Why does ice melt slower at higher altitudes? Why does ice melt slower at higher altitudes and how does Le Chatelier's principle play a role in this?

This question appeared right after this chapter in my text book and I've also read a very rough explanation from another source which cited this principle.

A simplified attempt.

For a simplified answer that attempts at a solution using Le Chatelier's principle, you'll need a constraint: Ice and water are at an equilibrium at $\pu{0^\circ C}$.

  $$\ce{Ice <=> Water}$$

Simple enough, now pull up the values of densities of ice and water, $\rho_{ice}= 0.9167~\pu{g/cm3}, \rho_{water}= 0.9998 ~\pu{g/cm3}$. Mass being the same, you can conclude that volume of ice is greater than that of water, so in accordance with Le Chatelier's principle, when you move at higher altitudes, the atmospheric pressure falls, thus the equilibrium is favored towards a side where the species occupy a higher volume, i.e. towards the formation of ice. This works antagonistically for $\ce{Ice->Water}$.

Alternatively, have you heard of phase diagrams? This might help you to understand things a bit more clearly after you go through its definition from other sources: https://chemistry.stackexchange.com/a/72732/43942

The phase diagrams for water are different though.

https://d2gne97vdumgn3.cloudfront.net/api/file/m1GFJVbQSBmwkDXRvglV

This, too indicates that for water, at lower pressures, the formation of solid (i.e. ice) is more favored.

A realistic insight.

To be more realistic, as MaxW pointed in the comments, the pressure difference is rather small. The summit pressure at Mt. Everest is $\pu{33.7 kPa}$, while at the sea level it is $101.3$, which gives us a difference of $\pu{67.6 kPa}$ which converts to a mere $\pu{0.676}$ Bar.

Although this is quite a significant pressure (see Jon's comment) for our perception, it doesn't affect freezing that much. For a comparison, this answer https://chemistry.stackexchange.com/a/8883/43942 states that the pressure required to solidify $\ce{CO2}$ to make it into dry-ice at room temperature, you'll need a pressure of about $\pu{56 bar}$.

Freezing at higher latitudes, requires us to also ponder on the ambient temperature. You probably wouldn't want to talk about latitudes as high as the Everest, because then the temperatures will be extremely low (averaging about $\pu{-19^\circ C}$ in summers!) and you'll have to apply an external pressure to adjust the melting point. Latitudes where the ambient temperature is (say) $\pu{3^\circ C}$ have a less temperature gradient when compared to a place (probably at the sea level) with an ambient temperature of (say) $\pu{25^\circ C}$. A higher magnitude of a gradient means a high rate of heat transfer, which hints at a slower melting of ice at higher latitudes.

Also, airhuff suggested that thermal conductivity of thin air at high latitudes may have a role to play in this. I searched this up, you might want to look at this paper: http://nist.gov/data/PDFfiles/jpcrd269.pdf


NH2- and Primary Alkyl Halides: Sn2 vs E2 When we mix 1-bromopropane and $\ce{NaNH2}$, what happens? Which reactions wins out to make the major product?

My textbook says that primary alkyl halides give mainly SN2 products unless the base is hindered ($\ce{t-BuO-}$) (where E2 would win out).

But it contradicts itself when you have for example 1,2-dichlorobutane and you add three molar equivalents of $\ce{NaNH2}$ and then $\ce{NH4Cl}$ to have an alkyne. Why is there no SN2 going on in that case?

IIRC, Sodamide is generally insoluble, except in liquid ammonia.  And why do people like to prepare amines with ammonia as a nucleophile?  The basicity complicates things, as does the fact that the primary amine initially formed will react further to give secondary and tertiary amines, even quaternary ammonium compounds.  It's not a reaction useful in preparative chemistry.

That said, the balance of nucleophilicity and basicity is also influenced by the solvent.  It's well known that KOt-Bu acts mostly as a base in THF and mostly as a nucleophile in DMSO, there's plenty of examples in the literature.  Hydrogen-bonding solvents tip the balance towards elimination.


Equivalent Line Angle Structures? Are the following equivalent structures for 2,2-dimethylpropane?

I think so, because my rendering simply rotates the cross 45 degrees!

Also, regarding 3,3-diethylpentane, are these both equally valid structures? My rendering seems to look way different but filling in the carbons and hydrogens, both appear correct:

a) yes, those are equivalent structures of 2,2-dimethylpropane

b) yes, those are equivalent structures of 3,3-diethylpentane


A monocyclic 6 carbon ring with 6 double bonds I would like to know if this molecule has ever been synthesized, if not why not, and what would its IUPAC name be?

There are several monocyclic $\ce{C6}$ isomers.  Three of them are pictured below

A - the compound you've drawn above, cyclohexahexaene
B - cyclohexatriyne
C - cyclohexa-2,5-diyne-bis-1,4-ylidene

While all of the carbons in these molecules are (more or less) $\ce{sp}$ hybridized, the molecules are different from one another due to differences in bond lengths and bond angles.  For example, all of the bonds in A are exactly the same length due to symmetry, whereas the triple bonds in B will be shorter than the adjacent single bonds.  Therefore, A-C are isomers, not resonance structures.

Ideally the atoms attached to an $\ce{sp}$ hybridized carbon form a linear arrangement (180° bond angle) with the central $\ce{sp}$ carbon, acetylene and the central carbon in allene serve as examples.

(image source)

(image source)

However in the cyclic $\ce{C6}$ compounds A-C, the $\ce{sp}$ hybridized carbons are forced by the ring to deviate from the desired 180° arrangement to something down around 120°.  Therefore, quite a bit of angle strain is incorporated into A-C making them extremely strained and unavailable through laboratory preparation to date.

As mentioned above, A-C have different bond lengths and bond angles therefore they are isomers, not resonance structures.  A fascinating question is,

Do A-C each sit in a potential well, separated from one another by an energy barrier and if so, which one is the most stable (energy minimum)?

or

Is there another monocyclic species lower in energy than A-C? Perhaps a molecule that could be described by resonance structures A'-C', where the prime denotes that all of the bond lengths and angles have been adjusted and made equal in A, B and C.


Why is the specific heat of water high? How is the specific heat of water so high?
I was reading the hyperphysics article on it but could not understand it properly.

Water has hydrogen bonding in it. Hydrogen bonding is some kind of intermolecular force (a tutorial and the wikipedia page) that is usually seen in molecules that have $\ce{OH}$, $\ce{NH}$ or $\ce{FH}$ somewhere in their structure.
How does it happen?
Hydrogen atom is really small (atomic radius: About 37 pm) When it bonds with some very electronegative species, like O atom, (in the case of water), the electron is highly absorbed by the more electronegative. That results in a relatively high concentration of positive charge. The other O atom in water is attracted to H and this is the basis of hydrogen bonding.
So...?
Hydrogen bonding in water is far too weak to be compared with intramolecular bonds like covalent or ionic bonds, but it's strong enough that to be broken, it will require lots of energy. That's why water boils at $100 ° C$.
Many know the reason of water's high specific heat to be hydrogen bonding. Partially, it's related to that: In giving heat to water, some heat is "spent" on loosening the hydrogen bonds rather than increasing water's kinetic energy. The argument is valid, and I haven't seen anyone bringing evidence as to its being wrong, but another reason, which was the one hyperphysics gave, relies on the motion of water.
Expanding the topic isn't necessary, and it will add up to confusion. In short, the water molecule is nonlinear. That will result in three rotational degrees of freedom that allows it to have an awful lots of rotational motions! (More types of motions are a cause of an additional bonus to the specific heat)
Liquid ammonia has a higher specific heat, just for the same reason.
Even a very easier reason to understand is the water's relatively low molar mass: $\frac{18 g}{mol}$. What should that mean? More moles of water per one kilogram.
As a summary, the reasons are:

The heated water will contribute much of the heat to loosening, bending or breaking the hydrogen bonds.
Water had three rotational degrees of freedom. In addition to vibration, rotations happen a lot to water molecules. This will result in a higher heat capacity.
Specific heat capacity is defined as the amount of heat required per unit mass to increase the temperature by a degrees celsius. The relatively low molar mass of water allows more moles of it to be there in a mass unit (either kg or g)

As a side note, the exceptional high specific heat is not the only weird property of water. See here for example.


Charge of an amino acid sequence I have the following problem:

  A polypeptide has the following sequence (amino end on the left)
  $$\mathbf{lysine - aspartic~acid-histidine-arginine-serine}$$

  Using the $pK_a$ information that follows, what would be the approximate charge on this molecule at pH = 3.3?

  $pK_a$ values for the various groups in the molecule are as follows:

  \begin{array}{|cc|}\hline
\rm Alpha~carboxyl & 2.2 \\ \rm Alpha~amino & 9.5 \\ \rm R-group~amino & 11.7 \\ \rm R-group~carboxyl & 4.5 \\\hline\end{array}

After my calculations I got +2 and my answer was wrong. How to solve it correctly?

My effort:

$\ce{NH3}$ has $+1$ when on and $0$ when off. $\ce{NH3}$ are Lysine, Histadine, and arginine.
$\ce{COOH}$ has $0$ when on and $-1$ when off. $\ce{COOH}$ are Serine and aspartic acid.
Based on the pH and $pK_a$ values the answer should be 2.
Lysine=+1 Aps=0 His=+1 Arg=+1 Serine=-1

When the pH is lower than the pKa, the group will be protonated, and when the pH is higher than the pKa, the group will be deprotonated. Your explanation is correct that basic groups will be positively charged when protonated and neutral when deprotonated and that acidic groups will be neutral when protonated and negatively charged when deprotonated. Shown below is the compound in it's fully protonated state with the given pKa's.*

The given pH is 3.3 which means that:

N-terminus: pH < pKa, protonated basic group, +1
lys-sidechain: pH < pKa, protonated basic group, +1
asp-sidechain: pH < pKa, protonated acidic group, 0
his-sidechain: pH < pKa, protonated basic group, +1
arg-sidechain: pH < pKa, protonated basic group, +1
C-terminus: pH > pKa, deprotonated acidic group, -1

The overall charge is +3.

*An aside, the pKa's for the side chains are a gross approximation here. It turns out they are sufficient to answer the question, but given a higher pH, the answer wouldn't be correct.


Calculation of volume required for titration with dibasic acid Here is a problem:

  Titrate $1 M$ sulfuric acid with $\pu{50 ml}$ of $1 M$ sodium hydroxide solution. What volume of sulfuric acid will be required for neutralization.

This is a simple problem. My reasoning is since both reactants are strong we will have $\pu{0.05 L}$ $\ce{OH-}$ ions and for neutralization we need an equal number of $\ce{H+}$ ions. Since $\ce{H2SO4}$ is a strong acid and ignoring the second dissociation which is weak, we get an equal number of $\ce{H+}$ as $\ce{H2SO4}$ ions. Since concentrations are equal, the volume is simply $\pu{50 mL}$.

Apparently this is wrong. I think the source of confusion is what is meant by "required for neutralization". If you write out the molecular equation for this reaction and do stoichiometry you get a different volume.

We first observe the balanced equation. $$\ce {H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$ we see the ratio of $\ce {NaOH}$ and $\ce {H2SO4}$ is 2:1. Given $ 0.050~\mathrm{L}$ of $\ce {1M~NaOH}$ we find moles of $\ce {NaOH}$ and use stoichiometry to find how many moles of $\ce {H2SO4}$ is needed. We find the volume to be $25~\mathrm{ml}$.


On the addition of HCl to a conjugated diene Is this allylic chloride formed by an ionic or radical mechanism? Is it the more stable isomer?

Benzylic carbocation is formed as it is stable due to resonance.
So Cl attaches to benzylic position.
More substituted alkenes are more stable so the terminal double bond changes its position.
(Also this isn’t free radical reaction)


Reaction Kinetics and Stoichiometry -- Mass conservation This question has been bothering me for some time, and I can't seem to find a good answer online.

Say I have four chemical species $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and these four react in the following ways:

\begin{align}\ce{
A + A &-> B\\
A + B &-> C\\
A + C &-> D\\
A + D &-> B + C\\
}\end{align}

The kinetic reaction equations for these four species should be:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2  -k_2[\ce{A}][\ce{B}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}

where $k_i$ is the Arrhenius coefficient for that reaction.

In my mind, the stoichiometry does not seem to add up right. It seems that when the system undergoes $\ce{A + A -> B}$, the concentration of $\ce{A}$ should decrease by twice as much as a reaction like $\ce{A + B -> C}$. Similarly, the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.

Therefore, stoichiometrically, I want to write:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -2k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2  -k_2[\ce{A}][\ce{B}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}

Is this wrong-headed? What conceptual issue am I missing?

Reaction rates should be defined in terms of the Extent of reaction ($\xi$) that corresponds to the number of moles (or the molarity for reactions in solution) of specie $i$ divided by the the stoichiometric number, $\nu_i$:
$$\xi = \frac{[i]}{\nu_i}$$
Therefore, considering that the first reaction should be:
$$\ce{A + A -> B ~=~ 2A -> B}$$
The corresponding reaction rates are:
$$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$
In this way, the reaction rate is always the same (positive) number, independently on the specie we are referring to.
The global rate equations are:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= −2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]−k_3[\ce{A}][\ce{C}]−k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= 2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]+k_4[\ce{A}][D]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}

It is conceptually wrong to assume that

  the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.

The reaction scheme you described is the "parallel reactions" model. The amount of $\ce{A}$ consumed by each reaction depends on the specific rate. The stoichiometry of the first step says that every time 1 molecule of $\ce{B}$ is produced 2 molecules of $\ce{A}$ are consumed. But maybe $k_1$ is very small and so all the other process will consume $\ce{A}$ much more rapidly than this one.

You cannot predict what will happen in a reaction system just looking at the stoichiometry of each step. If you have a large amount of $\ce{C}$, it will probably consume a lot of $\ce{A}$ in the third step, producing a lot of $\ce{D}$ that, in turn, will produce a lot of $\ce{B}$, etc. Stoichiometry is just one piece of information. But kinetics is much more that this.


2 electrons with the same spin in the same orbital When two electrons are in the same orbital and they have the same spin do we say that the electronic configuration is impossible or that it refers to an excited state?

It is impossible. Having two electrons with the same spin in the same orbital is a violation of the Pauli exclusion principle.


How to measure absorbance at 882 nm? I am trying to measure the absorbance of a solution at 882 nm, but this seems to be beyond the range of many UV-Vis spectrometers. Is this a job for the IR spectroscopy, or something else? I do not have much experience with IR spectroscopy.

No this is not a job for an IR machine. It all depends on the machine you are using, but provided the monochromators will extend that far all that may be necessary is to change the detector for one that is more red sensitive. If your machine will not do the job, and if its a one-off type of measurement  try to get time on another machine somewhere else. If this is not possible or you need to measure many samples over a long period you may need to experiment by using a lamp , interference filters and a detector etc and make your own machine to measure transmittance.


To what degree an N99 active coal filter can protect against products of burning paraffin oil? I sometimes do firedance and a part of it is waving a stick with burning endings made of Kevlar soaked in paraffin oil. I started to be a bit concerned about how healthy is it to inhale lot of paraffin oil fumes.
I was wondering if I can protect somehow against it and my idea is to wear a mask with N99 active coal filter. The vendor claims that it blocks dust of size 0,1 mm and more. What noxious products of burning paraffin oil can it block? Which of them it can't?

Activated carbon's forte is vapor adsorption of hydrocarbons due to the activated carbon's high surface area, and this mask is probably very good at adsorbing fumes from hot parafin.

However, activated carbon does not tend to make a good material for filtering out the most harmful components of that kind of smoke, namely particulates smaller than 2.5 microns. In fact, the specification on the mask for particulate removal is that "that it blocks dust of size 0.1 mm and more" (meaning 100 microns and larger).  The World Health Organization has classified particulate matter of 10 micron (one tenth of the size claimed to be removed by the mask) to be a Group 1 carcinogen $\mathrm{^{(1)}}$.

Your filters are likely very good at filtering the particulate size most relevant to a particular application. However, for your application (production of very fine smoke particulates), I suggest a mask that is rated to remove 95% of $\pu PM_{2.5}$.

1) Outdoor Particulate Matter Exposure and Lung Cancer: A Systematic Review and Meta-Analysis, Environ Health Perspect; DOI:10.1289/ehp.1408092


Why does the excitation and emission spectrum of a fluorescent molecule have overlap? Question is rather self-explanatory. Came up during a lecture without a concrete answer.

I understand that the differences in emission wavelengths is due to relaxation to the lowest energy level of S1, but why do fluorescent molecules necessarily overlap in their excitation and emission spectra?

Energy-Level Diagram
There is overlap because between the excitation and the emission spectra because there exists a Range of orbitals at different energies. You'll always notice that the energy for excitation is always higher, or the lambda value is always lower (these say the same thing since E=planck's constant * frequency).

I've attached a HE-Ne laser schematic from Verdeyen's Laser Electronics book. For a better visual. What you need to know is let's say your gain medium, in this case He-Ne was excited at it's lowest wavelength (we'll call it Lambda1) then it's emission wavelength may be only 2nm higher (or red shifted), Lambda2.

However, as you'll notice in this schematic there are many wavelengths at which this gain medium (He-Ne) can emit light. These other possible excitation orbitals can have lower energy levels that correspond to higher wavelengths and thus overlap on the spectrum.


Methods for Determining Partial Charges I want to run classical molecular dynamics simulations of a periodically replicated surface (rutile $\ce{TiO2}$ with grooves). In order to do so, I first need to solve for the partial charges residing on each atom of the surface. What electronic structure methods are available for determining the partial charges on a surface?

I am vaguely familiar with Mulliken population analysis, but my understanding is that this method is not the most accurate. What other methods are commonly used, and how accurate are each of these methods?

Finally, what are some good (and free) packages for performing such calculations? I've used SIESTA in the past, is that well-suited for determining partial charges of a surface?

There are lots, and I mean lots of methods and software programs to produce partial charges. See Wikipedia for a small (incomplete list)

Let's start with the basics. The idea of a partial atomic charge, while useful for concept, can not be defined uniquely. Quantum chemical methods (whether wavefunction or DFT) produce some sort of electron density. Different schemes divide up that electron density in different ways.

I like Cramer and Truhlar's categories:

Class I: Some sort of intuitive or empirical approach (i.e., non-QM). These include methods based on experimental dipole moments and atomic electronegativities, including Gasteiger-Marsili partial charges and other "electronegativity equalization methods" (EEM).
Class II: Partitioning using wave functions / orbital schemes (including Mulliken charges while very efficient to compute, are basis set dependent.
Class III: Partitioning the electron density or electrostatic potential (e.g., Hirshfeld, density-fitting schemes, etc.
Class IV: Semiempirical mapping from Class II or Class III (preferred) to match experimental data like dipole moments.

Almost any modern quantum package will include several schemes to fit partial charges. My personal preference is for something like Hirshfeld charges (i.e., from the electron density, so not basis-set dependent), or electrostatic potential fitting schemes like Merz-Kollman or CHelpG.

I think the latter (electrostatic potential fitting) will be better generally for molecular dynamics, since you're attempting to produce point charges that will best represent the quantum mechanical electrostatic potential.


Does the Location of Line-Dash-Wedge Notations Matter? In Line-Dash-Wedge notation, when illustrating that an atom or bond is directed towards or away from the viewer (solid triangle for coming forward, dotted lines for going back), does the location of these lines matter when doing illustrations?

For example, a THC molecule.

The two hydrogen atoms there. Could they be placed inside the carbon rings instead of appearing outside, like below?

I understand that there's a whole host of issues with that otherwise, but just asking about the triangle and dashes in particular.

Yes, both inner and outer placements are possible with some nuances.

From Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) [1, p. 1926]:

  ST-1.3.2: Stereogenic centers at ring fusion atoms

  Stereogenic centers at ring fusion atoms should be drawn with hashed wedged or solid wedged bonds to the exocyclic substituent at the fusion atom whenever possible. If necessary, an implicit hydrogen should be made explicit in order to provide an exocyclic substituent to bear the hashed wedged or solid wedged bond. When one of the fusion bonds is oriented vertically and the exocyclic substituent is graphically small (such as a hydrogen or a methyl group, or even such physically larger substituents as a phenyl group represented with a graphically small “Ph” group label), the exocyclic substituent is also
  preferentially oriented vertically and opposite to the vertical fusion bond. This orientation is preferred even if it results in the substituent being placed within the ring system, and is particularly common in the depiction of steroids and other natural products.

From Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) [2, p. 354–357]:

  GR-4.2.2 Fusion atoms with one substituent drawn

  Ring fusion atoms already have three bonds within the ring system. For external ring fusion atoms, the substituent should preferentially be positioned outside the ring system and oriented so that it bisects the angle between the adjacent bonds.

  [...]

  When two substituted fusion atoms are adjacent, placing both substituents outside the ring system will typically cause them to overlap. In those cases, it is better to draw one of the substituents within the ring system. A substituent drawn in this way should preferentially be drawn exactly vertically or horizontally, whichever direction would minimize overlap with other atoms and bonds.

  If a substituted fusion atom is adjacent to an unsubstituted fusion atom, there typically is enough room to orient the substituent either inside or outside the ring system. That substituent should be oriented outside the ring system if all substituents can be oriented outwards, but should preferably be drawn exactly vertically or horizontally if any other substituent is placed within the ring system to avoid
  conflict with an adjacent substituent.

  Small substituents on the C-8 and C-9 atoms of steroids should be drawn within the ring system in this fashion even if there are no substituents on the adjacent fusion atoms C-10 and C-14. Certain other natural products [23] also have preferred orientations that place substituents within the ring system.

  A substituent should not be oriented within a ring system when attached to a fusion atom that lacks any vertical or horizontal fusion bond. The substituent should be oriented outside the ring system, minimizing overlap as best as is possible. Sometimes, it might be possible to reorient the ring system so that the fusion bonds are vertical; the substituents should certainly be placed vertically if so.

  It is not acceptable to orient two substituents within the same ring if the ring has fewer than eight atoms. One or both of the substituents should be oriented outside the ring system, minimizing overlap as best as is possible.

  Substituents on interior fusion atoms have no option other than being drawn within the ring system. As above, such substituents should preferentially be drawn exactly vertically or horizontally, according to the direction that would minimize overlap with other atoms and bonds.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897. (Free Access)
Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277. (Free Access)


How do I make a paper white again? Suppose I have a white paper originally. Then it is placed in a trunk, where soda spills occasionally and gasoline gas/gasoline also spills due to no gas tank cover. Over time, this paper becomes discoloured.

How do I make it white again? What equipment do I use?

One approach would be to use various solvents. You would require solvents in which your various stains were soluble in. Alcohols / water would be suitable to dissolve the soda. Whereas an organic solvent would be suitable for washing away the gasoline. Though gasoline is typically colourless.

Another approach would be to expose the paper to copious UV irradiation, materials are coloured due to absorptions in the visible spectrum which trigger electronic excitation. These absorptions bias the transmission and reflection of otherwise white light and give the materials their colour. UV radiation is of higher energy than visible light radiation, and therefore the chemical excitations involved can prompt reactions which result in the destruction of the materials themselves.

A similar approach to invoke this chemical degradation of the coloured materials could be achieved through using a strong acid or base. This would be akin to the effect of bleach on household surfaces. Again, exposure to acids and bases can prompt reactions which result in the degradation of coloured materials.

It is worth noting that all of these methods will have some affect on the paper. Continued washing / drying of the paper with solvents will likely cause the paper to become brittle and affect the absorption of the paper, which may affect your ability to write / print on it. A similar effect may be noticed under UV irradiation, though I expect it wont be to the same extent.

A sufficiently strong acid or base may be able to completely degrade your paper as well as your coloured contaminants.


Arterial blood CO2 content, division and partial pressure My lecturer cites Boron and Boulpaep's Medical Physiology (2nd edition) for the following claims:

the total $\ce{CO2}$ content of arterial blood is $26.4$ mmol/L or $48\%$ v/v (not sure if these numbers convert)
including $90\%$ as bicarbonate, $5\%$
as dissolved gas and 5% as carbamino compounds (specifically
carbaminohaemoglobin)
the arterial partial pressure of $\ce{CO2}$ is $40$mmHg

How does one get from the arterial partial pressure value to determining these proportions and figures?

What I've tried: Henry solubility constant for $\ce{CO2}$ in water, equilibrium constants for the hydration of $\ce{CO2}$, ionisation of carbonic acid, assumed carbaminohaemoglobin formation went to completion and only consumed $\ce{CO2(aq)}$. Still don't seem to completely match my lecturer's figures.

Even when Douglas equation has been criticized is the best approach I know to calculate CO2 content. You can calculate plasma CO2 content and total blood CO2 content

Douglas 1988 (DOI: 10.1152/jappl.1988.65.1.473)


Why is time-dependent density-functional theory (TD-DFT) used to describe excited states? It seems that TD-DFT is a common choice for a computationally efficient method to describe excited states. But if DFT is a single-configurational method, and hence a ground state method, how does the TD-DFT method treat the excited states?

TD-DFT is used and developed because there is an excited state theorem, the Runge-Gross theorem, that is analogous to the Hohenberg-Kohn theorem for ground states.

(...) the Runge-Gross theorem shows that for a many-body system evolving from a given initial wavefunction, there exists a one-to-one mapping between the potential (or potentials) in which the system evolves and the density (or densities) of the system.
The Runge–Gross theorem provides the formal foundation of time-dependent density functional theory. It shows that the density can be used as the fundamental variable in describing quantum many-body systems in place of the wavefunction, and that all properties of the system are functionals of the density.

This means that any state of the system can be described by an electron density, not just the ground state. If I understand this correctly, the R-G theorem is more general than the H-K theorem, which can be derived from the R-G theorem by removing the explicit time dependence.
Here is a paywalled link to their original paper.
An interesting comment from the Wikipedia proof is

The proof relies heavily on the assumption that the external potential can be expanded in a Taylor series about the initial time. The proof also assumes that the density vanishes at infinity, making it valid only for finite systems.

In the case of applied fields, such as a slowly varying or oscillating electric field, these may be treated with perturbation theory rather than self-consistently; this is the same thing as the Taylor expansion. A consequence is that TD-DFT may not be valid in the presence of strong applied fields or a strong external potential.


Br- and I- ions give layer test, but Cl- ion doesn't. Why? One of the questions I found in an NCERT book. Interestingly I could not find the answer anywhere on the internet. Hence, stack exchange.

Organic Layer test is a test involving redox reactions among halides. This test is used to detect in the presence of bromide ions and iodide ions in a solution (e.g., Lassaigne’s test solution).  Layer test is carried out by adding organic solvent such as carbon disulphide or chloroform in the presence of nitric acid and chlorine water to the given solution. For example, Lassaigne’s extract is acidified with dilute nitric acid and chlorine water is added to it. When organic solvent is added to this aqueous solution and shaken the mixture vigorously but carefully, the liberated bromine or iodine liberated by redox reaction of bromide or iodide ions with chlorine) would be traveled to organic layer to give distinguished color. In case bromine is present, it will give an orange layer while it will give a violet layer in case of presence of Iodine. Relevant redox reactions are given below:

$$\ce{Br- + Cl2 -> Br2 + Cl-}$$
$$\ce{I- + Cl2 -> I2 + Cl-}$$

The key is bromine and iodine are much soluble in organic solvents than in water. Since we are using chlorine as a oxidant, chloride ions are negated here.


Identifying functional groups I've spent a few days googling functional groups and reading my text assigned on organic functional groups. I'm understanding very little and I just need a little clarification. I understand that at the top right and bottom left, the hexagon structure signifies an aromatic, right?

Top Left: My first guess was carboxylic acid, but with further googling, I now think it's an amide, because it contains nitrogen and oxygen?

Top Right: Since the hexagon is aromatic, I know it can only be the last three choices, but I'm going with C because the $\ce{-NH2}$ is not connected to an oxygen.

Bottom Left: Because of the $\ce{Br}$, I know it's a bromide and, assuming that my thinking is correct with aromatic, the answer should be B.

But the bottom right, I am not sure what exactly to look at here to even guess, besides that it is not an aromatic ring.

To show that I've tried the last one, all I can find is that an alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$ and, the aromatic being the hexagon structure, I'm unable to grasp/understand which one the last group would fall into.

Some pointers:

  the hexagon structure signifies an aromatic

Not quite. The hexagon structure depicted with alternating carbon-carbon double and single bonds is aromatic. It is called benzene and is very commonly encountered in organic chemistry. It can be represented by two resonance structures as shown below and is sometimes also represented with a circle inside a ring of singly bonded carbons to show the delocalised $\pi$ system.

However, the are other molecules depicted as hexagons which are not aromatic such as cyclohexane.

Additionally, there are other molecules which are aromatic which are not benzene, and many which are not hexagonal but you probably won't come across these at an introductory level.

  My first guess was carboxylic acid, but with further googling, I'm now thinking it's an amide, because it contains nitrogen and oxygen

You are correct that it is an amide. The amide functional group contains a nitrogen adjacent to a carbon-oxygen double bond. The carboxylic acid group is similar but the nitrogen is replaced by an $\ce{-OH}$ group. Amides can be primary, secondary or tertiary depending on the number of carbons that are bonded to the nitrogen.

  alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$

These are the simplest example of alkanes, alkenes and alkynes. They are called methane, ethene and ethyne respectively. However, there are more complicated molecules involving these functional groups. In general, alkanes have only carbons-carbon single bonds and carbon-hydrogen single bonds. Alkenes are distinguished by having a carbon-carbon double bond denoted by two lines between the two carbons which look like an equals sign. Alkynes have a carbon-carbon triple bond which is denoted by three lines between the carbons. Therefore, your example is an alkene.


Limiting fraction of association Show that the limiting fraction of association of a base in water, as
  the concentration of base approaches 0, is $$\alpha=\frac{10^7 K_b}{(1+10^7\times K_\mathrm{b})}.$$

I have tried this problem for a long time and have made little progress. Here is what I have tried:

We know that:

\begin{align}
K_\mathrm{b} &= \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}\tag{definition}\\
\ce{[HB+]} + \ce{[H+]} &= \ce{[OH-]} \tag{charge balance}\\
\ce{[H+][OH-]} &= 10^{-14} \tag{$K_\mathrm{w}$ at $\pu{25^o C}$}\\
\text{initial concentration of base added} &= \ce{[B]} + \ce{[HB+]} \tag{mass balance}
\end{align}

I know that the limiting fraction of association is
$$\alpha = \frac{\ce{[HB+]}}{\text{initial concentration of base added}} = \frac{\ce{[OH-]}}{(\ce{[B]} + \ce{[HB+]})},$$
but I'm confused about how to proceed from here. I tried putting $\alpha$ in terms of $\ce{[OH-]}$ to get a cancellation, but that proved difficult. How do I proceed?

Start from

$$K_\mathrm{b} = \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}$$

Define

$$\alpha = \frac{\ce{[HB+]}}{\ce{[B]} + \ce{[HB+]}} $$

or

$$\frac{\ce{[HB+]}}{\ce{[B]}}=\frac{\alpha}{1-\alpha}$$

so that

$$K_\mathrm{b} = \frac{\alpha\ce{[OH-]}}{1-\alpha}$$

or

$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+\ce{[OH-]}}$$

But in the limit of infinite dilution the concentration of $\ce{OH^-}$ is solely determined by the autodissociation of water (if the solution is pure unbuffered water) and

$$\ce{[OH-]} = \sqrt{K_\mathrm{w}}$$

so that

$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+K_\mathrm{w}^{1/2}}$$

But at RTP $K_\mathrm{w}^{1/2}\approx 10^{-7}$ resulting in the final expression

$$\alpha = \frac{10^{7}K_\mathrm{b}}{10^{7}K_\mathrm{b}+1}$$


What is the reaction mechanism of a 5-endo-trig cyclization? I am trying to solve this mechanism for a disfavoured 5-endo-trig cyclisation to generate the disfavoured product (4) from starting material (1). The favoured 5-exo-trig is given by (2), which I have got a mechanism for. My efforts take me to intermediate (3) as shown below, but I'm not too sure on how to generate the desired product without reagent reduction.

Your mechanism to the 5-endo-trig product (4) is very close.  Notice however that when you do the initial cyclization, the intermediate formed has a positive charge on nitrogen along with the negative charge on oxygen (as you showed).  This cyclized intermediate then just needs to move a proton from the nitrogen to oxygen to form the neutral enol which then converts to the (ester) carbonyl - which is your desired product.  No other reagents are required.


Why did we introduce the concept of oxidation number in chemistry? Isn't oxidation number a weird concept? It assumes complete transfer of an electron from a more EN to a less EN atom per covalent bond formed. This assumption isn't even true. We know that partial $δ^+$ and $δ^-$ charges develop on the atom. Why did we introduce such a weird concept? And what is it good for?

Oxidation number is a deliberate approximation to make keeping track of electrons in molecules easy
You are quite right to say that oxidation number isn't "true". It is clearly a simplified idea and not a perfect representation of "reality" whatever that is.
But there is a point to it. That point is to enable chemists to keep track of the electron count in molecules. Electron distributions may be infinitely variable in bonds but electrons are discrete and keeping count of them is important. In transition metal complexes, for example, it is important to make sure all the bonding electrons are correctly accounted for. Many transition metals can have variable oxidation states with very different properties and the (crude) approximation of where those electrons are helps us classify those differences even when the metal-ligand bonds are covalent and the electron distributions are not clear-cut or ionic.
So there is no intent to correctly represent the distribution of charge in bonds. But pretending the oxidation number is an integer helps keep track of things and is often very useful at explaining properties.


Why would water, and not saltwater, undergo electrolysis? I am attending a cram school and there was a question that goes:

  Electrolysis decomposes chemicals by passing an electric current through a liquid or solution containing ions. Which of the following can be decomposed by means of electrolysis?
  A. Water
  B. Cream
  C. Saltwater
  D. Carbonated drink

My answer was C since the first sentence explicitly stated that electrolysis decomposes chemicals by passing an electric current through a liquid or solution containing ions, but they said that the correct answer is A. Also, I was thinking that water in itself is not conductive and therefore will not facilitate electrolysis by itself. Is the anwer really Water? And if so, why water and not saltwater?

What is the "correct" answer

You are asking why the answer key has water and not saltwater as an answer. Decomposition is defined as:

  A reaction where a single compound breaks down into simpler compounds.

For salt water, the salt reacts at one electrode, and the water at the other, and you are making NaOH. For pure water, you are splitting water into the elements. Neither process matches the definition of decomposition that I quote exactly, but pure water is the better fit.

In your question, you write:

  I was thinking that water in itself is not conductive and therefore will not facilitate electrolysis by itself.

The half reactions could be written as:

$$\ce{2H2O + 2e- -> H2 + 2OH-}$$ and

$$\ce{H2O -> 1/2 O2 + 2H+ + 2e-}$$

So there are protons produced at one electrode and hydroxides at the other (You could also have protons be reduced at one electrode, and hydroxide oxidized at the other - the net effect is the same). The protons are very mobile in water and will quickly diffuse and neutralize the hydroxide. This takes care of the charge transport through the solution.

Is this a good question?

No, but most multiple choice questions aren't.


Why transition metal chloride complex have lower reduction potential than metal aqua ions? Why transition metal chloride complex have lower reduction potential than metal aqua ions? For example:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &-> Au(s)} &\quad E^\circ &= \pu{+1.52 V}\tag{1}\\
\ce{[AuCl4]-(aq) + 3 e- &-> Au(s) + 4 Cl-} &\quad E^\circ &= \pu{+0.93 V}\tag{2}
\end{align}
$$
Is it because the chloride complex formation constant is very high (higher than aqua ions) so the reduction of $\ce{Au^3+}$ is less favored?

The  anion  $\ce{AuCl4−}$ is  the  most  known  complex  of  gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we are dealing with:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &<=> Au(s)} &\quad E^\circ_\mathrm{cathode} \tag1\\
\ce{[AuCl4]-(aq) + 3 e- &<=> Au(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} \tag2
\end{align}
$$
If we substract the equation $(2)$  from the equation $(1)$, we get:
$$\ce{Au^3+(aq) +  4 Cl- (aq) <=> [AuCl4]-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag3$$
The equation $(3)$ is the reaction for the formation of $\ce{AuCl4−}$ with formation constant $K_f$. If you applied the Nernst equation for the equation $(3)$, then at $\pu{25 ^\circ C}$:
$$E^\circ_\mathrm{cell} = \frac{RT}{nF} \ln Q = \frac{RT}{nF} \ln K_f = \frac{0.0592}{n} \log K_f \tag4$$
This last solution of the equation $(4)$ (where $n = 3$) is because, when at equilibrium $Q = K$. If we know the $K_f$, we could have find the $E^\circ_\mathrm{cell}$, and hence $E^\circ_\mathrm{anode}$ (which is for $\ce{AuCl4−}$). Unfortunately, I couldn't find $K_f$ for $\ce{AuCl4−}$ ion formation. Thus, I'd show how big $K_f$ for $\ce{AuCl4−}$ by applying the given values of $E^\circ_\mathrm{cathode} = \pu{+1.52 V}$ and $E^\circ_\mathrm{anode} = \pu{+0.93 V}$ on the equation $(4)$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f \ \Rightarrow \log K_f = \frac{n}{0.0592} E^\circ_\mathrm{cell} = \frac{3}{0.0592} (1.52 - 0.93) = 29.9$$
$$\therefore \ K_f = 10^{29.9} = 7.91 \times 10^{29}$$
It is also in similar way you can find the $E^\circ$ for redox half-reaction of $\ce{PtCl4^2−}$ since we know its formation constant ($K_f = 1.0 \times 10^{16}$):
$$
\begin{align}
\ce{Pt^2+(aq) + 2 e- &<=> Pt(s)} &\quad E^\circ_\mathrm{cathode} = \pu{1.18 V} \tag5\\
\ce{[PtCl4]^2-(aq) + 2 e- &<=> Pt(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} = ? \tag6
\end{align}
$$
If we substract the equation $(6)$  from the equation $(5)$, we get:
$$\ce{Pt^2+(aq) +  4 Cl- (aq) <=> [PtCl4]^2-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag7$$
From equation $(4)$ where $n = 2$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f = \frac{0.0592}{2} \log( 1.0 \times 10^{16})  = 0.474$$
Since, $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.474 V}$, and $E^\circ_\mathrm{cathode} = \pu{1.18 V}$,
$$E^\circ_\mathrm{anode} = 1\pu{1.18 V} - \pu{0.474 V} = \pu{0.706 V}$$
$$\therefore \ \ce{[PtCl4]^2-(aq) + 2 e- <=> Pt(s) + 4 Cl- (aq)} \quad E^\circ = \pu{0.706 V}\text{ (calculated)} $$
This value is in good agreement with the experimental value, which is given as $\pu{0.755 V}$ in electrpchemical series of CRC Handbook of Chemistry and Physics.
Source of formation constant of platinum complex ion


Why are alkali hydrides strong reducing agents? My book says that alkali hydrides are strong reducing agents. It also says that this is especially true at high temperatures? Is there any particular reason behind it?

Just like I said in the answer to you broader question, it's all about elements and their individual preferences. Hydrides (of any metals, not necessarily alkaline) contain hydride ion $\ce{H-}$, which is basically a hydrogen having one extra electron (or, in other words, a hydrogen in oxidation state -1). Now, hydrogen's electronegativity is relatively low for a non-metal, and so is its electron affinity. It can be said that hydrogen takes that extra electron almost reluctantly, and is ready to give it away to any oxidizer, even a weak one. This makes the hydride ion (and any compound which it is a part of) a powerful reducing agent.

In due time you will learn about the redox potentials which are a quantitative measure of the compound's oxidizing or reducing power. (Yes, they may depend on temperature and other things.) I'd rather not touch on that right now, for it's a long story.


In search of a safe and inexpensive solvent that can dissolve depolymerized polypropylene waste plastic I work at a place where we depolymerize post consumer waste polypropylene waste plastic through a thermal catalytic conversion process. During process shutdown we have to clear up the vessels which becomes a really difficult job as the depolyermized material which we call wax hardens without heat and have to hammered and scraped out which takes a lot of time. Therefore to make this cleaning process easy I am looking for a chemical clean up technique. Is there a reasonable and economical solvent that can dissolve/soften this wax?

Short answer: Not really, unless you can do the cleaning still at high temperature. Solid hydrocarbons tend to only become soluble rather close to their melting point.

The only reasonable solvent is petrol ether, i.e. a mixture of alkanes. Polar solvents obviously don't work, chlorinated solvents are ecologically problematic and only have the advantage of an increased boiling point. Toluene or xylene might be alternatives.

So basically what you should do is to switch your feedstock from polyolefines to liquid alkane (petrol, basically) before shutdown. I would however be amazed if your process engineers don't do exactly that anyway. Maybe they could do that for a little longer, or maybe your process should be run (at least from time to time) at a slightly higher temperature so solidified (i.e. crystallised) residues in the reactor can melt and be washed out.

And whatever remains after shutdown and cooling, you will have to scrape off by hand, sorry.

(A sloppily designed or maintained reactor with bad flow profile, insufficient/broken insulation/heating devices of course might or might not be fixable.)


Can I make a container of (kind of) perpetually boiling water? It's a common school experiment to make water boil at room temperature by putting it in a container and reducing pressure.

I am wondering if it is possible to make a sealed sphere containing water where the pressure is carefully set to room temperature so that it boils "forever".

Obviously it can't boil forever, because it would eventually all turn to vapor. What I am thinking though is that minor fluctuations in room temperature will cause it to cycle between boiling and condensation without the need for any power.

Is this line of thinking correct? Has anyone attempted to make something like this? I think it would be a really cool desk ornament, like a little box of cycling boil-condense water.

You can buy a hand-heat powered boiler. But something must provide energy to keep the bubbles forming, and remove it to condense the gas back to liquid, or it would all be in the vapor state. In this toy, the heat of your palm provides the input energy, and ambient air cools the vapor in a condenser.

BTW, this demo won't work if the room temperature is exceedingly high, because the liquid (probably an ethanol-water mix) won't condense and the vapor pressure becomes too high for bubbles to form in the remaining liquid.


Why are some weak acid equilibrium approximations invalid in very dilute solutions? Using two approximations for the dissociation of a weak acid, we can write
$$K_\mathrm{a}= \frac{[\ce{H+(aq)}]^2}{[\ce{HA(aq)}]}$$
One of these two approximations we use is that the concentration of the acid HA has not changed upon dissociation, because the dissociation of a weak acid is so small compared to the starting concentration of the acid that is negligible.
However, I was taught that this approximation is invalid when the acid is stronger, but also when the solution is very dilute.
I don't understand why this approximation wouldn't work if the solution is very dilute. I have some ideas though:

The acid concentration decreases significantly when dissolved in a lot of water, so maybe the H+ would become significant then?

Maybe the H+ from water's dissociation affect the equilibrium of the weak acid's dissociation so that the dissociation becomes significant?

Are any of these right? Could someone explain why this approximation is invalid for very dilute solutions?

The above simplified formula considers 2 simplifying assumptions:

Concentration of hydrogen ions from water autodissociation can be neglected, so $[\ce{H+}] \approx [\ce{A-}]$ .
Concentration of an acid $c_0$ , in relation to its $\mathrm{p}K_\mathrm{a}$, is high enough for the fraction of ionized acid to be negligible, so $c_0 \approx [\ce{HA}]$ .

Especially the second assumption is broken for too dilute weak acid solutions. For more diluted solutions, higher fraction of an acid is ionized to match the equilibrium equation.
if we consider $$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$
then if we dilute the acid 4 times, concentration of the conjugate base decreases just 2 times, so the fraction of the ionized acid form increases 2 times as well.
It also means the stronger the acid is, the higher concentration must be . As for the same concentration, stronger acids are more ionized and $c_0 \approx [\ce{HA}]$ causes a bigger error.
The rule of thumb is, for the formula to be acceptable, this inequality should be true: $$ - \log {c_0} < \mathrm{p}K_\mathrm{a} - 2$$
so maximally about 1% of an acid is ionized.
The first assumption is broken mainly for too weak acids. E.g. if $\mathrm{p}K_\mathrm{a} = 10$, practically all $\ce{H+}$ comes from water autodissociation and the equation fails.


Are atoms really round? I'm not sure if this is a silly question, but I was sitting here with a cup full of cheezey poof balls thinking, "My goodness, it's like an amazing cheesey delicious liquid - huge water molecules!"

Of course my next thought was, "Wait a minute - water has two hydrogen atoms bonded to an oxygen, so that's not quite right. They wouldn't be round like this."

Then I started thinking about the diagrams we see in chemistry textbooks, etc., and how the atoms are always pictured as round balls. How do we know this is accurate? Is it possible for the atoms to be configured in more complicated shapes (e.g. not solid, crystalline, some type of lattice)?

It depends how you define the surface of an atom. Atoms maintain no surface in the normal sense; only regions of space where you have a better chance of finding electrons. So in fact it is not correct to say they have a true shape at all.
Shapes of Atomic Orbitals
However if you plot the region of higher probability of finding electrons in an atom you can obtain something like this:

These are the shapes of the first five atomic orbitals: $1\mathrm{s}$, $2\mathrm{s}$, $2\mathrm{p}_x$, $2\mathrm{p}_y$, and $2\mathrm{p}_z$ from Wikipedia. $1\mathrm{s}$ orbital is sphere-shaped but other orbitals have more complex shapes so atoms with many electrons have orbitals very different from a sphere.
Shapes of Molecular Orbitals
Molecules have more electrons and so even more orbitals. They can have very strange "shapes". I've calculated for you with GAMESS and Avogadro the LUMO 2$b_2$ water's orbital that is one of the molecular orbitals of water. This is the result:

Shapes of Atomic Constituents
For answering your question in the comment: In fact even protons, neutrons and electrons don't have a real shape due to wave-particle duality. However we can assume in many cases that neutrons are particles (so we suppose a spherical symmetry) but the  de Broglie hypothesis states that they have also a wavelength!


Why it is only carbons that have hydrogens attached (and not just hydrogens directly attached) to the charged carbon which result in hyperconjugation? For the following compound:

Looking at figure 2, my understanding of hyperconjugation is that carbon 2 is more electronegative than the hydrogens attached to it, and so pulls charge towards carbon 2, giving it a partial negative charge. This negative charge then stabilises the positive charge on carbon 1.

Figure 2

My book says, for the compoound $\ce{C+H2CH2CH2CH3}$ there is only one hyperconjugation centre (i.e. the $\ce{CH2}$ attached to the C+).

Why can't the hydrogens attached to the C+ act as hyperconjugation centres? E.g. won't these hydrogens be less electronegative than the C+ and so the C+ will attract the electrons closer to it , somewhat stabilising the positive charge on the cabron?

Hyperconjugation happens because of an overlap of orbitals.

Consider this: in (uncharged) 2-methylpropane, the central carbon has sp3 hybridisation (and a hydrogen bound to it).

When the carbocation of the image is formed, the hydrogen leaves, the hybridization becomes sp2 (hence, planar), and now there is a free pi orbital available.

Here is what causes hyperconjugation: pi orbitals are quite larger than how they are drawn in books, and they can overlap partially with the orbitals of the adjacent hydrogens, and get part of their charge.

In the example of butane, the thing is the same: when the carbocation is formed, the carbon and the two hydrogens attached become planar (sp2), and a perpendicular pi orbital is formed. It cannot interact with the hydrogens attached to the charged carbon because they are on its nodal plane. Instead, it overlaps partially with the orbitals of the hydrogens on the adjacent carbon.


Is there a substance that is hard (ideally abrasive) in solid form but would dissolve in water? doing some research for a dishwasher / dish cleaning product and interested in any compounds that could be fashioned into a brush or other abrasive form and then dissolve and wash away with the water cycle?

A usual cleaning product that does not pose any health hazard and that is easily washed away is sodium hydrogenocarbonate (aka baking soda). I use it myself, and it works a charm in the kitchen!


Why are full and half filled orbitals the most stable? Why are degenerate orbitals (restricted to a single spin) less stable when neither fully filled nor completely empty? Why, in most molecules, are half-filled and fully-filled shells more stable than partially filled shells? I understand that transition metals often promote electrons to their d orbital to make it half filled because this makes them "more stable." Why would this be the case? Relatedly, why are orbitals filled with single electrons in one spin before doubling up to form pairs and complete orbitals? Why must degenerate orbitals follow hund's rule? It seems like it would be more stable for electrons to pair up as they are added to an orbital rather than to become additional radicals in the orbital. Is hund's rule a result of maximizing stability, or is there some other principle at work?

In short, the so-called Fermi correlation which is a consequence of the Pauli principle and applies only to electrons of the like spin keeps such electrons further apart comparing to the case of electrons of unlike spin. Naively one might think that a larger average distance between electrons of like spin reduces electron-electron repulsion energy which is the reason for the stability of high spin states. And while it is true that for each and every pair of electrons electron-electron repulsion energy is decreased as the average distance increases, it was shown that in the whole electron-electron repulsion energy for the molecular system in a high spin state is actually greater that in a lower spin spate.

So, the modern explanation of the lower energy of high spin states is that a larger average distance between electrons also effectively leads to less screening of the nucleus, i.e. each electron experiences a larger nuclear effective charge. As a result the unpaired electrons can approach the nucleus more closely and the electron-nuclear attraction is increased. And this increase in the electron-nuclear attraction energy (which is negative) overcomes the above mentioned increase in the electron-electron repulsion energy (which is positive) leading to a greater stability of high spin states.

For details, see Boyd, R. J. Nature 1984, 310, 480-481, or any modern textbook on quantum chemistry.


Maximum observed optical rotation through a polarimeter Is there a maximum optical rotation of plane polarized light that can be observed using a polarimeter?

I understand that in a polarimeter, light passes through a filter that converts it into plane polarized light. This type of light only oscillates in one plane. When it is passed through an optically active solution, the plane is rotated due to interactions between the molecules. To measure this rotation, we use another a filter and try to orient it so that light passing through it appears brightest. The difference in angular orientations of the two filters is read as the observed angular rotation.

Here is an image I've taken from Solomon and Fryhle's Organic Chemistry textbook:

So is there a maximum value for this rotation? Surely after a 90 degrees rotation, the angle would be the same as another angle is we'd rotated in the opposite direction. For example, I could say that a +100 degrees rotation is the same as a -10 degrees rotation.
Could the angle be greater than 90 degrees?

The question might seem simple, but it's difficult for me to take into consideration multiple rotations of the plane polarized light before it passes through the second filter. Are the multiple rotations even possible?

This is something I discovered after an attempt at a more mathematically rigorous investigation of how to interpret polarimetry results. It's not that hard, but I don't think you'll ever find a similar analysis in a book because in practice the details are pretty much unnecessary for experimental purposes (though fun mathematically!).

Before we begin, it's very important to distinguish two types of optical rotation. The first is the real optical rotation, $\alpha_{real}$. This is the twist which a substance actually causes on the plane of polarized light as it travels through the optically active solution, and therefore is the rotation which is part of the specific rotation equation, shown further below. The numerical value of this angle can be any finite real number, in degrees or radians. There is no constraint on how much an optically active solution is "allowed" to twist a plane of polarized light; the fact that you can't directly see how many complete twists the plane of polarized light made is entirely an experimental limitation.

This experimental limitation is what gives rise to the other type of rotation: the observed or apparent rotation, $\alpha_{obs}$. This is an angle which is also real number, but which is limited between $0$ and $180$ degrees or $0$ and $\pi$ radians (or you could say between $-90$ and $90$ degrees or $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ radians, it makes no conceptual difference). The observed rotation is what you measure in a polarimeter, and in principle it cannot be directly put into the specific rotation equation.

Why is $\alpha_{obs}$ allowed only in the range $[0, 180]$ degrees? Because you can't observe a difference between an angle $\alpha_{obs}$ and $\alpha_{obs} + 180\times n$ (where $n$ is a positive or a negative integer). Look at a figure such as this (source) and try to convince yourself of this. If it helps, remember that two identical polarizers on top of eachother with an angle of $0^\circ$ let light through completely, but rotating one will block more and more light until they are perpendicular ($90^\circ$) and no light at all comes through, at which point increasing the angle will brighten the light until it reaches a maximum again at $180^\circ$ before once again starting to dim. This means that, in polarimetry the observed brightening or dimming by rotations of plane-polarized light are cyclic with period $180^\circ$, not $90^\circ$ as you suggest.

Turning the previous sentence into a simple equation, we obtain:

$$\alpha_{real} = \alpha_{obs} + 180\times n \ \ \ , \ \ n \in \mathbb{Z} $$

(The angles in this answer will always be treated in degrees, though for simplicity, in the equations I shall forget about the $^\circ$ unit and treat angles as pure numbers*).

Now we use the equation for optical rotation to find the specific rotation $[\alpha]^T_\lambda$, which characterizes a substance when measured at a temperature $T$ with a monochromatic light source of wavelength $\lambda$:

$$\alpha_{real} = [\alpha]^T_\lambda \times l \times c$$

, where $l$ is the optical path and $c$ is the concentration of the (enantiomerically pure) optically active substance. Substituting that into the previous equation:

$$[\alpha]^T_\lambda = \frac{\alpha_{obs} + 180 n}{lc}$$

So for each individual measurement of $\alpha_{obs}$ made, there are infinite solutions for $[\alpha]^T_\lambda$. This comes as no surprise.

But if we take several measurements, then we can find a value of $[\alpha]^T_\lambda$ as a unique solution to the equation, right? Speaking in strictly mathematical terms, no. Any finite amount of measurements, even with perfect accuracy and precision, will still yield infinite possible values of $[\alpha]^T_\lambda$. In fact, there are infinite levorotatory and dextrorotatory solutions, so in principle you can't even determine the direction of rotation by this method! See the graph below:

EDIT: In my previous graph, I incorrectly equated the values of $\theta$ with $\alpha_{real}$. The actual identity comes from $\mathrm{tan}\ \theta = \frac{d\alpha_{obs}}{dc} = \frac{d}{dc}(\alpha_{real}+180n)=\frac{d\alpha_{real}}{dc}=\frac{d}{dc}(lc[\alpha]^T_\lambda)=l[\alpha]^T_\lambda$, from which $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$. This however does not change the theoretical development of the answer.

Note that since the measurements are cyclic with period $180^\circ$, straight lines wrap around. The two black points are two measurements of $\alpha_{obs}$ at two different concentrations. There is an "obvious" red line connecting them, but it need not be the correct one. There are infinite lines which cut both points; I have represented two more perfectly valid solutions with the green and blue lines. Each line which cuts all points represents a possible solution, with an inclination given by the angle $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$ (see derivation above). From this, $[\alpha]^T_\lambda = \frac{\mathrm{tan}\ \theta}{l}$. Interestingly, the red and blue lines are represent dextrorotatory solutions to the problem, while the green line indicates a levorotatory solution for the same compound. There are infinite more solutions, characterized by steeper lines which wrap around more times. See the comments for information about how they come into play.

However, in practice, materials only have a finite and usually relatively small ability to perform optical rotation. Most of the more commonly encountered substances have a value of $[\alpha]^T_\lambda$ which tops out at a few hundreds. Some extreme examples are helicenes, for which specific optical rotations can reach as high as $\approx 10000^\circ$ in [13]helicene. After performing enough measurements (especially if you do them intelligently, by not measuring at different concentrations which are related to eachother by simple fractions), then you basically reach a situation where you can find an infinite family of solutions such as $[\alpha]^T_\lambda = -534 + 462785\times n$, in which case the true specific rotation is probably levorotatory with a value of just $-534^\circ$.

As a small additional comment, usually the filters in an optical polarimeter are rotated until the least amount of light comes through as the human eye is better at detecting small differences in light levels in darker conditions, which will make for a more accurate measurement of the observed rotation.

Originally I had answered this question by bringing up modular arithmetic, but while it's interesting it can possibly make the answer harder to understand at first. I have relocated my previous argument here.

The first equation shown is equivalent to:

$$\alpha_{obs} = \alpha_{real}\ \ (\mathrm{mod\  180})$$

The $\mathrm{mod\  180}$ part comes from modular arithmetic, which may sound scary and new, but it's just normal mathematics, with the small quirk that every time a number appears out of the range $[0, 180]$, you just add or subtract $180$ as many times as necessary for the answer to appear between $0$ and $180$.

* In reality some of the $\alpha_{real}$ and $\alpha_{obs}$ variables in certain equations above would have to be divided by $^\circ$, so that all equalities are dimensionally valid. Equivalently, the modulus could be expressed with the unit as $\mathrm{mod\  180^\circ}$, but I'm not sure if the modulus is allowed to have physical dimensions, even if I don't see any issue with it.


Where to find pre-optimized files for Quantum Chemistry Softwares? Many times Geometry optimization gets very time consuming ( even with access to super-computers ). I have seen some pre-optimized libraries for biomolecules like proteins that you can easily download machine optimized files from.

I use Avogadro,  GAMESS and McMolPlt for my quantum calculations.
I am thinking such libraries should exist for some complex chemical structures ( usable on softwares mentioned above ) but couldn't find any.

At this point I am specifically interested in " Hydroxyapatite Crystals ".
Does anyone know such resources ?

Short answer: we're building one for molecules

In your case, you don't actually want molecules. You're mentioning crystals. In that case, I suggest the Crystallography Open Database or if you have access, try the commercial Cambridge Crystal Structure Database. The COD is planning a theoretical crystal database, but I don't think it's online yet. These are experimental x-ray structures, but should be good starting structures for solid-state simulations.

For molecules, my group is building one. It's the Pitt Quantum Repository There are a few related efforts that I'm aware of.

With the PQR, you can either search through the website (even on mobile). The database is keyed by InChI keys, so if you have an existing structure you can generate the InChIKey and then get the structure, e.g.:

D-(-)-Morphine -> .mol2 file at http://pqr.pitt.edu/api/mol2/BQJCRHHNABKAKU-KBQPJGBKSA-N

The long answer is that some commercial software vendors consider such libraries as violations of the license. No joke, I've planned this for years to augment Avogadro, but the company behind a leading program told me that they would void my license to distribute output files in this fashion.

With the rise of community and open source quantum chemistry software, this is moot. Use programs that let you distribute output files. What remained was:

Gathering the 3D coordinates (i.e., a low-energy conformer) of X thousand/million molecules
Obtaining the computer time to optimize all of them
Devoting a server and disk space to serving them
Having a great website to help view, browse, and search them


X-ray fluorescence: where do expelled electrons go? I'm studying the XRF method cause I'm about to use a WDXRF spectrometer at college so I'm getting all the information about the physics and chemistry behind it.

I'm just curious about one thing, as you know when a beam of x-ray hits the electron of an inner shell if the energy is just about lower than the electron binding energy, that electron is expelled and the hole is filled with an electron from an outer shell and a photon (fluorescence) is emitted with the same energy as the difference between the shells.

What about the electron that has been expelled from the atom, where does it go and what happens to it?

There are a number of techniques that use excitation of inner shell electrons. Auger spectroscopy, and ESCA for example. But XRF is a "bulk" analysis method, not surface such as Auger. The excitation energy from the initial x-ray will be keV whereas the empty orbitals in the atoms of the sample will be a few eV. The excited electron rips trough the sample creating various ions until its energy is reduced enough so that the emitted electron can drop into some empty low energy orbital. (In general it would a "different" electron that drops into the vacancy and creates the fluorescent x-ray.)

In general the x-ray can penetrated deeper into the sample than an excited electron can escape. So the bulk of the excited electrons stay within the sample.


Nomenclature of inorganic compounds As depicted in books, $\ce{NO3-}$ and $\ce{NO2-}$ are called as nitrate and nitrite respectively. Where (-ate) suffix is used for higher oxidation state and (-ite) suffix is used for lower oxidation state (just like:- sulfate and sulfite).

Now since Nitrogen is 2nd period element, hence it can't expand it's octet (no vacant d-orbital). So I didn't get how (-ite or -ate) suffix can be used if nitrogen shows only 1 stable oxidation state i.e; (+3). Can anyone suggest?

Nitrogen can have a variety of oxidation states ranging from -3 to +5. For example:

  -3

The nitrogen atom in ammonia $\ce{NH3}$ has an oxidation state of -3 (each hydrogen is +1).

  -2

The nitrogen atoms in hydrazine $\ce{N2H4}$ has an oxidation state of -2.

  0

The oxidation state of the nitrogen atoms in $\ce{N2}$ is 0.

  Nitrate and nitrite

The nitrogen atom in the nitrate ion $\ce{NO3-}$ has an oxidation state of +5. Each oxygen atom has an oxidation state of -2, and the overall charge on the ion is -1.

The nitrogen atom in nitrite ion $\ce{NO2-}$ has an oxidation state of +3.

Nitrogen can have such a range of oxidation states because of its electron configuration, which is $\ce{[He]}2s^2 2p^3$ (5 valence electrons). Nitrogen can "gain" up to three electrons or "lose" up to five electrons.


Heat Capacity temperature dependence and Gibbs energy I got this problem in school as a tool to get ready for International Chemistry Olympiad, but I have some problems with this challenge.
So, the problem is to calculate as precise as possible the Gibbs free energy for $\ce{CaCO3}$ decomposition at 1200 K.

I know that $\mathrm{d}G=\mathrm{d}H-T\cdot\mathrm{d}S$.

But $\mathrm{d}S$ and $\mathrm{d}H$ are temperature dependant, so

$$\mathrm{d}H=\mathrm{d}H_0+\int_{T_0}^TC_p\mathrm{d}T=\mathrm{d}H_0+\mathrm{d}C_p\cdot\left(T-T_0\right)$$

and

$$\mathrm{d}S=\mathrm{d}S_0+\ln\left(T/T_0\right)\cdot\mathrm{d}C_p$$

But molar heat capacity at constant pressure is also temperature dependant, and the equation is

$C_p=a+b\cdot T+c/T^2$, where a, b, c coefficients for each compound at specific phase.

But the problem is with both solids, because this equation can not be used for them, because in 1200 K they are not gases and Einstein’s Quantum Mechanics and Debye models do not work, because Einstein’s approximates are maxed at $C_p\approx3R$, which is about 24.9, but $C_p$ at 298 K, 1 atm for both solids in tables are > 24.9.

Where is the problem? I would appreciate any possible lead to mistake/ correct answer.

At constant pressure, $dG=-SdT$.  Also, the definition of G is $G=H-TS$, so $S=\frac{H-G}{T}$.  Substituting,
$$dG=\frac{G-H}{T}dT$$
This equation can be manipulated into the form:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
So,
$$\frac{d(\Delta G/T)}{dT}=-\frac{\Delta H}{T^2}$$
I leave the rest up to you.

A math error was corrected on 10/16/15


Detection of chloride anion In a sample solution, containing $\ce{Cl-}$, while conducting the test for detection of $\ce{Cl-}$ ion first it is acidified with conc. $\ce{HNO3}$ then cooled and then $\ce{AgNO3}$ solution is added. A curdy white precipitate appears which is soluble in $\ce{NH4OH}$ but reappears on adding $\ce{HNO3}$.

My question: is the $\ce{HNO3}$ added dilute or concentrated?

To best understand, let us consider the reactions occurring.

First, we have $\ce{Cl-}$ ions in solution. Acidifying the solution serves to remove other anions that might give a false positive for the chloride test. Upon addition of $\ce{AgNO3}$, the following reaction occurs:

$$\ce{Ag+(aq) +Cl-(aq) -> AgCl (s)}$$

When enough ammonia is added to neutralize the $\ce{H3O+}$ in solution:

$$\ce{NH3(aq) +H3O+(aq) -> NH4+(aq) + H2O(l)}$$

The ammonia can then begin to form a metal complex with the $\ce{AgCl}$ that is soluble in water:

$$\ce{AgCl(s) +2NH3(aq)->Ag(NH3)2+(aq) +Cl-(aq)}$$

When $\ce{HNO3}$ is reintroduced, the ammonia ligand is protonated because it is more basic than water, and the metal complex is destroyed, reforming $\ce{AgCl}$:
$$\ce{Ag(NH3)2+(aq) +2H3O+(aq) +Cl-(aq) ->AgCl(s) +2NH4+(aq) +2H2O(l)}$$

Since you will likely be working with relatively small amounts, it is likely that you will be using concentrated acid in stoichiometric amounts to produce these products.


Michaelis menten constant motivation behind 1/2? So the michaelis constant is defined as the substrate concentration at half maximal „reaction-speed“.
I was wondering why 1/2 ?
I guess for higher values you need more substrate, which could be a problem. But would 1/3 also work ? Or does 1/2 optimize some other parameter?

To see why the $K_m$ is equivalent to the substrate concentration at which the rate is equal to $V_{max}/2$, we need to look at the derivation of the Michaelis-Menten, which is usually included in detail in any introductory biochemistry book. Here is a brief version.
We will consider an enzyme-catalyzed reaction described by the following scheme:
$$\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P},$$
where "E" is enzyme, "S" is substrate and "P" is product.
To get an expression for the rate, we start by noting that the rate of formation of product is $k_2[\ce{ES}]$, so we need to find an expression for $[\ce{ES}]$ using known quantities. Next, we make a steady-state approximation, setting $\frac{d[\ce{ES}]}{dt}=0$, so we have
$\displaystyle k_1[\ce{E}][\ce{S}]-(k_{-1}+k_2)[\ce{ES}]=0$
$\displaystyle \implies [\ce{ES}]=\frac{k_1}{k_{-1}+k_2}[\ce{E}][\ce{S}].$
To finish, we note that when the reaction has not consumed very much substrate, we can approximate $[\ce{S}]$ as equal to the starting substrate concentration $[\ce{S}]_0$. We also know that the total amount of enzyme in the reaction must be equal to the sum of the concentrations of E and ES, ie $[\ce{E}]+[\ce{ES}]=[\ce{E}]_{Total}$. We already have an equation for $[\ce{ES}]$ in terms of $[\ce{E}]$, so with some algebra, we can find that
$\displaystyle [\ce{ES}]=\frac{k_1[\ce{E}]_{Tot}[\ce{S}]}{k_{-1}+k_2+k_1[\ce{S}]}$.
Combining everything, we get
rate of formation of $\displaystyle \mathrm{P} =k_2[\ce{ES}]=\frac{k_2k_1[\ce{E}]_{Tot}[\ce{S}]_0}{k_{-1}+k_2+k_1[\ce{S}]_0}$.
In order to isolate $[\ce{S}]_0$ in the denominator, we divide the numerator and denominator by $k_1$:
rate $\displaystyle =\frac{k_2[\ce{E}]_{Tot}[\ce{S}]_0}{\frac{k_{-1}+k_2}{k_1}+[\ce{S}]_0}$,
and our final simplification is to define the unwieldy $\frac{k_{-1}+k_2}{k_1}$ as $K_m$ and $k_2[\ce{E}]_{Tot}$ as $V_{max}$:
rate $\displaystyle =\frac{V_{max}[\ce{S}]_0}{K_m+[\ce{S}]_0}$.
Getting back to your original question: Note that we never explicitly defined $K_m$ to be the substrate concentration at half-maximal rate. It simply resulted from our choice to isolate the substrate term in the denominator. However, that was not a completely arbitrary choice.
If we look at $K_m=\frac{k_{-1}+k_2}{k_1}$, we can see that if $k_{-1}$ is large and $k_2$ is small, $K_m$ approaches $\frac{k_{-1}}{k_1}$, which is the equilibrium constant for dissociation of the substrate from ES. For that reason, $K_m$ is often treated as an apparent measure of the affinity of the enzyme for the substrate, and that is an additional reason why we choose to define $K_m$ this way. Always remember that it is not actually the dissociation constant, and in some reactions will be quite far off from the actual $K_d$.


Reaction of barium peroxide forming barium carbonate I stumbled across this reaction:

$$\ce{BaO2 + H2CO3 -> BaCO3 + H2O2}$$

This reaction seems quite similar to the variant with sulfuric acid

$$\ce{BaO2 + H2SO4 -> BaSO4 + H2O2}$$

which is well known and works.

But as carbonic acid is a lot weaker than sulfuric acid, I started to question this reaction. Why does it even proceed in the first place? I thought that the sulfate/carbonate is really insoluble in water which is why no equilibrium forms (or does it?), however barium peroxide is itself pretty insoluble.

So, a) does the first reaction even happen and b) why?

Before the precipitation of $\ce{BaCO3}$ starts, the following equilibrium is arranged:

$$\ce{BaO2 + 2H-O-H <=> H2O2 + Ba^{2+} + 2OH-}$$

$\ce{BaO2}$ is but very slightly soluble in $\ce{H2O}$ because the then occurring hydrolysis produces $\ce{OH^-}$, which raises the pH, and hence stops hydrolysis (and dissolving) of further $\ce{BaO2}$. Simultaneously, by the formed $\ce{OH^-}$, the back reaction is fostered. However, these two blockades may be overcome by adding dilute acid, since this latter neutralizes at once the alkaline product $\ce{Ba(OH)2}(aq)$, and so the equilibrium may be continually shifted to the right side.

Hence, it doesn't matter whether the acid is more or less diluted (nor whether it is strong or weak), provided that acid is continually supplied to the extent that it is consumed during the dissolution/hydrolysis process of $\ce{BaO2}$. An aqueous solution of $\ce{Ba(OH)2}$ is an efficient gas detection or gas scrubbing method for $\ce{CO2}$; hence, by bubbling  $\ce{CO2}$-gas through it for a long time, the $\ce{Ba(OH)2}(aq)$ will easily be transformed into $\ce{BaCO3}(s)$. Of course, a gas frit will be used to spread the gas, and the reaction mixture should be constantly stirred.

In the textbook of Sarda/Handa/Arora$^1$ this process with $\ce{CO2}$ is shortly mentioned (without any explanations except that ice-cold water should be used, and that $\ce{BaCO3}$ may be removed by filtration, since the desired product is, of course, $\ce{H2O2}$).

$^1$ Sanda/Harda/Arora, Chemistry, part-2, New Dehli, 2016, p. 885.


Why is gadolinium specifically used in MRI contrast agents? Gadolinium(III) chelate complexes are routinely used as contrast agents in magnetic resonance imaging (MRI);1 the usual explanation is that paramagnetic species contain unpaired electrons, which cause relaxation of nearby $\ce{^1H}$ nuclei.

Why gadolinium, and not any of the other lanthanides?

In particular, $\ce{Tb^3+/Dy^3+/Ho^3+/Er^3+}$ have magnetic moments that are larger than that of $\ce{Gd^3+}$ (due to orbital angular momentum). At first glance, these should possess even higher relaxivity, since the rate of relaxation due to random fields is proportional to $\langle B_\mathrm{loc}^2 \rangle$.2

I do vaguely recall reading one sentence, about the fact that ground states that aren't S terms are not spherically symmetric, and there's some problem associated with this. But I can't find the reference anymore and I don't remember any other detail.

References

Werner, E. J.; Datta, A.; Jocher, C. J.; Raymond, K. N. High-Relaxivity MRI Contrast Agents: Where Coordination Chemistry Meets Medical Imaging. Angew. Chem. Int. Ed. 2008, 47 (45), 8568–8580. DOI: 10.1002/anie.200800212.
Hore, P. J. Nuclear Magnetic Resonance, 2nd ed.; Oxford University Press: Oxford, U.K., 2015; pp 65, 71.

The reasons may possibly be:

T1shortening
Relatively less adverse reactions
Increased contrast enhancement

  The most commonly used clinically approved contrast agents for MR
  imaging are gadolinium-based compounds that produce T1 shortening.

Tissue relaxation results from interactions between the unpaired electron of gadolinium and tissue hydrogen protons, which significantly decrease the T1 of the blood relative to the surrounding tissues.

Additionally, adverse reactions to this agent are far less frequent than those seen with iodinated compounds, with common reactions including nausea, vomiting, headache, paresthesias, or dizziness.

  T1weighted sequences are part of almost all MRI protocols and are
  best thought of as the most 'anatomical' of images, resulting in
  images that most closely approximate the appearances of tissues
  macroscopically, although even this is a gross simplification.

Figure 1-10. MR enterography on a patient with Crohn disease. A. Coronal T2-weighted image shows wall thickening
and stenotic ileum (arrow). B. Coronal gadolinium-enhanced fat suppressed T1-weighted image shows increased
contrast enhancement of one thickened segment of ileum (arrow).

Credits Basic Radiology Chapter 1 page 11

Update

Most of the answer above was based on Pharmacological/Medical aspects but some Chemistry aspects have been included below:

Also the article mentioned by@Linear Christmas contains valuable information on the subject

Contrast agents

  Contrast agents can be divided into two groups depending on whether
  they cause changes in either T1 (longitudinal relaxation – in simple
  terms, the time taken for the protons to realign with the external
  magnetic field) or T2 (transverse relaxation – in simple terms, the
  time taken for the protons to exchange energy with other nuclei)
  relaxation rates of the water protons, these being known as positive
  or negative agents respectively.

  The ability of an agent to affect T1 and T2is characterised by the
  concentration-normalised relaxivities r1 and r2respectively. These
  parameters refer to the amount of increase in 1/T1 and 1/T2
  respectively, per millimole of agent, and are normally quoted as a
  rate (mM21 s21).4 The values are used to determine the efficiency of a
  contrast agent, and they consist of contributions from both inner
  sphere and outer sphere relaxation mechanisms Negative contrast agents
  influence the signal intensity by shortening transverse relaxation
  (T2), thereby producing darker images as high T2 results in increased
  brightness of the images.

Gadolinium (III) vs other Lathanide (III) ions

  Gadolinium(III) reagents are commonly focused on, due to the coupling
  of a large magnetic moment with a long electron spin relaxation time
  of $\ce10^{-9}$s  at the magnetic field strengths used in MRI
  techniques.

Gadolinium (III) ion is unique in a number of respects:

It has the ability to shorten both the longitudinal and transverse relaxation times of water protons approximately to the same extent by relaxing all nearby protons. Other Ln(III) ions with large magnetic moments are less efficient in shortening T1.

  The Gadolinium ion is also unique among other Ln(III) because its
  symmetric seven electron ground state results in an electronic
  relaxation rate that is six orders of magnitude slower than the
  other Ln(III) ions. This in turn results in a strong magnetic
  anisotropy and fast electronic relaxation states with very short T1,
  on the order of $\ce10^{-13}$s.

Other Ln (III) ions tend to undergo Curie relaxation enhancement arising from the interaction of the nuclear spin with the thermal average of the electron spin. (The Curie-spin relxation effect is significant at lower temperatures and higher magnetic fields and for ions with a large magnetic moment).

It has been observed that the Curie spin relaxation effect affects the transverse relaxation more than the longitudinal relaxation. Consequently paramagnetic Ln(III) other than Gd (III) are less efficient T1 relaxation agents.

Use of other Lanthanide (III) ions as MRI contrast agents

Dysprosium(III) is another lanthanide ion that has been used in
MRI, being classed as a negative contrast agent. In the use of high magnetic fields,

  Gd(III) based contrast agents exhibit poor water relaxivity. As a
  result, the interest in dysprosium-based complexes is increasing as
  they display slow water exchange, due to the need to lengthen the
  residence time in order to optimise the r2 relaxivity. Dy(III) has a
  large magnetic susceptibility which induces local field gradients
  resulting in a lowering of T2.

Generally, dysprosium complexes where the water molecules have a long residence
time, possess potential application as negative contrast agents
at high magnetic fields due to their efficient transverse
relaxivity. However it has been demonstrated that lengthening the residence time of water
can actually be detrimental, because the transverse relaxivity
can then be limiting.

The overall conclusion from the use of dysprosium(III)
complexes as contrast agents is that due to the balancing of
factors required to optimise r2, design of a suitable molecular
structure is crucial. Fine tuning these residence times of the
water protons, and hence the relaxivity, may lead to promising
contrast agents for high field magnetic resonance imaging.

  Europium(II) analogues have been proposed as alternatives to Gd(III)
  because they are isoelectronic—each having seven unpaired electrons.

The only problem with this
complex was that the stability of the Eu(II) chelate was
107 times higher than that of the Eu(III) chelate, which
meant that the Eu(III) complex could be susceptible to
dissociation, possibly releasing toxic Eu(III) into the body.

Other strongly paramagnetic ions e.g Tb(III), Ho(III), Er(III) can have significant effect on line widths and such line broadening usually makes it difficult or even impossible to detect nuclei that are located within a certain distance of the lanthanide ion.

There is still promising use of other Lanthanide (III) compounds as contrast agents as alternatives to using T1 shortening contrast agents as contrast can originate from altering proton density or the total water signal that is detected (using another technique called Magnetisation Transfer (MT).In this case Gadolinium(III)cannot be used in this technique because the T1 of water would be too short, but other lanthanides, such as Eu(III), can be used because they have smaller magnetic moments, relaxing bulk water much less efficiently.

Other potential candidates are still being tested under laboratory conditions, Contrast agents used in the
body clearly should be biocompatible, but there are also other requisites need to be addressed as well. These include the requirements of rapid renal excretion, water solubility, stability in aqueous conditions, and a low osmotic potential when in solution for clinical work.

References

Basic Radiology Michael Y. M. Chen, MD
'Alternatives to Gadolinium-Based Metal Chelates for Magnetic Resonance Imaging' (Chem. Rev., 20102010, 110 (5), pp 2960–3018,
Lanthanides in magnetic resonance imaging. Melanie Bottrill, Lilian Kwok and Nicholas J. Long
Chemistry Society Reviews, DOI: 10.1039/b516376p


What compound is formed in this case? I read from a source that the aqueous solution of ferrous sulfate ($\ce{FeSO4}$) gives an unstable brown compound on treating with $\ce{NaNO2}$ and dilute $\ce{HCl}$. Well, it is quite understandable that it is the $\ce{HNO2}$ formed in situ that further causes the reaction. But what is the reaction and what is that unstable compound formed?

Iron(II) reduces nitrite to nitrogen monoxide which forms a brown pentaaquanitrosyliron(II) complex with the iron(II) ions.

$$\ce{NO2^{-} + Fe^{2+} + 2H+ -> NO + Fe^{3+} + H2O}$$

$$\ce{[Fe(H2O)6]^{2+} + NO -> [Fe(H2O)5NO]^{2+} + H2O}$$

Iron(II) can also reduce nitrate to nitrogen monoxide. This is used in the brown ring test to determine the presence of nitrate in aqueous solutions.


Why does saline have a pH of 5.5? I am a physician and administer 0.9% NaCl in water ("normal saline") to patients frequently. The bag says it has a pH of 5.5 (compared to the human body of 7.4).

I am asking why the pH is so low. It is because NaCl is the salt of HCl? I read on wikipedia that the pH of NaCl aqueous solutions is 7. I teach medical students and want to give an understandable explanation why the pH on a bag of "normal saline" reads pH 5.5 and when we give lots of saline to patients, they develop acidosis in the blood.

Medical saline has a $\mathrm{pH}$ of 5.5 do to the dissolved $\ce{CO2}$ within the solution as well as factoring in the degradation of the PVC packaging.
I reference the attached article from the International Journal of Medical Sciences from 2013:
Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-750 (doi:10.7150/ijms.5868)(PDF).


Precipitation of AgCl from the tap water solution of the group 2 chloride I had this question about an experiment where the group 2 chloride was dissolved in distilled water and excess $\ce{AgNO3}$ was added to the solution to form $\ce{AgCl}$ precipitate. One of the sub-questions read

  State how the amount of $\ce{AgCl}$ will change if tap water is used to dissolve the group 2 chloride instead of distilled water (will the amount of $\ce{AgCl}$ formed increase or decrease) and explain why.

Can someone please explain how the amount of $\ce{AgCl}$ formed will change if tap water is used? I assumed that tap water is chlorinated and so the amount of $\ce{AgCl}$ formed will increase. I don't know if this is right.

Yes the amount of $\ce{AgCl}$ formed will be more due to more $\ce{Cl-}$ ions present in the tap water. They are mainly because of some salts already present in water, and due to chlorination.


Do liquid O₂ and N₂ form ideal solutions? I suspect they do, but could not find data on this. (I mean at temperatures near those used in the distillation of liquid air.)

It's not perfectly ideal.  Try to get access to Excess enthalpies of the liquid mixtures nitrogen + oxygen... for quantitative data.


Nernst equation and equilibrium I am currently stuyding electrochemistry, and recently I've stumbled upon a problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate, but I obtained an unrealistic result and I became confused.

What I first did was write the half-reactions involved within the voltaic cell:
Anode (-): 5Fe2+(aq) ⇌ 5Fe3+(aq) + 5e- ε01= -0.77 V
Cathode (+): MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) ε02= 1.5 V
Notice the fact that I obtained the first reaction, the oxidation, by reversing the equation for the reduction of Fe3+.
By combining both half-cell reactions, one can obtain the equation for the whole process:
5Fe2+(aq) + MnO4-(aq) +  8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) E0= 0.73 V

A positive standard potential will yield a negative value for the Gibbs free energy, so the reaction is product-favoured. The problem is concerned with calculating the concentration at equillibrium, corresponding to the equivalence point of our titration, so the overall Gibbs free energy will be 0, just as the non-standard potential, which we can calculate using the Nernst equation (considering 298.15 K of temperature and 1 atm of pressure). I'm also assuming that the reaction quotient is equivalent to the equilibrium constant, since we are at the equivalence point.

\begin{align*}
    E=E^{0}-\frac{0.0592}{5}\log{K}
\end{align*}
\begin{align*}
    E=0, E^{0}=\frac{0.0592}{5}\log{K}
\end{align*}
I then obtained the mathematical expression for the equilibrium constant:
\begin{align*}
    K=10^{\frac{5E^{0}}{0.0592}}
\end{align*}
I then entered the values, and calculated K (substantially larger than 1):
\begin{align*}
    K=4.523*10^{61}
\end{align*}
However, this is the hurdle I have yet to overcome. I didn't quite understand what activities are in the Nernst equation (afaik they are not equilibrium concentrations), so I assigned a formula for K based on the equilibrium reaction:

\begin{align*}
    K=\frac{[Fe^{3+}]^{5}[Mn^{2+}]}{[Fe^{2+}]^{5}[MnO4^{-}][H^{+}]^{8}}
\end{align*}

The problem mentions the following concentrations for the titration:
Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
H2SO4: 1N (probably 0.1M for H+)
KMnO4: 1N (0.2M MnO4- following the same steps)

In my conception, the activities of the reagents in the equilibrium constant were their initial concentrations, and the activities of the products were their equillibrium concentrations. I then made an ice table, considering a concentration of x. By replacing those values into the equilibrium constant, I obtained:
\begin{align*}
    K=\frac{5x^{5}*x}{0.02^{5}*0.2*0.1^{8}}
\end{align*}
X was supposed to be the concentration of Fe^{2+}, however the answer I got is definitely not the right one.
\begin{align*}
    x=6726501.235
\end{align*}

Where did I go wrong?

problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate,

There are several things which may be problematic in your approach.

For the stated problem you do not need to invoke Nernst equation. Simply use the balanced equation.
When potentiometric titration curve are to be estimated, it is far more convenient would use the half-cell instead of a full reaction. I am assuming the redox electrode is dipping in iron solution.

\begin{align*}
    E=E^{0} -\frac{0.0592}{5}\log{[Fe^{2+}]/[Fe^{3+}]}
\end{align*}

where $E^{0}$ is the iron half cell value. You should not calculate equilibrium constants from potentiometric titrations.

See page 17 for a full example Potentometric Titration clculation

I do not know when people will stop teaching normality. It is obsolete (18th century concept). Your normality conversions are incorrect.

  The problem mentions the following concentrations for the titration:
  Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
  H2SO4: 1N (probably 0.1M for H+)
  KMnO4: 1N (0.2M MnO4- following the same steps)

1 N sulfuric acid will be half the molarity =0.5 M

1 N permanganate will be 1/5th the molarity = 1/5 M

Check the concept of normality in your textbook.


What is the difference between cofactors and coenzymes? I've just been wondering what are the differences between the two and I can't seem to find a formal delineation.

Enzyme cofactors are a collective name of all the chemical compunds or elements associated with the enzyme to increase its efficiency. There are mainly two types- inorganic ions and organic compounds. Inorganic ions are also known as enzyme activators (Cl ion in salivary amylase). Organic compunds are again divided into two- co-enzymes and prosthetic groups. Co-enzymes are loosely associated with the enzyme. They can be removed from the enzyme and undergo cyclic reactions to again combine with enzyme (EX: NAD, NADP). But prosthetic groups are tightly associated with enzyme molecules(EX : FAD). Hope this helps.