Preventing build-up of Diethyl Ether Peroxides I've never stored Diethyl Ether before but am thinking of producing minor amounts myself and storing said Ether inside of glass/PTFE bottles.

I am aware of possible pressure build-up due to vapors and storage temperature, but what always puts me off is the formation of explosive peroxides.

Now as the above Wikipedia article mentions:

  Diethyl ether hydroperoxide decomposes in the presence of sodium hydroxide and $\ce{Fe^2+}$ -containing salts.

This is obviously stating that I can stop the peroxide formation simply by adding in some $\ce{NaOH}$ which I have an abundance on hand.  However that above information is regarding hydroperoxides. The less nasty of the two peroxides.

Ethylidene peroxides are the explosive ones that detonates by simply touching it. This is what scares me. There's no information regarding the stopping of formation on the page, and frankly my google searches for "Safe storage of Diethyl Ether" and "Stop Diethyl Ether Peroxide formation" seem to come up with "put some iron wire in it" or Potassium Hydroxide.

So my question is: what is the most reliable, safe, and long-term solution to prevent building up of both of Diethyl Ether's peroxides, especially the Ethylidene peroxide, with the least amount of contamination possible.

There's no information regarding the stopping of formation [of ethylidene peroxide] on the page...

The required information is given implicitly. The wikipedia page states

  Diethyl ether peroxide, also known as ethylidene peroxide, [...] is a polymerization product of diethyl ether hydroperoxide

With other words:

If you prevent the formation of diethylether hydroperoxides by

storing diethylether in a brown bottle over sodium hydroxide

you will automatically prevent the formation of subsequent products, such as the ethylidene peroxide.

Note that there is a different approach, which is recommended in terms of workplace safety:

Substitute diethylether with methyl tert-butyl ether (MTBE), which does not form peroxides.


Why is the cathode of a lithium-ion battery metallic, which is the opposite of most other batteries? In most types of batteries, primary and secondary, the metallic electrode is the anode, and a non-metal acts as cathode.  Or a pure metal is the anode, and the cathode is a metal oxide.

Why is the opposite true in a lithium-ion battery?

This is not entirely true.  Wikipedia lists several possible cathode materials and they are oxides.  They may be "metallic" is the sense of being metallic conductors (some oxides are so), but they are ceramic compounds.
In secondary batteries the anode is typically Li-intercalated carbon (which is also a good conductor).  Lithium intercalates readily into carbon, allowing a safer and less side-reaction prone choice than pure lithium, whereas most other metals do not.


What is the n factor of h2o2 undergoing disproportionation reaction liberating oxygen There is a lot of confusion about the n-factor of hydrogen peroxide in its disproportionation into water and oxygen.
$$\ce{2H2O2 ->  2H2O + O2}$$
Some sources(1,2) say n-factor of $\ce{H2O2}$ is 2
and other say n-factor of $\ce{H2O2}$ is 1 for this reaction.
Actually my main problem is "the relationship between normality and volume strength of $\ce{H2O2}$". According to my respected teacher:
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
implying that n-factor of $\ce{H2O2}$ is 2 but I am calculating n-factor of $\ce{H2O2}$ as 1 because the definition of n-factor is "the no of electron gained  or lost per mole".
As we can see 2 electrons are participating in this reaction                            so electron exchanged per mole is 1 which is equal to n-factor. Please clarify.
Sources

https://socratic.org/questions/what-is-equivalent-weight-of-h2o2-in-decomposition-reaction#629354
https://www.meritnation.com/ask-answer/question/what-is-the-n-factor-of-h2o2-and-o2-in-the-reaction-h2o2-h2o/some-basic-concepts-of-chemistry/8526405

We can split the reaction
$$\ce{H2O2 -> H2O + O2}$$
into the respective reduction and oxidation half-reactions.
$$\ce{H2O2 -> O2 + 2H+ + 2e-}$$
$$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$$
Since the n-factor of $\ce{H2O2}$ for both these half-reactions is 2, the n-factor is:
$$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$
$$n_f = 1$$
Based on the comments under your post I believe you have gotten this far, but some sources claim that the n-factor for this reaction is 2. Well, how do you verify what is correct? In such a case, I'd simply write the balanced chemical reaction and verify whether the law of chemical equivalence holds or not.
$$\ce{2H2O2 -> 2H2O + O2}$$
The n-factor for $\ce{O2}$ should be $2$ as can be seen in the first half-reaction, and for $\ce{H2O}$ the n-factor should be $1$, since we have $2$ moles of $\ce{e-}$ in the second half-reaction and $2$ moles of $\ce{H2O}$, and n-factor is defined per-mole of the substance.
From the law of chemical equivalence, the number of equivalents of $\ce{H2O2}$, $\ce{H2O}$ and $\ce{O2}$ should all be equal. If the claim I make above that the n-factor of $\ce{H2O2}$ is $1$ is true, this should hold.
Plugging in the moles from the balanced chemical equation and using the n-factors described above,
Equivalents of $\ce{H2O2} = 2 \times 1$, $\ce{H2O} = 2 \times 1$, $\ce{O2} = 1 \times 2$ (moles $\times$ n-factor)
Thus the n-factor is indeed $1$. The n-factor of $\ce{H2O2}$ for the half-reactions is $2$.
(Note that all the n-factors above are with respect to the reaction $\ce{H2O2 -> H2O + O2}$ and they (most probably) will change for a different reaction)
Update (the above part answered the question before the edit):
From [1]:

The "volume strength" of hydrogen peroxide is defined as the number of times its own volume of oxygen a sample of hydrogen peroxide solution will evolve if decomposed naturally.
$$ \ce{H2O2 -> H2O + O2}$$

As you've mentioned in your question,
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
This does not match with the reasoning that the n-factor of $\ce{H2O2}$ for the above reaction is $1$, which would imply that the normality and molarity should be equal.
From [2]:

Now, equivalent weight of $\ce{H2O2} = \frac{34}{2} = 17$

A (N) $\ce{H2O2}$ solution contains 17 gms of $\ce{H2O2}$ per litre.

So in fact when we are relating the normality of an $\ce{H2O2}$ solution to it's volume strength, we are not referring to the normality of $\ce{H2O2}$ in the decomposition reaction. You would be right in saying that the normality of $\ce{H2O2}$ in the decomposition reaction is equal to its molarity.
One way of justifying the n-factor of $2$ of an $\ce{H2O2}$ solution (not undergoing any reaction) is due to the $-2$ charge on the $\ce{O2^{2-}}$ peroxide ion.
I can't really explain why this has been done, since it would make just as much sense to consider the n-factor of $\ce{H2O2}$ in the decomposition reaction, it is just a historical fact. Since n-factor is not a constant for a particular compound, it should be really important to specify context about normality when we relate it to volume strength, which is rarely done when the relation is taught which is what leads to this confusion. Both the n-factor's are correct in different contexts, so I would just recommend sticking to the n-factor of $2$ when talking about volume strength since that is what you have been taught.
[1]: Chemical Calculations by King A archive.org
[2]: An Introduction To Chemistry Vol. 1 Ed. 3rd by Das, Ranajit archive.org


How to assign priorities around the asymmetric carbon with its substituents having multiple bonds? How do I assign priorities to substituents with triple bonds?

If we take a look at the molecule above, the atom directly bonded to the asymmetric center is carbon for all three substituent.
Moving on to the atoms connected to those carbons, benzene side has carbons attached by a double bond and a single bond. The double bond is counted twice and the atoms count as $\ce{C, C, C}$.
For the sp carbon on the lower right, it is triple bonded to its neighboring sp carbon, so it counts as $\ce{C, C, C}$ as well.
The tert-butyl group is bonded to three carbons from methyl groups, so this gives another $\ce{C, C, C}$ making it a tie again.
If we look at the next set of atoms, the tert-butyl group has methyl groups so it's $\ce{H, H, H}$, giving it the lowest priority.
However, I'm not sure how to count the next set of atoms for benzene and the triple bond containing group. Judging from the R, S configuration the benzene group should have a higher priority, but I'm not sure how to figure that out. What should be my next step?

I would like to add a diagram to supplement @Raven's response and add a caveat about flipping structures to make R,S-assignments. The C's are the phantom carbons from the double and triple bonds. Working outward from the center of chirality, the first shell (in red) does not allow a decision to be made. At the second shell (in blue) the tertiary-butyl group drops out of the race and becomes the third priority group ahead of the hydrogen (4th priority). The third shell (in green) assigns the benzene ring as the top priority because the acetylene group has "run out of room" and must settle for second priority.
Flipping structures often result in drawing the enantiomer when one tries to place the lowest priority group in the rear. Leave the structure where it is and use your hands to determine chirality. Point your thumb from the chiral center to the lowest priority group. Curve your fingers so that they point in the direction 1 --> 2 --> 3. If you come up with 3 --> 2 --> 1, you are using the wrong hand! In this case your right hand (Rectus) will do the job.


What is the mechanism of functionalisation of nanoparticle using long chain alcohols How can long chain alcohols(not thiols) be functionalised on the surface of silver or gold nanoparticles?

I have read in an article that Polyvinyl alcohol is used as a capping agent for silver nanoparticle synthesis, but what is the mechanism for that?

What type of bonds form between nanoparticle and the long chain alcohol(covalent or ionic)?

Is there any relationship between chain length of alcohols(or thiols) to the functionalisation of nanoparticles?

The longer the alykylchains, the more likely you can form self-assembled monolayers (SAMs), which will hold themselves together by covalent interactions between chains.  The nanoparticle analog would be miceles.

However, alcohols are a fairly weak ligand for silver, so I'm not sure this is feasible.  SAMs are known to form on silver with carboxylate ligands.


Compressibility Factor Of Gases In the general equation of the compressibility factor $Z$, we define $Z$ as
$$Z=\frac{pV}{nRT}$$
Here, what is $p$? Is it $p_\text{real}$ or $p_\text{ideal}$?
Also, what is $V$? $V_\text{real}$ or $V_\text{container}$?

Both quantities are of the real gas.

Note that for the ideal gas behaviour, the compressibility factor $$Z=\frac{V_\text{real}}{V_\text{ideal}}=\frac{pV_\text{real}}{nRT}=1$$
 for any combination of $p, V, T$.

But real gases, in contrary to the ideal gas, have nonzero volume of molecules and there are intermolecular interactions.

The volume of molecules increases $Z$, as it makes the pressure higher, because the collision frequency is higher.

The cohesive forces between molecules decreases  $Z$. As molecule attraction, like hydrogen bonds or dipole interactions, decreases the effective number of molecules and therefore pressure.

This is reflected in the van der Waals equation - probably the simplest state equation of real gases, for not too high pressure:
$$\begin{align}
\left(p + \frac{an^2}{V^2}\right)\left(V - nb\right)&=nRT \\
\left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m - b\right)&=RT
\end{align}$$

$$\begin{align}
Z &= \frac{p\cdot V_\text{real}}{nRT} \\
Z&=  \frac{p\cdot V_\text{real}}{\left(p + \frac{an^2}{V_\text{real}^2}\right)(V_\text{real} - nb)} \\
Z&=  \frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \\
Z&=  \frac{1}{\left(1 + \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 - \frac{b}{V_{\mathrm m,\text{real}}}\right)}
\end{align}$$

For $|x|\ll1, 1/(1+x)=1-x$

For not too far from ideal behaviour, we can apply the above approximation.

$$\begin{align}
Z&=\left(1 - \frac{an^2}{pV_{\mathrm{real}}^2}\right)\left(1 + \frac{bn}{V_{\mathrm{real}}}\right) \\
Z&=\left(1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 + \frac{b}{V_{\mathrm m,\text{real}}}\right) \\
\end{align}$$

We can therefore also afford to neglect minor terms.

$$\begin{align}
Z&=1 - \frac{an^2}{pV_{\mathrm{real}}^2}+ \frac{bn}{V_{\mathrm{real}}}\\
Z&=1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}+ \frac{b}{V_{\mathrm m,\text{real}}}\\
\end{align}$$

To address clarified scenario, as a the same amounts of the ideal and real gas
are under the same temperature and pressure:

$$V_\text{ideal gas}=\frac{nRT}{p}$$
$$V_\text{real gas}=Z \frac{nRT}{p}$$

There is need to perform iteration for probably the easiest way to get the result.
With the $V_{\rm{ideal gas}}\ $ as the first approximation. The other option is to solve the root of the cubic equation what we probably do not want to.

$$\begin{align}
V_\text{real}&=Z(p, V_\text{real})\frac{nRT}{p}\\
V_\text{real}&=\frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \frac{nRT}{p} \\
V_\text{real}&=\left( 1 - \frac{an^2}{pV_{\rm{real}}^2}+ \frac{bn}{V_{\rm{real}}}\right) \frac{nRT}{p} \\
\end{align}$$

The van der Waals equation can also be expressed in terms of reduced
properties

$$\left( P_r + \frac{3}{V_\rm{r}^2}\right) \left( V_{\rm r}-\frac{1}{3}\right) = \frac{8}{3}T_\mathrm r$$

This yields a critical compressibility factor $3/8$.

The values of pressure, temperature and volume are divided by the
respective critical values of the given gas.


Reaction of Sulfur dioxide with Potassium What are the products of the reaction: $$\ce{K + SO2->}$$
Here, $\ce{S}$ is in $+4$ oxidation state. Hence, $\ce{SO2}$ has both oxidizing and reducing character. $\ce{K}$ is a stronger reducing agent and hence, it will reduce $\ce{SO2}$.

So, $$\ce{K + SO2->KO2 + S}$$ is a possibility.

But the answer given is $$\ce{K + SO2->K2SO3 + K2S2O3}$$

$\ce{KO2}$ is definitely out of question. True, it does form when $\ce{K}$ reacts with $\ce{O2}$, but that's another story. It is a peroxide (in a broader sense), so it contains oxygen in oxidation states other than -2. $\ce{SO2}$ contains oxygen in -2. Who is going to oxidize oxygen?

Turning down peroxides, we resort to the more realistic option of $\ce{K2O}$. Now, that's a basic oxide, and $\ce{SO2}$ is an acidic oxide. Surely some chemistry is going to happen between them.

As to where the reduced sulfur would go, it is not quite clear and depends on many things. With excess of potassium one might expect ending up with $\ce{K2O}$ and $\ce{K2S}$. Obviously that's not our case. Then we must consider a reaction between $\ce S$ and $\ce{K2SO3}$. Such reaction is known (though probably under different conditions than we have here) and indeed produces thiosulfate.

That being said, partially reduced compounds of sulfur are numerous and I see no reason to favor one above the other without some deep investigation. Have you heard about dithionites, for example? They are known to appear when reducing $\ce{SO2}$ with certain metals; why not here?


Relative rate of hydrolysis of the carbon chlorine bond in thioethers and ethers I am solving problems based on rates of hydrolysis of compounds but I don't have a general approach to such questions.

Which of the following compounds undergoes hydrolysis faster than the second under comparable conditions? It may be acid catalysed or base catalysed hydrolysis but the conditions must remain the same for each pair.

In my view since oxygen is more electronegative than sulfur, hence hydrolysis should be easier for second. But I am not sure about it.
 see part 2 here
see part 3 here

This is a classic case of neighboring group participation. The hydrolysis of the thioether is significantly faster because the sulfur can displace the chlorine to give a three member ring with a sulfur in it. The ring is quite reactive and will easily be opened by water.

For reference, you can also look at a compound like mustard gas. It has the same functional groups and is so reactive that is a war gas.


Need 28% acetic acid from 98% glacial acetic acid i need to make a solution of a 28% acetic acid, what i get is 98% glacial acetic acid. If someone could please guide me to make this solution.

I want to know the safety precautions i need to take when using 98% glacial acetic acid.

Also i would like to know safety precautions i need to take when using 28% glacial acetic acid.

Can i store it in PP narrow mouth bottles

How long will this solution stay good and also how to store it.

Thanks for your help.

As @Todd Minehardt already stated, a lot may be found in the internet.

Beside common sense (incl. appropriate equipment and protective clothing) and requesting help from trained persons (teaching assistant in lab class, chemistry teacher, to mention a few) in case of doubts:  If the chemical is well known and is (commercially) of importance, where available, chemical suppliers provide both the chemical as well as safety relevant data (in material safety data sheets, MSDS).  Often these data are provided in more than one language, too.  As two examples, look-up catalogues by Sigma-Aldrich or Acros Organics.  The wiki page mentioned above even mentions a dedicated search engine.

Sigma, by the way, compiled many useful data here, where the molarity calculator is one possible answer to your third question. In your case, the final volume of the diluted solution is 3.5 times the initial volume of the concentrated volume to be diluted in water.

Fourthly, the diluted acetic acid may be stored (at room temperature) in a bottle of PP.  Charts like this provide an overview. Do not forget to check it prior to use and then regularly for cracks or leaks, as for every container containing chemicals, and to store it safely according to your local law.


Nucleophilicity/basicity in protic and aprotic solvents In polar aprotic solvent, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

What I think: If I talk about their basicity, $\ce{F-}$ would be more basic due to instability from its high electron density. Is it applied same when talking about nucleophiliicty?
Why is an atom with high polarizability a strong nucleophile?

What I heard: an increased atom size means a high polarizability, and a high polarizability means a high nucleophilicty because the atom can give its weakly-held $e^-$ easier.

What I think: above sentence makes sense to me. But in another point of view, isn't it also right that the atom with properties like that would be stable, making it poor nucleophile?
When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?

What I heard: when deciding the nucleophilicty of an atom in a polar protic solvent, there are two factors that come into play: solvation and polarizability.

What I think: in a polar protic solvent, since the effect of solvation is significant, the polarizability shouldn't mean so much. Rather, its effect should be significant in polar aprotic solvent since there's no hydrogen bonding. Is this wrong?

Why is an atom with high polarizability a strong nucleophile?

An atom with a higher polarizability will share his electrons more easier since they are less attracted by the nucleus. The more the electrons are near the nucleus, the more it's hard the share them with an electrophile. $\mathrm{I^-}$ is thus more nucleophile than $\mathrm{F^-}$.

As you'll se below, the solvent can affect the nucleophilicity and reverse what I just said.

In polar aprotic solvent, why is $\mathrm{F^-}$ a stronger nucleophile than $\mathrm{I^-}$?

When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?

First, basicity and nucleophilicity are not the same at all. For instance, a strong base can be non-nucleophile, such as LDA for instance.

Then, let's talk about the polarizability of the $\mathrm{F^-}$ and $\mathrm{I^-}$ anions. The fluoride has a small polarizability: its charge is dense, very localized. It's called an hard anion. Iodide has a great polarizability with its dispersed charge. It's called a soft anion.

Using a protic polar solvent, the anions will be well solvated, even better if they are hard anions because the hydrogen bondings will be more efficient. The anion $\mathrm{F^-}$ is thus more solvated than $\mathrm{I^-}$ and less reactive.

In the case of an aprotic polar solvent, the anion is not so well solvated and the fluoride will be more reactive.


Why is dialysis tubing impermeable to sucrose? This question was posed a couple years ago but not answered adequately. You can view the older thread here. The cellulose membrane commonly used in high school and university labs is widely available from suppliers such as Wards, Carolina, Flinn, PASCO, etc. The specifications are either available online or were obtained by contacting the companies and in each case the molecular weight cut off (MWCO) was between 6,000 and 16,000 Da. Regardless of the specifications these products are used to conduct osmosis demonstrations by placing 0.2-1.0M sucrose on one side of the membrane and to create water potential. In addition to having conducted this exercise myself the results are validated in countless examples online. The mass/pressure results from osmosis and a benedicts test indicate sucrose is not moving across the membrane.

Considering Sucrose has a MW of ~342 Da the previous explanations seem insufficient;

1) The pore size of the membrane is variable. True, but according to specs at least 10% of the pores are larger than the specification and even with variation of 0-6,000 Da (worst case) it would still be like throwing a tennis ball through a garage door 10% of the time.

2) Interactions between solute molecules, shape, and solute-solvent causing a larger solvodynamic radius. Sucrose doesn't have any charges I'm aware of, it has a simple structure, and even covered in water molecules it would still be much smaller than the pores.

Any suggestions or alternative explanations?

Benedict's reagent detects reducing sugars (free aldehyde group) by reducing soluble blue Cu(II) to insoluble red Cu(I) oxide.  Sucrose aquoous solution is inert unless it is pre-hydrolyzed by heating with strong acid catalyst.

That certainly is a poser.  One possibility (SWAG, not fact) is regenerated crosslinked cellulose is overall not permeable to neutral solutes.  Its size selection is active toward dissociated solutes and proteins that are charged.  One can test this with anti-freeze.  The sodium fluorescene color should freely diffuse.  Tiny molecule ethylene glycol (EG) should be retained.

https://www.chemetrics.com/Glycol
  Test kit and how it works:  Iodate cleavage  of 1,2-diol to aldehyde, then aldehyde detection with Purpald.

One could iodate cleave permeated EG (or propylene glycol, glycerin, etc.) then use Benedict's solution.  Serial dilution of standard EG solution for control and calibration.  One expects any dialysis membrane to leak somewhat.  Major passage of uncharged 1,2-dihydroxylic solutes smaller and larger than than sucrose is the SWAG's falsifying observation.

http://chemsee-foods.com/antifreeze-detection
http://www.chemworld.com/Ethylene-Glycol-Test-Kit-p/k-4815.htm
http://www.hach.com/glycol-visual-test-kit-model-eg-1/product?id=7640218461
  EG in food and fluids tests.
http://www.astm.org/Standards/E202.htm
  Testing standards


Rates of Keto-Enol Tautomerism Are there any databases with rate constants for keto-enol tautomerism in small molecules? I'm looking for rate constants if possible.

This is not a database, and I don't know about how small molecules you are looking for, I found that article on Jacs Kinetics and thermodynamics of keto-enol tautomerism of simple carbonyl compounds: an approach based on a kinetic study of halogenation at low halogen concentrations

If you can access the file it will be better for you, if not I let you some matrix of values. I hope it can help ! I give you them in the same the order of the publication.

I added some other publication at the end.

Links

Tautomerization of 2-Acetylcyclohexanone. 1. Characterization of Keto−Enol/Enolate Equilibria and Reaction Rates in Water

Kinetic studies of keto-enol and other tautomeric
equilibria by flash photolysis (a lot of constants in this one)


What is the variation of atmospheric pressure at large heights? In an approximation (though not a very accurate one), temperature can be assumed constant throughout the atmosphere. The Earth's gravitational field can also be assumed constant (a better assumption).

For these approximations, what is the calculated pressure at the top of Mt. Everest ($8848\ \mathrm m$), given that the pressure at the top of Mont Blanc ($4807\ \mathrm m$) is $4.32\times10^4\ \mathrm{Pa}$?

  $T = 0.0\ \mathrm{^\circ C}$

  $\rm g = 9.81\ \mathrm{m\ s^{-2}}$

  $\ce{O2}$: $M = 32\ \mathrm{g/mol}$, percentage composition of atmosphere =
  $21\ \%$

  $\ce{N2}$: $M = 28\ \mathrm{g/mol}$, percentage composition of atmosphere =
  $79\ \%$

I thought about using $pV=n\mathbb{R}T$ for this question – perhaps using the difference in the volume of air above the different heights, but I couldn't come to an answer.

From the given composition of the atmosphere you can calculate the mean molecular mass $M$ of air.

Since Earth's gravitational field and temperature are assumed to be constant you can use the barometric formula

$$p(h_1) = p(h_0)~\mathrm{e}^{- \frac{M g}{R T} \Delta h}$$

to calculate the pressure at the top of Mt. Everest.


Heat energy vs. electrical energy in a galvanic cell or in an electrochemical reaction How can someone determine heat energy released during an electrochemical reaction?

Let's assume a simple galvanic cell at standard environment:

$$\ce{Zn(s) -> Zn^2+(aq) || Cu^2+(aq) -> Cu(s)}$$

Electrical energy can be derived from the electrode potential and electron flow, but how the heat release can be calculated? What's the heat release when $\ce{Zn(s)}$ gets solvated to $\ce{ZnSO4(aq)}$, and whats the heat release when $\ce{Cu^2+}$ ions gets plated to the electrode?

I'm using term "heat release", because I get confused about the enthalpy in electrochemical cells, since some of enthalpy seems to be connected to electrical work as well.

Operational electrochemical cells have $\Delta G\lt0$
And,
$$\Delta G=-nFE$$
$$\Delta G=\Delta H-T\,\Delta S$$
Knowing the entropy change, enthalpy change could be calculated
or
$$\Delta H=-nFE+nFT\frac{\mathrm d\Delta G}{\mathrm dT}\quad\text{at contant }p$$


Why a specific temperature for value of 1 calorie? Everything in chemistry often comes with a strong reason and facts behind it, but I stuck in definition of $1$ calorie.

My reference book define one calorie as amount of heat required to raise the temperature of one gram of water from $14.5$ degrees Celsius to $15.5$ degrees Celsius.

Why $14.5$ and $15.5$, why temperature such as  $0$ to $1$ or $19$ to $20$ degree Celsius is not chosen?

Please forgive me if this question is too easy and I can't figure out the logic related to it.

I shall be glad to have your suggestions.

why temperature such as 0 or 1 degree celsius is not chosen??

That has been chosen by some.
According to Physical Laboratory Experiments: Part III.--Heat, 4th edition (1912) :

The following different calories should not be confused:

ZERO DEGREE CALORIE:- The quantity of heat required to raise one gram of water from 0 [degrees] C to 1 [degree] C.

ORDINARY CALORIE: cal.-The quantity of heat required to raise one gram of water from 15 [degrees] C to 16 [degree] C, this being taken as the mean room temperature.

MEAN CALORIE OR ICE CALORIMETER CALORIE: The one-hundred part of heat required to raise one gram of water from 0 [degrees] C to 100 [degrees] C.

...

Heat capacity of liquid water is a continuously varying function of temperature itself, so a particular temperature or temperature interval must be specified.  Various intervals were chosen historically, and when a small interval near 15 degrees C is chosen it is because that was thought of as room temperature 100-200 years ago.


Why is the first member of group IVA and the second member of group VIA show stronger tendencies to catenate compared to the other group members? catenation is a process by which a molecule or an atom forms a straight or repeating chain with other atoms.in carbon it has 4 electrons in the outer most energy level hence can form chains repeatedly

The ability to catenate appears to be based on several factors:

The bond energy of the element to itself - stronger bonds promote catenation
Atom size - smaller atoms promote catenation due to lower steric demand and better orbital overlap (especially p-p)
Degree to which orbital hybridization is important in the structure of that elements compounds (for example, the bond angle in $\ce{NH3}$ is $\sim 108^\circ$ while that of $\ce{AsH3}$ is more like $90^\circ$
Electronegativity of that element - atoms with electronegativity in the middle of the scale appear to catenate best.

Here is a plot of bond energy and electronegativity for the first 36 elements. Note that there appears to be a sweet spot of electronegtaivity between ~2 and ~3 that produces high bond energies. In the box I drew are all of the elements that are known to have covalent compounds containing more than 3 catenated atoms of that element. In case you were wondering, I used bond energies for single bonds only so comparison was better.

Note that hydrogen is in this box. Hydrogen cannot catenate since it has only one valence electron, but if it had electrons in the second shell, it probably would catenate.


Can sodium acetate anhydrous be used as a desiccant? Sodium acetate trihydrate can be converted to sodium acetate anhydrous easily with an oven. I've been told to quickly store sodium acetate anhydrous in a sealed container because it takes water from the air and turns back into the trihydrate form. It's also non-toxic, and is even sometimes used as a food additive. For these reasons, would sodium acetate anhydrous be viable as a desiccant? Perhaps as a less-toxic replacement for silica gel?

No, not in most applications. While it may act as a dessicant, it is fairly volatile, so you'll get acetic acid evolving from it in the presence of water. Generally, you don't want to expose what ever you're keeping dry to acid. The two obvious issues when it comes to food uses (ie toxicity concerns) is smell (and taste) and acceleration of spoilage (although in the cases of acidic foods, it might not be an issue). You don't generally want your dessicant mixed into the product so keeping them separate is optimal but with the volatile acetic acid it could be a problem. If you wanted to use its acidity to enhance stability or taste (mixed into the food), then you'd have problems with changes in consistency as water was absorbed.


Why does heat solidify cake batter? Increased vibration causes the atoms in an object to spread out farther away. Thus, with added heat, and object goes form a solid to a liquid to a gas (and then to plasma, with enough heat). Why is it then, that when a heat something like cake batter, it solidifies?

It is caused by physical and chemical reactions in the cake batter of two types:

1) When heated and mixed with water, starches present in the flour will undergo a process called gelatinization in which water is absorbed in the structure. This will cause the batter to 'set' from a liquid into a solid.

2) Additionally, reactions of the proteins from the egg that you added can also work to 'set' the batter. At first the proteins will unfold (denature) after which they start to form bigger proteins through cross-linking reactions that turn the liquid into a solid.

The baking soda in the batter turns into sodium carbonate, water and carbon dioxide. The latter will make your cake nice and fluffy.


Regarding Analysis for CO2 Absorption in NaOH Background

Lab experiment, packed glass column (random packing of metal/ceramic material), countercurrent gas absorption, flowrates measured using rotameters, $rt \approx 27^\circ C$.

2 N NaOH flows downwards at 10-30 mL/s; Air + $\ce{CO2}$ (~10% $\ce{CO2}$ (v/v)) flows upwards at total rate of ~350-500 mL/s.

Samples were collected as (I) 10 mL + (II) 10 mL from the bottom of the column. An excess of $\ce{BaCl2}$ was added to (II). Both are titrated against 0.5 N HCl with phenolphthalein ($\ce{HPh}$) as indicator.

Expected Reactions

$$\ce{2NaOH + CO2 \to Na2CO3}$$
$$\ce{Na2CO3 + BaCl2 \to BaCO3 + 2NaCl}$$

Colour changes:

$$\ce{HPh + HO- \to Ph- + H2O}$$

$\ce{HPh}$ is colourless and $\ce{Ph-}$ is pink

The Problem

The first few readings were as expected and no irregularities were observed. The volumes of HCl added were $<$ 15 mL.

Later

On adding indicator and waiting for some time, the colour disappeared
(pink $\to$ colourless) for both (I) and (II). Shaking sped
up the colour change process.
On adding indicator and starting titration immediately, the colour
lightened (did not fade away completely) for a considerable volume of
HCl added, then darkened considerably, and after a few more drops,
suddenly became colourless (white in case of (II)). The volume of HCl added was $>$ 40 mL. Such a great change is not expected.

One Possibility:

$\ce{CO2}$ from the atmosphere reacts with the sample to increase the acidic nature of the solution and causes the disappearance of colour.
This would explain speeding up of colour change by shaking.

But, is the rate of absorption of CO2 into a still sample of $\ce{NaOH + Na2CO3}$ this fast?

Other possibilities

Formation of buffer: $\ce{Na+}$, $\ce{CO3^{2-}}$, $\ce{HO-}$ ions are present in (I). (II) additionally contains $\ce{Ba^{2+}}$, $\ce{Cl-}$. The tap water used might be a source of $\ce{Cl-}$ as well. $\ce{HCO3-}$ formation during titration (?)
The packing material might have reacted into the system somehow. This seems unlikely, and the packings did not show signs of fouling.
The chemicals used contain impurities causing these weird effects, in which case, which type of impurites could cause this?

Note: Do comment if you more information is needed.

Phenolphthalein becomes colorless again (from pink) in strongly basic medium with time. It is a known phenomenon called the alkaline fading of phenolphthalein. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=fading+of+phenolphthalein+in+alkaline+solution&btnG=

It is not a good choice to use phenolphthalein in the presence of sodium carbonate or bicarbonate. Methyl orange is typically used when NaOH and Na2CO3 are present together. Please consult the classic Vogel's "Quantitative Inorganic Analysis" for more details. Alternatively, you may try a potentiometeric titration (pH vs. volume of titrant). Measure the pH after every 0.5 mL addition of titrant and construct a titration curve.

The reaction of CO2 with NaOH is very fast.


Why is lutetium part of the lanthanide series? I’ve been looking at the periodic table and began to wonder - why is lutetium part of the lanthanide series? Or, for that matter, why is lawrencium part of the actinide series?

This isn’t readily apparent when looking at the traditional periodic table because the lanthanides and actinides are separate and below the main table. However, on the extended table, the two elements don’t appear that they should be in the groups that they are in: lutetium contains the $5d$ electron shell whereas the lanthanides do not, and lawrencium contains the $6d$ electron shell whereas the actinides do not.

The assignment of elements to groups or series in the periodic table is not based exclusively on the electron configurations concerned. Helium for example is an s-block element yet is placed above Ne in the p-block due its noble gas properties.

Lutetium is part of the lanthanide series, notwithstanding its electron configuration, because its properties and those of its compounds are similar to those of the other lanthanide elements.

Lawrencium is assigned to the actinides for basically the same reason. Experiments have so far failed to find any chemistry-based evidence for Lr having a p electron. While it may have a p electron in the gas phase it probably has a d electron in the condensed phase.

Whether either Lu and Lr, or La and Ac are assigned to group 3 will not make any difference to the composition of the lanthanides or actinides, since both of these are horizontal series rather than vertical groups.


IUPAC nomenclature: putting the number before or in the middle of the name Which is preferred according to "strict" IUPAC:

2-pentanone or pentan-2-one ?

The name “2-pentanone” and “pentan-2-one” describe the same compound. The name “2-pentanone” is in accordance with obsolete IUPAC Recommendations. The relevant rule of the IUPAC Nomenclature was changed in the 1993 IUPAC Recommendations:

R-0.1.2 Position of locants
Locants (numerals and/or letters) are placed immediately before the part of the name to which they relate, except in the case of traditional contracted forms (see R-2.5).

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

Examples:

hex-2-ene (not 2-hexene)
cyclohex-2-en-1-ol (not 2-cyclohexen-1-ol)
pentan-2-one (not 2-pentanone)
(propan-2-yl)benzene

Examples for traditional contracted names with locants at the front of the name:

2-naphthyl (not naphth-2-yl), the preferred name is naphthalen-2-yl
2-pyridyl (not pyrid-2-yl), the preferred name is pyridin-2-yl


Is this graphic of molecules correct? First off, huge fan of chemistry, not a chemist – do be nice :D

Would an image like this make sense as an infographic for the molecular composition of the air? I especially wonder if the different sizes of the molecules (or do they all have the same size?) and the constellation make sense in theory.

Many thanks

No.  Most molecules of air have two atoms.

So if you want to be quantitatively accurate, if you have 100 molecules in a graphic, it would show

78 dinitrogen (N-N) molecules

21 dioxygen (O-O) molecules

1 argon atom (Ar)

http://eesc.columbia.edu/courses/ees/slides/climate/table_1.html

And you could add some water (H-O-H) molecules if you want to depict humidity.


Why can't an enolate anion successfully displace a halogen from a benzene? Why can't an enolate anion successfully displace a halogen from a benzene? I am told that the geometry is incorrect; that such SN2 substitutions take place on $\mathrm{sp^3}$ hybridized (i.e. tetrahedral) substrates.

This reasoning is circular. So what's a better reason?

I can think of one reason but it is not satisfactory; it's still a bit circular. SN2 = backside attack; how would one attack a planar bromobenzene? From within the ring? Seems like there would be a lot of possibly unfavorable interactions with the p-orbitals for something to get within the ring. Also this doesn't explain why substitution reactions with enolate anions don't take place at other $\mathrm{sp^2}$ carbon atoms - i.e. bromoethene.

explain why substitution reactions with enolate anions don't take place at other sp2 carbon atoms

As explained in this earlier answer, $\ce{S_{N}2}$ reactions do occur at $\ce{sp^2}$ carbons, they're just higher energy and consequently, less common.  So the question becomes, "why is this pathway higher in energy?"

Since the $\ce{S_{N}2}$ reaction involves "backside" approach to the carbon undergoing substitution, steric considerations play a role.  However, the attacking reagent could alter its approach by 15 degrees or so lessening steric interactions (e.g. in benzene the attacking nucleophile could approach at an angle of 15 degrees above the plane of the benzene) and not increasing the activation energy too much.

One way to explain the high energy associated with $\ce{S_{N}2}$ attack at a vinylic carbon is as follows.  Consider the ethylenic double bond as a bent bond system, that is, we have a two-membered ring (instead of a traditional sigma-pi double bond) and each carbon in the ring is $\ce{sp^{4.3}}$ hybridized.  This bent bond description is equivalent to the alternate pi bond description in all regards including electronic (btw, it makes the reason [strain] for the high heat of combustion of olefins more palpable). If possible, build a model to help visualize the situation.  As the attacking nucleophile approaches and begins to interact with one of the ring carbons, that carbon will attempt to rehybridize to an $\ce{sp^2}$ carbon (just like in the traditional case of $\ce{S_{N}2}$ attack on an $\ce{sp^3}$ carbon).  In this case we're forcing the already highly strained bond angle in our two-membered ring to try and open further - towards 120 degrees. Hence, this is a high-energy pathway that is not often observed.

Alternately, this same concept can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state.  It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons in higher energy p orbitals, not offset by the energy lowering of the electrons shifting from $\ce{sp^2}$ bonds to $\ce{sp}$ bonds) thereby increasing the energy required to reach the transition state.

Another point to consider is that since the $\ce{S_{N}2}$ reaction proceeds with inversion of configuration, such a reaction in a ring system would lead to the formation of a trans-double bond.  This would be an extremely high energy process in the case of a small ring system such as benzene.

Finally, $\ce{sp^2}$ hybridized carbon is more electronegative than $\ce{sp^3}$ hybridized carbon due to the increased s-character.  This will increase the electron density around an $\ce{sp^2}$ carbon and, as a nucleophile approaches, the increased electron repulsion will make this a higher energy pathway that nucleophilic attack at a less electron rich $\ce{sp^3}$ carbon.


Intrinsic Reaction Coordinate - What does 'Connectivity' really mean? I am currently in the process of connecting transition state (TS) structures to other minima in order to determine a reaction pathway.  Now I understand the basic concepts of this method.  You start with your TS, run an IRC in both the Forward and Reverse directions, then you optimize the final structure in the successful IRC run.

My question is, what constitutes connectivity between a TS and a minimum energy structure?

Examples

1.) I run an IRC and I take the last resulting structure and optimize it. The optimization takes 4 steps and the geometry barely changed.  Is this structure connected to the TS?

2.) I run an IRC and I take the last resulting structure and optimize it.  The optimization takes 20 steps and the geometry changed fairly appreciably, however, connectivity of the atoms remains the same.  Is this structure connected to the TS?

3.) I have a TS where a water is being formed next to some other molecule.  I run the IRC and following the pathway shows you that the water forms and is being moved away from the molecule.  The IRC successfully finishes so I take the last structure and optimize it.  The optimization brings the water back towards the molecule to some particular spot where H-bonding occurs with the molecule.  Is this structure connected to the TS?

Note

These examples were purposely laid out in order to determine at what point TS->Minima are or are not connected.

Various criteria may be applied in order to ascertain the connectivity of structures. A common solution is a path generated by requiring it to be a Newton path on the potential energy surface (E):
$$
(\nabla\nabla^\ast E(x))^{-1}\nabla E(x) = -\frac{dx}{dt}
$$
Two points $x_0$ and $x_1$ are connected only if $\exists x:\mathbb{R}\to\Omega,\tau_0, \tau_1\in \mathbb{R}$ such that $x(t)$ is a Newton path and $x(\tau_i)=x_i$, i.e., starting and ending structures, where $\Omega$ is your space of structures (usually centered 3D coordinates). $x$ is then a parameterization of that path on $[\tau_0,\tau_1]\subset\mathbb{R}$.

Sometimes it is additionally required that there is at most 1 maximum on $E(x)$ between $\tau_0$ and $\tau_1$.

You may be interested in reading this page. In a literature search, you may want to read up on the work by Vanden-Eijnden, Bofill, Quapp, and Henkelman. This topic is too rich to discuss in this forum.

To address the specific examples, assuming the starting point is in the convex attraction basin of the nearest optimum and the used optimizer converges in such instances, the optimum is connected in the above sense. Most optimizers used in geometry optimizations are some variant of (quasi-) Newton-Raphson optimizer, so this criterion is given. There are several reasons why the IRC may not converge to the same optimum as the subsequently used geometry optimizer, including changes of basis sets, convergence tolerances, or in case of quasi-Newton methods, the Hessian guess.

The fact that the optimizer brings the water closer to the other molecule is no indication of lack of connectedness. Instead the IRC run did not end at an optimum. It is actually expected that the energy of two (neutral) molecules decreases as they approach each other from infinity even in the complete basis set limit (van der Waals forces). You just found the vdW/BSSE-bound pre/post-complex.


Explain isotopic uranium/perchloric acid digestion What is happening in this video? Can you explain this technically as well as in layman's terms?

I'll summarize it. I see a technician moving fuming mixtures of presumably uranium and organic compounds on and off hot plates. Perchloric acid is mentioned, and a large bottle is present.

If that is a perchlorate digestion, the technician needs more training.  The hood door must be lower despite the Official marker.  His cotton lab coat must NOT be exposed to vapors!  Sleeves should be inside long gloves.  He has a bottle of perchloric acid next to a running hot plate!  I'd have a 5-gallon bucket of water nearby if the gloves ignite.  As perchlorate digestions occasionally explode, and he's doing a lot of them, I'd wear goggles under the face shield and wear Dyneema or Spectra body armor under the lab coat.  One might have some concern for hands (longer pair of tongs) and neck, too.

He's apparently cooking off a sample of organics-contaminated uranium whatever for trace contamination or isotopic analysis.  Everything oxidizable will be gone, leaving uranium oxides.

That is NOT marked as a perchlorate fume hood!  Everything organic downwind of those fumes is a potential high explosive, and certainly a violently flammable material including its own oxidizer.  There will be massive corrosion of all metals.  You don't want to be standing, living, downwind of the exhaust stack.  OTOH, the operation is low cost/sample, and anybody off the street is a potential lab tech.


Structure of AsF3Cl2 Since the molecule has 5 bonding pairs and zero lone pairs, the shape should be trigonal bipyramidal. But one thing i am confused about is which atom(s) will occupy the axial and the planar positions, respectively, and why?

Thanks

The axial positions see 3 substituents at 90 degrees, while the equatorial positions only see 2.

To first order, it is generally favorable to place the smaller substituents in the axial positions. Unfortunately, when you have fluorine, higher order effects (like minimizing dipole moment) may overcome steric considerations, but I would still prefer to place the fluorines axially as my first guess.


Learning chemistry with software? I'm interested in learning organic chemistry, and am following a text (Organic Chemistry by Clayden) and an online course. I also have access to ChemBio3D.

I feel as though the way chemistry is taught conventionally is more difficult than it needs to be. Some online courses utilize industry software when teaching things (such as using MatLab in engineering), which makes visualizing and understanding a lot of the theories and concepts taught in books much more intuitive, since you can replicate the theory directly into a program.

Does a similar course exist somewhere for chemistry? I'm not talking about the little java applets that are available everywhere that mimic certain reactions or visualize concepts, I mean a course that utilizes a powerful software suite such as ChemBio in order to teach organic chemistry? Something that will teach both organic chemistry at a proper undergrad level and a good piece of industry software at the same time?

Do your chair parade.  See what the Kahn Academy has to offer.  Have more than one texbook.  If organic is not the best thing in the world pursued clothed, consider other options.  Employment at the bench will be under management that has other goals.  Grant funding considers discovery to be unacceptable risk.  Remember what they called the first designed degenerate Cope rearrangement - BULLVALENE.  It worked.  So did its two-step synthesis using no reagents.

LCAO for hybridization, geometry, resonance, and 1,3-shifts.  Nomenclature and chirality.  Woodward-Hoffmann comes later.  Carbocations, carbanions, radicals.  Carbonyl condensations.  Hard and soft acids and bases, Lewis and Brønsted acids and bases.  Reductants and oxidants; reagents in general.  Name reactions (that you already know as pedagogy).  Practice looking into stereograms.

NMR, IR, UV-Vis (not so much now), mass spec with database and software.  You should have some glassblowing under your belt - Schlenkware;  T-, butt, and ring seals; making an ampoule.  Own your own didymium glasses and Kennametal cutter.  Burn a pencil for its "lead."

Synthesis.  Work backwards via the synthon approach.  We used to read the new Aldrich catalog, and Fieser and Fieser, but both have has gotten out of hand.  Spend a weekend/month skimming the main journals - JACS, Angew. Chemie, JOC, Tetrahedron - and what interests you.  Never fill a reaction flask more than half full, because the universe hates you.  COTTON lab coats, because they don't melt onto your flesh.  Don't do anything stupid twice.  When in doubt - palladium.

Buy HyperChem Lite, oh yes indeed.  Remember to add hydrogens, assign atom types, then minimize well below the recommended cutoff for 10 times the recommended iterations, at least.  Note how the software is boggled when minimizing high symmetry cage ketones

If it isn't fun, you might be a theorist.  There is no shame in that - but you don't get to make crystals.


Concentrating sulfuric acid Since calcium oxalate is 3000 times less soluble than calcium sulfate, would it be possible to concentrate sulfuric acid from let's say $\pu{50\%}$ solution using following steps? :

adding calcium carbonate to cold dilute $\ce{H2SO4}$
filtering off resulting $\ce{CaSO4}$
mixing $\ce{CaSO4}$ with oxalic acid in 1:1 molar ratio
adding just enough water to submerge the solids
vacuum filtering off the now concentrated (hopefully) $\ce{H2SO4}$

My reasoning behind this is that even though at any given time there's only a small amount of $\ce{CaSO4}$ dissolved, the reaction producing sulfuric acid would eventually go to completion with high yield due to it's the massive kinetic favourability. The only problem I see is that the amount of water needed would limit the final concentration. Would adding merely enough water to wet all the solid still work, improving the final concentration?

Calcium oxalate solubility depends on acidity, as the weak oxalic acid is protonated in highly acidic sulphuric acid.

Additionally, oxalic acid would be dehydrated if sulphuric acid was concentrated enough.

If we for convenience mark oxalate anion $\ce{(COO)2^2-}$ as $\ce{Ox^2-}$, then

$$\ce{CaOx <<=> Ca^2+ + Ox^2-}$$

$$\ce{Ox^2- + H+ <=>> HOx^-}$$

$$\ce{ HOx^- + H+ <=>> H2Ox}$$

$$\ce{  H2Ox ->[H2SO4] H2O + CO ^ + CO2 ^}$$

Therefore oxalic acid has no chance to create concentrated sulphuric acid.


Food Grade Sodium hypochlorite decomposition/neutralization I recently purchased some 55 gallon blue barrels for rainwater harvesting.  Upon closer inspection of the barrels, I noticed they previously housed food grade sodium hypochlorite which was used to wash produce.  I was searching for food grade barrels but wasn't expecting bleach.  Because I will be watering plants and fruit tress with the rainwater, I don't want the water chlorinated which would kill the bacteria I'm trying to promote in the soil.  These barrels are empty but I was advised to rinse them when I initially purchased them.

Can I assume that just rinsing with water won't completely get rid of the bleach  and that the residual bleach which may be on the walls of the barrel will not decompose or neutralize by itself?

If I were to wash the barrels with water and Hydrogen Peroxide, will this suffice to remove the chlorine?

Several good tap water rinses should probably get rid of any significant remaining amounts of bleach, although it's possible some bleach residues might adsorb to the wall of the barrel and resist these efforts.

Per here and here, if you want to be very sure the residues are removed, dissolve a handful of vitamin C (ascorbic acid) tablets in water and use that to rinse the barrels until all bleach odor is gone.  The ascorbic acid will act as a reducing agent and should reduce the bleach levels to next to nothing.

Dmitry is right: as an oxidizer, $\ce{H2O2}$ will probably do little to help.


What are the mechanisms for the reduction of organic compounds employing red phosphorus and iodine? I was recently studying oxidation and reduction reactions.

I found red phosphorus and iodine reduce many different organic compounds.

Out of interest I would like to know the mechanism for these reactions. A concerted path is preferred.

I found Red P and Iodine reduce many different organic compounds.

Actually, it is typically used for the deoxygenation in benzylic positions, using benzylic alcohols or benzoins as starting materials.

Aliphatic alcohols yield iodoalkanes, propargylic alcohols often undergo elimination under these conditions whereas allylic alcohols are typically converted to complex mixtures without synthetic value.

The deoxygenation is often carried out with catalytic amounts of iodine but an excess of red phosphorous. In principle, catalytic amounts of $\ce{NaI}$ plus one equivalent of $\ce{H3PO3}$ in water would work as well.

When using iodine and red phosphorous, hydrogen iodide is generated in situ via the formation and subsequent hydrolysis of $\ce{PI3}$.

The formal replacement by a proton is not fully understood and is disputed whether the formation of a benzylic iodide is really involved.

In princpiple, elimination from water from the protonated starting material would yield a stabilized benzylic cation. This might get reduced by a iodide to a benzylic radical and, in a second step to a very shortlived benzylic anion which is instantaneously protonated.

Both reduction steps yield iodine radicals from iodide and in summary regenerate iodine.


How could you accomplish the following transformation in sufficient yield? I've considered Riley oxidation but the major product would have a hydroxyl group at the benzylic carbon to the left of the double bond instead of the methyl group as shown. Is there a way to make the reaction favor oxidation of the methyl group?

I'm not an organic chemist, but upon looking for similar systems I found a nice collection of various papers that all use selenium dioxide, so allylic oxidation as you mentioned it with various other functional groups. I copied the sections from the papers (therefore sorry for the quality and the Chinese in one paper) together with the preparations.

To me this looks like the most promising route here. According to that mechanism it should favor the allylic position here selectively and therefore and also due to sterics hopefully not oxidize the longer alkyl chain.


Change in Enthalpy in Free Expansion of Ideal and Real gas against Vacuum By free expansion, I am referring to gas allowed to expand freely against vacuum in a Joule Expansion.
If gas is ideal then change in Internal Energy '∆U' and change in Enthalpy '∆H' is zero.
(By ∆H = ∆U + nR∆T).
But I was wondering whether '∆H' would still be zero for real gas?
(I know that ∆U will be zero for real gas. I just want to ask about ∆H)

In the Joule experiment, $\Delta H$ is not zero for a real gas.  $\Delta H=\Delta U+\Delta (PV)$, and even if $\Delta U$ is zero, there is no physical reason to expect that $\Delta (PV)$ would be zero in this process.


What makes Coke acidic? What makes Coke and other soft drinks acidic?  These drinks are carbonated, so a freshly-opened can should have a lot of dissolved carbonic acid, but the ingredients also lists phosphoric acid.  Would flat Coke be significantly less acidic than carbonated Coke?

If you look at the list of ingredients, you will find that Coca-Cola and many other brands contain phosphoric acid (food grade - don't worry). This is probably the dominant cause of acidity. Citric acid is another common component. Of course, the carbon dioxide is also acidic, but weakly, so there should be very little difference between carbonated and flat.


Is work done by a system negative delta G or positive delta G? My textbook states that a process is spontaneous and can perform work only when it is moving toward equilibrium. Since the process is spontaneous, wouldn't that mean that the work done is attributed to negative free energy?

Another way of considering it is the question of what should the sign of the amount of work done on objects outside the system (or heat released from the system) should be when the reaction is exothermic or endothermic.

Depends what you mean by that.  I agree it can be a little bewildering to figure out exactly what all these words in thermodynamics mean -- energy and work and heat and free energy and God knows what else.

It may help to remember in all such cases we're just talking about energy.  That's it, that's all there is.  The various types of energy that we call enthalpy or free energy or whatever are just different ways to identify part of the energy with the system and part with the surroundings.  Most of the time, changes in enthalpy or free energy or energy are just statements that the energy got transferred under certain conditions, for example with the system at constant temperature and pressure, or maybe constant pressure and volume.

But the important thing to remember is it's all just energy, and you can use the same intuitions about it you'd use if you just used the word "energy" everywhere (up to a point, of course, because the distinctions aren't meaningless).

So for example in this case, I hope it makes intuitive sense to you that if a system does work in its surroundings (causes weights to be raised, springs to be compressed, batteries to be charged, whatever) that it has to lose energy for that to happen.  Energy has to flow out of the system and do the work.  So that would suggest the energy -- or enthalpy -- or free energy -- has to go down. And so they would, generally.  That's a good intuition to have, a good starting point, although there are certainly more complicated tricky situations where you'd want to refine your first guess.

Usually if you're talking about free energy, you're talking about a system at constant T and P.  For that system, to do work, it has to lose energy, which means G goes down, dG < 0.  On the other hand, if work is done on the system -- springs within the system are compressed, batteries charged, et cetera, then dG > 0 and the free energy goes up.  All just as intuition would suggest for what happens to the energy.


Six independent papers reporting epitaxial formation of silicene in 2012 According to a Nature news article cited below (info-box "The rise of silicone"), six independent papers were published reporting successful epitaxial formation of silicene.
Does anyone know which those were? I am just able to find the following two papers:

Fleurence, A., R. Friedlein, T. Ozaki, H. Kawai, Y. Wang, Y.
Yamada-Takamura (2012). "Experimental evidence for epitaxial silicene
on diboride thin films". Physical Review Letters 108(24)
Vogt, P., P. De Padova, C. Quaresima, J. Avila, E. Frantzeskakis, M.C. Asensio, A. Resta, B. Ealet, G. Le Lay (2012). "Silicene: compelling experimental evidence for graphenelike two-dimensional silicon". Physical Review Letters 108(15)
Nature (2015). "Graphene’s cousin silicene makes transistor debut". Nature news article. URL: http://www.nature.com/news/graphene-s-cousin-silicene-makes-transistor-debut-1.16839

Does anyone know where to find the other four?

Since this is a Nature News article, usually it is a way of promoting a scientific article in one of the current Nature publications - that's why it can get away with stating "Six independent reports" without six citations attached.  In this case, it is citing this 2015 article:

  Tao, L. et al. Nature Nanotechnol.
  http://dx.doi.org/10.1038/NNANO.2014.325 (2015)

There are many citations in that article from the 2010-2013 time period reporting silicene sheets and nanoribbons.  Keep in mind, the Fleurence et al. paper you mentioned might not be one of the six mentioned, because it uses a diboride substrate instead of silver.


Why does gaseous HCl not change dry blue litmus paper to red? I am doing a practical in lab, in which HCl is to be evolved. I found that when blue litmus is near the evolved gas, it doesn't turn red. But, when litmus is dipped in water and then brought near the HCl gas it does change to red. Why is this happening?

This is because gaseous HCl  does not dissociate to give up protons and chloride ions.
But, in aqueous solutions HCl does dissociate and exist in ionic form:

$$\ce{H2O(l) + HCl(aq) -> H3O+ (aq) + Cl−(aq)}$$

Thus it acts as a typical Brønsted acid (proton donor) in an aqueous environment. In this case, water is gaining a proton $\ce{(H+)}$, so it is a base, while HCl is giving one away, so it is an acid.

Therefore it is able to turn the blue litmus paper red (a behaviour typical of acids).

For related topic please see: How does the litmus pH indicator work? as suggested above in comments.


Why don't substances have a condensation point? All substances have a freezing point at which they transition from a liquid to a solid. They all also have a melting point at which they transition from a solid to a liquid. However, as far as I can tell a substance only has a boiling point at which it transitions from a liquid into a gas. There is no analogous concept of a "condensation point" at which a gas transitions into a liquid. Why not?

A "condensation point" is sort of a misdirection as to the physical process involved. "Point" is really a stand-in for "temperature."

Consider water in air. Let's say that the temp is 20 degrees C, but the "dew point" is 10 degrees C. So if the temperature drops to 10 degrees C then the air is saturated with water and and a further drop in temperature will result in the formation of dew. So being past the dew point is dependent on the amount of water in the air and the temperature, not just the temperature alone.


Why do we say there's no carbon dioxide present? Were were estimating dissolved $\ce{CO2}$ in water by  American Public Health Association method. It was a titrimetric method using phenolphthalein indicator.

Titrant used was $\ce{NaOH}$ and analyte was sample water.

Our instructor said that on adding phenolpthalein if the water sample turns pink prior to the experiment (addition of $\ce{NaOH}$) then it can be concluded that no free $\ce{CO2}$ is present . But how's it so?

You're doing a titration of the carbonic acid $\ce{H_2CO_3}$ which is the product of the dissolution of $\ce{CO_2}$ in water  by a base here the sodium hydroxide. According to wikipedia for carbonic acid, $\ce{pKa_1=3.6}$, $\ce{pKa_2=6.3}$ and $\ce{pKa_3=10.32}$.

When the carbon dioxyde is in water, the solution is acid then your phenolphtalein is colorless. During your titration while not all the carbonic acid as react then the solution will still be colorless because phenolphtalein's color change appears at around $\ce{pH=8.2}$ (Ref). Then when you finish you're titration the color of the solution will change.

You need to make some prediction to find the theorical pH at the equivalence.
In this case at the equivalence $\ce{[H^+]=\sqrt{Ka_1Ka_2}}=1.26 \times 10^{-10}$ then $$\ce{pH_{eq}=-\log_{10}(1.26 \times 10^{-10})=9.9}$$

So at this point the phenolphtalein must be pink and then no more acid (so carbone dioxyde) will be in your solution. You'll have only carbonate ion which will not react with sodium hydroxyde.


Reaction molecularity and order Question:

  A reaction involving two different reactants can never be a

  unimolecular reaction
  bimolecular reaction
  second order reaction
  first order reaction

The answer as per my book is 1. unimolecular.
But the nucleophilic substitution SN1 is unimolecular but it involves two different reactants. Let us take an example,
$$\ce{(CH3)3C-Cl + Br- -> (CH3)3C-Br + Cl-}$$ This follows SN1 mechanism. Yet, it involves two different reactants even though the R.D.S depends only on the formation of carbocation and the rate law expression being
$$\text{rate} = k[\ce{(CH3)3C-Cl}]$$ where $k$ is the rate constant.
So, is the answer that unimolecular reactions can never involve two different reactants wrong? Or am I missing something trivial?
Also it would be great if other options are discussed.

First of all,
Molecularity isn't defined for a complex reaction.
From Physical Chemistry by Ira. N Levine: (Page 531, Rate Laws And Equilibrium Constants For
Elementary Reactions  ):

The number of molecules that react in an elementary step is the molecularity of
the elementary reaction. Molecularity is defined only for elementary reactions and should not be used to describe overall reactions that consist of more than one elementary step. The elementary reaction $\ce{ A → products}$ is unimolecular. The elementary reactions $\ce{A + B → products}$ and $\ce{ 2A → products}$ are bimolecular. The elementary reactions $\ce{A + B + C → products}$, $\ce{2A + B → products}$, and $\ce{3A → products}$ are trimolecular
(or termolecular).

Atkins' mentions: (page 810, Elementary Reactions):

The molecularity of an
elementary reaction is the number of molecules coming together to react in an elementary
reaction.It is most important
to distinguish molecularity from order: reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism.

Now, from Wikipedia:

This difference can be illustrated on the reaction between nitric
oxide and hydrogen:
${\displaystyle {\ce {2NO + 2H2 -> N2 + 2H2O}}}$
The observed rate law is ${\displaystyle v=k{\ce {[NO]^2[H2]}}}$, so
that the reaction is third order. Since the order does not equal the
sum of reactant stoichiometric coefficients, the reaction must involve
more than one step. The proposed two-step mechanism is
$${\displaystyle {\ce {2 NO + H2 -> N2 + H2O2}}}$$
$${\displaystyle {\ce {H2O2 + H2 -> 2H2O}}}$$
On the other hand, the molecularity of this reaction is undefined,
because it involves a mechanism of more than one step. However, we can
consider the molecularity of the individual elementary reactions that
make up this mechanism:
The first step is termolecular because it involves three reactant
molecules, while the second step is bimolecular because it involves
two reactant molecules.

As far as your example of SN1 reactions is concerned, Peter Skyes'  acknowledges while discussing the kinetics of SN1:

The molecularity refers to the number of species that are undergoing bond-breaking or making in one step of the reaction, usually, in the rate limiting step.

Thus, when we say SN1 (S= substituition, N= nucleophilic, 1= unimolecular) we are essentially talking with respect to the RDS in which the carbocation is formed. Note that though most books identify 1 for unimolecular, Clayden's book refers to it as 1st order, which, in my honest opinion, is much more clearer.

The slope of the first graph is simply the first-order rate constant because rate = $k_1[t\text{-}\ce{BuBr]}$. But the slope of the second graph is zero. The rate determining step does not involve $\ce{NaOH}$ so adding more of it does not speed up the reaction. The reaction shows first-order kinetics (the rate is proportional to one concentration only) and the mechanism is called SN1, that is, Substitution, Nucleophilic, 1st order.

Conclusion:
The question is poorly framed. However, since molecularity is defined only for an elementary reaction, maybe after looking at the options, the question setters expect you to assume that they are talking about an elementary reaction. If this is the case, only then option 1 (but along with option 4) can be marked correct. However, because they haven't mentioned anything, the candidate attempting the question is bound to get confused. If they are talking about the RDS step, then your argument is valid. An SN1 reaction is indeed unimolecular wrt the RDS step (according to all standard books like Solomon Frhyle, Peter Skyes', etc.) which would deem all of the options incorrect.


Why is the boiling point of sugarcane juice lower than the boiling point of water? As the boiling point is a colligative property, one would expect that the boiling point of sugarcane juice should be higher than pure water because Sugarcane juice contains water (75–85%), non-reducing sugars (10–21%), reducing sugars (0.3–3%), organic substances (0.5–1%), inorganic substances (0.2–0.6%) and nitrogenous substances (0.5–1%)1. However, the boiling point of sugarcane juice is only 91-95°C [2]. Are there any chemistry theories that can explain this drop in boiling point?
[3]
1 Sreedevi, P., Jayachandran, L.E. & Rao, P.S. Browning and bioactive composition of sugarcane juice (Saccharum officinarum) as affected by high hydrostatic pressure processing. Food Measure 12, 1962–1971 (2018). https://doi.org/10.1007/s11694-018-9811-7
[2] Tiwari, G. & Prakash, Om & Kumar, Subodh. (2004). Evaluation of convective heat and mass transfer for pool boiling of sugarcane juice. Energy Conversion and Management. 45. 171–179. 10.1016/S0196-8904(03)00143-2.
[3]: Canilha, Larissa & Chandel, Anuj & Milessi, Thais & Antunes, Felipe & Freitas, Wagner & Felipe, Maria & da Silva, Silvio. (2012). Bioconversion of Sugarcane Biomass into Ethanol: An Overview about Composition, Pretreatment Methods, Detoxification of Hydrolysates, Enzymatic Saccharification, and Ethanol Fermentation. Journal of biomedicine & biotechnology. 2012. 989572. 10.1155/2012/989572.

A more plausible explanation is that it is not water that is boiling.  The  "organic substances" and maybe the "nitrogenous substances" in the juice could include some more volatile components that may pass selectively into the gas phase at a lower temperature than most of the solution.  Such a vapor would not be the pure organic or nitrogenous substance; it would carry off some water vapor as the vapor pressure of the water is a large fraction of an atmosphere.  The temperature at which such selective vapor begins to form is properly called the bubble point.
Of some relevance is the effect of the sugar itself on the boiling point.  In the absence of other solutes this effect is very small, cfailing to reach $5°C$ boiling point elevation until we have nearly 80% sucrose in the solution (by mass).  Sucrose has a large molar mass smd water has a high heat if vaporization, both of which factors attenuate the boiling point rise at concentrations relevant to sugar cane juice.  This lack of an effect from the main nonvolatile solute would enable small amounts of more volatile components to yield a bubble point lower than the boiling point of pure water.

From Ref. [1]
Reference
1.
Willism Heffner and Hinanshi Jin, "Building a Low Cost, Hands-on Learning Curriculum on Glass Science and Engineering using Candy Glass", MRS Online Proceeding Library Archive (January 2009) 1233,
https://www.researchgate.net/deref/http%3A%2F%2Fdx.doi.org%2F10.1557%2FPROC-1233-PP03-06.


Would cerium(III) in solution be oxidized to cerium(IV) in ESI mass spectroscopy? During ESI (assuming positive ion mode for now), we apply a strong voltage ($1.0{-}\pu{5.0 kV}$) to induce ionization.

Therefore, it is common to see oxidation of metal ions (e.g. $\ce{Cu^{+I} -> Cu^{+II}}$).

I wonder if a $\ce{Ce^{+III}}$ solution would be observed as $\ce{Ce^{+III}}$ or as $\ce{Ce^{+IV}}$ due to in-source oxidation.

I feel like there should be a systematic study on this kind of effect but could not find any references for it. Any hints would be appreciated.

I found one study that examined cerium(III) polyoxometalate salts and concluded that at least under the conditions of their study, $\ce{Ce(III)}$ did not oxidize under electrospray conditions.  The study is Bray et al. 2011 in the International Journal of Mass Spectrometry.

They say:

  The presence of the redox active $\ce{Ce}$ metal ion introduces additional complicating effects as it would be difficult to differentiate such ions as $\ce{HCe(IV)P2Mo22O75^3−}$ from $\ce{H2Ce(III)P2Mo22O75^3−}$. However, successive degradation of $\ce{H_{g}Li_{2−g}Ce(III)P2Mo22O75^3−}$ to eventually form $\ce{HCe(III)PMo7O26−}$ yields an ion that can be directly compared to $\ce{LiCe(III)PMo7O26−}$. In particular, the two ions $\ce{HCe(III)PMo7O26−}$ and $\ce{LiCe(III)PMo7O26−}$, are separated by ∼6 m/z (Fig. 6), supporting the proposition of substitution of H+ by Li+ and, therefore, the identity of $\ce{LiCe(III)PMo7O26−}$. These results also aid in assigning the proton count on $\ce{H2Ce(III)P2Mo22O75^3−}$, $\ce{HCe(III)P2Mo18O62^2−}$, and $\ce{HCe(III)PMo13O44−}$, thereby indicating the trivalent oxidation state of cerium throughout. Although Ce(III) would not necessarily expected to be oxidized to Ce(IV) under the conditions of these experiments, confirming that cerium remains in the Ce(III) oxidation state from solution throughout the electrospray and CID processes is a useful methodology for oxidation state determination of redox active metal-complexes with f-block elements such as Ce(IV) and Pu(III/IV).

So they didn't expect Ce to be oxidized by electrospray in general, although the basis for this expectation wasn't clear.  They also report hard data showing that at least for polyoxometalate salts, no oxidation was apparent under electrospray conditions.

In general, the thousands of volts applied during the electrospray process does not mean that individual ions are exposed to fields of thousands of volts, because electrosprayed solutions are conductive and field strengths in solution are lowered because of Debye shielding.


Is it practical to extract zymase from yeast for ethanol production? Suppose I wanted to make some ethanol fuel. Normally, I would get yeast, mix with water, add sugar and wait. However, there is an enzyme (zymase) in yeast that does this. Looking at the Wikipedia article for zymase, someone managed to extract it using only quartz and diatomaceous earth.

The advantages of this is that it could potentially produce large amounts of concentrated ethanol, whereas yeast could only produce less than 20% due to it dying. So, do people use such a method, and if not, why not?

So, do people use such a method, and if not, why not?

No, people do not use such a method.  There are several reasons:

In fact, conversion of glucose to ethanol requires many enzymes, not just one.  The crude extract originally known as zymase was not characterized until later to consist of so many different molecules.  The molecules that comprise zymase include all the enzymes of glycolysis (e.g. glucose-6-phosphate isomerase, triose phosphate isomerase, phosphofructokinase, triose phosphate isomerase, etc.)
The other component of "zymase" as it was understood historically included dialyzable components.  Later these were identified as coenzymes, and are now understood to consist of molecules such as NAD+, ATP, and coenzyme A.
Live yeasts can synthesize all the enzymes and co-enzymes they need to convert glucose to ethanol.  But yeast extracts such as zymase can't. As they convert sugar to ethanol, eventually cofactors are used up, lost to processes such as dilution, oxidative damage, or enzyme promiscuity.  When a zymase extract loses its activity because of these reasons, it can never get them back.  But live yeasts are always synthesizing more coenzymes (and enzymes) in a programmed response to any sensed shortage.
Even zymase would not go to 100% ethanol concentrations.  Enzymes denature and lose their effectiveness at high alcohol concentrations.  The concentration where this happens is probably higher than the concentration that is lethal for yeast, but I don't think it is close to 100%.


Determining the chemicals in potash and crystallizing them I have made potash by leaching wood ash in water. The potash in solution has a high pH indicating it is alkaline, and reacts with acetic and citric acid. Is there an easy way to determine the different chemicals in solution? The neutralization reaction creates an odorless gas. I believe it is carbon dioxide which would make the following equation work:

$$\ce{K2CO3 + 2 CH3COOH -> H2CO3 + 2 KCH3OO}$$

The carbonic acid ($\ce{H2CO3}$) would then decompose into water and carbon dioxide:

$$\ce{H2CO3 -> H2O + CO2}$$

Which leads me to believe the potash contains potassium carbonate (from the reaction) and potassium hydroxide (from the pH.) Is there any flaw in my reasoning and how can I determine and separate the other components? Potash can contain potassium hydroxide, potassium carbonate, and potassium sulfate, among other things.
Once I determine the chemicals, is there a way I can easily crystallize certain compounds out of solution. Thanks.

I think it would be quite difficult to separate all of the different compounds dissolved in the sample into their solids because they are all in the same phase. I do not think any method exists for this kind of separation.

However, determination is absolutely possible.

Firstly, regarding your deductions...

Although carbonates give carbon dioxide gas, which is colourless and odourless, it is nevertheless important to test that the gas is indeed carbon dioxide by bubbling it into limewater and looking out for white precipitate. If you have not conducted that test, then you cannot claim that the gas produced is carbon dioxide. Otherwise, you can conclude that there is carbonate in the sample.

The high pH is indeed indicative of high hydroxide ion concentration in the sample. That is for certain.

However, you have not proved that the cation is indeed potassium. Perhaps, you want to conduct a flame test or do analysis using atomic absorption spectroscopy (AAS). Procedure for flame test can be found here: https://www.soinc.org/sites/default/files/uploaded_files/flametest.pdf

If you would like to find out what other ions are present in the sample, carry out a simple qualitative analysis using some common lab reagents. Procedure can be found here: http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis

By boiling off the water, you would likely get some ionic solids but I believe that it should contain a lot of impurities.


Charge of polyatomic ions doesn't equal the charge of ions it consists of? I'm taking entry level collage chem. class, and we got to this topic that I don't really understand. I think I understood what an ion is (a atom whose amount of electrons is not equal to amount of protons, so it has positive or negative charge). But then polyatomic ions... Lets say peroxide; consists of two Oxygens, whose ions have charge of -2. Consequently, two of these anions are suppose to have charge of -4, but peroxide has charge of -2. How does this work? If polyatomic ions don't just sum the charge of ions they consist of, how else can I predict and understand their charge?

H-O-O-H.  The bond in the middle is covalent.  Deprotonate to $\ce{{HOO}^{-}}$ (hydroperoxide) then $\ce^{^{-}{OO}^{-}}$, peroxide dianion, $\ce{O2^{2-}}$.  You learn it by doing it, having examples, learning what to look for, and gaining understanding.  Both oxygens are in the -1 oxidation state.

Even good answers here have a legitimate diversity of explanation.  The world is complicated and dirty.  Unlike physicists (quantum gravitation, SUSY, dark matter), we can test our predictions with experiments.  Chemists are not too stinking proud to admit we were wrong, e.g., benzene displacements then palladium catalysis.  That is how we make real world progress, by discovering footnotes.

Introductory inorganic does not have wiggle room.


Does adding slices of lemons to water reduce its chlorine content? Often restaurants add slices of lemon to water. They seem to make the water taste better and perhaps reduce or at least cover up the chlorine taste.

I've heard that Vitamin C tablets can be used to remove chlorine from tap water.

Q: Does anyone know if lemons, being rich in vit C, will encourage the removal of chlorine from tap water?

Yes , a quick and cheap way to reduce chlorine residual is to add ascorbic acid
 (VITAMIN C), which if added in an equal molar amount to free chlorine would completely eliminate chlorine residual within 1 min.

I am quoting from ABSTRACT from a paper from, Int. Journal of Engineering Research and Applications , ISSN : 2248-9622, Vol. 3, Issue 5, Sep-Oct 2013, pp.1647-1651.

"Society’s preference for bottled water mainly comes from a dislike of the mildly pungent taste and odor of tap water. Such odor and taste may be caused by chlorine residual (in the form of free chlorine or chloramines), which is maintained in a water distribution system to prevent regrowth of microorganisms and is thus present in tap water.

"Two methods of reducing chlorine residual, boiling tap water and adding ascorbic acid to
tap water, were tested. The results showed that, for free chlorine, a quick and cheap way to reduce chlorine residual is to add ascorbic acid, which if added in an equal molar amount to free chlorine would complete eliminate chlorine residual within 1 min "

Reduction of Chlorine Residual in Tap Water with Temperature

source : https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf

As shown aboce, the chlorine residual decreased faster at higher temperatures.

Reduction of Free Chlorine and Monochloramine in Tap Water through Boiling

source : https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf

As shown in Figure 2, both residuals of free chlorine and monochloramine decreased with boiling time, and the residual of free chlorine decreased slightly faster than that of monochloramine.

The boiling time required to completely remove free chlorine was about 33 min.$\ce{^1}$

Effect of Ascorbic Acid on Free Chlorine and Monochloramine

Ascorbic acid added in a molar ratio of 1:1 (free chlorine to ascorbic acid) completely eliminated free chlorine. The DPD indicator(A DPD solution can also be used for the measurement of chlorine residual) did not produce a red color when added within 1 min,indicating complete elemination chlorine residual within 1 min.$\ce{^2}$

The reduction of free chlorine by ascorbic acid can be expressed as follows:

$\ce{C6H8O6}$ + $\ce{HOCl}$  →$\ce{C6H6O6}$ +$\ce{HCl}$ + $\ce{H2O}$

Ascorbic acid $\ce{(C6H8O6)}$ is oxidized to dehydroascorbic acid $\ce{(C6H6O6)}$ . Simultaneously, free chlorine $\ce{(HOCl)}$ is reduced to chloride ions (HCl).

In the experiment conducted it was determined that 10 mg of ascorbic acid is required to treat 1 L of tap water. and the quickest way to reduce the concentration of free chlorine in tap water is to add powdered vitamin C (ascorbic acid) tablets in less than 1 min.$\ce{^3}$

Thanks to N4v i will be using this cost effective method to remove chlorine from water.(wow).

Reference

https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf


Why does the pH of a weak acid not increase by 1 when diluted by a factor of 10? Strong acid pH increases by one unit when diluted by a factor of 10, but why do weak acids not?

In a weak acid  $\ce{HB}$ solution, with a nominal concentration $c$, a tiny amount of its molecules are dissociated into $\ce{H^+}$ and $\ce{B^-}$. Let's call this concentration $[\ce{H^+}]$ = $\ce{[B^-]}$ = $x <<c$, so that the following approximation can be made : $c - x$= $c$. The dissociation equilibrium constant $K_a$ of this weak acid $\ce{HB}$ can be approximated by$$K_a\ce{= \frac {[H^+][B^-]}{$c - x$} = \frac{$x$^2}{$c-x$} = \frac{$x$^2}{$c$}}$$ $$\ce{[H^+] = x = \sqrt{K_ac}}$$ $$\ce{$p$H = \frac{1}{2} (log$K$_a - log $c$)}$$ Look at the coefficient $1/2$ before the logarithm. If the concentration $c$ is multiplied by $10$, the log c increases by $1$, but the $p$H changes by $1/2$.


Naming an Ion with respect to its oxidation state When identifying compounds, where certain elements are in various oxidation states we write the oxidation state in Roman numbers after it. Does the same go for ions? What would the correct way of writing Cu$^{2+}$ in words be?

Does the same go for ions? What would the correct way of writing $\ce{Cu^2+}$ in words be?

The most common way of saying $\ce{Cu^{2+}} $ would be "copper(II) ion".  However keep in mind that in aqueous solution, various ligands bind to naked copper(II) ions forming various complexes.  So for example "hexaaquacopper(II)" ions are very common.  In these names the charge is inferred from (a) the oxidation state of the metal and (b) the ligands on the copper.  So copper(II) with four negative charged chloride ligands would be "tetrachlorocopper(II)", although since the latter is an anion, it is often called "tetrachlorocuprate(II)", e.g. on Wikipedia.


How will citric acid be deprotonated by pyridine? Why do we deprotonate that specific hydrogen as opposed to any other one? What information does the table on the right give us? It seems unrelated to citric acid - where does the fluorine come from?

The table on the right looks like a standard group of acids shown to illustrate the effect of inductive withdrawing on carboxylic acid pKa. The first two structures are relevant to the question. The table may have been copied from somewhere else because it had both structures on it. Two of the pKa values in the table are every similar to two of the pKa values given in the question. The fluoroacetic acid is a fine distraction; you are correct in thinking that it is not involved.

  As for figuring out which of citric acid's three carboxylic acids is deprotonated first...

Citric acid has three carboxylic acid groups of two different types. Can you see that? One of the acids is closer to the OH group than the other two. Which acid or acids look more like the first entry in the table (except jammed onto the citric acid structure)? Which acid or acids look more like the second entry on the table? Once you have made that distinction, you should be able to use the pKa data on the table to determine which of the three carboxylic acid groups in citric acid is the most acidic (goes with the 3.13 pKa).


In the addition of HBr to isoprene why does the less stable carbocation form? According to me, the answer is B as the tertiary allylic carbocation should be more stable. But the answer given is D and in the given solution, the tertiary allylic carbocation rearranges to primary allylic carbocation to yield Option D.

My question is why does rearrangement from tertiary allylic carbocation to primary allylic carbocation take place?

Source: JEE Advanced 2015 Paper 1 Q31

The carbocation will be formed as per stability rules but as it is conjugated system the cabocation resonates and as the less stable carbocation is more reactive  it is attacked by Br-


Why is pickle salty? How does salt get into pickle? I know this may look like some question easily answered with common sense, but please read.

If water is the solvent wouldn't the only thing that moves be water? (Osmosis)

Edit: this is basically what I think. If you imagine the skin of whatever you are pickling as a membrane that prohibit the flow of electrolytes but allows that if water. Then salt, NaCl, cannot move anywhere.

Your concept of a cell is a little vague. Although the cell wall is made from a diglyceride bilayer, it is not impregnable. In fact, there are many proteins that pass through the cell wall where dissolved material can pass. One of the proteins is called the sodium potassium pump. In a living cell, the protein uses ATP to move sodium out of the cell and drag potassium into the cell. Both of these ions are against the gradient. In a non-living cell, such as the pickle, sodium ions can move across the cell membrane and make the cells salty.


Find the molarity and molality We take 10mL of $H_2SO_4$ with a $96\%$ of mass and density $1.84g/cm^3$, and we add it into a flask that has $0.5L$ of capacity, filled up to half of the flask with distilled water. We shake the mixture and we add in more distilled water up to the level of $0.5L$. Find the molarity and molality.

$\mathbf{Attempt. }$ My problem comes up when trying to find molality, I've found molarity like this (and this meets with the book's solution):
Let $M$ be molarity, defined as $M=\frac{n_s}{V_{T}}$ where $n_s$ is the number of moles of the solute and $V_T$ the total volume of the mixture. As we know $V_s=0.01L$ and $d_s=1.84g/cm^3=1.84g/mL=1840g/L$. We can find the mass $m_s$: $$m_s=d_s\cdot V_s=18.4g$$ and then find the moles of the solute $n_s=18.4g\cdot\frac{1 mole\ H_2SO_4}{98g\ H_2SO_4}\approx 0.188 moles \ H_2SO_4$ and then the $V_T=0.5L$ so $$M=\frac{0.188moles}{0.5}\approx 0.38M$$ and this part agrees with my book now, the issue is with molality.
$\mathbf{Doubt/The\ part\ where\ I'm\ stuck.} $ With molality, which I'll denote $m$, defined as $m=\frac{n_s}{m_{solvent}(kg)}$, where $m_{solvent}$ is the mass of the solvent in kg. Okay well, I attempted this and it gave me $245.2m$ while the book's answer is $0.37m$. My attempt:
We know $n_s$, and $m_{solvent}=m_{mixture}-m_{solute}$, and from the percentage given of the solute $96\%$, we also know that $0.96m_{mixture}=m_{solute}\implies m_{mixture}=\frac{m_{solute}}{0.96}$, and we finally plug that in the previous equation: $m_{solvent}=\frac{m_{solute}}{0.96}-m_{solute}$, plugging $m_s=18.4g$ we get $m_{solvent}=0.767g$. Converting to kg, we get $m_{solvent}=0.000767kg$. And finally plugging into the molality: $$m=\frac{0.188moles}{0.000767kg}\approx 245.11m$$ where I have gone wrong so that it does not meet with my book's solution?

All in all this isn't a well stated problem. The molarity can be solved, but the molality can't really be solved.
The wording "with a 96% of mass" is odd. I'm not sure if this is your interpretation of the problem or if that is what is literately written in the book. I'll interpret the phrase to mean that the sulfuric acid was 96% sulfuric acid by weight.
For both problems the grams of sulfuric acid are:
$$\mathrm{grams\ }\ce{H2SO4} = 10 \times 1.84 \times 0.96 = 17.66$$
$$\mathrm{moles\ }\ce{H2SO4} = \dfrac{10 \times 1.84 \times 0.96}{98.079} = 0.1801 $$
Molarity
So the molarity, $M$, of the solution will be:
$$\mathrm{Molarity} = \dfrac{10 \times 1.84 \times 0.96}{98.079 \times 0.50} = 0.360 $$
Molality(1)
The problem with the molality calculation is that volumes are not additive. I'll admit that for two or three significant figures the calculation sort of works.
So for the 500 ml of solution, you must assume that 10 ml is sulfuric acid, the other 490 ml of volume is water. Water is assumed to be 1.00 kg/liter, so to figure out how many moles of sulfuric acid per kilogram of solvent we'll need to multiply the 0.360 moles per liter of solution by a correction factor of 1/0.980 which gives a value of 0.368 for the molality.
Molality(2)
As a second way to calculate the molality let's divide moles of $\ce{H2SO4}$ by the kilograms of water.
From above we know that the 500 ml of solution contained 0.1801 moles of $\ce{H2SO4}$.
Again we have to assume that the volumes are additive so that 10 ml of concentrated sulfuric acid and 490 ml of water yield 500 ml of solution. Given no temperature we can just assume that water is 1.00 g/ml, so 490 ml of water has a mass of 490 grams or 0.490 kilograms.
$$\mathrm{Molality} = \dfrac{0.1801}{0.490} = 0.368 $$

To perhaps beat a dead horse here let's back calculate the volumes for one liter of a similar solution. Steffen's Chemistry Pages give that a 4% solution of sulfuric acid by weight has a density of 1.0250 g/ml, 41.00 g sulfuric acid per liter and a molarity of 0.418 at 20°C.
To get 41.00 grams of sulfuric acid we'd need
$$\mathrm{ml\ conc\ sulfuric\ acid} = \dfrac{41.00}{1.0250\times 0.96} = \pu{41.7 ml}$$
But since water is 1 g/ml and the solution is 96% water
$$\mathrm{ml water} = 1000 \times 0.96 \times 1.0250 = \pu{984 ml}$$
$$41.7 + 984 = 1025.7$$


Why is methyl group more electron-donating than tert-butyl group? As title says, why is methyl group more electron-donating than tert-butyl group? The context behind this is stabilization of conjugate base. (http://www.khanacademy.org/science/organic-chemistry/organic-structures/acid-base-review/v/stabilization-of-conjugate-base-iii look at 10:51)

Methyl group is not more electron donating than tert-butyl group, as the video provided in the link suggests (11:20 of the video), which is true.

Inductive effect is a permanent effect which results in displacement of electrons along the chain of atoms in the molecule (here, carbon atoms).  The reason for this displacement can be worked out from the electronegativity difference between the atoms.

Consider the tert-butyl group.  In a C-H bond, the electron density is directed towards carbon.  Methyl group, which has three hydrogen atoms attached to a carbon, pushes the electron density towards $\ce{C2}$ of tert-butyl acetic acid.  So do the two other methyl groups.  The amount of partial negative charge accumulated on $\ce{C2}$ of tert-butyl group (has 3 methyl groups) is more than that on methyl group (only has 3 hydrogens) of acetic acid.

The more negative charge on $\ce{C2}$ in tert-butyl acetic acid destabilizes the conjugate base, compared to a simple acetate ion.  Hence acetic acid has less pKa than tert-butyl acetic acid.


The meaning of permanent dipole moment I was asked in one question that which of all given molecules have a permanent dipole moment?

And $\ce{XeF6}$ was given as a correct answer quoting that it is a distorted octahedral, and hence the lone pair results in the molecule having a dipole moment. But since we know that the lone pair goes at each face of the octahedral one by one, the net effect should cancel out. Hence, is it correct to say that the molecule has a PERMANENT dipole?

I am having similar doubt in the case of the very fundamental ammonia example. Since it undergoes inversion do we say that it has a permanent dipole?

Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me.

Indeed, $\ce{XeF6}$ has a distorted tetrahedral structure in all three phases. Two of solid state crystal structure modifications ($\bf{A}$ and $\bf{B}$) are shown in following diagram (Ref.1):

The abstract of Ref.1 tells them all:

  According to single crystal X-ray diffraction, neutron powder diffraction, solid state MAS NMR data, and differential scanning calorimetry, $\ce{XeF6}$ exists in at least six different modifications. Three of them are formed at temperatures above room temperature, one exists at room temperature, while two have been found at low temperatures. In the high temperature modifications $\ce{XeF6}$ forms a non-symmetric tetramer, better described as a cyclic trimer with a weakly associated monomer. The normal temperature modification is the previously described cubic phase IV, having disordered tetrameric and hexameric units. The low temperature modifications are regular tetramers. Only in presence of $\ce{HF}$ symmetric dimers are formed.

The tetramer $\bf{C}$ (ball and stick version of $\bf{A}$) represent the general version of all modifications, a cyclic trimer with a weakly associated monomer. The symmetric dimer $\bf{D}$ is the structure of $\ce{XeF6}$ in the presence of $\ce{HF}$ (highlighted by $\color{green}{\text{green}}$ ovals). Nonetheless, all of these modifications display distorted octahedral structure for individual $\ce{XeF6}$ molecule as represented by ball & stick 3D structure in the central box. None of $\ce{F-Xe-F}$ angles are equal to $90^\circ$, which is the normal angle of central atom with two adjacent ligands in regular octahedron. As a consequence, $\ce{XeF6}$ has net permanent dipole moment since all six $\ce{Xe^\delta+-F^\delta-}$ bonds are polar.

For your question about ammonia, look at Ivan Neretin's comment and here where there is an accepted answer.

References:

Sevim Hoyer, Thomas Emmler, Konrad Seppelt, “The structure of xenon hexafluoride in the solid state,” Journal of Fluorine Chemistry 2006, 127(10), 1415-1422 (https://doi.org/10.1016/j.jfluchem.2006.04.014).