What is the physical state of vitamin E acetate at room temperature and normal pressure Vitamin E acetate ((+)-alpha-tocopherol acetate) is sold as semi-solid/liquid by Sigma and their safety data sheet gives the melting point as 25 °C. On the other hand, internet databases such as Chemspider show melting points of either -28 °C or 28 °C. Supplements labeled either Vitamin E or Vitamin E oil are packaged as solid, gel capsules or liquid.
What is the normal melting point, and what is the physical state at room temperature (20 °C)?
Update
There are at least three products on the market, the single (+) stereoisomer (derived from biological sources, CAS 58-95-7), the racemic mixture, and the "all-racemic mixture" (8 stereoisomers, with some pairs being diastereomers and other pairs enantiomers). The latter is produced synthetically with non-stereospecific steps (CAS 7695-91-2). All have the formula $\ce{C31H52O3}$. To clear up any confusion, the physical state of the single isomer, of the (+/-) racemic mixture and the "all-racemic mixture" would be of interest.
Vitamin E, depending on the source, has different numbers of methyl groups on the aromatic ring; these forms are referred to as alpha, beta and gamma. The alpha form is more common in Europe while the gamma form is more common in the US (due to different diets).

Vitamin E acetate, or α-tocopheryl acetate, exists in two forms: D-form (solid/highly viscous at 20 °C) and a racemate (viscous liquid at 20 °C).
Interestingly enough, the difference in physical properties is well denoted in German Wikipedia article, whereas English Wikipedia omits this.
Also, I noticed many authoritative sources arbitrarily listing a melting point for one or another isomer without denoting the exact form, e.g. even CRC Handbook of Chemistry and Physics provides a single entry for vitamin E acetate with melting point −27.5 °C [1, p. 3-550], so no wonder online sources and sellers mix the things up a little.

More relevant details can be found in Handbook of Pharmaceutical Excipients [2, p. 32]:

  d-Alpha tocopheryl acetate

  Empirical formula $\ce{C31H52O3}$
  Molecular weight 472.73
  CAS number [58-95-7]
  Synonyms (+)-(2R,4'R,8'R)-2,5,7,8-Tetramethyl-2-(4',8',12'-trimethyltridecyl)-6-chromanyl acetate; d-α-tocopheryl acetate; vitamin E.
  Appearance A practically odorless, clear, yellow, or greenish-yellow colored viscous oil that may solidify in the cold.
  Melting point 28 °C
  Solubility Practically insoluble in water; soluble in ethanol (95%). Miscible with acetone, chloroform, ether, and vegetable oils.
  Specific rotation $[α]_\mathrm{D}^{25} = +0.25^\circ$ (10% w/v solution in chloroform)
  Comments Unstable to alkalis.

  dl-Alpha tocopheryl acetate

  Empirical formula $\ce{C31H52O3}$
  Molecular weight 472.73
  CAS number [7695-91-2]
  Synonyms (±)-3,4-Dihydro-2,5,7,8-tetramethyl-2-(4,8,12-trimethyltridecyl)-2H-1-benzopyran-6-ol acetate; (±)-
  (2RS,4'RS,8'RS)-2,5,7,8-tetramethyl-2-(4',8',12'-trimethyltridecyl)-6-chromanyl acetate; (±)-a-tocopherol acetate; α-tocopheroli acetas; all-rac-α-tocopheryl acetate; dl-α-tocopheryl acetate; vitamin E.
  Appearance A practically odorless, clear, yellow, or greenish-yellow viscous oil.
  Density $\pu{0.953 g/cm3}$
  Melting point –27.5 °C
  Refractive index $n_\mathrm{D}^{20} = \pu{1.4950 – 1.4972}$
  Solubility Practically insoluble in water; freely soluble in acetone, chloroform, ethanol, ether, and vegetable oils; soluble in ethanol (95%).
  Comments Unstable to alkali. However, unlike alpha tocopherol, the acetate is much less susceptible to the effects of air, light, or ultraviolet light. Alpha tocopherol acetate concentrate, a powdered form of alpha tocopherol acetate, is described in the
  PhEur 6.0. The concentrate may be prepared by either dispersing alpha tocopherol acetate in a suitable carrier such as acacia or gelatin, or by adsorbing alpha tocopherol acetate on silicic acid.

D-form may appear as a highly viscous oil, however it will become solid over time [3, p. 232] (emphasis mine):

  D-α-Tocopheryl acetate. CAS: 58-95-7. Obtained by vacuum steam distillation
  and acetylation of edible vegetable oil products. Occurs colorless to yellow, nearly
  odorless, clear, viscous oil. May solidify on standing, and melts at about 25 °C.
  Unstable in presence of alkalies.

  DL-α-Tocopheryl acetate. CAS:7695-91-2. Colorless to yellow or greenish yellow, nearly odorless, clear, viscous oil. Unstable in presence of alkalies.

  D-α-Tocopheryl acetate concentrate. Obtained by vacuum steam distillation and
  acetylation of edible vegetable oil products. May be adjusted by suitable physical
  or chemical means. Occurs as light brown to light yellow, nearly odorless, clear,
  viscous oil.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; CRC Press, 2016. ISBN 978-1-4987-5429-3.
Handbook of Pharmaceutical Excipients, 6th Ed.; Rowe, R. C., Ed.; APhA, (PhP) Pharmaceutical Press: London, 2009. ISBN 978-0-85369-792-3.
Food Additives, 2nd Ed., rev. expanded.; Branen, A. L., Ed.; Food science and technology; Marcel Dekker: New York, 2002. ISBN 978-0-8247-9343-2.


Do all reducing sugars show mutarotation? Do all reducing sugars show mutarotation?
I studied that for a sugar to be reducing it must possess a free aldehydo or keto group. Also, for a mutarotation interconversion between α- and β-forms of compound are required.
If there is a free aldehydo group or keto group in the compound must it show mutarotation?
Note: I am a high school student so please try to explain in a simpler manner.

The compounds having hemiacetal in their structure give mutarotation.
Generally, all monosaccharides (glucose, fructose, ribose, arabinose, …) and disaccharides (lactose, maltose, …) give mutarotation, except sucrose.
Sucrose, despite being a disaccharide, fails to give mutarotation because it doesn't contain hemiacetal.


Percent hydrogen and carbon A 1 g sample of an organic compound was burned in a combustion analyzer and yielded 1.42 g carbon dioxide and 0.872 g of water. Calculate the percent carbon and hydrogen in the sample. From the percentages, calculate the empirical formula for the compound.

Your initial compound weights 1g, and it is only made of C and H. The products are CO2 and H2O. In other words (the reaction is not balanced, but for our purpose we don't care):

CxHy + O2 --> CO2 + H2O

From the grams of CO2 you can get back to the moles (and therefore grams) of C in the initial compound, while from the grams of water you can get to the moles (and grams) of H.

moles of C = 1.42 g CO2 x $\frac{\text{1 mol CO2}}{\text{44.0 g CO2}}$  x $\frac{\text{1 mol C}}{\text{1 mol CO2}}$ = 0.0323 mol C
You then multiply by the molar mass of carbon, which is 12 g/mol, and you get 0.388 g.

We do the same thing for hydrogen:

moles of C = 0.872 g H2O x $\frac{\text{1 mol H2O}}{\text{18.0 g H2O}}$  x $\frac{\text{2 mol H}}{\text{1 mol H2O}}$  = 0.0969 mol H = 0.0989 g

Note that we did the same thing twice, but since we have two moles of H in water we multiplied by 2 in the second equation.
Now we just have to find the percentage:
You get

$\frac{\text{0.388 g}}{\text{1 g sample}}$ *100 % =  38.8 % C

$\frac{\text{0.0989 g}}{\text{1 g sample}}$ *100 % =  9.89 % H

I hope this helps!


How to determine whether two hydrogen are homotopic, diastereotopic, emantiotopic? I have been learning about nmr and I am confused about how to identify whether two hydrogen on different carbon atoms are homotopic, diastereotopic or emantiotopic. I can do identify this when two hydrogens are attached to same carbon atoms, but when it is different I am lost.I have searched lot on web, but they only include examples of two hydrogens on same atom, not on different ones. So can anyone help determine what the two hydrogen are in these compounds and why ? Thanks.

You determine by labeling. Take the two hydrogens in question and label them one at a time to get two molecules. What's the relationship between these two molecules? If the molecules are the same, then the hydrogens were homotopic. If they are enantiomers, the hydrogens are enantiotopic. Otherwise, if they are diastereomeric molecules, the hydrogens are diastereomeric.


In what compounds does oxygen exhibit +½ oxidation state? Theoretically it is said that the compound can show oxidation states in the interval $[n,n-8]$ where $n$ is the number of valence electrons.

Is there any compound in which oxygen shows +1/2 oxidation state. If so, then please provide an example.

Yes. The dioxygenyl ion, $\ce{O_{2}^+}$, is a rarely-encountered oxycation in which both oxygen atoms have a formal oxidation state of +1/2. It is formally derived from oxygen by the removal of an electron:

$$\ce{O_{2} → O_{2}^+ + e^{-}}$$

Source


how do i synthesize (E) 2-methyl-1-phenylpent-1-en-3-one? i need a way to synthesize the (E) 2-methyl-1-phenylpent-1-en-3-one
it's 2 days and still i can't find a solution.

Looks like a crossed aldol condensation between benzaldehyde and pentan-3-one.

In a classical approach, you would (under stirring)

Add your base to a dry solvent of your choice
Cool down (ice bath or dry ice/isopropanol)
Add pentan-3-one dropwise to make sure that it's all immediately converted to the corresponding enolate. You do not want self-condensation of the ketone!
Let it come to room temperature (and stir for a while)
Cool down again
Slowly add benzaldehyde. Freshly distilled benzaldehyde is recommended.
Let it come to room temperature again (and stir for a another while)

Check the progress of the reaction by TLC. You should have at least the aldol addition product by now. In order to facilitate elimination of water and thus complete the aldol condensation, it might be necessary to acidify the mixture and heat it.

There are other more modern (e.g. Mukayama aldol reaction, using the silyl enol ether of the ketone) and/or "greener" methods, but they are mostly a variation of the basic approach described above.


How were elements and molecular formulæ determined in the old days? Before modern chemical techniques, say in the 19th century and earlier, how did (al)chemists determine whether a substance was an element or a compound? If it was a compound, how did they determine its molecular formula and structure?

First thing, they couldn't tell elements and compounds apart. See Dalton's list of elements with their symbols.

Lime is right there--and it's not actually an element. Neither is Baryte(s). Soda and Potash probably refer to the oxides of Sodium and potassium, though terminology may have been different back then.

In those days, elements were things that could not be synthesized by addition reactions. If you have a reaction $\ce{A + B ->C}$, then $\ce{C}$ is a compound (can't say much about the other two). Alternatively, if the substance can be decomposed (via anaerobic heating/etc), then it's a compound.

Most probably they would stuff things into the list "innocent until proven guilty" (I don't know about this).

Molecular mass can be determined by osmotic pressure, ebullioscopy, Cryoscopy, etc. (Usually stuff delendant on colligative properties)

Determination of molecular formula is much harder.

For organic compounds, it can be done by burning in a controlled environment. The ratios of oxides produced lets us know the relative quantities of all elements but oxygen. The quantity of oxygen can be determined by finding the decrease in Oxygen content of the chamber, and subtracting it from the Oxygen content of the oxides. The molecular formula can be found if you use the emperical formula and determine molecular mass.

For inorganic compounds, they probably had various tests for ions/etc (like the silver nitrate test for halides). They also had flame tests (and the borax bead test), which could help identify certain ions. I'm not too sure here, though.


Is it possible to prevent methane combining with oxygen? The output from an anerobic kitchen waste digester is mostly $\ce{CH4}$, and $\ce{CO2}$.
If the methane is collected over water; algae/moss may use the $\ce{CO2}$ to grow. Yet oxygen is released ... which, together with methane, is a potent environment.

Will the oxygen, in this closed jar filled with water+algae+moss, recombine to form $\ce{CO2}$ and $\ce{H2O}$? Is there a means to prevent methane combining with oxygen released by the algae?

I doubt that the conditions in an Anaerobic digester (organic acids, ammonia (!), etc) allow algae to grow. If they do, the oxygen they set free will create oxic conditions and stop the AD - Methanogens are obligate anaerobe bacteria.
In nature, Methanogens and Algae inhabit different habitats - methanogens in the mud at the bottom of a swamp, algae (maybe) in the water above.

Processes for separating $\ce{CO2}$ from $\ce{CH4}$ usually work on the $\ce{CO2}$ (absorption in amines or water under high pressure), so it would be hard to pull the $\ce{CH4}$ from the headspace of your vessel - but given the biological points, this is a non-issue.


How do I find miller indices for a plane whose intercepts are fractions of the lattice constant? [I'm talking with respect to cubic lattices alone.]
For instance, if a plane has $x,y,z$ intercepts $a/2,a/2,a/2$ (where $a$ is the lattice constant) the miller index would be $[2\space2\space2]$. The book I'm referring to says that for fractional intercepts, the indices do not have to be reduced to the smallest whole numbers(hence $[2\space2\space2]$).
But, miller indices are supposed to represent a set of parallel planes and I can't think of any other plane with the index $[2\space2\space2]$
Then I saw this:

The image(3rd row, 3rd image) shows the set of $[2\space2\space2]$ planes.
I'm guessing that one of them has intercepts $a/2,a/2,a/2$. What are the intercepts of the other planes, and how are they all $[2\space2\space2]$?
Or, if my method of calculation is wrong, how else do I calculate miller indices for fractional intercepts?
PS: I've only just started to learn this concept, so it's possible that my understanding of miller indices is fundamentally flawed.

The plane closest to the origin has the intercepts you mentioned. The rest have intercepts that are integer multiples of this. In the figure, the origin is on the bottom left of each cell. As the figure shows, the [1 1 1] planes correspond to every second of the [2 2 2] planes because the latter is a multiple (by 2) of the former.

Higher indeces are associated with a lower distance of the planes and with a higher resolution (i.e. larger Bragg angle) of the corresponding reflection.


How does a hydroboration mechanism actually work? Let us say that you have 1-propene reacting with $\ce{BH3-THF}$ and then later with hydrogen peroxide and sodium hydroxide. I eventually get 1-propanol. The way it was explained to was that after reacting with $\ce{BH3-THF}$  we have $\ce{CH3CH2CH2BH2}$ and after the second set of reagents we get our product.

But, I never understood the mechanism, the explanation seems forced. $\ce{BH2}$  goes to the less hindered carbon which explains why its at the terminal. And then for some reason, $\ce{BH2}$  gets replaced by $\ce{OH}$.

I had to write the complete mechanism for you so that you can understand it better. See the image below. I have explained it in the simplest form. Some of the points I would like to discuss before you look at the mechanism are:

In BH3, Boron is less electronegative than H (i.e. hydrogen) , hence in the syn addition I have shown, hydrogen ends up being attached to a more substituted carbon atom.
For every bonding orbital we have an anti-bonding orbital which lies exactly opposite to the bonding orbital and each bonding-antibonding orbital pair can contain a maximum of 2 electrons in them.

Ok now lets head towards the mechanism which I wrote exclusively for you:

If you have any further doubts in the mechanism then you can ask in the comment section below my answer. I hope it helps.


What is the difference between steric strain and torsional strain? I know that in ethane, the extra energy present in the eclipsed conformer is caused by torsional strain.

In butane, the gauche conformation experiences steric strain. But the eclipsed conformation at 0 degrees has substantial amounts of both steric strain and torsional strain.

What is the difference?

TL;DR Torsional strain can be thought as the repulsion due to electrostatice forces between electrons in adjacent MOs. Meanwhile steric strain (also known as van der Waals strain) can be thought as the repulsion when two bulky groups which are not directly bonded to each other become too close to each other and hence there isn't enough space for them.

Here is the more detailed version.

Torsional Strain

Lets consider an ethane molecule. The C-C sigma bond is free to rotate and in principle there are an infinite number of possible conformations. However only 2 are significant, these are staggered and eclipse conformations. Different conformers are usually drawn as Newman projections as they can easily be compared with each other. Below are the Newman projections for the eclipsed and staggered conformer:

The staggered conformer is the most stable conformer while the eclipsed conformer is the least stable conformer. The staggered conformer is approximately $\mathrm{12~kJ~mol^{-1}}$ more stable than the eclipsed conformer. This energy difference between this maxima and minima is known as the torsional barrier.

So what is torsional strain? The reason for the eclipse conformer being higher energy than any other conformers is due to the destabilising electrostatic repulsion between the electron pairs of the C-H sigma bonds on the two carbons. Furthermore, there also exists a stabilising feature which is greater in the staggered conformer. In the staggered conformer there is a constructive orbital interaction involving the bonding and anti-bonding MOs of the adjacent H atoms. This results in hyper conjugation which stabilises the compound.

The two effects that I have mentioned above is what torsional strain refer to. So you can think of torsional strain as the strain that is the result of electrostatic forces.

Steric Strain

Now lets consider butane. Rotation of the $\mathrm{C_2-C_3}$ sigma bond also lead to infinite possible conformers. However there are 4 main conformers shown below:

Here there are 2 types of eclipsed conformers which are the least stable of the conformers. The most unstable one is known as the eclipsed syn form. The syn form is approximately $\mathrm{20~kJ~mol^{-1}}$ higher in energy than the staggered conformer. The reason for this can partly be attributed to torsional strain as there is repulsion between the electrons in the sigma bonding orbitals. However also it can largely be attributed to the repulsion of the two relatively bulky methyl groups as they become too close to each other and there isn't enough space for them.

This repulsion is known as steric strain. Therefore steric strain can be define as the repulsion which occurs when non-bonded groups which are not directly bonded approach each other too closely. This repulsion only exists for bulky substituents, such as methyl or ethyl groups. So in ethane there is no steric hindrance as the hydrogen atoms are not that bulky.


What does the intramolecular aldol condensation of 6-oxoheptanal form? I know that when 6-oxoheptanal is treated with NaOH, an intramolecular aldol condensation will take place. But will a five-membered ring or seven-membered ring be formed?

I know that the five-membered ring is more stable in comparison to the seven- membered ring, but as per the mechanism of aldol condensation the first step is removal of acidic hydrogen (which is also the slowest step i.e. the rate determining step {if I am not wrong}).

So if I remove the acidic hydrogen on C-7 (the terminal methyl group), then the (primary) carbanion formed will be more stable than if I were to deprotonate on C-5.

So I think that the major product should be the seven-membered ring.

Am I right? Or is my reasoning incorrect?

Intramolecular-Aldol reactions involves heat, as you have mentioned, and the base used (NaOH) isn't much affected by steric factors. Therefore, the preferred product should be thermodynamically controlled product. Moreover, as you commented about the formed carbanion being less stable in case of 2° carbon, actually the carbanion is converted to an enolate then the reaction proceeds further. So, the 5-membered cyclic ring should be the major product.


Why is the minus sign (-) not allowed in reaction equations? Recently in my chemistry class, the teacher told us that while writing "half equations" (such as a sodium atom losing one electron to form a sodium ion) we were not to use a simple minus sign. We were only allowed to use a plus sign. For example, this would not be allowed:

$$\ce{Na $-$ e- -> Na+}$$

The "correct" way of writing it would be:

$$\ce{Na -> Na+ + e-}$$

My question is: Is this conventional, or somehow only related to my school? Could any example of a chemical equation utilising the minus sign please be given for clarification?

Don't get too caught up in notational rigidity. You're "allowed" to use any notation you want, if it gets you to the correct answer. If you really want, go ahead and use $\ce{Na\ - \  e^{-} -> Na+}$, or go all out and use $\ce{-Na^{+}\ - \  e^{-} -> - Na}$. However, as Richard Feynman discovered while learning trigonometry, if you start using too much obscure/uncommon/personal notation, you get into trouble showing others your work, which makes doing science hard. Standardizing notation is just a way to save time and headaches later on.

The most basic chemical equations actually carry very little chemical information, to the point that it's easy to write reactions which are chemically outrageous. One of their most fundamental purposes is simply to establish atom conservation (which is why so much time is spent on teaching how to "balance" reactions). Thus, chemical equations are always equivalent to a system of linear equations, which is why you can treat them like such and add/subtract molecules to both sides, multiply by a number, add two different reactions together, etc. Using negative stoichiometric coefficients would thus be a very trivial extension of this behaviour, and the equation would remain completely valid.

Let's look at an example. Take the intermolecular dehydration of ethanol to produce diethyl ether:

$$\ce{2 C2H5OH -> C2H5OC2H5 + H2O}$$

The equation is just as valid if you subtract a water molecule from each side:

$$\ce{2 C2H5OH\ -\ H2O -> C2H5OC2H5}$$

Not many people would write the above equation. However, we do have another type of standardized notation which is often used, and which involves a negative stoichiometric coefficient. Using this notation, the equation is written as:

$$\ce{2 C2H5OH ->[\ce{- H2O}] C2H5OC2H5}$$

The water molecule has a legitimate negative stoichiometric coefficient of -1, and again the equation is exactly as valid as all of the ones above. Why is it that negative stoichiometric coefficients are rarely written in the reactants or products, but common when written atop the reaction arrow? I have no idea, but ultimately it makes no difference. That's just how notation happened to become standardized, and if everyone understands it, that's fine.


Does adding lemon juice to water chemically alter its nutrients and thereby negating any of its health benefits? An acquaintance who apparently has some chemistry understanding told me that eating lemon juice freshly squeezed to drinking water negates all of its health benefits because the water chemically alters the vitamins minerals and enzymes. I am very dubious because a lemon is 82% water and a human body is 75% water and of course lemon juice must be to a very great extent just water.
Also lemon juice is renowned as being a great morning drink because of its health benefits.
My acquaintance however emphatically stands by his statement.
Please clarify this situation for me.
Does water really chemically alter the nutrients in lemon juice rendering a lemon drink of no health benefit whatsoever other than the obvious water base?

Does water really chemically alter the nutrients in lemon juice rendering a lemon drink of no health benefit whatsoever other than the obvious water base?

No, of course not. Otherwise, humanity would have succumbed to scurvy a long time ago.

I encourage you and your friend to visit PubMed where you can peruse hundreds of refereed abstracts of studies that illustrate that nutrients from citrus fruits are bioavailable to humans and other species and necessarily require administration with water.

Here is a one-pager on lemon juice from the Cleveland Clinic, which is probably a reliable enough source in this case if you want to skip the more technical items at PubMed.


Acetic Acid % in vinegar? Given average molarity of acetic acid Question:

  The stockroom claims the percent acetic acid in vinegar to be $2.0\%$. The density of vinegar is $\pu{1.106 g {mL}-1}$. Using average molarity given ($\pu{0.2844M}$) calculate the mass percent acetic acid in vinegar for comparison to the stockroom claim.

Not sure how to approach the problem but this is what I've done:

$$\frac{\pu{0.2844 mol}}{\pu{1L}} \times (\pu{60.05g {mol}-1})\times \left(\frac{\pu{1L}}{\pu{1000mL}}\right) = \pu{ 0.01707822 g {mL}-1}$$

if you have $\pu{1000mL}$, you'll have $\pu{0.001106 g}\text{ Acetic Acid} \div \pu{0.0000170782 g}\text{ Vinegar}$ so that will give $\%$ of acetic acid? I could use some clarification in solving this

$1000$ mL of vinegar will weigh $1106$ g and contain $1000 · 0.017078$ g of acetic acid (= $17.078$ g). Then $17.078/1106 = 0.0154 = 1.54$% acetic acid.
The stockroom manager must be diluting the vinegar, perhaps for a specific experiment. The vinegar I buy is $5$% acetic acid (rarely $4$%).


Neutralisation Reaction Confusion What will happen when $\pu{1 mol}$ of $\ce{H2SO4}$ reacts with $\pu{1 mol}$ of $\ce{NaOH}$?

Will $\pu{1 mol}$ each of $\ce{NaHSO4}$ and water be formed?

Or will $\pu{0.5 mol}$ of $\ce{H2SO4}$ will remain yield $\pu{0.5 mol}$ of $\ce{Na2SO4}$ and $\pu{1 mol}$ of water?

I believe it will be the second way since H2SO4 is a strong acid and would again quickly react with the $\ce{NaHSO4}$ formed. But I am not quite sure how the mechanism will be and whether the Rate constant of the second reaction will be high enough to facilitate this.

What will happen and why?

First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are:

$$\ce{H2SO4(aq) + NaOH(aq) -> HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$

$$\ce{H2SO4(aq) + 2NaOH(aq) -> SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$

Your saying that reaction \eqref{rxn:2} might happen even when $\ce{NaOH}$ is limiting, and you would end up with some sulfuric acid and some sulfate. In a thought experiment, let's say that is what happened. Now, we can discuss whether sulfuric acid and sulfate, which are present in equimolar amounts, will react with each other according to:

$$\ce{H2SO4(aq) + SO4^2-(aq) <=> 2HSO4-(aq)}\label{rxn:3}\tag{3}$$

Sulfuric acid is a strong acid, and sulfate is a weak (very weak) base. This is like mixing hydronium ($\ce{H3O+}$) and hydroxide ($\ce{OH-}$) ions, which (mostly) form water. So just like water, which occurs mostly undissociated, the equilibrium of reaction \eqref{rxn:3} lies on the side of the products when the reactants are present in equimolar amounts.

Of course, if you add more $\ce{NaOH}$, you would start making a lot of sulfate.

If you wanted to know all of the products and their concentrations, you would have to consider water auto-dissociation as well. The argument I made is not meant to be quantitative, but to show why we don't expect sulfuric acid to be left over. In aqueous solution, there is never much sulfuric acid (even in the absence of $\ce{NaOH}$) because, as a strong acid, it dissociates.


Identification of types of alcohols An unknown compound was found to react with sodium hydride with the evolution of hydrogen. The compound could not be acetylated under the normal conditions for acetylation. The compound resisted oxidation under mild conditions, and under vigorous conditions yielded only products of molecular weight much smaller than the unknown. On the basis of these facts the compound is:

  a primary alcohol
  an aldehyde
  a secondary alcohol
  a tertiary alcohol
  a secondary amine

The correct answer is (4).

From the first statement, it is clear that this compound must be an alcohol (since alcohols react with sodium hydride and produce hydrogen gas in this reaction).  I could guess that the answer was a tertiary alcohol as secondary and primary alcohols would easily oxidize, but I'm not sure what the products of the oxidized tertiary alcohol would be?

Also, why wouldn't a tertiary alcohol undergo acetylation?  I don't see any reason why it shouldn't?

Reaction of tertiary alcohol with NaH

As all types of alcohol can easily reacts with NaH to produce hydrogen gas and sodium salt of alkoxide ion or alcohol. This shows alcohols have acidic hydrogen. It is type substitution reaction. The general reaction is as follows :

When tertiary alcohol say $\ce{(CH3)3COH}$ reacts with $\ce{NaH}$, hydrogen gas is released as follows :
$$ \ce{(CH3)3COH + NaH  ->  (CH3)3CONa + H2}$$

No acetylation of tertiary alcohol occurs under normal conditions

The acetylation means addition of acetyl group to a substrate. In the case of primary alcohol, acetylation is done under normal condition. Reactivity of secondary alcohol is somehow decrease for acetylation. In the case of tertiary alcohol, it is possible in extreme conditions. There is main reason behind.

As we know acetylation is nucleophilic substitution reaction. Here, the attacking nucleophile for acetylation of alcohol is alkoxide ion. More is nucleophilicity, more be reactivity of nucleophile to attack at carbon of carbonyl group of acetyl chloride.

Nucleophilicity order : $\ce{CH3O- > CH3CH2O- > (CH3)2CHO- > (CH3)3CO-}$

This is reason that tertiary alcohols are least reactive for acetylation. You can more understand by the following mechanism :

Oxidation of tertiary alcohol

Tertiary alcohols Undergo oxidation with oxidizing agents such as chromium tri oxide, potassium permanganate, potassium di chromate or nitric acid in neutral, acidic or alkaline medium to give the products. The oxidation may be partial or complete. Tertiary alcohols does not get oxidized by mild oxidizing agents either in aqueous, alkaline or under neutral conditions. They undergo oxidation under acidic condition.In this process of oxidation tertiary alcohols first undergo dehydration to form alkenes. The alkene thus formed then undergoes further oxidation and forms a ketone with one carbon atom less than the alkene. On further oxidation the ketone forms carboxylic acid with one more carbon atom less than the ketone. Thus the oxidation of tertiary alcohol gives a product with less number of carbon atoms than that of the original alcohol taken.


Size of hydrocarbons and boiling point Increasing the contract surface area between hydrocarbons raises boiling point, so hexane should have a higher boiling point than propane. This doesn't really make sense to me. What is the difference between pulling apart a pair of propane twice, versus one pair of hexane? Why would the total amount of energy needed differ? It's seems to go against the concept of pressure.

Some data on size of hydrocarbons and boiling point

I found complete data on the butane/octane pair but not the propane/hexane pair, so I am using those as illustration (I think they are taken at a temperature near the normal boiling point, respectively).

$$
\begin{array}{|c|c|c|c|c|}
\hline
\text{substance} &  \Delta H_{vap} & \Delta S_{vap}  & T_b & \text{ref}\\
\hline
\text{butane} & \pu{22.4 kJ/mol}  &  \pu{82.30 J/mol K}& 273 K & [1]\\
\hline
\text{octane} & \pu{39.1  kJ/mol}  &  \pu{98.3  J/mol K}& 398.7 K & [2]\\
\hline
\end{array}
$$

Ref. 1: https://webbook.nist.gov/cgi/cbook.cgi?ID=C106978&Mask=4

Ref. 2: https://file.scirp.org/pdf/OALibJ_2016091314302030.pdf

The enthalpy of vaporization is determined by measuring equilibrium vapor pressure over a range of temperatures (Clausius Clapeyron equation). Then, the entropy of vaporization is calculated from the boiling point and the enthalpy of vaporization ($\Delta G_{vap} = 0$ at the boiling point):

$$ \Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$$

  What is the difference between pulling apart a pair of propane twice, versus one pair of hexane?

As far as enthalpy is concerned, there is not much difference. You have to disrupt about the same dispersion forces removing one molecule of hexane from liquid hexane or removing two molecules of propane.

However, if enthalpy were the only aspect you had to consider, liquids would never turn into gases because you lose all interactions in the gas phase. Experience shows that if you increase the temperature sufficiently, at one point you disrupt the interactions between molecules in a liquid and get a gas.

  Why would the total amount of energy needed differ?

What determines whether a process happens or not is not the change in enthalpy alone, but the change in Gibbs energy. The Gibbs energy contains an enthalpy component and a highly temperature-dependent entropy component. The entropy change of a molecule from the liquid to the gas state is about the same for butane and octane (also see Trouton's rule), but the enthalpy change is different by a factor of two. That's why the two substances have different boiling points.

$$ \Delta G_{vap} = \Delta H_{vap} - T_b \Delta S_{vap} = 0$$

$$ T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}}$$

If you ask for the analysis for evaporating a kilogram each of butane and octane, the enthalpy change for the entire sample would be similar, but the entropy change for the entire sample would be off by a factor of about 2. This is because the molar entropy of vaporization is similar for butane and octane, but a kilogram of butane contains about twice the number of molecules compared to a kilogram of octane.

  It's seems to go against the concept of pressure.

This is a bit of a puzzling statement. There are two things that require energy when a liquid turns into a gas at constant pressure. The first is disrupting the intermolecular forces in the liquid. The second is pushing against the pressure to increase the volume of the sample as liquid turns into gas. The latter is abbreviated as P-V-work. If you look at internal energy changes, you have to consider it. If you look at enthalpy changes, however, it is taken out of the equation because enthalpy is defined as
$$ H = U + P V$$
As we are discussing Gibbs energy and enthalpy, we don't have to worry about P-V-work. ("It is not part of the bill").


Metallic bonding and covalent boding Why do metallic bonds have delocalized electrons, whereas covalent bonds only share electrons? Does it have something to with the atomic masses of the individual non/metals or something else? I'm really interested to know.

Why do metallic bonds have delocalized electrons, whereas covalent bonds only share electrons?

To be somewhat flip this is like asking why are redbirds red and blue birds blue?

In order to pigeon-hole bonding ionic, covalent, aromatic, and metallic bonds are "artificial" constructs used to predict behavior. But there are always exceptions, and no bond is purely in any one of the categories.

So in general a sample which has metallic bonds has delocalized electrons and hence will conduct electricity. Such elements would be metals. If the sample only has covalent bonds then there would not be any delocalized electrons and the sample would be a nonmetal.

For some elements the nature of the bonding yields different results. For example graphite conducts electricity, but diamond won't.

To add another wrinkle in this we can think of gaseous ionic or covalent molecules but in general being metallic is a "bulk" property. So we have to consider not only the bonding in the "molecule", but how "bonding" is distributed through the solid matrix of the sample. For example there are discrete water molecules in ice, but in a diamond there are no discrete carbon "molecules."


Why does a first order reaction only depend on the concentration of a single reactant? I am new to kinetics so please explain it from basic, If there is a reaction  -

$$\ce{A + B + C -> D}$$

In first order reaction rate is given by-

$\mathrm R = k[\mathrm A]$ or $\mathrm R = k[\mathrm B]$ or $\mathrm R = k[\mathrm C]$

my question is why cant $\mathrm R = k[\mathrm A]^{1/3}[\mathrm B]^{1/3}[\mathrm C]^{1/3}$
 even though order of reaction is still $1$,----- is it by definition that 1st order reaction always depend on concentration of single reactant, or is there any experimental proof for this?.

As a purely mathematical definition, a first order reaction is one in which the rate is directly proportional to concentration of reactants. The thing you should look for is physical examples of this. This only happens if you have reactions where one compound is decreasing and the other one is abundance (pseudo - first order) , or , radioactive decay where the decay rate is proportional to number of atoms at a point.
The reaction you have shown with three reactants would most probably me not first order unless two of the three are abundant. So, the solution to the confusion is to think of the physically possible scenarios which the equation models instead of creating hypothetical situations and force the math to model that.


Finding the charge passed in electrolysis of water using volume of hydrogen released During the electrolysis of water we have gotten $\pu{10 dm3}$ of hydrogen. Temperature was $\pu{0 °C}$ and pressure was $\pu{101325 Pa}$.
What charge has passed trough the electrolysis article/joint?

I don't know how to find the charge while knowing pressure, temperature and volume. How do I proceed with this question?

$PV = nRT$ and $\ce{2H+ + 2e- -> H2}$ are your best friends.
Step 1: Find the number of moles of hydrogen gas released
Use the formula: $PV = nRT$, where $P =$ Pressure exerted by the
gas, $V =$ Volume of gas, $n = $ number of moles of gas, $R = $ universal gas constant, $T = $ temperature of gas.
Using this formula with $P$ in atm and $V$ in liters we get:
$$n =\frac{PV}{RT} = \pu{0.0044 mol}$$
Step 2: Find the number of farads of charge used
We know that $$\ce{2H+ + 2e- -> H2}$$
This means that for every mole of hydrogen released, $\pu{2 moles}$ of electrons are used. A farad is the charge on $\pu{1 mol}$ of electrons and so number of farads of would be two times the number of moles of electrons. So charge would be $\pu{0.0088 F}$.
Step 3: One farad = 96500 coulombs
Therefore, $\pu{0.0088 F = 850.32 C}$ and that should be the answer.


What concentration is phenolphthalein indicator ? 1% or 0.1% I read some information about phenolphthalein indicator, some websites suggested phenolphthalein indicator of 1%, others mentioned phenolphthalein indicator of 0.1%. What is the difference of both of these concentrations when we use them in titrations?
I assume that 1-2 drops of phenolphthalein indicator of 1% can be used as an indicator but for phenolphthalein indicator of 0.1%, it needs to drop more with 4-5 drops to obtain the same effeciency.
Look forward to hear from you all

During titration of small amounts of acids, the molar amount of the indicator in 1-2 drops of $\pu{1 \%}$ indicator solution may not be negligible compared to the acid molar amount, affecting the result.
So for that cases, $\pu{0.1 \%}$ solution is used, to be able to dose smaller indicator amounts.
As the phenolphalein molar mass is about $M=\pu{318 g/mol}$, 1 drop of $\pu{1 \%}$ solution is equivalent about 3 drops of $\pu{0.01 M}\ \ce{NaOH}$.


Rate of reaction involving different stoichiometric coefficients Suppose I have a reaction (Say it is the Haber process):

$$\ce{N2 + 3H2 <=> 2NH3}$$

If we have to calculate the average rate of reaction of the above we follow the reasoning the rate of consumption of $\ce{N2}$ is one- third that of $\ce{H2}$ and half that of production of $\ce{NH3}$. We go ahead and divide the change in concentration of $\ce{N2}$ or $\ce{H2}$ or $\ce{NH3}$ by their stoichiometric coefficient. My doubt is why rate of reaction is taken as the rate of consumption of $\ce{N2}$? Why can't it be the rate of production of $\ce{NH3}$? Why can't we say that the rate of reaction is the rate of production of $\ce{NH3}$ or half the rate of consumption of $\ce{N2}$ or one and half times that of rate of consumption of $\ce{H2}$?

My doubt is why rate of reaction is taken as the rate of consumption of N2?

It's done that way because rates have units, and most people don't pay enough attention to the units.  Usually in chemistry, rates are units of moles per unit time (per unit volume or per unit catalyst mass or per unit surface area or something).  But you need to know moles of what?

By writing the reaction as $\ce{N2 + 3 H2 <=> 2 NH3}$, you are implicitly defining that a mole of "reaction" corresponds to one mole of $\ce{N2}$, two moles of $\ce{NH3}$, and three moles of $\ce{H2}$.  If you wrote the reaction as $\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3}$, then one mole of this new reation would correspond to one mole of $\ce{NH3}$, but only half a mole of $\ce{N2}$, etc.

  Why can't it be the rate of production of NH3? Why can't we say that the rate of reaction is the rate of production of NH3 or half the rate of consumption of N2 or one and half times that of rate of consumption of H2?

It could be any of those things.  The choice is up to you.  You just have to be consistent.  You might this related chem.SE question and answer helpful.

When you say "the rate", you have to specify what rate you are referring to.  And if you talk about rates "of the reaction", the agreed upon reference is that you are counting moles of species that have stoichiometric coefficients of 1.


What is the substance in the new water cooler filter that give bubbles and taste a bit like soap Recently the water cooler was replace with a new filter.

However, when I refill my water kettle with the water from the water cooler, there are many bubbles and the taste was a bit like soap.

What chemical cause such effects? (Is it related to the new filter?)

The most likely cause of a "soapy" taste and sudsing from filtered water is a reverse osmosis filter, for which the usual membrane preservative used to keep the filter from degrading before its first use is plain old glycerin, a common ingredient in and byproduct of traditional soap (though most cleansing products nowadays, even those marketed as soaps, are technically anionic surfactant "detergents", based on alkyl-sulfates). The glycerin is hygroscopic, meaning it will attract and hold water on the membrane surface to keep it from drying out.

An activated carbon filter sometimes has a similar preservative to improve the material's shelf life, or else a binder that allows the manufacturer to mold a block of the carbon rather than loading loose powder into filter housings, which reduces carbon dust in the factory (a major fire and health hazard of these types of plants).

In all cases, the manufacturer usually recommends flushing any new filter with between one and five gallons of source water to wash out these substances, along with any loose filter media or other byproducts of manufacturing. This obviously didn't happen in your case. The glycerin in the water won't kill you; it's an FDA-approved food additive used as a sweetener and "plasticizer" (it's used in cake frosting for instance, to avoid it hardening after it sets). In sufficient doses it's an irritant to the mucous membranes and the GI tract, which can cause nausea and diarrhea, but you'd have to eat almost a kilo of the pure stuff for it to become dangerously toxic.


Applicability of Enthalpy-Heat relation for a non-isobaric process Consider a process in which state variables changes from $(p_1, V_1, T_1)$ to $(p_2, V_2, T_3)$ . I've learned that to calculate the change in enthalpy for any process, we consider a hypothetical isobaric process (Since enthalpy is a state function, it doesn't depend on the path).
Should the hypothetical process have the same initial and final state variable or just the same initial and final temperature? Because if we consider the same initial and final state variables, then we can't form a hypothetical isobaric process since the pressure changes in the actual process and the formula $\Delta H = q_p = n \cdot c_p \cdot \Delta T $ fails here.

Disclaimer: The below discussion only holds to be completely true for reversible processes and for ideal gases.
I think the problem is you're thinking of it the wrong way, that is, you first think that $ q= nC_p \Delta T$ and for constant pressure process $q_p= \Delta H$ , and hence, $ \Delta H = nC_p \Delta T$
In actuality, it is the other way around. Similar to how internal energy is always $nC_v \Delta T$, the enthalpy is always $nC_p \Delta T$ for an ideal gas. A simple proof is shown below:
$$ \Delta H = \Delta U + \Delta (PV)$$
For $n$ fixed moles of an ideal gas,
$$ U = n C_v \Delta T$$
and,
$$ \Delta (PV) = nR \Delta T$$
Hence,
$$ \Delta H =n\Delta T [  R + C_v]$$
Now, from Mayers relation,
$$ C_p  = C_v + R$$
And hence,
$$ \Delta H = n \Delta T C_p$$
So, as seen above, if you can accept that $\Delta U= nC_v \Delta T$ for an ideal gas always then you must be able to accept a similar statement for enthalpy

Regarding state variables:
Yes, there are indeed many states for an ideal gas which correspond to the same enthalpy since enthalpy is a function of the only temperature. The state variables of pressure and volume are irrelevant for enthalpy calculations unless you don't have the final and initial temperatures. In such a case, you can use the ideal gas law to find expression for the temperature at different states.

Note:
A discussion of proving $q_p =\Delta H$ can be found in this post
You may think the above proof is circular reasoning since regularly Mayers relation is proven by using $ \Delta H = nC_p \Delta T$. However, there exists another way to prove it by considering path functions. Have a look at the first twenty minutes of this lecture by MIT OCW.


Nomenclature of amoxicillin I am trying to understand the nomenclature of amoxicillin:

The IUPAC name that Wikipedia has given is (2S,5R,6R)-6-{[(2R)-2-amino-2-(4-hydroxyphenyl)-acetyl]amino}-3,3-dimethyl-7-oxo-4-thia-1-azabicyclo[3.2.0]heptane-24-carboxylic acid.

I think I understand most of it but I have two problems:

Why is there a 24 before the carboxylic acid?
There is a $\ce{C=O}$ group between two nitrogens that doesn't appear in the name. I was expecting an amide, rather than amino or oxo.

To preface this, I'm not sure if this is indeed the correct IUPAC name as per the latest 2013 recommendations. It might well be, but I don't know. However, that's not crucial to your question, which is more about understanding the name that is given on Wikipedia. Here's the structure of amoxicillin:

The amide C=O is in "acetyl", which refers to the $\ce{CH3C(=O)}$ group (see e.g. acetyl chloride, $\ce{CH3COCl}$):

The '24' is probably a typo and should instead be '2'.

Following this numbering scheme you can see that the lactam C=O is in '-7-oxo'.


Can high-tin bronze be made sharper than low-tin bronze? I am not sure how sharpness is measured, but I do know that certain materials can be designated sharper than others, for example obsidian is said to be the sharpest material:

  The prismatic glass blade is infinitely sharper than a honed steel edge

So my question is, will a honed edge of high-tin bronze be sharper than a honed edge of low-tin bronze, and if so why?

Unfortunately I don't have a solid reference for this, but what I remember from my undergrad courses is that the limit to sharpening is basically determined by the porosity and tensile strength of the metal you are trying to sharpen. At some point you will just run into the microscopic roughness of the metal itself and that is where sharpening ends.

Low- and high-tin bronze are pretty rough classifications so I couldn't tell you which will result in a lower porosity or higher strength, but if I would have to guess than I think low-tin bronze can be made sharper, because copper has a higher strength than tin.

Maybe if you can get your hands on this book chapter, you can find out a more exact answer.


Prediction of products in reaction of dicarbonyl with alpha-haloketone Following is the question that was asked in my exam. a,b,c,d are the various options, one of which is the answer. I am familiar with the Reformatsky reaction, Aldol condensation, Knoevenagel's reaction and general base catalysed carbonyl reaction. 1 and 2 are my attempts at deriving the mechanism and hence getting at the products. But none of those succeeded to bring about the required products.

I need help in how to proceed through in this case and derive the products.

As you've already figured out, cyclohexane-1,3-dione is more acidic than chloroacetaldehyde, the pKa of the diketone in water should be around 5.3.
The enolate of 1 (sorry for the circle in the structure, i'm to lazy to fix it) and 2 undergo an aldol reaction yielding 3. With a bit of charge shifting, we get again a nicely stabilized enolate (4), which dehalogenates (you might call that an O-alkylation as well) to 5. Subsequent elimination under aqueous conditions results in the formation of 6. I'd say this is, more or less, an example of the Feist-Bénary reaction.

Btw, I created the drawings with BKChem.

A conceivable depiction for the "transformation" of 3 to 4 is given below, but please note that this is just a model to please our eyes and minds ;)


Is there a gas with metallic luster? We know that both solids and liquids can exhibit metallic luster. I wonder whether there is a gas that when filling a transparent envelope (such as a glass tube) would make it appear with metallic luster?

Good question... would a plasma qualify? Metallic luster is due to a sea of electrons, free to move and to reflect (well, re-radiate) incident radiation. A plasma can be created in a gas, such as the interstellar medium, or in Earth's atmosphere. The Kennelly-Heaviside layer is such a "shiny", reflective layer, causing skip signal propagation to extend the range of radio communication. So if you saw in radio wavelenghts, rather than the visual spectrum,  plasma is shiny!

Another possibility would be a Bose-Einstein condensate... but the act of illuminating it tends to destroy it.


If alloys are homogeneous mixtures, why can't we separate their components? An alloy is a material composed of two or more metals or a metal and a nonmetal. And, they are usually formed by heating the elements to their melting points, and then cooling them, so that the components mix. Now, why doesn't this works backwards i.e. if we heat the alloy again to melting point of their constituents, and they should separate?

If you melt the alloy and then use electrolysis, or if you use a redox reaction, you should be able to separate the components.

For example, if you used a basic acid, let's say hydrochloric acid, on a zinc-copper alloy, I'd imagine the zinc would be removed from the alloy and taking the hydrogen's place, forming zinc chloride, leaving hydrogen gas and copper metal.

This is based off of simple redox reaction memories and no personal experience. However, if you're curious, try it out!


Numbering heteroatoms and groups in benzene derivatives Can someone please explain why the carbon in 8-hydroxyquinoline is numbered 8, while in 1,10-phenanthroline the nitrogens are numbered 1 and 10? I'm counting carbon atoms and just cannot see why the hydroxy is in position 8 and nitrogen atoms are in positions 1 and 10.

Quaternary carbon atoms i.e. Those with no hydrogen atoms attached and in these examples, the ones that are common to both or all rings, are not counted. Try it with 1,8-dihydroxynaphthalene. Both OH groups should be pointing in the same direction.

In your examples, as you probably know, the heteroatom is numbered first.


Propagating uncertainty with the slope of a tangent line I have done an experiment that found the initial reaction rate for a reaction that created gas in $\pu{kPa s^-1}$ by finding the slope of the tangent line at time = 0. How do I propagate uncertainty for this? Is there any uncertainty propagation that has to be done?
$$\text{slope} = \frac{\pu{(99.726 ± 0.01) kPa} - \pu{(99.71753 ± 0.01) kPa}}{\pu{0.076 s} - \pu{0.061 s}} = \pu{0.56 ± 0.02 kPa s^-1}$$
This is what I have so far. My teacher said that the error for the time was so small that it is negligible (but not a perfect number). I do not think using percentage uncertainty would make sense here because you could measure the slope with however big or small numbers you wanted, depending on what two points you chose to use.
I got Vernier Logger Pro to do the tangent and slope calculations for me which is why there are more decimal places than uncertainty. Not rounding until the end. Is that wrong?
These are the instructions I was given:
only a simple treatment is required. For functions such as addition and subtraction, absolute uncertainties can be added; for multiplication, division and powers, percentage uncertainties can be added. If one uncertainty is much larger than the others, the approximate uncertainty in the calculated result can be taken as due to that quantity alone.

You have not propagated your errors properly. Note that the sum in your nominator evaluates to a small number with an error of roughly the same size (99.73±0.01 - 99.72±0.01 = 0.01±0.01). Also, error propagation cannot be done by simply adding up the errors.
Instead of calculating the error propagation manually (cumbersome, and prone to errors, hehe ;-) I think it's better to use suitable software.  Both R and Python (both are freely available, commonly used tools) can handle error propagation.
Here's how to do it with R (the code below should be reproducible, assuming you have installed R).
install.packages("errors")
library(errors)
options(errors.notation="plus-minus", errors.digits=1)
y1 <- set_errors(99.726, 0.01)
y2 <- set_errors(99.71753, 0.01)
t1 <- set_errors(0.076, 0)
t2 <- set_errors(0.061, 0)
(y1 - y2) / (t1 - t2)
# 0.6 ± 0.9

The value of the slope is 0.56 kPa/s, but since the error is 0.9 kPa/s, we should report just a single significant digit, so that makes it 0.6 ± 0.9 kPa/s.
And not rounding till the end is not wrong, in fact, it's good practice. As long as you handle your uncertainties correctly, too many significant digits (due to intermediate calculation steps or whatnot) become practically irrelevant, since your value cannot be more precise than the size of your uncertainty anyway.
Good luck!


Why do you need to convert copper (II) hydroxide to copper (II) oxide before converting to copper (II) sulfate? This is the protocol for a practical class I once took. The idea is that copper nitrate is converted to hydroxide, oxide, sulfate and then finally solid copper. I know that $\ce{CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O (l)}$, but can't you avoid converting the hydroxide to oxide altogether and just do $\ce{Cu(OH)2(s) + H2SO4(l) -> CuSO4(aq) + 2H2O(l)}$? Note that $\ce{NaCO3}$ is decanted and washed off before reacting copper (II) oxide with sulfuric acid, but surely you can also do the same thing with copper (II) hydroxide?

I would like to add to what Jason has already mentioned:

1.Copper Hydroxide is comparatively tougher to handle since it contains a lot more impurities ( when we are converting Cu(II)nitrate to hydroxide using NaOH, the impurities could be sodium nitrate, excess sodium hydroxide. )

Copper Hydroxide itself is blue, so the equivalence point (when exact amount of copper hydroxide has reacted with concentrated sulphuric acid) of the reaction would be hard to detect, and so yield would drop (since you would again have to neutralise the excess sulphuric acid before purification or conversion to copper metal).
In contrast to this CuO is a black precipitate and can be easily isolated (comparatively) by a Hirsch funnel (or any such equivalent apparatus). This improves the overall yield of further steps and eases the process of purification.

Hope this helps.


Reaction kinetics - Organic chemistry On an exam in organic chemistry I was asked why the reaction B proceeds faster than reaction A.

I answered that: in reaction B, I- is a better leaving group than Cl-, the reason being it is a weaker base.

The answer our teacher gave after correcting the exam was:
I- is a good nucleophile and at the same time a good leaving group.

I scored 1,5p/2p for my answer.

Are the two answers very different or is there some incorrect information in mine?

Your answer is simply missing something. Yes, I is a better leaving group, therefore an Sn reaction of an iodide proceeds faster than the one of the chloride. But your starting material in both cases is the chloride. You first need to exchange that chloride for an iodide and that only works well if the iodide is a good nucleophile.


Alpha hydrogen and hyperconjugation Why do we say that an alpha hydrogen is necessary for hyperconjugation? What exactly is an alpha hydrogen? From what I read online, it is the hydrogen attached to a functional group. Why is that required?

For example, in $\ce{CH3-CH=CH2}$, hyperconjugation takes place, but I don't see clearly any functional group that would facilitate it.

Alternatively, in $\ce{CH3-COOH}$ it does not take place.

Why is this?

To remember:

Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system. Further information.

The proposed systems to take into account.

Looking at the resonance structures (which is a synonym of mesomeric effect) of the propene

(source: tutorvista.com)

and acetic acid

(source: nonsibihighschool.org)

Therefore, in hyperconjugation analysis you need to look at the position of the electron density. Not the vicinal hydrogen.

Moreover and with the aim to give a complete answer, the hyperconjugation of the acetic acid can be very well understood at here.


Most stable chair conformation I am asked to draw the most stable chair conformation of compound 1.

After doing so I am a little uncertain which of the conformations is the most stable. The most stable conformation is the one where the most bulky group is positioned equatorial. However, do I prioritize Cl over the methyl- and isopropyl-group or are the two groups more prioritized due to them being bonded to Hydrogens which take up more space than a Cl-atom. From what I have read, I would say that the chair conformation on the top is the most stable, prioritizing the methyl- and isopropyl-group. Is that correct?

The lowest energy conformer for a cyclohexane ring with multiple substituents will be the one in which the largest group occupies an equatorial position.

Steric parameters are tabulated in many organic chemistry textbooks, the most common of which is the A-value. The A-value represents the difference in energy (ΔG) between a group being axial/equatorial on a cyclohexane ring, thus, as a group gets bigger, its A-value increases, reflecting the increasingly high energy needed to flip the ring.

Bulky substituents such as tert-butyl are often said to "lock the conformation", that is, with a tert-butyl group on a ring, the energy to flip is sufficiently high that ring is "locked" in the conformation with the tert-butyl group equatorial.

In your example, building a simple model with a kit will show you that having the iso-propyl group axial isn't terribly favourable (interestingly the iso-propyl group has a similar A-value to ethyl, due to the ability of the group to rotate, avoiding having a methyl group pointing into the ring).

I ran a few simple calculations (molecular mechanics followed by DFT optimisation - gas phase) on the axial and equatorial conformations, which showed that the two conformations are separated by around 3.4 kcal/mol.  The minimum energy conformation of your example is shown below:


How were weights of elements and compounds determined? I have browsed through many questions on this site but I still don't understand this:

How chemists in the past used to determine the weight of substances (which I believe they randomly called elements or compounds) without using modern techniques?

For example ... we know the molecular formula of water is $\ce{H2O}$

But how did chemists determine the composition of water?

How did they know hydrogen and oxygen, and not some other substance, made water and how could they determine the weight according to which combined?

What was the complete procedure behind it?

EDIT: Just like I said.... That answer still doesn't show us the complete picture... Moreover, the answerer has used phrases like (I don't know) and (Maybe) many times....

If you check out the Wikipedia page on Properties of Water, it has a very small history section which says Cavendish (not a chemist, arguably) determined the composition of water in 1781. Then following that lead to Cavendish's article, you'll learn quite a lot, unless you already knew it. As far as us putting "the complete picture" in front of you on a plate, yeah, that's not going to happen. We'd have to know what your background was, and even then what exactly would satisfy your concept of what the "complete" picture was. The up-side is that although satisfying yourself is up to you, the web has a lot of resources you can get help from. When it comes to a lot of scientific, mathematical, and technical subjects, though, you'll have to put in the study (and practice) time to learn and understand it, course work and textbooks will be difficult to replace with unorganized web research. OK, enough philosophy. I encourage you to put together a time-line of alchemy, natural philosophy, and science at around the start of the 19th Century. You probably know that the idea that matter is composed of small indivisible particles (atoms) was first seriously considered by the ancient Greeks, right? And that various technologists have used recipes (based on weights or volumes or both) to repeatedly create (or destroy) various materials of interest, from time immemorial. Alchemists were dedicated transmutation, so what they knew and when they knew it is important here. Proust didn't demonstrate his law of definite proportions until almost 20 years after Cavendish discovered the constituents. So first we knew what it was made of (oxygen and hydrogen, although the concept of elements was not yet firmly established.) And see also https://en.wikipedia.org/wiki/Oxygen#Early_experiments as well as the following discussion on the Phlogiston theory. Probably the reason you haven't gotten a clear answer here is because this isn't a technical question, it is a historical question. To get a clear complete "picture" you need to literally waste your time learning about all sorts of discredited theories (Phlogiston, and the earlier even more unfortunate Earth, Air, Fire and water "theory") and then learn how the various people overcame the blunders made before them to find the "right" explanations for the world around them. Needless to say, that is an ongoing effort but to recapitulate all of the dead-ends isn't something we're going to do in a short answer, frankly. Anyway, put together a time-line and you'll begin to see how the various strands of the story wove together. HTH.


Determine the structure for C6H13Cl using the following data. The H NMR for C6H13Cl states that there are 3 different types of protons:

6H, triplet
4H, quartet
3H, singlet

What will the structure look like?

Here's a useful hint:  when you have a triplet and a quartet with the area ratio given as above, it indicates the presence of ethyl groups.  Since the total area under those peaks is 10 H, there are two of these  groups equivalent to each other.  Only one isomer with the given formula does that.


Can we find 100% probability region in an orbital if the position of an electron is limited by speed of light? The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.

The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.

My question is, provided that we knew position of an electron at a give time, could we now draw a boundary surface diagram that encloses a region in which probability of finding the electron is 100% for some point in time in future?

This would be possible because the electron cannot travel faster than the speed of light, so we could draw a sphere of radius $3*10^8$ meters centered at the current known position and say that probability of finding the electron after one second will not be beyond the region enclosed by the sphere.

Or is this question invalid because we can't know the position of an electron ever, not even within a finite uncertainty that would allow us to draw such a sphere?

The (Wave)functional forms for the surface which contains the 100% probability for finding the electron extends to the literal edge of the universe.  In the context of 'wave' it is nonsensical to measure an electron's position, and hence its speed may not be determined.

In the context of particle, an electron, having nonzero mass, may not be accelerated to the speed of light.


Is there a real detection limit in NMR spectroscopy? From what I've read, my impression is that with a sufficient number of scans etc., one can get spectra from even very dilute samples (S:N ratio ~ $\sqrt{\text{NS}}$). Obviously, this is not always possible in practice and it is not necessarily efficient use of instrument time.

However, assuming that (1) we have infinite instrument time and (2) the sample is indefinitely stable, is there still a true detection limit? If there is, where does it arise from?

[Unrelated: this question originally came about because I was reading a paper by Gschwind on CEST (J. Am. Chem. Soc. 2018, 140, 1855), but I realise that that technique is dealing with short-lived intermediates – hence my insertion of assumption (2).]

This is an interesting problem which also appears in gamma spectroscopy. Both 1D NMR and normal gamma spectroscopy are related. A sample emits photons, the photons are then captured by a detector.

In the case of a modern FT NMR machine the nuclei are excited by a radiofrequency pulse, the nuclei in the sample then re-emit radiowaves and the sample relaxs back to the way it was before the pulse in the free induction decay. After capture of the signal a analouge-digitial converter (ADC) supplies it to a computer which then uses the fourier transform to convert it into a histrogram of intensity against frequency.

Gamma counting uses something similar a energy dispersive detector records a series of events, these events are then fed into a ADC whose output is recorded by the computer. Normally again as a histrogram of events at different energies.

In both NMR and gamma spectroscopy there are two sources of noise, I define noise as unwanted signals in a system or spectra.

If we use gamma spectrscopy as an example for a moment, we have random noise due to cosmic rays (things like muons) and effects like Compton scattering. This will cause the random appearance of events in the detector with random energies. This will have a similar effect on the appearance of the spectrum to the random noise which exists in the electronics of a NMR machine. If we were to lower the noise level of the amplifers in the signal path before the ADC in a FTNMR machine we would be able to record spectra faster.

The more times we run the experiment the smaller the effect of the random noise will be on the final spectrum. I once ran an experiment for three weeks to get a spectrum from a lump of trinitite I bought off ebay. It turned out to be genuine.

The question is about the measurement of a stable sample. When I was a PhD student I worked on some samples which were unstable. I suspect to this day that I managed to dissolve for a short time somethings in my NMR samples which then later formed a superinsoluble solid. Once I saw a peak appear in the FT-NMR spectrum at the expected chemical shift which then disappeared again. We can consider this side question also.

Now one problem I see in both systems is if some non random unwanted signal enters the spectrometer before the ADC.

Consider for a moment if we were to do NMR spectroscopy in a 1:1 mixture of CHCl3 and CDCl3. We would have a very strong chloroform peak, this would limit the sensitivity of the amplifiers which can be used before the ADC, by forcing the system to reduce the senstivity of the amplifiers before the ADC what will happen is that the effect of random noise in the ADC will become greater. This would make it harder to make the measurement.

Also with a NMR spectrometer we have a limit as to the number of bits in the computer memory which handles the output from the ADC. If we had eight bit memory. then when one of the points in the free induction decay was to reach 11111111 then the system should not attempt to pulse the sample to record any transient.

If we were to pulse the sample again and then try to increase the count in this point (in time) in the FID, then we would have a nonsensical situation which would cause mayhem with the fourier transform. I think it could then fail to work as intended

Also in gamma spectroscopy we would have the problem if we were to overflow the memory storing the output of the ADC we would start to get a distortion of the spectrum. I suspect that in gamma spectra the effect would be less. We would merely get the highest peak to become zero again. But I suspect that the rest of the spectrum would still be OK.


Which is better: iodine number or bromine number? The degree of unsaturation of fats can be determined with the help of two type of scale namely iodine number and bromine number.

Generally, iodine number is more preferred over bromine number (Is this statement true? Is it because the Wikipedia article of iodine number is more elaborated than that of bromine number?)

Iodine number is typically found out by using iodine monochloride, $\ce{ICl}$ where the iodine atom gets bonded with the carbon atom having double bond. The more iodine  used up to break the $\ce{C=C}$ bonds,  higher is the iodine number.

$$\ce{-CH=CH - + ICl -> -(I)CH-(Cl)CH - }$$

But in the 4th paragraph of the 1st page of this link, it is written that:

  Iodination
  is an endothermic process, resulting in vicinal diiodides that tend
  to revert to alkenes. Consequently, the most common applications of
  alkene halogenation are chlorination and bromination.

So, in this context, bromination is better and thus bromine number is better than iodine number. The statement contradicts the previous statement i.e. using iodide number.

H.P.Kaufmann method of finding iodine number uses bromine. Is this a better method than using $\ce{ICl}$ considering the disadvantage of iodination?

Bromine number (and Bromine Index) are based on theoretically good chemistry; however, that is where their virtues end. If these tests are run according to the current ASTM procedures, you are virtually guaranteed to get an incomplete reaction and low (perhaps as much as 50%) results. When properly applied, off the shelf equipment can result in exceptionally good accuracy and precision, but it does require a deviation from ASTM-accepted technology.


Enthalpy in adiabatic process A gas adiabatically expanded from $\pu{32atm}$ and $\pu{273K}$ to $\pu{1atm}$ and $\pu{251K}$, Calculate Joule-Thomson coefficient $\mu$ at $\pu{273K}$.

Answer:

$$\mu = {\Delta T \over \Delta p} = {-22 \over -31} = 0.7$$

I think the answer uses the definition of $\mu$ that is $\displaystyle\left({\partial T \over \partial p}\right)_H$, but this definition assumes the process to have constant enthalpy.

But in adiabatic process $dH = dU + Vdp + w_{ad}$, so for $dH = 0$ we need to have $dU + Vdp + w_{ad} = 0$ which is not possible since pressure is clearly changing and so $Vdp$ is not zero.

My questions are

Is $dH = 0$ for a adiabatic and I am missing something ?
If not then how did the author used $\displaystyle\left({\partial T \over \partial p}\right)_H$ to get the answer ?

The Joule-Thomson experiments occurs with no change in enthalpy.

Suppose that at the left of a porous plug there is a pressure $p_1$ and temperature $T_1$ and $p_2,T_2$ to the right of the plug, as $p_1>p_2$ the gas moves left to right. The experimental configuration must ensure that pressures remain constant and that the experiment is performed under adiabatic conditions when $q=0$.

If a volume $V_1$ of gas moves from the left to right the work done/mole is $W=p_1V_1-p_2V_2$. This is the difference between the work of compression on the left of the plug and work recovered on expansion on the right. If the gas were ideal then $w=0$, but real gases are not. The gas expansion is also adiabatic so that no heat leaves or enters then $q=0$ and the change in internal energy $\Delta U$  is equal to the net work $$\Delta U =U_2-U_1 = p_1V_1-p_2V_2$$ therefore $$U_2+p_2V_2=U_1 + p_1V_1$$
As $H=U+pV$, then  $$\Delta H = H_2-H_1=U_2 +p_2V_2-U_1 -p_1V_1 =0$$

The Joule-Thompson  coefficient $\mu$ is defined, as you write, $\left ( \partial T/\partial P \right)_H$ and this measures how much the intermolecular interactions make the gas differ from an perfect gas. Most gases cool when passing from high to low pressures at room temperature.

Notes:

The coefficient can be rewritten in other forms using
$$ \left ( \frac{\partial T}{\partial p} \right)_H \left ( \frac{\partial H}{\partial T} \right)_p \left ( \frac{\partial p}{\partial H} \right)_T =-1$$
then
$$ \mu C_p= -\left (\frac{\partial H}{\partial p}  \right)_T $$

as
$$ \frac{dH}{dP} = C_p\frac{dT}{dp} +V-T\left (\frac{\partial V}{\partial T}  \right)_p$$
then if $\alpha=(1/V)(\partial V/\partial T)_p$ is the coefficient of expansion then at constant T, $(\partial H/\partial p)_T= V(1-\alpha T)$ which means that the Joule-Thompson coefficient can be written as $\mu=(V/C_p)(\alpha T-1)$ (This last equation has been used as a way of measuring absolute temperature because $V,\mu$ and $ \alpha $ are all measurable quantities.


Determine the concentration of the original NaOH solution This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $NaOH$ solution to 90.0 mL of 0.400 M $HCl$.

The solution was then treated with an excess of $\text {nickel(II) nitrate}$, resulting in the formation of 1.06 of $Ni(OH)_2$ precipitate. Determine the concentration of the original $NaOH$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I  wrote first the equation that describes the first part of the reaction :

$NaOH +  HCl \rightarrow NaCl + H2O $

After that I wrote the second part of the reaction :

$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

I am going to need a unit for the 1.06. The question needs you to figure out the amount in moles of Nickel Hydroxide. Then multiply this by two as two moles of NaOH are required to form one mole of the Nickel precipitate. Then this is the number of moles that was surplus in the reaction with HCl (NaOH is added in excess - the key to the question). Find the number of moles of HCl using the volume and concentration. Add this number to the number of moles that were used to form the Nickel precipitate and then find the concentration using the moles and the volume.


Finding the Ka of a weak acid after addition of a strong base When 50.00 mL of an unknown 0.2461 M weak acid are titrated with 0.1968 M NaOH, it was found after the addition of exactly 16.00 mL of base that the pH of the solution was 3.79. From this information, calculate the Ka of the acid.

I have an idea of how to do this question, but looking in my book, I seem to be arriving at a wrong answer.

$$50 \text{mL of }0.2461\text{M weak acid} = 0.2461 \cdot 0.05 = 0.012305\text{ mol weak acid}$$
$$16 \text{mL of }0.1968\text{M NaOH} = 0.2461 \cdot 0.013 = 0.0031488\text{ mol NaOH}$$

$$\begin{array}{|c|cccc|}\hline
   & \ce{HA} & \ce{NaOH} & \ce{A^{-}} & \ce{H_{2}O} \\ \hline
  \mathrm I & 0.0123 & 0.0032 & 0 & -\\
  \mathrm C & -0.0032 & -0.0032 & +0.0032 & - \\
  \mathrm E & 0.0091 & 0 & 0.0032 & -\\ \hline
 \end{array}$$

$$\text{molarity } \ce{A^-} = \frac{0.0032~\textrm{mol}}{0.066~\textrm L} = 0.0485~\textrm M$$

from here, I work on the assumption that the weak conjugate base will associate in water.

$$\mathrm p\ce{OH} = 14 - \mathrm p\ce H = 14 - 3.79 = 10.21$$
$$[\ce{OH^-}] = 10^{-\mathrm p\ce{OH}} = 10^{-10.21} = 6.17\times 10^{-11}~\mathrm M$$

$$\begin{array}{|c|cccc|}\hline
   & \ce{A^-} & \ce{H_2O} & \ce{HA} & \ce{OH^-} \\ \hline
  \mathrm I & 0.0485 & - & 0 & 0\\
  \mathrm C & -6.17\times 10^{-11} & - & +6.17\times 10^{-11} & +6.17\times 10^{-11} \\
  \mathrm E & 0.0485 \text{ (negligible change)} & - & 6.17\times 10^{-11} & 6.17\times 10^{-11}\\
 \hline\end{array}$$

\begin{align}K_\mathrm {b} &= \frac{(6.17\times 10^{-11})^{2}}{0.0485} = 7.85\times 10^{-20}\\ & = \frac{K_\mathrm W}{K_\mathrm b}\\ & = \frac{1.0\times 10^{-14}}{7.85\times 10^{-20}} = 127388.54 \leftarrow\text{This is obviously wrong}\;.\end{align}

Where did I go wrong? Any help is much appreciated!

When you used the ICE method the second time, you forgot that HA does not have an initial concentration of 0. Instead, the initial concentration of HA is 0.0091 as you calculated previously. For simplicity's sake, you can just use Henderson-Hasselbalch equation after your first ICE method to determine the answer.


What is the energy package released to the outside world? Consider a chemical reaction where two different particles form another one

$$\ce{O + O_2 -> O_3}$$

I find it confusing how this can be an exothermic process. How to picture the release of $+\Delta H$ to the surroundings.

And am I right in viewing this energy value only as the net energy coming from an integration of the particle potentials as they get closer along the reaction coordinate?

Essentially yes - if you consider the two particles colliding into a potential well, the kinetic energy associated with each will (depending on the angular specifics) be apportioned into rotational, vibrational, and even occasionally electronic excitations. So, the two particles collide, and are then vibrating and/or rotating about each other with all the translational energy converted in the collision.

Non-reaction collisions with other ambient particles allow relaxation into lower-energy states, transmitting energy to those particles. After some collisions the new species formed is in equilibrium, transferring between low-energy rotational and vibrational states as it collides further, and the change in potential energy has been dissipated as rotational, vibrational, and translational energy in other particles.

(I guess I could really use some sort of animation for this.)


Similarities and Differences between Resonance and MCSCF treatments Recently, there has been a question by Voldemort concerning different resonance structures of $\ce{NCO-}$, requesting an explanation why one resonance structure would be more preferred than another.
There has been a little discussion on that between ron and Martin - マーチン that took my interest:

ron: "[...] structure #1 ($\ce{N#C-O-}$) is more stable (preferred) than structure #2 ($\ce{^{-}N=C=O}$)."
Martin - マーチン: When talking about resonance structures, I think stable is the wrong word. I agree with your analysis in terms of contribution. Therefore #1 should have the highest contribution of all.

Now while I agree with ron that structure #1 is the most preferred resonance structure (referring to electronegativities), I regret with him concerning his attitude towards the "stability" as Martin does.
But I mainly do this, because I used NBO6 with the Natural Resonance Theory (NRT) keyword on the Natural Bond Orbitals (NBO) produced of an MP2/6-311+G(2d) optimized wavefunction to find the contributions of all relevant resonance structures - $66~\%~(\ce{N#C-O-})$, $30~\%~(\ce{^{-}N=C=O})$ and $3~\%~(\ce{^{2-}N-C#O+})$ - and not because I have a superior understanding on all underlying theories.
For me, the resonance stuff lookes a lot like MCSCF treatment of some VB wavefunction, which means that the complete molecule is described best from it's contributing resonance structures.
Now on a more complete comparison, what are the main similarities and differences between both theories and as they don't exist but are rather part of an explanation, can resonance structures indeed be described as "stable"?

Great question - it engages with the difficult task concerning conceptualization of molecules, and how our models and jargon we use can often cause confusion while also helping to explain significant phenomena.

To begin, it is important to emphasize the fact that all these theories are models only (as you said, they are "part of an explanation") - we attempt to describe complex systems of quantum phenomena in a way that helps us explain chemical interactions in experiment (and/or what we would then expect from computational studies). Instead of focusing on all the minutiae of both resonance theory and MCSCF, I'm going to aim for using the differences between their orbital theory choice (VB or MO) to help with this distinction, and then approach the "stable" vs. "contribution" remarks. For further details on VB and MO theory, Martin - マーチン 's comment towards another question has a fantastically detailed answer, that I would recommend taking a look at here by hBy2Py.

Resonance theory relies on VB as a means to explain what is occurring in a molecular system, illustrated by the drawings seen in organic chemistry and later on with increasingly complex systems relying only on this idea of atomic orbitals (simple constructs localized around atoms). Resonance as a concept is applied to help us visualize the idea of electron delocalization in a way that also lets us select "better" structures to draw (based on electronegativities, geometric preference, etc...), which match up to chemical expectation/intuition. However, when drawn or used as the basis for a calculation, the orbitals are then limited to those around the atom, (with some hybridization) and how they overlap with neighbors, without much freedom to mold the orbital shapes (leading to computational difficulties due to non-orthogonality).

MCSCF, or other configuration interaction theories, build on the MO idea, which greatly expands the types of orbitals utilized when describing the bonding interactions between atoms, and captures the idea of delocalization much more effectively in doing so, a feature that applying the idea of "resonance" is supposed to recover in VB, but does not do quite as well as MO. Not perfect, but one of the best models so far, and continuing to be improved upon even today (or, built on, such as with new basis sets).

In essence, yes, those "resonance structures" do not exist by themselves, and thus there is no true "stability" to them, nor in that case much of a "contribution" as it is still not the whole picture. However, use of "contribution" helps to avoid implying that the resonance structures, drawn only to illustrate how VB theory handles delocalization, are localized/exact organizations of the electrons and atoms at a particular moment. In practicality, you will likely hear these interchanged, but it is best to keep in mind the idea trying to be captured: electron location is highly complex, with a range of local/delocal orbital characteristics that could be "best" depending on the system.

As further reading just in case, and as references for the topics discussed above:

Resonance Theory: if you go down to the section on misconceptions, it will show the common errors made when describing resonance, which you brought up nicely here in terms of computation as well. This is where use of "stable" may imply one of these misconceptions, and "contribution" may help a bit (but could also lead to the second misconception)


Why doesn't hydrogen peroxide decompose into hydrogen and oxygen? I'm trying to understand why hydrogen peroxide doesn't decompose into $\ce{H2}$ and $\ce{O2}$, but instead into $\ce{H2O}$ and $\ce{O2}$.

We have two theoretical decomposition pathways:

$$\ce{H2O2(l) -> H2O(l) + 1/2 O2(g)} \tag{1}$$

$$\ce{H2O2(l) -> H2(g) + O2(g)} \tag{2}$$

The required data (all at $298~\mathrm{K}$) is taken from the appendix of Atkins' Physical Chemistry, 9th ed:

$$\begin{array}{cccc}
\hline
\text{Species} & \Delta_\mathrm{f} H^\circ\mathrm{~/~kJ~mol^{-1}} & S^\circ_\mathrm{m} \mathrm{~/~J~K^{-1}~mol^{-1}} & \Delta_\mathrm{f} G^\circ\mathrm{~/~kJ~mol^{-1}} \\
\hline
\ce{H2O2(l)} & -187.78 & 109.6 & -120.35 \\
\ce{H2O(l)} & -285.83 & 69.91 & -237.13 \\
\ce{H2(g)} & 0 & 130.684 & 0 \\
\ce{O2(g)} & 0 & 205.138 & 0 \\
\hline
\end{array}$$

The short way

The standard Gibbs free energy change of a reaction ($\Delta_\mathrm{r} G^\circ$) can be calculated from the individual standard Gibbs free energies of formation exactly analogously to the enthalpy change:

$$\Delta_\mathrm{r} G^\circ = \sum_J \nu_J[\Delta_\mathrm{f} G^\circ(\ce{J})]$$

So, for reaction 1,

$$\begin{align}
\Delta G^\circ_1 &= [1(-237.13) + \frac{1}{2}(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\
&= -116.78~\mathrm{kJ~mol^{-1}}
\end{align}$$

For reaction 2,

$$\begin{align}
\Delta G^\circ_2 &= [1(0) + 1(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\
&= +120.35~\mathrm{kJ~mol^{-1}}
\end{align}$$

Clearly reaction 1 is favourable and reaction 2 isn't.

The long way

There's absolutely no need to do this, but you could use the "extra" data given above to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for both reactions. You'd get:

$$\begin{align}
\Delta H^\circ_1 &= [1(-285.83) + \frac{1}{2}(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\
&= -98.05 ~\mathrm{kJ~mol^{-1}} \\
\Delta H^\circ_2 &= [1(0) + 1(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\
&= +187.78 ~\mathrm{kJ~mol^{-1}} \\
\Delta S^\circ_1 &= [1(69.91) + \frac{1}{2}(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\
&= +62.879 ~\mathrm{J~K^{-1}~mol^{-1}} \\
\Delta S^\circ_2 &= [1(130.684) + 1(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\
&= +226.222 ~\mathrm{J~K^{-1}~mol^{-1}}
\end{align}$$

If you're only interested in the final result, you don't need to do all this at all. But it does give you some insight into the difference between the two reactions: reaction 1 is enthalpically favoured whereas reaction 2 is entropically favoured.

Generally speaking, enthalpy usually plays a larger role in determining $\Delta G^\circ$. However, it is not a very good idea to generalise this statement and compare reactions solely on the basis of their $\Delta H^\circ$.

Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, and taking $T = 298~\mathrm{K}$,

$$\begin{align}
\Delta G^\circ_1 &= -98.05 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+62.879 ~\mathrm{J~K^{-1}~mol^{-1}}) \\
&= -116.79~\mathrm{kJ~mol^{-1}} \\
\Delta G^\circ_2 &= +187.78 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+226.222 ~\mathrm{J~K^{-1}~mol^{-1}}) \\
&= +120.37~\mathrm{kJ~mol^{-1}} \\
\end{align}$$

consistent with our earlier findings.


Entropy as a state function independent of the path When entropy is a state function and the entropy change for a reversible process is

$$\mathrm dS_\mathrm{rev} = \frac{\delta Q_\mathrm{rev}}{T} = 0,$$

how can an irreversible process be calculated by the same formula? $\delta Q_\mathrm{irr}$ is not a state function, therefore $\mathrm dS_\mathrm{irr}$ cannot be a state function.

Another description I have found was that the irreversible path should be seen as divided in a infinite amount of reversible steps. But an infinite times zero is also zero.

Can you help me please?

Entropy is indeed a state function, and thus depends only on the state of the system.  Hence it doesn't matter how you get from state A to state B, the entropy change will be the same.  The analogy would be that it doesn't matter which path you use to get from the base of a mountain to the summit, your elevation change will be the same.  This is because altitude is a state function: your altitude depends only on how high you are (the state of your system), not how you got there.

But, since $dS= \frac{\text{đ}q_{rev}}{T}$, one can calculate the entropy change by integrating along a path connecting the two states, but that path must be reversible.  I.e., it doesn't matter if the way the system changed from A to B was reversible or irreversible.  $\Delta S$ will be the same.  However, to calculate $\Delta S$ for that change from A to B, you need to integrate along some reversible path that connects A to B.

Also, it is possible to sum the effects of an infinite number of infinitely small steps.  That's what happens when we do an integration.  [More precisely, in an integration we calculate the sum in the limit as the step size goes to zero and the number of steps goes to infinity.]  For a graphical representation of this, take a look at the graphics in Wikipedia's entry on Reimann sums (https://en.wikipedia.org/wiki/Riemann_sum), and imagine what happens in the limit as the bar width goes to zero.

The comments by Andrew, and Chet Miller, are both useful additions.  Further expanding on Andrew's comment: The system's net change isn't affected by the path.  This means that, to distinguish between the result of making a change in a system by a reversible vs. irreversible path, we need to look at differences in effects upon the surroundings.


What is the form of the Hamiltonian for solids? For atoms, and even molecules, I can  understand how the Hamiltonian would be constructed, but what of solids (as in, for electronic structure calculations)? Thank you for any help.

There are multiple possibilities. First of all, you can construct them like the ones for molecules, since a solid is also made from nuclei and electrons. So the Hamiltonian looks identical, but with much more terms.
On the practical side, you would do something different. If, for example, your system is periodic (think of a crystal), then you would choose a single elementary cell from the crystal and build the Hamiltonian for that (ignoring all the other cells), so you arrive at a small Hamiltonian.
Then to get the wavefunction for the whole system you do not change the definition of the Hamiltonian but the one of the wavefunction. While for molecules the wavefunction has to vanish at infinity, you now want the wavefunction to have special properties, called periodic boundary conditions:
For example, consider a one-dimensional crystal with unit cell from 0 to L. You now want the wavefunction $\psi$ at 0 and L to be equal, as well as the derivatives of $\psi$ to be equal at these points:
$\psi\left(0\right)=\psi\left(L\right), \frac{\partial\psi}{\partial x}|_{x=0}=\frac{\partial\psi}{\partial x}|_{x=L}$
With this information you can determine the wavefunction in this cell with any method you want. Now you could put this exact wavefunction into the cell from L to 2L and and the given wavefunction would be a solution for this second cell, since the cells are identical. This way you can get a wavefunction $\Psi$ for the whole system, by periodically putting the wavefunctions for the unit cells $\psi$ together:
$\Psi\left(x\right)=\psi\left(x-n\cdot L\right), x\in\left]n\cdot L,(n+1)\cdot L\right], n\in \mathbb{Z}$
The interaction of the wavefunction with the other nuclei in the crystal is taken care of by the periodic boundary conditions, so the wavefunction will be the real solution to the Problem.


Why are least electronegative elements usually in center of lewis structure Many sources state that one of the primary rules for molecular structures is that the least electronegative element is in the center. However, none of them explain why this is the case.

Is this because it is most conducive to sharing electrons and forming bonds, or is it the most stable, or is there a better reason?

Also, why are many molecules symmetric. Is there a reason why symmetry in nature and in chemistry is preferred over non-symmetry?

Why are least electronegative elements usually in center of lewis
  structure?

If there is a "central atom", then there are multiple (more than 1) bonds to the central atom.  If the central atom is relatively electropositive, then it will be better able to share its electrons and form bonds with other atoms, at least more so than an electronegative central atom would.  Also, electronegative atoms tend to carry multiple lone pairs of electrons.  If this electronegative atom and all of its lone pairs were at the center of the molecular structure, then we would have many more destabilizing (lone-pair - bonding pair) electron-electron repulsions, then if all of these lone pairs were on the periphery of the molecule.

  why are many molecules symmetric. Is there a reason why symmetry in
  nature and in chemistry is preferred over non-symmetry?

The symmetric arrangements (linear, tetrahedral, octahedral, etc.) minimize destabilizing bonding electron - bonding electron repulsions.  Remember methane is a perfect tetrahedron - that is the lowest energy arrangement because all of the bonding electrons are as far apart from one another as possible in this geometry.


General notation for one of the d-orbitals What is the general notation to represent the d-orbital with $l=2$, $m_l=0$, i.e. the orbital normally referred to as $\mathrm{d}_{z^2}$. To elaborate more, this orbital can be ordered in various ways in a crystal environment and can be written as: $\mathrm{d}_{3z^2-r^2}$, $\mathrm{d}_{3x^2-r^2}$ or $\mathrm{d}_{3y^2-r^2}$. So is there a general notation to represent all three, like $\mathrm{d}_{3A^2-r^2}$?

Generally you are free in your choice of coordinate system to describe a certain problem and calling the d-orbital with $l=2$, $m_l=0$  either $\mathrm{d}_{3z^2-r^2}$, $\mathrm{d}_{3x^2-r^2}$ or $\mathrm{d}_{3y^2-r^2}$ depends on this choice but by convention the $z$-direction is usually chosen to be the system's preferred direction, i.e. it points along the main symmetry of the system.

Let's move to $d$ orbitals in crystals: Usually, metal $d$ orbitals are not "properties" of the crystal itself but "belong" to the local metal centers. Thus, the $z$-direction along which a $\mathrm{d}_{3z^2}$ orbital points refers to the preferred direction of the metal ion in its local environment, e.g. in a Jahn-Teller distorted Perowskite $\ce{ABO3}$ the $\mathrm{d}_{3z^2}$ orbital of metal ion $\ce{B}$ always points along the direction of the elongated/compressed $\ce{B-O}$ bonds within the local $\ce{BO6}$ octahedron or quadratic bipyramid. So, even though in $\ce{LaMnO3}$ you have differently tilted $\ce{MnO3}$ octahedra that have different preferred directions and thus differently oriented $\mathrm{d}_{3z^2}$ orbitals, you still refer to all of them as "$\mathrm{d}_{3z^2}$" and everyone you talk to about this will know which orbitals you mean and where they will point in the different local octahedra in the crystal. That is pretty much the status quo.

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Now, what you want to do, is talk about the $\mathrm{d}$ orbitals not in the context of their local coordinate system but in the context of a more global coordinate system related to the whole crystal, e.g. declare a certain crystal plane to be the $x$,$y$-plane and name the $\mathrm{d}$ orbitals according to this choice to let their name reflect their orientation within the plane.

Of course, you can do this and it has been done here pretty much as you suggested  where the authors chose notations like "$\mathrm{d}_{3\sigma^2 - r^{2}}$ orbital ($\sigma = x, y , z$)" or "$\mathrm{d}_{3(x, y)^2 - r^{2}}$ orbitals". But be aware that this is a kind of notation that needs to be explained beforehand when you use it because it refers to an unconventional choice of coordinate system as a frame of reference for the $\mathrm{d}$ orbital labels.

TL;DR: Yes, you can use a general notation like you suggested but make sure to explain which coordinate system you refer to because otherwise you might cause confusion.


Separation of metal particles from fluid in ferrofluid I purchased commercial Ferrofluid, and when suspended, the water and/or alcohol turns into a brown sludge.  In 95% alcohol, all the metal particles separated from the "tar" Why, and how can I prevent this?

Assuming the ferrofluid is the one described at product description, then it is "an extremely fine powder, coated with a soap-like material called a surfactant, suspended in a mineral oil liquid base." From that description, it might be possible to dilute or extend the ferrofluid with mineral (or even vegetable) oil, but one would not expect ferrofluid to be compatible with water or with alcohol. It might be possible to dilute it with other aliphatic solvents.

There are directions for making your own water-based ferrofluid at Synthesis of an Aqueous Ferrofluid.