Calculate molar mass of a compound from a given molecular mass The mass of one molecule of a compound is $2.19\times10^{-22}\ \mathrm g$. What is the molar mass of the compound?

My attempt:
Using the formula moles=mass/molar mass, I found how to calculate molar mass using the formula molar mass=moles/mass.
Therefore, I was under the impression that, surely, $6.02\times10^{23}$ divided by the mass of the compound would have resulted in the molar mass. However, the answer was wrong.

To get the molar mass of your mystery compound, make use of the fact that there are $N_{\rm A}$ (Avogadro's number) of molecules of it in one mole of it.

To solve the problem, we multiply the molecular mass (given) by Avogadro's number (known) to get the mass of one mole of the mystery compound:

$${\rm molar \;mass} = N_{\rm A}\times 2.19\times 10^{-22}\,{\rm g} = (6.022\times 10^{23}\,{\rm mol}^{-1}) \times (2.19\times 10^{-22}\,{\rm g}) = 131.88\;{{\rm g}\over{\rm mol}}$$


How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock? As a non chemist I am most often charmed when visiting Wikipedia articles of chemical elements and see images of very pure specimens of element after element, proton by proton, and often also metal cube specimen made from smithing similar pure deposits.

The wiki article Periodic table allows me to do so easily; here are some elements I found looking almost the same and don't think I personally could distinguish between them without some instrument:

molybdenum and manganase
titanium and chromium
rutenium and cadmium
sodium and aluminium
silicone and germanium

My problem

Generally all examples are of synthetic specimens of about 99.9% purity but that me to wonder what test can a chemist make in a natural environment, say inside a natural cavemine, to determine which natural deposit contains what after the mining and positioning a seemingly multimineral natural deposit;
Of course, I understand that with natural deposits of mainly carbon or copper, or iron (water exposed and rusty) natural deposits, let alone with very pure specimens of these elements, it will be easy because of color, but I believe that generally or mostly color by itself won't be enough due to various color biases and I assume that today there are in-place very-accurate tests that can be done, especially for different parts of a multimineral deposits.

My question

How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock?

What you're describing is essentially the field of "qualitative inorganic analysis". That's a very broad subject, but a short description is that prior to the advent of advanced instrumentation, chemists developed extensive series of tests specifically to determine the elemental composition of samples.

For example, one would start with basic physical properties like color, density, and hardness, then look at reactions with heat, acids, bases, oxidants, etc. If the compound could be solubilized in any way, then one can look at how the solubility/color/etc is affected by addition of known cations and anions. If it was a geological specimen, then the location and the types of surrounding minerals would help inform the analysis as well.

A well trained inorganic chemist would have standard flow-charts of tests and could fairly confidently identify the elemental composition of common minerals and metals.


IUPAC name of cyclopentanol derivative Is the name of the following compound 3-(ethenyl)-5-(prop-1-enyl)-2-(prop-1-ynyl)-4-(prop-2-enyl)cyclopent-2-en-ol?

The problem I am facing is that how do I arrange 5-(prop-1-enyl), 2-(prop-1-enyl) and 4-(prop-2-enyl). In which order do I write these in the IUPAC name?

The relevant rule for the order of citation of detachable substituent prefixes in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-14.5 ALPHANUMERICAL ORDER
(…)
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
Alphanumerical order is applied as follows in organic nomenclature. Nonitalic Roman letters are considered first, unless used as locants or part of a compound or composite locant, for example, ‘N’ or ‘4a’ (see P-14.3), or in an isotopic descriptor. (…)

P-14.5.4 When two or more prefixes consist of identical Roman letters, priority for order of citation is given to the group that contains the lowest locant(s) at the first point of difference.

Therefore, the correct name for the compound that is shown in the question is
3-ethenyl-5-[(1E)-prop-1-en-1-yl]-4-(prop-2-en-1-yl)-2-(prop-1-yn-1-yl)cyclopentan-1-ol
since the order is ‘ethenyl’ > ‘propenyl’ > ‘propynyl’ and ‘prop-1-en-1-yl’ > ‘prop-2-en-1-yl’.
The fact that ‘4-propenyl’ is lower than ‘5-propenyl’ is irrelevant.


Are all organic molecules with odd-numbered electron count radicals? This question was triggered by another question on this site, about molecules with a total number of 13 electrons.

I couldn't think of any (except the boring answer of $\ce{Al}$), and the answers to the question are all about radicals, instead of 'normal' molecules. So that got me thinking: are all organic molecules with an odd total electron count radicals or radical ions? If so, is there any fundamental reason for this?

By IUPAC definition of radical, the organic molecules with odd number of electrons must have at least one of them unpaired and are therefore radicals. The fundamental reason is the definition.

Bear in mind, that other spin states must be taken into account, as simplest example the oxygen. It exists as a triplet, so two unpaired electrons.

The situation gets different, when you do have a metal in your compound, but it is out of the scope of the question.


Propagation of Uncertainties I have five values for the volume of sodium hydroxide needed to neutralise a fixed quantity of hydrochloric acid, each trial with an uncertainty of 0.05 mL. If I take the average of these five values, in what way would the uncertainty propagate? I can think of three of the following ways, but don't understand which would be correct:

Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5, we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.05 mL.
Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5 (which has no uncertainty), we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.25 mL.
Range divided by two is often another way to calculate the uncertainty of an average value, which for my case results in a value different to the above two for the uncertainty of the average.

Which one, and why, would be the correct one to state next to the average value (with a plus-minus sign), as an uncertainty of the average value?

None of your methods are correct. If you have a small data set - in this case, 5 values from a titration - then the absolute uncertainty for each individual value is half the range of data. For example, say you get the following results (in ml, from a burette with a precision of 0.05ml): $4.70$, $5.35$, $4.85$, $4.60$ and $5.20$. The range of these results is $5.35-4.60=0.75$. This gives an absolute uncertainty of $\frac{0.75}{2}=0.375$ for each individual piece of data e.g. $4.85\pm0.38$ml.

However, if the absolute uncertainty calculated from half the range is less than the precision of the instrument (in this case $<0.05$ ), then the precision of the instrument should be given as the uncertainty e.g. $5.00\pm0.05$. I think you may have incorrectly calculated the absolute uncertainties.

For small data sets ($<10$ pieces of data), the uncertainty in the mean is then given by $$\Delta {x}_{avg}=\frac{\Delta x}{\sqrt{N}}=\frac{R}{2\sqrt{N}}$$ where $\Delta {x}_{avg}$ is the uncertainty in the mean, ${N}$ is the number of times the measurement was repeated, $R$ is the range and $\Delta{x}$ is the uncertainty in one measurement.

For example, using the example data above, $\Delta {x}_{avg}=\frac{0.375}{\sqrt{5}}=0.168$. The average can then be stated as $4.94\pm0.17$ml


How to calculate the 15 geometrical isomers of an octahedral complex with all ligands different? Suppose we have a complex of type $\ce{[Mabcdef]}$ then we know it’ll show geometrical isomerism. According to my textbook (Cengage Inorganic Chemistry II, Page 7.37) there are 15 geometrical isomers possible.

I tried calculating it but ended up with different answers every time.

I think we have to use the knowledge of permutations and combinations for this calculation.

Assign "a" to one of the six equivalent positions.
Note that there are now 5 choices (b,c,d,e,f) for the opposing position.
You are left with four equivalent equatorial positions.  Once a ligand is assigned to one of these four, there are 3 ligands remaining to select for the opposing equatorial position.
Finally, there are 2 ways to assign the last two positions. The two possibilities are mirror images of each other.

So $5\times3\times2= 30$

30 isomers (15 pairs of enantiomers).


Why are there only 7 types of unit cells and 14 types of Bravais lattices? Why are there only 7 types of unit cells and 14 types of Bravais lattices?

I was reading about solid-state chemistry for the first time and this limitation made no sense to me.

I tried to do the math and realized that there could be many more possibilities. Usually, the standard unit cell is described on the basis of whether the sides are at 90 degrees to each other and whether the sides are equal or not. However, any angle other than 90 could be chosen as the reference angle. So why 90?

Is it because only these 14 systems are found in nature and there is no need to study others?

Moreover, why are there only 4 sided shapes or hexagonal shapes?
Why not pentagonal or trigonal?

Any reference from any site or textbook would be appreciated.

All quotes will be from Solid State Physics by Ashcroft and Mermin.

Bravais Lattice:

  A fundamental concept in the description of any crystalline solid is that of the Bravais lattice, which specifies the periodic array in which the repeated units of the crystal are arranged.  The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be."

Primitive Unit Cell:

  A volume of space that, when translated through all the vectors in a Bravais lattice, just fills all of space without either overlapping itself or leaving voids is called a primitive cell or primitive unit cell of the lattice.

Unit Cell; Conventional Unit Cell:

  One can fill space up with nonprimitive unit cells (known simply as unit cells or conventional unit cells). A unit cell is a region that just fills space without any overlapping when translated through some subset of the vectors of a Bravais lattice. The conventional unit cell is generally chosen to be bigger than the primitive cell and to have the required symmetry.

Crystal Structure:

  A physical crystal can be described by giving its underlying Bravais lattice, together with a description of the arrangement of atoms, molecules, ions, etc. within a particular primitive cell.

So, one comes up with 14 Bravais lattices from symmetry considerations, divided into 7 crystal systems (cubic, tetragonal, orthorhombic,monoclinic, triclinic, trigonal, and hexagonal). This comes solely by enumerating the ways in which a periodic array of points can exist in 3 dimensions.

Now, what is on those points is a unit cell, which will itself have some symmetry. Thus, the combination of Bravais lattice and unit cell symmetry can again be enumerated and one comes up with 230 space groups.

Now for some of your related questions:

All cubic-related Bravais lattices will have 90 degree angles because they are based on cubic symmetry.  The trigonal Bravais lattice has no 90 degree angles, but isn't talked about much in more basic textbooks because, well, it looks weird.

Why no pentagonal unit cells? Well, because you can't fill space with a 5-fold symmetric Bravais lattice. Quasicrystals, while they have 5-fold symmetry, are a tiling through space that does not obey the rules for a Bravais lattice.


Absorption spectrum of Hydrogen molecule in UV region 200 nm - 300 nm (not Atomic Hydrogen) Can somebody provide Absorption spectrum of Hydrogen molecule? I have searched days for absorption spectrum of Hydrogen molecule in 200 nm - 300 nm UV region and All I got is absorption spectrum of atomic hydrogen. I do find little about absorption spectrum of hydrogen molecule in the region 90 nm to 180 nm in some astrophysics research papers.

Being a homonuclear diatomic molecule, molecular hydrogen does not posses a pure rotational of vibrational (electric dipole) spectrum. The lowest excited singlet state is the B state (for historical reasons there is no A state in H$_2$) and transitions from this state to the X ground state are known as the Lyman band which starts around 120 nm, thus outside the range you are interested in. In principle, electric quadrupole transitions may be observed but these are several orders of magnitude weaker than electric dipole transitions. In addition, the dissociation energy of molecular hydrogen lies around 36000 cm$^{-1}$, so that only radiation longer than about 277 nm can result in bound-to-bound transitions. However, I expect that the Franck-Condon factors to these highly excited vibrational states are very poor.

So there reason that you cannot find a spectrum is that there is none.


What would be the most suitable solvent for a single-solvent recrystallization? The compound in question is acetanilide, and
the solvents in question are: water, ethanol, and ethyl acetate

I was thinking all three would be suitable since upon experimentation, acetanilide dissolved in all three solvents when heated.

Is it possible to rank the suitability?
As, well could someone provide an explanation as to why one is more suitable than the other ?

The ideal properties for a recrystallization solvent are that the compound of interest is soluble at high temperature but insoluble at low temperature. You have addressed the first part, so you just have to assess the second part.

Given that acetanilide is a non-polar compound, I would expect that it would have the lowest solubility in the most polar solvent. For this reason, water is probably the best, then ethanol. I would expect acetanilide to be readily soluble in ethyl acetate at room temperature, making it unsuitable for this recrystallization.


The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid? The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid?

What you meant is known as erosion. Erosion is the chemical loss of enamel due to acid.When acid continuously attacks teeth, they cannot repair themselves and will gradually begin to turn fuzzy and dissolve. Most soft drinks contain acids such as citric, phosphoric and malic or tartaric acids. A study was done where extracted human teeth were placed in softdrinks. Within two days, the teeth became very soft and the enamel surface lost much of its calcium.

Brush your teeth at least twice a day with a fluoride-containing toothpaste.A mouth rinse containing fluoride can help prevent tooth decay, according to the American Dental Association. For children, sealants can be used.


Hydrolysis of 3,4-dihydropyran In the 3,4-dihydropyran shown below will the oxygen or the pi electrons of the double bond attack the hydrogen cation during hydrolysis and why?

This is my attempt towards a solution: The oxygen lone pair clearly can attack the hydrogen cation better than the pi electrons being a better a nucleophile, but according to my textbook (T.W.Grahams Solomons) the pi electrons of the double bond attacks, but shouldn't the oxygen atom be a better nucleophile than the pi electrons?

The functional group in question is a vinyl ether, and it is a mistake to consider these groups as isolated.  The oxygen is in conjugation with the vinyl group, meaning that resonance favors the entire system attacking.  There is no way to leverage conjugation as an energetic driving force if the oxygen attacks by itself.


Why is nicotinoyl chloride available as its hydrochloride salt? How can we separate them? Nicotinoyl chloride is only available as its hydrochloride salt.

Why can't nicotinoyl chloride be synthesized in free form? How can we seperate nicotinoyl chloride from its salt form? Can we try any base for this purpose?

Unfortunately, i don't have access to the original reference from 1928 Über die Einwirkung von Thionylchlorid auf die Pyridin-monocarbonsäuren by Meyer and Graf, but it's cited in The Preparation of Nicotinyl Chloride from 1953.

In this later note, where dried and grounded potassium nicotinate was reacted with oxalyl chloride in anhydrous benzene and the resulting nicotinyl chloride eventually destilled off under reduced pressure (75-90 °C, 10-12 mm), the authors state:

  On exposure to the air, the oily acyl chloride soon solidified to nicotinic acid hydrochloride.

With other words, the combination of moisture and the basicity of the pyridine moiety results in the decomposition of the product. This is apparently prevented by deactivation of the pyridine upon conversion to its hydrochloride.


How to choose the correct name for calcium sulfate based on a calculations of the crystallisation water included? I have a test review question which i really need help on. It states :

  A student heated a sample of the hydrate $\ce{CaSO4 * xH2O}$ in an evaporating dish over a bunsen burner to remove all the water and recorded the following data:

  Mass of empty evaporating dish: $\pu{19.73g}$
  Mass of evaporating dish + hydrate: $\pu{21.50g}$
  Mass of evaporating dish + anhydrous $\ce{CaSO4}$: $\pu{20.80g}$

  Determine A) the value of $x$ and B) the most likely name/formula of the hydrate.

I already figured out $x$ by getting the mass of water then simplifying the ratio of anhydrous to water.
But the second question which asks to name the formula of the hydrate really confused me. I would appreciate if anyone can guide me how to solve B.

From my calculations

\begin{align}
x &= \frac{M(\ce{CaSO4}) \times \left[m(\ce{CaSO4 * xH2O}) - m(\ce{CaSO4})\right]}{M(\ce{H2O}) \times m(\ce{CaSO4})} =\\
&= \frac{\pu{136.14 g mol-1} \times (\pu{1.77 g} - \pu{1.07 g})}{\pu{18.02 g mol-1} \times \pu{1.07 g}} =\\
&= 4.94,
\end{align}

which gives $\ce{CaSO4* 5 H2O}$, calcium(II) sulfate pentahydrate.


Why ionisation enthalpy of early lanthanides are expected to be greater than early actinides? In my book Chemistry Part I (NCERT XII), a statement on p. 232 goes like this:

  It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids.

My thought on this matter is that $Z$ will increase and f-orbitals will shield electrons poorly and so ionisation enthalpy should increase. But I'm not able to understand what my book is trying to say. What is meant by less penetration to inner core of electrons? To be honest I wasn't able to understand all the above mentioned 3-4 lines. So, can anyone explain this properly to me, please?

This data is incorrct, and I am saying so because the Ionisation energy of the actinoid is a decent bit higher than the corresponding lanthanide

In each case IE 1 is significantly higher for lanthanoids, IE 2 is still higher for lanthanoids and IE 3 is comparable.

$\ce{Ce}$ $534.4$, $1050$, $1949$

$\ce{Th}$ $587$, $1110$, $1930$

These were two examples where i showed all 3 IEs

For, the sake of further clarification, here is a the IE1 of the first few

$\ce{Pr}$ $527$, $\ce{Pa}$ $568$

$\ce{Nd}$ $533.1$, $\ce{U}$ $597.1$

$\ce{Pm}$ $540$, $\ce{Np}$ $604.5$

$\ce{Sm}$ $544.5$, $\ce{Pu}$ $584.7$

(units used are $kJ/mol$)

The reason you stated in your doubt is absolutely correct, as even if we were to conclude at these aren't accurately determined, we see this phenomenon elsewhere in the periodic table also, where the values are accurately determined.

$\ce{Sn}$ is the lowest IE element of GP 14 and not $\ce{Pb}$.
$\ce{Fr}$ has a higher IE than $\ce{Cs}$.
$\ce{Ra}$ has a higher IE than $\ce{Ba}$.

All due to the same reason poor shieling, which increases the effective nuclear charge, which increases nuclear attraction, which in turn leads to a higher IE.

References

Wikipedia data page


What compounds do Nor1 and Nor2 citalopram refer to? In Goodman and Gilman's The Pharmacological Basis of Therapeutics 12th Edition pg. 406, two drugs, nor1-citalopram and nor2-citalopram are referred to but I have no idea what they are. Could somebody tell me a more commonly used name of these drugs or their IUPAC names if common names do not exist. Also please show me what exactly the nor1 and nor2 refers to. I'm guessing there's methyl groups missing in them at different positions because I know that norpethidine (or normeperidine) refers to the N-demethylated metabolite of pethidine (or meperidine).

Those names are referring to desmethylcitalopram and didesmethylcitalopram, which, as you correctly predicted, are the N-demethylated and N-didemethylated metabolites of citalopram. Per IUPAC conventions, the nor- prefix is used to refer to any structural derivative of some parent compound in which a single carbon and (possibly) its hydrogens have been replaced (typically by a hydrogen, but not exclusively). See the relevant Wikipedia page on the prefix for more detail.


Meaning of the symbol “$\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$” One of the back exercises (13(C)) from the book for practice provided at my school asked whether the following relation is true:

$\Delta_\mathrm{f}H^\circ(\ce{I2,g}) = \Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$ at $\pu{25 °C}.$

I know how to solve the question, but the answer would depend on the meaning of  $\Delta_\mathrm{sub}H^\circ[\ce{I2,s}].$ Which of the following reaction does the symbol $\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$ represent?
$$
\begin{align}
\ce{I2(g) &-> I2(s)}\tag{a}\\
\ce{I2(s) &-> I2(g)}\tag{b}
\end{align}
$$

$\ ΔH_{sub}(I_2,s)$ can be interpreted as the change in enthalpy when one mole of solid $\ I_2$  converts to gaseous $\ I_2$  at a constant temperature. So the answer should be reaction (b).


Why does F replace the axial bond in PCl5? Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$? I realize that it would be more stable there than at equatorial bond, but what is the reason of its stability? Similarly in $\ce{AB4}$ type of molecules with $\ce{sp^3d}$ hybridization (4 bond pairs and 1 lone pair) why is the geometry of the molecule that of a see-saw, where the lone pair is at equatorial position rather than at axial one?

My book states the reason to be "due to Bent's rule" but I have difficulty linking these two.
Also Bent's rule states that the hybridized orbitals of equivalent energy of central atom tend to give more %s character to the electropositive atom attached to it rather than the electronegative one. (e.g. - in $\ce{CH3F}$) Am I right? What else is more to this rule?

Recently, there has been a lot of discussion of Bent's rule (see for example "What is Bent's rule?") here in SE Chem. Simply stated, the rule suggests that $\mathrm{p}$-character tends to concentrate in orbitals directed at electronegative elements.

Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$?

In order to answer this question, we need to start by understanding the bonding in $\ce{PCl5}$ and its fluorinated isomers. In introductory courses and texts, it is usually stated that $\ce{AX5}$ type molecules adopt a trigonal bipyramid geometry and are $\mathrm{sp^{3}d}$ hybridized. As the comments by Martin and permeakra point out, and as is learned in more advanced classes, this hybridization scheme is likely incorrect. There are several reasons that argue against $\mathrm{sp^{3}d}$ hybridization including: 1) the fact that $\mathrm{d}$ orbitals are relatively high in energy compared to $\mathrm{s}$ and $\mathrm{p}$ orbitals and therefore it is energetically quite costly to involve them in bonding; and 2) $\mathrm{d}$ orbitals in non-metals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak.
A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved. Applying this concept to $\ce{PCl5}$ we would say that the central phosphorus atom is $\mathrm{sp^2}$ hybridized.  Thus, there would be 3 $\mathrm{sp^2}$ orbitals (these will be used to create the equatorial bonds) and a $\mathrm{p}$ orbital (this will be used to create our axial bonds) emanating from the central phosphorus atom.

The $\mathrm{p}$ orbital contains 2 electrons and will form bonds to 2 ligands (chlorine or fluorine in the case at hand). For simplicity, let's say that these ligands also use $\mathrm{p}$ orbitals for bonding (but it could be any type of orbital, $\mathrm{sp^3}$ or whatever is appropriate) and each of these orbitals contains one electron for bonding. This is our 3-center-4-electron bond and its MO diagram is pictured below. Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding M.O., so the bond order in this bond is reduced.

This reduced bond order in the bond using the phosphorus $\mathrm{p}$ orbital explains why the axial bonds in $\ce{AX5}$ type molecules are longer than the equatorial bonds.

Now that we understand the bonding in $\ce{PCl5}$ we can consider the case of $\ce{PCl4F}$ and how Bent's rule applies to the situation.  First, note that fluorine is more electronegative than chlorine.  As stated above, Bent's rule suggests that more electronegative ligands prefer to form bonds with orbitals that are high in $\mathrm{p}$-character. Why is this? $\mathrm{s}$-Orbitals are lower in energy than $\mathrm{p}$-orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more $\mathrm{s}$-character. The two electrons in the $\ce{P-F}$ bond will spend more time around the electronegative fluorine and less time around phosphorus. If that's the case (and it is), why "waste" precious, low-energy, $\mathrm{s}$-orbital character in an orbital that doesn't have much electron density to stabilize. Instead, save that $\mathrm{p}$-character for use in phosphorus hybrid orbitals that do have more electron density around phosphorus (like the $\ce{P-Cl}$ bonds). So, as a consequence of Bent's rule, we would expect phosphorus to use the orbital with lowest $\mathrm{s}$-character, the axial $\mathrm{p}$-orbital, to form the $\ce{P-F}$ bond; and the orbitals with more $\mathrm{s}$-character, the equatorial $\mathrm{sp^2}$ orbitals, to form $\ce{P-Cl}$ bonds.


Can ketones next to a nitrogen hetero atom tautomerise? I worry that the electronegative N will stabilise keto form so much that equilibrium will be so far towards keto it would be unreasonable to call the enolisiation process a factor

If I understand your question correctly, I think you're asking if an amide can enolize like a ketone.  The answer to that question is yes.

The enol content of most organic compounds is very low and special methods must be used to measure the enol content.  Perhaps the question becomes, "is there more or less enol content in an amide than in an ester, for example"?

We can draw resonance structures for an amide (or ester) showing that the nitrogen stabilizes the amide.  But above we also drew resonance structures showing how the nitrogen can also stabilize the enol.  We can draw similar structures for both an ester and its enol as well.  Although in the ester case we wind up putting a positive charge on oxygen in some of the resonance structures, so they might not count for as much as they do in the amide case - but both the starting ester and its enol would both be affected similarly.  So as a first approximation I might guess that the enol content in an ester and an amide would be similar.

Indeed the enol content in malonamide is similar to the enol content in diethyl malonate (see page 34 here)


Entropy of universe for non spontaneous processes When spontaneous processes are accompanied with an increase in entropy of universe and remains unchanged in a reversible process, what happens to entropy when one process is non spontaneous? Would entropy decrease??

Non-spontaneous processes do not occur. You will only see a decrease in the entropy of the universe in your theoretical calculations, never experimentally.


How to reduce Fe(III) impurity in Iron(II) sulfate? I have bought 50 kg $\ce{FeSO4}$, but it was stored on air, and it has turned to brown. Is there a cheap method to recover some $\ce{FeSO4}$?

I recommend you dissolve your iron sulfate in hot water, dissolving as much sulfate as possible in as little hot water as possible. While keeping it hot, dip some coiled iron wire in it as well as some steel wool, and stir gently. The metallic iron should reduce your iron III back to iron II.

Once the solution is clear green, remove the iron wool, but keep the iron wire in it, and let it cool down as slowly as possible, while protecting it from air the best you can. Finish the cooling by putting it in a fridge

You should be left with beautiful large, green crystals of highly pure iron(II) sulfate heptahydrate, and a clear solution of relatively dilute iron(II) sulfate.

Since you have 50 kg of it, you will likely have to proceed in several batches, so I recommend you keep that solution for the next batch to minimize product losses.

Once you are done recrystallizing your iron sulfate, you can heat up that solution to a boil, until some crystals start forming on its surface. Then, immediately stop the heating and cool down that solution again. You'll have some lower grade iron sulfate, but that should still be perfectly usable for gardening.

And if you want to keep some of the prettiest crystals you made in the process, I recommend you keep them in a sealed container with a small piece of moist cloth. Otherwise, they may tend to slowly dehydrate and turn into powder.


Why does N-phenylhydroxylamine give Fehling's test? I have studied that Fehling's test is given by aldehydes and alpha hydroxy ketones. I read in a book that it is also given by phenylhydroxylamine converting it to nitrosobenzene.
Does it gives the test as it has an alpha hydrogen or any other reason? Are there any more exceptions like this?

Quoting from "REDUCTION OF AROMATIC NITRO COMPOUNDS TO N-PHENYLHYDROXYLAMINES - A Comparative Study of the Yields Obtained when
Aromatic Nitro Compounds are Reduced, under Similar Conditions, to N-Phenylhydroxyiaminee by William Henry James"

  N-Phenylhydroxylamine is unstable. It is readily affected by light,
  especially white light. It is also affected by moist air, being
  readily oxidized to nitrosobenzene. It has strong reducing power,
  reducing almost quantitatively Fehling's Reagent to cuprous oxide, and
  causing the formation of a silver mirror when warmed with an
  aramoniacal solution of silver oxide.

So, I guess that its instability and its reducing power are the cause of the effect explained.

Reference

James, William Henry. Reduction of aromatic nitro compounds to N-phenylhydroxylamines. Thesis (M.A.), Boston University, 1932. Boston University Libraries OpenBU


Preparation of alcohols from Monosubstituted cyclic ethers Can methyl oxirane(epoxide) (a mono substituted 3 membered cyclic ether) form alcohols by reacting it with Grignard reagent? If yes, can you please explain the reaction with mechanism? If no, why?

Yes, Girgnard reagents can act as nucleophiles in the ring opening of epoxides. Nucleophilic attack occurs much faster at the less hindered carbon, so the product of this reaction will predominate. The reaction results in the formation of an alkoxide, and does not form the alcohol until treated with a protic acid. The mechanism for Grignard addition to epoxypropane (methyl oxirane) is:

$\hspace{4cm}$

For 2-methyl-2,3-epoxybutane, a more hindered epoxide, the addition still occurs at the less hindered carbon:

$\hspace{3.8cm}$


What is created when "Iron-Out®" is mixed with household bleach? I accidentally mixed bleach in a toilet bowl that had Iron-Out in it. The gas almost overwhelmed me, and left me gasping for breath for a good 15 minutes. What in the heck did I create with those two chemicals?

Good that you left for fresh air.

Assuming it is the same product I know, it calls here for caution:

The below mentioned data safety sheet available (SDS) here mentions for the very product as composition

and recommends as first aid (same source):

Upon the action of bleach, you likely generated sulfur dioxide, gas with pungent smell. Refraining from a remote diagnosis, but if you accident happened in a less ventilated space and you inhaled a lot, consultation of a physician may be adequate as sulfur dioxide ($\ce{SO2}$) is harmful and maybe you equally inhaled vapours of bleach, too.  Likely, mixing "iron out" and bleach together yielded a considerable heat, too.


Why does water react with H+? As we all know, if we put $\ce{H^+}$ ions in water then water reacts with them and produces $\ce{H_3O^+}$:

$$\ce{H_2O + H^+ <=> H_3O^+}$$

My question is, why does water react with $\ce{H^+}$ in the first place? It doesn't have an electron, then how can the reaction occur?

And again, it's already stable and adding one more $\ce{H^+}$ will, in my understanding, destabilize it as it will have one more proton in the molecule.

Water has two lone pairs. H$^+$ is a naked nucleus with a positive charge. It is extremely reactive and would bind to any neutral molecule. Water has two lone pairs of electrons and H$^+$ binds one of them to form H$_3$O$^+$.

In absence of water it would bind other things. For example, it would bind methane to form unstable CH$_5^+$ that breaks down to CH$_3^+$ + H$_2$.


Why does 1,3-dihydroxybenzene have a higher boiling point than 1,2-dihydroxybenzene? 1,2-dihydroxybenzene (catechol) shows intramolecular H-bonding. So due to strong bonding, it should be difficult to break the bonds in 1,2-dihydroxybenzene and therefore, 1,2-dihydroxybenzene should have a higher boiling point than 1,3-dihydroxybenzene. Why is it not so?

Being more volatile implies that the molecules should be set free from mutual interactions. As 1,2-dihydroxybenzene has intramolecular H bonding, it is easy for it to exist independently, but for 1,3-dihydroxybenzene there might be intermolecular H bonding (interaction between two molecules); hence, it is a bit more difficult to break it apart, thus having a higher boiling point than its counterpart.


How can both kinetic and potential energy increase simultaneously in a open substance? I have an open test tube containing a solid substance. In a school environment, we  heat it, which increases thermal (kinetic) energy, and this heat breaks its bonds, increasing its potential energy, turning it into a liquid. I understand in thermodynamics, there is a conversation of energy, yet since this test tube is open to other factors, such as air, can this occur?

Also, if this experiment were insulated what would occur?

You are adding energy by heating, this generally does not break chemical bonds (unless the temperature reaches hundreds of degrees) but instead the weak intermolecular potential energy between molecules (sometimes call van-der Waals interaction) and so the solid will melt, then the liquid will boil, obviously :)

The kinetic and potential energy of the sample has increased. Kinetic energy in translational motion of the molecules and internal potential and kinetic energy of the molecules has also increased as they have absorbed some of the heat as vibrational and rotational energy.  (A non-linear molecule with N atoms has $3N-6$ different ways of vibrating, e.g ethane has 18 ways).

The reason that the temperature increases rapidly is due to the fact that if you heat a pan on the stove, or a boiling tube with a Bunsen burner  is that convection of heat away into the air is very poor. So your boiling tube is effectively insulated. If it were completely insulated then less added heat would be needed to reach the same temperature. As an admittedly odd example, you will know that a plumber can join Cu pipes with solder using a blow torch, but only if they are dry. If they contain a even small amount of water more of the heat is carried away (and heats the water) and soldering becomes impossible.

If you heat a sealed tube, it is likely to explode violently due to increase in pressure; so don't do it! (To prevent explosion is why pressure cookers have a pressure release value).


Radon gas in earthquake prediction - Why Rn? I recently studied an article about predicting earthquakes and how correctly realizing that an increase in $\ce{Rn}$ gas is a sign for the earthquake saved a whole city. The following is derived from the wikipedia page for earthquake prediction:

  There are reports of spikes in the concentrations of such gases prior to a major earthquake; this has been attributed to release due to pre-seismic stress or fracturing of the rock. One of these gases is radon, produced by radioactive decay of the trace amounts of uranium present in most rock.

  Radon is useful as a potential earthquake predictor because being radioactive it is easily detected, and its short half-life (3.8 days) makes it sensitive to short-term fluctuations. A 2009 review found 125 reports of changes in radon emissions prior to 86 earthquakes since 1966.

However, wikipedia then attempted to prove that most of the assretions related to radon are false statements,

but all this leaves me with one question: Why radon? (Why isn't another chemical species used for identifying the danger of feasible earthquake?) Maybe things I studied were too technical for me to understand, so please be as simple as you can be.

(Feel free to edit tags)

Uranium is found in low-levels in all rocks and soil.  Radon is a gaseous radioactive decay product of uranium.  As the uranium undergoes radioactive decay, radioactive radon is generated and trapped in the rocks that contain the uranium.

The earthquake theory involving radon suggests that prior to the actual quake, there is some subterranean movement where rocks are crushed, soil is uncompacted and the trapped radon is released producing a pre-quake spike in radon concentration.

So radon presents the following attributes:

it is present in all rocks and soil
if a rock is broken or if soil is disturbed, radon will be released
it is gaseous and air currents / thermal gradients will carry it up to the earth's surface producing a detectable plume
it is radioactive, this makes detecting small amounts or small changes in radon concentration relatively routine due to well-developed and very sensitive methods for detecting and accurately measuring radioactivity.
radon has a very short radioactive half-life, a bit under 4 days.  Being a gas and having a short half-life is very useful in terms of measuring radon emissions.  If a radon emission spike occurs, the gas will dissipate quickly and after about 10 half-lives (40 days) normal background levels of radioactivity will return.


Is it safe to mix potassium alum powder with magnesium oil (made from water and magnesium flakes)? Will mixing these two together cause a reaction of any sort? If so, is it safe and is there an order in which one should be added to the other?

If you are not sure and you have to do the reaction, try it with small sample of $\ce{Mg}$ such as one flake. $\ce{Mg}$ flakes should react in water and will react with potassium alum.


Intuition for why effective nuclear charge increases down a column My textbook (Chemistry the Central Science, edition 12, Brown et al.) says that effective nuclear charge increases down a column. Indeed, this is consistent with the values provided by Clementi in the 1960s.

Is there intuition or some qualitative explanation for why effective nuclear charge ($Z_\mathrm{eff}$) for the outermost electrons increases down a column (as opposed to being roughly equal)?

Initially, let's take the electrons as probability density fields instead of classical particles.

The increase on the number of protons in the nucleus increases the electric force on all electrons, but the core electrons suffer higher attraction due to higher proximity and lighter electronic blindage, so their probability density field is forced nearer the nucleus.

Reminding of Electric Force:
$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$

Consequently, the outer electrons will suffer lower blindage force from these core electrons, parallelly to the nucleus attraction, allowing them to get closer to the nucleus as well.

As effect, of course, the outer electrons also force the inner electrons towards the nucleus, but their probability densities reach a minimum repulsion energy in such a distance that electrons on the same level are forced closer to the nucleus than on the atomic element above, going down a column in the periodic table.


Does simmering sparkling water change/decrease its mineral composition significantly? I know that heating sparkling water will cause it to lose $\ce{CO2}$.

Does this have any other effects on the water that remains, such as reducing its mineral content?  I understand when non-carbonated mineral water evaporates, its minerals stay behind and create a greater mineral concentration in the water that's left. Does sparkling water behave in the same way, or do other reactions occur that cause the water to change in additional ways?

No, simmering sparkling water should have negligible effect on its mineral composition.

You're exactly right that simmering it will drive off the $\ce{CO2}$. This $\ce{CO2}$ leaves as only/exactly $\ce{CO2}$ molecules, and any minerals that might have been associated with the carbonate/bicarbonate $(\ce{CO3^{2-}}/\ce{HCO3^-})$ will remain behind in the water.  Most such minerals will be present in dilute enough of concentrations that they will remain dissolved in the no-longer-sparkling water, as long as only a relatively small amount of the water itself is boiled away.


Compounds with aluminum and nitrogen atoms directly bonded to each other as part of an aromatic ring? Just hours ago, I asked a question about whether aromatic organoaluminum compounds with aluminum and carbon bonded together as part of an aromatic ring would be stable in water. The answer seemed to be that "it wouldn't even work at all since aluminum and carbon have relatively poor pi-overlap, and even then the resulting compound would still end up complexing with nitrogen- or oxygen-based protic solvents due to the high affinity for nitrogen and oxygen of aluminum", implying that aluminum-nitrogen bonds are reasonably stable, at least when they're alone by themselves.
My questions are- do aluminum and nitrogen have sufficient p-orbital overlap to be directly bonded to each other as part of an aromatic ring, and, if they do, how stable are they to different kinds of solvents (protic, aprotic, polar, non-polar etc), and is there any specific compound reported in the literature that is water-stable and in which there is a direct aluminum-nitrogen bond as part of an aromatic ring? (Note that these need not be organoaluminum compounds- that is to say, the aluminum atom in question need not be bonded to any carbon atoms)

Again no.  The aluminum-nitrogen bond is much stronger as a sigma bond than as a pi bond.

Borazine, with boron rather than aluminum in the six-membered ring, is known as an aromatic ring and even called "inorganic benzene".  Like its benzene counterpart, the borazine ring can be fused to give polymeric structures ultimately leading to hexagonal boron nitride.  But with aluminum nitride, where pi overlap is weaker and the bonding more polar, such a layered phase is not known, this nitride most commonly having a wurtzite structure.  In effect, the Al-N pi bonds are broken up by cross-linking with other nitrogen atoms even where the pi bonded structure is ostensibly aromatic.

Take a cue from this answer.  You are on firmest ground with aromaticity when the elements involved are a subset of boron, carbon, nitrogen, oxygen, sulfur.


Specific gravity of H2O2 bleached viscose What is the specific gravity of $\ce{H2O2}$ bleached viscose? In literature, I can find that this is $1.52~\mathrm{g~cm^{-3}}$, but for bleached viscose is it not findable. Can I presume $\ce{H2O2}$ treated viscose has the same density?

You can assume that the density remains unchanged.

Hydrogen peroxide simply oxidizes dyes, thereby destroying chromophores (groups that move the absorption of a molecule into the visible spectrum). The viscose structure should remain largely unscathed, and the change in density is so minuscule I would argue it is within the experimental error.


Relation between oxoanion acidity and number of oxygen double bonds https://en.m.wikipedia.org/wiki/Oxyanion

Was reading the above Wiki page when I came across the 'heuristic for acidity' section which mentions that the oxoanion's acidity can be predicted by the number of oxygen double bonds, with more double bonds making the oxoanion more acidic. My question is, why is this so? Can someone explain it to me by the definition of an acid, and is this prediction basis only limited to oxoanions or can it be extended to other molecules?

The strength of an acid is determined by how far the reaction

$$
\ce{HA} \rightarrow \ce{H^+} + \ce{A^-}
$$

proceeds. One of the products will always be the same for all acids ($\ce{H+}$), so no influence there. The other product is the anion ($\ce{A-}$) and the basic question is: "How stable is that anion?", or equivalently, "how well can that anion stabilize its negative charge?"

Many factors go into answering that question, the one under consideration in the rule alluded to in the question is: Are there atoms surrounding the central atom that can stabilize the charge effectively? Oxygen atoms bound only to the central atom (in the protonated form) are a prime example, because oxygen is a very electronegative element. (Another example not involving oxyanions/oxyacids is trifluoroacetic acid, which is more acidic than acetic acid.) The more such oxygens are in your acid, the stronger it will be - $\ce{H2SO3}$ is weaker than $\ce{H2SO4}$.

Jan commented on the question that double bonds between the central atom and oxygens only exist in the first row of main group elements (i.e. $\ce{C},\ce{N}$). This is true, however, the rather basic rule cited in the question does not go into this level of detail and simply assumes that the outdated picture of double bonds in say, $\ce{ClO4-}$, is correct.


cooling/refrigeration essentiality for carbonated softdrinks How does cooling/refrigeration help in maintaining the taste and shelf life of food products, especially CSD (Carbonated Soft Drinks)? For example, chilled drinks have a nicer taste than a normal drink and may have longer shelf life. Why is that?

It has a longer shelf life because the bacteria, (which is the reason it will go bad), cannot reproduce as fast in cold temperatures.

As far as taste goes with carbonated soft-drinks, I think it is mostly just about preference.


Ions and Ionic Bonds I've recently just started learning about Chemistry, and more specifically, Ions.

In class, we learned about how Valence Electrons are electrons located in the outer-most orbitals and how every atom (with the exception of Hydrogen and Helium), want $8$ valence electrons.

When possible, an atom will attempt to get rid, or add, an electron to get $8$ on its outermost shells. For example, Sodium and Chlorine.

We learned that it's easier for Sodium ($\ce{Na}$) to lose an electron to get the $8$ outer electrons it always wanted, and for Chlorine to receive that $1$ refuted electron. Then, they form an Ionic bond and you get Table salt!

But now, I'm wondering: what if that's not possible? What if you have an element such as Beryllium and Lithium? Would they simply not combine? Or would something else happen? I am confused.

The variation from ionic to covalent to metallic bonding is not clear-cut, but a matter of degree. For example, the bond between hydrogen and oxygen in water is mostly covalent (sharing electrons), but at room temperature, about one in 10,000,000 bonds break, with the hydrogen losing an electron to oxygen -- the ionic bond you discussed.

However, in the case lithium and beryllium, there is not enough difference between them to favor one getting an electron instead of the other. However, something more interesting happens: both are willing to let electrons go, so there is a sea of electrons, unattached to a particular atom! This is called a metallic bond, and it enables many metals to mingle to form alloys, such as bronze, amalgams and stainless steel.


How to calculate m/e value for the base peak in mass spectroscopy? My working out:

Propan-1-ol : CH3CH2CH2OH

This can be broken up into:

CH3 + CH2CH2OH
CH3CH2 + CH2OH
CH3CH2CH2 + OH

Base peak = most abundant fragment formed (highest peak)

How do i know which fragment is the "most abundant?"

After I know the fragment structure, i can work out the molar mass, and then deduce the m/e value on the assumption of a (+1 charge).

Here is the exam paper solution:

I would say that the question is not quite well formulated and that the answer that you quote is even further misleading in giving the proper answer.

Finding the answer

To start, let's consider fragmentation of primary alcohols, which is the question at hands here. First of all, can we rule out any of these answers?

  Propan-1-ol undergoes complete fragmentation in a mass spectrometer.

This sentence means that we can rule out the molecular ion $M^{\bullet +}$ as a base peak in the spectrum. Propan-1-ol has a monoisotopic peak at $m/z$ 60 so it rules out answer D.

Now we apply the standard fragmentation rules for alcohols: consider ionization on the oxygen lone pair, we have two possible $\alpha$-cleavage leading to $m/z$ 31 and $m/z$ 59. An additionnal side reaction could also be hydrogen transfer to the oxygen atom, followed by water loss. This would lead to an ion at $m/z$ 42.

In the options above, should we thus completely rule out $m/z$ 29? I think that in the spirit of those who asked the exam question, the answer is yes, and we should rule out answer A. But see below...

We are now left with two options which can be explained by the fragmentation chemistry. Which should we choose? Technically, the problem here is that now we rely on abundances and not on the simple occurence of these fragments anymore. This is much more tricky, as there are no absolute rules. If we have a look at good textbooks, we find that loss of the largest radical is favored. Therefore, we have here to compare loss of $H^{\bullet}$ vs loss of $CH_{3}CH_{2}^{\bullet}$. Thus the answer is B.

Is the above answer complete?

The problem in such a question (and in the answer that is given below) is that it assumes that in mass spectrometry one could correctly predict fragment ion abundances based on very simple rules. And, as I tell my students, currently, even if you were able to map out the complete potential energy surface of the fragmentation pathway and do some RRKM calculations on the transition states, you would probably be off by a factor of 2 from the experimental results in terms of abundances. So asking for predicting the base peak (i.e. the highest abundance peak in a spectrum) can be quite wrong when two fragmentation channels are quite close to each other.

Let's have a look at the experimental electron ionization spectrum. Well OK, the base peak is indeed $m/z$ 31, so we are not too bad.

But let's have a deeper look. Remember we crossed-out $m/z$ 29 as a possible base peak, and it happens to be the second most abundant peak in the spectrum, ahead of $m/z$ 42 and $m/z$ 59 which followed the basic rules!

$m/z$ 29 could be either $HCO^{+}$ or $C_{2}H_{5}^{+}$. The former could be a secondary fragmentation of $m/z$ 31 after $H_{2}$ loss, whereas the latter could be produced as a heterolytic bond cleavage of the C-C bond. (In fact it is the same bond elongation leading to $m/z$ 31, except that they electron remains on the opposite side.) Energetically, these two channels differ by the difference in ionization energy between the $C_{2}H_{5}^{\bullet}$ and $^{\bullet}CH_{2}OH$ radicals, i.e. 8.11 eV vs 7.56 eV. Although it amounts to almost 100 kJ/mol, with the fairly large energy distribution in electron ionization, this is still possible.

And for the second rule that we applied, is it always true that the largest radical if formed? This empirical rule works out quite well, even more so when the comparison is between a hydrogen radical (which is quite difficult to produce) and a carbon atom centered radical.

And how about the exam paper solution

Here I am very highly surprised by what they seem to affirm. OK as we just saw above, $m/z$ 31 is a quite characteristic peak of primary alcohols. And loss of water is also characteristic of alcohols having side chains longer than 3 carbons. But does it mean that these should be the base peaks?

Let's have a look at the series of small primary alcohols:

Alcohol              m/z 31        Water loss           Other base peak
Ethanol              100            10
Propan-1-ol          100            15
Butan-1-ol            95           100
2-methylpropan-1-ol   38            10                  m/z 43
Pentan-1-ol           38            50                  m/z 42

So I do not see the rational behind this answer. It starts failing as soon as the small alcohol sizes reaches four carbon atoms. Of course there is a rational behind each of these base peak (for instance, in the case of 2-methylpropan-1-ol, the ionization energy of the isopropyl radical is of 7.37 eV, i.e. lower than that of $^{\bullet}CH_{2}OH$ and thus it becomes the most abundant ion, with exactly the same bond cleaved), but you have to go through the mechanisms, the thermochemistry and sometimes even the transition states to get to the global picture.


Oily residue on old toy rubbery tyres I have some sets made by well-known Danish plastic toy manufacturer from my childhood which I've been reviewing recently. Most of the elements was in a condition I was expecting (some yellowing, a little bit damage), but not with the smaller tyres. Originally, I found them quite oily/greasy by touch with some even leaving a bit of, what seemed to look like, liquid residue when removed from plastic wheels. Oh, and they were also full of domestic dust.
Didn't pay too much attention as I was going to clean them with the rest of bricks in washing machine using mild detergent. Upon drying I've check affected elements again and found that plastic elements exposed to rubber tyres were clean. These small tyres however, while their conditions was much better, they still had a bit of oily feeling by touch.
Tyres having this issues are up to 30 years old and rather small. Bigger ones are from the same time period but seem to be made of different material and did not show any signs of being oily even prior cleaning.
Affected tyres look like that if it matters (pictures just show the form, not the condition):

So I'd like to know few this around this issue:

What is going on here?
Is it possible to clean rubber tyres of this residue and get them back to normal condition? How long would it last?
Is it safe for plastic elements made of ABS and poly-carbonate (two most common plastic types used by this manufacturer) to have contact with rubbery tyres exposing issue explained above? Any other possible hazards?
Is there anything I could do to prevent from such issue happening to similar tyres in the future?

Disclaimer. I have not included a brand name for couple of reasons. First, my interest is purely from science/practical point of view. Next, I am not looking to accuse any one of some issues in case there are any. And last reason - idea was to hint a company, as I thought some may have better knowledge of exact materials used in above elements.

The "rubber", or elastomer, likely has plasticizers to soften the material and make it more elastic and less brittle. These often migrate to the surface as parts age. Some of these plasticizers for elastomers include:

Sebacates, similar to animal fats
Adipates, based on adipic acid, which is found in beets
Phthalates, esters of phthalic anhydride
Some are innocuous, but others, such as bisphenol A (BPA) may have long-term toxic effects. Washing the parts, as you have done, should remove them from the surface. You might need to repeat washing, using different detergents.

Another possibility is depolymerization, actual breakdown of the plastic. At room temperature, this process is slow, though it could contribute to that sticky surface layer.
As for safety, you might check on rules at the time and place the toys were made. Modern toys are made with safety in mind, but plastics have had a bad rap in the past, such as causing liver damage to workers in vinyl production. Caveat emptor.


Is it correct to talk about an empty orbital? Professor A. J. Kirby mentions:

  The properties of an orbital are those of an electron contained in it. It is normal practice, illogical though it may sound, to talk of 'vacant orbitals'.The properties of vacant orbitals are those calculated for electrons occupying them.

Since an orbital isn't defined until it is occupied by an electron, would it still be correct to say that an empty orbital (such as a LUMO) can interact with other filled orbitals?

"The properties of an orbital are those of an electron contained in it. It is normal practice, illogical though it may sound, to talk of 'vacant orbitals'. The properties of vacant orbitals are those calculated for electrons occupying them."

I consider this a bit of a truism, especially the last sentence. IUPAC defines as an orbital:

  Wavefunction depending explicitly on the spatial coordinates of only one electron.

which as noted in this answer falls short of considering spin. So let's use that definition for spatial orbitals. (For a spin-orbital, the analogous definition would be: depending explicitly on the spatial coordinates and spin coordinate ...) At this point, let us point out that a spin-orbital is a one-electron wavefunction. It is not an observable. One can observe the density or even the spin-density by means of X-ray diffraction, indirectly by NMR, ESR etc. (Another truism: orbitals of a single-electron systems, such as the hydrogen atom, are an important, but also somewhat trivial exception.) The density can be calculated from a many-electron wavefunction (WF) and a popular way of obtaining such a WF is combining several orbitals. This involves a lot more theory and mathematics than introductory chemistry courses can show.

So one can decide that if no electron is present, there is no orbital and stand on sound mathematical ground. Then again, I can just put an electron there by means of mathematics. Depending on the "computation-chemistry-method", for instance when employing a basis set (which is by far the most common approach), one can calculate properties even for empty orbitals such as the orbital energy (not an observable), which can be used to (approximately) describe electronic excitations (an observable). Computational chemists have resolved to call the empty orbitals "virtual" to bridge the gap between the two opposing views and use them as mathematical tools to describe excited states or to improve the quality of the description of the ground state.

  Since an orbital isn't defined until it is occupied by an electron, would it still be correct to say that an empty orbital (such as a LUMO) can interact with other filled orbitals?

Let's consider the formation of a bond between two hydrogen atoms in a special way: reversing a heterolytic splitting, which we will briefly compare with one reversing a homolytic splitting.
In order to decide this question, one needs to split the non-existent hair dividing the following two views: a) Since there are no empty orbitals, in the heterolytic case, the proton will distort the orbitals on the anion until the bond is formed. Does the orbital on the former proton now magically appear? b) There is an empty orbital on the proton. However, at large distance, its effect on the anion can be simulated by an electric field, which should not carry orbitals (one can assume that the metal plates creating the field are far away and crank up the charge). If a situation involving an empty orbital cannot be distinguished from one where there isn't one, is it really there?

Of course, the end result of either way of looking at things is the same hydrogen molecule that we also get from combining two hydrogen atoms the standard way. I thus suggest to abandon the view of orbitals carried around by atoms (except in the computational chemist way, as will be outlined below). Rather, I suggest to think of the effective potential felt by a newly added electron - where would it go? Regardless of how the nuclei got to where they are now, where do the electrons go? "Unoccupied orbital" is then a useful shorthand for the relevant regions.

(At this point, I consider the original question answered. I will elaborate a bit on how I got here.)
The last question of the previous paragraph is one way of looking at the algorithm of the Hartree-Fock procedure (the first step of wavefunction-based quantum chemistry, which is comparable to density functional theory, DFT, in this regard). We have some "basis set" (and we do not care what is looks like right now) containing candidates for orbitals, nuclei (i.e. charges and positions) and a number of electrons. The first step is to evaluate the potential/the forces acting on the candidate orbitals, which is nuclei only at this point$^1$. One then linearly combines the candidate orbitals (that's LCAO right there) to form the best choice one can make. Obeying the aufbau principle, one fills the candidate orbitals until all electrons have got an orbital. One then updates the potential (which now also considers the electrons), combines again, fills the new candidates, updates the potential again and so forth until the changes between iterations are small. The result is then evaluated in terms of energy, and possibly electron density or other observables. By doing it this way, one arrives at the familiar MO picture without assuming an electron distribution or specific orbitals filled in a certain way on certain atoms.

Let's get back to the basis set. In the previous paragraph, no assumption was made on its nature. The initial candidate orbitals could be any shape (as long as they are reasonable from a mathematical point of view, for instance, they must be twice differentiable, square-integrable etc.) and many different choices exist. Experience shows that using functions resembling the exact orbitals of hydrogen are very useful for molecular chemistry$^2$. This is because they combine results of decent accuracy with computational efficiency and chemist-friendly starting points of interpretation - such as HOMOs and LUMOs that can be related to individual atoms, flawed though as that line of reasoning may be (as I have hoped to show with this wall of text).

Footnotes:
$^1$ A bit of a white lie: this is only the special situation of the so-called core guess. But done here for illustration.

$^2$ Solid state chemistry is a different matter, where standing waves are the norm.


How many stereoisomers are possible for 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane? How many stereoisomers (not counting different conformers) does 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane have?

I think the answer should be 3. There are two geometrical isomers (cis and trans) considering the two methyl groups on the ring, and the trans isomer will also show optical isomerism as it doesn't have a plane of symmetry. But in my book, the answer is given as 2. Is this a misprint?

In this structure the left hand methyl groups can be on the same side of the ring. This is one stereoisomer. This molecule is not chiral and there is only one isomer with the methyl groups on the same side of the ring.

Initially I assumed that there was only one other isomer: the two methyl groups on the opposite side of the ring. But building models sometimes wrecks intuition. The picture below shows the two possible version with trans (relative to the plane of the ring) methyl groups:

So my initial view was wrong: the molecule is chiral and that gives three total stereoisomers.


Which has maximum entropy Which state has a higher entropy — the
liquid–gas equilibrium or the gaseous state? I have learned that at equilibrium the entropy is maximised. But I also learned that the gaseous state has the highest disorder (entropy), if the latter is true then why does the equilibrium exist, why doesn’t the liquid transform into the gaseous state at equilibrium completely? What factor is responsible for this?

Whatever one observes (whether it is a liquid, a gas, or a mixture thereof) is the state of maximum entropy for the particular values of the internal energy and volume the system currently has. As OP noted, this is a consequence of the second law of thermodynamics. According to this law, a system with a fixed internal energy self-adjusts its free parameters in such a way that the entropy is maximized. The free parameter in OP's example is the fraction of molecules in the liquid (or gaseous) phase.

When we say that the entropy of the gaseous phase is larger than that of the liquid phase, we are referring to the specific entropies (i.e., entropy per mole) of these phases when they coexist. When the two phases coexist, they have a common temperature $T$ and a common pressure $P$. Furthermore, their specific Gibbs free energies are the same, that is,
\begin{equation}
0 = \Delta g (T,P) = g_{\mathrm{gas}} (T,P) - g_{\mathrm{liquid}} (T,P) = \Delta u(T,P) + P \Delta v(T,P) - T \Delta s(T,P).
\end{equation}
The gaseous phase has a higher specific internal energy ($\Delta u>0$) and a larger specific volume ($\Delta v>0$) as well as a higher specific entropy ($\Delta s >0$) in such a way that the effects of the three terms ($\Delta u$, $P\Delta v$, and $T\Delta s$) cancel out.

One may still wonder why liquid-gas coexistence is ever possible. As the gaseous phase has a higher specific entropy, why doesn't the system self-adjust to have all its molecules in the gaseous phase? The Answer: What we know for sure is that if the evaporation took place under a constant temperature and pressure, the entropy would increase (because $\Delta s >0$). But this process would also increase the internal energy and the volume ($\Delta u >0$ and $\Delta v >0$). Therefore, to have more molecules in the gaseous phase but stay at the same internal energy and volume, the system has to lower its temperature and raise its pressure. This results in a decrease in the entropy,1 which counteracts its increase that would have happened if the evaporation had taken place at constant temperature and pressure. Therefore, evaporation at fixed internal energy and volume doesn't necessarily lead to the increase of the total entropy. As stated in the first paragraph, if one observes liquid-gas coexistence, this should correspond to the state of maximum entropy for the given values of internal energy and volume.

1 For any (stable) thermodynamics system, the entropy decreases as the temperature is lowered. On the other hand, no such strict rule exists for the relation between the entropy and the pressure. Still, for most systems, entropy decreases as pressure is raised. [Notice the Maxwell relation $(\partial S/\partial P)_{T} = -(\partial V/\partial T)_{P}$, meaning that the entropy decreasing with increasing pressure is equivalent to the volume increasing with increasing temperature.]


Solubility of dimethyl ether in aliphatic hydrocarbons I'm having trouble finding information on the solubility of dimethyl ether in a hydrocarbon like butane, propane, or isobutane.

I've read that its soluble in methanol, ethanol, isoproanol, chlorinated hydrocarbons, and toluene but cant find much else.

Assuming 1 atmosphere of pressure and both chemicals are at a temperature of -11°F, would the then liquid isobutane be misicible with liquid Dimethyl ether?

Since dimethyl ether is a very uncommon solvent, it's not a big surprise that you have hard time finding its information just using Google. However, I found two relatively important parameters of dimethyl ether: Its Hildebrand Solubility Parameter is 7.3 and Kauri-Butanol index is 90 (INVENTEC). For comparision, Hildebrand Solubility Parameter of diethyl ether is 7.62 (Solubility Parameters: Theory and Application). According to this reference, if two solvents have about the same Hildebrand values, theoretically, such solvents should have solubility behaviors similar to each other.

Thus, dimethyl ether should behave similar to diethyl ether (in liquid phase, I believe).  Keep in mind that most of hydrocarbons dissolve in diethyl ether in all ratios. For example, we run column chromatography using such solvent systems (diethyl ether/hexanes) to isolate close nonpolar organic compounds from complex mixtures.


How derive g and u symmetry labels for orbitals? When asked whether a molecule has an inversion center, we "invert" the coordinates of all atoms; i.e. move each atom from its position through the center of symmetry and to a new position equidistant to the center of symmetry compared to its initial position, and see if the final configuration is indistinguishable from the initial configuration. Simple enough.

I am confused when trying to decide whether molecular orbitals have g or u symmetry. Should I move the atoms through the center of symmetry, should I move "each lobe" through the center of symmetry, or some other way? What is the most general way of describing the symmetry operation for orbitals?

In Figure 1, a bonding $\pi$ orbital is depicted. This is supposed to be antisymmetric with respect to the inversion center in the middle of the bond. It seems therefore I need to invert the lobes, somehow, or move in an "X" fashion across the bond. But this seems different than for the molecular case, where we simply invert atomic coordinates. Figure 2 should be symmetric under inversion, and inverting the lobe positions lead to this result.

And what when the two overlapping orbitals are not of the same type, e.g. a $\pi$ overlap between d and p orbitals? Will the MO ever have g symmetry?

I just fail to get the unifying link between the molecular case and the orbital case.

Edit: Something just hit me: Should I always think of it as "inverting all coordinates within the entity", regardless of whether the entity is an atom or an orbital?

Figure 1. Bonding $\pi$ orbital

Figure 2. Antibonding $\pi$ orbital

g and u labels are only used when the molecule itself possesses the inversion symmetry element. For a diatomic molecule, this means that both atoms have to be the same, i.e. it has to be a homonuclear diatomic molecule $(D_{\infty\mathrm{h}})$.

  How do I carry out the inversion?

$\hspace{25 mm}$

Let's consider the $\ce{O2}$ molecule.
Let's consider the $\pi$ bonding MO. (Or one of them, to be precise, since they occur in degenerate pairs.)
Find the point of inversion of the molecule. Clearly it has to be the midpoint of the O=O bond, i.e. exactly in the middle.
Draw a straight line from one lobe to the point of inversion.
Extend the straight line through the point of inversion. You will see that you have gone from a shaded lobe to an unshaded lobe, indicating that the phase has changed. Therefore, the $\pi$ bonding MO is ungerade, or u.

If you start from the bottom-left lobe, you would draw a line through the point of inversion and end up in the top-right lobe, and you would draw the same conclusion.

  And what when the two overlapping orbitals are not of the same type, e.g. a $\pi$ overlap between d and p orbitals? Will the MO ever have g symmetry?

The parity of such MOs cannot be classified as g or u. However, if we are talking about a homonuclear diatomic, then this wavefunction is not acceptable. You cannot have a wavefunction that has only d orbital character from one atom, but only p orbital character from the other. By symmetry, the contribution from both atom's d orbitals have to be equal.

This arises because the Hamiltonian is invariant to the permutation of identical particles. Essentially, the idea is that both atoms are exactly identical, so there cannot be any way for them to be distinguished from each other. If you had such a wavefunction, and if it was populated, then there would be different amounts of electron density on the two atoms, which would mean that the atoms could then be distinguished. This is quantum-mechanically forbidden.

Such MOs may arise in heteronuclear diatomics. However, heteronuclear diatomics $(C_{\infty\mathrm{v}})$ do not possess the inversion symmetry element, and accordingly their MOs cannot be classified as g or u. Since inversion does not preserve the molecule, there is no requirement that it should preserve the wavefunction.

  What if one atom does not have any d orbitals?

There is no such thing. Every atom has d orbitals.

Whether they have the same energy, or whether they are occupied, is another thing altogether. If the energies are different, or if one atom has occupied d orbitals and the other does not, then it is a heteronuclear diatomic and as I said earlier, the g and u labels are not appropriate in such a case.


How can aluminium oxide be called an acid? Aluminium oxide is amphoteric. It is easy to see that it is a Brønsted-Lowry base through the following reaction:

$$\ce{Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O}$$

The aluminum oxide splits and the oxygen accepts a proton, forming water.

But what about the reaction with a base? In my textbook, they say:

$$\ce{Al_2O_3 + NaOH \rightarrow 2NaAlO_2 + H_2O}$$

Now, the textbook claims that Aluminium oxide is an acid merely because it reacts with a base to form a salt and water, as is characteristic of a neutralization reaction.

But I'm not satisfied with this definition. I mean, acids aren't defined as 'things that neutralize bases', we have well-established definitions for them.

I tried to figure out for myself how this could be. Clearly, the Brønsted-Lowry theory cannot be applied here since the compound in question has no protons to donate. Therefore, the only alternative is the Lewis concept. I cannot see how that is applicable in this case.

The most basic definition of "acid" is that it is a proton donor (or one which accepts a lone pair)

All of this stuff is done in an aqueous medium, so we can assume that all aqueous ions and molecules are present. With this assumption (in this case, we are assuming that $\ce{OH-}$ is available to react), we get the following equation:

$$\ce{Al2O3 + OH- -> 2AlO2- + H+}$$

Similarly, we get:

$$\ce{Al2O3 + 6H+ -> 2Al^3+ + H2O}$$

where it is acting like a proton acceptor (base).


Why can't alcohols form hydrogen-bonded dimers like carboxylic acids? Carboxylic acids such as acetic acid are capable of forming dimers:

I'm wondering why alcohols like ethanol don't generally form dimers. In the diagram below, the oxygen atom on the left ethanol molecule should be capable of accepting a hydrogen bond from the partially positive $\ce{H}$ atom on the right ethanol molecule. Does it have something to do with the $\ce{-OH}$ bond angle?

The strength of a hydrogen bond somewhat depends on the $\ce{X-H\bond{...}X}$ angle that the hydrogen-bonding hydrogen forms with the two electronegative elements $\ce{X}$. In our case, carboxylic acids or alcohols, $\ce{X} = \ce{O}$ so the angle is $\ce{O-H\bond{...}O}$. The ideal angle for this fragment is $180^\circ$.

As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire $\ce{(R-COOH)2}$ feature as a benzene-like ring streched in a single direction. Most importantly, the acceptors and donors line up nicely, the carboxyl function features an angle of $120^\circ$ and the $\ce{C-O-H}$ angle (and the $\ce{C=O\bond{...}H}$ one!) is also close to $120^\circ$. These arae the theoretically predicted, unstrained angles.

For two alcohol (e.g. ethanol) molecules to attempt a similar arrangement, we would end up with only four atoms that have to form a rectangle with $\ce{O, H, O, H}$ at the four corners. Thus, the $\ce{O-H\bond{...}O}$ hydrogen bond angle would be much closer to $90^\circ$ — a very bad arrangement. Furthermore, this would put the two electronegative oxygens (which thus feature a negative partial charge) closer together with nothing in-between which would cause destabilisation due to like charges approaching. The hydrogen-bonding hydrogens cannot alleviate this unfavourable interaction since they are at the corners of the square.

The situation for alcohols is much better if they create a network of a number of molecules that allow for much more favourable angles. With four molecules, you could already create a cyclic structure of approximately $100^\circ$ $\ce{H\bond{...}O-H}$ angles and $170^\circ$ $\ce{O-H\bond{...}O}$ angles — much more favourable. Of course, in the actual solution this number will fluctuate strongly as hydrogen bonds are broken and reformed constantly and different ring sizes happen all the time.


What's the pKa (range) of a carbonyl functional group? Why is a $\mathrm pK_\mathrm a$ value for carbonyls never mentioned anywhere in textbooks I've use while other functional groups have these values?

As per comments you are asking for the $\mathrm{p}K_\mathrm{a}$ value of a hydrogen directly attached to a carbonyl carbon. This means, you are asking only for non-enolisable aldehydes as ketones have no hydrogens on the carbonyl carbon. Aldehydes that have a hydrogen in α-position will be deprotonated there generating an enolate ($\mathrm{p}K_\mathrm{a} \approx 20$) — more acidic than a carbonyl hydrogen. So all we have left is 2,2-dimethylpropanal (or pivaloylaldehyde). Its $\mathrm{p}K_\mathrm{a}$ is not given on SciFinder, but should be somewhere between $45$ (an alkane) and $25$ (an alkyne), probably just slightly higher than an alkene ($40$).


Possible Water peak in PMMA nmr spectrum solvent: $\ce{CDCl3}$

I'm unsure as to whether the large sharp peak at ~1.5 on my NMR spectra is water, I was under the impression water gave a broad peak in proton NMR, but may be confusing it with something else.

Can anyone confirm is deny?

Your peak at δ1.6 in $\ce{CDCl3}$ is indeed water. This is typical of the appearance of water in $\ce{CDCl3}$ at the sample concentration you present. Water can sometimes be broad; this depends on the overall sample concentration, acidity of the solution and relative concentration of other labile protons. This looks like a fairly new bottle of $\ce{CDCl3}$.

You will notice that the residual chloroform peak at δ7.26 has $\ce{^{13}C}$ satellites; intensity approximately 0.55% of the parent peak, with a splitting of ~210Hz (so about 105Hz either side of the parent $\ce{^{12}C}$ peak). You will then also see that the water peak, although of similar intensity, does not have any carbon satellites. You could also confirm this by adding a small amount of $\ce{D2O}$, and observing a change in chemical shift for this peak.


What is the concept behind the calculations for the dilution of solutions? I'm interested in the concept behind this computation and was wondering if anyone could give me some pointers or hints.
I've prepared $0.28~\mathrm{L}$ solution of $0.8500~\mathrm{M}$ of $\ce{Al(NO3)3}$.
I need a $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$ of $\ce{Al(NO3)3}$ made from my original solution.

For my dimensional analysis what I need should be up front right?
$50~\mathrm{mL}$ of the solution first converted to liters for the bridge conversion to 0.450 moles.

Once I have the molarity of my $50~\mathrm{mL}$ set to $0.450~\mathrm{mol/L}$ I divide my molarity by the molarity of the original solution I am taking this out of in the first place which gives me 0.026 liters of solution

$0.026~\mathrm{L}$ of solution is then what I need out of my $0.28~\mathrm{L}$ $0.8500~\mathrm{M}$ solution in order to come up with my 50 milliliters of $0.450~\mathrm{M}$ of solution? (I feel that something is missing)

My computations (conceptually explained above in my best effort)

$50~\mathrm{mL} \cdot 1~\mathrm{L}/1000~\mathrm{mL} \cdot 0.450~\mathrm{M}/1~\mathrm{L} \cdot 1~\mathrm{L}/0.8500~\mathrm{M} = 0.026~\mathrm{L}$ of $0.8500~\mathrm{M}$ solution needed to prepare $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$.

I'm not sure the exact question but I suppose you want to find the volume of 1st solution needed to come up with the 2nd solution. I always feel more comfortable with moles for this types of problem. Because mole is conserved (in a way).
So, to make the 2nd solution you need

$$50~\mathrm{ml} \times \frac{1~\mathrm{l}}{1000~\mathrm{ml}} \times 0.45~\mathrm{\frac {mol}{l}}=0.0225~\mathrm{mol}$$

That means you need $0.0225~\mathrm{mol}$ of $\ce{Al(NO3)3}$ from the first solution. You know the molarity and mole number. So the volume is
$$\frac {0.0225~\mathrm{mol}}{0.85~\mathrm{\frac{mol}{l}}}=0.0264~\mathrm{l}=26.47~\mathrm{ml}.$$
But you need $50~\mathrm{ml}$ of the 2nd solution, so you need to add solvent (Water or whatever you used). The volume of extra solvent required is $50-26.47=23.53~\mathrm{ml}$. Now you got the 2nd solution and to check that you have got everything right, you can find the molarity of your new solution.


How to determine the chemical state of a reactant or product in a chemical equation? I am asked this question:

Write a balanced equation, including state symbols, for this reaction:
$\ce{calcium + hydrochloric acid -> calcium chloride + hydrogen}$

I answered this:

$\ce{Ca(s) + 2HCl(aq) -> CaCl2(s) + H2}$

This was the correct answer:

$\ce{Ca(s) + 2HCl(aq) -> CaCl2(aq) + H2}$

I thought that the $\ce{CaCl2}$ was a solid precipitate (ppt.) instead of aqueous(aq).
Here are the known chemical states:

solid(s)
liquid(l)
gas(g)
aqueous solution(aq)

Is there any method to work out the chemical state of a reactant or product in a chemical equation, if I know the chemical states of all of the other reactants?

According to these basic solubility rules:

Calcium chloride (CaCl2) is soluble. Hence, you write CaCl2(aq) in the equation.


Boyle's law applied on mixture of gases Q1. If $\pu{1.00L}$ of oxygen gas at $\pu{1.00 bar}$ pressure is combined with $\pu{1.00L}$ of nitrogen gas at $\pu{1.00 bar}$ pressure in a $\pu{2.00L}$ vessel, what is the total pressure in the vessel?

a) $\pu{0.50 bar}$
b) $\pu{1.00 bar}$
c) $\pu{2.00 bar}$
d) $\pu{4.00 bar}$

Q1. If $\pu{1.00L}$ of oxygen gas at $\pu{1.00 bar}$ pressure is combined with $\pu{1.00L}$ of nitrogen gas at $\pu{1.00 bar}$ pressure in a $\pu{1.00L}$ vessel, what is the total pressure in the vessel?

a) $\pu{0.50 bar}$
b) $\pu{1.00 bar}$
c) $\pu{2.00 bar}$
d) $\pu{4.00 bar}$

My thought process:

I know that Boyle's law states that the volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure, so in these cases it is safe to assume that Boyle's law can be used. For Q1, would the answer be simply 2.00 bars (c)? Instinct tells me it is, but I can't seem to explain why. For Q2, I'd think the answer is 4.00 bar (d) because the volume is halved, so the pressure should be doubled? Again, I'm not too sure though. Any help would be much appreciated.

you have both answers wrong.

(1) b - The volume of oxygen and nitrogen are both 1 liter at 1 bar, but they are put into a 2 liter container. So the pressure stays the same.

(2) c - The volume of oxygen and nitrogen are both 1 liter at 1 bar, but they are put into a 1 liter container. Since the volume is halved the pressure must double.


Light scattering in sulfur colloids Recently I did a reaction in the lab (the reaction itself is not important) whose product was colloidal sulfur. As I watched the sulfur particles form I noticed something strange: At first the sulfur particles appeared white, but after a while the solution turned yellowish (probably when the size of the sulfur particles grew a little). I know that sulfur's colour is yellow by nature, because of its absortion soectrum, and I understand that the white colour is probably due to Mie scattering of light by the colloidal particles.

My question is: Why aren't the particles yellow? if the absortion spectrum has as a consequence a yellow colour I would expect that spectrum to remain valid at all dimensions of the particle, the particle can't just randomly decide to stop absorbing light of some wavelenght when it reaches a threshold size... can it?

I know that Mie scattering predicts that a particle will scatter all wavelenghts equally if its size is comparable or bigger than the wavelenght of light, but I would have thought that if the particle absorbed a specific wavelenght that wouldn't change with size.

There are a number of different effects:

Rayleigh scattering is the predominant effect for the tiniest particles, much smaller than the wavelength of light, i.e. before they've coalesced. It is dependent on the fourth (!) power of the frequency of the radiation, so short wavelength light is scattered toward the eye. Looking through the colloidal mixture, it should look redder.
Tyndall scattering applies to larger colloidal particles, and it too is dependent on the fourth power of the frequency of the radiation and preferentially scatter bluer light, so the yellow of the sulfur would be less apparent.
Finally, particles coalesce and scattering becomes less important than reflection from the larger particles.
Perhaps sulfur nano-particles behave a bit differently from bulk material due either to quantum effects or to surface effects. See Elemental Sulfur Nanoparticle Coarsening... If someone has experience with nanoscale sulfur, please feel free to comment or edit this.


How to numerically solve a reaction network with fast equilibrium steps? For example If I have to solve the following reaction numerically

\begin{align}
*\ce{ + A(g) &<=>[K1] [A*]} \tag{1}\\
\ce{[A{*}] &<=>[k1][k-1] [B{*}]} \tag{2}\\
\ce{[B*] &<=>[K2] [B](g) +}* \tag{3}
\end{align}

Where the first and third reactions are fast equilibrium steps.

[* ] represents vacnt site and [A*]  and [B*] represent adsorbed species.

The reaction is carried out in a closed container, therefore the concentrations of A(g) and B(g) are not constant.
Given initial A(g) and B(g) pressure and assuming zero coverage of reactants initially, how do I solve a system like this numerically.

For A adsorption:
$$r_1=0$$
because the reaction is in equilibrium.So,
$$K_1=\frac{[A]}{P_A \times [*]}$$
From this equation you can relate $[A]$ to $[*]$.Similarly, using the last reaction you can relate $[B]$ to $[*]$. Your [A] and [B] are connected via reaction 2.
$$\frac{d[A]}{dt}=-\frac{d[B]}{dt}=-k_1[A]+k_{-1}[B]$$
This is a first order differential equation, so you need only one boundary condition. You have three unknown and three boundary conditions $([A]_0=[B]_0=0$ and $[*]=1.00)$.


Why is carbon monoxide stable(compare with other flammable gas) in room temperature? I know carbon monoxide can burn with high temperature, but why doesn't it explode easily compared to other flammable or explosive gas?

What gives you the impression that carbon monoxide does not burn easily?

Below I will link to a youtube video of an accidental ignition of a large amount of carbon monoxide.  But be warned!  The video is a highly uncontrolled, dangerous experiment.  The person who performed it did not have appropriate safety equipment and acted very recklessly.  DO NOT TRY TO PERFORM A SIMILAR EXPERIMENT YOURSELF unless you know what you are doing and have

A CO detector / alarm
A buddy in the room with you
A fire extinguisher
Lots of experience synthesizing gases using laboratory apparatus

...

OK, all that said, here is the video.  As you can see, carbon monoxide is quite flammable.  And that guy is an idiot, at least in my book.

The autoignition temperature of $\ce{CO}$ is ~609 °C, only 100 °C or so higher than the $\ce{H2}$ autoignition temperature, and lower than the autoignition temperature for solid carbon. The LEL and UEL are 12.5% and 74%, respectively, which is a very wide range compared to most compounds.

Compared to other gases, $\ce{CO}$ explosions may be comparatively less violent because of the relatively smaller change in moles during the combustion reaction:

$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$

Thus, two moles of gas are formed from three moles of reactants, which means that the shock waves and pressure forces produced by a sudden $\ce{CO}$ combustion reaction / explosion are not as large as some gases such as say propane.


Assumptions made in calculating melting point from thermodynamic data I'm fine with doing these questions, such as:

For ammonia, enthalpy of fusion is 5.65 kJ/mol, and entropy of fusion is 28.9 J/K.mol

Hence find an approximate melting point:
Well $\Delta G = \Delta H - T\Delta S = 5.65*10^3 - T(28.9)=0$ Which yields $T=196K$. This is quite close to Google's value.

What I'm thinking, though, is that we shouldn't be able (in real life) to determine a melting point so easily, because $\Delta_r H$ and $\Delta_r S$ both vary with temperature, so in reality this isn't as simple as solving a linear equation for T?

I'd just like some confirmation here, as my textbook isn't giving much away on this point :)

In real life the melting point is determined by measuring the temperature. The entropy of fusion is usually determined by dividing the enthalpy of fusion by the melting temperature.

$$\Delta S_f = \Delta H_f / T_f$$

Your text book calculation just solves this equation for $T_f$ to recover the temperature that probably had been measured before.

Note that for constant pressure the temperature is constant $T_f$ as long as the phase transition takes place.


Solution to Raoult's Law for a volatile binary solution (no pun intended) Raoult's Law

In a closed container, $$P_A=P_A^\circ x_A$$ where $P_A$ is the pressure of compound in gaseous phase, and $x_A$ is the mole fraction of the liquid component A in the liquid mixture. If the liquid is alone in the container, $x_A=1$.

Problem I'm trying to solve

  $A$ moles of compound $A$ and $B$ moles of compound $B$ are stored in
  container closed with a piston. External pressure on the piston is
  maintained as $P_e$. Find moles of $A$ and $B$ that will be found in
  the gaseous phase at equilibrium. ($P_A^\circ$ and $P_B^\circ$ are
  known values)

Attempt at solving

Let $a$ moles of A and $b$ moles of B exist in gaseous phase. This means $A-a$ and $B-b$ moles of respective compounds are leftover in the liquid phase.

Now Raoult's Law gives us

$$a \frac{RT}V = P_A^\circ \left(\frac{A-a}{A+B-a-b}\right)$$
$$b \frac{RT}V = P_B^\circ \left(\frac{B-a}{A+B-a-b}\right)$$

(Ideal gas equation used implicitly)

Also total pressure of gases inside should equal $P_e$

$$(a+b) \frac{RT}V = P_e$$

These are not ordinary linear equations, so I'm not totally sure my calculations are correct, but I'm getting the solution as

$$P_A^\circ \left(\frac Aa -1\right) = P_B^\circ \left(\frac Bb -1\right) = \frac{AP_B^\circ (P_e-P_A^\circ)+BP_A^\circ(P_e-P_B^\circ)}{(A+B)(P_A^\circ+P_B^\circ-P_e)}$$

So far so good, but it turns out that on plugging in specific values for $A,B,P_A^\circ,P_B^\circ,P_e$, $a$ and $b$ attain negative values.

Mathematically it can be proved that this happens when the rightmost expression is negative. The rightmost expression can be rewritten as

$$P_e \left[\frac {\frac{\frac{A}{A+B}}{P_A^\circ}+\frac{\frac{B}{A+B}}{P_B^\circ}-\frac{1}{P_e}} {\frac{1}{P_A^\circ}+\frac{1}{P_B^\circ}-\frac{1}{P_e}} \right]$$

Let $N$ mean numerator and $D$ mean denominator of this expression. We can prove that $N<D$. This means that $\frac ND >0$ if and only if $N>0$ or $D<0$ holds.

Question 1

Assuming all the calculations are correct, this seems to be a rather arbitrary mathematical condition. Can someone explain the physical significance of this? Like a more intuitive way to understand why we must check such an arbitrary condition rather than something as simple as $P_e>P_A^\circ+P_B^\circ$?

Question 2

If $\frac ND <0$, is the solution always $a=A$, $b=B$ (which means no liquid evaporates)?

As you correctly pointed out, you can connect the moles of each compound in the gas phase, $a$ and $b$, to the composition of the liquid phase, $A-a$ and $B-b$, through their partial pressures, $p_A$ and $p_B$:

$p_A=a\cfrac{RT}{V}=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}$

$p_B=b\cfrac{RT}{V}=p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$

And, as you did, we can use the total number of moles in the gas phase to rewrite these equations in terms of $p_e$. Since

$p_e=p_A+p_B=\left(a+b\right)\cfrac{RT}{V}$

and therefore

$\cfrac{RT}{V}=\cfrac{p_e}{a+b}$

we can write

$p_e\cfrac{a}{a+b}=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}$

$p_e\cfrac{b}{a+b}=p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$

Note that this is a two-equation system with two unknowns, so, barring incompatible values of the constants $A$, $B$, $p_e$, $p_A^o$ and $p_B^o$, it's determined (as you essentially pointed out).

Let's find a relationship between $a$ and $b$. A straightforward way to do that is by dividing the equations:

$\cfrac{p_e\cfrac{a}{a+b}}{p_e\cfrac{b}{a+b}}=\cfrac{p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}}{p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}}$

$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}$

This expression is key; we will return to it later.

Clearing $a$:

$a=\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}\left(A-a\right)$

$a\left(1+\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}\right)=\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}A$

$a=\cfrac{\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}}{1+\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}}A=\cfrac{b}{\cfrac{p_B^o}{p_A^o}\left(B-b\right)+b}A$

We haven't made any additional assumptions in the derivation (beyond Raoult's law itself), so this is valid for all the situations where Raoult holds. Let's interpret the numbers to see what they mean.

With this expression you can check out various situations to get a feeling of how does Raoult behave. Some limit cases:

· If $p_A^o \approx p_B^o$, $\cfrac{a}{A} \approx \cfrac{b}{B}$ and both components are distributed to the same extent between the gas and liquid phase.

· If $p_A^o \gg p_B^o$, the vapour pressure quotient $p_B^o/p_A^o \ll 1$ and therefore $a \approx A$ (the more volatile component will be almost entirely in the gas phase).

· And, contrarily, if $p_A^o \ll p_B^o$, $a \approx 0$ (the less volatile component will be almost entirely in the liquid phase).

Let's return to the key expression above:

$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}$

If you ponder carefully what this means, it's telling you that for ideal mixtures (Raoult mixtures), the relationship between the components in the gas and liquid phases is always fixed as long as there is equilibrium, and the proportionality factor of that relationship is the vapour pressure ratio of the two components. So, if you have $p_A^o=2p_B^o$, the ratio of moles of A and B in the gas will always be twice the ratio of moles of A and B in the liquid. Or, equivalently, Raoult predicts that the gas phase is enriched in the most volatile component, precisely in proportion to the relative vapour pressures. The more volatile a component is, the more "additional fraction" of the gas phase it represents.

So, what values for $p_e$ are compatible with Raoult? Let's find out:

$p_e=p_A+p_B=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}+p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$

$p_e\left(\left(A-a\right)+\left(B-b\right)\right)=p_A^o\left(A-a\right)+p_B^o\left(B-b\right)$

$\left(A-a\right)\left(p_e-p_A^o\right)=\left(B-b\right)\left(p_B^o-p_e\right)$

$\cfrac{A-a}{B-b}=\cfrac{p_B^o-p_e}{p_e-p_A^o}$

Note that this is a particular case of the lever rule and, if all the quantities are positive, it demands that $p_A^o \leq p_e \leq p_B^o$ or $p_B^o \leq p_e \leq p_A^o$. A value of $p_e$ outside that range would correspond to a value of $a$ or $b$ outside of the range $0 \leq a \leq A$ or $0 \leq b \leq B$ - which doesn't make physical sense.

But is any value of $p_e$ between $p_A^o$ and $p_B^o$ valid? No, it depends on $A$ and $B$. Every value of $p_e$ corresponds to a certain composition - so you can't set both to arbitrary values.

Take for example $p_A^o=100$, $p_B^o=200$, $p_e=125$, $A=1$ and $B=4$. Applying the lever rule expression we just derived,

$\cfrac{A-a}{B-b}=\cfrac{p_B^o-p_e}{p_e-p_Ao}=\cfrac{200-125}{125-100}=3$

So, the ratio of moles of A to moles of B in the liquid in equilibrium is 3. In other words, 3/4 of the liquid molecules are of A.

Let's introduce this value in the key expression we derived earlier:

$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}=\cfrac{100}{200}3=1.5$

So, the ratio of moles of A to moles of B in the gas in equilibrium is 1.5. In other words, 3/5 of the gas molecules are of A.

All of this is compatible with Raoult's law... but it is incompatible with a total system composition of 1 mole of A and 4 moles of B!

Hope this helps.