Does NaOH react with K to form NaOK or KOH? I realized that the $\ce{-OH}$ group when it reacts with some metal like $\ce{Na}$ or $\ce{K}$ will replace the $\ce{H}$ element to form $\ce{-OK}$ or $\ce{-ONa}$:

\begin{align}
    \ce{H2O + Na &-> NaOH + H2}\\
    \ce{CH3OH + Na &-> CH3ONa + H2}\\
    \ce{H2O + K &-> KOH + H2}\\
\end{align}

So I am just wondering is that always true? Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$? Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?

The question: Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?

This is a redox reaction regardless of what species $\ce{K}$ would react with. The answer is two-fold: (a) Having potassium added to an aqueous solution of sodium hydroxide; and (b) Having potassium added to molten sodium hydroxide, therefore no water present (or trace amount).

Answer to situation (a) is potassium would essentially react with water (other ions in the solution are $\ce{Na+}$ and $\ce{OH-}$) based on standard potentials of following possible half-reactions.

$$
\begin{align}
\ce{K+ + e- &<=> K} &E^\circ &=\pu{-2.931V}&\\
\ce{Na+ + e- &<=> Na} &E^\circ &= \pu{-2.71V}&\\
\ce{2H2O + 2e- &<=> H2 + 2OH-} &E^\circ &=\pu{-0.8277V}&\\
\end{align}$$

Therefore, the only ionic species in the solution would be $(\ce{Na+ + K+ + OH-})$ in $\ce{H2O}$.

Answer to situation (b) is a little complicated than this. Nowadays, in electrochemical research, use of molten $\ce{NaOH}$ is the choice of the electrolyte. The choice of molten $\ce{NaOH}$ over molten $\ce{Na2CO3}$ is based on a number of advantages of the molten hydroxide electrolyte over others. The first advantage is its higher electrical (ionic) conductivity, which is important in the question at hand. The ionic conductivity of molten $\ce{NaOH}$ exceeds that of molten sodium carbonate by a factor of 1.5 [1]. Thus, it is safe to assume that molten $\ce{NaOH}$ functions as $\ce{Na+}$ and $\ce{OH-}$ ions in molten state. Additionally, $\ce{Ni}$ has already been shown to be a chemically stable material in a molten $\ce{NaOH}$ environment, and hence, nickel containers are used to produce anhydrous $\ce{NaOH}$. Concentrating an aqueous $\ce{NaOH}$ solution in $\ce{Ni}$ containers by heating up to $\pu{500 ^\circ C}$ will produce the desired $\ce{NaOH}$ melt.

According to [1], molecular oxygen is extremely oxidizing in fused hydroxides and reacts with hydroxide ions to produce superoxide and peroxide ions. The concentration of these ions is determined by the following equilibria (Note also that increasing the water vapor pressure would reduce the peroxide and superoxide ion content by favoring the equilibria to LHS, but it's not our concern here):

$$
\begin{align}
\ce{O2 + 4OH- &<=> 2O_2^{2-} &+& 2H2O }\\
\ce{3O2 + 4OH- &<=> 4O_2^{-} &+& 2H2O }\\
\ce{2OH- &<=> 4O^{2-} &+& H2O}
\end{align}$$

For example, see following Abstract:

  A thermodynamical discussion of the $\ce{Pt/O2}$ electrode in molten $\ce{NaOH}$ shows that the redox potential depends on $p_\mathrm{O_2}$, $p_\mathrm{H_2O}$, $a_\mathrm{O_2^{2-}}$ and $a_\mathrm{O^{2-}}$. Potential measurements have shown that the reaction $\ce{O2 + 2e− = O2^{2−}}$ is potential-controlling. From the electrode materials investigated—$\ce{Pt}$, $\ce{Au}$, $\ce{Ag}$, $\ce{Ni}$ and $\ce{Fe}$—the last one only exhibits a positive potential shift by comparison to the other metals; this shift has been interpreted in terms of an $\ce{O2^{2-}}$ impoverishment following corrosion.

  The peroxide which is required for the corrosion reactions can be formed only by chemical reaction of the melt with $\ce{O2}$ , not, however, by anodic oxidation, because peroxide oxidation essentially precedes the oxidation of hydroxyl ions yielding oxygen and water vapour. With anodic and cathodic polarization, peroxide is reacted with limiting current flow, the anodic reaction being less inhibited. Water vapour is cathodically reduced to $\ce{H2}$ and $\ce{OH-}$ —with limiting current flow—before $\ce{NaOH}$ is reduced. $\ce{Pt}$  corrosion is anodically enhanced by water vapour; in this context, current fluctuations are observed. In the pure $\ce{NaOH}$  melt the cathidic reduction may be assumed to involve formation of $\ce{H2}$ and $\ce{Na2O}$ ; not, however, $\ce{Na}$  deposition. [2]

Thus, in a molten hydroxide, one can expect ionic species such as metal cation ($\ce{Na+}$ in this case), $\ce{OH-}$, $\ce{O^{2-}}$, $\ce{O_2^{2-}}$, and $\ce{O_2^{-}}$ anions, and molecular $\ce{O2}$ present in this so-called basic melt in equilibria. Thus, when $\ce{K}$  reacts with any of these (most probably with $\ce{H2O}$ and $\ce{OH-}$ based on their concentrations at equilibrium) to oxidize to $\ce{K+}$, it will stay as mixture of $\ce{KOH}$, $\ce{K2O}$, $\ce{K2O2}$, and $\ce{KO2}$.

Example for ionic nature in molten salt solution molten $\ce{NaCl}$:

References:

Direct electrochemical power generation from carbon in fuel cells with molten sodium hydroxide: Chemical Engineering Communications, 2005, 192 (12), 1655-1670. DOI: 10.1080/009864490896241.
Elektrochemische messungen in NaOH-schmelzen: Electrochimica Acta, 1968, 13 (3), 625-643. DOI: 10.1016/0013-4686(68)87031-8.

KOH because K is highly reactive in the metal reactive series and metals don't react with metals as both are elctropostive


What are the different types of van der Waals forces? As per different websites on internet van der Waals force are of

Three types (Keesom force, Debye force and London dispersion force) @ Wikipedia
Two types (Dipole-dipole force/Keesome force and London dispersion force) @chemguide and chemwiki

While an expert website  also explained H-bonds to be a van der Waals force.

I'm confused. What is originally correct?

The definition of van der Waals forces according to IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8., is as follows:

  The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces. The term is sometimes used loosely for the totality of nonspecific attractive or repulsive intermolecular forces.

The phrase "other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules" would seem to imply that according to the IUPAC Gold Book, if a force is not an ionic or covalent chemical bond or electrostatic interaction, then it is a van der Waals force.

There are too many definitions for something that seems to be incompletely understood and it is best to look at it operationally.  There are forces of attraction between molecules. We know this because it is observed. Different types of molecules have different forces. I will only talk about the weakest of these, the "pure" van der Waals forces between simple molecules that shouldn't attract each other, but do, such as the inert gases especially He and Ne and the nonpolar diatomics such as H2 and the Halogens and possibly to the alkanes, N2 and O2. London dispersion forces arise because of the temporary dipoles in molecules by "motions" of the electrons about the nuclei. I have a problem with this explanation probably because I just don't understand it and have trouble understanding why it is only attractive.  The explanation I was taught was that as molecules approached each other the electron clouds were displaced more or less tangentially exposing the nonvalence electrons to the nucleus of the other molecule The distortion of the orbitals is of slightly lower energy than the attraction hence a positive attraction until the molecules got too close. A simple conversion of kinetic to potential energy. This has a problem with H2 and He because no inner electrons but I can imagine a distance where there is an attractive force. With alkanes the slight bond polarity will give some attraction and perhaps more induced dipoles, with actual polar bonds the dipole dipole attraction is evident. Hydrogen bonding is a great leap upwards in attraction and I wonder why these are considered types of van der Waals forces when they can stand on their own.  So the various definitions seem, to me at least, a bit arbitrary.


Does Br2/H2O oxidize all aldehydes including carbohydrates? In many books, it is mentioned that aldehydes are easily oxidized even by mild oxidizing agents. However, I have never seen bromine water explicitly mentioned. But in the carbohydrates chapter, it is mentioned that glucose etc. are oxidized by bromine water.

So is it that only carbohydrates are oxidized by bromine water or all aldehydes?

Also, do they react with bromine water under the conditions of the bromine water test for unsaturation? Or do they react only under vigorous conditions?

Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic brown color in the presence of aldoses, but will not in the presence of ketoses.

The general mechanism for the oxidation of an aldehyde under these conditions is:

Though ketones will be able to form a hydrate like aldehydes, no deprotonation can occur to eliminate the bromide ion as we see in aldehydes. Hence no further oxidation can occur.

it is a oxidising agent which oxidise aldehyde it is also used in testing for usaturation in organic compound as it decolourize by unsaturated compound. But can not to be used in benzene ring .


Why does O2NCl only have two resonance structures? Why is it that $\ce{O2NCl}$ has only two resonance structures? More specifically, why is the Lewis drawing where $\ce{N}$ is double bonded to $\ce{Cl}$ not considered a resonance structure along with the other two where $\ce{N}$ is double bonded to an oxygen?

$\ce{O2NCl}$ does not have two or three resonance structures but rather almost infinitely many that have different probabilities. Resonance structures will never bring you close to the actual picture; for the true picture to evolve you will need to do molecular orbital calculations. That said, resonance structures are often a somewhat good approximation if weighted accordingly.
This last expression in italics is important. Not all resonance structures are equal. People teaching resonance structures usually use expressions like most probable resonance structures. To determine whether a resonance structure is more probable than another one, check the following list:

does it give all main group atoms a valence electron octet?
are formal charges minimised?
are formal charges distributed in accordance with electronegativity (i.e. do electronegative atoms have more formal negative charge than electropositive ones)?

The list is ordered from most important to not so important although traditional teaching sometimes weighs maximised amount of bonds higher and thus arrives at hyperoctet structures.
Now let’s check the structures of $\ce{O2NCl}$ that you are talking about:

We see that all structures fulfill rule 1. However, rule 2 clearly shows a difference: The two structures on the left contain $+1$ and $-1$ formal charges each while the rightmost structure has $+2$ and $-2$. We don’t need to check rule three, we can see that the rightmost structure is unfavourable and thus less close to the truth. Finally, the formal charges are distributed correctly on all three structures, because oxygen is more electronegative.
Therefore, the left two structures are equally probable while the rightmost one is not so; oftentimes in exams, only the left two will be considered ‘correct’. But as I said, the true picture (or something really close to the truth) can only be determined by quantum chemical calculations.

There's nothing wrong with your suggestion. If $\ce{N}$ was double bonded to $\ce{Cl}$, then both $\ce{N}$ and $\ce{Cl}$ would have positive charges and both $\ce{O}$ atoms would have negative charges. That structure satisfies valence bond theory and octet rules, but it looks a bit strange being doubly zwitterionic. The structure with the $\ce{N=O}$ double bond $\ce{N}$ positive and the other oxygen is negative looks the most plausible.


How to dissolve PLA (polylactic acid)? I'm a maker and I've been interested in 3d printing for years.

I have found a technique for dissolve the PLA supports in a model printed in 3D in HIPS

I have printed the model in HIPS and the support in PLA
I have put the 3D model in a ultrasonic cleaner in the solution with water and sodium hydroxide

(1 liter H2O + 200 grams NaOH)

The PLA has dissolved after 3 hours with temperature 65° and 1 hour of ultrasonic sound.

I attach a youtube link with a similar experiment

https://www.youtube.com/watch?v=uWYn5wJ72zY

Is it possibile use another substance / solution less dangerous and less toxic?

In theory, the PLA is biodegradable. Is it correct?

Ps:
I want use this combination of materials HIPS for model and PLA for Support.

Thank you in advance for the answers.

Ethyl acetate disolves PLA - which is typically the ingredient in non-acetone nail polish removers.

https://en.wikipedia.org/wiki/Ethyl_acetate

According to Wikipedia:

  "If the filament is soaked in a small amount of ethyl acetate, it will dissolve and can be used to clean 3D printing extruder heads or remove PLA supports. The boiling point of ethyl acetate is low enough to also smooth PLA in a vapor chamber, similar to ABS and acetone."

PLA is biodegradable in the sense that it takes less time to biodegrade than other plastics. It normally takes around 80 years to biodegrade if I remember correctly, compared to a few hundred years for most other plastics. Acetone is able to partially melt PLA, but it won't dissolve; instead it will make the PLA sticky and soft, so I would normally use acetone for smoothing print layers.


Stability of tertiary carbocation Our teacher told us that a carbocation connected to 3 cyclopropane rings is among the most stable carbocations. But I am unable to understand how, as resonance (the most important factor governing stability) is absent. How did other factors bring so much stability?

Your teacher is correct, that

  a carbocation connected to 3 cyclopropane rings is among the most stable carbocations.

This because of various electrical effects, such as:

The Inductive Effect
Resonance
The Mesomeric Effect
Electromeric Effect
Hyperconjugation

Here's a link for you, to study them: Click Here.
Please make sure that you learn this correctly and from standard resources, in order to avoid confusion.

There may be a little confusion between mesomeric effect and resonance effect, so here's another link just to avoid confusion. : refer to the answer by Maria Dar

Please, do ask your teacher to help you understand this topic better :-)


How to explain the discrepancy between the calculated absorption wavelength within the particle in the box theory and the observed wavelength? The experimental observed wavelength for this molecule is 425 nm.

We are supposed to estimate this wavelength theoretically using this equation (energy change in a 1D box), where $N$ is the number of pi electrons.
$$\Delta E=\frac{h^2(N+1)}{8m_\mathrm{e}L^2}$$

$L$ is given as 11.84 angstroms.

I'm getting a wavelength of 200 nm. It seems to be too far off.

Molecule:

I'm counting 22 pi electrons and using that as N

This question is different from How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach? in that this question is specifically addressing the discrepancy between the theoretical and experimental wavelengths, whereas the previous question was specific to the number to use for $N$.

Also how does this even make sense? As conjugation increases the energy gap between HOMO and LUMO should decrease. With this relationship as $N$ increases as does the energy gap. To me it makes no sense, how can it be explained?

Luckily I found a source containing your molecule as an example.

You need your equation for the energy difference between the HOMO and the LUMO, which is (nearly) defined as you wrote:
$$\Delta E = \frac{(N+1)h^2}{8 m_e L^2}$$

As you want a wavelength, you need to convert the energy difference, which yields then:
$$\lambda = \frac{8m_ecL^2}{h(N+1)}$$

Now they state, that the length of the box can be guessed by the amount of carbons (p) between both nitrogens by using this formula:
$$L = (p+3+\alpha)l$$
... with $l$ being the length of typical benzene-C-C-bonds, thus being 0.139 nm and $\alpha$ is an empirical correction factor for polarizable end groups.

The amount of pi-electrons (N) that you need to inlude can be calculated by a similar equation:
$$N = p+3$$
... where the additional "+3" come from both nitrogens (would be actually +4, but it's +3 because one nitrogen atom had to form an ion).

This leads to their equation 4:
$$\lambda = 63.7\frac{(p+3+\alpha)^2}{p+4}$$
... where all the physical constants have been combined to give 63.7.

As you have three carbon atoms between both nitrogens, you come up with 328 nm and when you include their "correction factor for polarizable end groups" of $\alpha = 0.675$, you come up with 405 nm, which is more than close to the experimental result.

When I use your box length and use 12 electrons (3 from the "left" benzene ring, 4 from the two double bonds to the "right" nitrogen" and 1 from the lone pair of this nitrogen), I come up with about 356 nm, which is not that far away as it sounds. In eV this is an error of 20% (the 405 nm are wrong by about 5%), whereas your 200 nm would have been an error of 120%!

When one "arbitrarily" insists that the box is of a certain length, that is where any method fails.  The molecular geometry provides the boundary conditions on the standing waves for the electrons, so people fix things by adding an extra length (fudge factor) to account for the discrepancy in the model.  So, the Kekule structure(s) (Lewis structure(s)) people start with to model the box length are very approximate.

So, can we make a better approximation without having to account for every electron?  Yes, but we need to include those electrons in the molecule that are leading to the measured "characteristic length". What is that thing, but the "conjugation length".  You are gonna say, "the whole thing is conjugated", but look closer.

We all learn that conjugation is alternating double and single bonds right, but what is that here and how does it help?  Although the conjugation method is limited too, it does offer a better approximation to a "length" since it starts with a good fragment (stable "vinylic like CH2=CH-" chunks).

Use what you know about MO's from organic chemistry and try assigning phases to the Kekule structures and see what you need to do to make the longest length of alternating (++)_(--) units.  That is, try alternating phase pairs of the pi orbitals and see which configuration(s) gives the longest length.  You may need to change the number of alpha and beta phases that appear on one side of the molecule  to get the longest length (flip a pi-bond over to get its phase out of the way of conjugation path).  How many bonds long is that, and is that an integer multiple of the measured length (or a reasonably close approximation to an integer multiple of a bond length)?  Compare that best guess and shape to what the HOMO and LUMO look like.  What do you think conjugation is now and how does the end group affect conjugation?  Cheers.


Is graphene a combustible material? Various sources such as https://news.ycombinator.com/item?id=7679154 speculate that graphene will combust at higher temperatures.

Is this true? Is the flash point/ temperature of combustion for graphene known?

Graphene combusts at 620 K. See http://pubs.acs.org/doi/abs/10.1021/jp410044v.

It is postulated that it burns at this temperature "because of a large interlayer spacing of graphene sheets (i.e., 5.1 Å)".

Graphene is made from carbon atoms, just like diamond and graphite.
Just like diamond and graphite, it will burn in oxygen.


Express the maximum work from a voltaic cell The net cell reaction of an electrochemical cell and its standard potential is given below:
$$\ce{ Mg + 2Ag+ ->Mg^{2+} + 2Ag} \ \ \ \ \ \ \ \ E^\circ=3.17\:\mathrm{V}$$
The question is to find the maximum work obtainable from this electrochemical cell if the initial concentrations of $\ce{Mg^{2+}}=0.1\ \mathrm{M}$ and of $\ce{Ag+}=1\ \mathrm{M}$.

The solution just uses the Nernst equation to find the potential at this concentrations and uses $\Delta G=-nFE$ to calculate the Gibbs free energy change which is then equated to the maximum work obtainable.

But this is just the work obtainable per mole at this concentration only. As soon as the reaction proceeds, the concentration change and so does the value of $\Delta G$ per mole and therefore the maximum work obtainable is different for different concentrations.

Therefore, to find the maximum work obtainable, shouldn't we first calculate the equilibrium constant of this reaction (I calculated it to be $K_c=1.89\times10^{107}$), then the equilibrium concentrations(which is probably $\ce{Mg^{2+}}=0.6\ \mathrm{M}, \ \ce{Ag+}=0.178\times10^{-53}\ \mathrm{M}$) and then use something like:
$$
\int_{\mathrm{in}}^{\mathrm{eq}}\Delta G* \mathrm{d\,M}
$$
where $\Delta G$ is per mole and the $dM$ is a small amount in moles that the reaction proceeds at that concentration.

Please help me verify my method and to proceed further.

There is a simpler approach to this problem.

First, since both initial concentrations are not 1 M, you need to determine $E$ for the initial conditions, where $Q$ is the reaction quotient $Q=\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^{+}}]^2}$:

$$E=E^\circ-\dfrac{RT}{nF}\ln{Q}$$

The value of $n$ for this reaction is $n=2$, and assuming $T=298\ \mathrm{K}$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ K})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.1 \text{ M}}{(1.0 \text{ M})^2}}=3.22 \text{ V}$$

You can then go and convert this potential into free energy or maximum work or whatever else you want.

At equilbrium $E=0$ and $Q=K_c$, so we can calculate the equilibrium constant if we want to compare it to $Q$ and determine which direction is favored. We don't need to. Since $E_i>0$, this reaction procedures toward the products.

Now, as the reaction proceeds the concentrations of $\ce{Ag+}$ and $\ce{Mg^{2+}}$ will change. From the law of mass action:

$$-\Delta \ce{[Ag+]}= 2\Delta \ce{[Mg^{2+}]}$$

After some time $\Delta \ce{[Ag+]} = 0.25 \text{ M}$, so $\Delta \ce{[Mg^{2+}]}=0.125\text{ M}$ and $[\ce{Ag+}]=0.75 \text{ M}$ and $[\ce{Mg{2+}}]=0.225 \text{ M}$. The value of $Q$ has changed and so has the value of $E$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ V})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.225 \text{ M}}{(0.75 \text{ M})^2}}=3.18 \text{ V}$$

As the reaction proceeds, the value of $E$ decreases, which means the amount of work that can be done by the reaction also decreases. The maximum value of $E$ occurs at the initial state, and so the maximum work that this cell can do happens at the initial state. Since $E$ decreases as the reaction proceeds, so too does the work the cell can do at any moment.

If you want the total work, then you need to integrate with respect to $Q$, not worrying about the actual concentrations:

$$\int_{Q_i}^{K_c}E\cdot\mathrm{d}Q$$

It seems to me that your understanding of the question is flawed. You would require the integral suggested if you wish to calculate the total work done. The question asked is to calculate the maximum work at any given instant, which is obviously at the the initial state where the concentrations would be maximum and thereby the value of E cell. This would give you maximum value of work done at any particular instant until equilibrium is reached and not the summation of work done during the whole process until the cell dies.


How can over grinding lead to reduced API extraction in HPLC analysis? I am looking for some advice regarding particle size distribution (PSD) & how this can be affected by grinding a tablet.
During an assay by HPLC, the method requires grinding of a tablet before dissolving in order to extract the active pharmaceutical ingredient (API) into solution. Of the tablet, the API is only 0.5% m/m of the tablet. I have been told that over grinding the tablets can affect extraction of the API and this is supposedly related to PSD as the API is the smallest particle found in the tablet amongst other excipients.
Could someone explain how "over grinding" would cause a detremential effect? Thanks.

I sort of feel this is actually more of a mathematical/statistical/modeling related question than traditional chemistry.
In essence, one argument is that the sample derived effective API concentration is impacted by possible changes in the particle size distribution due to grinding. Now, there are, in effect, known statistical distributions that address particle size distributions (PSDs), and associated methods to assess parameters of said distributions relating to sample means and variability, and changes thereto.  This would seem to be a possible component of the analysis.
As background, I reference Wikipedia on the topic, which importantly notes, to quote:

The PSD of a material can be important in understanding its physical and chemical properties...It affects the reactivity of solids participating in chemical reactions, and needs to be tightly controlled in many industrial products such as the manufacture of printer toner, cosmetics, and pharmaceutical products.

as both surface area (a function of particle size) and concentration influence a reaction.
Also, some detail on PSD distributions:

The log-normal distribution is often used to approximate the particle size distribution of aerosols, aquatic particles and pulverized material.
The Weibull distribution or Rosin–Rammler distribution is a useful distribution for representing particle size distributions generated by grinding, milling and crushing operations.
The log-hyperbolic distribution was proposed by Bagnold and Barndorff-Nielsen[9] to model the particle-size distribution of naturally occurring sediments. This model suffers from having non-unique solutions for a range of probability coefficients.
The skew log-Laplace model was proposed by Fieller, Gilbertson and Olbricht[10] as a simpler alternative to the log-hyperbolic distribution.

Now, for example, in the case of the Log-normal distribution, the best estimate of the parent population mean particle size (which here can be used to estimate total concentration) is obtained by working with the log values of the sample particle sizes. Wikipedia on the Log-normal distribution notes that the sample mean (say 'm') and variance (${s^2}$) of these log-transformed values can be employed to more accurately specify the parent population (aka, the Log-normal) mean, which turns out to be ${Exp(m + 0.5*s^2)}$. Note, simply taking the straight average of the observed Log-normals size values is not recommended as a Log-normal random deviate can be very noisy being the ${Exp()}$ of a Normal random deviate, hence the use of a log-transform. The Weibull distribution is another example of where the parent population mean is a function of two of the distribution's to-be-estimated parameters.
Further, it may indeed be the case that excessive grinding affects particle size distribution and reaction rate. However, this should be verified experimentally to determine the potential impact. Perhaps using a proxy, that produces a more readily measurable product (as in gas evolution, or strongly colored compound) may be of assistance.
Finally, this educational source comments on the inter-connection between reaction rate, mass and concentration, noting an exception, to quote:

The concentration of the reactants determines the speed of the reaction. In simple reactions, an increase in the concentration of reactants accelerates the reaction. The more collisions over time, the faster the reaction can advance. The small particles have less mass and more surface area available for the collisions of other particles. However, in other more complex reaction mechanisms, this may not always hold true. This is often observed in reactions involving huge protein molecules with large masses and convoluted structures with reaction sites buried deep within them that are not easily approached by collision particles.

So, still some experimenting required here in an obviously specialty discipline with a possibly large molecule.


Schottky Defect in AB2 type crystals Are there any examples of Schottky defect in ${\ce{AB_2}} $ type crystals.
Does Schottky defect require that the cation and anion be of equal size?
According to me it is not necessary that the two ions are of equal size as even ${\ce{NaCl}} $ shows Schottky defect but ${\ce{Na+}} $ has radius is 95 pm and ${\ce{Cl-}}$ has radius of 181 pm which is a bit less than double of that of ${\ce{Na+}} $

Schottky defect occurs when equal number of positive and negative ions are missing from the lattice thereby maintaining net electrical neautrality and leading to a decreased density in the lattice (since there are now less number of ions) .
It is observed to be occurring in cations and anions of similar sizes.

NOW CONSIDER THAT AN ION IS AT THE LATTICE POSITION AND A DEFECT IS TO BE OCCURRED.

There are 2 ways stoichiometric conditions can be satisfied. - either by loosing this ion as well as another of opposite charge in equal ratio(SCHOTTKY) ,
OR not losing this ion at all , but just shifting it's position from lattice position to interstitial one.(FRENKEL)----
If we think logically ,

Since it's a stoichiometric defect it should maintain neautrality as well as loose cations and anions in a certain ratio.

If the sizes were very different , there are chances the smaller ion would occupy an interstitial space causing it to create a hole and still maintain neutrality and ratio without losing any ions. (Which is what happens in Frenkel defect) . Since however ions are of similar sizes , one can't occupy an interstitial space .. and therefore losing equal no. Of anions and cations is the only way to go.
You can think of these as 2 cases of defect... Yes it's not necessary the ions are of similar sizes for a defect to occur BUT IF THEY ARE...They are most likely to cause the Schottky defect and if not , then Frenkel Defect.
As for your NaCl doubt , I would suggest try finding the critical radius ratio for one ion to occupy the interstitial space . This way you can verify whether it can perform Frenkel defect or not ..Or why it's performing Schottky.
Hope this helps.

EDIT:- I can't find any examples of ${\ce{AB2}} $ type as such , but you can again try finding critical ratio of ions in ${\ce{AB2}} $ lattices to decide what type of defect they are likely to occur.


One half of a nail is exposed to vinegar. Un-exposed half develops a brown film. What is it? I have recently done an experiment for my chemistry EEI (Year 12) where we are testing what factors effect the rate of corrosion. I used vinegar (acetic acid) as one of the acids and Hydrochloric acid as the other.

In the one with the HCl, the results were as expected – the area exposed to the acid will corrode, the unexposed part won't (in the time frame I did).

However, for the vinegar, the results were different. The part exposed to the vinegar didn't show any rust however the part of it that wasn't exposed had a brown coating. Now I understand that the vinegar reacts with the rust, hence the reason we cant see it on the iron nail, but my question is: What is the brown coating on the top of the nail, and what is it caused from? Was there some sort of gas from the vinegar that caused it? If so, what is this gas called and how does it cause corrosion?

The nail is in two solutions. The lower solution is 5% acetic acid, saturated with air (20% $O_2$, 80% $N_2$). At 0.2 atmospheres, $O_2$ solubility in water is 40 mg/L x 0.2 = 8 mg/L, or about 8 parts per million.
The upper solution is a thin film on the nail (perhaps 10 microns thick - or less!) of $H_2O$ with some acetic acid (probably about 3-4%) and 8 ppm $O_2$. This film may have deposited from the lower solution by vapor deposition due to temperature differences, or by capillary action, as the meniscus crept up in a thin layer.
In the lower solution, $Fe^0$ goes into solution as $Fe^{++}$ and electrons go to $O_2$.
4$e^-$ + $O_2$ + 2 $H_2O$ --> 4 $OH^-$
The 8 ppm $O_2$ in the lower solution doesn't go very far, but the meniscus, and any film going higher up, will be continually refreshed with $O_2$, and will develop a higher pH and will precipitate iron hydroxide. Any iron dissolving in the lower solution will stay in solution because of the acidity, and the nail may not even show obvious signs of corrosion damage like pitting. The only obvious sign is the brown film of precipitated iron hydroxide.
This is an example of a differential aeration cell, where the upper solution is an oxygenated cathode, and the lower solution contains the anodic portion of the nail. http://www.corrosion-doctors.org/Corrosion-Factors-Cells/corrosion-cells-aeration.htm
We consider this corrosion in acetic acid to be an electrolytic reaction, and also the reaction of the nail with $HCl$, but the scale of cathodes and anodes in the case of $HCl$ is so small that we can't detect them. And in $HCl$ the $H_2$ is generated so fast that it bubbles off and doesn't need to be depolarized by dissolved oxygen.
Adding a picture suggesting the state of the nail:

https://www.ddcoatings.co.uk/2343/what-is-differential-aeration-corrosion.


What is the correct structure of sodium thiosulfate? Which structure of sodium thiosulfate is correct? One book says the second one (the one with co-ordinate bonds) and the other one says the first one?

According to me the structure 2 is correct as in that structure the octet of the middle S atom is complete whereas in the first one there's an excess.

Both of your Lewis structures are wrong.

In both of the shown structures, the solid line from sodium to oxygen/ sulfur implies that there is a discrete bond between these atoms. This is not the case.

The sodium ions and the thiosulfate ions form ion pairs (or ion triples) and are attracted to each other via diffuse electrostatic interactions. Usually that is what we consider ionic bonding.

In aqueous solution the ions will be very mobile. Sodium ions may coordinate to one or more thiosulfate ions, and therein to more than one site of the thiosulfate ion. At the same time thiosulfate ions may coordinate to more than just two sodium ion, or fewer. In such solutions, the solutes also coordinate to the solvent.
Ignoring any external field, the oxygen within the thiosulfate ion are equal.

The Lewis structure best describing the thiosulfate molecular ion is having a +2 formal charge at the central sulfur, and a -1 charge on each of the ligand atoms oxygen/ sulfur. (See below for details.)
One possible (and more accurate) Lewis structure is therefore:

(Note that the dashed lines do not represent bonds, but weakly coordinating interactions, but not necessarily all of them.)

More common are probably the following Lewis structure representations:

In solid state coordination of the ions depend on various factors, especially how much water is incorporated into the crystal structure. In any of these cases, it would be wrong to draw any bonds between the (molecular) ions. I'll include a few links to publications at the end of the post.

Additional thoughts of your input structures

  Which structure of sodium thiosulfate is correct? [...]

You noted correctly that in the first structure the central sulfur needs to "expand the octet". This would only be possible if d orbitals of said sulfur are involved. It has been proven many times that the contribution of d-orbitals in these hyper-coordinated species is minimal (usually below 1%).

The second structure is technically, or strictly speaking, not a Lewis structure, because dative bonds are not part of this description. However, if we allowed them as an extension, which has been proposed multiple times, then the arrow would need to point from the oxygen towards the central sulfur, as in those occurrences the more electronegative element would assume the higher electron density.

Disclaimer: The following section is a bit more advanced, and might not be appropriate to the casual reader, as another user pointed out that it might obfuscate the most important points about the structure of the anion.

Gas phase approach to determining the most likely Lewis structure

If we assume this molecule to be an ion-pair and calculate it in the gas phase, then we can draw a couple of conclusions, how a more accurate representation in the Lewis framework would look like.
I have performed a quick calculation on the DF-BP86/def2-SVP level of theory with Gaussian 09 and performed a natural bond order analysis (NBO 6.0) on it. The following picture shows the optimised geometry, alongside some inter-atomic distances.

As you can see, the sodium ions are coordinated to two oxygen, or one oxygen and a sulfur, respectively. The natural atomic charges reproduce well what I have initially stated, and reproduce well the Lewis structure I have suggested.

\begin{array}{rrr}
\text{Atom}& \text{No}& \text{Natural Charge}\\\hline
  \ce{S}  & 1 &   1.98 \\
  \ce{S}  & 2 &  -0.73 \\
  \ce{O}  & 3 &  -1.01 \\
  \ce{O}  & 4 &  -1.01 \\
  \ce{O}  & 5 &  -1.01 \\
  \ce{Na} & 6 &   0.91 \\
  \ce{Na} & 7 &   0.86 \\\hline
\end{array}

In terms of the NBO analysis, we can back this up, as it indicates no lone pair at the central sulfur (S1), three lone pairs at the oxygen (O3, O4, O5) and the ligand sulfur (S2), and four single bonds. There are vacant acceptor orbitals at the sodium ions (Na7, Na8). Here is an abridged version of the analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(skipping core orbitals)
  24. (1.98855) LP ( 1) S  2            s( 90.65%)p 0.10(  9.34%)d 0.00(  0.01%)
  25. (1.88728) LP ( 2) S  2            s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
  26. (1.84883) LP ( 3) S  2            s(  1.80%)p54.53( 98.09%)d 0.06(  0.11%)
  27. (1.97170) LP ( 1) O  3            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  28. (1.83558) LP ( 2) O  3            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  29. (1.82232) LP ( 3) O  3            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  30. (1.97170) LP ( 1) O  4            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  31. (1.83558) LP ( 2) O  4            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  32. (1.82232) LP ( 3) O  4            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  33. (1.96648) LP ( 1) O  5            s( 77.67%)p 0.29( 22.33%)d 0.00(  0.01%)
  34. (1.83753) LP ( 2) O  5            s(  0.02%)p99.99( 99.91%)d 4.78(  0.07%)
  35. (1.82147) LP ( 3) O  5            s(  0.00%)p 1.00( 99.93%)d 0.00(  0.07%)
  36. (1.95903) BD ( 1) S  1- S  2
               ( 56.47%)   0.7515* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 43.53%)   0.6598* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  37. (1.97694) BD ( 1) S  1- O  3
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  38. (1.97694) BD ( 1) S  1- O  4
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  39. (1.97758) BD ( 1) S  1- O  5
               ( 34.64%)   0.5886* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 65.36%)   0.8085* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)
 ---------------- non-Lewis ----------------------------------------------------
  40. (0.05958) LV ( 1)Na  6            s( 95.91%)p 0.04(  3.91%)d 0.00(  0.18%)
  41. (0.10019) LV ( 1)Na  7            s( 96.97%)p 0.03(  2.93%)d 0.00(  0.10%)
  42. (0.25853) BD*( 1) S  1- S  2
               ( 43.53%)   0.6598* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 56.47%)  -0.7515* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  43. (0.24584) BD*( 1) S  1- O  3
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  44. (0.24584) BD*( 1) S  1- O  4
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  45. (0.24610) BD*( 1) S  1- O  5
               ( 65.36%)   0.8085* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 34.64%)  -0.5886* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)

Key: LP = lone pair, BD = bond, LV = lone valency, BD* = "antibond"

Alternative coordinations

There are at least two other structures of the stoichiometry $\ce{Na2S2O3}$,  essentially with the same bonding motif, but different coordinations of the sodium atoms. The first is about $\pu{2.4 kJ mol-1}$ more stable than the previously shown.

The second is about $\pu{3.2 kJ mol-1}$ higher in energy.

These are electronic energies only, but they show that the sodium atoms are not discretely bound and may interchange positions quite freely.
Given that disodium thiosulfate probably only exists in a crystal lattice which cannot be described by Lewis structures, or in solution where we have to worry about other effects and equilibria, it is obvious, that there cannot be a single Lewis structure for this compound. It simply does not exist as a discrete molecule.

Literature (selection) of crystal structures

The crystal structure of anhydrous sodium thiosulphate. Sandor, E.; Csordas, L.; Acta Cryst. 1961, 14, 237-243. DOI: 10.1107/S0365110X61000802 Crystallography online: Crystal Structure of Na2 (S2 O3)
On Polymorphism of $\ce{Na2S2O3}$. von Benda, H.; von Benda, K.; Z. Naturforsch. B 1979, 34, 957-968. DOI:
10.1515/znb-1979-0716 Alternative direct link (pdf) Crystallography online: Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3)
Rerefinement of sodium thiosulfate pentahydrate. Prasad, S. M.; Rani, A.; Acta Cryst. E, 2001, 57, i67-i69. DOI: 10.1107/S1600536801011175 Crystallography online: Crystal Structure of Sodium thiosulfate pentahydrate
Structure of a new sodium thiosulfate hydrate. Hesse, W.; Leutner, B.; Böhn, K.-H.; Walker, N. P. C.; Acta Cryst. C 1993, 49, 363-365. DOI: 10.1107/S0108270192009399 Crystallography online: Crystal Structure of sodium thiosulphate 2/3-hydrate
The crystal structure of sodium thiosulphate pentahydrate. Taylor, P. G.; Beevers, C. A.; Acta Cryst. 1952, 5, 341-344. DOI: 10.1107/S0365110X52001027

Three cases:
In an aqueous environment, the anion $\ce{(S2O3)^2-}$ is surrounded by water as well as $\ce{Na+}$. In this case, none of the above.

In crystalline form, the $\ce{S-S}$ bond length is close to single $\ce{S-S}$ bond which implies a strong negative charge on the terminal sulfur. $\ce{S=O}$ bonds are shorter implying a primarily double bond character. Several pentahydrate and anhydrous isoforms are known. The sodium atom is interacting with several neighboring anions. This is a common situation in salt crystals. For example, in NaCl each $\ce{Na+}$ interacts with $\ce{6 Cl-}$ ions. In the case of the crystalline form, none of the above.

$\ce{NaS2O3-}$ was probably studied in a gas phase, but I haven't seen papers supporting this. All the forms are probably present in the gas phase, but I don't know the exact distribution. Given that the $\ce{S-S}$ bond in the crystal structure has a single bond character, structure 2 should be more commonly found in the gas phase.


Solubility Product and Buffers The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and solubility product for silver benzoate is $2.5 \times 10^{-13}$. How many times silver benzoate is more soluble in a buffer of $\mathrm{pH}$ 3.19 compared to its solubility in pure water?

Equation used to calculate solubility of silver benzoate in water:

$$\ce{C6H5COOAg <=> C6H5COO- + Ag+}$$

I've calculated the solubility of silver benzoate in water , it is $5 \times 10^{-7} \pu{mol/L}$ (since that is all I calculated, this is the only equilibrium equation I know).

But I cannot understand what to do next.

As ion benzoate is a base in the couple $\ce{C6H5COOH /C6H5COO- }$, we can write
$$s=\ce{[Ag+] =[C6H5COO- ] + [C6H5COOH ]}$$
$$s=\ce{[Ag+] =[C6H5COO- ] (1+\frac{[C6H5COOH ]}{[C6H5COO- ] })}$$

$$s=\ce{[Ag+] =[C6H5COO- ]} (1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} })$$
On the other hand,$$K_\mathrm{sp}=\ce{[Ag+] \times [C6H5COO- ]} $$
$$K_\mathrm{sp}=\frac{s^2}{(1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} })} $$
$$s=\sqrt{K_\mathrm{sp}} \times \sqrt{{(1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} })}} $$

So, the solubility in the acidic buffer solution is increased by the factor:$\sqrt{1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} }}$


Does the amine group participate in resonance in aniline? When we draw resonance structures of aniline, 8 electrons participate in resonance, making it antiaromatic. But in our school textbook, it is said to be aromatic. Does the lone pair of $\ce{NH2}$ participate or not? If it doesn't, then it would make sense.

The lone pair of nitrogen is also part of the extended π-system; the molecule itself is still aromatic. As I already explained in the question about fulvene, aromaticity is nothing you can determine based on simple rules, especially not counting electrons.
Structural indicators like bond lengths and angles give a good idea how well established resonance is. Additionally you can draw a qualitative molecular orbitals diagram and look if it is similar to benzene. Eventually the determination of aromaticity is an experimental one, for example with NMR spectroscopy.

From the crystal structure (M. Fukuyo, K. Hirotsu, and T. Higuchi, Acta Cryst. 1982, B38, 640-643. DOI: 10.1107/S056774088200363X) we know that the carbon-carbon bond lengths are all similar and between 136 and 141 pm. The NMR spectrum clearly shows the protons have aromatic shift, see chemicalbook.com.

The molecular orbitals of aniline (calculated at DF-BP86-D3BJ/def2-SVP; Gaussian 09 D.01) show that the nitrogen is part of the π-system. They are also very similar to benzene. Here are the orbitals of the π-system in decreasing energy. Virtual (unoccupied) orbitals are red/yellow, and occupied orbitals are blue/orange.

For aromaticity, you need several rules to be satisfied:

1) The compound must be cyclic;

2) It must be ring-conjugated (no cross-conjugation!);

3) The number of electrons on π-orbitals in ring conjugation must obey Huckel's rule and be equal to a number of (4n+2) electrons, where n is a whole non-negative number.

Al that means that we need to count only those electrons, which are in aromatic cycle, but we do not care for "exo" electrons.

You can read more about quinone non-aromaticity here, here, and some general information about aromaticity here.

Back to your question:

Resonance forms 2-4 do not have any kind of ring conjugation, as when the aminogroup is involved into resonance, it gives its non-bonding pair for bond formation, and makes the compound cross-conjugated. Since there is no ring conjugation the forms 2-4 are non-aromatic.

Aminogroup does participate in resonance, but it has little influence on aromaticity.


Will this beta-branched secondary alkyl halide undergo SN1 or SN2? What is the correct mechanism here? Would SN1 occur here? I chose SN2 because we have a secondary alkyl halide with a weak base $\ce{CN-}$.

However, the solution says SN1 is major. Can you explain why?

My friend got full credit for this:

Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.

Full mechanism:

I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.

I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required


how are colloids and suspensions solutions? my definition of solution is a homogenous mixture of a solute and solvent.

and in my textbook they call colloids and suspensions solutions . how is this possible if both of them are heterogenous mixtures.

The definition of solution is a little bit cloudy. IUPAC define a solution in this way:

  A liquid or solid phase containing more than one substance, when for
  convenience one (or more) substance, which is called the solvent, is
  treated differently from the other substances, which are called
  solutes.

And indeed this is a quite general approch that takes in account that we usaually call solutions systems (such as metallic solution) that are not compose of single molecules or ions dissolved in a solvent. Besides this approch chemists generally call solution a system with the folowing charachteristics:

It is stable.
It has only one phase.
It has to be an homogeneous mixture of two or more substances.
The solute can't be distinguished from the solvent (no scattering, can't be filtered)

And so I would not define colloids and suspension solution because they have a dispersed phase and often they are not stable.

Reducto ad absurdum, all matter is discrete, none is "homogeneous". So, there are no "homogeneous" solutions, since not only are there no solutions without particles present, but there are no solutions which contain exactly the same number of atoms in every cubic volume element. You may be thinking that my argument is silly, and you wouldn't be wrong. OTOH, your definition uses the words "solute" and "solvent" which is quite useless (circular definitions are really not appropriate nor necessary here). The only information you give us in it, is that the mixture must be homogeneous - which you don't define. Did you know that you can take a solution of some heavy metal salts, and centrifuge them and concentrate the salts at the bottom? So, the idea that there is a single criteria you can use to separate dispersions from solutions is incorrect - it changes with context. To be perhaps a bit clearer, some would consider colloidal mixtures, and even suspensions to be "homogeneous enough" to be "solutions". Generally, a colloid has a lot of interactions between its surface and the solvent, so that it makes sense in some contexts to call it a solution. I'm less enthused about calling a suspension a "solution", I'd call it a dispersion. Bottom line here is that "phases" are macroscopic things, and both suspensions and colloids are microscopically distributed, so a mixture of either may for some purposes be considered homogeneous. This is about terminology. I won't say its a bad question, but I will point out that this type of question ultimately depends on definitions, so may not be useful to worry about. (I decided not to "go there" as far as inquiring what qualifies you to say "my definition" - generally, you should be widely published in the appropriate area, and even experts may use terminology differently. This question isn't about things, its about words. It would be more interesting if YOUR definition was obtained from the text book, in which case it would seem to be inconsistent. I have no problem with the fact that the usage in the textbook is different than what you've seen elsewhere.


Which has higher priority according to Cahn-Ingold-Prelog priority rules, vinyl or secondary butyl? According to my textbook, vinyl has higher priority than isopropyl, because an atom which is connected by a double bond can be treated as two identical atoms (one of them is just an imaginary atom).

How about vinyl and secondary butyl? Should the imaginary atom be treated as that nothing else is connected to it? Does secondary butyl have higher priority?

The assignment of CIP priorities has some intricate details when it comes to multiple bonds.

First, the general procedure (Wikipedia):

  If there is a tie, we must consider the atoms at distance 2 from the stereocenter — as a list is made for each group of the atoms bonded
  to the one directly attached to the stereocenter. Each list is
  arranged in order of decreasing atomic number. Then the lists are
  compared atom by atom; at the earliest difference, the group
  containing the atom of higher atomic number receives higher priority.
  If there is still a tie, each atom in each of the two lists is replaced with a sub-list of the other atoms bonded to it (at distance
  3 from the stereocenter), the sub-lists are arranged in decreasing
  order of atomic number, and the entire structure is again compared
  atom by atom. This process is repeated, each time with atoms one bond
  farther from the stereocenter, until the tie is broken.

Now, concerning double bonds in particular:

  If an atom A is double-bonded to an atom B, A is treated as being
  singly bonded to two atoms: B and a ghost atom that has the same
  atomic number as B but is not attached to anything except A. In turn,
  when B is replaced with a list of attached atoms, A itself is excluded
  in accordance with the general principle of not doubling back along a
  bond that has just been followed, but a ghost atom for A is included
  so that the double bond is properly represented from both ends.

I have constructed a table of these atom lists for the vinyl, isopropyl and sec-butyl rests:
$$\small
\begin{array}{llll}
\hline
\text{Dist} & \text{Vinyl} & \text{Isopropyl} & sec\text{-Butyl} \\
\hline
1 & \text{C} & \text{C} & \text{C} \\
2 & \text{C} & \text{C} & \text{C} \\
  & \text{(C)} & \text{C} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
3 & \text{(C)} & \text{H} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
  & \text{H} & \text{H} & \text{H} \\
  & & \text{H} & \text{H} \\
  & & \vdots & \vdots \\
\hline
\end{array}
$$

The sec-butyl rest has thus the highest priority of the three, followed by vinyl and then isopropyl.

The ghost atom at distance three is the ghost carbon at distance one, as seen from the ghost carbon that was doubly bonded to it. If the distance 4 was to be considered, no more ghost atoms would appear, as the ghost atom at position 1 does not have any more ghost atoms bonded to it (atom 1 however is bonded to the ghost atom at position 2).

It's wrong. Vinyl loses at position 2 of distance 2 against both isopropyl and sec-butyl and therefore gets priority 3. sec-butyl beats isopropyl at the first position of distance 3 and therefore gets priority 1 thus isopropyl gets priority 2.
Phantom atoms are less worth than real atoms but more worth than real atoms with lower atomic number.


How do I lower the surface tension of water without using a detergent? For many years I've used a Rainbow vacuum, which uses a reservoir of water as the filter.   While I'm quite pleased with it, I've found that when trying to vacuum cement dust or wood ash not all of the  particles are captured in the water.   I believe a remedy to this would be to lower the surface tension of the water.  Yes, I can always be sure to use hot water, but I'd like to go further.  Obviously I can't use a suds-producing detergent because the vacuum produces lots of turbulence in the reservoir.

What commonly available chemicals could I use that would give me the greatest reduction of surface tension? So far I've tried to research trisodium phosphate, sodium carbonate, and a product called Haemo-Sol which is a non-ionic surfactant... but I can't find surface tension numbers for any of these much less the concentration to use.

The manufacturer suggests adding a tablespoon of vinegar, which I think is
rather lame.

Can anyone make a good guess or point me in the right direction?

Your goal is to reduce the surface tension of the water so that it does not support the formation of large bubbles or inhibit the whetting of the dirt particles.

When the bubbles are small, the dust inside will be more likely to make contact with the liquid surface of the bubble within the lifetime of the bubble. Also, when it does make contact, it will be much more likely to irreversibly adhere to the water if you have reduced the surface tension.

The word "surfactant" is a blend of "surface active agent"$^1$, and this is what you need to interfere with the surface tension of your water.  I believe that you may have written off detergents too hastily. I think the best solution will be to add a small amount of laundry detergent, or possibly non-foaming carpet cleaner. Each of these products are designed to foam to a lesser degree than, say, dishwashing detergent or most hand soaps.

I suspect that you will have to experiment a bit, but that something on the order of a single cc per liter may be sufficient to increase the effectiveness of dust removal without causing excessive suds. Your goal should be to see how much detergent you can add without adversely affecting your vacuum, and decide whether this has resulted in a decreased throughput of dust particles.

As to your title question, "How do I lower the surface tension of water without using a detergent?", what I'm really saying is 1) that most anything that does what you want could probably be deemed a detergent and 2) that a detergent is likely the most practical solution to your problem.

1) Surfactants and Interfacial Phenomena, Fourth Edition, Rosen and Kunjappu, John Wiley and Sons, Inc., 2012

Ethanol, ionic liquids, dyes, protein, reduce surface tension of water, however these chemicals increase viscosity so it is better to deal with friccohesity. The surface tension and viscosity determinations need density data measurements but friccohesity does not. The deepest mechanistic changes are accurately resolved by friccohesity. Acetic acid, sodium acetate and others reduce the surface tension of water.


Anomalous electronic configuration of platinum Why is the electronic configuration of platinum
$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$

and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or
$\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$?

The Madelung energy ordering rule gives the energy of the orbitals approximately:
$ \mathrm{1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d <
5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p}$
That would speak for $\mathrm{[Xe] 4f^{14} 5d^{8} 6s^{2}}$ if you follow the Aufbau principle (from German Aufbau = setup). But $\ce{Pt}$ is an exception (as there are some). A rule of thumb is that half-filled shells are stabilized. So that means in the case of $\ce{Pt}$, $\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$.
The real answer is much more complicated. It comes from relativistic effects, electron correlation and shielding effects. There is an interplay between the attraction of nucleus and the electrons, and the electron repulsion between all electrons. The heavier atoms become, the more important relativistic effects become, since the inner electrons are moving much faster as they are in a stronger electric field from the higher charge of the nucleus. Often it is said that the outer orbitals are less compact than the inner ones, which is true if one calculates and analyses them, but a stronger effect is that outer electrons are shielded and therefore the nucleus attraction is weaker. So there is no simple way, learn the exceptions or solve the Schrödinger/Dirac equations.
Also be careful when speaking about orbitals. An orbital is a one-electron wavefunction and they are more a chemical concept than reality in multi-electron atoms.

A quicker, less lengthy answer to your question:

Because it is in Group 10, Period 6 and all elements in that group are required to end with the 5d and 6s orbitals. And since it MUST use 6s, it is natural to put 1 electron in it (simply to show it exists, based on the element's placement on the periodic table). Also, because of the Aufbau principle, the other arrangements you mention simply cannot be.


How to calculate the molar mass of a protein from its osmotic pressure? I am unsure how to go about solving this question. I know how to get the osmotic pressure, but where do I go from there?
I can't use the gas law since the protein is not a gas.

$\pu{2.38 g}$ of a protein is dissolved in $\pu{100 mL}$ of water at $\pu{25^\circ C}$. The solution rises to a height of $\pu{8.5 cm}$ in an osmometer (a device for measuring osmotic pressure). What is the molar mass, in grams per mole, of the protein?
Data: assuming the water ls pure, then the osmotic pressure $\pi = \rho gh$, where $h$ is the height of the water, $\rho$ is the density of water ($\pu{1 g/cm^3}$),and $g$ is the acceleration due to gravity ($\pu{9.81 m/s^2})$.

UPDATE: I've spoken with a masters student (my TA) about this and it turns out that the way to solve this is to use an osmotic pressure equation which is identical to the ideal gas law, but only because it is an approximation. If this question is unlocked I would be happy to share my answer.

Use osmotic pressure $Π=iMRT$
From this get your concentration and calculate molarity of solution.
By the way if you are looking more on this topic then you must consider the chapter on solutions and their colligative properties in your chemistry textbooks

It seems to me that you should use Morse equation $\mathrm{\Pi=iMRT}$ - from your data you should get  $\mathrm{M}$ - molarity and then molar mass of the protein. You can skip $\mathrm{i}$ - it's not a problem if it's a dimer or oligomer.


Why opposite of electrophilic is not Protophilic In organic chemistry , considering definitions of  electrophilic and nucleophilic:

Electrophilic:having an affinity for electrons : being an electron acceptor.
nucleophilic : having an affinity for atomic neuclei (also can we say proton?) : being an electron donor.

So both being opposite to each, why dont we say nucleophilic as protophilic?

Because not all "nucleophilic" agents react equally well with all types of relatively exposed nuclei.  Phosphines are very unwilling to extract a proton but quite nucleophilic if the available electronic vacancy is in carbon or a transition metal.

Nucleophile-  nucleus loving .
but it means the one which donates electron or has a negative charge on itself.

The name/ term is given as nucleophile because it would try to give away its electron pair to molecule which has less electron density possibly an electrophile.
like in nucleophillic addition reactions the nucleophile attacks that part for example say in : carbonyl compounds on the carbon which has a less density of electron as the $\pi e^-$ land on oxygen .
So was that a proton site or nucleus site  or a site where electron density is less .


Difference between electropositivity and electronegativity Sodium is more electropositive than hydrogen although they are in the same group.

Why is the electronegative value of hydrogen greater than sodium (metal)?

Background:
Electropositive  implies having tendency to lose electrons to form positive ions. Electronegative implies having tendency to gain electrons to form negative ions. Electronegativity is a concept used to describe the tendency of an atom to attract electrons in a covalent bond.

  Why is the electronegative value of hydrogen greater than that of sodium?

This is because sodium after losing an electron gets its octet configuration, but a hydrogen atom after losing an electron will not get the configuration of helium but it can get helium configuration if it accepts an extra electron or forms a covalent bond.

So this implies that sodium is more electropositive that hydrogen, even though they belong to same group.

Sometimes "more electronegative" is used to describe higher electronegativity and "more electropositive" is used to describe lower electronegativity. In this sense, elements are less electronegative (or more electropositive) as you go down any group in the periodic table. However, the difference in electronegativity between hydrogen and sodium is exceptionally large, and correlates with the very different chemistry these two elements show.

An electropositive element is an element which has a propensity to lose an electron. (e.g. metals in rancidity)

An electronegative element is an element which has a propensity to gain an electron. (e.g. non-metal in oxidation)


How to know if computational calculations are correct I've recently been modelling some transition states of a reaction I've been doing (read: struggling to do).

Whilst I have a lot of technical support in doing this (both step by step help and also academics who research in the field of computational chemistry), one thing that bugs me is the difficulty in knowing whether the calculation is meaningful.

In synthetic chemistry, I can do an experiment, then take an NMR/crystal structure, and be confident that i've got what I was intending to do. Whilst different bases may promote different reactions, its always possible to work out whats happened.

In computational chemistry, I do a calculation and get an answer, however depending on the methods used the answer might change.

This can be minimal: isomer A is favoured over isomer B, but the extent changes depending on the basis set/experiment. Or can be significant: isomer A and B change their order.

I was wondering if there is any way of telling whether a calculation is relevant. Obviously experiment can do this, but the reason I'm doing the calculation to begin with is to try and rationalise the difficulties in reaction .

Is there a concrete way to quantify the 'meaningfulness' of a computational result, or is it always a case of just fitting the computational data to experiment and then working backwards.

The only guaranteed way to achieve absolute correctness is to use asymptotically better methods and see the asymptotic behavior yourself. This, however, ofter requires prohibitively complex approaches.

Outside of that, one needs an experiment to check your data against. A good way (with limits) is to perform vibration frequency calculations and see if that fit the experiment. (making correction of unharmonic behavior). Another way is to run through similar well-studied cases and fit computational results. Most probably, a systematic error would be found, that can be easily eliminated. Luckily, most probably you are not the first to work with your particular kind of systems, so using google scholar it is usually possible to find a review or an establishing work that would list what work and what does not. Most practical methods have known 'glitches', and the ways to deal with them were considered. In particular, DFT cannot catch dispersion interactions, requiring empirical correction schemes.

Based on your question I guess you are doing some sort of quantum calculations on small molecules, most likely DFT. (Please, correct me if I am wrong. Also, I'm not a fan of DFT, so my answer can be a bit biased...) In this case, knowing if the "numbers" you get are correct it's not too easy.

What I would check is how sensitive are your results on the setting of the calculation. In the case of the two isomers, for example, does the order of stability change if you use bigger basis sets or if you change the DFT functional? If you get always the same answer then you can be reasonably sure that answer is correct. If instead, you get a set of unreasonable results, then you should question all results, and try to understand (i.e. search the literature) which method is best for your particular problem.

For other computational methods such as ab-initio methods, semiempirical methods, or classical force fields, very different ways of checking the results exist, which is quite difficult to list them all.


Why do we call a benzene ring attached to an hydroxyl functionality an alcohol and not a benzene? But if an alkyl group is attached to a benzene ring, we call the parent chain the benzene ring, no matter how many carbons the alkyl group has. Why? For example: heptylbenzene.

For another example: when 1-propenyl-benzene reacts with sulfuric acid we get 1-phenyl-1-propanol. We don't call this product 1-propanol-1-benzene.

In your first example, heptylbenzene, the benzene ring is the parent structure for the preferred IUPAC name because the ring is always the parent structure if the ring and the chain have the same number of substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name). The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

In heptylbenzene, neither the benzene ring nor the heptane chain has any substituents that could be cited as the principal characteristic group. Therefore, the ring is selected as the parent structure.
In 1-phenylpropan-1-ol, however, the propane chain has an $\ce{-OH}$ group, which is the principal characteristic group of this compound and expressed as suffix (‘-ol’). Thus, Rule P-52.2.8 cannot be used. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of substituents cited as the principal characteristic group.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the propane chain is the parent structure in 1-phenylpropan-1-ol.

To the best of my knowledge, I think it is because the hydroxyl group (-OH) when connected to the benzene ring overrides the basic chemical properties of the benzene, displaying it's own more prominently.  Thus it takes the priority in naming.
But in the case of alkyl groups, they don't bring about much change to the basic properties of benzene and the chemical properties of the benzene remain more prominent. Thus benzene takes the priority in naming.
However I believe this may change if the alkyl chain is relatively large and overrides the benzene's behavior in the molecule.
I haven't cross checked anywhere so I'm not entirely sure but that's what I think makes the most sense. Hope it helps.


Will there be change in electronegativity difference in C-O and C=O? Is there any change in EN difference in C-O and C=O. If yes, why?
Does EN difference change if its bond is changed to single, double, or triple?

I think that you didn't understand the answer to this question.

A dipole is created when two equal (only in magnitude) and opposite charges are separated by some distance. The dipole moment depends on the magnitude of the charges and the distance between them. Electronegativity is just a concept that helps us to get an idea of the magnitude of the charges. So, in order to find the dipole moment, you should focus on the magnitude of the two charges rather than the electronegativity difference.

Look at the hybridization of the atoms in the single and double bond. That will tell you that both the atoms should be more electronegative after forming a double bond, but won't help you comment on the electronegativity difference.


Calculate the rate of decomposition of H2O2 at 25 °C when [H2O2] is exactly 10.0 mmol/L I'm really confused?

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

$$\begin{array}{ccc}\hline
\text{Experiment no.} & c(\ce{H2O2}) & \text{rate} \\
\hline
1 & 14.45 & 3.35 \times 10^{-3}\\
2 & 28.90 & 6.65 \times 10^{-3}\\
3 & 57.80 & 13.35 \times 10^{-3}\\
\hline\end{array}$$

So from previous parts of the question, I found the order of reaction (1st), the rate equation for the reaction ($r = k[\ce{A}]$), the rate constant ($2.3 \times 10^{-4}\ \mathrm{s}^{-1}$, which is correct according to the answers), and now I need to find the rate of decomposition of $\ce{H2O2}$ when $\ce{[H2O2]}$ is exactly $10.0\ \mathrm{mmol\ L^{-1}}$.

I thought it's just $2.3 \times 10^{-4} \times 10.0$, which is $2.3 \times 10^{-3}$, but the answers provided is double that (so $4.6 \times 10 ^{-3}$), so I must have gotten the rate equation wrong?

It is a matter of definition:

The rate of the reaction is defined as: $$r=-\frac{1}{2}\frac{d[\ce{H2O2}]}{dt}$$
The rate of the decomposition of $\ce{H2O2}$ is defined as:$$r_{\mathrm{decomp}}=-\frac{d[\ce{H2O2}]}{dt}= 2r$$

The correct answer depends on what who wrote the problem had in mind about the values of rates  in the  last column in the data table:

If he considers the  rate is  rate of decomposition of $\ce{H2O2}$      , then  the answer is $2.3\times 10^{−3}$

If he considers the  rate is  rate of formation of $\ce{O2}$, then  the answer is $4.6\times10^{−3}$,

How we answer if we write the balanced equation in this way:

  $\ce{H2O2 -> H2O + 1/2 O2}$


Are there any other elements that get less dense in their solid state? I was learning about how water, because of its hydrogen bonds, actually gets less dense as it goes into its solid state - I was just wondering, what other elements do this? Are they similar to water?

P.S. I want to make sure that I understand why ice is less dense than water in its liquid state, so please correct me if I am wrong.

In water, the hydrogen bonds have the ability to move around and bounce with other hydrogen bonds. They also have the ability to break and reform. Their ability to move around and bounce around allow them to be more 'closely knit'. But, as they get cooler, they, the molecules, loose their kinetic energy and thus stop moving around and become more framed and rigid. They loose the ability to come closely together. And thus, ice is less dense.

What I don't understand also, is why making water cold makes the hydrogen bonds spread out. Why don't they just freeze in the position they are in, instead of spreading out?

Your terminology is a little off: water isn't an element.

There isn't really a simple single explanation of why some solids are less dense than their associated liquids. The general explanation (which doesn't really explain much) is that the solid has a structure and, sometimes, that structure takes us more space than the average structure that occurs in the liquid.

In water, ice has a definite structure where each oxygen sits at the approximate centre of a tetrahedron with two bonds to hydrogen and two hydrogen bonds to the hydrogens of other water molecules. In liquid water the structure is far more fluid and averages out to be slightly more dense that the (more ordered) solid. Exactly why this is true in this case requires a lot more than could fit into an answer here. If water was frozen extremely quickly you might be able to freeze the structure in the liquid; normal cooling allows enough time for the bonding to become more organised.

But the phenomenon isn't unique to water, though the specifics will be very different for other substances. Among elements Gallium, for example, also expands on freezing (don't store it in glass bottles: see this other answer here).

Other examples of elements are germanium and silicon. Acetic acid also forms a less dense solid. In all cases the substances form a relatively ordered crystal structure which just happens to be less dense than the (less ordered) liquid phases.

The bonding of one elements atoms into solid, or even a number atoms giving way to a chemical like water depends a lot on how the solid is presented to us.  It my feel like a solid piece of iron or copper, but in reality although the pure solid form is more dense we never see it like that.  There are many small air bubbles or other impurities in it.  These are burned away in liquid form and all you have in liquid is on atom bonded to the next.  This is also why liquids cannot be compressed.  Yet solids can be compressed. That is to say only until their atoms remain (this heat and pressure may cause the solid into a liquid anyway).


Determining the order of molar conductivity This was a question which came in a objective type examination.

Which of the following have a $\lambda^\infty$ (molar conductivity at infinite dilution) larger than $\ce{KCl}$?
A. $\ce{CH3COOH}$
B. $\ce{HC}$
C. $\ce{NaOH}$
D. None of the above

I was not provided any experimental data to arrive at the answer.I know that $\ce{KCl}$ is a strong electrolyte and so my guess was that it must have greater molar conductivity than at least $\ce{CH_3COOH}$. But the answer key gives all A, B, and C. Is there a definite rule here?

I would agree with the answer key. The trick here is it's limiting molar conductivity, the molar conductivity at infinite dilution. For a neutral electrolyte compound, I was taught the notation $\Lambda ^0$, and $\lambda ^0$ was reserved for individual ions. (The $0$ superscript represents zero concentration, equivalent to $\infty$ dilution.)

Anyway, limiting molar conductivity is an interesting property because it does not represent a physically possibly scenario (a solute conducting electric charge when no solute is present). It can be extrapolated by the Kolhrausch Law, where it $\Lambda^0$ is the y-intercept of molar conductivity $\Lambda$ vs $\sqrt C$, concentration (note this relationship is only valid for strong electrolytes).

The degree of ionization of a weak electrolyte depends on the concentration, and tends to unity (100% dissociation) in the limit of concentration approaching zero. Limiting molar conductivity $\Lambda^0$ is defined (only) for this exact limiting scenario.

So when we consider limiting molar conductivity, we are doing so in a (abstract) condition where both strong and weak electrolyte fully dissociate. Thus for $\Lambda^0$ it makes no difference whether the electrolytes being compared are "strong" or "weak" at finite concentrations (for we are comparing them at "zero" concentration, i.e. infinite dilution). So we should have no problem accepting that a "weak" electrolyte might have a higher limiting molar conductivity value than a "strong" one, as in this case.

Why might $AcOH$ (organic chemist abbreviation), $HCl$, and $NaOH$ have higher $\Lambda^0$ values than $KCl$? Well, the $H^+$ ion is better at transporting its charge through an aqueous medium than $K^+$ because it can "water hop" via the
Grotthuss mechanism. This makes acidic compounds very effective electrolyte conductors. $OH^-$ is also an extremely effective charge carrier by an analogous "deprotonation-chain" mechanism. Because they use water itself to "tunnel"* through the bulk solution, protons and hydroxide have extremely high ion mobility in aqueous solutions and thus provide abnormally high molar conductivity.

So in general acids and bases would be expected to out-conduct a pH-neutral salt ($KCl$), and in the limit of zero concentration/infinite dilution, even "weak" acids and bases will have higher limiting molar conductivity values.

*Not in the quantum sense.

The key is WRONG. There is NO WAY acetic acid has a lambda° (don't know where you got ∞ from, but a rose by any other name will smell the same) that is greater than KCl's. At 0.001 m/L KCl's Λ is 146, HCl's is over 400, NaOH's is 245 and HOAc's is 49 (S cm²). For obvious (I hope!) reasons the $\ce{H+}$ and OH(-) ions' mobility in water is different from all other ions. The amount of energy "available" to an ion in solution depends mostly on the temperature of the solution, which means the kinetic energy of any two ions will be (on average) the same at the same temperature. Since K.E. = ½mv² we reach the approximation v = SqrRoot(2KE/m) for an ion of mass m. Obviously, as m increases, v decreases. So, you'd THINK that $\ce{K+}$ being larger than $\ce{Na+}$, that it would be slower, and this is correct - so far as it goes. It is more complicated, because in real solutions, ions are never naked. That is, they are always surrounded by solvation layers. This means that as charge of an ion increases, the ion will (in polar solvents) surround itself with more and more solvent molecules, and it will have to "carry" these along as it moves. This complicates the picture, since now it's not just the mass of the ion we have to consider, but also the mass of the solvation sheath - which depends on the electric field [which is greater for smaller spheres (with the same charge).] The good news is that your 'ions' are all mono-valent. So, its just their relative size that matters. The bad news is that $\ce{AcOH}$, and $\ce{HCl}$, and NaOH all dissociate to ions which can move really fast in water, so you're not just comparing Cl with, say, Br or F. For $\ce{AcOH}$ (acetic acid) we might assume that at infinite dilution it will be completely ionized, but the mass of the acetate anion still substantially slow down any charge movement. In other words, its like a team race, the speed can't be any faster than its slowest runner. Neither NaOH nor HCl have that problem, so we'd definitely expect them to move faster than KCl in aqueous solution. Chemistry has both theoretical logic, and pragmatic - it is great for people who both like to understand why and who like to get their hands dirty by DOING stuff. Unfortunately, some of the ways it is taught are all about "rules" and formulas.


Positive charge on a zwitter ion at pH=7? I faced this question. But I am having some problems understanding the solution to this:

How can the negative charge on the peptide vanish on $\pu{pH = 7}$? Suppose, we have some $\ce{H+}$ donor on either $\ce{R1}$ or $\ce{R2}$. Then the $\ce{H+}$ donor can make the carboxyl radical neutral. But by donating a $\ce{H+}$ ion, the donor becomes negatively charged. Maybe, the donor can take the $\ce{H+}$ from the $\ce{NH3+}$ radical. But the result, a completely neutral tetrapeptide, doesn't satisfy the condition of the question.

Need a hint.

To sum up: of the side chains listed, only the aminoalkyl ones will be positively charged at pH 7, so four peptides -- 4, 6, 8 and 9 -- will have a net positive charge at pH 7.


Unknown surface coating Forgive me if this isn't the best forum for this question, and guide me elsewhere if so.
I have two identical smartphones, both made by Sony (Xperia z2 to be exact). One I purchased new and one second hand. I want to affix an adhesive-backed mounting system to the back of both of them.
With one phone (that I had from new) there's no trouble, everything (e.g. adhesive-backed Velcro) sticks really well. But with the other phone nothing will stick . I've checked that there's no protective overlay sheet and can only assume it's been treated with some sort of chemical coating. Every solvent I've tried to remove this coating just forms a meniscus and runs straight off. Water, detergent, isopropyl, acetone, WD-40 - nothing makes any difference. Any suggestions?

Very likely, the title of your question already hits the nail.

Since mobile phone tend to be "smart", filled to the rim with computing power and data, users wish to protect their companions with a coating that repells water.  Often, the devices are not only coated on the front side, which would at least hamper accumulation of residues of gliding fingers, but all around, including the back.  As a result, you might get a cell phone that may drop into a bowl of water, and still remain functional.

Among the different technologies available (a short presentation by CNN, back 2013 is found here), there are polymer coatings, or a mixture of anastase / $\ce{TiO2}$ and PFTE / teflon (like the coating of your frying pan) particles with some perfluorinated liquid (example).  For some, you need a larger installation that appears like a scilab, for others, you may prepare it at home with a spray can (example).  Hence the results in terms of performance, maintenance, and longevity may vary, too.  As a result, if applied both on front and back of the phone, not only water, but the components of "normal glue" (referring to what backs a velcro tape) may be repelled more or less efficiently, too.

Your observation that the coating appears little to non-affected both by water as well as iso-propanol (which for some coatings is a no-go, as it would tear the layer off), as well as for WD 40 (which may be summerised as a blend of apolar hydrocarbons (ref)) allows the speculation that the coating might be either polymer-based, or the triple mix of anastase + teflon + perfluorinated binder.  To discern between the two, remotely, is not possible.

If a mechanical, pocket-shaped mount like

(source)

or a pouch, to be bound around your upper arm while riding a bicycle, like

(source)

is not what you may deploy (or alter a little to fit your needs), and if the coating consists of the "polymer-only" variant, then maybe it is possible to establish a connection between the back plate of the phone, and either the velcro tape or something already connected to the velcro tape with the help of cyan-acrylate based gorrilla super glue.  These glues often bind very strong and to polymers, however partially dissole the polymer a little.  This chemical approach however comes with many "ifs" and uncertain outcome.

Well, the worst coating would be a perfluorinated one. You're not likely to get much to stick to that. Silicones are more common (in general) and much more likely to be used.  MEK (methylethylketone, aka 2-propanone) works pretty well at swelling silicones, if you allow soaking. Acetone would be similar if you could allow that adequate contact time. The problem is that you should soak the surface for an hour (or more). So how do you slow the solvent's evaporation? In this type of application, using a powder, like flour (finer the better) to slow down evaporation...you may also need to cover it with some sort of film or if that's not feasible keep on replenishing the solvent. I'd definitely soak it for 20 minutes and then try rubbing the area you want cleaned with a solvent moistened rag. Methylene chloride is another good solvent for attacking silicone, but afaik it's no longer available. I don't know what kind of paint strippers you can purchase (it varies by local environmental regulations), but these are also a possibility. Note that it is possible that these could damage the finish and you definitely don't want any of these nasties to leak into your phone or onto the face! An alternate to using flour (or even tissue paper) to "hold" the solvent is to construct a dam around the area (like a circle of toothpaste or cake frosting or sealant - if you've tried a sealant and it doesn't stick very well, then perhaps you could make a dam of than and when done just pull it off. Basically, the dam would be acting as the sides of a pool the solvent goes into. Toluene is another not too bad solvent. Proper safety procedures always when working with chemicals. WD-40 is an awful idea, unless you think an adhesive will stick better to a slippery surface...


Can I react a Grignard reagent (ethylmagnesium bromide) with water and then ammonia? I came across this question where I had to convert $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$.
If I need to do so can I just react the given compound with water to get ethane and then react it with $\ce{NH3}$.

If the answer is no then why can't I?
Is there any set of principles in organic chemistry which govern as to which substances can be replaced with what?
Kind of like the reactivity series of metals?
For e.g.: If I react an alkane with a halogen to get a haloalkane and then react this back with hydrogen to get the alkane then I don't think so it would be correct to do so. And I don't think that organic chemistry works that way.
But how do I understand while solving a question whether I am going in the right direction.
Please help.

NB: This is not a homework question its just a question I stumbled across while my preparation for a college entrance test. (edit the title maybe)

I came across this question where I had to convert
  $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$. If I need to do so can I
  just react the given compound with water [...]

Always be cautious about unwanted, side reactions that may occur. Proton exchanges are one huge class of possible side reactions. Protons transfers occur readily in many circumstances because the protons is small and can be shuffled around without much to get in its way, physically.

So in the case of your hypothetical reaction between a Grignard reagent and water - realize that Grignards are strongly basic and therefore will deprotonate water. Unfortunately protonated Grignards are no good in terms of reactivity. They just fall back apart into hydrocarbons and a basic halide salt.

Further reading regarding Grignards - It is tempting to use Grignards to create carbon-carbon bonds, but anything that is acidic as water or more acidic will be deprotonated by the Grignard before any carbon-carbon bonds are created. When preparing Grignards, glassware must be perfectly dry. Not just wiped dry with a paper towel but heated dry. Grignards also live in ether solvents since ethers are only very weakly acidic.

Grignards may be used to attack aldehydes and ketones in creating carbon-carbon bonds, but realize that if the aldehyde or ketone contains a functional group that is about as acidic as water or more so - then the first equivalent of Grignard you add isn't going to the aldehyde or ketone carbon - it's going to perform a proton-transfer.

  Is there any set of principles in organic chemistry which govern as to
  which substances can be replaced with what?

In general, it is good to keep in mind the motif nucleophile attacks electrophile in introductory organic chemistry.

Exceptions to motif: nucleophile attacks electrophile

Other heuristics you can use in predicting what reactions might commence ...

Bond energies. This is mainly useful for studying radical reactions - to which the motif nucleophile attacks electrophile doesn't apply as well. Are we going from weaker to stronger bonds? That's a favorable move, enthalpically.
Polarities. This falls under nucleophile attacks electrophile.
Kinetics. It's good to learn the difference between thermodynamics and kinetics early in organic chemistry. Sometimes chemical entities may be thermodynamically unstable but kinetically stable. Maybe a carbon is particularly electrophilic, but it's too well "hidden" by neighboring groups (steric hindrance). So will a nucleophile attack this carbon? Probably not to a great extent.

Yes, a Grignard reagent can react with ammonia like the following:

$$\ce{RCH3Mg + NH3 -> R-H + MgNH3OH}$$


Can a solvent be a solid? Can a solvent be a solid? I do not think so. But is it possible for a solvent to be a solid?

When most people think of the term solvent, a liquid medium comes to mind; however, in the technical sense of the definition, this does not have to be the case. Here is the definition of a solution present in the IUPAC gold book:

  A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

The definition of a solvent is incorporated into the definition of a solution. This definition states that the substance present in the larger quantity is the solvent while that present in the smaller quantity is the solute. Based on this definition, it is entirely possible to have a solvent that is in the solid phase. For example, brass is a mixture of 64% copper and 34% zinc. According to the IUPAC definition, a solution can be in the solid phase. Because copper is present in a larger amount, it is considered the solvent -- a perfect, everyday example of a solvent in the solid phase.

Both your question and the answers seem to rely on a common linguistic/mental shortcut which regards phase as some kind of inherent property of a compound, but I would argue that it becomes a confusing oversimplification when regarding this specific question.

I would say that the proper statement here would be "During the process of dissolving, the solvent has to be in a liquid state", or maybe "in a fluid state" depending on how you like to define the semantics of a solution. The reason for this is that the solvent has to get "between" the molecules of the solute to do its job, and that can only work when the solvent flows.

As for the answers which discuss solid solutions, they are correct, under the implicit assumption that the phase change happened after the dissolving process was finished. Sure, there are some kind of edge cases of diffusion between metals, but really, nobody makes brass by putting a block of copper and a block of zinc in a press at room temperature and waiting a century, to then harvest a fraction of a millimeter thick alloy. So, if we discount such unrealistic edge cases, the statement as I formulated it above is still true, and absolutely compatible with the statement "There are solid solutions" which could be rewritten more explicitely as "A liquid solution can change its phase to solid". (The opposite is of course also true - you can melt an alloy - but not especially relevant to the apparent contradiction).

The comments under M. Farooq's answer do a good job of describing the point about gases. In that case, you have a simple linguistic/categorization problem. The prototypical process of dissolving something involves a liquid, and the process of mixing two gases is in some ways similar and in some ways different. So it is up to the individual language user to "decide" (typically not a conscious decision, but a gut feeling that can be rationalized post-hoc) whether they consider this process to be similar enough to deserve to be called "dissolving" or not. Depending on that decision, you can have a gaseous solvent (at the time point of creating the solution) or not. The above point about phase changes still apply - if you would mix, say, water vapor and ethanol vapor and then cool it down to a liquid, I am pretty certain that everybody would term it "a solution of ethanol in water" (or water in ethanol, depending on ratios) even though there will be a dispute whether the process of dissolving happened when the mixture was still in a gas state.

A separate linguistic case which deserves a mention: there is a widespread usage of "dissolving" among laypeople that is actually incorrect by scientific standards. It is used to describe a situaion in which you have a chemical reaction by a corrosive gas or liquid with a solid. If you pour liquid acid over a basic salt, chances are that you will get a chemical reaction - but many people don't recognize that detail and, seeing the crystal disappear, will state that the acid dissolved the substance (they are also unlikely to call it "the salt", btw, reserving that name for sodium chloride only). Even if one of the two reactants happened to be in a gaseous phase, I would say that it is incorrect to consider this an example of a solution with a gaseous solvent, even though you might meet people (without advanced scientific education) who would classify it as such.