What does the amount of acetic acid in vinegar depend on? I have different types of vinegar and I'm wondering what factors influence how much acetic acid is present in different vinegars (e.g wine vinegar/white vinegar/apple cider). I was thinking maybe it has something to do with different fermentation processes. Any thoughts?

Vinegar comes from French "vin aigre", aka sour wine. The transformation of wine into vinegar happens when some (possibly unwanted) aerobic bacteria transform the alcohol into acetic acid by oxidation. So any alcoholic mixture which allows growth of those bacteria can be turned into vinegar, like cider, sake...

The amount of acetic acid depends on the alcohol content which in turn depends on the initial concentration of sugar. It is also dependent on the time allowed for fermentation.

In fact it is much more complicated than that. Have a look at Wikipedia for more info.


Can five membered ring form through Friedel-Crafts alkylation? Our teacher told us that five membered ring can’t be formed through Friedel-Crafts alkylation but acylation is OK. Why is that? Is it because alkylation is an unfavored 5-endo-trig reaction but acylation is 5-endo-dig, thus favored?

Maybe your teacher means to say that because further reactions would happen due to the further activation of the ring due to the alkyl group as opposed to the deactivation due to the acyl group.
Also further catalyst induced polymerization reaction can occur apart from possible polyalkylation

In general alkylation on rings are not preferred because of the above reasons as well


How much heat is required to raise the temperature of 230 g CH3OH(l) from 22.0 to 30.0 ∘C? How much heat is required to raise the temperature of $230\ \mathrm g$ $\ce{CH3OH(l)}$ from $22.0\ ^\circ\mathrm C$ to $30.0\ ^\circ\mathrm C$ and then vaporize it at $30.0\ ^\circ\mathrm C$? Use enthalpy of vaporization $\ce{CH3OH(l)} = 38.0\ \mathrm{kJ\ mol^{−1}}$ and a molar heat capacity of $\ce{CH3OH(l)}$ of $81.1\ \mathrm{J\ mol^{−1}\ K^{−1}}$.

I Used the equation $Q_\text{total} = Q_\text{heat} + Q_\text{vap}$. My answer was $88.9\ \mathrm{kJ}$.

However, it is wrong.

The first thing to be sure of is that you're doing all the conversions correctly. Molar mass is the number of grams in a mol. It's found by adding the molar mass of all the elements in a compound times the number in the compound. For example the molar mass of $CH_4$ would be $$\textrm{Molar Mass of }C + 4\times\textrm{ Molar Mass of }H$$
Once you've found the correct molar mass it should have units of mass per mol - most commonly $\textrm{grams mol}^{-1}$. Thus if you have the mass of a substance and the mass required to have a mol, you can divide mass by molar mass to find moles. Then you simply:
$$Q_{heat} = \Delta T\times \textrm{ number of moles } \times C_p$$
Where $C_p$ is heat capacity. To find the heat of vaporisation again you just multiply the per mol value by the number of moles:
$$Q_{vap} = \textrm{ number of moles } \times \Delta_{vap}H^\circ$$
Where $\Delta_{vap}H^\circ$ is the heat of vaporisation. Be sure you got all these steps right and be aware of the difference between kJ and J! ($1\textrm{ kJ} = 1000\textrm{ J}$) Using this method I got a very different answer to yours. Feel free to post your own working for us to look at!

Note: Sometimes enthalpy of vaporisation values are actually the energy to vaporise from 298K (24.5C) but this can't be your problem because, as Jason Patterson said, even with just the enthalpy of vaporisation, no heating, your value is too low.


Can you electrolyse a single H2O molecule? I don't know why I'm asking or why you'd want to but I'm just curious.

Every resource I read about electrolysis says the reaction is $\ce{2H_2O (l) \rightarrow 2H_2 (g) +O_2 (g)}$.

Can you electrolyse a single molecule?

In a traditional experimental setup this would never happen, as there are always several water molecules. But lets just think about it for the sake of it.

If you wanted to design an experiment where you only have one water molecule, you would have to get two electrodes extremely close together. Lets say you find a way to levitate the molecule between the two electrodes. You might be able to then apply a voltage and tear the molecule apart, forming one H$_2$(g) molecule at the cathode. The oxygen radical would then most likely immediately react with the anode to which it is attracted by the Lorentz force and form an oxide.

So the result of electrolysing several single molecules of water might result in gaseous hydrogen and no oxygen, but a buildup of oxide at the anode.

However, having two electrodes so close together would probably create a series of problems. Not only to hold the single water molecule in place (which could be achieved by optical levitation) but because quantum mechanical effects such as tunneling currents would occur. Probably there are several more complications that would by all means make it nearly impossible to actually observe this. The knowledge gain derived from such an experiment would likely not justify the effort and expenses.

EDIT

I just revisited this question and thought about it again:
it's not absolutely necessary that the two H atoms would form an H$_2$ molecule. Depending on the way the cathodes are formed, the electric field might not pull the charged particles together, but to a collision with something else in the vicinity; the cathode or any ambient gas.
Furthermore, my notion above that "H$_2$(g)" would form is a bit nonsensical, as a single molecule cannot really be attributed a state of matter (i.e., being gaseous in this case).


What happens when a symmetry element is added or removed from a point group? What new point group is formed if symmetry element i is added to C3 point group?

Inversion $i$ is equivalent to a two-fold improper rotation $S_2$. Introducing a $2n$-fold improper rotation axis to the existing $C_n$ point group will result in a new $S_{2n}$ point group, e.g. in your case $C_3 ⊗ i = S_6$.


How can I turn Pine-Sol into an oil? I'm wanting to make  a Pine-Sol scented candle.  However, in candle making you have to add a fragrance oil.  People make them all the time by adding vanilla, etc and canola oil, vegetable oil, etc.  I'm wondering what proportions I need to do this with Pine-Sol since it has chemicals in it??  I'm using the Original Pine-Sol - it contains 8.7% Pine oil in it if that helps.  That's all I got from the label.

Thanks for your help!
Mandy

I can answer the chemistry part of your question... the candle-making part is better suited to a candle making forum.

There is no easy way of converting the other 90% of (older) pine-sol into pine oil and newer pine-sol has 0% pine oil. Best to purchase pure pine oil.

Pine oil is cyclic:

The other pine-sol ingredients are esters and other linear compounds. So using pine-sol to make pine oil would be ridiculous.

I am not certain how much pine oil is needed for a candle to be scented though. It may just be enough. For that you will have to just experiment... or ask on a candle forum.


Why are potassium organic salts ever preferable to the sodium equivalents? $\ce{KOH}$, $\ce{KCN}$, $\ce{KNO3}$, $\ce{K2CO3}$, and plenty of other potassium salts are frequently used in industry.  But, as far as I can see, the sodium equivalents are also widely used, often interchangeably.  Since Na weighs only 60% of K it seems like the sodium salts should always be preferred because they carry more of whatever base is attached per unit weight (i.e., the sodium salts have "more of what you want"). $\ce{Na}$ is also slightly more abundant (on earth) than K.

So what other factors favor the industrial use of K-salts?  E.g., are K-salts just more abundant in natural sources?  Are they more reactive because the ionization energies of K are lower than those of Na?  Are the behaviors of these simpler salts actually different between the K- and Na- variants, and if so, is there some other elemental characteristic that suggests this?

Going the other way, it seems like if K is preferable to Na, Rb should be even more so (because it appears that in every elemental characteristic in which K exceeds Na, Rb exceeds K).  Would it be if it weren't so relatively scarce?

The differences in the properties of the alkali cations are subtle, and the rationale for choosing one over the other will depend heavily on the application. I suspect you're right, that to some extent natural abundance of, e.g., potassium salts relative to sodium ones in certain mineral contexts explains the industrial preference for one or the other -- but, really, variations in abundance are just further examples of these subtle differences in the properties of the cations!

One specific example that I recently came across is the Hofmeister series (see also the Wikipedia article), where the choice of cation makes a demonstrable difference in the solubility properties of proteins.  Speaking also from personal experience, the choice of, e.g., $\ce{Na2SO4}$ versus $\ce{K2SO4}$ versus $\ce{(NH4)2SO4}$ as a component of the supporting electrolyte in electrodeposition of copper from acidic $\ce{CuSO4}$ matters significantly to the properties of the deposit obtained.


Meaning of the term "Activation of alkene" While reading organic texts, I have come across authors referring to "Activation of alkene" what does that mean !? Does it mean to include the alkene in resonance or what else exactly ?
I saw this notation, while reading about Heck reaction:

The palladium-catalyzed C-C coupling between aryl halides or vinyl halides and activated alkenes in the presence of a base is referred as the "Heck Reaction"

I also observed this while reading about nucleophilic conjugate addition:

Ordinary nucleophilic additions or 1,2-nucleophilic additions deal mostly with additions to carbonyl compounds. Simple alkene compounds do not show 1,2- reactivity due to lack of polarity, unless the alkene is activated with special substituents.

Activation of an alkene just means that the double bond has a higher electron density than that of a normal isolated double bond. That is, the electron density in the double bond is greater than the one observed in ethene $\ce{CH2=CH2}$.
Activation, in organic chemistry, generally means the compound displays a greater nucleophilic nature than it normally should due to increased electron density. For example, $\ce{-OCH3}$ group activates benzene when it forms toluidine. This makes the benzene ring more electron rich and so make it easier to react in nucleophilic reactions.


Calculating entropy: why consider a reversible path? I am reading up on entropy in a textbook and I got confused by this:

It says that to calculate the entropy for an irreversible process using heat flow, one must imagine the reversible process in which the initial and final states are the same as for the irreversible process. Why is this so and why do we need to think of it in the reversible way?

There are two parts to this answer.  The first is that the entropy change is defined only for reversible processes.  The second is that the entropy is a state function.  Many, but not all, the functions in thermodynamics (including the internal energy) are state functions.

A state function is one that is independent of path.  That is, if one goes from state $A$ to state $B$, the change in internal energy is independent of how the change takes place.  The same is true for the entropy.

So the entropy change in going from state $A$ to state $B$ is independent of how the change takes place.  But we can calculate the change ONLY for reversible changes.

So we mentally find a reversible path from state $A$ to state $B$ and calculate the entropy change.  Since entropy is a state function, that must also be the change for a non-reversible path.


Why is enthalpy and not heat released by system used while calculating entropy of surroundings? While finding the sum of change in entropy of the universe and thus defining Gibbs free energy, why is the change in entropy of surrounding the negative of enthalpy of the system divided by the temperature?
I mean why are we considering the pressure to be constant ( thus considering enthalpy) in all systems while calculating Gibbs free energy ? Shouldn’t a more general form of entropy of surroundings be negative of heat released by system divided by the temperature?
I was thinking that maybe it is because all closed and open systems are constant pressure systems and change in entropy of surroundings is zero for isolated system but I don’t know, I am just guessing.
So, could someone please help me in understanding why enthalpy and not just heat in general is considered for any system while calculating Gibbs free energy?
Any help will be greatly appreciated!

Because $q=\Delta H$ only if only expansion work is done and the pressure of the system is constant. In that case, for an isothermal process, in which T is the same for system and surroundings, the entropy of the surroundings can be equated with the change in enthalpy divided by T. Which is why the applicability of the integrated formula $\Delta G = \Delta H -T \Delta S$ is restricted to constant p and T. You could write $q_p$ in place of $\Delta H$ the equation but only when work is restricted to pV work and when T and p are constant. It would not hold generally since sometimes work other than expansion work is performed. In general $\Delta H = q_p+w_\textrm{non-pV}$ at constant T and p.
To show this more explicitly, for a small (differential) change:
$$dU=dq+dw$$
$$dU=dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}}$$
$$dH = dU+d(pV) = dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}} + pdV+Vdp$$
At constant pressure of the system $dw_{\textrm{pV}}=-pdV$, $Vdp=0$ and
$$dH = dq+dw_{\textrm{non-pV}}$$
Further derivation is possible* to show how this relates to $G$, by assuming constant T and considering reversibility, but this should suffice to show that $dH$ and $dq$ (or $\Delta H$ and $q$) are not in general the same.

*For a reversible process $dq=TdS$ and
$$dH =  TdS+dw_{\textrm{non-pV,rev}}$$
Finally the differential expression for $G$ at constant $T$ follows (since then $dG=dH-d(TS)=dH-TdS$):
$$dG=dH - TdS = dw_{\textrm{non-pV,rev}} $$


Predicting a precipitate based on Ksp $\pu{105 mL}$ of $\pu{0.10 M}$ $\ce{AgNO3}$ is added to $\pu{125 mL}$ of $\pu{0.35 M}$ $\ce{K2CrO4}.$ Will a precipitate form? $(K_\mathrm{sp}$ for $\ce{Ag2CrO4}$ is $\pu{1.12E-12}.)$

Answer:

   Yes, $Q = \pu{4.0E-4}$ which is greater than $K_\mathrm{sp}.$

Can someone please help me out with this question? I know that I have to find $Q_\mathrm{sp}$ and compare this to the $K_\mathrm{sp}$ given, and if $Q_\mathrm{sp}$ is bigger, then a precipitate will form. However, I do not know how to find $Q_\mathrm{sp}$ when two substances are mixed together.

To obtain $Q_\mathrm{sp}$ for

$$\ce{Ag2CrO4 <=> 2 Ag+ + CrO4^2-}$$

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CrO4^2-}]$$

$$Q_\mathrm{sp} = c(\ce{Ag+})^2\cdot c(\ce{CrO4^2-})\label{eqn:1}\tag{1}$$

you need to use new concentrations of ions in solution $c(\ce{Ag+})$ and $c(\ce{CrO4^2-})$, which can be found from the corresponding values for initial concentration $c_0$ and volume $V_0$:

$$
\begin{align}
c(\ce{Ag+}) &= \frac{c_0(\ce{AgNO3})\cdot V_0(\ce{AgNO3})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &= \frac{\pu{0.10 mol L-1} × \pu{105 mL}}{\pu{105 mL} + \pu{125 mL}}\\
  &= \pu{4.57E-2 mol L-1} \label{eqn:2}\tag{2}
\end{align}
$$

$$
\begin{align}
c(\ce{CrO4^2-}) &= \frac{c_0(\ce{K2CrO4})\cdot V_0(\ce{K2CrO4})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &= \frac{\pu{0.35 mol L-1} × \pu{125 mL}}{\pu{105 mL} + \pu{125 mL}} \\
  &= \pu{1.90E-1 mol L-1} \label{eqn:3}\tag{3}
\end{align}
$$

Finally, plug in the concentrations of ions from \eqref{eqn:2} and \eqref{eqn:3} in \eqref{eqn:1} and you find a value for $Q_\mathrm{sp},$ which indeed exceeds $K_\mathrm{sp}$:

$$Q_\mathrm{sp} = (\pu{4.57E-2 mol L-1})^2 × \pu{1.90E-1 mol L-1} = \pu{4.00E-4 mol3 L-3}$$


Alcoholic KOH mechanism with alkyl halides Does alcoholic $\ce{KOH}$ only show $E_2$ mechanism with alkyhalides?

I have studied with two organic chemistry teachers, one told me that there is always $E_2$ mechanism, except in the case of tertiary alkylhalide and the other told me it can be any depending on the substrate with $\alpha$ hydrogens more than 4 or if it has resonance it will show $E_1$.

Which one is true?

With a strong base $E2$ will always predominate over $E1$.

Both elimination mechanisms have a preference for tertiary substrates over secondary substrates over primary substrates. This behavior is on contrast with the substitution reactions, where steric factors are important for $S_N2$. Steric factors are less important for the proton transfer necessary for $E2$.

Since $E2$ and $E1$ have the same substrate preference order, you may feel that you cannot decide which is more likely. However, $E2$ is usually faster because $E2$ is bimolecular. Consider the rate laws:

$\mathrm{rate}_{E2}=k\mathrm{[substrate][base]}$
$\mathrm{rate}_{E1}=k\mathrm{[substrate]}$

If we make the assumption that the rate constants are of similar magnitude, then increasing the concentration of the base increases the rate of $E2$.


Electrocatalysis of Hydrogen Evolution Reaction with metal oxides I'm new to the topic of electrocatalysis so I have a few questions. Also, I don't have much knowledge about thermodynamics, but I will try to ask anyway.

I'm looking for a material to catalyze the Hydrogen Evolution Reaction for $\ce{H2}$ production. So, for pure metals we have the Volcano curve:

Then, with this curve, I found in literature that we can mix metals from the different branches (positive and negative slopes) and get a new alloy with an intermediate M-H energy strength (which is desirable to improve the catalyst activity.)

But, I want to know if there exists the same curve (and the same logic) for metal oxides, because I need to predict if an alloy of rhenium oxides with copper oxides are going to be a better catalyst than only rhenium oxides.

Is there some method of theory to predict this before going to the lab and doing some experiments?

You can use a similar, DFT-based approach to estimating the electrocatalytic activity of oxides, as has been done with the oxygen evolution reaction, but since most oxides are not stable at reducing conditions present in HER, there's not as much literature on that particular material-reaction pairing.  As an aside, a popular theory of metal-adsorbate bond formation, the d-band model was formulated specifically for metals, so the question as to whether or not there's a model with comparable simplicity that includes oxides (or other binary compounds) is still somewhat of an open question.


Right Raman Spectrum of Cyclohexane In this site, https://www.princetoninstruments.com/userfiles/files/assetLibrary/Technical%20posters/Poster-Aberration-Free-Raman-Spectroscopy-with-the-IsoPlane.pdf the raman spectrum of cyclohexane shows higher peak at the right.

While in this site, http://group.nanovietech.com/products/advantage-raman.html, the raman is showing higher peak at the left:

Both are raman spectra of cyclohexane. My questions relate to the 2853 cm-1 peak and 802 cm-1 peak. Based on structure of cyclohexane. Which of them should be higher? Ignore the other peaks which I wasn't concerned.

I found this too:

Should the 802 cm-1 peak be higher than the 2853 cm-1 peak based on the molecular structure of cyclohexane?

First of all, you are extremely lucky in asking this question for cyclohexane: cyclohexane serves as reference material for checking your intensity calibration, so it is one of the few substances for which integrated band intensity ratios are available in the literature.

Integral intensity 2567 - 3068 cm$^{-1}$ is higher than integral intensity 700 - 900 (i.e. 801 cm$^{-1}$ band). At 532 nm excitation, the ratio is roughly 10.3 : 1.

You always have the fundamental $\nu^4$-dependence of Raman scattered intensity. If you look closely, that's $\nu_0 (\nu_0 - \nu_j)^3$ for the band with Raman shift $\nu_j$ for photon count and  $(\nu_0 - \nu_j)^4$ for power. Thus, the relative band area differs depending on excitation wavelength.
For dispersive Raman instruments, the sensitivity is highly wavelength-dependent and this varies from instrument to instrument and for instruments with moving grating also depending on the grating position.

Thus, for intensity comparisons across different instruments, you'll need to calibrate the intensity axis for the spectrometers (and possibly fold the higher-resolved spectrum with the difference in line shape if that isn't negligible due to integration of broad spectral ranges)
However, when training chemometric models (calibration, quantitation, classification, ...) for one instrument (or maybe very similar instruments: say, same manufacturer & model), as long as that instrument is stable intensity calibration is not necessary as it will automatically become part of the model.
Which is why many Raman spectra are shown as raw (uncalibrated) intensity.

This doesn't have much to do with "highly sophisticated" Raman spectrometer rather than doing sophisticated calibration or not. And among the highly sophisticated Raman spectrometers, you have tradeoffs that can make such a calibration more or less stable. In fact, different instruments are highly sophisticated in different aspects (e.g. spatial resolution, spectral resolution, signal stability/drift, versatility/possibility to adapt to different experimental setups, ...) some of these influence how easy intensity calibration is and/or how often it needs to be re-done.

Literature:

McCreery: Photometric Standards for Raman Spectroscopy, in: Handbook of Vibrational Spectroscopy, Wiley 2006
McCreery: Raman Spectroscopy for Chemical Analysis, Wiley 2000


Measurement of cell constant Why is cell constant value measured through indirect methods (like using a solution whose conductivity is already known)? Isn't it easier to  measure the length and area of cross section of a conductivity cell directly?

The mathematical description of the cell constant ($l/A$) does not include factors like fringing fields (the ionic current does not travel precisely in a well-defined path between the electrodes) and the actual surface area for rough surfaces.
The cross section of a round electrode may have different fringing fields depending on its separation from its companion. And if the two electrodes are of different geometry, you have to decide how to average the areas out. I think there was more standardization in the choice of electrode materials, size and spacing than has been carried over into the beautifully simple formula $l/A$, so when the formula is used outside of certain narrow bounds (and I don't know what they are), it's only an approximation, and resistance or conductance can be measured more precisely than the cell constant from geometry. So you might as well calibrate against a known cell or solution.


Does water ionically bond to chloride ion? This user absolutely insists that when a chloride ion is present in water (for example, when $\ce{NaCl}$ dissolves in water) that $\ce{Cl-}$ ion is ionically bound to the hydrogen atom:

https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55324_34449

I then insist that this would form $\ce{H2OCl}$, but that's impossible due to the lack of valence electrons of $\ce{H}$:

https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55386_34449

Who is right?

The most common mechanism for when $\ce{NaCl}$ dissolves is as folows:

As water is a polar molecule, as per the UC Davis ChemWiki page All About Water, shown below:

The caption reads:

  the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule.

So, when $\ce{NaCl}$ salt is dissolved in water, the sodium and chloride ions are pulled apart by the water molecules with the result explained in the image below:

Image source

You'll note, that there is no ionic (nor covalent) bonding present, rather an ion-dipole interaction between the water molecules and the sodium and chloride ions (University of Waterloo).

The UCLA page Illustrated Glossary of Organic Chemistry has a very nice definition of the ion-dipole interaction:

  A noncovalent attraction between one pole of a bond dipole and an oppositely-charged ion. The δ+ end of the bond dipole is attracted to an anion, and the δ- end of the bond dipole is attracted to a cation.

There is a neat YouTube clip showing the process here.

The ion $\ce{H2OCl+}$, does actually exist according to NIST, but I could not find any references that link it to the dissolution of $\ce{NaCl}$ salt.


Degradation of sulfuryl fluoride SO2F2 Recently, the houses next to us were fumigated with "Vikane", which apparently is $\ce{SO2F2}$. To my surprise, the gas was just pumped out in the atmosphere without further ado at the end of the procedure. I was told by the pest control people that this gas is "degraded by light" and that there is no hazard whatsoever for humans living close to the house. However, I've also read that sulfuryl fluoride is a potent greenhouse gas staying in the atmosphere for up to 40 years which it probably wouldn't do if it were deteriorated easily by light. Which is it??

I think this pretty much comes down to three things:
1) The estimates of the atmospheric half-life of $\ce{SO2F2}$ has changed over time.
2) The current estimate of 30-40 years isn't a particularly long time relative to some other players (i.e. methane's atmospheric lifetime is around 6-8 years and that of carbon dioxide is around 50 years, though there is poor agreement on that one).
3) The emissions relative to the other greenhouse gas players is small enough to offset the fact that it is "potent" in this aspect.

From the sulfuryl fluoride Wikipedia page:

  Based on the first high frequency, high precision, in situ atmospheric
  and archived air measurements, sulfuryl fluoride has an atmospheric
  lifetime of 30–40 years, much longer than the 5 years earlier
  estimated.

That addresses my first item.  From the same page regarding my second and third items:

  Sulfuryl fluoride has been reported to be a greenhouse gas which is
  about 4000-5000 times more efficient in trapping infrared radiation
  (per kg) than carbon dioxide (per kg). It is important to
  note, however, that amounts of sulfuryl fluoride released into the
  atmosphere (about 2000 metric tons per yr) are far lower than the
  amounts of $\ce{CO2}$ released by hydrocarbon-burning vehicles, industry, and
  other processes (about 30 billion metric tons per year).

That means that annual carbon dioxide emissions will trap 3000 times as much heat as will sulfuryl fluoride.  So the pest control people may not have worded it quite right, but the bottom line is that sulfuryl fluoride emissions have a negligible effect on the planet's heat budget.


How to find the mass spectrum of Cl3+? The setting of this problem is on the moon of Jupiter: Io.

  Given that chlorine exists in two isotopes: $\ce{{}^{35}_{17}Cl}~(33.3\%)$ and $\ce{{}^{37}_{17}Cl}~(66.7\%)$, what is the mass spectrum of $\ce{Cl3+}$?

What I know now is that there are 3 peaks in the mass spectrum of $\ce{Cl2+}$ because there are 3 possible combinations ($\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{35}_{17}Cl}$ / $\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$ / $\ce{{}^{37}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$). Therefore, there are 4 possible combinations and 4 peaks for $\ce{Cl3+}$.

However, I do not know how to make a mass spectrum graph with only this much information.

Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a $\ce{{}^{35}_{17}Cl}$ is $a=33.3\%\approx\frac13$ and the probability for finding a $\ce{{}^{37}_{17}Cl}$ is $b=66.7\%\approx\frac23$. Since you have three positions to fill, this will result in a binomial formula of third order, $(a+b)^3$. You can solve this easily with Pascal's triangle, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. You spectrum should therefore look like this table:
\begin{array}{lrr}\hline
   & \text{m/z} &\text{abundance}/\%\\\hline
\ce{[{}^{35}_{17}Cl_3]^{+}}                & 105   &  3.7  \\
\ce{[{}^{35}_{17}Cl_2{}^{37}_{17}Cl]^{+}}  & 107   & 22.2  \\
\ce{[{}^{35}_{17}Cl{}^{37}_{17}Cl_2]^{+}}  & 109   & 44.4  \\
\ce{[{}^{37}_{17}Cl_3]^{+}}                & 111   & 29.7  \\\hline
\end{array}

The following snowball diagram should further demonstrate this:


Why do significant figure rules contradict each other? So I came across a problem in my chemistry textbook that brought up another question. The problem had to do with calculating the mass% composition of chlorine in a certain molecule.

So to go about solving this, you could calculate the mass in grams of 1 mole of the molecule and divide the mass in grams of the corresponding amount of moles of chlorine by that number and multiply that by 100 to get a percent.

The molecule had 3 Cl molecules, so I would look and see that the molar mass of 1 mole of Cl is 35.45g. So I would multiply that by 3 to get 106.35g of Cl, but I would have to round that to 4 significant figures since 3 has infinite significant figures, and the mass of Cl was given in 4. Which means the mass of 3 moles of Cl is 106.4g.

Which poses my question... Isn’t multiplying the molar mass of Cl by 3 the same as adding the molar mass of Cl 3 times? And if so, wouldn’t that mean that if you were to add the mass 3 times, then the significant figure rules for adding would tell us to round to the number in the problem with the least decimal places, which is 2 decimal places? So when you add the mass of Cl 3 times you get 106.35g so then you round to 2 decimal places and get 106.35g.

Why does this give a different answer? How come when you multiply by a number with infinite significant figures, the number of significant figures in your rounded answer can be different than if you were to add that something to itself that same amount of times?

Can’t you be certain to the tenths place in this example, and have the hundredths place by your uncertain digit since when you add, you don’t have to round any numbers off? If not, why not?

Bear in mind the significant digit rules are just  simplifications of the rules of absolute (+, -) and relative (*,/) variance propagation.

The rule of significant digit numbers for summing is for 2 statistically independent variables. Their  correlation coefficient = 0 and the error is $$  s_\mathrm{C}= \sqrt{(s_\mathrm{A})^2+(s_\mathrm{B})^2}$$
and therefore about at the same order as for less precisely known variable.

Multiplying a variable by number means summing the same variable $N$ times, with the correlation coefficient 1 and the error is growing linearly by $N$. As here applies the multiplicative  rule for propagation of relative variance.

$$ s_\mathrm{C,r}= \sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{B,r})^2}\\=\sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{3})^2}=s_\mathrm{A,r}$$

I agree with the comment to follow the rule of significant digits on the final result only, to avoid cummulative rounding errors.

On the other side, it can be hard to track the digits but for simple formulas. Applying the variance propagation rules may be easier way to go.


How to visualise the total electron density from a Gaussian 16 calculation? I am trying to generate a 3D visualisation of the total electron density of a molecule. Using Gaussian 16, I have optimised this structure, with DFT, and formatted the resulting .chk file into a .fchk using the formchk utility. I have used the cubegen utility:

cubegen 0 fdensity=scf molecule.fchk molecule.cube -5 h <0.20_3d_cube.xyz

to generate a .cube file with 0.20 angstrom spacing about a cube of the molecule.

However, the resulting .cube file fails to visualise any total electron density in visualisation software. Gaussview 6 fails to create a surface, prompting an error:

  Surface build failed! Processing mesh, No data to plot. IsoValue may be out of range

VMD allows the visualisation of an isosurface, but visually nothing appears.

When GaussView gives you the error

  Surface build failed! Processing mesh, No data to plot. IsoValue may be out of range

it means that the data in your .cube file doesn't have values of density at that isovalue. So when it tries to plot a surface, it cannot plot anything.

The IsoValue (default is 0.00040, I believe) means that you construct a surface at every place in the density where the density equals the IsoValue ('iso' = 'same'). So for the default, it opens up your .cube file and looks for the cube coordinates that have density at (or near) 0.00040, and plots a surface that connects these points (this is your 'mesh').

Most likely, your density values are too small large to plot an isosurface at this value. Try making the default density isovalue in GaussView to be something smaller larger and go from there. (Practically, I'd keep shifting an order of magnitude until a surface builds, e.g. try a value of 0.004 next).

Even better, you should look into the .cube file itself to see what your volumetric data looks like. The first few lines define the grid spacing and dimensions of the cube, but then you will see a bunch of floating point values. Here's an example .cube I took from this site (also a good reference):

CPMD CUBE FILE.
 OUTER LOOP: X, MIDDLE LOOP: Y, INNER LOOP: Z
    3    0.000000    0.000000    0.000000
   40    0.283459    0.000000    0.000000
   40    0.000000    0.283459    0.000000
   40    0.000000    0.000000    0.283459
    8    0.000000    5.570575    5.669178    5.593517
    1    0.000000    5.562867    5.669178    7.428055
    1    0.000000    7.340606    5.669178    5.111259
 -0.25568E-04  0.59213E-05  0.81068E-05  0.10868E-04  0.11313E-04  0.35999E-05
      :             :             :           :            :            :
      :             :             :           :            :            :
      :             :             :           :            :            :
        In this case there will be 40 x 40 x 40 floating point values
      :             :             :           :            :            :
      :             :             :           :            :            :
      :             :             :           :            :            :

Look at the value of your floating point numbers at the end. As a starting point, pick a number for your isovalue that is about an order of magnitude less than your largest values, and adjust from there.

If your values are all at or near zero (so, practically speaking, on the order of 1E-8 or lower) then you need to revisit how you are generating the .cube file, since your cube is not capturing any density---then it's a cubegen issue not an isosurface issue.


How can I compute molar concentration if I am given parts per trillion? A question in my book asks...

You have a gas X that is found in quantities of $261~\mathrm{PPTV}$ (parts per trillion by volume). Compute the molar concentration at $283.15~\mathrm{K}$ and $101325~\mathrm{Pa}$.

Would you just find the volume using $\frac{n\mathcal{R}T}{P}$ for $1~\mathrm{mol}$ of gas, then multiply this answer by $261/10^{12}$?

I have the feeling, that this is a superbly stupid question. And I also have the feeling that it is quite possible, that my answer will be wrong.
First of all the parts-per-notation should be avoided as it is not compliant with SI and highly ambiguous.
Secondly, it generally may only refer to a unitless number - a molar concentration has a unit ($\mathrm{mol/L}$).

The here used unit may refer to a couple of things. In general its meaning should most likely refer to $\mathrm{PPTV = 10^{-12}L/L}$. Sometimes the use of $\mathrm{PPTV = 10^{-12}g/L}$

From here you can probably make your best guesses, determining how many moles of compound do you have:
\begin{align}
pV &=n\mathcal{R}T\\
n  &= \frac{pV}{\mathcal{R}T}\\
n  &= \frac{101325~\mathrm{Pa}\cdot 261\cdot10^{-12}~\mathrm{L}}{8.314~\mathrm{\frac{J}{mol\cdot K}}283.15~\mathrm{K}}\\
n  &= 1.12\cdot10^{-8}~\mathrm{mol} &\implies c&=  1.12\cdot10^{-8}~\mathrm{\frac{mol}{L}}
\end{align}

If you have the mass given, then you need to know which gas it is, as $n=\frac{m}{M}$.

Another possible use could be the particles by volume notation, so it could also refer to $\mathrm{PPTV = 10^{-12}/m^3}$. (Apparently this is somewhat common in ecological air measurements.) Here you simply have to use $n=\frac{N}{\mathcal{N}_\mathrm{A}}$ to figure out the concentration in $\mathrm{mol/L}$.


Basic and nucleophilic character of NaOH (aq) Sodium Hydroxide (aq) is very effective in plucking off an acidic hydrogen. How effective will it be in substituting chlorine of an alkyl hallide (let us assume that the alkyl hallide is primary) and why ?

Even if, The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factor.

Nucleophilic substitution reaction has $\ce{SN2}$ reaction mechanism.

For $\ce{SN2}$ favors the reaction with primary substituted alkyl halide more than secondary than tertiary, because its effected by steric hinderance. The more crowded it is, the more difficult it is for a nucleophile to attack.

So, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution.

For a given halogenoalkane, to favor elimination rather than substitution, use:

higher temperatures
a concentrated solution of sodium or potassium hydroxide
pure ethanol as the solvent

To favor substitution rather than elimination, use:

lower temperatures
more dilute solutions of sodium or potassium hydroxide
more water in the solvent mixture


Acceptable levels of HCl? For a science/engineering project we are taking pool water (disinfected), filtering to 0.1 microns, and pumping to the kitchen.  Of course there is still 5 ppm of chlorine present so we want to neutralize the chlorine.  Asorbic acid in the form of orange juice neutralized the chlorine,  but I understand that it also creates some amount of HCl. This method is used to neutralize chlorine in water being introduced into ponds, rivers, etc. so that the chlorine does not kill the fish.

Could the asorbic acid actually create enough HCl to be dangerous,  how can I test this,  and what is acceptable?  I used a standard pool testing kit to test the chlorine.

pH of lemon juice and cola is around 2-3. As long as you don't go below this point, drinking should be fine.

It is different if you consider fish living in such water. pH from several ppm would not drop below 5, which is close to rainwater, but significantly lower than normal river or sea water (6-7 and 7-8 respectively). You should consult specialized literature in the case.


Application of limiting reagents and acid-base stoichiometry I'm having a bit of trouble with the latter questions of this problem.

  $\pu{15.0 mL}$ of $\pu{1.4 M}$ $\ce{HCl}$ was mixed with $\pu{1.00 g}$ of $\ce{CaCO3}$ until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $\pu{200 mL}$ with water. A $\pu{20.0 mL}$ portion of this solution was then neutralized by $\pu{8.50 mL}$ of a $\pu{0.100 M}$ $\ce{NaOH}$ solution.
  Calculate:

  Amount of substance in excess of $\ce{HCl}$ in the $\pu{20.0 mL}$ portion
  Amount of substance in excess of $\ce{HCl}$ in the $\pu{200.0 mL}$ portion
  Amount of substance in excess of $\ce{HCl}$ which reacted with $\ce{CaCO3}$

My approach

$$\ce{2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O(l) + CO2(g)}$$

Finding the amount of substance of each reactant and hence the limiting reactant,

For $\ce{HCl}$:
\begin{align}
M[\ce{HCl}] &= \frac{\mathrm{mol}}{\mathrm{L}}\\
\mathrm{mol}[\ce{HCl}] &= \mathrm{M}\times \mathrm{L}\\
\therefore \mathrm{mol}[\ce{HCl}] &= 1.4\times (15.0\times 10^{-3}) = 0.021
\end{align}

For $\ce{CaCO_3}$:
\begin{align}
\mathrm{mol}[\ce{CaCO_3}] &= \frac{\mathrm{m}}{\mathrm{mm}}\\
\mathrm{mol}[\ce{CaCO_3}] &= \frac{1.00}{100.0869 }\\
\therefore \mathrm{mol}[\ce{CaCO_3}] &= 9.9913\times10^{-3}
\end{align}

Therefore (after dividing by stoichiometric coefficients) $\ce{CaCO_3}$ is limiting.

From here I'm not sure if what I did next is correct. I found the amount of substance of $\ce{CaCl_2}$ produced the remaining $\ce{HCl}$ in excess. Which were,

\begin{align}
\mathrm{mol}[\ce{CaCl_2}] &= 9.9913\times10^{-3} \text{ and,}\\
\mathrm{mol}[\ce{HCl}] &= 1.1009\times10^{-2}
\end{align}

(Meaning that we have $9.9913\times10^{-3}\mathrm{mol}$ of $\ce{CaCl_2}$ and $1.1009\times10^{-2}\mathrm{mol}$ of $\ce{HCl}$ in in $\pu{15 mL}$ of solution?)

From here on I'm stumped.

I understand the procedure but not the questions. So there are two reactions:

$$\ce{2HCl(aq) + CaCO3(s) -> H2CO3(aq) + 2 Cl-(aq) + Ca2+(aq)}$$
$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Cl-(aq) + Na+(aq)}$$

The first reaction has reactants in non-stoichiometric amounts. The second reaction, a titration, is done with stoichiometric amounts, yielding a neutral reaction mixture.

The strange thing is that the problem is over-determined: I can calculate the amount of HCl not neutralized by calcium carbonate either by looking at the amounts of the two (assuming the reaction goes to completion) or by the titration (assuming that neutralizing means to add sufficient NaOH to be at stoichiometry ratio with the remaining HCl).

There is also a technical detail: it would be better to boil off the carbonic acid made by protonating the carbonate. Otherwise, the analyte will contain strong and weak acid, and buffer will form during the titration (with the endpoint of the titration ill-defined: is it when the HCl is depleted, or when all acid is depleted?).

However, it seems like the question wants us to make an argument about amounts starting with the titration. We will assume that the titration and the reaction between $\ce{CaCO3 and HCl}$ go to completion, and chalk up (get it?) any inconsistencies to impure $\ce{CaCO3}$.

Amount of NaOH

$$c_{\mathrm{\ce{NaOH}}} = 0.100\  \mathrm{M}$$

$$n_{\mathrm{\ce{NaOH}}} = c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH}}}$$

$$\ \ \ =0.100\  \mathrm{M} \cdot 8.50\  \mathrm{mL}$$

$$\ \ \ =8.50\times 10^{-4}\  \mathrm{mol}$$

Amount of HCl

The amount of HCl in the 20 mL sample was $\pu{8.50e-4 mol}$ (from the titration). So the amount in the entire 200 mL sample after adding carbonate was $\pu{8.50e-3 mol}$. The amount used up by reacting with carbonate is the difference between the original amount, $\pu{0.021 mol}$ or $\pu{21e-3 mol}$, and what is left after the first reaction.

$$n_\mathrm{reacted} = \pu{21e-3 mol}  - \pu{8.50e-3 mol} = \pu{12.5e-3 mol}$$

Amount of $\ce{CaCO3}$

As the OP writes, there is a 2:1 stoichiometry (HCl vs $\ce{CaCO3}$), so

$$n_\ce{CaCO3} = \pu{12.5e-3 mol} / 2 = \pu{6.2e-3 mol} $$

On the other hand, if the sample were 100% and everything has 100% yield and we had boiled off the carbonic acid, we would expect a higher amount in 1.00 g of calcium carbonate, i.e. $\pu{9.99e-3 mol}$ as correctly pointed out by the OP.

Why the discrepancy?

Without more information, there is no way of knowing.


How is "dietary cholesterol" processed in a human body? Person eats an egg that is high in animal cholesterol ("dietary cholesterol").
What do chemical reactions play a role in breaking down it in stomach?

Most lipids, such as cholesterol (and triglycerides), are nonpolar and therefore very hydrophobic molecules. In order to be transported in watery blood, such molecules first must be made more water-soluble by combining them with proteins produced by the liver and intestine.

Lipids and proteins associate noncovalently to form lipoproteins (globular micelle like particles that consist of a nonpolar core of triacylglycerols and cholesteryl esters surrounded by an amphiphilic coating of protein, phospholipid, and cholesterol):

Lipoproteins have been classified into five broad categories on the basis of their functional and physical properties:

1. Chylomicrons, which transport exogenous (dietary) triacylglycerols and cholesterol from the intestines to the tissues.

Chylomicrons therefore function to deliver dietary cholesterol and triacylglycerols to muscle and adipose tissue and dietary cholesterol to the liver.

2–4. Very low density lipoproteins (VLDL), intermediate density lipoproteins (IDL), and low density lipoproteins (LDL), a group of related particles that transport endogenous triacylglycerols and cholesterol from the liver to the tissues (the liver synthesizes triacylglycerols from excess carbohydrates).
5. High density lipoproteins (HDL), which transport endogenous cholesterol from the tissues to the liver.

Circulating HDL acquires its cholesterol by extracting it from cell-surface membranes and converts it to cholesteryl esters through the action of lecithin-cholesterol acyltransferase (LCAT), an enzyme that is activated by apoA-I. HDL therefore functions as a cholesterol scavenger.

This enzyme transfers a fatty acid residue from atom C2 of lecithin to cholesterol with the concomitant formation of lysolecithin

The cholesteryl esters are hydrolyzed by a lysosomal lipase to yield cholesterol, which is subsequently incorporated into the cell membranes. Any excess intracellular cholesterol is reesterified for storage within the cell through the action of acyl- CoA:cholesterol acyltransferase (ACAT).

References

Principles of Anatomy and Physiology 13e Tortora, Derrickson
Biochemistry Voet and Voet: Lipoproteins


Redox reactions of halides in hexane When we have two test tubes and we add:

In the first test tube: hexane, aqueous solution of $\ce{NaCl}$ and aqueous solution of $\ce{I2}$
In the second test tube: hexane, aqueous solution of $\ce{NaI}$ and aqueous solution of $\ce{Cl2}$

We are asked to write the reactions that take place in these experiments. Do we write only this:

$$\ce{NaCl + I2 → no reaction}$$

$$\ce{2NaI + Cl2 → 2NaCl + I2 (displacement)}$$

Or should hexane appear in the reactions too?

Hexane is the solvent for the two halogens. It does not react with either so leave it out of your answer.


Change in volume of two beakers on reaching equilibrium Two beakers are placed side by side in a sealed container. The first beaker contains 100 ml of pure water. The other an equal volume of an aqueous solution of sugar. What should be the volume of the liquid in the two beakers when all physical change stop, that is when the equilibrium is reached?

I know that the vapour pressure of pure water is higher than the vapour pressure of the sugar solution. But is it anyway going to help me in solving this question.

Okay, so you know that the vapor pressure of the pure water is some value, $p_w$. And you know from Raoult's Law that a solution with a nonvolatile solute of mole fraction $\chi_s$ will have a vapor pressure $p=p_w \chi_w$, where $\chi_w = 1- \chi_s$, so $p=p_w(1-\chi_s)$.

Since $\chi_s <1$, then $p < p_w$. How do you use this information?

Well, as the system approaches equilibrium, the two beakers will want to exchange water through the vapor phase to equilibrate the vapor pressures. What will that look like when it happens?  Can it happen?

By this logic, you can determine where the water is in your system. However, you cannot determine the volume(s) in the beakers without the following additional information:

Volume of the enclosing vessel - helps determine what mass of water remains in the gas phase
Temperature of the system - determines initial vapor pressure of pure water
Concentration of the sugar solution.
The non-ideal behavior of the sugar-water solution's volume.


Structures present in liquid water I know this may be a too generic question, but here it goes anyway.

It is already know that solid water (ice) organizes its molecules in a lattice (in a variety of 16 different crystals and one amorphous phase). See, for instance, http://en.wikipedia.org/wiki/Ice .

Now, what about liquid water? Is it the case that there are a (limited) number of local structures, like stable clusters or lattices configurations, combining and dissociating over relatively short periods of time, that dominate the liquid phase, loosely resembling the organization of a solid phase (the principal difference being that the hydrogen bonds would be weaker and would break more easily because of thermal agitation)? If so, what are those frequent structures?

Yes, there are various water clusters leading to local order on the scale of a few molecules holding stabilized orientation with one another for measurable timescales.  This article(1) is a pretty good introduction to the major varieties in liquid water.  This article(2) talks more about various structures available in any phase, including the ices, which gives some pretty good context.

(1) Keutsch, Frank N., Saykally, Richard J., Water clusters: Untangling the mysteries of the liquid, one molecule at a time, 2001, PNAS, 98, 19, 10533–10540, doi:10.1073/pnas.191266498

(2) Ludwig, Ralf, Water: From Clusters to the Bulk, 2001, Angewandte Chemie International Edition, 40, 10, 1808-1827, doi:10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1


How to synthesise N,N,3‐trimethylbutanamide from 1‐chloro‐2‐methylpropane? Suggest a synthetic route to amide C from the alkyl chloride B (reaction mechanisms are not required). Standard reagents and solvents may also be used.

In this question immediate reaction is to make a grignard reagent with B by reacting it with magnesium and have it attack $\ce{COX(N(Me)2)}$ to form C.
I'm not sure how it would be possible to make $\ce{COX(N(Me)2)}$ from standard reagents.
Are there any obvious synthesis ideas I'm missing in making $\ce{COX(N(Me)2)}$ as surely if I started from $\ce{COXY}$ where $\ce{X}$ and $\ce{Y}$ are Leaving groups, I would end up with $\ce{CO(N(Me)2)2}$ on reaction with $\ce{NH(Me)2}$ which would give an alcohol as the end product, as the amine isn't a good leaving group?
Also would $\ce{COXY}$ i.e. $\ce{COCl2}$ even count as a standard reagent? I have no idea what that really entails.

Phosgene, $\ce{O=CCl2}$, is a potent toxin and in no way would I consider it a safe or even standard laboratory reagent (at least on an undergraduate level).

Allow me to lead you down the garden path of organic synthesis.

We have a haloalkane. What do we know about such compounds; they have a polarised $\ce{C-X}$ bond ($\ce{x}$ is a halogen) and are prone to nucleophilic attack/substitution. In this case, we have a nice primary, haloalkane, hence, any nucleophilic substitution is likely to be SN2.

Now, the target is an amide, but what do we note is different about the target versus the starting material (SM)? There's an extra carbon present (and yes there's no halogen).

Warning bells should be going off at this time. What nucleophilic, reagent do we know of that could extend a carbon chain?
I can think of one: $\ce{CN-}$ (cyanide ion).
So, first react with $\ce{KCN}$, second react with conc. $\ce{H2SO4}$ to give a carboxylic acid, third react with $\ce{NH(Me)2}$ to give the product.
Now, once you've isolated the carboxylic acid, you can react it 'neat' with $\ce{NH(Me)2}$ and molecular sieves. The molecular sieves simply ensure a dry reaction and very high temperatures are needed (approx $\pu{160 ^\circ C}$). You could convert the carboxylic acid to an acyl chloride and react the amine to ensure a very high yield and 'fast' albeit more 'vigorous' reaction; in this case any ordinary organic solvent should suffice.


What will be the possible major product(s) on heating this primary alcohol i.e. 9-fluorenylmethanol with (conc) H2SO4? When the above compound is heated with (conc.) $\ce{H2SO4}$ the $\ce{OH}$ will probably be protonated and it will leave, producing a carbocation. After that possibly there will be a hydride shift. Now, after this step what could be the possibilities. I was thinking of ring expansion (which could potentially form phenanthrene) but again I doubt that a negative charge on a $sp^2$ carbon of the benzene ring would attack the positive charge after C-C bond cleavage. Or another possibility is formation of a double bond to form a structure like:

Which is the correct (major) product formed?

What I feel is that forming phenanthrene would be possible because of its high stability due to resonance. Initially, $\ce{-OH2}$ leaves out forming a 1°carbocation. Now the $\ce{\beta C-\gamma C}$ cleaves so as to make way for the better 2°carbocation which is also thermodynamically favourable.


Would atomic oxygen and atomic hydrogen result in O2 and H2 only, or some H2O too? Let's assume we have atomic oxygen and atomic hydrogen in a closed chamber, the temperature and pressure of which we're able to monitor (and control). Number of hydrogen atoms are twice that of oxygen.

Would this result in spontaneous formation of:

$\ce{O2}$ and $\ce{H2}$ only?
$\ce{O2}$ and $\ce{H2}$ with some $\ce{H2O}$?
and how would the process vary with temperature and pressure?
And I assume creating mostly $\ce{H2O}$ is the hardest (though I would be interested in a comment on this as well).

and how could one carry out analysis on paper to figure out what would happen? (I'm from a physics background, and my chemistry knowledge is a bit rusty. Though I do have a good background on thermodynamics, enthalpy, Helmholtz free, Gibbs free energies, etc, etc).

For the sake of brevity and focus of this question, we can ignore the question of how atomic oxygen and hydrogen came about in the first place. (This assumption is even more acceptable in the special case of carrying out molecular dynamics computer simulation, something I'm interested in, because we don't have to worry about how to form atomic $\ce O$ and $\ce H$)

edit: I guess I should clarify that I'm interested in a first principles analysis of this problem (since I'm mainly interested in MD). Also what if we had two other species of atomic gases (Ar and Ne, Ar and O, Ar and H, etc, etc). I think I'm more interested in it from a physical chemistry point of view (though without quantum mechanics if possible).

It would ignite and form water vapor. It's not complicated to figure out. The conditions you have described with a mixture of monoatomic hydrogen and oxygen is what is formed when a spark passes through a mixture of hydrogen and oxygen gases.


Are there ion combinations which can't be separated using the ion exchange resin method? I am currently investigating the uses of ion exchange resins (both cation and anion resins). I have not come across any ion combinations that cannot be separated using this method, are there any exceptions or does this method work for all ion combinations? If there are ions that can not be separated using the resin, what methods would work to separate them?

Ion Exchange resins are not selective.  You are essentially just statistically overpowering the situation by providing many more times the concentration of the ion you want compared to the ion you are 'extracting'.

There are some situations where chelation is so strong you can't exchange because the other ion is less stable in said position, say for example in crown ethers.

So the answer to your question, is no, there are not specific ions that can't be exchanged as a general rule, but there are ion pair combinations which won't exchange.


Why does capillary action occur better in narrow tubes than in wide tubes? We all have seen the meniscus of a small graduated cylinder in the lab. The reasons for the occurrence of capillary action are clear, but why does it occur more profusely in smaller-rimmed containers?

You have to look which forces act where. You have a force that tries pull the liquid up the capillary, which stems from the gain in energy due to adhesion. At the same time, however, gravity acts against this force. In equilibrium, both forces have the same value:

$$F_\text{g} = F_{\text{surf}}$$
$$ mg=\rho Vg= \rho R^2\pi h g=2\pi R\sigma_s \cos(\varphi)$$

Which leads to the formula presented by Babounet in another answer:

$$ h = \frac{2\sigma_s\cos(\varphi)}{R\rho g}$$

So, to boil it down, if you lower the radius $R$, you lower the force due to gravity $F_g$, but the force due to surface energy $F_\text{surf}$ does not fall as fast, which leads to a higher height $h$.

(source, $W=F_g$ here, naming isn't perfect, $F_{\text{surf}}$ is not drawn into)


Determining Rate Law and Reaction Mechanism from Method of Initial Rates The data collected from using the method of initial rates is tabulated above.
Suggest a rate law and a mechanism consistent with the data. State any assumptions.

My Attempt
Obviously B and D has no impact on the rate of the reaction. From the first table it appears that the reaction is first order in A, however this doesn't make sense when you examine the last two rows of the second table. Since the rate law is first order in A and is independent of B, decreasing the concentration of A by 4 and increasing the concentration of B by 4 should result in a fourfold decrease in the rate. However the rate of the reaction decrease by 2.5 times. So then my logic is somehow wrong.
Furthermore I am even more confused about the order of C. I know that concentration of C is inversely proportional to the rate of the reaction. However the reaction still occurs without the presence of C (evident by the first table). Doesn't this imply that the rate law is independent of the concentration of C?
From the comments below, I have realized that the rate order of B is in fact not $0$. However now I am even more confused. How can it be that the rate is not dependent on B when there is no C, but it is dependent on B when there is C present? What would its rate law be like?

Empirical rate law is:

$$\mathrm{Rate} = \frac{[\ce{A}][\ce{B}]}{[\ce{B}] + [\ce{C}]}$$

In the first block of data you can notice that $\ce{C}$ and $\ce{D}$ are not necessarily required for reaction to proceed because otherwise the rate would be zero. You can also deduce that their concentration must appear as a sum in the rate law for the same reason. The sum will appear in denominator because $\ce{C}$ slows down the overall reaction.

Next step is to guess elementary reactions. From the law's denominator sum you can conclude that intermediate will competitively react with $\ce{B}$ and $\ce{C}$.

$\ce{A ->[k1] I}$ (Probably some kind of excitation)

$\ce{I + C ->[k2] D}$

$\ce{I + B ->[k3]Product}$

Then you derive the overall rate via steady state approximation (assumption).

$$\mathrm{Rate} = \frac{k_1k_3[\ce{A}][\ce{B}]}{k_3[\ce{B}] + k_2[\ce{C}]}$$

and ${k_1}={k_2}={k_3}$


What is the conversion rate in a chemical reaction? Consider the following chemical reaction

$$\ce{X <=>[w_1][w_2] Y}$$

where $w_1$ is called the conversion rate of $X$ to molecule $Y$. What is the exact meaning of this conversion rate? Is it the same as probability?

The reference paper is:
Ref

The paper you linked to pertains to birth-death processes. An example of a birth-death process in the physical chemistry context that you're interested in  would be a set of molecules where each molecule is in one of two states, $X$ or $Y$, which are isomers of each other. The model assumes that due to the molecules' environment, each isomer can turn into the other isomer at times that are modeled as being random.

In the paper you provided a link to, $w_1$ is the probability per unit time that within a short period of time $\Delta t \ll 1/w_1$, a given molecule that's in state $X$ will transition to state $Y$. I.e., if a given molecule is in state $X$ at time $t$, then there is a probability of $w_1 \Delta t$ that the molecule will have randomly transitioned to state $Y$ by time $t+\Delta t$.


Energy cost of C−N rotation in hydroxamate group I am looking for a literature reference on the energetic cost of C−N rotation in hydroxamate group such in LC5 in water. I have been googling for a while.

Reference 1 reports the isomerization barrier for N‐methylacetohydroxamic acid. From the abstract:

  We have now determined the activation barrier to C–N rotation from major to minor conformation of a small secondary HA in $\ce{D2O}$ to be 67.3 kJ/mol. The HA rotational barrier scales with solvent polarity, as is observed in amides, although the HA barrier is less than that of a comparable tertiary amide in aqueous solution.

N‐methylacetohydroxamic acid (image from PubChem):

References

Stefanie P. Sippl, Paul B. White, Charles G. Fry, Sarah E. Volk, Lingxiao Ye, Heather L. Schenck. Kinetic and thermodynamic characterization of C–N bond rotation by N‐methylacetohydroxamic acid in aqueous media. Mag Res Chem 54 (2016), 46-50. DOI: 10.1002/mrc.4318.


Would this be a reasonable mechanism for the formation of a diazomethene? I am mostly concerned about the first step. Are there any species in this mechanism that look too reactive for this mechanism to make sense? Any thoughts would be greatly appreciated.

The accepted mechanism is here: 1

KOH attacks the carbonyl of the urea, the tetrahedral intermediate breaks down with Me-N=N-O- leaving. This picks up a proton from solvent to give Me-N=N-OH. The second step is OH- removing a proton from the Me group of Me-N=N-OH which then loses OH- giving diazomethane.


Reaction between acrolein and hydrazine When acrolein is heated with hydrazine we get dihydropyrazole. But I can't see how. First I think hydrazine should add on acrolein to form corresponding hydrazone, but what will happen next? Also is my first step correct?

Your suggestion of resulting corresponding hydrazone by the reaction between acrolein and hydrazine is correct. It is normal hydrazine-hydrozone reaction you perform in idetifocation of aldehyde and ketone carbonyl groups. Usually, E-isomer is preferred (Ref.1). E- to Z-isomerization can be occurred in presence of trace amount of acid (Ref.1). I expect that may be the case during constant heating. As a result, Z-isomer can undergo intramolecular Michael addition as depicted in following mechanism to give dihydropyrazole:

References:

Shigehisa Uchiyama, Masanori Ando, Shohei Aoyagagi, "Isomerization of aldehyde-2,4-dinitrophenylhydrazone derivativesand validation of high-performance liquid chromatographic analysis," Journal of Chromatography A 2003, 996(1-2), 95–102 (https://doi.org/10.1016/S0021-9673(03)00542-9).


Can singlet oxygen act as a catalyst for a (2+2) cycloaddition reaction? Can singlet oxygen facilitate a (2+2) cycloaddition reaction, for example between two alkenes, by reacting first with the pi-electrons in one alkene and then lowering the energy barrier for the second alkene to attack (given the instability of 1,2-dioxetane)?  Are there any examples of this sort of thing?

I'm not aware of any such examples.  In the reactions of singlet oxygen with olefins only products from cycloaddition to form a dioxetane and \ or products from the ene reaction are observed.  There are no reports of olefin dimerization.


Effect of pH on volatility (odor) of organic acid Based on two assumptions:

Ionic solutes can generally be considered nonvolatile
As the pH of a solution of organic acid is increased, an increasing percentage of it is deprotonated, or ionized (Henderson-Hasselbach eqn).

I think I should be able to decrease the volatility of an organic acid solution by increasing pH. However, in several rough experiments with acetic acid, I don't observe a very strong reduction in odor with increasing pH. Even a theoretically neutralized solution retains a significant vinegar odor.

Any ideas why this might be?

Acetic acid is a weak acid (pKa 4.8). Even if you neutralize it, some acetate ions will react with water to form acetic acid and hydroxide ion (hydrolysis). Since acetic acid has a quite strong smell, we are able to detect its presence in neutral acetate solutions. If you will increase pH to a high value with a strong base, the concentration of acetic acid will drop (Le Châtelier's principle).


How to separate sodium carbonate from sodium hydroxide? I got a drain cleaner in dust form, which consists of $\ce{Na2CO3}$ and $\ce{NaOH}$. I wanted to extract the $\ce{NaOH}$ so I tried to heat the mixture in order to just melt the sodium hydroxide, but it didn't really work as expected. Is there any easy way to do it?

There is a forum which discusses the similar problem of separating sodium hydroxide and sodium carbonate from drain cleaner. The solution is taken from there(slightly edited):

  This is relatively easy because sodium carbonate is practically
  insoluble in 50% sodium hydroxide solution so a glass frit type filter
  is needed to remover the insoluble sodium carbonate. Most commercial
  sodium hydroxide contains only a little sodium carbonate from
  absorption of atmospheric $\ce{CO2}$. So for carbonate free sodium
  hydroxide solution, make up one litre at a time of say 60% solution.
  To do this, add about 300g of sodium hydroxide pellets of flakes to
  500ml of ice-cold water in a large beaker and stirr until practically
  dissolved, cover and leave to cool . Allow it to cool to about 20-25 C
  and the add another 300g of sodium hydroxide and stirr until most has
  dissolved (much less heat is generated on the second addition), dilute
  to one litre, pour into a 1 L thick-walled polythene bottle and leave
  it to stand for about 1 month in a cool place and then decant the
  clear sodium hydroxide solution.

  If the material is say 50% sodium carbonate do the following:

  Add 500g of mixed sodium salts to 500ml of ice-cold water, stirr the
  slurry periodically for a couple of hours and then filter through a
  glass frit funnel (a vacuum filtration set-up will be needed) and wash
  the cake with a little water, say 50-100ml, to remove most of the
  remaining caustic soda; some sodium carbonate will dissolve but this
  will precipitate on the next addition. The cake will be almost pure
  sodium carbonate but you can always treat it with saturated sodium
  bicarbonate solution to ensure that it is.

  The clear filtrate should be placed in a calibrated vessel and another
  500g of material added and again stirred periodically for a couple of
  hours then filter and wash as before. Hopefully there should be about
  700-800ml of very strong sodium hydroxide solution. Place it in a
  polythene bottle and let it stand of a month so the sodium carbonate
  can crystallize out, decant the clear solution and dilute as required.

  NOTE:

  60% caustic soda solution is rather viscous and will not
  filter without suction even on a fairly coarse frit, 50% is still
  pretty viscous and difficult to filter.
  The sodium carbonate that
  crystallises out sinks very slowly hence the long standing time.
  Sodium carbonate dissolves rather slowly in water form some reason so
  with vacuum filtation the residence time of the wash water will
  actually dissolve very little sodium carbonate, too much wash water
  will dilute the solution below 50% and make the carbonate more
  soluble.
  Hot 50% caustic soda solution is extremely aggresive
  towards skin! When handling strong sodium hydroxide solutions, keep a
  bowl full of dilute citric around in case of emergency.

  There are other method such as treating a more dilute solution with
  slaked lime, filtering (difficult because the precipitate is fine
  grained) and evaporating down.


What are protecting groups in organic chemistry? My textbook does not give a very good explanation of what protecting groups are, how they are made, or how they function in Grignard reactions and others. It just gives a couple of examples of protecting groups and that's all.

So my questions are:

How are compounds chosen to be protecting groups? What are the criteria?
I noticed that protecting groups tends to be large molecules, is that always the case?

Functional groups become protecting groups out of practicality.

I would define an ideal protecting group (PG) as a functional group that is

easy to introduce/attach to a given, other functional group (the protection target)
stable/non-reactive, i.e. protecting, under a wide range of reaction conditions
easy to remove by well-defined and mild reaction conditions
with high yields throughout

Clayden/Greeves/Warren/Wothers have referred to PGs as functional groups with an Achilles' Heel.[1] Obviously, the usefulness of a given PG is always dependent on the target molecule and the reactions it will undergo. The "easy" part of the definition also encompasses factors such as price and toxicity.

As a corollary, one can introduce the concept of orthogonal PGs. Since the removal conditions can be very different for different PGs, one can successively attach PGs to, say, the hydroxyl groups of a sugar. By later removing them one by one, a chemist is then able to target them one at a time.[2]

Why do protecting groups tend to be large?

I am not convinced this is entirely the case, because I have seen applications of the methyl group as a PG. However, a larger group can introduce more steric hindrance or a point of attack for the removal that is different from the point of attachment to the protection target.

[1] Clayden, Greeves, Warren, Wothers: Organic Chemistry. Oxford University Press, 2004 (page 635).

[2] Werz, Seeberger, Angew. Chem. Int. Ed., 44, 6135, (2005). DOI:10.1002/anie.200502615.


Sulfuric acid's effect on rocks I am no chemist, nor did I pick up much understanding of chemical reactions in school. But I would like a basic understanding of how "reactive" and "potent" sulfuric acid is on rock; i.e., whether or not, and if so, how fast it would dissolve or otherwise affect the rock. Now, I know there are various types of rocks, and I'm sure this question could be too broad if not narrowed to some extent. So let me give some specifics.

My questions

Assuming a highly concentrated sulfuric acid (98%; but weaker, if perchance that would be more potent against the particular rock):

Will the sulfuric acid affect (dissolve or otherwise) it at all (and the ideal concentration to do so)?
If it affects it, then:

how fast would it dissolve/otherwise a solid piece of rock (not powder), say 1 cubic inch (though I'm looking for rate answers, so whatever the evidence gives as a rate works)?
Is rate affected by pressure (i.e., the rock "sitting" in the acid versus a pressurized spray of the acid on it)?
How much faster(?) would it dissolve it if the rock was powdered?
What are the general products/results left after the effect?

Igneous Rocks

Basalt
Gabbros, which apparently has a large amount of pyroxene and plagioclase
Granite

Sedimentary Rocks

Limestone
Dolomite

Metamorphic Rocks

(Perhaps the chemically diverse make-up of these rocks makes answering not feasible.)

Marble (perhaps reacts similar to limestone, "Geologists use the term 'marble' to refer to metamorphosed limestone"?)
Slate

While I have no problems with you giving in your answer the chemical reactions involved, I desire at least a "laymen's" summary of what occurs to each.

(Also, feel free to re-tag if there are some viable tags I was not aware of.)

This question probably cannot be answered correctly without conducting any experiment.

However, I would not expect that any of the rocks given would be dissolved by highly concentrated sulfuric acid. None of them has a composition that may completely result in the formation of soluble salts when reacting with sulfuric acid.

Presumably (hot?) dilute sulfuric acid may be able to dissolve parts of the rocks if they are powdered. E.g. $\ce{Na2O}$, $\ce{K2O}$, $\ce{MgO}$, and up to a small amount $\ce{CaO}$ and $\ce{Al}$.

Hydrochloric acid would do better since it forms soluble salts not only with Na, K, Mg, Fe, and Al but also with Ca. E.g. it could almost completely dissolve limestone, dolomite, and marble.


Why does the cyclization of open-chain glucose occur via the C5 hydroxyl group? In converting glucose from its acyclic (open-chain) form to either cyclic form (alpha or beta) why is the C5 hydroxyl group used instead of the bottom-most hydroxyl group?

Also, are these structures considered isomers? All sources I've seen only say "form" or "representation."

Due to the internal ring strain the best possible conformation is a six-membered heterocycle in this case called pyranose . A five-membered heterocycle may in rare cases also be formed, called furanose. (From left to right: α-D-Glucopyranose, β-D-Glucopyranose, α-D-Glucofuranose, β-D-Glucofuranose; courtesy of wikipedia)

Including the linear chain form they are all isomers (wikipedia), as they have the same chemical formula. More specifically the relationship between $\ce{linear <-> pyranose <-> furanose}$ is called constitutional or structural isomer.
The $\ce{\alpha-D <-> \beta-D}$ forms are diastereomers, or more specific diastereomeric conformers.


What are the constraints on the two electron integral matrix? In semi-empirical methods, the two electron integral matrix is simplified and the entries are replaced by actual values instead of integrals to evaluate. However, while doing that, the variational principle does not seem to directly hold; in other words, you can put values in the two electron integral matrix that make the energy go down infinitely. To make the variational principle hold, what are the contraints that the constants you put in the two electron integral matrix obey?

General Answer:

The basic premise of semi-empirical methods is to replace the computationally expensive integrals with values fit to experimental data or, in some cases, to high level ab initio calculations of the relevant quantity [1]. You are completely correct to point out that making this substitution will break the theoretical validity of the variational principle.

The short answer to your main question then is that the constraints are given by the experimental values to which one fits. That is to say, if you fit the values of your integrals to experimental data and all of a sudden you get out energies which are unbounded, discontinuous, or have some other undesirable and un-physical feature, then the problem is either in the fitting or in the corollary between the experiment and integral you are fitting. So, as in many physical contexts, the answer is circular. Sure, we could have any value for the integral, but the answers which experiments give live in a very small region of all possible answers to the question.

Approximations:

First of all, there are tons of semi-empirical methods which differ slightly in how they do the fitting, what approximations are made, etc. One important thing to note is that no semi-empirical method is really claiming to get the physics of these interactions correct for the right reason. They all rely on cancellation of errors, but this is built into the method on purpose, so this cancellation tends to be rather robust. One can achieve this because the behavior of the integrals arising in electronic structure is quite systematic, so if you are careful to know what you neglected in one part, you can try to neglect something of the opposite sign in another part. Reading ref. [1] is a good place to start to see how this is done in practice.

DFTB:

I should also point out that density function tight binding (DFTB) is often described as a semi-empirical method, but DFTB does obey a variational principle[2, 3]. That is, one can expand the total energy from regular density functional theory with respect to variations of the charge density (the central object in DFT). To zeroth order, one gets a non-iterative theory which depends only on properties of the individual atoms in the molecules and then corrects for interatomic repulsions. This is trivially variational.

If one includes terms up to second-order in the density variations, then you get self-consistent-charge DFTB (SCC-DFTB). This still defines an approximate Hamiltonian, so one can variationally solve for the energy. So, you might ask why this is considered semi-empirical. I'll just quote from ref. [2] as they give a nice summary.

  Comparing SCC-DFTB with MNDO-type methods, the
  underlying philosophy and the derivation of the applied approximations seem quite different at first sight. If one focuses
  on the actual working equations and their implementation, there
  are, however, many similarities: All these methods are valenceelectron SCF-MO treatments with a minimal valence AO basis
  set, only one-center and two-center terms are included, the twocenter two-electron integrals are represented by damped Coulomb interactions with the correct limit at large and small
  distances, and there are repulsive atom-pair terms that correct
  for deficiencies in the formalism. The methods are semiempirical
  in the sense that they employ empirical parameters which are
  adjusted to reproduce selected reference data, i.e., DFT energies
  and potential curves in the case of SCC-DFTB, and experimental
  heats of formation and other experimental data in the case of
  the MNDO-type methods.

Hopefully I've addressed your main concerns in the original question and left you with some resources so you better approach further questions.

References:

[1]: Thiel, W. (2014). Semiempirical quantum–chemical methods. Wiley Interdisciplinary Reviews: Computational Molecular Science, 4(2), 145-157.

[2]: Otte, N., Scholten, M., & Thiel, W. (2007). Looking at self-consistent-charge density functional tight binding from a semiempirical perspective. The Journal of Physical Chemistry A, 111(26), 5751-5755.

[3]: Seifert, G. (2007). Tight-binding density functional theory: an approximate Kohn− Sham DFT scheme. The Journal of Physical Chemistry A, 111(26), 5609-5613.


Conversion of oxalic acid to malonic acid The reagents I have used here are $\ce{LiAlH_4}$ which reduces the carboxylic groups into alcohol.Used $\ce{HIO_4}$ which breaks up the molecule into methanal. Then  I applied Grignard reagent which consisted ethyl for the alkyl part.Then I tried to oxide the terminal carbon into $\ce{COOH}$  but could not find a suitable reagent.

What is a better approach and a method to solve this question.

The Arndt-Eistert procedure suggested above seems very applicable to this problem. To use it we need ethyl oxalyl chloride. This can be prepared in 2 steps from oxalic acid by first esterification to form diethyl oxalate  procedure here then reaction with PCl5 to form ethyl oxalyl chloride procedure here. This can undergo the Arndt-Eistert mechanism here with excess diazomethane then either irradiation or treatment with silver oxide to give monoethyl malonate. Simple base hydrolysis then gives the required.


What are the CIP rules for cyclic substituents? Here's what I read on Wikipedia (section "Cycles"):

  To handle a molecule containing one or more cycles, one must first expand it into a tree (called a hierarchical digraph by the authors) by traversing bonds in all possible paths starting at the stereocenter. When the traversal encounters an atom through which the current path has already passed, a ghost atom is generated in order to keep the tree finite. A single atom of the original molecule may appear in many places (some as ghosts, some not) in the tree.

I don't understand how to assign CIP priorities to cyclic systems, for example, consider this:

How do we decide the CIP priorities? What does that paragraph I found on Wikipedia really mean? Could someone please give a detailed explanation?

I presume you want to know the E/Z geometry of the double bond. You need to determine on each end of the double bond which ring has priority. Intuitively one may argue that cyclohexane is larger than cyclopentane and cyclobutane is larger than cyclopropane. In this simple case, this is true and the double bond has the Z-configuration. However, digraphs are used in more complex cases. Your structure below has four colored dots. Each attached ring bond is "cut" and stretched out until one returns to the original atom. The three red atoms from the cyclohexane ring are identical. The same is true of the other three rings. Now treat the digraph as you would any acyclic alkene. The digraph has the Z-configuration.


calculate cell potential for the voltaic cell where its involving the two cobalt species with Copper Calculate the cell potential for the voltaic cell that results when the following two half-cells are connected at $\pu{25 ^\circ C}$:
(1) A platinum electrode inserted into a solution of $\pu{0.10 M}$ $\ce{Co^3+}$ and $\pu{0.0010 M}$ $\ce{Co^2+}$
(2) A copper electrode inserted into a solution of $\pu{0.010 M}$ $\ce{Cu^2+}$ ions

Answer. $\pu{1.66 V}$

how it's calculated?
I know how to calculate the second cell half but what about the first half?
Edit 1
will I need more than this equation?
Cathode
$\ce{{Co^{3+}} + e− ⇌ {Co^{2+}}  ->   1.92V}$
Edit 2 from Nicolas Answer
$\ce{E=E^0 -0.059*log \frac{[Co^{2 +}]}{[Co^{3 +}]}}$
$\ce{E=1.92 -0.059*log \frac{0.001}{0.1}}$
$\ce{E=2.038}$
Now how to use the Nernst equation again, which concentration I will use for Co?

I Thought I will use it for the cell, not just half cell which requires two concentrations

Edit 3
Applying Nernst to the second half
anode:
$\ce{Cu^{2 +}(aq) + 2e- → Cu(s) ->  +0.34}$
$\ce{E=E^0 - \frac{0.059}{2}*log \frac{1}{[Cu^{2 +}]}}$
$\ce{E=0.34 -\frac{0.059}{2}*log \frac{1}{0.01}}$
$\ce{E=0.281}$
E cell = E cathode - E anode = 2.038 - 0.281 = 1.757
The answer is 1.66 did I make everything right and That's a book's mistake?
Edit 4
trying to solve it in one step
Cathode
$\ce{{2Co^{3+}} + 2e− ⇌ {2Co^{2+}}  ->  1.92V}$
Anode :
$\ce{Cu ⇌  Cu^{2 +} + 2e- ->    0.34V}$
Cell:
$\ce{{2Co^{3+}} + Cu ⇌  Cu^{2 +} + {2Co^{2+}} ->    1.58V}$
$\ce{E=E^0 - \frac{0.059}{2}*log \frac{[Co^{2 +}]^2[Cu^{2 +}]}{[Co^{3 +}]^2}}$
$\ce{E=1.58 -\frac{0.059}{2}*log \frac{0.001^2 * 0.01}{0.1^2}}$
E = 1.76
So I think it's a book's mistake written 1.66 instead of 1.76

The equation noted corresponds well to that of the first half-cell: on the other hand, the potential indicated is not the good one but just that of the standard potential.
The Nernst relation is written $\ce{E=E^0 +0.06*log \frac{[Co^{3 +}]}{[Co^{2 +}]}}$: the potential therefore depends on the concentrations
You can therefore calculate the potential of this half-cell.
You do the same thing with the second to arrive at its potential: from the 2, you can calculate the potential of the cell


Which is the strongest base? When I was in lower grades, I was taught $\ce{NaOH and KOH}$ are the strongest bases. But I realised later that there can be more stronger ones. After a bit of research on the internet, I found these three to be spoken more about.

Lithium diisopropylamide (LDA)
Methyllithium (MeLi)
Butyllithium (BuLi)

Wikipedia on LDA says that the $\mathrm{p}K_\mathrm{a}$ of its conjugate acid is $36$, which means the $\mathrm{p}K_\mathrm{b}$ value at $\pu{298K}$ is $-22$. (Since $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$).
Likewise, for BuLi and MeLi I found the $\mathrm{p}K_\mathrm{b}$ value to be $-36$ and $-34$ respectively from the $\mathrm{p}K_\mathrm{a}$ values of their conjugate acids. So, can we conclude that BuLi is the strongest base? Is this absolute basic strength defined?

Sources for $\mathrm{p}K_\mathrm{a}$ values:

Wikipedia-LDA
pKa of MeLi and BuLi

Looks like it's neither of the three; ortho-diethynylbenzene dianion has the highest calculated proton affinity.

From Wikipedia

according to calculations, ortho-diethynylbenzene dianion is the strongest base, with a proton affinity of $\pu{1843 kJ \cdot mol^{−1}}$.

This base claimed the title of the strongest base 2016 ish, another base may have grabbed the spot but I haven't been able to find it.

B L J Poad et al, Chem. Sci., 2016, DOI: 10.1039/c6sc01726f


When is it necessary to include solvent effects? When modelling a process like keto-enol tautomerization or SN2 ring opening reaction, I am unsure as to what factors need to be considered when deciding whether solvent effects need to be accounted for. Because if they are important, a solvation approach, implicit or explicit is chosen to model the process.
For example, in a keto-enol tautomerization, solvent effects are important as the equilibrium is maintained by the solvent.

What I was thinking is that using the explicit water model, as few as  1 to 2 water molecules can be used to model the tautomerization. But many studies are showing that continuum solvent models (implicit) are the benchmark for keto-enol tautomerizations. I don't understand why and what factors determine this?

Well, first of all, explicit is more expensive than implicit. So, most people do not go for the former if they can help it. So, in my experience, you want to use explicit solvent only when you have to, like in the formation of $\ce{H2SO4}$ from $\ce{SO3}$ and $\ce{H2O}$. Here, extra water molecules act as catalyzers [1]. As for the case keto-enol tautomerization, I guess the results are in good agreement with the experiments using implicit solvent.

The bifunctional catalytic role of water clusters in the formation of acid rain. Chem. Commun. 2017, 53, 3516-3519.


Why does zinc react with sodium hydroxide? If zinc is less reactive than sodium, then why does it still react with sodium hydroxide?

Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction won't take place

$$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$

Remember, however, that $\ce{Zn}$ is amphoteric, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate

$$\ce{Zn + 2 H2O + 2 NaOH -> Na2Zn(OH)4 + H2}$$

P.S. I forget a lot of good ol' chemistry, and I have just realized that usually we do not call metals themselves amphoteric, but rather use this term for their oxides/hydroxides. For metals we just say that some of them (such as $\ce{Zn}$) dissolve both in acids and in bases. The point is that $\ce{Zn}$ does not react as an acid in the reaction above since the reaction is not an acid-base one, but rather oxidation-reduction. So, the answer then is that $\ce{Zn}$ reacts with aqueous solution of sodium hydroxide because it can be oxidized in these condition in accordance with the reaction mentioned.


Assigning Cahn-Ingold-Prelog priority numbering rules for 2 groups I was asked to draw the perspective formula of

  $$\ce{(3R,4R)CH2=CH-CH(CH3)-CH(CH3)-CH2-CH3}$$

Then I had to assign priority between the two groups:

  $$\ce{-CH=CH2}$$
  and
  $$\ce{-CH(CH3)-CH2-CH3}$$

Actually, they are bonded to the same atoms until the end of the first group, but one of them should have priority over the other.
Can please someone help?

When you've reached the end of your first group (after resolving the double bond), you're next comparing no atom (on the ghost C connected to C2) to a hydrogen atom (on the 4-methyl group), and any atom has a higher priority than no atom, so your second group gets the higher priority.


Does the stoichiometry effect the value of the equilibrium constant of an equilibrium reaction? Will be there any change in the $K_c$ values for same reaction, if we write stoichiometry as a fraction or as a integer? For example, assuming that we calculate $K_c$ at same temperature:

$$\begin{align}
&(1)  &&\ce{1/2A + 3/2B -> C}\\
&(2)  &&\ce{A + 3B -> 2C}\\
\end{align}$$

If we calculate $K_c$ for above two equilibrium reactions, do we get different answers?

$$K_c = \mathrm{products \over reactants}$$ So this is done by multiplying the concentrations of the products and dividing by the concentrations of the reactants. The concentrations are put to the power of their coefficient. So in your problems at equilibrium $K_c$ would be:
\begin{align}
  K_c &= \frac{[\ce{C}]^1}{[\ce{A}]^{\frac{1}{2}}[\ce{B}]^{\frac{3}{2}}}\tag1\\
  K_c &= \frac{[\ce{C}]^2}{[\ce{A}]^1[\ce{B}]^3}\tag2\\
\end{align}

Note that $[\ce{A}]$ means concentration of $\ce{A}$.


What's the relationship between dew point and temperature? From a Ph.D dissertation, the writer change the land surface from nature earth to urbn and built up to compare the difference involving with this land use change.

Due to the urban heat land effect, the temperature of lower atmosphere above the surface increase, and the dew point decrease.

Dew point is a parameter represent the absolute moisture. When the temperature increase, how to analysis the dew point increase?

There are two separate effects:

The average temperature above urban land is greater than above rural land. This could be because of the cooling effect of plants evaporating water, or because of the higher albedo of vegetation than pavement and roofing.
The absolute humidity decreases; that is, the total amount of water vapor in a given volume of air decreases, perhaps due to the capture and run-off of rainwater into storm drains, rather than evaporating.


Titration color change duration Why is it that in titration a color change occurs initially but then the solution quickly reverts to its previous color?

Thanks so much!

As an example, let's say you're titrating a strong acid with a strong base using phenolphthalein as an indicator. As you initially add the base to the acid, you may not see the color change occurring because the acid so readily neutralizes the added base. This is because of the large concentration of H+. As you approach a basic pH, however, concentration of H+ decreases and the acid is not able to neutralize the base as quickly. As the base disperses in the solution more slowly, the phenolphthalein will have time to react and show a slight color change.


Does a strong acid's conjugate base act as Brønsted-Lowry base? Why is the conjugate base of a strong acid not considered a Brønsted-Lowry base? Is it because the base is so weak that it essentially has no ability to accept a proton?

(Quote from my lecture notes: "$\ce{Br-}$ is the conjugate base of a strong acid ($\ce{HBr}$), so it will not act as a Brønsted-Lowry base").

In theory, a Brønsted-Lowry base is a chemical substance which can accept a proton.

But in the case of a strong acid, the acid will lose a proton and form a weak conjugate base. Say there is a strong acid HA.

$$\ce{HA + H2O <=> A- +H3O+}$$

So if we take a certain quantity of HA, most of it will undergo the forward reaction since HA is a strong acid. A- is a weak base and will not want to accept the proton and thus only a small quantity of the conjugate base will do undergo the backward reaction. Hence practically, you could say that the conjugate base of a strong acid is not a Brønsted-Lowry base because of its reluctancy to accept the proton.


How to calculate the abundances of certain gases from the peak areas, GC-MS In an experiment we had a series of air samples which were analysed through GC-MS, before using TargetLynx software to integrate the mass spectrum obtained.

So I now have the specific peak areas for a number of compounds (for example CFC-12 has an area of 28335299.35m^2) and a calibration standard for each specific gas investigated. For example CFC-12 has a mixing ratio (ppt) of 539.894.

With this information how could I determine the actual abundance of CFC-12 from the air sample?

Apologies for any confusion!
Thanks!

If you only have a single calibration standard concentration for each compound as your question implies, then you will simply need to calculate a calibration factor for each compound then multiply your measured peak area by the calibration factor.

This is also know as a single point calibration and it is important that the peak area of the standard is greater than that of your sample or you might be out of the linear response range of the instrument.  Also, a single point calibration assumes that the instrument response is linear from zero up to the concentration of the standard, or at least that the concentration of your sample is very close to that of your standard.

In your example you would calculate the calibration factor as:

  $\mathrm{539.894\ ppt / 28335299\ m^2 = 1.905\ x\ 10^{-5}\ ppt/m^2}$

Then if you had a sample with a peak area of 18335299.12, the concentration of CFC - 12 would be:

  $\mathrm{18335299.12\ m^2\ x\  1.905\ x\ 10^{-5}\ ppt/m^2\ =349\ ppt}$

Note: The calibration factor is often calculated in the inverse manner of my example above.  If calculated in that way, then of course you would divide your peak area by the calibration factor.


What is meant when people say a molecule has high multi-reference character? I have read some other posts explaining what multi-reference and multi-configuration are with regards to the wave-function (such as What exactly is meant by 'multi-configurational' and 'multireference'?).

However, what does it mean for a molecule specifically to exhibit high multi-reference character? Does it just mean the results from calculations will be strongly dependent on if the level of theory incorporates multi-reference?

When doing a CI calculation based on Hartree-Fock orbitals, then usually the HF configuration has a very high weight in the CI vector (>90%). The next largest weight is then very small (for example only 1%), and there is a very large number of those very small configurations.
Such cases are called weakly or dynamically correlated. Because of the large weight, it is reasonable to optimize the orbitals for the leading configuration only and then run some truncated CI calculations (typically CCSD) on top.

If there are multiple configurations with similar (or same) weight, then it is no longer reasonable to optimize the orbitals for just a single configuration, because this will introduce a bias towards this configuration. The number of these configurations is rather small.
Additionally, those configurations are not necessarily related by Single or Double excitations, so CISD may not include all. However, they are commonly generated by an active space of just a few orbitals.
This is called strong or static correlation, and is typically found in systems with partially occupied close-degenerate orbitals, e.g. d orbitals in transition metals.

In FCI this orbital bias would not matter, because the basis is (numerically) complete. But in truncations like CISD or CCSD this leads to a qualitative error. Some important configurations are poorly represented (due to the bias), other are completely missing (due to the truncation).

A CASSCF calculation resolves the bias by considering all important configurations of similar weight (by chosing an appropriate active space) and balancing the orbitals to all of the considered configurations. Since only some configurations are required for static correlation, it is often feasible to optimize both (CI and orbital coefficients) simultaneously.

The term multi-reference then refers to a calculation where all (or most) of the CASSCF configurations are used to generate excitations from.


Calculation of Atomic weight percentage in AAS I am currently doing doping studies in a system similar to this ($\ce{A_{x}B_{1-x}}$) and I have to find atomic weight percentage of $\ce{B}$ element using atomic absorption spectroscopy(AAS). I have prepared different PPM of solutions according to the weight percentage of $\ce{B}$ element.

Example to prepare 10 atomic weight percentage of $\ce{B}$ in $5~\mathrm{ppm}$ solution, I have used this equation to calculate molecular weight:
$$\text{Actual mol weight}=\text{mol.wt}(A)\cdot x + \text{mol.wt}(B) \cdot (1-x).$$

In this case $x = 0.9$ and $1-x = 0.1$.

From this, I have prepared $5~\mathrm{ppm}$ of a solution of the doped material assuming that $10\%$ of $\ce{B}$ is doped completely (but in reality only very low percentage is doped).

I got the below data from AAS:

B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)
    10          5                 1.5
    20          5                 0.9

From the above data, I have to calculate actual atomic weight percentage of $\ce{B}$.
My colleague has suggested this method in order to find $\text{Actual atomic weight percentage}$

$$\text{act.At.Wt.%} =\frac{\text{Obs.AAS.Conc(µg/ml)}}{\text{Sol.Conc(ppm)}}\cdot \text{At.Wt.percentage.of.}\ce{B}$$

Using this I got the results below.

B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)   Actual B (At wt %)
10                5                  1.25            2.5
20                5                  0.9             3.6

But I have doubt in this method as you can see in the table the observed AAS concentration for 10 and 20 At wt % $\ce{B}$ is 1.25 and 0.9 (µg/ml) respectively. You can see that the AAS concentration of 20 At.wt % is lower than 10 A.wt %. But when we calculate actual $\ce{B}$ At.wt.% we can see that the 10 A.wt %  is lower comparing 20 A.wt %. Can you tell me whether the equation suggested by my colleague is correct or not? If not can you suggest any other method in order to find the actual atomic weight percentage?

Relating mass fraction (ppm) with concentration (μg/ml) without knowing the density of the solution is impossible. You have to create your solutions with given concentrations (μg/ml). You have to weight a given mass of your substance and fill with water to a certain volume. You have to use volumetric flasks therefore. Consider the rules for applying volumetric flasks for getting correct results.