Unique carbons on rings In rings in general, how do I determine how many unique carbons there are? In this example, I am thinking that there are 2 unique carbons (one on top and one with a double bond) since this molecule is symmetrical. Any feedback is appreciated.

Yes, one would try to find points or lines of symmetry in general to check for equivalent ring atoms. It immensely helps to have a model and a lot of practice.

In your specific example, notice that you have two lines of symmetry in your (idealized) diagram: one passing through the sp3 carbons, and one crossing the two double bonds. You can thus "fold" the diagram into "quarters", leaving you with two non-equivalent atoms.


Is varnish with a flash point above 40 celsius safe in a microwave oven? I've painted regular white mugs and want to varnish them too.
But my varnish has the following specs:

Gloss varnish based on alkyd resin.

Flash point: Above 40 degrees celsius.

Is it safe to varnish on some sections of the painting on the outside of a mug? Or will it catch fire if microwaved to boil water or something?

I have no background in chemistry... took me a while searching around before I finally decided to just ask.

The flash point is for the solvent, i.e. for undried paint. Dried paint will have a considerably higher flash point, well over 100 degrees Celsius.

You should also consider toxicity. I don't think alkyd based varnish is approved for use in items you put in you mouth (like mugs).


Heat lost in evaporation at constant temperature I ran an experiment evaporating salt water at different temperatures and concentrations. I’m trying to figure out how much energy is needed to maintain water at a specific temperature even when its evaporating. I have tried a calculating the heat loss by evaporation but since the mass is decreasing during evaporation, both mass and heat would change, but temperature would have to remain constant. I am going crazy trying to figure this out.
Using the equation here would give me watts
https://www.google.ca/amp/s/www.engineeringtoolbox.com/amp/evaporation-water-surface-d_690.html
but it doesnt account for change in mass or temperature of the water.

Not a complete answer, but this should help clarify things for you:

You wrote "since the mass is decreasing during evaporation, both mass and heat would change".  But note that the change in mass per se does not affect the needed heat flow.  For instance, consider pure water.  Here, the change in mass would not affect the rate at which heat is lost due to evaporation (assuming other conditions, like surface area, temperature, the partial pressure of water vapor in the air above the water, wind speed, etc., don't change), and thus would not affect the heat flow needed to maintain the temperature.  This is because you're not trying to change the temperature of the water (in which case its mass would be relevant), you're just trying to maintain an energy balance of heat flow in and heat flow out.

Hence, regardless of the volume of water, for a specific evaporation rate, the heat flow needed to maintain constant temperature would be the same, since the latter is determined only by the rate at which heat is lost due to evaporation.

However, since you are working with salt water solutions, their concentration would increase with time, which would decrease the evaporation rate and thus decrease the heat flow needed to maintain the temperature.

So you have two choices:

(1) Just calculate the heat flow as a function of concentration (i.e., for each solution, determine the initial heat flow needed).  If the volume of water is large relative to the surface area, treating the concentration as a constant could be a good approximation for short periods of time.

(2) Use the evaporation rate as a function of concentration to determine the concentration as a function of time. I.e., derive a time-dependent expression. This is trickier.

I'm not sure from your question which you want to do, since your concern seemed to be the change in the total mass of the water, not the concentration. Note also that the equation you found in the linked article doesn't include a variable for concentration.


Why does barium sulphate not dissolve in HCl while barium sulphite does? Intuition tells me that the ionic bond in the sulphate must be stronger than that in the sulphite.
Is this true?
If it is then why? The charge on the barium ion is 2+ in both cases and the charge on the sulphate and sulphite ions are both 2-.
The sulphate has an extra oxygen atom. It's presence is somehow making the ionic bond stronger. But how/why?

$\ce{BaSO_3}$ has the same property as $\ce{BaCO_3}$ :
When reacting with an acidic solution, these two substances produce a gas which can get out of the solution, and displaces the equilibrium to the right  : $$\ce{BaCO_3 + 2 H^+ ->  Ba^{2+} + CO_2(g) + H_2O}$$
$$\ce{BaSO_3 + 2 H^+ -> Ba^{2+} + H_2O + SO_2(g)}$$ This allows these two substances to be soluble in $\ce{HCl}$ solutions. As a contrary $\ce{BaSO_4}$ has not this possibility to produce a gas in an acidic solution : it is and remains insoluble in $\ce{HCl}$ solutions.


Are nucleophilic substitutions reversible? When $\ce{OH-}$ interacts with a halogenoalkane, for instance $\ce{CH3CH2Cl}$, it acts as a nucleophile and undergoes $\mathrm{S_N2}$ mechanism, which results in the following reaction:

$$\ce{CH3CH2Cl + OH- -> CH3CH2OH + Cl-}$$

Since $\ce{Cl-}$ has a free pair of electrons, can it act as a nucleophile that will reverse the previous reaction?

$$\ce{CH3CH2OH + Cl- -> CH3CH2Cl + OH-}$$

Since an $\ce{OH}$ group is more electronegative than $\ce{Cl}$ group, the carbon atom to which it attaches would have a greater partial positive charge ($\delta+$), does this mean it's more prone to nucleophilic substitution?

In general, yes, $\mathrm{S_N2}$ reactions are reversible. But not in this particular case.

Rates of $\mathrm{S_N2}$ reactions depend on several factors: the nucleophile, the leaving group, the alkyl group undergoing substitution, and so on. In the context of your question, the leaving group ability is quite possibly the most important. Iodide, $\ce{I-}$, is an excellent leaving group. Fluorine, $\ce{F-}$, is an extremely poor leaving group. Again, there are several reasons which dictate leaving group ability, but at the simplest level, it has to do with the stability of the negative charge on the leaving group. This broadly correlates with basicity: iodide is hardly basic at all (its conjugate acid, $\ce{HI}$, is a very strong acid) and is quite happy to exist as $\ce{I-}$. Conversely, fluoride is reasonably basic (its conjugate acid, $\ce{HF}$, is a weak acid) and is not all that happy existing as $\ce{F-}$.

Using this logic you will find that hydroxide ion $\ce{OH-}$ is a very poor leaving group. Consequently, it's almost impossible to get ethanol (or other alcohols) to directly react in an $\ce{S_N2}$ fashion. So in the particular case you have raised, no, the reaction is effectively irreversible.*

The way around the leaving group problem is to convert the hydroxyl group to something that leaves better. Because alcohols are common, lots of people have devoted effort into making sophisticated ways of doing this. But by far the simplest way is to simply protonate it:

$$\ce{ROH + H+ <=> ROH2+}$$

The leaving group now is not hydroxide $\ce{OH-}$; it's now water $\ce{H2O}$, which is a much, much better leaving group. Consequently, if you treat alcohols with $\ce{HCl}$, you have a much better chance of getting an alkyl chloride, as opposed to if you were to just add chloride ion (in the form of e.g. $\ce{NaCl}$). The $\ce{H+}$ protonates the alcohol, and the $\ce{Cl-}$ does the actual substitution.

For an example of a reversible $\mathrm{S_N2}$ reaction, the best example is one where the nucleophile is the same as the leaving group. For example, consider a chiral alkyl iodide $\ce{R1R_2R_3CI}$. Let's say we have an enantiomerically pure sample of this alkyl iodide. Since $\mathrm{S_N2}$ reactions occur with inversion, if we treat it with sodium iodide, can we effect an $\mathrm{S_N2}$ reaction which will convert it entirely to the opposite enantiomer ?

The answer is no, and that's precisely because the reverse reaction will also occur. Both the forward and reverse reactions have the same nucleophile and the same leaving group, so in this case there aren't any considerations which would make one less effective than the other.

* "effectively" because technically speaking every reaction is reversible.


¹H NMR and MS analysis of products obtained by Claisen condensation reaction between methyl 2‐methylpropanoate and sodium methoxide A student attempted to perform the Claisen condensation below and  analyzed their isolated product by 1H NMR and MS.
Propose structures for  both the expected Claisen product and the observed product of this  reaction.
What could the student change to obtain their desired product?

This question was asking me to predict the product, both expected and observed.

Looking at your comment, I believe the main source of confusion may be understanding the underlying reason why the Claisen condensation requires the two α protons.
The first one, obviously, is required for the enolate to form. The second one is arguably equally important because it is the removal of this proton in the condensed product that prevents a retro-Claisen. The Claisen condensation is equilibrium driven to the product by that last deprotonation. Under the conditions listed here, you cannot deprotonate the product to generate the "real" product, so a retro-Claisen will quickly take place to provide the original starting material.
The "desired" product is the bis-methylated β-keto ester. You cannot use the conditions for a Claisen, but, as mentioned in comments, alkylation of the unmethylated β-keto ester should work. If you use a strong electrophile, like methyl iodide, you shouldn't need any harsh basic conditions, especially given the fact that a β-keto ester is quite acidic.


How important are those fields of chemistry in biotechnology and medicinal chemistry At my university there is some fields of chemistry that i don't know if i should focus on them...

How important are in biotechnology and medicinal chemistry those fields of chemistry? :

Quantum chemistry
Physical chemistry
Computional chemistry

Your question is broad and these chemistry which you have pointed out in your question is interlinked (Quantum ,Physical and Computational chemistry). The best way which I can explain this question is by giving links to main research carried out using these chemistry in biological science.

The study of quantum mechanics in photosynthesis is one such area and this nature paper(Nature 446, 782-786 (12 April 2007) is about that research

the highlights of the paper is the evidence for wavelike energy transfer through quantum coherence in photosynthetic systems

  two-dimensional electronic spectroscopy investigations of the FMO
  bacteriochlorophyll complex, and obtain direct evidence for remarkably
  long-lived electronic quantum coherence playing an important part in
  energy transfer processes within this system. The quantum coherence
  manifests itself in characteristic, directly observable quantum
  beating signals among the excitons within the Chlorobium tepidum FMO
  complex at 77 K. This wavelike characteristic of the energy transfer
  within the photosynthetic complex can explain its extreme efficiency,
  in that it allows the complexes to sample vast areas of phase space to
  find the most efficient path.

Other useful links

Nature 431, 256-257 (16 September 2004)

Nature 446, 740-741 (12 April 2007)


Why is hydrogen peroxide non-planar? I had my exam this week and one question asked whether $\ce{H2O2}$ was planar or not. I though that since the oxygen atoms are both sp³ hybridized and have two lone pairs each, the molecule should have a bent shape. I compared it $\ce{H2O}$ which also has the same shape and I knew was planar. But it turns out $\ce{H2O2}$ is not planar. Does anyone know the reason behind this?

Because the O atoms are sp3 hybrid and as a result they adopt a tetrahedral geometry, now the lone lairs also repel each other and also the bond pairs so according to VSEPR theory to minimize the repulsion and attain a confirmation of minimum possible energy, the molecular structure is non planar, although it deceptively looks planar in a flash! The actual structure is like that of an open book


Are the units of rate constants affected if we have more than one substance in the rate law $$A+B \rightarrow C$$ $$rate = k[A] [B]$$
[A]= x mol of A/L
[B]= y mol of B/L
So why the unit of the rate constant is merely $$L/M$$ not having mol of A or mol of B?

The heart of your problem is that you're assuming that there exist different types of the unit mole.

Consider the unit of length one meter. Take two meters of cotton thread, and two meters of jute thread. Place them together and you get a nice square of $1\times1 \text{ <unit of area>}$. Now what should the unit of area be? Should it be $=\text{cotton meter} \times \text{jute meter}$?

Well, it should NOT be $\text{cotton meter} \times \text{jute meter}$ because a meter, being a unit, is same for all substances being measured. A meter of jute is exactly the same length as a meter of cotton. Hence, the unit of area should be $\text{meter}^2$, which it always is.

Similarly, in your solution, you are assuming there exist two different types of the unit mole, one for $x$ and one for $y$. However, there is only one fundamental unit, and it measures the same amount of substance for both $x$ and $y$.

Hope it helps!


Why does aluminum chloride have a higher lattice energy than aluminum fluoride? From the table below (source: McMurry's Chemistry [1, p. 212]), it is evident that $\ce{AlCl_3}$ has a higher lattice energy than $\ce{AlF3},$ even though $\ce{F}$ is smaller than $\ce{Cl}$. Why is this so?

However, the melting point of $\ce{AlF3}$ $(\pu{1291 °C})$ is larger than that of $\ce{AlCl3}$ $(\pu{192.4 °C}).$ I think the huge melting point difference is because $\ce{AlCl3}$ is more covalent, while $\ce{AlF3}$ is more ionic. That being said, does lattice energy account for the covalent and ionic character of compounds?

References

McMurry, J.; Fay, R. C.; Robinson, J. K. Chemistry, 7th ed.; Pearson: Boston, 2015. ISBN 978-0-321-94317-0.

Lattice energy is a function of the radius ratio, which when tends to 1, improves the packing efficiency of the molecule. Here the radius ratio of the chloride ions and aluminium ions is much closer than the fluoride ions.
You could think of this from the born habers cycle also. Where the electron affinity of chlorine is higher, which leads to overall higher lattice energy. However, the radius ratio logic is much more convincing as the habers cycle logic would require the other parameters like the heat of formation also


Diluting a sample of weak base before titration. What happens to equivalence point pH of diluted vs undiluted? A student titrated a 10.00 mL of 0.20 M methylamine ($\ce{CH3NH2}$) and determined that 10.00 mL of $\ce{HCl}$ was needed to reach the equivalence point.
If you diluted your original sample of methylamine by adding 10.00 mL of water to your sample before you started adding $\ce{HCl}$, which of the following statements would be true?
a) You now only need to add 5.00 mL of $\ce{HCl}$ to reach equivalence.
b) The equivalence point pH would be the same as in the undiluted titration.
c) Both (a) and (b) are true.
d) Neither (a) nor (b) are true.

The answer is d) Neither (a) nor (b) are true.
I always thought that both the $\mathrm{p}K_\mathrm{a}$ and equivalence $\mathrm{pH}$ of an acid stay the same no matter what the concentration. Except where it is so dilute, ionization from water would have an effect. But I don't understand why or how it would change. I can't find anything in the textbook or online, can someone please explain this?

At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$.

If we neglect $\ce{H+}$ from water dissociation and consider $c_\ce{A-}\ll c_\ce{HA}$, then $c_\ce{A-} = c_\ce{H+}$
Therefore

$$K_\mathrm{a} = \frac{c_\ce{H+} c_\ce{A-}}{c_\ce{HA}} = \frac{c_\ce{H+} c_\ce{H+}}{c_\ce{HA} + c_\ce{A-}}$$

$$\begin{align}
\mathrm{pH} &= \frac12(\mathrm{p}K_\mathrm{a} - \log c) \\
  &= \frac12(14 - \mathrm{p}K_\mathrm{b} - \log c) \\
  &= 7 - \frac12(\mathrm{p}K_\mathrm{b} + \log c) \\
\end{align}$$

where $\mathrm{p}K_\mathrm{b}$ is the methylamine basicity constant;
$c$ is summary concentration of methylamine and it's a conjugate acid.

Therefore, if there is final volume 30 instead of 20 ml, the final

$$\Delta\mathrm{pH} = -0.5\log \left(\frac{\pu{20 ml}}{\pu{30 ml}}\right) = +0.088$$


Spreading of Chemicals (chemistry lab) Is it possible to spread a chemical or the danger of a chemical if the chemical has dried up on a piece of paper? For example if in a lab I have gotten a certain chemical on my lab paper but the chemical has dried up does that lab paper now pose a threat to my health and the health of others? Also if after touching the lab paper I touch my jacket and then after touching my jacket I eat something later on would I be in danger?

The chemical I am referring to is copper sulfate.

I have not experienced any irritation or burns (maybe some weird stomach or digestive problems).

Thank you

There is no sensible single answer to this: it depends on the chemical (and you really should read the safety notes before using anything or you will cause severe trouble.)

Consider these scenarios.

You are working with a solution of common salt and you spill it on your lab coat. No problem at all; you could eat your dinner off the lab coat as it is the same slat you might sprinkle on food. But you spill a solution containing some mercury salts on your coat, get it cleaned and decontaminated immediately as it will eventually poison you even if you don't use it to wrap your dinner.

Or you spill some ether on your lab coat. Don't stand near naked flames until it has evaporated, but once it has gone, it has gone with no lingering effects. Spill some nitric acid/sulphuric acid mixtures on it and even if you wash it off before wearing it again you might still have the problem that it has created some nitrocellulose, which can spontaneously combust, taking you with it. Equally badly, you spill some tellurium salts on your coat. They might not kill you but you will be shunned for weeks as you will smell strongly of something like garlic, but more offensive.

Bottom line: read the safety instructions and act accordingly: there is no general rule.


Finding the average rate of consumption The question is:

If $2.0 \cdot 10^{-4}$ moles of dye in $\pu{50 mL}$ of solution is consumed in 188 seconds, what is the average rate of consumption of dye in $\pu{mol L-1 s-1}$?

I am not sure if I have done this correctly. I use the rate of reaction formula which is $\text{Rate} = -\frac{\mathrm{d[A]}}{\mathrm dt}$ There is a negative sign because the dye is being consumed and it is disappearing.
Given:

[dye] = $2.0 \cdot 10^{-4}$ moles in $\pu{50 mL}$

volume of solution = $\pu{50 mL}$ = $\pu{0.05 L}$

$t = \pu{188 s}$

Since we have $2.0 \cdot 10^{-4}$ moles of dye per $\pu{50 mL}$ of solution, that means in $\pu{1 L}$ of solution, we get $2*(2.0 \cdot 10^{-4})$ moles of dye = $4.0 \cdot 10^{-4}$ moles
\begin{align}
\text{Rate of consumption/disappearance} &= \frac{-4.0 \cdot 10^{-4} \, \pu{mol}}{\pu{188 s}} \\
 &= -2.1 \cdot 10^{-6} \pu{mol L-1 s-1}
\end{align}
Is my answer correct? Please point out my mistakes if I did it incorrectly.

Since we have $2.0 \times 10^{-4}\text{ mol}$ per $50\text{ mL}$ of solution, to get the amount of substance per liter, you multiply by:

$$ \frac{1000\text{ mL}}{50\text{ mL}}=20$$

So we get $4.0 \times 10^{-3}\text{ mol L}^{-1}$ as Klaus says.

You're being asked to find a rate of dye consumption. This should be a positive number, if the dye is being consumed - which it is. If dye was being consumed at a negative rate, more dye would appear over time!

Imagine we begin with $c = 1.0\text{ mol L}^{-1}$. We end with $c = 0.5\text{ mol L}^{-1}$. Thus we might say that $\Delta c = -0.5\text{ mol L}^{-1}$. We see that the change is negative, signifying that dye has been consumed. If this occured over the course of 180 seconds, we might write:

$$\text{average rate of consumption} = -\frac{\Delta c}{\Delta t} = -\frac{-0.5\text{ mol L}^{-1}}{180\text{ s}} = 2.77\times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}$$

So even though the change in dye concentration is negative, the rate of dye consumption should be a positive number.


Why doesn't aromatic substitution add at multiple positions in one reaction? When you do aromatic substitutions, you usually see the major products with one substituent added. My question is why can't you add multiple substituents in one reaction? For example, the image below shows nitrobenzene reacting with $\ce{Br2}$ and $\ce{FeBr3}$ to add $\ce{Br}$ to the meta position. Why can't the reaction continue and add to the other meta position and thus make 1,3-Dibromo-5-nitrobenzene?

The rate of Aromatic substitution depends upon the activity of the aromatic system, because when the collision happens the aromatic system has to donate electrons to an electrophile.
In the example above you have used nitro-benzene, which is very strongly deactivated due to the $\ce{-NO2}$ group and thus the ortho and para positions are completely blocked. The meta position of nitro-benzene is not good but also isn't as bad as the para and ortho positions, in terms of electron density. So, we can say that the meta position of nitro-benzene is just a smidgen less electron rich than normal benzene, and thus first bromination takes place. But, after the addition of one bromine the ring becomes even more electron deficient which just makes further bromination even harder and slower. The reaction slows down to such an extent that it practically ceases to proceed.
On the contrary, if you would use a strongly activated aromatic system like phenol or aniline. It would provide such electron density to the para and ortho positions that they would undergo complete bromination, forming a 2,4,6-tribromo product. The electron density provided by $\ce{-OH}$ and $\ce{-NH2}$ is simply so high that it overcomes the electron deficiency created by the added bromines.


The Combining of Carbon Dioxide and Hydrogen How does this work exactly? Does it need some sort of kick in order to come together, or does it come together when put in a pressurized chamber?

I believe you are referring to the Sabatier process. $$\ce{CO2 + 4H2 -> CH4 + 2H2O}$$

The reaction uses a $\ce{Ni}$ catalyst and is done at $300-400\,^{\circ}{\rm C}$ and high pressures. The reaction is exothermic, $\Delta H = −165.0~\mathrm{kJ\,mol}^{-1}$, but requires initial heating in order to start the reaction off due to the activation energy.

Read more on Wikipedia.


Bond energies database I’m looking for a public database of bond energies of biochemical compounds. Does one exist?

Otherwise, which is the standard computational methodology to calculate them? Any references would be very appreciated.

The Comprehensive Handbook of Chemical Bond Energies might be able to help.  It walks through bonds in different functional groups.  It can be found at http://www.ebah.com.br/content/ABAAABH2cAD/comprehensive-handbook-of-chemical-bond-energies-0849373662


Sulfur Trioxide - Ionic Character I am told that the sulfur trioxide molecule exhibits charge separation because of poor p-orbital overlap. Sulfur's 3p orbitals are much bigger than oxygen's 3p orbitals, and thus the $\ce{S=O}$ bond is best described as $\ce{S+ - O-}$.

So is the author's argument suggesting that sulfur trioxide is unhybridized, and that it uses pure s orbitals to form sigma bonds and pure p orbitals to form pi bonds? Because that's the only way I can see sulfur as having three empty p orbitals to utilize. If we were to assign hybridization through counting the number of sigma pairs of electrons (3) we would arrive at sp2 hybridization, and there aren't enough p-orbitals to go around.

Also while we're on this topic, Wikipedia says that sulfur trioxide is best represented as having one $\ce{S=O}$ bond and two $\ce{S+ - O-}$ bonds. Now, does bond length differ between an  $\ce{S+ - O-}$ and an  $\ce{S=O}$ bond? Wikipedia suggests not, as only one bond length is given. If bond length is equivalent between the two, why?

Sulfur trioxide is planar with 120 degree O-S-O bond angles.  If the molecule were unhybridized and we used the 3 orthogonal p-orbitals for our sigma bonds, then we would expect the O-S-O angle to be 90 degrees.  Further, in the unhybridized scheme, there are not enough s orbitals to form 3 sigma bonds that are pure s. Just considering s and p orbital involvement, we could say that the sulfur is $\ce{sp^2}$ hybridized.  This would produce 3 $\ce{sp^2}$ orbitals for sigma bonds from sulfur to oxygen and leave one p orbital for one pi bond.  We can draw 3 resonance structures for this arrangement, so we would expect all bond lengths to be equivalent.


Why do some element symbols contain a single letter while others contain two letters? As the title says, why do some element symbols contain a single letter while others contain two letters?

There is a bit of history behind this.

In the website Development of the chemical symbols and the Periodic Table, it is stated that the Swedish chemist Berzelius updated Dalton's previous method.  He established a protocol for the symbols, still in use today:

  In the class which I call metalloids, I shall employ the initial letter only, even when this letter is common to the metalloid and some metal.
  In the class of metals, I shall distinguish those that have the same initials with another metal, or a metalloid, by writing the first two letters of the word.
  If the first two letters be common to two metals, I shall, in that case, add to the initial letter the first consonant which they have not in common:

  for example, S = sulphur, Si = silicium, Sb = stibium (antimony), Sn =
  stannum (tin), C = carbonicum, Co = cobaltum (cobalt), Cu = cuprum
  (copper), O = oxygen, Os = osmium, &c

The basis of the symbols are the Latin word for the elements.

A notes as what Berzellius was refering to when he used the term 'metalloid', from the Wikipedia page Origin and use of the term metalloid:

  In 1811, Berzelius referred to nonmetallic elements as metalloids, in reference to their ability to form oxyanions.

.


Are food calorie values really integers? According to what I've seen, fat has a listed caloric value of 9 Cal/g, while carbs and protein have listed caloric values of 4 Cal/g.* Are these numbers exact, or have they been rounded? And if they have been rounded what are the exact numbers? One hit said that the numbers are rounded, but not by how much.
*These are "food calories" (aka "large calories" or "Calories" with a capital "C").
1 food calorie = 1000 "small calories" or "gram calories" = 1 kcal = 4.184 kJ
See: https://en.wikipedia.org/wiki/Calorie

They're not exact numbers. These numbers aren't exact for three reasons:

Each type of carb, protein, and fat has a different caloric value.  These are overall averages for each class.

Even if you were dealing with a single pure compound, the value couldn't be exact because there is individual variation in how much of that compound is metabolized based on digestion, absorption, etc.

Even if #1 and #2 were not issues, the number still couldn't be exact, because it is a measured value, and measured values are never exact.  Only counted numbers, and values that are exact by defintion (e.g., the speed of light) are exact.

To get a better idea of how these values are determined, and the issues and uncertainties associated with measuring them, I'd recommend reading through:
Food and Agriculture Organization of the United Nations. FAO food and nutrition paper. Chapter 3:  Calculation of the energy content of foods - energy conversion factors. Food and Agriculture Organization of the United Nations, 1979.
Here is a direct link to chapter 3:
http://www.fao.org/3/Y5022E/y5022e04.htm


Free Open Source Gas Chromatography Software Collecting data is one thing, however sometimes one needs to plot or analyse data at a later date when access to the original software may no longer be feasible or possible.

Open Source software is at times a good alternative to commercial products. In this case, please list open source and/or free software for gas chromatography.
(Having said that, the databases for mass spectrometry are not open as far as I am aware, though suggestions are also welcome.)

One software that I have come across and used is OpenChrom https://www.openchrom.net/

At least for basic tasks it performs well as far as I can tell.

For those seeking pretty chromatograms, OpenChrom can be used in conjunction with R. Open a plot in OpenChrom, export it to a csv file and then plot the data using R.

A script for plotting two overlapped chromatograms in R is below:

filename = "Dodecane-Neat"
filename2 = "Dodecane-Stressed"
PlotTitle = "Dodecane Neat Chromatogram - Pre and Post Test"

infilename = paste0("./",paste(filename,"csv",sep="."))
infilename2 = paste0("./",paste(filename2,"csv",sep="."))
outfilename = paste(filename,"pdf",sep="-rect.")

## Read Data -> File is called 14mielczarek11.csv
undiluted <- read.table(infilename, sep=";", header=TRUE, nrows=10000)
undiluted2 <- read.table(infilename2, sep=";", header=TRUE, nrows=10000)
## Work out Total Peak (want the Chromatogram, not the mass spec)
undiluted$peaks <- rowSums( undiluted[,3:404] )
undiluted2$peaks <- rowSums( undiluted2[,3:404] )

GCxx <- c(undiluted$RT.milliseconds./1000/60, rev(undiluted$RT.milliseconds./1000/60))
GCyy <- c(rep(0, nrow(undiluted)), rev(undiluted$peaks)/max(undiluted$peaks))
GCxx2 <- c(undiluted2$RT.milliseconds./1000/60, rev(undiluted2$RT.milliseconds./1000/60))
GCyy2 <- c(rep(0, nrow(undiluted2)), rev(undiluted2$peaks)/max(undiluted2$peaks))

library(survival)
library(Hmisc)

# A4 is 8.27 × 11.7
pdf(outfilename, height=5, width=8)
par(mar=c(5,5,3,2))

plot("",
tck=-0.02,lab=c(10,10,4),las=1,
xaxs="i",xlab="Time (min)",
xlim=c(0,15),
yaxs="i",ylab="Detector Count Fraction",
ylim=c(0,0.15)
)

polygon(GCxx2+0.017, GCyy2, col= rgb(0,0,1,0.5), border=NA) #Blue - Shift the blue ever so slightly to properly overlap
polygon(GCxx, GCyy, col= rgb(1,1,1,1), border=NA) # White (so we don't get violet)
polygon(GCxx, GCyy, col= rgb(1,0,0,0.6), border=NA) # Red

lines(undiluted2$RT.milliseconds./1000/60+0.017,undiluted2$peaks/max(undiluted2$peaks),
type="l",col=4,lwd="0.05",lty=1) #Blue
lines(undiluted$RT.milliseconds./1000/60,undiluted$peaks/max(undiluted$peaks),
type="l",col=2,lwd="0.05",lty=1) # Red

title(PlotTitle)
minor.tick(nx=5, ny=5, tick.ratio=0.3)

dev.off()

And the results:


How do you find the pH of lime juice through titration? I am doing an experiment on the effect of time on the $\mathrm{pH}$ of Lime juice.
I was wondering how you would calculate the $\mathrm{pH}$ through titrations using $\pu{1 Molar}$ $\ce{NaOH}$. I have been racking my brain on this for a few hours and searching online.
Sorry if this has already been answered. Thank you for your time.

As said, titration and the concept of $\mathrm{pH}$ is very different. The process of titration is used to figure out an estimate of how much of acid $\ce{A}$ (of come concentration) will neutralize base $\ce{B}$ in an acid-base reaction (of come concentration) (also the reverse). The acid and base reaction is solely dependent on the coefficient of a chemical equation regardless of the $\mathrm{pH}$.

e.g., One mole of aqueous $\ce{HCl}$ will react always with one mole aqueous of $\ce{NaOH}$.

The $\mathrm{pH}$ in this case only indicates the "acidity" of either of the reactants based on the amount of $\ce{H3O+/OH-}$ ions that dissociated form them:
$$\mathrm{pH} = - \log[\ce{H3O+}]$$
You can of course figure out the concentration of a base/acid in titration if you know the basicity of one of the solution and the mole of acid/base in the other. You can do this by:

Let's say you want to know the concentration of a solution of ethanoic acid ($\ce{CH3COOH}$), we will have to prepare a suitable base to use as a titrant. We mix a universal indicator with the solution of $\ce{CH3COOH}$ and add it to a flask. Dissolving one mole of $\ce{NaOH}$ into x volume of water and put it in the buret. Run the buret until the resultant solution is neutral.

If the original volume of the $\ce{CH3COOH}$ solution is y and the amount of $\ce{NaOH}$ solution deposited is k.
Concentration of original acid is given by:
$$[\text{acid}] = {\frac{k}{x*\text{(original basicity of acid)}*y} }$$
as in
\begin{align}
 \\ [\text{acid}] &= {\frac{(\text{mole of}\, \ce{NaOH}\,\text{used})\cdot(\text{volume of}\, \ce{NaOH}\,\text{used})}{(\text{volume of}\, \ce{NaOH})\cdot(\text{original basicity of acid})\cdot(\text{volume of acid})} }
\end{align}
However, we still wouldn't be able to determine the $\mathrm{pH}$ of the original acid as we do not know how many of the acid in the original solution actually ionized:
$$[\text{acid}]*(\text{basicity}) \neq [\ce{H3O+}]$$
Thus we do not know the $[\ce{H3O+}]$ and the $\mathrm{pH}$ cannot be determined.


Why are dendrites formed during dilute CuSO4 electrolysis black? During the weekend, I wanted to conduct a kid's experiment - the copper dendrite formation. I was (partially) successful and now I would like to know what went wrong.

The procedure was taken from the weird science page (unfortunately in polish only).

The procedure was straight forward: Two, thin copper conductors on a microscope slide, connected to DC supply (voltage "over a dozen volts") and few drops of dilute $\ce{CuSO4}$ (according to the procedure, the solution should be transparent or maximum pale blue). It seemed pretty easy and the circuit was built in a few minutes. According to the description (and included pictures), I should see beautiful copper dendrites on the cathode, where $\ce{Cu^2+}$ reduces to $\ce{Cu}$, thanks to electric current (the reaction $\ce{Cu^2+ + 2e- -> Cu}$)…

In fact, the reality was slightly different (see attached figure). Below is my interpretation, but I am aware that it could be wrong:

At the beginning, bubbles on the cathode started to show. I assume it was hydrogen – high voltage (approx. $\pu{12 V}$) caused water electrolysis and $\ce{H2}$ evolution. No gas was evolving on the anode – I presume that the only reaction was oxidation of the copper wire $\ce{Cu-> Cu^2+ + 2e-}$.
A while later green flocs showed on the cathode, progressively moving towards the anode – I presume that $\ce{Cu(OH)2}$ was precipitated, due to increased $\ce{OH-}$ concentration near the cathode, caused by $\ce{H+}$ consumption.
Finally, the dendrites started to show on the cathode – but to my surprise, these were black, not orange like copper.

Why the dendrites are black? It does not look like copper but rather like $\ce{CuO}$. I’ve found the reference stating that $\ce{Cu(OH)2}$ transforms into $\ce{CuO}$ in basic solutions at room temperature, which could be an explanation (Cudennec and Lecerf, The Transformation of $\ce{Cu(OH)2}$ into $\ce{CuO}$, Revisited.). But why are these growing on the cathode instead of pure $\ce{Cu}$? And is it $\ce{CuO}$ or am I wrong?

To quote a source as you also noted:

  Copper hydroxide Cu(OH)2 is metastable. It easily transforms into copper oxide CuO more stable, either in the solid state by a thermal dehydration or at room temperature, in aqueous basic solutions. In the solid state, the transformation is performed at a relatively low temperature, 423 K.

In the current case, a basic solution develops and copper hydroxide breaks down into CuO and water.

However, there may also be heat from applying the overpotential, and this could speculatively be the source of CuO observed.

More precisely related to the cathode, start by noting per this source (Lumen Learning), for example, to quote:

  Electrolysis of a solution of sulfuric acid or of a salt, such as NaNO3, results in the decomposition of water at both electrodes. Hydrogen will appear at the cathode and oxygen will appear at the anode.

Assuming on the cathode the formation of an intermediate species (like $\ce{H*}$ mentioned on StackExchange in an advanced discussion of the underlying mechanic on the electrolysis of water) upon self-reaction, to quote:

  In an alkaline electrolyte they propose the following mechanism, which only slightly deviates from the above. The intermediate species are the same, that's why they did not need to treat them separately. On the cathode they just summarised as $$\ce{2H2O + 2e- -> 2{}^{-}OH + H2}$$ as the mechanism is bound to be the same.

Next, the action at the cathode of created OH- on Cu++ forms Cu(OH)2 which, in the basic conditions, results in CuO.


Acids of noble gases Are there any other acids of noble gasses besides xenic acid? Why other ones seem to be very unstable (like perxenic), or completely unknown (acids of any other noble gases). Particularly are there fluoroxenic acids, for example $\ce{HXeF5}$?

Yes there is and it has a very simple structure and is the strongest acid known: HeH$^+$. You might find it interesting to know that HeH$^+$ is also most likely the first molecule to have formed after the Big Bang.


How does one classify aqueous solutions of ions as acidic, basic and neutral? The following is a homework question:

  Classify the aqueous solutions of the following ions as acidic, basic or neutral:
          $\ce{C6H5NH3+, Cl- , Fe^3+ , Mg^2+ , HSO4- , NO3- , NH4+ , Al^3+ , C5H5NH+, H2PO4-, CH3CO2-, Cu^2+, Na+, Ag+, ClO4-, PO4^3-, CN-, SH-, Ba^2+}$

I am looking for an approach to solve it.

My thought process so far:

If any of the radical is highly reactive with $\ce{H+}$, then the solution will be basic as $\ce{OH-}$ is left. Similarly, if it reacts with $\ce{OH-}$, the solution will be acidic.
However, this is where I am confused. Take $\ce{Fe^3+}$ for instance. It is known that it reacts to form $\ce{Fe(OH)3}$, which is a strong base. But, according to my logic above, the solution will be acidic.

Can someone help me with this and possibly give me hints at solving the rest?

Your approach thus far is correct. $\ce{Fe^3+}$ is acidic; $\ce{Fe(OH)3}$ is not a strong base. Indeed, if $\ce{Fe(OH)3}$ were a strong base, it would dissociate and form $\ce{Fe^3+}$, but, as you've mentioned, the opposite process happens.

Your confusion may arise from the fact that $\ce{Fe^3+}$ is a Lewis acid, and produces protons in a "different" way than Arrhenius acids:
\begin{align*} \ce{HCl &<=> H+ + Cl-}\\ \ce{Fe^3+ + H2O &<=> Fe(OH)^2+ + H+ }\end{align*}

Either way, your thought process (1.) will work for both cases.


What is the complete reaction for Cyanide ions reacting with Thiosulfate? I was recently studying about the properties of Sodium Thiosulfate, where I came across the fact that $\ce{S2O3^2-}$ reacts with $\ce{CN-}$ ions to produce $\ce{SCN-}$ ions. This was quite useful in treating cyanide poisoning.

I have tried to find the complete reaction, but with no success. I believe this must be a redox reaction, since $\ce{S2O3^2-}$ has an average oxidation state of $+2$, while $\ce{SCN-}$ has a state of $-2$ (correct me if this is wrong). Can someone give me the complete reaction?

Source 1(If you consider $\ce{Na+}$ as counterions)

$$\ce{Na2S2O3 + NaCN -> Na2SO3 + NaSCN}$$

  Sodium thiosulfate reacts with sodium cyanide to produce sodium
  sulfite and sodium thiocyanate.

Source 2(If you consider $\ce{Ag+}$ and $\ce{Na+}$ as counterions for cyanide and thiosulfate ion respectively)

$$\ce{AgCN + 2Na2S2O3 → Na3[Ag(S2O3)2] + NaCN}$$

  Silver(I) cyanide react with sodium thiosulfate to produce sodium
  dithiosulfatoargentate(I) and sodium cyanide.

In this case, thiocyanate does not form, rather a complex forms.

Explanation of this reaction can be found by observing the related reactions given in the Wikipedia article of sodium thiosulfate.

  Silver halides, e.g., AgBr are typical components of photographic
  emulsions, dissolve upon treatment with aqueous thiosulfate:

  $$\ce{2S2O3^{2−} + AgBr → [Ag(S2O3)2]^{3−} + Br^{−}}$$

Since, $\ce{CN-}$  and $\ce{Br-}$ hold same charge,  you can easily replace it with silver bromide with silver cyanide  to get  cyanide ions as products.

Furthermore:

  Sodium thiosulfate is a component of an
  alternative lixiviant to cyanide for extraction of gold to form a
  strong soluble complex $\ce{[Au(S2O3)2]^{3−}}$. The advantage of this
  approach is that thiosulfate is essentially not toxic and that ore
  types that are refractory to gold
  cyanidation (e.g. carbonaceous or Carlin-type ores) can be leached by
  thiosulfate. Some problems  include the high consumption of
  thiosulfate, and the lack of a suitable recovery technique, since
  $\ce{[Au(S2O3)2]^{3−}}$ does not adsorb to activated carbon, which is
  the standard technique used in gold cyanidation to separate the gold
  complex from the ore slurry.

This process is also suitable for silver extraction using sodium thiosulfate and silver cyanide to form corresponding silver complex, $\ce{[Ag(S2O3)2]^{3-}}$. Just replace gold cyanide with silver cyanide and it will work pretty much well.


What is the "‡" symbol meaning in a reaction mechanism? I was studying hydroboration from Clayden's Organic Chemistry [1, p. 1281] and the transition state had “‡” symbol in the upper right corner of the activated complex:

  We know that this is not the whole story because of the stereochemistry. Hydroboration is a syn addition across the alkene. As the addition of the empty p orbital to the less substituted end of the alkene gets under way, a hydrogen atom from the boron adds, with its pair of electrons, to the carbon
  atom, which is becoming positively charged. The two steps shown above are concerted, but formation of the C–B bond goes ahead of formation of the C–H bond so that boron and carbon are partially charged in the four-centred transition state.

What does this symbol mean? I have seen it before as a superscript at the Gibbs energy symbol.

References

Organic Chemistry; Clayden, J., Ed.; Oxford University Press: Oxford; New York, 2001. ISBN 978-0-19-850347-7.

The symbol is called "double dagger" (sometimes also "double cross") and is used to denote transition state (a maximum in an energy diagram; also often denoted with "*" or "TS") or a related physical property.
Note, however, that a transition state and an intermediate are two different terms.

The symbol has the peculiar origins: as written by H. Eyring, it's entirely the merit of the secretary of the department, Miss Lucy D’Arcy [1, p. 9]:

  The symbol, ‡, first appeared in the second paper dealing with the activated complex.
  In that manuscript a star was used to designate the activated state but
  Miss Lucy D'Arcy, the departmental secretary, lacking a star used a plus and minus sign.
  Believing the typesetters would interpret this as a star it was left and by
  their option the cross, ‡, has become the almost universally used sign for the activated state.

Also, note that according to IUPAC recommendations [2, p. 1080]:

  In accordance with previous IUPAC recommendations (IUPAC QUANTITIES (1988)) the symbol ‡ to indicate transition states ("double dagger") is used as a prefix to the appropriate quantities, e.g. $Δ^‡G$ rather than the more often used $ΔG^‡.$

In the future, I suggest to use services like Shapecatcher or Detexify to find the name of the symbol.
Finding its usage in chemistry afterwards is normally a trivial task.

References

Eyring, H. Models in Research. International Journal of Quantum Chemistry 1969, 3 (S3A), 5–15. DOI: 10.1002/qua.560030705.
Muller, P. Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). Pure and Applied Chemistry 1994, 66 (5), 1077–1184. DOI: 10.1351/pac199466051077. (Free Access)


Configurations of Reactants in Second Order Nucleophilic Substitutions The question asks me to draw the product for the reactions. However, I am stuck and do not understand how to draw the reactants with configurations correctly.

(a) (S)-2-Chloropentane and NaSH

I thought the reactant I drew was in S configuration. Also, I thought since it was a chirality center, inversion of configuration will occur. Can someone please explain this in detail?

I tried to pinpoint you in the right direction through comments, but apparantly I failed. The bottom line is that the molecule you drew and the one in the textbook answer are identical. Here’s why:

In the top row, you can see the (S)-2-chloropentane, which I have drawn as you have on the left, slowly rotating around an axis equivalent to the reaction arrows to give you the same molecule drawn in a different orientation.

Now, if you do an $\mathrm{S}_{\mathrm{N}}2$ reaction, all you need to do is replace the chlorine atom and invert the stereochemistry which I have done for both cases. You can now see that both sides are identical.

It does not (and must not) matter whether you draw the 2-chloropentane with the substituted carbon to the right or to the left. Both ways are identical. Usually, you will chose an orientation that is consistent with further or earlier steps of your synthesis, but if you only have one one reaction that point is moot. (It matters a lot more in total synthesis, where it helps to draw reactands in a way that show their conformation in the final molecule.) Furthermore, there is a IUPAC style guide out there somewhere, which probably says one of the two ways is to be preferred if no other reasons speak for the other … but that’s more relevant for journal publications, not for practising or for exam answers.

So just for total consistency’s sake: Everything you did was correct; just remember that molecules can be drawn with different orientations, and learn to turn them around before your inner eye.


How do I solve this Ionic Equilibrium problem? Problem :

  Equal volumes of $\pu{0.02M}$ $\ce{AgNO3}$ and $\pu{0.02M}$ $\ce{HCN}$ were mixed. Calculate
  $\ce{[Ag+]}$ at equilibrium assuming no cyano-complex formation.
  $K_{sp}(\ce{AgCN})= \pu{2.2\times10^{-16}}$, $K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}$

My solution:
Concentration of $\ce{AgNO3}$ and $\ce{HCN}$ before reacting when mixed will become $\pu{0.01M}$. Since $\ce{AgNO3}$ is a strong electrolyte it dissociates completely. So, initial concentration $[Ag^+]$ is $\pu{0.01M}$.
$$\ce{HCN<=>H+ + CN-} $$
After reaction let
$[\ce{HCN}]= \pu{0.01-x}$, $[\ce{H+}]=\pu{x}$, $[\ce{CN-}]=\pu{x}$.
Now $\ce{CN-}$ reacts with $\ce{Ag+}$
$$\ce{Ag+ + CN^-<=> AgCN}$$
Initial $[\ce{Ag+}] = \pu{0.01}$, $[CN^-]=\pu{x}$.
After reaction let $[\ce{Ag+}] = \pu{0.01-y}$, $[\ce{CN-}]=\pu{x-y}$, $[\ce{AgCN}] = \pu{y}$.
Now, $K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \frac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}=\frac{x(x-y)}{0.01-x}\approx\frac{x(x-y)}{0.01} $.
And $K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}] = (0.01-y)(x-y)$.
Solving these two equations I get $\pu{(0.01-y)= 8.9\times10^{-11}=[\ce{Ag+}]}$.
But the solution in my book says the answer is $\pu{5.9\times10^{-5}}$.
I don't where am I going wrong.

Chemistry
AgNO3 is a strong electrolyte it dissociates completely.
$\ce{AgNO3 -> Ag+ + NO3-}$
If it was just HCN in solution, then most of the $\ce{CN^-}$ species would be as HCN. But since $\ce{AgCN}$ precipitates, more HCN dissociates. So essentially the overall reaction is:
$\ce{HCN + Ag+ -> AgCN}\,\ce{v + H+}$
The $K_{sp}$ of $\ce{AgCN}$ is such that it is reasonable to assume that for all practical purposes that the precipitation is essentially complete.
Assumptions
(1) I assume that the correct answer should have two significant figures which are all that the equilibrium constants have. However I will carry 4 significant figures in the intermediate calculations to try to avoid significant rounding errors.
(2) $\ce{[H^+]} \approx 0.01000$ molar and $\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$ (In other words about 1 ppt error for both)
Mass balance
If we consider 1 liter of the mixture of solution then we also have the following two mass balance equations:

The volume doesn't matter. Thinking of it as 1 liter just makes it simpler to get to the final result.

$0.01000 = \ce{[CN-] + [HCN]} +\ \mathrm{moles\ \ce{AgCN}\ce{v}\tag{1}}$
$0.01000 = \ce{[Ag+] +\ \mathrm{moles}\ \ce{AgCN}\ce{v}\tag{2}}$
$\therefore \ce{[Ag+] = [CN-] + [HCN]}\tag{3} $
Equilibrium equations
$K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}\tag{4}$
$K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}]\tag{5}$
dividing (4) by (5) we can get:
$2.819\times10^6 = \dfrac{\ce{[H+]}}{\ce{[Ag+][HCN]}}\tag{6}$
Rearranging (6) and substituting $\ce{[H+] = 0.01000\ \mathrm{molar}}$
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}  }{\ce{[Ag+]}}\tag{7}$
Rearranging (5) we get:
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}}\tag{8}$
Substituting (7) and (8) into (3) we get:
$\ce{[Ag+] =} \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}} + \dfrac{3.547\times10^{-9} }{\ce{[Ag+]}}\tag{9} $
Solving:
$\ce{[Ag+]^2} \approx 3.547\times10^{-9}\tag{10} $
$\ce{[Ag+] =} 5.955 \times10^{-5}\tag{11}$
Rounding to two places... $\ce{[Ag+] =} 6.0 \times10^{-5}$
Solving for other species
Substituting for $\ce{[Ag+]}$ in (8) gives
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{5.955 \times10^{-5}} = 3.694\times 10^{-12}\tag{12}$
Substituting for $\ce{[Ag+]}$ in (7) gives
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}   }{5.955 \times10^{-5}} = 5.956\times10^{-5}\tag{13}$
Argentocyanide Complex
The problem has ignored the argentocyanide complex. Is that reasonable?
$K = 5.6\times10^{18} =\dfrac{\ce{[Ag(CN)2^-]}}{\ce{[Ag+][CN-]^2}}\tag{15}$
rearranging
$\ce{[Ag(CN)2^-]} = \ce{[Ag+][CN-]^2}\times(5.6\times10^{18})=4.794\times10^{-9}\tag{16}$
So it is reasonable to ignore the additional complication.
Checking

You always check your assumptions on this sort of problem. It is easy to go down the wrong rabbit hole.

with $K_\mathrm{a}\ce{(HCN)}$
$\pu{6.2\times10^{-10}} = \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]} = \dfrac{(0.01)(3.694\times 10^{-12})}{5.956\times10^{-5}} = 6.202\times10^{-10}\tag{14}$
mass balance
It makes sense chemically that in the acidic solution $\ce{[HCN]} \gg \ce{[CN-]}$ so that $\ce{[Ag+]} \approx \ce{[HCN]}$. Making that assumption earlier using equation (6) would have reduced the work for this problem.
[H+] = 0.01000 molar and 0.01000 moles AgCN ppt per liter of the mixture
Also assuming that $\ce{[H^+] = 0.01000\ \mathrm{molar}}$ is certainly correct to two significant figures. However I tried to carry 4 significant figures in the intermediate calculations to avoid rounding errors. The assumption isn't that good.
!! PROBLEM !!
The assumptions that: $\quad\quad\ce{[H^+]} \approx 0.01000$ molar  $\quad\quad\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$
are not good to 4 significant figures. It would have been better to use:
$\quad\quad\ce{[Ag+]} \approx \ce{[HCN]}$
$\quad\quad\ce{[H^+] \approx 0.01000 - [HCN] \approx 0.010000 -[Ag+]}$
and get a quadratic equation when simplifying equation (6). Then $\ce{[Ag+]} = 5.938\times10^{-5}$ and $\ce{[H+]} = 9.941\times10^{-3}$.
So the value $\ce{[Ag+]} = 5.956\times10^{-5}$ is off by +0.303%, or 3 ppt, which I think is acceptable, but it is marginal.


Determining overtones bands of degenerate modes I have a question of a more mathematical nature on the mathSE (Symmetric Direct Product Distributive?) that received a good answer, but I think an answer more oriented to chemists would be a useful resource here.

I'm trying to determine the symmetry of the second overtone band of the degenerate $\Pi_u$ bend of $\ce{CO2}$. I'm told I need to use the formula$$\chi_v(\hat{R})=\frac{1}{2}[\chi(\hat{R})\chi_{v-1}(\hat{R})+\chi(\hat{R}^v)]$$
where $v$ is the number of quanta in the mode, $\hat{R}$ is an operation and $\chi_{v}$ is the character of that operation for the $(v-1)^{\text{th}}$ overtone. I arrive at the correct answer, which is that the second overtone has symmetry $\Pi_u\oplus\Phi_u$.

But this formula seemed complicated so I wanted to see if I could get the result just from taking the symmetric direct product of the state three times like so:
$$\Pi_u\otimes\Pi_u\otimes\Pi_u=(\Sigma^+_g\oplus\Delta_g)\otimes\Pi_u=2\Pi_u\oplus\Phi_u$$

Following the direct product tables I seem to get this, which has an extra $\Pi_u$.

I know this is wrong because the representation should only be quadruply degenerate, corresponding to the 4 different ways of distributing the quanta between the original degenerate modes: $(3,0) (2,1) (1,2) (0,3)$.

Why does this approach fail? Does the symmetric direct product not distribute over the direct sums?

The issue with this approach is that the symmetric direct product doesn't commute, but it is a bit more subtle than that. Summarizing in somewhat less mathematical terms the answer given to my question on MathSE, the issue stems from the fact that once you move past doing a single direct product, there is no longer just a symmetric and an antisymmetric part, but rather $n+1$ parts, where $n$ is the number of direct products being taken. So, trying to drop any antisymmetric terms that arise at each step will still leave terms that aren't in the symmetric part of the representation. (To say this in mathematical language for those interested, the representation for $n$ direct products of a particular irreducible representation of a group can be decomposed into $n+1$ parts by projecting onto the symmetric group $S_{n+1}$).

Physically, the reason for needing only the symmetric part rather than the whole direct product is that we are filling quanta into one particular mode. For example, if we wanted to find a combination band composed of two different $\Pi_u$ modes, we could just take the direct product to find the representation. But, filling two into a single $\Pi_u$ mode limits the number of possible distributions of the quanta to only those given by the symmetric direct product.

As to why the recursive formula gives the symmetric direct product, I give a summary of the argument from Wilson, Decius, and Cross's Molecular Vibrations for the case of a doubly degenerate mode.

Let $Q_a$ and $Q_b$ be the normal coordinates of a doubly degenerate mode. We choose these coordinates such that for an operation $R$      $$Q_a \xrightarrow{R}R_aQ_a$$
$$Q_b \xrightarrow{R}R_bQ_b$$
While the same coordinates will not diagonalize each $R$ in the group, we can find a set of coordinates for each $R$ that will. The character of $R$ for the $(v-1)^{\mathrm{th}}$ overtone, $\chi_v(R)$ can then be determined by summing the contribution of $R$ acting on each possible distribution of quanta between the modes $$\chi_v(R)=R_a^v+R_a^{v-1}R_b+...+R_aR_b^{v-1}+R_b^v$$
We can see that recursion formula holds by looking at the form of the terms on the right hand side
$$\chi(R)\chi_{v-1}(R)=(R_a+R_b)(R_a^{v-1}+R_a^{v-2}R_b+...+R_aR_b^{v-2}+R_b^{v-1})=R_a^v+2R_a^{v-1}R_b+...+2R_aR_b^{v-1}+R_b^v$$
and
$$\chi(R^v)=R_a^v+R_b^v$$
Averaging these two terms, we obtain the recursion formula from the question.
$$\chi_v(\hat{R})=\frac{1}{2}[\chi({R})\chi_{v-1}({R})+\chi({R}^v)]$$

By a similar argument, we can obtain a recursive expression for triply degenerate states
$$\chi_v({R})=\frac{1}{3}\bigg[2\chi({R})\chi_{v-1}({R})+\chi({R}^v)+\frac{1}{2}\big(\chi({R^2})+[\chi({R}]^2\big)\chi_{v-2}({R})+\chi({R}^v)\bigg]$$


Acid in Imine Formation Why is a strong acid like $\ce{H2SO4}$ used for imine formation here instead of a weaker acid like TsOH? I thought that if the pH was too low, the amine molecule would get protonated to an ammonium, which are not nucleophilic?

You need to protonate the oxygen in order to actívate the molecule towards nucleophilic attack by the nitrogen of Ethyl amine. pH plays a very important rule here. Too low pH gives an efficient protonation of the oxygen, but it also protonates the nitrogen of the Ethylamine(As ethylamine is weak base due to the lone pair of electrons on nitrogen in it), so the production of the imine results inefficient due to less amount of the desired reagent. On the other hand, at higher pH values, the oxygen does not get protonated enough to proceed the reaction as the electronegativity of oxygen dominates it from giving electrons to a low concentration of protons( As pH is high). So there is an optimum pH between pH 4.5 and 5, for this reaction to take place. So dil H2SO4 can give this pH level while weaker acids cannot. Also when you look at the mechanism for we need to dehydrate to get imine. Dil H2SO4 is good dehydrating agent.

For all the factors above diluted strong acids are better than weak acids.


How can we predict the colour of transition metal complexes? I am facing problems in indentifying colour of complex compunds just by seeing the their molecular formula. Is there any method or concept to predict the colour of complex compunds?

The basic principle of coloured compounds is that there is some kind of electronic transition whose energy difference corresponds to a photon whose wavelength is in the visible region ($\pu{400nm}<\lambda<\pu{700nm}$). Thus, to determine the colour of a compound we should always be looking at the molecular orbital scheme; for reference, I have attached the MO scheme of a typical octahedral $\ce{[ML6]^n+}$ compound in figure 1.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

The most common electronic transition corresponding to visible wavelengths is the transition between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$, marked with $\pu{10Dq}$ in figure 1. In simple terms, these five orbitals correspond to the central metal’s $\mathrm{d}$ orbitals meaning that the metal’s electronic configuration plus the strength of the $\ce{L\bond{->}M}$ interaction will determine the colour.

The problem with attempting to predict colours ab initio is that you cannot really predict the difference in energy levels between the ligand orbitals (right-hand side) and metal orbitals (left hand side) — it is mostly the metal and its oxidation state that determines this, a higher oxidation state corresponding to lower-lying metal orbitals, therefore a greater value of $\pu{10Dq}$ and therefore a blue-shift in colour.

However, there are still a few shortcuts that you can use assuming you have memorised the ‘general colour’ of a certain metal in a certain oxidation state. For example, you may consider stronger and weaker field ligands. When going from a slightly weaker ligand to a slightly stronger ligand — e.g. going from $\ce{[Cu(H2O)6]^2+}$ to $\ce{[Cu(NH3)4(H2O)2]^2+}$ — $\pu{10Dq}$ is predicted to increase and thus the absorbed wavelength predicted to decrease.

Unfortunately, the generality of this scheme is strongly limited. It fails as soon as:

the complex transitions from a high-spin state to a low-spin state (e.g. $\ce{[Fe(H2O)6]^3+}$ to $\ce{[Fe(CN)6]^3-}$)
the geometry of the complex changes from (distorted) octahedral to something else like tetrahedral (e.g. $\ce{[Co(H2O)6]^3+}$ to $\ce{[CoCl4]-}$).

So you might just be better off memorising the colours of certain complexes rather than attempting to learn the pattern and remember the exceptions.


Does the distance increase as the polarity increases? According to the dipole moment's definition, as polarity increases the distance between the centres of positive charge & negative charge increases.In the ionic bond ,as the polarity is maximum, distance between the two charged centres is maximum.Is this concept correct? If not what is the relationship between distance and polarity. I m not getting any intuition.

I'm not sure if the guideline actually works out in empirical observations. There are at least two factors affecting bond length - atomic radii - and polarity of the bond. A more polar bond will have a less even distribution of electrons. The electron density will generally be closer to the more electronegative atom. This isolation of electrons may have the effect of "stretching" the bond.

Consider the following two molecules, both of which are linear and polar:

From polarity considerations alone, we might expect the $\ce{H-F}$ bond to be longer than the $\ce{H-Cl}$ bond because fluorine is more electronegative than chlorine. However, chlorine is a bigger atom than fluorine by nature of having additional sublevels of electrons.

Also, consider the below data: bond length actually increases with decreasing polarity (quantified through dipole moments).

Fluoromethane Bond length 136pm, Bond strength 108 kcal/mol, Dipole moment (D) 1.85

Bromomethane Bond length 178pm, Bond strength 84 kcal/mol, Dipole moment (D) 1.87

Chloromethane Bond length 193pm, Bond strength 70 kcal/mol, Dipole moment (D) 1.81

Iodomethane Bond length 214pm, Bond strength 56 kcal/mol, Dipole moment (D) 1.62

The unfortunate thing about your guideline is that I thought that it would work at least across periods - in which atomic size stays relatively constant. However, the facts do not fit the case. Consider $\ce{H-N}$ and $\ce{H-O}$. Oxygen is more electronegative than nitrogen; we'd expect the $\ce{H-O}$ bond to be more polar, and thus longer than the $\ce{H-N}$, at least according to your guideline. The opposite is actually true; the $\ce{H-O}$ bond is actually shorter than the $\ce{H-N}$ bond.


Does ozone (O₃) gas have a color? Does ozone have a color or is it colorless? I have heard that ozone has a bluish color.

Could someone explain this observation scientifically?

Ozone gas does indeed have a slight blue color. It is dark blue in a liquid form and purple-black in solid form.

The color comes from the fact that it absorbs all wavelengths of light other than those in the blue region. As the phase changes, the molecules get closer together, shifting the area of absorption, changing the color of the material.


Can luciferin + ATP → luciferyl adenylate + PPi be reversed? Reading about the light emitting reaction of luciferin + ATP, catalysed by luciferase I am confused about the following.

On this webpage the partial formula is written as:

luciferin + ATP → luciferyl adenylate + PPi

which seems to me like a one way reaction.

But on searching on Google "luciferase ATP equilibrium" I found websites like this one and this one which seem to indicate that the reaction is reversible, as there is an equilibrium.

Is it possible for this reaction to take place in reverse? And what about the link on books.google.nl showing a reversible reaction with light involved? What about that?

Thank you for clarifying.

EDIT 2014-02-25:
This article: "Kinetics of inhibition of firefly luciferase by dehydroluciferyl-coenzyme A, dehydroluciferin and L-luciferin" by Luis Pinto da Silva etc (Link) shows the following reactions:

Where Luc is firefly luciferase, LH2 is luciferin, L-CoA is dehydroluciferyl-coenzyme A, L is dehydroluciferin, ATP is Adenosine triphosphate, AMP is  adenosine-5`-monophosphate, PPi is inorganic pyrophosphate, O2 is oxygen, CO2 is carbondioxide.

Looking at the last equation, it seems it is possible to form ATP as a product, as it is on the right side of the arrow. Remarkable, I would say. still I do not see any luciferin (LH2) on the right side of any equation, so the answer to question of this topic remains no thus far.

In theory every chemical reaction is reversible.

The reaction that you mention does not emit any light yet, that comes later with the following reaction:
$$ \text{Luciferyl adenylate} + \ce{O2 ->} \text{Oxyluciferin} + \text{AMP} +h\nu $$

Since the reaction you mentioned occurs under loss of pyrophosphate ($\ce{PP_{i}}$) which is where the energy was stored beforehand, I guess it's somewhat hard to get it back.

It's almost like hoping that
$$ \ce{2CO2 + 3H2O -> C2H5OH + 3O2} $$
occurs under room temperature. Which it doesn't, because $\ce{CO2}$ is such a thermodynamical black hole. I think the same applies to your question.


Is readily available ammonia a substitute for ammonia water? I am an artist attempting to duplicate a recipe for a painting medium.  The source says I can use powdered ammonia or ammonia water.  The recipe is from the 1940-50 and I am no chemist.

So, my question is can I use the readily available ammonia in place of "ammonia water" ... are they the same thing?

Ammonia at standard temperature and pressure is a colourless gas. So when your recipe says "powdered ammonia" there must be some kind of mistake. It might refer to a salt, that can be produced with ammonia and hence serve as a source of it. One of such salts is ammonium chloride.

In household stores available ammonia is a solution of the gas in water. Depending on the product there might be other ingredients as well, for example surfactants if used as a cleaner. In any case there should be some information about the concentration of the solution. Often given in percent and it strength may vary, but I doubt it contains more than 10% in any case.

In conclusion, the available ammonia is ammonia water.

Lastly a word of warning. As an artist you are probably handling paint thinners and other chemicals all the time. Treat ammonia just as carefully as these. And keep your place well ventilated. You will probably do this anyway, because the smell is just horrible.


SN1 (lim) and SN2 (lim) I noticed (lim) is sometimes added to the SN1 or SN2 notation. I suppose it means limiting, but still do not understand it completely. Can someone please explain how is SN1 and SN2 different to SN1 (lim) and SN2 (lim)?

Text from the first row:

Existence of an intermediate with a lower coordination number

No intermediate

Existence of an intermediate with a higher coordination number

Text from the first collumn:

Stoichiometric mechanism

intimate-mechanism

Ingold-Definition

Dependance of the speed on the nature of the entering group

Text from the last row:

The speed does not depend on the entering group.

The speed does depend on the entering group.

I think long story short: the difference between (lim) and not is that the dominant step drives the limiting case more strongly than the non-limiting case.

There are two aspects to a substitution reaction:

Stoichiometric mechanism: the sequence of elementary steps that take us from reactants to product

Associative (A): the bond formation between the metal and the entering ligand i
Dissociative (D): the bond breaking between the metal and the exiting ligand is the important step, i.e. the intermediate has a decreased coordination number
Interchange (I): there is no evidence of an intermediate

Intimate mechanism: what affects the rate constants of particular steps?

The reaction rate is affected by the entering group (a)
The reaction rate is affected by the leaving group (d)

You can see that SN1 and SN1 (lim) both have dissociative intimate mechanism, which means that the breaking of the bond between the metal and the leaving group is the rate determining step. In SN1 (lim), the leaving ligand bond breaks first forming an intermediate with a lower coordination number; the entering ligand then forms a bond with the metal. However, SN1 is an interchange stiochiometric mechanism which means that there isn't a true intermediate where the leaving group has totally left. In this case you could imagine that the leaving metal-ligand bond merely weakens or 'stretches' before the incoming ligand bonds to the metal.

Similar principle applies to SN2 (lim) and SN2. In SN2 (lim), an intermediate with higher coordination number is formed when the entering ligand bonds with the metal. For SN2, no such intermediate step is observed. You could say that the incoming ligand simply begins to form a bond to the metal before the leaving group bond starts breaking.

(http://nptel.ac.in/courses/104106063/Module%205/Lectures%208-10/Lectures%208-10.pdf for image above)


How small a peak could one identify by GCMS I'm trying to characterize the impurity profile of an organic liquid sample that is already described as 99.7% pure (Boiling Point 230 C).

Would a GC-MS be a viable way to identify the smaller peaks? e.g. How small a peak could one expect to get a decent mass spec. pattern for? Say, a peak with 0.01%  of total peak area, would it be viable to identify it on a GC-MS?

I'm assuming the smaller peaks can be resolved satisfactorily. I don't need quantification of the peaks just identification. I can match against the NIST library of spectra but also have a clue based on the manufacturing process about what the likely impurities may be.

You can get GCMS peaks for very small components of the overall mixture.

However, most GCMS patterns are electron impact, and hence probabilistic, so if you don't have enough sample, the fingerprint won't be reliable.

In my experience 0.3% produces spectra which are 'good enough'... as long as you don't inadvertently overload your column.


What is the density of nitrogen at standard conditions? I find diverging answers to what the density of nitrogen at standard conditions is.
By standard conditions I mean T=273.15 K and P=1 atm (=1.01325 bar).
Different sources state:

Wikipedia : 1.2506 g/L
Fluidat : 1.250 g/L
NIST : 1.2504 g/L
When I use ideal gas law : 1.2498 g/L (using molar mass as 28.0134 g/mol and gas constant as 8.31446261815324*10^3 J/(K mol)

Which one is more correct and why ?

One possibility for differences in measured density of an element is isotopic variation. For example, sulfur isotope abundance may vary, depending if the element is derived from soil, from sulfide minerals or as the native element from a volcanic vent. Nitrogen has two common stable isotopes: 14N and 15N. It is possible that the measurement of density varied because of natural variation.

However, it is possible the density varied because the definition of "Standard Temperature and Pressure" (STP) has changed over time (e.g., 101.325 kPa pre-1982, 100 kPa afterwards).

It is possible the density measurements varied because the definition of mass has changed over time.

It is possible the density measurements varied because of experimental error.

So if the discrepancies are a cause for concern in your work, check the source given for each measurement to help explain the differences.


How are $\sigma$ and $\sigma^\pm$ determined in Hammett plots? The Hammett plot is commonly invoked in organic chemistry to reason about the plausibility (or implausibility) of various reaction mechanisms. The vertical axis is essentially the logarithm of an equilibrium constant (or rate constant) measured relative to a hydrogen functional group, which I understand. However, I am at a complete loss for understanding what is being plotted on the horizontal axis, which conventionally are denoted $\sigma$ and $\sigma^\pm$. What are these parameters and how are their values determined for various functional groups? In practice it seems that people look them up in tables, but where do the values in these tables come from?

Hammett wanted to find a way to quantify the effects electron-withdrawing (EWG) and - donating (EDG) groups have on the transition state or intermediate during the course of a reaction. Initially he took the $pK_{\mathrm{a}}$-values of benzoic acids carrying the respective functional group in para or meta position (ortho acids and aliphatic acids weren't used because steric effects would overlay with the electronic effects) as a guide and plotted them against the logarithm of the reaction rates.
So, he based his scale on the following reaction reaction:

Later Hammett decided not to use the $pK_{\mathrm{a}}$s themselves for his correlation but defined a new parameter, which he called $\sigma$.
This $\sigma$ shows how electron-donating or -withdrawing a group is relative to $\ce{H}$ as a ratio of
the $\log K_a$s or the difference of the $pK_{\mathrm{a}}$
between the substituted benzoate and benzoic acid itself.
If the acid required to determine $\sigma$ for a new substituent was not available, $\sigma$ could be determined by correlation with other reactions.
A different value of $\sigma$ for any given substituent was needed for the meta and para positions (sometimes called $\sigma_{\mathrm{m}}$ and $\sigma_{\mathrm{p}}$, respectively).
The equation for $\sigma$ is

\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ K_{\mathrm{a}}(\ce{X-C6H4COOH}) }{ K_{\mathrm{a}}(\ce{H-C6H4COOH}) } \right) = pK_{\mathrm{a}}(\ce{X-C6H4COOH}) - pK_{\mathrm{a}}(\ce{H-C6H4COOH})
\end{equation}

or in short notation

\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ K_{a} (\ce{X}) }{ K_{a} (\ce{H}) } \right) = pK_{\mathrm{a}}(\ce{X}) - pK_{\mathrm{a}}(\ce{H})
\end{equation}

where $\ce{X}$ is the functional group, i.e. the EWG or EDG, whose effect on the reaction rate shall be evaluated.
It is also possible to determine the $\sigma$ values via the logarithm of the ratio of the rate constants for the aforementioned reaction with substituent $\ce{X}$ (in short notation: $k_{\ce{X}}$) and with $\ce{X}=\ce{H}$ (in short notation: $k_{\ce{H}}$)

\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ k_{\ce{X}} }{ k_{\ce{H}} } \right)
\end{equation}

Update

AcidFlask's comment reminded me that I forgot to say something about $\sigma^{+}$ and $\sigma^{-}$ values.
As mentioned above the "normal" $\sigma$ values are based upon the difference of the $pK_{\mathrm{a}}$ between a substituted benzoate and benzoic acid itself.
But for benzoates one cannot draw resonance structures that delocalize the negative charge onto the benzene ring via the $\pi$ electron system.
Yet, many reations of interest create negative or positive charges that can be stabilized by delocalization via resonance with the substituent.
For these reactions, one finds that Hammett plots using $\sigma$ values have considerable scatter.
Therefore, two new substituent effect scales were produced, one for groups that stabilize negative charges via resonance ($\sigma^{-}$), and one for groups that stabilized positive charges via resonance ($\sigma^{+}$).
The $\sigma^{-}$ scale is based upon the ionization of para-substituted phenols (I've found different accounts for this. According to the German Wikipedia the $\sigma^{-}$ scale is based upon the ionization of para-substituted anilins.),

for which groups like nitro can stabilize the negative charge via resonance.
The $\sigma^{+}$ scale is based upon the heterolysis ($\mathrm{S}_{\mathrm{N}}1$) reaction of para-substituted cumyl chlorides (phenyldimethyl chloromethanes),

for which groups like amino can stabilize the positive charge via resonance.
The $\sigma^{\pm}$ values can then be evaluated by the aforementioned equation

\begin{equation}
\sigma_{\ce{X}}^{\pm} = \log \left( \frac{ K_{a}^{\pm} (\ce{X}) }{ K_{a}^{\pm} (\ce{H}) } \right) = \log \left( \frac{ k_{\ce{X}}^{\pm} }{ k_{\ce{H}}^{\pm} } \right)
\end{equation}

but with the rate constants $k_{\ce{X}}^{\pm}$ and $k_{\ce{H}}^{\pm}$ (or the equilibrium constants $K_{a}^{\pm} (\ce{X})$ and $K_{a}^{\pm} (\ce{H})$) measured for their respective defining reactions.


The Verb for Polymorphism How can you elegantly describe that a polymorphic substance changes its structure?

  It polymorphs
  It morphs
  It undergoes polymorphic transition
  It experiences a morphic change

Or what would be even better terms?

It "transitions from one polymorph to another"

It "transitions to another polymorph"

It "transition to polymorph beta"

It "transforms to another polymorph"

It "transforms to polymorph beta"

I like "transitions" better than "transforms", but I've seen both


What are the byproducts in a Suzuki reaction? I am looking precisely at a coupling of a trifluoroborate with an aryl bromide. I know the mechanism of the Suzuki reaction, yet I cannot find any mention of the side products or degradation products that one could obtain if the coupling fails. Maybe those products are obvious for those who know well how palladium reacts, but this is not my case. I tried to find a book that would give me a feel for what might happen and could find anything close enough.

I'm expecting a protodeboration as a side reaction, yet I can't find any source nor mechanism. Could anyone shed some light on the subject?

It is postulated that the trifluroborates are not the transmetallating specis, rather than to be the source of slow hydrolysis to yield boronic acids:

(source 1, slide 41; citing Molander and Biolatto, 2003 (Ref.1))
which then enter the catalytic cycle:

(preview in Kürti & Czakó, page 448 bottom (Ref.2)
Organic Reactions (vide infra) explicitly states that complete hydrolysis of the trifluoroborates is necessary to allow successful transmetallation.
In addition, the fluoride anion is said to interact in the following way with the palladium catalyst:

(source 2)

References:

Gary A. Molander, Betina Biolatto, "Palladium-Catalyzed Suzuki−Miyaura Cross-Coupling Reactions of Potassium Aryl- and Heteroaryltrifluoroborates," J. Org. Chem. 2003, 68(11), 4302-4314 (https://doi.org/10.1021/jo0342368).
László Kürti, Barbara Czakó, In Strategic Applications of Named Reactions in Organic Synthesis: Background and Detailed Mechanisms; Forwarded by E. J. Corey, Introduction by K. C. Nicolaou; Elsevier Academic Press: Burlington, MA, 2005 (ISBN-13: 978-0-12-429785-2).


Should bromine water be called a solution? Bromine water is a reagent which is used to test for unsaturation in organic compound. It is $2.8~\%$ bromine in water. In many places, it is refer to as bromine solution.

But it is observed that adding bromine to water gives hypobromous acid ($\ce{HOBr}$) and hydrobromic acid ($\ce{HBr}$) via a disproportionation reaction.

$$\ce{Br2 + H2O <=>> HOBr + HBr}$$

So, if bromine reacts with water to form a mixture of acids, how can it be called a solution? Components of solution do not react with each other.

Much like with chlorine or iodine, the disproportionation of bromine can occur in pure water, but it is not fast. As you can see, the reaction generates acids, so without any base present the equilibrium is pretty close to the bromine side. You do need bases to disproportion bromine in any meaningful amount — much like chlorine which is generally quenched with $\ce{NaOH}$ solution to generate hypochlorite solutions.

So if you just put bromine and water together, they may react very slightly, but not much, and most of the bromine will sink to the bottom (assuming you used a lot) while the lesser part will dissolve in the water first. Since the concentration of $\ce{Br2}$ is much higher (and we’re talking orders of magnitude) that those of $\ce{Br-}$ and $\ce{OBr-}$, it is best to consider this a bromine solution and not a bromide/hypobromite one.


Any chemicals combines readily oxygen? I am designing an industrial process which can remove oxygen from air.
This chemical should combine readily with oxygen in the presence of heat.

In particular, I am looking for a chemical with the following properties:

Combines readily with oxygen, thus removing the oxygen in air
Inexpensive
Environmentally friendly

For point #1, heating is available if it is an endothermic reaction.

If $\ce{CO2}$ is acceptable in the feed gas, along with $\ce{CO}$ and a bit of left-over oxygen, then carbon (coal, charcoal etc.) might do. It's ignition temperature is ~1,000 K.

Magnesium is more expensive, but leaves less oxygen and has no gaseous byproduct. It's self-ignition temperature is ~800 K.

Many other materials could be used; it really depends on how the gas is to be used.


Gibbs energy and chemical equilibrium : false proof of spontaneous generation, but why? Let’s assume we have a reaction medium containing only simple molecules such as $\ce{H2O}$, $\ce{CH4}$, $\ce{NH4+}$ (and maybe some more).

Now we consider the following equation ;

  $\Delta G_\mathrm r = \Delta G_\mathrm r^{\circ} + RT \ln Q_\mathrm r$

were $\Delta G_\mathrm r$ is the Gibbs energy change associated to a chemical reaction occurring in this reaction medium, $\Delta G_\mathrm r^{\circ}$ the standard Gibbs energy change of this reaction and $Q_\mathrm r$ the mass action ratio of the reaction in the reaction medium at a given time.

Let’s consider the reaction consisting of the synthesis of a very complex molecule from the simple molecules of the reaction medium (for example, a protein). Whatever the complex molecule is, it is so complex that we are sure that it is absent from the reaction medium, i.e. its concentration is exactly 0. Also, the standard Gibbs energy change of this reaction must be very high (positive), but not infinite.

If the concentration of the complex molecule starts at 0; then $Q_\mathrm r$ is 0; then $RT \ln Q_\mathrm r$ is $-\infty$ ; as $\Delta G_\mathrm r^{\circ}$ is finite, $\Delta G_\mathrm r$ is $-\infty$. So in these conditions, the synthesis of at least one copy of our very complex molecule is infinitely exergonic. This reasoning could be applied to any molecule whose concentration in the reaction medium is 0 at a given time and whose synthesis from the reaction medium is theoretically possible. In the thermodynamics courses I attended, a negative Gibbs energy change was associated with the “favourableness” of a reaction, so those reactions would be “infinitely favourable”.

Of course, this is absurd. But why?

Within the context of the equation in question, $\Delta G$ refers to the change in free energy between two specific thermodynamic equilibrium states:

State 1:  Stoichiometric quantities of pure reactants in separate containers at temperature T and individual specified pressures

State 2:  Corresponding stoichiometric quantities of pure products in separate containers at temperature T and individual specified pressures

In the case of $\Delta G^0$, the specified pressures are all 1 atmosphere.  This means that Q = 1, and $\Delta G =\Delta G^0$.

If the specified pressures in states 1 and 2 happen to correspond to the partial pressures of an equilibrium mixture of reactants and products, $\Delta G$ will be equal to zero.  In this case, if one had an equilibrium mixture of the reactants and products, the partial pressures in the equilibrium  mixture could also be held in equilibrium through individual semipermeable membranes with the gases in the separate containers of reactants and products (with no tendency for the reactants or products to enter or leave the reaction mixture through the semipermeable membranes).


Why do we not include water in equilibrium expression calculations? Why do we not include water in equilibrium expression calculations? And can you go into detail on why pure liquids and solids have a constant concentration as well, please and Thanks.

Usually when you have water in a reaction it is taken in excess, i.e in a much larger quantity compared to the other reactants. Ultimately the change in concentration of water is negligible. As an example you can consider the hydrolysis of ethyl acetate to ethanol and ethanoic acid. For better understanding consider the initial amount of ethyl acetate to be 0.01mol and that of water as 10 mol.
This however is valid only when water is taken in excess. If the amount of water and the other reactants is comparable, then water has to be included in the equilibrium expression.


Two different containers full up with pure water and salt water in osmosis. What happens? I would like to understand what happens when there are two containers which are connected together and separated by only semipermeable membrane. Then in one container I fill up (all the way to the top) pure water and in the second container I put water with a lot of salt ($\ce{NaCl}$). Now In container 2 it said to be an high osmotic pressure.

Then my question is: Is the water in container 1 will drain out / will be pushed out by the osmotic pressure in container 2? (see sections a,b,c below).

■  My question popped up after reading the following things in a book called "RBS - physiology" (6t edition p. 5):

  Osmosis is the flow of water across a semipermeable membrane from a
  solution with low solute concentration to a solution with high solute
  concentration.

Example of osmosis:

  a. Solutions 1 and 2 are separated by a semipermeable membrane.
  Solution 1 contains a solute that is too large to cross the membrane.
  Solution 2 is pure water. The presence of the solute in solution 1
  produces an osmotic pressure.

  b. The osmotic pressure difference across the membrane causes water to
  flow from solution 2 (which has no solute and the lower osmotic
  pressure) to solution 1 (which has the solute and the higher osmotic
  pressure).

  c. With time, the volume of solution 1 increases and the volume of solution 2 decreases.

In your described scenario, there will be osmotic flow in an attempt to equalize the chemical potential of the solvent (water in this case, assuming the membrane is only permeable to water). The effect of the osmotic flow will be that pure water from container 1 will flow into container 2. Since this is a closed system, one will observe the liquid level rise in container 2 and fall in container 1, and the amount of displacement will be a function of how different the chemical potential of the water is between container 1 and container 2. There is no classical force that will push or pull the liquid from one container to another, osmotic flow is an instance of diffusion which is governed by the concentration gradient that is present.


Can a negative pH exist? I am programming a small model to simulate diffusion of $\ce{H+}$ ions in human tissue. It's quite a simplistic model based on a naive algorithm, and I'm mostly doing it to familiarize with the concept of agent-based models.

My space is an $n\times n$ grid where each position $(x;y)$ hosts $h$ $\ce{H+}$ ions.

I understand that $\ce{pH}$ can be calculated based on $h$ as $-\log(h)$. As I want my initial $\ce{pH}$ to be stable at average human $\ce{pH}$, I'd set $h$ across the matrix to be $10^{7.5}$. (In fact, as I don't want to simulate $10^{7.5}$ protons I decided each particle will represent a bulk of $10^5$ protons and made my equation $-\log(10^5\cdot h)$. )

However, I do not understand the equation, wouldn't it at some point yield a negative $\ce{pH}$ as $-\log(x) < 0$ for $x > 1$?

Yes, pH can be negative.

The official definition of pH is: negative log of H+ activity, not negative log of H+ concentration, however.  See pH Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+] J. Chem. Educ., 2006, 83 (5), p 752.

By either definition pH can be negative in concentrated solutions of strong acids.  For example 5M HCl would have a negative pH.


Addition of oxygen molecule to given aromatic compound I did figure out that the hydrogen atom is reactive as it has three electron donating groups attached to it, I know it reacts. But the answer given is that 6 different compounds can be formed (minor and major; including stereoisomers). But I need some help figuring them out.

The three hydrogen bond at -OCH3 would be reactive and
-CH3 at attached directly to ring would react so 6 products possible.


Would a salt bridge work through which the ions could not diffuse into the half cells? Instead of covering the ends of the salt bridge in an electrochemical cell with a membrane, one could also plug them on both sides with a conductive non-porous material, for example gold, which doesn't go into solution. Would the accumulation of the ions at the plugs and their electrostatic force field be enough to keep the cell running?

It is a very good thought experiment as to why we don't cap the salt bridge ends with gold plugs. By capping, you are essentially providing an interface which can source and sink for electrons in the cells, thus each end will act as an electrode in the cell. One end of the salt bridge will act as a cathode and one end will act as an anode (strange as it may sound!)

As to your comment above "I want them to be non porous and able to let the electric field through that the accumulating charges have. I think a closed glass tube would also work". No a close tube will not work. You need the ions or electrons to move in order to observe current flow. A closed glass tube will have an infinite resistance and hence no current can flow!


Why do laws of thermodynamics "outweigh" any other law ever formulated? "It is the only physical theory of universal content concerning which I am convinced that within the framework of the applicability of its basic concepts, it will never be overthrown." - Albert Einstein
"Nothing in life is certain except death, taxes and the second law of thermodynamics". - Seth Lloyd
“If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell's equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation." - Arthur Eddington
"“Every question or effect has the right to exist if it does not contradict the second law of thermodynamics”. - Boris Pavlovich
“A good many times I have been present at gatherings of people who, by the standards of the traditional culture, are thought highly educated and who have with considerable gusto been expressing their incredulity at the illiteracy of scientists. Once or twice I have been provoked and have asked the company how many of them could describe the second law of thermodynamics. The response was cold: it was also negative. Yet I was asking something which is the scientific equivalent of: Have you read a work of Shakespeare's?” - C.P. Snow

I find these particularly interesting but there are many others which mark the supremacy of the laws of thermodynamics over any other law ever formulated by man.
I have just begun to learn the laws of thermodynamics and it is indeed very different and "interesting" but so was Newton's laws of motion, Faraday's laws, Heisenberg's uncertainty principle, Einstein's theory of relativity, then why are laws of thermodynamics considered so immutable and supreme to other laws. What is it that makes all these great people hold them as so impactful and perpetual and makes these laws outweigh others?

The laws never outweigh each other, nor do they have to. Newton's laws of motion, Maxwell's equations of electromagnetism, Einstein's theory of relativity are all equally fundamental, in that they've never been overthrown within their respective frameworks of applicability. Sure, these frameworks have their limits: Newton fails at near-light speeds where Einstein takes over, Maxwell and Einstein fail at quantum limit. So what? Thermodynamics has its limits as well. If you don't believe this, then good luck trying to apply it to a single molecule.

If anything, the laws of thermodynamics are less fundamental, in that they are true only statistically, that is, with less precision than the rest. With relatively little effort, you can measure the mass of a body to six-digit precision. Can you do the same to its temperature?


Quantitative measurement of ring strain While looking for data about ring strain in cycloalkanes, I came across two types of data:

Ring strain per methylene group (let's denote this quantity by X): The difference between heat of combustion of the cycloalkane per methylene group and the reference value of $\pu{658.6 kJ/mol}$.
Total ring strain (let's denote this quantity by Y): The difference between molar heat of combustion of the cycloalkane and the reference value of $\pu{658.6 kJ/mol}$ multiplied by number of carbon atoms in the cycloalkane.

From what I have read till now, it appears that the first quantity is the better representation of ring strain. (I say this on the basis that Y has the same value for cyclopentane as well as cycloheptane but still cycloheptane is considered to be less strained as it has lower value of X).

I want to know why X is a better measure for ring strain than Y. When we talk about a ring, we consider it as a whole; then why do we calculate ring strain per methylene group?

Let's look at following table from a Yale website:
$$
\begin{array}{c|ccc}
 & \ce{(CH2)3} & \ce{(CH2)4} & \ce{(CH2)5} & \ce{(CH2)6} & \ce{(CH2)7} \\
\hline
\Delta H_\mathrm{Combusion} \ (\pu{kcal/mol}) & -499.8 & -656 & -793.5 & -944.6 & -1108.3  \\
\Delta H_\mathrm{Combusion} \text{ per } \ce{-CH2}-  \ (\pu{kcal/mol}) & -166.6 & -164 & -158.7 & -157.4 & -158.3 \\
\text{Ring strain per } \ce{-CH2}-  \ (\pu{kcal/mol}) & 9.2 & 6.6 & 1.3 & 0 & 0.9 \\
\text{Total ring strain}  \ (\pu{kcal/mol}) & 27.6 & 26.3 & 6.5 & 0 & 6.3 \\
\hline
\end{array}
$$
Now, just concentrate on ring strain values for cyclopropane and cyclobutane only:

Ring strain per $\ce{-CH2}-$ (X-values) are $9.2$ and $\pu{6.6 kcal/mol}$, respectively.
Total ring strains (Y-values) are $27.6$ and $\pu{26.3 kcal/mol}$, respectively.

According to Y-values, cyclopropane and cyclobutane are equally unstable (relatively speaking), while according to Y-values, cyclopropane is significantly less stable than cyclobutane. Which valuse should be more reliable? Let's look at experimental findings:

Although cyclopropanes are far less reactive than alkenes, they react with chlorine and bromine under polar conditions to form corresponding addition products, 1,3-dihalopropanes, as the predominant product (Ref.1). To my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.
Cyclopropane also reacts with HCl and HBr to form corresponding addition products, 1-halopropanes (Ref.2; Reactions of bromocyclopropane with hydrobromic acid has given mixture of 1,1-, 1,2- and 1,3-dibromopropanes). Again, to my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.

So which one you think should be more reliable and why?
References:

Joseph B. Lambert, Erik C. Chelius, William J. Schulz, Jr., Nancy E. Carpenter, “Polar bromination and chlorination of cyclopropane,”  J. Am. Chem. Soc. 1990, 112(8), 3156–3162. (https://doi.org/10.1021/ja00164a043).
Choi Chuck Lee, Bo-Sup Hahn, Kwok-Ming Wan, D. J. Woodcock, “Reactions of cyclopropane with hydrochloric acid and of bromocyclopropane with hydrobromic acid,” J. Org. Chem. 1969, 34(10), 3210–3211. (https://doi.org/10.1021/jo01262a097).


Finding pH of tri-protic acid I am studying for my final and I am given this problem:

  A solution of sodium phosphate is made from 10.5 g sodium phosphate in
  150 mL of water. What is the pH of this solution

Given: Ka1, Ka2.

The solution says that we can simply calculate this pH with the following eqution:
$$pH=\frac{pK_{a1}+pK_{a2}}{2}$$
However this makes absolutely no sense to me, as this is the pH at the equivalence point.  What am I missing as towhy this equation can be used?

amphoteric

  A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.

from the IUPAC goldbook.

For ampholytes like $\ce{NaH2PO4}$ the $\ce{pH}$ is independent on the concentration (in first approximation). Consider the reactions of that are possible:
\begin{align}\ce{
H2A-    + H3+O &~<=> H3A     + H2O  \tag1\\
H2A-    + H2O  &~<=> HA^{2-} + H3+O \tag2\\
HA^{2-} + H2O  &~<=> A^{3-}  + H3+O \tag3\\
}\end{align}

As the first approximation, we will assume that everything happening in equation $(3)$ is negligible, as is will only very little influence the concentration of $\ce{HA^{2-}}$. Therefore we can focus on the acidity constants $K_{a_1}$ of $(1)$ and $K_{a_2}$ of $(2)$:
\begin{align}
K_{a_1}  &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})}  \tag4\\
K_{a_2}  &= \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \tag5\\
\end{align}
We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels.
\begin{align}
K_{a_1} \cdot K_{a_2} &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})} \cdot
  \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \\
K_{a_1} \cdot K_{a_2} &= c^2(\ce{H3+O})\cdot  \frac{c(\ce{HA^{2-}})}{c(\ce{H3A})} \tag7\\
\end{align}
Now another major assumption is that the reactions $(1)$ and $(2)$ are happening to the same extent and therefore we can approximate
$$c(\ce{HA^{2-}})=c(\ce{H3A})\tag8$$
and rewrite $(7)$ as
\begin{align}
c^2(\ce{H3+O}) &= K_{a_1} \cdot K_{a_2}\\
c(\ce{H3+O})   &= \sqrt{K_{a_1} \cdot K_{a_2}}\\
\ce{pH}        &= \frac12\left(\mathrm{p}K_{a_1}+\mathrm{p}K_{a_2}\right).
\end{align}

However, this equation only works as a first approximation. Usually the coupled equilibria are very complex and in most instances a computer needs to be involved to calculate it accurately. More information on polyprotic acids.


Can all NMR spectrometers perform special techniques such as COSY, NOESY, HETCOR,..? Do you need special spectrometers for:

probing nuclei other than hydrogen or carbon?
performing 2D spectroscopy such as COSY, NOESY, HETCOR, HSQC,..

Or are these techniques usually available on standard NMR spectrometers?

It depends on a lot of things, software-wise

some of these techniques are simply a software option
you can implement pulse sequences yourself, if that´s possible with your control software
you need to have the right software to transform and evaluate the data
... or write it yourself (not exactly rocket science if you know your way with Matlab/Octave etc.)

and also regarding your hardware

you can have several probes for one spectrometer

they can be single channel, or dual ("HX" usually), triple ... channel
the "X" channel can be tuneable over a wide(r) range, or you are limited to e.g. carbon

of course you need a transceiver with enough channels, which can be used at the wanted frequencies
and a suitable preamp for every channel
some techniques need more special hardware, like CPMAS, pulsed gradients, cryo etc.
low field spectrometers on the market (afaik) have no user exchangable parts, so you are stuck with the hardware you bought

But to answer your immediate question: I guess with basically any living (as in "not undead") spectrometer, all those you named above should be possible, provided you can tune the X channel to the het nucleus you are after. They are the basic standards.

If you want to know a lot more, 200 and More NMR Experiments by Berger and Braun is probably your book.