Relation between ionic resonance energy and polarity The ionic resonance of the bonds are given as follows:

  $$
\begin{array}{cr}
\ce{A-B} & 24.3 \\
\ce{A-C} & 50.6 \\
\ce{A-D} & 102.3 \\
\ce{A-E} & 105.9
\end{array}
$$

  Which of the bonds is the most polar?

  a) $\ce{A-B}$
  b) $\ce{A-C}$
  c) $\ce{A-D}$
  d) $\ce{A-E}$

I thought that more the resonance energy, more is the electron delocalisation and less polar is the bond, but this is not the case with the answer given.
I was getting a), but the answer is d).

I'm not too familiar with resonance theory, but I'll try:

According to modern VBT, the general idea is that the state of the molecule is not described by a single state wave function, but rather a linear combination ("superposition" if you will) of multiple different state wave functions. In your particular case, the resonance is between a "dominant" covalent structure (involving a completely equal sharing of electron density) and a "minor contributor" ionic structure (with unequal electron density).

Since resonance energy is defined as the amount of energy needed to convert the true delocalized structure into that of the most stable contributing structure (which can be interpreted as the amount of stabilisation provided by the linear combination of the state wave functions), a higher resonance energy would mean a greater contribution (and hence weighing coefficient) from the ionic structure. This means that there is a greater ionic character to the bond (and it is more polar).

Hope this helped.


How can I Identify Elements in Objects? Since all objects on earth are composed of elements, how can I identify the fundamental make-up of objects such as trees, glass, bricks, wood, plastic, shoes, and other entitys?

There is a complete science behind your question called Analytical Chemistry. However the goal is not to identify what a given object is made of, people usually look for targeted compounds or elements of interest. Take chocolate for example, there are thousands of compounds there. Analyzing it completely from scratch may take more than a year! There you will learn plenty of techniques to identify elements. If you just wish to know what this glass is made of, one may use a technique called X-ray fluorescence, it is good for quickly detecting most of the elements except the lighter ones.


Possible erratum in solutions to Orbital Interactions in Chemistry 2ed I was attempting the problems in Chapter 1 of Orbital Interactions in Chemistry, and found a slight problem.

ftp://ftp.wiley.com/public/sci_tech_med/orbital_interactions_2e/Chapter%201%20-%20Answers.pdf

For Q1 (d), I obtained the same secular determinant as the question. However, my answer was slightly different as solving the determinant yielded completely different answers for me.

For the question, H12 (resonance integral) was given as -14.18 eV, S12 (overlap) was given as 0.596, and the energy of the 1s hydrogen atomic orbital was given as -13.60eV. However, plugging these numbers into the secular determinant yielded me orbital energies of -18.26eV, -13.6eV and 41.07eV (the answer given was -33.657eV, -13.6eV, 6.457eV).

Here's the question, for reference. Is my answer incorrect, and if so, how is it incorrect? Thanks!

Here's the question.

Any help would be appreciated.

I ran the calculations through the MATLAB command window and obtained the same answers as you:

>> H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60]
>> S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1]
>> energies = eig(H,S)

energies =

  -18.2615
  -13.6000
   41.0717

On the other hand, if you assume that $S_{12} = 0$ i.e. $S = I_3$ (3x3 identity matrix), then you get the same result as the authors:

>> energies = eig(H)       % or eig(H, eye(3))

energies =

  -33.6535
  -13.6000
    6.4535

I find it slightly telling that the authors explicitly comment that the bonding orbital is stabilised by exactly as much as the antibonding orbital is destabilised. If you solve the secular equations without explicitly substituting in numbers, you should find that this can only be true if the overlap matrix is set to the identity matrix.

The same is true of part (e); you only get the authors' answer by setting $S = I_3$.

I would consider this an erratum in the provided answers.

To find the eigenvalues, use instead [energies, MOs] = eig(H,S). You can try this block of code online. Note that this website has Octave instead of MATLAB, so energies and MOs need to be reversed:

H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60];
S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1];
[MOs, energies] = eig(H,S);
disp(energies);
disp(MOs);

energies is a 3x3 matrix with the eigenvalues as the diagonal elements. MOs is a 3x3 matrix, where each column vector is the MO corresponding to the relevant eigenvalue (i.e. the left-most column is the MO corresponding to the first eigenvalue in energies).


Does gravity affect the rate of a chemical reaction? For example, on planets with lower gravity, would the rate of reactions be lower because the reactants are slower to mix with one another?

It sort of depends what you mean. If by "affect the rate of a chemical reaction", you mean affect the individual steps of some overall reaction as to slow them down, then no, gravity definitely has no effect at all.

This is simply because the gravitational force is so much weaker than the electrostatic force, which will dominate all interactions relevant in chemical reactions, that it does not have sufficient time to make any appreciable difference in the behavior of molecules or atoms.

A kind of standard way of demonstrating this is by imagining a situation in which it is guaranteed that electromagnetism will be stronger than gravity, which is all normal situations in chemistry, and then plugging the numbers into Coulomb's law or Newton's law of gravity.

This, however, is contrived because there are situations where gravity is stronger than electromagnetic interactions. Just never in chemistry. This article does a pretty good job of explaining this, and pointing out that the ratio of forces can really be anything at all.

The key reason gravity will never be relevant to individual chemical processes, and hence never affect the rate, is that the external gravitational field will affect the molecules equally once they are close enough to interact electromagnetically. And once they are interacting electromagnetically, it is guaranteed this force will be much, much stronger than the gravitational interaction between them because both follows the same $1/r^2$ law, so it is just a matter of comparing the proportionality constants which go in front of the law, and inputting some masses and charges characteristic to the situation. The article I linked does provide a context in which the gravitational force dominates the electromagnetic force, so this is possible.

So, no individual rates will not be affected. It's fun to imagine how gravity could affect macroscopic equilibria though, so let's think about that.

Given sufficient time, gravity is very good at organizing things in a particular way. For instance, if you imagine two planets which have atmospheres that at some time are exactly identical, but one planet is much more massive than the other one, I don't think anyone would contend that these atmospheres will look exactly the same after a long time. For instance, the escape velocity of an object is $\sqrt{2rg}$, $r$ being the radius of the planet and $g$ the gravitational constant of acceleration. If $g$ is increased significantly, then you would not lose nearly as much hydrogen from the atmosphere, which could affect the location of certain chemical equilibria in the atmosphere. The atmosphere as a whole would be kept much closer to the planet, which could also be relevant.

All of this is kind of contrived, however. The point is that gravity will only really affect chemistry in how it organizes materials, but cannot affect the individual rates.


Confusion about direction of dipole arrow in alpha-helices and other molecules I understand that molecular dipoles are electric dipoles. And electric dipole moment vectors point from the negative to the positive charge.

In class we learned to draw these special molecular dipole arrows (with a "plus" at the beginning) that point from the positive to the negative partial charge. I also see these arrows all over the internet, e.g.:

But for alpha-helices, the dipole moment points from the C to the N terminus. And the detail always shows the arrow pointing from the oxygen (negative) to the hydrogen (positive):

Should I just accept that even though an alpha-helix is a molecule like water, the "standard" way to draw its molecular dipole arrow is in the opposite direction than what people normally draw for water and other small molecules?

Edit: Another image, from a recent publication, but without macro-dipole arrow...:

I (accidentally) stumbled upon the following statement in Atkins' "Elements of Physical Chemistry" (p378):

  We represent dipole moments by an arrow with a length proportional to $\pmb{\mu}$ and pointing from the negative charge to the positive charge (1). (Be careful with this convention: for historical reasons the opposite convention is still widely used.)

Unfortunately he does not go into more detail. And I know this does not really answer your question.

The definition from the IUPAC is the same as the one used by Atkins:

  electric dipole moment, $\mathbf{p}$
  Vector quantity, the vector product of which with the electric field strength, $\mathbf{E}$, of a homogeneous field is equal to the torque.  $\mathbf{T} = \mathbf{p} \times \mathbf{E}$. The direction of the dipole moment is from the negative to the positive charge.

The source quoted there is from 1993, so you can probably understand my surprise, when I did a little more searching and found in C. Párkányi's "Theoretical Organic Chemistry" (1997, p239):

  [...] in organic chemistry the positive direction of the dipole moment is normally defined as the direction from the center of the positive charge towards the center of the negative charge. This convention prevails in physical organic chemistry and in inorganic chemistry. However, while the dipole still points from the positive charge to the negative charge, in physical chemistry and in chemical physics the positive direction of the dipole moment is defined in the opposite way, i.e., from the negative charge to the positive charge.

There are also a few sources given, but I currently have not enough time to look them up. I find this statement (definition) highly confusing and it also does not give another reason why this is the direction it uses. Please forgive me for not going into more detail here, I really don't want to add any more to the confusion.

Conclusion
Don't add to the confusion. Use
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
as your definition from now on. Popular use does not make it right. Help using it consistently in the correct way and flush out the historic remnants that are still being taught. However, keep Atkins' warning in mind when you read books, paper, etc.. In the literature you will find both versions.

If anyone wants to argue with you about this, give the following derivation. The dipole moment operator $\mathbf{P}$ is a vector operator  that is the sum of the position vectors $\mathbf{r}$ of all $N$ charged particles weighted with their charge $q$.[goldbook]
$$\mathbf{P} = \sum_i^N q_i \mathbf{r}_i$$
For a molecule (neutral by definition)
we find
$$\begin{align}
  0& =\sum_i^N q_i &\Leftrightarrow&&
  0&=\sum_{i_+}^{N_+}q_{i_+} + \sum_{i_-}^{N_-}q_{i_-} &\Leftrightarrow&&
  \sum_{i_+}^{N_+}q_{i_+} &= - \sum_{i_-}^{N_-}q_{i_-} &.
\end{align}$$
This can be transformed into
$$\sum_{i_+}^{N_+}|q_{i_+}|\mathrm{e} =
- \sum_{i_-}^{N_-}|q_{i_-}|(\mathrm{-e}).$$
Therefore we can write
$$\begin{align}
  \mathbf{P} &= \sum_{i_+}^{N_+} q_{i_+} \mathbf{r}_{i_+}
             + \sum_{i_-}^{N_-} q_{i_-} \mathbf{r}_{i_-}
             &\Leftrightarrow&&
  \mathbf{P} &= \mathrm{e}\left(
                \sum_{i_+}^{N_+} |q_{i_+}| \mathbf{r}_{i_+}
              - \sum_{i_-}^{N_-} |q_{i_-}| \mathbf{r}_{i_-}
                \right)&.
\end{align}$$
In the parenthesis the first term is a linear combination of vectors for all positive charges with the resulting vector $\mathbf{r}_+$ and the second term is a linear combination for all negative charges with the resulting vector $\mathbf{r}_-$. The dipole operator is therefore equivalent to
$$\begin{align}
  \mathbf{P} &= \mathrm{e}\left(\mathbf{r}_{i_+} - \mathbf{r}_{i_-}\right),
\end{align}$$
which is a linear combination of vectors that points from negative to positive.


Why do some gases liquefy easily under pressure and normal room temperature? Is this a combination of Boyle's and Charles' Laws or simply one of them?

I'm confused as to which applies. I don't feel I fully understand the implications of both laws individually and I'm wondering why it's easy to take a gas like butane and liquefy it under pressure at "normal" temperatures and why other gases need to be cooled to extremely low temperatures or very high pressures to liquefy.

I understand that temperature, pressure and volume are all related, I just don't know which rules apply and how and how they all interact. How do these laws apply to changes of state from gas to liquid?

I'm sure this is stuff I forgot from 10th grade but I'd love to hear an explanation again.

The gas laws you describe, Boyle's and Charles' Laws, can be combined into one relationship known as the Ideal Gas Law. According to this law,

$$PV=nRT$$

Where

$P$ is the pressure of the gas ($\ce{Pa}$),

$V$ is the volume of the gas ($\ce{m^3}$),

$n$ is the number of moles or amount of gas ($\ce{mol}$),

$R$ is a constant ($R\sim 8.31 \ce{\frac{J}{mol.K}}$), and

$T$ is the temperature of the gas ($\ce{K}$).

To see how changing one property affects another property of the sample of gas, hold the other values constant and see what must happen.

For example, according to the law, an increase in Pressure will result in a decrease in Volume, if Temperature is held constant.
But if the Volume is instead held constant, then the Temperature will increase instead.

In effect, the ideal gas law is the combination of the other laws affecting a gas and its properties.

But this does not explain why some substances are liquids at room temperature while others are not and require cooling and/or a change of pressure.

To describe this, an understanding of what are known as Intermolecular forces must be expressed.

Between molecules of a given substance, there exist attractions. These attractions vary in strength and are what determine the pressures and temperatures needed to liquify specific chemicals.

The (usually) weakest kind of attractions are known as London Dispersion Forces. These forces are generated randomly due to the random movement of electrons about a chemical. Since the movement is random, there come times where the electron density is more negative on one side of a molecule than another. This results in a very slight charge, which can induce charges onto neighboring molecules and make them attract each other.

The next class of attractions are known as Dipole-Dipole Interactions. This is a result of a permanent charge density difference across a molecule. In these molecules, one or more atoms are much greater in their affinities for electrons and will draw the density away from the atoms with lesser affinity. This makes for a permeant slight charge on one or more sides of a molecule that are positive, while the other sides become negative. The negative ends of one molecule will attract the positive ends of neighboring molecules, resulting in a usually pretty strong attractive force.

The strongest by far class of interactions are known as Hydrogen Bonding. This is when hydrogen is bonded to en extremely electronegative atom, like oxygen or fluorine. Take water for example. The Oxygen atom greatly skews the electron density, making the oxygen end much more negative in charge than the positive hydrogen ends. This makes for very strong intermolecular attractions.

With these forces understood, you can now see that chemicals vary in the strength of their intermolecular attractions.

All that said, the weaker the intermolecular attractions, the easier it is for the molecules to break their attractions and become gaseous. Such chemicals with weak attractions require so little energy to become gaseous that at room temperature and standard pressure, the molecules act in the gas phase. To make these chemicals turn into liquids, one must either increase the pressure to make the formation of attractions more favorable or lower the temperature to lessen the energy of the molecules so that they don't break their attractive forces.

Some chemicals have extremely weak attractive forces, meaning that the temperature will have to be very low or the pressure very high to force the chemical into its liquid state.

Other chemicals, like water, exhibit such strong forces that the temperature has to be raised or the pressure lowered to vaporize them.

So yes, the gas properties of pressure, temperature, and volume do affect changes of state, but its the strength of the intermolecular attractions that are the most important and are the reasons that chemicals behave so differently under the same conditions.


How is the ln k term in the linearised Arrhenius equation acceptable? In my syllabus we only study about rate laws under the topic "Rates of reactions", but as I was interested in it, I started to search about that topic. And then I found the Arrhenius equation

$$\ln k = \ln A - \frac{E}{RT}.$$

But according to this equation rate constant $k$ doesn't have units because we are taking the natural logarithm of it. According to rate laws, rate constants have units and they depend on the reaction.

Are we talking about two different rate constants here? I searched this on internet and all the books I could refer but I couldn't find an answer.

The reason is because of sloppy notation. This sort of logarithmic plot is widespread in chemistry, though, and isn't limited to the Arrhenius equation, so do not expect it to change very much in the near future. You will have to get used to it.

With the Arrhenius equation

$$k = A \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

you cannot directly take logarithms on both sides, because of the inherent units in $k$ and $A$. First you would have to divide through by a constant which we could call $k^\circ$:

$$\frac{k}{k^\circ} = \frac{A}{k^\circ} \, \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

Now we can write

$$\ln \left( \frac{k}{k^\circ} \right) = \ln \left( \frac{A}{k^\circ} \right) - \frac{E_\mathrm a}{RT}$$

and as long as $k$, $k^\circ$, and $A$ are expressed in the same units, the resulting arguments of the logarithms will not have units. Note that the actual value of $k^\circ$ does not matter: you will of course get a different y-intercept, but the final values of $A$ and $E_\mathrm a$ that you get out of the Arrhenius plot will not be affected.

The division by $k^\circ$ is often implicit. When people write $\ln k$, what they usually mean is $\ln(k/k^\circ)$ where $k^\circ$ has the numerical value of $1$ and the units of whatever $k$ is in. Once you have the y-intercept, you take the exponential of that and tack the units back on to get $A$.

Although it is not rigorous, from a practical standpoint, this generally does not lead to much confusion as long as the author is consistent with their usage of units. Another reason for omission is that the actual value of $k^\circ$ has no physical effect, as previously discussed, and an overly pedantic focus on writing $k/k^\circ$ and $A/k^\circ$ can distract one from the chemical principles behind these equations, especially for more complicated equations which have many more symbols.


Pyridine synthesis by tin(II) chloride reduction of 5-nitronorbornene What is the reaction mechanism here?
I guess the $\ce{SnCl2}$ reduces $\ce{NO2}$ to $\ce{NH2}$ but then what happens?

This reaction was conducted by Ho and Liao1. Nitro compound 1 is reduced to the nitroso compound 2 which undergoes a [3+3] electrocyclization to form 3 (4). Tautomerization of 4 to 5 is followed by a retro-"Diels-Alder" reaction to open to 6. Conjugate addition of the nitrogen to the unsaturated aldehyde moiety of 6 to afford dihydro pyridine 7. Tautomerization of 7 to 8 is followed by the loss of acetaldehyde. [Structures are from the paper; my numbering.]

ADDENDUM: The mechanism provided in the paper does not address a number of stereochemical issues. Nitro compound 1 is obviously formed as an endo-Diels-Alder product between (E)-(2-nitrovinyl)benzene and 1,3-cyclopentadiene. The bicyclic compound 3 (4) must have a cis-fusion given the mechanism of the [3+3]-electrocyclization (2 → 3) (vide infra). However, the stereochemistry arising from the retro-Diels-Alder reaction would not be expected to kinetically form the (Z,Z)-diene 6 but rather the (E,Z)-isomer 6a. Clearly, 6a is incapable of cyclization as prescribed. But 6a is a masked, vinylogous malonaldehyde that is susceptible to enolization and isomerization to 6b or 6c. While 6c can cyclize by the reported procedure of conjugate addition, there is also the probability that ring closure occurs by electrocyclization of 6b.

Tse-Lok Ho and Po-Yau Liao, Tetrahedron Lett., 1994, 35, 2211. DOI:10.1016/S0040-4039(00)76799-2


What is a chemical element? I would like to know what is the difference between a chemical element and a type of atom? Can  anyone give the definition, please!

From a scientific standpoint, there is no difference.  Elements=Atoms=the 100 or so basic building blocks of chemistry.

This is purely conjecture, but I suspect that the term/concept "elements" pre-date "atoms" (at least atoms as we know them today).  I could see "elements" referring to the macroscopic description, e.g. a jar of carbon (graphite), but it may not have been realized back-in-the-day that these were aggregations of many carbon atoms.

"Atom" strikes me as quantum scale, and "Element" strikes me as something on the macroscopic scale, perhaps hearkening back to the pre-atomic days.  But really, there is no difference.


Why we don't we save water by boiling the contaminated water to get the pure steam? I see lot of drinkable water wasted in household and industrial activities, water usually gets contaminated with detergents(and other household materials),chemical effluents released from factories .
Why we don't just boil off the contaminated water to get the steam and again condense it to get pure water safe for drinking?

I get this idea from the fact that when we boil out salted water we get all the salt as precipitate thus distilling the water.I don't know much about distillation but I think this process can be economically feasable when done on a large scale.

By this method of treatment,I am considering countries that cannot afford water treatment plants on a large scale and with problem of water scarcity,as this would be quick way to get drinkable water.

Boiling a water mixture doesn't always produce pure water.

Boiling a solution of salt and water (or evaporating the water) will separate the salt from the water, but those are compounds with very different properties.

In the making of distilled alcohols (Ethanol), the vapor phase has more alcohol than water.  Repeated distillations can increase the concentration of the alcohol.  But, the leftover undistilled solution is not pure water.  Ethanol and water are very "sticky" to each other.

Ethanol is just an example.  Household 'grey' water (waste water that doesn't include human waste) will have a whole mess of compounds.  Some you can isolate through boiling, some, like ethanol, will 'come along for the ride'.  Wastewater treatment is complicated.  Look at the system that had to be engineered for the International Space Station.  Actually, a version of this system has been tested for use in the types of communities you are describing.

Purifying grey water isn't a simple process.  There are, however, plenty of things that can be done with grey water.  Using grey water to flush toilets is perfectly fine.  With some simple amount of treatment, grey water can be perfectly suitable for watering crops.

So, since it's cheaper and easier to repurpose grey water, and harder to purify it back to drinking-water, reuse is more efficient than trying to completely treat it at home.


What chemicals are used in receipt paper? Most "invisible inks" turn brown when heated and they take a while to transform.  I would like to make (or obtain) an "ink" that responds to the heat of an iron and permanently turns black or a similarly dark color - instantly.  Thermal receipt paper does this where the paper is white until it is heated with almost instant response times AND the "ink" has a permanence of sorts (e.g. sticking a receipt in a freezer won't revert the reaction).  What chemical combinations in what amounts will do an equivalent transformation?

Bonus question:  What chemical combinations would create similar effects for standard palette colors?  That is, transparent/hidden until heat from an iron is applied, whereupon it instantly appears.

My goal is to use such an ink with paintbrushes and fountain pens on canvas.  Black is probably sufficient, but knowing how to make standard colors would dramatically expand my options.  The ink ideally shouldn't be toxic to touch after drying nor emit toxic fumes when heating with an iron at a "Cotton" setting.  It can be semi-toxic while wet but nothing that will melt nitrile gloves, fingers, or human lungs.  I will, of course, wear sufficient protection at all times as I don't really need any chemicals leeching through my skin.  Having a drawing/painting appear instantly in front of an audience from a seemingly blank canvas would be very dramatic and painting with invisible ink will be an interesting adventure.

Have a look at wikipedia article on Thermal Paper. The whole trick is in the three components, which are mixed as solid substances and applied as coating to the paper. When you heat the paper, the solid substances melt and the coloring reaction occurs.

For this you need fine pulverized chemicals, stirred in a form of suspension in some volatile solvent and apply them on the canvas. It should not be a big problem, see e.g. patent on Thermal Paper.

If all this effort is for an artistic purpose, I would suggest as a first try to approach the companies manufacturing the thermal paper (there should be lot of them) and ask them for a bottle of the prepared suspension, telling them your plans. The success rate would not be high, but hopefully non-zero and it saves you 99% of work.

Additionally, the painting itself could be quite fun, I would expect the components to be UV-active, so the black light could help you.


Why Na/C2H5OH not preferred in the reduction of carboxylic acid, despite being used for the reduction of esters why is Na/C2H5OH  not preferred in the reduction of carboxylic acid to alcohols, despite being used for esters?

As the comments point out: the sodium in ethanol generates sodium ethoxide, just as sodium in water generates the hydroxide.  The ethoxide ion, acting as a strong base, deprotonates the acid to a salt and thwarts the reduction.  Get rid of the easily removed proton by using an ester and you unleash the reduction reaction you want.


Determination of mercury in AAS Its known that the only method to determine mercury in atomic absorbtion spectroscopy is the cold vapor method.

What is the reason that mercury can't be determined with other methods, such as flame, electrothermal AAS or even emission spectroscopy ?

Mercury is an element with poor sensitivity when analyzed by traditional flame atomic absorption. This happens because the element is extremely volatile (remember mercury is a liquid at room temperatures and has significant vapor pressure at room temperature). Cold vapor is a widely used method, that's EPA approved, and shows a significant increase in sensitivity (nearly four fold).
Source

There's many different devices out there that can be used to do elemental analysis, but from a personal experience, if you're working with dry solid material (ex. soil), then X-ray fluorescence (XRF) could be worth looking at. It's relatively cheap and does multi-element analysis. With mercury however, you're looking at a LOD of about 5 ppm, which is... okay. It depends on your experiment.


Why do only certain acids disassociate in water? I understand that most acids are weak acids, i.e. they do not fully disassociate in solution. However, certain acids, such as $\ce{H2SO4}$, do. I have been unable to find out what the defining characteristic is that differentiates strong acids from weak ones.

For example, HCl is strong, yet both HF and HBr are weak (even though they're based on halogens as well). Shouldn't HF be even stronger?

What characteristic makes only a select few acids fully disassociate (and hence belong to the strong acid group)?

There are many factors that determine the strength of an acid:

electronegativity of the non-metal atom: an electronegative atom can attract more easily  the electrons of the hydrogen an so generate the proton.
atomic radius greater the radius of the conjugate base greater the acidity, this because the electrons are dispersed around the atom and there is a less attraction between the positive proton and the negative anion.
stability of the conjugate base if the conjugate base is stable due to some particular electronic configuration (e.g. stability due the resonance structure) the species are more prone to be dissociated.


What physical properties are manifested when an electron leaves Na and moves to Cl to form an ionic bond? When an ionic bond is formed between Na and Cl, the lone electron in Sodium's outer shell leaves, and completes Chlorines outer shell.

Are there any physical characteristics that I can use as an observer to see this movement?

When the reaction begins (as seen in this video https://www.youtube.com/watch?v=oZdQJi-UwYs) a lot of light and heat is created. Is this the electron moving from one atom to the other? If not, what is causing it?

Secondary to this. Why does the reaction need help (by either adding water in the video above, or pre-heating the Sodium before dropping it in the Chlorine gas) before it gets started?

The light and heat are created by the exothermic nature of the reaction.  If you look up "Born Haber cycle sodium chloride" you will see information on why the reaction is exothermic.

The reaction "needs help" because of the fact that the sodium will likely be coated with a layer of sodium oxide or hydroxide that will not allow chlorine to collide with the sodium metal.  Even then, with pure sodium metal the reaction will be faster if heated as any reaction needs activation energy to proceed.  Adding water causes an exothermic reaction that can provide activation energy for the reaction with chlorine.


What is LE (plant contents)? I am reading the contents of the oatstraw:

  .. high in vitamins A, C, B complex, LE, G..

I am unable to find on the Internet what LE refers to.

My answer will unfortunately have almost nothing to do with chemistry.
The quoted statement is BS, there's no vitamin LE.
This claim appears to appear on several web pages, the common source seems to be a post from an nutrition/herbal oriented mailing list from 1998:[1]

…
OATSTRAW
Oatstraw as any livestock breeder will confirm, builds the strongest possible
body, with the greatest resilience. Used extensively in European cultures
throughout time for health and beauty, our country has let this valuable herb
slide into disuse.
Contains: calcium. high in Vitamins A, C, B complex, LE, G, phosphorus,
potassium, proteins.
Oatstraw contains Avenin, an amorphous alkaloid which is highly nutritive to
cells, improving normal cellular reproduction.
…

(also note e.g. that avenin is not an alkaloid but protein)
I have an evolution theory how the author could have made this typo, when (hopefully) compiling a literature:

… vitamins A, B complex, C, E, …
… vitamins A, B complex, C E, …
… vitamins A, B complex, CE, …
… vitamins A, B complex, LE, …
… vitamins A, C, B complex, LE, …

(Some intermediate forms might be purely hypothetical. An intermediate form “… vitamins A, B complex, C E, …” can be seen alive at the page mentioning oatstraw, at a different product though[2])
References:

http://www.geocities.ws/chdigest/herbs/pages/apr1998
https://www.healthylife-solutions.com/blog/2017/5/15/womens-herbal-tea-recipe


Why is sugar syrup so sticky? On the festival of Diwali, Indians exchange sweets and share their love. I did just that this very morning, but things don't always go the right way. I dropped a box of sweets and the sugary syrup spread all over the floor. Now while cleaning up the floor, a question popped up in my mind: Why is sugar syrup is so sticky? Is there any chemical phenomenon responsible for its stickiness?

Since I'm a high school student and not a chemist, I can't seem to think of any possible explanation. I shall be thankful if someone could direct me the right way here.

The "sugar" you refer to is sucrose, which is a disaccharide (a fancy term we use for a type of sugar, composed of two other bits of sugar called monosaccharides; which in this case, are glucose and fructose). I felt I ought to have brought up the sucrose specification, since when you talk chemistry, "sugar" can refer to a whole lot of things.
Anyways have a look at this diagram:

Observe carefully, and you'll see that there are 8 -OH groups (hydroxyl groups) in a molecule of sucrose. Now why did I direct your attention to these particular groups?  It's because they're capable of taking part in a fascinating phenomenon called hydrogen bonding (I believe you already know a bit about it, so I won't go into details).
Now when you add a solute to water, that is capable of H-bonding, it sorta causes all the water molecules and solute molecules to 'clump' together. The greater the amount of H-bonding the solute displays with water, the greater the tendency for all those interaction to put up some resistance to the flow of water. This ability to 'resist flow', that is inherent to every real fluid (to different degrees of course), is what you call viscosity.
Now 8 hydroxyl groups, can result in some pretty extensive H-bonding. To give you a rough idea as to how much that contributes to the viscosity of the sugar solution, think of glycerine (glycerol). Rub some of it between your fingers, it's pretty viscous right? Now the thing is, a glycerol molecule has only 3 hydroxyl groups. So imagine how much more viscous/sticky the sugar solution ought to be, when it's got nearly 3 times as many hydroxyl groups! [No, I'm not saying H-bonding is the only factor you should consider]. Of course, the more concentrated the sugar solution, the more the sucrose molecules and hence more the H-bonding; so it'll get stickier and stickier.
Now another thing you ought to consider is the geometry of the sucrose molecule. Now it isn't as important as the H-bonding bit, but it's still noteworthy. What you must know is that the sucrose molecule is not a planar molecule. It isn't some 2-dimensional structure. It's got all sorts of groups sticking out above and below the plane of the carbon rings. Now at low concentrations this isn't really an awfully important thing, but the story changes when you're dealing with concentrated solutions. That's when the 'jagged' physique of the sucrose molecule comes into play. It serves to compound the effect of H-bonding and further obstructs flow.
Want an analogy for that bit on the importance of the molecule's geometry?
Imagine you've got this large sack of coins. Dump them onto the floor and they spread out so well, that you wouldn't be wrong if you said they flowed. But suppose you had nothing better to do, and decided to glue a couple of screws on both faces of each coin. Put all your coins back in the sack, and proceed to dump out the whole lot. You can easily picture how well this 'modification' obstructs the flow of the coins, and probably now, you can appreciate the importance of molecular geometry in this regard better.


Is a microwave oven dangerous? If we take a look at the electromagnetic specter, microwave wavelength is higher than infrared, which means its energy is very small. To break  chemical bonds, the energy required is 100-1000 kJ/mol, and this kind of energy can be afforded only by UV and VIS. Due to this, we can conclude that microwave radiation has only a vibrational effect on molecules of food and does not break any bonds ( the food remains the same ).

Considering those facts, why is the microwave avoided in concern for health?

If there is no concern over the quality of the food, I'm guessing the question is more specific to whether the microwave oven is a safe piece of equipment to use.

The U.S. Food and Drug Administration says:

  Manufacturers must certify that their microwave ovens comply with strict FDA emission limits. The emission limits are well below the threshold for risk to public health.
  Most injuries related to microwave ovens are the result of serious thermal burns from hot containers, overheated foods, or exploding liquids.

  There have been extremely rare instances of radiation injury due to unusual circumstances or improper servicing.

If we're not talking about the oven itself, and rather the nature of chemicals involved with its use, then we could talk about microwaving with plastics with foods. There are sources online that raise concerns over plastics leeching dangerous chemicals into the food.

Harvard School of Medicine says:

  If you’re concerned about plastic wraps or containers in the
  microwave, transfer food to glass or ceramic containers labeled for
  use in microwave ovens.
  Don’t let plastic wrap touch food during
  microwaving because it may melt.
  Wax paper, kitchen parchment paper,
  white paper towels, or a domed container that fits over a plate or
  bowl are better alternatives.
  Most takeout containers, water bottles,
  and plastic tubs or jars made to hold margarine, yogurt, whipped
  topping, and foods such as cream cheese, mayonnaise, and mustard are
  not microwave-safe.
  Microwavable takeout dinner trays are formulated
  for one-time use only and will say so on the package.
  Old, scratched,
  or cracked containers, or those that have been microwaved many times,
  may leach out more plasticizers.
  Don’t microwave plastic storage bags
  or plastic bags from the grocery store.
  Before microwaving food, be
  sure to vent the container: leave the lid ajar, or lift the edge of
  the cover.


How much potassium permanganate is needed to completely oxidize ferrous oxalate? Amount of substance of $\ce{MnO4-}$ required to oxidize one mole of ferrous oxalate completely in acid medium will be:
(a) 7.5 moles
(b) 0.2 moles
(c) 0.6 moles
(d) 0.4 moles

In acidic medium, 1 mole of $\ce{MnO_4^-}$ accepts 5 moles of electrons. Oxalate is dibasic so will give 2 moles of electrons when oxidized to $\ce{CO2}$.
$\ce{Fe^2+ -> Fe^3+ + e-} $ so total 3 moles of electrons.
$\text{Amount of substance of } \ce{MnO_4^-}=\frac{3}{5}=0.6$
However, the given answer is 0.4 moles indicating that only oxalate is considered to be oxidized.

Does  $\ce{Fe^{2+} -> Fe^{3+} + e^-} $  actually occur in this reaction?
What would be the net reaction when $\ce{KMnO_4}$ is used(not just the ions)?

See here.

The "green salt" is $\ce{K3Fe(C2O4)3.3 H2O}$.
\begin{align}
\ce{MnO4- + 8H+ +5e- &-> Mn^2+ + 4 H2O}\\
\ce{C2O4^2- &-> 2 CO2 + 2e-}\\
\ce{2 MnO4^- + 16 H+ + 5 C2O4^2- &-> 2 Mn^2+ + 8 H2O + 10 CO2 (g)}\\
\end{align}

One expects all the oxalate is oxidized to $\ce{CO2}$ and $\ce{Fe(II)}$ is oxidized to $\ce{Fe(III)}$.


Are acids and bases always aqueous? I am tying to determine if an acid is a liquid or aqueous solution? Please can some one enlighten me?

There are varying definitions of"acid" (or "base") depending on solvent system; some do not involve water at all. For example, there are Brønsted–Lowry acids (proton donors) and Lewis acids (electron acceptors), though the Arrhenius definition is based on aqueous solutions dissociating into $\ce{H+}$ (acid) or $\ce{OH-}$ (base) ions.

Of course, many acids are not liquids, such as solid citric acid, which may be anhydrous or the monohdrate.


Acidities of amines and carboxylic acids I have been told that the first of the two acids is the stronger acid. According to this person,deprotonation for both occur at the nitrogen.The resulting conjugate bases include,respectively a neutral amine group,and a negatively charged amine group.Because negatively charged nitrogens are highly unstable from an acid/base standpoint (consider lithium diisopropyl amide)... the first must be the stronger acid (because of its more stable conjugate base).

The problem I have with this line of reasoning is that it's based on a faulty premise - that the nitrogen is the site of deprotonation.I see a much better site of deprotonation for both - and that is the carboxylic acid site. A negative charge here will not only be resonance stabilized but also inductively stabilized by oxygen.Plus, the Ka of ammonium ion is 5.6*10^-10, while the Ka of acetic acid is 1.8*10^-5.In other words - orders of magnitude of difference. I can't see why something that is structurally analogous to ammonium ion (much less ammonia) would be deprotonated preferentially over a carboxylic acid.

Compound 1's conjugate base IS compound 2. The positive NH3 loses a proton to become the neutral NH2, but deprotonation of the NH2 group is extremely unlikely. Compound 2 will lose a proton at the carboxylic acid group first, so your friend is wrong to say deprotonation will occur at the nitrogen in both.

This will likely form a zwitterion, where the nitrogen is NH3+ and the carboxylic acid is COO-. This compound will have 2 pKas, one for the carboxylic acid, 1 for the amine. The carboxylic acid will be the lower of the 2 pKas. So if we start at very low pH and add base, the compound will start like compound 1. As we approach the pKa for the carboxylic acid, it will deprotonate to form the zwitterion. Then, as we approach the pKa of the NH3+, it will also deprotonate, forming a compound with an NH2 and COO-.


Why does the titanium go to the tungsten filament in the crystal bar process? I've tried to read a report by the US atomic energy commission [1], but couldn't really find the answer, or maybe I missed it.
Why is the gaseous titanium deposited on the tungsten wire/filament during the van Arkel–de Boer process?
Reference

Petersen, A. W. Preparation of Metallic Titanium by Film Boiling (Thesis); UCRL-2523, 4393936; United States, 1954. DOI: 10.2172/4393936.

By "gaseous titanium", do you mean "titanium tetraiodide ($\ce{TiI4}$) vapor"?
Van-Arkel De-Boer process is a purification process of titanium and zirconium. Basically, what happens is that the impure metal, let's say titanium is heated in iodine environment at a temperature of $\ce{250 ^\circ C}$ to form volatile titanium tetraiodide ($\ce{TiI4}$) vapor. The impurities are left behind, as they do not react with iodine. This vapor is then  passed over a hot tungsten filament at $\ce{1400 ^\circ C}$ for which the vapor gets decomposed and pure titanium is deposited on the filament and is removed. The iodine is reused. The overall reaction is:
$$\ce{Ti _{(impure)} + 2I2 ->[250 ^\circ C] TiI4 ->[1400 ^\circ C][W filament] Ti_{(pure)} + 2I2}$$
The thermodynamics of this reaction is discussed in Buttonwood's comment.


Dead-body Ashes: biodegradable or not? In many cultures, human dead bodies are burnt to ashes, which is then thrown in rivers and some time buried in earth.
Are human ashes after burning dead bodies biodegradable?

Note: Earlier I posted this question here
But I was suggested to post it here.

According to one source, the major elemental constituents of cremation remains are calcium and phosphorus, presumably from the hydroxylapatite in bone (which consists primarily of calcium phosphate). The rest is mainly assorted trace metals, which exist in relatively minute concentrations in the human body and don't vaporize at the temperatures typical of a crematorium furnace.

Virtually all organic matter, comprising mainly oxygen, carbon, hydrogen, nitrogen, and sulfur, is oxidized into a variety of simple compounds and is removed as a mixture of gases in the furnace fumes (and perhaps some finely dispersed, light solid particulates).

Consequently, what remains is largely inorganic and non-calorific in nature. While the mineral and metal content of the ash can be dissolved and subsequently absorbed by various organisms in trace amounts, there's probably little or no organic material left that can be metabolically decomposed by bacteria, fungi, or plants as a direct energy source. Of course, there are chemical processes that can further break down the ash and isolate specific components via redox reactions (dissolution in strong acids to give salts, electrolysis, etc.), but obviously that's distinct from biodegradation.


What is the range of values of n in [Cu(NH3)4(H2O)n]SO4? My textbook represents the copper ion as "tetraaminecopper(II) sulfate-1-water", so I assumed that this complex ion has only 1 water of crystallization.

Then, in the Wikipedia article, they represent the ion as $\ce{[Cu(NH3)4(H2O)_n]SO4}$. The same formula is used but the number of water of crystallization is variable, denoted by $n$. So $n$ must have a range of values.

My question is what determines the value of $n$? What condition would cause $n$ to be larger or lower?

Tetraamminecopper(II) Sulfate or cuprammonium sulfate can be synthesized as its monohydrate (Ref.1):

  A solution of $\pu{50 g}$ of finely divided $\ce{CuSO4 . 5 H2O}$ in $\pu{75 mL}$ of conc. ammonia and $\pu{50 mL}$ of water is filtered and precipitated by slow addition of $\pu{75 mL}$ of alcohol. After standing for several hours in the cold, the crystals are filtered on a Büchner funnel, washed with a mixture of alcohol and conc. ammonia (1:1) and then with alcohol and ether, and dried by suction.

$$\ce{CuSO4.5H2O + 4NH3 -> [Cu(NH3)4]SO4.H2O + 4H2O}$$

The properties of tetraamminecopper(II) Sulfate monohydrate is listed as follows (Ref.1):

  Deep blue crystals, $\mathrm{d}^{20}_4$: 1.81. Solubility ($\pu{21.5 ^{\circ}C}$): $\pu{18.5 g} / \pu{100 g}\: \ce{H2O}$. Decomposes in air. Loses $\ce{H2O}$ and $\ce{2NH3}$ on heating to $\pu{120 ^{\circ}C}$; the remaining ammonia is evolved at $\pu{160 ^{\circ}C}$.

The crystallographic study of tetraamminecopper(II) Sulfate monohydrate was done several times, and most recently by Mazzi (Ref.2). In the abstract of Ref.1, the author states that:

  $\ce{Cu(NH3)4SO4.H2O}$ is orthorhombie, $Pmcn-D^{16}_{2h}$, with four molecules in the unit cell. The determination of the structure was carried out by means of Patterson and Fourier projections on $(001)$ and $(100)$. The structure consists in layers parallel to $(010)$ formed by $\ce{Cu(NH3)4}$ squares and $\ce{SO4}$ tetrahedra held together by $\ce{H2O}$ molecules. Copper atoms, which lie on symmetry planes, have four nearest $\ce{NH3}$ neighbours ($\ce{Cu-N} = \pu{2.04-2.06 Å}$) and two more $\ce{H2O}$ neighbours ($\ce{Cu-O} = \pu{2.59-3.37 Å}$), which complete the distorted octahedral co-ordination group. Each $\ce{SO4}$  tetrahedron ($\ce{S-O} = \pu{1.40-1.43-1.46 Å}$) has two oxygen atoms linked with hydrogen bonds to adjacent $\ce{H2O}$ molecules ($\ce{OH...O} = \pu{2.67 Å}$).

All of these data directed to the comment of @andselisk who states that tetraamminecopper(II) sulfate is a monohydrate but, Probably why Wikipeda article suggesting it as  $\ce{[Cu(NH3)4(H2O)n]SO4}$ that this compound may be hygroscopic, and thus, its water content may deviate (or vary). The crystal structure has confirmed it is a monohydrate.

References:

Georg Brauer, Ed., Handbook of Preparative Inorganic Chemistry – Volume 2; 2nd Edn, P. G. Stecher, Translation Ed., Translated by Scripta Technica, Inc.; Academic Press: New York, NY, 1965, “Section 19: Copper, Silver, Gold, by O. Glemser & H. Sauer,” p. 1021.
F. Mazzi, “The crystal structure of cupric tetrammine sulfate monohydrate, $\ce{Cu(NH3)4SO4.H2O}$,” Acta Cryst. 1955, 8, 137–141 (https://doi.org/10.1107/S0365110X55000571).


Finding the chemical name of SnMg₂O₄ Can somebody help me in identifying the name of $\ce{SnMg2O4}$?
Other examples are $\ce{ZnAl2O4}$ (zinc aluminate), $\ce{ZnIn2S4}$ (zinc thioindate).

Please kindly provide the reference if possible.

$\ce{Mg2SnO4}$ Magnesium Orthostannate

$\ce{MgSnO3}$ Magnesium Metastannate


How does the inverted Marcus region explain chemiluminescence? The invert Marcus region offers one explanation of chemiluminescence, the process by which light is one of the products of a chemical reaction.

I tried reading one of Marcus's original papers on the subject [1], but I don't really get his argument. I get confused about which of the potential energy surfaces in his Figure (copied here, hopefully that's okay) that refer to the ground state reactant, ground state product, excited reactant, and excited product.

[1] R.J. Marcus (1965), "On the Theory of Chemiluminescent Electron-Transfer Reactions", The Journal of Chemical Physics 43(8):2654-2657

Marcus theory of electron transfer shows that it is possible for a reaction with a large negative free energy ($-\Delta G$ ) to be slower than one with a less negative free energy. The region where this happens is called the 'inverted' region and is shown in figure (c) in your diagram. You can see that the energy of the products is far below that of the reactants but that there is still a substantial activation energy, that is the energy from the minimum of the R curve to the crossing point of the R-P curve on the left.

In fig (A) this is called the 'normal' region and the free energy is close to zero and there is still an activation barrier, in (b) the reaction is slightly exothermic has an activation energy and is still in the 'normal' region. What is missing is the case when the reaction is activationless, this lies between figs (b) and (c) and occurs when the P curve crosses the R curve at its minimum. In this situation the reaction has its maximum rate constant.

The figure shows some data illustrating the rate constnats for electron transfer in the normal(left side) activationless (top) and inverted inverted (right) regions vs $-\Delta G$. The data is for electron transfer from bases in DNA to different intercalated dyes. The fitted Marcus curve is the solid line one from classical Marcus theory the other from a quantum calculation.

Hopefully this will description allow you to follow the paper you have been reading. (You may also want to  look at BBA v811, p265, 1985)

It was not until late 1980's that Marcus theory was confirmed experimentally, first in the solid state and then with chemically linked donor and acceptor molecules as this prevents diffusion, which had hitherto greatly confused interpreting data, and also meant that the inverted region could not be observed.

The energetics for an electron transfer reaction

$$\ce{D^* +A \rightarrow D^+ + A^-  }$$ is

$$\Delta G = E^o_{D^+/D} -E^o_{A/A^-}-E_{s_1} +\epsilon_r$$

for an excited state $\ce{D^*}$ with energy $E_{s_1}$ and electrode potentials $E^o$. Marcus in the paper also includes a solvation term in a solvent, of dielectric constant $\epsilon$ to account for change in redox. This term can often be neglected but is

$$\epsilon_r = \frac{\Delta e^2}{4\pi \epsilon_o}\left (\left(\frac{1}{r}-\frac{1}{2a}\right)\frac{1}{\epsilon}-\frac{1}{\epsilon_ar}\right)$$

where $\epsilon_a$ is the dielectric constant of acetonitrile, $2a$ is radius of donor + acceptor and r their separation.

The rate constant is calculated from
$$ k_\mathrm{et} = k^0\exp\left(- \frac{(\Delta G +\lambda)^2}{4\lambda k_\mathrm{B}T}\right)$$

where $\lambda$ is the reorganisation energy and is the vertical (positive) energy between the minimum of the acceptor potential and the donor potential. Its is determine experimentally from the maximum in the rate vs $-\Delta G$ plot as is $k^0$. In the plot $\lambda \approx 1~\mathrm{eV}$ and $k^0 \approx 10^{13} \pu{s^{-1}}$.

The chemiluminescent reaction can be considered as the reverse of the example given.


When is equilibrium analysis appropriate? What is a good rule of thumb to use to determine if equilibrium is the appropriate assumption?  I assume it is mainly to do with the time scales of the problem, meaning if the reaction time scales are very short, we could probably assume that we get to equilibrium very quickly.  Is there an accepted metric for determining if the equilibrium assumption is valid?

It depends on the context.

The accepted threshold is "when everything else is slow by comparison." For example, if you have a reaction that takes place in two parts, a Jones oxidation taking a secondary alcohol to a ketone and Diels-Alder (which is far slower), you can assume for modeling purposes that the oxidation will be in equilibrium whenever you're looking at how the Diels-Alder will interact with the oxidation.

By contrast, if you're reacting something like a $\ce{CN7}$ anion (which will react very very quickly, detonating with no provocation), the Jones oxidation can't be considered in equilibrium by comparison.


What's the reaction here? I have found a gif photo on the internet and I am very curious what is happening there? What reaction is the reason for that?

$\ce{KI}$ with $\ce{H2O2}$ with a tiny bit of dishwashing liquid. It is a rather violent redox reaction wherein $\ce{H2O2}$ is decomposed to $\ce{H2O}$ and $\ce{O2}$. You can actually see that $\ce{I2}$ is being formed by the brownish-yellowish color in the beginning of the reaction.

There is a lot of information about this reaction. I performed this reaction quite a few times myself for my students. It is a lot of fun. Some people call it making elephant toothpaste. :-)


What is the wavelength of an electron? The kinetic energy of an electron is $1.67 \times 10^{-17}\mathrm{J}$. Calculate the wavelength ($\lambda$) of the electron.

I know the formula $\lambda= \frac{h}{mv}$; where h is Planck's constant.

How do I find $\lambda$ without $v$ (velocity) and $m$ (mass)?

The mass of electron is never given in question you should remember it. Mass of electron is $9.109 \times 10^{-31} \mathrm{kg}$.

Now read the question Kinetic Energy is given, so from that you can easily find velocity of electron.

$$K.E.=\frac{1}{2}mv^2$$

so You will get $v= 6 \times10^6 \mathrm{m/s}$

Now you can substitute value of m and v in formula $\lambda = \frac{h}{mv}$

You will get wave length of electron as $0.12 \times 10^{-9} \mathrm{m}= 0.12 \mathrm{nm}$


How to compute the residence time for given flow rate in standard cubic meters? I'm given a flow rate of a gas mixture of $3\ \mathrm{Nm^3/h}$ in standard cubic meters through a cylinder of length $50\ \mathrm{cm}$ and cross section $70\ \mathrm{cm^2}$, can I compute how long a particle will stay inside of the cylinder (which would equal the time the particle can react inside the cylinder)?

And what if the temperature changes during the reaction?

If the flow rate would be given in volume per time, I had an idea to compute this, but the use of standard cubic meters is baffling to me.

The N stands for "norm" and should be avoided (in fact, the German Wiki page lists the Nm³ as wrong abbreviation) as it can create confusion.(qed)

So your flow should be written as "3 m³/h under standard condtions".
If you can guess German, have a look at the German Wiki page. It is much better than the English one.

Standard conditions vary a bit between different norms

T = 0 °C or 15 °C or 20 °C
p = 1.01325 bar (sometimes 1 bar)

The gas is supposed to be dry.


How does hyperconjugation lead to stability? I know that hyperconjugation leads to delocalization of electrons between filled and unfilled orbitals and thus greater stability (that's what most textbooks said).

However, I think that it reduces the bond strength, because by saying that the electrons delocalized would mean that the electron density between two atoms decreases and hence the 'glue' (bond) that hold the two atoms together becomes weaker.

Any misconception here? Because my statement seems to contradict with the fact that hyperconjugation increasing stability. Why reduction in bond strength increase stability? (or does it actually decrease the bond strength?)

What you are ignoring is the magnitude of the effect.

While hyperconjugation does decrease the electron density of a localized bond and slightly increases the energy, the overlap with a high-energy, empty orbital is at least very poor. Think that the orbitals are very far from each other and have different symmetries. The total electron density of this bond would remain largely unaffected.

On the other hand, the difference between zero electrons and partial electron density for an empty orbital of high energy is significant, and it increases (to a moderate extent) the stability of, say, a carbocation. Besides, a single orbital often receives electron density by hypercongujation with more than one bond.


Why did HF disolve this lightbulb at the surface rather than more submerged points? I watched this video where a lightbulb is suspended in a solution of HF.  The result is that the HF dissolves the glass, but it seems like the part of glass on the surface reacted faster than fully submerged parts.  Here is an image of the lightbulb after the experiment:

The glass breaks at approximately the 2:50 mark, and it looks like at that point the bottom portion was cut from a clean line.  Perhaps the larger portion dissolved afterwards or was damaged during cleanup, I'm not sure.

Is there a reason why the glass closer to the surface would react faster than more submerged parts?  Why would the very tip not dissolve first?

My best guess is that the specific line happened to be the thinnest glass.  I find it unlikely that the specific line is the thinnest part.  Wouldn't it be much more likely for there to be random holes in the submerged portion from the thinnest part of the glass?

Near surface of the liquid the glass is a bit hotter. All glass of the bulb gains comparable amount of heat energy from tungsten wire, but parts above the liquid are cooled only by air, so they are much hotter. Part of this heat is transferred into slightly submerged part, so it is the hottest part that is under liquid, and all reactions are quickened by increase of temperature.

However, this exact experiment is a bit ambiguous, as glass here is under extreme thermal stress, and shattering near surface of the liquid may be addressed to that.


How do you know when the first compound is completely distilled and how do you know when the next compound starts Did a distillation between two solvents. Would the answer have to do with temperature?

There is no such point when the first solvent is completely distilled and then the next one starts. From the very beginning, both solvents boil together, though the vapor initially is relatively enriched with the lower-boiling one. The cutoff point is set somewhat arbitrarily. And yes, the answer surely has something to do with temperature, because temperature is the only parameter we can control.

Like NotNicolaou said, if the boiling points of your solvents are reasonably far away, you may separate the mixture almost completely by collecting the fraction that distills near the boiling point of the low-boiling one. Otherwise, you might not be able to get pure compounds, unless you use an industrial rectification column.


Can eggs boil indefinitely? At breakfast someone asked if the eggs were boiling. While obviously they meant the water for the eggs, but it lead to a discussion regarding if you can make an egg boil without it coagulating or even if you can heat a coagulated egg to the point that it starts boiling under special circumstances such as in an oxygen free environment to prevent combustion.

So would it be possible to make an egg boil without coagulation or otherwise destroying it? The shell would break either way so we can remove that for the experiment.

Boiling means dynamic evaporation of liquid.

Infinite boiling would mean to have infinite amount of that liquid and therefore the egg would have infinite mass.

Such an egg would collapse into a black hole and we would not be able to continue in our observation.

In reality, in inert atmosphere, with overlapping stages:

protein coagulation
water boiling
organic matter thermal dehydration
general organic matter decomposion.
carbon stage

If we kept temperature just above $\pu{100 ^\circ C}$ to keep boiling,  stage 3-4 would be very slow and we would not reach carbon.


Is my reasoning for resonance structures correct? given this picture, which resonance structure is correct? I'm thinking 1 and 3. You can't have 2 because the carbon which originally had a negative charge would be making 5 bonds, and you can't have 4 because the carbon on the right side of the new double bond would also have 5 bonds.

So is this reasoning correct? I'm just having a bunch of trouble drawing resonance structures.

Well, your answer seems to be correct but not your reasoning. In 2 and 4 there are 2 extra pi electrons in the system which seem to have come out of nowhere. The best way to solve is to draw all the resonance structures and see which ones match with the options.


Definition of an "equivalent" What does it mean to add "1 equivalent" of something? I looked it up on google and couldn't really find a good explanation for it.

Thanks!

An equivalent is the factor of the number of molecules you need, at minimum, to carry out a reaction. The number of equivalent depends on your choice of what you consider is your main reactant.

Simply put, if you take a chemical equation where A + B → C, you need 1 eq. of B compared to A, i.e. you need one molecule of B for each molecule of A.

But sometimes, you can get something like 2A + B → C, and you want to know the "number of equivalent of B you should add to A". The quantity of B is one half that of A. If you consider your main reactant to be A, then you need to add only 1/2 (aka 0.5) equivalent of B.


Calculate the pressure of the gas samples as indicated by the manometer Part A) Given a barometric pressure of $\pu{764.0 mmHg}$, calculate the pressure of gas sample (a) as indicated by the manometer.

Part B)
Given a barometric pressure of $\pu{764.0 mmHg}$, calculate the pressure of gas sample (b) as indicated by the manometer.

For Part A) I tried to subtract $764.0-40.0$ to get $724$ but the answer is wrong.
For Part B) I tried to add $764.0+40.0$ to get $804$, but the answer is also wrong.

In part (A), ${\rm P}_{\rm gas} > {\rm P}_{\rm atm}$, as evidenced by the column of mercury being "pushed up" in the manometer on the side of ${\rm P}_{\rm atm}$ - in this case, you'll need to add (the amount it's "pushed up") to atmospheric pressure (i.e., 764 mm Hg). You state that you tried to subtract 40.0 from 764.0: Look at the graphic again (I hold it's 7 cm - or 70 mm - Hg) and try adding that number to 764 mm Hg instead of 40 mm Hg.

For part (B), ${\rm P}_{\rm atm} > {\rm P}_{\rm gas}$, so reverse the process above. I believe what you state is your answer to part (A) is actually the correct answer to part (B).


Why is O2 the supporter of combustion? It's there in every high-school textbook: $\ce{O2}$ is the supporter of combustion. Without $\ce{O2}$ combustion cannot take place. Why? And why only $\ce{O2}$? Why not some other element?

And, what happens when a combustible gas burns in air? Say $\ce{H2}$?

Oxygen is not strictly needed, it just happens to be very, very good at what it does!

Specifically, combustion requires three things:

1) Fuel: The thing that burns. This is is often a hydrocarbon, or other organic molecule. The simplest possible fuel is pure hydrogen gas.

2) Energy: What gets the reaction started, which is true of most chemical reactions.  Note that this so-called activation energy is usually much less than the energy ultimately released from combustion.  This is like rolling a boulder some distance in order to roll it down a hill, which releases much potential energy.

3) Oxidizer: The molecule that accepts electrons.  It turns out that combustion requires the fuel to be oxidized, that is, it donates electrons.  So we need something to accept the electrons, and that's the oxidizer, which is then reduced.

Oxygen is a great oxidizer because it is so electronegative, which means it really wants to accept electrons.  Only fluorine is more electronegative, and fluorine is a superb oxidizer: blow fluorine gas at nearly any substance and it bursts into flames.  For example, see this video: Fluorine.

Oxygen is in many ways a perfect oxidizer for supporting life: it doesn't set things on fire like fluorine, but otherwise can oxidize very many things.  When you eat, oxygen is oxidizing the food (fuel) you ingested, to generate energy you need to live.  So there is "combustion" going on inside you!


How do I find the moles of NaOH required to reach the equivalence point? For each trial with the unknown acid, calculate the following:
  a) Moles of $\ce{NaOH}$ required to reach the equivalence point.

I need help with part A. I have already solved and figured out the concentration of $\ce{NaOH}$ ($.125$M). I've been reading my notes and it says moles of acid will equal moles of base. I've tried solving for both using the volumes and molarity and using the $M_1V_1 = M_2V_2$
I think that's wrong and I'm not sure how else to solve it.

Do I even have the correct equation?

Assuming you are titrating a monoprotic acid, the moles of acid and base should be the same at the equivalence point. To find the moles of base used to reach equivalence, you simply need to multiply the volume of base you added to reach equivalence by the molarity of the base (which you determined from your standardization runs). You also will then know the moles of acid, since they have to be the same at equivalence.


Reaction Order Greater than 1 What is the chemistry behind why reactions of order >1 exist? And what role does the number of activated complexes formed have in determining reaction rate?

The following answers might be vague or overtly simplified or basic, but the questions by nature are fairly open-ended, aren't they? feel free to inquire further if something isn’t clear enough.

Take $$\ce{2NO2 \to 2NO + O2}$$ You can (extremely basically) think of this reaction as two $\ce{NO2}$ particles colliding, (forming a complex), then coming out as $3$ molecules, $2$ being $\ce{NO}$ and $1$ being $\ce{O2}$. But you needed two $\ce{NO2}$ molecules to collide. Collisions are unlikely, first the two molecules have to find each other, then they have to have enough energy, and be in the right orientation.
I’m not sure about multiple activated complexes, but I would only imagine that the reaction becomes even less likely if more than one activated complex is necessary for it; that basically means you need a second reaction from the first activated complex, which means a second collision. (likely slower rates? Someone else feel free to correct if I’m wrong)
Three molecules colliding at once ($3$ rd order reaction): way less likely, and that’s why it’s not studied so much, it’s much harder getting $3$ molecules to collide (especially in the gas phase without) in the right orientation with enough energy.
By general rate law do you mean finding the constant $k$? Or deriving its form entirely? Finding the orders for each reactant and the constant can only be done experimentally by using different concentrations and seeing what comes out.
I think there is more than one way to derive the general rate law’s form. One way is to take the steady state approximation. It has a useful problem that might explain more clearly. The steady state approximation is a commonly made assumption in various situations. Essentially, it assumes that the intermediate is consumed as soon as it’s produced (so there’s no net accumulation of intermediate – which makes sense for many situations). Using the info gained from this approximation you can use the intermediates and the relations that result to form the general rate law. Of course, it turns out the $k$ is composed of multiple constants.
It’s true because they told you it was. This gives you details that you wouldn’t have necessarily been able to deduce just by having the reaction equation there. Importantly, it’s there to tell you straight up that the equation you have can’t be broken down to further reactions... Again you can think of it like this: $\ce{a A}$ molecules have to collide with $\ce{b B}$ molecules to produce a single activated complex. That’s one step. If they didn’t specify that, maybe there could have been intermediate steps that could change the rate law.
Maybe it'll be easier to see if you consider extremes. Say Reaction $R$ can be divided into $4$ reactions, $r1, r2, r3, r4$. Now, say $r1, r2$, and $r4$ happen in microseconds, while $r3$ takes $100$ seconds. Already you're at the point where you can basically ignore the times of $r1, r2$, and $r4,$ because they're so small compared to the time $r3$ took (comparing $1\mathrm E-6$ to $1\mathrm E+2$ here). So as you can see in this example, $r3$ was literally the rate determining step.

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html


How to go about deciding the acidic strength of a functional group in an aromatic compound? The question requires me to identify the decreasing order of the acidic strength of the following functional group. But I'm afraid I don't know how to distinguish one from the other since all the hydroxyl groups would be in resonance. What would be the next step in this case? I felt like the answer should start with (i) as there is a carbonyl carbon near and the delocalisation would be stabilised.

This question is, at best, flawed. The most acidic protons on the structure provided are not any of the labeled OH protons. Rather, the most acidic position is the one I have drawn a rectangle around below.

Deprotonation at this position produces an anion with a resonance contributor in which both rings are aromatic. The circa 150 kJ/mol resonance stabilization for each ring will vastly outweigh any delocalization that might occur for the oxyanions formed by deprotonating any of the OH groups.


Why is nylon durable and how does it have such exceptional strength? I was wondering how the chemical composition of a synthetic polymer such as nylon results in strength and durability.

Polyamides (such as Nylon) are molecules containing a large number of amide functions:

In those functions, the $\ce{C=O}$ bond is polarized (due to the difference in electronegativity between C and O) : $\ce{^{\delta+}C=O^\delta-}$.
The N-H bond is polarized, too: $\ce{^{\delta-}N-H^\delta+}$.
This leads to H-bonds between the chains:

While H-bonds are weak interactions, their high number (as there are a lot of amides in polyamide chains) leads to strongly bonded chains, which results in increased strength and durability.


electron - nuclei recombination within stars Nuclear fusion within stars combine atom nuclei, so the atoms are ionized and electrons run for free.
My question is, when or how these electrons recombine with the nuclei to form neutral atoms? inside the star? outside?

Nuclear fusion in stars produces huge amounts of energy that raise the temperature of the star (e.g. our sun has a surface temperature of about 6000K). At such high temperatures, a lot of atoms possess enough energy to become ionized. However, what percentage of atoms at a particular temperature has what amount of energy is determined by the Boltzmann distribution. So, there will always be some atoms which are not completely ionized (i.e. maybe the electrons are excited from the ground energy level, but they have not completely 'left' the nucleus).
The upshot of all of this is that as the temperature rises, more and more atoms get ionized. Atoms with lower ionization energies like Na, Mg and other metals lose their electrons first, whereas atoms with high ionization energies like He hold onto their electrons until the temperature reaches ~30,000K. Elements like C, N, O, H come somewhere between them. Also, the Boltzmann distribution means that as the temperature rises, more number of a certain atom become ionized. (This page contains a nice diagram showing which ions form at which temperature)
Conversely, if you lower the temperature, you start to form more neutral atoms, and even molecules after a point. However, that happens only below ~3000K, where it starts to look less and less like a star, and more and more like a huge gas giant planet (see Brown dwarf).
As you lower the temperature even more, you can get big complicated molecules like the ones we have on Earth. All of this happens outside of stars, of course*. (*It requires elements like C,N,O,P all of which were synthesized in a pre-existing star, see how planets form )

As an aside, there is another way electron and nuclei can 'recombine' in stars. At the end of the life of a heavy star (> 8 times the mass of sun), the nuclear fusion (that provides an outwards radiation force) suddenly stops. The star then collapses inward on itself due to the heavy gravitational force. This force is so massive that it crushes the electrons and protons together to form neutrons and neutrinos and a huge amount of energy (a type of nuclear reaction). That type of rapid nuclear reaction causes the star to detonate in a supernova.


Fluorination of Carbonyl Compounds using Molybdenum Hexafluoride What is the mechanism of halogenation of carbonyl compounds using molybdenum hexafluoride? How does it work?

Here's a reaction I came across:

Can we use it for all carbonyl compounds? Are there any limitations to the method?

Molybdenum(VI) fluoride can be used as a fluorinating agent though not as popular as the selective DAST and Deoxyfluor as @Waylander noted. Molybdenum(VI) fluoride fall under the category of electrophilic fluorinating agent where it acts as a Lewis acid. The reaction is run in an inert solvent like $\ce{CH2Cl2}$ and a catalyst like boron trifluoride is used. The following is the reaction procedure taken from an e-book$\ce{^{[1]}}$. Also yield of various gem-difluorocarbonyl compounds from corresponding carbonyl compounds are listed:

  A solution of $\ce{MoF6}$ (102g, 0.49g) in anhydrous $\ce{CH2Cl2}$(500mL) is cooled at 20-30 C. While passing a gentle flow of $\ce{BF3}$ through the solution, the carbonyl compound(0.5g) is added in small portion. The mixture is stirred while warming upto rt. Anhyd. $\ce{NaF}$ and silica gel were added. After filtration, the solvent is distilled in vacuo. The residue is distilled/purified by chromatography.

Further, it is written that:

  In general, the yields are not that high due to the sensitivity of the
  reaction towards electronic and steric factors. Yield increase with
  the elctronic withdrawing group of the substituent of the substrate
  and decrease with the steric hindrance of the carbonyl group.
  $\ce{MoF6}$ attacks ether, amino, hydroxy, carboxy as well as
  $\ce{C=C}$ bonds. Ester, amide, cyano, nitro are inert towards the
  reagent. Halo substituent are not exchanged for fluorine.

The general reaction is given by$\ce{^{[2]}}$:

If a cyclic compound is used then R1 and R2 will be fused together.

If fluorination is proceeded at high temperature then catalyst is not needed. Aromatic carboxylic acids are converted into into gem-trifluoro derivatives without any catalyst.
Also, Aryl trifluoromethyl derivatives can also be obtained under less drastic conditions by fluorination of aromatic acid chlorides.

Now, the same reaction can be achieved by using DAST or Deoxylfluor and they are the more popular fluorinating reagent than $\ce{MoF6}$. Their reaction mechanism has been studied extensively. The mechanism is more or less similar than the mechanism achieved by $\ce{MoF6}$. The following is an explanation of the mechanism$\ce{^{[3]}}$(emphasis mine):

  The electronic effect of the substituent exerted a significant
  influence on the reactivities of benzophenones. The presence of an
  electron-withdrawing group is believed to facilitate the nucleophilic
  attack of the fluoride anion to the carbonyl group, causing more
  facile cleavage of the carbon–oxygen double bond. Among the
  benzophenone substrates examined, 4-nitrobenzophenone exhibited the
  highest reactivity and the corresponding gem-difluorinated product was
  obtained in 91% yield[...]

Notes and References

Houben-Weyl Methods of Organic Chemistry Vol. E 10b/1, 4th Edition Supplement: "Organo-Fluorine Compounds - Synthesis of Fluorinated Compounds I, Transformations of Fluorinated Compounds" by Bernd Baasner and Georg Thieme Verlag, 14-May-2014 (link)
http://reag.paperplane.io/00002051.htm#2
Ying Chang, Amit Tewari, Avi-Izak Adi, Chul sung Bae "Direct nucleophilic fluorination of carbonyl groups of benzophenones and benzils with Deoxofluor" Volume 64, Issue 42, 13 October 2008, Pages 9837-9842
DOI: https://doi.org/10.1016/j.tet.2008.08.009
https://www.alfa.com/media/docs/FluorinatingAgents.pdf
Direct carbonyl group attack can be avoided and α-fluorination of ketones can be achieved by using highly selective fluorinating agent, NFSI. α-fluorination of cyclohexanone has been observed.

Enantioselective Organocatalytic α-Fluorination of Cyclic Ketones
Piotr Kwiatkowski, Teresa D. Beeson, Jay C. Conrad, and David W. C. MacMillan
Journal of the American Chemical Society 2011 133 (6), 1738-1741
DOI: 10.1021/ja111163u


How to dissolve beeswax on the underside of a porous wood structure Honeybees, like most insects, communicate through smell. They are attracted to the smell of beeswax. My profession is to rescue and relocate honeybees that have taken up residence inside walls, attics and chimneys. New bees should be discouraged to return by removing all beeswax residue so I recommend scraping down to bare wood and then encapsulating with oil based spray paint. The scraping process is laborious, time consuming and inferior since wood is soft and porous. It would be ideal if a chemical could be applied that would actually dissolve the beeswax without harming the structure. Occasionally bees build over electrical wire and galvanized hurricane ties with hundreds of 8 mm dimples making cleaning very difficult. Heat and a wire brush may work but I sense that heating beeswax only makes the oil spread out. Remember this is on the underside of studs, etc so letting solvent sit to dissolve will not work unless in gel form. Any chemist have suggestions?

Esters are good for dissolving wax, but use a long carbon chain that is not flammable.
Isopropyl myristate is sold online to people that want to make their own cosmetics. It is relatively cheap. It will leave an oil film that can be removed with liquid detergent in warm water.

For thick wax, warm with a blow dryer and scrape with a plastic spatula. Then try above.


How do I compute the mean dissociation time from the dissociation constant, assuming instantaneous binding reaction rate? There is a solution with two chemical species, A and B. Given the equilibrium dissociation constant
$
K_D=\frac{[A][B]}{[AB]}
$
how do I relate the mean dissociation time $t_{off}$ to $K_D$, and initial reagent concentrations, assuming that the reaction time it takes for A and B to bind is much lower than the diffusion time from one to the other?
I know that $k_D = t_{off}/t_{on}$, so here I'm looking for $t_{on}$ in the approximation of instantaneous binding reaction time, which I assume should be the time required for a particle of a species to diffuse through the solution from a particle of the other species to the next. How do I do that to solve the question in the previous paragraph?
EDIT: changed notation to fix  typo and to incorporate Poutnik's feedback.

Note that convention usually used Capital K for TD equilibrium constants and small k for kinetic rate constants.
Thermodynamic equilibrium constants say nothing about kinetic reaction rate constants, but about their ratio. There can be 2 reactions accidentally with the same equilibrium constant, but one with both forward and backward reaction rate very fast, the other very slow.
If there is reaction $\ce{AB <=> A + B}$ with the equilibrium constant $K_\mathrm{c}=\frac{[A][B]}{[AB]}$
with the forward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = - k_\mathrm{f} \cdot [\ce{AB}]$ and the reaction halftime $t_{1/2}=\frac{\ln{2}}{k_\mathrm{f}}$
and with the backward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = k_\mathrm{b} \cdot [\ce{A}][\ce{B}]$
then $K_\mathrm{c}=\frac {k_\mathrm{f}}{k_\mathrm{b}}$
All is assuming the reaction has trivial simple reaction mechanism.
Credit to @Porphyrin:

If $\ce{ A + B -> AB}$ is diffusion controlled, the rate constant would be (for equal sized molecules or approximately so ) $k_\mathrm{d} = k_\mathrm{b} = 8000 \frac {RT}{3η}$
where η is the solvent viscosity,
$k_\mathrm{d}$ evaluates to $≈\pu{5⋅1010dm3/mol/s}$ in low viscosity solvents such as ethanol.

Then :
$$kf=k_\mathrm{b} \cdot K_\mathrm{c}=K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)$$
and the reaction halftime is:
$$t_{1/2,f} = \frac {\ln 2}{k_\mathrm{f}} =\frac{\ln {2}}{K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)} =\frac{\ln {2} \cdot 3η}{  8000 \cdot K_\mathrm{c} \cdot RT} $$


Is there a terminology contradiction about whether the conjugate of a strong acid is a "weak base"? (Please note, I am not simply asking, "Is the conjugate of a strong acid a weak base?"  I'm asking about the contradictory ways those terms seem to be used.)

I was revisiting strong/weak acids/bases to help some high school chemistry students and found a lot of apparent contradictions in the way that sources use the terms.  One is whether the conjugate acid of a strong base is considered a "weak acid".

I understand that the strongER the base, the weakER the conjugate acid, and all sources use the relative terms consistently.  The contradiction is in the use of the absolute terms "strong" and "weak".

There seem to be two conflicting conventions in use.  In Convention I, the conjugate of a strong base is indeed called a weak acid (whereas weak bases and weak acids can also be conjugates), so you have these conjugate pairings:

\begin{array}{cc}
\hline
\text{acid} & \text{base} \\
\hline
\text{strong} & \text{weak} \\
\text{weak} & \text{weak} \\
\text{weak} & \text{strong} \\
\hline
\end{array}

In Convention II, the conjugate of a weak acid is a weak base, and vice versa; the conjugate of a strong base is too weak to be considered an acid, and the conjugate of a strong acid is too weak to be considered a base:

\begin{array}{cc}
\hline
\text{acid} & \text{base} \\
\hline
\text{strong} & \text{not a base} \\
\text{weak} & \text{weak} \\
\text{not an acid} & \text{strong} \\
\hline
\end{array}

You can find examples of both just by googling "conjugate of a strong acid", "conjugate of a weak base", etc.  Many sources complete the sentence "conjugate of a strong acid [or base]" with "is a weak base [or acid]" (Convention I).  On the other hand many sources complete "conjugate of a weak acid [or base]" with "is a weak base [or acid]" (Convention II).

I just want confirmation different sources really are using these terms in ways that contradict each other, so I can pass that on to my students and move on.

So, which of the following is true?

There is no contradiction; I'm reading it wrong.  (This seems unlikely since you can literally google "conjugate of a weak acid" and get what seem to be multiple contradictory answers.)
One usage of these terms is correct, and the sources that say otherwise are using the terms incorrectly.  (In particular, the more reliable sources seem to skew towards saying that the conjugate of a weak acid is a weak base, i.e. Convention II.  Maybe the sources that say "the conjugate of a strong base is a weak acid" are wrong, and they're just garbling the (correct) statement that the strongER the base, the weakER the conjugate acid.)
There is no agreed-upon definition of "weak acid" and different sources use it in contradictory ways, but real chemists don't care because they just use Ka values.

To avoid horrible confusion, please be clear in your answer if you're voting for #1, #2, or #3, or something else not listed here.

(Would rather have made this a comment than an answer but still new. So please don't hang me.)

Through high-school, I was also bugged by this. No, you are not going mad, there are many different views on this matter when you google it. The problem is that like many other scales humans create, the way we define 'strong' or 'weak' is a rather relative concept, especially when it gets to the mid-ranges. It is only a way for us to explain what we observe and get a handle on things.

One thing all chemists would agree on is that you should rather look at how strong an acid or base is relative to another (based on Ka or Kb values). In the end, these are the things you would actually use. However, as you need an answer in a school context I would suggest the following:

At school (at least where I come from) we mainly dealt with Bronsted-Lowry acids and bases. Here the strength of an acid depends on how easily it donates ("accepts" for bases) protons. If your students already know a bit about Ka values then anything with a Ka>1 can be considered (reasonably) strong: on a school level, this is usually limited to HI, HBr, HCl, HNO3, H2SO4 and sometimes HClO3 and HClO4. You might want to look at the relative strengths of these 'strong acids' (I seem to recall an exam question about why HI is a stronger acid than HCl (how stable are the ions) and why HF isn't considered a strong acid - the answer is H-bonding). Anything with Ka<1 can be seen as weak.

As to conjugates. Your statement in 2 about 'The stronger the Acid the weaker the conjugate base' and its inverse applies. Here you also work comparatively rather than absolute. You could try explaining it by saying if something gives away a proton easily (acid strength) why would it want to take that proton back later on? Mathematically this can be expressed by relating the Ka value to the Kb value of its conjugate Ka = Kw/Kb (Kw = 1e-14). You would notice that very few 'weak bases' (Kb<1) will give you a Ka>1, from there the confusion about "the conjugate of a weak base is a weak acid". eg. H2PO4^- (Ka = 6.23e-8) is neither a particularly strong base nor acid, but if it reacts as a base and forms H3PO4 this conjugate still only has a Ka = 7.52e-3 - which is still very weak compared to HI.


Major product of a cyclopropanol with silver ions I'm having trouble predicting the major product of the reaction of 1-(chloro(cyclohexyl)methyl)cyclopropan-1-ol and silver ions:

I thought that in presence of $\ce{Ag^+}$, the three-membered ring would open forming a ketone with chlorine atom in the $\alpha$-position. Then I thought that there might also be some migration of the ketonic group to the ring, but I'm not sure.

A common test for haloalkanes (save fluoroalkanes) is reaction with $\ce{AgNO3}$, which initially yields the corresponding silver halide and a carbocation intermediate. The formation of the insoluble salt is the driving force for the reaction:

On a primary haloalkane this carbocation would be quenched with water, but adjacent to a three-membered ring this carbocation is primed for ring expansion. In addition to reducing ring strain, the resultant carbocation is even further stabilized by the oxygen atom:

Deprotonation gives the final substituted cyclobutanone product.


How are Grignard reagents stable in an ethereal solution? I have heard of the leveling effect, which mentions that highly basic ions cannot exist in a solvent if it's acidity is high enough to be deprotonated by the base.
Considering that, check out the Grignard reagent. I'll use a methylgrignard for this purpose, generated in a diethylether solvent.
This Grignard reagent furnished methanide ions ($\ce{CH3-}$). This ion can abstract protons from any compound thats more acidic than methane. Diethyl ether possesses $\beta$-hydrogens that are easily abstracted due to the inductive pull of the oxygen atom:

The inductive pull may not be much, but it certainly makes it more acidic than a regular hydrocarbon hydrogen.
However we know that Grig reagents do work in ethereal solvents. How is it so?
Edit: This isn't a duplicate of the mentioned question. My question deals with acidity of $\beta$ hydrogens.

We tend to assume that acid-base reactions are instantaneous. But they need to go through an activated complex of higher energy, like every other reaction.

These reactions are usually done at - 78°C. As long as the energy provided to the system is below that which is required to reach the activated complex there should be no reaction.

I know for sure that tetrahydrofuran is fully deprononated by nBuLi at room temperature. The color turns deep yellow, if I recall correctly.


Fluorination with xenon difluoride Is this an error? Why is the nitro group reduced as the benzene ring is oxidized? I don't see any possible reducing agents here..

Also, what is the mechanism of aromatic fluorination by xenon difluoride? This appears analogous to electrophilic aromatic substitution in which benzene acts as a nucleophile and attacks an electrophile because of the expected meta-substitution pattern given the presence of the nitro group.

Is fluorine really electrophilic here? How so?

The reaction as pictured above is incorrect, the nitro group is not reduced.  As you suggest this is (predominantly) an electrophilic aromatic substitution reaction.

Usually $\ce{HF}$ is used as a catalyst in these reactions and the following equilibrium is established.

$$\ce{XeF2 + HF <=> XeF+ + HF2^-}$$

The electrophilic $\ce{XeF^+}$ is the reactive species forming a complex with the aromatic ring.  Xenon is expelled from the complex and the reaction proceeds to yield the meta-fluoro isomer as expected for an electrophilic aromatic substitution.  Some other products, such as substituted biphenyls, are usually formed in small amounts suggesting that some minor pathway (radical-cation?) is also in play.


The quantum mechanics behind periodicity of elements Especially in high school/first-year undergraduate chemistry courses, we learn with great dedication the periodicities along groups and periods. There are various useful and interesting trends.
I wanted, however, to find the theoretical framework behind it which explains the trends through mathematical analyses. I skimmed through the book Physical Chemistry by Peter Atkins, but couldn't find much information.
So I was also considering the possibility that the trends were mainly discovered and investigated through experiments, and quantum mechanics was there only to provide some rough theoretical yet non-quantitative interpretations. (For example, explaining the fact that the $4s$ orbital has a lower energy level than $3p$ orbital using what's called the penetration effect, which says that there is a comparably significant value of $|\psi|^2$ for $4s$ than $3p$ for smaller radius despite the probability densities at higher radius or expectation values of radius)
Is there any qualitative analysis of such periodic trends using quantum theory? If so, where may I find it?

As you surmise, quantum mechanics was developed to help explain observations such as that periodicity, spectral lines and the photoelectric effect.
However, the solution of the Schrödinger equation is difficult, and not complete beyond hydrogen. For example Wikipedia states, "Unlike for hydrogen, a closed-form solution to the Schrödinger equation for the helium atom has not been found." Matrix mechanics can also predict some properties, such as spectral emissions. Neither, approach, though, is likely to give a clear insight into chemical behavior for heavier elements, particularly where relativity affects inner electron behavior for the actinides.
So you've raised a good question, and as mathematics and computational speed increase, exact solutions for other elements will be found.
[To me, it seems a bit odd (no pun intended) that there is one pair of s electrons, three pairs of p, five pairs of d and seven of f, as if it's numerology. But that's my limitation. Sigh... or ψ.]


Why is o-halophenol more acidic than m-halophenol? Why is o-halophenol more acidic than m-halophenol?

My thoughts: m-halophenol has only the -I effect which favors acidity while o-halophenol (which has greater -I) but also has +r which should reduce the effect caused by -I.

It's because the halogen is highly electronegative. Therefore, the closer the halogen is to the alkoxide, the more it helps in stabilizing its negative charge after it's been deprotonated. Conversely, the further away the halogen is, the less it is able to help stabilize the negative charge. The more stable the conjugate base, the stronger the acid.

This question is basically testing induction, which is covered pretty well here:

http://orgomadesimple.com/organic-chemistry-acids-and-bases/


Why do liquids vaporise spontaneously at temperatures below boiling point despite ∆G being positive? We say that for the equilibrium
$$\ce{A(l)<=>A(g)}$$ the ∆G is zero at its boiling point (the thermodynamic definition of boiling point), and becomes positive at temperatures below the boiling point. However every liquid has a characteristic vapour pressure at all temperatures. How does this happen even though the ∆G has become positive thus the reaction has lost spontaneity?

Start with a closed container with a quantity of liquid water, and no vapor. Maintain the container at room temperature. What will happen? If the temperature does not change, the overall Gibbs free energy will not change in equilibrium. Now, $G = H - TS$, and at constant temperature $\Delta G = \Delta H - T\Delta S$.

Converting liquid (take as the reference state) to vapor costs enthalpy, the $\Delta H$. But, by allowing vapor to occupy the empty space in the container you have gained a great deal of entropy, $\Delta S$. The first water molecule to go in to the vapor gains a huge amount of entropy - it has the entire volume to roam around in - overwhelming any enthalpy required. The reaction of liquid -> vapor is, in fact, highly spontaneous because of this! But, the next little bit of vapor gets a smaller entropy gain - there is already vapor occupying the head space in the closed vessel. And the next little bit gets even less entropy. Yet, it keeps costing the same amount of enthalpy to convert the liquid to vapor.

As more and more liquid goes in to the vapor phase, the additional entropy gain decreases, while the enthalpy cost remains the same. Eventually the reaction reaches an equilibrium point when $\Delta G = 0 = \Delta H - T \Delta S$ for the overall reaction. The $\Delta H$ is the total enthalpy required to make the equilibrium vapor concentration, and the $\Delta S$ is the total entropy gained by doing so. You have converted $x$ moles of liquid in to vapor, and it is now in equilibrium with the remaining $y$ moles of liquid.

Perhaps that gives you a better perspective.


I'm trying to verify the integrity of my cheap "24k gold" leaf In what acid would a combination of gold and copper only dissolve the copper, but leave the gold?

I've been googling around, and so far they follow very close to the same solubility rules - as in, it would react with both the copper and gold, and both were semi soluble.  If the compound is brightly colored that would help too.

EDIT: as pointed out by Stian Yttervik, my previous answer (below) is probably too dangerous for someone without prior experience with handling hazardous chemicals.

I would therefore recommend that you use nitric acid: simply add a few drops of clear fuming nitric acid to a test tube containing your gold leaf and stir with a glass rod.

The solution will then go green/blue if it contains copper. Only do this with very small quantities of gold leaf and when outside as one of the products of the reaction with copper is nitrogen dioxide - an extremely toxic brown gas.

I would recommend concentrated $\ce{H2SO4}$, as it is easily available and produces a brightly coloured salt when reacted with copper, thus making it easy to tell whether your gold leaf is 24K or not without any expensive weighing equipment.

Make sure you do this outside or in a very well-ventilated area, as it will give off small amounts of $\ce{SO2}$, a toxic gas. Wear thick gloves and eye protection and make sure you have a water supply nearby in case of an accident.

Pour 10 ml of concentrated sulfuric acid (98%) into a boiling tube (or any other narrow Pyrex container you might have) and heat the liquid with a Bunsen burner until it starts to boil. Avoid the fumes as these are corrosive. Without removing the boiling tube from the flame, add a few small sheets of gold leaf and stir the solution with a glass rod for a few minutes.

If you find that the solution is completely colourless, there certainly was no copper in the gold leaf. However, even the slightest tinge of blue/green would indicate copper sulfate - formed by reacting copper with sulfuric acid.

When you're done, make sure to dispose of the sulfuric acid by emptying it into a litre or so of water (acid in water, not the other way around) and then neutralising this with bicarbonate of soda


Naming substituted branched alkanes with different projections Hi all, I would like some help with naming these two compounds.

For compound 1, I know it is a 2-methyl-pentane, but I am unsure how to name the 3,4 $\ce{OCH3}$ compounds. I know that in the final name, it would be 2-methyl-3,4-di[$\ce{OCH3}$]-pentane (or vice verse depending on alphabetical order.).

For compound 2, I am absolutely lost and would appreciate any help. At first I though it would be a bicyclic cycloalkane, but I steered away from that idea because of the three bridgeheads. Is it possible to have three bridgeheads? I just learned this content yesterday in class and have scoured through my textbook but have seen no mention of a compound like this one. If you could at least give my a general idea of how the 3D structure looks, that may help me. Thank you!

You are right about the first one, it's called 2,3-dimethoxy-4-methylpentane. The $\ce{-OCH3}$ group is called methoxy. The second compound is a
newman projection
. It is basically the view that you will see when you are looking down the $\ce{C-C}$ bond axis. The second one should be called 2,3-difluoro-2,3-dimethylbutane.

Here's what it looks like (http://www.chemspider.com/Chemical-Structure.57499056.html?rid=29f2de9b-b762-46d4-a2ca-7452b8a5d6b6). In the newman projection you are looking down the the C2-C3 axis, and the arms show the substituents bonded to them.