When are all cannabinoids infused into grape seed oil using non-decarboxylated, dried plant matter? I am unsure which is the correct terminology, infusion or extraction? For all practical purposes I will continue using infusion here. (What is the difference between the two?)

When making the infusion1 I'm using a ratio of 2 oz of dried, but not decarboxylated plant matter to 1 quart of grape seed oil. Any brand will work. The plant matter mainly consists of the buds and colas. These have gone through the curing phase already and are dry.

I put the plant material in the oil inside of a small 1.5 qt crockpot. The last time I did this, I let the crockpot sit on high for 8 hours before switching to low for 3 days (72 hours).
A lot of tutorials and online videos will say leave it sit for 4, 8, 24, or all the way up to 72 hours and more.
How long will it actually take until most cannabinoids are infused into the oil, without infusing other, unwanted compounds like chlorophyll?

I'm assuming that there is a breaking point where the plant matter is done decarboxylating, and the oil has extracted all of the cannabinoids.

I've done this kind infusion using coconut oil and I didn't get the same potency as when I did the infusion with grape seed oil. What is the reason on a molecular level that allows grape seed oil to infuse cannabinoids more easily?

1 Legality is not an issue since I live in a state where I am allowed to grow up to 6 plants of varying maturity and also produce my own grape seed oil infusions.

The time taken (before there are diminishing returns) of extracting the herb is totally dependent on the solvent used, or in your case, the oil. If you were to use water, the answer might be days. If you used heptane, a couple of hours.

The decarboxylation and the extraction - a catch all chemical term for removal of components or a component from a particular solid liquid or gas phase, as opposed to infusion which seems to be specific to botanical extractions (so both are correct in this instance) - should be decoupled. If you start with the cannabinoidal acid and heat in solvent, you'll start extracting components at different rates, while also decarboxylating and extracting the cannabinoid. This reduces the selectivity of the extraction.

Again, depending on your chosen oil, the rate at which your desired components and undesired ones are pulled into your oil will differ. You seem to have a handle on  which of your two oils is better. I would try two experiments:

Heat the dry herb to decarboxylate; less than two hours should be ample and may even be too much. Then extract/infuse your oil. Bear in mind, the more oil you use also leaves you open to extracting more rubbish you don't want, as does extending the time. You're better cutting your time down first and monitoring potency (as extracting more undesired components will lower your potency in the final extract, so potency should increase as you bring your time down) and then experimenting on reducing the oil content until your potency decreases. A decarboxylation should take about one hour at around 110C, but it depends on the amount of herb you have and if it's whole leaf,  shredded,  milled etc. I'm talking fine material at kilo scale. Extraction under ideal conditions is done in six hours, depending on solvent.
The second option is to extract at cooler temperatures (cool temp to 50-60C absolute maximum) first to get the cannabinoidal acids. If you have the facility, add base (diluted sodium hydroxide is ideal and would take minutes . Sodium carbonate dissolved in water may work but would be slow (days?) and you'd have to monitor carbon dioxide evolution). The acids you want will form salts and extract into the water.  The neutral stuff you don't want (terpenes, waxes - long chain alkanes) stay in the oil. Separate the oil and water. You have a choice here.

Add acid to your water layer - this may cause your cannabinoid acid to precipitate, so filter it off.
Otherwise add fresh oil to the water,  then acid and your cannabinoid acid will now be in the oil after mixing.

Whatever you choose, heat your cannabinoid acid to over 100 degrees Celsius and you should be done in an hour or two. Over 120 degrees Celsius may be too much. As I mention, some experimenting will be needed but try to ensure your herbs are consistent and not mixtures of different varieties each time.

As for your oils. One is more like a liquid "fat" - the coconut oil. The grapeseed oil is still a mixture, but is primarily "polyunsaturated" ( it's actually di-unsaturated) and about 50% bigger than the main coconut oil component. These oil mixtures are not usually used in chemical extractions as they're not pure. I assume you use digestible, non-toxic solvent oils. Otherwise I'd suggest limonene.


Which compound reacts faster in the Cannizzaro Reaction? Which reacts faster in the Cannizzaro Reaction?
  a) $\ce{OHC-C6H4-NO2}$
  b) $\ce{OHC-C6H4-OCH3}$

Obviously, a better hydride releasing group will react faster. Therefore my answer was b, as $\ce{-OCH3}$ shows -I (inductive) effect as well as +R (resonance) effect, as it's in the para position. This increases the electron concentration in the $\ce{C}$ atom of $\ce{CHO}$ and due to the excess electrons, it is a better donor of an electron to the $\ce{H}$ attached to it. Thus, leaving a  $\ce{H-}$.

But the answer given is a. The answer states that $\ce{C}$ can only give electron to $\ce{H}$ when it accepts an electron from the $\ce{O}$ attached to it. For that the $\ce{C}$ should be electorn deficient. Therefore an electron withdrawing group (in this case $\ce{NO2}$, which shows -R effect at ortho and para positions) better releases hydride.

But I don't think that happens. That's because if $\ce{C}$ is electron deficient, then it wouldn't give an electron to $\ce{H}$ (less electronegative) in the first place, even if it takes an electron from oxygen (which is also tough).

Can anyone give a proper explanation to the whole problem?

The mechanism of the Cannizzaro reaction is illustrated below.

The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate.  Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon because it is now $\ce{sp^3}$ hybridized. So we need to look at the step leading to the tetrahedral intermediate - the first step.

However, if we examine some of the resonance structures for the starting carbonyl compounds (where the substituents on the aromatic ring can interact with the carbonyl through resonance) we see that the nitro compound places positive charge adjacent to the already positive polarized carbonyl carbon - a destabilizing situation (remember, destabilizing something means making it higher energy, more reactive).  On the other hand, the methoxy compound places negative charge adjacent to the already positive polarized carbonyl carbon - a stabilizing situation.

The destabilized carbonyl in the nitro compound will be more reactive towards nucleophilic attack.


Lattice energy of NaCl and MgCl In an exam paper it says that the lattice energy of NaCl is more exothermic than MgCl as the Na+ cation has a smaller ionic radius than Mg+.
Here are the resources given:

What is the reasoning for for Na+ having a smaller ionic radius even though Mg+ has a higher proton number?

Thanks.

We actually do mean $\ce{Mg^+}$ here.  The idea is to show what factors are involved when an alkaline earth metal forms stable compounds with $\ce{M^{2+}}$ ions whereas an alkali metal favors $\ce{M^+}$.  In this case, as pointed out in some of the comments, $\ce{Mg^+}$ is bulked up by the electron that remains in the relatively diffuse, loosely held $3s$ subshell.  You need to remove that electron, making $\ce{Mg^{2+}}$, to get a compact ion that gives good lattice energies in ionic crystals.


Why does electronegativity create a stronger acid, but so does greater electron distribution? I just learned that:

Acids get stronger with:

Increasing electronegativity of atom bonded to $\ce{H}$.
Size of atom bearing negative charge.
Hybridization of negative charge (more s character, $\ce{e-}$ closer to nucleus, more acidic).
Resonance stabilization (like size, greater distribution of electrons).

So basically, the  more concentrated the negative charge (think $\ce{F-}$, $-1$ and very small), the tighter it's going to "hold on" to the proton. With resonance and large negative groups, the charge is less concentrated, so it's not going to "hold on" to the proton as well.

However, electronegativity is the tendency for an atom to "hog" electrons. Therefore, the more electronegative, the more electrons there are in a similarly sized space.

More concentrated electrons = more stable base = weaker acid.

But electronegativity = more concentrated electrons (even though size trumps it).
And electronegativity = stronger acid.

Feel free to edit if you see what I'm trying to say but can say it better!

The main point about strong acids is ‘how stable is the conjugate base?’

The better stabilised the negative charge on a conjugate base of a neutral acid is, the happier the species will be to loose a proton and thus the more acidic the proton is. All of the points you noted just emphasise that.

Therefore, the only wrong assumption you are making is:

  more stable base = weaker acid.

The correct way is:

  more stable base = stronger acid.


What was the lithium concentration in 1940's 7-Up? I just came across this article in the New York Times, where lithium (or more correctly, lithium salts), a known anti-depressant had been added to beverages in the 1940's. They also mention that a study found negative correlation between the suicide rate of a county and the amount of lithium in its tap water, despite the fact that the concentration is relatively low ($\pu{<0.170 mg L^-1}$), especially compared to the doses administered in therapy.

Now I was wondering what the lithium content in those beverages was (obviously, they are not sold anymore), but couldn't find anything?

The article mentions that you can still buy lithia water ($\pu{0.50 mg L^-1}$), but I'm more interested in the 1940's 7-Up mentioned, as those probably had different food standards at the time they were made.

Given that the formula was a trade secret and that lithium was removed from the product ~70 years ago, the likelihood of this question being answered by providing a detailed formula seems remote. Note in addition that the formula might have evolved to reduce cost or improve market reception. You might therefore want to settle for an estimate of the upper limit on the lithium citrate concentration, which is relatively easy to provide.

A basic estimate can be made based on (1) the historical limits of sodium concentration in sodas; (2) the current composition of 7-up; and (3) the relative saltiness (taste) of lithium citrate versus sodium.

First one can attempt to determine what range of sodium concentration makes for a flavorful soda. Here is a quote:

  The amount of sodium in soda varies depending on the type. A well-known brand of cola has 46 milligrams of sodium in a 12-ounce can. The same sized can of root beer has 71 milligrams; a can of lemon-lime soda has 64 milligrams; a 12-ounce can of all-natural soda contains 36 milligrams. These are just examples, and the amounts in particular brands of soda may differ. It's best to check the label of your favorite soda to make sure.

This leads to a range of ~150-200 mg/L sodium concentration.

Next, consider the current formula for 7-up (in the USA):

  in 12 fl oz (355 mL):
    40-50 mg  sodium,
    60 mg potassium

The current formula contains ~110-140 mg/L sodium and ~170 mg/L potassium.

Finally, consider the saltiness of lithium citrate versus sodium. Lithium citrate was the salt form of lithium in the 7-up recipe  according to a number of sources including the Wikipedia:

  The soft drink 7Up was originally named "Bib-Label Lithiated Lemon-Lime Soda" when it was formulated in 1929 because it contained lithium citrate. The beverage was a patent medicine marketed as a cure for hangover. Lithium citrate was removed from 7Up in 1948.[Ref 4]

and  this one (which cites Ref 1):

  As to why “7Up,” C.L. Grigg never explained how he came up with the cryptic name. Several theories exist about its origin:
     "7Up was the product of seven ingredients. (Which, in a way, was at least true with regard to the classes of ingredients in that original formulation: sugar, carbonated water, essences of lemon and lime oils, citric acid, sodium citrate, and lithium citrate.)"

Lithium and potassium are among few cations with a salty flavor similar to that of sodium, albeit weaker. Citrate reduces the saltiness further. According to Ref 2, the saltiness of lithium and potassium can be estimated to be ~0.4 and 0.6 that of sodium, repectively, while Ref 3 explains that citrate has an inhibitory effect.

Consider now the potassium concentration (~170 mg/L) in the current formula of 7-up. Potassium is added as the citrate salt, and could be replaced with the lithium salt with little change in taste perception, according to what is reported on the relative saltiness of the cations and citrate salts. This suggests that if the taste of the beverage has remained reasonably constant since inception, it is not unreasonable to estimate an upper bound of ~200 mg/L in the concentration of lithium citrate (replacing the potassium salt) in the orginal recipe.

Note this is based entirely on the taste perception and not on any medical implications of consuming lithium at such concentrations. Lithia water (a brand of spring water sold in the USA) contains only ~0.5 mg/L of lithium. For reference, an online medical advice site suggests that a dose for treatment of depression is ~60 mg of lithium daily. So a dose of ~60 mg lithium citrate per serving certainly seems high relative to lithium-rich spring water but is just about a typical clinical dose.

References

Rodengen, Jeffrey.   The Legend of Dr. Pepper/7Up.   Fort Lauderdale: Write Stuff Syndicate, 1995.   ISBN 0-945903-49-9.
McLaughlin, S., Margolskee, R.F.. The Sense of Taste. American Scientist, 82, 538-545 (1994).
C.M.D. Man. Technological functions of salt in food products. in Reducing Salt in Foods, Woodhead Publishing, 2007. ISBN 978-1-84569-018-2.
Gielen, Marcel; Edward R. T. Tiekink (2005). Metallotherapeutic drugs and metal-based diagnostic agents: The use of metals in medicine. John Wiley and Sons. p. 3. ISBN 0-470-86403-6.


How to calculate the amount charge transfer from a natural population analysis? I am trying to reproduce the quantity referred to as the degree of charge transfer reported in this publication by Zhu et al. [Ref. 1] for the F4-TCNQ molecule. I am interested in section 3.2 in the main text about using the bond length as an estimate for the charge transfer amount.

In figure 3b, the authors show the correlation between the degree of charge transfer, CT$_{pop}$ from their natural population analysis compared to the approximation of the bond lengths of F4-TCNQ from equation 1, shown below.

$$\rho = \frac{\alpha_{CT}-\alpha_{0}}{\alpha_{-1}-\alpha_{0}}$$
with $\alpha_i = l_3 (l_2+l_4)$ where $l$ is the bond length, and subscript $i=0, -1, CT$ denote the netural molecule, the anion, and the compelx, respectively.

It's not clear to me, how the authors determined the values on the y-axis in Figure 3b, and there are no further clarification or references to a methodology for the values within the text or supplement information. Both axes in Figure 3b use the F4-TCNQ anion as an arbitrary reference.

I can calculate $\rho$ from the above equation, but I want to compare $\rho$ to the CT$_{pop}$ at the level of theory I am employing, verifying the accuracy of my calculations.

Through reading and googling, I believe the CT$_{pop}$ refers to a population analysis such as a Mulliken or natural population (AKA Lowdin population), referenced in Mendez et al. [Ref. 2] (bottom of page 9 in the Theory footnote). I do not know how to obtain the degree of charge transfer from the resulting population analysis.

I have one reference from Joo et al. [Ref. 3] who gives more details on a similar method, writing on page 2, left column:

The calculated atomic charges were then used to obtain the degree
of charge transfer between the donor and acceptor molecules,
which may be defined as the sum of all atomic charges on the
donor part of the complex.

The above text leaves me to believe the method of determining the degree of charge transfer is as follows:

Calculate a population analysis of the DA (donor and acceptor) complex to determine the atomic charge
Determine the atomic charge of each of the fragments D and A by summing the atomic charge for the D and A fragments, respectively
Report the atomic charge of fragments as the degree of charge transfer.

The last aspect is Joo et al. [Ref. 3] reports the value for the electron donor while Zhu et al. [Ref. 1] indicates the value for the electron acceptor, which is F4-TCNQ.

Does this procedure sound reasonable? Any thoughts or comments would be appreciated.

TL;DR
How can I reproduce the values shown in Figure 3b on the y-axis, the degree of charge transfer from a natural population analysis?

References

Zhu L; Kim E.-G; Yi Y; Bredas J.-L. Charge Transfer in Molecular Complexes with 2,3,5,6-Tetrafluoro-7,7,8,8- Tetracyanoquinodimethane (f4-Tcnq): A Density Functional Theory Study. Chemistry of Materials 2011, 23 (23), 5149–5159 DOI: 10.1021/cm201798x.
Méndez Henry; Heimel, G.; Winkler, S.; Frisch, J.; Opitz, A.; Sauer, K.; Wegner, B.; Oehzelt, M.; Röthel Christian; Duhm, S.; et al. Charge-Transfer Crystallites As Molecular Electrical Dopants. Nature Communications 2015, 6 (1) DOI: 10.1038/ncomms9560.
Joo B; Kim EG. Model-Independent Determination of the Degree of Charge Transfer in Molecular and Metal Complexes. Chemical Communications (Cambridge, England) 2015, 51 (81), 15071–15074 DOI: 10.1039/c5cc05274b.

I will evade answering the main question because I am not feeling expert enough, but I want to point out that Natural population analysis (NPA) is very different (and better) from the Loewdin one. Loewdin is computed from orbitals of a special basis set (although it comes from the density, it is still free-atom-derived basis set which is used to obtain the density matrix). NPA (part of NBO) is from density matrix which is adjusted to the particular molecular environment, hence better representing the "real" atomic population. NPA is available in the commercial NBO program, old version of which is also included with Gaussian package (maybe also in other packages, I don't know). You can also try AIM charges which are absolutely basis-independent (if Your density is good enough).

Personally I would think the analysis You proposed is fine with NPA or AIM charges, but I state again that I am not an expert in the topic.

See https://www.q-chem.com/qchem-website/manual/qchem43_manual/sect0061.html and also http://www.huntresearchgroup.org.uk/teaching/teaching_comp_chem_year4/L7_bonding.pdf .


How to predict (know) the products of the reaction SnO+NaOH? I would like to know how to complete the reaction $\ce{SnO}+\ce{NaOH ->}$, how can I form a salt and water from this?

I've seen that the complete reaction is:

$$
\ce{SnO + 2NaOH ->Na2SnO2 + H2O}
$$

How to know what will be formed?

I've discovered the answer, at least I guess...

The given reaction is

$$
\ce{SnO + 2NaOH -> Sn^{2+} + O^{2-} + 2Na+ + 2OH-}
$$

The hidroxy breaks in $\ce{H2O + O^{2-}}$. This released oxigen will get together with the oxigen from the $\ce{SnO}$, to form $\ce{O2^{2-}}$.

$$
\ce{SnO + 2NaOH -> Sn^{2+} + \underbrace{O2^{-2}}_{\textrm{nox = -4}} + 2Na+ + H2O}
$$

The oxidation state of this molecule is -4. Performing the double replacement reaction we get

$$
\ce{SnO + 2NaOH -> SnO2^{-2} + 2Na+ + H2O}
$$

And finally,

$$
\ce{SnO + 2NaOH -> Na2SnO2 + H2O}
$$

What is already balanced.


What does this three-number isotope notation mean? What does the superscript $3+$ mean?
$$_{21}^{45}\mathrm{Sc}^{3+}$$

I understand that $45$ is the atomic mass number and $21$ is the number of protons. Is the $3+$ to indicate that it is a particular isotope, or does it have some other meaning?

The superscript '$3+$' denotes a normalised electric charge. Charges are quantised, that is they take on only certain discrete values.

Neutrons do not have a net charge ($q = 0 $).
Electrons and protons have the same absolute charge[footnote]. They have opposite, arbitrary signs. By historical convention, protons are assigned positive charges.

Hence,

$$\text{electric charge} = \text{no. of protons} - \text{no. of electrons.}$$

For shorthand purposes, we usually normalise the electric charge by dividing the total electric charge of our agreed system with the elementary charge $e$.[footnote] This provides us an opportunity to operate with integer values in chemistry; there are fractional charges for quarks which were discovered later but these particles cannot be observed separately by current understanding.

Note also that we write '$3+$' instead of '$+3$' since we are, again by convention, counting charges (that is $+ + + $), not denoting the mathematical normalised charge itself.

As of 2011, this is verified up to essentially $1$ part in $10^{21}$. The absolute charge of an electron (or charge of a proton) is called the elementary charge $e$, and

$$e=1.6021766208(98) \cdot 10^{-19}\ \mathrm{C}$$

where the parentheses designate a standard uncertainty (assuming a normal distribution, $\approx 0.68$ probability, coverage factor $k =1$).

ref: Mohr, P. J.; Newell, D. B.; Taylor, B. N. 'CODATA Recommended Values of the Fundamental Physical Constants: 2014'. Reviews of Modern Physics, 2016, 88 (3). DOI: 10.1103/RevModPhys.88.035009, freely available on arXiv here, also archived link.

  Disclaimer: originally, I claimed that proton–electron charge symmetry is verified up to $1$ part in $10^{26}$. This was wrong, and based on a naive approach. Again, see How accurately is it known that protons have the same charge as electrons? for a correction.


Calculating the molar mass of a solute from freezing point A $\pu{4.305 g}$ sample of a nonelectrolyte is dissolved in $\pu{105 g}$ of water. The solution freezes at $\pu{-1.23^\circ C}$. Calculate the molar mass of the solute. $k_f$ for water is $\pu{1.86 C/m}$.

My steps:

$$1.23 = 1.86 \times \frac{x}{1.05},$$

where $x = y / 4.305$.

$$0.66 = \frac{x}{1.05}$$
$$x = 0.69$$
$$0.69 = y / 4.305$$
$$y = \pu{2.989 g}$$

But I know this is not correct.

The freezing point depression is based on the molal concentration (moles of solute per kg of solvent).
$$\Delta T_f = -k_f \cdot m$$
You know the freezing point depression of the solution and the cryoscopic constant, so you can calculate the molality:
$$m = -\dfrac{\Delta T_f}{k_f}=-\dfrac{\pu{-1.23 ^\circ C}}{1.86 \frac{\textrm{kg}\cdot ^\circ \text{C}}{\textrm{mol}}}=\pu{0.66 molal}$$
Since molality is the number of moles of solute divided by the mass of the solvent in kg, you can calculate the moles of the solute:
$$n_\textrm{solute}= \pu{0.66 m} \cdot \pu{ 0.105 kg}= \ ...$$
Divide your mass of solute by the moles of solute and you will get the molar mass (and not 2.989 g).


Why is sulfur the central atom in the Lewis structure for SO2? So apparently the Lewis structure of $\ce{SO_2}$ is

It was my understanding that the central atom is the one that is more electronegative. And an atom is more electronegative the closer it is to Fluorine (top-right). Oxygen is definitely closer to fluorine than sulfur is. Then, why is sulfur the central atom?

Also, one thing that has been bugging me for a while: if there are two or more atoms for the central element, how do you make the Lewis structure?

I think you messed up something. Because it is exactly opposite.

"Central part of atom is normally least(not the most) electronegative atom i.e. S in this case.


Carbocation Stability: Tertiary versus Secondary Benzyl for Hydrobromic Acid Addition I haven't been able to find anything concrete or consistent on this topic.
For the following reactions, which product would be favored, assuming that they both have to go through a carbocation intermediate?

What I've gathered so far:
The electron withdrawing group (-NO2) would destabilize the benzylic carbocation (through loss of electron density for stabilization), and would at least help push the reaction in the direction of the tertiary carbocation.

Thanks for any and all help.

In the first instance, bromine will attach itself on the benzylic carbon as the benzylic carbocation formed in this case would be more stable than than the tertiary carbocation due to the mesomeric effect.

However in the second case, the benzylic carbonation formed would be destabilized due to the -M effect of the nitro group, like you pointed out. So bromine would favor the the tertiary carbocation which in this case would be more stable due to hyperconjugation and inductive effects.


Is there a (carbon) heterocyclic compounds named from number three to infinity? For example, oxepane is a 7-atom ring. Can there be 8-atom ring or 9-atom ring … or million-atom ring to infinity?

To my knowledge, the Hantzsch-Widman nomenclature defines stems for ring sizes up to 10 (ecine for unsaturated, ecane for saturated heterocycles).

Oxepane is not the general stem for a seven-membered heterocyclic ring, but the suffix for one containing an oxygen atom.


Removing diamonds from a 14 karat gold ring with acid I have a 14 karat gold ring with diamonds. I want to weaken the metal (to get the diamonds out without damaging them).

If I dip the ring in sulfuric acid, will I be able to dissolve the base metal in the alloy leaving the gold in place?

I think that asking a jeweler is the best thing to do because gold is not something to throw away. However talking about the chemical concern, sulfuric acid can't oxidize gold, if the base metal (you should specify where and how it is) is iron theoretically you will be able to dissolve it with sulfuric acid:
$$\ce{Fe + H2SO4 ->FeSO4 + H2 ^}$$

Something similar can happen with copper and tin but I think that temperature of the acid is a determinant parameter for these reactions.

If the base metal is inside the alloy is quite unlikely that you can affect it because gold is not a porous material, so you can only interact with the superficial layer.


Is it possible to produce sulfuric acid at large scale by oxidizing and hydrating pyrite? I was reading The Knowledge by Lewis Dartnell, which is a book about recreating our technology from scratch should we experience a civilization-ending apocalypse. The author thinks that acquiring vanadium catalyst for contact process is too difficult, so his recommendation to produce sulfuric acid is to: (1) bake pyrite, (2) react the sulfur dioxide produced with chlorine gas (electrolyzed from brine), using activated carbon as the catalyst, (3) then react the resulting sulfuryl chloride with water to produce sulfuric acid.

I don't know how hard acquiring vanadium is, but I think it will be easier to replicate the same process happening in acid mine drainage:
$$\ce{2 FeS2 + 7 O2 + 2 H2O → 2 FeSO4 + 2 H2SO4}$$

The iron(II) sulfate can then be baked to produce sulfur trioxide, which can also be processed into sulfuric acid.

I think this method would have been simpler. Am I correct? Is there another simple method of producing sulfuric acid?

Per a source, there is an iron bacteria path acting on FeS2 to create sulfuric acid:

Thiobacillus spp. are participants of active oxidative corrosion. They oxidize a range of sulfur compounds to sulfuric acid, and the acid released from the cells may attack alloys.

My experience, however, suggests that the easiest way to produce H2SO4 is by the action of a concentrated solution of Oxalic acid (used in commercial products) on magnesium sulfate (FeSO4 may also work, but likely leads to an iron impurity).
Per a source (old Wikipedia commentary) to quote:

Relationship to oxalic acid
The dissociation of protons from oxalic acid proceeds in a stepwise manner as for other polyprotic acids. Loss of a single proton results in the monovalent hydrogenoxalate anion HC2O4−. A salt with this anion is sometimes called an acid oxalate, monobasic oxalate, or hydrogen oxalate."

As such expected reactions:
$\ce{MgSO4 + 2 H2C2O4 -> Mg(HC2O4)2 + H2SO4}$
$\ce{MgSO4 + Mg(HC2O4)2 -> 2 MgC2O4 (s) + H2SO4}$
Be careful, the created Sulfuric acid could be very strong depending on the level of hydration.
Warning: Attempting (in the presence of H2C2O4) to produce fuming sulfuric by boiling down the solution can lead to a spontaneous ejection of the acid (reported in an old thread of mine).


What is LSD-1, …, LSD-24, LSD-26, …? LSD (lysergic acid diethylamide) was formerly often referred to as ‘LSD-25’.
If it really is a (Albert Hofmann / Sandoz)'s code name (and not a combination of ‘LSD’ and usual LSD (Delysid) dosage numeral ‘25’ μg; see also Quora: What does the "25" in "LSD-25" mean?), are some other substances from the series LSD-1, LSD-2, … etc., being at a time probably a part of industrial secret, currently already known?
UPDATE
Summary
Identify at least one another Sandoz LSD-x, where x ≥ 1 and x ≠ 25 (with good sources or at least proven insider knowledge).

The number "25" indeed comes from the fact that it was the 25th compound of the lysergic acid amide series or lysergamides. I found an article* by Frank J. Ayd, Jr. & Barry Blackwell who accounted Alexander Hoffman's investigation in his LSD synthesis:

A number of pharmacological experiments were carried out by Professor
Ernst Rothlin with this new compound, which was given the laboratory
code name LSD-25 because it was the twenty-fifth compound of the
lysergic acid amide series.

You can check the other compounds of the series in the Wikipedia article linked above. Most of them have their own specific names and not LSD-1, LSD-2 etc.

I also did find something in an e-book(Ref) which says that the number has something to do with date:

Dr. Hoffmann described the results of accidental ingestion of
LSD. It was in Switzerland in 1943, five years after Dr. Hoffmann and
an associate has synthesized lysergic acid diethylamide tartarate
(from ergot). The substance was labeled "25" because of the date: "May
2, 1938" (the date when LSD was synthesized(?)).

Ref.: LSD-25: a Factual Account: Layman's Guide to the Pharmacology, Physiology, Psychology, and Sociology of LSD, Louise G. Richards, U.S. Bureau of Narcotics and Dangerous Drugs, 1969
ಠ_ಠ

*I was researching about something else i.e "serendipity in Chemistry". While I was surfing the Internet, I stumbled upon this article and from that, I came to know about LSD, searched for the sources of their information and from there I got access to the above article and then proceeded to compile this answer without any prior intentions. What a beautiful irony!


Why are excited Slater determinants used to describe electron correlation? Usually the first step in describing the electronic wave function of a molecule or atom is to describe it with a single Slater determinant. I get that this is an independent particle approximation, and that in the context of Hartree-Fock, each electron moves in the average field of the other electrons. That's fine.

But when we correlate the electrons, this is done by generating excitations out of the Slater determinant, and then taking weighted combinations of them. For example, configuration interaction mixes the ground state Slater determinant with singly, doubly, triply, etc excited determinants. A fancier way of doing this is coupled cluster, where cluster operators generate excitations out of the Slater determinant.

Hermann Kummel says that we can think about electron correlation this way:

  "The first thing one may imagine happening is that two particles mutually interact, thereby lifting themselves out of the Fermi sea, so that after the interaction both are in unoccupied orbitals"

Kümmel, Hermann. "Origins of the coupled cluster method." Theoretica chimica acta 80.2-3 (1991): 81-89.

Why can we (or should we) think about electron correlation that way?

It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!

The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra chemistry that we want to add is electron correlation. The extra maths we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.

(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)


Double Replacement Reactions form Precipitates Why do most all double replacement reactions result in a precipitate and an aqueous molecule? Is it because the individual solute molecules always form stronger bonds such as how in single replacement the molecule with more energy bonds with the molecule of the opposite charge?

Precipitation of silver chloride is a common example of a double replacement reaction. e.g. $$\ce{AgNO3~_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3~_{(aq)}}$$

Initially all four ions are present in solution. When the solutions are mixed the silver and chloride ions bond together and precipitate, leaving the sodium and nitrate ions remaining in solution along with a small number of silver and chloride ions because the precipitation is an equilibrium so a small number of dissolved ions will still be present (thanks to @Martin for pointing this out). The reaction happens because the precipitation of silver chloride removes these ions from the solution equilibrium, driving the reaction to completion.

Silver chloride is insoluble because the $\ce{Ag-Cl}$ bond has a high degree of covalent character due to the high polarizing ability of the silver cation and the high polarizability of the chloride anion and the similar size of both ions which allows a shorter bond length and so a stronger bond requiring more energy to break apart in order to dissolve the solid. See this question for more detail on this point.


Round Bottom Flask with Inner Coating I am using a rotary evaporator to separate DCM (dichloromethane) from some nasty tars.
Before the evaporation is completed, I stop the process and collect the remaining slurry.
The slurry is then positioned in a constant temperature oven, for a secondary evaporation. After cooling in a desiccator, the tar is measured.

The problem is that some of the tars remain deposited on the flask's wall. I can easily remove them with a soft brush, however this is a problem for 2 reasons:

Some of the tar is lost, retained on the brush itself ( I require a very high precision since I work with samples of 100 or so mg)
The amount of DCM for secondary evaporation is increased (I am trying to reduce that amount as much as possible)

Is there a round bottomed flask that has a TEFLON like coating on the inside, that can prevent the tar deposition on the walls?

There seem to be a lot of products that are referred as " Flask with Plastic Safety Coating". On which side is the coating applied (I assume is the outside)

I've seen a couple of companies making flasks entirely from PTFE or PFA, which might be suitable. They may not hold up to heating with internal vacuum though, for example, check out the notes here.

And yes, the typical safety coatings are on the outside to hold the glass together on fracture.


Why don't gases of elements with negative electron affinities exist as ions in nature? Fluorine has an electron affinity of about -300 kJ/mol. Lithium has an electron affinity of about 60 kJ/mol.

As I understand it when $\ce{Li^-}$ loses an electron about 60 kJ/mol is released but when $\ce{F}$ gains an electron about 300 kJ/mol is released.

Furthermore, when $\ce{Li^+}$ gains an electron about 520 kJ/mol is released because that is the first ionization energy of lithium and when $\ce{F^+}$ gains an electron about 1700 kJ/mol of energy is released because that is the first ionization energy of fluorine.

Shouldn't the cases that release the most energy be $\ce{Li}$ and $\ce{F^-}$ and so shouldn't those be the states that gases of those elements occupy?

Is it the case that $\ce{2F^-}$ is less stable than $\ce{F2}$ or something? Or is it the case that $\ce{F^-}$ could only exist in a highly artificial situation such as a magnetic contraption that trapped ions?

Why don't gases of elements with negative electron affinities exist as ions in nature?

It totally does not work like that. You can't say that $\ce{2F^-}$ is less stable (or more stable) than $\ce{F2}$; you can't compare them at all. People would say "B is more stable than A" when there is a spontaneous reaction $\ce{A\to B}$, but this is not the case here. A reaction must be balanced in charge and in all elements, otherwise it is just a bunch of letters that makes no sense. There is no reaction $\ce{F2<->2F^-}$ (either way), much like there is no reaction $\ce{F2->Au}$.

There is indeed a reaction $\ce{F2 + 2e- -> 2F^-}$, and it is quite favorable energetically (as you already know), but that's another story. You need the electrons for it, and they are just not there. Fluorine can't wrestle those electrons from another fluorine, as explained by IT Tsoi. And if you have lithium around, why, of course they would react to form $\ce{LiF}$, but that's a chemical compound, not just $\ce F$ by itself; also, at room temperature it is solid.

There is more to it. A gas (as well as a liquid or solid, for that matter) formed of negative ions alone is very much impossible. The ions will repel with such force that no material in the world will hold them even for a second. (I mean macroscopic amount, not a thin trickle of ions inside a mass spectrometer.) So you can't have fluorine in this state even if you have a stash of electrons which you are willing to share. $\ce{F-}$ ions do exist, of course, and there is nothing exotic about it (arguably, almost all fluorine in the world is found in this form), but they exist as a part of various ionic compounds, and not by themselves.


Molecular orbitals of allylic cations, anions and radicals I was reading the explanations of electronic configurations of the Allylic Radicals, Cations and Anions in my textbook. I completely understand everything, but there is one thing that confuses me. Namely, it is said that the  antibonding pi3* orbitals of allylic cations, radicals and anions have no electrons. Now my questions is, where are they? Regardless of the type of bonding, there should be 2 pi electrons in an allylic cation, 3 pi electrons in allylic radical, and 4 pi electrons in allylic anion. So the electrons should be somewhere. For example I would think that the allylic radical of pi3* is a molecule with 1 pi electron in each carbon, but because of the absence of orbital overlap they are not shared between the carbons, whereas in the case of pi2 nonbonding they are partially shared and fully shared in pi1 bonding orbitals.

In allylic species 3 orbital wave functions are interfering to form 3 molecular orbitals among which the pi antibonding molecular orbital will have the max energy. The first two mo can accomodate 4 electrons. Since pi antibonding are at a higher energy level the first 4 electrons will remain in first two mo only. Similarly in cation and radical electrons will remain in these orbitals hence the pi antibonding will always remain empty.


Why is Vapor Pressure of Liquid Equal to Ambient Pressure at Boiling? What does phase change from liquid to gas have to do with equality of vapor pressure and ambient pressure? I understand that at higher pressures more kinetic energy is needed for molecules in liquid to spread out to form bubbles of vapor since vapor has lower density than liquid which means that higher temperature is needed. But why should equality of vapor pressure and ambient pressure determine when phase change takes place?

When the vapor pressure equals the applied ambient pressure the liquid can form vapor in the bulk liquid hence "boiling".  To accurately measure this care must be taken to mix the liquid well and to prevent overheating. At lower vapor pressures evaporation is only from the exposed surfaces.


How many possible nucleotide types can be created? In the RNA world hypothesis, it has been argued that RNA is made of 4 nucleotides and it can act as a template to create another copy of itself.
And it's been claimed that we only have 5 nucleotides, 4 of which is used in RNA.
And I also searched and realized that a Nucleotide is made of a phosphate groul, a sugar, and a nitrogenous base.
My question is, in the world of possibilities, how many types of nucleotids can be created? And if it's more than 5, then why nature has only created 5 of them?

I don't know why RNAs with 5 nucleotides became so prominent in our biosphere but you can bet that it is probably because it was better (even slightly so) than other RNA models it may eventually have competed with. So in order to arrive at a nice hypothesis about why this thing exists, we need to understand what was the competition like and how would our 5-symbol code benefited against rivals.
Consider the setting we are discussing. We have self-replicating RNA strands competing for the building blocks in the primordial soup (and maybe in each other)  in order to use those to make more of itself. The single most important resource that these replicators will have is time and speed. RNA strands aren't terribly stable and will hydrolyze after a certain window of time, meaning that our critter have a lifespan. If an RNA strand is to work as a self-replicator it must be able to assemble copies of itself within its lifespan otherwise its code will die with it.
We know that the supply of nutrients must have been quite limited because if you put mindless replicators that can eat each other in a soup of free dinner they reproduce at an exponential rate until the supply and demands are met and the population reaches and equilibrium appropriate to the free energy available. A population in this level leaves very little free nutrients around.
So in any given pool teeming with RNA critters, life must have been hard and dinner difficult to come by, because the time of bounty, when life was rare and food was abundant was followed by "that time when everyone ate like nutjobs until all the food was in their bodies rather than out there".
We also have reason to accept that there must be an upper limit to the length of the self-replicating strands. Longer strands are more prone to hydrolysis, oxidation and radiation. Environmental hazards that may lead to mutations and the rupture of the strand, leading to the corruption of the code and potentially the loss of its self-replicating quality, effectively resulting in its death. Longer chains have a shorter lifespans and take longer to copy than longer chains.
These two things generate selective pressures that might help us explain why the amount of nucleotides isn't larger. On one side, individuals with too few symbols (shallow code) will require longer chains to hold their code. This will shorten their lifespans and make their code harder to copy, since it is longer. So going too shallow will hamper an individual's ability to reproduce by increasing the time required to make copies and shortening the lifespan. Eventually, there must be a specific chain size above which the average copy/lifespan is less than 1. With a shallow code only the simplest individuals can reproduce reliably and that leads to low genetic diversity and less adaptability.
On the other hand, individuals with too many symbols (deep code) can have very short chains which are, in theory, quicker to copy but must compete for a more varied assortment of resources and must search harder and longer to meet their nutritional needs. If the code gets too deep, the time needed to procure the neccessary building blocks will exceed the critter's lifespan, meaning that, on avarage, an individual using this code will die before it can make a single copy of itself. With deep codes we can have rich, dense complexity but its maintenance cost can be prohibitive for replicators of this type.
So here is a possible explanation. Maybe 5 nucleotides is the best bang for our buck. The longest code that would have granted our RNA ancestors enough genetic diversity to survive crises that may have wiped shallower rivals while keeping their average copies/lifespan greater than their deeper rivals.
While this might look like an appealing explanation, I urge you not to take this at face value. I have zero evidence to corroborate with any of this. Since your question was so broad, I do believe that some speculation is in order. However the example here may help you learn how to think these questions through and arrive a satisfactory hypothesis to test in a suitable laboratory. Send the link up here when you publish it, cheers.


What are the reactions occurring during the titration of a carbon dioxide contaminated water sample with diluted sodium hydroxide solution? During estimation of dissolved $\ce{CO2}$ in water, we took a water sample, added few drops of phenolphthalein and titrated it against dil $\ce{NaOH}$.

Questions

What are the reactions that happen:

initially  in the water sample?
after addition of phenolphthalein (if any)?
after addition of NaOH?

My speculations

Inside the sample water

I think dissolved $\ce{CO2}$ reacts with $\ce{H2O}$ to form $\ce{H2CO3}$. But then does all $\ce{CO2}$ get converted into $\ce{H2CO3}$?
After addition of phenolphthalein

In an acidic medium the equilibrium shifts to the left and the weak acid phenolphthalein ($\ce{HIn}$) mostly remains undissociated and hence colourless.

$$\ce{HIn <- H+ (excess) + In-}$$

After addition of $\ce{NaOH}$

According to my book after $\ce{NaOH}$ is added to the sample the following occurs:
\begin{align}
\ce{2NaOH + CO2 &= Na2CO3 + H2O}\\
\ce{Na2CO3 + H2O + CO2 &= 2NaHCO3}\\
\end{align}
From my understanding
\begin{align}
\ce{NaHCO3 &= Na+ + HCO3-}\\
\ce{HCO3- + H2O &= H2CO3 + OH-}\\
\ce{OH- + H+ + In- &= H2O + In-}\\
\end{align}
As the $\ce{H+}$ concentration decreases equilibrium shifts to the right
$$\ce{HIn -> H+ + In-}$$

This $\ce{In-}$ ion imparts the pink colour which at pH 8.3 becomes susceptible to our eyes.

At the beginning some of the dissolved $\ce{CO2}$ got converted into $\ce{H2CO3}$. They remain beyond the scope of estimation, isn't it?

Are my speculations correct?

Your thinking is in principle correct, let me just add a few more things, that may help you further.

What you are effectively titrating is carbonic acid, or more precise the hydrogen carbonate ion. Carbonic acid is not very stable in water, an most of the carbon dioxide stays the way it is, i.e. dissolved and interaction via van der Waals forces. The main equilibria are as follows (and derivations thereof):
$$\begin{align}
\ce{CO2 + 2H2O &<=> HCO3- + H3O+}\tag1\label{major}\\
\ce{HCO3- + H2O &<=> H2CO3 + OH-}.\tag2\label{minor}\\
\end{align}$$
In a contaminated water sample $\eqref{major}$ is by far the more predominant reaction.

When you now add sodium hydroxide solution you shift $\eqref{major}$ to the right side, as you are removing hydrogen carbonate ions via
$$\ce{HCO3- + OH- <=> CO3^2- + H2O}\tag3\label{conversion}$$

As you correctly identified, phenolphtalein is a weak acid by itself and the colour change happens via
$$\ce{H2In + 2OH- <=> In^2- + 2H2O}.\tag4\label{indicator}$$

Once you have added enough sodium hydroxide solution to completely shift the equilibrium $\eqref{conversion}$ to the right, you will start deprotonating the indicator molecule via $\eqref{indicator}$.

It is important to understand, that all these reactions happen at the same time and they keep on happening. This is the basic characteristic of a dynamic equilibrium. Since you are changing the composition of these equilibria the overall composition changes and the system adjusts, see Le Chatelier's principle.


Is it wrong to call carbon dioxide "methanedial"? Since dialdehydes have a carbonyl group on either end of an alkane, is it reasonable to call carbon dioxide "methanedial", or is it that ethanedial (glyoxal) is the first dialdehyde?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

Examples:
pentanal ($\ce{CH3-CH2-CH2-CH2-CHO}$)
pentanedial ($\ce{OHC-CH2-CH2-CH2-CHO}$)
In this context, the term ‘aldehyde’ is unambiguously defined:

P-66.6.0 Introduction.
The class name ‘aldehyde’ traditionally refers to compounds containing the $\ce{-CH=O}$ group attached to a carbon atom. However, nomenclature for aldehydes has been extended to describe a $\ce{-CHO}$ group attached to a heteroatom.

Since carbon dioxide ($\ce{O=C=O}$) does not contain a $\ce{-CH=O}$ group, it is not considered an aldehyde and thus Section P-66.6.1.1.1 does not apply to carbon dioxide.
Therefore, the shortest dialdehyde is ethanedial ($\ce{OHC-CHO}$). (Note that the preferred IUPAC name is ‘oxaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.)


How does an energy input actually change a molecule? I'm reading an online biochemistry textbook for fun, and a question that has bothered me on chemistry before popped up again. The question here is not specific for the chemical process described, but I needed an example:

By adding energy to a molecule of formic acid ($\ce{CH2O2}$), it kicks out an O and becomes formaldehyde ($\ce{CH2O}$). More energy makes it methanol ($\ce{CH3OH}$), and more energy makes that methane ($\ce{CH4}$).

I get that energy allows the molecule to become a different molecule of a higher energy state; it essentially stores the energy as new or altered chemical bonds. What I never understood is why? It is natural for the molecule to try to go to a lower energy state, releasing energy in the bonds, as it does under oxidation/"burning". So why doesn't it just immediately release the added energy, rather than store it in bonds?

What prevents the formaldehyde from saying "I see your energy, but I'll just do what's easier for me and release some of my own, and become formic acid"? After all, no matter how much energy I use to blast a block of stone into the air, it's going to fall down again, unless someone up there catches it. What's "catching" the higher-energy molecule and preventing it from "falling back" into the more stable, lower-energy molecule?

The answer seems like it might be leaning on atomic theory more than chemistry, but I thought I'd give it a shot here.

It is about transition energy barriers.

Imagine a mountainous landscape with lots of valleys of different depths. Stable molecules are large balls than sit in the bottoms of the valleys. Add some energy and the ball might rise up the valley floor but it will fall back. Add enough energy and the ball can rise high enough to get over the barrier to the next valley and will fall into a different place with a different depth (and the analogy falls down as it also becomes a different ball). So, whatever the relative depths of the valleys, there is a big barrier to getting from one valley to another. Even when one valley is much lower than another (so moving from one to another releases energy) you still have to put energy in to cause the transition.

Chemical reactions are often controlled by those transition barriers between one molecule and another rather than by the absolute energy difference between the molecules. Methane and oxygen will burn to give various compounds and will release energy, but there is a big barrier to starting the burning process. This means that the deoxygenation reactions you describe are possible and won't spontaneously reverse even though they would release energy if they did.


What is the area that 1 litre of gasoline would spread above the sea? I am studying for a general chemistry test with Atkins and Jones "Principles of Chemistry" 5th edition.

In the Chapter 18, page 743, section 18.3, there is this funny question:

  How would you estimate the area that 1 litre of gasoline would spread
  above the water surface?

Observation: I translated it from the Brazilian Portuguese edition. The English version may have different units, but the idea should be the same.

Let's say that the water body is the sea. But it could be a big lagoon, too. It doesn't really matter.

It got me thinking and now I'm stuck. How can we do that?

The hint here assumes a layer thickness of (crude) oil of about 1 mm. If I remember the pictures in the News around Deepwater Horizon correctly, and compare what I saw with own eyes at the local petrol station, petrol has a lower viscosity; likely spreads out wider than crude oil.

From here we learn the layer thickness is much lesser than 1 mm (millimetre), namely 168 nm (nanometre); hence all these colourful interference patterns in visible light.

$$ 1 \text{ L} = 1 \text{ dm}^3 =  10^{24} \text{ nm}^3$$

Deviding by 168 nm provides with googles built-in calculator an area of about $$5.95 \times 10^{21} \text{ nm}^2 = 5952 \text{ m}^2$$

To set this into a context, UEFA and FIFA agreed on the size of European Football / Association (soccer) field ($ 105 \times 68 \text{ m}$) which equals $7140 \text{ m}^2$.

Addendum:  According to this, an American Football field measures "$120.0 \text{ yd } (109.73 \text{ m}) \times 53.3 \text{ yd } (48.74 \text{ m})$ or $6400 \text{ yd}^2 (5351.2 \text{ m}^2$)".


Physical interpretation of the Gibbs-Duhem equation For a binary system at constant temperature and pressure, the Gibbs-Duhem equation can be reduced to
$$\mathrm d\mu_B=-\frac{n_A}{n_B}\mathrm d\mu_A$$
How does one make sense of the Gibbs-Duhem equation in terms of intermolecular forces and lattice packing? Why does the chemical potential of one component decrease (relative to the other) by a factor of $n_A/n_B$?

If you write down the equation in terms of the very definition of $\mu_i=\frac{\rm dG}{{\rm d}n_i}$ and impose ${\rm d}n_\text{totoal}=0$, it's just the conservation of the energy potential ${\rm d}G=0$.

Say you own a theme park where, every hour, each adult has to pay ${\rm d}\mu_a=\$7$ (money=energy) and each kid has to pay ${\rm d}\mu_k=\$3$.
We assign positive signs to when they put money out of their pockets.

If there are $n_a=10$ adults and $n_k=5$ kids in the park, and if this number of people stays fixed (${\rm d}n_\text{totoal}=0$ over some time), then they together have to spend

$$n_a\cdot {\rm d}\mu_a + n_k\cdot {\rm d}\mu_k = \$70 + \$15 = \$85$$

Now say you, the park owner, are not allowed to actually make any money (${\rm d}G=0$) and you relax the condition that the kids have to pay any money at all. That is, now just the adult have to pay ... and since the money has to go somewhere, the kids are on the receiving end. Then the ten parents still spend $\$70$ per hour and now the five kids will split that money among each other. Each kids receives
$$-{\rm d}\mu_k = \frac{1}{n_k}\cdot n_a\cdot {\rm d}\mu_a = \frac{1}{5}\cdot \$70 = \$14$$

The comparison with "money per time" lacks in that here people don't bring in money just from coming to the park (as it's the case with particles coming into the system and energy). But I hope this clear up the meaning of factor $\frac{1}{n_k}$ clear.


Why is Graphene So Strong? There has been a lot of news about Graphene since its discovery in 2004. And as we are all told it is a revolutionary material which is very strong, conductive and transparent; even in some cases it has revolutionized especial laws of physics (see this page).

But what is it about the structure of Graphene which makes it so strong?

If I understood what strong means, then graphene has high strength because of the 2-D layer structure of carbon atoms. The strength can also be attributed to its $sp^{2}$ hybridisation.

Any crystalline material will have a lattice structure. When we mean strength of a material its usually the ability to break apart the bond between two lattice atoms. But the observed strength is usually 2 or 3 orders lower than theoretically predicted strength due to flaws (voids, dislocations etc). But when graphene has a near flawless 2-D structure we can reach its theoretical strength. Though its is a very expensive process to actually realize "flawless". Even with flaws its strength is really high compared to steel.

So its essentially the 2-D one atom layer structure that can be responsible for its high strength.

One important thing is though graphene has high strength and elasticity (ability to get stretched)  it is relatively brittle (fracture toughness ~4 MPa$\sqrt{m}$).


Calcium sulfate soluble in water In the volumetric estimation of calcium in a given solution as calcium oxalate. We convert the calcium oxalate to oxalic acid by dissolving the former in hot (~70 °C) 2 N sulfuric acid solution. This is titrated against $\ce{KMnO4}$.

The reaction would be
$\ce{H2SO4 + CaC2O4 -> H2C2O4 + CaSO4}$

The calcium sulfate formed is normally not soluble in water. But here we get a clear solution which we titrate against $\ce{KMnO4}$.
Why does the calcium sulfate not precipitate in the solution?

The overall reaction is really better given as: $$\ce{2H2SO4 + CaC2O4 -> H2C2O4 + Ca^{2+} + 2HSO4^{-}}$$ since at that strong of a pH the predominate sulfuric acid species is $\ce{HSO4^{-}}$.

Better yet the reaction could be written as  $$\ce{2H^+ + CaC2O4 ->C[{0.2 N \ H2SO4, 70 °C}]\ H2C2O4 + Ca^{2+}}$$

As was already pointed out in the comments, calcium sulfate has a fair solubility 0.2 g per 100 ml of water at at 20 °C, and the $K_\text{sp} = 4\times10^{-5}$.

The solubility varies with temperature. The solubility of calcium sulfate dihydrate increases from 0.223 g/100 ml at 0 C to about 0.265 g/100 ml at 40 C, then decreases to 0.205 g/100 ml at 100 C.


Does alumina become transparent at high temperatures? Consider a normal opaque alumina crucible. I've heard that they become transparent at higher temperatures and therefore, are not appropriate for high precision DSC (caloric) measurements. The reason is, that there might be a lack of energy due to radiation losses (in addition to the bad thermal conductivity).

I was not able to find at least one quotable source for this claim. So my two questions are:

Is this actually right?
Does someone know of a quotable source for this (scientific papers, textbooks, etc...)?

Thanks!

Edit: Based on my research, the absorption rises with temperature, so the transparency should degrease, what is a somewhat contradictory to A.K. answer, isn't it?

Well, alumina is sapphire so stand to reason that it is transparent at a low temperature too.  I think what is meant by what you heard is that the higher intensity of infrared light is transmitted by the alumina causing thermal loss.

Alumina is transparent to UV, visible and infrared light. To our visible light vision, the scattering and reflection of polycrystalline alumina make it appear as opaque white. In the IR spectrum though it may appear transparent. This is because scattering is proportional to the inverse of the wavelength to the fourth power ($\alpha _{\text{scat}} \propto \lambda ^{-4}$).  Thus a wavelength just twice as long a red light (the longest wavelength of visible light; $\lambda = \pu{750nm}$ ), in this case, $\pu{1.5 \mu m}$, which is near infrared, would scatter one-sixteenth ($6.25 \% $) as much as the red light would. A wavelength twice as long as that wavelength ($\pu{3\mu m}$) would be in the thermal infrared spectrum and be scattered one-two-hundred-fifty-sixth as much as the red light ($0.39\%$) This reduced scattering allows more transmission. At high temperature when infrared radiation is more intense, the energy loss due to transmission becomes more significant, but the alumina was always transparent.


Book recommendations for protein structure refinement I’m an undergraduate Chemistry student and I’m currently writing my dissertation about semi-empirical methods, particularly PM7, applied to protein structure refinement from X-ray structure. Could you recommend me any book/article about it?

I found an article here about PM6/7 which might be useful.


Choice of base for malonic ester synthesis Most reaction mechanisms use $\ce{EtO-}$ as a base to deprotonate the alpha hydrogen in the malonic ester sythesis. But in the equilibrium there are still esters that haven't been deprotonated. Why won't the enolate do a nucleophilic attack on the ester, just like in the Claisen reaction?

Why isnt LDA used instead, which would deprotonate all the ester molecules, preventing Claisen reactions?

Understanding how this reaction is actually performed in the lab is key to answering your question.

Typically in the malonic ester synthesis, a full equivalent (or 2 equivalents if you are carrying out a dialkylation) of base is used, so little, if any, ester remains with the $\ce{\alpha}$-proton still in place.

(image source)

Also, keep in mind that both the alkylation and Claisen reaction have an activation energy. This means that it is usually necessary to heat the solution to bring about reaction (see here for a lab example).

The reaction is carried out by first forming the deprotonated malonate ester at room temperature and then dripping it into a refluxing solution of the alkylating agent ($\ce{RX}$ in the above diagram). This procedure achieves  the following:

concentration of non-deprotonated ester is very low
deprotonated ester is formed at room temperature, so even if any non-deprotonated ester remains there is not enough energy (at room temperature) to make it over the energy barrier and react
the deprotonated ester is dripped into the solution of refluxing alkylating agent; therefore, now that there is enough thermal energy to bring about reaction, the relative concentration of alkylating agent (the entire solution) is much higher than the relative concentration of deprotonated ester (added drop by drop) making the desired alkylation reaction much more probable than reaction of deprotonated ester with protonated ester (which, for reasons explained above, is in very low concentration in any case).


In a neutralization reaction between a strong acid and weak base, can I assume the weak base is totally ionized? Here's a Chemistry question from the infamously terrible Nelson textbook.

  10.0 mL of $\ce{NH3(aq)}$, with a concentration of $\mathrm{5.70\dot\ 10^{-2}}$ mol/L, is titrated to the endpoint (by which I assume they mean equivalence point) with $\mathrm{2.85}$ mL of $\ce{HBr}$ solution. What is the concentration of the $\ce{HBr}$ solution?

The answer they give is $\mathrm{0.200}$ mol/L which assumes that all of the moles of $\ce{NH3}$ reacts with the acid. I am unsure of this because I thought weak bases only ionized partially, and the amount they ionize has to be gleaned from the $\mathrm{K_b}$ table.

Why, if it is at all, correct to assume all of the base reacts?

You're never going to puzzle this out without some guidance so here goes...

  Why, if it is at all, correct to assume all of the base reacts?

All of the base will never react no matter how much acid you add. If you setup the equilibrium equation you can solve for $\ce{[NH3]}$ for any given pH. Thus there will always be some small amount of $\ce{NH3}$ in an acidic solution.

The gist is that there is $5.70\cdot10^{-4}$ moles of base. If you add "exactly" that much acid then the necessary assumption is that the acid balances the base and the solution is neutralized.

One other thing to consider in all of this is the significant figures. So +/- $1\cdot10^{-6}$ moles of acid doesn't make any real difference. That is enough to swing the final pH from 6 to 8.

Just like for the acid if $\ce{[NH3]} < 1\cdot10^{-6} $ then the free ammonia is negligible to the amount of $\ce{[NH4+]}$. Thus, considering the significant figures, the assumption that all of the ammonia was converted is reasonable, even though it is not actually absolutely true.


Can freezing break up sugar polymers? I recently freezed a huge bunch of apple slices, and thawed them again a few days later. After thawing, I noticed a few things compared to a batch that I had put in the fridge (no freezing)

The apples turned brown far quicker
After pureeing, they tasted sweeter

I know that slow freezingg can destroy cell walls in animals due to crystallization. My guess what has happened here is that the same mechanism also broke up long chain sugar polymers. Thus the sugars are chemically more active, oxidize more easily after contact with air, and taste stronger than in the other batch.

But is my guess plausible? Do polymers actually break under those mechanical stresses? Is this or a similiar effect observed elsewhere?

Freezing breaks the cell structure of an apple, not individual chains of short polysaccharides you can feel the taste. What happens is probably enzymes which the apple contains in cell structures go out of these structures and split long polysaccharides (which have no taste) into short sugars which you feel as sweet.

In other words, the apple digest itself by its own enzyme systems after breaking the cell structures.


Can I use Sunflower oil to make soap? When reading about making glycerine soap, the recipes suggested to melt animal fat and mix it with $\ce{NaOH}$.

Could I use Sunflower oil instead? Or a olive oil for nice flavor? Is heating necessary in that case?

You can use any triglyceride to make soap.  The process is called saponification.  In this process fatty acid molecules are cleaved from glycerol (glycerine).  In order for the reaction to proceed it is heated.  Regardless of the fatty acid source, the soap will probably not taste very good.


How to clean out a contaminated aluminum automotive radiator I have a very expensive custom made automotive aluminum radiator that has had a mouse or mice nesting inside the top tank.  I have removed most of the nesting material through the top inlet but I want to clear the radiator tubes that may be plugged up with the nesting material (looks like cotton of some sort) and be sure the radiator is flowing water through it as it should. As it is an aluminum radiator I am told they cannot remove the tank and "rod out" the radiator tubes.  The only solution offered by the company is replacement.  The radiator cost about $1200.
I have thought about an enzymatic drain cleaner but am not sure that it will not harm the aluminum radiator internally.  What ideas do you have?

Well I actually tried a small jar of the Enzymatic product with a sample of the nesting material I already removed as well as small piece of .023 6061 aluminum... Overnight the aluminum was completely decomposed.  The nesting material seemed unfazed.  So much for that idea!


Can glutamic acid and arginine form H-bond at physiological pH? I was wondering if, say, glutamic acid and arginine can form H-bonds at physiological pH?

According to the figure arginine has a $\ce{NH3+}$ group and glutamic acid a $\ce{COO-}$ group at physiological pH.
I know hydrogen bonding is possible amongst $\ce{NH2}$, $\ce{HF}$ and $\ce{OH}$. But $\ce{-NH2}$ is in the protonated form and the $\ce{-COOH}$ in the deprotonated form.

What will happen:

Will $\ce{-NH3+}$ donate its $\ce{H+}$ to $\ce{-COO^-}$ to form $\ce{-NH2}$ and $\ce{-COOH}$ and then form hydrogen bonds?
Will they attract each other because of the opposite charge, with resulting H-bonding
No H-bonding will occur.

(maybe the first and second option are combined, I didn't really know how to formulate it)

UPDATE
How is the H-bond possible in the following figure (I would think this is just an electrostatic interaction, this figure suggest it would a combination of both):

source of picture: Wikipedia

Yes, glutamate and arginine can form a hydrogen bond at physiological pH. Remember of course, that as soon as you enter aquaeous media, everything will be saturated by hydrogen bonds from the surrounding water molecules. But inside a protein an arginine—glutamate hydrogen bond is perfectly viable. Arginine will donate the hydrogen bond using an $\ce{N-H}$ bond and glutamate will receive towards a carboxylate.

Note, by the way, that your picture gives a wrong structure of protonated arginine. The correct protonation of the guanidine group is given in the image below (taken from Wikipedia).


If heavy metals bind to sulfur why isn't sulfur used as a safer way to chelate them? I was wondering about sulfur as an alternative life-threatening chelators. Why isn't sulfur used as a chelator if heavy metals have an affinity for it?

Sulfur-containing molecules are indeed used in chelation therapy: prime examples are dimercaprol and penicillamine, both of which contain thiol functional groups.


How to extract pure caffeine from tablets? I have some caffeine tablets and I want to extract the pure caffeine from the rest of the ingredients (caffeine 200 mg. Inactive ingredients in order are corn starch, dextrose, dicalcium phosphate, D&C yellow #10 aluminum lake, magnesium stearate, microcrystalline cellulose, silica gel).

I assume the result would be aqueous solution but any method which works would be great.

The best option would be an organic extraction with dichloromethane. All the other ingredients (except possibly D&C yellow #10 aluminum lake) will be insoluble and therefore unaffected. Filter, let the dichloromethane evaporate and you should have your caffeine.

EDIT: I agree with J M. This caffeine IS NOT EDIBLE. You could easily become intoxicated by the small amounts of dichloromethane that will remain. Also be careful when handling the dichloromethane. Vapors are also toxic and if the liquid is heated or shaken in a closed recipient it might explode. Use caution.


Mechanism of synthesis of allene I would like to know the mechanism of this reaction. In my first attempt to the question, I had replaced the two $\ce{Br}$s by $\ce{CH3}$s by $\ce{S_N2}$ of the organometallic compound but the answer given was the above allene. I think my reaction isn't possible because of too much hindrance. So if that's the case, then what is the mechanism of the actual reaction?

This is the Skattebøl rearrangement, the first step is of which is a lithium-halogen exchange to form a carbenoid. A mechanistic study of the rearrangement of monosubstituded lithium bromocyclopropylidenoids found that allene formation can occur either in a concerted (Scheme 2) or step-wise (Scheme 3) fashion depending on the ability of the substituent to stabilize the intermediate carbene$^{[1]}$.

The study found that electron donating groups (like the methyl groups in your example) lower the barrier to allene formation. For several EDG-substituted species, the free carbene structure 14 could not be found as an energy minimum and all attempts to locate these structures lead directly to allenes. This means the stepwise mechanism does not play a role in the Skattebøl rearrangement for EDG-sbustituded species$^{[1]}$.

$[1]$ Azizoglu, A.; Balci, M.; Mieusset, J.-L.; Brinker, U. H. The Journal of Organic Chemistry 2008, 73 (21), 8182–8188.


Van 't Hoff equation In thermodynamics many definition has been made from partial derivatives at constant conditions for instance:
Let $U$ be $U:f(S,V,m_)$ then:
$$T = \left( \frac{\partial U}{\partial S} \right)_{V, m_{k}};  p = \left( \frac{\partial U}{\partial V} \right)_{S, m_{k}}; \mu_k = \left( \frac{\partial U}{\partial m_k} \right)_{V, S}$$
In a particular case, heat of reaction and van 't Hoff relation
Being $H:H(T,p,\xi)$
$$\mathrm dH= \left( \frac{\partial H}{\partial T} \right)_{p, \xi}\,\mathrm dT + \left( \frac{\partial H}{\partial p} \right)_{T, \xi}\,\mathrm dp + \left( \frac{\partial H}{\partial \xi} \right)_{T, p}\,\mathrm d\xi$$
Where $C_p$ is the specific heat at constant pressure, $h_T$ is the heat compressibility and $r_{T,p}$ is the heat of reaction at constant temperature and pressure:
$$C_p = \left( \frac{\partial H}{\partial T} \right)_{p, \xi} ; h_T = \left( \frac{\partial H}{\partial p} \right)_{T, \xi}; r_{T,p} = \left( \frac{\partial H}{\partial \xi} \right)_{T, p}$$
In terms of variation of affinity with temperature, the heat of reaction is given by (where $K$ is the equilibrium constant of a chemical reaction):
$$r_{T,p} = -RT^{2}\frac{\partial }{\partial T} \ln K(T,p)$$
So the question is: How can I define an equation where the left side needs constant temperature, but the right side implies the derivative of temperature?

The T that is held constant on the LHS of your equation has a different meaning from the T that is allowed to vary on the RHS.

The van 't Hoff equation answers the following:

Suppose I have a reaction that takes place at const T, p at T = T1.  Now I change to a new temperature, T2, and allow that reaction to again take place at const T, p, except now T = T2.  How does the equilibrium constant change?

Now that you understand this, you can see there is no inconsistency between "constant T" and "changing T", since they refer to two different things.

I.e., the fact that the reaction is constrained to take place at constant T in no way prevents me from testing the reaction at a different, but still constant, T.

The analogy I'd give is this.  Suppose we want to determine the effect of altitude on 100 m sprint times.  The 100 m is always run on a level surface (no change in altitude during the race).  Yet this does not in any way prevent us from comparing 100 m run times at different altitudes (e.g,  sea level vs. 10,000 feet).

So your question is like asking:  "How can we determine the effect of changing altitude on 100 m run times, given that the 100 m run is always done at constant altitude?"  Well, we can, of course.


Theoretical plates in Plate theory of chromatography In "Chromatography theory" by Scott and Cazes,
The plate volume is given as ($v_m+Kv_s$) where $v_m$ and $v_s$ are the volume of stationary and mobile phase in the column and $K$ is the distribution coefficient (basically the equilibrium constant of the solute concentration in stationary and mobile phase).
They give the definition of plate volume as

The plate volume is defined as that volume of mobile phase that can contain all the solute in the plate at the equilibrium concentration of the solute in the mobile phase.

The plate concept is introduced to explain the equilibrium attained by the solute in distributing between stationary and mobile phase. We can't assume this equilibrium in the whole column as mobile phase continuously flows during the process and disturbs the equilibrium, so column is considered to be made of number of theoretical plates in which equilibrium is attained for some finite time.
Basically, I am not able to understand the definition of plate volume. Because as the name suggests one can interpret as the column to be made of number of plates which is made of the volume of mobile phase and stationary phase in which the solute is distributed between the two phases. So, I initially misinterpreted the formula of plate volume as $v_m+v_s$.
Please explain the definition of the plate volume. It seems somewhat confusing.

I shall take the example of the mixture ethanol-water, first without chromatography. If I start distilling 250 mL of a mixture 10% ethanol and 90% water, the solution boils at 96°C, and the vapor contains 55% water + 45%  ethanol. This vapor is condensed by passing through the cold condenser. So the first fraction of maybe 10 mL condensed vapor contains 45% ethanol. Suppose 250 mL of this first mixture (45% ethanol) is heated in another distilling apparatus. At maybe 85°C, the liquid boils and the first vapor will contain 70% ethanol and 25% water. If this vapor is condensed, the liquid so obtained contains 70% ethanol. These distillation - condensation operations can be repeated, with the following results.
1st distillation. Original liquid : 10% ethanol. Distilled liquid : 45% ethanol
2nd distillation. Original liquid : 45% ethanol. Distilled liquid : 70% ethanol
3rd distillation. Original liquid : 70% ethanol. Distilled liquid : 83% ethanol
4th distillation. Original liquid : 83% ethanol. Distilled Liquid : 90% ethanol.
etc.
nth distillation. Original liquid : 96% ethanol. Distilled liquid : 96% ethanol.
Now suppose all these distillations are carried out in the same apparatus. Instead of having the vapor go through a cold tube and be recovered in a flask out of the condenser, imagine that the vapor is sent into a long vertical tubes situated just above the first container half filled with the mixture 10% ethanol. Suppose this vertical tube contains a lot of successive floors or plates, punched with one or two small holes. In the beginning the vapor made of 45% ethanol will pass through the holes of the first floor or plate, and will be condensed just above this level and may partly go down to the bottom of the setup. But a fraction of this liquid will now boil again and produce a vapor made of 70% ethanol. This vapor passes through the second floor and be condensed just above the second floor. On this second floor the liquid (containing 70% ethanol) will later on be evaporated and produce a vapor containing 83% ethanol. And so on. After n floors the vapor and the liquid will contain 96% ethanol. As a result, the separation ethanol-water improves if the number of floors or plates increases. And you may imagine that the height of the room between two floors is critical. If it is too small, it will be immediately filled. If it is too big, some secondary condensation will happen on the wall of each room.
Now ethanol is not the best example for my purpose because the successive distillations cannot get to pure ethanol. Suppose for the rest of my description that after n distillations, a pure amount of a solvant X can be obtained. Suppose that to get a better that 99% separation you need 8 floors, which are all 5 mm high.
Suppose now that instead of having used a distillation apparatus, you are using of chromatographic column. In order to get a better than 99% separation of pure compounds you need a column measuring 60 cm. The plate volume is : 60 cm / 8 plates = 7.5 cm. Of course, if you use a longer column, the separation might be 99.9% or better.
Of course this explanation is an over-simplification of the reality. But it may help you understand what is a plate volume.
Did you follow me ?


Finding identities of diatomics from Spectroscopy and Mass Spec Data Let's say I have a sample of 2 (assumingly different) diatomics A and B.

Through spectroscopy I found the data below:

For molecule A:
$$\begin{align}
B&\approx2.17690\ \mathrm{cm^{−1}}\\
D&\approx4.79000\times10^{-5}\ \mathrm{cm^{−1}}\\
I&=1.28590\times10^{-46}\ \mathrm{kg\ m^2}\\
\tilde ν_e&=928.15\ \mathrm{cm^{−1}}
\end{align}$$

For molecule B:

$$\begin{align}
B&\approx1.51695\ \mathrm{cm^{−1}}\\
D&\approx7.1049\times10^{-6}\ \mathrm{cm^{−1}}\\
I&=1.84533\times10^{-46}\ \mathrm{kg\ m^2}\\
\tilde ν_e&=1401.86\ \mathrm{cm^{−1}}
\end{align}$$

Through Mass Spec, I found 2 strong spectral lines at $30.0077086\ \mathrm{g\ mol^{-1}}$ and $41.9766928\ \mathrm{g\ mol^{-1}}$. How do I deduce which is which and further, how do I determine the bond length. [FYI, I'm thinking the former mass is the molecule NO and the latter NaF just by guessing]

I'm thinking that the heavier molecule will have the larger inertia value. And from the inertia equation $I=\mu R^2$ I'll derive the bond length $R$. but how do I find the $\mu$ from the actually mass of the molecule from the mass spec data?

Is this problem conceptual, based on hypothetical data?

  Through [m]ass [s]pec, I found 2 strong spectral lines at $30.0077086\ \mathrm{g\ mol^{-1}}$ and $41.9766928\ \mathrm{g\ mol^{-1}}$.

These mass-to-charge ratios or $m/z$ values are implausibly precise.  The significant figures you show imply a precision of about 30 parts-per-billion (ppb) mass accuracy.  That is well outside the specification of any commercial instrument I am aware of.

Also, if you are right about the first ion being, $\ce{NO}$, the value is inaccurate.  The neutral $\ce{^14N^16O}$ molecule has a mass of 29.99799 Da.  The positively charged ion formed by electron loss from the neutral species would have a mass of 29.9974 Da.  This latter value is off from the value you quote by about 341 parts-per-million.  So if the mass spectrometry data really is precise to within 30 parts-per-billion, and it is accurate, then the molecule can't be $\ce{NO}$.

In fact, with mass spectrometry data that precise (and accurate), we shouldn't need any other data to identify the diatomics, other than the fact that they are diatomics.  We can add the mass of every possible nuclide to every other possible nuclide and see which give the right mass.

Here's how to do that using the R packages ecipex and dplyr.

require(ecipex)
require(dplyr)
nistiso %>%
    mutate(foo = 1) %>%
    left_join(., ., by='foo') %>%
    filter(mass.x > mass.y) %>%
    mutate(diatomic_mass = mass.x + mass.y) %>%
    filter(diatomic_mass < 30.00771, diatomic_mass > 30.00770)

The result is that only the $\ce{^11B^19F}$ diatomic molecule has the requisite mass. Even if you loosen the mass tolerances to allow for e.g. for electron loss or gain, this is still the only possible diatomic with the right mass.

You can repeat this exercise for the other data point, and find that $\ce{^40Ca^2H}$ is the only possible ion consistent with the data.  ($\ce{^40Ar^2H}$ is close in mass but is more than one electron's mass too light.)

I'll leave it to you to decide whether these ions are chemically meaningful or possible for you, and if they are consistent with the rotational spectra.


Calculate the concentration in milimoles per litre (mM) of a solution that is 20% H2SO4 by mass and that has a density of 1.198 g/L Calculate the concentration in milimoles per litre ($\pu{mM}$) of a solution that is $\pu{20\%} \ce{H2SO4}$ by mass and that has a density of $\pu{1.198 g/L}$.

For the answer, please provide two decimal places.

I calculated an answer but I'm just checking to see if I am correct...

I got $\pu{61.7 mM}$

First, I calculated the molar mass of $\ce{H2SO4}$ which is $\pu{98.086g/mol}$ and then I multiplied it by $\pu{20\%}$ which then got me $\pu{19.6172g/mol}$.

Second, I divided density by mass to cancel out the grams to get $\pu{mol/L}$ which then came out to $\pu{0.06106M}$ and then I divided it by $1000$ to get $\pu{mM} = \pu{61.07mM}$

If someone could check if this is correct that would be great, thanks!

the solution has a density of 1.198 g/L.

This has to be wrong. I assume that the density is 1.198 g/ml.

Thus a liter has a mass of

$$1000\text{ ml/L} \times 1.198\text{ g/ml}= 1198\text{ g/L}$$

20% of the mass is $\ce{H2SO4}$, thus the mass of $\ce{H2SO4}$ in 1.000 L of solutiom is:

$$0.20 \times 1198\text{ g/L} = 239.6 \text{ g/L}$$

$\ce{H2SO4}$ has a molecular mass of $98.079\text{ g/mol}$ so the number of moles is

$$\dfrac{239.6 \text{ g/L}}{98.079\text{ g/mol}} = 2.443\text{ moles/L}$$

Now converting to millimoles

$$2.443\text{ moles/L}\times 1000\text{ mM/mole} = 2443\text{ mM/L} \ce{->[rounding]} 2.4\times 10^3\text{ mM/L} $$


Associated occupation numbers of frozen natural orbitals (FNOs) from MP2 method The paper Frozen Virtual Natural Orbitals for Coupled-Cluster Linear-Response Theory gives the following (simplified) description of frozen natural orbitals (FNOs).

In the MP2 method, the unrelaxed one-electron density matrix can be written in terms of spin orbitals

$\gamma_{pq}=\langle\Psi^{1}|\{a_{p}^{\dagger}a_{q}\}|\Psi^{1}\rangle$

where $|\Psi^{1}\rangle$ is the first-order correction to the Hartree-Fock wavefunction.
In the MP2 based NO method, the virtual-virtual block is constructed:

$\gamma_{ab}=\frac{1}{2}\sum t_{ij}^{ac}t_{ij}^{bc}$

and then diagonalised:

$\gamma V=nV$

The eigenvectors V are the virtual NOs, and the eigenvalues n are the associated occupation numbers.

If the eigenvalues are found from a virtual-virtual (unoccupied-unoccupied) block, how can there be an associated occupation number?

Virtual or unoccupied refers to the Hartree-Fock solution - the virtual orbitals are thus unoccupied in Hartree-Fock, i.e. have occupation number 0, whereas the occupied orbitals have occupation number 2 (spin-traced occupation in restricted HF) or 1 (for a given spin in unrestricted HF). The first-order MP2 correction to the wave-function will have different occupation numbers - the occupied orbitals will become slightly less occupied, whereas the virtual orbitals become slightly occupied by the same total amount as a result of electron correlation. This doesn't change the nomenclature: occupied and virtual orbital still refers to the HF reference wave-function.


problem with loops over basis sets in psi4 using python I've recently started using the psi4 quantum chemistry software. I wanted to compare a couple of different basis sets (mainly pople style) on an amino acid. So I've written a python code inside the input file to iterate over the different sets.But only the first loop runs and the second fails miserably giving an error code:

PSIO_ERROR: Attempt to write into next entry: 35, SO-basis Overlap Ints
PSIO_ERROR: unit = 35, errval = 18
PSIO_ERROR: 18 (Incorrect block end address)
Traceback (most recent call last):
File "<string>", line 42, in <module>
File "/usr/share/psi/python/driver.py", line 617, in energy
procedures['energy'][lowername](lowername, **kwargs)
File "/usr/share/psi/python/proc.py", line 1231, in run_scf
scf_helper(name, **kwargs)
File "/usr/share/psi/python/proc.py", line 1457, in scf_helper
e_scf = psi4.scf(precallback, postcallback)
RuntimeError: PSIO Error

here is my code:

basis_sets = ["6-31G", "6-31G*", "6-31G**"]
for basis_set in basis_sets:
    set basis =  $basis_set
    energy('scf')

I've tried to run it without the code and there was no problem.
I copied the code from the psi4 tutorial page for python scripting
Thanks for the help.

Here is an input file that will reproduce this error:

molecule {
    b      0.0000        0.0000        0.0000
    h      0.0000        0.0000        1.0000
}

basis_sets = ["6-31G", "6-31G*", "6-31G**"]
for basis_set in basis_sets:
    set basis = $basis_set
    energy('scf')

The problem is that Psi4 sees there are files remaining in scratch that it can use, starting with the spin orbital-basis overlap integrals, but because the basis set has changed, the file is the wrong size. You'll need to clean these up after the energy call by adding clean() within the loop.


What are the products of the haloform reaction of acetylacetone? I searched a lot to find the products of the haloform reaction of acetylacetone (pentane-2,4-dione),  but I didn't find any results.

What are the products of the haloform reaction of acetylacetone?

Because acetylacetone is a 1,3-dione, the carbonyls are very enolizable.  This makes the hydrogens attached to carbon 3 on acetylacetone very acidic.  When introducing the conditions for a haloform reaction, base is added.  The base will take carbon 3's hydrogens, in the presence of some halogen, they will be replaced, this happens twice.  Now, with the product 3,3-dihalopentane-2,4-dione, the reaction will end.

If you insist on a haloform reaction occurring, a much stronger base will be needed and the reaction vessel would probably need to be heated.  It will proceed as usual and create first 2,2-dihalo-3-oxobutanoic acid (2,2-dihaloacetoacetic acid). The reaction will probably stop at this point. Upon heating though, this compound may decarboxylate to give 1,1-dihaloacetone and carbon dioxide. If you wanted to continue with this, you could do another haloform reaction to first generate perhaloacetone, and if you wanted to, this may be able to become trihaloethanoic acid and haloform.

But realistically, the reaction would end at the 2,2-disubstituted dihalo acetoacetic acid (2,2-dihalo-3-oxobutanoic acid)


Can bromide ion ever act like a base? I was reading MECHANISM OF α-HALOGENATION OF ALDEHYDES AND KETONES:

As you can see in the third step. Bromide ion is picking up a base from protonated acetone. Doesn't that step mean that protonated acetone is stronger acid than Hydrobromic acid? Is that even possible?

Come on, you can figure out this one on your own! How do you compare acidities? Here's a helpful hint.

Armed with that information, I have to add something. I believe the mechanism as drawn above is not universal: it sure depends on the conditions and solvent used. Some other proposed mechanisms involve the enol, rather than the protonated ketone, which might be more plausible in mild conditions. In water, anyway, HBr is fully dissociated and it shouldn't be “reformed” like that. The proton will return to be solvated in the solution.


Will cinnamaldehyde undergo self-aldol condensation? Will cinnamaldehyde undergo self-aldol condensation?

My book says it won't, but since it has an α-hydrogen, I'm confused.

Not every α-hydrogen automatically means that a ketone is enoliseable. It would need to be an α-hydrogen capable of aligning with the π system of the $\ce{C=O}$ bond such that removing the hydrogen automatically generates the required enolate anion system. In the case of cinnamaldehyde — your compound — there is conjugation across the phenyl ring, the double bond and the aldehyde meaning that this compound is entirely planar. Thus, there is no way for the α-hydrogen to align favourably for deprotonation to occur. If it did, it would generate an allene-olate $\ce{Ph-CH=C=CH-O-}$ which is not particularly favourable.

In general, α,β-unsaturated ketones and aldehydes are deprotonated vinylogously, meaning that the double bond shifts away upon deprotonation. For example crotonaldehyde $\ce{H3C-CH=CH-CHO}$ is deprotonated in the following way:

$$\ce{H3C-CH=CH-CHO ->[][- H+] H2C=CH-CH=CH-O-}\tag{1}$$

I.e. the γ-hydrogen is abstracted in place of the α-hydrogen.


Why is pressure negative in force and work The equation to measure work is $$w=F\cdot D$$

However, the equation for force is  $$F=(-P \cdot A)$$

Plugging this into the work equation we have:

$$w=-P\cdot A\cdot D$$

My question is, why is pressure negative in force?

In this context we are talking about the work done by a gas.

Rather than representing it as $w = -PAD$, think of the formula as representing an area (that the gas is pressing against) moving a distance $D$. Since this a change in volume, we can set $w = -P\Delta V$

If you are putting work into the system (say pressing a piston into a cylinder containing a gas at a given pressure), the volume is becoming smaller, so $\Delta V$ is negative.  The negative of the pressure times a negative $\Delta V$ is a positive value, which matches with the sign convention for putting work in.  Vice versa, an expansion of a compressed volume at a given pressure will push the piston back out of the cylinder and allow the gas to do work, so $w$ is negative which follows the sign convention for work output.


Separation of 2 compounds ( Powder ) Suppose I have a mixture of 2 pure binary compounds, a white powder. What is the best method to separate the 2 compounds?

What I have thought of so far is to dissolve powder in DI water and carry out filtration if one compound is soluble and the other is not. However, I have not been able to think of any other methods to separate the 2 compounds if both are either water-soluble or water-insoluble. I considered fractional distillation but the compounds may become anhydrous. Furthermore, the boiling points of the compounds are not given and they could end up being too close for fractional distillation to work.

Is there any efficient, and preferably simple, method to achieve the above-mentioned desired result?

Edit: Is there any general method of approach for separating such 2 compounds as there is for separating water-soluble and water-insoluble compounds?[Filtration]  Or does it have to be achieved purely by trial and error of various methods?

Your question is about as broad as chemistry itself. Like you said, you may try dissolving them in water, if one is soluble and the other is not. If it's not working with water, chances are it will work with ethanol, or acetone, or... see, there are many solvents on the shelf. No luck? Maybe by now you've found something that dissolves them both, so you may try fractional crystallization. Still no luck? Maybe the compounds differ in density or selective adhesion of something, so flotation might help. Still no luck? Chromatography, by any chance? No? Distillation, like you said (provided they both would survive until boiling point), or maybe sublimation (good for $\ce{NH4Cl + NaCl}$)? No? Well, there is a multitude of chemical processes; some of them might change (that is, destroy) one compound or both, but probably in such a way that they can be regenerated later.

And if everything else fails, you can always look at the mixture under a microscope and separate the crystals manually.


Solution , Solubility and Colligative Property? Which of the following  0.1 M aqueous solution will have the lowest freezing point ?

1) Glucose
2) Urea
3) NaCl
4) K2SO4

How to relate Molarity with freezing point ?
Should I use the relation
Kf = MR(Tf)2/1000Hv

Let's say we have some ice.

Beautiful structure. Ice stays solid at some temperatures because of its neat, tight lattice. Ice stacks this over and over until there's no more ice left to stack.

If I were to put a lot of ions between this stack, the stack might bend or even break; it isn't as reinforcing as it once was. The stack isn't as durable anymore, so we would expect it to melt a lot easier - there's too much structural stress for it to be a solid.

The freezing point of solutions depends, at least in this case, on how much dissociation occurs. If we can push more ions into solution, it's going to have a harder time making a sturdy lattice - a harder time freezing.

$\ce{C6H6O6 -> C6H6O6}$ It doesn't dissociate at all.

I'm not sure how much urea dissociates, but I assume it would be like glucose.

$\ce{NaCl -> Na+ + Cl-}$ It dissociates into 2 ions.

$\ce{K2SO4 -> K+ + K+ + SO4^{2-}}$ It dissociates into 3 ions.

$\ce{K2SO4}$ dissociates the most, so it's freezing point is lowest. Hope this helps!


Anti elimination from vinylic halides An $\mathrm{E2}$ (or $\mathrm{E1cb}$) mechanism requires antiperiplanar configuration. In case of restricted rotations, like in cycles, where rotation is not possible to achieve the antiperiplanar configuration, elimination occurs to favour the Hofmann product instead of the Zaitsev, if antiperiplanar configuration is possible at the less substituted β C.

In menthyl chloride, antiperiplanar configuration is only seen with the red $\ce{H}$, so 2 menthene is the only product, although it is not the most substituted.

In the molecule above, due to restricted rotation, the leaving group $\ce{Cl}$ and the base attacking the $\ce{H}$ will be on the same side. So will elimination be possible in this case?

An E2 reaction will not occur with this reagent.

An E1 reaction, on the other hand, could do so (though quite slowly: E1 is an all around pathetic reaction mechanism, and this molecule does nothing special to stabilize it). The chlorine will leave and there will be a carbocation on the end of the alkene, whereupon the proton from the other side of the double can be extracted with a base in a second step, leaving you with an alkyne.

That said, if you want the alkyne, it would be infinitely infinitely easier to start with the cis isomer of the molecule you have and just run it through very strong base. E2 is far far better than E1 in terms of reaction rate and yield.


Using AC voltage for electroplating As I know, in electroplating method, the type of voltage used is DC voltage. My question is, does replacing DC current supply by AC one affect the functioning of the cell? And why?

The difference is quite huge. Firstly, the deposition will occur on both electrodes (if possible). Secondly, if the frequency is high enough, there will be no deposition at all as only polarization and depolarization effect take place. The frequency effects mass, transport and electrical double layer. Depending on the electrolyte composition, the dissolution (if possible) will reduce the deposition. Usually pulse electroplating is used.  The problematic is huge, please see some articles for more info:

An Overview of Pulse Plating - Norman M. Osero

and

Pulse and pulse reverse plating—Conceptual, advantages and applications, M.S. Chandrasekar, Malathy Pushpavanam


Find pH and Kw for water with given values of conductivity At $\pu{25 ^\circ C}$, $\lambda_\infty (\ce{H+}) = \pu{3.5 \times 10^{-2} S m2 mol-1}$ and $\lambda_\infty (\ce{OH-}) = \pu{2 \times 10^{-2} S m2 mol-1}$. Determine $\mathrm{pH}$ and $K_\mathrm{w}$.

Given: Specific conductance $= \pu{5.5 \times 10^{-6} Sm-1}$ for $\ce{H2O}$

My Attempt:

I know I need to find the concentration of $\ce{H+}$ or $\ce{OH-}$ for $\mathrm{pH}$. $K_\mathrm{w}$ can be found subsequently. However, I have no clue as to how do I find that with the given information.

I am also aware of the relation between conductivity, Specific conductance, and molarity. It is easy to find the molarity from the given information, but that would be of no use as we need the concentration of the individual ions rather than the molarity of the solution.

Any help would be appreciated.

We know,

$$\lambda_c = \frac{(\text{Specific Conductivity})}{1000 \times c}$$

$$\alpha\lambda_\infty = \lambda_c$$

(where $c$ is concentration in moles per litres of the electrolyte and $\alpha$ is degree of dissociation of the electrolyte)

$\lambda_{\infty}$ of water (the electrolyte) is the sum of $\lambda_{\infty}$ of $\ce{H+}$ and $\ce{OH-}$.

From equations above, one can easily find  the the value of $c\alpha$ is $10^{-7}$, which is the concentration of $\ce{H+}$.