Can hydrobromic acid perform substitution with propane? I am a student in the 11th grade, we recently had a chemistry exam, I lost marks because of a question that asks about the reaction between $\ce{HBr}$ and $\ce{C3H8}$. According to our teacher  $\ce{HBr + C3H8}$ is going to react and form $\ce{C3H7Br + H2}$. In my opinion I don’t think that a reaction is going to happen, because $\ce{HBr}$ is an acid and acids can't react with alkanes. Can you please state your opinion and justification?

I am afraid that you cannot support your argument. Here are two reasons why as I expand on my comment above. If this process were to be a radical chain reaction, the following process should be considered. Hydrogen bromide has a weaker bond then any of the C-H bonds in propane. An initiation step (heat, light?) would be expected to generate atoms of hydrogen and bromine. If you use a bromine atom as a reactant in propagation step 1 (P1), you will produce a hydrogen atom at the end of P2. So we start with a hydrogen and abstract a secondary hydrogen (primary hydrogen is higher in bond strength but it will only make the analysis less favorable) to produce a secondary propyl radical. The bond dissociation energies (BDE's in kcal/mol) are listed over the bonds of interest. The first step is good news. Exothermic by -5 kcal/mol. But P2 is endothermic by +14 kcal/mol. If you add P1 + P2 you get the overall reaction that has a heat (enthalpy) of reaction of +9 kcal/mol. The reaction is endothermic and uphill. The initiation step, the hydrogen atom and the propyl radical do not participate in the computation. The second method is to use heats of formation (in red). Using Hess's Law (ΔHorxn= ΔHfoproducts - ΔHforeactants) gives the same result, give or take a little. However, the reaction will be exothermic in the reverse direction given the right conditions. Sorry not to have better news.


Can chemicals be identified by their molar mass? Wouldn't it be possible to identify a chemical by measuring its molar mass? How much accuracy does the measured molar mass need to be to guess what chemical it could be?

EDIT:
@ortho makes an important point about isomers, but organic chemistry is much harder ;). Let's just focus on inorganic chemistry (though there are still isomers, usually there aren't).

It depends.

This was pretty much the way that the elements were discovered.

For inorganic chemicals in general the problem is really one of specificity. So I want to work with a chemical sample between 1 gram and 20 grams and I have an analysis good to 1%. That only yields about 300 statistically significantly different numbers $1.01^{300} = 20$. So for various inorganic salts this wouldn't work too well.

To take this into another field think of geology. Hardness and specific gravity are simple tests used to help identify minerals. Neither is very good by itself, but combined they sort out the minerals into a lot more bins.


Rate-limiting in transition state theory Suppose an initial state $A$ can transition to either state $C$ or state $D$. Suppose further that both of these two processes are rate-limited by a transition to the same intermediate state $B$, as follows:

Note that $\Delta G_{AB}> \Delta G_{BC}>\Delta G_{BD}$.

According to transition state theory, rates are determined by the rate-limiting step. So the rate of each process will be the same,

\begin{align}
r_{A\to C}&=r_{A\to B} \\
r_{A\to D}&=r_{A\to B}
\end{align}

from which it follows that the number of transitions made during some large $\Delta t$ will also be the same,

\begin{align}
N_{A\to C}&=r_{A\to B}\Delta t \\
N_{A\to D}&=r_{A\to B}\Delta t
\end{align}

However, if we were to consider the two pathways collectively then we would predict more transitions to be made to state $D$ than $B$ since $\Delta G_{BD}<\Delta G_{BC}$.

How do we resolve this apparent discrepancy?

In transition state theory (TST, goldbook) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us
\begin{align}\ce{
A &<=> [AB]^{$\ddagger$} -> B\\
B &<=> [AB]^{$\ddagger$} -> A\\
A &<=> B,
}\end{align}
and in addition to this, also
\begin{align}\ce{
B &<=> [BC]^{$\ddagger$} -> C & B &<=> [BD]^{$\ddagger$} -> D\\
C &<=> [BC]^{$\ddagger$} -> B & D &<=> [BD]^{$\ddagger$} -> B\\
B &<=> C                      & B &<=> D,
}\end{align}
as well as
\begin{align}\ce{
C <=> [BC]^{$\ddagger$} -> & B <=> [BD]^{$\ddagger$} -> D\\
D <=> [BD]^{$\ddagger$} -> & B <=> [BC]^{$\ddagger$} -> C\\
C <=> & B <=> D,
}\end{align}
and
$$\ce{
C <=> A <=> D\\
[AB]^{$\ddagger$} <=> [BC]^{$\ddagger$} <=> [BD]^{$\ddagger$}
}.$$

You can use TST to estimate rate constants.
$$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$

Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence:
$$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} =  \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$

The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics.

For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.


Can you eliminate cigarette smoke odor between walls? If holes are made in dry walls is there a substance that can be sprayed in the walls to eliminate cigarette  odor?

Getting rid of odor is a difficult task in general, particularly since molecules stick to surfaces and often stick in surfaces.

There's a good reason that "Febreze" works well. It contains a cyclodextrin ring that binds many odor-causing molecules. So the molecules are still there, but bound in a form you can't smell anymore.

(I remember an anecdote that they had to develop special fragrances to work with Febreze, but I can't find that anywhere.)

I suspect spraying paint primer inside the walls could work. This is often used to fix smoke damage from fires.


Neither Raman nor IR Active vibrational modes Is there any molecular vibrational mode (or activity) that is neither Raman nor IR active?

In other words. Is there any activity in the molecules that can't be captured by either Raman spectrometer (all wavelenght) or IR absorption spectrometer (all wavelenght)? What are those?

Yes it is possible. While no normal modes can be both IR and Raman active in molecules with a centre of symmetry, as mentioned in a comment molecules with silent modes exist.

The easiest example is  probably the HCH out-of-plane twisting of ethylene:

Source: http://www.shodor.org/succeed-1.0/compchem/labs/vibrations/ethylene.jpg

Due to the symmetry of the molecule, when the HCH twist out of plane the dipole and polarizability of molecule is unchanged.

Other examples can be found here

https://www.chem.purdue.edu/jmol/vibs/c6h6.html

where the modes of benzene are sketched. The silent modes easily seen are kind of distorted breathing of the ring, that due to the C6 axis of symmetry again let the molecule unchanged both for the electric dipole moment and its polarisability.


Internal energy calculation The temperature of 1 mole of a liquid is raised by heating it with 750 joules of energy. It expands and does 200 joules of work, calculate the change in internal energy of the liquid.

I want to use the expression:
$$\Delta U = \Delta Q + \Delta W$$
so that:
$$\Delta U = 750\,\mathrm J- 200\,\mathrm J = 550\,\mathrm J$$

but it strikes me that it can't be that simple (first year college exam paper). Also, of what significance is the '1 mole' of liquid?

Your calculation is correct. The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is

$$\Delta U=Q+W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur). Therefore, heat transferred to the system is assigned a positive sign in the equation
$$Q=750\ \mathrm J$$
whereas work done by the system on the surroundings during the expansion of the liquid is assigned a negative sign
$$W=-200\ \mathrm J$$
Thus, the change in internal energy is
$$\begin{align}
\Delta U&=Q+W\\
&=750\ \mathrm J-200\ \mathrm J\\
&=550\ \mathrm J\\
\end{align}$$

However, the question is a bit flawed since the given values are not typical for a liquid. By way of comparison, realistic values for water are shown in the following table.

$$\textbf{Water (liquid)}\\
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\
\text{Volume} & V & 18.0476\ \mathrm{ml} & 18.0938\ \mathrm{ml} & 0.0462\ \mathrm{ml} \\
 & & 1.80476\times10^{-5}\ \mathrm{m^3} & 1.80938\times10^{-5}\ \mathrm{m^3} & 4.62\times10^{-8}\ \mathrm{m^3} \\
\text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\
 & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\
\text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 29.9560\ \mathrm{^\circ C} & 9.9560\ \mathrm{^\circ C} \\
 & & 293.1500\ \mathrm{K} & 303.1060\ \mathrm{K} & 9.9560\ \mathrm{K} \\
\text{Internal energy} & U & 1\,511.59\ \mathrm{J} & 2\,261.58\ \mathrm{J} & 749.99\ \mathrm{J} \\
\text{Enthalpy} & H & 1\,513.39\ \mathrm{J} & 2\,263.39\ \mathrm{J} & 750.00\ \mathrm{J} \\
\hline
\end{array}$$

When $1\ \mathrm{mol}$ of water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting expansion is actually only
$$\begin{align}
\Delta V&=V_1-V_0\\
&=18.0938\ \mathrm{ml}-18.0476\ \mathrm{ml}\\
&=0.0462\ \mathrm{ml}\\
&=4.62\times10^{-8}\ \mathrm{m^3}
\end{align}$$

The corresponding pressure-volume work is
$$\begin{align}
W&=p\Delta V\\
&=100\,000\ \mathrm{Pa}\times4.62\times10^{-8}\ \mathrm{m^3}\\
&=0.00462\ \mathrm J
\end{align}$$
which is clearly below the value given in the question $(W=200\ \mathrm J)$.

The values given in the question are appropriate for a gas. For example, realistic values for nitrogen are shown in the following table.

$$\textbf{Nitrogen (gas)}\\
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\
\text{Volume} & V & 24.3681\ \mathrm{l} & 26.5104\ \mathrm{l} & 2.1423\ \mathrm{l} \\
 & & 0.0243681\ \mathrm{m^3} & 0.0265104\ \mathrm{m^3} & 0.0021423\ \mathrm{m^3} \\
\text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\
 & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\
\text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 45.7088\ \mathrm{^\circ C} & 25.7088\ \mathrm{^\circ C} \\
 & & 293.1500\ \mathrm{K} & 318.8588\ \mathrm{K} & 25.7088\ \mathrm{K} \\
\text{Internal energy} & U & 6\,081.06\ \mathrm{J} & 6\,616.83\ \mathrm{J} & 535.77\ \mathrm{J} \\
\text{Enthalpy} & H & 8\,517.87\ \mathrm{J} & 9\,267.87\ \mathrm{J} & 750.00\ \mathrm{J} \\
\hline
\end{array}$$

When $1\ \mathrm{mol}$ of nitrogen with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting pressure-volume work is

$$\begin{align}
W&=p\Delta V\\
&=100\,000\ \mathrm{Pa}\times0.0021423\ \mathrm{m^3}\\
&=214.23\ \mathrm{J}
\end{align}$$
The corresponding enthalpy balance
$$\begin{align}
\Delta H&=\Delta U+W\\
750.00\ \mathrm{J}&=535.77\ \mathrm{J}+214.23\ \mathrm{J}
\end{align}$$
is quite similar to the values of the question $(\Delta H=Q=750\ \mathrm J,$ $\Delta U=550\ \mathrm J,$ and $W=200\ \mathrm{J}).$


Calculating molar mass of an oxidant by iodometry I am stuck on a question from the Australian Chemistry Olympiad involving molar mass, shown here:

BTW, this whole paper is available online for free, and is distributed by its owners for free. Just search up: Australian Chemistry Olympiad Papers and look at the past papers.
Anyways, in this question, I am able to solve part a and part b, but part c is a little tricky. I understand how to solve for molar mass: it is simply the mass of the substance divided by the number of moles in the same amount of substance. However, in the solution they use the number of moles of the metal cation M, as well as the subscript variable x. Can someone please describe in detail what this means. If there are any more questions, feel free to comment below. Thanks in advance.
P.S. The answers are not my own, but those from the answer sheet.

The molar mass of a substance, as you said earlier, is:

  simply the mass of the substance divided by the number of moles in the same amount of substance

In this question, you can see that as per part b), we found that there were $7.553\times10^{-4} \text{ moles}$ of $\ce{M^{n+}}$ present.

Hence, the salt, i.e., $\ce{M_xA_y.ZH2O}$ contains $\ce{M_x}$ = $7.553\times 10^{-4} \text{ moles}$ $\ce{M^{n+}}$.

=> $\ce{M}$ = $ \frac{7.553\times 10^{-4}}{x} \text{ moles}$ of $\ce{M^{n+}}$

$\therefore$ taken moles of $\ce{M_xA_y.ZH2O}$ = $\frac{7.553\times 10^{-4}}{x} \text{ moles}$

But we know that the taken moles have a mass of $0.2642 g$.

Now, divide $0.2642 g$ by $\frac{7.553\times 10^{-4}}{x} \text{ moles}$ to get the molar mass, in terms of x, which is $349.8x$


How can you manage SCF convergence problems? "Failure to locate stationary point, SCF has not coverged"

This is the most annoying message after Optimization run ( even with a good primary start geometry like UFF we ay see that) .
The only solution I know is to use the output log file for another run but it is very time consuming.
Is there any way to force GAMESS to automatically do that ?( I mean makes another automatic run on current output results ).

There can be many reasons for the SCF to not converge.

The initial guess wasn't great, and the convergence is slow. Increasing the number of SCF iterations can help here.
The SCF oscillates. This can happen if there are two orbitals close in energy. Using damping, mixing (i.e., the next iteration is 50% the old and 50% the updated iteration), Fermi broadening, etc. can help here.
The SCF strategy (e.g., DIIS) is causing problems. Usually DIIS can greatly accelerate convergence, but in some hard-to-converge systems, you need a lot of iterations with another strategy (e.g., Newton-Raphson)

Without more information from your particular file, it's impossible to know how to help. Many times, you may simply want to increase the maximum number of SCF iterations and it'll work fine.


Different solutions in cyclic voltammetry CV is usually used to monitor corrosion (from what I have learned) because, if it is a reversible reaction, we can see the cathodic and anodic peak in the voltammogram. I know that the scan rate influences the intensity of the current peak and I also know why that happens. However, I just can't seem to find the reason why different solutions result in different current peaks in the voltammogram. Especially if the solutions get more acidic the current peak changes.
So my question is why do the peaks in the voltammogram change when we change the solution?

Recording a CV assess both so-called half cell reactions.  If balanced individually, some of them only need only the transfer of an electron, e.g. for
$$\ce{Fe^{2+} -> Fe^{3+} + e-}$$
Other half-reactions include either $\ce{H+}$ or $\ce{OH-}$, and their consumption / generation alters the pH value of the solution, e.g.
$$\ce{MnO4^- + H2O + 3 e- -> MnO2 + 4 OH-}$$
or
$$\ce{MnO4^- + 4 H3O+ + 2 e- -> Mn^{2+} + 6 H2O}$$
Since each of them are characterized by a standard electrode potential, altering the pH value of the solution thus may alter the reaction pathway accessible; the reaction of permanganate to manganese dioxide is typical for a reaction in a basic aqueous reaction, the one to yield $\ce{Mn^{2+}}$ typical for an acidic one. Indeed pH electrode is an application of this pH dependence which may be quantified by the Nernst equation.
While the examples shown deal with small inorganic ions, the same principle is valid for recording a CV for organic molecules, too.  For the couple quinone / hydroquinone, the variation of the pH value thus may yield a plot like the following:

(reference)
The changes for the redox couple, recorded in different solutions of different pH value equally were plot with the cell potential in function of said variation of the pH value of the analyte:

Note: compare the half reactions with each other.  If both the number of electrons, as well as the number of protons exchanged are equal, the potential of the cell will not be affected by a change of the pH value.
reference: pH Dependent Redox Couple: An Illustration of the Nernst Equation, Walczak et al., J. Chem. Educ. 1997, 74, 1195-1197; doi 10.1021/ed074p1195


Best way to dissolve/emulsify beeswax and carnauba wax in water-based solution What I can use to dissolve beeswax and carnauba wax and keep it in an aqueous solution? I do not want to use ethanol or acetone as they are too harsh.

I am adding about $\pu{5\%}$ wax to a $\pu{16oz}$ aqueous solution that also has some oil in it. I need to stop it from coagulating and be fine enough or in solution to pass through the nozzle of a spray bottle without clogging it ?

Try Oelic acid and TEA with Waxes.
1. Take 3% Oelic acid
2. Add 3% TEA in Oelic acid and stir it with continuous heating up to 50C.
3. Take 100ml Water in Above solution.
4. Take 20grm carnuaba wax and dissolve it in above solution with continuous mixing and heating up to 60C.
5. Again add 40 to 50ml water in above solution.


the particle in 1-D box momentum and total energy(P.E + K.E (x)) Employing the particle in 1-D box can momentum and total energy(P.E + K.E (x)) can be measured simultaneously?

As this is obviously a homework problem, I'll just give you a hint:

Observables can be measured simultaneously with arbitrarily small uncertainties if their operators commute. Compute the commutator of the momentum and particle-in-a-box Hamiltonian operators; if the commutator is zero, the answer to your question is "yes".


Which double bond is more electrophilic : carbon-carbon or carbon-oxygen? Question

Attempt

It seems to me that the carbon-oxygen bond would be more electrophilic due to the higher electronegativity of oxygen. Hence I believe $\ce{CH3-}$ should attack the carbonyl site. But this doesn't seem to be the case.

Is it because that the intermediate carbanion can undergo resonance with oxygen, had $\ce{CH3-}$ attacked carbon- carbon double bond?

What can be a valid reason?

step-1: cleavage of C=O bond:

the C=O bond undergoes heterolytic cleavage to give positive charge on carbon and negative charge on oxygen just like almost all other organometallic reagents react with a ketone, for example a grignard reagent (not exactly, but similar to it).

step-2:(the real step which caused deviation from your answer)-resonance of intermediate:

the positive charge on the carbon of the erstwhile C=O group now undergoes resonance with the double bond at the allylic position to give positive charge at carbon no. 3 instead of carbon no. 1
in this sort of reaction is 1,4 addition occurs instead of 1,2 addition. For more information, you can visit this link: https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/1%2C4-Addition

step 3:(as usual):-attack of CH3 group:

now on as usual, the CH3 group from our organo copper reagent ( called as gilmann reagent, used in Corey-house synthesis), just like a grignard reagent, attacks the new carbocation formed at carbon no.3 and the water molecule in the second step donates it's proton to the remnant oxygen of our old C=O group and it becomes an enol. tautomerisation takes place and we bet 3-methyl cyclohexanone as a result.


Solve for wavelength of photon given threshold frequency The threshold frequency of a potassium metal plate is $\pu{4.35\times10^14 Hz}$. Determine the wavelength of a photon required to reach $\pu{2.50 eV}$ of the maximum kinetic energy of an emitted electron.

Options:
A. $\ \ \ \pu{2.88\times10^2 nm}$
B. $\pu{-4.96\times10^2 nm}$
C. $\ \ \ \pu{9.50\times10^-21 nm}$
D. $\ \ \ \pu{6.69\times10^2 nm}$
E. $\ \ \ \pu{2.02\times10^15 nm}$

From what I understand, I used the equation $KE = h\nu + h\nu_0$ and replaced $\nu$ with $\frac cA$ because wavelength and frequency are inversely proportional. Then I plugged in the given components, $\nu_0$ and a fraction of $KE$ to solve; however, I don't understand how to approach this as I'm only given $2.5 \pu{eV} (4.005 \times 10^{-19} \pu J)$ of maximum $KE$, not an actual value for $KE$. The answer I'm getting is $\pu{1.14 \times 10^{-12} nm}$ if I just use $\pu{2.5 eV} = KE$ max, which isn't an answer choice, and the given answer is A. $2.88 \times 10^2 \pu{nm}$. Please help.

Always work in energy not wavelength. You are given $\nu_0$ in Hz ($\mathrm{s}^{-1}$), multiply this by $h$ to convert it to Joules; ($h$ has units J s). You have the kinetic energy in eV, use the conversion  $1\mathrm{eV} = 1.602\cdot 10^{-19}$ J to get into joules. Now you have only to find energy $h\nu$ by subtraction. Finally convert energy to frequency using $E=h\nu$ and then to wavelength using $c=\lambda \nu$. Make sure $c$ is in m/s and as you answer will be in meters then convert to nm from metres.


Why does 3-bromopropanoic acid have least pKb value when compared to 3-chloropropanoic acid, 3-fluropropanoic acid and fluoroacetic acid? Shouldn't 3-bromopropanoic acid have the least pKb due to the reduced inductive effect and electronegativity, leaving the carboxylate anion with more electric charge on it?
Also, what will be the decreasing order of pKb?

The correct order of PKb is
fluoroacetic acid> 3-fluoropropanoic acid>3-chloropropanoic acid>3-bromopropanoic acid.
To find the order of pKb, first find the order of pKa which can be found by checking the stability of negative charge on carboxylate anion once it donated proton.
Halogens are electron withdrawing groups. The halogen which will withdraw the most electron density will stabilise the anion the best. Fluorine, being the most electronegative withdraws the most electron density thus reducing the charge on oxygen and stabilising the anion, therefore having the least pKa and maximum pKb.


Simple Explanation of orbitals Imagine I was your little brother, how would you explain an orbital to him? (assuming he knows what atoms electrons and neutrons are, and the rest of the basics till that point.)

I have been reading up on this and can't make heads out tails of it.
As a side how do the different energy levels relate to them.

I hope I'm clear, I'm doing my best. I read several books but can't get a clear bearing on this.

Well, the first step is to stop thinking electrons as being very small balls that orbit around the nucleus in a circular path. This is known as the Bohr model. Despite this model being an excellent model at an introductory level, it doesn't tell the entire story.

There are several issues with the notion that electrons travel in a circular orbit around the nucleus, the main one being that they should eventually undergo orbital decay and hence eventually slow down and crash into the nucleus. Of course, this doesn't happen. So how do electrons move around the nucleus?

This is where quantum theory came to the rescue in the early 20th century. It stated that electrons could not be treated as a classical particle and didn't not have a definite position and momentum. So this basically means that we can't exactly know an electron's position and its momentum at the same time. This relationship is given by the Heisenberg Uncertainty Principle.

I won't go into detail about it since to keep this explanation basic, but if you are interested check out this video which provides a really good explanation.

So basically what quantum theory tells us that we actually can't specify where and what a path an electron is going to take. However the Bohr model assumes that you can precisely plot out where an electron is going to be, but we now know that due to the Heisenberg Uncertainty Principle you can’t precisely know the exact position and momentum of an electron. So we actually can't construct an orbit and say that the electron is always going to follow that path.

Therefore, it makes it impossible to plot an orbit for an electron as how can you? If you don’t know where the electron is going to be next, there is no way you can predict its path.

So how can we get around this problem? Well, if we can’t draw the orbit for an electron the next best thing we can do is to plot an electron density probability orbital for it. Also known as atomic orbitals.

An orbital is a well defined region of space. Therefore an atomic orbital is a region of space that shows where the electron will be 95% of the time (usually we take 95% but it could be any number really such as 90% or 75%).

A simple way of thinking of orbitals is that imagine that you have a magic camera that can take a sequence of photos of an electron in a hydrogen atom. The electron appears as a dot. Now if we superimpose all of these images, you will see something like this:

Since we generally take atomic orbitals to be the region of space that shows where the electron will be 95% of the time, the very outer sphere of the above image is said to be the orbital of the electron.

Hopes this helps you understand orbitals. If you have any question, feel free to ask.


Why do the pennies in the pre/post 1982 penny experiment sometimes dissolve completely? USA pennies were nearly pure copper pre-1982.  Post 1982 they are copper clad zinc.  A common experiment in US high schools is to notch pennies with a file and add them to an HCl solution.  The old pennies remain untouched and the zinc of the new pennies dissolves, leaving the copper cladding.

Sometimes the copper cladding also dissolves and sometimes a considerable amount of the old penny dissolves.

Why does the copper dissolve?

I propose that the copper in the penny is reacting with atmospheric oxygen, forming copper oxide that dissolves in the acidic solution.

Another proposal is that the formation of a complex ion between copper and chloride ion drives the equilibrium, allowing the copper to react in HCl.  This is analogous to how aqua regia dissolves gold.

Another proposal is that the copper oxide layer on the penny reacts with copper metal in a comproportionation to form copper (I) ion that then reacts with oxygen to make copper (II) and dissolves?

Which is right?  Why does the copper dissolve?

Your conclusion about formation of the Copper Oxide is correct, but not about the reaction with atmospheric oxygen.

The origin of the Oxygen is from the hydrochloric acid solution. The solution does not mean is pure HCl , it means is a dissolve in water.  There is a redox reaction with water

I leave to you to find the redox reaction.


Origin of the 15% difference limit between solute and solvent atom radii in the Hume-Rothery rules According to the Hume-Rothery rules for the substitutional solid solutions the atomic sizes of the the solute and solvent must not differ by more than 15%.
Why should it be so?

Possibly, there are two extrema to account:
a) the atomic radii of metal A and metal B do not differ significantly, and thus there are chances both metals blend statistically with each other.  Which would be covered by the $< 15\%$-rule, a situation similar to mixing apples and oranges in one crate.
b) the atomic radii differ a lot.  If we assume atoms as hard spheres, even the closest arrangement of spheres of one sort does not occupy fully the space available (see e.g., fcc and hcp packing).  If the diameter of the other metal(s) is/are much less than the one of the host, it is possible for some of these pockets to be filled more, and for others to be filled less. Overall, the distribution of metal B and metal A in the blend of (A .and. B) no longer is even; the smaller atoms may segregate.  As if you would transport a crate of filled with melons and hazel nuts.  This non-regular demixing is possible because -- contrary e.g., to spinels -- you may assume absence of ions charged oppositely charged.
For the purpose of illustration, both the examples as well as the figure below intentionally exaggerate the contrast between case a) and b).


Why does milk clot while making yogurt? Why does milk clot when it becomes yogurt? Is it only because of the decrease in the $\mathrm{pH}$ of milk? If it does so, what is happening when I increase the $\mathrm{pH}$ level? And by decreasing the $\mathrm{pH}$ level can I clot any type of liquid like milk?

The clotting in a yoghurt is indeed caused by the lactic bacteria producing lactic acid and therefore lowering the pH of the milk. However in this particular case (yoghurt making) this is not due to the denaturation of the milk protein, they are in fact intact.
When the pH drop in the milk, it get near the isoelectric point (IEP) of the casein (milk protein) therefore making them partialy neutral (or completly if the pH reach it).
It weaken the electrostatic repulsion between the casein.
Consequently when the pH get closer to the IEP, casein solubility decrease and often precipitate.

And this is reversible, if you add caustic soda or another base, the yoghurt or any sour milk will go back to being liquid (well not perfect in some case because the bacteria often produce exopolysaccharide which thicken the milk).


What are the limitations of a waters of hydration experiment? When a hydrated salt such as $\ce{CuSO4.5H2O}$ is subjected to heat, an anhydrous salt ($\ce{CuSO4}$) is produced. However the theoretical molar ratio of $\ce{H2O:CuSO4}$ and experimental molar ratio of $\ce{H2O:CuSO4}$ are not the same. Instead there is a percentage error, why does this occur? And how can we minimise this in this experiment?
The hydrated salt was left on hot plate at 70 degrees celsius for 25 mins.

You can mention experimental details and how you achieved 1:4 ratio, which may explained if we know the maximum temperature of heating.  CuSO4.5H2O loses water of hydration in steps as a function of temperature showing that those five water molecules are not equivalent. In this figure from German Wiki, follow the green line (ignore blue). The y-axis is % loss in weight and x-axis is the temperature. It clearly shows that water is lost in steps. The labels read Mass change (loss of crystal water). If your heating temperature is not enough you are left with CuSO4.H2O.


What is the enthalpy of formation of 1 mol of sulfuric acid? In the contact process, the commercial process of producing $\ce{H2SO4}$, several reactions occur before the final product is produced:

  \begin{align}
\ce{S8(s) + 8O2(g) &-> 8SO2(g)} &\Delta H_\mathrm{f} &= \pu{-70.9 kcal mol^{-1}}\tag{1} \\
\ce{2SO2(g) + O2(g) &->[V2O5] 2SO3(g)} &\Delta H_\mathrm{f} &= \pu{-94.5 kcal mol^{-1}}\tag{2} \\
\ce{SO3(g) + H2O(g) &-> H2SO4(g)} &\Delta H_\mathrm{f} &= \pu{-193.9 kcal mol^{-1}}\tag{3}
\end{align}

  Given that $\Delta H_\mathrm{f}$ for $\ce{H2O(g)}$ is $\pu{-68.3 kcal mol^{-1}}$, what is the enthalpy of formation of $\pu{1 mol}$ of $\ce{H2SO4}$?

What I've tried to do:

Summing all the enthalpies, I arrived at to $\Delta H_\mathrm{f} = \pu{-427.6 kcal mol^{-1}}$.
Using the data from equation $(3)$, I conclude that the answer is $\Delta H_\mathrm{f} = \pu{-193.9 kcal}$.

Since enthalpy is a state function, all that is needed is the $\Delta H_{\mathrm{f}}$ of each reactants that form $\ce{H2SO4}$. Multiply the equations through by the correct fractions (including the enthalpies) to result in the correct coefficients and enthalpy for the desired reaction:

$$\ce{SO3{(g)} +H2O{(g)} -> H2SO4{(g)}}~~~~~\Delta H_{\mathrm{f}}=-193.9~\mathrm{kcal~mol^{-1}}$$

$$\ce{H2{(g)} + \frac{1}{2}O2{(g)}->H2O{(g)}}~~~~~\Delta H_{\mathrm{f}}=-68.3~\mathrm{kcal~mol^{-1}}$$

$$\ce{SO2{(g)} + \frac{1}{2}O2{(g)} \xrightarrow{\ce{V2O5}} SO3{(g)}}~~~~~\Delta H_{\mathrm{f}}=-47.25~\mathrm{kcal~mol^{-1}}$$

$$\ce{\frac{1}{8}S8{(s)} +O2{(g)} -> SO2{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-8.8625~\mathrm{kcal~mol^{-1}}$$

Combining these equations yields:

$$\ce{\frac{1}{8}S8{(s)} +2O2{(g)} +H2{(g)} -> H2SO4{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-318~\mathrm{kcal~mol^{-1}}$$

(according to the equations and numerical values provided)


Caffeine dissolution I am looking to add caffeine powder in pure form to water that has a pH level of 7 or higher. What would the pH level be if I do this? Is Caffeine itself acidic?

Caffeine is not acidic. In fact, it is basic. It will raise the pH of water to some degree. It is a weak base, so the free base will exist in equilibrium with its protonated counterpart.

Due to its basicity, it will be more soluble in low pH aqueous solutions than in high pH solutions. This is because in low pH solutions, caffeine would exist completely as a protonated salt, making it more hydrophilic. Despite this, caffeine is very polar and is still quite soluble in water at higher pH levels.

The six-membered pyrimidinedione can exist in an aromatic, zwitterionic form where both amide nitrogens have formed double bonds to respective adjacent carbonyls. This six-membered ring had been observed as being planar. Therefore, it has been concluded that caffeine does indeed exist primarily as its zwitterionic resonance form, with two positive charges and two negative charges. The negative charges exist on the two carbonyl oxygens. These may be slightly basic, too, but significantly less so than a structurally similar hydroxyphenol (e.g. resorcinol) due the electronegative nature of the attached aromatic ring. Therefore, the basic imidazole nitrogen likely has largest effect on the pH.

The fact that there is possibly more that one weakly basic atom on the caffeine molecule complicates calculations as far as determining a theoretical pH. Also, the calculation would require the quantity of caffeine you will add. If you need an exact number, you will have to measure it. You can be sure, however, that the pH of the water will rise as caffeine is added.


Aromatic Character in Percentage? I was just searching whether Uracil is aromatic or not and in this link$^1$ it is said that (from the abstract):

High-level computational methods were tested for the enthalpy of hydrogenation of uracil, but only with G3 was possible to obtain results in agreement with the experimental ones. It was found that uracil possesses 30.0% of aromatic character in the gaseous phase.

What is the meaning of this sentence? I have only basic knowledge on Aromaticity limited to Huckel's rule and the names of a few aromatic compounds. I didn't expect Aromaticity to be expressed in percentages. Can someone explain why?
Reference

Tiago L. P. Galvão, Inês M. Rocha, Maria D. M. C. Ribeiro da Silva*, and Manuel A. V. Ribeiro da Silva. J. Phys. Chem. A 2013, 117, 28, 5826–5836.
https://doi.org/10.1021/jp404938u

G3 is generally a term used to describe a particular computational model used in calculating molecular bonding in quantum chemistry.  The G refers to Gaussian and 3 is just a third iteration of including additionally complex terms to the molecular orbital calculations being used.  There's a brief discussion on wikipedia, but you'll find more extensive discussion of these types of calculations in Quantum Chemistry text books or directly from the researcher's works just an example, a G3 Theory site.
Aromaticity can actually be measured quantitatively via Aromatic ring current and calculated with [NICS] nucleus-independent chemical shift.  The wikipedia article included a link to a paper describing the quantitative scale.  It's just a way of saying how aromatic is the molecule compared to another (call it a standard / perfect example molecule.)  The problem with such relative standards is there may always be one that exceeds the standard, so you could potentially end up with results larger than 100%.


What could cause one brand of carbonated water to be superiorly carbonated compared to another? I buy a lot of carbonated water, its pretty much all I drink. There is something I've noticed that just made me curious though.

The usual brand I buy is the cheapest one available, 14 cents per 1,5l. This one almost always tries to give me a surprise shower when I open it, no matter how still I keep it prior to opening. The bubbles are large and plentiful at first but it goes flat pretty quickly, usually within the day but thats fine since I'll drink it in a day anyways.

The other brand I buy when the first one isn't available, its the second cheapest one and is 35 cents per 1.5l. Opens up pretty calmly, the bubbles are much smaller but lots of them around. The interesting bit is that it doesn't go flat nearly as quickly and I can expect some carbonation to be left even a few days later. Another point to make is that the second one behaves much more like various other brands.

The question is, what could cause such difference in the bubbling behavior?

The bottles are the same kind of clear plastic, the ingredients aren't listed which should mean that its just water + $\ce{CO2}$. Over here, if they put any other chemical in it, they are required to name it on the bottle, many water bottles have chlorine in them for example.

Could it be the carbonation process? Or the natural water composition? Maybe the transportation or storage?

Carbonated water is supersaturated in carbon dioxide, and the only thing preventing all that excess $\ce{CO2}$ from bursting out all at once is the lack of nucleation sites for $\ce{CO2}$ bubbles to form. These nucleation sites can be almost any solid (including the sides of the container to some degree) from undissolved salts to soil particles or tiny particles of most anything.

The most likely reason for the formation of such a copious amount of bubbles forming in the cheap water is the presence of a large amount of undissolved solids relative to the more expensive waters. Depending on the product, one step of the processing of carbonated water is filtering in order to make the product appear clearer and more sparkly (and maybe to prevent rapid formation of $\ce{CO2}$ bubbles). Of course this process costs money, and it is likely that the cheaper brand either skips the filtration process or uses a much less efficient (and cheaper) process.


How to draw Lewis dot structures? What is the most logical and easy to remember step by step method for drawing Lewis dot structures?
I'm having trouble drawing a simple one for $\ce{CO2}$.

Calculate the total number of valance shell electrons ($n_1$) in all constituent atoms or ions. For example, in $\ce{NH_4^+}$, $n_1$ will be 8 (5 of N, 4 from all H's and -1 because the molecule has +1 charge).
Now, calculate the total number of electrons that are required to fill valance shell of every constituent atom or ion ($n_2$). In other words, calculate the total number of electrons that will be required to fill valance shell of each atom or ion individually. We can use following formula to find $n_2$$n_2 = [\text{# of Hydrogen atoms} \times 2] + [\text{# of all other atoms except Hydrogen} \times 8]$
In case of $\ce{NH_4^+}$, $n_2$ will be 16 ($[4 \times 2] + [1 \times 8]$).
Now, calculate the total number of electrons that atoms or ions will need to share among each other in order to complete their octets ($n_3$).
$n_3 = n_2 - n_1$
For $\ce{NH_4^+}$, $n_3$ will be 8.
Now calculate the number of bonds that will be present in our molecule
$n_4 = \frac{n_3}{2}$
For $\ce{NH_4^+}$, $n_4$ will be 4.
Now let's find the number of unshared electrons, $n_5$
$n_5 = n_1 - n_3$For $\ce{NH_4^+}$, $n_5$ will be 0.
Now find the number of lone pairs, $n_6$, which is nothing but $n_5$ divided by 2._For $\ce{NH_4^+}$, $n_6$ will be 0.
Now find the central atom. Central atom is the one which is either least in count or more electropositive.
Always remember that Hydrogen and Fluorine can never be placed at center.
So in our case, Nitrogen gets to be the central atom.
After allocating the central atom, place the surrounding atoms such that the molecule is symmetric.
In our case, geometry of molecule will be tetragonal.
Distribute the bonds between atoms. Remember that the total number of bonds, that you distribute, should be equal to $n_4$.
In our case, we simply place a bond between Nitrogen and each Hydrogen atom.
Now use the lone pairs to complete the octet of atoms. Remember that, by now, every atom will already have 2 electrons in their valance shell because of the bond.In our case, which is $\ce{NH_4^+}$, we don't have any lone pairs. OOPS!
Now comes the Formal Charge.

  Formal Charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in a chemical bond are shared equally between atoms, regardless of relative electronegativity. It is not an actual electrical charge but just a mathematical construct.

  $\text{Formal Charge of an atom} = \text{# of valence shell electrons (in Ground state)} - \text{# of unshared/non-bonded electron pairs or lone pairs} - \text{# of bonds}$

Please note that you must place formal charge on each and every atom when it is a non-zero value for that atom.
This step is little confusing so let's get back to our example. Nitrogen has 5 electrons in its valance shell in ground state. It made 4 bonds in our molecule and has none unshared electron pairs or lone pairs (please note that a single unpaired electron doesn't count here). So formal charge on Nitrogen will be 1 (5 - 0 - 4), whereas formal charge on each Hydrogen atom will be 0 as every Hydrogen atom has 1 electron in its valance shell and made one bond in our molecule.

In a sense, you actually obtain the structure in $10^{th}$ step but Formal charge, if a non-zero value, should be represented for each atom in molecule in order to complete its Lewis Dot Structure.


Balancing complex haloform redox reaction I just completed a haloform reaction experiment, using acetophenone and sodium hypochlorite to form benzoic acid and chloroform.

My lab report has a question regarding balancing the equation of the reaction.  I have been working on it for quite a while, and I can't figure out how to do it.  I've tried looking up methods but too many atoms are reliant on too many molecules!  Can someone help with this question and explain how to do it so I can really understand this process?

I need to be able to identify the limiting reactant as well, which relies on the molar ratio of the reactant to the product, so I really need to know how to do this.

As you noted, you need to find the limiting reactant. This is most of the time done by taking the one with a 1:1 ratio of educt:product.

So in this case, the $\ce{ROCH_3}$. So you take that once. Now you see immediately that you need 2 $\ce{Na}$ on the right side, so you can put a 2 in front of the $\ce{NaOH}$. You have two $\ce{X}$ on the left, but 4 on the right, so put a 2 in front of the left one. Now count all your atoms. Subtract them from each other:

You have $a * \ce{H}$ on the left and $b * \ce{H}$ on the right. if $a-b=0$ its balanced. Do that for every reactant and you're done.

There are of course more strictly formulated methods, but in my time I noticed that a straightforward approach with a little thinking can be applied most universally without having to memorize some formulas.


Which of the following alkenes would produce two products when they undergo an addition reaction with hydrogen bromide? Which of the following alkenes would produce two products when they undergo an addition reaction with hydrogen bromide?

Ethene
But-1-ene
But-2-ene
Hex-3-ene

The correct answer (according to my textbook) is But-1-ene but I do not understand why this is.

Well as mithoron said.its about symmetry.Let's see how but-1-ene comes out to be the answer

On addition of HBr markownikov rule is followed but lets draw both product.

1.)CH2=CH2 + HBr--->Br-CH2-CH3 + CH3-CH2-Br  (both products are 1-bromoethane)

2)CH2=CH-CH2-CH3 + HBr--->Br-CH2-CH2-CH2-CH3 + CH3-BrCH-CH2-CH3 (notice the                name of both the product. 1st one is 1-bromobutane. 2nd one is 2-bromobutane)

3)CH3-CH=CH-CH3 + HBr---->CH3-BrCH-CH2-CH3 + CH3-CH2-BrCH-CH3 (both products are same i.e both are 2-bromobutane)

4)CH3-CH2-CH=CH-CH2-CH3 + HBr--->CH3-CH2-BrCH-CH2-CH2-CH3 +
                    CH3-CH2-CH2-BrCH-CH2-CH3 (here also both the compounds are same i.e both are 3-bromohexane)

So what distinguishes the 2nd one from these three is that the second molecule  is unsymmetrical.You can find its unsymmetrical by looking at the no:of carbon to both side of double bond.If they are equal then molecule is symmetrical.


Inductive effect of alkyl ammonium groups Why $\ce{-NR3+}$ has more -I effect than $\ce{-NH3+}?$
And what about order of $\ce{-NHR2+}$ and $\ce{-NH2R+}?$

My attempts
I got idea about electronegativity of carbon atom in $\ce{R}$ is more than hydrogen atom, hence positive charge gets more concentrated on nitrogen atom and it exerts more electronic attraction. But this didn't worked out in latter case when -I effect of $\ce{-NHR2+}$ is less than $\ce{-NH2R+}.$

The inductive effect is both distance-dependent and χ-dependent. N-H bonds are 1.01 Å, yet the χ of H is only 2.1. N-C bonds are 1.42 Å, yet the χ of C is 2.5. At the extremes, where there are only N-C or N-H bonds, the ordering of -I is easy to rationalize.

However, the two, short N-H bonds in -NH2R+ outweigh the higher χ of N-C bonds, giving -NH2R+ a higher -I than -NHR2+.


Calculating the heat of formation of propane given its enthalpy of combustion Given the equation $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$ and that enthalpies of formation for $\ce{H2O (l)}$ is $\pu{-285.3 kJ/mol}$ and $\ce{CO2 (g)}$ is $\pu{-393.5 kJ/mol}$, and the enthalpy of combustion for the reaction is $\pu{-2220.1 kJ/mol}$, I need to find the heat of formation of propane.

My initial idea was to use Hess's law, and I got
$[3(-393.5) + 4(-285.3)] - [-2220.1] = \pu{-101.6 kJ}$

I then doubted myself because Hess's law gives the $\Delta H^o_\text{rxn}$ which is different from the $\Delta H^o_\mathrm{f}$ for  $\ce{C3H8}$ that we are trying to find.

So I used another approach using the individual equations for the formation of $\ce{H2O, CO2}$, and $\ce{C3H8 + 5H2O -> 3CO2 + 4H2O}$ and combining the equations to get the equation for the formation of $\ce{C3H8}$ (which is $\ce{3C + 4H2 -> C3H8}$). After manipulating the equations, I got $+2220.1 - 1141.2 - 1180.5 = \pu{-101.6 kJ}$.

I got the same answer, and I want to understand why Hess's law still works even though the equation I used in Hess's law had nothing to do with the formation of $\pu{1 mol}$ of $\ce{C3H8}$. I understand my second method, but why does the first method work?

The equation I had earlier in my first method was wrong. Using Hess's law, we know the change in enthalpy of combustion to be $\pu{-2201.1 kJ/mol}$. Thus:

$$\Delta H^\circ_\text{rxn}=-22201.1 = [3(−393.5) + 4(−285.3)] − [5(0) + x]$$

where $x$ is the heat of formation of propane.

Solving the equation, we get $x = \pu{-101.6 kJ/mol}$.


Finding the kp with density, temperature, and pressure Consider the following reaction
$$\ce{3O2(g) <=> 2O3(g)}$$
At $\pu{175°C}$ and a pressure of $\pu{128Torr}$, and equilibrium mixture of $\ce{O2}$ and $\ce{O3}$ has a density of $\pu{0.168g/L}$. Calculate $K_p$ for the above reaction at $\pu{175°C}$.

I used the ideal gas law $pV=nRT,$ $K_p=K_c(RT)^n$, and $\rho=M \times c$ but I am not certain what to do after finding the total $c$ of the reaction. I was thinking of using the ICE chart to find the concentration of reactant and products which will lead to the $K_c$ and then $K_p$ value but since I don't know the initial concentration of $\ce{O2}$ I end up with an equation with two variables.
I was thinking of using the density to find the molar mass of the reaction and then somehow the  weight of $\ce{O2}$ and $\ce{O3}$ in grams but it seems like a stretch.
$$pV=nRT$$
$$T=175\ \mathrm{^\circ C}=448\ \mathrm K$$
$$p=128\ \mathrm{Torr}=0.1684\ \mathrm{atm}$$
$$0.0821=\frac nV\times\left(0.0821\ \mathrm{\frac{atm\ l}{K\ mol}}\times448\ \mathrm K\right)$$
$$0.0821=\frac nV\times36.78\ \mathrm{\frac{atm\ l}{mol}}$$
$$\frac nV=0.00457\ \mathrm{\frac{mol}l}=c$$
$$ \begin{bmatrix}
  & \ce{3O2} & \ce{2O3} \\
  \mathrm I & y & 0 \\
  \mathrm C & -3x  & 2x \\
  \mathrm E & y-3x & 2x
 \end{bmatrix}0.00457\ \mathrm{mol/l}=(y-3x)+2x$$

You can use the formula for the ideal gas
$$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$

As pressure, temperature and density are given, you get the mean molar mass.

$$M_\mathrm{mean}=\frac{\rho RT}{p}$$

From that, you get molar fractions.

$$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$

$$x_{\ce{O3}}=1-x_{\ce{O2}}$$

From that, you get partial pressures.

$$p_{\ce{O2}}=p\cdot x_{\ce{O2}}$$

$$p_{\ce{O3}}=p-p_{\ce{O2}}$$

From that, you get $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$


Model of London forces and Hydrogen Bonding; the question is carried on forward, partially, as a personal question, from another question Does Model of London forces and Hydrogen Bonding are themselves complicated and require multiple sub-models and assisted theories to explain its correctness.

I asked this question in the form (ironically) of a question, but I came up a new doubt later on. Just read below:

["
Lighthart says, "The actual magnitude is somewhat difficult to measure (and frequently dynamically changing due to molecular motion), but their existence is used to describe non-covalent bonding interactions between molecules such as London forces and Hydrogen Bonding." So, does it implies that the Model of London forces and Hydrogen Bonding are themselves complicated and require multiple sub-models and assisted theories to explain its correctness.
Can the statement "Now with partial charges the water molecule can have a dipole moment. ... So the gist is that ionic bonds and covalent bonds are labels that we put on molecules to predict behavior. Such models are useful in limited circumstances. But if we try to overextend the model it leads to wrongful conclusions." (quoted from text written as answer by MaxW.) can be used to strengthen my question's argument."]

The link to the site is: Partial charges in a covalent bond

If i underatabd correctly,
Now a days when in molecular dybamics simulation or docking studies, to calculate everything requires precise knowledge of polar, hydrophobic, ionic and non-bonding interactions. We have models to deal with each, and each has limitations. With polar groups the issues are not so much the models, which have distance dependance and orientation dependance, but computer power. This is really what limits the models. Often solvent effects and hydrophobic effects are ignored or estimated and simplified.

If there is a model missing is a hydrophobicity model. There isn't anything that is super.


Can there be more than one combination of reactant/product concentration for a specific equilibrium constant? For an elementary reaction like $$\ce{A + B -> AB},$$ $$K_{c} = \frac{\ce{[AB]}}{\ce{[A][B]}}$$ and $$\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A][B]}.$$ If I let the reaction reach equilibrium, and at this point, without altering temperature, suck away $\ce{A}$ and add $\ce{B}$ at the same rate (so $-\mathrm{d}\ce{[A]}/\mathrm{d}t = \mathrm{d}\ce{[B]}/\mathrm{d}t$) and alter both concentrations slightly, once I am finished, will the system still be in equilibrium or try to shift?

What if the reaction is not an elementary reaction such that $\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A]}^n\ce{[B]}^m$?

A single "elementary reaction" does not involve an equilibrium.

There must be a forward reaction and a reverse reaction to have an equilibrium.  This is at least two elementary reactions.

Considering an equilbrium reaction: Just as a square encloses the most area for a rectangle of given perimeter, if the sum [A] + [B] is kept constant,  [A][B] is maximal when [A] = [B].  Deviation from this maximum will shift the equilibrium away from the product and toward the reactants.  In other words, if the system is at equilibrium with [A] = [B], and [A] is decreased and [B] increased, keeping [A] + [B] constant, there will be net reverse reaction until equilibrium is again reached, and [AB] at equilibrium will be less than orginally.


Why are osmoles not considered SI units? The Wikipedia article for osmoles is pretty clear on what they are and how they are used. Right off the bat, though, it states that the osmole is not an SI unit, and indeed the usual lists of derived SI units don't mention it. This seems weird to me - it looks like a pretty reasonable thing to be talking about and it looks reasonably uniquely defined for each solute to me.

Is there some fundamental reason why this unit did not make the cut? Is it down to historical accident? Or did the unit stop being useful by the time the SI was designed? Or is it simply not that well defined, changing with respect to exactly what situation one is talking about?

The unit osmole (symbol: $\mathrm{osmol}$) is a coherent derived unit with a special name and symbol like the similar unit equivalent (symbol: $\mathrm{eq}$). Both units have not been accepted for use with the International System of Units (SI).

The main reason why these units have not been adopted is probably that such use is neither necessary nor convenient. All concerned quantities have the same dimension as the amount of substance $n$ $(\dim n = \mathsf{N})$. Therefore, all values can be expressed in terms of the base unit mole (symbol: $\mathrm{mol}$).

Note that, according to the definition of the mole,

  The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
  When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

Thus, the amount of substance can always be expressed in terms of the unit mole, regardless of the considered elementary entities. For example,

amount of mercury atoms: $n(\ce{Hg})=1\ \mathrm{mol}$
amount of hydrogen molecules: $n(\ce{H2})=1\ \mathrm{mol}$
amount of chloride ions: $n(\ce{Cl^-})=1\ \mathrm{mol}$
amount of electrons: $n(\ce{e-})=1\ \mathrm{mol}$
amount of equivalent entities $\ce{1/2H2SO4}$ corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction: $n(\ce{1/2H2SO4})=1\ \mathrm{mol}$
amount of equivalent entities $\ce{1/5MnO4-}$ corresponding to the transfer of an electron in a redox reaction: $n(\ce{1/5MnO4-})=1\ \mathrm{mol}$
amount of solute that contributes to the osmotic pressure of a solution: $n_\text{solute}=1\ \mathrm{mol}$

Nevertheless, certain coherent derived units with special names and symbols have been adopted for use with the SI. The main reason is convenience. The special names and symbols are simply a compact form for the expression of combinations of base units that are used frequently, for example “one kilogram metre squared per second squared” may be written “one joule” $(1\ \mathrm{kg\ m^2\ s^{-2}}=1\ \mathrm J)$. However, his reason does not apply to the units osmole and equivalent, since “osmole” and “equivalent” are not shorter than “mole” and $1\ \mathrm{osmol}$ and $1\ \mathrm{eq}$ are not more compact than $1\ \mathrm{mol}$.

In many cases, coherent derived units with special names and symbols may also serve to remind the reader of the quantity involved. For example, a value expressed in the unit newton may help to remind the reader that the considered quantity is a force. However, the unit symbol should not be used to provide specific information about the quantity, and should never be the sole source of information on the quantity. Therefore, even if the special name “osmole” or “equivalent” is used, the special nature of the considered quantity would still have to be explained in the text.

There are, however, a few important exceptions. For the quantity plane angle, the unit one is given the special name “radian” (symbol: rad), and for the quantity solid angle, the unit one is given the special name “steradian” (symbol sr). For the quantity frequency, the unit reciprocal second is given the special name “hertz” (symbol: Hz), and for the quantity activity, the unit reciprocal second is given the special name “becquerel” (symbol: Bq) in order to emphasize the different nature of the quantities. For the quantity absorbed dose $D$, the unit joule per kilogram is given the special name “gray” (symbol: Gy), and for the quantity equivalent dose $H$, the unit joule per kilogram is given the special name “sievert” (symbol: Sv) to avoid the grave risks of confusion between these quantities (e.g. in therapeutic work). Apparently, the concept of “osmotic concentration” (formerly known as “osmolarity”) is simply not important enough to justify a similar exception for the unit osmole.


How to calculate vibrational spectra of organic molecules? I watched a ted talk (Luca Turin: The science of scent) and was wondering if anyone had any examples of how to calculate vibrational spectra of molecules (used as fragrances)?  Would love to figure out how he computed these charts.

The usual approach is:

Find the equilibrium geometry (within the Born-Oppenheimer app.).
Expand the energy in Taylor series till second order term. The first order will be 0 because of the minimum energy condition.
Include this expression in the Hamiltonian.
Generate a coordinate transformation to get normal coordinates.
Relate second order derivative of the potential energy with the force constant of Hook law (Using second order term we can identify it with a harmonic oscillator). Then get the harmonic frequencies.
The software employed should give also the intensity for each frequency.
Plot intensity vs. frequency/energy/wave length. You'll get only "points". To transform them to a continuous function you can use a Lorentzian like function.

Details can be found in most computational quantum chemistry books.


Comparing adamantane carbon types for accepting OH using MM ForceField I want to compare two type of carbons accepting OH in adamantane which I've shown below.

I know the blue position is preferred. But I wanna prove it with energy optimization and MM Forcefield. I thought it is related to transition state and corresponding carbocations. But MM+ ignores all charges.

How can I test it using ForceField softwares ?

One thing to try would be just to build the two isomers and compare their total energies .. that would at least show which position was preferred thermodynamically (i.e. more stable).  It sounds like you are more interested in the kinetic consideration though (i.e. which reacts faster) ... in that case, looking at the relative energies of the transition states will give you some basis for rationalizing the preference.  I am not sure how good MM is at dealing with transition states, however ... you would need to rely rather heavily on your chemical intuition to find the correct structures, and even then, the bond angles and bond lengths may be distorted to such an extent that your forcefield is not well parameterized to deal with them.


Determining the amount of chlorine in a swimming pool I need to determine the amount of chlorine in a swimming pool. How do I get started? What kind of experiment would I need to carry out?
What equipment would I need for this experiment?

If you need a no-frills solution, get some quick-test paper, such as this.

A slightly more complex solution would be a test kit like this, which makes use of the formation of Bandrowski's base from 1,4-phenylenediamine in the presence of oxidants, such as chlorine.


mechanism of mechanically triggered explosion There are several well known explosives which can be mechanically triggered. I am curious about the mechanism of such a reaction, though.

I guess one can give an easy answer when the explosive is sensitive to everything, flames and mechanical activation, too. What puzzles me is that several of these compounds, e.g. nitroglycerin or ammonium-nitrate much more difficult to trigger by simple flame. How mechanical activation is different?

There are differing propagation mechanisms for an explosion:

Thermal: a deflagrating explosive such as gunpowder has a comparatively slowly propagating wave-front, depending on heat propagating through the bulk material to set fire to the next layer. This is similar to a candle slowly burning down.
Compression: some brissant (but opaque: see 3. below) explosives create a shock wave. "[High explosives] detonate to produce a defining supersonic over-pressurization shock wave."
Light: fuel-air mixtures, glyceryl trinitrate and nitrogen triiodide explosion can be initiated by light (auto manufacturers are looking at laser "spark plugs" to improve internal combustion engine efficiency). In these types of explosives, light from the explosion itself might propagate a supersonic blast front.

So initiating an explosion through heat alone may initiate just a deflagration, without enough over-pressure to form a blast wave.


Why is there no effective alternative to metal salts in antiperspirants? Metal salts, especially aluminum compounds are used in antiperspirants to prevent sweating. I found out aluminum chloride hexahydrate, which is found in most antiperspirants, undergoes an acid-base reaction with water which causes the proteins to denaturate, thus these proteins are precipitated out as a gel blocking the sweat glands. Additionally, aluminum is able to bind water molecules.

However, I would like to understand the biological and chemical processes in more detail as I am not quite sure why there is no effective alternative to metal salts to prevent sweating. There are many compounds that could undergo an acid-base reaction with water. There are also different metal salts which are able to bind several layers of water.

My specific questions are:

Which biological and chemical processes underlie the effect of antiperspirants that make metal salts so effective?
Why are aluminum salts used most commonly?

Which biological and chemical processes underlie the effect of    antiperspirants that make metal salts so effective?

Aluminium-based complexes present in the antiperspirants react with the electrolytes in the sweat to form a gel plug in the duct of the sweat gland. The plugs prevent the gland from excreting liquid and are removed over time by the natural sloughing of the skin.  From a biological point of view, it interact with the keratin fibrils in the sweat ducts and form a physical plug that prevents sweat from reaching the skin’s surface.

Why are aluminium salts used most commonly?

Aluminium salts have a  astringent effect on the pores causing them to contract, further preventing sweat from reaching the surface of the skin. Aluminium salts are also natural antimicrobial agents, so they control bacteria on the skin, reducing unpleasant odors.
Some of the aluminium salts used in antiperspirants:

Aluminium chloride
Aluminium chlorohydrate
Aluminium zirconium tetrachlorohydrex gly (an aluminium-zirconium compound)


How can a nonlinear field gradient be generated in NMR? I am working on the topic of optimal control in NMR, where spins undergo non-unitary evolution caused by gradient field pulses at different time intervals. I am studying the effects of this on the spin dynamics.
For my project I need a particular kind of gradient field as given by the equation $-z\exp(-z^2)$, where $z$ is the the spatial position of the region from center of the sample. All the books I had referred to (Levitt's Spin Dynamics and Keeler's Understanding NMR Spectroscopy) talk about constant gradients. So I have no idea if it is possible to create even an approximation of a spatially varying gradient as given here.
In particular, I need my field due to the gradient to look like a delta function in space. So I thought of using a derivative of a Gaussian with a small standard deviation as an approximation, which is the equation given above.

Regular off-the-shelve pulsed field gradient (PFG) systems only give linear gradients, obviously. However you can produce higher-order gradients with your shim coils, normally to compensate similar opposing unwanted gradients in the field of the main magnet.
So, by manipulating the shim set, you can create any odd gradient field that you can sufficiently approximate with the terms that your shim system supplies (usually up to 5th order in $z$). To find out what exactly the terms are (coordinates and absolute amplitudes), you can image them using the regular PFG system.
There are two hitches: You cannot switch the shim system quite so fast as the dedicated PFG set, and the "room temperature" shim system isn't nearly as powerful as a PFG, so you are limited in terms of gradient strength. (The "cryoshims" are stronger, but you cannot switch them, as they are superconducting.)
So, if you're OK with a static gradient of low amplitude, you can go straight ahead. If you need it to be pulsed, give it a try. It's been done before, switching shim currents during a pulse sequence. Very good chance to really get to know the insides of your spectrometer system.

If you're secretly an engineer also, check my other answer. ;)


Does the hybridization model gives us any further insight on Molecular Geometry? It doesn't seem as though the hybridization model adds anything useful to the discussion of molecular geometry as predicted by the VSEPR model. It's just another way of labeling linear, trigonal planar, and tetrahedral geometry (as $sp$, $sp^2$, and $sp^3$, respectively). Considering that orbital hybridization may not be the most accurate view of molecular bonding, what is the utility of having it around?

I find orbital hybridization of very limited use.  Orbital hybridization is a mathematical manipulation of atomic wavefunctions, but that that is the only relationship to a quantum mechanical description of an atom.  Presently, we have no evidence that supports the claim that orbital hybridization happens.  The reason why I am not in favor of using orbital hybridization to describe chemical phenomena is because it is not a predictive tool.  We only know the hybridization of an atom after we know the structure.

VSEPR, however, is a predictive tool, although it has no significant quantum mechanical basis.  The VSEPR model is based upon the idea that pairs of electrons (whether bonding or not) will repel one another.  Geometrical factors are considered to determine how far away pairs of electrons can be from one another (e.g 180 degrees apart for two pairs, and 120 degrees apart for 3 pairs).  The model accurately predicts the H-C-H angle in methane and also predicts that the H-N-H angle in ammonia is closer to 109 degrees than 120 (as is the case in a molecule such as $BF_3$.  Because VSEPR assumes ideal conditions, the model starts to fall apart when considering complex molecules.  For example, while the VSEPR model correctly predicts that $SF_4$ is seesaw shaped, it can only predict that the structure won't be precisely seesaw (with F-S-F angles of 180, 120 and 90 degrees).  It only allows us to suggest that the lone pair on the sulfur will be a "space hog" in a manner of speaking and repel the bonding electrons.

Now, on to the last part of your question why do we still use it?  (This is somewhat speculative and opinion-based.) We have to blame the organic chemists here.  orbital hybridization terminology ($sp^2, sp^3$, ...) provides a lexicon for describing the bonding environments of carbon atoms that is compact and visually constructive for organic molecules.  If I state that there is an $sp^2$ carbon in a molecule, you can begin to visualize what that part of the molecule looks like, that it is a region of planarity, that it cannot be a chiral center, that it likely has a double bond and therefore may undergo certain types of chemical reactions, etc. So while orbital hybridization doesn't provide any insight into molecular geometry, it provides a basis for terminology that chemists can use to transmit meaningful information.


Rationale for titant standardization in alkalinity measurements [ISO 9963-1:1994] I work in a environmental chemistry laboratory and I will start to do some alkalinity measurements in my project. My research group is pretty big on following any and all standardization possible, which most often mean compliance according to ISO-standards.

In the ISO-document Water quality - Determination of alkalinity - Part1: Determination of total and composite alkalinity (ISO 9963-1:1994), you need to prepare your titrant, in this case $0.1 \text{M HCl}$ and a standard solution $0.025 \text{M Na}_{2}\text{CO}_{3}$. Then you need to standardize your titrant using either a potentiometric detection (titrate a dilute standard solution to pH 4.5 and note the volume used) or visual endpoint detection (titrate a dilute standard solution with bromocresol green-methyl red indicator until color change). And lastly, a blank using water as the titrand instead of the standard solution.

Using these two acid volume consumption you then recalculate your $0.1 \text{M HCl}$ to it's actual concentration. And if you keep your stock acid solution for a longer period you should perform this standardization every week. Then you can perform the actual titration and determination of the alkalinity in the sample.

My question is, why is this necessary? Why does the ISO stipulate this standardization?

I understand that if I prepare an acid it won't be super precise since there are effects that influence the preparation such as my handling, age of chemical, pipetting, graduated cylinder error intervals etc. But the same would apply to my preparation of the sodium carbonate. In principle I am shifting the error effects from the acid preparation to the sodium carbonate preparation which I am standardizing against.

According to this listing, $\ce{Na2CO3}$ is on of the primary standards used in volumetric analysis, i.e. is easily available in a reproducible form and concentration, has a high molecular mass, is -- once the container is opened -- not (so much) hygroscopic; and is stable.
I speculate, a dilute solution of HCl (0.1 mol/L), if stored over weeks in a bottle recurrently opened, maybe stored on a window bench (seen so frequently in the undergrad's labs) subject to ventilation and underneath heating by the radiators does not have this resillience.
Using a cartrige to prepare a standard solution of HCl that you rinse into a volumetric flask and fill up with deionised water may be seen valid as alternative for a secondary standard by convention. (ISO standards are settled agreements; similar to ASTM standards, if you see things differently, you may contact them.)  Obviously, alteration of a single step within a whole chain of procedures may invalidate the protocol's fit into ISO; a less severe issue in a teaching lab than for a GMP-accredited lab.
Addendum
"How is the diluted solution of HCl obtained?" brought an other element into play that I did not consider earlier: the use of concentrated hydrochloric acid. If purchased and used freshly, it may be a batch of concentrated (37%) or even "fuming HCl" (about 40%), and both are hygroscopic.  Small errors about the concentration of the stem solution of this corrosive reagent may indeed yield larger systematic errors, down the road of repeated dilutions.  (Obviously not so important if the sole intend is to fill up again the bottle of 1 M or 2 M HCl for the extractive workups ...)


Why lanthanides and actinides are shown separate from standard Periodic Table layout? In the standard Periodic Table layout , all the elements up to 56 are in order i.e are in the same layout table. However, lanthanides and actinides are always shown separately from the layout like in this layout:

What is the reason behind this structure?

Is this standard layout or can I represent it like this too?

The Periodic Table arranges elements in blocks as each type of orbital fills with electrons - $s,p,d,f,g,h$.  Alkali metals and alkaline earths are $s$-block filling (but could be one $s$-block slot).  $p$-block six electrons to fill are trelides, tetralides, pnticides, chalcogenides, halides, inert gases (but could be one $p$-block slot).  Transition metal $d$-block is ten elements (but could be one $d$-block slot).  Filling the $f$-block are 14 elements, lanthanoids and actinoids.  That gets sloppy to print and the elements are (or at least were) overall obscure.  They get condensed.

Representing the Periodic Table has become an an art form.  The plain vanilla variety is terse and useful.

http://en.wikipedia.org/wiki/Alternative_periodic_tables


How to figure out the state of aggregation of an unknown substance via properties of a chemical equilibrium in a closed system? This is a question from the College Entrance Test in China.

In a closed, constant temperature container, reaction $\ce{A(s)<=>B +C(g)}$ reached chemical equilibrium. If the volume of the container is reduced, when the equilibrium is reached again, the concentration of $\ce{C}$ equals the concentration of $\ce{C}$ before the reduction.
Is it possible that $\ce{B}$ is in gas state?

The official answer is impossible.
And the official explanation to this is:

If $\ce{B}$ is in gas state, we have $K=C_\ce{B} \cdot C_\ce{C}$
Otherwise, if $\ce{B}$ is in liquid or solid state, we have $K=C_\ce{C}$
Since the concentration of $\ce{C}$ does not change after the reduction, in order to remain the $K$ value unchanged, $C_\ce{B}$ must not be involved in this process. So $\ce{B}$ is impossible to be in gas state.

But I don't think it makes sense.
Consider a container with a volume of $V$, $b$ mol of $\ce{B(g)}$ and $c$ mol of $\ce{C(g)}$ is inside it.
If   $K_{\text{pre}}=K_{\text{post}}$
, we have $C_{\ce{B},\text{pre}} \cdot C_{\ce{C},\text{pre}}=C_{\ce{B},\text{post}} \cdot C_{\ce{C},\text{post}}$.
When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{\text{post}}}{b_{\text{pre}}}=\dfrac{c_{\text{post}}}{c_{\text{pre}}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.
Conclusion: $\ce{B}$ can be in gas state.
So, is the official answer wrong or I just didn't make the right conclusion?

Distinction: you are only considering Kp, which is an equilibrium constant that doesn't take concentration into account. You're going wrong here because the question asks for concentration, which has a unit of $L^{-1}$, not mole fraction, which is just in units of moles.
$$Kc=[C]c[D]d / [A]a[B]b$$
$$Kp=(C)c(D)d / (A)a(B)b$$
Think about their answer to the question. It states that $K=Cc_{pre}$ and $K=Cc_{post}$. The only way for this to make sense is if $Cc$ is in units of concentration because then as the total moles of c go up, the fact that the volume is decreasing doesn't change the value of Kp.

When you say,

  If   $K_{pre}=K_{post}$
  , we have $C_{\ce{B},pre} \cdot C_{\ce{C},pre}=C_{\ce{B},post} \cdot C_{\ce{C},post}$

  When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{post}}
{b_{pre}}=\dfrac{c_{post}}{c_{pre}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.

you do not take concentration into account, or the fact that $Cc_{pre} = Cc_{post}$

For then you cannot restate your equation in terms of concentration: $\dfrac{Cc_{post}}{Cc_{pre}}=\dfrac{1}{2}$


Which atoms in a given amino acid are able to form hydrogen bonds with water? Can anyone help me with this question I have tried everything.
I know hydrogen bonding is with $\ce{F}$, $\ce{O}$, or $\ce{N}$, but every time I select those is says it's wrong.

The positive end of water can hydrogen bond with N, O, or F, as you said, but only if A) the geometry of the atom allows the water and N/O/F to get close to one another and B) the N/O/F has a lone pair of electrons.  This happens at each of the green circled nitrogen and oxygen atoms.  The red circled nitrogen is a decoy, as it is neither geometrically available nor does it have the appropriate configuration of electrons.

The negative end of water can hydrogen bond with hydrogen atoms that are attached to the N/O/F atoms, including the ones that are attached to the N atom that was not correct for the first part.


What is the Kovat's Index used for? As mentioned in the title, why do we use the Kovat's Index in gas phase chromatography? It seems a little arbitrary to compare everything to the order that n-alkanes (such as n-pentane, n-hexane, and n-heptane) emerge from the column, and I am curious to know reasons why we use such an index.

Why do we use the Kováts index?

The Kováts index is used to normalize GC data. This Wikipedia page(though not very long), sums up why you would want to do so.

Retention times of the same compound on even two different versions of identical instrument with the same temperature and pressure program using the same column from the same manufacturer might not be the same! Retention time is dependent on so many things:

The identity of the analyte
The identity of the stationary phase
The length of the column (did you know that columns sometimes get trimmed as they age to remove deposits?)
The diameter of the column
The thickness of the stationary phase film
The flow rate of the carrier gas
The temperature or temperature program of the oven

With so many variables, gas chromatography (indeed all chromatography) can seem like a forbidding black art in which one has no hope of being able to compare results from one instrument to the next or one lab to the next.

Fortunately, we're smarter than that. We can always compare our analyte's retention time to some retention times of well behaved standards. If we know the elution order (will not change) of a series of well-behaved standards on several instruments, and we know when our analyte elutes in relation to those standards on one instrument, we have a good starting point for predicting the retention time of that analyte on another instrument.

Ervin Kováts took this idea a step farther and created a quantitative index relating the retention time of an unknown analyte to the retention times of known standards in a way that produced system-independent constants describing the elution behavior of compounds. The system-independent part is important, given all of the factors listed above that can influence retention time. Every analyte should have (nearly) the same Kováts index value on any GC system under any set of parameters. Thus, if we determine the Kováts index of analyte $\ce{A}$ on one GC, we can know something about how $\ce{A}$ will behave on other instruments or with a different temperature program, etc.

  Why the linear alkanes?

The choice of the n-alkanes was probably not arbitrary, but I cannot say for certain. Of the two links from the Wikipedia article that might contain a rationale, one is dead and the other is in German behind a paywall. However, any well-understood series of standards would have worked. My guess would be that Kováts chose the linear alkanes for the following reasons:

They form a homologous series increasing one carbon at a time.
A great many of them are (were - Kováts developed this index in the 1950s) available commercially in high purity (high enough purity that they can be easily purified further). At Sigma Aldrich, you can buy all of the n-alkanes at least up to eicosane $\ce{C20H42}$.
The first 20 linear alkanes cover a wide range of boiling points (and thus retention times): Methane $\text{BP}=-161.5\ ^\circ \text{C}$, pentane (the first liquid at room temperature) $\text{BP}=35.9\ ^\circ \text{C}$, decane $\text{BP}=173.8\ ^\circ \text{C}$, and eicosane $\text{BP}=343.1\ ^\circ \text{C}$.
There is a relatively reliable linear relationship between $\log(t_r)$ and the number of carbon atoms in the chain.

  How do you determine and use the Kováts index?

Lets say we have an analyte $\ce{A}$ that has a retention time of 193 seconds on some system. First, we identify the linear alkanes that have retention times on either side of the analyte. Lets say that heptane has a retention time of 170 seconds and octane has a retention time of 200 seconds in this system.

The formula for the index looks like this (for non-isothermal programs):

$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$

$I$ is the Kováts index
$n$ is the number of carbon atoms in the alkane with the lower retention time
$N$ is the number of carbon atoms in the alkane with the higher retention time
$a$ is the analyte
$t_r$ is retention time, so $t_{r,a}$ is the retention time of the analyte

For our example:

$$I=100\times \left[7+\left(8-7\right)\dfrac{193-170}{200-170} \right]$$
$$I=100\times \left[7+\dfrac{23}{30}\right]=777$$

Now, we want to analyze $\ce{A}$ on a different instrument, where the retention times of heptane and octane are 217 and 289 seconds, respectively. We can plug values back into the equation for the index to predict the retention time of $\ce{A}$ on this system:

$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$
$$777=100\times \left[7+\left(8-7\right)\dfrac{t_{r,a}-217}{289-217} \right]$$
$$7.77=\left[7+\dfrac{t_{r,a}-217}{72} \right]$$
$$0.77=\dfrac{t_{r,a}-217}{72}$$
$$55.2={t_{r,a}-217}$$
$$t_{r,a}=272$$


Energy bands in Liquid In solid, energy bands means the 1s orbital with spin up and down become 3 states when there are 3 atoms.
But then. If you look at hydrogen atoms, the allowed bound states have energies equal to En=Ry/(n2) and n is the main quantum number ranging from 1 to infinity. In other words, there is an infinite number of available electronic states in a single hydrogen atom.
When we talk about energy bands. Which of the above is usually referred to? The few states orbitals or the almost infinite allowed bound states?
For the latter, how do you characterize the energy bands in liquid like
water? What kind of bands are formed in such cases?

When we talk about energy bands, we are talking about mixing of orbitals on different atoms (AOs) with similar energies. First, assume simple models. In the solid, the structure is very regular and all atoms have AOs with identical matching energies. This makes the number of orbitals available for mixing in solids very large, and mixing results in bands. In a liquid, on the other hand, structure is highly irregular. Orbitals on different atoms often have energy or overlap mismatches, but occassionally match and mix, and this results in much local mixing (say due to hydrogen bonding) but not extended delocalized orbitals. Because of energy mismatch you can also end up with bands in liquids but their origin (in this model) is different than in solids.
The difference between the origin of bands in liquids and solids is similar to the different cause of homogeneous and inhomogeneous broadening. In both cases you get bands (a distribution of energies), but in one case broadening involves delocalization over resonantly matched orbitals on different atoms/molecules and in the latter it involves variations in the local environment due to randomly fluctuating perturbations caused by neighboring molecules  ("the solvent"). Extensive orbital mixing and Pauli's exclusion principle results in homogeneous broadening in the first case. Motional jostling in the case of a liquid results in inhomogeneous broadening.
Energy levels of individual molecules in liquid water are "perturbed" by neighboring molecules. These perturbations are small compared to the ionization energy of the valence electrons. The stability conferred by an H-bond in water is in the order of 20 kJ/mol. Compare that to the ionization energy, ~1200 kJ/mol. Therefore electrons remain bound to local waters even as their energies are slightly perturbed by interactions with other molecules. H-bonds do contain covalent character, that is, involve sharing of electrons between two molecules (specifically between H-bond donor and acceptor atoms). However this does not imply electrons in liquid water are extensively delocalized beyond local bonds, certainly not on the spatial scale possible in solids.
As a note of interest ice under pressure exhibits metallic phases.
Note perturbations due to the solvent are evident in the difference between vibrational spectra obtained in different phases.


What all information can be gathered qualitatively from the radial distribution functions of a molecule? Apart from knowing the radius of coordination shells and the probability of finding, is there any other thermodynamic  information we can gather by looking at radial distribution functions of a molecule?

For instance, If I plot the radial distribution functions for 'O' atom of H2O, once for pure H2O and then once in the presence of ethanol. Will i be able to relate the excess molar volume through the radial distribution functions?
What other information can I get?

You can get alot from the RDF... an amazing amount actually. It is true though that people like printing out the RDF and thinking their paper now has substance, but there is so much more to the pair correlation function.

Integral equations are based on g(r). Kirkwood-Buff solution theory is too. Kirkwood-Buff solution theory will make you dizzy with what it can calculate.

See wikipedia for more on Kirkwood-Buff solution theory. See Arieh Ben-Naim's Molecular Theory of Solutions for the real deal explanation. What a good read. I get excited everytime I get to link it. Ben-Naim, amazing book


Why is black-body radiation curve smooth without a sharp cutoff? Planck's law is able to predict a graph that is consistent with experimental observation:

In essence, unlike Rayleigh-Jeans law that assumes equipartition theorem to hold (that each mode of motion shares equal energy at thermal equilibrium, so all modes excited at thermal equilibrium ⇒ ultraviolet catastrophe), Planck's hypothesis proposed that oscillators of frequency v will be excited only if they can acquire energy of at least hv , where h is Planck constant.1
So based on this argument, I am not sure why the shaded region of the graph will slope in a smooth manner when wavelength is being decreased. Say if you are only able to supply energy of hv, then all oscillators with frequency > v should not be excited and hence will not contribute to energy density ⇒ ρ = 0:

I expect the behaviour to be same as that in photoelectric effect: if frequency of incidence wave is below threshold frequency, you will never observe photoemission. But the actual observed graph is the first one. Where did I go wrong? I am not a physicist so a not-too-complicated explanation will be appreciated.

1 Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 247.

The key to understanding the curve is thinking about how energy is distributed among components of the system
The mathematics that produce the curve involves some not-very-simple statistical mechanics some of which Planck didn't understand when he first developed his theory.
But it isn't that hard to get the intuitive idea. Consider the case of a gas where molecules have some kinetic energy. The individual molecules don't all have the same energy: some are moving faster than others. The temperature of the gas is function of the average kinetic energy of the molecules. But individual molecules are exchanging energy very frequently as they bang into each other. But those random collisions exchange random amounts of the kinetic energy between the molecules. Sometimes one molecule will gain a great deal of energy so it ends up with far more than the average amount. But the probability of a series of collisions giving a single molecule a very, very large kinetic energy is low and the higher that energy, the lower the probability. There is no sharp cut-off just an exponentially decreasing probability of getting a higher and higher energy.
Take that picture and apply the appropriate statistical probability theory and you get the overall Planck distribution: a curve where the average kinetic energy is given by the temperature but where individual molecules have some probability of having much lower energy (with slowly decreasing probability) and others have some probability of having much higher energy (but with a sharply decreasing probability). There is no sharp cutoff at high energies just sharply decreasing probability of reaching those levels.
In reality the picture is far more complex as molecules have vibrational and electronic energy as well as kinetic energy, but these details don't matter much for the intuitive picture.


Are wrinkles\folds in paper reversible by heat, pressure or something else? I ran into this question on Lifehacks.SE, regarding how to remove wrinkles from a crumpled piece of paper. Suggestions included ironing the paper or placing it under some heavy books for a few days.

I am wondering about the chemistry of these processes and whether they are actually effective. This Reddit thread gives some general background about the molecular structure of paper and what happens when you fold paper with enough force to permanently create folds, i.e. to break bonds (as opposed to rolling it within the "elastic boundaries", as it says there). See also this excellent infographic from Compoundchem.

So, can ironing\applying pressure really straighten out a folded paper back to factory level? If not, is there any other way to unwrinkle a paper?

I'd appreciate explanations in the molecular level.

Once you fold a paper or crush a paper, then it was found experimentally that the internal geometry of paper developed ridges throughout inside it and these ridges inside the folded or crushed paper holds the energy you imparted by folding the paper. This energy imparted by you by folding the paper gets stored in the ridges of the paper in form of bending energy (and this bending energy is also the reason that a crumpled paper ball cannot be further compressed beyond around 90% air).

The crumpling of a piece of paper leaves localised permanent marks. This shows that, by some mechanism, the crumpling has focused the stress at some points, exceeding there the limit yield of the material and leading thus to irreversible deformations.

Indeed, a piece of crumpled paper does not usually return to its planar shape even if you iron it or apply pressure to it because along the folds, the material is often strained beyond the level of reversible elasticity.


What happened to this stainless steel tool when exposed to vinegar? Based on an article I found online about how to disinfect hygiene tools I've been soaking nail clippers and similar items in a mixture of white vinegar, dish soap, and warm water for a few hours in a plastic sandwich container.

So far it's been fine, but recently did it with a stainless steel cuticle nipper and after taking it out and rinsing it and drying it off, the metal had turned a dark gray color it's giving off a foul odor.

What happened to the metal, and is the tool still safe to use?

For reference, here is a photo of the tool (on top) and a different one showing what it used to look like (bottom).

What grade of stainless steel was the clipper made of? Was it labeled stainless steel, or is this an assumption because it looked shiny or similar to some other corrosion-resistant object (which may itself be stainless steel or not)?
Test with a magnet: if they are non-magnetic, the clippers could be stainless steel. Without knowing more about the source of the clippers, I could guarantee you that they are not stainless steel - after, they were not stain-less even in vinegar.
It is difficult to imagine any kind of heat treatment or surface treatment (short of plating with a more active metal) that could result in this kind of generalized corrosion.


Enthalpy change of gas expansion against constant external pressure A certain mass of gas is expanded from $\pu{1 L}$ at $\pu{10 atm}$ to $\pu{4L}$ at $\pu{5 atm}$ against a constant external pressure of $\pu{1 atm}$. If initial temperature of gas is $\pu{300 K}$ and heat capacity at constant volume of preocess is $\pu{50 J °C-1}$, find enthalpy change of process.

Here are my thoughts:

$$∆H = nC_p\,\Delta T = nC_V\,\Delta T + p\,\Delta V$$

where $p$ is external pressure.

$$\Delta U = c\,\Delta T$$

with $T_1 = \pu{300 K}$ and $T_2 = \pu{600 K}$ (which I got from gas equation), so

$$\Delta U = \pu{50 J K-1} \times \pu{300 K} = \pu{1.5 kJ}$$

Still I am not getting the correct answer. I believe my methodology is right.

Assume an ideal gas
I think we have to assume an ideal gas, otherwise it is not clear how to get the final temperature of the gas. In the OP's comments, the final temperature is determined as 600 K.
The difference between change in energy and change in enthalpy
Enthalpy is defined as:
$$H = U + PV$$
If both P and V change from state one to state two:
$$H_1 = U_1 + P_1 V_1$$
$$H_2 = U_2 + P_2 V_2$$
we can relate the change in enthalpy to the change in energy by subtracting the two equations:
$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1$$
We could write $\Delta(PV)$ for $P_2 V_2 - P_1 V_1$, but maybe that is confusing.
Change in energy of the gas
In the question, the temperature, the volume and the pressure of the gas all change as we are going from initial to final state. Because energy and enthalpy are state functions, we can decide in which order these changes happen. It makes sense to use steps from initial to final that are easy to calculate. I will change temperature first, then volume (the pressure will change as well, and we will see how that plays into the calculation).
Initial state: $V = \pu{1 L}, T = \pu{300 K}$
Intermediate state: $V = \pu{1 L}, T = \pu{600 K}$
Final state: $V = \pu{4 L}, T = \pu{600 K}$
The change in energy from the initial state to the intermediate state is given by the OP:

$$\Delta U = C_V \Delta T = \pu{1.5 kJ}$$

Taking that intermediate state (V = 1L) to the final state (V = 5L) does not involve any change in internal energy if the gas is ideal, so $\Delta U = 0$ for this step. We could expand into a vacuum, or against pressure. In the latter case, the temperature would drop and we would have to transfer heat into the system to the same amount as the work it did on the surrounding.
So for the entire process, $\Delta U = \pu{1.5 kJ}$.
Change in enthalpy of the gas
$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1 = $$
$$ = \pu{1.5 kJ + 5 atm 4 L - 1 atm 10 L} = $$
$$ = \pu{1.5 kJ + 10 atm L = 1.5 kJ - 0.101 kJ = 1.4 kJ}$$
What about the external pressure?
It does not matter for the calculation because the internal energy of a gas (assuming ideality) does not depend on it. In other problems, where the volume of the gas is not constrained, it might be important, but here it is not.
You could solve this problem going along another path, and keeping track of volume work and heat transferred to get at the changes in energy (using the first law of thermodynamics). This would be more complicated, though.


Difference between ΔQ and dQ What is the difference between $$\Delta Q$$ and $$\mathrm dQ$$ in thermodynamics? I think one is used when the change is very small and the other with bigger changes, but I am not sure.

This is not about mere arbitrary notation.  It is about fundamental concepts in classical thermodynamics, and the notation flows from these concepts.
So: q and w are not state functions; rather, they are path-dependent. Hence their differential forms are inexact rather than exact. The dyet, $\text{đ}$, is used to indicate that (though typically it is an angled slash rather than a horizontal bar; I don't know how to make that with MathJax).
Since the differential forms of heat and work are inexact, any integration with respect to them must be a path integral. While not all physical chemistry/thermodynamics texts use this specific notation for inexact differentials, I've never seen a physical chemistry textbook that doesn't carefully use some sort of notation to distinguish exact from inexact differentials, because of the conceptual and pedagogical importance of distinguishing between state functions and path functions.
[After seeing the comment about Atkins, I went back and looked at my old Atkins pchem book and, sure enough, he doesn't use distinguishing notation for inexact differentials.  I didn't know this because Atkins' thermo treatment was considered poor, so we only used him for kinetics.  We used Castellan for thermo.]
Hence we have, for a closed system:
$d\text{U} = \text{đ}q + \text{đ}w$
And:
$\Delta \text{U} = q + w$
Or, alternately:
$\Delta \text{U} = \text{Q}  + \text{W}$
Heat and work are path functions that reveal themselves only during a change in state—during a process.  Hence, for a given process, there will be a certain amount of heat flow, q, and work flow, w.  One should not use "$\Delta q$"  or "$\Delta \text{Q}$", as this is conceptually incorrect: It implies that heat is a state function, and thus you could have a change in heat.  No.  There was (or could have been) a change in internal energy, along with some heat flow; there was no "change in heat".
More explicitly:
$$\int_{state 1}^{state2} dU = U_2 - U_2 = \Delta U$$
But:
$$\int_{state 1}^{state2} \text{đ}q \ne q_2 - q_1 = \Delta q$$
because there is no q for state 1 or state 2—q is associated with the path only, not with the state of the system.
Instead:
$$\int_\limits {\text{path from state 1 to state 2}} \text{đ}q = q$$
The only context in which it might make sense to use $\Delta q$  (or $\Delta \text{Q}$) would be if you were comparing two different processes, and calculating the difference in heat flow between those two processes.  This is acceptable, but perilous, since such usage risks people falling into the common misconception of thinking that it is acceptable to use $\Delta q$ to characterize individual processes, which it is not.
Adding something from one of my comments: A key property of exact differentials that both connects them with state functions, and distinguishes them from inexact differentials, is that their cyclic integrals must always be zero. Hence:
$$\oint dU = 0= \Delta U_{cyclic}$$
Conversely, it is not generally the case that:
$$\oint \text{đ}q = 0$$
Interestingly, if we carefully define the path to be reversible, and apply an integrating factor $(\frac{1}{T}) $, we can convert the inexact differential $\text{đ}q_{rev}$ into one of the most recognizable exact differentials in thermodynamics, as follows:
$$\oint \frac{\text{đ}q_{rev}}{T} = \oint dS =0$$