Find gas pressures of injection knowing mixture pressure without knowing volume Let's say we have a gas 1 moving inside a pipe (we don't know the geometry) at mass flow rate $Q_1$ and temperature $T_1.$ We inject a gas 2 at some point inside the pipe with gas 1 at mass flow rate $Q_2$ and at temperature $T_2.$
We know that the resulting mixture is moving at the mass flow rate $Q_\mathrm{mix} = Q_1 + Q_2$ at the temperature $T_\mathrm{mix}$ and is at pressure $p_\mathrm{mix}.$ All gases are ideal gases.
Is it possible to find the gas 1 pressure $p_1$ and the gas 2 pressure $p_2?$ And if so, what is the relation?
So far I haven't found any clues to solve this problem.
I thought about partial pressure, but I feel wrong to say that $p_1$ and $p_2$ are respectively equal the partial pressures of gas 1 and gas 2 in the gas mixture because there is no mention of volume in this problem.

Of course it is a pipe and it has a infinite volume hypotactically,
To use ideal gas equations is a bit tricky with mass flow rate need to convert to molar flow rate of each gas thus need to divide by $\ce{M_w}$ molecular weight respective to each gas.
Consider $\ce{1}$ time unit where inside of pipe gets
$$\frac{Q_1}{M_{w_1}}$$
$$\frac{Q_2}{M_{w_2}}$$
moles of gas to one volume section of pipe
At one time unit according to the ideal gas equation
$$\ce{PV=nRT}$$
Gas 1 will consume $\ce{V1}$ amount of volume
$$V_1 = \frac{Q_1RT_1}{M_{w_1}p_1}$$
Gas 2 will consume $\ce{V2}$ amount of volume
$$V_2 = \frac{Q_2RT_2}{M_{w_2}p_2}$$
Total of $\ce{V}$ volume per each time unit
$$V = V_1 + V_2$$
But $\ce{V}$ is just a section of the pipe it just shifts to what ever direction the gas is flowing and keeps going for ever in the infinite pipe.
But as further the gases go away just like you said turbulence will stabilize and final temperature and pressure inside pipe will be almost equal to $\ce{T_{mix}}$ and $\ce{p_{mix}}$
When it is stabilized,
Gas 1 will consume $\ce{V'_1}$ amount of volume
$$V'_1 = \frac{Q_1RT_{mix}}{M_{w_1}p'_1}$$
Gas 2 will consume $\ce{V'_2}$ amount of volume
$$V'_2 = \frac{Q_2RT_{mix}}{M_{w_2}p'_2}$$
Total of $\ce{V'}$ volume per each time unit
$$V' = V'_1 + V'_2$$
The important factor here is that number moles of gas will shift to next stage will be the same thus molar fraction for each gas is a constant. Think as $\ce{V'}$ stage in pipe
Molar fraction can be easily calculated
$$\ce{X_1 = \frac{(\frac{Q1}{M_{w_1}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})}}$$
$$\ce{X_2 = \frac{(\frac{Q2}{M_{w_2}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})}}$$
Using the partial pressure equation
$$\ce{P_A = X_A P_{tot}}$$

what is the relation?

Refer here!
$$\ce{p'_1 = \frac{(\frac{Q1}{M_{w_1}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})} p_{mix}}$$
$$\ce{p'_2 = \frac{(\frac{Q2}{M_{w_2}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})} p_{mix}}$$
There you have it final answer but consider that its one time unit thus take only the magnitude with respective unit of mass flow rate * unit time $\ce{|Q1| |Q2|}$ symbol ` is there to show gases are stabilized.


Is boat form of cyclohexane non polar? In chair form of cyclohexane the C-H dipole moments cancel out each other but in boat form i don't think all dipole moments will be cancelled out. Also if boat form is polar then what will we say overall about cyclohexane is it polar or nonpolar?

Being nonpolar does not require having the zero dipole moment.

Enough is being much closer to having a zero dipole moment than to having the a strong dipole moment. It applies also to parts of the molecule.

E.g. carbon dioxide has zero permanent dipole moment, yet liquid carbon dioxide is rather polar solvent due its polar bonds.


How do I calculate the temperature of hot air in kelvin? The question simply says “calculate the temperature of hot air in kelvin” on a data calculation section for this lab. We recorded:

volume change on cooling = 170 mL
original volume of hot air = 20 mL
volume of the cooled air = 150 mL
temperature of heated water = 100 °C
temperature of cool water = 20.4 °C

and that is all.

The lab was focused on Charles’ law, namely to predict how much air should contract when cooled from 100 °C to room temperature. We took a flask and inserted a rubber stopper with glass tubing through the center. We then heated the flask in a water bath until the water boiled, then keeping a finger over the top of the glass tubing we inverted the flask in a “cool” water bath and held it there for 10 minutes. Then we equalized the pressure within the flask with the atmospheric pressure by raising the flask until the water level inside equaled the outside.

You already know the temperature of the hot air in Celsius: 100 degrees

Using the equation Kelvin = Celsius + (273.15 degrees), you would get 100 degrees + 273.15 degrees, or 373.15 degrees.

I am assuming that this is a preliminary step that is required in order to find the temperature to plug into Charles' law (which must be in Kelvin), in which case you can use the formula I mentioned above for your other calculations as well.


Comparing acidity of anilinium ion and para-fluoro anilium ion Which of these compounds - anilinium ion and para-fluoro anilium ion - is more acidic?

I thought the answer would the be first one since $\ce{-F}$ is an ortho-para activating group, so the electron density at the carbon at which $\ce{-NH3+}$ is attached would be more. I know that $\ce{-F}$ has a strong inductive effect, but since this is para with respect to $\ce{-NH3+}$, I thought I could say +M dominates. However the answer given was the second one. Why is this so and when can I neglect the inductive effect of halogen substituents in benzene to favour the mesomeric effect caused by them?

The best way to compare acidities of organic compounds is by drawing out the conjugate bases and figuring out which one is more stable. Molecules with more stable conjugate bases are more acidic, so doing the same here:

The conjugate bases are aniline and para-fluoro aniline respectively. The $\ce{-NH2}$ group shows +M (mesomeric effect), so on drawing out the resonance structures we find that the second resonance structure is more stable as the resulting negative charge is inductively withdrawn by fluorine group.

For halogens the inductive effect dominates over their positive mesomeric effect, especially in the case of fluorine (because it is small and the lone pairs experience a high effective nuclear charge,in case of large atoms like chlorine, bromine and iodine, the outer electrons are held weakly)


Electrochemical Cell Now in an electrochemical cell , we make use of the fact that whenever there is a potential drop , electrons will flow from lower to higher potential as for a potential drop , there must be an electric field .

Now suppose there are two single electrodes , what creates the electrostatic potential on each of them in the first place ?

And just to confirm , a redox reaction is used just that we can maintain a steady flow ?And so why is it that we always put in that electrolyte that has one of the ions same as the electrode element ?

Creating a Galvanic Cell

To simplify, let's forget galvanic cells for a moment. Instead, just picture an oxidation/reduction reaction in aqueous solution.  If we had a solution of $\ce{Cu^+}$ and $\ce{Co^{3+}}$, the metal ions would exchange an electron, and we would end up with $\ce{Cu^{2+}}$ and $\ce{Co^{2+}}$. This happens because of the relative stability of the two ions; the electron is at a lower energy with the cobalt than with the copper (at these oxidation states). Now, if we separate the two metals into two different solutions, insert an electrode into each, and connect the two with a conductive wire, we can still transfer an electron. The potential between the electrodes comes from the same source as the mixture: the chemical potential between the two ions.

A salt bridge is also necessary. The salt bridge balances charge in the two solution. Otherwise, we would have a net positive charge in the copper solution and a net negative charge in the cobalt solution. This would create a potential difference and the reaction would eventually and stop.

Galvanic Cell using Non-ionic Species

Galvanic cells do not have to use species that are ionic in the oxidized and reduced states (or either, for that matter). Consider the cell pictured below. In this example, zinc metal is being oxidized to $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$ is being reduced to copper metal. In other words, the zinc electrode is getting dissolved, and the copper electrode is growing. The anodic salt solution is used to create what is called a poised electrode. If a zinc electrode were inserted into pure water, according to the Nernst equation, the system would have an infinite potential.

$$\mathcal E = \mathcal E^0 - \frac {RT} {nF} \ln 0 \rightarrow -\infty$$

As soon as even a few molecules per liter are in solution (which would happen very quickly) the potential drops to around 50 V, and would continue to drop as more zinc was oxidized (although there are certainly deviations from the Nernst equation in an observed system). The salt solution puts the system into a steady state. In terms of the cathode, the metal does not matter. Anelectrode made from any conductive material would behave the same as a copper electrode in a solution of $\ce{Cu^{2+}}$; copper would plate onto the electrode as it is reduced. The system potential is not altered because the activity of a pure solid is unity. It is normally just convention (and for learning purposes) that the electrode material is the same as the oxidant.


What is the number of stereoisomers for hex-4-en-2-ol? For the following compound, find

  Total number of stereoisomers
  Number of optically active stereoisomers
  Total number of fractions on fractional distillation of all stereoisomers

There is a chiral carbon marked with "*" in the compound. The compound is thus chiral and is optically active. The configuration at the chirality centre can either be R or S. There is also a double bond in the compound and cis-trans (geometrical isomerism) is possible here. So using basic permutations there are totally $2\times2=4$ stereoisomers (optical and geometrical).

Since there is a chirality centre the compound exists in right-handed and left handed systems and thus is optically active. Thus all the 4 stereoisomers are optically active.

Out of R-cis, R-trans, S-cis, S-trans isomers there are two pairs of diastereomers - R-cis, S-trans and R-trans, S-cis - so there are 2 fractions obtained due to fractional distillation.

Final Answer

Total number of stereoisomers = 4
Number of optically active stereoisomers = 4
Total number of fractions on fractional distillation of all stereoisomers = 2

Are my approach and my answers right?

The approach is right, and your answers are right, but there are a couple of things to look at in your explanation.

First, the presence of a chiral centre does not in itself make a compound chiral.

It is in this case, but a molecule is chiral if, and only if, it is not superimposable on its mirror image.  You could have a molecule with more than one chiral centre, which was superimposable on its mirror image, and was therefore not chiral.

Second, a chiral molecule is not necessarily optically active.  Its specific rotation may be zero (or at least unmeasurable), in which case it is said to exhibit cryptochirality.

Thirdly, the R- and S- form of each of the geometric isomers are enantiomers (ie non-superimposable mirror images) of one another, not diastereomers.  Diastereomers occur when a molecule has (for instance) two chiral centres and rather than inverting both (giving the enantiomer, its mirror image) you only invert one, which is actually a different compound, with different physical and chemical properties.

Personally I think using the term stereoisomer to refer to the cis and trans versions is misleading, even if strictly geometric isomers are a subset of stereoisomers.  I'd prefer to say that the R- and S- forms of each geometric isomer are enantiomers of one another, and leave it at that, rather than lumping all four together as stereoisomers.  But you don't get to choose the wording of the question, I guess.


How to prepare potassium cyanide from potassium ferrocyanide How do we prepare potassium cyanide from potassium ferrocyanide?

It is something which I am curious about, but cannot find the answer. I am studying inorganic chemistry but this reaction cannot be found.

Until about 1900, potassium cyanide was produced by melting dry potassium carbonate with dry potassium hexacyanidoferrate(II):
$$\ce{K4[Fe(CN)6] + K2CO3 -> 6KCN + FeO + CO2}$$
However, the product contains potassium cyanate, which is approximately formed as follows:
$$\ce{K4[Fe(CN)6] + K2CO3 -> 5KCN + KOCN + Fe + CO2}$$

In a small scale, dry potassium hexacyanidoferrate(II) is melted in a covered porcelain crucible. It is heated as long as nitrogen escapes.
$$\ce{K4[Fe(CN)6] -> 4KCN + N2 +FeC2}$$
The iron carbide sinks to the bottom and the molten potassium cyanide can be poured off.


Concentration from Absorbance vs Time I am given a protein, BAPNA, and the initial concentration of the protein. The experiment involves reaction rates of varying protein concentrations. Each sample cuvette is inserted into a spectrometer, 100% transmittance is set, has the enzyme inserted, and then has transmittance measured every 20 s for 600 s.

I understand absorbance is given as: $A = 2 - \log (T\%)$ and that's no issue.

The problem I run into is trying to solve for the concentration or molar extinction constant. I understand that Beer-Lambert law, $A = \epsilon bc$  is used for this purpose. The only given concentration is the initial concentration. Assuming $b = 1$, we can solve for $\epsilon$ or $c$, but having 100% set in the beginning causes $A = 0$, meaning initially $\epsilon = 0$ if $c \neq 0$.

This is where my calculations run into a dead end. I'm not given $\epsilon$ or concentration at any other point, but I'm supposed to be able to calculate the final concentration and maximum reaction rate. I've only a set of transmittance numbers over time, and that initial concentration.

Is there something I am misunderstanding or not seeing?

The only given concentration is the initial concentration [...]

Unless there is a major misunderstanding on my side, you have all the data you need.

Given the initial concentration $c_0$ and knowing the path length $d$ of your cuvette, you measure the extinction $E$ at this point and use - as you have correctly stated in your question - the Lambert-Beer law to calculate the molar extinction coefficient $\epsilon$.

$$E = \epsilon\cdot c\cdot d$$
$$\epsilon = \frac{E}{c\cdot d}$$

Did I mention units? Units are your friend!

Write the Lambert-Beer law with units!
Percent (%) is NOT a proper unit for the initial concentration of your protein.

  but I'm supposed to be able to calculate the final concentration and maximum reaction rate

Having determined $\epsilon$, you can now correlate at any point along your reaction the measured extinction with the actual concentration of your sample, including the final concentration.

From the slope of the curve of extinction $E\ $ vs concentration $c$, the rate of your reaction can be obtained.

If the relation is not a straight line, logarithms are your friend.


Why do the protons in pentacarbonylmethylmanganese show up at –0.21 ppm? The $\ce{^1H}$ NMR of $\ce{Mn(CO)5CH3}$ shows a resonance at –0.21 ppm, which has to be due to the methyl group. Why are the protons so shielded?

I would have thought that all the pi backdonation from the Mn to the carbonyl would reduce the electron density on the $\ce{CH3}$, hence increasing the chemical shift.

Unfortunately, you don't give an indication of what you expect the chemical shift for the -Me to be, other than higher than is observed. However, you are correct in your assumption; the π back donation from the Mn centre to CO ligands does decrease the shielding, and therefore does increase the chemical shift, but this is a good case of how chemical shifts are all relative.

The dominating factor in shielding in d-orbital transition metal complexes comes from the paramagnetic ring currents within the incomplete valence d-shell. The circulating electron in the d-orbital produces a much greater angular momentum than for a p-orbital complexes (typically organic molecules, say). The influence of the d-orbital ring current induces a very large diatropic shift for ligands close to the metal centre. These can by very large for hydrides; up to -50ppm in some cases, and still significant for methyl ligands.

So, your starting point for the chemical shift of the -Me ligand is something that has significant shielding, and would be largely negative. What you would expect, based on your argument, and indeed what is observed by experiment, is that as you increase the number of π accepting ligands around the metal centre, you get an increase in the chemical shift of your -Me ligand. For example, as you go from M(PR3)5(CO)1 to M(PR3)1(CO)5, there would be a general trend to deshielding and an increase in the observed chemical shift.

Relative changes in chemical shift in transition metal/ligand chemistry are a great tool for determining stereochemistry and relative positioning of other ligands; a trans substitution will have a greater influence on the shielding than cis, for instance, however the overall dominating factor in the local shielding of your observed -Me still remains the diatropic shift from the metal centre.


Reasons for weakness of London dispersion force in graphite? The layers of graphite are held together via London dispersion force. This force is weak. I know carbon having a smaller atomic radius and few electrons is one factor for this force's weakness.

But, the free electron is weakly bound with the adjacent three atoms. Does this constrains the free electron's ability to disperse, making the London force weaker?

P.S. This might be a duplicate question. Inform me if you find similar one(I haven't found any)

We need to look at the orbitals, where are the electrons localized?
because the hybridization of the carbons in grpahite are sp2 we have one free p orbital with one electron, which is denoted as the pz orbital.
The free pz electron then forms the resonance bond with the adjacent carbons. Therefore there are no non bonding valence electrons to contribute much for the London dispersion interactions.


Does potential difference or current drive electrolysis? Would a high voltage ( of the order of a few score kV ) drive electrolysis?

Or, does it require a large current and low voltage? Alternatively, does electrolysis require both large voltage & large current?

For any electrochemcial reaction, electrons are transfer from the oxidant to the reductant. By definition, any flow of charge per unit time period is an electric current.

However, in the case of electrolysis, the electron transfer is not spontaneous. An external energy source is required for the reaction to take place. The energy (work) provided per unit charge is called the voltage.

For any electrolysis reaction to occur therefore requires a supply of electrical energy which means both a voltage and electric current are needed.

Specifically, the amount of work (energy) resulting from a voltage, $V$ and a current, $I$ applied for a time $t$ is given by:

Work produced by electrical energy, $W = V\cdot I\cdot t$

For a given electrolytic reaction to proceed, the required minimum voltage which must be applied is determined from the Gibb's function,

$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}$

Where $E^\circ_\text{cell}$ is the minimum voltage needed for the electrolysis reaction to occur, $\Delta G^\circ$ is the change in Gibb's free energy under standard conditions, $n$ is the number of electrons transferred and $F$ is Faraday's constant (96,485 coulombs per mole).

For example, if pure water is placed in an 'electrolytic cell' with two non-reactive electrodes (eg: platinum), electrons forced into the 'cell' by an electrical source (eg: battery) will react with water molecules at the cathode forcing them to lyze (split) into hydrogen ions and hydroxide ions.

$\ce{H2O(aq) ->[\text{elect}] H+(aq) + OH- (aq)}$

At the surface of the 'anode', hydroxide ion will be oxidised (dontate electrons) to form oxygen gas.

$\ce{4OH- (aq) -> 2H2O + O2 + 4e- }$ (anode half-cell reaction)

Meanwhile, at the surface of the 'cathode', hydrogen ions will accept electrons to form hydrogen gas.

$\ce{2H+(aq) + 2e- -> H2(g)}$ (cathode half-cell reaction)

The overall cell reaction is therefore:

$\ce{2H2O(aq) -> 2H2(g) + O2(g)}$ ; $\Delta G^\circ=237.2\ \mathrm{kJ/mol}$

The positive Gibb's function for this reaction indicates that the reaction will not occur sponteneously but requires an external energy source (eg: electrical energy), which is reflected in the negative cell potential.

The minimum (theoretical) voltage required is:

$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}=\frac{-237.2 \times 10^{3}}{2\times 96485}=-1.23\ \mathrm{V}$

However, in practice, a slightly larger voltage of 1.48 V is required since the  enthalpy (heating) of the products results in slightly lower efficiency which is manifested as an overpotential of about 0.25 V.

Once this critical voltage level is exceeded, the electrolysis reaction proceeds at a rate determined largely by the the current, since the current represents the rate at which charge is delivered to the system. Basically, the higher the current, the more molecules will react (electrolyze) and the more products (hydrogen gas) will form per unit time.


Cooling Curves differences between Solution and Pure Solvent To put it short, why is the cooling curve for a solution slanted during the phase change? The pure solvent plateaus at a fixed temperature, but for a solution, the phase change occurs over a range of temperatures. What explains this phenomenon?

During the phase change, the composition of the solution phase is changing from that of the original solution as the solidifying component is removed.  This doesn't happen for a pure liquid.  In a simple system, the slant continues until both components begin to solidify simultaneously - the eutectic point.  The temperature then flattens out because the composition of the liquid phase doesn't change.  It stays flat until no liquid remains.  This can be complicated if there are other solid phases or formation of new chemical components.


What is the formally correct notation of t-Bu, i-Pr, p-NO2, n-Bu, a-HCl, etc.? In chemical notation, can I safely assume that the notation of

a single lower-case latin character
followed by a hyphen
followed by an Element upper-case latin character (and possibly more characters)

should always be typeset as

italic lower-case character
hyphen  (i.e. never is a bond)
element following characters

This question is regarding the further improvement of mhchem.

I know that not all of them stand for the same concept, but I assume they share the same typographical representation. Beware, I am no chemist. And because I don't know the name of these notations, I could not find them in the colored IUPAC books.

The notation is quite rare. There were only 21 usages in 43k \ce calls extracted from chemistry.SE:

(p-NO2Ph)_2CO
t-BuO-
i-Pr
t-Bu
R=i-Pr
n-BuNH2
(c-Hex)2BCl

a-HCl  // should this be \alpha?

1s-H  // what is this?

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), single italic lower-case Latin letters are used in some stereodescriptors (such as ‘c’, ‘m’, ‘p’, ‘r’, ‘s’, and ‘t’), in descriptors for fusion sites (e.g. dibenzo[c,g]phenanthrene), in locants (‘o’, ‘m’, and ‘p’ in place of ortho, meta, and para, respectively; however, this usage is strongly discouraged), and in other structural descriptors (e.g. s-indacene).

However, single lower-case Latin letters that are used for numbering are not italicized (e.g. naphthalene-4a,8a-diol).

All these symbols are typically used in names (which are usually not typeset using MathJax/mhchem) and do not necessarily appear in formulas.

Furthermore, common abbreviations for organic structure fragments may be used in formulas, but not in text. According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the permissible abbreviations include:

isopropyl: iPr
isobutyl: iBu
sec-butyl: s-Bu
tert-butyl: t-Bu

However, the ACS Style Guide recommends different abbreviations:

isopropyl: i-Pr
isobutyl: i-Bu
sec-butyl: sec-Bu
tert-butyl: t-Bu

Furthermore, authors may create their own abbreviations, but any abbreviations not included in the recommendations shall be defined clearly when they are used.

Also note (since the question mentions letters followed by an element) that several structural abbreviations in current usage are identical to an element symbol:

Ac (acetyl, actinium)
Ar (aryl, argon)
Cm (carboxymethyl, curium)
Nb (p-nitrobenzyl, niobium)
Np (p-nitrophenyl, neptunium)
Pr (propyl, praseodymium)


Distillation: temperature not reaching azeotrope boiling point I ran a simple distillation of $\ce{HBr}$/water mixture because I wanted to purify it. Boiling point of azeotropic hydrobromic acid is approx. $\pu{124 ^\circ C}$, however my distillation ran almost all the way through at vapor temperature of $\pu{112\! -\! 113 ^\circ C}$. Atmospheric pressure is normal. It is around $\pu{8 ^\circ C}$ where I'm working though. What could be the reason the temperature never reached $\pu{124 ^\circ C}$?

I think your major mistake is using a simple distillation method, which may not give accurate reading for azeotropic distillation. The early distillate may not have enough cooling to get to expected azeotrope temperature of aqueous $\ce{HBr}$. If you have followed the work of Carriére et Cerveau (Ref.1), you may able to assume approximate composition of your fractions. Although their work cannot be regarded as of high precision as the composition was determined by titration (Ref.2), they have given an insight to $\ce{HBr}$ distillation. Their work has clearly showed aqueous $\ce{HBr}$ has been yielding a "high-boiling" azeotrope (boiling temperature of azeotrope is higher than boiling points of either of pure components). For example, the $\ce{HBr}$ composition of the early distillate collected at $\pu{110 ^\circ C}$ was reported as 50.8%. The temperature had increased to $\pu{126 ^\circ C}$ maximum (47.5% $\ce{HBr}$), and then declined passed  $\pu{111.5 ^\circ C}$, at which, the $\ce{HBr}$ composition was found to be 30.8%. Since your initial $\ce{HBr}$ concentration was around 30%, I’d say your distillate is already at downward slope and never have able to reach constant boiling azeotrope temperature. If you had used fractional distillation setup with long enough column (and measured atmospheric pressure accurately), you may have able to collect some azeotrope. For your convenience, I listed published data of azeotrope boiling point, density, and composition at different pressure (measured by pressure control apparatus):

$$
\begin{array}{ccc}
\text{Reference} &\text{Pressure} & \text{Boiling Point} & \text{Density at}\: \pu{25 ^\circ C} & \text{Composition, % HBr}\\\hline
\text{Carriére et Cerveau}^1 & \pu{760 mmHg} & \pu{126 ^\circ C} & - & 47.5\\
\text{Ewing and Shadduck}^3 & \pu{760 mmHg} & \pu{125 ^\circ C} & - & 47.795 \pm0.033\\
\text{Bonner et al.}^2 & \pu{760\pm0.5 mmHg} & \pu{124.3\pm0.02 ^\circ C} & - & 47.63\pm0.01\\
\text{Bonner et al.}^2 & \pu{100 mmHg} & \pu{74.12 ^\circ C} & 1.5116 & 49.80\\
\text{Bonner et al.}^2 & \pu{200 mmHg} & \pu{90.35 ^\circ C} & 1.5030 & 49.28\\
\text{Bonner et al.}^2 & \pu{300 mmHg} & \pu{99.91 ^\circ C} & 1.4961 & 48.83\\
\text{Bonner et al.}^2 & \pu{400 mmHg} & \pu{107.00 ^\circ C} & 1.4908 & 48.47\\
\text{Bonner et al.}^2 & \pu{500 mmHg} & \pu{112.94 ^\circ C} & 1.4866 & 48.19\\
\text{Bonner et al.}^2 & \pu{600 mmHg} & \pu{117.82 ^\circ C} & 1.4832 & 47.95\\
\text{Bonner et al.}^2 & \pu{700 mmHg} & \pu{122.00 ^\circ C} & 1.4802 & 47.74\\
\text{Bonner et al.}^2 & \pu{800 mmHg} & \pu{125.79 ^\circ C} & 1.4775 & 47.56\\
\text{Bonner et al.}^2 & \pu{900 mmHg} & \pu{129.13 ^\circ C} & 1.4752 & 47.40\\
\text{Bonner et al.}^2 & \pu{1000 mmHg} & \pu{132.12 ^\circ C} & 1.4733 & 47.27\\
\text{Bonner et al.}^2 & \pu{1100 mmHg} & \pu{134.80 ^\circ C} & 1.4716 & 47.14\\
\text{Bonner et al.}^2 & \pu{1200 mmHg} & \pu{137.34 ^\circ C} & 1.4700 & 47.03\\
\end{array}
$$

References:

MM. E. Carriére and Cerveau, Compt. rend. Acad. Sci. 1923, 177, 46-48 (https://gallica.bnf.fr/ark:/12148/bpt6k3130n/f50.image).
W. D. Bonner, L. G. Bonner, F. J. Gurney, “Azeotropic Hydrobromic Acid Solutions at Pressures of 100 mm to 1200 mm,” J. Am. Chem. Soc. 1933, 55(4), 1406-1409 (DOI: 10.1021/ja01331a012).
D. T. Ewing, H. A. Shadduck, “The Composition of a Constant-Boiling Solution of Hydrogen Bromide in Water,” J. Am. Chem. Soc. 1925, 47(7), 1901–1904 (DOI: 10.1021/ja01684a017).


Why isn't auto ionisation of water considered when we find the pH during salt hydrolysis? $\ce{H+}$ concentration from the auto-ionisation of water is $10^{-7}$
If we have $\ce{HA}$ as weak acid and $\ce{BOH}$ as weak base having $\mathrm pK_\mathrm{a} = 3.2$ and $\mathrm pK_\mathrm{b} =3.4$ respectively, we are get a salt $\ce{AB}$.
When we find $\ce{H+}$ concentration for this salt using the formula,
$$\ce{H+} = \sqrt{\frac{K_\mathrm wK_\mathrm a}{K_\mathrm b}}$$
We get $\ce{H+}$ concentration in the solution to be $10^{-8}$.
In such a scenario, why don't we add $\ce{H+}$ concentration from the auto-ionisation of water to the $\ce{H+}$ concentration?

What happens in salt hydrolysis is enhancement of water ionization with salts. Try asking yourself where do $\ce{H+}$ and $\ce{OH-}$ come from in the solution of $\ce{AB}$, and you'll find that they come from $\ce{H2O}$. Or equivalently, hydrolysis of $\ce{A-}$ can be written as
$$
\ce{A- + H+ <=> HA,H2O <=> H+ + OH-}\stackrel{\text{add}}{\Longrightarrow}\ce{A- + H2O <=> HA + OH-}.
$$
In a word, ionization has been taken into consideration in your calculation. There are multiple equilibria in the system, so you cannot simply add $10^{-7}$.


Water electrolysis - what is happening to an iron anode? So I made an experiment to find a good electrolyte for a water electrolysis.

I tried citric acid, which turned out to not produce any gases at the anode, I tried sodium hydroxide, which turned out to be the very best, and I tried sodium carbonate.

The $\ce{Na2CO3}$ solution was about 2 g/l.

As my electrodes I decided to try some nails, very likely made of iron or some iron alloy.

So I hooked up my power source, just two nine-volt-batteries in series, and started the electrolysis.

The kathode (negative terminal) bubbled vigorously, indicating the production of hydrogen gas.

The anode (positive terminal) however bubbled very little and started decomposing, gray-green flakes started falling off and sinking to the bottom. This decomposition started as soon as the power was supplied.

This struck me, because I was expecting oxygen to be liberated at the anode!

When I took my anode out, it was at first a dirty green or green-gray, but it quickly turned a rusty brown.

So in conclusion, the kathode produced hydrogen with no sign of wear when on the other hand the anode made little gas and decomposed into probably some iron compound, which turned into probably iron oxide (I'm judging by the color here) on prolonged contact with air.

So, what could it be? I thought it might be iron hydroxide, but that'd be brown. This however was green.

On an anode made of iron, some of the removed electrons may be used oxidizing iron metal to ferrous iron(II) ions, or oxidizing iron(II) ions to iron(III), rather than bubbling off oxygen gas.  Pushing more current through the cell is likely to result in gas evolution alongside the iron oxidation, since all three of these processes will run simultaneously.

$$\ce{Fe(metal) -> Fe^{2+}(aq) + 2 e^- }$$
$$\ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^- }$$
$$\ce{2H_2O(liquid) -> O_2 (gas) + 4H^+ (aq) + 4e^- }$$

Ferrous iron(II) ion solutions are a pale green, but rapidly oxidize in air to vivid red-brown ferric iron(III).  Iron oxides and hydroxides form precipitates and slimes in a variety of colors including black (magnetite), red (iron(III) oxide), and yellow (iron(III) hydroxide).


Is it possible to cut yourself on pyrite? I am planning an RPG adventure where the characters might end up getting a large cubic lump of "gold", which in reality, as you might have guessed, is in fact pyrite.

I also wanted to plan for the contingency that the group wouldn't notice that it's not the real deal, so I thought I could maybe let one of the characters cut himself on the crystal. A quick google search yields images such as this:

(Source)

This looks pretty sharp and possibly blood-drawing, but is there a way to know for sure? Any recommendations as to how to handle sharp minerals? Any first-hand experiences?

The Wikipedia page for Mohs scale shows that pyrite has a hardness of 6-7 (according to the page on pyrite, it's 6-6.5) while plate glass has a hardness of 5.5 and a pocketknife has 5.1. I've cut myself on both knives and glass before, so I'd wager that pyrite should be able to slice.

The catch is that I've never cut myself on a cube of glass or a cube of steel before--a 90 degree edge just doesn't generate enough pressure for that. So either your pyrite must have very nice serrations on the edge or it must have an edge sharper than 90 degrees.

Is that possible? I don't know. I suppose at this point you could invoke the leeway granted to you by fiction.


¹H-MRS Detection of Dopamine Why is it difficult to detect dopamine using $\ce{^1H}$ Magnetic Resonance Spectroscopy (MRS) in the brain (in vivo)?

The concentration of dopamine is low
The concentration of other protons is very high
The magnetic field is not homogeneous.
The brain cannot be rapidly rotated the way a laboratory solution state NMR sample is to average magnetic field inhomogenity
Compare to laboratory solution state NMR, the viscosity in the brain would be high causing further line broadening.

If you look at the spectra in 1H NMR Probe for in Situ Monitoring of Dopamine Metabolism and Its Application to Inhibitor Screening J. Am. Chem. Soc., 2012, 134 (30), pp 12398–12401, you'll see that even in tissue samples, there are too many other stronger signals for conventional 1H NMR to be successful, so C-13 labeling is used in combination with 1H NMR.


Is temperature double-counted in the Gibbs free energy equation? Complete amateur here. When looking at one form of the equation deriving the change in Gibbs free energy for a given chemical reaction, the terms change in enthalpy, and the negative product of temperature (in kelvin) and change in entropy:

$$\Delta G = \Delta H - T\,\Delta S$$

This seems to double count how temperature will influence the spontaneity of the reaction if temperature is already considered when deriving entropy, $S$.

Any guidance would be appreciated.

One reason we write equations this way is because all of the parameters on the right-hand-side can be measured or computed independently.

Consider an analogous (but more intuitive) equation, the Stokes-Einstein relation for the diffusion coefficient $D$ of a particle in a liquid:

$$D = \frac{k_\mathrm{B} T}{6\pi\eta r}$$

where $\eta$ is the dynamic viscosity, $r$ is the radius of the particle, and $k_\mathrm{B}$ is Botzmann's constant. Here we see an explicit dependence of $D$ on $T$, but there is also an implicit dependence on $T$ through $\eta$ and possibly also $r$. Is this a problem? No, assuming the equation holds and we can determine the parameters accurately at each $T$ of interest. Whether that is a simple thing is another question.

Also, it is fair to wonder to what extent the parameters are fundamentally independent, or, whether by digging into the gory details of atomic structure of matter, we might not find $r$ and $\eta$ to be connected in some other way. But that is another question.


How many atoms of helium are in the chamber? A $22.7 \,\mathrm{L}$ helium-filled chamber at $298\,\mathrm{K}$ has a tiny hole of area $1.63 \times 10^{-5}\,\mathrm{mm^2}$ punched in it through which is leaking helium at a rate of $2.25 \times 10^{15}$ atoms per second. How many atoms of helium are in the chamber?

I'm personally not too sure how to answer this, I'm thinking that it probably uses the root-mean-square speed or collisions with the wall equation ($Z_\mathrm w$). But I'm sort of lost when it comes to substituting the values and so on. (This is also a question on last years Gas Laws test, and I was planning on solving it for practice questions)

What I tried doing was using the Collisions with the wall equation:
https://en.wikipedia.org/wiki/Kinetic_theory#Collisions_with_container
then solved for $N$, but the answer is supposed to be $N = 9.98 \times 10^{21}$, and the closest I've gotten was getting $3.38 \times 10^{22}$, so I'm sort of stuck on the procedure of solving this question.

You already mentioned the equation for the rate of collisions of gas molecules with a section of wall:

$$Z_\mathrm w=\frac14\frac NV\sqrt{\frac{8RT}{\pi M}}A$$

We know the values of
rate $Z_\mathrm w=2.25\times10^{15}\ \mathrm{s^{-1}}$,
volume $V=22.7\ \mathrm l=0.0227\ \mathrm{m^3}$,
molar gas constant $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$,
temperature $T=298\ \mathrm K$,
molar mass of helium $M=4.003\ \mathrm{g\ mol^{-1}}=0.004003\ \mathrm{kg\ mol^{-1}}$, and
area $A=1.63\times10^{-5}\ \mathrm{mm^2}=1.63\times10^{-11}\ \mathrm{m^2}$.

Thus, the number $N$ of helium atoms is

$$\begin{align}
N&=\frac{4Z_\mathrm wV}{\sqrt{\dfrac{8RT}{\pi M}}A}\\[6pt]
&=\frac{4\times2.25\times10^{15}\ \mathrm{s^{-1}}\times 0.0227\ \mathrm{m^3}}{\sqrt{\dfrac{8\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm K}{\pi \times 0.004003\ \mathrm{kg\ mol^{-1}}}}\times1.63\times10^{-11}\ \mathrm{m^2}}\\[6pt]
&=9.98\times10^{21}
\end{align}$$


Why don't alcohols undergo nucleophilic substitution with ammonia? Alcohols can undergo nucleophilic substitution with $\ce{PCl5, POCl3, HCl}$.  Halogenoalkanes can undergo nucleophilic substitution with $\ce{NH3}$. The $\ce{C-O}$ (in alcohol) and $\ce{C-Cl}$ (in chloroalkanes) bond energies are comparable. Why don't alcohols undergo nucleophilic substitution with $\ce{NH3}$?

To build on ssavec's answer: Nucleophilic substitution reactions require two species: a nucleophile (a Lewis Base) and a substrate with a leaving group. Nucleophiles are often generically represented as $\ce{Nu}$ and leavings groups as $\ce{LG}$.

All leaving groups are not created equal. The leaving group ability of a leaving group is defined as the relative rates of a particular analogous series of substitutions. For example, the relative leaving group ability might be based on the following reaction, with rates determined for various leaving groups:

$$\ce{EtOH + PhCH(LG)CH3 -> PhCH(OEt)CH3 + H-LG}$$

This is an $\mathrm{S_N1}$ substitution, so the first (and rate determining) step of the mechanism is loss of the leaving group (and is independent of the nucleophile):

$$\ce{PhCH(LG)CH3 -> PhCH+CH3 + LG-}$$

The relative rates of this reaction are influenced by the stability of the $\ce{LG-}$ anion (see the Hammond Postulate, which proposes that the transition state of an endothermic process resembles the products).

The stability of the $\ce{LG-}$ anion can be predicted from the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of $\ce{LG-}$, which is controlled by the same ability to stabilize a negative charge.

The conjugate acid of $\ce{Cl-}$ is $\ce{HCl}$, which is a strong acid. Using some Hess's Law trickery, the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCl}$ has been predicted to be around $-8$.

The conjugate acid of $\ce{OH-}$ is $\ce{H2O}$, which has a $\mathrm{p}K_\mathrm{a}$ around $+16$.

$\ce{H2O}$ is a much weaker acid than $\ce{HCl}$, which means that the negative charge on  $\ce{OH-}$ is less stabilized than $\ce{Cl-}$. Thus, $\ce{OH-}$ is a poorer leaving group than $\ce{Cl-}$ (by a lot).

Another good discussion of leaving groups (and the $\mathrm{p}K_\mathrm{a}$ data used above) come from this site.

In your example reaction (ammonia + ethanol), the product of the reaction has a better leaving group ($\ce{NH3}$, conjugate base of $\ce{NH4+}$, which has a $\mathrm{p}K_\mathrm{a}$ of $+9.75$) than the $\ce{OH-}$ leaving group in the reactant, so the reaction will also run in reverse, and the equilibrium will strongly favor the reactants.

$$\ce{CH3CH2OH + NH3 <=> CH3CH2}\color{red}{\ce{NH3+}}\ce{+ OH-}\ \ K_\mathrm{a} \ll 1$$

As for your confusion about bond energies...

Bond dissociation energies are defined for homolytic bond cleavage, in which each atom keeps one electron when the bond breaks (see image).

Substitution reactions involve heterolytic bond cleavage, in which one atom gets both electrons:


Are there any reducing agents for reduction of nitrobenzene to aniline other than Sn/HCl? I was taught that Sn/HCl should be used for reduction of nitrobenzene to aniline. Can I ask if there are any alternative reducing agents (for example, LiAlH4?) to do the same task?

Hydrogenation of aromatic nitro groups (over, for example, $\ce{Pd/C}$) usually results in reduction to the corresponding aniline and is probably easier to carry out and work up as compared to $\ce{Sn/HCl}$.

Wikipedia has a list of several conditions for reduction of $\ce{ArNO2}$ to $\ce{ArNH2}$.  It also states that $\ce{LiAlH4}$ reduces it to the azo compound $\ce{ArN=NAr}$, which is corroborated by March's Advanced Organic Chemistry (7th ed.).

In any case, from a practical perspective, you would probably want to avoid using $\ce{LiAlH4}$ if there's a simpler alternative. It's dangerous and the workup is not fun.


Conversion of CO2 to methane Suppose one could generate an enzyme with arbitrary conditions at the active site. What conditions would then be necessary to cause carbon from dissolved $\ce{CO2}$ in low $pH$ water to combine with the free hydrogen and make methane?

Let's take a close look at what the reaction entails. You want to convert $\ce{CO_2}$ to $\ce{CH_4}$ using $\ce{H^+}$. It is possible that the hydrogen ion will co-ordinate with $\ce{CO_2}$ to produce $\ce{[CO_2H]^+}$.

Unfortunately it is very unlikely to have a second hydrogen ion attaching to the same oxygen to produce $\ce{[CO_2H_2]^{2+}}$. Now for the impossible step of water leaving and producing $\ce{[CO]^{2+}}$. This step would require a large amount of energy and $\ce{[CO]^{2+}}$ would be extremely reactive. Since it is highly charged, you would not have any hydrogen ions approaching the moiety.

At this point, any further progress towards $\ce{CH_4}$ is fantasy. To stabilize the reaction, you need to add electrons to the carbon (which you don't have a source).


Determining equilibrium concentrations from initial conditions and equilibrium constant The equilibrium constant for the following reaction at $\pu{600^{\circ}C}$ is 4.0.  Initially, two moles of $\ce{CO}$ and one mole of $\ce{H2O}$ were mixed in a $\pu{1.0 L}$ container.  Determine the concentration of all species at equilibrium.

Attempt #1:

$$\ce{CO (g) + H2O (g) -> CO2 (g) + H2 (g)}$$

\begin{array}{|r|c|c|c|}
\hline
\text{Initial}~(M)     & 2.0      & 1.0     & 0  & 0 \\ \hline
\text{Change}~(M)      &  -x      & -x      & +x & +x \\ \hline
\text{Equilibrium}~(M) & 2.0 – x  & 1.0 - x \\ \hline
\end{array}

  Plug $x$ into the equilibrium expression and solve for $x$. $4.0  = 2[x]$, so $x = 0.85 [1.0 – x][2.0 – x]$.

  Determine concentrations: the equilibrium values become 2.0 – 0.85, 1.0 - 0.85, 0.85, and 0.85, giving 1.2, 0.1, 0.85, and 0.85.

Where does the 0.85 come from? Could it be cross multiplied some way or is there another way this is done?

Furthermore, I want to make sure that I've done the 3.00 correctly for $\ce{Fe(SCN)2+}$.

  Consider the reaction represented by the equation: $$\ce{Fe^3+ (aq) + SCN- (aq) -> Fe(SCN)2+ (aq)}$$

Attempt #2:

\begin{array}{ccc}
\text{Initial}     & \pu{6.00 M}\ \ce{Fe^{3+} (aq)} & \pu{6.00 M}\ \ce{SCN^{−} (aq)} \\
\text{Equilibrium} & \pu{? M}\ \ce{FeSCN2^+ (aq)}, \ K = 0.33
\end{array}

\begin{align}
\frac{1}{3}x^2 - \frac{124}{25}x + \frac{297}{25} &= 0 &
 x_{(1;2)} &= \frac{\color{red}{\pmb{-}}(-\frac{124}{25})\pm\sqrt{\left(\frac{124}{25}\right)^2 - 4\cdot\frac13\frac{297}{25}}}{\frac{2}{3}} \\
&& x_{(1;2)} &= \frac{124/25\pm\sqrt{\frac{15376}{(625)}-\frac{15601}{1000}}}{\frac{2}{3}} \\
&&           &= \left(124/25\pm{3}\right)\cdot\frac{3}{2} \\
&& x_{(1;2)} &= 12;3
\end{align}

The text

  4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x]

appears to be an error. The proper equation is:

$$4  = \frac{x^2}{(2-x)(1-x)}$$

The 0.85 is one of two roots of the quadratic equation you'll solve, using

$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$

after you simplify the first expression, above.

As for the rest of your question, please expand: What does

  the 3.00

mean?


Which of the following is most likely to undergo hydride shift? For the following question the answer is given as option (a) . However both,option (a) and (c) has alpha H atoms and both will form 3° carbocations after rearrangement. So what makes (a) more favorable than (c)?

Besides Avinsh's comment, there is another factor.  When you have a carbocation center next to a cyclopropane ring or cyclobutane ring, the "bent bonds" of the ring partially overlap the "vacant" orbital on the cation center.  To the extent that this occurs, the formally primary ion (c) is something like a more stable tertiary one.  Ion (a) does not have this factor because the cation center is already inside the ring instead of next to it.


How to find the ionization constant of an unknown weak acid? A $\pu{0.45 M}$ solution of a weak acid, $\ce{HX}$, has a $\pu{pH}$ of $4.5$.  What is the ionization constant, $K_a$, of the acid?

$\ce{[HX]}$ is already given as $\pu{0.45 M}$.
$\ce{[H+]}$ is given by $10^{-4.5}$.

Since $\ce{pH}$ is $4.5$, $\ce{pOH} = 14 - 4.5 = 9.5$. Then we can get $\ce{[OH^-]}$ by computing $10^{-9.5}$.

Then computing $K_a$, I get
$$\frac{10^{-4.5} \cdot 10^{-9.5}}{0.45} = 2.22 \times 10^{-14}.$$
But that is apparently not the right answer. What did I do wrong?

Look at the answer given here and modify it:
How to set up equation for buffer reaction?

In your case the concentration of base $c_B \approx 0$ and $[\ce{OH^-}] = K_\mathrm{w}/[\ce{H^+}] \approx 0$, so that the equation reduces to $$K_\mathrm{a} = \frac{[\ce{H^+}]^2}{c_\mathrm{a}-[\ce{H^+}]} \approx \frac{[\ce{H^+}]^2}{c_\mathrm{a}},$$
as $c_\mathrm{a}$ is so large.


How is a NMR spectrum obtained? I am reading about NMR, and from what I'm understanding it should give information on the transition energies in the spectrum of the nuclear spin in a magnetic field.

What I don't understand is how this information is accessed during the experiment.

The NMR measurement is usually described as: a magnetic field $H_0$ polarizes the sample along a certain direction; then a short pulse of an auxiliary magnetic field alters the direction of the magnetization, which then starts to precess around $H_0$. The changing magnetization induces a current the coils of the NMR machine which is measured and gives the precession frequency. The magnetization gradually aligns with $H_0$, leading to a diminishing current intensity. The decrease in intensity gives the nuclear relaxation rate.

A first question is: is the above description of the measurement correct?

A second question is: How do we extract the information on the energy levels $E_m$, or their separation, from the precession frequency? If not from the precession, how is the energy spectrum obtained?

The description is simplified, but actually fairly accurate. You can find all the details in a specialised NMR book (other general texts usually do not go into much detail).

Quantum mechanically, you can show that the precession frequency (Larmor frequency), denoted $\omega_0$, is related to the energy difference between the up and down spin states of a spin-1/2 nucleus. The energy levels are

$$\begin{align}
E_\alpha &= +\frac{1}{2}\hbar\omega_0 \\
E_\beta &= -\frac{1}{2}\hbar\omega_0 \\
\end{align}$$

and so the transition occurs at the frequency $|E_\beta - E_\alpha|/\hbar = \omega_0$, so if we can identify how much precession is happening at the frequency $\omega_0$, that tells us how strong the corresponding signal should be in the spectrum.

The initial data that you get is in the form of a free induction decay (FID), which plots the amount of detected magnetisation against time. The process of extracting the frequencies from the raw time-domain data is accomplished by a Fourier transform.

Feel free to ask if you would like any part of this to be elaborated. I intentionally did not go into great detail, partly because that is the role of a textbook, but also partly because it can get pretty complicated very quickly.

(There is also a complicated sign convention to do with the definition of $\omega_0$, and different books define it differently. Occasionally the definition even depends on the sign of the gyromagnetic ratio $\gamma$. I would advise to not worry too much about plus/minus signs.)


How are polarizabilities measured experimentally? Most common spectroscopies that produce either a full spectrum, a tensor, or a scalar value have a specific instrument associated with them that is relatively self-contained and not custom. For example, (linear) IR spectra come from benchtop FT-IR spectrometers, NMR chemical shifts use a FT-NMR spectrometer (which is capable of many experiments), and so on. One spectroscopic property that serves as a benchmark for electronic structure theory and is important for designing optical materials is the (electric dipole) polarizability, which "is defined as the ratio of the induced dipole moment $\mathbf{p}$ of (a molecule) to the electric field $\mathbf{E}$ that produces this dipole moment":

$$
\mathbf{p} = \alpha \cdot \mathbf{E}.
$$

Physically, this corresponds to how easy it is to shift a system's charge density or electron cloud around. More explicitly, it is a symmetric rank-2 tensor:

$$
  \begin{bmatrix}
    p_{1} \\ p_{2} \\ p_{3}
  \end{bmatrix}
  =
  \begin{bmatrix}
    \alpha_{11} & \alpha_{12} & \alpha_{13} \\
    \alpha_{21} & \alpha_{22} & \alpha_{23} \\
    \alpha_{31} & \alpha_{32} & \alpha_{33}
  \end{bmatrix}
  \begin{bmatrix}
    E_{1} \\ E_{2} \\ E_{3}
  \end{bmatrix},
$$

as the applied electric field may have 3 independent Cartesian coordinates, and the system being measured may have some anisotropic response to this field. In some cases, calculating the tensor may be time-consuming, but formulation is well-understood and there are standardized methods for doing so. However, it is not obvious to me how it is measured experimentally. There is the Lorentz-Lorenz equation (rearranged),

$$
\alpha = \frac{n^2 - 1}{n^2 + 2} \frac{3}{4\pi N},
$$

which relates the refractive index $n$ of a system or medium to its polarizability. There is also the Clausius-Mossoti relation, which relates the dielectric constant $\epsilon$ to the polarizability, as the refractive index and (complex) dielectric constant are related:

$$
n \propto \sqrt{\epsilon}.
$$

This implies to me that if you can measure the refractive index with a refractometer, then that is the preferred way of measuring the polarizability.

Is the refractive index the most common method of experimentally measuring the polarizability? Can it be measured directly, or must be through a relationship such as those above?
Does this relationship hold for oriented (anisotropic) polarizabilities, or only averaged (isotropic) ones?
The refractive index and the polarizability are frequency-dependent properties. How is the static (frequency-independent) polarizability measured? Or, is it even possible without extrapolation from multiple frequency-dependent data points?

The polarizability of molecules determines how a molecule interacts with other molecules (intermolecular interactions and chemical reactions), how it interacts with electromagnetic radiation (optical properties), and how it interacts with electrical fields (electrical properties). Thus, there are multiple ways to compare the polarizability of two molecules or samples, and to quantify it.

[OP] Is the refractive index the most common method of experimentally measuring the polarizability?

That is hard to say without a systematic search of all the polarizability data. A starting point for such a search might be the polarizability data provided by NIST, which includes references that describe methods used. In general, there are electrical techniques (static electrical field or slowly oscillating) and optical techniques (electromagnetic field). Light scattering depends on the electric polarizability of the medium, so any technique that is based on light scattering (Raman spectroscopy, X-ray diffraction, refraction) in principle could be used to determine polarizability.

[OP] Can it be measured directly, or must be through a relationship such as those above?

The refractive index can be measured directly. The polarizability, while it has direct impact on molecular properties like the strength of dispersion forces, is determined from other measurements.

[OP] Does this relationship hold for oriented (anisotropic) polarizabilities, or only averaged (isotropic) ones?

As written in the question, the Lorentz-Lorenz equation is for the isotropic case. The literature linked to below also shows the more general equation.

[OP] The refractive index and the polarizability are frequency-dependent properties. How is the static (frequency-independent) polarizability measured? Or, is it even possible without extrapolation from multiple frequency-dependent data points?

The refractive index is frequency-dependent while the polarizability is not. You measure the refractive index at multiple wavelengths and extrapolate to infinitive wavelength. Alternatively, you can do a non-linear fit to the respective theoretical dispersion function.
Measurements of molecular polarizability are done at low concentrations in the gas phase, and yield the isotropic polarizability. Here is one paper that compares the experimental data to computational results for the iodine molecule. More complicated experiments involving polarized light are required to figure out the anisotropy, if any. For example, this paper from the same two labs compares the anisotropy of the polarizability of butadiene to computational results. An example of a quantumchemical computation of polarizability is here, for the hydrogen atom. In the solid phase with oriented molecules, the anisotropy of polarization is easier to measure (for example with cross-polarized filters as in the example below showing a plastic lens under mechanical stress):

Source: https://commons.wikimedia.org/wiki/File:Stress_analysis_of_plastic_lens.png


Order of rate of acid catalysed dehydration of alcohols The relative rate of acid catalysed dehydration of following alcohols would be:

According to me, we have to arrange them in order of their carbocation stability.
Q will be the most stable as it is stabilized by Resonance and Hyperconjugation. P should come next. After that I think it should be S>R as in case of R there is no hyperconjugation and hyperconjugation is a “stronger” effect than the Inductive effect.
But in my Book R>S is given. Is Inductive effect stronger here? Or is there some other reason to it?

Your basic assumptions are correct.
It can be observed as such:

In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest.
Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also greater in P than in R. Hence, it forms carbocation faster.
Both R and S form primary carbocations. Inductive effect is more prevelent in R than in S, while S has 3 hyperconjugating structures.
But R offers capacity to rearrange and stablise the carbocation, by methyl migration to produce highly stable tertiary carbocation, also stablised by resonance over the adjoining phenyl ring.
But, S also forms a carbocation which is primary, but offers no stablising effect other than 2 +1 or 3 hyperconjugating structures.

Similarly, this rearrangement may also be attributed to the comparison of P and R, where hydride shift and methyl shift occur. The migration amplitude of hydride, being lower, leads to rearrangement of positive charge quickly and hence is kinetically also favoured.
Thus finally the order may be:
Q>P>R>S


Mass number, (relative) atomic mass, average mass What is the difference between mass number, atomic mass and average atomic mass?
I know the mass number is the amount of protons + the amount of neutrons. The average mass is the weighed average of the isotopes that occur in nature. But then what is the (relative) atomic mass?

It’s pretty simple:

The mass number is the integer you get if you count (and add up) the neutrons and the protons of a given element. Thus, a hydrogen atom of the $\ce{^1_1H}$ isotope has a mass number of $1$ (only proton), $\ce{^12_6C}$ has $12$ ($6$ protons, $6$ neutrons) and $\ce{^81_35Br}$ has $81$ (of which $35$ are protons, the remaining $46$ neutrons).
The atomic mass is what these atoms actually weigh in atomic mass units. For reasons that boil down to $E = mc^2$ (or so I believe) and the nonzero mass of an electron, this is not an integer except for one exception:

$\ce{^1H}$’s atomic mass is $1.007825032 1(4)$
$\ce{^12C}$’s atomic mass is exactly $12$. This is because $1~\mathrm{u}$ was defined as exactly $1/12$th of the mass of a carbon-12 atom.
$\ce{^81Br}$’s atomic mass is $80.916289(6)$

The average mass takes into account an elements different isotopes and their natural abundance and calculates an overall average. Thus, this is no longer defined on an isotopal basis but on an elemental one. The average masses of the elements discussed above are:

$\ce{H}$ has $1.00794 (7)$ (this is larger than the atomic mass of $\ce{^1H}$ due to the heavier isotopes.)
$\ce{C}$ has $12.0107 (8)$
$\ce{Br}$ has $79.904 (1)$ (this is lower than the atomic mass of $\ce{^81Br}$, because about half of the naturally occuring bromine atoms are $\ce{^79Br}$.)


What happens when oxidizing Ni(OH)2 with a strong base? In an aqueous solution, Nickel oxyhydroxide is oxidised with a 1:1 ratio of aqueous Sodium hydroxide. My goal is to produce Nickel oxyhydroxide ($\ce{NiOOH}$).

Strong Base: $\ce{NaOH + H2O \leftrightharpoons Na+ + OH-}$
Add that to some stirred $\ce{Ni(OH)2}$:

$$\ce{ Ni(OH)2 + OH- \leftrightharpoons NiOOH + H2O + e-}$$

This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make $\ce{Na (s)}$?:

$$\ce{ Na+ + e- -> Na (s) }$$

But then, $$\ce{Na(s) + H2O \leftrightharpoons NaOH }$$

That just brings everything back to the beginning.

I guess the real question is: will the initial oxidation of the $\ce{Ni(OH)2}$  happen? If not, how can I force it? I initially thought of just grounding the solution, as odd as that sounds.

Thanks for your help!

P.S. I'm a 12th grade high school student.

This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make
  $\ce{Na (s)}$ ?: $\ce{Na+ + e- -> Na (s)}$

This reaction is, you might be knowing, the reverse of one of "the best" known favourable reaction, as the data suggests:
$$\ce{Na->Na+ +e-}\tag{$E^\circ=2.71\;V$}\\
2.71\;V\equiv-261.475\;\text{kJ/mol}\equiv1.0938\times10^{50}(\text{rate})$$
Just imagine the reverse reaction.

  But then,
  $\ce{Na(s) + H2O -> NaOH }$

This then would never actually then happen. Intersetingly see this


When does hydrogen peroxide act as a reducing and when as an oxidising agent? My chemistry books details out both, the oxidizing and reducing properties of hydrogen peroxide, $\ce{H2O2}$.

  Its reducing properties:

  It reduces $\ce{Cl}$ to $\ce{HCl}$
  It reduces $\ce{Ag2O}$ to $\ce{Ag}$
  It reduces $\ce{K3[Fe(CN)6]}$ (potassium ferricyanide) to $\ce{K4[Fe(CN)6]}$ (potassium ferrocyanide)
  Etc.

  Half equations for reduction:

  In acidic medium: $\ce{H2O2 -> 2H+ + O2 + 2e-}$
  In basic medium: $\ce{H2O2 + 2OH- -> 2H2O + O2 + 2e-}$

  Its oxidizing properties:

  It oxidizes sulphites to sulphates
  It oxidizes arsenites to arsenates
  It oxidizes nitrites to nitrates
  It oxidizes potassium ferricyanide to potassium ferrocyanide
  Etc.

  Half equations for oxidation:

  In acidic medium: $\ce{H2O2 + 2H+ + 2e- -> 2H2O}$
  In basic medium: $\ce{H2O2 + 2e- -> 2OH-}$

I was unable to find any pattern in all of the reactions I studied above (except for potassium ferrocyanide, for which the medium triggers the electron addition/removal) to recognize when hydrogen peroxide behaves as an oxidising agent and when it behaves as a reducing agent.

Let us say that $\ce{H2O2}$ reacts with some ion $\ce{Y+}$. If $+1$ is yttrium's highest oxidation state then it will have no other option but to to get reduced to one of its lower oxidation states, and hydrogen peroxide acts as reducing agent. If this is not the case it will get oxidized to some higher oxidation state, and hydrogen peroxide will act as oxidizing agent. This is because $\ce{H2O2}$ is both an oxidizing and reducing agent.


Which will give faster SN2 reaction In $\ce{H2C=CH-Br}$ and $\ce{H3C-CH2-Br}$, which will react faster towards a $\mathrm{S_N2}$ reaction?

According to me, as double bond exhibit −I effect, hence the 1st should do a faster reaction. Am I right, or is there any other reason?

There is no sense of comparing the reaction rate when one of the compound doesn't show SN2 mechanism. And, that compound is $\ce{CH2=CHBr}$, due to the fact that it is a vinyl halide.

$\ce{CH2=CHBr}$ will hinder the approach of nucleophile due to the presence of pi-electron cloud around double bond.

And, therefore the obvious answer is $\ce{CH3-CH2Br}$


Coming up with a Mathematical Model to Predict how long a redox reaction can last The description of my problem is as follows:

I will like to find out how can I obtain a mathematical equation which could give me a surface plot of the current output as a function of both concentration and time from an electrochemical cell which uses the following redox equation:

O2 + 2Pb -> 2PbO

I have the following information:

Mass of Pb - 10g
A fixed concentration of oxygen gas - 20.9% over volume of air in room

Assumptions made:

Unlimited supply of oxygen
The gas is supplied at constant room temperature and pressure (25 degree Celcius and 1 atm)

Tentatively, my plan was to use Farday's Law of Electrolysis (https://en.wikipedia.org/wiki/Faraday%27s_laws_of_electrolysis) to calculate the current which flows as a result of the transfer of electrons from the redox reaction. My hypothesis is:

The magnitude of the current generated by the following reaction will be proportional to the rate of oxygen consumption.
The magnitude of the current generated will decrease over time as the rate of reaction decrease (from the decrease in concentration of Pb).

My problem is that the plot I obtained shows me that the current vs time relationship is a reciprocal one under fixed concentration of oxygen gas.

This does not agree with the experimental results I obtained which shows that current output is steady and starts decreasing after a long period of time (a few months at least). Can someone help point out my the flaw in my argument? What should I do instead to obtain the plot I described?

In addition to the redox reactions the transport of material towards the electrodes also needs to be considered. This involves using Fick's equations for diffusion to calculate the concentration of species at distances away from the electrodes as reaction starts and reaches constant conditions. The steady state equations are of the form

$$ D_{ox} \frac{\partial C_{ox}}{\partial x}= k_cC_{ox}-k_aC_{rd} \\
   D_{rd} \frac{\partial C_{rd}}{\partial x}= -k_cC_{ox}+k_aC_{rd}$$

These lead to the Cottrell equation for the limiting flux (current density)  $j=-nFC^o_{ox}\sqrt{D_{ox}/\pi t}$. To get what you want you will probably need to consult some specialist text books on electrochemistry.


Why is ice less dense than water? I know the traditional explanation, which says that ice has large spaces between $\ce{H2O}$ molecules because hydrogen-bonding gives it an open structure. But what does the open structure have to do with hydrogen-bonding? Why isn't a similar phenomenon observed in other species which exhibit hydrogen-bonding, like $\ce{HF}$ or $\ce{NH3}$?

The following is an image of the hexagonal crystaline form of ordinary ice (Ice I$_h$) taken from S.S. Zumdahl, Chemistry, 3rd ed., copyright © 1993 by D.C. Heath and Company:

Note that the dashed lines represent hydrogen bonds. Liquid water actually has a similar "open" structure also due to hydrogen bonding. But in the case of liquid water, the hydrogen bonds are not rigid and semi-permanent as in ice. So imagine that in the image above, the hydrogen bonding network collapses. This is what happens when enough thermal energy is present to break the rigid hydrogen bonds resulting in melting. Clearly, once this crystaline structure is no longer forced into place by the rigid hydrogen bonding in ice, it can collapse into itself, resulting a greater density of water molecules.

Thus the liquid form of water, although engaged in transient hydrogen bonding, is not as open and expanded as when held into it's solid form by the rigid, semi-permanent hydrogen bonding.


Must a sample of a deuterated compound be isotopically pure for kinetic isotope experiments? I'm trying to do a project on the primary kinetic isotope effect on the dehydroxylation of $\alpha-FeOOH$, goethite. The goethite consists of edge-sharing double octahedra of 3 oxyanions and 3 hydroxyanide. The double octahedra are joined at corners by the oxyanion or hydroxide, leaving empty channels across which hyroxides and oxyanions form hydrogen bonds.

The dehydroxylation mechanism is believed to consist of proton diffusion from the bulk of the material to the surface (likely facilitated by a series of protonations down the hydrogen-bonding channels), then binding of hydrogens to surface hydroxyl group, forming water desorbing. The proton diffusion is believed to be the rate-determining step.

I thought I could lend more evidence to whether this is the case by performing the dehydroxylation on a deuterated compound and a compound of natural isotopic abundance. However, I cannot find the deuterated compound for sale and will have to synthesize it. The most facile synthetic method given time constraints involves potassium hydroxide. We could use deuterated hydroxide, but that would be more expensive. Grey et. al.** were able to do this synthesis with the undeuterated hydroxide, but they didn't report an H/D ratio.

All the papers I've come across on kinetic isotope effect experiments work under the assumption that their deuterated compounds are isotopically pure. It may just be a matter of convention, but would I actually need to do this experiment on an isotopically pure sample? Could I use a sample with, say, an isotope ratio halfway between pure deuterated and natural abundance? Would the calculated kinetic isotope effect (defined as hydrogen rate constant/deuterium rate constant) then be twice what I measure with this partially-enriched sample?

*Surface and Bulk Reactivity of Iron Oxyhydroxides - A Molecular Perspective
Xiaowie Song
Umea University

**2H MAS NMR Studies of Deuterated Goethite (α-FeOOD)
Kathryn E. Cole,†,§, Younkee Paik,†,§, Richard J. Reeder,‡,§, Martin Schoonen,‡,§ and, and Clare P. Grey*,†,§
The Journal of Physical Chemistry B 2004 108 (22), 6938-6940
DOI: 10.1021/jp0486090

There are two approaches to measuring kinetic isotope effects.

One is to have two completely separate, isotopically pure systems, one heavy, and one light. In each system you can measure the rate of the reaction however you want.  You don't have to be able to measure the rate in an "isotopically resolved" way, just measure the bulk rate.  Each of your systems is isotopically pure, so the rate of each system is 100% reflective of the isotope composition of that system.

The other way is to have a single system with variable/unknown isotopic composition.  You measure the rate of your reaction, but you have to be able to measure (e.g.) the rate of depletion of each substrate isotopomer separately.  (Or the rate of formation of each product isotopomer, but that sounds very difficult in your system.)  For example, if you had a mixture of $\ce{CH4}$ and $\ce{CD4}$ and you were interested in the kinetic isotope effect of some kind of combustion reaction of those gases with oxygen, you could measure the ratio of $\ce{CD4}$ to $\ce{CH4}$ over time, e.g. using mass spectrometry.  It would increase as the $\ce{CH4}$ is preferentially depleted by the combustion reaction.  You could also measure the rates of formation of $\ce{H2O}$ and $\ce{D2O}$.  The rate of formation of $\ce{H2O}$ would be much higher than that of $\ce{D2O}$.  (The exact numerical values here would depend strongly on the nature of your combustion system, e.g. temperature, catalysts used, etc.)

So iff you can measure $\ce{FeOOH}$ and $\ce{FeOOD}$ separately in the same system, then you don't need an isotopically pure material.  I'm not an expert in these types of measurements, but it sounds very hard to me.  Probably if you don't already know how to do it, it will be cheaper just to buy $\ce{KOD}$ and get on with the synthesis of an isotopically pure material so you can use the first approach.


Grignard reaction on electron-poor benzyl bromides I am trying to prepare a Grignard reagent from 4-nitrobenzyl bromide

but the reaction just doesn't work. My conditions are:

10 mL of THF (anhydrous);
7.04 mmol of 4-nitrobenzyl bromide;
7.04 mmol of Mg;
Argon atmosphere.

Water is not a problem since I left all my glassware in the oven overnight and flame-dried the round bottom flask + stir bar before adding Mg and the bromide. I also did everything under argon. I know the reaction is not proceeding because virtually all of the Mg turnings I added are still there and the solution is still clear. I tried adding two crystals of iodine but it didn't help either.

Is there something about this particular bromide that prevents/slows down the formation of the Grignard? I was wondering if the fact that the nitro group is electron withdrawing could make it more difficult to form the carbanion-like structure of the Grignard.

This question has aged a bit with no other answers, so here I go.

The same nitro group that can oxidize Grignard reagents could be oxidizing the magnesium too, thus keeping it passivated.  What you need is not so much a milder EWG but one with less oxidizing potential.  Try your reaction with nonoxidizing EWGs like acetyl (watch for side reactions), a $\ce{CF3}$ group, or a chlorine atom attached to the aryl group.


Composites of Immiscible Solids - Aluminum and Lead Example The $\ce{Al}$-$\ce{Pb}$ (aluminium-lead) phase diagram clearly depicts that when cooled from the liquid state, aluminium and lead are not miscible in the solid state, i.e. the crystal structure of only one of the components prevails while the other remains a liquid or does not interact in any way with the other in its own solidification process.

What does this have to do - if anything - with the fact that a solid combination of $\ce{Al}$ and $\ce{Pb}$ is not classified as a composite?

You are correct that solid Al (fcc) and Pb (fcc) are (surprisingly for both being fcc) not miscible in each other (well, there is always some miscibility as driven by entropy). Heck, there is even an miscibility gap in the liquid phase as well. Al and Pb really don't like each other.

So, as you cool the liquid from above 1418C (the top of the liquid miscibility gap), you will phase separate in the liquid phase. At the aluminum melting point, the aluminum will solidify, and the Pb will remain liquid. When the Pb melting temperature is reached, the Pb will solidify. The exact specifics of the solidification will depend on how it is done (the kinetics), not the thermodynamics. This means you can't predict the final micro-structure without  details of the process.

The final result will be solid Al and Pb, likely intermixed. It is not an alloy, but a two-phase mixture. I would not call it a composite as I would generally reserve that name for actual composite materials.


Can you predict the resistance of conjugated polymers as a function of degree of polymerization (DP)? Electrical properties of very simple systems can be calculated using NEGF. Is there a way to calculate the resistance (and other electrical properties) of a conjugated polymer, as a function of degree of polymerization (DP)?

Short answer: No.

First off, the degree of polymerization does not directly relate to the charge transport properties of conducting polymers (more below). You can find highly conductive oligomers.

Secondly, in my opinion, NEGF is not a suitable method to understand the charge transport in polymers. It works for nanoscale devices and molecular wires, but most polymer devices are on the ~100 nm to micron scale.

That's not to say that different oligomer and polymer lengths don't have different transport properties. Absolutely, different molecular weight polymers have different charge mobility, etc. The effects are usually due to a number of hard to characterize factors:

Concentration of defects and traps
Morphology of the film (e.g., higher MW might yield smoother or more ordered films)
Polydispersity (e.g., broader MW distribution is often problematic)
Effective conjugation length
Packing

etc.

The electronic structure of oligomers and conjugated polymers can be simulated fairly well using first-principals methods, including DFT. This can often predict components like the energy levels, band gaps, band dispersion, etc. for crystals and parameters for amorphous films such as the Marcus reorganization energy which is often related to the charge mobility in many of these materials.

What you usually find is that as a function of chain length, charges become more delocalized and the reorganization energy drops (increasing mobility). In reality, this trend should saturate, although most DFT methods incorrectly delocalize charges completely.

Also, most devices with conjugated polymers are amorphous or polycrystalline, making device performance predictions difficult. Transport is not usually band-like, but some form of variable-range hopping between chains or sites within a chain.


Are resonating structures real or not? We say that hybrid structure is real. It has distinct properties as compared to its Resonating Structure. Is it because that all the experiments we carry out are on a sample, lots of molecules are present. Are they hybrid or are they lots of Resonating Structures in an equilibrium state that they collectively show properties of the hybrid. Also, some structures are said to contribute more to a hybrid structure. What does that mean?

Also, in charged molecules like carbonate ion, the charge is fractional. Won't that mean that electrons get "divided"? I know that partial charges can arise in covalent bonds due to a difference in electronegativity, but this seems different from that.

Resonance is a concept that had to be invented by man because the previous concepts to describe reality were insufficient. Resonance is sometimes wrongly taught as swinging back and forth between the different resonance structures, but that is an idea that people thought up and that has little to do with the actual reality.

It is very hard if not impossible for humans to actually understand the reality of chemistry at a molecular or atomic level. At these scales, quantum mechanical descriptions must be considered and any observed object can be both a wave and a particle — I think, at one point interference patterns could be observed with Buckyballs, confirming the wave-particle dualism up to $\ce{C60}$; just to give you an idea. Considering this, it is very understandable that concepts are hard to grasp and frequently misinterpreted, even by nobel laureates.

The key to correctly understanding resonance is that there are not many resonance structures that co-exist, but that overlaying all the resonance structures together into one gives us a better clue of the chemical reality at atomic scales. Taking the benzene molecule, we do not have double bonds on one side half the time or for half the molecules we’re observing. We have six identical bonds with a bond order of $1.5$. Localised double bonds, if present at any time, should react with halogens in an electrophilic addition reaction, yet that never happens with benzene. (Other experimental proof can also be used.) It is the fact that writing this $D_\mathrm{6h}$ symmetric structure on paper with our Lewis model formalisms is impossible that inspired chemists to think of resonance structures to better explain what is going on.

You also mention fractional charges and the possibility that ‘electrons get “divided”.’ Rest assured that fractional charges are possible even without electrons getting divided. A computationally well-studied system is the allyl anion[1]. Two canonic resonance structures as shown below are typically drawn.

$$\ce{H2C=CH-\overset{-}{C}H2 <-> H2\overset{-}{C}-CH=CH2}\tag{1}$$

Each of the terminal carbon atoms has an effective charge of $-\frac12$. How is this possible, you may ask? Molecular orbital theory does the best job of explaining it. The key orbital is the π HOMO which looks as shown in the image.

Notice that it is a wave function; it has four lobes, two centred on each of the terminal carbon atoms. If you try to understand this MO with conventional mechanics, you may be inclined to propose some kind of tunneling mechanism for the electrons to get from one lobe to the other. That would be wrong. They are not in one lobe and then go somewhere else, the two electrons are in all four lobes simultaneously. So we can have fractional charges (happens if you add up the electrons on each carbon; proof left to the readed) even without having to split electrons simply because of the wave-like nature of electrons.

Many fascinating observations await you as you enter quantum chemistry and

  [attempt to] understand whatever
  Binds the world’s innermost core together[2]

Be prepared to forget what you think you may know if you decide to dive into this world.

Note and Reference:

[1]: Taken from Professor H. Zipse’s webpage (LMU Munich). For some reason, the allyl anion system shows N/A. Instead, I linked to the allyl cation which is identical except for one orbital without electrons.

[2]: J. W. von Goethe: Faust. Der Tragödie erster Teil. 1808.


$\beta$-mercaptoethanol molarity I'm writing up a practical and $\beta$-mercaptoethanol was used, with no concentration given. I've found I don't need the BME molarity since I'm giving percentage (v/v) concentration, however I'm slightly confused as to why there are different values for its concentration.

Sigma-Aldrich provide BME concentration 14.3 M (pure liquid), yet the openwetware.org protocol for SDS sample buffers gives a molarity of 14.7 M. I can't see precisely how this is derived from its density, or why there'd be different molarities floating around.

Not that it matters for my writing here, but since I don't know the molarity I'm just a bit puzzled now as to what it may have actually been, and would rather understand now than have to work it out further down the line. Any answers greatly appreciated!

From the Aldrich data:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}}\ \text{at}\ 25\ \mathrm{^\circ C}$

Converted to a more convenient unit:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot \frac {1000\ \mathrm{mL}}{1\ \mathrm{L}} = 1114\ \frac{\mathrm{g}}{\mathrm{L}}$

Given the molar mass (also from Aldrich) of:
 $M = 78.13\ \frac{\mathrm{g}}{\mathrm{mol}}$

We learn that one liter has:
 $1114\ \frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{78.13 \ \mathrm{g}} = \frac{1114}{78.13}\ \frac{\mathrm{mol}}{\mathrm{L}} = 14.26\ \frac{\mathrm{mol}}{\mathrm{L}}$

I can't speak to why your other source has a different number.


heating methane in a vacuum chamber Using radio waves to heat, and electromagnetic fields to contain....could methane be heated to much higher temperatures than via simple oxygen/combustion, thus producing a much higher specific impulse for, say, a rocket engine?

Not talking plasma drive temps here (millions of degrees), but several thousand degrees above normal combustion temperature.

The problem is where the energy for heating comes from not the gas or the mechanism of heating

Rocket engines often use burning fuels to produce thrust because chemicals are a very good store of energy and burning often produces hot gaseous products.

Yes, you can produce hot gases by other means (there are electrically driven plasma rockets and you could, maybe, heat gases with microwaves or similar radiation). But the problem you have is where to get the energy for that process? A self-contained rocket has to have an energy source and that source has to be compact and a slight as possible (a long electrical cord to an electric socket isn't practical in orbit or in space). Just about the best way to store energy is various combinations of reactive chemicals (eg oxygen and kerosine) and the easiest way to turn that into thrust is burning. The only other approach know to work well is to put a nuclear reactor on the rocket as an energy source but that has obvious disadvantages for other reasons.

Rocket scientists have spent a lot of heroic effort to optimise mixtures of (often dangerous) chemicals to achieve better thrust. For a good summary read John D Clark's Ignition where he traces the history of trying to make use of extreme oxidising agents like chlorine trifluoride to get better performance.


Finding the pOH of a buffer from the scratch without the Henderson Hasselbalch equation I found this question on the internet that said:

Find the pOH of the solution obtained by mixing $\pu{0.1 mol}$ of $\mathrm{NH_4 OH}$ $(K_\mathrm b = 10^{-5})$ and $\pu{0.1 mol}$ of $\ce{(NH4)2SO4}$ in a $\pu{500 mL}$ solution.

What my approach was to do it without applying the Henderson Hasselbalch equation and doing it from the scratch using the "common ion" effect. Here is how I did it:
$\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
$\alpha$= The dissociation constant per mole
$$\ce{NH4OH <<=> NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH4)2SO4 -> 2 NH4+ + SO4^2-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{dissociation} &0 &0.2 & 0.1\\
\hline
\end{array}
Now, we can say that
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
$$10^{-5}= \frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1}{0.5}}$$
As, $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$.
Doing the further calculations, we get:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
$$\implies \boxed{\ce{[OH-]} = 0.1 \alpha = \frac{5}{2} \times 10^{-6} \frac{\pu{mol}}{\pu{500 mL}}}$$
Now, $$\mathrm p\ce{OH}=\log[\ce{OH-}] = -[\log(5)-\log(2)-6] = \boxed{5.6}$$
But if we directly apply the Henderson Hasselbalch equation for the buffer, we get:
$$\mathrm p\ce{OH} = \mathrm pK_\mathrm b + \log\left(\frac{[\mathrm{salt}]}{[\mathrm{base}]}\right)$$
$$\mathrm p\ce{OH} = 5 + \log\left(\frac{\dfrac{0.2}{0.5}}{\dfrac{0.1}{0.5}} \right) = 5+ \log(2)=\boxed{5.3}$$
[.] = molarity
Where did I go wrong in the calculation or what mistake did I do while solving this question?

I found out where the mistake was in my method while solving the question. Here it goes.
• $\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
• $\alpha$= The dissociation constant per mole.
$$\ce{NH4OH <<=> NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH_4)_2SO_4 -> 2 NH_4^{+} + SO4^{2-}}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{dissociation} &0 &0.2 & 0.1\\
\hline
\end{array}
Now, we can say that:
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
Plugging in the values mentioned and derived,
$$10^{-5}= \left(\frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1(1-\alpha)}{0.5}}\right)$$
As $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$ and $(1- \alpha) \approx 1$. Doing the calculations above, we get that:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
Which further implies that, there are $0.1 \alpha$ moles of $\ce{OH^{-}}$ ions in the 0.5L solution.
$$ \therefore \ \boxed{\ce{[OH-]} = \frac{0.1 \alpha \ \text{mol}}{0.5 \ \text{L}} = \frac{0.1}{0.5} \cdot \frac{5}{2} 10^{-5} = \frac{10^{-5}}{2} \frac{\text{mol}}{\text{L}}}$$
Now, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$


Effective diffusion coefficient of a species in a heterogeneous mixture I want to calculate the effective diffusion coefficient of hydrogen radical in a mixed-phase consisting of metal oxide, metal hydride and metal. I consider the interfaces to be diffuse.

How can I calculate the effective diffusivity or effective diffusion coefficient of a species (ion or radical) in a heterogeneous mixture consisting of 3 different components? Can I weight average the diffusivities or is it more complicated than that?

This greatly depends on macro structure. If the phases were say arranged in a sheet-like structure (like a diffusion front) then the low diffusivity phase would be the dominating determinate of the diffusion rate. For this macrostructure, if we assume steady state with no change in phase or length from diffusion or time, then mass transport of the bulk can be approximated by:

$$\frac{\mathrm dn}{\mathrm dt} = \frac{A \times \nabla C}{\sum\frac{L_i}{D_i}}$$

If the high diffusivity phase is continuous then that could be the dominating determinate of the diffusion rate. That does not mean it is easily calculable, just noting that dominating factors can change. If no phase is continuous or other phases are continuous then this also becomes convoluted. For non-sheet-like structures calculating diffusion requires a computer model of the grain structure and finite element analysis to resolve an approximate solution.


how to manufacture PTC ink I am trying to create a PTC ink similar to Henkel ECI8120 , I researched PTC rubber and found that it was Polydimethylsiloxane (PDMS) loaded carbon nanoparticles, I was wondering if anyone here has some input on how to go about this.

I was thinking of dispersing carbon nanoparticles into the PDMS , but how do I get this to cure/solidify ?

I want to express that I am just asking for experimentation pointers/tips and not a complete solution.

It would have been helpful if you have said what kind of polymer-based nanocomposites you would like to create. Yet, I found a good article for the fabrication of poly dimethylsiloxane/carbon nanofiber (CNF)-based nanocomposites. The schematic diagram of experimental procedure is illustrated in following figure (from the Ref. 1).

The experimental section in Ref.1 describes the step-wise procedure for the fabrication. It also provides a useful few more references for polymer-based nanocomposite formulations. I also included another reference (Ref.2), which may be helpful to you.

References:

S. Chowdhury, M. Olima, Y. Liu, M. Saha, J. Bergman, T. Robison, Poly dimethylsiloxane/carbon nanofiber nanocomposites: fabrication and characterization of electrical and thermal properties, International Journal of Smart and Nano Materials, 2016, 7(4), 236-247 (https://doi.org/10.1080/19475411.2016.1269027).
H. Azimi, F H. Tezel, J. Thibault, Effect of embedded activated carbon nanoparticles on the performance of polydimethylsiloxane (PDMS) membrane for pervaporation separation of butanol, J. Chem. Technol. Biotechnol., 2017, 92(12), 2901–2911 (https://onlinelibrary.wiley.com/doi/pdf/10.1002/jctb.5306).


Can flexible plastics have reflective coating (vacuum metalized)? Is it possible to achieve an optical mirror coating to a flexible plastic part?

If vacuum metalization is possible, would the reflective coating layer be durable enough to withstand the bending of a flexible plastic such as polyurethane or a thermoplastic elastomer, without cracking over time?

There are many techniques for metal-coated polymers, as linked by permeakra Generally, vacuum deposition is an issue because the metal is hot. Cold deposition techniques (e.g., sputtering, ink coating) often work better.

Common, commercially-available materials include:

Pyralux - copper-coated polyimide from DuPont.
Mylar - DuPont's name for biaxially-oriented polyethylene terephthalate (BoPET)

As mentioned in other answers, adhesion is an issue, and to my knowledge, the processes to bind Pyralux and metabolized Mylar are proprietary.

Chemistry.About.com does mention vapor deposition of metals into Mylar, though.


Why does the beryllium 2+ ion have a larger radius than a helium atom? $\ce{Be^2+}$ has an ionic radius of $\pu{45 pm}$, while $\ce{He}$ has a radius of only $\pu{31 pm}$. If they are isoelectronic (they both have two elections in the $\mathrm{1p}$ orbital), and $\ce{Be}$ has more protons to attract them, why doesn't $\ce{Be^2+}$ have a smaller radius?

You are mixing apples and oranges.
Or, to be more precise, an ionic radius for $\ce{Be^2+}$ with coordination number (C.N.) 6 and van der Waals radius of $\ce{He}$.
To make it clear I compiled data for van der Waals and covalent radii [1, p. 9-58] as well as ionic radii [1, p. 12-13]:

$$
\begin{array}{lccccc}
\hline
\text{Element} & R_\mathrm{vdW}/\pu{Å} & R_\mathrm{cov}/\pu{Å} & \text{Ion} & \text{C.N.} & R_\mathrm{i}/\pu{Å} \\
\hline
\ce{Be} & 1.52 & 0.99 & \ce{Be^2+} & 4 & 0.27 \\
  &  &  &  & 6 & 0.45 \\
\ce{He} & 1.40 & 0.37 &  &  & \\
\hline
\end{array}
$$

Apparently, there is no experimental ionic radii for helium to compare with as helium, well, represents one of the most non-reactive elements.
Also, all radii set are determined using various methods and cannot be directly compared.
From the foreword [1. p. 9-58]:

  There are many scales of van der Waals radii, but they are not fully consistent with one another. The van der Waals radii determined by Bondi [...] from x-ray diffraction data, crystal densities, gas kinetic collision cross sections, critical densities, and liquid-
  state properties are the most widely used values. [...]

  The covalent radii tabulated here are recommendations for single covalent bonds,
  and they are based on a comprehensive evaluation of experimental data [...].

Ionic radii are estimated first and foremost from the experimental crystal structure data, averaged and supplemented by empirical and theoretical calculations (extrapolations, e.g. using Zachariasen method).

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


Are there stronger oxidizing agents than fluorine gas? Are there stronger oxidizing agents than fluorine gas, so it could oxidize fluoride to fluorine?

Also, in case of oxygen, fluorine gas can oxidize oxygen gas to the exotic dioxygenyl ion. Can anything oxidize fluorine gas to $\ce{F+}$ ion?

There are stronger oxidization agents than fluorine gas. I can name two in particular. But I don't think either would actually oxidize fluorine gas in it's pure state, for obvious reasons that you'll see shortly.

Dioxygen Difluoride. Affectionately named 'FOOF'. This is an extremely, extremely potent oxidizer. It also lights ice on fire. In fact, it lights essentially everything on fire or blows up on contact with said things.

Chlorine Trifluoride. Arguably, worse. This gas burns through practically everything, except for a few specific structural metals, but only because these form a very thin passivization layer. If this scratches, the tank burns. A notable incident with this gas, a 900kg spill of it burnt through 30cm of concrete, then 90cm of gravel underneath that concrete. It burnt over a meter into the ground.

As a foreword: I have no idea why you'd want to get within three miles of these things. Please don't light yourself on fire. It's not fun. (Trust me).


How can i detect Formalin in food easily I am trying to detect formalin in different types of food. Formalin $(\ce{CH2(OH)2})$is an aqueous solution of formaldehyde, which is $\ce{CH2O}$ or $\ce{H2CO}$. I used many chemicals to do this, but every time I fail. Please help me to do this. I got some results in my test by using NFPA 704 and ChEMBL, but that is not enough to save all. I would like to make a simple and easy food formalin detector that can save all the people. Please help me or give me some direction.

Simple commercially available tests contain formaldehyde dehydrogenase. This enzyme selectively catalyzes the chemical reaction of formaldehyde:

$$\ce{CH2O + NAD+ + H2O -> HCOO- + NADH + H+}$$

The enzymatic reaction is typically indicated by reduction of a tetrazolium salt to a coloured formazan dye.


Connecting a computer to Genesys 10S Spectrophotometer Has anybody ever connected a computer to a Thermo Scientific Genesys 10S Spectrophotometer without their (costly!) software?  I'd like to pull absorbance data off of it to perform analyses.  There are several ways I could think of to do this:

A USB flash drive could be plugged into the front of the instrument, but none of my drives are recognized by the spectrophotometer.
The PC could be connected as an external printer, but I haven't found a way to receive the data without ThermoFisher's software.
The PC could be connected as an external printer, but set the instrument to believe it is a PCL format printer.  Could the data be intercepted and converted to a usable form?

I appreciate your input.

This is a long comment....
The webpage specification says that there are three USB connections.

USB Type A port for USB memory device (front panel)
USB Type B port for optional computer connectivity (rear panel)
USB Type A port for external printer (rear panel)

The printer uses the HP PCL 3.0 specification via a USB cable. You should be able to get the HP specification and emulate that to get the data. The spectrometer doesn't know what it is communicating with...
The type A port on the front is supposedly for a USB stick for data storage. I'd guess that the Thermo Scientific software formats the stick in some screwy manner. You'll need the Thermo Scientific  specification to duplicate that. You might be able to get someone with the software to format some USB sticks for you. No idea what format the data format would be. I'd also guess that the Thermo Scientific software manual doesn't detail the format of USB data.
The USB Type B port on the back must have some bidirectional communication protocol. Unknown if Thermo Scientific would release details of the communications specification even if you bought the program. If you know someone that has such a computer instrument setup then you could install a sniffer in the USB communications line and figure it out. A lot of work.


How to detect if a beverage has Sucralose or aspartame? There is a drink that claims to be sweetened with stevia, but I think I taste Sucralose in there even though it is not listed on the ingredients.

Is there any easy way I could test for the presence of Sucralose to confirm my hunch?

How about Aspartame?

By easy, I mean not needing access to something like a gas chromatograph.

Here's the structure of sucralose

and here is the structure of aspartame

They are very different one from the other and should be easily separable.  You can do a "poor man's" separation using paper chromatography.  Aldrich sells both of these materials, but I'm sure you could find them much cheaper at a local Health or Health Food store.  Get the samples and some filter paper.  You'll need some solvents, a tank (preferably glass) with a lid, some filter paper coated with a fluorescent material (as an easier, but more expensive alternative, you could always buy some TLC plates from a chemical supply house) and a black light.  Put one of the solvents in the tank so that it is 1/8 - 1/4" deep.  Put the top on the tank and let it equilibrate. Make dilute solutions of your known samples using the same solvent that's in the tank.  Using the tip of a needle or a toothpick place a small spot of each of the dissolved materials on the filter paper, put the spots about 1" apart and high enough above the bottom edge of the filter paper so that they will be above the solvent level when you place the filter paper in the tank.  Give the spots a moment to dry and then put the spotted paper in the tank.  Watch as the solvent level climbs up the filter paper (or tlc plate) and remove the filter paper when the solvent is about 3/4 of the way up the paper or plate.  Let the paper or plate dry and then examine it with the black light.  The paper should fluoresce under black light except where the two spots have migrated up to.  The spots cover the fluorescent material and will appear as dark spots under the black light.  If you got a good separation between the two spots stop here and repeat the experiment putting a spot of your diluted drink on a new piece of filter paper along with spots of the two known materials.  Rerun it in the tank and see if there are any spots in your drink that match either of the spots from the two known compounds.  If there is no match, then sucralose or aspartame are not in your drink.  If there is a match, then whichever spots matched might be in your drink - more advance experiments will be needed to further analyze the situation.  If you didn't get a good separation between the spots for the two known compounds to begin with, start over and repeat with different solvents until you do get a separation.