CAS #          % by Wt

Trisodium phosphate   7601-54-9           1-5
Sodium Carbonate       497-19-8           < 1

I'm guessing the third might be water.
Question: What might the term "tri-salt polymer" refer to in general, and if possible to answer, what might it be in this particular case?
I did find a patent in a google search but I'm not sure of it's relevance and I don't understand it.

CA2504014C, Non-toxic water soluble inorganic anti-microbial polymer and related methods

Beside the three-step deprotonation  of (ortho) phosphoric acid, $\ce{H3PO4}$, along

$\ce{H3PO4 <=> H2PO4^- + H+ <=> HPO4^{2-} + 2H+  <=> PO4^{3-} + 3H+}$

where each of the oxygen-bound hydrogen atoms may be substituted by a cation, phosphoric acid may dehydrate partially to yield polyphosphates.  Which seems, reading a related Canadian patent CA2504014 (A1) (deposit in English), the working principle here:

(loc. cit., p. 6; cations generalized)

The know-how includes tuning the proccessing as such that the composition of these blend polyphosphates is obtained reliably, since the complete dehydration simply would yield the waxy polyphosphoric acid used as acidic catalyst in the lab.


How to rationalise the major product formed in a nucleophile promoted epoxide cleavage reaction? The Problem is to find major product of reaction as given below (the star is a labelled carbon):

The two possible reactions that come to my mind are:

Nucleophile promoted epoxide cleavage reaction, ① in image below
SN2 substitution of chloride as a strong nucleophile $\ce{C2H5O^-}$ is being used, ② in image below

I thought that both these products should be formed, but to my surprise the answer is given as product ③ (in image), in which it appears that after ring opening of epoxide to ① they they have substituted the chlorine by using $\ce{OH-}$.

So my questions are:

Why is the product ③ major product, why not ①?
Is the product ② also formed and if no, then why not? If yes, then on what basis is product ③ the major product? Shouldn't product ② be also formed proportionately as in the final structure both product ② and product ③ appear very similar to me.
Can anyone please give me hints on how to solve such types of question when to form epoxides as in product ③ (even though epoxide is a ③ membered ring with ring strain also if $\ce{C2H5O^-}$ is left in solution it may again try to open the epoxide ring) and when not to form epoxide as in product ① which doesn't have any ring strain.
It would be great if anyone could also suggest the approximate percentage of products ①, ②, ③. However it is not necessary a simple explanation is sufficient.

Let's do this step by step and figure your questions out.
First, what a nucleophile needs is positive terminal in the compound.

The other carbon attached to oxygen is not as positive as the one where the attack has been done due to inductive effect of $\ce{-CH2}$ adjacent to it. The product (1) is not formed as there is no source providing $\ce{H+}$ to form alcohol. If possible a very small amount  may be formed by $\ce{H+}$ extraction from $\ce{EtO-}$. Next the intramolecular nucleophilic substitution takes place.

I do agree that product (2) is formed, but as a minor product. Here what we observe is that there is neighboring group participation in the mechanism. Generally these neighboring groups increase the rate of the reaction by switching on abilities like dipole in the bond. Whenever we observe compounds with such epoxides or sometimes even alcohols, we better think of such consequences which can help us predict the product.


What determines how favorable (exothermic) the electron affinity of a certain element is? (Favorable means how exothermic a reaction is, more favorable= more exothermic)

There are some abnormalities in the trends of how favorable the electron affinity is.

From Al to Cl, the
Electronegativity is increasing, but the energy releasing from the electron affinity of P is smaller than that of Si. Why? Is it because of how the electrons are arranged?

There are many determinants of how favorable electron affinity is. For example, oxygen and sulfur. Generally, we will think oxygen to have a more favorable electron affinity, due to its stronger electronegativity. Both have the same electrons config, except for the number of shells. But sulfur appears to have a more favorable electron affinity because of its larger size. The smaller size of oxygen atom produces more electron-electron repulsion when one or more electrons are added.

All I ask for is a list of determinants. I tried to find one but couldn't.

The electron affinity (EA) of an atom (A) is defined as the minimal energy that is required to free an electron (e$^-$) from the associated anion (A$^-$)

$$
\text{EA: }\qquad\text{A}^-\text{(g)}\rightarrow \text{A(g)}+\text{e}^-,
$$
in other words, the electron affinity is the ionization energy of the associated anion.

A large positive EA means that the anion A$^-$ is stable, while a negative EA indicates that the anion ${\text{A}^-}$ is unstable (such as He$^-$). Halogens have the largest electronegativity as the associated anion obtains a completely filled shell.

In general, if an atom A obtains a full/half full shell/sub shell by addition of an additional electron, then A will have a large EA, while A will have a small EA if it already has a full/half full shell/sub shell.

In addition there are some competing effects along a group in the periodic table. Consider the EA of the halogens for example:

Atom:      F       Cl      Br      I
EA /eV     3.40    3.61    3.36    3.06

(i) The smaller the principle quantum number $n$ of the highest sub shell, the larger is the attraction between the nucleus and electrons and the larger is EA. This would predict the trend to be EA(F) > EA(Cl) > EA(Br) > EA(I).

(ii) The smaller the principle quantum number $n$ of the highest sub shell, the larger is the electron repulsion the smaller is EA. The expected trend is EA(F) < EA(Cl) < EA(Br) < EA(I).

As these effects are in competition with each other, there is a maximum in the observed trend for the halogens around chlorine. Where the maximum occurs, is unfortunately difficult to predict.


Structural isomers of hexaammineiron(III) hexacyanidocobaltate(III) Find the number of structural isomers of $\ce{[Fe(NH_3)_6][Co(CN)_6]}$

This was a question solved by my teacher as follows:

The $\ce{CN}$ will show linkage isomerism while $\ce{NH_3}$ and $\ce{CN}$ will show coordination isomerism.

We will now consider cases of coordination isomerism and then use linkage isomerism.

$\ce{[Fe(NH_3)_6][Co(CN)_6]}\rightarrow 7$ isomers(due to linkage isomerism of $\ce{CN}$)
$\ce{[Fe(NH_3)_5(CN)][Co(CN)_5(NH_3)]}\rightarrow 2\times 6=12$
$\ce{[Fe(NH_3)_4(CN)_2][Co(CN)_4(NH_3)_2]}\rightarrow 3\times 5=15$
$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}\rightarrow$ not counted.

The further $3$ cases are similar to the first $3$ and will result in the same number of isomers. Hence the answer is $2\times(7+12+15)=68$.

Why is $\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$ "not counted"?

I feel that isomers due to $\ce{CN}$ showing linkage isomerism are perfectly valid. Please correct me if I'm wrong and sorry if the question is too trivial.

I have found the answer:

  In the following case:
  $$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$$
  Both the constituent entities are neutral and hence will exist independently, and thus will not form isomers of the original compound.


Is there a mathematical formula for electronegativity? If we lost all the numerical values we have for the electronegativity of elements, is there any mathematical formula that can be used to re-calculate them based on proton (and neutron) numbers alone? For example, if I had no way to look up the electronegativity of Carbon, could I use some existing formula to calculate it from scratch? Or would I have to go actually measure it?

Electronegativity cannot be calculated from the number of subatomic particles in an atom. However, using Ionization Energy and Electron Affinity, one can calculate the Electronegativity of an atom using the Mulliken Electronegativity Equation.

Ionization Energy : The amount of energy required to remove the
electron that is least attracted to the nucleus in a gaseous state.
Electron Affinity : The amount of energy an atom gains when an
electron is added.

To find the Ionization and Electron Affinity values of an atom, one must reference a chart.  In the Mulliken Electronegativity Equation, Ei stands for Ionization Energy and Eea stands for Electron Affinity.  Plug in your values and solve.

Equation for Energies in Kilojoules/Mole :
ENMulliken = (1.97×10^−3)(Ei+Eea) + 0.19
Equation for Energies in Electronvolts :
ENMulliken = 0.187(Ei+Eea) + 0.17

Lithium Example in kJ/mol :

The Ionization Energy for Lithium is 520 kJ/mol
The Electron Affinity for Lithium is 60 kJ mol^-1
ENMulliken = (1.97×10^−3)(Ei+Eea) + 0.19
ENMulliken = (1.97×10^−3)(520 + 60) + 0.19
ENMulliken = 1.143 + 0.19 = 1.333

NOTE : There are multiple scales for measuring Electronegativity, including the Pauling Scale, Sanderson, Allen, and more.  This example measures Electronegativity using the Mulliken scale, and will compute an answer that will differ from other models.


Would magnesium and calcium hydroxide undergo a single replacement reaction? Can calcium hydroxide react with magnesium (if given sufficient activation energy) to produce magnesium hydroxide and calcium, in the same way that NaOH + $\ce{Mg} \ce{->} \ce{Mg(OH)2} + \ce{Na}$ ?

Or, is calcium's electronegativity (or some other factor) prohibitive of this?

The key part of the question is given sufficient energy.  The difference in electronegativity between Ca and Mg is ~0.3 eV. The relationship of eV:K is about 1:12,000. Given that, you should be able to calculate a temperature at which the reaction $\ce{Ca(OH)2 + Mg -> Mg(OH)2 + Ca}$ is favored over the reverse.

However, temperature is based on the random motion of particles, so the energy distribution follows a Gaussian curve. Some Mg atoms are sure to have enough energy to knock loose some Ca atoms at room temperature.

Furthermore, you can force the suggested reaction by making use of relative solubilities. Find a solvent (not water, since magnesium and calcium are far more electronegative than hydrogen, and would decompose water to form hydroxides of both metals) in which $\ce{Ca(OH)2}$ is more soluble than $\ce{Mg(OH)2}$. Precipitation will "lock up" the $\ce{Mg(OH)2}$ as it forms, little by little, forcing the reaction to proceed even though it is endothermic.

Of course, you could also force this to proceed in molten salts by driving it with an electric current.


Availabilty of magnesium ions in phosphate buffer Phosphate buffer is a common buffer in biological applications, it's especially popular for NMR. Magnesium ions are necessary for many biological systems to work, though the amount of magnesium ions you can add in a phosphate buffer is limited by the relatively low solubility of magnesium phosphate.

What I'm curious about is whether the phosphate buffer can affect the availability of magnesium ions even in concentrations where no visible precipitation occurs? Common phosphate buffer concentrations are around 10-100 mM, and magnesium ions are usually 5 mM or below. At those concentrations I see no visible precipitation, but is the magnesium actually fully available or is the phosphate sequestering it and thereby lowering the effective magnesium ion concentration in the solution?

Even just calculating the solubility of magnesium salts in phosphate buffer is not straightforward, as there are multiple species of magnesium phosphate and possibly even other salts like $\ce{MgKPO4}$.

Does phosphate buffer affect the availability of magnesium ions in solution compared to a non-interacting buffer? If that is so, is there a way to calculate the size of that effect?

If you're still interested in this, fluorescence may be a good method to test the phosphate dependence on magnesium activity. First record a fluorescence spectrum of a solution of magnesium and fluorescent dye, then add small volumes of concentrated phosphate and note any changes in the fluorescence. Of course, you would have to do a control experiment to see if the phosphate is directly perturbing the fluorescence. This write-up describes some magnesium-chelating dyes that you could use, though if you don't want to spend a lot of money on a dye, you may already have some chemicals in your lab that will work. Phenol looks like promising alternative as it fluoresces, is decently water soluble (83 g/L), and forms a complex with magnesium which breaks planarity, which would certainly change the fluorescence spectrum.

It would only take a day to get accurate quantitative results if you have access to a fluorometer.


Equilibrium constant calculation from concentrations Consider the following equilibrium: $\ce{A2\ _{(g)} + 2B_{(g)} <=> 2AB_{(g)}}$

A chemist puts 8.0 moles of $\ce{A2}$ and 10.0 moles of $\ce{B}$ into an evacuated 1.0 L container. The system reaches equilibrium at a certain temp according to the process outlined above. When the system is at equilibrium the chemist determines $\ce{[AB]}$ is 8.0 M. What is the $K_c$ for this process at this temperature (278 K). I set up a nice chart and everything and I cant figure this out. My answer choices are 10, 3.0, 1.0, 4.0, and .50

Initially I thought that The Kc was [AB]^2/([A2][B^2]) but I put in the values and didn't get any of the answers...

Then i set up an ice chart

It's hard to show on here but I then ended up with [8+2x]^2/([8-x][10-2x]^2) but that didn't really get me anywhere

Well, I haven't done that in a while, but we seem to agree that the equilibrium constant is expressed as

$$ K_c = \frac{[AB]^2}{[A_2][B]^2}$$

What else do we know?

Initially, $[A_2]$ = 8.0M, $[B]$ = 10.0M
In equilibrium, $[AB]$ = 8.0M
For every mole of $B$, one mole of $AB$ is produced.
For every mole of $A_2$, two moles of $AB$ are produced.

With just one cup of coffee so far, I get

$$ K_c = \frac{8^2}{(8 - \frac{8}{2})(10-8)^2} = 4$$


Binding energy question The electrons in the 1s orbital of chlorine have a binding energy of 273 MJ/mol, but the 1s electrons in sulfur have a binding energy of 239 MJ/mol. Why is this?

Chlorine has 17 protons and Sulfur has 16. In the 1s orbital there is no ineer electron to provide shielding effect of repulsion. So greater the positive charge, greater is the attraction and better is the binding.


extraction/isolation of Curcumin from Turmeric using ethanol I want to extract/isolate Curcumin from Turmeric using ethanol solvent, already know paramaters such as optimum particulate size, optimum temp (140 degrees F) etc, I need to know how to safely set up a basic distillation to recover the ethanol for reuse.  I have basic lab glassware incl and heat plate/stirrer, condenser.
Thank you

In their open access article Optimization of the extraction of curcumin from Curcuma longa rhizomes, published in Revista Brasileira de Farmacognosia, 2013, 23, 94-100, the authors systematically examined the effect of various parameters on the extraction yield, namely

drug to solvent weight ratio (1/6 and 1/4 g/g)
agitation speed (30 and 70 rpm)
extraction time (12 and 24 h)
extraction temperature (50 and 80 °C)
ethanolic strength (70 and 90 % v/v)

Based on their analysis, the following optimum conditions were seemingly predicted:

drug to solvent weight ratio: 1/6 g/g
agitation speed: 30 rpm
extraction time: (12 and 24 h
extraction temperature: 80 °C
ethanolic strength: 70% v/v

Please have a look at the full article for further details.

Edit

  I need to know how to safely set up a basic distillation to recover the ethanol for reuse.

That is a rather fundamental question which suggests that you don't have a lot of experience in organic benchwork.

You might want to have a look at the English wikipedia page on distillation for the general setup.

Note that you will not be able to fully separate ethanol and water by destillation. Instead, you will obtain an azeotrope, a mixture that has a boiling point lower than that of ethanol and water. In order to check the density of your destillate and to dilute again to an ethanolic strength suitable for extraction, a hydrometer might be a suitable tool.

In terms of lab safety, the following points should be considered:

Do not work alone!
Ethanol vapour is flammable!

Keep open flames and electric sparks away.
Run your still in a well-vented place.
Have a fire extinguisher at hand.
Have a work space with an unblocked emergency exit.

If you use standard lab glassware, a stirrer and an oil bath to heat the still pot, make sure that the tubes for the cooling water that cools your condenser are properly secured by wire or clamps. Remember that the heating bath is hot! The last thing that you want is cooling water splashing into the hot silicone oil.

On a lab scale, I would would not set up a classical distillation apparatus, but remove (and collect the ethanol) at reduced pressure using a rotary evaporator with a water bath!

In any case, use suitable eye protection!


Are molecules chiral if a rotation leads back to the same compound? I understand that if you rotate a molecule and it has the same configuration as its enantiomer, then it is indeed the same compound as the enantiomer (meso), and hence is achiral (despite having chirality centers).

If compound is rotated and the resulting Fisher projection has exactly the same configuration after its rotation, what does that mean in terms of chirality? (I'm not talking about enantiomers here,) It will have a different enantiomer (and hence not be meso), and will be polarizable, but the compound itself has exactly the same perspective if you rotate it. Does this hold any signifance chemically or stereochemically, or is it meaningless?

Take for example D-idaric acid (2S,3R,4R,5S)-2,3,4,5-tetrahydroxyhexanedioic acid).

In the below example of Idaric acid, the enantiomer appears to be a different compound. When the compound is rotated 180 degrees, it looks exactly the same.

In the below example of galactaric acid, when the compound is rotated 180 degrees, it is a different image. This is despite it being meso.

Obviously, being meso has a lot of chemical significance (no polarization of light). But does a compound possessing the property of being rotated 180 degrees, and having the same image (in Fischer form) present any chemical significance compared to a compound that doesn't do this?

The property that you are describing for D-idaric acid is called C2 symmetry. This means that when you rotate a molecule (or any object) by 180 degrees, it appears identically to the original orientation. In D-idaric acid, the C2 symmetry axis is halfway along the bond between C3 and C4.

Having a C2 symmetry axis doesn't tell us if the compound is chiral or not. Achiral compounds have an (internal) plane of symmetry or a point of symmetry. D-Idaric acid does not meet either of these requirements and is a chiral molecule.

Galactaric acid does not have a C2 symmetry axis but does have a plane of symmetry. Galactaric acid is achiral. Since galactaric acid has stereocenters but is not chiral, we describe it as meso.


Why use potassium iodide instead of copper iodide in Sandmeyer reaction? The Sandmeyer reaction is an extremely useful reaction for the functionalization of aromatic rings through a diazonium intermediate. For conversion of a diazonium intermediate to the corresponding chloride, bromide, or cyanide, the copper(I) salt is used. Surprisingly (to me at least) for iodination, potassium iodide is the reagent of choice, as described in Maitland Jones's Organic Chemistry (and elsewhere I'm sure), but no rationale is given. SciFinder searches also show that potassium iodide is most frequently used.

Is there something special about copper(I) iodide that makes it unsuitable for use in the Sandmeyer reaction?

Sandmeyer reactions using iodides, water, and thiols do not require the copper(I) ion as a catalyst for the reaction to proceed whereas your other nucleophiles do (though not necessarily stoichiometrically).

As you can see the copper(I) ion is important in reducing the diazonium to nitrogen and forming a radical.  Iodine, oxygen, and sulfur must be able to be oxidized by diazonium and therefore do not require copper.


Aluminum powder from aluminum sulfate? I recently did the experiment where you drop a piece of iron in an aqueous solution of copper sulfate. After a while, the piece of iron gets coated with a fluffy coating of copper that can be dried and mashed into a powder. Apparently this happens because iron is more reactive than copper, according to the reactivity series.
Can I do the same thing by making an aqueous solution of aluminum sulfate, then dropping a piece of magnesium metal in? It should work because magnesium is above aluminum on the reactivity series, right?

Metallic aluminum cannot be obtained from a solution of its salts. It is because pure aluminum reacts with water, a little bit like sodium. Of course a piece or a sheet of aluminum does not usually react with water, because the metal is covered by a thin, colorless and waterproof layer of aluminum oxide, which prevents the contact between aluminum and air. And if you remove this layer with a knife, the oxide layer will immediately be remade by air.
The only way of showing that aluminum reacts with water is to dip the aluminum piece in a solution of Mercury chloride. There will be a surface reaction like $$\ce{2 Al + 3 HgCl2 ->  3 Hg + 2 AlCl3}$$ Then the mercury Hg will make an alloy with aluminum. And this alloy cannot be covered by the alumina shield. So such a piece of amalgamated aluminum will be quickly oxidized in air, and will produce long white filaments of $\ce{Al2O3}$ or $\ce{Al(OH)3}$ if abandoned in air for a couple of minutes due to one of the following reactions $$\ce{4Al + 3 O2  -> 2 Al2O3}$$  $$\ce{4 Al + 3 O2 + 6 H2O ->  4 Al(OH)3}$$ This production of alumina filaments is very interessant to observe. Of course it is not allowed any more today, because of the toxicity of mercury.


Why does magnesium bromide transfers from a carbon atom to nitrogen in piperidine? In this reaction, why does $\ce{MgBr}$ goes to the $\ce{>NH}$ group and form a bond with nitrogen by removing $\ce{-H}$.
Also where does the negative charge go?

There is no step 1 and step 2 in this reaction. Both are consecutive and second one is fast because it is an acid-base reaction:

Note that approximate $\mathrm{p}K_\mathrm{a}$ values of ring $\ce{C-H}$ and $\ce{N-H}$ are 50 and 40, respectively. Thus, as soon as Grignard reagent is formed, it exchanges relatively acidic protons of $\ce{N-H}$ fast (Remember, Grignard is also a strong base). To visualize the exchange process, I included schematic representation at the bottom of the image.


How do I make a solution of spinach to titrate with pottasium permanganate? My teacher said something about straining it but I didnt quite get it

The best approach is to ask the teacher again after doing some "homework" (i.e. reading and searching). Extracting and then titrating oxalic acid/oxalate is not that easy from spinach. You will need a lot of spinach to extract a titratable amount. Oxalic acid can be extracted by boiling spinach in 0.1 M HCl. However, tons of other redox active substances will also appear in the extract. Direct titration with KMnO4 will give an extra-ordinary large percentage, which is of course erroneous. After extraction, you need to precipitate oxalic acid via some cation. After isolation of the precipitate you would need to bring back oxalic acid into solution, one can do a redox titration. In real-world nobody would use this approach but as an academic project, it could be a good exercise. Again ask your teacher if she/he is willing to go to this extent.


Using a gas stove instead of Bunsen burner for a flame test? In all chemistry labs we use the Bunsen burner. What's the difference between it and a good old gas stove used in home? Can the stove also be used to get positive flame tests for some cations (like $\ce{Ca^2+}$)?

Yes, you can use a common stove to test for cations.
But a stove is designed to minimize the risk of incomplete combustion (which could lead to production of carbon monoxide), hence its flame always appears as an intense blue flame.
Such color contamination could be problematic when testing for cations.
In contrast, the combustion (and the color of the flame) can be regulated in a Bunsen burner.

If you want to experiment at home, an old (but gold) method is that of placing a wick through the metal lid of a closed glass vial, thus creating a burner. You can put alcohol in it, and light the wick.

You can adjust the flame by moving the wick.

Plus, as reported in
[1],
a solution of the salt in alcohol produces a long lasting colored flame in that setting.

References

Dragojlovic, V.; Richard F. Jones. Flame Tests Using Improvised Alcohol Burners. J. Chem. Educ. 1999, 76 (7), 929. https://doi.org/10.1021/ed076p929.


Why do oxidation numbers work? I understand that oxidation numbers are a method for keeping track of electrons in a reaction and how they are generally assigned (electrons in a bond are assigned to the more electronegative atom).

Why does this method work? Why does assigning these arbitrary numbers based on electronegativity work in predicting and balancing reactions?

First of all they are not really arbitrary.

But the main point, regarding balancing the equations is that: You can keep track of the total number of electrons per species, which is a well defined value. Like, for one permanganate ($\ce{MnO4-}$) you have 25+4*8+1 electrons, for one $\ce{Mn(II)}$ you have 23 electrons, for one water you have 10 electrons, for one $\ce{H+}$ you have 0 electrons. So for this half-equation, you must have 5 electrons on the left side:

$$\ce{8H+ + MnO4- + 5e- -> Mn(II) + 4H2O}$$

or you can say that, here I assume all H atoms are +, all oxygens are -2, and for consistency, Mn in $\ce{MnO4-}$ is +7. I can dismiss H and O's since all are same, and all change is because of Mn, which goes from +7 to +2 and difference is 5 electrons.

You could do this like that as well: O is -4 everywhere, so H's in $\ce{H2O}$ should be +2 and they were +1 on the left side. They donated 8 electrons. Mn in $\ce{MnO4-}$ should be +15 and it is +2 on the right side, which means it accepted 13 electrons. And the difference is 5 electrons, again.

Bottomline: It is all about consistency. If all elements in each and every species sum up to the charge of this species, whatever you say their formal charges are, it will work -because this just another way of keeping track of total electron count.  So it is better to treat most of them constant like, O is -2, H is +1 almost always.


Calculating the pH of a 'buffer' solution? Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$.

The given solution:
Concentration of acid: $\frac{0.030}{0.300} = \pu{0.10 mol dm-3}$
Concentration of base/salt: $\frac{0.020}{0.300} = \pu{0.067 mol dm-3}$
$$\ce{[H+]} = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times 0.10}{0.067} = \pu{1.9 \times 10^{-3} mol dm-3}$$
$$\mathrm{pH} = 2.72$$
Now, here is why I am confused: this calculation is for a buffer solution, where $\ce{HA}$ and $\ce{A-}$ are a conjugate acid/base pair. However, chloroethanoic acid and sodium hydroxide are not conjugate acid/base (are they?) - so why does this calculation work? What is the intuitive way to think about this?

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$.

I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of chemical reactions and buffers. That's probably why OP seems so confused of the solution. The answer elsewhere explains to certain level but has not explore the total picture. Hence, I thought it would be beneficial to OP if I explain it a little further:
$\ce{CH2ClCOOH}$ is a weak acid (about 100 times acidic than acetic acid) with $K_\mathrm{a}$ of $1.3 \times 10^{-3}$ (which is $1.8 \times 10^{-5}$ for acetic acid). On the other hand, $\ce{NaOH}$ is a strong base, which reacts with strong or weak acids irreversibly. When add $\ce{CH2ClCOOH}$ and $\ce{NaOH}$ together, following reaction would take place:
$$\ce{CH2ClCOOH (aq) + NaOH (aq) -> CH2ClCOONa (aq) + H2O (l)} \tag1$$
The net reaction is:
$$\ce{CH2ClCOOH (aq) + OH- (aq) -> CH2ClCOO- (aq) + H2O (l)} \tag2$$
The acid to base mole ratio of the reaction is 1:1. However, added amounts of acid and base are:
Weak acid, $\ce{CH2ClCOOH}$: $\pu{0.50 mol dm-3} \times \pu{0.100 dm3} = \pu{0.050 mol}$
Strong base, $\ce{NaOH}$: $\pu{0.10 mol dm-3} \times \pu{0.200 dm3} = \pu{0.020 mol}$
That makes $\ce{NaOH}$ the limiting reagent (recall your general chemistry knowledge). Thus, $\pu{0.020 mol}$ of $\ce{NaOH}$ reacts with $\pu{0.020 mol}$ of $\ce{\ce{CH2ClCOOH}}$ to give $\pu{0.020 mol}$ of $\ce{\ce{CH2ClCOONa}}$ or $\ce{\ce{CH2ClCOO- + Na+}}$, if you look at the net reaction.
The unreacted amount of $\ce{\ce{CH2ClCOOH}}$ in the mixture ($\pu{0.300 dm3}$ in total volume) is: $\pu{0.050 mol} - \pu{0.020 mol} = \pu{0.030 mol}$.
Therefore, the concentration of acid, $\ce{\ce{CH2ClCOOH}}$: $\frac{\pu{0.030 mol}}{\pu{0.300 dm3}} = \pu{0.10 mol dm-3}$
Simillarly, the concentration of salt, $\ce{\ce{CH2ClCOONa}}$ or $\ce{\ce{CH2ClCOO-}}$: $\frac{\pu{0.020 mol}}{\pu{0.300 dm3}} = \pu{0.067 mol dm-3}$
Since, $\ce{\ce{CH2ClCOOH}}$ is a weak acid, it would ionize in aqueous solution as follows:
$$\ce{CH2ClCOOH (aq) + H2O (aq) -> CH2ClCOO- (aq) + H3O+ (aq)} \tag3$$
Suppose, $\alpha$ amount (in $\pu{mol dm-3}$) of $\ce{\ce{CH2ClCOOH}}$ ionizes at $\ce{\ce{CH2ClCOOH}}$, the equilibrium concentrations of $\ce{\ce{CH2ClCOOH}}$, $\ce{\ce{CH2ClCOO-}}$, and $\ce{\ce{H3O+}}$ are $0.10 - \alpha$, $0.067 + \alpha$, and $\alpha$, respectively.
$$K_\mathrm{a} = \frac{\ce{[H3O+]}\ce{[A-]}}{\ce{[HA]}} $$
$$\therefore \ \ce{[H+]} = \alpha = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times (0.10  - \alpha)}{(0.067  + \alpha)} \approx \pu{1.9 \times 10^{-3} mol dm-3}$$
$$\mathrm{pH} = -\log\alpha = -\log (1.9 \times 10^{-3}) = 2.72$$
On the other hand this can be easily solved using the Henderson-Hasselbalch equation as directed in the other answer elsewhere:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac {\ce{[A-]}}{\ce{[HA]}}\right) = 2.89 + \log \left(\frac {0.067}{0,10}\right) = 2.72 $$


Photochemistry and steady state approximation I am unsure whether to ask this in physics or chemistry as there is overlap but I am taking this in a chemistry module.

System A undergoes photophysical pathways:
$$\ce{A + hv -> A\mathrm{*}}$$
$$\ce{A\mathrm{*} -> A + hv   \tag{k1}}$$
$$\ce{A\mathrm{*} + B -> A + B\mathrm{*} \tag{k2}}$$
$$\ce{B\mathrm{*} -> B + hv      \tag{k3}}$$
$$\ce{B\mathrm{*} -> C           \tag{k4}}$$

It wants me to show that the quantum yield of energy transfer is equal to the quantum yield of fluorescene B and the quantum yield of photochemical conversion from B to C.

I am quite stuck on this problem. I assume the steady state approximation needs to be employed in order to determine quantum yields for the processes asked for.
I am wondering what the general strategy would be to answer this question?

The processes are absorb light into A, and then A* has only two pathways, i.e. fluoresces and reacts with B to form B*.

The B* has only two pathways, it  fluoresces and converts to C. All B* molecules must therefore follow one of these paths, so the sum of yields of B* (paths fluorescence + forming C) must be the same as the yield to form B* in the first place, thus you don't need to do a calculation.

In general if you have to find a yield of any process x this is, $\displaystyle \phi_x = \frac{\text{rate of process x}}{\text{rate of absorption}} \equiv \frac{\text{rate const x} }{\text{sum of all other rate constants from x}}$.

In your example $\displaystyle \phi_{ET}= \frac{k_{ET}[B]}{k_f^A+k_{ET}[B]}$ where superscript $A$ refers to fluorescence from $A^*$.

To work all this out write down $d[A^*]/dt = +k_{abs}[A]- k_f^A[A^*]-k_{ET}[A^*][B]$ and use steady state.


Where does the pH come from in the equation of Standard Hydrogen Electrode? $$\mathrm{E=-{2.303RT \over F}pH - {RT \over 2F}\ln {p_{H_2}/p^0}}$$
This is a equation on Wikipedia page of Standard Hydrogen Electrode, but I have no idea where the $pH$ come from.
Since $\mathrm{a_{H^+}=f_{H^+} C_{H^+} /C_0}$, how does the $\mathrm{f_{H^+}}$ disappear?

There are a few logarithm rules that you need to apply here.
\begin{align}
E &= \mathrm{RT \over F}\ln {a_{H^+} \over (p_{H_2}/p^0)^{1/2}}\\
  &= \mathrm{RT \over F}\ln a_{H^+} - \mathrm{RT \over F} \ln \left({p_{H_2} \over p^0}\right)^{1/2}\\
  &= \mathrm{\ln(10) RT \over F} \lg a_{H^+} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)\\
  &= -\mathrm{2.303 RT \over F} \mathrm{pH} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)
\end{align}

You need a few logarithm rules:

$\log\left({a \over b}\right) = \log(a) - \log(b)$
$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$
$\log_b(x^r) = r \log_b(x)$


What are quantifier and qualifier ions in mass spectrometry data for GC-MS or LC-MS? I've read that one should pick an ion peak that is unique to an analyte for quantification purposes and then pick at least 2 qualifier peaks and use their ratios to identify the analyte (or avoid misidentification thereof). How exactly does this work?

This is a general "identification" rule which should not be restricted to mass spectrometry.
In chromatography you may be analyzing very closely related analytes whose masses are very close or even identical. Think of structural isomers (low resolution MS is blind to enantiomers, isotopomers, isotopologues etc.). Structural isomers may very closely related mass spectra and their molecular ion peaks would be identical in mass, but their fragmentation patterns might be slightly different. So the "good" advice is pick not one but more than one line in order to positively identify a target analyte.
Think of the legal consequences of false positives. A sportsman or women may loose their career if the tests show a false positive or a criminal may go scot free. Therefore analytical chemists have a huge responsibility on the society in terms of having a very high degree of confidence in identifying or quantifying anything.
Although most of the people think that mass spectrometer is a "perfect" detector in chromatography, it is nowhere close to being one. Many compounds do not ionize well and hence go undetected. It is extremely useful though.


Stoichiometric reaction of hydrochloric acid, potassium nitrate, and copper I have the following reaction involving these elements:

$\ce{HCl}$ (hydrochloric acid)
$\ce{KNO3}$ (potassium nitrate, a nitrate salt)
$\ce{Cu}$ (copper)

The reaction should produce nitrogen dioxide ($\ce{NO2}$).

The chemical equation should look like this:

$$
\ce{HCl + KNO3 + Cu -> NO2}
$$

But I'm clearly missing something. Can anybody help me equalize and complete this equation?

My apologizes, I found the answer. The full equation is :
$$\ce{Cu + 4KNO3 + 4HCl -> Cu(NO3)2 + 2NO2 + 4KCl + 2H2O}$$
(I'll detail this later)


Question about calculating enthalpy change $$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$

$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$

Both $\Delta H$'s given by the above formulas should be equivalent.
However, I found one special case that the above 2 formulas give 2 different results.
$$\ce{2 H+ + 2 e- → H2}$$
$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)= 0-0=0$$
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)<0$$ because there is only bond formation, no bond break.
How to explain this difference? Can I say the heat is released during this process, because $$ \Delta H $$ is smaller than 0?

The approach to estimating the enthalpy change in a reaction by using bond dissociation energies assumes homolytic cleavage of the bonds: each atom gets one electron from the covalent bond.

The bond dissociation energy of the $\ce{H-H}$ bond is reported as 436 kJ/mol.  The relevant reaction is:

$$\ce{H2(g) -> 2H.(g)}$$

Note that this reaction is not the same as $\ce{H2 -> 2H+ + 2e-}$. The homolytic cleavage   reaction produces hydrogen atoms, not hydrogen ions. If we look up the standard enthalpy of formation for hydrogen atoms, it is +218 kJ/mol. Since the standard enthalpy of formation for $\ce{H2}$ is defined as zero, using the enthalpies of formation we also get $\Delta H = +436$ kJ/mol for this reaction.

It is generally hard to use bond dissociation energies for heterolytic reactions (reactions that produce ions from molecules).


Why do principal energy levels in an atom get closer together as n increases? The title says it all. Reasons that I can supply include: increased nuclear charge increasingly catches up in terms of influence to the increasing shielding and proof by contradiction in that if the space remained the same there would be negligible nuclear pull on the valence electrons and so this must not be the case. What is a comprehensive explanation for why this phenomenon is so?

Thank you!

The energy levels for an atom are similar to a "particle in a box", except that the walls have a different shape. When you analyzed the "particle in a box" situation, you used a rectangular box. An atom is spherical, so the walls of the atom open outwards at the same rate as the sphere increases in surface area. You should have noticed that the energy levels for a "particle in a box" decreased in distance at higher energy levels. The surface area of the sphere brings the energy levels even closer.


Mechanism for reaction of NOCl (Tilden reagent) with primary amines According to my textbook, the reaction between $\ce{NOCl}$ and $\ce{CH3CH2-NH2}$ is as follows:

  $$\ce{CH3CH2-NH2 + NOCl -> CH3CH2Cl + N2 + H2O}$$

But how exactly does the reaction proceed?

The first step (what I think) could be that $\ce{NOCl}$ dissociates into $\ce{NO+}$ and $\ce{Cl-}$ ions and then the $\ce{Cl-}$ ion attacks the amine group and replaces it by $\mathrm{S_N2}$ mechanism.

But according to this site, a diazonium product is produced. Even though a diazo product is formed, how does it produce an alkyl halide according to my textbook?

Why would the chloride ion attack the electron rich amine? The amine is the best nucleophile is the system and will attack the NO+. This gives a RNH-NO nitroso species which is then protonated on Oxygen, loses water and gives the diazonium cation. This adds loses nitrogen rapidly and adds Chloride to give the primary alkyl chloride.

More can be read about this process here (which I where I sourced the reaction scheme) or by searching the term "nitrosation of primary amine"


What is the functional difference between a radical anion and a nucleophile? Radical Anions are hundreds if not thousands of times more reactive than nucleophiles (right?). A radical anion is just a molecule with an unpaired electron. Where as a normal radical is missing an electron and an atom. If an acidic organic compound is deprotonated to form a nucleophile by removing a proton, then is there not an unpaired electron availible? (which makes the compound nucleophilic)

Why aren't nucleophiles as reactive as radical anions, since they both have "free electrons"? What is the functional difference?

Radical Anions are hundreds if not thousands of times more reactive
  than nucleophiles (right?).

Radical anions are very reactive, but some nucleophiles (such as $\ce{CH3^{-}}$) are also very reactive.

A radical anion is a molecule that has had an electron added to it.  For example, treating naphthalene with sodium in liquid ammonia transfers an electron from sodium to naphthalene forming the naphthalene radical anion, the electron is unpaired.

A radical arises from the homolytic breaking of a bond.  This process leaves an unpaired electron with each of the two molecular fragments.

  If an acidic organic compound is deprotonated to form a nucleophile by
  removing a proton, then is there not an unpaired electron availible?

Removal of a proton $\ce{H^+}$ (just a proton, no electrons) would leave 2 paired electrons (and a negative charge) on the remainder of the original molecule.

Nucleophiles come in a lot of flavors.  They can be neutral, like the nitrogen in an amine, or they can be negatively charged.  If negatively charged, the nucleophile could be a radical anion or a carbanion.  So radical anions just belong to a sub-class of nucleophiles.  It's probably fair to say that carbanions and radical anions are usually better nucleophiles than a neutral nucleophile.  Comparing the nucleophilicity of a carbanion and a radical anion is harder to generalize since they are both so reactive, but the carbanion with 2 electrons would probably be more nucleophilic than the radical anion with one electron..


pH Curve in Acid/Base Titrations So I came across a question on the equation of the pH curve, and I wanted to know how the user here arrived at the matter balance equation. Any help would be appreciated!

Here is the question:

The reason behind the steep rise in pH in the acid base titration curve

Nicolau already outlined it quite clearly in his answer.

  Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ($\ce{HA}$ or $\ce{A−}$) is simply $n_A = C_A V_A$

so you have $n_{\ce{HA}} + n_{\ce{A-}} = C_A V_A$

  We will assume that volumes are additive (total volume $V_T = V_A + V_B$)

and given that $[\ce{HA}] = n_{\ce{HA}}/V_T$ and $[\ce{A-}] = n_{\ce{A-}}/V_T$ it should be fairly clear that

$$[\ce{HA}] + [\ce{A-}] = \frac{C_A V_A}{V_A + V_B}$$

as desired.


Can water electrolysis detect the water impurities? A salesman came to my house to demonstrate his water purifier machine which is a RO Purifier. To show the difference between his purifier machine and the machine I own, he took some water from my machine in a glass. Then he pour water from a bottle he bought in another glass claiming that this was a purified water from his machine (little suspicious). Then he put a small electrolysis machine (that one we used for experiments in school) on both glass and started it. After a minute, there was lot of uhh... particles looking like mud floating in my water, while in his glass, there were very few in dissolved state.

So he said that those particles were salts and bad impurities which my purifier can not filter out, and his purifier can, thus having few particles in his glass.
So my question is, can this method detect the impurities in water, or is it just some dirty sales tactic? Because as I remember, i was taught in school that electrolysis does not has to do anything with water, it just transfer cathode to anode or something like that (sorry for my bad chemistry knowledge :P)

Please help me out!

[…] can this method detect the impurities in water […]

In principle, yes. The electrolysis can, as an example oxidize iron ions ($\ce{Fe^2+ -> Fe^3+}$) The latter forms a brownish, voluminous oxide hydroxide, $\ce{FeO(OH)·n H2O}$.

The WHO suggests an upper limit of 0,2 mg/l iron in drinking water.

Am I right that your drinking water comes from a well? Did you perform any analyses on the the water quality or consult local authorities concerning possible health risks and their prevention?

  […] or is it just some dirty sales tactic […]

I can't comment on the dirty part, but salesmen need to sell. We don't know whether the reference sample was actually produced by reverse osmosis or just a sample of distilled water (which you would not want to drink over a longer time).

Even if reverse osmosis can reduce the mineral content in your drinking water, you might want to check whether the additional costs for operation are really worth it.

Again, try to obtain some information from local authorities.


How does life choose it's nominal temperature preferences? If all life relies on water, and water freezes at 32F and boils at 212F.

Then why isn't the nominal temperature for life 122F?

122F will literally kill some animals, including us.

Do our biochemical reactions need "cold" (most people prefer ~72F) for some reason?

Here's a link to a brief Scientific American article that presents the results from a study of this question.  Basically, for every $\ce{1^{o}C}$ increase in body temperature, 6% fewer fungal species can survive - so the hotter, the better (the longer you're likely to survive).  The downside of "hot" is the increased metabolic rate required, which translates into more time spent consuming food.  If you were to plot the effect of these two competing effects on longevity versus body temperature you would get something like an inverted (upside-down) parabola.  The study found a maximum longevity at 98.1 F, not far off from the observed average body temperature of 98.6 F.


Exercising to use up energy consumed Suppose I eat a whole $450~ \rm{g}$ chocolate Easter rabbit. The enthalpy of combustion of chocolate is $−22.4~\rm{ kJ/g}$. One kilogram of body fat contains $39000 ~\rm{kJ}$ of energy. Walking for an hour uses up $690 ~\rm{kJ}$ of energy.

For how many hours do I have to walk to use up the energy consumed?

So far I have calculated the energy found in $450 ~\rm{g}$ of the Easter bunny by multiplying $450~ \rm{g}$ by $-22.4 ~\mathrm{kJ/g}$. This gave me $-10080~ \mathrm{kJ}$.

I’m thinking my next step would be dividing $-10080 ~\mathrm{kJ}$ by $690 ~\mathrm{kJ}$ to find the answer. But I feel this is incorrect because I don’t understand the relevance of $1~\mathrm{kg}$ of body fat containing $39000 ~ \mathrm{kJ}$.

As binary geek said, that information is not required in the calculations. Its probably given to make people realize that by gulping chocolate and walking for just 5-10 mins wont make you reduce your weight. Standard Satire has been employed!


What is the relationship between the stability constant of a copper (II) electrolyte complex ion and the output voltage Ecell? As a part of a school investigation I constructed a voltaic cell experiment where for each run I added a different ligands to the copper (II) sulfate half-cell electrolyte in a daniell cell to see if there was a change in the output voltage Ecell. When plotting a graph of this, it seemed that Ecell increased as the stability of the ligand increased (indicated by the stability constant Kc) however I am struggling to explain why. I presume its because the standard potential at the cathode Erhe increases with ligand stability due to greater oxidation.

The redox potential of Copper does not depend directly on the stability of the complex Cu-ligand. It depends on the concentration of the residual $\ce{Cu^{2+}}$ ion in solution. And in any solution of $\ce{CuSO4}$, adding a ligand produces a decrease in concentration of the free ion $\ce{Cu^{2+}}$. So, according to Nernst's law, the potential $\ce{Cu^{2+}/Cu}$ becomes less positive.
For example, the standard redox potential of $\ce{Cu^{2+}/Cu}$ is + $0.34$ V. Let's consider the ligand chloride $\ce{Cl-}$. If enough chloride ions $\ce{Cl-}$ is added to a $1$ M $\ce{Cu^{2+}}$ solution so that at the end the excess of $\ce{Cl}$- ions is $1$ M, the complex $\ce{CuCl4^{2-}}$ will be formed, with a concentration $1$ M. The measured residual [$\ce{Cu^{2+}}$] concentration is $2.5 ·10^{-6}$ M. In this case, Nernst's law gives the following $\ce{Cu^{2+}/Cu}$ redox potential : $$\ce{E_{Cu^{2+}/Cu} = + 0.34 V + \frac{0.0592}{2}log(2.5·10^{-6}) = + 0.34 V - 0.0296·5.6 = +0.34 V -0.166 V = +0.174 V}$$ This +$0.174$ V is lower than the standard redox potential (+$0.34$ V).
The stability complex of the $\ce{CuCl4^{2-}}$ ion is : $\ce{K = \frac{[CuCl4^{2-}]}{[Cu^{2+}][Cl^-]^4} = \frac{1}{2.5·10^{-6}·1^4} = 4·10^5 M^{-4}}$
If another ligand is used, which makes a stronger complex with copper $\ce{Cu^{2+}}$, the stability constant is higher, and the redox potential will be still lower.


Will mixing methyl 2-cyanoacrylate and cellulose make the cellulose harden quickly? The other day I was gluing together cotton fabric with methyl 2-cyanoacrylate (superglue) and the fabric started heating up and then eventually began smoking. I have been working on making a 2-part plastic-like substance for personal reasons for some time now, and I know that ionic liquids pull the individual polymer chains that make cellulose apart and that once they are separated they go into solution easily. This in mind, could I get the dissoluted cellulose and add superglue to create a cheap plastic-like material?

The working principle of super glue, based on 2-cyano acrylate, is an anionic polymerisation reaction which in principle triggered by any nucleophile.  Water conveniently may be such a nucleophile (the basic anion of water, hydroxyde $\ce{OH^-}$ were even better) which is why the container's mouth originally is not only closed by a cap, but by a thin metal plate.

Cotton consists mainly of cellulose, and cellulose itself

(source)

posesses many hydroxyl groups which like water may trigger the polymerisation.  Since there are so many that are easily accessible, the process is very fast, and proceeds under liberation of much heat (i.e., an exothermic reaction).  Because of this and because the acrylate may generate a strong 3D polymer, one should not wear cloths made of cotton-based fabric, nor gloves of such.

While it is new for me to learn that ionic liquids are able to dissolve cellulose, it might be that such an unmoderated polymerisation "runs away"; i.e. already proceeds at a high rate and, because of the liberated heat and high local concentration of the nucleophiles and monomers, will become even faster, eventually is no longer manageable.  Beside potential splatter of liquid-as-such (thermal threat), the polymerisation reactions could proceed further even on and with the skin of an operator (chemical threat) if not protected properly.  Be warned, as each of the two are potentially dangerous, in addition to the potential hazard that working with ionic liquids, and chemicals in general, represent.

Wikipedia hints to this potentially relevant Critical Review doi 10.1039/C2CS15311D in Chem. Soc. Rev., 2012, 41, 1519-1537.


What is the difference between TIC and TICC? I have picked up a textbook on mass spectrometry, and I am finding it did not clearly distinguish between total ion chromatogram (TIC) and total ion current chromatogram (TICC) in the section (1.3 Ion Chromatograms) I just read. I suspect there was some sort of typo, but I don't have the experience to tell yet.

Here is an abridged excerpt from the textbook that has confused me:

  The total ion current (TIC) can either be measured by a hardware TIC
  monitor before mass analysis (nA to uA range), or its equivalent can
  be reconstructed or extracted after mass analysis. [...] Thus, the TIC
  represents a measure of the overall intensity  of ion production of
  ion production or of mass spectral output as a function of time,
  respectively. The TIC obtained by means of data reduction, [...], is
  termed total ion chromatogram (TIC). The term total ion current
  chromatogram (TICC) refers to a chromatogram obtained by plotting the
  total ion detected in each of a series of mass spectra recorded as a
  function of retention times of the chromatographically separated
  components of a mixture (which is essentially implicated by: TIC).

I would like to an explanation of the difference between total ion chromatogram (TIC), and total ion current chromatogram (TICC).

Ok, I had to read the actual text on pages 11 and 12 of the textbook link to understand.

TIC - In the olden days of mass spectrometers there was a way to measure the ion current from the sample (all masses). This total current is the current of all masses that is being injected into the mass spectrometer. Since the current in detector was also being measured, the spectra is plotted as the ratio of currents vs the mass (scan time).

TICC - The current from the sample is ignored and the spectra is just plotted as current in the detector vs mass (scan time).


Is “amount of substance” the same thing as “number of moles”? Background:
Amount of substance is a fundamental physical quantity which has mole (mol) as it's SI unit. Therefore all expressions for amount of substance should have the unit mole on simplification.
If $A$ is the amount of substance (normally in moles), $m$ is the mass of the substance in a particular unit (normally in grams), $M$ is the mass per unit amount of substance (normally in grams/mole), then
\begin{align}
A &= \frac{m}{M}\tag1\label1
\end{align}
If the substance is a molecule then $M$ is known as molar mass of the substance.
But in my textbook and in many websites on the internet, I have encountered the phrase number of moles. I think it refers to the amount of substance or maybe the numerical part in the amount of substance. The formulas given there were strange.
If $n$ is the number of moles, $m$ is the mass of the substance in grams and $x$ is the atomic weight (for atoms of elements) or molecular weight (for molecules of elements and compounds), then,
\begin{align}
n &= \frac{m}{x}\tag2\label2
\end{align}
This equation is not dimensionally correct if I am right.
My questions:
First of all I would like to ask whether amount of substance and number of moles refer to the same thing, or is it that amount of substance has a unit along with a numerical value whereas number of moles does not have a unit and represents the numerical value in the magnitude of the amount of substance.
Secondly, between equations $\eqref1$ and $\eqref2$, which one is completely correct (both in meaning and dimension)?
Thirdly, since in some place I have encountered gram atomic/molecular mass in place of molar mass, I would like to know what are the differences between both and do they have the same units or different units?

"Amount of substance" is one of the seven base quantities of the SI [1]. Its dimension is noted with the symbol N, and its base unit is the mole (mol). Similarly "length" is another base quantity with dimension noted L and with base unit the meter (m).

"Number of moles of [substance X]" rigorously means the numerical value (thus dimensionless) associated with the corresponding "amount of [substance X]" expressed in moles, but from my experience it is often misused for "amount of [substance X]" (thus of the dimension of an "amount of substance"). - Think about how you would respond to the question: "How many meters are there between your house and the nearest bakery?" You could answer "35" or "35 meters", both would be grammatically correct; but strictly speaking the question expects just a numerical answer.

Note that according to the SI, a number of entities (molecules, atoms,..) is dimensionless and the unit of a dimensionless quantity is simply the number one, 1 [1]. My way of reconciling these two views (an "amount of substance" of dimension N and of unit the mole vs. a number of entities, dimensionless and of unit 1) is that the "amount of substance" refers to a combination of an amount (a number) and a given type of entity of a given substance, whereas a number of entity simply refers to the number itself. Why not consider an amount of substance as a dimensionless quantity? I think it is a reminiscence of history, at the time when chemists had to measure amount of substance without knowing how this substance was structured (but I can be wrong).

Regarding the equations, I agree with this other answer by @desc (equation 1 is correct, equation 2 isn't). Importantly: "atomic weight" is dimensionless (it is a ratio between a mass and a reference mass) and should not to be mistaken with "atomic mass". [2]

Finally, regarding the units "gram-atom" or "gram-molecule", these are outdated units which, according to SI were connected with "atomic weights" and "molecular weights" [2]:

  Following the discovery of the fundamental laws of chemistry, units
  called, for example, "gram-atom" and "gram-molecule", were used to
  specify amounts of chemical elements or compounds. These units had a
  direct connection with "atomic weights" and "molecular weights", which
  are in fact relative masses. "Atomic weights" were originally referred
  to the atomic weight of oxygen, by general agreement taken as 16. But
  whereas physicists separated the isotopes in a mass spectrometer and
  attributed the value 16 to one of the isotopes of oxygen, chemists
  attributed the same value to the (slightly variable) mixture of
  isotopes 16, 17 and 18, which was for them the naturally occurring
  element oxygen. Finally an agreement between the International Union
  of Pure and Applied Physics (IUPAP) and the International Union of
  Pure and Applied Chemistry (IUPAC) brought this duality to an end in
  1959/60. Physicists and chemists have ever since agreed to assign the
  value 12, exactly, to the so-called atomic weight of the isotope of
  carbon with mass number 12 (carbon 12, $^{12}$C), correctly called the
  relative atomic mass $A_r$($^{12}$C). The unified scale thus obtained gives
  the relative atomic and molecular masses, also known as the atomic and
  molecular weights, respectively.
  [2]

https://www.bipm.org/en/publications/si-brochure/chapter1.html
https://www.bipm.org/en/publications/si-brochure/mole.html


Excitation of an electron of H-atom from n=2 to any $n_2 $ I got taught that if an electron de-excites from any $n_2 $(where $n_2 >3$) to $n=2$, the  energy change can be represented by $$\Delta E=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \mathrm{eV}$$
What if an electron excites from $n=2$ to $n_2 $(where $n_2 >3$)?

Do we have to reverse the signs in the brackets?

Also, if it excites like I said, will it be absorbing visible light?(Because in de-excitation, it emits visible light)

The excitation of electrons are caused by photons while their de-excitation is causes the emission of photons of the same energy as the energy differemce between two states. The general expression for energy of n-th excited state of H-atom is given by
$$E_n=-\frac{13.6 eV}{n^2}$$

Hence by the conservation of energy, the energy released/absorbed is given by the formula you have written.
$$h\nu=\Delta E=E_{n_2}-E_{n_1}=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) eV$$

Hence your initial state is always $n_1$ and final state is always $n_2$. The energy will be positive if photon is absorbed and negative if it is released.

So, for specific values of $n_1$ and $n_2$ ($n_1<n_2$), if the photon released while going from $n_2$ to $n_1$ is in visible range, the photon absorbed for corresponding excitation will have same frequency, and hence it has to be in visible range.

Please note that there are selection rules for the transitions, determined by the
transition moment integral. Please go through the book in refernece for details.

References:

D.J. Griffiths, Introduction to Quantum Mechanics , Pearson, 2 ed,  2004. ISBN: 9332542899


Global warming kinetics A rule of thumb in chemistry is that a reaction doubles in rate for every 10 °C increase in temperature. Considering the impact of a 10 °C increase in global temperature, would the same rule apply to kinetic weather processes – for example, would the average wind speed around the world double?

That is a good rule of thumb.  It is based on the Arrhenius equation $$k = A {\operatorname{e}^{\frac{-{E_{\text{act}}}}{RT}}}$$ You'll find that the "temperature doubles every 10 degrees" rule  only applies if the following assumptions are valid

The reaction must obey the Arrhenius equation

it must have an entropic pre-term based on collision probability
the reaction must be an activated process (e.g. it must have an activation energy)

The Arrhenius equation works well for simple unimolecular reactions like the ring opening of cyclobutene to buta-1,3-diene.

Now you'll need to ask a weather physicist or a meteorologist if weather can be treated as an activated process.  My guess is that the answer is "no", but that's just a guess.  It seems to me that there is no minimum energy required (activation energy) for the rain to fall or for the wind to blow.


Why are conformational isomers considered as isomers? I was taught that geometric (cis-trans) isomers are considered isomers because there is a high energy barrier to breaking the double bond for rotation (to convert into the other isomer).
 Given that there is a low energy barrier to rotation about a single bond, why are conformational isomers considered as isomers?

Gaurang's answer deals well with the theory part I just wanted to give a visual aid.

(Taken from Atropisomerism of naphthyl alcohol)

Let's call the rotating substituent on the naphtahlene a fan.
The fan has two black fins one red fin.
Now let us consider two conformations

  Red fin at 3'o clock

And

  Red fin at 9'o clock

If you were to take snapshots of both the conformations they look different.
Now these two isomers are different and naming them as 9'o clock and 3'o clock  is obviously very inefficient.
That's why conformational isomers and their nomenclature exists, for us to point out specific configurations of a compound.


Mechanism for reaction of aryl diazonium with potassium iodide I’m working on a synthesis problem that requires me to draw the arrow mechanism for the second half of the Sandmeyer reaction. Basically it’s the halogenation of an aryl diazonium compound. Typically it’s done using a copper catalyst, but it can also be done using just potassium or sodium iodide. Problem is, I can’t find any reference showing how this works in an arrow pushing mechanism. Does anyone know where I could find an example to show me how it works?

It follows a radical mechanism (SET  = Single Electron Transfer)


Looking for a non-volatile liquid less viscous than water I have tried searching everywhere for a stable, non-volatile, bio-safe, low viscosity (< .8cST) fluid. Everything I have come across that is low viscosity is inflammable - acetone, ether, silicone oil, fuel, etc.

I have found some confusing things in my research. Acetone has a viscosity 3x less than water but weighs 2.5 times more. I would like to know how water weighs less but flows slower. Google says viscosity is from low intermolecular forces, but I do not know how to quantify those forces or how to find chemicals that match.

I am trying to make a science project that shows the principle of siphoning, but I want a lower viscosity liquid to show how viscosity effects flow rate. I am working with glass right now and PVC tubes, but I can change my materials based on whatever low viscosity liquid you can find.

Also if anyone knows anything about siphoning, viscosity or material chemistry please let me know if there is a database I can search or if you know chemical that is less viscous than water and meets my specifications.

Is there something I can add to water to lower the viscosity aside from heating? I do not want to have boiling water in my hands. I hear there are surfactants that lower the viscosity of water, but google says surfactants increase viscosity.

I guess you could go with Peppermint essential oil. Its density is 0.898 g/mL at 25 °C. It obviously is volatile to some extent if left uncovered however its viscosity is considerably lower than water. Hope this helps you out!


Are aliphatic amines SABRE amenable? Hyperpolarization of molecules via the SABRE-SHEATH technique is a fairly new idea. Does anybody know if hyperpolarizing aliphatic nitrogens is possible? The articles I've read mostly use N-heterocyclic compounds like pyridine or imidazole as substrates. If not, why?

There have been a number of interesting papers just this year on this. Simon Duckett from York is leading the charge, and has recently demonstrated successful hyperpolarisation of a number of amines, including ammonia, as well as a number amides. Previously, a limiting factor in the success of the SABRE method has been the ability of the substrate to coordinate to the catalyst centre. Recent results show that this can be overcome using a relayed polarisation transfer method using a co-ligand.

For a couple of examples, check out:

Chem. Sci. 2018, 9 (15), 3677–3684
Sci. Adv. 2018, 4 (1), eaao6250
anything else by Simon Duckett in recent years


Why note the solvent level on chromatography paper when it is removed from the chamber? Why is it important to mark solvent level on chromatography paper when you remove it from the chamber?

Building on my comment, marking the distance that the solvent traveled allows us to calculate the retention factor $R_f$ (or apparently retardation factor, according to IUPAC meddlers who need to rename things with perfectly good names that everyone else uses). The retardation factor is the ratio of the distance traveled by the spot to the distance traveled by the solvent:

$$R_f=\dfrac{d_{spot}}{d_{solvent}}$$

If you are only performing one thin-layer chromatography experiment (TLC) ever, then you may be okay not marking the solvent distance. Since everything is on one piece of paper or one plate, you can compare readily:

From the above TLC picture, we can easily tell that unknowns A and B are probably the same compound as the standard, but unknown C is something else.

Now, what if we have two TLC plates/papers?

We can draw the same conclusions about the relationships between standard 1 and unknowns A, B, and C. Likewise, we can make some conclusions about the relationship between standard 2 and unknowns D, E, and F. However, we cannot easily determine if unknown F is the same compound as unknown A.

Determining the $R_F$ value lets us do that. Using the same solvent system, the distance traveled by the solvent and the distance traveled by a compound are both proportionally related to the length/height of the plate/paper. The solvent can travel twice as far on a paper that is twice as long. It turns out that the compound being analyzed also travels twice as far on a paper that is twice as long. The the ratio, $R_f$ is constant.

Let's drop in some numbers:

Now we can calculate the $R_f$ values for each spot:

After calculating the $R_f$ values, we can make some conclusions (assuming there are only two compounds in the entire experiment):

Unknowns A, B, D, and E are the same compound as Standard 1.
Unknowns C and F are the same compound as Standard 2.


Why is atomic orbital one electron wave function? Why distance from centre is proportional to angular wave function? Consider the following statements:

  An atomic orbital is one electron wave function $\psi(r,\theta,\varphi)$ obtained from the solution to the Schrödinger equation.

There two electrons in an atomic orbital, so shouldn't atomic orbital be two electron wave function? Or since we get three eigenvalues from Schrödinger's equation, not four, so atomic orbital is one electron wave function?

  In angular wave function plots, the distance from the centre is proportional to numerical values of $\Theta\Phi$ in that direction and is not the distance from centre of the nucleus.

Distance is from which centre if not from centre of nucleus? Also why distance from centre is proportional to $\Theta\Phi$?

Lastly, I guess, it is impossible to plot the variation of $\psi$ with $r,\theta,\varphi$, because it would require 4 axis (or four-dimension graph). Am I right?

OK, let's address these one by one.

Yes, it should be a two-electron wave function, i.e., the one that depends on 6 spatial coordinates. Just imagine the amount of trouble required to simply write it down, let alone produce a meaningful plot or do anything else with it. So we go with two one-electron functions instead (see Hartree-Fock method). That's an approximation, of course, but it would suffice for now.
Yes, in angular wave function plots, it is the numerical value of $\Theta\Phi$ that is actually plotted by means of the distance from the origin. There is no physical reasons behind this. It is not a physics question at all. It is a question about drawings made by humans on pieces of paper. We may plot whatever we like against whatever else we like. If you find the type of plot which gives you better intuition of what the spherical harmonics look like, you are welcome to use those. (Hint: people have been using these for quite a while now.)
The "not the distance" part of the quote is cryptic to me. I can only guess it is a botched attempt to point out that the plot is not related at all to the radial part of the wave function, and hence to the distance from the electron to the origin.
Yes, that would require a four-dimensional graph, which we can't do on paper. Wait, we can't even do a three-dimensional graph on paper. That's why we resort to cross-sections and other weird plots in the first place.

So it goes.


Which plastic type is used inside dishwashers? We have a drinking bottle made partly from polypropylene that wraps when washed in a dishwasher. We would like to change the plastic type so that it can withstand in dishwasher.

I know for a fact that my dishwasher at home has a lining inside made from polypropylene. After googling this  I have found a few sources that suggest that dishwashers would use a filler material in the PP called "calcium carbonate".

I would like to hear from someone who actually with confidence can tell us:

What is the usual plastic type used inside dishwashers? How is this plastic different from "normal" PP? Will we be able to use the same mold for this kind of plastic as "normal" PP?

One possible reason for the different behaviour of the drinking bottle, compared to the lining of the dishwasher may be indeed the different composition of the polypropylene deployed.  The part of poly indicates that many hundreds and thousands smaller propylene molecules are brought together to constitute the polypropylene.  Yet there are different ways this may happen, as this picture (source) depicts:

Along the polymer chain, there a methyl groups (marked as $\ce{CH3}$) occupying more volume than hydrogen atoms (marked by $\ce{H}$). It is imaginable that all of these methyl groups are attached on one side of the chain only (the upper picture), or consistently "once on one, the next time on the opposite side" (the lower picture).  Not shown here, there may be a randomized distribution, too, where you can't predict on which side the next methyl group will be attached (not shown here) as an intermediate situation.

Such alterations in the molecular structure may alter how individual polymer molecules interact with each other.  It influences if the polymer becomes more crystalline, or more glass-like; and by the same token, it influences the melting point of the material.  Indeed, the reference states melting points of 130 degree Celsius (266 Fahrenheit), as well as 171 degree Celsius (340 Fahrenheit).

Now think about the production of the bottle: it is done by (blow) molding.  A manufacture aims to produce a high number of bottles per unit of time, and preferentially at a low cost to spend, for example to heat the polymer that it becomes flexible enough to form it.  So it is better to select a polymer with a lower melting point, and by admixing / blending with other polymers to polypropylene, the melting point probably may be lowered further.

On contrast, a manufacturer of a linings of dishwashers equally takes into account the processes while the dishwasher is used: chemicals (detergent), hot water during the washing, and heating to eventually dry the tableware.  So the selection prefers a polymer that tends to be more crystalline and hence exhibits a higher melting point.

You may encounter situations showing you that this comes with an disadvantage, too:  The more crystalline, high melting polypropylene liner often is more brisk than the more glassy, low melting polypropylene of your bottle.


Computed partial charges for protonated amine: discrepencies between Gasteiger and other methods Consider the following molecule:

SMILES: COC(=O)[C@@H](c1ccccc1)[C@@H]1CCCC[NH2+]1

I would expect the computed partial chage on the amine to be positive, due to its protonation state.

When using Gasteiger charges calculator, it is affected a positive charge of 0.2422.

However, when using more sophisticated methods (AtomicChargCalculator or Antechamber's AM1-BCC implementation) it is affected a negative charge (-0.649), in  these results for example.

It seems the charges computed for the other atoms are roughly similar, as shown in this graph (x-axis: AtomicChargCalculator charge, y-axis: Gasteiger)

Does this amine really have a negative partial charge ? Why the discrepency ? Is there something I am missing ?

tldr - atomic partial charges can be non-intuitive

Formal charges in a Lewis structure don't always match up with partial charge assignment methods.

Typically, one wants 'accurate' partial charge models, for some definition of 'accurate' - but what that means can vary considerably. I explained some of this in a related question.

While I haven't worked extensively with AM1-BCC, most of the common partial charge methods either attempt to divide the electron density, assign charges based on best fits to the electrostatic potential at the van der Waals surface, or approximate one or both of these concepts. The electronegativity equalization method (EEM) server you linked mentions fits to electrostatic potential models.

I think you have two parts to your question:

1) Why does Gasteiger give a different answer than AM1-BCC and EEM?

This one is easy. The Gasteiger model uses the molecular connection table (i.e., the valence bond diagram) and thus considers the initial formal charge on the nitrogen. AM1-BCC begins with a quantum calculation, and EEM is attempting to approximate a quantum calculation. You can't specify a formal charge on an atom to quantum mechanics -- the best you can do is express that the molecule is a cation and let the electrons go wherever.

2) Why do AM1-BCC and EEM methods give a negative charge to the nitrogen?

This is less clear to me, but consider that most of these models are attempting to fit point charges at an atomic center to best fit the electrostatic potential on the molecular van der Waals surface. In this molecule, the nitrogen is buried beneath four bonds - two to external hydrogen atoms that are assuredly positive partial charges.

There's ample literature that assigning the point charge to such a buried atom is very difficult because the nitrogen doesn't really have exposed VdW surface.

I haven't taken much time to look at other charge models, but I can tell you that the N+ in your structure does have a more positive charge  than the corresponding neutral N when you strip off the extra proton (i.e, B3LYP RESP gives -0.194 for NH2+ and -0.755 for NH).

That's a very quick, rough calculation, but it should indicate that relative partial charges are often better considerations. In other words, this nitrogen is more positively charged than the neutral analogue.


Green Potassium Manganate and Sodium Hydroxide Equation When a sample of green potassium manganate (VI), $\ce{K2MnO4}$, is added to aqueous sodium hydroxide, a brown black solid A and a purple solution of B are obtained.

I know that the brown black solid A is $\ce{MnO2}$, but I can't seem to write out the entire equation.

Can someone help me with this?

Thanks! Really appreciated.

$$\ce{ 3K2MnO4 + 2NaOH + 2H2O -> MnO2 +2NaMnO4 + 6KOH }$$

As you said the brown solid A is $\ce{MnO2}$ while the purple solution is due to the permanganate ion .


Calculate pH of a weak acid and weak base neutralization reaction I'm an industrial pharmacist needing some help with some chemistry basics.

I'm trying to calculate the theoretical pH of an effervescent preparation containing 2 mmol citric acid ($\ce{H3C6H5O7}$), 4 mmol sodium bicarbonate ($\ce{NaHCO3}$), and 0.4 mmol sodium carbonate ($\ce{Na2CO3}$). These powders are dissolved in 100 ml of pH 6.0 water.

Citric acid: $\mathrm{p}K_\mathrm{a1}$ = 3.13, $\mathrm{p}K_\mathrm{a2}$ = 4.76, $\mathrm{p}K_\mathrm{a3}$ = 6.39
Sodium bicarbonate: $\mathrm{p}K_\mathrm{a}$ = 10.329, 6.351 (carbonic acid)
Sodium carbonate: $\mathrm{p}K_\mathrm{b}$ = 3.67

I don't know how to calculate pH in a case like this, where both acid and base are weak and polyprotic. If someone can do an example calculation, I'd really appreciate it.

You have a system of an acid, an ampholyte and a base. As you are interested in the pH value, your main question is what does "produce" or "consume" them. The proton concentration for your specific system is (citric Acid, sodium Bicarbonate, sodium Carbonate):
$$\ce{[H+] = [OH-] + \underbrace{[H2A-] + 2[HA^2-] + 3[A^3-]}_{citric acid's bases} \underbrace{+ [B^2-] - [H2B]}_{bicarbonate's deprotonated\\ and protonated form} \underbrace{- [CH+] - 2 [CH2^2+]}_{carbonate's sinlgy and\\ doubly protonated form}}$$

The multipliers arise from the fact, that a n-times deprotonated/protonated ion yields/consumes n protons.

Now the question is, where you get these concentrations from. It is by solving a linear equation system for each subsystem.

Citric acid, to be deprotonated three times:
\begin{align}
\ce{[H3A] + [H2O] &<=> [H2A-] + [H3O+]\\
[H2A-] + [H2O] &<=> [HA^2-] + [H3O+]\\
[HA^2-] + [H2O] &<=> [A^3-] + [H3O+]}
\end{align}
This gives you the following linear equation system
\begin{align}
k_{a1} &= \ce{\frac{[H2A-][H3O+]}{[H3A]}} \\
k_{a1} \ce{[H3A]} &- \ce{[H2A-][H3O+]} = 0\\
k_{a2} &= \ce{\frac{[HA^2-][H3O+]}{[H2A-]}} \\
k_{a2} \ce{[H2A-]} &- \ce{[HA^2-][H3O+]} = 0\\
k_{a3} &= \ce{\frac{[A^3-][H3O+]}{[HA-]}} \\
k_{a3} \ce{[HA-]} &- \ce{[A^3-][H3O+]} = 0\\
\ce{[H3A] + [H2A-] &+ [HA2-] + [A3-] = [H3A]0} = c_1\\
\end{align}
Which solves to:
\begin{align}
\ce{[H3A]} &= \frac{c_1 x^3}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[H2A-]} &= \frac{c_1 k_{a1} x^2}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[HA^2-]} &= \frac{c_1 k_{a1} k_{a2} x}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[A^3-]} &= \frac{c_1 k_{a1} k_{a2} k_{a3}}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3}
\end{align}

Sodium bicarbonate, to be protonated or deprotonated:
$$
\ce{[HB^-] + [H2O] <=> [B^2-] + [H3O+]\\
[HB^-] + [H2O] <=> [H2B] + [OH-]}
$$
\begin{align}
k_{a4} &= \ce{\frac{[B^2-][H3O+]}{[HB-]}} \\
k_{a4} \ce{[HB-]} &- \ce{[B^2-][H3O+]} = 0\\
k_{b1} &= \ce{\frac{[H2B][OH-]}{[HB-]}} \\
k_{b1} \ce{[HB-]} &- \ce{[H2B][OH-]} = 0\\
\ce{[H2B] &+ \ce{[HB-] + [B^2-] = [B]0}} = c_2\\
\end{align}
Which solves to:
\begin{align}
\ce{[H2B]} &= \frac{c_2 x k_{b1}}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
       = \frac{c_2 x k_{b1}}{x k_{b1} + k_W + k_{a4} k_W/x} \\
\ce{[HB-]} &= \frac{c_2 x k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
       = \frac{c_2 k_W}{x k_{b1} + k_W + k_{a4} k_W/x} \\
\ce{[B^2-]} &= \frac{c_2 k_{a2} k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
        = \frac{c_2 k_{a2} k_W/x}{x k_{b1} + k_W + k_{a4} k_W/x}\\
\end{align}

Sodium carbonate, to be doubly protonated:
$$
\ce{[C] + [H2O] <=> [CH+] + [OH-]\\
[CH+] + [H2O] <=> [CH2^2+] + [OH-]}
$$
\begin{align}
k_{b2} &= \ce{\frac{[CH+][OH-]}{[C]}} \\
k_{b2} \ce{[C]} &- \ce{[CH+][OH-]} = 0\\
k_{b3} &= \ce{\frac{[CH2^2+][OH-]}{[CH+]}} \\
k_{b3} \ce{[CH+]} &- \ce{[CH2^2+][OH-]} = 0\\
\ce{[C] &+ [CH+] + [CH2^2+] = [C]0} = c_3\\
\end{align}
Which solves to:
\begin{align}
\ce{[B]} &= \frac{c_3 (k_W/x)^2}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\
\ce{[BH+]} &= \frac{c_3 k_{b2} k_W/x}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\
\ce{[BH^2+]} &= \frac{c_3 k_{b2} k_{b3}}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2}\\
\end{align}

Putting it all together (it looks a bit different, though, as it is copied from a simplified form of Mathematica):
$$x=\frac{k_W}{x} +
c_1 \frac{3 k_{a1} k_{a2} k_{a3}+2 x k_{a1} k_{a2}+x^2 k_{a1}}{k_{a1} k_{a2} k_{a3}+x k_{a1} k_{a2}+x^2 k_{a1}+x^3} +
c_2 \left(\frac{k_{a4} k_W}{k_{a4} k_W+x^2 k_{b1}+x k_W}-\frac{x^2 k_{b1}}{k_{a4} k_W+x^2 k_{b1}+x k_W}\right) -
c_3 \frac{2 x^2 k_{b2} k_{b3}+k_W x k_{b2}}{x^2 k_{b2} k_{b3}+k_W x k_{b2}+k_w^2}$$

Solving this little monster for $x$ and calculating its negative decadic logarithm gives you the wanted pH of your system, which is 5.65.

As you mention, further thinking about the problem makes it possible to simplify it a litle bit.
\begin{align}
\ce{[H+]  &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [B^2-] - [H2B] - [CH+] - 2 [CH2^2+]} \\
\ce{[H+] &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [CO3^2-]_B - [H2CO3]_B - [HCO3-]_C - 2 [H2CO3]_C} \\
x &= \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 - c_3 x k_{a4} + 2 c_2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\
\end{align}

This does not change the resulting pH value and thus it is not more accurate in my eyes, it only simplifies (admittedly quite a bit) the equation to be solved.

If we assume your initial pH value to be exactly 6 and that is caused by CO2, then we can add it to the equation.
$$x = \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 + (c_4 - c_3) x k_{a4} + 2 (c_2 + c_4) k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} $$

Where $c_4$ can be calculated by rearranging the proton concentration equation for solely carbonic acid:
\begin{align}
x &= \frac{k_W}{x} + c_4 \frac{x k_{a4} + 2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\
c_4 &= \left(x - \frac{k_W}{x}\right) \left(\frac{x^2 + x k_{a4} + k_{a4} k_{a5}}{x k_{a4} + 2 k_{a4} k_{a5}}\right) \\
c_4 &= 3.21119 \cdot 10^{-6}
\end{align}

Without initial pH 6: 5.6467
With initial pH 6: 5.6465


Column length for baseline separation I have a GC-chromatogram with 4 retention times and peak widths and a column length.

How can I calculate the column length that I need in order to get baseline separation for all components in the sample?

I know I can calculate the resolution between the peaks and that it should be equal or greater than 1.5 for baseline separation, but in what way should I use the equation of Rs in order to calculate the best column length, using all retention times for the components?

Thanks in advance!

Resolution depends on three factors (i) efficiency N (ii) alpha or selectivity $\alpha$ and (iii) retention factor k, which the same as the capacity factor in older texts.

Guess which factor is dependent on the column length? It is only the efficiency. If you recall the resolution equation:

$$R_s=\frac{\sqrt{N_2}}{4}\left[\frac{\alpha-1}{\alpha}\right]\left[\frac{k_2}{k_2+1}\right]$$

you need the desired resolution, say 1.5, you already know $\alpha$ and ks from the experimental retention times. All you need is to solve N. You would also know that N scales linearly with the column length. With these hints you can estimate the necessary column length.

Other ways to increase resolution, if it were a research problem: You would use a smaller thermal gradient, change temperatures, and lastly change the column length.

In the real world, people would use brute force, i.e. they would change the column, because it's the selectivity which affects the resolution the most. For example, pharmaceutical companies test about 8-16 columns at once (HPLC). The column which separates a given sample with their fixed mobile phases is chosen for further optimization.


What does gases at constant pressure mean? My teacher told me that if we keep some weight on a piston containing 2 different gases, both gases will be at constant pressure, and hence the volume occupied by each gas will be proportional to number of moles of each gas according to the ideal gas law ($pV=nRT$)

My doubt is that won't the 2 gases exert different pressures and the sum of the two different pressures will be equal and opposite to the pressure exerted by the weight? Also, if the two gases exert different pressures, then how can we say that both gases are at constant pressure? Do gases at constant pressure exert equal pressure?

Without a clearer setup it's difficult to say whether someone misspoke, whether you misunderstood precisely what your teacher was saying, or whether your teacher was incorrect.

If the setup is that there is a single piston that contains two gases, and a weight is placed upon it, then:

The volume of each of the gases in that piston is equal to the entire volume of the piston.  The volume of any gas is the entire volume of its container.  If the gases are ideal, then the volume of the piston will be the sum of the volumes of the piston if each of the gases were kept in it individually.
The pressures of the two gases sum to the pressure exerted on the piston by the weight (and the outside atmosphere.)  The pressures will be constant as long as the amount of gas present, the applied pressure, and the temperature are kept constant.

If the setup is that there are two pistons that each contain a gas, and an identical weight is placed upon each (or a single piston filled with two gases, one at a time) then:

The volume of the two pistons will be proportional to the number of moles of ideal gas that it contains, as long as both are also at the same temperature (we're already assuming the same pressure due to the identical weight.)
The pressures of the gases in the two pistons will be equal, because they are opposing the same weight + outside atmospheric pressure.

As to your other questions:
Gases at different pressures do indeed exert constant pressure, assuming things like constant temperature, constant volume, and constant number of moles of gas.  The pressure of a gas is the result of huge numbers of gas particles striking the walls of their container every second.  The rate at which those particles strike and rebound and the force with which they do so determines the pressure.  Because there are sooooo many gas particles in any reasonable quantity of gas, the fluctuation in the rate and intensity of those collisions is nearly constant (to many, many decimal places.)

Gases at constant pressure do not necessarily exert equal pressures.  If I pump up the front tire of a bicycle to twice the pressure as the back tire, each has a constant pressure, but they are different.  If I put two moles of gas A and one mole of gas B into a sealed container, the pressure due to A will be twice that of the pressure due to B, but both will be constant, as will their sum, the total pressure of the container.


Conjugate addition vs Electrophilic addition I was just doing an organic chemistry problem from an online source recently, on the topic of conjugate additions (or 1,4-additions). Below is an image illustrating the problem.

I chose the upper route while the solution manual claims that the answer is the lower route. My rationale for choosing the upper route was that perhaps the alkene is not nucleophilic enough to react feasibly in an electrophilic addition with $\ce {HCl}$. This is due to the presence of the electron-withdrawing carbonyl. Thus, it is important that we first get rid of the carbonyl by using the ethyl Grignard. Then, we can easily add the $\ce {HCl}$ for the electrophilic addition.

Upon reconsidering my reasoning, perhaps I was too narrow-minded in saying that the $\ce {HCl}$ must add in the fashion of electrophilic addition. Perhaps, conjugate addition could also take place. And also, by choosing the upper route, I cannot assure that my chlorine atom would end up in the desired position due considerations of the stability of the carbocation intermediate. Thus, I agree that my choice is incorrect.

The correct choice is the lower route and this does ensure that the $\ce {Cl}$ is in the right place in the final product, as it adds via a conjugate fashion. But how is the writer of the solution so sure that the $\ce {HCl}$ would add in a conjugate fashion, instead of via an electrophilic addition?

Grignards unless modified with Cu(I) salts add 1,2 not 1,4 so the EtMgBr will add to the ketone to give the t-alcohol and leaving the double bond intact.

Consider then the HCl step - if it is strong enough to protonate the double bond leading to Cl addition then it will protonate the t-alcohol in preference leading to elimination or t-Cl formation. So the answer is to do the reaction on the double bond first then the Grignard addition. Protonation will occcur at the carbonyl oxygen but you have to remember that the double bond is in conjugation with the carbonyl system. This polarises the double bond, leading to Cl- attack at the end of the system.

The Grignard is more nucleophile than base so you will get majority addition to the ketone though some deprotonation/elimination may occur depending on the reaction conditions.