H NMR of 4-chloroacetophenone For the H NMR of the compound I am wondering why peak (A) is the farthest out of all of them? I think that it is due to the electro negativity of oxygen. Am I correct? If so, why would the presence of oxygen push the peak to the left?

The further to the left a peak in an NMR spectrum is, the more deshielded it is. Deshielding or shielding are terms to describe how much a certain nucleus is shielded from the outer magnetic field that is applied during NMR measurements. Shielded nuclei will experience less of a magnetic field and thus resonate at a lower frequency meaning lower ppm-values; and vice-versa. In general, shielding can be explained by more electrons being in the vicinity of an atom because the electrons — in layman’s terms — can be thought to create an electric current induced by the outer magnetic field, which in turn generates an opposing magnetic field and lowers the effective magnetic field at a nucleus.

If we have an electronegative atom close by, that atom will draw electrons towards it, reducing the electron density around the nucleus we are observing and thus deshielding it. That not only works for electronegative atoms alone, but also for electron-withdrawing groups like the carbonyl group.

In your specific example we have two electron-withdrawing groups that will deshield neighbouring protons. However, the deshielding that the acetyl group is responsible for is stronger than that of the chlorine atom. So both pairs of protons are moved downfield (to higher ppm values) when compared to toluene or para-xylol, but the protons closer to the acetyl group are deshielded more.

It ultimately boils down to oxygen’s electronegativity in combination with an $\ce{sp^2}$ carbon that is not attached to any hydrogen, but it may be easier to think of carbonyl groups being electron-withdrawing per se.


Why is the value of "a" is positive instead of negative in Van der Waals equation? Higher value of a means higher atttraction force between molecules. Attraction force causes the pressure to be less than expected. Now the actual pressure becomes $P + \frac{n^2a}{V^2}$ .
Which means if a increases the neat pressure increases too. But isn't it the other way around?
(I assume here P is the pressure exerted on the wall of the gas container considering there is no intermolecular force)

You understand the chemistry: intermolecular attraction forces cause the pressure to be less than expected. The problem here is with the math. Though unintuitive at first sight, a positive $a$ leads to a negative correction to pressure. Let's look at a numerical example.

  Assume that the pressure inside a certain container is $50$ atm with the ideal gas approximation and $45$ atm with the van der Waals equation of state. Then $P_\text{ideal} = P_\text{vdW} + 5$.

In other words, the corrected pressure is $P_\text{vdW}$, not $P_\text{vdW} + a\frac{n^2}{V^2}$.


Additive to an aluminum sulfate and water solution to lower pumping pressures I am currently looking for a surfactant (powder form) that I can add to a solution of aluminum sulfate and water.  My current solution is comprised of 34 gallons of water and 4 (four) 50 lbs bag of aluminum sulfate.  It has a specific gravity of 1.275

It is used as an additive to a gypsum based fireproofing product to lower the density as well as help with a quick set of the fireproofing material.  It is injected into the fireproofing material at the base or a 1" hose that is 20 feet long.  I was hoping to find an additive (surfactant?) that would reduce the pumping pressure of the fireproofing material thru the 1" hose without affecting the chemical nature of the aluminum sulfate and water combination.

The aluminum sulfate and water are mixed in a 50-gallon drum with a high propeller blade that is fairly aggressive.  I worry that a normal surfactant will foam up too much and make the product unusable.

If a non-foaming agent can be added to the water that will offer to lower the pumping pressures of the material thru the 1" hose that would be what I am looking for.

The needed pressure depends on liquid viscosity. The significant decrease of water solution viscosity cannot be achieved by adding of minor additives.

It could be achieved by methods that may not be applicable:

Decreasing concentration of the solute
Decreasing viscosity of solvent by different solvent composition
Increasing temperature of solution. A rule of thumb says viscosity decreases by 3% by increasing temperature by $1 ^{\circ}\mathrm{C}$.


why bond energy of hydrogen molecule is twice the chlorine molecule? why Bond energy of Hydrogen molecule is greater than chlorine and HCl molecule?
Here's a link

Because there are shielding electrons, meaning that the atoms can't come as close to each other. The outermost electron is the one being shared to fill a p orbital in chlorine where as it fills the s orbital in the hydrogen


Order of reactivity of carbonyl compounds to Nucleophilic addition reaction One of the questions in my textbook was to arrange the following in order of their reactivity for nucleophilic addition reaction.

$\ce{(p-NO2Ph)_2CO}$
$\ce{(CH3)_2CO}$
$\ce{(p-NO2Ph)COH}$
$\ce{CH3COH}$

I know there are two factors in play here. One is the steric factor (4>3>2>1) and annother is the electropositive nature of the carbonyl carbon, which is the site of nucleophilic attack (1>3>4>2). I am unable to decide which of these factors shall overpower the other and what shall be the final order?

My instinct is that there are probably too many variables here for a definitive and accurate answer to be given strictly from first principles. When comparing molecules in which there is only one key difference, especially when that difference leads to steric and electronic effects which are essentially additive, it's fairly easy to draw accurate conclusions. This question, however, is a bit thornier.

Comparing acetone, $\ce{(CH3)2CO}$, to acetaldehyde, $\ce{CH3COH}$, is quite straightforward: acetaldehyde should be the more electrophilic of the two. Firstly, the greater steric bulk of acetone's methyl group slightly hinders nucleophilic attack. Secondly, the additional methyl group is electron-donating by hyperconjugation, comparatively reducing acetone's electrophilicity. Here, both steric effects and electronic effects are synergistic.

The other comparisons are not quite as straightforward, as they require (in the absence of empirical data) qualitative judgment calls to be made.

Consider p-nitrobenzaldehyde. Sterically, reactivity should be reduced due to the obvious bulk of the aromatic ring. Electronically, the picture seems to me to be more complex. From the perspective of valence bond theory, the $sp^2$ carbons may be electron-withdrawing by induction, particularly in concert with the electronegative nitro group. On the other hand, the carbonyl group is also highly electronegative and should remove electron density from the ring, primarily via the $\pi$ system, and I would expect that effect to be greater. Finally, from the MO theory perspective, considering the frontier orbitals, conjugation of the carbonyl to the aromatic ring should result in lowering the energy of the $\pi^*$ LUMO, which ought to enhance reactivity. That, however, may be mitigated by any overall increase in electron density around the carbonyl group. By comparison to acetaldehyde, the steric effects are clearly unfavorable, while the electronic effects are slightly murky.

Finally, for the 4,4'-dinitrobenzophenone, the analysis of electronic effects is roughly the same as for p-nitrobenzaldehyde, but here sterics is undoubtedly going to be the overriding factor, and should result in massively diminished reactivity.

Overall, I would predict the reactivity order to be acetaldehyde >> p-nitrobenzaldehyde > acetone >> 4,4'-dinitrobenzophenone. I don't claim this is in any way authoritative, so I hope others weigh in as well. If I can find some relevant literature containing, e.g., relative rates, I'll update this answer with a link.


Is OH the same as O-H? When 2-hydroxylpropanitrite undergoes hydrolysis by H20, it will become 2-hydroxylpropanoic acid. In the structural formula, I wrote the O-H that are connected to the carbon with double bond as OH. Is it still the same thing or it is incorrect?

$\ce{-OH}$ and $\ce{-O-H}$ are the same.

The single bond between oxygen and hydrogen is implied in the first representation and this representation is the most commonly used because of it's brevity.


Strength of substituted Carbon Double Bonds and C-halo bonds The question asks me to identify the compound which has a stronger double bond. I came up with two kinds of reasoning but they do not agree upon a common answer.

1: $sp^2$ carbons are more electro-negative and they would like to have electron donating groups sitting next to them. So in the compound one , the extra methyl group should increase stability and therefore, one would expect the bond strengths to increase. Thus decreasing the bond length (smaller bond length is more stronger? stronger bonds imply more stability)

2: Since the methyl group partially donates electrons to the $sp^2$ carbon, the $sp^2$ carbon's tendency of pulling the double bond electrons towards itself should decrease. Thus increasing the bond length.

So where am I going wrong?

If you have any good resource/material on strength of bonds, please let me know.

EDIT (another Q):

Which double bond has higher disassociation energy?

Substituted alkene is more stable than unsubstantiated alkene. So I is more stable which means it has higher bond disassociation energy so II must be having lesser bond disassociation energy.
Due to hyper conjugation, the bond order in compound I decreases so it must have lower bond disassociation energy.

Which one is correct among the above two reasoning?

If the electro-negativity difference between the atoms is larger, then the bond will be more ionic and hence should be stronger?

Out of the 2 bromo compounds, the first one has has stronger $\ce{C-Br}$ bond, because the lone pair of $\ce{Br}$ is in conjugation with $\ce{C=C}$.
In first case I think Compound 1 is more stable for the reason that it has more substituted $\ce{C=C}$. In reasoning 2 I think when you think about decrease in bond order, the tendency of pulling electrons of $sp^2$ carbon has not decreased. Me- group itself donates electrons because Carbon is $sp^2$ hybridised. The tendency will not increase or decrease. I think it will remain in the form of slight polarisation.


How can one optimise the formation of copper acetate? Mixing solid copper, $5\%$ vinegar, and hydrogen peroxide ($\ce{H2O2}$) causes copper acetate to form. The process will occur very slowly without hydrogen peroxide. Adding $\ce{H2O2}$ speeds the formation of copper acetate.

How can I test the solution to see if all of the acetic acid has been converted into copper acetate?

If possible, provide answers that employ only household items or readily available devices.

Your question and title do not correspond, IMHO. I'll answer one of both:

Vinegar is mostly water, with some acetic acid ($5\%$). A classic experiment, performed by a lot of kids at school, is to mix vinegar with baking soda. The baking soda neutralizes the acid and produces carbon dioxide. The carbon dioxide is what you're looking for: it's the sign that the reaction is taking place.

Pour a pinch of baking soda inside your solution. If it makes bubbles on contact, your solution still contains a lot of vinegar. If you don't see any bubbles, even when stirring, most of the acid has been consumed.


Why are the reactants in the solubility product constant multiplied by their coefficients and raised to the power of their coefficients? It is probably easiest for me to explain the question in an example. Consider the following chemical reaction:

$\ce{MgF2(s) -> Mg^{2+} (aq) + 2F- (aq)}$

Plugging this into the formula for the solubility product constant, $K_{sp}=[M^{y+}]^{x}[A^{x-}]^{y}$, we get:

$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$.

The solubility product constant for magnesium fluoride is $5.16\times10^{-11}$. To determine molar solubility, the equation to solve for magnesium and fluoride is $5.16\times10^{-11}=X\times (2X)^{2}$.

$X$ is magnesium and $(2X)^{2}$ is fluoride. This is because there are two moles of fluoride for every one mole of $\ce{MgFl_{2}}$, and one mole of magnesium for every one mole of $\ce{MgFl_{2}}$. (Please correct me if I'm wrong anywhere.)

But here's whats confusing to me. Why is fluoride denoted as $(2X)^{2}$? I know that there are two moles of fluoride per one mole of magnesium fluoride, but why does it need to be denoted by being both multiplied to its coefficient and raised to the power of that coefficient?

Sources: Magnesium Fluoride Solubility product constant

The reason why in the equilibrium constant expression, the concentration is to the power of coefficient is because the above equation can be written like this:
$$\ce{MgF_2(s) -> Mg^{2+} (aq) + F^- + F^- (aq)}$$
Hence when you plug this into the equilibrium constant expression, you get this:
$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}][\ce{F-}]$$ which becomes the equivalent to:$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$$ So really the equilibrium constant is proportional to the square of the concentration of fluoride. So that is the reason for the square. However, the concentration of fluoride is twice as much as the concentration of Mg, hence that is the reason for the multiplying the concentration of fluoride by 2. So that is why fluoride is denoted as $(2X)^2$.


Kinetic isotope effect - why transition states taken as isoenergetic I am working on the derivation of the equation below for the difference in activation energies for the two reactions (1) and (2).

I've done the derivation, but I am unsure on some of the approximations I have used. When considering the activation energies I have just used the zero point energy for the C-H and C-D bonds - I am not sure why I can do that, but I assume its because the rest of the zero point energies cancel out.

Secondly I am confused why I can consider the transition states as isoenergetic, as to me they look like they would different energies.

Thank you for your help.

When you deal with energies, absolute energies never matter. Energy only makes sense when you consider it relative to some reference point. For example, for standard values, we consider standard states. You're perfectly fine declaring that both TS's have the same energy because you're not comparing them to each other; you're comparing them to two ground states.


How to calculate the half time of a unimolecular reaction given the Arrhenius coefficient and the activation energy? Consider the following unimolecular reaction:
  $$\ce{H2O2 + M -> OH + OH + M}$$

  The high pressure limit unimolecular rate coefficient for this reaction is $k_\mathrm{uni} = A\cdot \mathrm{e}^{\frac{-E_\mathrm{a}}{R \cdot T}}$, where $A = \pu{3E14 1/s}$, and $E_\mathrm{a} = \pu{305 kJ/mol}$. Calculate the half-life of this reaction at $\pu{1 atm}$ and $\pu{1000 K}$.

Source: Question 14.1 from John W. Daily: Statistical Thermodynamics: An Engineering Approach. Cambridge University Press, 2018. ISBN: 1108244645, 9781108244640. DOI: 10.1017/9781108233194.

I started by getting that $k_\mathrm{uni}$ through the formula, got $\pu{0.03515 1/s}$ and then, since it's unimolecular:
 $$t_{\frac{1}{2}} = \frac{\ln(2)}{k_\mathrm{uni}}.$$
But I got it wrong, I don't have the solutions manual to this book to check what is wrong. I also tried using the Troe form, using the $k_\mathrm{uni}$ from the starting formula as the $k_{\inf}$ but still got it wrong.

Okay, I finally found the solutions manual. And saw that the answer on it is completely wrong, so there was no way that introducing what that math gave would work. The halftime is actually 19.7 s, but in the book it is 9 $\mu$s, which doesn't make sense.

You calculated the high-pressure-limit half life, where the rate of collisions with M is so high that it has no influence on the rate constant. Now all you have to do is to consider that the pressure is 1 atm instead of at the high-pressure limit. That slows down the reaction such that the half-time is in the second instead of the micro second range.
Something similar happens with ozone. It is not very stable at sea level, but has a substantial half-time in the upper atmosphere (unless there are certain man-made substance up there).
To solve numerically, you could assume a Lindemann mechanism (see https://en.wikipedia.org/wiki/Lindemann_mechanism) and that $k_1$ is diffusion-limited (i.e. no activation energy, every collision leads to activated complex).
The textbook cited by the OP has:

Strangely, in an earlier table they show $k_0$ and $k_\infty$ as (along with pre-exponential temperature dependence $T^n$ and ?activation energy? E - units unknown):


What is the hybridisation of titanium in the hexaaqua- and hexaamminetitanium(III) complex? I want to check if their hybridizations are as follows:

$\mathrm{sp^3d^2}$ in  $\ce{[Ti(NH3)6]^3+}$ and $\mathrm{d^2sp^3}$ in  $\ce{[Ti(H2O)6]^3+}$.

Immediately stop using the hybridisation concept for transition metals. Don’t even take it a step further. It is not helpful here at all.

Both complexes have essentially identical electronic structures except that the relative orbital energies differ slightly. The following image — originally taken from Professor Klüfers’ internet scriptum for his coordination chemistry 1 course (LMU Munich; in German) — shows a typical octahedral coordination compound and only includes the ligand–metal σ interactions.

You can see the metal’s orbitals on the left-hand side; from bottom to top they are 3d, 4s, 4p. 4d would be even higher in energy, possibly even higher up than 5s. These virtual energies can be really hard to calculate or measure accurately. On the right-hand side, you can see the six donor orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. (For some reason, the u is missing.) The ligands’ p-orbitals that can take part in π bonding would transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, and would therefore form another bonding-antibonding interaction with the metal’s $\mathrm{t_{2g}}$ orbitals. The ligand orbitals would again be bonding and the metal’s antibonding. If you check out the link and look for the image you’re seeing, you can also see pictures of the metal orbitals, and further down the scheme that includes the π interactions (only the relevant part highlighted).

The metal orbitals that interact with ligand orbitals in a σ fashion are two 3d, 4s and three 4p. The remaining three 3d orbitals interact in a π fashion. So if you really, really, really must use hybridisation in this case (did I say it is not recommended?) then you should report titantium’s hybridisation as $\mathrm{d^2sp^3}$ and not the other way round: 4d orbitals do not play any role.

The only difference between the hexaaquacomplex and the hexaamminecomplex is that the former has lower-lying ligand orbitals than the latter so the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ will be slightly different. That’s all.


Pericyclic Reaction Mechanism We've been asked to draw the mechanism for the following pericyclic reaction:

I know that the tin tetrachloride will act as a Lewis acid and activate the one of the carbonyl groups of the ethyl glyoxylate (I'm not sure which one). I've been trying to see where the cyclic transition state could come from but when I've been drawing the curly arrows I can't seem to get a cyclic transition state (which is required for a pericyclic reaction).

If someone could show me the initial steps of the mechanism then I would be very grateful.

We don't need to consider the stereochemistry for this reaction.

Although you probably know the answer by now, here is the cyclic transition state. It is an oxy-ene reaction. The Lewis acid lowers the activation energy of the reaction.


Formation of tetrahydrofuran in the reaction of n-butanol with lead tetraacetate When n-butanol reacts with lead tetraacetate it forms tetrahydrofuran.

And lead tetraacetate is a good oxidising agent so why doesn't it form n-butanal?

So when does cyclisation take place and why?

The reaction of lead tetraacetate, $\ce{Pb(OAc)4}$ with alkanols, $\ce{R-OH}$, is believed to proceed via an alkoxy-lead acetate, $\ce{R-O-Pb(OAc)3}$.

In this intermediate, the bond between the $\ce{Pb}$ atom and the oxygen of the alkanol cleaves homolytically, this yields an alkoxy radical, $\ce{R-O^.}$ and $\ce{Pb^.(OAc)3}$.

If the alkoxy radical has a hydrogen atom in the $\delta$ position, intramolecular hydrogen atom abstraction may occur.

By this $\ce{CH3CH2CH2CH2O^.}$ is converted to a carbon-centered radical, $\ce{^.CH2CH2CH2CH2OH}$.

The later is oxidized by $\ce{Pb^.(OAc)3}$, which leads to $\ce{Pb(OAc)2}$
and the cation $\ce{^+CH2CH2CH2CH2OH}$.

This cation undergoes cyclization to furnish tetrahydrofuran.


What is the Franck-Condon region What precisely is the Franck-Condon region? My understanding is that the FC region is the "bonding" or "equilibrium" region in a potential energy surface, i.e. near the minimum. But when a transition takes place from S0 to S1, it is said that molecule is excited to the FC region of the S1 state - but there is no guarantee that going vertically from an S0 minimum will lead to a near S1 minimum. So something is probably incorrect with my current understanding.

Also, in theoretical minimum energy path calculations, we start from an optimised high energy state, often in a "Franck-Condon region" of an excited state.

So, what is the precise description of the Franck-Condon region?

The Franck–Condon region is a region of the final potential energy surface which is accessed by the vertical transition from the initial state. Usually, the initial state is the ground vibrational state of the ground electronic state and the final is some vibrational state of an electronically excited state. For instance, the Franck–Condon region is shaded below on the Figure from Wikipedia article on the Franck–Condon principle.

  But when a transition takes place from S0 to S1, it is said that
  molecule is excited to the FC region of the S1 state - but there is no
  guarantee that going vertically from an S0 minimum will lead to a near
  S1 minimum. So something is probably incorrect with my current
  understanding.

Nope, what you are saying is correct. Indeed, according to the Franck–Condon principle vertical transitions are the most probable. The minimum of the upper potential energy surface might lie outside of the Franck–Condon region, so a vertical transition from the ground vibrational state of the ground electronic state to the excited electronic state might indeed result in one of the vibrationally excited states like it is shown in the picture above. Once a molecule is excited in such way it quickly relaxes to the lowest vibrational level of the electronically excited state from which it can decay then back to the electronic ground state. And again, on this way back to the ground state the most probable transition is the vertical one, so that photon emission can lead to a vibrationally excited state of the ground electronic state, not necessarily to the ground vibrational state of it.


Ether cleavage with conc HBr I've been asked to draw the mechanism by which benzene with $\ce{Me}$ on position 1 and $\ce{OMe}$ on position 2 goes to benzene with $\ce{Me}$ on 1 and $\ce{OH}$ on 2 by reaction with concentrated $\ce{HBr}$.

I'm really unsure as to what is going on, I think its probably most likely that $\ce{Br}$ acts as a nucleophile and attacks the $\ce{C-O}$ $\sigma^*$ and then $\ce{OH}$ attacks the $\ce{C-Br}$ $\sigma^*$ but it doesn't seem like a likely pathway as $\ce{OMe}$ has a higher pKaH than $\ce{Br}$ and substitutions don't occur at $\mathrm{sp^2}$ centres.

Cleavage of ethers using strong acids (classically $\ce{HI}$ but $\ce{HBr}$ should work as well) can proceed by either an $S_N1$ or $S_N2$ mechanism. In this case the cation formed would be primary or phenyl, neither of which are stable, and so the reaction proceeds via an $S_N2$ mechanism.

The oxygen is protonated in an equilibrium reaction. The bromide ion then attacks at the least hindered carbon and the $\ce{C-O}$ bond breaks in a concerted $S_N2$ mechanism.

(source)


Why bother to standardize a solution? I have been preparing lessons for some of my students about preparing acidic and basic solutions, and I keep finding pre-made labs in which the students "standardize a basic solution" by titrating it with an acid of "known" concentration.

The problem is that the students create a solution of NaOH by measuring a certain mass and adding it to a volumetric flask, filling with water to the desired volume.  But the students will do the same thing with an acid (potassium hydrogen pthalate, for example).  They will measure out a specific mass, do the calculations to determine the molarity of the KHP solution, and then do the titration to determine the molarity of the NaOH solution.  But if the students took the mass of the NaOH in the beginning, can they not just use that information to determine the approximate molarity of the solution?

As the title says, why bother to standardize?  If our NaOH solution had some room for error in measurement and calculation, surely titrating against another solution of "known" concentration would only introduce more error, no?

Usually, if we're doing these for basic application or experimentation, standardization doesn't really matter. But when it comes to anything analytical where you start to involve calculations, standardization is a must. This is done with NaOH because it's hygroscopic and readily sucks up the moisture in the air. So what is being weighed isn't totally NaOH, but also the moisture that it has absorbed. So almost always the concentration will be lesser than what is sought to be prepared because of this. Tedious as it may be (because you also have to heat up the KHP for around an hour before titrating), standardization brings you closer to the true concentration, but not exactly on the mark. At least in the end, calculations using the standardized concentrations will be analytical. This is why they never introduce standardization in general chemistry courses, I only got to grips with it when I took up Analytical Chemistry, and back then it was really tedious to do.

For example, I actually experienced this before where we had to prepare a 1M stock of NaOH, but after standardization it turned out to be around 0.8 (maybe due to the humidity in the lab at the time).


Why do the boiling and melting points decrease as you go down group 1 and vice versa for group 7? I used to think that because an alkali metal needs to lose one electron to complete its outer shell, when the atom increases in size (atomic radius), the electron would be easier to lose as the attraction between it and the nucleus would be lower, resulting in increased reactivity as you go down the group (the opposite being true for halogens as they need one more electron and so need to retain their electrons).

Now the trouble I am having is that why would the melting point also increase, what does the ability to lose an electron have to do with the intermolecular forces between several atoms?

The group 1 elements are the so-called alkali-metals. The bonding between the atoms is caused by the interaction of the nuclei with the delocalized electrons. With increasing number of electrons and protons, the atomic radii get bigger and hence this interaction becomes weaker as the average distance between nuclei and electrons increases as you go down in group 1.
The group 7 elements are the so-called halogens. They exist under normal circumstances in their molecular form ($\ce{F2}$, $\ce{Cl2}$ and so on). In contrast to the group 1 elements, the dominant intramolecular force here is London dispersion (or van der Waals forces if you prefer). This attraction is caused by the correlated motion of electrons. With increasing amount of electrons, there can be more correlated motion and hence there is a stronger interaction between the molecules and an increasing melting / boiling point when you go down in group 7.


How to find out oxidation state when compound has multiple electronegativities How do I solve for oxidation state in something like CH3-O-F? The fluorine is more electronegative than the oxygen which is more electronegative than the carbon which is more electronegative than the hydrogen. In this case would everything except the fluorine be oxidized?

You can assign oxidation states with a method similar to assigning formal charges.

Draw the Lewis structure.
Assign each atom all of its lone pair electrons.
For bonding electrons assign all of the electrons to the more electronegative element.
For each atom perform the following calculation:
number of electrons in the atom in its elemental state - assigned electrons = oxidation state.

In your example, the oxidation number of each $\ce{H}$ is +1 and of $\ce{F}$ is -1.  The oxidation number of $\ce{C}$ is -2 and of $\ce{O}$ is 0.


How long must a sample be irradiated before all 59-Co atoms are converted to 60-Co? For the interaction,  n$^0$ + $^{59}$Co $\to$ $^{60}$Co

At a constant neutron flux, this is a first-order reaction:

$ - \frac{{{\text{d}}\left[ {{}^{59}{\text{Co}}} \right]}}{{{\text{d}}t}} = k \cdot \left[ {{}^{59}{\text{Co}}} \right]$

$\left[ {{}^{59}{\text{Co}}} \right] = {\left[ {{}^{59}{\text{Co}}} \right]_0} \cdot {\operatorname{e} ^{ - k \cdot t}}$

Theoretically, $\left[ {{}^{59}{\text{Co}}} \right]$ will never become zero.


EPDM tube resistant to ozone on the long run? I am working with ozone produced with an ozone generator and pure oxygen.

The resulting O2 / O3 gas mixture is probably about 80-85% O2 and up to 15-20% O3.

The generator does produce about 500g of O3 per hour.

Currently I am using PTFE tubes to bring the O3 where it is needed.

However I am not happy with the mechanical properties of the PTFE tubes, as they are stiff and as a result they are uneasy to work with.

I have found now that EPDM tubes are way easier to work with, but I have doubts about their resistance to O3.

While many datasheets and Wikipedia do tell that EPDM is resistant to O3, I question if it is really resistant to extreme high concentrations of O3 on the long run.

Switching from PTFE to EPDM in the future would be a good thing as EPDM is also cheaper and easier to work with.

Does anyone here have experience with high concentrated O3 and EPDM?

The problem with EPDM is that it has tertiary carbon sites (and some double bonds for cross-linking). Forget the latter as let's assume the material is already cross-linking.
Normally carbon backbone polymers with substitution need anti-oxidants to prevent the C-H, where there is a side group, from oxidising. These delay the oxidation process and define the lifespan of the material. So after time, the anti-oxidants will be consumed and the polymer will be prone to oxidation, which leads to fragmentation.
Ozone is a great oxidant as you know. When we used high pressure ozone,  we always used fluorinated everything. By using fluorine saturated polymers, there's no sites for oxidation to occur.


Why is the pH scale only from 0-14? Why can't there be more hydrogen ion concentration in acids or lesser hydrogen ion concentration in bases? Why should there even be any hydrogen ion concentration in alkalis? (I know pH is calculated by the negative log of H+ ion concentration.)

It isn't.  We're just used to working in the lab where solutions are generally 1 M or less in concentration.  So we get no more than 1M (solvated) hydrogen ions where pH is 0, or at the other end no more than 1 M hydroxide ions corresponding to pH 14.

These 1 M limits are, however, just for our convenience and not any law of nature.  Right in our own labs we likely have concentrated acid stock solutions, well above 1M, where if we attempted to measure pH we would get less than zero.  Acid pickling solutions in the steel industry are concentrated enough to do the same thing, as manufacturers are more interested in getting hot mill scale cleaned off than in a convenient pH scale.  Bases like sodium hydroxide also have concentrated stock solutions with pH above 14.


What is the activation energy of creating molecular hydrogen from atomic hydrogen? If molecular hydrogen is dissociated, or two hydrogen atoms in space collide, how much energy (in $\pu{eV}$, or perhaps $\pu{kJ/mol}$) does it take for an $\ce{H_2}$ molecule to form?

As porphyrin pointed out, two hydrogen atoms coming together from infinite distance will tend to bond, if the energy of the bond can be dissipated. This is unlike the situation where two magnets click together and stay together (because that interaction hits an absolute minimum, rather than a roller-coaster upswing as in the diagram. Ref 1). Two hydrogen atoms coming together from infinity would be attracted, gain kinetic energy from the potential energy of their separation, come too close, then repel each other - to infinity again. Over and over!

Perhaps instead of thinking of an activation energy, it is more relevant to consider a deactivation energy, which is all the energy that the bond needs to dissipate. You could ask whether it is possible for the two atoms to come together and emit a photon of the exact energy needed to quell the oscillation. Or even a lower energy photon, just to keep the molecule bonded, even if still in a high energy state. Alas! Selection rules seem to prevent this or at least minimize its likelihood, and recombination of hydrogen atoms is reported to occur thru three-body interactions. Ref 2
Can you do this any other way? Perhaps instead of observing them in free space, you could attach both atoms to a metal surface and let them find each other in small steps. Some metals, like zinc, evolve H$_2$ in acidic solution. Very very pure zinc reacts only very slowly with acids. Ref 3 My interpretation of that situation is that when the first evolution of hydrogen occurs, the hydrogen is in atomic form, and is attached (not strongly bonded) to the surface, whether that is actually metallic zinc or a nanothin layer of ZnO, perhaps as a ZnO  -  H moiety. A more stable state would be for two of these loosely attached hydrogen atoms to bind together to form H$_2$, but movement is inhibited (by a high activation energy). In less pure zinc, the presence of cathodic sites (like a little copper) provides a better site for H$_2$ evolution, lowers the activation energy and the process spreads over the rest of the zinc to allow rapid corrosion. Less pure zinc could be coated with a thin film of mercury to provide another surface for hydrogen atoms to be relatively immobilized on, and this also gives a high overpotential (a high activation energy)for hydrogen evolution. Ref 4
Some metals, noble metals which have no oxide layer, like Pt, Au and Pd, enable rapid migration of H$_2$ across their surface (thus not only a low, near zero activation energy, but also connection with a massive metal substrate that can accept some vibrational energy when two hydrogen atoms get together and dance for a while before settling down. They are good surfaces for evolving H$_2$.
A number of organic reactions were done with atomic hydrogen, using high activation energy to prevent  formation of molecular hydrogen; this was classified as due to nascent hydrogen, a term now in disfavor. Ref 5
Ref 1. http://www1.biologie.uni-hamburg.de/b-online/library/newton/Chy251_253/Lectures/LewisStructures/Dihydrogen.html
Ref 2. I. Amdur,  J. Am. Chem. Soc. 1938, 60, 10, 2347–2355
Ref 3. https://en.wikipedia.org/wiki/Zinc
Ref 4. https://en.wikipedia.org/wiki/Overpotential
Ref 5.https://en.wikipedia.org/wiki/Nascent_hydrogen


Role of Hg2+ as catalyst in the reaction of addition of hydrogen halides to alkynes $\ce{Hg^2+}$ can be used as a catalyst. But what is its role? How does it increase the rate of reaction.

I think that the vacant orbitals in Hg would help in the process of ionisation of hydrogen halides. Is this concept correct? please help

The main use of $\ce{Hg}$$^{\text{2+}}$ is to polarize the electron cloud of the alkyne.
Else the attacking specie wont be able to approach the bond


Curious as to what CorA1 is and what it does? First time Chemistry SE poster,

Was wondering what actually CorA1 is and what it actually does?

I ask because I found out I was allergic to a incredibly long list of items and this popped up to be a high allergen as I work through the list.

Google defines it as a: Magnesium transport protein CorA and it contributes to Ion transport?

You see, CorA and CorA1 are two different things. CorA is a protein, while CorA1 is the gene that codes for that protein.

It is the genes that when translated give rise to the protein.

This gene was isolated from the Rhizobium meliloti, a bacterium commonly known for forming leguminous nodules in certain plants.

As for it's function, I'm not really a biologist, but I know that magnesium is a component of chlorophyll (It's coordinated at the center of the porphyrin complexes).

So you could say magnesium is rather important for plants. The $\ce{Mg^2+}$ ion needs to be transported from the soil to the leaves, and this protein facilitates this transport.


How much more salt can I dissolve in hot water compared to cold water? Hot water increases not only the speed of solubility but also the amount.

I don't know whether this is for all materials in water, apart from salt.

But, if this is the case, what causes hot water to dissolve more salt (or other solutes)?

There seems to be two parts to your question. First, not all substances have increased solubility as water temperature rises. Gasses for example have decreased solubility in this case.

As for the second part of the question, increased heat adds energy to the water molecules which makes it easier to overcome the bonds between $\ce{Na}$ and $\ce{Cl}$. According to Wikipedia, the solubility of $\ce{NaCl}$ in nearly freezing water is $35.65 \, \mathrm{g} / 100 \, \mathrm{mL} \, \ce{H2O}$ versus boiling water which is $38.99 \,\mathrm{g} / 100 \, \mathrm{mL}\, \ce{H2O}$. Hope this helped.


Why can't amylase digest glycogen? Amylase is an enzyme that breaks down starch in the form of amylopectin and amylose. Both amylose and amylopectin are formed by alpha glucose joined together by (1-4) and (1-6) glycosidic bonds. Glycogen is no exception, just that it has more branching.

However, why is it that a Google search shows that it is hydrolyzed by glycogen phosphorylase rather than amylase? Also, how can amylase digest both (1,6) and (1,4) glycosidic bonds?

Amylase can’t digest glycogen because of its inability to attack the branching (1→6) linkages.

Perhaps, another very important reason is controlling the rate of glycogen metabolism through glycogen phosphorylase. Just like any other biological system, regulation of metabolic substrates and/products is crucial to maintaining the balance (homeostasis) so to prevent excess glucose production from glycogen metabolism or to little according the needs of the organism.

Tl;dr

There are  actually three forms of amylases:

α-amylase (an endoglycosidase, which can hydrolyze a glycosidic linkage anywhere along the chain to produce glucose and maltose).
β-amylase (an exoglycosidase that cleaves from the nonreducing end of the polymer).
gamma amylase

Now starch consists of two main components amylose and amylopectin;

Amylose is a linear polymer of several thousand glucose residues linked by (1→4)
bonds:

Amylopectin consists mainly of (1→4)-linked glucose residues but is a branched molecule with (1→6) branch points every 24 to 30 glucose residues on average.
The primary structure of glycogen resembles that of amylopectin(not amylose), but glycogen is more highly branched, with branch points occurring every 8 to 14 glucose residues:

Digestion

The digestion of starch, the main carbohydrate source in the human diet, begins in the mouth. Saliva contains α-amylase, which randomly hydrolyzes all the (1→4) glycosidic bonds of starch except its outermost bonds and those next to branches.
On the other hand glycogen being a highly branched molecule, it is evident that amylase won’t be a good enzyme to digest it due to physical structure constraints.
 Substrate specificity of the enzymes ensure that the appropriate substrate which is highly optimized for the active site will be broken down.

Although glycogen posseses parts similar to starch's amlyopectin its structure is not optimized to be broken down by amylases. Note: even these amylases only partially break down starch to different oligosaccharides


Lambert Beer's law deviations Can somebody explain as concrete as possible, how the change of temperature and the change of pH  leads to deviations of this law?

A compound S follows the Lambert-Beer law if the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.

\[E_\lambda = \left(\frac{I_0}{I} \right) = \epsilon\cdot c \cdot d \]

A "deviation" is observed when the real concentration $c$ is different from the expected/assumed.

This may happen if

S is an acid or a base:

variation of the pH leads to (de)protonation and forms another species with a completely different absorption spectrum, or at least a different $\epsilon$ at the wavelength under observation

S undergoes a reaction to S' (with different optical properties) at elevated temperatures (or change of pH)
Decrease of temperature or change of pH leads to the formation of a turbid solution from which light is scattered into the detector


Difference between organic and inorganic chemistry This question may seem vague at a first glance, and even lead to the thought: "come on, just go Google it!"

But it's a real question because it does not make enough sense to me (even after Googling it, a lot).

I was just presented to the words, but not to their definition.

Like: "Hey, now we will speak about organic nomenclature".

At some point, books and videos start talking about inorganic and organic as if it was a common sense thing.

A frequent rule I found while trying to understand the difference:

Organic: has carbon
Inorganic: does not have carbon

But then, some inorganic compounds do have carbon too, and there may even be some  compounds that some call organic, and others call inorganic, like $CO_2$.

As I have felt it, in my learnings so far, it's like inorganic chemistry is the default chemistry and organic chemistry goes a step beyond.

But I don't quite grasp the difference.

What is the real semantics behind the word "organic"?

For example, we humans are made of loads of water, and that's a pretty organic thing to me. But then, water is inorganic.

Diamonds are the carbon top of the cake, and do not transmit the idea of being an "organic" thing.

Another very confusing thing are polymers, chanins of loads of carbons with other elements, in many shapes and textures. To me, a piece of "plastic" is not a very organic thing, but indeed, they are!

That brings the semantics into an even more confusing level.

And of course, there must be historical reasons for those chosen words.

Could someone please point out where this distinction comes from and why it is important?

With all my respect to science and the people who made chemistry a useful thing. This question is not about critics, it's about not knowing the facts, so of course I am the ignorant here.

Related and useful: What is the definition of organic compounds?

IUPAC is the International Union of Pure and Applied Chemistry, they make recommendations on the nomenclature. IUPAC mentions that the difference between organic and inorganic is not distinct. To quote "The boundaries between ‘organic’ and ‘inorganic’ compounds are blurred." in Brief Guide to the Nomenclature of Inorganic Chemistry R. M. Hartshorn, K.-H. Hellwich, A. Yerin.

Since the terminology of organic vs. inorganic is all human classification, it is not a binary system 0 or 1. What we can say now is that traditionally, all organic compounds do contain carbon. It can come from natural sources or purely synthetic. There is no such restriction. Plastic is an organic compound because it contains a lot of carbon chains.

Note that this word organic, as used in chemistry, has nothing to with the buzz word used in marketing of organic food, organic fruits, organically grown stuff. The word organic comes from French organique designating the jugular vein, hence related to organs or living beings.

It is a myth that Wohler started the distinction of organic vs. inorganic. There is a nice French article. The 16-paged paper is titled “Les origines de la chimie organique au-delà du mythe fondateur”  (The origins of organic chemistry beyond the founding myth), Comptes Rendus Chimie, 2012, 15, 553–568.

Read the table at the end, and the abstract.

It clearly shows that Berzelius coined the term organic chemistry between 1807-1813.

Use Google Translate or DeepL to translate it.


Applying Raoult's Law in volatile solids–liquid solutions My textbook explains applying Raoult's Law for the two cases:

liquid–liquid solutions, in which the vapour pressure of the solution is due to both the components
non-volatile solids–liquid solutions, in which the vapour pressure of the solution is only due to the liquid component of the solution

I am curious as to what will happen (how things change for the application of Raoult's Law) if the solution consisted of some volatile solid like camphor, in some liquid (possibly some liquid hydrocarbon).

Certainly, the solid now also contribute to the total vapour pressure of the solution, but does it still block the escaping of the solvent molecules into vapour phase as much as the non-volatile solid blocks the liquid solvent molecules?

If yes, then how can I apply Raoult's Law for the case to find the total pressure of the solution?

Raoult's Law is only valid when the solute is non-volatile. Therefore, you cannot apply it.


Refrigeration for Cryochemistry −35 °C I need to cool a tube containing liquid to −35 °C for a cold chemical experiment that will last about a week. The glass vessel I need cooled is 500 mL in volume and will be giving off $\ce{H2S}$ gas. I currently only have a fume hood. Most freezers do not seem capable of this low temperature. Most cryochemistry refrigeration seems designed for much lower temperatures, and is needlessly expensive. I realize that refrigeration is relatively simple: 1 compressor, 1 expansion valve, two heat exchangers, and electronic control/monitoring. My first impulse is to modify a freezer, but I am completely new to this. At this point I would be very happy with some general advice and perhaps a reference for custom designing these sorts of cooling devices.

I would suggest a Frigorific Mixture of calcium chloride (often sold as DampRid) and Ice in a cooler.  You will need quite a bit of each and to moniter at least twice a day, but it wont break the bank and can reach the desired temperature.


Does the third law of thermodynamics apply to all compounds and elements? According to wikipedia, the third law of thermodynamics is "The entropy of a perfect crystal, at absolute zero (zero kelvins), is exactly equal to zero.".

Then, theoretically, could all substances, including molecular compounds, attain absolute zero? Another way of thinking of this question is whether all compounds could become perfect crystals.

I strongly suggest you read this. From there:

  The laws of thermodynamics dictate that absolute zero cannot be reached using only thermodynamic means, as the temperature of the substance being cooled approaches the temperature of the cooling agent asymptotically. A system at absolute zero still possesses quantum mechanical zero-point energy, the energy of its ground state. The kinetic energy of the ground state cannot be removed.

And:

  An even stronger assertion is that It is impossible by any procedure to reduce the temperature of a system to zero in a finite number of operations. (≈ Guggenheim, p. 157)

And:

  Perfect crystals never occur in practice; imperfections, and even entire amorphous materials, simply get "frozen in" at low temperatures, so transitions to more stable states do not occur.


Reactivity of alkyl halides I am unable to understand how to decide reactivity of alkyl halides based on inductive effect in a nucleophilic substitution reaction. I have 2 explanation for deciding. Consider 2 alkyl halides:

(A) 2-chloro-2-methylbutane
(B) 2-chloro-3-methylbutane

First explanation:

In (A) we have a $3^\circ$ carbon so if we break the $\ce{C--Cl}$ bond a $3^\circ$ carbocation will be formed, which is stabilized by +Inductive effect of surrounding  groups. Therefore we can say that as  breaking of $\ce{C--X}$ bond in (A) leads to the formation of a more stable carbocation than that formed in (B){$2^\circ$ carbocation}. And due to this the reactivity will be more of (A).

Second explanation:

As in (A) $3^\circ$ carbocation is formed, it will get more stabilized than $2^\circ$ carbocation formed in (B). So this means that $2^\circ$ carbocation in (B) becomes more positive than $3^\circ$ carbocation in (A), making it easier for a nucleophile to attack on (B)'s carbocation. This would make (B) more reactive.

Which one of them is correct?

Assuming your concern is on $\mathrm{S_N1}$ type nucleophilic substitution reactions, you are in the correct path with your first explanation. However, everything has gone south from that point on. As you found out, the stability of carbocations in $\mathrm{S_N1}$ reactions are: $3^\circ \gt 2^\circ \gt 1^\circ$. However, reaction rates of $\mathrm{S_N1}$ reactions do not depend on electron density (or lack of it) on carbocation, because rate determine step is rate of carbocation formation step:

As depicted in the diagram, $3^\circ$-carbocation from 2-chloro-2-methylbutane forms faster than $2^\circ$-carbocation from 2-chloro-3-methylbutane because the stability of $3^\circ$-carbocation is much larger than that of $2^\circ$-carbocation as you predicted in your first explanation. Also, you need to understand that in these reactions, 1,2-hydride (or methide) transfer to give more stable $3^\circ$-carbocation from $2^\circ$-carbocation is possible if there are available $\alpha$-hydrogen or $\alpha$-methyl group in the sought $2^\circ$-carbocation, alike in the case of 2-chloro-3-methylbutane here. Regardless of this rearrangement, the rate of carbocation formation of 2-chloro-2-methylbutane is still faster than that of 2-chloro-3-methylbutane, because it is strictly the stability of $3^\circ$- versus $2^\circ$-carbocations.

The second reaction (attack of the nucleophile to available carbocation) is very fast compared to chloride dissociation, thus it is not affecting the overall rate of the reaction. Thus, the product formation from 2-chloro-2-methylbutane is faster than the product formation from 2-chloro-3-methylbutane, and hence, 2-chloro-2-methylbutane is more reactive than 2-chloro-3-methylbutane.


Approximating strength and hardness of alloy I do not have any formal education in metallurgy, but I am trying to make a game mechanic based on it.

I need to be able to alloy together different metals and alloyable materials such as carbon in different percentages, and get an approximation of the strength and hardness of the resulting alloy. If anyone has any ideas how what I would need to do to calculate such a thing I would be very appreciative.

Look at the free resource provided at MatWeb.

Under the Physical Properties tab, you can choose up to 3 of them from a rather extensive list, including hardness. For example, I chose "Carbon" and "Hardness, Rockwell R," and 25 results were returned.

Under the Alloy Composition tab, you can choose up to 3 elements and percentages for your alloy, and see if you get a hit. I chose "Nonferrous Metal" and carbon (minimum 5%, maximum 75%) and silicon (minimum 5%, maximum 75%), and got one hit.

I believe for your purposes, you can assemble a list of materials by hardness and strength (I'm not a metallurgist either, so you might have to do some research on which of the hardness and strength metrics you'll need to choose) and/or by selecting by composition (again, you might need to research which 2- and 3-element combinations make sense and which ones don't).


Reference Books on Electrolysis and Chemical Titration process I'm in a General Chemistry course in college. We have laboratory, and then we are supposed to make a report on what we did, etc.

I'm not entirely sure what the last laboratory was about, but it was called Salt and Water Electrolysis

To sum up the procedures -two general procedures-, here's what we did:

1st Procedure:

We prepared 50 mL of a $\ce{NaCl(aq)}$ solution at 0.1 M (molarity) in a 100 mL beaker
We made a DC (direct current) circuit at 5V and 700 mA, with graphite electrodes
We turned the circuit on for roughly 20 minutes
We employed a magnetic stirrer to stir the solution
We added 3 drops of Phenolphthalein in the solution, and took notes of the phenomena. (I guess it kinda turned pink)

And that was that for the 1st procedure.

2nd procedure:

We measured the solution's pH with a pH meter
We measured roughly 15 mL of $\ce{HCl}$ (Hydrochloric Acid) at 0.5 M with a test tube
We poured the solution of $\ce{HCl}$ in a 25 mL burette.
We prepared the titration equipment
We were supposed to add the $\ce{HCl}$ (drop by drop) to the $\ce{NaCl}$ solution and at the same time be measuring the pH until it got to 7

And that was that for the second procedure.

Of course we kind of messed up, as we overshot the end point. The excess $\ce{HCl}$ brought it's pH to 3 point something.

The $\ce{NaCl}$ solution lost its pink color, and then turned yellowish.

So.. I guess I am asking for references to the equipment involved -I kind of have references for the pH and the properties involved, still, if you know of some references, let me know!-, I've looked for electrolysis processes in books but don't find any. Or they're too technical. Also, haven't found titration procedures in books.

Can you help me out, just so I can finish my report?

Any good general chemistry textbook has a chapter on electrolysis. If you wish to add more scholarly work check the Journal of Chemical Education. https://pubs.acs.org/action/doSearch?text1=Electrolysis&quickLinkYear=&quickLinkVolume=&field1=Title&type=within&publication=346464552

In the first part you electrolyze a solution of salt. Check the half cell reactions from the text: What happens at the cathode and what happens at the anode?

Water is reduced at the cathode and -OH ions are produced. This makes your salt solution basic. In the second part, your lab requires you to estimate how much base was produced.*

Essentially you do a titration of NaOH with standardized HCl. The end point is estimated by a pH meter. Look up any general chemistry textbook again for titration of a base with HCl using a pH meter. The technique is called pH titration. You did a very rough version of a pH titration.

Depending on the conc. of NaCl, either chlorine or oxygen is produced at the anode.


How can I free base cystamine dihydrochloride? I am using cystamine (free based) to conjugate to a phosphate group on an oligonucleotide. I tried several keyword searches on Google for neutralizing the dihydrochloride at the ends of the cystamine. But, I am unable to find a protocol. I had once found a procedure for the process, but I did not reference it and hence I am unable to find it again.

The procedure I followed was:

Dissolve the cystamine.2HCl in water.
Transfer to a separation funnel and add chloroform.
To this add 1N NaOH.
After shaking vigorously collect the chloroform layer which contains the cystamine.
Repeat extraction couple of times using more chloroform.
Pool fractions - dry with anhyd. MgSO4 - transfer to an RBF.
Evaporate on a RotaVap
Dry on schlenk line and store in an inert environment at 4 deg. C.

However, the extraction was not very efficient. Can anyone refer a protocol for the same?

Thanks a lot for your help!!!!

Maybe the MgSO4 is absorbing your product. You may want to try another drying agent Try Na2SO4, zeolites, or azeotropic distillation (e.g. with benzene). Maybe NaOH will work, but it may react with cysteamine.


Resolution in Mass Spectrometry I have been reading about resolution in mass spectrometry and there are a few things I do not understand.

The resolving power is determined by $m/(m_2-m_1)$ at full width half height of a peak. The higher the value obtained the better. This is where I struggle to form a link between the resolution of the mass spectrometer and the resolving power. In image (a) below the resolution of the mass spectrometer is 1000 and would have a very low mass resolution value, but in image (b) the resolution is 5000 and the nominal peaks masses are separated, therefore providing a high mass resolution value.

Why can a mass spectrometer with a resolution of 5000 resolve the nominal masses and the mass spectrometer with a resolution of 1000 cannot? And how does the mass resolution come into play? Also, what do the values of 1000 and 5000 mean?

I'm not sure I completely understand what exactly your problem is, but:
You want to resolve to signals at approx. m/z = 1060.2 and m/z = 1061.2.
So $\Delta m = 1$ The required resolution is: $\frac{m}{\Delta m} = \frac{1061}{1} \approx 1060$: The spectrometer in (a) has only 1000 which is too low. The spectrometer in (b) has 5000 which is quite more than needed.
Maybe it is easier to think of the reciprocal: $\frac{\Delta m}{m}$ which tells you how far two signal need to be apart so your spectrometer can resolve them.
In (a) this is $\frac{\Delta m}{m} = \frac{1}{1000} = \frac{1.06}{1060}$ two signals can be resolved if they are at least 1.06 m/z apart at m = 1060. I.e. if they were slightly further apart than they are.
The spectrometer in (b) can resolve signals that are at least  $\Delta m/z =  0.21$ apart at  $m/z = 1060$.
FWHM and resolution
The FWHM is basically part of the resolution. The $\Delta m$ is the m/z difference between two signals that are just resolved.
Have a look at the IUPAC's definitions of resolution in mass spec. There are several approaches to specify what "just resolved" means.
The FWHM can be used directly as the $\Delta m$ of the "peak width definition".
If instrument A has $5\times$ the FWHM than instrument B, then resolution of A  $\approx \frac{1}{5}$ resolution of B.
I'm more used to the "10 % valley" which roughly equals the full with at 5 % height: you specify how much signal is allowed at the minimum between the two peaks, e.g. 10 %.
It is important to realize that these two definitions differ by almost a factor 2!
Here are some results of "simulated" peaks at different resolutions and $\Delta m/z$ (all gaussian, which of course doesn't need to be the case in reality). I use the full width at 5% = 10 % valley definitions for the resolution.
The first row is resolution 5000 with the, second row is resolution 1000. You see that the two signals are much better resolved than in your example picture above. Which means that above another resolution definition was used, possibly $\Delta m$ = FWHM. In any case, a resolution of 1000 should be almost sufficient to resolve m/z 1060.2 from 1061.2.
The third row shows two signals of equal height with $\Delta m/z $ = 1 = FWHM. According to the resolution definitions above, this is a resolution of ca. 520. Note that while you can say that there are at least two peaks, you cannot take the two maxima as the m/z of the two underlying signals (grey).
The fourth row is almost the same, but I multiplied the second signal by 56% to account for the frequency of having bradykinin with one $^{13}$C. Looks quite similar to your picture.

Also, have a look at M. P. Balogh: Debating Resolution and Mass Accuracy, LC•GC Europe, 17(3), 152–159 (2004).


Why does chlorine have a higher electron affinity than fluorine? Since fluorine has its valence electrons in the n=2 energy level, and since chlorine has its valence electrons in the n=3 energy level, one would initially expect that an electron rushing towards fluorine would release more energy, as it would land in the n=2 energy level, whereas in chlorine, the electron would land only in the n=3 energy level, and would then not release as much energy.  Thus, one would expect fluorine to have a greater electron affinity than chlorine.  However, why is it that chlorine has a higher electron affinity (349 kJ/mol)  than fluorine (328.165 kJ/mol)?

Fluorine, though higher than chlorine in the periodic table, has  a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

See this.


Titration - Just need your judgement A student investigated the volume of sulfuric acid that reacted with 25 cm3 sodium hydroxide. Describe a titration method the student could use in this investigation. {6 marks}

My answer:

Use the pipette and pipette filler to safely add 25 cm3 of $\ce{NaOH}$ to a clean conical flask.
Add a few drops of indicator and put the conical flask on a white tile (so you can see the colour of the indicator more easily).
Fill the burette with $\ce{H2SO4}$ and note the starting volume (record from bottom of the meniscus at eye level).
Slowly add the acid from the burette to the alkali in the conical flask dropwise, swirling to mix.
Stop adding the acid when the end-point is reached (the appropriate colour change in the indicator happens). Note the final volume reading of $\ce{H2SO4}$ (record from the bottom of the meniscus at eye-level).
Repeat steps 1 to 5 until you get consistent readings

Have I missed anything that will prevent me from getting 6 marks?

As titration is based on the relationship of molar quantity ($n$), volume  ($V$) and concentration ($c$) of the reagents

$$ c = \frac{n}{V}$$

and as your titration is to determine the concentration of NaOH in a known volume of solution of analyte, you would need to know both the volume as well as the concentration of the $\ce{H2SO4}$ solution deployed by addition via the burette.  The later is missing in your list.

Knowing the stoichiometry of the neutralisation is equally of importance, since each equivalent of $\ce{H2SO4}$ reacts with two equivalents of $\ce{NaOH}$. (This differs from a determination of, say, of / with $\ce{HCl}$, or $\ce{H3PO4}$.)  Your current list neither accounts for a chemical equation, nor this stoichimetric factor.


What is the geometric formula of CaI2 and AlCl3 Well, I think $\ce{CaI2}$ is the same as $\ce{BeH2}$ and $\ce{AlCl3}$ the same as $\ce{NH3}$, but I'm not sure since $\ce{I}$ and $\ce{Cl}$ have their own pair electrons and I don't know if this affects the structure...

$\ce{CaI2}$ is mostly ionic in the solid and aqueous phases.  In the gas phases it is a linear molecule.  See Ionic or covalent? Can some first hints be derived from the solid state
structures of alkaline earth metal halide adducts?

$\ce{AlCl3}$ has an octaheral structure in the solid phase.

In the liquid phase:

Trigonal planar monomers also exist at high temperature.

Source: http://en.wikipedia.org/wiki/Aluminium_chloride


What causes the "Gd break" in the trend of lanthanide-EDTA formation constants? Smith and Martell obtained a series of data for the binding of trivalent lanthanide ions, $\ce{Ln^3+}$, with various carboxylic acid ligands (amongst them the well-known EDTA).1 A graph of the formation constants is attached ($K_\mathrm{f} = [\ce{Ln(edta)-}]/[\ce{Ln^3+}][\ce{edta^4-}]$):

Due to the contracted nature of the 4f orbitals, the lanthanide ions characteristically do not exhibit covalency in metal-ligand bonding, which is strongly reflected in their chemistry. Therefore, the almost linear increase in $K_\mathrm{f}$ going from $\ce{La^3+}$ to $\ce{Lu^3+}$ is entirely to be expected. The ionic radius decreases going across the 4f block, and consequently, the interactions with the ligand (primarily electrostatic in nature) become stronger.

However, as the title suggests, I'm interested in the slight blip at gadolinium ($\ce{Gd^3+}$). The value of $K_\mathrm{f}$ seems to be lower than would be expected, and the astute reader will notice that this applies not just to the EDTA complexes but also for some of the other ligands investigated.

Is there a reason for this anomaly?

There is a review by Moeller et al. from 1965,2 which wrote:

  For all ligands which have been studied, the gadolinium complex is less stable than would be expected from the simple electrostatic model. This behavior has been called the “gadolinium break” and cannot be explained, as was originally attempted, by assuming a steric effect, since it is still apparent in ligands for which there should be no steric interference.

Of course, this is more than half a century ago. I would be grateful if anybody could provide either a sound theoretical explanation for why the Gd formation constants should be smaller or a newer reference that explains this effect.

References:

Smith, R. M.; Martell, A.E. Critical stability constants, enthalpies and entropies for the formation of metal complexes of aminopolycarboxylic acids and carboxylic acids. Sci. Total Environ. 1987, 64 (1–2), 125–147. DOI: 10.1016/0048-9697(87)90127-6.
Moeller, T.; Martin, D. F.; Thompson, L. C.; Ferrús, R.; Feistel, G. R.; Randall, W. J. The Coordination Chemistry of Yttrium and the Rare Earth Metal Ions. Chem. Rev. 1965, 65 (1), 1–50. DOI: 10.1021/cr60233a001.

If you consider the solvent extraction of lanthanides by reagents such as H-DEHPA, then to a first approximation the distribution ratio is given by the equation.

$$\ce{DLn}= k[\ce{H-DEHPA}]^3[\ce{H+}]^{-3}$$

If you plot k as a function of atomic number for DEHPA dissolved in most organic solvents (solvent extraction term is diluent) then the value goes up as a function of Z. But the line has bumps on it.

The reason is that the extraction reaction is

$$\ce{Ln(H2O)_n^3+(aq) + 3H-DEHPA(org) -> [Ln(DEHPA)3](org) + H^+(aq)}$$

As you go from one side of the lanthanide group to the other the binding of the lanthanide to the anionic DEHPA ligands is likely to become stronger, but the value of $n$ changes in several steps. This will give bumps on the graph of $K$ for both the extraction of the lanthanides and also for their binding to ligands such as EDTA.


Entropy change in mixing a gas at constant temperature and pressure The expression given on Wikipedia is (https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwj-_pDA2IfSAhUFto8KHa2kBAwQygQIIDAA&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FEntropy_of_mixing%23Mixing_of_ideal_materials_at_constant_temperature_and_pressure&usg=AFQjCNFVf8mAoquhiL05HeVXC1H3cw54xA)
But If I evaluate it as following I get a different answer .
$$\Delta S=\int q_{rev}/T=\int pdV/T $$
as p and T are constants $$p(V_1+V_2)/T $$
What is wrong with my method ?

The reference you gave involves mixing two different chemical species together at constant temperature and pressure.  The problem you seem to be solving is mixing the same chemical species with itself at constant temperature and pressure.  For that process, the change in entropy is zero.  If you really intended to determine the entropy of mixing two different chemical species, then the path you identified is not a reversible path for transitioning between the initial and final states of the system.


Is there any free GAMESS input file maker for Mac OS? My main purpose is to simulate excited state calculations on a molecule with Time Dependent DFT.

I am looking for free software that makes input files for GAMESS. I have tried Avogadro and it works fine (in Extensions> GAMESS > Input Generator > Advanced tab) but it can not make TD-DFT input codes.

I also checked the recommended software by the Gordon Group (here) and tried installing some of them but couldn't install Gabedit at all (I was unable to compile the binary, maybe because I am using Mac OS X Yosemite).  I also tried WebMO but it is available just on iTunes for iOS.

What I need is a software that makes the input code interactively to avoid possible errors.

If you can give me the requisite keywords for TD-DFT in GAMESS, it's not hard to add that to Avogadro

I've never done TDDFT in GAMESS because I use other programs, but given a link to the docs, it's easy enough to enable more drop-down menus.

That said, you can interactively edit the text in the Avogadro input for any quantum program (not just GAMESS).

If you don't want to use Avogadro, the "native" program for GAMESS is MacMolPlt which was developed in the Gordon group itself. I haven't checked, but I'd assume it supports TDDFT options, since it enables pretty much everything else in GAMESS.


How to report a triplet of septets in NMR tabulation? Does anyone know the proper format to report an iso-butyl group, eg. $\ce{R-CH2-C\color{red}{H}(CH3)2}$, in $\ce{^1H}$ NMR tabulation? The proton in red should be a triplet of septets, so I've written:

δ 2.72 (d, J = 5.5 Hz, 2H), 1.80 (tsep, J = 5.5, 5.3 Hz, 1H), 1.02 (d, J = 5.3 Hz, 6 H).

Is the use of "tsep" acceptable here?

I personally use a space, so "t sep" as long as both "t" and "sep" are defined. I think it's slightly clearer than "tsep", but I don't think there's any official adjudication on what's a good or bad acronym.

Also, of some interest to me is how you manage to distinguish 5.5 and 5.3 Hz. If it looks quite like an nonet, I would probably prefer reporting it as an apparent nonet, although different people may have different opinions... I know some groups will have "in-house" rules / guidelines for reporting NMR spectra, so it might be a good idea to check with your supervisor on this front.


Why isn't ethane used for cooking? We commonly use methane and propane for cooking (and home heating), but not ethane. I would expect ethane to be suitable for this, being in between the two, but I've never heard of anyone using it for this purpose. Why is that?

On a related note, why is butane used for cigarette lighters and basically nothing else (in ordinary life, I mean)?

Probably the biggest drivers behind using methane as a fuel is that it is abundant in natural gas and is (currently) mostly useless as a chemical feedstock.
Ethane makes up a few percent of natural gas and can also be obtained as byproducts of petroleum refining, but the big difference from methane is that ethane is extremely useful in chemical synthesis (mostly to make polyethylene).
In fact, household natural gas often contains a bit of ethane, which may vary depending on the current demand for ethane as a feedstock.
In short: it burns just as well, but ethane has other important uses.

Propane has some use as a feedstock, but it is an attractive fuel because it is easily stored as a liquid without requiring huge pressures.


Why is atmospheric composition compared by volume? The composition of the Earth's atmosphere by its constituent gases (nitrogen, oxygen, argon...), is often given by percent volume. (See this Wikipedia page, for example). Why are these compositions given by volume? I understand that a volume composition shows the mole fractions, which is the real point of interest. Volume compositions compare the volume taken up by different amounts of gases under the same hypothetical pressure and temperature. However, in any "container," each gas occupies the same volume as another; what differs is the partial pressure of each gas. Thus, why not base compositions off of partial pressures? It is physically correct to say that oxygen accounts for 21% of the pressure of the atmosphere, rather than saying that oxygen accounts for 21% of the volume of the atmosphere (which is thankfully not true). I understand that comparing pressure under equal volume and comparing volume and equal pressure both show the same thing, but why not use pressure if it is more physically correct?

The short answer is that (a) partial pressures have units of pressure, and (b) pressure changes with altitude and/or temperature.

  The composition of the Earth's atmosphere by its constituent gases (nitrogen, oxygen, argon...), is often given by percent volume. (See this Wikipedia page, for example). Why are these compositions given by volume? I understand that a volume composition shows the mole fractions, which is the real point of interest.

You are right that the mole fractions are the real points of interest.  You may additionally be interested to know that many techniques that measure the percent volume of a certain gas actually measure its mole fraction.  Thus, reporting the results in percent volume is correct only if the gas is ideal.

  Volume compositions compare the volume taken up by different amounts of gases under the same hypothetical pressure and temperature. However, in any "container," each gas occupies the same volume as another; what differs is the partial pressure of each gas. Thus, why not base compositions off of partial pressures?

Partial pressures have units of pressure.  So at 1 bar of pressure at 298 K, oxygen has a partial pressure of 0.21 bar and nitrogen has a partial pressure of 0.78 bar.  Let's say you had a closed rigid container of gas initially at this pressure and temperature.  Let's now say that it's a hot day so the temperature is 313 K.  Now the total pressure is $1\times\frac{313}{298}=1.05$  bar, and as a result the oxygen partial pressure is 0.22 bar and the nitrogen partial pressure is 0.82 bar.  But meanwhile the volume fraction (or equivalently the mole fraction) of each gas is unchanged.

  It is physically correct to say that oxygen accounts for 21% of the pressure of the atmosphere, rather than saying that oxygen accounts for 21% of the volume of the atmosphere (which is thankfully not true).

I don't fully understand this sentence but do not believe it is correct.  Oxygen does account for 21% of the volume of the atmosphere.

  I understand that comparing pressure under equal volume and comparing volume and equal pressure both show the same thing, but why not use pressure if it is more physically correct?

It is not any more physically correct than using volume fraction.  It additionally is complicated that partial pressures have units and will change with temperature or total pressure.


How do oxide formations affect voltaic cells? I asked a question previously on here but still have some confusion. To briefly summarize, I made a voltaic cell for a school project with 0.1 M copper nitrate and varying concentrations of aluminum, zinc and nickel nitrates. My results were fairly consistent with the theoretical values from the Nernst equation for zinc, however, the aluminum and nickel cells produced substantially less voltage than their theoretical values. The answer I got from my previous question explained how the low voltage in the aluminum cell was a result of its oxide layer.
However, I still had some confusion over why there were differences between aluminum and nickel, and zinc. Firstly, it's my understanding that a metal's ability to form an oxide layer is related to its reactivity. This would explain aluminum's tendency to form such a strong oxide layer however, zinc has a higher reactivity than nickel so wouldn't it also form an oxide layer and its outputted voltage would be lowered similarly.
My next explanation for the differences between aluminum and nickel, and zinc was related to the ability of the oxides to either react or dissolve in the nitrate solutions, however, I found that nickel oxide is soluble in acids and aluminum oxide reacts with acids.
I was also wondering how I even got a voltage in the aluminum cell if the oxide layer does not allow any aluminum ions to be transferred.
I would appreciate some guidance from the chemistry wizards because at this point I am stumped.

There are several questions in your text.
First, you imagine that "a metal's ability to form an oxide layer is related to its reactivity". No. It is not. Aluminum is rather reactive with water. But sodium is more. And metallic sodium is not protected by any oxide layer. On the contrary. Sodium reacts dangerously with water. Aluminum is a strange metal, as it forms an oxide that has the same density as the metal. So this oxide can perfectly cover the surface of the metal.
Then you ask about the difference of reactivity between zinc and nickel. I cannot answer this question, as I don't know enough of the nickel chemistry.
Then you say that aluminum oxide reacts with acids. It is right, but this reaction is extremely slow. If some aluminum piece or foil is put into a solution of $\ce{HCl}$ made by mixing the concentrated solution ($36$%) by an equal amount of water, you will have to wait a couple of minutes before the oxide layer is dissolved according to : $$\ce{Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2}$$ And this last reaction is so slow that the expected reaction producing hydrogen gas happens only two or three minutes later according to $$\ce{2 Al  +  6 HCl  ->  AlCl3  +  3 H2}$$ In this case, the production of gas is violent and highly exothermic, but it only happens with delay.
At the end, you ask why "I even got a voltage in the aluminum cell if the oxide layer does not allow any aluminum ions to be transferred". This is a good question, for which I don't have a real answer. It may be due to some default of the alumina layer.


Does latex protect against acids? I heard somewhere that some plastics are pretty good at resisting decently strong acids. To what degree is this true? Does latex protect against corrosives? Are plastics good at "resisting" bases or other strong solvents?

I'm not a chemist, so I apologize if my question seems dumb. I couldn't find much info on the subject of latex vs corrosives via Google.

Also, I don't plan on handling any dangerous chemicals; I'm just curious.

There are all sorts of resources for picking the best gloves for handling any given type of chemical. This resource describes latex as being good for protecting against bases, alcohols, dilute water solutions, and fair against aldehydes, ketones.

Indeed if you check the Glove Type and Chemical Use table towards the bottom of the page, you can see that latex fairs pretty well against most acids. These ratings give you an idea of how long it takes to penetrate the glove when exposed to that chemical. As you can also see from this table, neoprene gloves are slightly better for handling acids, and though it's not shown on the table, PVC gloves are also good.


Why does covalent nature in ionic solids increase the lattice energy? I have observed in my textbooks that ionic solid that have a greater covalent character tend to have higher lattice energy. Can this be generalised as fact?

(First of all I have to note that I'm just a student, not a specialist in solid body physics or chemistry)

Let's consider chemical bond in general, starting with diatomic molecule (our findings for this case we will generalize to crystalline structure later):

When we say that chemical bond is formed, we mean that for electrons of each atom by atomic interaction (interaction of atomic orbitals of each atom to each other) become available to be situated on molecular orbitals(MO) (defining as superposition with different contributions of atomic orbitals(AO)). Now we must say that depending from atomic orbital energy and its symmetry, result of interaction with another AO may be different.
If we have two interacting AOs with big (bigger than stated value) difference in energies it leads to formation of two MOs: one of them, bonding MO, must have energy closer to such AO that is lower by energy, another relatively must have energy closer to another AO (see picture). In this case we have low energy of stabilization (system weakly feel energy changing before and after interaction).
If difference as big as whole system looks like we give both electrons to one of AOs it means that electron density of bond is shifted to one of atoms. In this case we say that we deal with IONIC bond.
Now COVALENT bond: energies of interacting orbitals are close. This way we have big stabilization energy (see picture). Because of it we can say that covalent bond is stronger then ionic.

Simplifying: ionic bond is provided by Coulomb interactions (in approximation), but covalent bond is provided not only by them, and this is the reason why it's stronger.

Generalization to crystalline lattice: now we must consider group of interacting orbitals, but main findings remain the same: bond forming by AOs with same energy is stronger.

And, as lattice energy is energy of interacting particles forming lattice, it's clearly determined by MO's energy.

That's why I think yes, we may generalize your finding as a fact.


What is the ground state electron configuration of isoprene? Also the electron configuration of isoprene at its singlet state and also that of the triplet state?

The ground state is a singlet, when excited by UV light a $\pi\pi^*$ singlet excited state can be produced. (Singlet has electron spin paired with one electron in $\pi$ HOMO orbital one in LUMO [or higher orbital for other excited states]). The singlet may then form a triplet or decay to the ground state by emission/ internal conversion/reaction as per normal. In some polyenes the lowest excited state may be symmetry forbidden from the ground state, you will have to look into the details for each molecule.


Is melting of Tar for road construction a reversible change? In India roads are made up of a black liquid which is prepared by melting a black material(probably coal tar). After cooling the liquid becomes very rigid. Can this solid be converted back to its previous liquid state by heating?

P.S: This is a question in my book. I think that after getting solidified the chemical composition of Tar may change so it can't be converted back to its original state.

Tar
 is mainly made up of a collection of very heavy hydrocarbons, with
rather high melting temperatures. Commonly, it is mixed with some stones
 or gravel to make a harder surface for roads. It is possible to heat
this up to soften or melt this; you can try it yourself by heating it in
 a metal can over a stove (don't use a container that you plan to keep
clean) or in a wood or (char)coal fire.

EDIT: As MaxW pointed out, it would be a bad idea to do this inside the house because it would create a strong smell, similar to that created by whatever machine they used to make the road.


Belousov-Zhabotinsky reaction: questions about rate determining step, k and activation energy In one of our lab courses, we performed a variation of the BZ reaction, the experimental procedure is outlined below:

Experimental Procedure

10 mL of 3.0 M $\ce{H2SO4}$ was added to test tube T1.
Test tube T1 was placed on a stand and immersed in the water bath maintained at 35 °C
5.0 mL of 0.25 M $\ce{KBrO3}$ was added to test tube T2.
2.5 mL 0.02 M $\ce{Ce(NH4)4(SO4)4}$ in a 1.0 M $\ce{H2SO4}$ aqueous solution was added to test tube T3.
5.0 mL 1.0 M $\ce{CH2(COOH)2}$ was added to test tube T4.
Test tube T2 and T3 were added to test tube T1 and then the contents were swirled.
Test tube T1 and T4 were put into the water bath for 10 minutes.
While test tube T1 was still in the water bath, the contents of test tube T4 was put into test tube T1 and the timer was started.
Changes were observed while the tube T1 was vigorously shaken in the water bath.
The length of the induction period was recorded.
The reaction was allowed to oscillated 10 times, and the length of the period of oscillation was recorded.
The experiment was repeated at $\pu{40 ^\circ C},~\pu{45 ^\circ C},~ \text{and}~\pu{50 ^\circ C}$.

Also, here's the data set that I obtained:

Now, based on the instructions in our lab manual I plotted the following curves:

Now, my questions:

The manual instructs us to calculate activation energy using the period of induction and the period of oscillation from the Arrhenius plots above, which I calculate as $\pu{69.3 kJ/mol}$ (oscillation period) and $\pu{51.4 kJ/mol}$ (induction period).
But in doing so, we utilise the fact the $k = \frac{1}{\mathrm{period}}$ (i.e., $\ln(K)$ corresponds to $\ln(\frac{1}{\mathrm{period}})$. We were just given this information with no explanation. I can intuitively guess that for a first order reaction $k$ has the dimension $\mathrm{s}^{-1}$, but is there a stronger reason for making this selection?
I would guess the rate-determining step occurs during the induction period (since it is the slowest), but then the activation energy is lower. I would think that the energy barrier of the rate-determining step should be higher. Am I seeing a relation where none exists?
I noticed that the period of the reaction changed when the reaction mix was stirred more vigorously (temperature  and starting concentration being the same). Is that possible? If that is the case, then does that mean that the period isn't a reliable measure?

I realise your homework was probably due a LONG time ago but I just stumbled across this.

In answer to (3) yes, the reaction is notoriously sensitive to a wide range of conditions and stirring is one of them...when the system is stirred faster (probably generating a vortex) oxygen is more readily available to react with radical species in the reaction - effectively these will react with oxygen rather than the bromide. As such the bromide in the system is above the threshold level for oscillations to occur - if you were to slow this down again then you'd see the oscillations resume.

Also I don't get why they'd make you use cerium sulfate...everyone knows ferroin is better.


Relaxed scan of Cp-M torsion angles of piano stool compounds As part of a conformation analysis of a library of piano stool complexes to find global minima and TS, a relaxed scan where the legs of the piano stool complex rotated about the Cp-M bond need to be performed (Here a $\ce{[(Me-Cp)RuClH(CO)]}$ is given here to illustrate the task in question) :

However it was found that Gaussian 09 or 16 does not allow dummy atoms in redundant coordinates and cannot build dihedrals where one of the atoms is a dummy atom as far as Generalised Internal Coordinates (GIC) goes, because atoms are not considered an internal coordinate. Also unlike this question, since the rotation has to be done about the Cp-M bond,the workaround in that question will not work.
In addition, none of the existing dihedrals can be added together to give a dihedral where the middle bond will be aligned to the Cp-M axis. This means, the dummy atom for the centre of the Cp is necessary.

What are the possible workarounds that can achieve an outcome equivalent to the relaxed scan of the Cp-M dihedral?

Are there other chemistry packages that allow relaxed scan using dummy atoms or (better) user defined rotation axes?

I would use the dummy, define the 3 X 10 11 (16 and 18 too) dihedrals in the connectivity matrix, run the optimization and later write a shell script for running subsequent calculations in separate files replacing the value of the previously defined dihedrals. So I guess my solution is only slightly above doing it by hand but so far I haven't found any software that does this and G09 has no workaround.

I hope this helps


How to determine the iron content of a screw? In Analytical Chemistry, I was asked to determine the iron content of a screw.

I thought it could be solved in the following way:

Treat the screw with a solution of sulfuric acid or hydrochloric acid.
As the iron in solution can be found in the form of $\ce{Fe^{II}}$ or $\ce{Fe^{III}}$, I would reduce the possible $\ce{Fe^{III}}$ ions with $\ce{SnCl2}$ or Walden's reducer.
Titrate the solution with potassium permanganate.
From the volume spent, the amount of iron present could be determined.

I don't know if it will be OK, it is the first time I have faced a problem of this type.

There is another problem with this.

Potassium permanganate is a very strong oxidizing agent, which is capable of oxidizing Chloride ions to Chlorine. This will result in additional consumption of $\ce{KMnO4}$ solution, which makes you think there is more iron in your screw than there actually is.

To solve this, you should look up the titration of Iron using the Reinhardt-Zimmermann solution. It contains sulfuric acid, Manganese (II) sulfate and phosphoric acid.

The $\ce{MnSO4}$ changes the redox potential so the acidic $\ce{KMnO4}$ solution can no longer oxidize Chloride ions. Because Iron(III) gives an intense yellow solution, the addition of Phosphoric acid is necessary since it will react with Iron(III) to colorless Iron phosphates. That makes it easier to detect the end point.

So the way I did this back in my first semester was to make sure we have all Iron as Fe(II) by adding an excess of $\ce{SnCl2}$. The excessive $\ce{SnCl2}$ was quenched using $\ce{HgCl2}$. Then the Reinhard-Zimmermann solution was added and titration with $\ce{KMnO4}$ could start.


What is the limiting value of band gap that makes a material semiconductor? What is the limiting value of band gap (in eV) that makes a material semiconductor? Or what is the value of band gap that separates insulator and semiconductor?

The transition of electron from the valence band to the conduction band is a probabilistic event.  Electrons get this excess energy equivalent to the band gap ($\Delta E$) from the thermal energy ($kT$, where $k$ is the Boltzmann constant and $T$, temperature in Kelvin). Hence, the probability of such an event will be proportional to $e^{-\Delta E/kT}$.

Answering your question, although there is no standard defined value of band gap, you can see since it's an exponentially falling smooth curve that one can define a cutoff $\Delta E$, saying that for band gap $> \Delta E$ there are practically no electrons transiting to the conduction band, qualifying the material as an insulator (i.e., exhibiting insulator-like properties).


Why is thionyl chloride preferred for preparing alkyl chlorides from alcohols? Why is thionyl chloride ($\ce{SOCl2}$) preferred for preparing alkyl chlorides from alcohols, as opposed to say $\ce{PCl5}$ or $\ce{PCl3}$ ?

Thionyl chloride is preferred for preparing alkyl chlorides from alcohols because the by-products formed in the reaction are $\ce{SO2}$ and $\ce{HCl}$ which are in gaseous form and escape into the atmosphere leaving behind pure alkyl chlorides.

$$\ce{CH3CH2-OH  + SOCl2   →  CH3CH2-Cl  +  SO2 ↑ + HCl ↑}$$


Why is 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) often considered a prototypical room temperature ionic liquid? It seems from a brief search of the literature that 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) is a prototypical room temperature ionic liquid (RTIL) that has been studied extensively.  However, numerous -- almost limitless -- combinations of cations and anions that give rise to ionic liquids are possible.  Why does it seem that the imidazolium cations are so extensively studied?  Is one possible reason the fact that $\text{EMI}^+$ is aromatic and thus affords the ionic liquid a high degree of stability?

Similarly, why does it seem that $\text{BF}_4^-$ (along with, perhaps, $\text{PF}_6^-$) is the prototypical ionic liquid anion?  I have seen studies of other anions (for example, $\text{CF}_3 \text{SO}_3^-$ and $(\text{CF}_3 \text{SO}_2)_2 \text{N}^-$), but $\text{BF}_4^-$ seems to be one of the most popular.

This is a very good question. These compounds have been studied for long enough that most people working with them "know" the answers to these questions, but the answers are buried in physical chemistry articles from 20 or more years ago. The imidazolium tetrafluoroborate salts are popular RTILs because 1) they have the "right" properties, and 2) they are easy to make.

Concerning these compounds having the "right" properties:
Most RTILs require two characteristics in the their constituent ions to minimize the intermolecular order which would cause them to solidify. First, they need a cation-anion pair that cannot associate strongly, and second they need to have lower symmetry. Immidazolium tetrafluoroborate (BF4-) salts have both of these. The aromaticity of imidazole is important. The positive charge of the imidazolium cation is distributed through the resonance of the aromatic pi system. The negative charge in BF4- is distributed inductively over the four fluorine atoms. Additionally, imidazolium is planar and BF4- is spherical. Both the diffuse charge distribution and poor shape match prevent these two ions from having a strong association. The two different alkyl groups (ethyl and methyl) on the imidazolium ring are enough to break the symmetry of the system and prevent more efficient packing, which would lead to the substance being solid.

Concerning their ease of synthesis:
The imidazolium salts are easy to prepare from imidazole and various alkyl halides, all commercially available and inexpensive. The predominance of the BF4- anion is also a result of the synthesis. The synthesis of an imidazolium salt RTIL begins with imidazole and and an alkyl halide, say chloromethane. The product of this step is 1-methylimidazolium chloride. This salt is reacted with another alkyl halide, say chloroethane, in the presence of a base, say triethylamine. The products of the this step are 1-ethyl-3-methylimidazolium chloride and triethylammonium chloride. 1-ethyl-3-methylimidazolium chloride is a solid, because the chloride ion is small, with concentrated charge. The tetrafluoroborate salt is prepared by salt metathesis. 1-ethyl-3-methylimidazolium chloride is dissolved in water and an aqueous solution of silver tetrafluoroborate (AgBF4) is added. Silver chloride, AgCl, is virtually insoluble in water (Ksp = 1.77 x 10-10) and precipitates. The solution is filtered, and the filtrate is concentrated to produce 1-ethyl-3-methylimidazolium tetrafluoroborate (EMIM BF4-). AgBF4 and AgPF6 are commercially available, so you will see those ions more frequently. The other ions, while they have good properties for ionic liquids, are not usually available as silver salts, so those reagents have to be prepared. You are already doing three reactions to get to EMIM BF4-. Who wants to do more synthesis just to make their solvents?


What does the subscript n on methylene groups in a cyclic alkene mean? I've recently come across an unfamiliar notation on a cyclic alkane. One carbon had parentheses around it with a subscript $n$. Does this refer to rings of variable sizes (i.e.  one or more $\ce{CH2}$ groups are located in this position)? If not, what does this refer to?

Yes, n denotes the number of $\ce{CH2}$ groups between the parentheses.

Example:

n = 1, 2 or 4 means cyclopentane, cyclohexane, or cyclooctane.


Why laboratory glassware shatter in big pieces? What are laboratory glassware made up of? This question came up in mind because once I accidentally dropped a test tube and it shattered into big pieces. I observed that it didn't broke into pieces on first impact to the floor rather it bounced and broke into pieces on second impact. Any ordinary glass would shatter into small pieces On the very first impact to the floor. Does this type of glass possess elasticity?

Kitchen glassware is made to contain large internal stresses, so it shatters into small, non-dangerous pieces. That is hard to achieve with Duran etc., because that stuff has a small thermal expansion coefficient, just so that it does not break under large temperature changes.

I expect that the kind of kitchen glassware that is supposed to withstand large temperature changes also breaks into rather large pieces.

And yes, glass is highly elastic. Elasticity is effectively the only mechanical property it does possess, before breaking. ;-)


Why alcohols don't behave as acids? If I understand everything correctly, carboxylic acids behave like acids because oxygen is much more electronegative than hydrogen, so it takes his electron and releases H+. The problem is that alcohols have an OH group too, so why don't they behave like acids?

The figure below illustrates these two ionization reactions.  The key difference between the two is that when the carboxylic ionizes the carboxylate anion is formed.  The carboxylate anion is much more stable than the alkoxide anion formed when the alcohol ionizes.  The carboxylate anion is much more stable because we can draw resonance structures to describe it.  This resonance stabilization of the carboxylate anion explains why carboxylic acids are so much more acidic than an alcohol.


How do I write the complete and net ionic equations for Insoluble compounds? So, I was doing my homework and had to write out the complete equation and net ionic equation for a combination of Lead (II) Iodate and Sodium Hydroxide.

So my equation is given as:

$\ce{Pb(IO3)2_{(aq)} + NaOH_{(aq)} -> ?}$

However my dilemma is that both of these compounds seem to be insolvable and I'm not really sure that I can either write the complete equation or even the net ionic equation, since I have no idea what to expect. If I do it on Wolfram Alpha I get the resulting products as:

$\ce{H2 + I2 + 2Na + 4O2 + Pb}$

However I have no idea if this is right/and or what's even going on here?

While lead(II) iodate is indeed sparingly soluble in water, this is not true for sodium hydroxide (111 g/100 mL at 20 °C). The reaction that occurs is a precipitation:

$$\ce{Pb(IO3)2_{(aq)} + 2NaOH_{(aq)} -> Pb(OH)2\downarrow + 2NaIO3_{(aq)}}$$

Both lead salts are not readily soluble in water; however, $\ce{Pb(OH)2}$ has a lower solubility ($K_{sp}=1.43 \times 10^{-20}$) than $\ce{Pb(IO3)2}$ with $K_{sp}=3.69 \times 10^{-13}$. Adding $\ce{NaOH}$ to an aqueous solution of lead iodate should therefore precipitate lead hydroxide, which dissolves again with excess $\ce{NaOH}$ under formation of hydroxoplumbates.