How to calculate the mass of iron with a temperature of 325 °C needed to melt ice and raise the temperature of water from 0 to 22 °C? A large well-insulated container holds a mixture of $75\ \mathrm g$ of ice and $100\ \mathrm g$ of water at $0\ \mathrm{^\circ C}$. Using the data given below, calculate the mass in grams of solid iron, $\ce{Fe}$, at $325\ \mathrm{^\circ C}$ that you would have to add to this mixture in order to melt all of the ice and raise the temperature of the resulting $175\ \mathrm g$ of water to $22\ \mathrm{^\circ C}$.
Specific heat capacity of water is $c_p(\ce{H2O})=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$
Specific heat capacity of iron, $\ce{Fe}$, is $c_p(\ce{Fe})=0.45\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$
Molar enthalpy of fusion (melting) of water is $\Delta_\text{fus}H_\mathrm m(\ce{H2O})=6.01\ \mathrm{kJ\ mol^{-1}}$
$M(\ce{H2O})=18\ \mathrm{g\ mol^{-1}}$

My answer was that you needed $25041.67\ \mathrm J$ to melt the ice and $16108.4\ \mathrm J$ to raise the temperature to 22 degrees Celsius.
Therefore $mc \Delta T(\ce{Fe}) = -(25041.67\ \mathrm J + 16108.4\ \mathrm J)$ and dividing this by the mass of iron and the change in temperature results in a mass of $301.8\ \mathrm g$.
Is this correct?

I can not see anything wrong with your calculations at all.  For posterity, the following is a rough method to confirm:

Melting ice

Enthalpy of fusion $\Delta H_\text{ice} = 6.01\ \mathrm{kJ\ mol^{-1}}$
Mass of ice $m = 75\ \mathrm g$
Molar mass of $\ce{H2O}$ $M = 18\ \mathrm{g\ mol^{-1}}$

Hint: usually, when you are given mass and molar mass, then the amount of the substance is required.

Note that the Enthalpy of fusion ($\Delta H_\text{ice}$) is measurement in $\mathrm{kJ\ mol^{-1}}$, and as the ice is measured in grams, and you are provided with the molar mass, you'll need to use the formula:

$$n = \frac{m}{M}$$

hence:

$$n_\text{ice} = \frac{75\ \mathrm g}{18\ \mathrm{kJ\ mol^{-1}}}$$

$$n_\text{ice} = 4.1666\ \mathrm{mol}$$

Therefore, the enthalpy of fusion:

$$\Delta H_\text{fus} = 6.01\ \mathrm{kJ\ mol^{-1}} \times 4.1666\ \mathrm{mol} = 25.04167\ \mathrm{kJ} = 25041.67\ \mathrm J$$

Heating water

Specific heat capacity of water $c_p = 4.184\ \mathrm{J\ g^{-1}\ ^\circ C}$
Mass of water $m = 175\ \mathrm g$
Temperature change $\Delta T = 22\ \mathrm{^\circ C}$

So:

$$Q = mc_p \Delta T$$

$$Q = 175\ \mathrm g \times 4.184\ \mathrm{J\ g^{-1}\ ^\circ C} \times 22\ \mathrm{^\circ C} = 16108.4\ \mathrm J$$

Total energy used by the water/ice mixture

$$Q_\text{total} = 25041.67\ \mathrm J + 16108.40\ \mathrm J = 41150.07\ \mathrm J$$

Iron

Specific heat capacity of iron $c_p = 0.45\ \mathrm{J\ g^{-1}\ ^\circ C}$
Mass of water $m = X$ (unknown)
Temperature change $\Delta T = 325\ \mathrm{^\circ C} - 22\ \mathrm{^\circ C} = 303\ \mathrm{^\circ C}$ decrease

As, according to conservation laws:

$$Q_{\ce{H2O}} = Q_{\ce{Fe}}$$

Then:

$$41150.07\ \mathrm J = X \times 0.45\ \mathrm{J\ g^{-1}\ ^\circ C} \times 303\ \mathrm{^\circ C}$$

Or:

$$41150.07\ \mathrm J = 136.35\ \mathrm{J\ g^{-1}}\times X$$

Then:

$$X = \frac{41150.07\ \mathrm J}{136.35\ \mathrm{J\ g^{-1}}}$$

$$X = 301.8\ \mathrm g$$


pH of a weak base calculation Calculate the $\ce{pH}$ of a $100$ $\ce{mL}$ solution containing $0$$.0375$ moles of the weak base, Sodium Benzoate, $\ce{C6H5COONa}$.

  $$\ce{C6H5COO + H2O -> C6H6COO- + OH-}$$

$$\frac{0.0375 ~\mathrm{mol}}{.100~ \mathrm{L}} = 0.375~M$$
$$\ce{pOH} = 0.43$$
$$\ce{pH} = 13.57$$

Is this right? Thank you.

Welcome to Chemistry SE!

To do this specific problem you need to use the $K_{b}$:

$$\ce{\frac{x^{2}}{.375-x}}=K_{b}$$

Finding the $K_{b}$ and solving for x will give you the concentration of $\ce{OH-}$. Remember to take the $-log$ of that and solve for the $pH$.

Also next time please show some work and take your time to format these questions so people can try to reply back to you.


Electron configurations for d-block and p-block Question 1: Why is it that when you get to the transition metals (or the d-block), the energy of the $\ce{3d}$ orbitals becomes slightly less than that of $\ce{4s}$, making $\ce{4s}$ have the electrons with the most energy?

Also, is it correct to write electron configurations for transition metals with the $\ce{4s}$ orbital being the outermost one? For example like this $\ce{Sc = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$ instead of this $\ce{Sc = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$

Question 2: If my first thought is true, then do you write electron configurations for the p-block just like the d-block, with the $\ce{4s}$ orbital being the outermost one? Or do you follow the regular pattern of letting the $\ce{3d}$ block comes after the $\ce{4d}$ block?

For example, would it be like this $\ce{Ga = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^1}$ or would it be like this $\ce{Ga = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^1}$

According to google, if you search up "scandium electron configuration" it would say my first one). If you search up "gallium electron configuration", you would see my second one)

I did not completely understand you question here, but I am gonna give it a try, so maybe I might answer your question.

I do not think that it matters how you write them, but its just depends on how you fill them. The order for filling orbitals is as follows:
$$\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s}$$

In high school, I just used to write the $(n+1)s$ before the $nd$. Because with some exceptions, the rule says that you will always fill the $(n+1)s$ orbital before you fill the $nd$ orbital. Meaning that you fill $\ce{4s}$ before you fill $\ce{3d}$, $\ce{5s}$ before you fill $\ce{4d}$, and so on.

The reason is that the $\ce{s}$ actually has a lower energy than the $\ce{d}$ orbital. You wrote that wrong in your question. That is the only you fill the s before the d, and when you make ions, you will have to remove the electrons from s before you do from $\ce{d}$.

But like I said, you can write them like you want as long as you have filled them correctly in the right order. Likewise once you have completely filled your $ns$, only then you can move to fill the $(n-1)d$ orbital and then again proceed to fill the $\ce{np}$ orbital. Its just how we fill them.

In some books, you will find that they write the s as the outermost but some write d as the outermost. Frankly, in my book and myself have learned to write $\ce{d}$ at the end. But I do not think that writing it matters.

For example, take a look at this site. They write $s$ orbital at the end of the configuration and also note the energy diagram. But in the following Wikipedia website, they have written the $\ce{d}$ at the end instead of $\ce{s}$. So basically, the writing of the orbitals doesnt matter, but the order of filling matters.


Quaternary Ammonium salts and peroxy acid I came across this question in my test:

  This compound

  is treated with $\ce{CH_3CH_2I}$ and then with $\ce{H_2SO_5}$ and heated.
  The product is:
  A)

  B)
  $\ce{CH_2CH_2}$
  and two other impossible options.

I first don't understand the role of $\ce{H_2SO_5}$ in this. I know about Hoffmann elimination of quaternary ammonium hydroxides but in that we use silver oxide to form the hydroxide from the iodide. And then heating is done for elimination.

Secondly the answer given is option B but I am not able to understand why is that. The second example in this page says otherwise Quaternary ammonium salts
Does this have anything to do with the $\ce{H_2SO_5}$?

My third doubt is a general one. What happens in general when different alkyl groups are attached to the the nitrogen in a quaternary ammonium hydroxide. Which one undergoes elimination first? Are there any rules to determine that?

I believe this is an example of a Cope elimination mechanism here

The EtI is making the tertiary amine. The persulfate is taking the tertiary amine to the N-oxide. Heating this produces ethene and hydroxylamine of the starting amine.

This is consistent with the Cope as "The sterically demanding amine oxide function reacts preferentially with the more easily accessible hydrogens"


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

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

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

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


Direction of external cell when voltage is supplied by it to Daniel cell to oppose cell potential The above diagram is from our textbook for a Daniel cell.

This is a diagram from our textbook showing a Daniel cell when an external voltage opposing the cell potential is a applied. Here ${E}_{ext}$ is less than cell potential.

This diagram show the Daniel cell when ${E}_{ext}$ is greater than cell potential. What I do not understand is why has the direction of the cell been reversed in the last diagram?
Is this an error? If it is not an error, then why are the $\ce{Cu^2+}$ ions in the solution NOT accepting the electrons provided by the cell and getting reduced? Why is the Copper electrode acting as an anode?

Suppose two identical Daniell cells are connected with the zinc electrodes electrically connected through an amperemeter, and the two copper plates also electrically connected. Of course no currant will be measured by the ammeter. But the tension between the plate is independent on the connections. Now if one of the cell has a higher nominal voltage, maybe because of aa lower zinc ion concentration, this "stronger cell" imposes its capacity of producing electrons to the other one. Electrons will be flying through the ammeter. So the weakest cell has to reverse the sense of its reactions on its plates. The cell will work as in an electrolysis. Zinc ions $\ce{Zn^{2+}}$ will be reduced into metallic zinc in this cell. Slowly the two zinc ion concentrations will tend to equalize themselves in both cells.


How to distinguish palmitic acid from lauric acid? I'm reacting either palmitic acid or lauric acid ($\ce{C16H32O2}$ or $\ce{C12H24O2}$) with $\ce{NaOH}$, sodium hydroxide.

I'm wondering if there's any way to tell which one of those two acids it is after I've already reacted it with NaOH. It has to be a relatively short process, as I won't have much time. I'm doing this for a lab.

In case you feel uncomfortable about what you may call "helping me cheat", please note that my assignment states that I can have online help:

  You may look up information to complete your flow-chart, just be sure to cite your sources...

Note: it doesn't have to be NaOH, it could be any base, I just picked that one. If you think another one would work better, recommend that as well, please.

Consider the reaction which happens, what is produced and then what is different about the products? The key has to be in their properties.

The saponification reaction which takes place will produce a salt, in this case it will make sodium palmitate and sodium laurate. These two salts have very different solubilities and this makes a marked difference in the lathering ability. Sodium laurate is higher and thus lathers more readily.

For more details :

Solubility data

Saponification

Scientific Soapmaking: The Chemistry of the Cold Process
By Kevin M. Dunn


Will trenbolone acetate in the presence of methanol yield trenbolone acetate if distilled water is subsequently added thereafter? I have 2 grams of trenbolone acetate. I was told to add it to 20 ml of methanol. Then wait until it is fully dissolved. Then distilled water is to be added yeilding a precipitate that is supposed to still be trenbolone. Can anyone shed light on this?
Here is the link.

The method will not convert trenbolone acetate to trenbolone, the ester is not cleaved under these conditions.

It is however possible to separate the steroid from a mixture with other, water-soluble compounds this way.

Unless you are a veterinarian and use trenbolone or its acetate as a growth promoter for cattle, you might want to consider the legal situation in your country. As far as Germany is concerned, the possession of as much as 2 g might lead to real trouble. Here, the non-punishable, non-significant amount is in the 150 mg range (for its use in sports).


Question on organic chemistry where we are required to find enol content Which of the following compounds is expected to  have maximum enol content in ethanol?
(A) $\ce{CH3COCH2CO2CH3}$
(B) $\ce{CH3COCH2CO2C2H5}$ ✔(I think this is the correct answer)
(C) $\ce{CH3COCH2COCH3}$
(D) $\ce{CH3COCH(CH3)COCH3}$

I think the ketoesters will have greater enol content than the diketones because they have stronger $-I$ effect resulting in greater positive charge on the hydrogen atoms of the $\ce{CH2}$ group in between the ester and ketone group. This leads me to believe the hydrogen will be lost easily. Now between option one and two I think option (A) makes no sense because it will be converted to option (B) via trans esterification, so I have marked option (B). Please let me know where I am going wrong and the correct concept involved if what I think is not right.

Thermodynamically, the greater enol content can be associated with greater acidity. Thus, the question can be rephrased as ‘which is the strongest acid of these four?’ We can ignore the interconversion of the methyl ester A into the ethyl ester B — it will happen because we are in ethanolic solution, but extremely slowly as we do not have any strong bases present. (And even if we had strong bases, these would deprotonate A more further reducing the susceptibility of nucleophilic attack on the carboxy group.)

For reference and posteriority, these are the compounds we are talking about:

methyl acetoacetate (A)
ethyl acetoacetate (B)
acetoacetone (C)
3-methylacetoacetone (D)

Simply by looking at the names we can see how highly related they are and how much subtle differences must matter. We can predict all four of them to be rather acidic since if the inter-carbonyl hydrogen is released and an enolate is formed, this enolate will always be in contact with an electron-withdrawing π system on the right-hand side: either another keto-group or an ester group.

That immediately gives us our first hint to compare the four: an ester group is less electron-withdrawing than a ketone, so we predict A and B to be less strong acids than C and D. This is convenient as it means we do not have to concern ourselves with the acidity difference of a methyl versus ethyl ester.

The difference in acidity between C and D is also subtle but more obvious. In both cases, the corresponding enolate anion would be planar. However, the methyl group is much larger than a hydrogen atom meaning that the enolate of D is subject to 1,3-allylic strain — which cannot be lessened without destroying planarity and thereby reducing conjugation. Thus, the enolate of C is more stable than the enolate of D.

This is backed by acidity data from the Bordwell tables — see table.

$$\textbf{Table 1: }\text{acidity data of the four compounds}\\
\begin{array}{lcccc}\hline
\text{solvent} & \mathrm{p}K_\mathrm{a}(\textbf{A}) & \mathrm{p}K_\mathrm{a}(\textbf{B}) & \mathrm{p}K_\mathrm{a}(\textbf{C}) & \mathrm{p}K_\mathrm{a}(\textbf{D}) \\ \hline
\ce{H2O} & -/- & 10.7 & \phantom{0}9.0 & -/-\\
\text{DMSO} & -/- & 14.2 & 13.3 & 15.1\\ \hline\end{array}$$


3 chemicals to produce light I'm actually looking for 2 chemicals that will produce a really small exothermic reaction. The 3rd chemical would be there to separate the chemicals - like oil,  water, and air.

Ideally, the 3rd chemical would be a nonreactive oxide of the other 2.

Also, they should all stay liquid on a hot summer's day.

As stated, they should produce a small exothermic reaction - I'm looking for mostly light that won't melt a glass bottle.
Melting the bottle isn't ideal, but I can find a stronger material too.

Also, the chemicals should be as nontoxic as possible. They don't have to be though.

As discussed in the comments... The two component chemiluminescence reaction:
(Solution 1) To 50 ml of deionized water add 10-50 mg of luminol, 10-20 ml of 10% ammonia solution and 1-5 ml of 3% hydrogen peroxide. (Solution 2) To 100 ml of deionized water add few milligrams of copper salt (e.g., sulfate) or few drops of blood or fresh horseradish juice. These are approximate quantities (they should work), but you can experiment with the concentrations, after all experimenting is the most fun part of chemistry. Now take the solution 2 (whole) and add few milliliters of solution 1 to it and shake it. You should be able to observe a faint glow. A brighter emission requires organic solvents (DMSO, DMF) and corrosive chemicals (KOH). Adding a third liquid that will separate the two other (as the OP suggested) is not a simple thing to do. I would do this otherwise: add some neutral ingredient to solution 2 to increase its density (possibly sugar, but I'm not sure if that will not quench the chemiluminescence) and place the solution in a container (preferably a long tube). Then carefully pour the solution 1 on the surface of the solution 2 (so that the solutions won't mix). When you shake the container the solutions will mix and produce glow. Another idea: place one solution in a container and layer it with a hot paraffin from a candle and allow it to cool down. Then you can pour the second solution on the solid paraffin. To start the reaction, simply break the thin paraffin layer by shaking (you can place some glass marbles in the container).


Why is the boat form of paraldehyde not theoretically possible? Wikipedia says the following about paraldehyde:

  Theoretically, four stereoisomeric structures are possible. The
  structures (1) and (2) are known as cis- and trans-paraldehyde. The
  structures (3) (a conformer of (2)) and (4) (a conformer of (1)) don't
  exist for steric reasons.

If this is correct, it seems that the boat conformer is not theoretically possible. I would expect the boat conformer to be possible but be much less common than the chair conformer (as is the case with glucose and cyclohexane for example), but that does not seem to be the case.

Wikipedia says that conformers 3 and 4 don't exist for steric reasons, however they are still theoretically possible. To me this sounds like the nonexixtence of boat conformation is not just because of steric reasons, but there is a more fundamental reason for that.

So, my questions are:

Is it true that the boat conformer is theoretically impossible?
If yes, why is it impossible and what makes paraldehyde different from glucose and cyclohexane in this regard?

A $\ce{CH_3}$ group is quite big. Steric hindrance is the reason.

Its not theoretically impossible, but it is energetically a lot less favourable than (1) and (2).

You wouldn't find it in nature so to speak.

to 2: A $\ce{OH}$ group is sterically a lot smaller than a $\ce{CH_3}$ group.


Why are Terminal Alkynes less reactive than internal alkynes? I got in Paula Bruice under the section Mechanism for Mercuric ion catalysed Hydration of Alkyne

Terminal alkynes are less reactive than internal alkynes towards addition of water. Hydration will take place if Hg²+ is added to acidic mixture.

Why are Terminal Alkynes less reactive than internal alkynes?

The mechanism of hydration is an electrophilic addition and requires polarization of the carbon-carbon triple bond. This task is more difficult compared to case of alkene due to it having more %s character
In case of internal alkynes the carbocation intermediate formed ( since this mechanism is electrophilic addition the intermediate is a carbocation) is stabilized by the hyperconjugative effect and the inductive effect of the alpha carbons.
This support is taken away in case of terminal alkynes as their is no alpha-carbon. Now terminal alkynes can still be hydrated but only in presence of $\ce{Hg^2+}$ ion in acidic medium because of $\ce{Hg^2+}$'s high polarizing power(because it has a pseudo inert configuration). This polarization succeeds in catalyzing the reaction.
Ref.: Morrison and Boyd, sec 12.10 (hydration of alkynes)


Affinity to mobile or stationary phase during column chromatography I have carried a column chromatography and obtained two compounds. The first one is liquid (ester) and the second one is solid (nitro) yet I know that the second compound is less polar than the first one. Why does the result differ from the knowledge about polarity (the more polar the molecule is, the slower the compound)?

I don’t know in which universe you would consider a dinitrobenzene less polar than an ethyl ester but this universe is not the correct one.

As your experimental data (the $R_\mathrm f$ values of the two eluted compounds) clearly show, the dinitrobenzene is more polar than ethyl benzoate.


Does liquid ammonia react with hydrogen gas? This may seem like a basic question to educated chemists, but I failed to get this answer on Internet searches.  I keep getting more complicated stuff like the Haber process which is not relevant.

At this website it describes how the absorption cooling process works in a propane powered refrigerator.  The system uses ammonia, water, and hydrogen in closed loop system.  In the evaporator section, it says liquid ammonia reacts with hydrogen gas thus vaporizing it and this process of ammonia boiling in hydrogen absorbs heat from the fridge box. It doesn't mention what the end products of the ammonia and hydrogen reaction are.

Is there a chemical reaction between hydrogen and ammonia here?  Or is there no actual chemical reaction, just partial pressure vaporization of ammonia liquid in hydrogen gas?

No, it does not. I do not know what they described but I am pretty sure liq. ammonia doe snot react with hydrogen gas. I believe they use hydrogen gas as a carrier but I found that to be questionable as well, since hydrogen in residential setting can be pretty dangerous.


Can gaseous hydrogen and gaseous oxygen be compressed without reaction? I'm looking into building a clean Bunsen-style burner (or rocket engine - I haven't decided yet). I will be using electrolysis to extract hydrogen and oxygen from water, but the apparatus I am thinking about building will render gaseous hydrogen and gaseous oxygen, but in a mixture. Preferably I would store them compressed, and not have spontaneous combustion.

My question is this:
Will gaseous hydrogen and gaseous oxygen react with each other if compressed in the same chamber?If so, how much pressure is required before it blows up?

The study Explosion Characteristics of Hydrogen-Air and Hydrogen-Oxygen Mixtures at Elevated Pressures includes data for pressures up to 200 bar.

Data were collected for Hydrogen-Oxygen mixtures at both 20 and 80 degrees C at pressures ranging from 1 to 200 bar.  A high voltage spark was still needed to cause explosion.

Storing a compressed mixture of hydrogen and oxygen still seems extremely unsafe to me.  I definitely wouldn't do it.


Wittig versus Corey–Chaykovsky reaction The Wittig and Corey–Chaykovsky reaction both involve an ylid, phosphonium ylid in the former versus a sulfonium ylid in the latter, attacking a carbonyl compound.

In the Wittig reaction, a concerted [2+2] cycloaddition leads to direct formation of a four-membered ring, but this does not happen with the sulfur ylids. Instead, nucleophilic addition to the carbonyl occurs to give a betaine, which then decomposes via intramolecular nucleophilic substitution by the oxyanion. Thus, in the Corey–Chaykovsky reaction, we end up with an epoxide, while the Wittig reaction affords an alkene.

Possibly formation of the strong $\ce{P=O}$ bond is a key driving force for the wittig reaction (the P=O bond strength in $\ce{Ph3PO}$ is $\pu{529 kJ mol-1}$ vs. the $\ce{S=O}$ bond strength in $\ce{Ph2SO}$ which is $\pu{367 kJ mol-1}$).

I am looking for an answer that explains the differences between the outcomes of these two reactions, and their respective intermediates. It would be great if the answer can be supported with references, and/or calculations.

Overview of mechanisms

There have been recent excellent summaries of the reaction profiles of the Wittig1 and the Corey–Chaykovsky2 reactions. For the purposes of answering this question, there's no need to go into exquisite detail about each individual step (a lot of the research has been poured into understanding stereoselectivities which isn't relevant here).

The general mechanisms are shown below, which should suffice; however, the interested reader is encouraged to consult the articles above.

Both reactions are considered to be essentially irreversible. As such, an argument based on the thermodynamic stability of the products cannot be valid.

For a kinetically controlled reaction, one should look at the activation energies of betaine vs. oxethietane or oxaphosphetane formation from the ylid and carbonyl.

Phosphorus ylids

The mechanism of the Wittig reaction has been studied very thoroughly, and the general consensus is that betaines are not intermediates. See: Which is the currently accepted mechanism of a Wittig reaction?

As described in the answers there, true betaines have never been observed as intermediates in "salt-free" Wittig reactions, and have been rigorously disproven in some cases. Betaines can be obtained by deprotonation of their conjugate acids, called β-hydroxyphosphonium salts. However, the only observed products of this deprotonation are either oxaphosphetanes3 or the alkene + phosphine oxide products4. There is a large body of evidence for a four-membered transition state leading directly to an oxaphosphetane.1

As for computational studies, Harvey and Aggarwal5 found a betaine intermediate at 14.6 kcal/mol higher than the cis oxaphosphetane for the reaction of $\ce{Me3P=CHMe}$ with acetaldehyde, but could not find a corresponding transition state. For the reaction with $\ce{Ph3P=CHMe}$, the betaine was found to be 16.2 kcal/mol higher than the oxaphosphetane, and the TS leading to the betaine 3 kcal/mol higher than the TS leading to the oxaphosphetane.

If the phosphorus betaine is never formed, the next step to form an epoxide (intramolecular nucleophilic substitution) cannot occur. Hence, the Corey–Chaykovsky reaction will not occur with phosphorus ylids.

Sulfur ylids

Very few cases of oxathietanes are known. A rare case was provided by Okazaki in 1996,6 characterised by NMR and crystallography:

However, the oxathietanes possibly formed in a typical Corey–Chaykovsky reaction differ from this in structure. For the model reaction of $\ce{H2S=CH2}$ with formaldehyde, an early computational study found a local minimum corresponding to an oxathietane with S–O bond length of 2.10 Å,7 and postulated that a Wittig outcome (alkene formation) did not occur because of the high activation energy to pseudorotation about sulfur as well as [2+2]-cycloreversion. However, these studies also did not take solvation effects into account, which will stabilise the highly charged betaine.

In a more recent DFT study, Harvey and Aggarwal2 were able to reproduce the molecular geometry of Okazaki's oxathietane, proving that there was no inherent computational bias against such structures. However, attempts to find a stable oxathietane from the addition of $\ce{Me2S=CH2}$ to formaldehyde, or from the addition of $\ce{Me2S=CHPh}$ to benzaldehyde, failed. In particular, the S–O distances in the optimised intermediates were 2.81 and 2.48 Å respectively (the full geometries are in the supporting information). Geometry optimisation starting from an oxathietane structure also yielded nothing but betaines.

Likewise, if an oxathietane cannot be formed, then the decomposition to alkene and sulfoxide cannot occur. Hence, sulfur ylids do not undergo the Wittig reaction.

Why the difference?

Harvey and Aggarwal2 wrote of their inability to find oxathietanes:

  Clearly, the energetic gain associated with bonding between these atoms [sulfur and oxygen] is more than offset by the ring strain it would entail.

Evidently, the key to understanding the different behaviour lies in bond strengths: not the double bonds in the product, but the single bonds in the oxetane intermediate. Kutzelnigg8 has found binding energies for $\ce{H2P-OH}$ (relative to $\ce{PH3 + O}$) and $\ce{HS-OH}$ (relative to $\ce{SH2 + O}$) to be 114 and 59 kcal/mol respectively. A qualitative explanation is not easy, and I am not really inclined to try.

Extra: Phosphorus betaines

As mentioned above, phosphorus betaines can be independently generated by deprotonation of β-hydroxyphosphonium salts. If we ignore the immediate cyclisation to form oxaphosphetanes, it is theoretically possible that these betaines could undergo intramolecular nucleophilic substitution to yield epoxides:

In a 2005 study, Aggarwal and Harvey again showed that such a process was much more energetically unfavourable for phosphorus than for sulfur.9 This was mainly rationalised by the poor leaving group ability of $\ce{PR3}$ compared to $\ce{SR2}$. Firstly, the intrinsic leaving group ability of phosphorus is lower (this is a purely kinetic measure of leaving group ability, based on the activation energy of the reaction $\ce{X + CH3X+ <=> X+CH3 + X}$). Secondly, the C–P bond is stronger than the C–S bond, which adds a further thermodynamic cost to epoxide formation. The difference in the energy profiles is striking:

For $\ce{X = PMe3}$, epoxide formation from the betaine is associated with an activation energy of 28.3 kcal/mol. Compared to an essentially barrierless cyclisation to the oxaphosphetane, it is easy to see why such a process cannot compete.

For $\ce{X = SMe2}$, on the other hand, the activation energy is only 5.4 kcal/mol. In fact, as the authors had noted in their earlier paper,2 the major challenge in the Corey–Chaykovsky reaction is getting the betaine into the correct conformation for intramolecular $\mathrm{S_N2}$ (i.e. going from 4-syn to 4-anti in the scheme above)!

References

Byrne, P. A.; Gilheany, D. G. Unequivocal Experimental Evidence for a Unified Lithium Salt-Free Wittig Reaction Mechanism for All Phosphonium Ylide Types: Reactions with β-Heteroatom-Substituted Aldehydes Are Consistently Selective for cis-Oxaphosphetane-Derived Products. J. Am. Chem. Soc. 2012, 134 (22), 9225–9239. DOI: 10.1021/ja300943z.
Aggarwal, V. K.; Harvey, J. N.; Richardson, J. Unraveling the Mechanism of Epoxide Formation from Sulfur Ylides and Aldehydes. J. Am. Chem. Soc. 2002, 124 (20), 5747–5756. DOI: 10.1021/ja025633n.
Vedejs, E.; Marth, C. F. Mechanism of Wittig reaction: evidence against betaine intermediates. J. Am. Chem. Soc. 1990, 112 (10), 3905–3909. DOI: 10.1021/ja00166a026.
Maryanoff, B.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108 (24), 7664–7678. DOI: 10.1021/ja00284a034.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.
Ohno, F.; Kawashima, T.; Okazaki, R. Synthesis, Crystal Structure, and Thermolysis of a Pentacoordinate 1,2λ6-Oxathietane: An Intermediate of the Corey−Chaykovsky Reaction of Oxosulfonium Ylides? J. Am. Chem. Soc. 1996, 118 (3), 697–698. DOI: 10.1021/ja953533u.
Volatron, F.; Eisenstein, O. Wittig versus Corey-Chaykovsky Reaction. Theoretical study of the reactivity of phosphonium methylide and sulfonium methylide with formaldehyde. J. Am. Chem. Soc. 1987, 109 (1), 1–14. DOI: 10.1021/ja00235a001.
Wallmeier, H.; Kutzelnigg, W. Nature of the semipolar XO bond. Comparative ab initio study of H3NO, H2NOH, H3PO, H2POH, H2P(O)F, H2SO, HSOH, HClO, ArO, and related molecules. J. Am. Chem. Soc. 1979, 101 (11), 2804–2814. DOI: 10.1021/ja00505a002.
Aggarwal, V. K.; Harvey, J. N.; Robiette, R. On the Importance of Leaving Group Ability in Reactions of Ammonium, Oxonium, Phosphonium, and Sulfonium Ylides. Angew. Chem. Int. Ed. 2005, 44 (34), 5468–5471. DOI: 10.1002/anie.200501526.


Why is 2s more stable than 2p, if 2p experiences greater effective nuclear charge? I'm referring specifically to this graph, which appeared in a chemistry lecture but which the lecturer could not explain. We know that 2s is lower energy than 2p. But surely those electrons in 2p orbitals, which apparently experience greater effective nuclear charge for Z > 10, would have lower energy! Can someone reconcile my understanding of orbital energy levels and this diagram?

One big issue here is that we generally don't talk about different ionization energies of anything but the outermost orbitals. Once electrons are in "core," we usually don't care since they generally do not participate in any kind of everyday chemistry.

We generally consider the $2s$ orbital to be lower in energy than the $2p$ because the $2s$ has some density inside the radial node that we think of as "nuclear penetration," i.e., the $2s$ electrons are able to get closer to the nucleus through the $1s$ core electrons. Now, once you add a $3s$ electron, I wonder if the innermost radial probability regions of the $3s$ actually repel the $2s$ electrons more than the $2p$ electrons which are generally farther away. The differential repulsion would have to happen in such a way so that it more than compensates for any nuclear penetration effects of the $2s$ electrons. This would certainly align with the observation that the $2p$ electrons are more stabilized only when we add a $3s$ electron.


Why is the water solubility of CO worse than that of CO2? I tried my best to ask SciFinder, Google, and Google Scholar the wrong questions ...
I found many paper on the fitting of the solubility to some equations to be able to calculate the amount of $\ce{CO}$ in water, but I could hardly find any explanation on the low solubility. Only one paper [1] dealt with it.
Why is $\ce{CO}$ way less soluble in water than $\ce{CO2}$, despite, e.g., the first having a dipole moment and the latter doesn't? What am I missing?

[1] H. Sato, N. Matubayasi, M. Nakahara, F. Hirata, "Which carbon oxide is more soluble? Ab initio study on carbon monoxide and dioxide in aqueous solution," Chemical Physics Letters 2000, 323(3-4), 257–262 (https://doi.org/10.1016/S0009-2614(00)00508-X).

The fact that $\ce{CO2}$ does not have a dipole moment does not really mean it's nonpolar.  What $\ce{CO2}$ has is a quadrupole, two opposing dipoles that occupy different portions of space (opposite sides of the carbon atom).
Ordinarily we expect an electrostatic interaction with a quadrupole to be relatively weak, but in $\ce{CO2}$ the component dipoles are strong and at close range the water molecules can "home in" on one end of the molecule and interact selectively with the corresponding component dipole.  Thereby the electrostatic interaction between water and the $\ce{CO2}$ quadrupole is stronger than the interaction between water and the weak $\ce{CO}$ dipole.


What do positive, negative and zero overlap of atomic orbitals mean? I am confused about positive, negative and zero overlaps.

Do they represent the extent of overlap or are they related to the bonding and anti-bonding orbitals as described in Molecular Orbital Theory?

The terms in the lecture slides you refer to describe the situation in a certain, very specific subset of molecular orbital theory: the idea of linear combining single atomic orbitals to form bonding and antibonding orbitals. While one can arrive at the same bonding/antibonding orbitals with elaborate molecular orbital theory through the localisation of orbitals approach, the simplified view these lecture slides (and many others) present is not consistent with molecular orbital theory. (However, a subset of the result can be obtained by a proper symmetry analysis.)

Let’s consider a rather simple case such as $\ce{HCN}$. It is obvious that the carbon atom’s atomic orbitals will in some way interact with the atomic orbitals of both hydrogen and nitrogen to form bonds. In the method presented in the lecture slides, you would look at certain pairs of orbitals and then determine what type of interaction would be between. I will only present a few representative examples. For consistency, I will define the molecule axis to be my coordinate system’s $z$ axis.

The $2\mathrm p_z$ orbital of carbon and the $2\mathrm p_z$ orbital of nitrogen.

These two p-type orbitals are aligned in the same way. Without taking their phases into account, we can see that the $-z$ lobe of the nitrogen orbital and the $+z$ lobe of the carbon orbital can interact. Depending on whether the phases match or do not match, we would arrive at a bonding and an antibonding orbital, respectively. The resulting bonding orbitals would be a $\sigma$ and $\sigma^*$ orbital.
The $2\mathrm p_x$ orbital of nitrogen and the $2\mathrm p_x$ orbital of carbon.

These orbitals are parallel and still well aligned. However, at any one time both lobes of one will interact with both lobes of the other. There is a nodal plane that completely contains the bond axis. If the $+x$ lobe of both orbitals feature the same phase, the interaction is bonding. Otherwise it is again antibonding. The bonding orbitals are $\pi$ and $\pi^*$, respectively.
The $2\mathrm p_x$ orbital of carbon and the $1\mathrm s$ orbital of hydrogen.

Now, we are trying to mix an orbital that only has one phase (the s orbital) and one that has two phases (the p orbital). This would be okay if only one of the two phases of the p orbital were pointing towards the s orbital; however, because the p orbital is perpendicular to the bond axis, both its phases would have to interact at the same time. Thus, we have one interaction on the $+x$ half of the coordinate system which is bonding and at the same time an interaction on the $-x$ half which is antibonding. Mathematically, the bonding and antibonding contributions cancel out perfectly. Thus, there is no interaction between the two orbitals. This is zero overlap.
The $2\mathrm p_x$ orbital of carbon and the $2\mathrm p_y$ orbital of nitrogen.

Now, we are attempting to mix two different p orbitals. However, they are not oriented in the same direction. The $+x$ direction of one orbital and the $+y$ direction of the other have the same phase while the $-y$ and $-x$ directions have the other. If we start at $+x$ and go around in a circle, we will start in an area of no interaction (the $\mathrm p_y$ orbital has a nodal plane in the $xz$ plane). Then, turning towards $+y$ we reach an area of bonding overlap. After crossing $45°$, the bonding overlap decreases until we reach zero again at $+y$. From there to $-x$, we experience an increasing and decreasing antibonding interaction. We then rinse and repeat what we just saw for the remaining $180°$.

It would probably not be impossible to consider this interaction a curly bond that rotates around the central axis by $90°$ between the two atoms. However, traditional analysis would note that at the same time there is something like an antibonding second part displaced by $90°$ to this. Therefore, the analysis would consider these two orbitals nonbonding with zero overlap.

In a proper MO or LCAO approach, molecular orbitals will stretch across the entire molecule. Thus, the analysis of whether an orbital is bonding or antibonding with respect to a certain bond reduces itself to the question of ‘does this orbital feature a nodal plane along the bond axis or does it not?’. From the $+,-,0$ picture we had above, we arrive at a black and white picture. This is because nonbonding interactions are typically separated out before the molecular orbitals are even calculated by a symmetry analysis. If you do that, you would realise that the $2\mathrm px$ orbital and the $1\mathrm s$ mentioned in case three are representative of two different irreducible representations. (Strictly speaking, $2\mathrm p_x$ and $2\mathrm p_y$ both belong to the same degenerate irreducible representation.) Only orbitals that belong to the same irreducible representation can be combined in bonding or antibonding orbitals.


How to find the mass of a solute given a mass fraction? How many grams of $\ce{CuSO4 * 5 H2O}$ are needed to prepare a $20\%$ (w/w%) solution of $\ce{CuSO4}$?

In a $20\%$ solution of $\ce{CuSO4}$ we have $\pu{20 g}$ of $\ce{Cu}$ and $\pu{80 g}$ of $\ce{H2O}$. Then

in every $\pu{259.6 g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{159.6 g}$ $\ce{CuSO4}$,

so for every $x~\pu{g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{20 g}$ $\ce{CuSO4}$

$$x = \pu{31.87 g}$$


How to rationalise the protonation of a compound with multiple double bonds? In the following reaction, why is the molecule protonated at the indicated carbon and not on one of the other double bonds?

The key is delocalization of the positive charge in carbocations.

Most stable carbocation that you can get from the protonation of the compound above. Reasons for it being the most stable:

The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)
In one of the resonance forms, the positive charge is at a tertiary carbon atom

Second most stable carbocation.

The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)

When protonation happens on the third pi bond. No delocalization - not as stable as the carbocations shown above.

There are other possible ways in which the compound could have been protonated (below), however none of them are as stable as (1) or (2), because there is no delocalization.


How deep does a counter electrode need to penetrate the solution in electrochemical experiments? I ordered platinum-wire counter electrodes that are shorter than I was expecting. How much should it be submerged in my solution (right now it is only the tip that is submerged)? My experiment is hydrogen permeation using an H-cell with counter and reference electrodes in each side and my working electrode in the middle. My solution is 1 M NaOH. Should I order longer electrodes? How do I know if it is submerged enough?

TL;DR As long as you are not getting overloads on your instrument, it is submerged enough.

Longer answer:
Your potentiostat will increase the potential on your counter electrode until it achieves the desired potential across the working electrode(WE) reference electrode(RE) circuit. The exact potential applied on the counter will depend on a lot of factors like solution resistance, amount of current flowing on the working at the potential desired, material your counter is made of and finally, the area of the counter electrode.

If you end up with a situation where your potentiostat hits its electronic limits before the potential difference WE-RE is the desired value, you will start seeing overloads.

If you are curious, you can always put a voltmeter between your WE and your counter electrode to see how much of a potential your instrument applies.


How can I tell if a liquid is (not) emitting light? I found an image of Nuka Cola Quantim, a soft drink from the Fallout video game series, which depicts the bottle as glowing, but it doesn't feel quite real:

Then I also saw this video of a chemiluminescent reaction of luminol:

https://youtu.be/TJj7t2mrpxc

Experienced chemists and/or photo editors might tell right away which is real and which is fake. What hints should I look for when determining if a liquid itself is glowing or is being lit by a different light source?

In a photograph, a light source is always brighter than any surface it is illuminating.

Therefore, if the liquid is the actual light source, the liquid should be significantly brighter than the rest of the bottle. See for example this picture by David Mülheims CC BY-SA 2.5, via Wikimedia Commons:

In the given picture, however, the parts above and below the water level have a similar brightness.


Is matter really divided in two categories? Page 1 of almost every Textbook start with the ordinary divison of matter in 2 categories: Substances and Mixtures.
Substances, or Pure Substances sometimes, are defined

as a form of matter that has constant chemical composition and
characteristic properties.It cannot be separated into components
by physical separation method.

This  is according to Wikipedia, at least.
Mixtures always citing wikipedia, instead are

materials made up of two or more different substances which are mixed
but are not combined chemically....
Mixtures are one product of a mechanical blending or mixing chemical
substances such as elements and compounds, without chemical bonding or
other chemical change, so that each ingredient substance retains its
own chemical properties and makeup.

Is this separation so radical?
Is there something in between? What about alloys?

Matter....

Matter is generally categorized as either mixtures or pure substances.  Pure substances are elements or compounds.  Mixtures are either heterogeneous or homogeneous.  For heterogeneous mixtures one should be easily able to discern the different components, while if the mixture is homogeneous it appears uniform.  Solutions and alloys are homogeneous mixtures.

A concept map which I give to my students: https://c7chemistry.wikispaces.com/file/view/CW%20matter%20concept%20map%20new%20rev%20091211.pdf/300804074/CW%20matter%20concept%20map%20new%20rev%20091211.pdf


Which hydrogen halides react with group 14 oxides? $$\ce{6HF +SiO2->H2SiF6 +2H2O}$$

$\ce{HF}$ reacts with $\ce{SiO2}$ to form $\ce{H2SiF6}$.

  Do other hydrogen halides like $\ce{HCl}$ and $\ce{HBr}$ react with group 14 oxides in the same way?

All reactions taken from this site.
First some maths: 4 hydrogen halides can react with 5 Group 14 monoxides and dioxides. So, there should be 40 different chemical reaction combination possible. But not all reactions are feasible.
HF

With $\ce{SiO2}$ (already mentioned)

$$\ce{GeO2 + 4HF → GeF4 + 2H2O}$$

$$\ce{SnO + 2HF ->[333 K] SnF2 + H2O}$$

HCl

$$\ce{GeO2 + 4HCl(g) ->[450-500 C] GeCl4 + 2H2O}$$

$$\ce{GeO2 + 4HCl(l) ->[170-180 C] GeCl4 + 2H2O}$$

$$\ce{SnO + 3HCl → H[SnCl3] + H2O}$$
$$\ce{SnO2 + 6HCl → H2[SnCl6] + 2H2O}$$
$$\ce{PbO + 2HCl → PbCl2 + H2O}$$
$$\ce{PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O}$$

$\ce{HBr}$ and $\ce{HI}$ do not show such type of reaction.


Why do I have to use a 1:1 ethanol:water solution to hydrolyse aspirin? I understand that water itself would not work as aspirin is not very soluble in water but Why not just use 100% ethanol?

Hydrolysis means to break something apart (e.g. chemical bonds) with water.

If there's no water, there's no hydrolysis.


Will nitrate be able to oxidise mercury? We have the following data:

\begin{align}
\ce{NO3- + 3 e- &-> NO} &\quad E^\circ_1 &= \pu{0.96 V}\\
\ce{Hg^2+ + 2 e- &-> Hg} &\quad E^\circ_2 &= \pu{0.86 V}
\end{align}

By the reasoning given in my book, nitrate oxidises mercury:

  ... of two substances, the one whose reduction $E^\circ$ value is greater, will be able to oxidise the other substance the other.

Then I tried calculating $E^\circ$ for overall reaction also taking into account number of electrons in each reaction. Or that we calculate $\Delta G$ for

$$\ce{NO3^- + Hg -> NO + Hg^2+}$$

$$\Delta G^\circ = -0.96 \cdot 2F + 0.86 \cdot 3  = 0.66F$$

So, $\Delta G^\circ$ is positive and so nitrate does not oxidise mercury. Which is correct?

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

\begin{align}
\ce{NO3− (aq) + 2 H+ + e− &<=>  NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\
\ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\
\ce{Hg^2+ + 2 e− &<=> Hg (l)} &\quad &E_0 = \pu{+0.85 V} \\
\ce{2 Hg^2+ + 2 e− &<=> Hg2^2+} &\quad &E_0 = \pu{+0.91 V} \\
\ce{NO3−(aq) + 4 H+ + 3 e− &<=> NO (g) + 2 H2O (l)} &\quad &E_0 = \pu{+0.958 V} \\
\end{align}

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.


Rate expression for the final product Let's say we have:

  $$
\begin{align}
\ce{A &-> B}&\quad&\text{(fast equilibrium)}\label{rxn:1}\tag{1}\\
\ce{2 B + C &-> D}&\quad&\text{(slow)}\label{rxn:2}\tag{2}\\
\ce{D + E &-> F}&\quad&\text{(fast)}\label{rxn:3}\tag{3}
\end{align}
$$

  Find the rate expression for the formation of $\ce{F}$.

Since I am writing the rate law for $\ce{F}$, should I write it as

$$\text{rate} = k_3[\ce{D}][\ce{E}]$$

because that's the reaction step that actually includes $\ce{F}$ as the product, or do I always have to write is in terms of the slow step i.e.

$$\text{rate} = k_2[\ce{B}]^2[\ce{C}]?$$

What happens with the leftover $[\ce{B}]$ and $[\ce{A}]$ in the steps \eqref{rxn:2} and \eqref{rxn:1}, respectively — do I include those concentrations as well?

I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that

$$k_2[B]^2[C]=k_3[D][E]$$

This approximation is why the teacher is always saying that only the slow reaction is important

Fast equilibrium means that the equilibrium is quickly governed by thermodynamics laws and so the constant

$$K = \frac{[B]}{[A]}$$

is not variating, so you can express $B$ in function of $A$

$$\frac{d[F]}{dt} = k_2K^2[A]^2[C]$$

I don't know how far you need to go but if you really want to solve for $[F]$ you need to write an equation of conservation

$$[N] = [A] + [B] + [C] + [D] + [E] + [F] = cst$$
$$d[A] + d[B] + d[C] + d[D] + d[E] + d[F] = 0$$

$$d[D] = 0$$
$$d[E] = 0$$

You can also say that $d[C] \approx 0$ because it looks like it's just a partner of collision, (it's the solvent) because the solvent is in much larger quantity you can say the concentration stay pretty the same

$$d[A] + d[B] = - d[F]$$

$$[A] - [A]_0 + [B] = -[F]$$

$$[A](1+K) = [A]_0 - [F]$$

$$[A]^2= \frac{([A]_0 - [F])^2}{(1+K)^2}$$

$$\frac{d[F]}{dt} = K_0\frac{([A]_0-[F])^2}{(1+K)^2}$$

with $K_0 = k_2K^2[C]_0$


Why is carbon dioxide a Lewis acid? I know this has been asked before, but can anyone explain me in simpler terms? Why is $\ce{CO2}$ a Lewis acid? The carbon atom doesn't have any vacant orbitals to accept a lone pair. How can it be a Lewis acid?

Suppose one of the doublets of one C=O bond moves to one of these two atoms. It will not be to the Carbon atom. The Oxygen atom is a better choice. And now this Oxygen atom would be negatively charged. As a consequence the Carbon atom is positively charged and has a vacant orbital able to accept a lone pair, for example from an $\ce{OH-}$ ion or from water. This will produce the ion $\ce{HCO3^-}$ (from $\ce{OH-}$) or $\ce{H2CO3}$ (from $\ce{H2O}$).


Process of transferring kinetic energy in regard to photons. Now I know that temperature is the average kinetic energy of atoms, molecules and compounds. Photons increase the kinetic energy/temperature of molecules since photons are pure energy. Hence, wouldn't the transfer of kinetic energy mean that photons need to be transferred from a higher temperature molecule to a lower temperature molecule?

You may be experiencing a very common misconception, which I blame on Star Trek and Star Wars and all their talk of pure energy that always comes as bright lights, with the implicit assumption that energy is some kind of fiery bright substance, like a glowing fluid or something.  It isn't.  Energy is a property of matter, and of the rather peculiar substance we call electromagnetic radiation, chunks of which are called photons.  Think of energy as kind of like an adjective that describes a noun (some kind of matter, or radiation).  An atom can be energetic or not energetic, and so can a photon, just as a cat can be furry or not furry.  But you would  never mistake "furriness" as some kind of object itself.  In the same sense, you should not mistake energy as some kind of substance itself.  It's a property of substances, just like furriness is a property of cats.

So energy has to be transferred from a hot substance to a cold one, and indeed it could go by way of photons.  One particle could emit a photon, transferring part of its energy to the photon, and maybe slow done, and then this photon could be absorbed by another particle, which would receive the transferred energy, and maybe accelerate (go faster).  That's definitely one way it could happen.  But it doesn't have to happen that way, because energy can be transferred other ways.  For example, the first particle could simply run into the second, bang, and the impact itself could transfer the energy directly, no photons involved.  Or the first particle could run into a third particle, or the wall of a container, and the second particle could pick up the energy from the third particle, or the wall of a container.  And so on.


Electron pairing in atoms A question asks:

  This question refers to isolated gaseous atoms. In which atom are all electrons paired? Ba, Br, S or Si?

I thought it was Ba as its electron configuration is [Xe] 6s2.

However, the answer says Si, which seems incorrect to me.

Could someone please confirm the answer for me?

Thanks

The configuration for the free atoms is:

Ba - $\mathrm{[Xe]\ 6s^2}$
Br - $\mathrm{[Ar]\ 3d^{10} 4s^2 4p^5}$
S - $\mathrm{[Ne]\ 3s^2 3p^4}$
Si - $\mathrm{[Ne]\ 3s^2 3p^2}$

Thus:

Barium has no unpaired electrons.
Bromine has one unpaired electron in 4p subshell.
Sulfur has two unpaired electrons in the 3p subshell.
Silicon has two unpaired electrons in the 3p subshell.


Why does the hydroxide ion have a negative charge? I've been studying the roles of hydroxide & hydronium in acids and bases, and it was mentioned that a hydroxide ion (OH-) has a negative charge. Can someone give me a layman's explanation of what causes this charge?

Since hydrogen and oxygen are sharing a covalent bond, is the charge negative because an electron is donated from oxygen, thus giving hydrogen 2 electrons?

A water molecule is charge neutral because there is the same number of positive
charges as there are negative charges.

In this diagram, called a Lewis structure, the dots represent electrons while the lines or dashes represent a covalent bond of two electrons.

When water ionizes one of the hydrogen atoms absconds with itself and leaves
it's electron behind, giving us the hydroxide ion. The extra electron gives hydroxide a net charge of -1.

The brackets indicate that this is an ion, charge is denoted at top right.

To go deeper down the rabbit hole on this one I recommend reading up on the Octet rule and Electronegativity.


What is the correct structure for trans-3,4-dichlorocyclopentene? I have to draw the structure for trans-3,4-dichlorocyclopentene.

What position should I draw the double bond at? Should I draw it as molecule 2? Because there's a trans.

Or should I draw it as shown in molecule 1 because it didn't state?

I feel that molecule 2 in the diagram is correct. What do you think? Where are my mistakes?

The issue here really is that the name 'trans-3,4-dichlorocyclopentene' is not really correct, as it is unspecific. It lacks stereo-descriptors, and the locant for the double bond.

Assuming that the double bond starts at the first carbon, which is usually the case if it is omitted, there are two enatiomers still possible:

(3R,4R)-3,4-Dichlorocyclopent-1-ene

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=12430197.
(3S,4S)-3,4-Dichlorocyclopent-1-ene

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=131015828.

So the structure you have drawn as 1 is (3R,4R)-3,4-dichlorocyclopent-1-ene.

The structure you have drawn as 2 doesn't really exist, as the double bond is planar, but ignoring that it would be 1,2-dichlorocyclopent-1-ene.

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=10606793.


Is there any molecule visualizing software for Python? I am looking for a free and open-source software written or wrapped for Python that would simply allow me to generate bunch of figures (in eps, jpg, or pdf format) from a set of xyz files, e.g. water clusters of different sizes.

I am aware of PyMol, but it seems a bit complicated to use (it has its own script language) for such a simple task and I guess it does not recognize xyz files, so I'd first need to convert to pdb format, etc.

In case you really need it has to be written in python probably you want to try Atomic Simulation Environment (ASE). It has a GUI that "allows users to visualize, manipulate, and render molecular systems and atoms objects".

Let me add (even when you are asking explicitly for codes written in Python), just in case that what you need is to create images from .xyz files (regardless of the language), Jmol probably will suits you, as it reads .xyz files straightforwardly. For saving your image once it is loaded, you will need to open a console in Jmol and type:

write image 1200 720 jpg 90 "filename.jpg"

where:

1200 and 720 are the width and height of your image (choose the ones you want to)
90 is the quality of the compression


Gibbs of vaporization of a substance at a non-standard temperature I'm trying to solve for the gibbs energy of vaporization of ethanol at a temperature lower than its boiling point. I am given the heat capacities of liquid and gaseous ethanol, the standard enthalpy of vaporization, and the standard boiling point of ethanol.

I start with the form:

$\ce{dH=(C_{p,g}-C_{p,l})dt}$

Following integration, I get the form:

$\ce{\Delta H=\Delta H_{vap} +\Delta T(C_{p,g}-C_{p,l})}$

This form gives me the new enthalpy of vaporization at a non-standard temperature. However, When I try to enter this into the form:

$\ce{\Delta G=\Delta H - T\Delta S}$, with $\ce{\Delta S=\frac{\Delta H}{T}}$

This solution always results in an answer of zero (equilibrium). Intuitively, I understand that phase changes are typically reversible, however using another representation:

$\ce{\Delta G=-RTln(K)}$, with $\ce{K=\frac{P}{P_{0}}}$, the ratio of vapor pressure to standard pressure of the liquid (liquid activity = 1).

My question is, why do these formulas apparently conflate?

Alright, this is kind of fast but after thinking about it I think the issue with the problem is the definition of $\ce{\Delta S}$. My second try to compute this value involves starting with the form:

$\ce{dS = \frac{C_{p,g}-C_{p,l}}{T}*dT}$

Following integration, I arrive at the form:

$\ce{\Delta S = (C_{p,g}-C_{p,l})ln\frac{T}{T_b}+\frac{\Delta H}{T}}$ where $\ce{\Delta H}$ is determined in the way as in the question above.

Placing everything into the form $\ce{\Delta G = \Delta H-T\Delta S}$:

$\ce{\Delta G = \Delta H_{vap} +(T-T_b)(C_{p,g}-C_{p,l}) - T((C_{p,g}-C_{p,l})*ln\frac{T}{T_b}+\frac{\Delta H}{T})}$ where $\ce{\Delta H}$ is the non-standard enthalpy of vaporization.


What exactly makes a carbon atom "α" in a protein residue? I'm a computer scientist and have no biochemical background, but was working on a project recently that had me going through a lot of protein residues for one reason or other. Feel free to correct me from here on, but my understanding is that proteins, also called "peptides" also "subchains" when a part of a larger macromolecule are made of amino-acids(central dogma) which are also frequently referred to as "residues".

Upon closer (programmatic) inspection, each residue (amino-acid) turns out to be a bunch of atoms like so:

I was wondering what is the significance of that CA there in each of the residues, and whether i identified it correctly as "alpha-carbon". Protein Data Bank docs refer frequently to a residue's alpha-carbon. If it is indeed the alpha-carbon, i have two questions which are probably dead-simple to anyone with biochem background:

Why is that one(in space) considered alpha and not the carbon atom next to it?
Is it a given that each residue out there has only one alpha-carbon? Seems pretty obvious, but please.

Thanks a ton!

This nomenclature is due to the fact that amino acids are carboxylic acids.

Near the carboxylic acid moiety, the carbon chain is unbranched and simple, so the positions are named like an unbranched, simple aliphatic carboxylic acid.

The carboxylic acid ($\ce{-CO2H}$) is not indicated with a position. But the carbon immediately next to it is $\alpha$. The one next to that is $\beta$. That should be enough to explain the $\alpha$ in your context because that carbon is the one immediately next to the carboxylic acid. In a protein or polypeptide, the carboxylic acid is most frequently converted to an amide in a peptide bond, but the Greek letter nomenclature is unchanged.

This nomenclature is used in other cases for example, with an $\alpha$,$\beta$-unsaturated carbonyl.

The most publicly visible use of this nomenclature is when we use $\omega$ to denote the end of the carboxylic acid chain. This is the origin of the term $\omega$-3 [unsaturated] fatty acid.


Which would have a more intense transition: a low spin d6 complex or a high spin d5 complex? I have an assignment question for second year inorganic which asks to rank the intensity of d-d transitions for a number of complexes. Two are $\ce{[Fe(OH2)6]^3+}$ and $\ce{[Fe(CN)6]^4-}.$ Both will have Laporte-forbidden transitions, but which will be 'more forbidden', so to speak?

You see that both cases are Laporte-forbidden, now how about spin-forbidden?

Assuming the complex vibrates so it's no longer exactly octahedral, can you make a spin-allowed transition for $\ce{[Fe(CN)_6]^{4-}}$?  How about $\ce{[Fe(H_2O)_6]^{3+}}$?  When you answer this you will answer your question.

   You should find that $\ce{[Fe(H_2O)_6]^{3+}}$, with all orbitals occupied by electrons having the same spin, has only spin-forbidden d-d transitions.  Therefore it will undergo weaker transitions than $\ce{[Fe(CN)_6]^{4-}}$ which has some spin-allowed transitions.


Which chemical properties make a substance explosive? I know that some chemical substances are used as explosives. Amongst the most famous nitroglycerin in dynamite and cyclonite in C-4. Which chemical properties (I suspect that physical properties have less of an effect) make a substance a good explosive? If we know, for example, the formula and its geometry, is it easy to predict how explosive it can be?

I'm going to assume by 'good explosive' you mean 'high energy to mass ratio', which by no means implies anything about whether the explosive is 'good' in the 'won't suddenly explode when thought about' sense.

This is not intended to be a full answer as it doesn't touch on the terrifying explosivity of organic peroxides, however it's a start. The presence of nitro ($\ce{R-NO2}$) or azide ($\ce{R-N=N=N}$) groups in a molecule is strongly suggestive of explosivity. This has a lot to do with the large negative free energy change of formation of dinitrogen and (I guess) oxides of nitrogen upon decomposition of these molecules. $\ce{N2}$ has one of the strongest bonds known and releasing a lot of $\ce{N2_{(g)}}$ entails a large increase in entropy. Inorganic nitrates and azides are also likely to be explosive.

The state of the art in explosive research seems to involve cramming as many consecutive nitrogen atoms as possible into a molecule, as showcased on Derek Lowe's blog:

$\ce{C4H4N8}$

$\ce{C2N14}$ (what?!)

Alternatively, you can take a highly strained polyhedrane of some description and stick nitro groups all over it. The ur example of this sort of philosophy is octanitrocubane.

Bonus: tetranitrotetrahedrane

Disclaimer: Explosives are deadly and people who prepare them are either highly trained experts with exquisite safety infrastructure, or morons. Nothing in my answer should be taken by any person as an endorsement to attempt to synthesise anything described.


Reverse washing soda buildup on stainless steel In a home brewing project, I use washing soda (Na2CO3) to remove labels and glue from previously used bottles. In doing this, my flip-top bottles (which have a stainless steel flip top arm) often get unruly buildup of residual soda, which tarnishes and blemishes the steel. Is there a chemical mixture I can use to reverse this process and return the steel to a near-new appearance?

Here are a couple of photos of the bottles and buildup: the left shows a typical amount of residue; the right shows my attempt to wipe it clean with white vinegar (which partially works, but is still blemished).

If you use tap-water together with Sodiumcarbonate, chances are that the residue is Calciumcarbonate (unless your tap-water is very soft).

Calciumcarbonate is significantly less soluble in water, but can be dissolved in acid. I would try to put a bottle/arm in a hot 5% solution of acid (acetic acid is fine, personally I would prefer citric acid) to see if any (small) gas-bubbles are formed at the residues. If it does, it is carbonate and can be removed in an acid bath.

If not, the residues could be glue from the labels, and getting rid of this is probably more complicated.


Which argon compounds exist, supported by experiment or computation? Just as $\ce{HArF}$ was synthesized...Could something like $\ce{Li-Ar-Li}$ exist with AXE geometry $\ce{AX2E3}$? Any other argon compound proposals out there?

I also know that there are some posible excimers like $\ce{LiAr}$ and $\ce{KrF}$ (maybe $\ce{KrF2}$). Can they be considered covalent compounds, even if explained via excited dimers/excimers or the OM theory (recall the technical name for lithium-doped argon matrices in the chemical literature)?

ArBeO and ArBeS are experimentally known.

ArAuF, ArAgF and ArCuF are also experimentally known.

CUOAr$_n$ and related complexes are known.

ArBeNCN and ArBeNBO are predicted by theory.


Dissolvant for the black spots around the rear mirror of the front window of my car I make pictures for Mapillary.
I would like to place my dashcam as high as possible, but those black spots get in the way.  Scraping them off does not work.
But I can feel their relief.
If no one has a better idea I will begin the trials with a sticker remover.

I do not find the exact word for those spots any more.

The part of the windshield that you are talking about is called the frit. This is typically made up of some form of high-heat enamel, and as far as I was able to find, it is intended to increase the surface roughness of the inside of the glass to increase the ability for the adhesive to stick to it.

I was not able to find any specific mention for what type of enamel frits are actually made of - that is whether it is an enamel glass or an enamel paint -  and as such I cannot give any specific recommendations for what solvent to use. If it is enamel glass, as I would hazard to guess it is - given that some sources indicate that it is applied prior to the windshield being bent - then you will likely not be able to remove the frit without also damaging the glass. If it is an enamel paint then you may be able to remove it with an organic solvent. In general, I like to start with acetone as a general organic solvent.

All of this said, given both the purpose of the frit, and its proximity to the adhesive that holds your windshield on, I strongly recommend that you consider moving your dashcam rather than attempting to dissolve the frit.

Best of luck.


Best solvent for crystallization + mass recovered from crystallization + purity of the product I have a chemistry final tomorrow and this question on my sample exam through me for a loop. I'm not entirely confident I can work out problems B and C, but those questions rely on the answer to A.

If you could please especially explain how I'm supposed to read the solubility chart to understand which solvent is the best, that would be awesome.

Expanding on LDC3's comment a little.

The information in the table (irrelevant of how good or bad the method proposed is) tells you what the saturation points are for solvent mixtures at given temperatures for each compound.

Fundamentally, you want to separate your desired product from your undesired product. In this case, you are trying to recrystallise the 3-phenylacrylaldehyde, leaving the benzaldehyde in solution. You know this because for all temperatures and solvent mixtures it has lower solubility (this cannot be assumed for a more complex solvent array).

You need the minimum of hot solvent, here 100 degrees, to dissolve all of your crude compound, and you want to maximise the return of pure compound on cooling. This explains why you want the biggest ratio of hot to cold solubility as LDC3 indicated.

You also want to minimise the amount of your starting material that crashes out with your product. So, calculate the same ratios for the starting material, but here you want to minimise the return on cooling, so you want the smallest ratio.

You now have a table of solubility ratios, columns being 'solvent mix', 'rP' (ratio of product solubilities) and 'rSM' (ratio of starting material solubilities). The largest value of rP/rSM gives you the best solvent mixture.


Was nitroglycerin invented as an explosive? Although I have researched this as carefully as the web allows, it is not clear to me if the inventor was planning to make an explosive and if so, why he thought the process he undertook would yield one. Were there similar compounds created with acids that were also explosive? I understand that in the early 1800s, there was a lot of apparently random experimentation.

Probably ... but he doesn't actually say

$Nitroglycerin^1$ was discovered by Dr. Ascanio Sobrero in 1847. Both Sobrero and Alfred Nobel were students of Prof. Théophile-Jules_Pelouze.

It is rather unfair to describe chemistry of this era as "random". 1847, of course, was mid-1800s rather than early 1800s, and it was some two decades after Wöhler's famous demonstration that organic chemicals could be synthesized from inorganic ones. Organic chemistry of this era was more in its adolescence than its infancy and was being pursued in a systematic scientific manner with publications in peer reviewed journals.

Some of these journals have now been digitised and are searchable, so we have some idea of Pelouze's work, and his students. Much of it involved preparation and study of nitro-esters of simple carbohydrates. They don't seem to say why, but in the context of the era, it is very likely they were looking for practical explosives. Braconnot had discovered nitrostarch in 1832; Pelouze apparently first prepared nitrocellulose in 1838, but didn't recognise it as a new substance. Then in 1846 -- just one year before Sobrero nitrated glycerin -- no fewer than 3 chemists prepared and described nitrocellulose in the same year, with Otto being first to publish.

For a long time, it appeared that nitrocellulose was not a practical material as it is very difficult to purify, and extremely unstable when impure. So it is not surprising that a year later, we find one of the early nitrocellulose investigators now conducting a program of testing nitrates of a range of simple organic compounds, especially carbohydrates. (Cellulose itself is a carbohydrate.) They don't say exactly why in anything I have read. It is possible that they are using nitrations to investigate fundamental chemistry; for example, such a process does help to identify the structure of simple sugars. However, they test them all for explosive properties and refer to them all as "fulminant" (i.e. explosive.)

Sobrero's paper on his discovery of nitroglycerin has been digitised. (Note: article is in French.) He mentions earlier work with sugars and notes that glycerin is not a carbohydrate, but doesn't seem to say what inspired him to try to nitrate it except that it is "analogous" with other materials that his group had nitrated. Incidentally, in this paper he describes purification methods at some length, and also discusses the toxicity:

  ... for it is sufficient to take a very small quantity (which can be taken
  by slightly wetting the tip of the little finger) on the tongue to
  feel a powerful migraine for many hours. This action on the human body has
  been observed by many people in my laboratory ...

However, he doesn't mention its extreme shock sensitivity. Later, Sobrero came to the conclusion that nitroglycerin was too dangerous to use, and it was, of course, his colleague Alfred Nobel who did the considerably longer and more complicated studies to find a way to render it practicable. (The same situation occurred with nitrocellulose; discovered by Schönbein, Böttger and Otto, but it was Abel who did the real spade work to make a practical industry for it.)

Note that whilst many of the substances in this answer were historically called nitroX where X is the original moiety, none of them are nitro compounds in the modern sense. Rather, all explosives discussed in this answer are nitrate esters. So nitroglycerin, for example, should really be called glyceryl trinitrate.


Hydrohalogenation vs Hydration of Alkenes So, I am doing an investigation involving the hydration and the hydrohalogenation of ethylene. Essentially, to produce the ethylene, I heat ethanol, in the presence of concentrated sulfuric acid, to obtain ethylene gas, which I plan to bubble into a solution of hydrochloric acid at varying concentrations.

Now, when the ethylene is bubbled in, I imagine that it will react in the solution and there are two possible products it could potentially form; Ethyl Chloride (from the hydrohalogenation mechanism) and ethanol (from the hydration mechanism). The problem is that I want to make a prediction of which product would be found in greater proportion, though I'm not very sure how I might go about doing this.

I'm thinking that, because the HCl dissociates to form hydronium ions and chloride ions, there are more hydronium ions than HCl molecules and therefore, hydration is likely to take precedence.

However, I've also thought that, in both mechanisms, the first step is protonation. In that case, it would also be possible that the chloride might add to the positively charged ethyl carbocation to a greater extent than the water molecules. Again, this is because both the carbocation and the chloride are charged and attract more easily. The oxygen in water is only partially charged and therefore, does not act as a good nucleophile in this case. However, I'm not sure how valid my arguments are.

Could someone help me out with pointing out flaws in this scheme of work?

"I'm thinking that, because the HCl dissociates to form hydronium ions and chloride ions, there are more hydronium ions than HCl molecules and therefore, hydration is likely to take precedence."

Definitely don't go by the that logic. The first step of hydrohalogenation requires the dissociation of $HCl$ and produce $H^+$ (/hydronium), so that the $H^+$ can activate the alkene forming the carbocation intermediate. This is also the first step of the competing hydration mechanism, as you ended up realizing.

The key here is the second step, where a nucleophile attacks the carbocation to form the neutral product. The nucleophilicity of water and chloride aren't extremely different. One might argue that "in a vacuum", chloride is a better nucleophile due to the charge (though that charge is quite stabilized over a large volume). But, practically speaking, water tends to attack instead of chloride in most situations. This is because when both are present, water tends to be in much higher concentration which increases the rate of the water attack pathway.

So consider your conditions, and if water is in much higher concentration than chloride, I would assume the hydration pathway will dominate.


What are the key chemical characteristics that determine the octane rating of fuel? The octane rating of petrol (gasoline) is a measure of the how much compression a fuel air mixture can have before detonating (which in petrol engines is a bad thing as you want the mix to combust only when the spark plug arcs).

What are the key chemical characteristics that determine this rating? What is the molecular-level explanation? And how do the various non-hydrocarbon additives work to modify the result?

Combustion
 is (usually) chain radical reaction with heavy branching. More
branching means faster combustion and less branching is slower
combustion. And detonation is wave of super-fast combustion.

So, if you can suppress branching or deaden some branches, you can
increase durability of air-fuel mixture. Actually, this means, that some
 radical trap is needed. The oldest trap was tetraethyllead, but it is
currently dropped (lead oxides as fine dust is not a health friendly
substance). Non-saturated hydrocarbons of different types also can trap
active radicals, transforming them into more stable allyl radicals, so
fuel companies tries to keep as much of them in fuel as possible. Some
other compounds, like ferrocene, also can be used to increase octane
rating.


Preparation of an aromatic Grignard reagent I was tasked with preparing 4-fluorophenylmagnesium chloride from the corresponding aryl chloride. But I've heard that aryl Grignard reagents are very difficult to prepare. I think that the optimal solvent for the reaction would be THF. Any advice would be greatly appreciated!

Also what are the downsides of using magnesium powder instead of magnesium turnings?

I have prepared many aryl Grignards in my career. Generally the aryl bromide or iodide is easier than the chloride, but it can be done. THF is certainly fine for this, run under inert atmosphere, of course. Magnesium powder may prove to have more of an oxide film than turnings and hence be less reactive. Most Grignard preps use turnings. It might be helpful to give your $\ce{Mg}$ turnings a brief wash with $\pu{2N}$ $\ce{HCl}$, then rinse with $\ce{EtOH}$ and $\ce{Et2O}$ just before you use them.

To initiate the reaction - do not stir the reaction but either hotspot it by using a heatgun to heat the region immediately around where the turnings sit at the bottom of the flask or put the whole flask in an ultrasound bath. A crystal of iodine in the reaction mixture will often serve to initiate the reaction, when it decolourises you know the reaction is proceeding. Depending on scale the heat of the reaction may bring the solution to reflux so you'll need a condenser on it.

Finally if all else fails at couple of mls of Red-$\ce{Al}$-$\ce{H}$ solution should start it.

If all fails with $\ce{Mg}$ metal, consider running a Knochel exchange with $\ce{iPrMgCl}$.


When is rotation around double bonds to form cis-trans isomers allowed? Textbook Question:

The rhodopsin molecules in the retina contain a derivative of vitamin A called cis-retinal (Figure 6). On exposure to light, cis-retinal becomes an isomer called trans-retinal.

This process helps cause an electrical impulse to be transferred to the brain. Around which numbered bonds does the isomerism rotation occur?

The answer is bond 11 (shown in red).
What I don't understand is: I thought rotation is not allowed around double bonds.
So when light hits the structure in Figure $6$, how does it cause rotation around bond $11$ to form the structure in Figure $7$?

In rhodopsin this normally occurs only after the 11 cis in the chromophore (a retinylidene schiff base) is placed into an electronically excited singlet state after absorbing a visible photon. This excited state has one electron promoted from the HOMO orbital into the LUMO (antibonding) orbital of the conjugated $\pi$ system. Most data has been obtained from the protein bactoeriorhodposin, which acts a  proton pump, as it is easier to use and has a similar chromophore to rhodopsin.

In the protein the rhodopsin molecule is somewhat strained by adjacent (tryptophan) and other amino acids. The isomerisation takes about 200fs so is extremely fast; among the fastest reaction ever studied. After that there is a series of at least 6 intermediates that slowly returns the trans molecule to the cis form and these thermal reactions take about a second or so.

The initial reaction is thought to proceed as follows; just after the electronically excited state is formed calculations indicate that the cis bond is rapidly stretched (~20 femtoseconds) and experiments show that a  small barrier exists between the minimum energy of the excited state and the pathway to isomerisation. At or close to this barrier rotation & oscillation of the groups either side of the 'double' bond occur during isomerisation. (Kobayashi et.al Nature, 2001, v414,p531).

Many experiments have ben performed on trans/cis stilbene isomerisation which shows similar behaviour. Trans stilbene is the most stable isomer and can be converted into the cis also by photo excitation. The cis ground state has a relatively low barrier to the trans at room temperature (the opposite is not true) and so reverts to the trans spontaneously. This does not happen in rhodopsin  so the protein has to go through a complex set of reactions.


Taking kinetic readings for the Briggs–Rauscher oscillating reaction I am current doing the IB diploma program where we are required to do an extended essay which is a form of a research paper on a specific subject. I chose chemistry.

I am analysing the kinetics of the Briggs–Rauscher reaction, adapting my experiment from the paper by Schmitz and Furrow [1].

The paper appears to use a spectrophotometer to measure iodine concentrations which I have tired, however the readings failed since the reaction needs to be continuously stirred. Another alternative they have used is to use an ion selective electrode to determine iodide ions in the solution however I don't have access to this.
I do have an AgCl reference electrode and was considering trying method from this YouTube video, however I lack a potentiostat and the ion-selective electrode. I was wondering if anyone could tell me any tricks on how to build this electrode if possible as I couldn't find anything online.

Alternatively, is there any other recommended way to determine the kinetics of this reaction. I would like to not have to use a stopwatch to clock the oscillations since that does not provide me with much detail.

References

Schmitz, G.; Furrow, S. Kinetics of the Iodate Reduction by Hydrogen Peroxide and Relation with the Briggs–Rauscher and Bray–Liebhafsky Oscillating Reactions. Phys. Chem. Chem. Phys. 2012, 14 (16), 5711–5717. https://doi.org/10.1039/C2CP23805E.

The comment by K. T. gives me a nice idea.

Set up your reaction in an environment with stable illumination and against a white background or far from any specific one. Then you can take ordered photographs that you print on transparent slides.

You can also set a cuvette in a white box and fill it with sample from the reaction vessel, to take better photographs.

After you cut them in stripes with the cuvette face size and by inserting them in the slits of the spectrophotometer you get the absorbance spectrum of a sample that mimics your reaction.

You should easily get stripes of the cuvette size, if not you can always arrange something to be measured against the appropriate background (baseline).

You can also make standards or at least checkpoints using a single component of the reaction, and in all cases measuring at a relevant wavelength.

The only hypothesis is that the A of the stripes scale linearly with that of the reaction solution, and this can be confirmed as in the above paragraph.

I never thought before of this especially because my work required some accuracy. But I could have tried it indeed when the kinetic study apparatus was already in use (just to identify a change in colour, for instance). For your essay, it might be more than enough.

If a student of mine would circumvent the given limitations in experimental set-ups using a trick as I have described, I'd be rather impressed.

edit I must add that once, for fun, I used the smartphone to photograph my finger at different levels of blood oxigenations, and after emphasising the differences looked for a trend, which I have found.  This made me consider possible colour picker applications or photography suites as a tool to quantify the color (absorbance) without even going to a spectrophotometer. Digging Internet with this in mind I have found that my idea isn't original anymore and found project similar to that in the Q, and indeed conducted using a smartphone. Examples can be easily found and downloaded searching for "smartphone as a colorimeter" and similar search keys.

https://pubs.acs.org/doi/abs/10.1021/ed300567p?src=recsys&journalCode=jceda8

is an introduction in J. Chem. Ed.


Ion/Counter ion layers in the colloid precipitate of silver chloride You use a precipitation method to determine [Ag+] in an unknown sample using Cl- ions. Reaction is conducted in presence of HNO3 in excess (relative to Ag+) Which statement is correct concerning the colloid precipitate at the start of the rxn (rxn not yet complete).
A) The ion in Primary Absorption layer is Cl-
B) The Colloid Particle is (-) charge
C) Ag+ is the major ion in Counter-ion Layer
D) Most Cl- ions added are in the precipitate form.

So I think the correct answer is D through process of Elimination and not chemistry knowledge. I know B is incorrect, B/C the precipitate must be AgCl, or possibly AgNO3 since there's Nitric acid here. Both of these are neutrally charged.
I'm pretty sure that Primary absorption layer is the "opposite" of the counter ion layer. So whichever ion goes into the Primary, the opposite ion goes in the counter. Therefore I eliminated A and C, since one cannot be correct without the other being correct as well. (although my first instinct was to put Ag+ in the primary ion layer).
I still have doubts about D. Since the reaction is not complete, and we are adding Cl- to a solution of Ag+, its safe to assume that any Cl- added will react. But HNO3 is a strong acid which dissociates completely, that means there's an excess of NO3- ions floating around. I'm not given any precipitation constants. Can I safely assume Ag+ will bind with Cl- before making AgNO3?
Also could you please confirm or correct my assumptions about Primary/counter-ion layer? I can't find any info it besides vague definitions.

It isn't absolutely clear from the wording, but it seems you're adding $\ce{Cl^{-}}$ to a $\ce{Ag^{+}}$ solution.

As some of the $\ce{Cl^{-}}$ is added, $\ce{Ag^{+}}$ is still in excess in the solution so virtually all of the $\ce{Cl^{-}}$ precipitates. $\ce{Ag^{+}}$ will be the primary adsorbent on the precipitate, and the colloidal particles will have a positive charge.

Since the colloidal particles have a positive charge, the ion in the counter-ion layer around the particle needs to be negatively charged. There isn't any  $\ce{Cl^{-}}$ in solution so it has to be $\ce{NO3^{-}}$ as the anion.

Thus the correct answer is (D).


ESI & MALDI Mass Spectrometry: particle mass vs. measurement error For MALDI or ESI mass spectrometry of large macromolecules (e.g. in the $\approx 100$ kDa to $\approx 10 - 100$ MDa mass range), I've always how measurement error scales with particle mass.  Not having any hands on experience with mass spetrometry, and calling anything build in the last 5 - 10 years "modern", are there any published tables of general rules of thumb for how measurement error scales with particle mass (the two must surely be positively correlated)?  If we're interested in measuring the mass of large intact nucleic acids or proteins, at what single-molecule particle masses does it make sense to use ESI or MALDI?

I'd guess that at higher mass ranges (e.g. in the MDa range) ESI might become preferable since it can produce multiply charged species, allowing for smaller $\frac{m}{z}$ ratios?

The answer to your question is not an easy one, as multiple parameters need to be taken into account to give an answer. At the high masses that you quote, most of the litterature trends come to their limits in terms of extrapolation, as the major mass ranges in mass spectrometry covers an $m/z$ range 10 - 4 000.

As you indicate in your question, one should separate the question into two parts: how does mass accuracy change with high $m/z$ values and how does mass accuracy change with high mass values, considering a high number of charges, as can be achieved in ESI.

Dependence with ion charges

I will first tackle the second part of the question, because this will be needed to answer the first part. If one considers a mass analyzer in which measurement accuracy ($\Delta m/z$) is directly proportional to the measured $m/z$ result, we have:

$$\frac{\Delta m/z}{m/z} = Constant = \frac{\Delta m}{m}$$

Therefore for such a mass analyzer there is no direct advantage in using multiply charged ions (as produced in ESI) compared to singly charged ions as produced in MALDI. From this point, it comes that on this criterium alone, one should consider if measurement accuracy increases faster or slower than a proportional increase to decide whether multiply charged ions come as an advantage or a disadvantage.

Dependence with mass analyzers

Let us now consider a series of classic mass analyzers and consider how they would perform in terms of high mass measurement.

Quadrupole filters and quadrupole ion traps

I group these in the same subtopic, as when one is concerned about high masses they have fairly similar properties. Commercial instruments are usually limited to $m/z$ 4,000 (or 32,000 for the specially devised systems). So ESI is a requirement for high mass species.

Quadrupole mass filters are usually operated in the constant mass width mode ($\Delta m/z = C$, with $C$ on the order of 0.5 to 1.3 amu), which means that the mass error will be roughly only dependent on $z$ and that there is an advantage to work on the high end of the $m/z$ range, as the overall error will be lower than on the low end. One point worth notice though: if you consider a 2 MDa system at $m/z$ 1,000, it would mean $z = 2000$. Thus spacing between neighbouring charge states is of only 0.5, lower than quadrupole resolution, which means that a deconvolution of the charge state distribution will not be doable and that is will be globally not possible to determine the mass of the molecule with this technique.

Time-of-flight analyzers

These are probably the mostly used analyzers for high mass compounds, both in MALDI and ESI modes. In an ideal TOF, the mass accuracy would depend only on the digitizer (and thus scale as $1/\sqrt{m/z}$), but it also depends on the applied potentials and other factors leading to a mass accuracy that scales as $\sqrt{m/z}$. This still means that considering the above points, mass accuracy will be better achieved at low $z$ values. The mass range limit for TOF analyzers is in theory infinite, but it is in fact limited by the detector (see below).

Orbitrap mass analyzers

Technically an Orbitrap mass analyzer could benefit from working at high $m$, low $z$ conditions, as mass accuracy in an Orbitrap scales as $\sqrt(m/z)$. The instrument is nevertheless limited by the possibility to inject ions within the trap, which is currently limited to 4,000 (or maybe 8,000, I did not check the most recent developments in the field). Thus ESI is required for such high mass systems, but the highest mass accuracies would be achieved on the high end of the achievable mass range.

FT-ICR mass analyzers

In FT-ICR mass analyzers, mass accuracy scales as $m/z$. Thus there is no effect of the charge on mass accuracy. Technically, FT-ICR instruments always include a quadrupole (trap or filter) between the ion source and the FT-ICR mass analyzer and the $m/z$ range is thus limited by the range of this filter (4,000 - 8,000).

Other issues of high mass measurements

One of the difficulties that arises when using high mass ions is also ion detection: devices using a secondary electron emission (such as MCP's or conversion dynode) have efficiencies that depend on the linear velocity of the impacting ion. Therefore the standard detectors are not well adapted to high $m/z$ ions. Some detectors have been developped to circumvent this limitation, but this remains an issue for masses in the 2 - 10 MDa range that you quote in your question.

One other issue is ion desolvation: as size increases, adducts (with salt, solvent or matrix) tend to increase also, leading to changes in the total mass of the ions and a broadening of the peaks. Mass accuracy will also be reduced because there is no direct rule for removing the masses of these adducts from the measured mass to lead to the mass of the molecule.


Electroplating a spoon (electrolytic cell) using a electrochemical voltaic cell I drew a rough picture of the two cells, and marked each position with a distinct letter:

The whole reason I am asking this question is that I am bit confused and unsure about some of the chemistry and physics concepts. It will be extremely helpful if you guys can tell me if what I did is correct and why.

What I think is correct: to make this whole thing work.

Connect alligator clip x to position A, right on the zinc anode not the wire.

And connect clip y to B, on the copper cathode plate.

From what I read, the electron is forced from Ag plate, anode, through the wire (with two alligator clips on each end), then reduces $Ag_{+}$ ions to $Ag$ metal on the spoon cathode. But from the picture, it looks more like all the electrons came from the zinc anode of the voltaic cell and all the electrons from the $Ag$ anode of the electrolytic cell go to $Cu$ cathode of the voltaic cell.

Some physics aspect of this: if x clip was to be connected to position C and y clip to position D, is it true that there will be no potential difference?(unless the wire itself has a high resistance)

Let's start with the last question: The points A and C have the same potential, because we  generally neglect the resistance of the wire. The same argument is valid for points B and D. But according to your scheme, you have a short circuit (as C and D are connected).

Now, to electroplate the spoon, you want the silver to be oxidized at the silver electrode. So, silver electrode is the anode, and must be connected to pole (+) of the voltaic cell (point B):
$$\ce{Ag -> Ag+ + e- }$$
The spoon is the electrode where ion silver will be reduced. So, it's the cathode and must be connected to pole (-) of of the voltaic cell (point A). The electrons enter the cathode and reduced the silver ion according to the equation:
$$\ce{Ag+ + e- -> Ag}$$

So, you're right!


How did Scientists Capture The First Images of Molecules Before and After Reaction? Using a state-of-the-art atomic force microscope, the scientists have taken the first atom-by-atom pictures, including images of the chemical bonds between atoms, clearly depicting how a molecule's structure changed during a reaction. Until now, scientists have only been able to infer this type of information from spectroscopic analysis.  — Source

Are there any papers how this thing actually works?

There are a ton of papers on how AFM works, but I'll have a go at explaining it anyhow.

Atomic Force Microscopy (AFM) is a venerable technology for imaging nanostructures down to individual molecules by physically contacting atoms on surfaces. The whole process is very similar to a finger reading Braille.

What does it mean to physically contact an atom? It means that the AFM works by detecting some kind of repulsion between an atom and a very, very sharp probe due to their respective electron clouds repelling. The forces between surfaces on the nanoscale are a complicated mixture of attraction due to Casimir-Polder/Van der Waals interactions, dipole-dipole interactions, electrostatic repulsion, Pauli exclusion, etc. etc., but the key point is that if two objects get too close together they will experience a rapidly increasing repulsive interaction, which the AFM picks up.

This repulsion is detected through the deflection of a cantilever (a thin, long, elastic beam of material). The probe is mounted on the end of this cantilever, and the cantilever bends when the probe is pushed up against a substrate. The deflection of this cantilever is measured by bouncing a laser beam off the end, the change in position of the reflected beam indicating the force.

The AFM probe has an atomically sharp tip that can localise single atoms on a surface, and the tip is moved around using exquisitely precise actuators so that the tip can be scanned back and forth over a surface, picking up the bumps and valleys of the sample that indicate the positions of atoms. I've seen ads for AFM probes that actually have flared tips - these can obtain even more 3D information about nanostructures as they can hook into nanostructures from the side and measure the texture and relief parallel to the surface.


Law of reciprocal proportion Is it necessary that the valency of an element must have the same valency?

For eg. If 2 elements A and B combine with the same mass of C, is it necessary that they will have the same valency as they would have when combining AB?

The following are ladders of oxidation states for two elements. Simple examination shows there is no hard and fast rule for valence and stoichiometry overall. Valence need not be integral, either, for example, the Creutz-Taube ion.

Fe(-2), $\ce{Na2[Fe(CO)4]}$, Collman's reagent
Fe(0), $\ce{Fe(CO)5}$
Fe(0), $\ce{Fe3(CO)12}$
Fe(+2), $\ce{FeCl2}$
Fe(+3), $\ce{FeCl3}$
Fe(+4), $\ce{K4[FeO4]}$
Fe(+5), $\ce{K3[FeO4]}$
Fe(+6), $\ce{K2[FeO4]}$

V(-2), $\ce{[Na(diglyme)2][V(CO)6]}$
V(-1), $\ce{(Cp)V(CO)4}$
V(0), $\ce{V(CO)6}$

Chemistry seeks to establish rules.  Then the fun starts - breaking the rules.


How does pH change during the electrolysis of water? How does the pH change during electrolysis of a water and  magnesium sulphate solution? Also, will this pH change happen every time? Does the pH change differently when there is just water in the anode and cathode?

The pH will decrease, because magnesium sulphate is acidic.

As the electrolysis proceeds, the magnesium sulphate becomes more concentrated.