How do derive the equation for relative velocity of gas molecules The relative velocity is $$v_{rel}=\sqrt2v_{mean}$$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$$$v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$$ because the reduced mass of two of the same particles is half the mass of one. In general: $$v_{rel}=(\frac{8KT}{\pi \mu})^{\frac12}$$

However, my question is: what is the need for reduced mass? If $v_{rel}=\sqrt2v_{mean}$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$. Wouldn't that just leave you with: $v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ Where does reduced mass come into it?

Your equation $v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$ is incorrect.

$v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ (your last equation) is also incorrect.

Instead $v_{rel}=(\frac{16KT}{\pi m})^{\frac12}$ and $v_{rel}=(\frac{8KT}{\pi \frac m2})^{\frac12}$ would both be correct.

The need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms.

$\mu = m_1m_2/(m_1 + m_2)$ by definition.  If $m_1 = m_2$ then $\mu = m/2$

When there are two different masses, there are two different mean velocities.

To see where the equation with the reduced mass comes from, look at appendix 1.4 of Kinetic Theory of Gases, along with pages 1-30 to 1-34.


Difference between coordination entity and coordination sphere What’s difference between coordination entity and coordination sphere? Can they be same? Here in my book same example is given for both coordination entity and coordination sphere.

Please help.

Source:http://ncert.nic.in/ncerts/l/lech109.pdf

Section 9.2 Definitions of Some
Important Terms Pertaining to Coordination Compounds

Page no. 240-241

Coordination entity is the part of complex compound having central atom surrounded with the number of secondary linkage.
But,if this entity is written along with charge on it ,then it is called coordination sphere
For example  [Fe(CN)6]4- is a coordination sphere while Fe(CN)6 is the coordination entity


Silicone for use in casting Bismuth I am trying to cast bismuth, which melts at 271.4 °C. Silicone rubber seemed like a good material to use for a mold, given its flexibility and the ease with which it can be cast itself, as well as its high (~300 °C from some sources) heat tolerance.

I have had the opportunity to test casting bismuth in this way once so far (will have the opportunity to try again this weekend) using a mixture of acetoxy silicone and cornstarch as the mold, for ease of molding and quicker curing.

While the mold itself seemed to stand up well to the heating, aside from discoloration, there was a large amount of bubbling as the bismuth cooled, resulting in deformation of the final structure.

My best guess as to the cause of the bubbling is that the silicone had not finished curing, and trapped, evaporating water and acetic acid were the source of the bubbles. Is this likely to be correct, and if so, is there a way to avoid this in future casts, such as possibly by letting the silicone cure longer? Alternately, are there different formulations/materials that may perform better, without being significantly harder to make?

Acetoxy silicone (I assume this is silicone caulk) is a bit of a different thing to the RTV silicone used for low-melt metal casting.

Really you want to find some high temperature mould silicone, this is fairly widely available from sculpture and model making suppliers and is used for things like casting white metal miniatures and jewellery.


Rutherford gold foil experiment This question may sound silly.
What actually I understood from this experiment is that the positively charged nucleus which is in the center is responsible for the bending and reflection of the alpha rays.Here is my question:Why cant there be any small holes in the gold foil which might have caused this kind of behavior?

Reflect means to bounce back, as in a ball bouncing off a wall. The average density of an atom is very low, so the observed reflection was startling... like throwing many billiard balls at a mass of fluffy cotton  candy and every now and then a ball comes bouncing back!
The explanation offered was that there must be some "hard", dense object inside that fluff... the nucleus. Electrons are effectively spread out in their orbitals, and tend to act individually (bound by electromagnetic forces). An alpha particle, massing $\pu{4 \times 10^{9} eV}$, could not be reflected by an electron massing ~$\pu{5 \times 10^{5} eV}$ -- that would be like bouncing a baseball off a ping-pong ball.
Since both alpha particle ($\ce{He}$ nucleus) and gold nucleus are positively charged, and the gold nucleus is held together by powerful nuclear forces (so all its particles' masses are effectively combined to ~$\pu{2 \times 10^{11} eV}$), it explained the reflection. Since reflection occurred only rarely, it implied that the nucleus was a small, dense target.


Enthalpy and Internal Energy in Constant Volume Calorimeter It is my understanding that enthalpy differs from internal energy only in that the change in enthalpy takes into account any pressure-volume work done on the system (or by the system) in a reaction, and thus it is equal to the heat transferred to the system, at least in constant pressure.

Many sources say that in a constant pressure, "coffee cup" calorimeter, $q_p = \Delta H$. I understand that. However, they also say that the $q_v$ in a constant-volume calorimeter is equal to the change in internal energy, and not necessarily the change in enthalpy. My thought is that since $\Delta H = \Delta U + P\Delta V$, with $\Delta V$ being zero (it is by definition constant volume), is not $\Delta H = \Delta U = q_v$?

The definition of enthalpy is
$$H = U + pV$$

which gives

$$\begin{align}H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
\Delta H &= \Delta U + \Delta(pV)\end{align}$$

The equation you cited is only true at constant pressure: if $p_1 = p_2 = p$,

$$\begin{align}
H_2 - H_1 &= (U_2 + pV_2) - (U_1 + pV_1) \\
\Delta H &= \Delta U + p(V_2 - V_1) \\
&= \Delta U + p\Delta V
\end{align}$$

The condition for that equation is that the pressure is constant. That equation is not applicable for constant-volume (isochoric) processes unless the process is also constant-pressure (isobaric).

However, if you do have a process that is both isobaric and isochoric, you are right to say that $\Delta H = \Delta U$.


Visualizing PDB file in VMD, how to upload molecules? Problem: When I load PDB-file of my protein of interest in VMD, nothing happens. The molecules is uploaded, but when I push "D" the screens with with big "VMD" letter freezes instead of presenting the structure of the protein in the PDB file. This is what is looks like:

Any Ideas what causes this? Should the PDB file be in a specific folder? I am a complete novice at this, so I wont understand very complicated answers.

I have downloaded the mac version of VMD version 1.9.2, I have Mac OS X version 10.11.2.  Here is the PDB file that I want to open.

I found a shortcut, and that is if I go to (click on):

file
New Molecule
Enter in the "browser" window the direct PDB accession code "3WD5" (for my protein) then VMD retrieved the protein structure form the PDB database on its own, and is showing it as it should in the visualization window.

However, this does still not resolve why it will not load a downloaded PDB file, but I suspect its most likely due to some problem with downloading of the files or installing the program its self that has gone wrong.


Looking at the equation of a reaction, how do we know it is endotermic or exotermic For example, $MnCO_3 → MnO + CO_2$ is endotermic, and occurs at 473 K

the polymerization of ethylene $H_2C=CH_2$ to $n(H_2C=CH_2)$ is exotermic

I suspect it has something to do with the standard enthalpy of formation of the different terms of the equation (right side - left side = $\Delta H$ if negative, the reaction is exotermic and releases $\Delta H$ to the environment as heat, if positive, it is endotermic and needs $\Delta H$ to happen.

Is this correct?

If not, What am I missing?

You are correct. If you have enthalpies of formation of each reagent and each product then you sum up enthalpies of each product multiplied by stoichiometric coefficient and subtract enthalpies of each reagent multiplied by stoichiometric coefficient. If the result is negative then the reaction is exothermic and vice versa.


If acyl chlorides are reacted with aqueous ammonia, will carboxylic acids also be formed? As part of my A-level chemistry course, I have been taught that an acyl chloride will react violently with water to produce a carboxylic acid and hydrochloric acid. I have also been taught that an acyl chloride will react with an aqueous solution of ammonia to produce an amide and ammonium chloride. My question is, would the acyl chloride also react with the water in the aqueous ammonia solution? If so, this could decrease the yield of the amide, because some of the acyl chloride will react to produce the carboxylic acid.

I would appreciate any additional relevant details about the reaction, such as answers to the following questions (assuming that the acyl chloride does react with the water):

Is there an easy method to separate the amide from the
carboxylic acid?
Would it be best to use a concentrated ammonia solution, so that
there is plenty of ammonia for the acyl chloride to react with
instead of reacting with the water?
Is the reaction with water negligible compared to the reaction with ammonia?

Thanks in advance!

Acyl halides react with water but usually not violently. The acyl halides are normally not soluble in water so the reaction is slow. Then, when the reaction proceeds, acid is formed and the pH decreases, reducing the reactivity of water.

Ammonia and amines are much more nucleophilic and react much faster with the acyl halide. This reaction is so much faster that it can be carried out in water. In this context it is called the Schotten–Baumann reaction, which normally proceeds with high yields.

An excess of ammonia is required at least to absorb the HCl formed, but normally a huge excess is used. The acyl halide is more expensive than ammonia so we want to have it completely transformed to the amide. The excess of ammonia will also help to dissolve the acid, if it is formed in the reaction.

The other points:

To separate acids and amides, extract with alkali. The acid will go to the aqueous phase and the amide to the organic phase.  Or, if they are small, recrystallize it. The amides are less soluble than acids.
Huge excess of ammonia or simple amines is normally used.
The reaction of water is negligible under these conditions.


Coefficients and Parameters for contracted Gaussian basis sets I'm trying to write a program to calculate fixed-point Hartree-Fock level energies of molecules (for my amusement) and everything makes sense but this. I've been agonizing over this for almost 3 hours now. I've tried pretty much every Google search I can think of (mostly returning results that were either too vague or without enough detail) and looked in almost every book/ebook I own (pretty much always too vague). Any help whatsoever would be extremely appreciated.

As far as I understand, an STO-NG contracted Gaussian basis function has the following form:

$$\phi_{\mu}^{\textrm{CGF}}(\vec{r}) = \sum_{p}^{N_{\textrm{PGF}}} d_{p\mu} (x-X_A)^{i_{p\mu}} (y-Y_A)^{j_{p\mu}} (z-Z_A)^{k_{p\mu}} e^{-\alpha_{p\mu}|\vec{r}-\vec{R_{A}}|^{2}}$$

where the contraction coefficients, $d_{k\mu}$ and exponents, $\alpha_{k\mu}$ are chosen such that $\phi_{\mu}$ provides the 'best fit' to a Slater-type orbital having a Slater exponent $\zeta$, $\vec{R_{A}}=(X_A,Y_A,Z_A)$ is a fixed reference centre (usually a nucleus) and $l_{p\mu} = i_{p\mu} + j_{p\mu} + k_{p\mu}$ defines the angular momentum of the primitive Gaussian function:

$$\phi_{p\mu}^{\textrm{PGF}} = (x-X_A)^{i_{p\mu}} (y-Y_A)^{j_{p\mu}} (z-Z_A)^{k_{p\mu}} e^{-\alpha_{p\mu}|\vec{r}-\vec{R_{A}}|^{2}}.$$

I'm having problems interpreting data files documenting contraction coefficients and exponents for various basis sets.

For example, please consider the following snippet from the Gaussian 94 STO-3G basis set file documenting contraction parameters for Carbon:

****
C     0
S   3   1.00
     71.6168370              0.15432897
     13.0450960              0.53532814
      3.5305122              0.44463454
SP   3   1.00
      2.9412494             -0.09996723             0.15591627
      0.6834831              0.39951283             0.60768372
      0.2222899              0.70011547             0.39195739
****

I have read in various places that there are enough coefficients here to describe 5 contracted functions, but I see only enough for 3 maximum (if the exponents are the same, but the coefficients change between 2S and 2P). I'd be eternally grateful if someone could explain for me, precisely how one would determine the following information from such a basis set data file:

1) Which EXACT basis functions appear within each contracted function?

2) How many contracted Gaussian basis functions there are?

If possible, would you be able to do the same for the following snippet, again for Carbon, Gaussian 94, but this time for the more complicated 6-31++G** basis set?:

****
C     0
S   6   1.00
   3047.5249000              0.0018347
    457.3695100              0.0140373
    103.9486900              0.0688426
     29.2101550              0.2321844
      9.2866630              0.4679413
      3.1639270              0.3623120
SP   3   1.00
      7.8682724             -0.1193324              0.0689991
      1.8812885             -0.1608542              0.3164240
      0.5442493              1.1434564              0.7443083
SP   1   1.00
      0.1687144              1.0000000              1.0000000
SP   1   1.00
      0.0438000              1.0000000              1.0000000
D   1   1.00
      0.8000000              1.0000000
****

Take a look at the carbon STO-3G first. It means for each AO, you form it with three Gaussians. For carbon, you have a 1S, 2S and (three) 2P. 5 functions when finished.

The top line says you are defining the basis for carbon; no big deal.

Next line says you are creating an S shell, with three primitives, with a 1x scale factor.

Then you list your three rows of primitives.

The first column is the Gaussian exponent. The next column is the S expansion coefficient. These are the three Gaussians that make up your 1S "Slater" atomic orbital.

In the next section, you define SP. What this is, and I quote from the Gaussian website now: "Note that 2SP requests the best least-squares fit simultaneously to S and P slater orbitals and is not equivalent to separately specifying the best S and the best P expansions."

S and P (and all higher angular momenta, actually) can be related by a recursion relationship. It's an efficient way of computing integrals. I believe the Obara-Saika recursion takes advantage of this, though in Gaussian the integrals are formed using the PRISM algorithm.

The first column is the exponent for your Gaussian, the second column is the S expansion coefficient, and the third column is the P expansion coefficient. They just share the same exponent. Thus you have 4 + 1 (from before) = 5 AOs total formed with three primitives each.

To jump ahead to your last example, the pattern simply continues. The first section gives six primitives to form one 1S (the 6 in 6-311++G**). The second section gives three Gaussians for you valence S and P, the third section is one Gaussian for valence SP. This is the 31 in 6-31++G**. You have a split valence description, one with three primitives, one with a single primitive.

The next section is the diffuse function, hence the really small exponent in the first column. This is the ++ part of the 6-31++G**. The small exponent gives a really wide spreading Gaussian. Helpful for describing electronic activity far from the nucleus. (It's helpful to compare with the huge exponents in the first section, which gives really tight functions that describe the core electrons).

Finally, you have the D functions which are polarization functions. Carbon doesn't have a D valence shell, but we include them to describe polarization of electronic density in molecules.

Note that depending on your program, you either have 5D AOs or 6D AOs. This is the difference between spherical and cartesian representations, and is more of a computational trick than anything else. Gaussian uses 5D by default, except in some cases.


Expansion work by a gas when pressure is not explicitly constant? How is it possible to calculate the work done by a gas when pressure is not explicitly stated to be constant? The question states:

  Calculate $\Delta H$, $Q$, and $W$ when $1\ \mathrm{mol}$ of He expands from $V = 5\ \mathrm L$ at $T = 298.15\ \mathrm K$ to $V = 10\ \mathrm L$ at $T = 373.15\ K$.

I am not told that the process is isobaric, adiabatic, or even to treat He as an ideal gas (although I know it roughly approximates one).

If I treat it as an ideal gas, I can find $\Delta H$ by using the equations $$H = U + pV$$ $$p = nRT/V$$ and $$U = (3/2)RT$$ but I have no ideas about finding $Q$ or $W$ since I cannot make any other assumptions. Is there an implicit assumption I should make about He gas (or ideal gases in general) expanding, or is there simply not enough information to solve this problem?

You can calculate the work done and Heat exchange if you assume that the process is quassi-statically performed and it is reversible and most importantly done in a single step.

Assume that the process through which the $\ce{He}$ is expanded obeys the equation, $PV^x = K$(constant). This equation can also be alternatively stated as, $TV^{x-1} = K$. Now, suppose the gas is transformed to the state $(T_2,V_2)$ from the state $(T_1,V_1)$. So, we can thus write, $$T_2V_2^{x-1} = T_1V_1^{x-1}$$ so, putting the given conditions, we can find $x$ as, $$x= 1+ \frac{\log\left(\frac{T_2}{T_1}\right)}{\log\left(\frac{V_1}{V_2}\right)} \approx 0.676$$ Thus, after knowing $x$ ,you can calculate the work done by the gas as,  $$W= \int_{V_1}^{V_2} P\mathrm{d}V = \int_{V_1}^{V_2} \frac{K}{V^x}\mathrm{d}V = \frac{P_2V_2 - P_1V_1}{1-x} = \frac{R(T_2-T_1)}{1-x}$$ We know all the value of $T_1, T_2 $ & $x$, and work done can be easily calculated. and we already know the change in Internal energy  and thus heat exchange can be easily calculated and also $\Delta H $ can be calculated   ($Q = (C_V + \frac{R}{1-x}) \Delta T$ and $\Delta H = C_P \Delta T $) and thus all the quantitites can be known.


In a Koppers-Totzek reactor, why do I get CO and H₂ instead of CO₂ and H₂O? A Koppers-Totzek reactor (there seems to be only a German Wikipedia page) is an entrained flow gasifier that blows coal dust, oxygen and water (as steam, of course) into a burning chamber where the coal burns at about 1600 °C. The output contains around 60% $\ce{CO}$ and 33% $\ce{H2}$, plus some $\ce{CO2}$ and (very little) $\ce{CH4}$.

When burning coal, I would expect $\ce{CO2}$ and $\ce{H2O}$ as output, is the reactor simply providing too little oxygen, or is there something about the process that produces the output observed?

The reaction occurs in two stages: in first $\ce{C}$ oxidized to $\ce{CO2}$ and then $\ce{CO2}$ is reduced to $\ce{CO}$.

The process of having carbon react with oxygen and water vapor is called steam-oxygen conversion. It is used, because in reality, due to the high reaction temperatures, the allothermic principle is hard to provide.

$$\ce{C + O2->CO2}$$
$$\ce{C + 2H2O->CO2 + 2H2}$$
$$\ce{C + H2O->CO2 + H2}$$
$$\ce{C + CO2->2CO}$$

The equilibrium concentrations of $\ce{O2}$ and $\ce{H2O}$ in the gasification products at all temperatures are negligible. Steam-oxygen gasification products are a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$.

Before carrying out the reaction of the shift
$$\ce{CO + H2O->CO2 + H2}$$ from a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$ must be separated $\ce{CO2}$. To this must be expended work division (again burn some $\ce{C}$).


Energy loss in the process for exothermic reactions Some chemical reactions while producing the product molecules releases heat to the surroundings and such reactions are called “exothermic reactions”. Is it true that such systems must be showing energy loss in the process meaning “ΔE ” must be negative?

No.  If means that to maintain the temperature of the products equal to the temperature of the reactants, you need to remove heat, and, form the first law, if you do this, then $\Delta E$ is negative.  However, if you don't remove the heat, the temperature of the products will be higher than those of the reactants, and $\Delta E$ of the system will then be zero (as required by the first law).


Why won't this reaction work? I'm trying to do a reaction with phenyl piperazine, triphosgene, and 4-bromophenol. However, every time I do this reaction I seem to get the carbonyl di-4-phenylpiperazine as a major product (based on the LCMS masses). I tried to reduce the triphosgene step to 5min, which seemed to yield trace amounts of the intended product, but I still can't get my intended product as the major product. This should be such a simple reaction - what am I doing wrong?

You are forming the urea because there is always an excess of phenylpiperazine present to react with the carbamoyl chloride you are forming. You need to add the phenylpiperazine to the triphosgene. The paper by Varjosaari et al. [1] has a procedure (described on Organic Chemistry Portal) that should work.

@Mathew Mahindaratne's suggestion will also work:

  Try reversing the order. First, add 1: 1 triphosgene to 4-bromophenol (with stronger base) and then add phenyl piperazine.

References

Varjosaari, S. E.; Suating, P.; Adler, M. J. One-Pot Synthesis of O-Aryl Carbamates. Synthesis 2016, 48 (01), 43–47. https://doi.org/10/gf4wh2.


Under what circumstances would azeotrope exist in non ideal solutions? (when would the dew point curve and the bubble point curve meet?) [solved. For anyone who's interested in this topic, please view MaxW's comment under his/her answer. A thousand thanks to Poutnik and MaxW.]
I looked up several books on thermodynamics, and they all states that for non ideal solutions, the dew point curve and the boiling point curve sometimes meet at some point. Azeotropes forms there.
What I couldn't figure out was, under what circumstances would the two curves meet? Do they always meet if both of them have a local minimum? What is the logic under this phenomenon?

Azeotropes do not exist for all non-ideal solvent solutions. All solvent solutions are more or less non-ideal, but not all solvent combinations form azeotropes.

E.g. ethanol and isopropanol form azeotropes with water, but methanol does not.

Azeotropes exist if there is large enough deviation from the Raoult law. It means if the azeotrope point of the diagram is pushed enough away from the endpoint connections.

For near ideal solutions, like n-pentane + n-hexane. there is no bottleneck and onle one convex/concave chart segment.

Back to the meeting of boiling and condensation curves, they meet at the pure solvent points and eventually at the azeotrope point, if it exists.

Existance of such minimums ( or maximums ) ob the vapour diagram means there is an azeotrope. An azeotrope behaves like if it were the 3rd substance, with  the same composition of liquid and vapour above it, so the curves meet there.

Minimums meet if and only if liquid and vapour compositions are the same. That happens if and only if either the liquid is pure solvent either it has the composition of the azeotrope ( if it exists ).

You can create simulations yourself by defining nonideal vapour pressure equations as $p_i(T)<>p_{i,0}(T) \cdot x_i$. It cannot be exactly calculated, as neither the mixture of pure solvent and azeotropic mixture has the ideas behaviour.


What kind of solvents favor elimination reaction? This question is confusing and I can’t seem to find a answer to this. The question is what solvent promotes E2 and E1 reactions? My teacher says that methanol as solvent with strong base favors E2 over Sn2. But she never explained why. Also she keeps using methanol as the example which is confusing. Can anyone please elaborate on this?

This website gives good rules of thumb: http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

And so does this website: http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch08substitutionvselimination.html

In addition, according to page 583 of "Physical Organic Chemistry" by Anslyn and Dougherty:

  Almost all bases are also nucleophiles, and hence we expect competition between elimination and substitutions. In both $\ce{S_N2}$ and $\ce{E\text{2}}$ reactions, the nucleophile or base reacts in a single rate-determining step with the reactant. In both $\ce{S_N1}$ and $\ce{E\text{1}}$ reactions, the nucleophile or base reacts in a step after the rate-determining heterolysis...

  An $\ce{E\text{2}}$ reaction requires the addition of a base, and can be performed in solvents of low or high ionizing power. Since most strong bases are also good nucleophiles, $\ce{S_N2}$ reactions are important competitors. However, elimination will dominate if the carbon with the leaving group is not susceptible to nucleophilic attack, such as a tertiary R group. Examples of this trends are given in Table 10.6.

  $\ce{E\text{1}}$  reactions involve carbenium ion intermediates, and therefore are facilitated by all the factors that stabilize carbenium ions. These are the same factors that facilitate $\ce{S_N1}$ reactions. Strongly ionizing solvents and substitution of electron donating groups on the carbon undergoing heterolysis are necessary.

  In highly ionizing solvents and with R groups that readily form carbenium ions, the ratio of substitution to elimination is typically independent of the leaving group. This evidence supports the notion that the substitution and elimination products are formed b branching from a common intermediate, and therefore the two reactions share a common rate-determining step. In contrast, in solvents of lower ionizing power, the ratio of substitution to elimination products does depend on leaving group, and indication that the two reactions do not share a common intermediate. Instead, contact ion pairs are formed, and either the leaving group or the solvent can remove the proton to give elimination products. The formation of a contact ion pair also has an influence on the stereochemistry of the elimination.

Table 10.6

Reactant            Solvent   Base      Elimination
Tert-Butyl Bromide  Ethanol   NaOEt     100.00%
Tert-Butyl Bromide  Acetone   NBu4Cl    96.00%
isopropyl Bromide   Ethanol   NaOEt     75.00%
isopropyl Bromide   Acetone   NBu4Cl    0.00%
Ethyl Bromide       Ethanol   NaOEt     9.00%
Ethyl Bromide       Acetone   NBu4Cl    0.00%


Why do we use the external pressure to calculate the work done by gas I read in a textbook that in the case when we have a gas in a cylinder fitted with a massless frictionless piston being held with an external pressure $p_1$, and when the pressure is reduced to become the value $p_2$, the gas pushes up against the piston and then the work done by the gas for a small change in volume is calculated by:

$$\mathrm dW=p_2\,\mathrm dV$$

Here, is what I don't conceptually understand. If the gas's molecules was under some pressure $p_1$ which is equal to the external pressure in the static state then after the external pressure became lower than the internal pressure, shouldn't the work done by the gas be the difference between the two pressures?

If the piston is frictonless and massless, then, if you do a force balance on the piston, you must have that the force per unit area that the gas exerts on the inside face of the piston will always be equal to the external force per unit area that one imposes on the outside face of the piston.  The sudden drop in pressure on the outside face of the piston causes the gas to undergo an irreversible expansion.  During an irreversible expansion, the local pressure within the cylinder becomes non-uniform, so that the average pressure of the gas differs from the force per unit area at the piston face.  As a result, the ideal gas law (or other equation of state) cannot be applied globally to the gas in the cylinder.  In addition, during an irreversible expansion, there are viscous stresses present in the gas that allow the force per unit area at the piston face to drop to the new lower value while requiring that force to match the external force on the outer face.  So the work done by the gas on the piston is equal to the external force per unit area times the change in volume:  $$W = \int{P_{ext}dV}$$  This equation is always satisfied, irrespective of whether the expansion is reversible or irreversible.


What is the unit of molar conductivity? Could you explain what is unit of  molar conductivity along with derivation

Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.
It's unit in S.I. is: $$\ce{S.m^2.mol^{-1}}$$


What chemical properties make LSD so psychoactive? Lysergic acid diethylamide (LSD) is a well known hallucinogen and (usually) illegal psychoactive drug with the structure below:

The drug is active is doses measured in micrograms in contrast to many other active compounds where the dose is milligrams or even grams. The effective dose of Mescaline, a related psychoactive, is something like 200mg (more than 1,000 times more than for LSD).

Why is LSD so active in small doses? Do we know which features of the structure that contribute to its activity?

It's the indole portion of the molecule. the lowest two rings...if you remove those, the compound has no such properties. Indoles, the side chains of tryptophan , an amino acid, are suitable for binding to receptors that affect emotion, perception, cognition, sleep/wake cycles etc. You can see how the indole looks similar to both DMT (dimethyltryptamine--another hallucinogen) and the benzene ring bears similarity to psilocin, the major constituent of psilocybin (the active principle in magic mushrooms). Tryptamines are common hallucinogens.


Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart? Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart? E.g., is it possible to change dextrodopa to levodopa?

Yes,this process is called stereoconversion. Both SN1 and SN2 reactions can invert the chirality of a carbon center. In the case of the SN1 mechanism, departure of a leaving group generates a prochiral intermediate which is then open to nucleophilic attack on either face. If both faces are equally acessible to the nucleophile this will result in racemisation of the molecule. In the case of the SN2 reaction the nucleophile will enter a trigonal bipyramidal intermediate with the molecule. Assuming the nucleophile has the same stereochemical precedence as the leaving group, the stereochemistry will be inverted.


When reactants are in their transition state, can they go back into their reactants form instead of progressing to form the products? When reactants are in their transition state, can they go back into their reactants form instead of progressing to form the products? I don't see any reason why the reaction has to keep progressing forward.

In the gas phase the reactants have kinetic energy as they move towards one another, so you would think that they would just cross the transition state, but molecules also have vibrational energy which means that it is possible to reach the transition state and then turn back. See the figure below which shows a simple LEPS calculation of the O+H2 reaction.

The solid black line shows the position and vibration of the H2 molecule as it approached the O atom and then 'reflects' at the transition state. It has just enough energy to cross, but the vibrational motion and position on the energy surface prevents it.  The axes show the separation of H and H2 (mid point) and of H and OH products.

In solution (or high pressure gas phase) the situation is a little different because collisions with may solvent inhibit or enhance the motion of the reactants towards one another. The species diffuse together and also diffuse along the transition state pathway so that the transition state can be crossed and recrossed many times before reaction occurs or the reactants are reformed. This approach is named after Kramers and is the most important one to understand when thinking about transition states in solution. (Strangely, it is rarely mentioned in Chemistry textbooks even though it was developed more than 70 years ago.)


How do ligands and coordination complexes work in this case with copper (II) chloride? What happens when Ammonia  is added to an (aq) solution of copper II chloride?

$\ce{CuCl2 + NH3 ->\; ?}$

I know that ligands bond with metals to form ion but I'm not sure about how they work in a reaction.

When solid $\ce{CuCl2}$ is dissolved in water, aquo complexes are formed.

$\ce{CuCl2_{(s)} + H2O -> [Cu(H2O)6]^2+ + 2 Cl-}$

Addition of ammonia results in a ligand exchange reaction to give tetrammine complexes.

$\ce{[Cu(H2O)6]^2+ + 4 NH3 -> [Cu(NH3)4(H2O)2]^2+ + 2 H2O}$


How does water freeze and crystallize on a cold superhydrophobic surface? Superhydrophobic surfaces are nano-structured to "repel" liquid water and there has been a lot of work done to understand the microscopics of wetting of these nanostructures.

What I currently am interested in understanding is wetting and freezing of water on a cold, below-freezing superhydrophobic surface.

I would expect that the ice would still have mechanical adhesion to the surface, but I don't know to what extent.

So my question is the following: How does water freeze and crystallize on a superhydrophobic surface?

I know this is a general question that would depend on the nature of how superhydrophobicity is achieved, so feel free to specifically treat a certain branch of superhydrophobic surfaces.

As acknowledged in your question, there are a large number of factors at play here. The most basic of these is that there are three fundamental mechanisms by which freezing of a droplet can initiate when in contact with a superhydrophobic surface: 1) heterogeneous nucleation where an ice freezing nucleus (IFN) is present in the water droplet, 2) heterogeneous nucleation where the superhydrophobic surface serves as the IFN, and 3) homogenous freezing of the water droplet.

An excellent demonstration of the first or third scenario above (it is not clear from the article which is actually the case) is given in video format from an article published in Nature$^1$. In this experiment, the water droplet is evaporatively cooled until freezing initiates at the water/air interface, and continues without adhering to the aluminum-based superhydrophobic surface.

Whether or not freezing will initiate at a superhydrophobic surface depends upon the degree of supercooling, the presence or absence of potential IFN within the water droplet, and the potential for the specific material (the superhydrophobic surface) to serve as IFN. It would seem that the latter of these would be unlikely, except that a surface being hydrophobic doesn't mean that it is equally icephobic. Basically, the more the surface resembles the crystalline structure of ice, the more likely it is to act as an IFN.

Please don't hesitate to ask for any clarifications in the comments below.

1) Thomas M. Schutzius, Stefan Jung, Tanmoy Maitra, Gustav Graeber, Moritz Köhme & Dimos Poulikakos, Spontaneous droplet trampolining on rigid superhydrophobic surfaces, Nature 527, 82–85 (05 November 2015) doi:10.1038/nature15738


The structure of 1-(butan-2-yl)-4-tert-butylbenzene is displayed below. What is -yl on this compound and when do I use it? I can't figure out how to use -yl on organic nomenclature and how do I use it in nomenclature. Please explain everything about the use of -yl in nomenclature.

The "yl" refers to removing a hydrogen atom from a molecule making a univalent group.

Butane is

$\ce{CH3-CH2-CH2-CH3}$

To make butan-2-yl take a hydrogen atom off carbon 2:

$\ce{CH3-CH-CH2-CH3}$

Then butan-2-yl X would be

$\ce{CH3-CH(X)-CH2-CH3}$.


Why is it important to use a salt bridge in a voltaic cell? Can a wire be used? I was learning about voltaic cells and came across salt bridges. If the purpose of the salt bridge is only to move electrons from an electrolyte solution to the other, then why can I not use a wire?

Also, will using $\ce{NaCl}$ instead of $\ce{KNO3}$ in making the salt bridge have any effects on voltage/current output of the cell? why?

Plus if it matters, I'm using a Zinc-Copper voltaic cell with a tissue paper soaked in $\ce{KNO3}$ as salt bridge

There's another question related to salt bridges on this site.

The purpose of a salt bridge is not to move electrons from the electrolyte, rather it's to maintain charge balance because the electrons are moving from one-half cell to the other.

The electrons flow from the anode to the cathode.  The oxidation reaction that occurs at the anode generates electrons and positively charged ions.  The electrons move through the wire (and your device, which I haven't included in the diagram), leaving the unbalanced positive charge in this vessel.  In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell.  A similar (but reversed) situation is found in the cathodic cell, where $\ce{Cu^{2+}}$ ions are being consumed, and therefore electroneutrality is maintained by the migration of $\ce{K+}$ ions from the salt bridge into this half cell.

Regarding the second part of your question, it is important to use a salt with inert ions in your salt bridge. In your case, you probably won't notice a difference between $\ce{NaCl}$ and $\ce{KNO3}$ since the $\ce{Cu^{2+}}$ and $\ce{Zn^{2+}}$ salts of $\ce{Cl-}$ and $\ce{NO3-}$ are soluble. There will be a difference in the liquid junction potential, but that topic is a bit advanced for someone just starting out with voltaic/galvanic cells.


How can we measure the rate of reaction of EDTA chelation with zinc carbonate? I am performing an experiment in order to find the activation energy of EDTA chelation. Consequently, I need to measure the rate of reaction of the process. I am high school student, and so I do not have access to sophisticated equipment, although they might be unnecessary. However, I have been attempting to measure the volume of gas produced, but I am not achieving any result. So, can you suggest any other method to measure the rate of reaction? I would greatly appreciate the help since there is not a sufficient amount of information on the internet to help me get a clear answer, I have been working on the reaction for more than a month and a half.

I think that the reaction will be too fast for you to measure with the tools in a high school lab. I think that you are not measuring the zinc binding to EDTA but instead making a measurement of the dissolution rate of zinc carbonate in an aqueous solution of EDTA.

If you were to want to measure the rate of the binding rate of zinc by EDTA the best way might be to use a radioactive tracer. I would start with radioactive zinc in an anionic EDTA solution. I would bring this into contact with a non-radioactive solution of zinc. I would then keep on pulling out small samples from the mixture and filtering them through an ion exchange resin which is able to remove either the anionic EDTA complex or the cationic form of zinc.

I would use zinc nitrate or some other form of zinc which does not tend to form anionic complexes when in solution.

I know that this experiment has been done as a teaching lab experiment in a nuclear chemistry unit with radioactive europium, but it has a reputation as a difficult experiment to do.

I suggest you read Williams, K.R.; Choppin, G.R., 1974,  J. Inorg. Nucl. Chem., v. 36, no. 8, pp. 1849-1853. This has details of an experiment on the kinetics of metal exchange between EDTA complexes and "free unsolvated metal" for trivalent metals.

An alternative non radioactive method would be to use stopped flow experiments with a metal senstive indicactor present in the mixture. If we assume that the exchange of zinc between the solution and the indicator is very fast then this method could be of use.

Another method could be to take a solution of the zinc complex of EDTA and then treat it with C13 KCN, you could measure the C13 NMR spectrum again and again if it is a relatively slow process. But I think for you that you will be unable to deal with this problem with the chemcials and tools you have in a school lab. Sorry for having to pour some cold water around.


EDTA titrations of singly charged cations It is well known that EDTA can be used for titrations involving complexations of highly charged metal cations, with the most well known being those of the alkali earth metals and transition metals. However, would it be possible to analyze metal cations such as the alkali metals or singly charged transition metals?

Titrations involving chelating of metal ions with ethylenediaminetetraacetic acid (EDTA) is an extremely useful analytical tool and a very well studied system in analytical chemistry. It can be bind to virtually any metal ion, evidently even to alkali metals. For instance, sodium ion complexes with EDTA in certain situations (Ref.1). The abstract of the reference states that:

The crystal structures of tetrasodium (1,2-ethanediyl-dinitrilo)tetraacetate pentahydrate and disodium dihydrogen (1,2-ethanediyldinitrilo)tetraacetate dihydrate have been determined. $\ce{Na}$ ions prefer to act as bridges between different $\mathrm{edta}$ ligands and only in the tetrasodium salt is the $\mathrm{edta}$ ligand hexadentate to one $\ce{Na}$ ion. $\ce{Na}$ ions display seven-coordination, when coordinated to one $\mathrm{edta}$ ligand (distorted pentagonal bipyramid), and six-coordination, when linked to several $\mathrm{edta}$ ligands (from distorted octahedral to trapezoidal bipyramidal geometry). An $\ce{Na}$ ion in the tetrasodium salt shows five-coordination, because the sixth site is blocked by the $\mathrm{edta}$ ligand. $\ce{H}$ atoms in the disodium salt are linked to amine groups. Only these two salts can be obtained on varying the $\mathrm{pH}$ of an aqueous solution containing $\ce{Na+}$ and ethylenediaminetetraacetic acid. The first is obtained around $\mathrm{pH}$ 2-5 and the second around $\mathrm{pH}$ 12.

Nonetheless, there are no evident in the literature that EDTA titrations can be done to determine alkali metal concentration (agreeing with M. Farooq in his comment). As a matter of fact, disodium ethylenediaminetetraacetate (dihydrate) salt is the EDTA source it has been almost always used in most of EDTA titrations to determine  a large number of metal cations including alkaline earth and transition metals. Keep in mind that $K_f$ values of alkali metal complexation with EDTA are very law compared to other metals (Ref.2):
$$
\begin{array}{|c|c|c|c|}
\hline
\text{Cation} & K_f & \text{Cation} & K_f & \text{Cation} & K_f & \text{Cation} & K_f\\
\hline
\ce{Li+} & 6.17 \times 10^2 & \ce{Be^2+} & 1.58 \times 10^9 & \ce{Cr^2+} & 3.98 \times 10^{13} & \ce{Sc^3+} & 1.26 \times 10^{23}\\
\ce{Na+} & 4.57 \times 10^1 & \ce{Mg^2+} & 6.17 \times 10^8 & \ce{Mn^2+} & 7.41 \times 10^{13} & \ce{Y^3+} & 1.23 \times 10^{18}\\
\ce{K+} & 6.31 \times 10^0 & \ce{Ca^2+} & 4.90 \times 10^{10} &\ce{Fe^2+} & 2.09 \times 10^{14} & \ce{La^3+} & 3.16 \times 10^{15}\\
\ce{Ag+} & 2.09 \times 10^7 & \ce{Sr^2+} & 5.37 \times 10^8 & \ce{Co^2+} & 2.04 \times 10^{16} & \ce{In^3+} & 1.00 \times 10^{25} \\
\ce{Tl+} & 3.47 \times 10^6 & \ce{Ba^2+} &  7.24 \times 10^7 & \ce{Ni^2+} & 4.17 \times 10^{18} & \ce{Bi^3+} & 6.31 \times 10^{27} \\
\hline
\end{array}
$$
This lack of complexation of alkali metals with even strong chelator such as EDTA can be attributed to their lower Lewis acidity (EDTA is a strong Lewis base because it contains both amino and carboxylate groups).
Apart from alkali metal cations, singly charged transition metal cations form  complexes with chelates and can be determine by relevant titrations. For example, concentration of silver(I) ions ($\ce{Ag+}$) can be determined by cyanide titration (a monodentate):
$$\ce{Ag+ + 2CN- -> [:\!N#C-Ag-C#N\!:]-}$$
The fairly high $K_f$ values for those singly charge cations (e.g., see the values for $\ce{Ag+}$ and $\ce{Tl+}$ in the table above) other than alkali metals suggest that they were able to involve in complexation titrations (They are compatible with alkaline earth metals).
References:

M. Font-Bardia, X. Solans, M. Font-Altaba, “Sodium ion complexes with ethylenediaminetetraacetic acid,” Acta Crystallographica Section C 1993, C49, 1452-1456 (https://doi.org/10.1107/S0108270193001350).
Arthur Earl Martell, Robert M Smith, Editors, In Critical Stability Constants, Volume 1; Plenum Press: New York, NY, 1974, pp. 204-211 (ISBN-13: 978-0306352119).


What are the products of [Ti(H2O)6]Cl3 + H2SO4 + K2Cr2O7 ---> I am slightly lost on how to proceed with this equation. I know potassium dichromate is a strong oxidizing agent but I am unsure of how to deal with it along with the titanium complex. I was wondering if I could get tips on how to handle these types of equations.

I know potassium dichromate is a strong oxidizing agent

You have already identified one partner and its role in a redox reaction. Good start!

So which component will be oxidized to what?

Can we assume that the reaction is carried out in water? If so, let's have a look at the ions in solution.

We'll have $\ce{H3O+}$, $\ce{K+}$, $\ce{Cr2O7^{2-}}$, and "that complex".

Let's have a look at $\ce{[Ti(H2O)6]Cl3}$.

Apparently, it's a complex of titanium with 6 neutral molecules of water as the ligands and 3 $\ce{Cl-}$ as the counter ions. So, what is the oxidation state of titanium in the complex when it is neutral with 3 $\ce{Cl-}$ as the counter ions?

Let's look again: $\ce{[Ti(H2O)6]^{3+}}$. OK, it is obviously $\ce{Ti(III)}$ and it seems that we have found the victim of our stong oxidant in the system.

We remember that most titanium compounds in nature are $\ce{Ti(IV)}$ oxides and conclude that we will have to balance our redox reaction to account for a $\ce{Ti(III) -> Ti(IV)}$ oxidation.

We could now continue to conclude that the reduction of $\ce{Cr(VI)}$ to $\ce{Cr(III)}$ is accompanied by $\ce{[Ti(H2O)6]^{3+} -> [Ti(H2O)6]^{4+}}$ but then we remember that $\ce{Ti(IV)}$ - tiny and higly charged - is a hard Lewis acid and will not exist as a naked cation, barely covered with just six neutral ligands in a hexaquo complex!

Instead, the likely $\ce{Ti(IV)}$ species is the titanyl cation, $\ce{TiO}^{2+}.$

Time to collect the puzzle pieces and write the redox reaction.


Are molecular optimizations in quantum chemistry really in vacuum? I know that Gaussian (or other quantum chemical packages) assumes the molecule is alone in vacuum, that is, the molecule doesn't interact with anything, when optimizing the molecule.

But I've found that optimization at HF or DFT levels results are almost the same as experimental data, which were from the solid state by X-ray diffraction. Everyone knows that the solid state is quite different from the vacuum, for molecules interact with each other in the solid state.

Then why do the two generate nearly the same results? Or have I looked at so few samples that I neglect occasions when the two give quite different results?

You raise an excellent question and the short answer is "yes, quantum chemistry calculations are intrinsically in vacuum".

Approximate methods like semiempirical (AM1, PM6, PM7) and hybrid DFT methods (e.g., B3LYP) use experimental data, which often comes from crystal structures.

For many years, single-crystal x-ray diffraction studies were the gold standard of "what is the geometry of a molecule." Performing gas-phase measurements of bond angles and bond lengths are extremely difficult, since they typically require a molecular beam of the intended target and then very high resolution IR,  rotational spectroscopy, or gas electron diffraction.

I once heard a talk by an extremely talented theorist, who asked the question "is an experimental crystal structure accurate enough?" He showed that in many cases, the answer is "no." (Consider that establishing the position of hydrogen atoms is difficult if not impossible with many single crystals.) The method he was proposing (geometry optimization with CCSD(T) and a basis set limit) isn't really appropriate for large molecules.

Until there's some huge database of gas-phase molecular geometry measurements and CCSD(T) calculations, most methods are compared with crystal structures.


Why do impurities lower the melting point of an isolated substance? It is known that impurities in a desired isolated product lower the melting point of the mixture, even if the impurities' melting point is much higher than the desired product. Why is that so?

It's a very general statement, but it's not always true. I'll explain why it's often true, and give a counter-example at the end.

Your majority component B and the impurity (let's call it A) form a binary system. In most cases, such binary mixtures exhibit a solid–liquid phase diagram as follows:

(image taken from these lecture notes).

This binary phase diagram has pure A on the left, pure B on the right. A and B form, somewhere, a eutectic. It is the point here at concentration e and temperature y. Because the existence of a eutectic point is guaranteed for any A/B binary system, and because the eutectic corresponds to a lower temperature, your liquidus curve decreases with increasing impurity concentration, and the impurity thus lowers the melting point.

However, not all binary mixtures form a eutectic. In the words of Wikipedia:

  Not all binary alloys have a eutectic point; for example, in the silver-gold system the melt temperature (liquidus) and freeze temperature (solidus) both increase monotonically as the mix changes from pure silver to pure gold.

The corresponding phase diagram is as follows:


Calculus and physical chemistry Is calculus and differential equations really that useful in physical chemistry? How does knowing that helps me in physical chemistry? And how much of calculus must I know in order to do physical chemistry topics such as thermodynamics? Because I am currently a chem major and am wondering what maths courses would be relevant. I currently completed single variable calculus and intend to go on to multivariate calculus as well as differential equations. I wonder how useful that'd be.

Calculus is the basis of physical chemistry. Have a look to Maxwell relations,Clapeyron equation. How stated Uncle Al you should take a book of physical chemistry and look into, furthermore some books (see Atkins) have an appendix showing you the mathematical skills required to understand the content of the book. Here what you should know only for thermodynamics in my opinion:

algebra
logarithms
vectors
calculus
Partial derivative
differential equations

If you go deep into the atoms more mathematical skills are required.


Does anisole react with concentrated nitric or sulfuric acid Can we add concentrated $\ce{H2SO4}$ and concentrated $\ce{HNO3}$ to anisole?

I heard that we can't add acids into phenol and aniline which make them blast.

Anisole (methoxybenzene) is an electron-rich arene that will react with sulfuric acid or nitric acid in an electrophilic aromatic substitution.

I suggest that you look the reaction up. Then grab a pencil, some sheets of paper and start to draw. Draw the arenium ions (Wheland intermediates) for the addition of the electrophiles at the different positions of anisole, have a look at the possible resonance structures and enjoy chemistry.

UPDATE

Btw, you can do the same reaction with aniline or phenol without causing an explosion.

My comment on aniline and phenol only works on paper but not in real life and Mithoron's comment on aniline and phenol is correct!

Nitroanilines and nitrophenols are synthesized in a different way.
In principle, nitration of aniline is possible with a mixture of nitric and nitrous acid, or after protecting the aniline as its formanilide.

Technically, 4-nitroaniline is produced by reacting 4-chloronitrobenzene with aqueous ammonia at elevated temperatures in an autoclave.

Details can be found in Ullmann's Encyclopedia of Industrial Chemistry:

  In the alternative continuous process, molten 4-chloronitrobenzene and 40 % ammonia liquor in a molar ratio of 1 : 17 are pumped at 20 MPa through a steel reaction coil held at 237-240 °C by oil bath heating. Inflow is controlled
  to give a contact time of ca. 10 min, and outflow is directed into an expansion chamber with a tangential flow of hot water to give, in quantitative yield, a suspension of product that is readily isolated and washed.

The same is true for the formation of 4-nitrophenol. Again, the starting material is 4-chloronitrobenzene, which is reacted with strong aqueous $\ce{NaOH}$ to yield the phenol.

These two reactions are examples for nucleophilic aromatic substitutions!


NMR and pseudorotation I think it has something to do with the fact that at high temperature it interconverts and so all the coupling constants are the same hence quintet but at lower temperatures the frequency of the rotation is less than the sample rate and so there are different coupling constants for the planar and axial fluorine atoms. We have two structures to consider; 3 planar-1 axial and 2 planar-2 axial. If my line of thought it correct the P-31 NMR spectra for each of these should superimpose to give the spectrum shown.

But I'm lost because when I draw my diagrams they don't match the spectrum shown, I end up with a much more complicated splitting pattern.

You are certainly thinking in the right direction.  Fluorine strongly prefers the axial position (for reasons described here), so at low temperature, the isomer that you've drawn on the right (2 axial and 2 equatorial fluorines) is the only isomer present.  As a result we have an $\ce{A2B2X}$ spectrum. The 2 equivalent equatorial fluorines (spin = 1/2) split the phosphorous signal into a triplet.  Each line of this triplet is further split into a triplet by the 2 equivalent axial fluorine's. BTW, the same reasoning that explains fluorines axial preference, also explains why generally $\ce{J(P-F)_{equatorial}}$ > $\ce{J(P-F)_{axial}}$

At room temperature, pseudorotation is rapid (more thermal energy is available and the barrier to pseudorotation can be surmounted) and all fluorines become equivalent on the nmr timescale. Therefore, at room temperature and above a quintet is observed in the $\ce{^{31}P}$ nmr spectrum.


Ab initio calculations Why are quantum mechanical calculations so computationally expensive? What exactly is the process that makes it so time taking?

What I mean is, one can perform classical dynamics simulations on a large system like proteins with no difficulty unlike ab initio calculations. Why? What type of calculations are to be performed in ab initio that makes it expensive? Matrix diagonalization?

It basically comes down to scaling with the size of the system, which will correspond to the number of electrons (that is, occupied valence orbitals) and the size of the basis set (which must have a sufficient quality, that is, size to yield meaningful results).

Edit: I should note that not all quantum chemists consider DFT to be ab-initio. Given the prevalence of such calculations, I have included it in my answer.

Let's consider density-functional theory (DFT) and Hartree-Fock (HF) theory first. These are a ground-state methods that describe a system with a single "wave function" called a Kohn-Sham or Slater determinant, respectively. This is valid for many, but not all systems - especially when there are several d-block metals closely interacting, this can be a bad approximation. For DFT, the cheapest useful specific methods (called functionals) in the most common implementations (i.e. programs) will scale as $\mathcal{O}(N^3)$ (where $N$ is some measure of the system size) due to the calculation of the electron-electron interaction.$^1$ At some point, a matrix diagonalization is performed, which has the same scaling. For HF, the scaling is worse, because of the calculation of a term called exchange (that arises from the specific form of the wave function ansatz), which scales as $\mathcal{O}(N^4)$.

At this point, I should mention that efforts to lower the prefactor of the computational cost function of these workhorses of quantum chemistry are ongoing and some have reached maturity. For "small" systems, the cost is then often dominated by terms other than the highest scaling ones. However, you will eventually hit the scaling wall. There are also efforts to lower the scaling behavior by introducing more approximations, smarter discarding of near-zero terms etc. Again, you may find out the hard way, that in the system you wish to calculate, the approximations do not hold or are not effective.

For excited states, one can assume a $+1$ scaling behavior and a worse prefactor. When considering the ground state with more accuracy, one then turns to post-DFT methods such as double-hybrid density functionals or post-HF methods such as Configuration Interaction, Møller–Plesset perturbation theory, Coupled-Cluster theory (e.g. CISD, MP$n$, CCSD(T)). In the standard implementations, they scale at least as $\mathcal{O}(N^5)$, but can go up to 6 and 7. Again, smart selection of the involved terms can lead to significant speed-ups, leading to an empirical linear or near-linear scaling at the expense of increased prefactors and less certainty.

The comments to the question mention classical dynamics of proteins. The scaling wall even for a single-point energy is hit well before approaching 100 $\ce{CNO}$ atoms (in my personal experience, which is a bit dated. Make it 200, then). The necessary derivatives with respect to the nuclear coordinates also tend to add to the prefactor or the scaling in a significant fashion.

$^1$ This scaling applies if a particular additional approximation called density fitting is made, which is usually a very good one. Note that better DFT methods will have worse scaling.


Does the mass of sulfur really decrease when dissolved in water and increase when burnt? I was going through a bunch of interesting science 'facts' and one entry went this way:

  Name an element whose mass decreases when it is dissolved in water and increases if it is burnt.

I tried guessing, but that didn't help. So I had a look at the answer: it was Sulfur

There are two questions that struck my mind after I saw the answer:

What does mass decrease/increase as stated here mean?(Does it mean that mass of sulfur along with other elements combined with it changes/or anything ambiguous such as change in actual mass is taking place) )
Is sulfur really the correct answer to that question? If yes, then why does sulfur show this anomalous property? Are there elements that display a similar property?

Any suggestion is welcome.

Upon reading the answers on Quora (thanks S007 for pointing that out) I realized this trick question is a lousy play upon two somewhat peculiar features of sulfur:

When submerged in water (not "dissolved", mind you, for sulfur does not dissolve), its apparent weight becomes less, thanks to Archimedes and the buoyancy force. This is indeed true of any element and compound, but sulfur has density of about 2, so its apparent weight decreases about twofold.
When burnt, sulfur forms a compound with oxygen which (containing all that sulfur plus oxygen) naturally weighs more than sulfur alone. Again, this is true of any element or compound that can be burnt, which are numerous. If anything, sulfur is "special" in that its atomic weight is 32, and the molecular weight of $\ce{SO2}$ is 64, so the increase is exactly twofold.

The way it stands now, though, the question is quite meaningless. Mass of any element never changes (barring nuclear reactions), that's the law of conservation. If you count compounds, then the mass would increase in any reaction (presuming we started from the pure element) and stay unchanged otherwise. If you count all compounds except gases, then the mass would sometimes increase and sometimes decrease, so there might be an answer, but then sulfur fails both conditions: it does not dissolve in water at all, and it all transforms to gases (that is, "vanishes") when burnt.


Can I use my extech pH 60 meter to measure soil pH if so how? Can I use my extech pH60 meter to measure soil pH if so how?
I included a link to the meter.

Extech pH 60

Take a soil sample and mix it with distilled water. The most common ratio used for soil-water pH is 1:1 soil:water.
Stir the soil and solution and wait 15 minutes to 1 hour to set the slurry.
Measure the pH (check the temperature first).


What does heterogeneous mean? We say that a homogeneous sample consists of the chemically same type of molecules whereas heterogeneous ones vary chemically, i.e. the molecules throughout are not same and also not uniform.
Why do we treat a (water + water vapour) system as heterogeneous in spite of the presence of the same kind of molecules? Also if there exists an equilibrium in the two phases would we still not call them homogeneous?
Does heterogeneity have a chemical basis or physical basis?
I do not seem to get a clear idea as to what defines heterogeneity. The definition has kept on changing since my high school and later on.

Any classification like chemical or a physical change, homogeneous or heterogeneous substance, will eventually fail at one point. From an analytical or physical chemistry's perspective, homogeneous substance is a substance, in a single phase, whose chemical composition is uniform. A heterogeneous substance has different composition from place to place like your garden soil.

Why do we treat a water+water vapour system as heterogeneous in spite of similar molecules present? Also if there exists an equilibrium in the two phases would we still not call them homogeneous?

It is a bad example to label water+ water vapor system as heterogeneous- but with respect to what? It is better to call it a two phase system. You have to define heterogeneous with respect to some property.
For example, if we take tap water in a glass, we can say the liquid (water) is homogeneous with respect to its chemical composition. Let us put some ice cubes in water, now it is a heterogeneous phase, although chemically it is all water.
More formally, from the International Union of Pure and Applied Chemistry's Gold Book:

homogeneity (in analytical chemistry): The degree to which a property
or a constituent is uniformly distributed throughout a quantity of
material. A material may be homogeneous with respect to one analyte or
property but heterogeneous with respect to another.

You can see the subtilty above in italics.


Simple alternative to UNIFAC to prove that a mixture containing 1ml of x and 5 litres of propylene glycol is not flammable? Section 2.6.1 of the REGULATION (EC) No 1272/2008 (CLP) defines a flammable liquid as:

  Flammable liquid means a liquid having a flash point of not more than
  $60 ^{\circ}$C.

So data on flash point are needed for the classification of a liquid.
The CLP goes on to state that data can be determined by testing, found in literature or calculated.

In the case of mixtures the flash point need not be determined experimentally if the calculated flash point of the
mixture, using the method described in Gmehling and Rasmussen (Ind. Eng. Fundament, 21, 186, (1982))], is at least 5°C greater than the legally defined threshold/s. This assumes that the liquid is homogenous and we know certain details about the components.

I have had a look at this method which appears to be called the UNIFAC method. As a non-scientist, it looks overly complicated for situations where you just intuitively know that a mixture is not going to be flammable.

I have a hunch that if I pour 1 ml of food flavouring into a 5 litre bottle of propylene glycol (PG) the resulting mixture is not going be classed as 'flammable' (where the flash point of PG is deemed to be 104°C ).

Is this hunch one of those areas where science and intuition conflict?

Does it make sense to perform the whole UNIFAC process to confirm this conclusion or is there a simpler (but still scientific) 'rule of thumb' that we can use to discount having to perform a UNIFAC method?

For example, if I have a mixture containing:

1 ml of pure water, which as far as I can tell has no flash point, and
2 litres of PG (flashpoint 104°C)

Is the UNIFAC process the simplest way to determine the flammability of this mixture or is there a simpler, yet reliable, method to determine the flammability of this mixture?

UNIFAC is not a procedure, unless by chance in a similar field two things got the same name, which is unlikely.

UNIFAC is used by activity coefficient models for predicting liquid properties. It is often coupled with a vapor phase equation of state, and quite often simply the ideal gas law...

A flash point calculation is a procedure, and must be done in whole. Note: you can do either an experimental flash calculation, or a computational one!. UNIFAC is used in the latter. It combines the fact that at equilibrium a specie in both liquid and vapor must have the same chemical potential (gotten through via activity coefficients in the liquid phase and fugacity coefficients in the vapor typically) coupled with mass balance. You cannot do half of it and have an answer... like all calculations it must be done fully and correctly. It is not that difficult, but it should be done by someone who is a scientist or engineer... This is the type of thing that would be expected of a chemical engineer.

Most chemical engineering Thermodynamics textbooks have a decent section covering flash and bubble/dew point calculations.

There are many programs that can do flash calculations for you. VMGSim, Aspen/HYSIS to name just two of the many commercial/industrial simulators plus a plethora of small programs online made by people like me (mine isn't online ;) ).

I just did a quick google and this looks like it would work https://checalc.com/solved/multi_flash.html

Note, you do not need to use UNIQUAC in the liquid, but, it usually by default will use UNIFAC to fill in unknown interaction parameters, so it is basically a safe activity coefficient model. Depending on what is in the vapor, the peng-robinson or even ideal gas equations of state are probably good enough. If there is association or the molecules are polar, then I recommend the Haydon O Connell.

The link I sent looks like it uses the Peng-Robinson equation of state for liquid and vapor. It is probably okay for what you are doing. Actually the link I sent doesn't look like it has PG in its database... this one may... https://www.daoftx.com/usingflsc.htm

But you know what to look for now... These premade programs just need you to tell them the species, initial composition, T, P and then you just hit calculate. You just need to find one that has your molecules in its database.


How to find the angle of emission of an X-ray during impact of accelerated cathode rays? X rays are produced when accelerated cathode rays strike a metal target or a metal anode.
How can we find the angle at which the X-ray is emitted from the metal anode or the target when the cathode ray strikes the metal perpendicular to the surface?
Does anyone know how?

The angle of resulting X-ray radiation, product of the impact, is related to the amount of energy or momentum of the cathode ray. Bremsstrahlung relation will amount how much energy the produced photon has.
$$hf = \frac{hc}{\lambda} = K - K'\tag{1}$$
$$\lambda_\mathrm{min} = \frac{hc}{K} = \frac{hc}{e\,\Delta V}\tag{2}$$
Back scatter the possible photon angles with the Compton effect equation,
$$\lambda' - \lambda = \frac{h}{m_\mathrm{e}c}(1 - \cos\theta)\tag{3}$$
A difference in kinetic energy is as difference in wavelength. Precise results should consider a vacuum and use imaginary numbers; also probability distribution on superposition.
For more details refer to the chapter 3 of Krane's Modern Physics [1].
Reference

Krane, K. S. Modern Physics, 3rd ed.; Wiley: Hoboken, NJ, 2012. ISBN 978-1-118-06114-5.


What's the smallest Hoberman sphere possible? What is the smallest Hoberman sphere possible? Could it be possible on an atomic or molecular level?

Image courtesy of Hoberman Associates, Inc

I realize the issue with this will be the scissor joints. I am wondering how small these could be made and arranged. Would it be possible to create a functional Hoberman sphere with carbon nanotubes?

I am also curious if anyone has any opinion or knowledge about if it is possible to make a Hoberman “molecule” or rather a Hoberman sphere made of atoms? If so, what elements would make for the best Hoberman molecule? Could scissor joints be made functional at the molecular level both expanded and closed using something like redox reactions?

Thanks in advance for any answers!

Hoberman sphere shares the same topological features being an icosidodecahedron with a family of polyoxometalates called Keplerates $\ce{\{(\color{teal}{Mo|W})_72\color{orange}{M}_30\}}$ ($\ce{\color{orange}{M} = Fe,Cr,V}$, anionic residues and alkali metal cations are removed for clarity) ([1, 2]):

Also, there is a cluster $\ce{[\color{blue}{La}_20\color{green}{Ni}_30(IDA)30(CO3)6(NO3)6(OH)30(H2O)12](CO3)6 * 72H2O}$ (IDA - iminodiacetate $\ce{C4H5NO4}$) consisting of an icosidodecahedron of Ni(II) Ions
encapsulating an inner La(III)-dodecahedron [3]:

The problem is, these structures only resemble the symmetry, but not the expandability of Hoberman sphere. Currently there are many attempts of creating nanomachines with mechanically active molecular assemblies, but I didn't find any working models close to Hoberman structures (yet).

Also, some "spherical" viruses such as Cowpea chlorotic mottle virus (CCMV) with a structure of truncated icosahedron, which is significantly larger than Keplerates ($\pu{28 nm}$ (CCMV) vs. $\pu{3.2 nm}$ (average Keplerate)), can undergo swelling (pH < 6.5) and expansion (pH > 6.5), and show pH and metal ion dependent polymorphism [4, 5]

  as a result of deprotonation of carboxyl moieties at the inter-capsomere contacts, leading to electrostatic repulsion that opens out the structure but falls short of complete disassembly through preservation of interwoven
  carboxyl/protein links between capsomeres.

Don't forget that "there's plenty of room at the bottom", so Hoberman sphere can possibly be a real working entity at molecular scale.

Bibliography

Botar, B.; Ellern, A.; Hermann, R.; Kögerler, P. Angewandte Chemie International Edition 2009, 48 (48), 9080–9083. DOI 10.1002/anie.200903541.
Schäffer, C.; Merca, A.; Bögge, H.; Todea, A. M.; Kistler, M. L.; Liu, T.; Thouvenot, R.; Gouzerh, P.; Müller, A. Angewandte Chemie 2009, 121 (1), 155–159. DOI 10.1002/ange.200804496.
Kong, X.-J.; Ren, Y.-P.; Long, L.-S.; Zheng, Z.; Huang, R.-B.; Zheng, L.-S. J. Am. Chem. Soc. 2007, 129 (22), 7016–7017. DOI 10.1021/ja0726198.
Speir, J. A.; Munshi, S.; Wang, G.; Baker, T. S.; Johnson, J. E. Structure 1995, 3 (1), 63–78. DOI 10.1016/S0969-2126(01)00135-6.
Kovacs, F.; Tarnai, T.; Guest, S. D.; Fowler, P. W. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 2004, 460 (2051), 3191–3202. DOI 10.1098/rspa.2004.1344.


Why is the speed of ideal gases mass dependent while the kinetic energy is not? The rms speed of an ideal gas is $v_{\text{rms}}$ = $\sqrt{\frac{3RT}{M} }$ and the kinetic energy is $E_\text{k} = \frac32RT$. From this, it is concluded that the speed is mass dependent, while the kinetic energy isn't.

This doesn't make sense to me. I know the speed and kinetic energy for things other than ideal gases are surely mass dependent. The heavier an object is, the slower it is and will have a bigger kinetic energy. Why isn't this true for ideal gases?

As the mass increases, the velocity decreases, yes that is true. But how does the kinetic energy remain the same?

The increment in the mass is balanced by the decrement in the velocity. In other words, the velocity changes as the mass changes in such a manner that the kinetic energy of the molecules of the gas will remain to be same no matter what. It's something like the multiplication of two variables always giving the same constant value. Keeping in mind that this is true in an ideal environment.


Are phenols good antioxidants? Are phenols good antioxidants? If so what makes them good? I think it is because they oxidize easily but I'm not sure how that helps.

Phenolic compounds, or in general aromatics actually, has electrons which many oxidants can easily abstract or bind to molecule.

http://www.springer.com/cda/content/document/cda_downloaddocument/9781441978257-c1.pdf?SGWID=0-0-45-1106848-p174056970

From the above link in page 17,

as you see the highest rates, order of magnitude or even more higher ones come from aromatics, or at least pi-bond between carbons. It can reasonably assumed that if you have these compounds in a mixture, the possibility of any oxidant introduced reacts with them is higher than the other ones. In short, they are good anti-oxidants in that way.


Why does the amount of CO2 produced decrease with more baking soda? I'm in chemistry class and we have an assignment and they gave us a video and I took down the data they gave us. They fist measured out 10mL of vinegar for each trial and used different amounts of baking soda in all 8 trials. They put the baking soda in a beaker then poured the vinegar into the beaker and capped it with a one hole stopper and a syringe in the hole. It was about a 1-1.5 second delay between poring the vinegar and capping it. I don't understand why the amount of CO2 produced wouldn't increase at the beginning then flat-line when the vinegar becomes the limiting reactant. I posted a graph of the data we received. All trials had the same amount of vinegar.

This graph is a very good example of experimental errors which may give rise to a pseudo-phenomenon i.e., an observation which apparently looks like a scientific phenomenon but it is an artifact of the experimental design.

First of all work out the moles of acetic acid (5% acetic acid) and moles of NaHCO3. As you can see vinegar has become a limiting agent. And it should have become a flat curve only if your system was a closed system.

So where is the culprit in the experimental design?

As you stated "It was about a 1-1.5 second delay between poring the vinegar and capping it."

By the time, someone puts a syringe, a large volume of gas has already escaped when acetic acid was poured on a heap of sodium bicarbonate. How big is the heap of 3 g NaHCO3!

There is a nice trick to avoid these problems. You put the sodium bicarbonate in an open plastic pouch so that acid does not immediately encounter the salt. You can stopper and then shake to mix (quite well).

So your hypothesis is right for a flat curve, sadly there is an experimental design issue.


How does lime prevent soda lime glass from dissolving in water? How does lime addition as an ingredient prevent soda lime glass from dissolving in water? Does it form calcium silicate in the process of melting?

The addition of lime does indeed form calcium silicate, but it doesn't serve the purpose you mentioned. Silica of itself isn't water soluble, nor soda glass.

The difference is in the crystalline structure. Silica ($\ce{SiO2}$) consists of silicon atoms bridged by oxygen atoms, forming a periodic crystal with no net charges on any atom.

However when some alkali is introduced ($\ce{Na2O}$), the additional oxygen atoms remain charged, and cannot link silicon atoms together; those form silicate anions. The silica breaks and depolimerizes, and the remaining $\ce{Na+}$ fill holes in the structure to maintain charge balance. Thus the solid melts at lower temperature with reduced viscosity, it's "more fluid".

Lime contains calcium, which bonds more strongly to silicate anions than sodium. It also decreases viscosity of molten glass because of the non bridging oxygen, but in addition increases hardness and tensile strength of the glass in solid state.


Why is the formation of double helix between non complementary sequences disfavoured? The textbooks starts with the phenomena that double helix is an expression of the rules of chemistry.

  Hydrogen bonds are important in determining the formation of specific
   base pairs in the double helix. However in single staranded DNA, the hydrogen bond donors and acceptors are exposed to solution and can form hydrogen bonds with water molecules.
  When two single strands come together, these hydrogen bonds with water are broken and new hydrogen bonds between the bases are formed. Because the number of hydrogen bonds broken is the same as the number formed, these hydrogen bonds do not contribute substantially to driving the overall process of double helix-formation. However, they contribute greately to the specificity of binding.

I see the point that some neucleotides which form stable links are the complimentary bases (purine and pyramidine). Some bases do not form stable links. I did search my question online, I am not able to get near anything that's directly related or comprehensible at my level. So following is what I do not understand.

  Suppose two bases that cannot form Watson-Crick base pairs are brought together. Hydrogen bonds with water must be broken as the bases come into contact. Because the bases are not complementary in structure, not all of these bonds can be simultaneously replaced by hydrogen bonds between the bases. Thus, the formation of a double helix between noncomplementary sequences is disfavoured.

What's the specificity mentioned here? What does it mean by non complementary in structure? An example is appreciated.

If you take a look at the base pairings, as shown below, you can see that adenine and thymine form two hydrogen bonds when they are matched, while guanine and cytosine form three hydrogen bonds when they are matched. If you try to match an adenine with cytosine or guanine with thymine, the orientation of hydrogens and lone pairs will not line up.

For example, guanine has two hydrogens and one lone pair available for hydrogen bonding, while cytosine has one hydrogen and two lone pairs available for hydrogen bonding. They fit! However, if you try to match adenine (with one hydrogen and one lone pair available for hydrogen bonding) with cytosine, the elements will not match or line up, resulting in an energetically less-favorable interaction.

Summary: A base on one strand of DNA requires another, specific base to complement the exposed hydrogens and lone pairs to form the most most energetically favorable interactions.


How do you separate enantiomers? There are some stereochemical reactions that result in the presence of enantiomers. When moving forward with a practical organic synthesis, how does one usually separate them in order to continue with one of the enantiomers?

There are several ways that enantiomers can be separated, but none of them are particularly simple.

The first way to separate them is chiral chromatography.  In chiral chromatography, silica gel is bonded to chiral molecules to form what is called a chiral stationary phase.  The enantiomers will then separate as they run down the column because one of the enantiomers will interact more strongly with the column and "stick" in place.  Chiral sugars (ex. cellulose) are frequently used in chiral chromatography.

The second common method is to react the enantiomers with another chemical to form diastereomers.  While enantiomers are identical in terms of chemical properties, diastereomers are not.  Diastereomers can be created by reacting a mixture of both the enantiomers with another chiral molecule, such as s-brucine, which is commonly used because it is cheap.  Diastereomers have different chemical properties (for example melting points), so it is much easier to separate them.  Then, after separation, the enantiomers can be recovered from the single diastereomer.


Why is gold unreactive when only one electron is in the outer shell? I've been trying to answer my (high school) daughter's questions about the periodic table, and the reactivity series, but we keep hitting gaps in my knowledge.

So I showed that the noble gases have a full outer shell, which is why they don't react with anything. And then over the other side of the periodic table we have potassium and sodium, which have only one electron in their outer shell, which is what makes them so reactive, and at the top of our reactivity list. (And the bigger they get, the more reactive, which is why we were not allowed to play with caesium in class...)

But then we looked up gold, which is at the bottom of the reactivity series, and found it also has only one electron in its outermost shell (2-8-18-32-18-1).

Is there an easy explanation for why gold doesn't fizz like potassium when you drop it in water?

(This question could be rephrased as "What properties of each element decide their ranking in the metal reactivity series?" if you prefer; that was the original question we were trying to answer.)

First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer

The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).

A nice summary of some work by Jens K. Norskov can be found here:
http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490

  In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms.

I think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions.

You can, of course, react gold with aggressive reagents like aqua regia, a 3:1 mix of $\ce{HCl}$ and $\ce{HNO3}$.

If properly done, the product is $\ce{HAuCl4}$ or chloroauric acid.


Why isn't the silicon analogue of benzene flat? Why isn't the silicon analogue of benzene flat? As far as I understand, silicon and carbon make similar structures and it doesn't make sense that its benzene analogue should be substantially different.

Quoted from the article:

  Chemists in the UK have constructed a structural analogue of benzene made from silicon atoms. The molecule is not flat like benzene, but it reveals a new type of aromatic stabilisation.

This 'new' type of aromatic stabilisation is said to occur only around the seat of the chair structure.

I am thinking that silicon's bonds are too weak to be kept in a double bond, and therefore it doesn't have enough energy to change the shape of the bonding and is thus quite unstable.

The linked article is a short summary of a Science article: "A Tricyclic Aromatic Isomer of Hexasilabenzene" by Kai Abersfelder, Andrew J. P. White, Henry S. Rzepa, and David Scheschkewitz; Science 2010: 327 pp. 564-566.

They propose a new term, "dismutational aromaticity":

  The highly dispersed $\ce{^29Si}$ nuclear magnetic resonance shifts in solution ranging from +125 to −90 parts per million indicate an inhomogeneous electron distribution due to the dismutation of formal oxidation numbers as compared with that of benzene. Theoretical analysis reveals nonetheless the cyclic delocalization of six mobile electrons of the $\pi$-, $\sigma$- and non-bonding type across the central four-membered ring.

They point out that while $\ce{Si=Si}$ bonds were ignored for a long time, multiple compounds have been prepared, even including delocalization, like 1,2-disilabenzenes.

They prepared $\ce{Si6R6}$ with "Tip" or 2,4,6-triisopropylphenyl groups.

  The surprisingly stable 3a can be exposed to air for hours as a solid or for minutes in solution without detectable changes.

Long story short, there is delocalization around the central 4-center ring:

Note that there is a 1,2-shift occurring, and that 2 $\ce{Si}$ atoms have two substituents, 2 have one substituent, and 2 have no substituents.

The paper argues that a fully planar hexasilabenzene might be possible, but:

  the choice of precursor may be important in any synthetic approach.

Presumably the choice of substituent is also relevant to prevent the rearrangement here.


quantum number - question about the ms-value I've got the quantum number $\ce{(3,2,2,-1/2)}$, and the question is which electron belongs to this quantity. The answer is orbital $\ce{3d^1}$. But I don't understand why. I would have understood it, if the ms-value was $\ce{+1/2}$. But how can you have $\ce{-1/2}$?

As porphyrin clearly explained, the first three quantum numbers usually are enough to determine the orbital of a given electron.

You probably know this but I'd like to let it here for everyone else.

Each different value of the principal quantum number $n$ yields a different wave function and a different state of the electron. In atoms, $n$ defines the order of all states according to energy.

The second and third quantum numbers $l$ and $m_{l}$, arise from the quantization of rotation where $l$ gives the shape of an orbital and its angular distribution and $m_{l}$ determines the number of orbitals and their orientations in a subshell.

The spin quantum number of an electron is denoted $s$ and it has the value of $s=1/2$. The spin magnetic quantum number is denoted $m_{s}$ gives the projection of the spin angular momentum on the z-axis. Therefore, one electron can have only one orientation: either $m_{s}=+1/2$ which we commonly denote up-spin or $\alpha$ and $m_{s}=-1/2$ denoted as down-spin or $\beta$.

What I see in the comments are partial answers because to associate an electron (not an orbital) to these numbers you need to use $m_{s}$. Then, you have a spin-down or $\beta$ electron in a $3d$ orbital.

Answering your question in the comments. To have a $3d^{2}$ you'd need two pairs of four quantum numbers: $(3,2,2,+1/2)$ and $(3,2,2,-1/2)$. Here, two electrons occupy the same region in space but they have different orientations with respect to their spin angular momenta.


Oxidizer/reducer in decomposition of potassium chlorate I'm in the process of tutoring a close friend of mine in basic redox chemistry, and this problem shows a little bit of my chemistry atrophy (It's been two years since I've been active in chemistry work).

The problem stated is the decomposition of potassium chlorate into potassium chloride and oxygen gas, so

$$\ce{2KClO3 -> 2KCl + 3O2 ^}$$

Obviously, the half-reactions of this process can be written as the following:

$$\ce{2Cl^5+ + 12e^- -> 2Cl^-}$$
$$\ce{6O^2- -> 3O2 + 12e^-}$$

The issue I'm encountering is in successfully demonstrating the identity of the reducing agent and oxidizing agent. It's clear from the half-reactions which entity is oxidized versus reduced (and the fact that $\ce{K^+}$ is a spectator ion), but my friend has prior worked examples that appear to show that the identities of the reducing/oxidizing agent are that of the reactants. If so, can $\ce{KClO3}$ be both the reducing and oxidizing agent? If not, would this information be stated as the individual ionic entities listed above?

If I remember my introductory lab course correctly, there are two reactions going on, one after the other. First, $\ce{ClO3-}$ disproportions according to the following equation:

$$\ce{4 ClO3- -> Cl- + 3 ClO4-}$$

Which has these two half-equations:

$$\ce{ClO3- + 6 e- + 6 H+ -> Cl- + 3 H2O}\\
3(\ce{ClO3- + H2O -> ClO4- + 2 e- + 2 H+})$$

Where the formally required water is merely spectator.

Only upon further heating does the perchlorate anion decompose according to:

$$\ce{ClO4- -> Cl- + 2 O2}$$

Which again can be broken down to:

$$\ce{ClO4- + 8 e- + 8 H+ -> Cl- + 4 H2O}\\
2(\ce{2H2O -> O2 + 4 e- + 4 H+})$$

And again formally requires water as a spectator.

Note that it is not at all unusual for one entity to be both; the simplest example is probably the disproportion of hydrogen peroxide according to:

$$\ce{2 H2O2 -> O2 + 2 H2O}$$

Where one molecule is reduced by another of the same type.


When a bond is formed between an electophile and a nucleophile, they both lose energy and become more stable right? This is a section from clayden's organic chemistry textbook. I do not understand why it says energy gain and not energy loss

In the context of the picture provided, it seems just a matter of language choice. When energy is lost, it is gained somewhere else because of conservation of energy. So a change in energy corresponding to a gain is at the same time a loss elsewhere. The author of that diagram could have used interchangeably energy loss without loss of meaning.


Does O2 have a color in the gas phase I have noticed that liquid $\ce{O2}$ (I clarify it as $\ce{O2}$ because oxygen exists in several other forms which have different colors in the liquid state than $\ce{O2}$) has a light blue color to it.

In the solid state it has a light blue color like it does in the liquid state and then finally a red color if enough pressure is applied. It can also be pink, black, or metallic again depending on the pressure and temperature.

Does it also have a light blue color in the gas phase?

Background

$\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of solid oxygen with color ranging from pale blue to red to black.  In the liquid phase it has a light blue color.  This color is due to light absorption by the ground state triplet according to the following equation
$$\ce{2 O2(^3\Sigma_{g}) ->C[{h\nu}]\   2 O2(^1\Delta_{g})}$$
This absorption requires light from the red region of the spectrum (~630 nm).  If red light is absorbed , then blue light is transmitted or reflected and this gives rise to the blue color associated with liquid oxygen.

Answer

Note that the light absorption requires 2 molecules of $\ce{O2}$ and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller.  Hence, the probability of photon absorption in the gas phase is much reduced.  To the human eye oxygen gas will appear colorless (at normal and reduced pressures).  However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption.


General justification for steady state approximation I want to understand the assumptions that have to be fulfilled in order for the steady state approximation to work. I did some research myself and was able to deduce assumptions of the form $k_1\gg k_2$ for some systems, by solving the system exactly and taking the limiting case in which the exact system reduces to the approximative system. (Later I found this answer doing the same thing Meaning of steady state (kinetics)) This approach will not work for most systems though, as no closed exact solution exists for most non-linear differential equation systems.

I want to develop a clear chemical intuition in which case it is justified to use the approximation. Most literature seems to just tell the reader that it's fine to use the approximation in the reaction thats discussed. This is unsatisfying. From a mathematical standpoint this approximation is all but trivial.

No very satisfying justifications are available unless you do a case-by-case analysis. But PSSA does not come from common sense. In general, pseudostationary state approximation (PSSA) is justified by the theory of differential equations.

In fact, the pseudostationary state concentrations represent the zeroth approximation
to a set of particular solutions of the correct chemical kinetic equations, which are called principal solutions. After the initial period, the concentrations of the intermediates approach their principal solution rather than their pseudostationary state values.

Therefore, the accuracy or usefulness of the pseudostationary state assumption depends upon the magnitude of the differences between the principal solution and the pseudostationary state concentrations.

OK, that sounds useless. But after doing some maths (like finding the principle solutions), some (obviously very ambiguous) rules are indeed justified mathematically.

  Induction period should be short. (As you can see PSSA solutions cannot satisfy the initial conditions)
  Destruction of the intermediates should be fast.

For example, in $\ce{A ->[$k_1$] B ->[$k_2$] C}$ if $k_2\gg k_1$ then the conditions above are met while if $k_1\gg k_2$, they aren't.

By the way, PSSA originates from Michaelis-Menton theory for enzyme kinetics. PSSA in it is well justified while the conditions above are not necessarily met. Consider

$$
\ce{E + S <=>[$k_1,k_{-1}$] ES ->[$k_2$] P}.
$$

Let $[\ce{S}]=s,[\ce{E}]=e,[\ce{ES}]=c,[\ce{P}]=p,e+c:=e_0,s+p+c:=s_0$. Furthermore, let $\tau=k_1e_0t,u(\tau)=s/s_0,v(\tau)=c/e_0$. The kinetic equations can be written as

$$
\begin{cases}
\frac{\mathrm du}{\mathrm d\tau}=-u+(u+K-\lambda)v\\
\color{red}{\varepsilon}\frac{\mathrm dv}{\mathrm d\tau}=u-(u+K)v
\end{cases}
$$

where  $\lambda=k_2/\left(k_1s_0\right),K=\left(k_{-1}+k_2\right)/\left(k_1s_0\right),\color{red}{\varepsilon=e_0/s_0\ll 1}$. Therefore, PSSA can be applied (i.e. setting $u-(u+K)v=0$, justified by perturbation methods) without loss of much accuracy. This example indicates that more often than not, a case-by-case analysis is required.

Finally, I think to use PSSA safely, one should do some experiments to check (numerical simulations, real experiments, etc.).


should charge balance be based on activities? We have a fluid with $\pu{0.01 M}~\ce{HCl}$ and $\pu{0.02 M}~\ce{K2HPO4}$, so if we assume activity coefficients are $1$ throughout we can find pH as $6.800012$ and confirm that at that pH there is approximately zero balance ($6.104839\!\times\!10^{-13}$) between positive and negative charges calculated as

$$\mathrm{  K^+ +H^+ – OH^- - Cl^- -}\ce{H2PO4-}\mathrm{ - 2\times} \ce{HPO4^2-} \mathrm{\ -\ 3\ \times}\ce{PO4^3-}$$

Now we can find the ionic strength and via Debye-Hückel/Davies find activity coefficients for absolute charge 1-3. This allows correction of the thermodynamic dissociation coefficients for phosphoric acid (and water dissociation) so we can correct the concentrations and find a new pH to minimize charge balance. After $4$ iterations we have pH 6.436864 at a charge balance of $1.429412\!\times\!10^{-15}$. This procedure allows reproduction of the results in a recent paper (Schell et al J Chem Eng Data 2018; 63: 2151-2161) with correlation $>0.9999$.

Now the question: these results requires the charge balance being formulated strictly in concentrations multi-plied by the relevant charges. I simply wonder why we should not for the charge balance equation multiply the found activity coefficients on the charged species so making the balance to minimize a function of $f_3\times 3\times [\ce{PO4}]$ and so forth with $f_3$ being the trivalent activity coefficient from Davies?

No, charge is not really a activity based property.  It is a number.  Just as the activity coefficients don't change the number (or moles) of potassium atoms or phosphate anions, they don't change the charge either.

If you told me that one liter of a 1.0 molar potassium chloride solution had activity coefficients of 21 for potassium and 186 for chloride, then the amount of charge in the solution is still one mole of potassium cations and one mole chloride anions.


Does there exist a black coloured metal? Most of the metals are just painted/coated black. Does there exist a black coloured metal, either natural or artificial, such that even when we scratch it, it will still remain black?

Highly and degenerately doped semiconductors and semi-metals can be at least dark in visible light in many cases yet have good electrical conductivity. However it would be a stretch to call them either "metals" or "black" without qualifying those labels.
They appear dark in visible light for (superficially at least) the same reason that obsidian or black marble appears dark; they mostly transmit light entering the surface rather than reflect it, then slowly absorb it internally over microns or millimeters. The same is true for many semiconductors, but since their index of refraction in visible wavelengths is in the 2.5 to 5 ballpark their specular Fresnel reflection can be 20 to 50%! Compare to glass with an index of 1.4 or 1.5 and Fresnel reflection of ~4%.
See all the excellent answers to Why are most metals gray/silver? in Physics SE.
There's a short discussion of plasma frequency and semiconductors in this answer which begins:

Metals have a high density of free carriers that can be though of as a plasma. The plasma frequency of metals is usually in the ultraviolet (except for Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?), meaning any electromagnetic wave below the plasma frequency would be rapidly absorbed and re-emitted (fancy words for "reflected" by the metal). Just like Earth's ionosphere reflecting HF frequencies and lower but transmitting most VHF and higher, a metallized film that reflects visible light and infrared light (heat) would also reflect all RF radiation; anything below the plasma frequency.
$$\omega_{pe}=\sqrt{\frac{n_e e^2}{m^* \epsilon_0}}$$

A polished surface could look different depending on what is being reflected, but a roughened or ground surface will look grayish.
This piece looks fairly smooth and so it has a shiny appearance due to Fresnel reflection, but it only reflects about 40% of the light, whereas a polished metal can be 95% or higher in some cases.


Why is the Friedel-Crafts alkylation of nitrobenzene disfavoured? Suggest a synthetic route to m-nitrotoluene, starting from benzene.

The conversion can be achieved in two ways:

$\hspace{10 mm}$

or:

However, the first route is not possible as the Friedel-Crafts reaction does not yield a meta-substituted product. Why is that so?

The nitro group is so deactivating that yields are really poor for Friedel-Crafts reactions with nitrobenzene; not even the meta-substituted product is found in good yield. In fact, nitrobenzene is actually used as a solvent in Friedel-Crafts reactions.