How many molecules are there typically in items made of plastics? In objects made of plastic (and other polymers), how many molecules would one generally expect to find?

Intuitively, it seems like, say, a plastic shopping bag or PET bottle couldn't be a single molecule, but on the other hand, that's one of the strengths of polymers.

So the question is, I suppose, do items made of plastic consist of a single molecule, or many molecules fused together by way of glues or mechanical action?

Edit: Following a comment, I realise that the core of my question could be clearer, so here's an alternate phrasing:

If I chose an atom in a typical plastic object and gave it a colour, and then gave every atom connected to it by way of covalent bonds the same colour by flood fill, and then chose a colourless atom in the object and repeated the process until no more colourless atoms existed in the object, how many colours would a typical plastic object have?

The short answer is a very large number and far more than you could reasonably count

Polymerization reactions typically produce very large molecules but not so large that a single molecule would make up as much material as a whole plastic bag or item of any significant size. This source on LDPE tells us that the typical molecular weight of the industrial product is 90,000 (some polymers are much bigger than this).

For non-chemists molecular weights are (simplifying slightly) the weight relative to a single hydrogen atom. Carbon is 12 on this scale and ethylene (the monomer from which the plastic is made which consists of 2 carbons and 4 hydrogens) is 32. This mean that a typical molecule of the plastic is (crudely) made from a little under 3,000 ethylene molecules. The other fact you need to know is that there are about 6 x 1023 hydrogen atoms in a gramme of hydrogen (see the wikipedia definition of Avogadro's number). Simple mathematics tells us that a gramme of LDPE will contain roughly 6.7 x 1018 molecules so that is the order of the number in small plastic item which, I expect, weighs just a few grammes.

So, the bottom line is that even in a small plastic item consisting of a long chain polymer (LDPE is relatively short but other polymers might have chains 10-100 times more at most) there will be something like 1016 to 1018 molecules of polymer, which is a lot.


Which functional group is -CSSH? If we replace two oxygen in $\ce{-COOH}$ what is the name of the compound we get?

Is a dithioic acid.

For example, a benzene with two -CSSH hubstituents in 1, 2 positions is a benzene-1,2-dicarbodithioic acid.


Stronger reducing agent Why H$_2$Te is a good reducing agent than H$_2$O .

I think it should be opposite as in water , oxygen has only tendency to release electrons .

According to this Wikipedia entry:

"[electronegativity] is a chemical property that describes the
tendency of an atom to attract electrons (or electron density) towards
itself.

then according to an electronegativity scale called the Allen electronegativity:

"...neon has the highest electronegativity of all elements, followed
by fluorine, helium, and oxygen."

By any scale, oxygen is among the most electronegative elements. More to the point, electronegativity decreases as you go down a period, so  $\ce{Te}$ is much less electronegative, much more able to give up its electrons, than oxygen.  Thus, $\ce{H2Te}$ makes a better reducing agent than $\ce{H2O}$.


What is the charge of the amino acid lysine at pH 2? What is the charge of the amino acid lysine at $\mathrm{pH}=2$? $\mathrm{p}K_\mathrm{a}$'s are carboxylic: 2.18; amino: 8.95; side chain: 10.79.

The side chain of lysine is an amine group, so I assume that when fully protonated it has positive charge ($\ce{NH3+}$).

At $\mathrm{pH} = 2$, everything will be protonated because their $\mathrm{p}K_\mathrm{a} > \mathrm{pH}$.

This gives a net 0 (carboxy) + 1 (amino) + 1 (side chain) = +2 charge. Why do sites such as this say that at $\mathrm{pH} = 2$, lysine's charge is only +1, not +2?

There's probably something really simple I don't know. If anyone can help point it out, that would be great.

At $\mathrm{pH}=2$, the charge will be:

(charge near neutral pH, 2 groups positive, one negative) + (the fraction of the pKa = 2.18 group that is protonated)

=1 + (the fraction of the $\mathrm{p}K_\mathrm{a} = 2.18$ group that is protonated)

$=1 + 1/(1 + 10^{\mathrm{pH}-\mathrm{pK}}) =1 + 1/(1 + 10^{2-2.18})=1.60$

So the charge is +1.60, which is closer to 2 than 1, as you are saying.


Sodium cyanoborohydride and iminium reductions I understand that the LAH will reduce the nitrile into a primary amine. The addition of acid will likely destroy the acetal mask and leave us with ethylene glycol and a molecule with both a ketone and an amine functional group.

However, what does the sodium cyanoborohydride do here with the acid and heat? I'm again thinking this is a red herring; $\ce{NaBH3CN}$ only reduces iminium ions as far as we are concerned... and we have an amine. Amines don't exist in equilibrium with imines.

I believe the reaction effected in the third step is a reductive amination. In this case, the ketone and amine moieties are both present within the same molecule, so the reaction proceeds intramolecularly. The $\ce{TsOH}$ is an acid, presumably serving to protonate the carbonyl oxygen and activate the carbon toward nucleophilic attack by the amine. As the iminium group forms (with concomitant loss of water), it is rapidly reduced to an amine by the $\ce{NaBH3CN}$. After workup, the final product is:


Relationship between the first and the second quantum number Does the secondary quantum number tell how many subshells a specific principal quantum number shell has?

E.g., if the principal quantum number is $n$, there are ($n-1$) subshells.

Does the secondary quantum number tell how many subshells a specific
  principal quantum number shell has?

No, it does not. It is the principal quantum number $n$ which does it (as you actually mentioned yourself later in the question). The second quantum number $l$ describes the shape of the orbitals in a particular subshell. Say, $l=0$ for $\mathrm{s}$-subshell of any shell and all $\mathrm{s}$-orbitals have a spherical shape.

  If the principal quantum number is $n$, there are $(n−1)$ subshells.

Why $(n−1)$? Think about the most trivial example: how many subshells the first shell ($n=1$) have? One, not zero, so $(n−1)$ is obviously wrong formula. The situation is in fact simpler: $n$-th shell has $n$ subshells.


Strength of bases and acids Is it possible that strong acids like $\ce{H2SO4}$ behave as weaker ones like $\ce{CH3COOH}$ when the concentration is really high and vice versa for weak acid? And the same question for bases.
By this I am generally referring to pH values.

Why do concentrated acids react more than dilute ones? The degree of ionization of concentrated acids are lower than diluted acids, so they are supposed to react less. Where am I wrong?

In the case of weak acids dissociating like strong ones when they are highly dilute, there is a limit to how far this can go.  If you were to try to measure acetic acid dissociation at $10^{-n}$ molar for $n=4,5,6,7,8,...$ at ambient temperature you would at first rise nicely towards $100$% as you might expect, but eventually level off at only about $99.4$%.
What stops you is the autoionization of water.  As you approach infinite dilution the hydrogen ion concentration approaches not zero but the autoionized value of $10^{-7}$ molar.  So the ratio of dissociated acid to undissociated acid levels off at $K_a/10^{-7}$.  Accordingly the limiting degree of dissociation is
$1/[1+(10^{-7}/K_a)]=K_a/(K_a+10^{-7})$
For acetic acid with $K_a\approx1.8×10^{-5}$ this limit is about $99.4$%, but for a much weaker acid such as a typical phenol or hydro-cyanic acid the dissociation is much lower even at extremely low concentrations.  When $K_a<10^{-7}$ the limiting dissociation is less than $50$%; you need a stronger base than water to coax dissociation as a major reaction in aqueous solution.


Will "soft nucleophiles" undergo 1,4-conjugate addition even if the β-carbon is sterically hindered? In general soft nucleophiles do 1,4 conjugate addition to a carbonyl carbon under thermodynamic control. Will soft nucleophiles like organocuprates still preferentially undergo 1,4 addition if the β-carbon is more sterically hindered than the carbonyl carbon?
For example, if I consider the following case

should I just accept it as a rule of thumb that these "soft nucleophiles" undergo 1,4-nucleophilic addition to the carbonyl group, or does it vary depending on the structure and the situation?
In addition to the example above, what happens if there is a methyl group at C-2 of the ring?

Your line of thought possibly is influenced by the Walden inversion which is characteristic for the nucleophilic substitution mechanism of $\mathrm{S_N2}$ where steric hindrance affects the rate when this umbrella-like structure flips upon entry of the nuclophile and departure of the nucleofuge around a formally $\mathrm{sp^3}$ hybridized center:

(credit)
The reaction shown by you however is an addition on to a carbon with molecular orbitals formally hybridized as $\mathrm{sp^2}$.  Similar to the Bürgi–Dunitz trajectory for the entry of a nucleophile on to a $\ce{C=O}$ bond a) it is plausible that the angle of attack for the 1,4-addition b) equally is tilted, rather than an attack from the back.

(composed with a picture from here)


Balancing the equation when MnO₂ is reduced to MnO by H₂(g) I am studying for a final tomorrow and I was going through some problems on older exams and came across one which I wasn't really sure how to solve.

The problem is to write the balance equation when:

managnese(IV)oxide is reduced to manganese(II) oxide by hydrogen gas.

I took a few approaches to this, because at first I did not realize it say reduced "by" hydrogen gas. This part threw me off because I initially added $\ce{H2(g)}$ to the product side.

So far I have:
$\ce{MnO2(s) + H2(g)-> MnO(s) + H2O(l)}$ which is not balanced yet. I noticed in this case the conventional type of balancing the equation does not work. Would this be a redox rxn in which I find each half reaction and then balance?

I am completely stuck at this point, and I don't think my formula is correct. Does anyone have any insight or good examples ?

This is balanced: I count the same number of elements on each side, which is the definition of balanced. What is not explicitly stated is that your hydrogen goes from $0$ to $+1$, and is oxidized, thus reducing your manganese from $+4$ to $+2$. But this all balances!


Having problems running DFTB using dftb.org parameter files on Gauss 09 (G09)? I ran into an issue trying to run dftb.org parameter files on G09 and found questions asked on this but not answers.  I thought I would post the solution here to help anyone else who may run into this problem.

First, at least for C.01, dftb=readparameters, in the route section, does not work dftb=read does but is not necessary.

Skip a line after molecular descritption and include your parameter files there. Make sure you include all pair interaction files, not just bonded atoms.  It should look something like this:

@C:/G09W/C-C_mio.skf
@C:/G09W/C-H_mio.skf
@C:/G09W/H-C_mio.skf
@C:/G09W/H-H_mio.skf

(I included the mio because we are running more than one parameter set, not a requirement just a name)

The real glitch comes from my final hint.  The DFTB parameters generally have more than 500 points, but the first line in the code is 500 (not 519, or 520, etc).  G09 Rev. D.01 doesn't care about this, B.01 and C.01 do.  You simply need to change 500 ro reflect the correct amount of points and add in the correct order for your atomic numbers (this part is explained on Gauss' website) and you are good to go.

Hopefully this helps anyone trying to run DFTB on G09 for the first time, and helps save some time.


Is metallic sodium a strong base? Chemical properties: Sodium metal and sodium amide ($\ce{NaNH2}$ ) are strong bases. They react with ethyne to form sodium acetylide with the liberation of hydrogen gas.

How can a metal considered a base (that too strong)? To act as base it should either donate a pair of electrons (in this case it's single electron) or accept proton or give $\ce{OH-}$ ion but here it's not qualifying any one. So is that statement wrong?

Formation of monosodium ethynide:
$$\ce{HC#CH + Na -> HC#C^-Na+ + 1/2H2}\tag{13.59}$$
Formation of disodium ethynide:
$$\ce{HC#C^-Na+ + Na -> Na+C^{-} #C^-Na+ + 1/2H2}\tag{13.60}$$

Source: NCERT 11, Hydrocarbons, Page 386

No, stricly speaking metallic sodium is not a strong base. Strictly speaking, metallic sodium is not a base at all. (Sodium amide clearly is one, though: due to the amide part.)

Not everything that reacts with an acid is a base. If an acid reacts to form hydrogen and there are no hydrides present, that is a very strong indicator that the reaction was not an acid-base reaction but a redox reaction. And that is what metallic sodium is: a strong reducing agent.

Just because reducing agents can react with acids (proton donors) to generate hydrogen radicals and thus nascent hydrogen gas does not make them bases. For the reaction partner of an acid to be a base, it must accept (and keep!, at least to a certain extent) the proton donated by the acid — or, in the Lewis theory, it must donate an electron pair.


Help understanding the synthesis of HOCl My goal is to add acetic acid to sodium hypochlorite to reach a ph of between 3.8-5.5 and produce hypochlorous acid. I’ll be using bromocresol green as a ph indicator. I was reading about how to synthesize it from an article here by the acidification of hypochlorite, which was noted to be the safest and most controlled way. My initial question was how do I calculate how much acid is needed to reach the specified pH range. However, other users noted that this may dangerously produce chlorine gas, in which case this experiment should be executed under a fume hood with care and proper safety precautions.
Figure 2 from the article notes the production of various byproducts of the reaction at different pH levels. It appears that chlorine gas is produced at both low pH and at high pH, but it is a bit confusing since it is overlapping with another line making it hard to understand.

Ideally, I'd like to reduce any production of Cl$_2$. Is there a way to do that? Anything to help me better understand this reaction is appreciated.
For the moderators,
I don't understand why this question was so poorly received and closed. It clearly lies within the allowed topics listed in the help center. Even though my question was about a potentially dangerous experiment, it is an important one nonetheless. I was able to learn an important detail concerning the safety of the experiment from this post. If anyone else was to follow down the same path as me, they should be able to google about it, see this question, and learn of the safety concerns as well. I think this question should be reopened to educate people instead of being heavily downvoted because OP is an inexperienced chemist trying a dangerous experiment, because that is not the case.

Chlorine bleach is commonly produced via the chloralkali process as a highly alkaline mixture of sodium hypochlorite (NaOCl) and hydroxide (NaOH) along with some unreacted NaCl as well.
It's usually dangerous to lower the pH of chlorine bleach, as you may inadvertently generate chlorine gas
$$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) \rightleftharpoons Cl2(g) + H2O (\ell)} $$
$$ \ce{4H+(aq) + 4OCl- (aq) \rightleftharpoons  2Cl2(g) + O2(g) + 2H2O (\ell)} $$
And that is why it is kept at a high pH, to help favor the left side of the above equilibria.


Thermodynamics heat, enthalpy, Gibbs We know the basic Thermodynamic equations are:
$$ dU=-PdV+ TdS + \mu dN$$
$$ dH=dU+d(PV)= TdS+ \mu dN + VdP$$
we typically assume constant Pressure for a reaction carried out in air and neglect the dP term
$$dG=dH-d(TS)=  \mu dN - SdT+ VdP$$
once again we neglect the dT and dP terms
but for heat of reaction we use the enthalpy not just the TdS term. When we have a battery though the electrical energy is the Gibbs free energy with the heat of reaction being TdS and other energy losses also contributing to the heat. When the reaction is not separated (not an electrochemical cell) then the Gibbs is part of the heat.
For a typical reaction where the chemicals are in contact why do they use the enthalpy to determine the heat of the reaction produced?

When a reaction occurs at constant pressure and no work other than expansion work  (such as electrochemical) is performed, then it can be shown, from the definition of the enthalpy and the first law of thermodynamics applied to a closed system, that the change in enthalpy is equal to the heat:
$$\begin{align}dH&=dU+d(pV)\\&=dw_{pV}+dw_{other}+dq+pdV+Vdp\\&=-p_{ext}dV+dw_{other}+dq+pdV+Vdp\\&=dq+dw_{other} \end{align}$$
The last equality follows when the pressure is constant and equal to the external pressure. When only pV work is done then clearly heat and enthalpy change are equal. Readily measured heats of reaction provide a very convenient way to determine values of the enthalpy state function which can be tabulated and employed in further thermodynamic computations.


Can you create a temporary bond in between an atom and a filter by electrically charging the filter? I want to be able to turn a filter on and off by electrically charging the filter so that it creates a temporary bond that blocks up the filter then, by discharging the filter, releasing the temporary bonds and allowing particles through again. Is this possible?

In a sense, you can increase and decrease flow through an osmotic filter with electricity. The process is called electro-osmosis. This can even be used to create a pressure differential in the electroosmotic pump.


Etymology of alanine Oxford dictionary online gives etymology of alanine as:

  Coined in German as Alanin, from aldehyde + -an (for ease of pronunciation) + -ine.

But I see no resemblance to the aldehyde structure in alanine. Is this etymology incorrect, or am I mistaken about the resemblance?

In the original German paper [1] Adolf Strecker used Aldehyd-Ammoniak or aldehyde-ammonia as a precursor, that's where the name derives from:

  Vor einigen Jahren habe ich gezeigt, daſs Aldehyd-Ammoniak und Blausäure beim Erwärmen mit verdünnter Chlorwasserstoffsäure sich zu einer schwachen Basis, Alanin genannt, vereinigen [...]:
  $$\ce{\underset{\text{Aldehyd-Ammoniak}}{C4H4O2 * NH3} + HCl + \underset{Blausäure}{C2NH} +2 HO = \underset{Alanin}{C6H7NO4} + NH4Cl}$$

As David Richerby mentioned in the comments, Strecker's brutto-formula ($\ce{C6H7NO4}$) deviates from the moden one ($\ce{C3H7NO2}$), also the reaction scheme is a bit different.

Strecker, A. Annalen der Chemie und Pharmacie 1854, 91 (3), 349–351. DOI 10.1002/jlac.18540910309.


What does the reaction of chromium with ozone form? I know almost nothing beyond 10th grade chemistry. I'm trying to name a web application, and was wondering what the products are of a reaction between the element Chromium and the gas Ozone (ozonolysis of chromium?).

Ozone is a strong oxidizing agent, even stronger than molecular oxygen. But it is still a donor of oxygen atoms. Thus, depending on conditions, you'll have different chromium oxides. The most stable and common chromium oxide is chromium (III) oxide $\ce{Cr2O3}$. It has numerous nicknames. Other oxides are much harder to get.

Substances named ozonides do exist and in two different flavours.

Inorganic ozonides are salts of big, inoxidizable cations, like $\ce{Cs}$ and have formulas like $\ce{KO3}$. I'm not sure which cations have ozonides that have been already produced, but to my knowledge the only reasonably easy-to-produce ozonides are $\ce{KO3}$, $\ce{RbO3}$ and $\ce{CsO3}$ by reaction of the corresponding hydroxides with ozone.

Another type of ozonides is organic ozonides that are quite weird heterocyclic compound, which are product of reaction between alkenes and ozone.

If you need a name based on word games, you may consider that chromium is a derivative of an ancient word for 'color' and ozone, as a strong oxidizer, has bleaching and antiseptic properties.


What happens when C17H35COO-, a soap, reacts with an acidic solution? My book says that-

  In acidic water, the negative fatty acid ions become fatty acids.

However, they did not include a chemical equation to demonstrate this, so I made my own:

$$\ce{C17H35COO- (aq) + HCl (aq) -> C17H35COOH (aq) + Cl- (aq)}$$

But according to this equation, the fatty acid ion is a base, since it gets a proton. So this must be wrong. Can anyone point out my mistakes (if any), and explain what happens when a negative fatty acid ion comes in contact with acidic water, and becomes a fatty acid?

Your equation is correct. Hydrochloric acid ($\ce{HCl}$) is a stronger acid than stearic acid ($\ce{C17H35COOH}$),i.e. it is dissociated into $\ce{H+}$ and $\ce{X-}$, the anionic rest, to a greater degree. Therefore, it protonates the stearate ion. One can also say that the conjugated base of stearic acid (the stearate ion) is a stronger base than chloride, the conjugated base of hydrochloric acid.


Why is pyrrole more reactive than pyridine and benzene for an electrophilic substitution reaction? According to me, pyridine should be the most reactive as it has got an extra pair of electrons on the nitrogen atom which is further localised and readily available for electrophile.

From Peter Sykes:

  Pyridine (62), like benzene, has six at electrons (one being supplied bynitrogen) in delocalised it orbitals but, unlike benzene, the orbitals will be
  deformed by being attracted towards the nitrogen atom because of the latter's
  being more electronegative than carbon. This is reflected in the dipole of pyridine, which has the negative end on N and the positive end on the nucleus:

  Pyridine is thus referred to as a $\pi$-deficient heterocycle and, by analogywith a benzene ring that carries an electron-withdrawing substituent, e.g.$\ce{NO2}$ (p. 6.22), one would expect it to be deactivated towards electrophilic
  attack. Substitution takes place, with difficulty, at the 3-position because this leads to the most stable Wheland intermediate (63); the intermediates
  for 2- and 4-attack (64 and 65, respectively) each has a canonical state in which the charge is located on divalent N a highly unstable,i.e. high energy, state.

  There are certain formal analogies here to m- attack or
  p. 6.24), byt pyridine is very much more difficult to substi
  Thus nitration, chlorination, bromination and FriedelCraf
  really be made to take place usefully, and sulphonation on
  ing with oleum for 24 hours at 230°C, with an $\ce{Hg^2+}$ catalyst attack is due partly to the fact that pyridine has an available nitrogen, and can thus protonate (66), or interact with an electrophile (67):

  The positive charge will clearly further destabilise any of the $\sigma$-com
  plexes for electrophilic substitution, as did a substituent such as $\ce{-NR3^+}$, on the
  benzene nucleus (p. 6.24), but the destabilisation will be much more marked
  than with $\ce{-NR3^+}$; as the charge is now on an atom of the ring itself and not merely on a substituent.
  Pyrrole (68) also has 6 electrons in delocalised orbitals, but here the
  nitrogen atom has to contribute two electrons to make up the six (thus be
  coming essentially non-basic in the process, cf. p. 3.26), and the dipole of
  pyrrole is found to be in the opposite direction to that of pyridine, i.e. with
  the positive end on nitrogen and the negative end on the nucleus:

  Pyrrole is thus referred to as a π- excessive heterocycle and behaves rather like a reactive benzene derivative, e.g. aniline (p. 6.25), undergoing very ready electrophilic attack. This may be complicated by the fact that in
  strongly acid solution protonation (69) is forced even on the weakly basic
  pyrrole (it takes place on the 2-carbon atom rather than on N. cf. p. 3.26)

TLDR

Pyridine is a $\pi$-deficient heterocycle owing to nitrogen's more electronegative nature than carbon. Pyrrole on the other hand is $\pi$ excessive since nitrogen donates its lone pair to the ring.
The non-delocalized lone pair on pyridine tends to interact with the electrophile/$\ce{H+}$. This further destabilizes any of the $\sigma$ complexes formed in electrophilic aromatic substitution (EAS).


Does silver oxidise in a pH 1 solution? Consider submerging a plate of pure silver in a solution with $\mathrm{pH} = 1$. Does the redox reaction with hydrogen happen spontaneously?

  The composition of the solution is not given.

My attempt:

I considered the standard potential for $\mathrm{pH} = 0$. Since the silver must be oxidised, the reaction happens at the anode.

$$
\begin{align}
\ce{Ag+ + e- &-> Ag} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{2 H+ + e- &-> H2} &\qquad E^\circ &= \pu{0.00 V}
\end{align}
$$

At this point, it is obvious that the reaction cannot be spontaneous since the potential for this cell is less than zero. In fact:

$$\Delta E = E_\mathrm{red} - E_\mathrm{ox} = \pu{-0.80 V}$$

From Nernst equation, the potential for the hydrogen at $\mathrm{pH}=1$ must be negative. Thus the reaction is all the more non spontaneous ($E = \pu{-0.059 V}$).

However, this seems to be wrong. I must be applying the $\Delta E$ wrong, because the solution for this problem states clearly that it is given by:

$$\Delta E = E_\mathrm{red} - E_\mathrm{ox} = \pu{-0.059 V} - (\pu{-0.80 V}) = \pu{0.741 V}$$

This is driving me crazy. Where does the minus sign come from?

The answer given in the solution was wrong. It has been corrected, and it eventually proved my point.


Oven safe dessiccant? Im sure there is a chemist/lab tech that knows a simple solution to my problem, as I cant seem to find a solution.  When trying to strip and re-season cast iron cookware, I occasionally have problems with rust.  I live near the ocean, so there is a lot of humidity in the air.  I strip the cast iron by putting in the oven on the self-cleaning setting for an hour.  This normally works perfectly.  However, sometimes at the end of the hour, I open the door and see the pan has developed surface rust.  Obviously, heat + iron + water = rust.  I am sure there is a way to absorb the moisture in the oven, but I cant find anything.  I looked at desiccants, but they release moisture in the oven when heated - at least the ones I saw.  Is there a desiccant that works at oven temperatures?  Or is there another solution?  Remember, food will be cooked on the cast iron, so the solution has to be food safe in the end.

I like the suggested treatment suggested here which is to basically clean/scrub the pan, dry and coat with vegetable oil, invert the pan, place over aluminum foil on an oven rack and bake. To quote:

Now that you’ve returned your skillet to its base layer, you’ll need to season it for use. Drop a tablespoon of vegetable oil on the pan and spread it around with your fingers or a paper towel. (Take care not to burn yourself on a warm skillet!). Oil the outside and the handle, too. Next, take a folded napkin and wipe off the excess oil. You’ll want the skillet to be lightly greased. Set your oven to 350 degrees and put some aluminum foil on the bottom rack. Then pop the skillet in, upside down, on the upper rack for about 1 hour. Turn the oven off and leave the pan inside until it’s cool. The oil will bake into the pores of the pan, providing a non-stick finish.
Repeat this step a second time and you’ll have an almost glassy surface of seasoning.
Ta-da! Your rusted cast-iron skillet is good as new.

The created carbonized layer is safe, chemically inert and durable.


What are modern approaches to describe and/or quantify aromaticity? I know that Schleyer described an approach for evaluating aromaticity using nucleus-independent chemical shifts (NICS, see e.g. Chem. Rev. 2005, 105 (10), 3842–3888) quite some time back. I also somehow have the feeling that there should be a more direct way to describe aromaticity than via chemical shifts.

What are modern approaches to describe/quantify aromaticity? Could anyone point out some links/references on recent developments in the field?

I'm more familiar with the Electron Localization Function approach to aromaticity commented by Quantum AMERICCINO, it's been proposed that the average value of the $\text{ELF}_{\sigma}$ and $\text{ELF}_{\pi}$ bifurcations would be used for a general scale of aromaticity of a molecule or molecular cluster. The points of bifurcation can be interpreted as a measure of the interaction among the different basins or, simply, electron delocalization. Although the ELF depends on the density (i.e. no separate information of $\pi$ or $\sigma$ bond), a topological analysis on a separated ELF formed only by the $\pi$ orbitals and other formed by the $\sigma$ orbitals can be done (also you can do a separation for the spin $\alpha$ and $\beta$ contributions to the ELF). [J. Chem. Theory Comput. 2005, 1 (1), 83–86.]

There is another approach based on how a bond responds when its electron density is perturbed by a finite amount. Specifically, they used the interaction coordinate (IC) to see this. Quoting

  If the delocalization in the aromatic system is more extensive, for constraining a bond by unit displacement from the equilibrium geometry, it requires more energy. So the response of the other internal coordinates will be greater for constrained optimization and is proportional to the effective delocalization and hence related to the aromatic stabilization. The IC is a measure of the response of a bond (its electron density) for constrained optimization when another bond or angle is stretched by one unit (its electron density is perturbed by a measured amount).

Instead of taking the full molecule, they take only the aromatic skeleton to account the electron density perturbation on the system. Also, it says that this is very general yet, so the IC approach can take new directions.
[J. Phys. Chem. A 2016, 120 (18), 2894-2901.]

Another approach is making a relationship between the amount of the electron density (due to ring size, number of $\pi$ electrons, etc) and the non-nuclear magnetic shielding (cf. NICS).
[Phys. Chem. Chem. Phys. 2010, 12 (39), 12630–12637.]


Catalysts to accelerate foamy polyurethane reaction This is the reaction that I am referring to:  https://www.youtube.com/watch?v=EoRUSo0BWC0

I am interested in finding catalysts to accelerate this fun chemistry demonstration. I have heard that some amide catalysts can accelerate CO2 production leading to faster foam rise, but I have not found any sources confirming this. Does anyone know of any catalysts to accelerate this fun chemical demonstration?

In those preparations, the foam arises from the reaction between water and isocyanate, which leads to the production of CO2 via formation of a carbamic acid, which upon decarboxylation generates the blowing agent, CO2.

The trick would be that of making water more reactive towards the isocyanate, without (as expected by a catalyst) partecipating actively in the reaction.

Several compounds of this kind have been reported and are sold, for instance: https://patents.google.com/patent/WO2009048807A2/en mentions N,N,N',N',N",N"-pentaethyldiethylenetriamine and morpholine derivatives, and commercially (I won't link any product, but googling "Blowing Catalysts" might help you) catalysts based on  N-methylmorpholine, Bis[2-(N,N-dimethylamino)ethyl] ether, DMEA and PMDTA are available.

This said, for practical applications the formulations have probably the best ratio of reactants, surfactants and catalyst for obtaining a "good" foam (with good physical properties, and possibly with negligible smell), but if your goal is that of making the reaction go faster, maybe an excess of (any?) tertiary amine, and possibly some drops of water in excess should work in making things faster (with the possible drawback of obtaining a low quality foam, maybe with "bubbles" larger than intended, or with bad "texture").

Additional note: if speed is what you are looking for, gently heating the mixture should work without adding anything else


Density of saturated n-octane in air at 300 K? I am trying to calculate (if possible) the gas density of air saturated with n-octane at 300 K. Any suggestion of how to do this, or better, where I can look up the data?

Thermodynamic parameter values for many substances (not just water and steam) can be found using so-called steam tables. In the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). For a simplified estimate, however, we only need the value of the vapour pressure of octane.

For simplicity’s sake, we may assume that the gas phase (containing air and octane) is an ideal mixture of non-reacting gases and that the liquid phase contains only pure octane. Thus, for the calculation of parameter values for octane, we may ignore the air and only consider the equilibrium in a system of pure liquid octane and pure gaseous octane.

Using tabulated data, we find that the saturation point for octane at a temperature of $T = 300\ \mathrm K$ corresponds to a pressure of $p_{\ce{C8H18(g)}} = 2060\ \mathrm{Pa}$.

The corresponding density can be estimated using the ideal gas law:
$$\begin{align}
p\cdot V &=n\cdot R\cdot T \\[6pt]
\frac nV &= \frac p{R\cdot T} \\[6pt]
\frac m{M\cdot V} &= \frac p{R\cdot T} \\[6pt]
\frac mV &= \frac {M\cdot p}{R\cdot T} \\[6pt]
\rho &= \frac {M\cdot p}{R\cdot T} \\[6pt]
&= \frac {114.23\ \mathrm{g\ mol^{-1}}\times 2060\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&= \frac {0.11423\ \mathrm{kg\ mol^{-1}}\times 2060\ \mathrm{kg\ m^{-1}\ s^{-2}}}{8.314462618\ \mathrm{kg\ m^2\ s^{-2}\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&= 0.09434\ \mathrm{kg/m^3}
\end{align}$$

By way of comparison, the corresponding density given by REFPROP is $\rho_{\ce{C8H18(g)}} = 0.094649\ \mathrm{kg/m^3}$.

Assuming that the total pressure is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, we may calculate the partial pressure for the remaining air:
$$\begin{align}
p_\text{tot} &= p_\text{air} + p_{\ce{C8H18(g)}} \\
p_\text{air} &= p_\text{tot} - p_{\ce{C8H18(g)}} \\
&= 101325\ \mathrm{Pa} - 2060\ \mathrm{Pa} \\
&= 99265\ \mathrm{Pa}
\end{align}$$

The density of dry air at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1770\ \mathrm{kg/m^3}$. The corresponding density at a partial pressure of $p_\text{air} = 99265\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1531\ \mathrm{kg/m^3}$.

Thus, the total density of the gas mixture is
$$\begin{align}
\rho_\mathrm g &= \rho_{\ce{C8H18(g)}} + \rho_\text{air} \\
&= 0.094649\ \mathrm{kg/m^3} + 1.1531\ \mathrm{kg/m^3} \\
&= 1.2477\ \mathrm{kg/m^3}
\end{align}$$

By way of comparison, I made a quick simulation using REFPROP assuming that air is a mixture with the mole fractions $x_{\ce{N2}}=78.12\,\%$, $x_{\ce{O2}}=20.96\,\%$, and $x_{\ce{Ar}}=0.92\,\%$. I added an excess amount of octane $x_{\ce{C8H18}}=10\,\%$ (in order to obtain a liquid phase) and renormalized the composition to $100\,\%$.
The resulting density of the vapour phase at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{g} = 1.2483\ \mathrm{kg/m^3}$.


Why are some elements more abundant than others in the universe? Hydrogen is the most abundant element in the universe. But what makes it so? At the time of big bang, what made certain elements more abundant than the others?

I don't find this order of abundance too predictable. What makes neon more abundant than nitrogen?

WHAT MAKES HYDROGEN ABUNDANT IN UNIVERSE:

After few minutes of creation of the universe, protons and neutrons began to react with each other to form deuterium, an isotope of hydrogen. Deuterium, soon collected another neutron to form tritium. Rapidly following this reaction was the addition of another proton which produced a helium nucleus.

Sources say that there was one helium nucleus for every ten protons within the first three minutes of the universe. After further cooling, these excess protons would be able to capture an electron to create common hydrogen. Consequently, the universe today is observed to contain one helium atom for every ten or eleven atoms of hydrogen.

WHAT MAKES NEON STAND 4TH ON THE ABUNDANCE LIST:

The star in its first stage is powered by a series of nuclear fusion reactions that convert hydrogen to helium. The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two γ rays, and a great amount of energy. It then takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. The after reaching high temperature the helium-4 nuclei begin to fuse, producing beryllium-8.

The neon production:

  Beryllium-8 has low neutron-to-proton ratio makes it very unstable. So
  it reacts with a third helium-4 nucleus to form neon, magnesium and
  carbon. Neon production doesnt stop here. Again from the reaction of
  carbon with helium-4 to initiate a series of reactions, neon, oxygen
  and some more elements are produced. So in the stages of stars, neon
  is produced multiple times in huge amounts making it one of the
  abundant element in the universe.


Formamide and Hoffmann Bromamide Reaction I encountered a problem to identify the compound which does not give Hoffmann bromamide degradation reaction. One of the options was formamide, and it was given as the answer.

Why can't formamide give Hoffmann bromamide reaction? I feel that the answer is wrong (it should produce ammonia).

I have spent a lot of hours to find the evidence to support OP's idea of formamide may undergoes Hoffmann degradation reaction, but I can't find any. However, I find at least one publication, which may suggest formamide may decompose to formic acid and ammonia (ironically) in hot alkaline medium, before exchanging amide $\ce{N-H}$ to $\ce{N-Br}$ (please keep in mind that $\mathrm{p}K_\mathrm{a}$ of $\ce{HC(O)NH2}$ is about 24 at $\pu{25 ^{\circ}C}$ in DMSO, Evan's pKa Table):

  Ref.1 has described the rates of decomposition rates of acetamide and formamide, which have been studied by individually decomposing them in pressurized hot water. According to the studies, while apparent decomposition rates for both amides were more represented by the second order rather than the first order reaction kinetics, the rate constant for formamide was more than 10 times higher than that for acetamide (Ref.1).

Even though, the work has been done in high temperature and high pressure, the authors observed the decomposition without using acids or bases, and throughout wide range of concentrations. They also found that rate increases with increasing concentrations. Thus, one can expect that in highly alkaline media used in Hoffmann degradation, some amide may hydrolyzed before rearrangement. According to a mechanistic study of the rearrangement, hydrolysis of even aromatic amides under alkaline conditions has been observed:

  It has been found that in the presence of excess alkali at $\pu{30 ^{\circ}C}$, p-nitrobromobenzamide gives approximately as much p-nitrobenzoic acid by hydrolysis as p-nitroaniline by rearrangement; at $\pu{90\!-\! 100 ^{\circ}C}$, however, a $90\%$ yield of p-nitroaniline is obtained.

All of these evidence directed to the theory of hydrolysis of formamide (which is aliphatic and most susceptible to hydrolyze under the conditions due to its size) before it converted to N-bromformamide. Further more, the Ref.3 described a theoretical study of the Curtius rearrangement, which also believed to be going through same mechanism as Hoffmann degradation. Although authors have discussed various pathways of rearrangement including going via oxazirene ion, and identified the most stable intermediate is isocyanic acid during the rearrangement, their conclusion wasn't clear on the hydride ion shift.

References:

M. Okazaki, T. Funazukuri, “Decomposition of acetamide and formamide in pressurized hot water,” Journal of Materials Science 2006, 41(5), 1517–1521 (DOI: 10.1007/s10853-006-4616-1).
C. R. Hauser, W. B. Renfrow, Jr., “The Removal of $\ce{HX}$ from Organic Compounds by Means of Bases. III. The Rates of Removal of Hydrogen Bromide from Substituted N-Bromobenzamides and their Relative Ease of Rearrangement in the Presence of Alkali. The Hofmann Rearrangement,” J. Am. Chem. Soc. 1937, 59(1), 121–125 (DOI: 10.1021/ja01280a029).
A. Rauk, P. F. Alewood, “A theoretical study of the Curtius rearrangement. The electronic structures and interconversions of the $\ce{CHNO}$ species,” Canadian Journal of Chemistry 1977, 55(9), 1498–1510 (https://doi.org/10.1139/v77-209).


Preferred IUPAC name of Protoporphyrin IX I was looking up protoporphyrin IX and was surprised to find that no two websites gave the same IUPAC name for the compound. Here are some examples:

3,3'-(7,12-Diethenyl-3,8,13,17-tetramethylporphyrin-2,18-diyl)dipropanoic acid (EPA)
3,3'-(3,7,12,17-Tetramethyl-8,13-divinyl-2,18-porphyrindiyl)dipropanoic acid
(CHEMSPIDER)
3-[18-(2-carboxyethyl)-8,13-bis(ethenyl)-3,7,12,17-tetramethyl-22,23-dihydroporphyrin-2-yl]propanoic acid (PUBCHEM)

Anyone know if an IUPAC name is agreed upon among scientists or if there is a preferred IUPAC name?

I think both EPA and ChemSpider names are acceptable. However, I don't agree with dihydroporphyrin part in PubChem name. This is because the Gold Book gives following framework for porphyrin nucleus:

According to the numbering the IUPAC name of protoporphyrin IX is 3,3'-(7,12-Diethenyl-3,8,13,17-tetramethylporphyrin-2,18-diyl)dipropanoic acid. This name based on least locant numbers for ethynyl group, which have higher priority before methyl groups (3,3'- is based on positions of propanoic acid). The same name is given to protoporphyrin IX in Ref.2, without 3,3'- locants.

References:

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
IUPAC-IUB Joint Commission on Biochemical Nomenclature (JCBN): Nomenclature of tetrapyrroles (Recommendations 1986); Prepared for publication by Gerard P. Moss, Queen Mary College, London, Eur. J. Biochem. 1988, 178(2), 277-328 (https://doi.org/10.1111/j.1432-1033.1988.tb14453.x) (PDF).


Using solubility product to determine the mass of the precipitate of a reaction I mix together $\pu{100 mL}$ of an aqueous $\ce{NaCl}$ solution at $\pu{0.5 M}$ and $\pu{100 mL}$ of an aqueous $\ce{AgNO3}$ solution at $\pu{0.3 M}$. Assuming that the solubility of $\ce{NaCl}$ in water is $\pu{360 g/L}$, the solubility of $\ce{AgNO3}$ in water is $\pu{2340 g/L}$, the solubility of $\ce{NaNO3}$ is $\pu{921 g/L}$ and the $K_\mathrm{sp}$ of $\ce{AgCl}$ is $1.77 \times 10^{-10}$, what is the mass of the precipitate ($\ce{AgCl}$) formed?

I know that the answer to this problem is $\pu{42.1 g}$, I solved it using stoichiometry. However, I want to know how to solve it using the solubility product and the other solubility figures given.

This is awfully closed to a homework question. Since you already have an answer (real answer is actually $\pu{4.3 g}$, see following calculations) and still insisting on how to use the solubility product information, I'd like to give some insight. Since all components are in solutions, solubilities of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ should not be in concern. Solubility of $\ce{AgCl}$ is the only one should be concern:

$$
\begin{array}{lccc}
\ce{&2 AgNO3(aq) & + NaCl(aq) &<=> &AgCl(s) &+ & NaNO3(aq)} \\
\text{Initial}, \pu{mol} & 0.3 \times 0.10 = 0.03 & 0.5 \times 0.10 = 0.05 && 0 && 0 \\
\text{Final}, \pu{mol} & 0 & 0.05 - 0.03 = 0.02 && 0.03 && 0.03 \\
\text{Final}, \pu{mol/L}& 0 & \frac{0.02}{0.2}= 0.1 && \pu{0.03 mol} && \frac{0.03}{0.2}= 0.15 \\
\end{array}
$$

Thus, amount of $\ce{AgCl}$ precipitated $= \pu{0.03 mol} \times \pu{143.3 g/mol} = \pu{4.3 g}$

Since $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2}$ is such a small number, it is safe to say almost all of $\ce{AgCl}$ is in solid form. However, the solubility product come to play a role if there is a question about what is the final $\ce{[Ag+]}$ in solution. Now you have to consider following equilibrium:

$$\ce{AgCl(s) + H2O <=> Ag+ + Cl-}$$

Hence, $K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\ce{Cl}] = \pu{1.77 \times 10^{-10} mol^2L^-2}  \tag1$

Suppose the solubility of $[\ce{Ag+}]$ is $\alpha \ \pu{ mol\:L-1}$. Then, $[\ce{Cl-}]$ is $\alpha \ \pu{+ 0.1 mol\:L-1}$ (common ion effect). Yet, $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2} $ is such a small number, the $\alpha$ amount from the dissolution of $\ce{AgCl}$ can be considered negligible.

Thus, from equation $(1)$:

$$K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\pu{0.1 mol L-1}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \\ \Rightarrow [\ce{Ag+}] = \frac{\pu{1.77 \times 10^{-10} mol^2L^-2}}{\pu{0.1 mol L-1}} = \pu{1.77 \times 10^{-9} molL-1} $$


What's the difference between graphite, nanotubes, and fullerenes? I understand that all of these are made up of carbon atoms but I don't understand what differentiates them. Since they are carbon they all must have covalent bonds, right?

All three carbon modifications indeed contain covalently bond carbon atoms. The $sp^2$-hybridized C atoms are connected by $\sigma$ and delocalized $\pi$ bonds in all cases. It is the geometric arrangement of the atoms that differentiates the modifications: Stacked planes in graphite, fullerenes: ball-shaped molecules (hence the name buckyball), and for carbon nanotubes, the name is quite self-explanatory.

A single layer of the graphite structure:

A $\ce{C60}$ fullerene molecule:

And finally some carbon nanotubes:


Clarification of textbook concepts relating to "perfect", "ideal", and "real" gases My textbook, Atkins' Physical Chemistry, 11th edition, by Atkins, de Paula, and Keeler, says the following:

  The resulting expression

  $$pV = nRT \tag{1A.4}$$

  is the perfect gas law (or perfect gas equation of state). It is the approximate equation of state of any gas, and becomes increasingly exact as the pressure of the gas approaches zero. A gas that obeys eqn 1A.4 exactly under all conditions is called a perfect gas (or ideal gas). A real gas, an actual gas, behaves more like a perfect gas the lower the pressure, and is described exactly by eqn 1A.4 in the limit of $p \to 0$. The gas constant $R$ can be determined by evaluating $R = pV/nT$ for a gas in the limit of zero pressure (to guarantee that it is behaving perfectly).

  A note on good practice Despite 'ideal gas' being the more common term, 'perfect gas' is preferable. As explained in Topic 5B, in an 'ideal mixture' of A and B, the AA, BB, and AB interactions are all the same but not necessarily zero. In a perfect gas, not only are the interactions all the same, they are also zero.

My questions here relate to this part:

  A note on good practice Despite 'ideal gas' being the more common term, 'perfect gas' is preferable. As explained in Topic 5B, in an 'ideal mixture' of A and B, the AA, BB, and AB interactions are all the same but not necessarily zero. In a perfect gas, not only are the interactions all the same, they are also zero.

It seems that the authors here are implying that a "perfect gas" and an "ideal gas" are synonymous, but my research indicates that they are actually different concepts. Are these two concepts the same, or are they different? How should one think about (distinguish) these two concepts?
What do the authors mean when they say that "in an 'ideal mixture' of A and B, the AA, BB, and AB interactions are all the same but not necessarily zero"? And what is meant by the interactions being "zero" here?
What do the authors mean when they say that "in a perfect gas, not only are the interactions all the same, they are also zero"? And what is meant by the interactions being "zero" here?

I would greatly appreciate it if people would please take the time to clarify these points.

The authors attempt to explain that in many sources the term "ideal gas" is used in place of "perfect gas" to indicate a gas following the ideal gas law and which has the property that the molecules do not interact. They equate the absence of an intermolecular interaction with the statement that the "interactions are zero". In other words, there is no intermolecular potential between the idealized gas molecules in a "perfect gas". These idealized molecules lack a volume, so that even steric (excluded volume) interactions expected for hard spheres are absent. They are "point particles".

The authors further explain that it is preferable to use the word "perfect" rather than "ideal" because the word "ideal" is used in other contexts where the condition of no interaction is not observed, ie during ideal mixing. Ideal mixing occurs when different molecules have the same volume and intermolecular potential in the pure and in the mixed states.

Opinion-based aside

BTW I've always been partial to the term "ideal gas". IMHO the term "perfect" seems to imply that the molecules excel in some way, whereas "ideal" simply means that it is the product of human minds, an abstraction, and not necessarily a real thing.

IUPAC does not to have any qualms about continued use of the term "ideal gas", so I don't see reason to stop using this term.


What is the link between hydrogen production and potassium oxide? Is there an industrial process that produces hydrogen with potassium oxide as a catalyst?

Or is there a process for the production of potassium oxide that has hydrogen as a by-product?

When you react potassium and potassium hydroxide, you will produce potassium oxide and hydrogen as a byproduct.

$$\ce{2K + 2KOH ->[\Delta] 2K2O + H2}$$

  Potassium react with potassium hydroxide to produce potassium oxide
  and hydrogen. This reaction takes place at a temperature near 450°C.(Chemiday)

The Wikipedia page of potassium oxide states:

  Potassium hydroxide cannot be further dehydrated to the oxide but it
  can react with molten potassium to produce it, releasing hydrogen as a
  byproduct.


How to calculate the height of an hcp lattice? A hexagonal closed packing (hcp) unit cell has an ABAB type of packing. For calculating the packing fraction we require the volume of the unit cell.

Volume of hcp lattice = (Base area) $\cdot$ (Height of unit cell)
Each hexagon has a side = $2\cdot r$
Base area = $6$ (Area of small equilateral triangles making up the hexagon)
$$=6 \cdot \frac{\sqrt{3}}{4}\times(2r)^2$$
$$=6 \cdot \sqrt{3} \cdot r^2$$

Hence, volume $= 6 \cdot \sqrt{3} \cdot r^2 $ (Height of unit cell)

This is the point where I am stuck. How do I find out the height of the unit cell?

I searched in textbooks and found out that height $= 4r \cdot \sqrt{\frac{2}{3}}$. Can you please explain why is this so?

To calculate the height of a unit cell, consider a tetrahedral void in an hexagonal closed packing arrangement. It can be imagined as a 3 solid spheres touching each other and at the center-point, you have another sphere stacked over them. An interactive version can be viewed on this site. The situation looks like this:

If you join the centers of these four spheres, you'll get a tetrahedron. That's basically a pyramid with a triangular base. I'm assuming each edge of our tetrahedron to be equal to $a$.

Now, you have a pyramid ($ABCD$), with an equilateral base ($\Delta BCD$), I would like you to drop a perpendicular from the highest point ($A$) to the center ($G$) triangular base. If you're following me correctly, you'll have a figure like this:

All we have to do now is to calculate the length $AG$. For this, simply use the Pythagorean theorem in $\Delta AGD$.

$$ \begin{align*} AD^2 &= AG^2 + GD^2 \tag{1} \end{align*} $$

Although we know that $AD=a$, the side $GD$ remains unknown. But that's easy to calculate. The point $G$ is the centroid of $\Delta BCD$. Thus, the length $GD$ equals $a/\sqrt{3}$. Pluggging in the values in our first equation, we get $AG=a \sqrt{\frac{2}{3}}$. But note, this is half the height of our unit cell. Thus, the required height is $2a \sqrt{\frac{2}{3}}$.


How to calculate the percentage yield for enantiomers in free radical halogenation? How many mono-chlorinated products of 2-methyl butane will be formed upon chlorination in the presence of sunlight?
  Find the percentage of major product formed.

Six products were formed: two pairs of enantiomers and two achiral chloroalkanes.

Until here my answer was correct, confirmed by my teacher.

For the second part that is the percentage yield I had an issue with the enantiomers. I calculated the relative yield ratios by multiplying probability factor and rate factor.

For cases A and B, where enantiomers are being formed, I further multiplied $p_\mathrm{f} \times r_\mathrm{f}$ by half since there's an equal likelihood of each of the enantiomers to form. My teacher said this is wrong, as we do not multiply it by half.

So, my question is why?

I'll say that this is just a matter of convention, and not really worth worrying about. Your underlying method and concept is 100% correct. Your final question boils down to:

  Are enantiomers separate chemical compounds?

To elaborate, if the question was:

  Find the percentage of major product formed, counting the enantiomers separately.

Your approach would have without doubt been correct.

But since the question only says:

  Find the percentage of major product formed.

It is difficult to ascertain what the question wants. Enantiomers themselves have different chemical reactions - so they may be interpreted as different chemical products. But, your teacher has interpreted them as a single compound, presuming that optical isomers are just one compound.

Again, I guess this is just convention, because I haven't been yet able to find references to favor either definition.

Finally, to satisfy both the parties involved, and write a foolproof answer that guarantees a complete score, you can frame your results like so:

  Percentage of (R)-2-chloro-3-methylbutane = $x$
  Percentage of (S)-2-chloro-3-methylbutane = $x$
  Total percentage of 2-chloro-3-methylbutane = $2x$

where $x$ is your calculated value for a single enantiomer.


What is called a salt? While I was doing experiments in my school days, they gave me different substances with the combination of base and acid radicals, for example ammonium nitrate and zinc chloride and they seldom called it a salt.

But actually a salt is a NaCl as far as I know.

So My Question is: Actually, what is called a salt?
Why do they call the acid and basic radical a salt?

I think what confuses you  is that 'salt' has two meanings: In the every day sense it means table salt (which has the formula NaCl) but in the chemical sense it means a certain way for particles to hold together and form a tangible substance.
I'd define the latter as any periodic arrangement/lattice (see e.g. here ) of particles held together by ionic bonds.
The positive ion is almost always a metal, the negative one can be anything (so usually either a halogen (those lack only one valence electron to have a full shell, see e.g. Cl) or a (smallish) molecule-ion like SO4^2−)

  base and acid radicals

I really don't see much connection between salts and either bases/acids and radicals. Do consider that not all ions are bases/acids and all radicals are highly reactive. It would rather be an edge case to have a salt with a radical in it; most salts have ions with energetically favourable number of electrons (as a rough rule this means all "electron shells" are either full or empty.)
NaCl is a good example.

Because different sized salt crystals still have extremely similar behaviour, the salt formulas express only the ratio of anions and cations. Realize that on the whole salts are neutral.

  Why do they call the acid and basic radical a salt?

A single kind of particle is never called a salt, salt means an arrangement of two ions of different polarity. I guess some lab 'recipes' have sloppy language if something is provided as a salt, but only one component is important for the reaction.
Perhaps I can give a better answer I understood what you mean by radical in this context.


Where to find Atmospheric data for various cities Hi im searching for atmospheric data on the following:

Air quality, Dust Particle Concentration, Pollution, Average Temperature etc

for various cities such as Frankfurt, London, Hong Kong, Bangkok...

Are there any websites that can give me this data (or similar atmospheric measurements) for each city?

Most cities used to have own pages about pollution data. But there are other sources. For example:

London: http://www.londonair.org.uk/LondonAir/Default.aspx

Bangkok: http://www.pcd.go.th/airquality/bangkok/

If you are looking for a single page, try this:

http://aqicn.org/city/all/


Is the subtrahend in the Nernst equation related to ohmic drop? Let us consider the Nernst equation $$E= E^\circ -\frac{2.303RT}{nF} \log Q$$ which once at $\pu{298.15 K}$, is the equivalent  of $$E= E^\circ - \frac{0.059}{n} \log Q .$$

Does the part of the Nernst equation $-\frac{0.059}{n} \log Q$ address Ohmic drop? In other words, is $-\frac{0.059}{n} \log Q$ related to solution (electrolyte) resistance of an electrochemical cell?

In short: No.

The Nernst equation describes the thermodynamic equilibrium state. By definition, there is no current flowing through the cell. Therefore, there is no ohmic drop in any term within the Nernst equation.

The Q term relates to the activities of the reactants and products at equilibrium.


XRD Peaks in databases I had a look at the JCPDS card of BiOCl and saw that there were very small peaks (such as 0.1 percent relative intensity) that I think wouldn't be observed in XRD pattern, they would just blend in with the noise, so I'd like to know how do they calculate the XRD peaks and know their relative intensities so well (this card referenced a 1935 paper).

Many thanks

short answer: The record is based on the old record by Bannister 1934 as first description of this phase . The fine intensity scaling 1:1000 on the record card is based on a more recent re-investigation.

recording the data:

Beside the point detector / Geiger counter technique mentioned in Jon Custer's comment -- still highly valuable in powder diffraction analysis -- a common method was to record the diffraction pattern on a photographic film.

In the case of powder samples, you used for example Guinier cameras:

(source)

with a stripe of a film placed in the inner of the housing.  After the experiment, the film was developed; and the more diffraction intensity, the darker the band on the developed film.  The following image depicts a simultaneous analysis of four samples:

(source).

In case of single crystals, one method was to place the oriented sample in front of a film, for example with a Buerger precession camera

(source)

to yield clichés like

(source)

where the location of the spots provides information about symmetry distance relationships of the lattice.  (Relative intensity of the spots equally has a meaning, but beyond the scope of this question here.)

In both cases, the intensity of the signals were attributed by eye (e.g., strong, medium, weak), or later by densitometry and densitometers in numbers. While these film methods today are superseeded in convenience by positional detectors and CCD arrays (no wet chemistry to develop the films, quick numeric read-out), you still find these noted in volume C of the International Tables of Crystallography, section 7.1 «Detectors for X-rays».

(source)

your phase:

You find the first entry about $\ce{BiOCl}$ as entry #1011175 published 1934 in Nature, and #9011782 published 1935 in Mineralogical Magazine by the same author as indexed by the COD with their .cif files. You may use these to predict how a powder diffraction pattern of this phase should look like:

And indeed, there are some signals expected to be much weaker, than others, e.g., reflection $(103)$ (blue rectangle) would be weak and maybe not spot.

Interestingly, the AMS database equally offers access to the Bismoclite data, including a copy of Bannister's publication in Mineralogical Magazine, too (here).  And table 2 (loc. cit.), based on the photographs and their visual readout lists only 16 entries.  You find all of them in the subsequent re-investigation in 1983 (here, view text file next to download diffraction data), now with a total of observations equal to 37.

Thus, it is plausible the record card you accessed refers either to this set (1983) of data (or an  even more recent study) to list $(h,k,l)$ and their diffraction intensities, but credits Bannister for the first description of the very same phase in the 1930s.


Why is the maximum % for a gold alloy 92%? The assignment question is asking us to explain why you can't get more than 92% gold in a gold alloy, and I can't seem to find out why it's specifically 92% in either the textbook or online.

Any help is appreciated. Thank you.

It's not that you can't, it's that you usually don't.

The majority of Gold goes into jewellery with some in coinage (more in the past, we make gold coins now but they don't circulate much). But pure gold is too soft to be useful in most applications (coins have to last, rings have to not scratch too much and they aren't much use if they bend too easily). It is also expensive. So the most used alloys have to make a compromise between usability and cost. There are also some that try to achieve a particular aesthetic effect like a different colour.

This imposes limits of practicality. At the cheap end alloys that look like gold might be mostly other metals. But at the high end the issue is to get the correct aesthetic effect while retaining the appeal and value of gold without having something  too soft or weak to be practical. Many coloured golds are only about 75% gold (18 karat) because that gives a wide range of colours. But top end "pure" gold objects can be made at 22 karat (92% isn). We could make richer alloys, but they would be too soft.

So, mostly we don't. We can make many alloys with more gold, including many that would be useless as jewellery like the AuBe alloy mentioned in another answer (it is far too brittle), but you will rarely see them as they are not useful for anything.


Reaction between n-butylithium and either an aromatic bromide or an alkylbromide What exactly happens? Does the butylithium chain add to the carbon where the bromine is attached? or do the 2 undergo halogen metal exchange?

It's quite a long time that I did myself, but I used BuLi to transmetalate. I.e. $\ce{MesBr + BuLi -> MesLi + BuBr}$


Why is it the middle ring of anthracene which reacts in a Diels–Alder? I'm wondering why maleic anhydride adds to the middle cycle of anthracene, and not the outer two.

I would have expected that a Diels–Alder with the outer ring would be better, because I expected a naphtalene part to be lower in energy than two benzene parts (more resonance stabilisation).

I guess it has to do with reactant based arguments that the atomic coefficients for the two center carbon atoms (C-9 and C-10) are higher than from the outer cycle (C-1 and C-4). I ran a calculation using http://www.chem.ucalgary.ca/SHMO and the coefficients on C-9 and C-10 were 0.44, whereas those on C-1 and C-4 were only 0.31.

To provide a reason for the observed regioselectivity, it is helpful to draw anthracene's aromatic π-electron system in alternance of single and double bonds.  In this instance, it is more beneficial than "the ring" symbolizing the delocalised electron system, as this helps you to account for the precise number of π-electrons before the reaction (starting materials), during the reaction (the mechanism), and after the reaction (the product).

For the Diels–Alder reaction, you may imagine two different pathways.  I invite you to draw the mechanisms by yourself:

A reaction that involves carbon atoms #1 and #4 (or #5 and #8).  Possible, by mechanism.  There are five double bonds remaining in conjugation, and you count one six-membered ring in the state of "a benzene ring" (the very left one).  In the very right six-membered ring, there is only a single double bond, too.
Alternatively, a Diels–Alder reaction with carbon atoms #9 and #10.  This is more favourable then the former example, because both the very left as well as the very right ring of the product are in the state of "a benzene ring", each of these rings contain 6 π-electrons, the prerequisite of a Hückel aromatic compound.  It is so much favourable to the former, that this is the reaction observed.

It may be helpful to add that benzene, naphthalene and anthracene are of course Hückel-aromatic compounds; with 6, 10 or 14 π-electrons they fit into the rule of $(4n + 2)$.  Yet gradually, as experimentally found, in this group of three, benzene is the most, anthracene the least aromatic compound.


What does it mean to calculate hardness in terms of ppm of calcium carbonate? PPM means per part million or milligram per litre of solution.Calculation of ppm for a given compound say $x \pu{gms}$ of $\ce{Ca(HCO_3)}$ in 1 liter of water would be done like

$$ppm= \frac{1000\cdot x}{1\pu{1L}}$$

But I do not understand by what's meant by in terms of $\ce{CaCO3}$ is it equivalents or weight? This was asked in an exam and the formula is not covered in my textbook.

It is meant as molar equivalent. $\ce{CaCO3}$ has the molar mass about  $\pu{100 g / mol}$, $\ce{Ca(HCO3)2}$ has the molar mass cca $\pu{ 162 g / mol}$. Therefore $\pu{100 g}$ of the former is equivalent to $\pu{ 162 g }$ of the latter. The similar for the concentration expressed in ppm. E.g. if water contains $\pu{1.62 ppm}$ of $\ce{Ca(HCO3)2}$, it  is equivalent to $\pu{1 ppm}$ of $\ce{CaCO3}$


What does the acid do in this reaction? Which of the following is the product of the series of reactions shown above?

The correct answer is (C).

So the first step of the first reaction meant to create an enolate (deprotonating the hydrogen sandwiched between the two carbonyls), which upon addition of the substrate, forms:

.

I think the second step, potassium hydroxide is meant to hydrolyze the ester to a carboxylic acid.

Finally, the third step is meant to decarboxylate the acid by emitting $\ce{CO2}$.

However, I'm unsure as to what role the acid plays in the last step.  The only thing I could think of is that the acid helps the enol (the actual product from decarboxylation) tautomerize, but I don't think acid needs to be present for tautomerization.

My guess is that the acid is just there to protonate the carboxylate salt and allow decarboxylation. Hydrolysis of the ester with $\ce{KOH}$ generates the potassium carboxylate salt:

Since the mechanism for thermal decarboxylation involves a cyclic transition state including the carboxyl $\ce{-OH}$ group, I expect that the carboxylate salt must be protonated before decarboxylation can take place, hence the addition of aqueous acid.


For a 1st order reaction, how do you derive the integrated rate law for a product versus time? For a chemical reaction in which some stoichiometric quantity of arbitrary chemical $A$ decomposes into products $B$ and $C$,

$$a \ce{A} \ce{->} b \ce{B} + c \ce{C}$$

or more simply:

$$a\ce{A} \ce{->} \text{Products}$$

The depletion of substrate A at a concentration [A] can be modeled:

$$\frac{{\rm d}[\ce{A}]}{{\rm d} t} = - k[\ce{A}]^n$$

If this is a first order reaction then n = 1:

d[A] / dt = -k * [A]

This is a separable differential equation. When integrated from time 0 to time t and initial concentration [A]0 to [A]t which is the concentration at time t, the integrated form is this:

ln([A]t) = ln([A]0) - k * t

or in terms of [A]:

[A]t = [A]0 * e-k * t

This is useful if you want to know how the concentration of A as [A] changes with respect to time. But if you want the concentration of products, I think you could consider the total concentration:

[A]0 = [A]t + [P]t

Substitution into the the last equation yields a form in terms of products:

[P]t = [A]0 * (1 - e -k * t)

This kind of makes sense to me. The substrate is being exponentially depleted and the products are being exponentially produced to a maximum concentration which is [A]0. The following assumptions were made that the reverse reaction is negligible and that the reaction proceeds to completion which isn't always true. In fact, the time coordinate where these two curves intersect is actually the half-life!

If [P]t = [B]t + [C]t

Solving the equation for [B]t yields:

[B]t = [A]0 * (1 - e -k * t) - [C]t

This doesn't make sense because at the start of the experiment (time = 0),
negative concentration of [B] would be observed. Am I doing something wrong?

I have never seen integrated rate laws modeling the concentration of a single product versus time, only the depletion of reagent concentration versus time.

If you are doing this on a molar basis, you have to be careful about assumptions like: $$[A_o] = [A]_t + [P]_t$$
In the same way that you have defined a component balance for the reactant system, you can also do so for the product system. Let's look at the concentration of B in your reaction equation $\ce{aA -> bB + cC}$.

Defining the component balance,
$$\frac{d[B]}{dt}=k[A]$$
$$[B]_o = 0, t = 0$$
This is essentially saying that the generation of $B$ is proportional to the inverse depletion of $A$ (as defined in your component balance). We know that from the solution of the A component balance that, $$[A](t) = [A]_oe^{-kt}$$ Plugging in this expression to the differential equation for $[B](t)$ yields,
$$\frac{d[B]}{dt} = k[A]_oe^{-kt}$$ Solving this differential equation will yield the following expression for $[B]$,
$$[B](t)=-[A]_oe^{-kt}+C_1$$
Using our intial condition that $[B]=0, t = 0$ we solve for our constant $C_1$ and end up with,
$$[B](t)=[A]_o(1-e^{-kt})$$
Note that this is similar to the form you had already derived, however... this is only for $B$. Since the kinetic expression will be the same assuming no initial mols of $B$ or $C$ and same stoiciometric coefficients (i.e $b = c$) then we can reasonably assume that $$[P]_t = 2[A]_o(1-e^{-kt})$$


Calculating volume of gases in different conditions Doctor Who has a secret laboratory on Venus where atmospheric pressure is $92$ times that of Earth and room temperature is $\pu{735 K}$. There, Doctor Who mixes $\pu{4.00 g}$ of $\ce{CaCO3}$ with $\pu{40.00 mL}$ of $\pu{1.00 M}$ $\ce{HCl}$. What volume of carbon dioxide gas at VSTP (Venusian Standard Temperature and Pressure) is produced?

I know that the chemical equation is

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

Since there are $\pu{0.04 mol}$ of $\ce{CaCO3}$ and $\pu{0.04 mol}$ of $\ce{HCl}$, $\ce{HCl}$ is the limiting reagent; thus, $\pu{0.02 mol}$ of $\ce{CO2}$ are produced.

This is equivalent to:

$$V_\ce{CO2} = \frac{nRT}{p} =\frac{0.02 \times 0.08205 \times 735}{92} = \pu{0.013 L}.$$

However, the answer key says $\pu{26.2 mL}$ of $\ce{CO2}$ is produced, which is double my answer. Where did I go wrong in my reasoning?

I haven't found any flaws in your math either, and I can only join those who recommended to include units in your calculations. By googling around I found this problem appeared in 2012 Australian Science Olympiad Exam, and $\pu{13.1 mL}$ was one of the options there. Maybe your source of the answer key is incorrect?

Now to the fun part. I'm pretty sure you were expected to use ideal gas law, but come on, it's Venus and no way under the pressure of $\pu{92 atm}$ any gas would behave even remotely as the ideal one. In fact, $\ce{CO2}$ on Venus with its critical temperature of $\pu{304 K}$ and critical pressure of $\pu{7.39 MPa}$ is going to be in supercritical state!

You have to use Van der Waals' equation:

$$
pV_\mathrm{m} = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2},
$$

where $p$ – pressure; $V_\mathrm{m}$ – molar volume; T – temperature; R – molar gas constant; $a$ and $b$ – characteristic parameters related to the critical temperature $T_\mathrm{cr}$ and pressure $p_\mathrm{cr}$ (reduced form):

$$
a = \frac{27R^2T_\mathrm{cr}^2}{64p_\mathrm{cr}};
\qquad
b = \frac{RT_\mathrm{cr}}{8p_\mathrm{cr}}
$$

Using the above-mentioned values coefficients $a$ and $b$ can be found:

$$
a = \frac{27\cdot (\pu{8.314 J mol-1 K-1})^2\cdot (\pu{304 K})^2}{64\cdot\pu{7.39e6 Pa}} = \pu{3.65e-1 J2 mol-2 Pa-1}
$$

$$
b = \frac{\pu{8.314 J mol-1 K-1}\cdot \pu{304 K}}{8\cdot \pu{7.39e6 Pa}} = \pu{4.28e-5 J mol-1 Pa-1}
$$

Now you need to plug both values in the first equation and solve quadratic equation for $V_\mathrm{m}$. Wolfram Alpha claims $V_\mathrm{m} = \pu{2.56e-2 m3 mol-1}$, so

$$
V(\ce{CO2}) = \pu{2.56e-2 m3 mol-1}\cdot\pu{2e-2 mol} = \pu{5.12e-1 L},
$$

which, of course, deviates quite a lot from the value found using ideal gas law.


How can we separate a mixture of two sulphide ores using froth floatation method? As we know froth floatation method makes use of the wetting properties of mineral particles.

  Mineral particles (sulphide) become wet with oils, while gangue particles get wet by water

According to my textbook (NCERT Class 12), we can use the froth-flotation method to separate a mixture of sulphide ores also using depressants.

How do these depressants actually work in this process?

In the froth floatation process, the role of the depressants is to separate two sulphide ores. They do so by  preventing one ore from forming froth.

In this process, we want some substances to settle down and some to float. Now ideally, we want the sulphide of the substance that we wish to extract tonfloat and the gangue to sink. However, sometimes the ore we are using also has other sulphides present in it as impurities. For example, lead sulphide generally has zinc sulphide also present as impurity which will also float along with the lead sulphide. To make sure that lead sulphide floats but zinc sulphide do not float, we add depressants.
Commonly used depressant is $\ce{NaCN}$.
To separate two sulphide ores ($\ce{ZnS and Pbs}$), $\ce{NaCN}$ selectively allows $\ce{PbS}$ to come with froth, but prevents ZnS. This happens because of the reaction of $\ce{NaCN}$ with $\ce{ZnS}$ to form $\ce{Na2[Zn(CN)4]}$.
The complete reaction is,
$$\ce{4NaCN + ZnS  →  Na2[Zn(CN)4] + Na2S}$$


Why does internal combustion engine smoke appear blue? Could there be "blue molecules" in it? Four-stroke and two-stroke gasoline engines as well as diesel engines can all emit dense clouds of blue-tinged smoke under various transient loads or incorrectly tuned conditions.

Raleigh scattering - in this case probably from particulates of a limited range in size - strongly favors short wavelengths, so there is a possible physical explanation.

But I would like to know if there are any chemistry effects that could lead to or contribute to the slightly blue color sometimes seen in engine smoke under these conditions.

Could there be large "blue molecules" in the smoke? I don't think so; usually gases are transparent, though there are a few notable exceptions like molecular iodine and chlorine, but I'm not sure.

above: Iodine gas, from here

above: Chlorine gas, from here

below: snapshot sitting on a bicycle, looking at adjacent scooter while holding breath and waiting for the light to change.

below: screenshot from the YouTube video Big Engines Starting Up.

I can't really tell you the chemical difference but usually blue smoke indicates that oil is burned, too. Like some of the oil cycles get into contact with the engine and it is burned as well. And then probably heavier components of your gasoline once you start the cold engine. My Prof (lecture on Petrochemistry) once told us that there is indeed a difference between premium gasoline and the cheap one although the octane number could be similar. For the cheap ones they add a lot of stuff with a very low octane number and finally only a small amount compounds with higher octane numbers to get to 95. But it makes a difference in burning it. Maybe if the engine is still cold some of the "harder to combust" compounds don't burn that well and cause the blue color, too.

It's probably due to Rayleigh scattering as you mentioned. The sky is blue and clouds are white, the normal exhaust gases should be water and CO2 so similar to clouds and the blue particles are smaller rests which appear blue.


Why is only bromine used in the degradation reaction? Why specifically is bromine used in the degradation reaction (Hoffmann bromamide)? Why can't the same be performed using chlorine or iodine?

In other words, what makes bromine a more convenient reagent as compared to the other halogen gases? For instance, during halogenation of alkanes, we don't use fluorine due to the explosive nature of the reaction, and we don't use iodine because of the slow rate. Is the reason somewhat analogous to this?

I thought because of kinetic factors, maybe somewhat similar to how monochlorination and monobromination are much more easier than monoiodination and monofluorination

It isn't.

Under alkaline conditions, the Hofmann rearrangement of primary amides to amines can be performed with chlorine too. But why work with chlorine when bromine is much more convenient.

Choosing bromine over chlorine when possible is an obvious decision for everyone with practical lab experience!

Under standard conditions, bromine is a liquid and can be added using a dropping funnel, while chlorine is a gas, which makes adding stoichiometric amounts much more tedious.

Hypohalites have been used too: in the presence of alkanols, rearranged carbamates are formed. Under acidic conditions, hypervalent iodine compounds have been shown to be effective.


What decides whether a reaction releases light or heat I understand the reason that a chemical reaction would create photons (physics is more my strong point), but why would it create heat instead? My only guess is that the light released somehow causes the molecules around it to gain kinetic energy through some process and thus create heat, but I don't know how this would occur. What decides whether a given reaction produces photons or heat?
I'm a chemistry noob so try to explain using noob-friendly terms. Also, I don't know what tags suit so please put appropriate tags if my tags aren't.

The amount of energy released by a reaction is what most often creates light. Burning, the energetic reaction usually caused by oxidation of something, creates light because the amount of heat released is large enough to heat the reactants or products to temperatures where the radiation emitted contains a lot of visible light (coals in a fire glow red because they are hot enough that black-body radiation has a significant component in the visible).

Most exothermic chemical reactions don't generate this amount of heat and, although they release heat, you don't see it. Hand warming pads, for example, often contain supercooled sodium acetate which can be made to crystallise by popping a small metal insert. The exothermic reaction (crystallisation) releases heat to warm your hands, but not enough to see visible light.

A very small class of chemical reactions produce light directly as the reaction leaves molecules in an excited state which decays with the emission of visible light. Glow sticks are an example of this.

Overall, though, most reactions produce heat because the products have lower energy than the reactants and the released energy has to go somewhere. And it usually just ends up as heat. No photons are usually involved as the energy passes directly into the vibrational and translational energy of the products. So the excess energy in the reactants' bonds ends up as vibrational or translational energy in the products.


Why were elements discovered "out of order" after 1950? I'm referring to the fact that there is one element that corresponds to one proton in an atom, a second that corresponds to two protons, up to, I believe, an element with 115 protons. These elements are listed in the periodic table discovered by Mendeleev in the 19th century, or modern expansions thereof.

By "out of order," I'm referring to a situation where elements were not discovered consecutively by number of atoms, e.g. where an element with 103 protons was discovered before an element with 102 protons, and again with 109 protons before 108, according to this table. I use the qualification "after 1950", that is, the "modern era," because this would have given scientists time to sort out Mendeleev's Table after it gave us a road map to atoms. Perhaps more to the the point, the scientific process is a lot more organized than it was in Mendeleyeev's time.

What were the circumstances of these out-of-order discoveries? Isn't it exponentially harder to discover elements with 103 protons than 102 or 101, etc.? Or wouldn't scientists concentrate on creating the atom with the "next" number of protons, e.g. 116 using advanced technology nowadays? Because this is a process that is different from "randomly" coming across atoms commonly found in nature.

The synthetic trans-uranic elements (the "modern era" elements as you call them) are synthesized by bombarding a certain isotope of one element with a certain isotope of another element with a lot of energy in order to get nuclear fusion. The reason they are "out of order" is that the building blocks of these elements have to be very specific isotopes (extra rich in neutrons for example, somewhat abundant for another). This limits the possible sources considerably and so the heavy elements synthesized (and hence discovered) earlier were those where there was a convenient pair of smaller elements to combine to form the larger ones. You don't just bombard each successive element with protons to make the next one.

A lot of the early discoveries were made by bombarding sources with alpha particles, so there are several discoveries that are increases by 2 protons (e.g. Cm->Cf, Am->Bk). Indeed, the table you linked gives more examples of the pairs of elements used to make the heavy actinides and beyond.


molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF Here was my attempt #$1$ at a solution.

$K_{sp}=[Ca^{2+}][F^-]^2$

Using the information at pH=7, I estimated that $K_{sp}\approx3.7\cdot10^{-11}$, which means that barely any HF formed.

I assumed that only $HF$ forms and not $CaOH$ because it's insoluble.

I got $\displaystyle \frac{[H^+][F^-]}{HF}=10^{-3.17}$

We have $  \displaystyle \sqrt{\frac{3.7\cdot10^{-11}}{[Ca^{2+}]}}$ fluoride ions.

We rewrite $K_a=10^{-3.17}=\displaystyle\frac{10^{-3}[F^-]}{2[Ca^{2+}]-F^-}$ and plug in the concentration of fluoride ions in terms of calcium ions.

This yields $[Ca^{2+}]=3.7*10^{-4}$, which gives a multiplier of 1.76.

The correct answer is 1.83, which I guessed based on my answer. How do I arrive at that?

The $K_{sp}$ of $\ce{CaF2}$ which you found out as $3.7 \times 10^{-11}$ is indeed correct, and after that essentially you have to solve two equations.

Now, suppose at $pH =3$, (i.e. $\ce{[H+] = 10^{-3}}$),  $\ce{CaF2}$ dissociates as $\ce{Ca^2+}$ and $\ce{F-}$ and their concentration in the solution are as follows, $$\ce{CaF2 <=> Ca^2+ + 2F-}$$ $\ce{[Ca^2+ ] = S }$ mol/L. and $\ce{[F-] = 2S -}$ $x$ mol/L. ,as, this $x$ mol/L. of $\ce{F-}$ is in equilibrium with $\ce{HF}$, as, $$\ce{H+ + F- <=> HF}$$ This equilibrium satisfies the equation $$\ce{\frac{[H^+][F^-]}{[HF]}= 10^{-3.17 }   => \frac{10^{-3}(2S - x)}{x} = 10^{-3.17}  ...(i)}$$.

And another equation which you have to solve is just regarding $K_{sp}$, i.e $$S(2S-x)^2 = 3.7 \times 10^{-11} .....(ii)$$ From first equation, you will get $\frac{2S}{x} = 1.676$ ; if you substitute for $x$ in the second equation by this, you can now solve for $S$ and the answer will be $S= 3.845 \times 10^{-4}$.

Thus , the factor will be $\frac{3.845 \times 10^{-4}}{2.1 \times 10^{-4}} = 1.83$


How can I have iron powder without a fast reaction with oxygen? I've seen this video in which fine iron powder reacts immediately with oxygen when exposed to the air, turning into rust.

Now, will this reaction always occur with iron powder? Or is there something special with this specific sample?

I'm writing a steampunk story where people are liberating iron powder in the air. Now, I know that iron powder exists without creating sparks, but I don't want to make some huge mistake, so I'd like to understand a little more about this reaction.

Also, the presence of coal combustion smoke or water steam would have some influence in the reaction?

All is matter of how fine the powder is ( surface per mass ratio ) and what is the level of prior surface reaction with oxygen.
If the powder is very fine ( I mean very ) and the surface is "naked iron", it would burn in a small firework of sparks.
If the powder is course or the surface is already passivated by oxygen because of the powder history of preparation, it would not provide noticeable reaction.
I remember some old book of chemical experiments, preparing pyrophoric powder by pyrolysis of ferrous oxalate,
but it was not iron powder, but pyrophoric ferrous oxide $\ce{FeO}$:

Preparation
$\ce{FeO}$ can be prepared by the thermal decomposition of iron(II) oxalate.
$$\ce{FeC2O4  -> FeO  +  CO2  +  CO}$$
$\ce{FeO}$ is thermodynamically unstable below $\pu{575 °C}$, tending to disproportionate to metal and $\ce{Fe3O4}$:
$$\ce{4 FeO(s) -> Fe3O4(s) + Fe(s)}$$

With the following oxidation by air;
$$\ce{3 Fe(s) + 2 O2(g)-> Fe3O4(s)}$$
respectively;
$$\ce{6 FeO(s) + O2(g)-> 2 Fe3O4(s)}$$


Based on Molecular Orbital Theory , how is it decided that the double bond in O2 molecule has one sigma bond and one pi bond? My Chemistry textbook says this in regard to a C2 molecule:

  It is important to note that double bond
  in C2 consists of both pi bonds because of the
  presence of four electrons in two pi molecular
  orbitals.

Is it possible to account for the double bond in an Oxygen atom consisting of a sigma bond and a pi bond in the same way, based on MOT?

$\ce{O2}$ MO diagram
From the MO diagram, we can see that it has 2 electrons in a $\sigma$ orbital, 4 in $\pi$ orbitals, and 2 in $\pi$* orbitals. Taking this together gives a bond order 2, with 1 coming from the sigma electrons and another 1 being the net value for the $\pi$ and $\pi$* electrons.


How many electrons are transferred in this redox reaction? I have the following reaction:

$$\ce{CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2 O(l)}$$

How do I determine how many electrons are transferred in this reaction? More specifically, I need to know the value of $v_e$ which I am supposed to plug into the Nernst equation, which my textbook explains as the coefficient of the number of electrons transferred in each half-reaction. How does one go about splitting this into two half-reactions?

The carbon atom must have an oxidation state (O.S.) of -4 on the left hand side since each hydrogen bonded to it must have an O.S. of +1. On the right hand side, the carbon atom has an O.S. of +4 since each oxygen bonded to it has an O.S. of -2. The carbon's oxidation state increased by 8. This is a hint that 8 electrons were transferred. To verify, examine the oxygen atoms. On the left, we only have free oxygen, therefore the O.S. is 0. On the right, we have 4 total oxygen atoms, each with an O.S. of -2. Thus our net change in oxidation state is 0 - 8 = -8. We can almost conclude that all these electrons were transferred from carbon, but we must check the oxidation states of the hydrogen atoms. On both sides, each H atom has an O.S. of +1, so there is no net O.S. change. Thus we can confirm that 8 electrons were indeed transferred in this reaction.


What does it mean for a higher glass transition temperature in terms of crystallinity? The glass transition temperature is associated with amorphous regions in a polymer. If I have observed in a lab experiment that polymers with a higher glass transition temperature tend to have less crystalline structures, how should I interpret this?

A higher glass transition indicates less energy is required for the bonds of the glass to relax their strain. This is usually due to a slower cooling rate. In a slower cooling rate regime, more time is allowed for crystallization to occur.  So while you are correct to observe a correlation, one does not cause the other but rather they have a common cause of the slower cooling rate.


Calculating Vapor Pressure from Delta G/ Kp For the process, $\ce{CH3OH(l) -> CH3OH(g)}$
  $\Delta G^\circ = 4.30\ \mathrm{kJ/mol}$ at 25 °C. What is the vapor pressure
  of $\ce{CH3OH(l)}$ at 25 °C in mmHg?
  (A) 0.176 mmHg
  (B) 14.0 mmHg
  (C) 134 mmHg
  (D) 759 mmHg

This question is from the 2012 National Chemistry Olympiad (US).

Attempt at solution:

$\Delta G^\circ = -RT\ln K$
$4.3\times10^3\ \mathrm{J/mol}=-8.314\ \mathrm{J/(mol\cdot K)}\cdot 298\ \mathrm{K} \cdot\ln K$
thus
$0.1763\ \text{(unitless)} = K$

Now, I’m supposed to know that $K$ is in units of atm and convert to mmHg accordingly. How would it know that this unitless quantity is atm and not pascal or mmHg?

This topic came up the other day but in regards to the evaporation of water.  You might find my answer to that question helpful.

Here's how I would solve the problem.

Explicitly note standard state used and assumptions.  I will assume:

Methanol vapor is an ideal gas
Standard states for all species are 1 bar of partial pressure 25 °C.  (This assumption matters a lot.  The meaning of the $^\circ$ symbol in $\Delta G^\circ$ can't be understood without it.
The free energy of methanol liquid does not vary with pressure, i.e. $G_{liq} = G_{liq}^\circ$ at all pressures.  I'm fairly confident that this assumption is equivalent to assuming methanol liquid is an incompressible fluid.

Note the equilibrium condition.  At equilibrium,

$$\Delta G = 0 = G_{vap} - G_{liq}$$

Substitute the pressure-dependent free energy of an ideal gas

$$ 0 = G_{vap} - G_{liq} = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}$$

Use the assumption of incompressibility of liquid methanol that we made above

$$0 = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}^\circ$$

Rearrange and substitute

$$0 = \left [G_{vap}^\circ - G_{liq}^\circ \right ] + RT \ln{\frac{p}{p^\circ}}$$

$$0 = \Delta G^\circ  + RT \ln{\frac{p}{p^\circ}}$$

$$0 = 4.3 \frac{\textrm{kJ}}{\textrm{mol}} + RT \ln \frac{p}{p^\circ}$$

Solve for $p$

$$\ln{\frac{p}{p^\circ}} = \frac{-4.3 \frac{\textrm{kJ}}{\textrm{mol}}}{RT} $$

$$p = p^\circ e^{\frac{-4300 \textrm{ J/mol}}{RT}}$$

$$p = p^\circ e^{\frac{-4300}{8.314\cdot 298}}=  p^\circ e^{-1.73} = 0.176~p^\circ  $$

Figure out what $p^\circ$ is.  Just like it does for $\Delta G$, adding a $^\circ$ indicates that we are referring to a standard state.  Thus $p^\circ$ refers to the partial pressure of methanol vapor at its standard state.  This is almost always defined as one atmosphere (or in more modern times, one bar, which is nearly but not exactly the same thing) of pressure. Assuming that the standard state is the usual one, then $p^\circ$ is one bar, which is equal to 750 mmHg.  This is what we have assumed in step 1.  If that is the case, then $p = 0.176~ p^\circ=0.176\cdot750~\textrm{mmHg}=134~\textrm{mmHg}$.

  How would it know that this unitless quantity is atm and not pascal or mmHg?

Hopefully my answer above helps you answer your question.  It has to do with the common use of $\circ$ to indicate a standard state of one bar of partial pressure for all gases.


Equilibriums and Overshooting If we have some reaction $\ce{A(aq) + B(aq) -> AB(aq)}$  and right now we have less $\ce{A}$ and $\ce{B}$ then we would have at equilibrium, is there ever a time where the amount of $\ce{A}$ and $\ce{B}$ would overshoot the equilibrium of $\ce{A}$ and $\ce{B}$ if the reaction proceeded without any new sources of energy or reactants. In other words, will the concentrations of $\ce{A}$ and $\ce{B}$ always increase towards their equilibrium concentrations but never actually reach those points or will they sometimes overshoot that concentration a large amount (not just an amount that can be ascribed to chance) but then fall back towards the equilibrium concentration.

Not really. Overshooting doesn't happen, rather the concentrations gradually approaches the equilibrium concentration as an asymptote.

You'll soon see what I mean.

As an example, for simplicity, lets consider:

$$\ce{P <=>[k_1][k_2] Q}$$

I will assume here that the forward and backward reactions are first order, so we get the rate to be:

$$\text{Rate}_{\text{forward}}=k_1\ce{[P]}$$

and

$$\text{Rate}_{\text{backward}}=k_2\ce{[Q]}$$

As a result the net rate would be:

$$\frac{d\ce{[P]}}{dt}=-k_1\ce{[P]}+k_2\ce{[Q]}$$

Let's take the initial concentration of $\ce{P}$ to be $\ce{[P]_0}$, and the initial concentration of $\ce{Q}$ to be zero.

Suppose at any time $t$, the concentration of the reactant is $\ce{[P]}$. Thus, $\ce{[Q]} = \ce{[P]_0} - \ce{[P]}$. Now, we get the rate law to be:

$$\frac{d\ce{[P]}}{dt} = k_2\ce{[P]_0} - (k_1 + k_2)\ce{[P]}$$

Integrating this from $\ce{[P]_0}$ to $\ce{[P]}$:

$$\int_{\ce{[P]_0}}^{\ce{[P]}}\frac{d\ce{[P]}}{k_2{[P]_0}-(k_1 + k_2)\ce{[P]}}=\int_0^tdt$$

Solving the integral, we get:

$$\frac{-1}{(k_1 + k_2)}\log_e{\left(\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}\right)}=t$$

$$\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}=e^{-(k_1 + k_2)t}$$

Finally, on simplifying it, we get our integrated rate equation as:

$$\ce{[P]}=\ce{[P]_0}\frac{k_2 + k_1e^{-(k_1+k_2)t}}{k_1+k_2}$$

Let's plot this on a graph to understand what this is actually saying:

(purple-$\ce{[P]}$ and orange-$\ce{[Q]}$)

You can clearly observe that the concentrations never overshoot the asymptotes (black dotted line). The asymptotes denote the equilibrium concentrations, so it's evident that the reaction tends toward the equilibrium concentration.


Question about change in pressure and change in temperature for system at equilibrium I was reading my Chemistry textbook and I noticed that it stated that for a system at equilibrium:

At constant temperature, reducing the volume of gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. Increasing the volume will shift the system in the direction of more moles of gas.
If the temperature is increased at equilibrium the system will shift in the direction to consume the heat. If the temperature is decreased, the system will shift in order to produce excess heat.

My question is why does this occur?

Please answer my questions in terms of how the molecules in the system interact, instead of just saying that what my textbook states is true.

Also, explain (if possible) using the chemical reaction of N2 (g) + 3H2 (g) ↔ 2NH3 (g), or a similar one.

Why would increasing the pressure cause it to shift in the direction of fewer moles? What, physically, makes the reaction that produces fewer moles favorable?

Also, why would increasing the temperature cause the system to shift to the more endothermic side? What, physically, makes the endothermic reaction more favorable than the exothermic reaction?

For a general idea, see Le Chatelier's principle

  When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure,
  (1) the system changes to a new equilibrium and
  (2) this change partly counteracts the applied change.

For the changes of $p, V$, read about equilibrium constant $K$ and Reaction quotient $Q$ and their relation.

Notice, how the Reaction quotient changes with increased pressure/decreased volume, if the reaction increases the molar amount of gases.
And how it changes if it decreases them.

For the former case, increasing pressure increases Q and the equilibrium shifts backwards to make Q  equal to K again.

For the latter case, increasing pressure decreases Q and the equilibrium shifts forwards to make Q equal to K again.

For the temperature effect, see van't Hoff equation:

  $$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

Endothermic reaction direction has higher activation energy than the exothermic direction.

As the consequence,

the reaction rate of the endothermic reaction direction

raises with temperature faster

than the reaction rate of the exothermic reaction direction.

Faster endothermic then exothermic reaction means the equilibrium constant shifts at higher temperature in favour of the endothermic reaction.


Identity a metal based on its properties A metal X can be cut with a knife. When placed in water, metal X floats and melts. What is X? The options are calcium, copper, magnesium and potassium.

I think it can't be potassium because it will react vigorously with water, So I believe it's calcium. But the only problem is that calcium is denser than water so it will sink.

Copper is much denser than water and it does not react with it. Magnesium is also denser than water and reacts only very slowly. Calcium will also sink in water, although it will react with it producing hydrogen, so small pieces of the element may float on the surface. However, it would definitely not melt. The correct answer is potassium - it is a very soft metal, less dense than water and reacts with it producing enough heat to melt it (potassium melts at only $\pu{63.5^\circ C}$).


Geometrical isomerism in cycloalkenes Why doesn't cyclohexene have a cis-trans isomer whereas cyclodecene has one?
Is strain a factor?

You simply can't make a transoid molecule of cyclohexene because there is not enough "length" in the carbon chain.

I don't know if it is "strain" per se--but a putative transoid cyclohexene would have extraordinarily exotic bonding arrangement--thus being very high in energy.