text:
Base-promoted rearrangement of β-ketoester

Propose a reasonable mechanism for the following transformation:

I guess an enolate probably forms first, but then what does it attack to give something like the product in the question?

Retro-Claisen then a forward Claisen, sorta similar to this but Claisen instead of aldol.

This problem has probably been on everyone's mechanistic chemistry exam at some time or another. Does the alkyl (allyl) just "jump" from one side of the carbonyl group to the other? In most examples, the allyl group is a saturated alkyl group. Under such circumstances I am in full agreement with @orthocresol as to the mechanism of the process (1 --> 9). The reaction is not catalytic in methoxide and requires sufficiently concentrated alkali to assure sufficient ester enolate, lest the reaction stops at the diester 4. Notice that the red hydrogens in β-ketoester 1 retain their position in the product 9.

While @Carlson Miller suggested the formation of an enolate, his proposal brings up an interesting possibility in this particular example. The ensuing mechanisms are improbable in protic media but imagine running the reaction with a kinetic base ((NaH, (TMS)2NNa, LDA) in an aprotic solvent is intriguing. There are two possible mechanisms for the reaction. First, enolate 10 can fragment to a ketene-ester enolate that must lose methoxide and generate diketene 11. At this point the reaction is stoichiometric in diketene and methoxide. These two species lead to enolate 12. Bimolecular hydrogen exchange affords the more stable enolate 8. Although I am not particularly enamored with this option, it is nonetheless possible in spite of the limited amount of base relative to the protic reaction. Incidently, any product of intramolecular ketene dimerization would be susceptible to opening by methoxide. This mechanistic pathway, as with the protic reaction, retains the red hydrogens in their original location.

However, the more likely pathway for enolate 10 is via an oxy-Cope-like rearrangement. Although enolate formation is the driving force for the oxy-Cope-like rearrangement (13 --> 14), the rearrangement of enolate 10 gives the more stable enolate of the β-ketoester 8. In this case, the red vinylic hydrogens of ketoester 1 are now located on the methylene group of the allyl residue. Indeed, the allyl group does "jump" in the oxy-Cope.
Enolate 16 is generally formed by alkylation of the dianion of an acetoacetic ester with an allyl halide. Is anyone in SE-land aware of the transformation 15 --> 16?

The proper way to differentiate would be check with a labeling experiment.

Agreed, obviously!

https://www.sciencedirect.com/science/article/pii/S0040403900733429 has some examples of 15 -> 16, albeit with a different mechanism proposed. & a Cope rearrangement of the silyl enol ether: http://www.nrcresearchpress.com/doi/10.1139/v87-115


text:
Experimental bond lengths and binding energies of diatomic molecules
Is it possible to find the experimental values of the bond length and the binding energy of the diatomic molecules $\text H_2$, $\text{Be}_2$, and $\text{Ar}_2$ in one source?
I found a lot of chemistry handbooks and databases, but non of the provided both, the bond length and the binding energy for all three molecules.
I guess the experienced chemists know some good sources?

IMHO, Be2 and Ar2 hardly deserve to be called molecules.

Yes, could be right, because the binding energy of Ar2 is about -12 meV and of Be2 about -100 meV. Meybe this is the reason why I can not find data...

Might not have all the information you are looking for, but if it has an experimental (or computational) bond length, it will probably be here: https://cccbdb.nist.gov/diatomicexpbondx.asp. Diargon is no dice, but they actually do have a value for diberylium.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Reactivity of different isomers of N2F2
$\ce{ N2F2}$ exists in two forms cis and trans. Today my chemistry teacher tells me that among the two only cis form takes part in reactions and that it is more reactive. But I have read elsewhere that trans is less stable. As far as I understand less stable compounds are usually more reactive. So why this discrepancy?

According to Dinitrogen Difluoride Chemistry Inorg. Chem., 2010, 49 (15), pp 6823–6833:

This increased reactivity of the cis-isomer can be readily
  explained. In both $\ce{N2F2}$ isomers, each nitrogen atom
  possesses a sterically active free valence electron pair.
  The transformation of $\ce{N2F2}$ to $\ce{N2F+}$ requires the conversion
  of the $\ce{N=N}$ double bond into a $\ce{N#N}$ triple bond
  by an overlap of electrons from these two free valence electron pairs. Obviously, this overlap is greatly facilitated
  in cis-$\ce{N2F2}$, where these free pairs are on the same
  side of the molecule.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
UV–Vis spectra of hexaaquacobalt(II) ion
I observe the UV–Vis spectrum of $\ce{[Co(H2O)6]^{2+}}$ and there is a little peak at $\sim\pu{640 nm}$. By observing the Tanabe–Sugano diagram I see that the spin allowed transitions are:
$$
\begin{align}
\mathrm{^4T_{1g}(F)} &\to \mathrm{^4T_{1g}(P)} \\
\mathrm{^4T_{1g}(F)} &\to \mathrm{^4A_{2g}} \\
\mathrm{^4T_{1g}(F)} &\to \mathrm{^4T_{2g}}
\end{align}
$$
The answer about why there is this little peak is that it is explained by the $\mathrm{^4T_{1g}(F)} \to \mathrm{^4A_{2g}}$ transition and it has to do with a two-electron transition. I can't understand why it is explained by this transition. Also, how do I know that it has to do with two electrons?

The transition seems to go from $t_{2g}^3e_g^4 \to  t_{2g}^5e_g^2$ thus the photon shifts two electrons but the overall transition is still spin allowed. You might expect that such a transition would be very weak as the matrix element probably involves some product of terms involving a  transient configuration.

@chemistrylove all of these allowed transitions could only be two photon or vibronically allowed since for one photon, g<->g is not allowed

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Precipitation of CaCl2 and FeCl3 in Na2CO3
Why does $\ce{CaCl2}$ gives precipitation of $\ce{CaCO3}$ when $\ce{Na2CO3}$ is added, but $\ce{FeCl3}$ does not give the precipitation of $\ce{Fe2(CO3)3}$ in $\ce{Na_2CO_3}$  ?

If it gives a precipitate then it isn't/is slightly soluble.

@KianStevens ... I wanted to know the reason ... If I'm correct or not ...

What about the down vote ?? Is it not related .?

Obviously, $\ce{FeCl3}$ does not give the precipitate of  $\ce{CaCO3}$, because, er... why would it? There is no  $\ce{Ca}$ in it, to begin with.

@Ivan ...  ... By precipitate , I meant $Fe_2CO_3$ ...

Man ... What's the reason for down vote ? Is it not related .?

Well, then here's another correct (if probably useless) answer for you: $\ce{FeCl3}$ contains  $\ce{Fe^3+}$, and $\ce{FeCO3}$ contains  $\ce{Fe^2+}$, and the former would not just up and transform into the latter (unless this is a redox, which it isn't).

@Ivan ... I've edited my question .. I hope you would go through it ...

Now we've come to the heart of the matter. See, $\ce{Fe2(CO3)3}$ (yes, you've got the formula right this time around) is unstable due to hydrolysis. For all practical purposes, it doesn't exist.

Yes, iron carbonate does not exist. See this post: https://chemistry.stackexchange.com/questions/64331/is-iron-iii-carbonate-stable-in-water

@Ivan ... Is it something like $Fe_2(CO)_3$ is unstable due to decomposition for polarization ...?

I don't know what do you mean by "decomposition for polarization". Other than that, yes, it is unstable.

@NilayGhosh You've just sent the OP to an unanswered post with pretty much the same question. I don't think this is going to help.

@Ivan Guess that would be a duplicate...

@NilayGhosh You need to have either an accepted or upvoted answer in the previously asked question to close this as a duplicate,

@Pritt I am aware of that fact. I was just joking.

This must have been a very good question for a geochemist. They'd give you a complete answer with details of why $\ce{Fe2(CO3)3}$ does not but $\ce{Fe2O3}$ exist in our surrounding. Yet, I would like to give only a simple answer for the question: No, $\ce{Fe2(CO3)3}$ does not form (in sort) when you added $\ce{Na2(CO3)3}$ to an aqueous solution of $\ce{FeCl3}$. The reason?
An aqueous solution of $\ce{FeCl3}$ is acidic because $\ce{FeCl3}$ is a salt of a strong acid $(\ce{HCl})$ and a weak base $(\ce{Fe(OH)3})$. Thus an aqueous solution of $\ce{FeCl3}$ is in following equilibrium:
$$\ce {Fe^{3+} + 2H2O <=> [FeOH]^{2+} + H3O+ }$$
Thus, $\ce{H3O+}$ ions in solution can react with added $\ce{CO3^{2-}}$ ions to form carbonic acid $(\ce{H2(CO3)3})$,which can decomposed to form $\ce{CO2}$  and $\ce{H2O}$. Remaining $\ce{Fe^{3+}}$ will combine with left behind $\ce{OH-}$ ions to make $\ce{Fe(OH)3}$, which would decomposed to $\ce{Fe2O3}$. Thus, when $\ce{Na2CO3}$ is added to an aqueous solution of $\ce{FeCl3}$, one can expect the following sequence of reactions to be taken place:
$$\ce {FeCl3 (aq) + Na2CO3 (aq) -> [Fe2(CO3)]^{3+} (aq) -> Fe2O3 (s) + CO2 (g) + H2O (l)}$$

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Catalytic poison in Pt catalysed manufacture of H2SO4
I came across a question that goes as:

In manufacturing of sulfuric acid in presence of platinum catalyst, which metal impurity acts as a catalytic poison?

Fe
Cr
Cu
V

I searched everywhere I could but did not find anything that talks about this.
I read about $\ce{As2O3}$ being a catalytic poison in this reaction and recommendations for using the Contact process using $\ce{V2O5}$ instead of this reaction.
I can understand that this is more of a fact based question but reasons or explanation (if any) should please be included.

$\ce{Fe}$ and $\ce{V}$ are used in the contact process.
$\ce{Cu}$ is used in the preparation of chlorine gas.
Thus $\ce{Cr}$ can be used as a catalyst poison.

The question specifically asks for more explanation and/or reasons. It is best to avoid writing one line answers in such cases. Please comply by expanding your answer. Kindly also consider good formatting in future posts, it makes them more readable! -From Review


text:
Ideal gas Law best describes the properties of which of the following gases at 0 degrees Celsius and 1 atm?
a. $\ce{PH3}$
b. $\ce{HBr}$
c. $\ce{SO2}$
d. $\ce{N2}$
I believe that it is $\ce{N2}$, but I'm not too sure, since technically all of them would act the same at STP.

They would act the same if they all were ideal, which they aren't.

One of the defining characteristics of ideal gases is that ideal gas particles do not interact with each other, and there are no intermolecular forces between them. If a certain type of gas molecule has a molecular bipole, then each molecule will have ends with opposite partial charges. In a collection of gas particles, if the particles have a molecular dipole, these particles will attract and repel each other and thus will experience intermolecular forces. Because the two nitrogens in $\ce{N2}$ are equally electronegative, the bond and each $\ce{N2}$ gas particle will have minimal partial charges and thus minimal intermolecular forces. Thus, $\ce{N2}$ behaves very ideally.

Ideal Gas Law disregards intermolecular interaction. So polar molecules will most likely differ from ideal gases to a higher degree than nonpolar molecules


text:
Hazards of trivalent chromium and hexavalent chromium
I am examining a material safety data sheet that reports on trivalent and hexavalent chromium in chemical films. It covers each type where it is used in a product individually.
Is there a practical difference in safety between the two, such that they should be reported separately?

Sure, and a great one. Chromium (III) is just another light metal cation, almost as harmless as iron (II, III). Chromium (VI) is acutely toxic.

Hexavalent chromium is a genotoxic carcinogen. Due to the similarity of the chromate(VI) and sulfate ions, it is transported around the body via the sulfate channels. Due to this it is a major health hazard. Trivalent chromium on the other hand is not generally regarded as a health hazard due to its poor solubility in water. With this information you should report both separately.

Thank you for providing these details. I'm now curious why trivalent chromium would even be mentioned in a hazardous materials report.

Possible irritant effect maybe?

Since you're referencing an SDS, I'm assuming your talking about the occupational environment, in which case it makes sense to talk about occupational exposure limits. The American Conference of Governmental Industrial Hygienists (ACGIH) has established non-enforceable Threshold Limit Values (TLVs) for different oxidation states of Chromium:

Chromium VI - 0.0002 mg/m3
Chromium III - 0.003 mg/m3
Chromium metal - 0.5 mg/m3

These values refer to the maximum recommended 8-hour time weighted average exposure concentration for a given substance. A lower number means someone is allowed less exposure; a higher number means someone is allowed a greater amount of exposure.
Depending on your product and process, it may make sense to look at measuring exposure to each of the different oxidation states. Particularly for chromium VI, because the exposure limit is so low, it's doesn't take much to overexpose someone.
Note also that if you're in the US, OSHA has a limit for Chromium VI of 0.005 mg/m3. If you exceed this limit, there are a ton of requirements that kick in. See 29 CFR 1910.1026 for more information.


text:
Does salinity affect seawater's pH?
If so, how? It seems like excessive salinity within seawater has the same destructive effect on marine organisms just like excessive acidity. But in real terms, is there any relation between salinity and the acidity of seawater?
P.S. I don't take chemistry. This is for my biology depth study. Please try to respond in a way that is eay to understand. :)

Short answer: no.

Well, sea salt (NaCl) isn't acidic nor basic. So yeah... no.

Drinking cyanide has the same destructive effect on human life as being run over by a truck, but it doesn't follow that all trucks contain cyanide or vice versa.

I am going to say the opposite. Short answer: yes.

Yes, salinity affects the pH of seawater.
See RECOMMENDATONS FOR THE
DETERMINATION OF pH IN SEA WATER
AND ESTUARINE WATERS Pure & Appl. Chem., Vol. 60, No. 6, pp. 865-870, 1988.
An approximation of pH as a function of temperature and salinity is given:
pH = (2559.7 + 4.5 S)/T - 0.5523 - 0.01391 S
Where "T" is temperature in kelvins and "S" is salinity in units of parts per thousand.
As seen above, in the temperature range of Earth's oceans, increasing salinity increases pH.
See also pH of seawater Marine Chemistry Volume 126, September 2011, pages 89-96.
Fig. 2 plots pH versus salinity at 25 degrees C and 0.000334 atmospheres of CO2 and shows pH increasing significantly with salinity.

Is it actually more acidic? A different pH doesn't mean that the water is more acidic, it just means the [H+] is different, like how pH decreases as temperature increase even though water doesn't get any more acidic.

Excess salinity does have detrimental effect on marine organisms, but that is for an entirely different reason; it has nothing to do with the pH of the saline water. When the water is highly saline, then water from the cell cytoplasm tends to move out of the body by osmosis. Thus the cells are plasmolysed. Moreover, ions like $\ce{Na+}$, $\ce{Mg^{2+}}$, $\ce{Cl-}$ etc. which are present abundantly in the saline water, tends to diffuse into the body. For removing these ions, the kidney starts to produce more and more urine. The end result is that the organism cannot survive in saline water.
When the level of salinity is not high enough (like in oceans), then the aquatic plants and animals can use various physiological modifications or adaptations to stay alive in the salty water. In some fishes, the ions are pumped out of the cell cytoplasm and blood by active transport (ATP operated proteins). Many fishes also drink sea water to counteract the loss of water by osmosis. Some other fishes may also have dissolved compounds in blood to increase its concentration and prevent osmosis.
When the salinity is really high, then neither of these countermeasures become effective enough. So, marine organisms cannot survive. (Think about dead sea)
But to answer your original question, increasing the salinity does affect the pH of the water (see below), but the osmosis is more likely cause of death than change of pH.
The main ions in the sea water are $\ce{Na+}$, $\ce{Cl-}$, $\ce{SO4^{2-}}$, $\ce{Mg^2+}$, $\ce{Ca^2+}$ and $\ce{K+}$ [1]. As you can see, the acids and bases which produce these ions in solutions are:  $\ce{NaOH}$, $\ce{HCl}$, $\ce{H2SO4}$, $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{KOH}$. Out of these, only $\ce{Mg(OH)2}$ is not a strong base. So, change in the concentration of $\ce{Mg^2+}$ will make the water more acidic[2]. Also, sea water contains an appreciable amount of $\ce{CO3^2-}$. Increase in concentration of this ion will make the water more alkaline. So, it all depends upon the specific salt which is the cause of excess salinity at the region.

Notes:

Britannica article
The conjugate acid of a strong base is a weak acid. The conjugate acid of a weak base is a strong acid. So, the other ions do not affect the pH. (Read the Wikipedia page and the Chemistry LibreTexts page for further information)


text:
Do Biuret test always means proteins presence in a substance?
I am a student and I got this answer I cant really say for sure if I understood correctly :
"A substance reacts positively to Biuret's test. ¿ Can you say with 100% accuracy that such substance is a protein? Justify"
Is this correct to say that the substance is a protein?
Is there a chance that the test reacts but the substance is NOT a protein?
Additionally, I am being asked why we are using whole milk instead of skim milk in this test.

Biuret test react with a peptide bond, so nylon can react with.( https://en.wikipedia.org/wiki/Biuret_test, https://twitter.com/jane_essex/status/921359485939134465)


text:
What is the basic concept of positive and negative charge of an atom?
In ionic compounds, e.g. NaCl, chlorine 'steals' one electron from sodium. Because of this, sodium's proton number is increased by one, and the number of electrons in chlorine is increased by one. This makes sodium positively charged and chlorine negatively charged; opposite charges makes them attract to each other.
My question is that the nucleus of the sodium ion is still surrounded by electrons, and so does the nucleus of chlorine. Do those electrons repel, even though we say opposite charges attract? What do the electrons do while they attract?

Please don't edit your question to ask more questions. Also, 'core electrons' is an ambiguous term. In a sodium atom, there's only 1 valence electron. But in a sodium ion, there's 8; those 8 electrons which are now valence electrons were the 'core electrons' in the sodium atom. Refer to them as just 'electrons' for this sake to avoid confusion.

Basically you are asking if someone would be so nice to give you a comprehensive introduction to quantum mechanics for chemical compounds. While one should encourage people asking questions like you do, it's a bit much to ask. If you are really interested, there are entire libraries written on that topic. I'm sure your teacher (or whatever) will give you literature if you ask him/her nicely

Sodium's proton number doesn't change else it'd become a different element.
In basic terms, sodium donates an electron to chlorine's outer electron shell; there is no electron 'stealing'. As much as both ions' nuclei are surrounded by electrons, the charges of the ions are based on the ratio of protons to electrons. For example, in an atom, the number of protons and electrons are equal to give no overall charge.
However, in an ion it's different; the sodium ion has 11 protons but only 10 electrons, so has a relative charge of 1+, as there's one more positive charge (proton) than there are negative charges (electrons). On the flip side, the chloride ion has 17 protons and 18 electrons, so has a relative charge of 1-, as there's one more negative charge than there are positive charges.
This then causes the electrostatic attraction between the two ions, which we call ionic bonding.

IMHO, the transfer of electron can be described as stealing all right.

@Ivan Neretin Meh. I don't really like to imagine atoms 'stealing' electrons from other atoms... Whatever works for us.

The first ionization energy (the minimum energy required to eject an electron out of a neutral atom in its ground state) of sodium is about 496 kJ/mol; therefore, sodium is not donating its electron voluntarily. However, it’s also not chlorine that is stealing the electron since the first electron affinity (the energy released when an additional electron is attached to a neutral atom) of chlorine is only 349 kJ/mol and thus not enough to steal the electron from sodium. It’s actually the large lattice energy of NaCl that provides the energy for its formation.

@Loong I was going to talk about ionisation energies and electron affinities but I decided to keep it simple.

The overall, or so to speak average charge of an ion is what attracts an ion of opposite charge.
The electrons within an atom do repel each other alright. That is why the electron hull is not a uniform structure but consists of so called orbitals allowing the electrons to distribute themselves with maximum distance to each other.
Besides the electric charge the electron has another property called spin. Electrons with different spin can occupy the same spot in spacetime. Electrons with same spin repel each other. This is why electrons inhabit oribtals pairwise.
In your example $\ce{NaCl}$ the orbitals just align so perfectly that the attracting Sodium and Chloride ions fit into a very dense grid. But not denser, because, as you said, the electron hulls repel each other.

I want to add that what I explained above is just a model of the reality. A model that happens to describe the phaenomena at hand well enough and to be academically accepted.
My chemistry teacher in school used to hold up a matchbox car and to say
"This is not a car. It's a model of one".


text:
Mechanism for the formation of an alkyl halide from alcohol and phosphorus trihalide
I was reading about synthesis of alkyl halides from $\ce{PX3}$. I want to know the mechanism for this reaction. According to Solomons and Fryhle, no rearrangement takes place in case of $\ce{PBr3}$ , which leads me to think that the mechanism is SN2. But what about phosphorus bromides and iodides?
A Google search confirmed that the mechanism for reaction with phosphorus tribromide is indeed SN2, but I could not find anything on trichlorides and iodides. I need help for those.
Thanks.

Why do you think they may be different?

HBr undergoes AMK addition to an alkene in the presence of peroxide whereas other hydrogen halides do not. I think that anything is possible in chemistry and I do not want to make a mistake if a tricky question based on this appears in my exam.

The HBr is quite different from the phosphorus halides as there is no protonation step on the alcohol

https://chemistry.stackexchange.com/questions/69063/if-socl2-reacts-with-alcohols-via-sni-why-doesnt-pocl3

Thanks for the links @Mithoron ! The reaction with $PCl_3$ looks like S_N^2 or S_N^i to me but I can't quite tell which. Which one is it?

I would assume that since the other answer suggests PCl3 gives the inverted product, it occurs via SN2.

see https://www.masterorganicchemistry.com/2015/03/20/pbr3-and-socl2/ (see step 4)


text:
Calculating base dissociation constant given the molarity and pH

At $\pu{25°C}$, a $\pu{0.100 M}$ aqueous solution of $\ce{C_6H_5NH_2}$ has $\mathrm{pH}=8.80$. Find the $K_{\mathrm{b}}$.

I figured I would have to use an ICE table, but I am not too sure how to set it up.
I can find the equation, which is $$\ce{C6H5NH2 -> C6H5NH3^+ +OH-}$$ but I don't think it helps. I only have an elementary knowledge of acid/bases, so I am hoping for a simple answer.

I've never used $K_{b}$ before, but it sounds like $K_{a}$. So what you could do is approximate the numerator in the  $K_{b}$ equation to $[OH^{-}]^{2}$. Then find $[OH^{-}]$ by calculating $[H^{+}]$ and using $K_{w}$. Finally, use the data you've calculated (and been given) to calculate $K_{b}$.

@KianStevens $K_b$ is $(1.0\cdot10^{-14})/K_a$, and I'll try that thanks

Your given equation is incomplete. It needs a water molecule in left hand side.

have a read of this answer and the link. https://chemistry.stackexchange.com/questions/95681/calculating-ph-is-my-answer-correct/95740#95740 It explains how to do these type of questions.

$\ce{C6H5NH2}$ (aniline) is a weak base, and therefore, in its aqueous solution, it takes a proton from water and forms $\ce{C6H5NH3^+}$ and $\ce{OH-}$, which are correct as you have written.
Suppose the extent of protonation is $\alpha$. After taking up the proton, $[\ce{C6H5NH2}] = c (1 - \alpha)$ and $[\ce{C6H5NH3^+}]= c \alpha = [\ce{OH-}]$.
So, we can write the following, from the above mentioned reaction:
$$
K_{\mathrm{b}} = \frac{[\ce{C6H5NH3^+}][\ce{OH-}]}{[\ce{C6H5NH2}]} = \frac{c^{2}\alpha^{2}}{c(1 - \alpha)} \approx c \alpha^{2}
$$
As the degree of protonation is very less than 1 ($\alpha << 1$), we get $\alpha = \sqrt{K_{\mathrm{b}} / c}$. Now,
\begin{align*}
\mathrm{p}\ce{OH} &= -\log[\ce{OH-}] \\
& = -\log(c\alpha)=-\log(c\sqrt{K_{\mathrm{b}}/ c})=-\log(\sqrt{cK_{\mathrm{b}}}) \\
&= \frac{\mathrm{p}K_{\mathrm{b}}}{2} - \frac{\log c}{2}
\end{align*}
(as $-\log K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{b}}$).
In the question, $\mathrm{pH} = 8.8$, so, $\mathrm{pOH} = 14 - 8.8 = 5.2$, and $c = 0.1$. Putting these values in the equation we will get
$$
\mathrm{p}K_{\mathrm{b}} = 2.(5.2 + \frac{\log(0.1)}{2}) = 9.4 \implies K_{\mathrm{b}} = 10^{-9.4} = 3.98 \times 10^{-10}
$$
Thus we get the $K_{\mathrm{b}}$ of aniline as $3.98 \times 10^{-10}$.
Note that, if you have any aqueous solution consisting of only weak base, you can apply the direct formula as $\mathrm{pOH} = \frac{\mathrm{p}K_{\mathrm{b}}-\log c}{2}$, and easily get the $K_{\mathrm{b}}$.

Thanks! I'm not a huge fan of memorizing formulas but I'm sure I can derive it pretty quickly now


text:
Is carbon suboxide planar?
I was asked to comment on the planarity of the carbon suboxide molecule. I thought that the Lewis structure of the molecule looks somewhat like an Allene and must have similar bonding. The two terminal pi bonds must be out of plane and hence the molecule would be non planar. But the answer given in my book states that it is planar.
I couldn’t find a satisfactory description of its structure anywhere. The molecule would be linear, but that doesn’t imply that the molecule should be planar.

When we talk about a molecule being linear or planar, we only consider the positions of the nuclei and ignore the electron orbitals. Sure, π-bonds might be out of plane, but that does not count. If it would be otherwise, we would have to consider that any atom except hydrogen has p-orbitals, and some of those are out of plane, and so no molecule ever would have the right to be called planar.
So yes, $\ce{C3O2}$ is kinda like an allene, with all its carbons having sp-hybrid orbitals, which inevitably makes the molecule linear and hence planar.
One might point out that the molecule is relatively easy to bend, but that's another story.

What if it polymerizes?

What of it? A polymer is a different thing with different structure and different everything.

One would have to imagine quite some bending for the molecule to be non-planar.

Lay it obliquely onto a cylinder, maybe?

I wonder how that translates to the vibrational quantum states.


text:
Confused about enolate chemistry
I'm not so sure how this mechanism is supposed to go. Could anyone provide some pointers?

The answer is supposed to be:

In particular, I'm confused about how to break up the two esters.

Google “malonic ester synthesis”, there might be a few helpful pages.

What do you know about the stability of 1,3-dicarboxylic acids?

Possible duplicate of https://chemistry.stackexchange.com/questions/19410/decarboxylation-under-acid-water-and-extreme-heat


text:
Which would be more acidic, benzene or cyclohexane?
If benzene were to donate a hydrogen ion, then could the negative charge still be delocalised around the ring? Wouldn't that break aromaticity and thus destabilise the molecule?

See: Relative acidities of alkanes, alkenes, and alkynes. Essentially the same explanation holds here, as benzene carbons are sp2-hybridised.
The newly formed lone pair is in a σ-type orbital (blue) which does not overlap with the π system (red), hence no delocalisation is possible.


text:
reimer tieman reaction only one ortho product
why we cannot have two reimer tiemann substitutions on the same phenol
 as CHO- group is electron withdrawing it wont disturb electron concentration at its meta which is phenol's ortho- product

But it deactivates the system overall

If you look at the electron flow, you'd see the electron density on  $\ce{O}$ in $\ce{Ar-OH}$ is not completely available after the first substitution.

why am i getting down votes? you guys find it difficult to help

There are some literature evidence that shows to have two Reimer–Tiemann substitutions on same phenol ring could be possible. For example, salicilic acid is a Reimer–Tiemann product from phenol and carbon tetrachloride. Yet, it had undergone another Reimer–Tiemann reaction to give  5-formyl-2-hydroxybenzoic acid with $17\%$ yield (para-substitution) but no 3-formyl-2-hydroxybenzoic acid (ortho-substitution) when chloroform was replaced by trichloroacetic acid (J. Chem. Soc., 1933, 496-500).
For more examples: Read, Chem. Rev., 1960, 60(2), 169–184.
J. Chem. Soc., 1933, 496-500.


text:
Reaction of alkyl bromide in presence of piperidine

What are the products of this reaction? I believe there'll be E2 elimination to give the products. Not sure. Help would be greatly appreciated.

The above is the product which I thought might be possible due to E2 elimination
In case there is carbocation formation, this might be the intermediate:

Could there be nucleophilic substitution?

How likely is it that the secondary cation will rearrange to a more stable species?

@Waylander I thought so, but given that reaction type isn't mentioned I wasn't sure about it. Also, Piperidine being a strong base might abstract Hydrogen for E2 Elimination

Could the cyclopropane ring behave like a double bond, making this the analogue of an allyl bromide, and yielding R2NCH2CH2CH=CH-tBu by attack on a cyclopropyl CH2 and ring opening? Especially given the steric hindrance near the carbon bearing the leaving group.

@user6376297 Won't that require a Carbocation formation? But a protic polar medium hasn't been mentioned

@User6376297 raised a question:

"Could the cyclopropane ring behave like a double bond, making this the analogue of an allyl bromide, and yielding $\ce{R2NCH2CH2CH=CH-tBu}$ by attack on a cyclopropyl $\ce{CH2}$ and ring opening?" Especially given the steric hindrance near the carbon bearing the leaving group.

Indeed, that is the process going on in this case when the reaction was taken place in piperidine either at refluxing conditions or at room temperature. When (1-bromo-2,2-dimethylpropyl)cyclopropane (the alkyl bromide in question) was reacted in refluxing piperidine, it has yielded $88\%$ of (E)-1-(5,5-dimethylhex-3-en-1-yl)piperidine as a major product, which corresponds to the homoallylic analogue of the well-known SN2 process observed with allylic halide.
In addition, $8\%$ of 1-(1-cyclopropyl-2,2-dimethylpropyl)piperidine was also yielded as a minor product (Ref. 1 & 2). The authors claimed the formation of the minor product was not compatible with simple SN2 displacement due to the great steric hindrance imposed by the neighboring $\ce{t-Bu}$ and cyclopropyl groups, but the tight ion-pair mechanism (not separated by solvent) may explain the presence of SN2 product very well.
The percent yields of SN2' and SN2 products for the reaction at $\pu{25^\circ C}$ had been $73\%$ and $8\%$, respectively.
References:

Reaction of amines with cyclopropylcarbinyl halides: SN2' or
solvolysis?: J. Org. Chem., 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).
Homoallylic substitution reaction of piperidine with 1-bromo-1-cyclopropylalkanes: Tetrahedron Lett., 1983, 24(46), 5031–5034 (https://doi.org/10.1016/S0040-4039(00)94033-4).


text:
Differences between hexaaqua cobalt (III) complex and hexammine cobalt (III) complex in aqueous solution
Why is $\ce{[Co(H2O)6]^3+}$ a strong oxidizing agent and $\ce{[Co(NH3)6]^3+}$ stable in Aqueous solution?
I think this has got to do with Pairing Energy and Ligand strengths of $\ce{H_2O}$ and $\ce{NH_3}$ but I haven't been able to get a definite answer so far, neither by my friends nor by google.


text:
Approximated formula of acid dissociation constant
$K_\mathrm{a}$ can be expressed as $K_{\mathrm{a}} = \frac{[\ce{H^{+}}] [\ce{A^-}]} {[\ce{HA}]}$, however, it can be approximated that $K_{\mathrm{a}} = \frac{[\ce{H^{+}}]^{2}} {[\ce{HA}]}$.
I usually use the approximated equation when I'm dealing with questions that don't provide [$\ce{A-}$]. What I don't understand is why/how can we approximate it to this?
To approximate to $\ce{[H+]^2}$, then $\ce{[H+]}$ and $\ce{[A-]}$ must be pretty much equal, and so this must assume that the acid completely dissociates. However, in practice, do acids dissociate to the point that $\ce{[H+]}$ and $\ce{[A-]}$ are equal? If not, then how can we approximate to $\ce{[H+]}^2$, and are there any mathematics that demonstrate this approximation?

As an approximation, it's valid when 1) the amount of $\ce{H3O+}$ from the dissociation of water is negligible compared to that from the dissociation of the acid, and 2) dissociation of the acid is almost complete ($\alpha \approx 1$). So in general it works for relatively concentrated ($1 < \mathrm{p}C < 5$) solutions of moderately weak ($3 < \mathrm{p}K_\mathrm{a} < 8$) acids.

The acid dissociates in equal amounts of $[\ce{H^+}]$ and $[\ce{A^-}]$ hence we can say they are equal quantities.
We can approximate it like this because we assume the $[\ce{H^+}]$ from the dissociation of water to be negligible compared with the $[\ce{H^+}]$ from the dissociation of the acid.
We also assume the $[\ce{HA}]$ at the start is equal to $[\ce{HA}]$ at equilibrium because the dissociation of weak acids is small.
Here is some mathematics that you requested:

$K_{\mathrm{a}}$ can be calculated using equilibrium concentrations:
  $$K_{\mathrm{a}} = \frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{equim}}=\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}$$

Taken from my OCR A Level Chemistry Textbook (The specification under which this year's exams will be.)

Here is an extension to fully show how the two equate. Taken from the same textbook:

$$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}\rightarrow\frac{[\ce{H^{+}(aq)}]_\text{equim}^2} {[\ce{HA(aq)}]_\text{start}}$$
  Therefore, $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]^2} {[\ce{HA(aq)}]}$$


text:
Calculating the pH of a solution of sodium carbonate
I've seen similar questions on this site, but I'm having trouble as my book said something else. For a $\pu{0.09989 M}$ solution of $\ce{Na2CO3}$, my book said:
$$\ce{Na2CO3 + 2H2O -> H2CO3 + 2Na+ + 2OH-}$$
So , $\mathrm{pH} = 14 + \log(2 \times 0.09989) = 13.3006$ .
The problem is that I haven't been given a value of $K_\mathrm b$ as I saw in other questions on the site. We usually don't use any of them if they're not supplied in the question. What should I do?

You should apply the correct formula  of hydrolysis of salts( here the salt is of strong base and weak acid) which is, $$\ce{pH = 7 + \frac{pK_a(HCO_3^-)}{2}  + \frac{log (c)}{2}}$$ You should look  for the correctness,  justify by your own which should be correct. You should be confident enough to say the thing wrong if it seems wrong by all possible logics even though it is written in book. By this formula you will find that the answer is close to $11.66$.

If you are given, say $K_b$, then you need to know that $K_b$ and $K_a$ are related as $K_w=K_aK_b$ .

https://chemistry.stackexchange.com/questions/99012/how-to-calculate-ph-of-the-na2co3-solution-given-ambiguous-ka-values/99814?sfb=2#99814


text:
Catalyst with zero order kinetics
I've read in a book that a catalyst may be included in the rate equation even though it is never used up. It was written as having zero order kinetics however I thought catalyst concentration will always have an effect on the rate, especially since in a lot of biological enzyme-controlled reactions, enzyme concentration is a key element of the rate of a reaction. Is this only the case if the catalyst is above a certain concentration such that adding more will have no effect?

Enzyme reactions usually follow the Michaelis-Menten scheme $\ce{E + S <=> ES -> P + E}$ where E is enzyme, S substrate and P product. Overall the enzyme is unchanged but the initial rate  is first order in enzyme concentration. Your intuition about rate is confirmed in this case.

What makes the reaction a zero-order reaction, is it being dependent only on the concentration of one of the elements.
In a system made by a reagent and a catalyst, we expect the reaction to behave in a zero-order fashion when the concentration of the reactant varies in a linear fashion during the course of the reaction.
You could write an equation in the form: $rate = k$ and, implying that only the reactant A varies in concentration, $[A]=[A_0]-kt$.
Now, during a reaction you know that the concentration of the catalyst remains constant.
What you are asking is: what happens if I vary the concentration of the catalyst? Assuming that you are still in zero-order conditions and you run another reaction with a larger amount of catalyst, then you will obtain another equation rate (another value for k).
But that will not make the curve [A] vs t non linear (that is, non zero-order).
More catalyst means a larger rate constant, but doesn't imply a variation in kinetic behaviour of the reaction.
Still, if you were to evaluate the variation of rate in respect to the catalyst's concentration, then you would obtain a reaction rate that is still of order 0 in respect to A, and of order non-zero in respect to the catalyst.
Maybe your book was just pointing out that a set of reactions in which the same amount of catalyst is used, could be studied with a zero-order approach.


text:
When calculating enthalpy changes of reaction using, why is only the amount of substance of one reactant used?
When calculating the thermal energy released during an acid-base neutralisation using the equation $Q = mc\,\Delta T$, and then using $\Delta H = - Q/n$ to calculate the enthalpy change of neutralisation, why is the amount of substance of one reactant used ($n$ in the second equation) rather than the total amount of substance of acid and base reacting together?

Consider the following reaction ($\Delta H_{\text{rxn}}$ is the enthalpy change for one mole of the reaction, or the amount of moles of each substance found in a balanced equation: in this case, 1 mole $\ce{A}$, 1 mole $\ce{B}$):
$$\ce{A->B}\qquad\Delta H_{\text{rxn}}=~?$$
You are given 5 moles of $\ce{A}$, and you measure the change in heat to be $\pu{-100 kJ}$. However, this $\pu{-100 kJ}$ is actually just the $\Delta H$ of this reaction: $$\ce{5A->5B}\qquad\Delta H=\pu{-100~kJ}$$
Looks the same, right? That's because it is the same. However, if you want to calculate the enthalpy change of the first reaction, you are going to have to divide the measured enthalpy change, $\pu{-100 kJ}$, by 5. Thus, the $\Delta H_{\text{rxn}}=\pu{-20kJ}$.
Now consider this:
$$\ce{A +B->C}\qquad\Delta H_{\text{rxn}}=~?$$
If you were again given 5 moles of $\ce{A}$, along with 5 moles of $\ce{B}$, and you measured the $\Delta H$ to be $\pu{-100 kJ}$, you would get something similar to the first scenario:
$$\ce{5A + 5B -> 5C}\qquad\Delta H=\pu{100kJ}$$
Looks familiar right? Now, if you wanted to calculate the $\Delta H_{\text{rxn}}$, you wouldn't divide 100 by 10; you would divide it by 5, since the second reaction is just $5x$ the first one.
Side-note: This same reaction would be written if we have more than 5 moles of $\ce{B}$. Since $\ce{A}$ and $\ce{B}$ react in a 1 : 1 ratio, adding more $\ce{B}$ than $\ce{A}$ would mean that there is excess $\ce{B}$.


text:
How does the cathode of an electrolytic cell drive the anode reaction?
Consider this electrolytic cell:

In this cell, electrons from the anode of the battery travel to the cathode of the electrolytic cell, where hydrogen ions are reduced into hydrogen gas. This drives the oxidation reaction at the anode, where hydroxide ions are oxidized into oxygen gas.
My question is: How does the reduction reaction at the cathode drive the oxidation reaction at the anode? They aren't connected by a wire or anything, so there is no electron flow from the cathode to the anode like in a galvanic cell. Also, since this is an electrolytic cell, the reaction is non-spontaneous.
Some possible hypotheses:

The galvanic cell just gave some of its electrons away, so it is negatively charged and sucks in the electrons from the anode of the electrolytic cell, forcing the oxidation reaction to occur. However, assuming the galvanic cell has a salt bridge, shouldn't ions be able to compensate for this?
The loss of positively charged ions at the cathode causes a buildup of negative charge in the cell; the anode tries to compensate for this. However, how do electrolytic cells involving the reduction of water, which has neither a positive or negative charge, into hydrogen gas occur?


text:
Can Hückel's rule be applied to inorganic compounds?
Hückel's rule helps us know which molecular orbital, bonding or antibonding contains more electrons, and it is derived from the same. And on the basis of this, we can find out the stability of the compounds.
But, this rule was specifically introduced for cyclic organic compounds to know if they are aromatic or not. But again, the reasoning remains the same across all compounds, that is, if more electrons are in BMO than in ABMO – stable; otherwise not.
So, I was wondering why we couldn't use the same rule for cyclic inorganic compounds? I don't know, but I might be missing out somewhere. Any help would be appreciated.
Also: This pdf here, agrees that Hückel's rule can be applied to inorganic cyclic compounds, but with a 'maybe'.

Whereas organic
  compounds need to be planar to be aromatic, no such limitation applies to inorganics. An inorganic compound may be expected to be aromatic if it satisfies a “counting rule”; the familiar
  Hückel rule applies to planar aromatics and the Hirsch rule works for spherical analogs.

Huckel's rule can be applied to "inorganic" cyclic molecules but only if they have a delocalised pi system. They also have to have flat (or close to flat) rings.
Recently some N5 rings which 6 pi electrons have been made, these are "inorganic" aromatic rings.
Also borazine has been the subject of considerable debate, it has been discussed on stack exchange before. It is the subject of a interesting paper: Islas, R.; Chamorro, E.; Robles, J.; Heine, T.; Santos, J. C.; Merino, G. Borazine: to be or not to be aromatic. Struct Chem 2007, 18 (6), 833–839. DOI: 10.1007/s11224-007-9229-z.

This rule can also be applied to all-metal compounds, which due to the electron deficiency can have both pi- and sigma-aromaticity. See, e.g.: Wikipedia article on Metal Aromaticity or the paper: Boldyrev, A. I.; Wang, L. All-Metal Aromaticity and Antiaromaticity. Chem. Rev. 2005, 105 (10), 3716–3757 DOI: 10.1021/cr030091t.

Welcome to chemistry.se! It would be nice, if you include human readable citations, instead of just links to a paper.

The tetrasulfur dication, $\ce{S_4^{2+}}$, has six pi electrons in its four-membered ring and shows the symmetry expected of an aromatic ring (reference: Elemental Sulfur and Sulfur-Rich Compounds II, edited by Ralf Steudel). It has even been obtained as a salt, $\ce{S4(SO3F)2}$, suggesting enhanced stability.


text:
Nitrosonium ion in the nitration of phenols
Peter Sykes mentions in the book Guidebook to Mechanism in Organic Chemistry (6th ed.) on p. 137 that:

Highly reactive aromatic compounds, such as phenol, are found to undergo ready nitration even in dilute nitric acid, and at a far more rapid rate than can be explained on the basis of the concentration of $\ce{^⊕NO2}$ that is present in the mixture. This has been shown  to be due to the presence of nitrous acid in the system which nitrosates the reactive nucleus via the nitrosonium ion, $\ce{^⊕NO}$ […]

However, there was no mention as to why the nitration of more reactive aromatic compounds would result in there being nitrous acid in the system. I understand that the conventional explanation for why the nitration of activated benzene derivatives is more rapid is that there is a stronger attraction between the electrophile and the nucleophilic benzene ring due to the greater amount of electron density in the ring. This explanation using the presence of the nitrosonium ion seems interesting but I don't see how nitrous acid could be generated due to there being a more reactive aromatic compound being nitrated. Could someone enlighten me on this?

Related but not a dupe:Why does a mixture of nitrous acid with concentrated nitric acid cause nitration of benzene instead of nitrosation?

What is the "dilute nitric acid" diluted in?  Nitrations are commonly carried out in a sulfuric acid solvent, not water, where nitric acid is actally a Bronsted-Lowry base.  In such a solvent a dilute nitric acid solution still presents a fair concentration of nitryl ions.

I believe the solvent should be water, if I interpreted correctly.

Possibly the electrophile is molecular $\ce{HNO3}$. Nitric acid is strong but not super-strong ($\mathrm{p}K_\mathrm{a} = -1.4$), so solutions on the order of a mole per liter have a significant amount of molecular $\ce{HNO3}$. As $\ce{NO2(OH)}$ this can transfer its nitryl ($\ce{NO2+}$) ion moiety to the aromatic ring in exchange for a proton to make the nitrated ring and water. Obviously molecular nitric acid is less electrophilic than the straight-out nitryl ion, so an activated aromatic ring is needed for this to work.

But the book mentions specifically that NO+ is generated

Not sure how that is, unless some of the organic compound is oxidized first.  And nitrogen(III) species would disproportionate or be oxidized by air, no?  Molecular nitric acid is in the solution with no side reactions, ergo a more parsimonious explanation.


text:
What explains the color in clear sigma versus green pi complexes formed by methylbenzene?
Peter Sykes mentions in Guidebook to Mechanism in Organic Chemistry (6th ed.) on p.132 that there would significant differences in physical properties of the resultant mixture when we:

React methylbenzene with $\ce {HCl}$ to form a $\sigma$ complex
React methylbenzene with $\ce {HCl}$, with $\ce {AlCl3}$ as Lewis acid catalyst, to form a $\pi$ complex

One of the differences he mentioned was that the former would remain colourless while the latter would give a green colour. However, he did not make clear what gives rise to the colour difference. Is it due to formation of the $\pi$ complex?

Yes as it brings in new levels for which transition energies fall within the Vis part of the spectrum

However, these complexes must be very different to one another as one has lower energy levels than the other. Most probably they are completely different species with the coloured one involving Al possibly complexed to the benzene via its delocalised orbitals.

The examples are probably wrong, HCl can't protonate benzene to significant degree, and HAlCl4 acts as Bronsted acid not Lewis, so there's no π complex created.

This page of experimentalists talks about the reaction of $\ce{HCl + AlCl3}$.

Check out this video if you see what you get.


text:
Does the mixture of a ligand exchange reaction contain all the complexes obtained in the part reactions?
A ligand exchange reaction involves many part reactions. So, does the mixture contain all the complexes from those part reactions?
Quoting an example from ChemGuide:

This can be written as an equilibrium reaction to show the overall effect:
$$\ce{[Cu(H2O)6]^2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]^2+ + 4H2O}$$
In fact, the water molecules get replaced one at a time, and so this is made up of a series of part-reactions:
$$
\begin{align}
\ce{[Cu(H2O)6]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)}(H2O)5]^2+ + H2O}\\
\ce{[Cu\color{blue}{(NH3)}(H2O)5]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)2}(H2O)4]^2+ + H2O}\\
\ce{[Cu\color{blue}{(NH3)2}(H2O)4]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)3}(H2O)3]^2+ + H2O}\\
\ce{[Cu\color{blue}{(NH3)3}(H2O)3]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)4}(H2O)2]^2+ + H2O}\\
\end{align}$$

And if yes, then which complex has the highest concentration?

The answer is pretty much given in the link you provide. Each equilibrium constant is large , hundreds to thousands, so pretty much complete reaction and as each species is formed it then reacts to form the next. The vary majority of species will be the last one in  your scheme.

Related: Why is ligand substitution only partial with copper(II) ions and ammonia?

Yes, all those complexes from all those reactions will be present, and some more. Essentially Loong did all the mathematics in his answer to: Why is ligand substitution only partial with copper(II) ions and ammonia?
Let's shorten this a little bit and write the general reaction equation:
\begin{align}
  \ce{[Cu(H2O)_{(7-n)}]^2+ + n NH3 &<=> [Cu(NH3)_n(H2O)_{(6-n)}]^2+ + n H2O};&
  n&\in\{\mathbb{N}; 1,\dots,6\}
\end{align}
Therefore we can write the general equilibrium constant:
\begin{align}
  K_n &=
    \frac{{\left[ \ce{[Cu(NH3)_n(H2O)_{(6-n)}]^2+} \right]}}
   {{\left[ \ce{[Cu(H2O)_{(7-n)}]^2+} \right]\left[ \ce{NH3} \right]^n}}\\
%  K_\mathrm{B} &=
%    \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}
%    {{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}}
%  &&= \prod_{n=1}^{6} K_n
\end{align}
Since all reactions are in equilibrium, all possible species will exist, although they might not be quantifiable. The overall equilibrium depends on the concentration of the involved species, especially ammonia.
If $[\ce{NH3}]$ is low, then a complex with smaller $n$ will be more stable, if you go to higher concentrations, larger $n$ will be preferred.
In all of this it should be mentioned that there are also complexes with vacant coordination sites $\square$, like anything in between
$$\ce{[Cu(H2O)5\square_1]^2+},\dots,
  \ce{[Cu(H2O)3(NH3)2\square_1]^2+},\dots,
  \ce{[Cu(NH3)5\square_1]^2+},\\
  \left(\ce{[Cu(H2O)4\square_2]^2+},\dots,\ce{[Cu(NH3)4\square_2]^2+}\right),$$
and possibly others.

Thanks Martin! :D "In all of this it should be mentioned that there are also complexes with vacant coordination sites □, like anything in between" are they just defective complex compounds? Because they don't seem to be present in any of the reactions.

" if you go to higher concentrations, larger $n$ will be preferred."
But, according to my logic, when $\ce{[NH3]}>\pu{1M}$, then $\ce{[NH3]}^n$ would increase with increase in $n$, and thus $K_n$ would decrease. So, the tendency for a complex with higher $n$ to form would decrease. Why is this incorrect?

@GaurangTandon There is nothing defective about complexes with empty coordination sites, they exist in equilibrium with the other complexes. I have just not written down all the equations. Your logic is wrong, because with an increasing concentration of one of the reactants, the equilibrium constant won't change; more product will be formed instead.


text:
Phenylalanine reduction to phenylalaninol
Since phenylalanine is reduced to phenylalaninol in 16-18 hours under reflux with sodium borohydride and sulfuric acid in dry tetrahydrofuran. My question is, to reduce $\pu{100g}$ of phenylalanine, how much sodium borohydride and sulfuric acid are needed?
Also, what is the quantity of dry THF used in the reflux? After the reflux can I remove the sodium borohydride with a filtration, as well as the THF/sulfuric acid/phenylalanine batch that is neutralised with NaOH? I also want to extract the phenylalaninol with toluene.

How is there sodium borohydride plus sulfuric acid?  I thought sodium borohydride breaks down with strong acid.

I am not familiar with the Sodium Borohydride/sulfuric acid procedure, however this paper https://pubs.acs.org/doi/abs/10.1021/jo000242h contains a procedure with 2eq LiBH4 and 4eq TMSiCl in THF for the reduction you want. Based on this method you will need 600ml THF.
There is also this paper that uses NaBH4 and NiCl2 in water and claims an 83% yield for the redution of Phenylalanine.  http://journals.tubitak.gov.tr/chem/issues/kim-99-23-2/kim-23-2-2-98058.pdf


text:
What to do with (large) imaginary frequencies for constrained minimum structures?
I am performing DFT calculations using ORCA 4.0.1 on an enzyme active site model. The model contains 89 atoms including the substrate (see Animation 1), five of which are fixed in space (the spherical atoms in Animation 1). I am using dispersion-corrected B3LYP with 6-31g(d,p) basis set, with the "RIJCOSX" approximation, and with a CPCM model for the solvent (dielectric constant $\epsilon = 4)$. From what I have seen in the literature, this method is often applied to enzyme models with success, and this is the reason for using it. Once I start computing reaction paths, I plan to try different functionals and basis sets to see if this give drastically different results. My geometry optimization tolerances are
Energy change: 5e-6;
Max gradient: 2e-3;
MRS gradient: 5e-4;
Max displacement: 4e-3;
RMS displacement: 2e-3

The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient). Additionally, it has been absolutely necessary to compute a better guess for the initial Hessian for the optimization, done at the B3LYP/3-21G-rijcosx level.
The fixed atoms are kept fixed to maintain the orientation of the active site residues. Only atoms where the side chain or backbone has been truncated are kept fixed. Without fixing these atoms, the residues will most likely  deviate way too much from the crystal structure, and no longer represent the enzyme environment we try to simulate.
Now, due to the constrained atoms, I expected imaginary frequencies to occur. However, for one of my presumably converged structures, I have quite strong imaginary frequencies. The list of frequencies start as follows: -220.645934, -69.117105, -65.597557, -61.314719, -48.720497, -40.439949, -32.562415, -21.395791, -16.637113, 6.126145. The two strongest negative frequencies are mainly a sigma bond rotation of a methyl group (of which the carbon is fixed in space during the optimization).
It is difficult to find any good literature on what to do with and how to interpret such negative frequencies in constrained optimizations. Is it "safe" to disregard the frequencies that mainly involved fixed atoms? Should I converge my geometry tighter to come closer to the true minimum? That's not so easy, I find, as the molecule is quite flexible. It took me quite some time to come up with a method that converges, which involve computing a good guess for the initial Hessian used in the optimization (computed at the B3lYP/3-21G level). I hope the imaginary frequencies are artefacts of the constraints, and that I can safely disregard them. However, they seem to have quite higher frequencies than what other people report in the literature (many say that their imaginary frequencies are around 20 cm-1).
This has to be a common challenge for people doing electronic structure calculations on large and constrained systems, but very few address this issue specifically in their publications.
I did read through this question, but it addresses not quite the same issue that I face.

Animation 1. The active site model. A total of 89 atoms.

Animation 2. Imaginary mode with frequency -220.645934 cm$^{-1}$.
What I am trying right now
I have three jobs waiting in the queue right now:

I have made a small rotation of the methyl groups strongly involved in the strong imaginary mode
I have started another geometry optimization of the structure given above (exhibiting the imaginary frequencies), but by reading the computed Hessian (the "exact" one, from same level of theory as optimization)
I have restarted an identical optimization, but by computing the "exact" Hessian from the start, giving a much better guess of which direction to go in.

I can add here later what these attempts result in.

Could you comment on why those atoms have to be fixed? I assume these are known positions from the crystal structure of the protein; how did you decide to cut there? Is QM/MM an option for you? Do you simulate a solvent in any way? Is there any reason why you are using this terrible level of theory? As far as I know ORCA has a plenitude of better (and way cheaper) functionals available; as well as a decent basis (def2-SV(P) with optimised aux. basis). I would also recommend a tighter convergence for the SCF and a larger grid.

I have updated the question with answers to your questions. So you recommend using something other than B3LYP/pople basis sets for such calculations?

I would never, ever recommend using a Pople basis set, except for training students and immediately tell them what not to use in the future. I also generally avoid B3LYP like a bad hangover. So yes, I would totally advise to something more advanced, always. Plus on the performance side: You are starting a large calculation with a hybrid DFA, either you have resources like the god of computation, or you have time like the doctor. Use a GGA/meta-GGA 'pure' functional instead, options: PBE, BP86, M06-L, ... Basis set: def2-XVP has especially optimised aux basis sets, use this!

Thank you for your insight! I will look into this. Do you know of any literature on "best practices"/benchmarks for DFT for organic reactions? While very useful for me, maybe this is a small digression for the actual question. Or do you think the imaginary frequencies are artefacts of a poor computational method?

This is a common occurrence when you fix atoms. Many programs (e,g. Gaussian) allow you to fix certain atoms during the vib freq calculation, which would prevent spurious imag modes. If the fixed atoms actually are expected to be fairly rigid, this is also more physically reasonable when calculating the vibrational modes from the Hessian. I don't think you can fix atoms during the freq analysis in ORCA though. Before Gaussian had this feature, I know some people deleted the vib modes of the fixed atoms, although in reality the remaining vib modes should change a bit by having atoms fixed.

May I ask what you used to generate the 2nd animation? Thanks!

@AndersSteenChristensen Animated the normal mode with Avogadro, and used GIPHY capture software (Mac only) to make the gif file.

I'd add dispersion corrections. The molecule is large, floppy, and there are non-covalently bound parts. I'd maybe even suggest PBEh-3c as an overall composite method (which includes -D3(BJ) and gCP), if B3LYP is too expensive.

Large imaginary frequencies, while you are looking for a local minimum, are usually an indicator that your guessed starting geometry is not good. Values larger than $10i~\pu{cm-1}$ are only observed for transition states, and only if you have one of these modes, it is actually meaningful.
When you are fixing atom coordinates, then you are (in a practical sense) employing an external potential. If the program you are using does not account for that constraint, it is not suited for these kinds of optimisations.note
At the very least, your thermochemistry will be wrong. If your aiming for anything else than a rough guesstimate, then this is not good enough (if you are contempt with that, why bother with frequencies).
In general there are a few things you can do to improve convergence. One of the most important ones and often underrated is the SCF convergence. If you have a sloppy converged SCF, then also gradients will be sloppy, which leads to poor performance of the geometry optimiser. Tighten it to 1E-7 to achieve better results.
Often numerical noise is a problem in calculations, make sure that you are using a tight (tighter, the tightest) grid for DFT calculations. In principle the same as above applies; if it is to coarse, your SCF will converge sloppy, etc.
This should at least let you converge the geometry on default cutoffs (gradient, displacement). If you then have imaginary frequencies, you should use path following algorithms to get rid of these; and if they are not numerical noise, this should not be a problem.
Use dispersion corrections. There is no excuse not using them (except missing parameters, but then use a different method).
From my experience, solvent models can drastically influence the convergence, in many cases it may lead to oscillation. Try optimising without it first, turn it back on later.
You have stated that this is a part of an enzyme. My first guess seeing this large imaginary modes, was that you chose to cut the moieties at the wrong position, substituting more demanding groups with hydrogen atoms. While this is common practise for many computational models, if you have problems with the molecular structure, this is a sign that you cut away something important. Increase your cut-out; don't replace with hydrogen, but with methyl instead (in doubt fix those, too).
Leaving something out of the calculation can lead to all sorts of artefacts. You should have a look at a methodology where you can describe the system in a whole. There a various applications developed just for this case. Something along the lines of DFTB+ might do wonders for your system. Another possibility is using QM/MM, and ORCA is capable of that.
If you want to describe reactions, in solution, sooner, rather than later, you need to look at these options. Aqueous media are especially challenging.
B3LYP is a terrible functional. This is probably more often true than not. It is not only one of the slowest performers (for hybrid functionals), it also has been shown to produce unreliable results (just search B3LYP failure). Starting with a hybrid functional is overkill for this system size in any case. Always use a non-hybrid functional (BP86-D3(BJ), B97D3, M06-L, etc.) first, it saves a lot of time, resources, brain cells. Work your way up Jacob's ladder, not down.
Use def2- basis sets. The close cooperation with Ahlrich's group has many of the great features developed there (re: turbomole) readily implemented in ORCA. This includes the extremely well performing basis set, which do already come with especially prepared auxiliary basis sets for density fitting (RI).
TL;DR: Tighten SCF; tighten grid; use dispersion; use a cheaper functional; use better basis; increase system size (you can fix adjacent moieties).

Note: If the program does not correct for these constraints, then the derived gradients won't be good enough to achieve convergence.

B3LYP is not a terrible functional per se. It strongly depends on your system and what properties you are interested in, as with every functional so far.

The first I noticed is you said:

The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient).

At first sight, the structure of your molecule should not leads to serious problems, but the above is a bad signal, and turns the results unreliable (at least doubtful). The constrains must be considered in the calculation of vibrational frequencies for they make sense.
Maybe no one can give you a foolproof steps series to follow. My suggestions are below.
What to do/check

If comments are right, and Orca does not care about fixing atoms for
the calculation of the vibrational frequencies, then you need to
change the software for this task (because in this case results are not meaningful) (This also answer the question "Is it "safe" to disregard the frequencies that mainly involved fixed atoms?"). The simpler way to check this is by looking at the number of freedom degrees reported (or deduced form the output). I was referring directly to vibrational f.d. because low frequency modes may be treated like rotations. I doubt that Orca lacks of this basic functionality. In such case I bet that this is described their User Manual.
The step you are doing are reasonable. My guess is that maybe you are far from a true minimum, in some unstable geometry, and that is why you have problems with accuracy changes. You said: "Should I converge my geometry tighter to come closer to the true minimum? ". Yes, it is always advisable to get good frequencies. So I would use not only tighter geometry convergence criteria but the TIGHT convergence option for the SCF part.
In the same lines, you can try using modern semi-empirical methods and then make the fine tuning with DFT. Even using a simpler DFT functional.
I agree with comments that you are not using the best basis set out there, but I think that this is not a problem here.
If you want to discard weird behaviors you can try with another functional, but I do not think that this is the problem (because it is known that B3LYP performs well in compounds with the elements involved here).

In brief. I would check if Orca compute correctly the frequencies, if not: change the software. If yes: discard those results, try to converge the structure using cheaper methods (if you are lack of computational resources) from different initial conditions, store those results and start new geometry optimizations and vibrational frequencies calculations starting from those previous converged states. The last step using the tight convergence criteria for SCF and normal convergence criteria for geometries, and using the target basis set (for energy and its derivatives).
Good luck
Note:

Once I start computing reaction paths, I plan to try different
  functionals and basis sets to see if this give drastically different
  results.

Why do you expect that something like this would happen? You can save time by using a little large basis set and a better functional once you get the B3LYP/6-31g(d,p) results. It should not take too much time.

EDIT DUE TO COMMENTS
This answers generated some discussion about B3LYP or not B3LYP?
In that respect, my viewpoint is as follows:

B3LYP is not the most accurate functional (there should be no doubt about that).
It is reasonably good for geometries in systems like yours. Errors in geometries can be larger due to small basis set than due to B3LYP, or at least of the same order.
While there are functional with better results for geometries, you also must take care of how large basis sets those functional needs to be a better option. I mean, maybe using 5$\zeta$ basis sets they are better, but not using a small basis set.  (and maybe not)
The above discussion can be very long, but in any case, you are using approximations that (in my opinion) are much stronger than the errors due to  B3LYP. For example, the permittivity value, or the geometry of the constained parts, etc..
Also, even if you could perform a FullCI/CBS you shoud notice that the simulated system is different to the real system, so extremely exacts results in the simulated system may no improve your knowledge about reality.
So, if the functionals are accurate enough, what it is important here is obtaining the corresponding results for the functional/basis set. The functionales can be more or less problematic. B3LYP should not be problematic for your system. My guess is what I said in the answer.
Because of the above, I think that B3LYP is reasonable and enough for your case.

"I agree with comments that you are not using the best basis set out there, but I think that this is not a problem here." Anecdotally, I have had issues with the PES provided by such poor basis sets, leading to imaginary frequencies I couldn't get rid of. The diffuse versions are even worse. But it's true that it probably isn't the issue here.

@pentavalentcarbon Interesting. I will definitely try something else, PBE of maybe M06.

@Yoda I said basis set, not functional. The use of Pople basis sets is more problematic than the use of B3LYP. Though B3LYP-D3(BJ) is supposed to be quite good, at least for energetics.

I must have had two thoughts at once writing that comment.  I will try different basis sets!

+1 For modern semi-empiric; -1000 for saying that B3LYP performs well (for anything).

@Martin-マーチン B3LYP (or at least the dispersion corrected forms) did pretty respectably in Martin Head Gordon's recent benchmarking of density functionals https://www.tandfonline.com/doi/full/10.1080/00268976.2017.1333644?scroll=top&needAccess=true

@Tyberius Thanks for the link, I'll give that a read in due time. I would be a fool to dismiss B3LYP, and not to acknowledge the huge impact it had on the development of DFT. I still believe there are better choices, and better performing functionals; especially for a first choice. The remark was meant to be read with a ;) as I still up-voted the answer.

I do many calculations on the active site model, but using Gaussian instead. Like what you have also mentioned in the question, the constraint on the truncated atoms caused some imaginary frequencies. However, I was greatly surprised by the -200 cm-1. Try to optimize the structure and calculate the freq at the level B3LYP/6-31**. I would expect much much smaller imaginary there.
update: I thought the calculations were done at the level of B3lYP/3-21G. Sorry about that. Have you found the solution now?

That is exactly what I have done, what you suggested.


text:
Probabilities of Electrons in Orbitals
I've heard both of the following:

An orbital is a region where there's a 90% probability of finding an electron
An orbital is a region where an electron spends 90% of its time

Are these essentially the same things? Or which one is correct and why?
Edit- I've heard both from veritable sources and only need a confirmation not a precise definition.

essentially the same thing

If an electron spends 90% of its time in an orbital, then the probability of finding an electron there is 90%; they're interchangeable in my opinion. Orbitals are where we think electrons exist anyway.

@KianStevens Like every statement in your comment is incorrect. :/

@Mithoron, can you please tell me what's correct then ?

https://chemistry.stackexchange.com/questions/39438/what-exactly-is-an-orbital

This is not the definition of an orbital. It is not even a volume/region dimension.


text:
Mechanism for reaction of enol ether with HBr

Please refer to the image for the question. I tried two different mechanisms for the reaction (1 and 2) and ended up with two different products. Now I can't decide which of the two mechanisms would give the major product.
I think that mechanism 1 should give the major product as it involves the formation of a double bond between $\ce{C}$ and $\ce{O}$ atom which is thermodynamically favorable. Is this reasoning correct?
The answer to the question had given product of mechanism 2 as major product. But I don't know if it is really correct or not.

The last s step in Mech. 2 is reversible allowing liberate bromide to participate in Mech. 1. In Mech. 1 the cation in the upper right may undergo SN2 vs. SN1 reaction with bromide. In all, you did well!

@user55119 Thanks! That's another point to consider. One of my friend said that mechanism 1 involves cleavage of a C-O bond and so the reaction should not occur through Mech. 1. Is what he is saying correct?

Both mechanisms involve cleavage of a C-O bond at some point.


text:
Will vegetable oil and kerosene mix?
Is it possible to mix vegetable oil with kerosene?
If not, what other solvents can be used to reduce the viscosity of cooking oil/vegetable oil.

Asking for answers and not providing enough relevant details, purpose, context or background of questions are contradictory decisions.

Cooking oil and kerosene will mix well, just do not cook with it afterwards !
It is important to keep in mind that there is no such thing as a "standard kerosene". I tend to use two different kerosenes for solvent extraction, I like the solvent 70 which is aliphatic kerosene and solvesso150 which is an aromatic kerosene.
You can regard an aliphatic kerosene as a longer chain version of hexane, it will be similar to nujol.
One thing you may notice is that when you mix some cooking oil with kerosene is that the water (if there is any in the cooking oil) may separate from the organic liquids. This may make it go cloudy. But I think that the organic part of cooking oil (fatty acid glycerol esters) will be miscible with kerosene.


text:
Effusion rate 1.4 times faster SO2(g)
How do I solve this-
Which diatomic gas would effuse at a rate that is 1.4 times faster than the rate at which so2(g) effuses at any given temperature? This is high school chemistry.
I did: (using graham's law) 1.4= sqroot of 64gmol-1 / xgmol-1 => 1.96= 64/x => 1.96x=64 => x=32.65g/mol. 32.65/2 (bc its a diatom) => 16.325g/mol which is close to oxygen.
At first, I found Urms of SO2 using 273k for T, 0.064kg for M and 8.314 for R, I got 326. Then multiplied that by 1.4 and got 456. Then I put 456/326= sqroot of 64gmol-1/ xgmol-1 and solved for x. I realized the 456/326 equaled 1.4, so I figured I didn't need to find Urms in the first place.

See Graham's law of effusion

would O2 be the correct answer?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

@Chrissy we will tell you whether oxygen is correct if you tell us how you arrived at that possibility.

@OscarLanzi sorry didn't know about the policy. I updated the post.

@chrissy now it can be told, you're good.

The rate of effusion is proportional to the mean speed of the molecules which in turn is proportional to $ 1\sqrt{M}$ (at constant temperature).  From this and by using ratios you can work out what the M for the unknown is.


text:
The sign of enthalpy of formation of magnesium oxide
I'm currently doing a lab to calculate the enthalpy of formation for $\ce{MgO}$. However at the moment me and my lab partner are having a disagreement. We've both calculated and agreed upon the same heat released in the reactions ($\pu{4.2\frac {kJ}{mol}}$ (reaction of $\ce{MgO +2HCl})$ and $\pu{7.5\frac {kJ}{mol}}$ (reaction of $\ce{Mg +2HCl}$)). Where we disagree is over how to calculate the enthalpy of reaction.

My argument:
Enthalpy is the amount of energy put into the reaction, what we measured is the heat released thus the equation should be:
$$\Delta H=-\frac{Q}{n}$$
Also, this would make sense considering our reactions were blatantly exothermic, and using this equation provides negative values.

Her argument:
The equation is simple and the teacher agreed with her results:
$$\Delta H=\frac{Q}{n}$$

My question: Who is correct?

She is right, and the Q should be negative.

I'm pretty sure you are correct. Q is the change in heat for the surroundings. Thus, you have to put a negative sign in front of it to determine the change in heat for the system. If the system loses heat, the surroundings gain heat, vice versa.

@Frank, That is not correct.  Q represents the amount of heat you have to add to the system so that the final temperature of the products is the same as the initial temperature of the reactants.  So Q represents heat added to the system, not heat added to the surroundings.

@ChesterMiller I'm pretty sure this experiment is carried out in a calorimeter, and Q is calculated using Q = mcΔT. I feel like that would be the most reasonable method to carry out such an experiment. Also, why would you want to add heat to make the products the initial temperature instead of just measuring the temperature change?

@Frank you are correct we conducted in the reactions in a styrofoam cup, and measured the temperature change. We then used Q=mcΔT to calculate heat.

The calorimeter experiment is just a vehicle used for determining the quantity I described.  You only measure a small temperature rise because of the large thermal inertia of the calorimeter, and you use the measured rise to determine the amount of heat you would have had to add to result in the temperatures of the products being equal to the temperatures if the reactants.  (For an exothermic reaction, this would be negative)  When you learned how to use a calorimeter to measure the heat of reaction, didn't they teach you this stuff?  They should have.

I think we're talking about the same thing in opposite ways. What I'm talking about is, when you measure the temperature change in the calorimeter, and you convert that to heat, you are calculating the heat added/removed from the surroundings. In a calorimetry experiment, you measure the temperature change of water. OP even noted that the calculated Q (heat) was positive, and since the reaction was exothermic, the Q he measured was obviously the heat added to the surroundings. I think what you are talking about is the inverse of this, although I'm not completely sure what you mean.

@Frank  I stand by what I said in my earlier post.  For an exothermic reaction, the change in enthalpy is negative, and heat is removed from the system (negative Q).

Exactly. He has positive Q values from his calculations and must multiply them by -1 to get the actual ΔH, which is negative.

In thermodynamics, the convention is that Q represents the heat transferred from the surroundings to the system.  So, for an exothermic reaction, both Q and $Delta H$ are negative.  There is no need to multiply by -1 if the convention is followed properly.

I guess you've been answered in the comments already, but anyway. You are correct. In thermodynamics, enthalpy is the energy supplied to the system at constant pressure. If a reaction is accompanied by the release of heat, i.e. if it is exothermic, its $\Delta H$ is negative. If $x$ kJ is released into the calorimeter upon dissolving 1 mol of substance, resulting in the rise of calorimeter temperature, then $\Delta H_{diss} = -x$ kJ/mol.
As for the heat, depends on the type of the calorimeter, i.e. on the direction in which the heat flows. For conventional "coffee cup" calorimeter, where $Q=mC\Delta T$, $Q$ is the energy released, so $Q = -\Delta H$. For some power-compensated calorimeter, where you vary the heater power so as to maintain the temperature, $Q$ would be equal to the energy supplied. Since you have to supply excess energy ($Q>0$) to offset the endothermic reaction ($\Delta H>0$), $Q = \Delta H$.
I've heard that a long time ago, at least in some countries, there were two systems of signs for enthalpies, so-called "thermodynamic" and "thermochemical," with reversed signs in the latter. However, I haven't seen "thermochemical" one in real life, and it seems to be long since abandoned. It is not even mentioned in modern textbooks on thermodynamics. So only the thermodynamic sign system, where $\Delta H>0$ for endothermic processes, remains.


text:
Pipette pumps durability
According to your experience, in terms of durability and longevity of the tools, which one of these types of pipette pumps is better and why?

Tri-valve pipette bulb
Syringe pipette pump
Electronic pipette pump

These pipette pumps are mentioned in the YouTube video Bio-Rad Laboratories — Using Serological Pipets.

As long as the chemical is not very corrosive, medical syringes with needles are easier to use than pipettes with pumps. Medical syringe AS a pump (with help of a short silicone pipe as connector) also works good enough, but is less precise than other methods. Still, I have several years of experience using it in this capacity. It worked fine.

Which syringe are you talking about? The regular syringe?

@Rexita Yep, plain old regular medical one-use plastic syringes.  They are good enough for surprising amount of cases, and are OK as pumps for pipettes. Furthermore, medical syringes work great with https://www.plasticoid.com/sleeve-stoppers.html for works in argon atmosphere.

You mean this syringe: https://en.wikipedia.org/wiki/Syringe

No screws like this, they prevent use with glass pipettes https://www.nicvape.com/syringes-blunt-tip-needle ||.  This one is better https://www.amazon.com/Pack-BDTM-Dispensing-Syringe-needle/dp/B00FUO2X06 |||You can fix a short silicone tube on it and plug it onto your pipettes. It works well enough, is cheap and actually reliable and durable (I saw one in intense use for several years as a pump).

Syringe pipette pump.
Tri-valve is not as easy to use as syringe one. And the little metal sphere in it can be thrown while using. The syringe is more comfortable, it is just required to have 3 types for different volumes.
Additionally, when the pipette is broken at the top or has similar problems, you can somehow adapt the syringe, like inserting pipette deeper or tightening the syringe. It is also not expensive and easy to find from others when you need more.

"tightening the syringe" like at the end part of the syringe where the pipette is inserted? Isn't that part made of some kind of rigid plastic or is it rubber made? If it's rigid plastic how would you tighten that?

"when the pipette is broken at the top or has similar problems, you can somehow adapt the syringe, like inserting pipette deeper". Is it not possible to do that with the tri-valve pump?

" It is also not expensive and easy to find from others when you need more." I think this statement doesn't apply to me..

-For your first comment, it is a screwed plastic, you can loosen and tighten for your need. It is important as for instance pipetting DCM is troublesome in any other way, it just leaks without control.
-Trivalve is not as rugged as syringe ones, it will be not as effective; replying to your second comment.
-I don't want to sound like a supplier of syringe pump but it is both more convenient and has more lifetime, it can be cheaper on that scale. In Turkey, this is the situation.


text:
Difference in application between Standard Type Bulb Pipette Filler and Universal Type?
I want to ask what is the difference in terms of application between the HHH (Heinz Herenz Hamburg) Bulb Pipette Filler Standard Type (1081801) and the HHH Bulb Pipette Filler Universal Type (1081802)?

They both basically have same functions with same operations. However, universal type is longer than standard type with an adapter for pipette head (see the picture below: standard type is to your left).

Note: The bulbs shown are Deutsch-Neumann, but they are similar to HHH.

What's the advantage of one over the other?

Are they used for the same  pipette with the same measurement capacity or they provide different capacity of suction? Example : one is for pipette of X ml and the other can be used for pipette with higher ">X" ml.


text:
How to find the number of optically active and optically inactive isomers for a spiropentane derivative?
What will be the number of optically active and inactive isomers 1,2,4,5-tetramethylspiro[2.2]pentane?

The question asked for the difference between number of optically active and optically inactive isomers. When I drew them, I could find 6 optically active, and none optically inactive isomers. So, according to me answer should have been $6-0=6$. But the given answer is 5. Apparently, one optically inactive isomer is possible, but I don't see how.
I am assuming the answer given to be correct. Though, there is very slightest possibility of it being incorrect too, but I don't have any source to confirm, thefore asking on Chemistry SE. :)

There is one center of symmetry, giving rise to one meso compound which is optically inactive.


text:
Can DIBAL-H reduce carboxylic acid?
Question : Can DIBAL-H (Diisobutylaluminium hydride) reduce carboxylic acid?
According to "March's Advanced Organic Chemistry" (by Jerry March),  acid can be reduced upto aldehyde using DIBAL-H. But on the other hand, "Solomons & Fryhle Organic Chemistry" and "Organic Chemistry by Jonathan Clayden" this isn't possible (though not said explicitly).
I am confused now. Can someone please confirm that whether this reduction is possible or not?

But still there is no explanation and answer for this, I will thank full if any one can give proper explanation for it

Please share the relevant page numbers from the books you have mentioned so that users who have them can check it too. Thanks.


text:
is rydberg constant and rydberg unit are same?
i want to ask a question that.is rydberg constant and rydberg unit are same. UNIT OF RYDBERG CONSTANT IS m^-1, right? Then what is rydberg unit.

Read this article.


text:
Why does ammonia accept a hydrogen to form ammonium when its octet is already complete?
Why does ammonia accept a hydrogen to form ammonium when its octet is already complete as $\ce{NH3}$ and it should be, seemingly, stable as $\ce{NH3}$?

It is the same reason why $\ce{H3O+}$ exists (formed by $\ce{H2O}$ accepting a $\ce{H+}$ from stronger acid).

It is stable and happy as NH3. The problem is that H+ is very unhappy.

"Hydrogen" means different things at different times. It can mean $\ce{H+}$, $\ce{H.}$, or $\ce{H2}$. (Not to mention deuterium and tritium!)
A bare proton ($\ce{H+}$) does not bring an electron to the filled octet on nitrogen, but attaches to the lone pair. The positive charge is now shared over 5 atoms, and the energy of formation of a new bond stabilizes the ammonium cation. The octet on nitrogen is still only an octet.
Interestingly, a hydrogen atom ($\ce{H.}$) can also attach to an ammonia molecule to form ammonium radical. Electrolysis of ammonium salts with a mercury cathode  yields ammonium amalgam, which rapidly decomposes to $\ce{NH3}$ plus $\ce{H2}$. In this case, the extra electron in $\ce{NH4.}$ is also shared among the 5 atoms, and is more stable than when it is attached to just one proton. But not that much more stable, and the bond between two hydrogens is even stronger.


text:
What is the relation between the following two optical isomers?

I think they should be diastereomers, but the answer to this question is enantiomers, if it is enantiomers then how?

They are mirror images. Look at the $(S,R)$ configuration of each chiral center.

Rotate one of them by 180-degrees. Compare.

you can check this out it is not exactly the same but Cl is replaced by OH


text:
Explaining the colour change in the potassium permanganate titration of iron(II) ions
In the redox titration:
$$\ce{MnO4- + 8 H+ + 5 Fe^2+ -> Mn^2+ + 4 H2O + 5 Fe^3+}$$
the colour change which occurs is purple to colourless, because of the decreased concentration of permanganate ions. But doesn't this fail to account for the visible iron ions? $\ce{Fe^2+(aq)}$ is green and $\ce{Fe^3+(aq)}$ is brown.
Anyone have an explanation for why the end result is colourless and not brown solution?

It's permanganate, not manganate. As for your question, consider the intensity of color. Compared to permanganate, both Fe2+ and Fe3+ are almost colorless.

Fe(3+) is very pale yellow, hardly brown

Furthermore, note the common addition of $\ce{H3PO4}$ whose dissociated anions give colourless complexes with $\ce{Fe^{3+}}$.

I think this is because $\ce{MnO4-}$ ions have a very distinct colour. At the end of the reaction, the solution is colourless. I think that the $\ce{Fe^3+}$ ions give the solution a brown colour, but it is in a very low concentration. The brown $\ce{Fe^3+}$ ions are too few to give the whole solution a brown colour.


text:
Difference between equivalent conductance and equivalent conductivity
I find questions in my book which ask for the equivalent conductivity and sometimes for the equivalent conductance. And then when I look at the solution, they've used the same formula for both. This has confused me a lot.

Conductance and conductivity are distinct, although related, magnitudes.
Conductance, $G$, is a measure of the total electric current, $I$, that can pass through a conducting element (such as a wire, a space through which a charged beam flows, or an electrolytic solution) when a given voltage, $V$, is applied through it:
$G = \cfrac{I}{V}$
Correspondingly, the IS unit for conductance is the siemens (S), equivalent to an ampere per volt. Since conductance is the inverse of resistance, a siemens is also the inverse of an ohm.
However, in general, conductance does not only depend on the nature of the material; it also depends on the geometry of the conducting element. For instance, the conductance of copper wires of different length and/or different section will be different. Conductivity, $\sigma$, is defined in order to compare the contribution to conductance of materials, isolating it from shape effects.
In general, for a linear electric flow across a conducting element, conductivity increases with cross-section (as more flow is able to pass through simultaneously) and decreases with length (as more resistance is found). Conductivity is thus defined as
$\sigma =  G \cfrac{l}{A}$
which, ideally, should be identical for all conductors made of the same material (in practice, there are a number of effects that can make the relationship between conductance and geometry non-linear, but for macroscopic ohmic conductors it holds fairly well). This allows direct comparison of the electric properties of materials, effect of doping, temperature, etc.
The IS units of conductivity are therefore siemens per metre (S/m).
Note that unlike conductance, which is an extensive quantity, conductivity is an intensive quantity which doesn't depend on the size of the conducting element. According to 19th century naming conventions, conductivity can also be seen as a specific conductance (in the same way specific gravity is an intensive measure for mass, or specific heat is an intensive measure for heat capacity) and indeed that name was widely used in the early days of both electrochemistry and electromagnetism. While in electromagnetism that standard has changed and currently conductivity is the almost universal name for $\sigma$, in electrochemistry that name lingered and conductivity measurements in solutions were reported as "specific conductance" for longer.
Compound the facts that conductivity measures in electrochemistry are usually made with standard-sized cells in which geometry isn't a factor, and that specific values of conductivity (such as molar conductivity or equivalent conductivity) need an additional qualifier, and these magnitudes were often called "molar conductance" or "equivalent conductance" short for "molar specific conductance" or "equivalent specific conductance". However, you'll find that these magnitudes are always reported in units of conductivity (typically $\mathrm{\mu S/cm}$ or $\mathrm{mS/cm}$).
The use of specific conductance to mean conductivity is now considered obsolete; see for instance the IUPAC Gold Book's page for conductivity.


text:
Dehydration of carboxylic acids using P2O5
Why isn’t concentrated $\ce{H2SO4}$ used for the dehydration of carboxylic acids to anhydrides? What is special about $\ce{P2O5}$?

OK, thanks. Now to the point: you may use H2SO4, but how would you separate it from your anhydrides?

well i dont know

The advantage of P2O5 is that it is solid. Also, it is a stronger dehydrating agent than H2SO4.

$\ce{P2O5}$ is a solid. Hence, it is easy to separate from reaction products. Moreover, $\ce{P2O5}$ is a stronger dehydrating agent than $\ce{H2SO4}$

Intermolecular forces are the reason. Carboxilic acids have capability of forming a dimer by hydrogen bonding. That enables a very stable arrangement. The hydrogen bonding stabilise the molecule, enabling the organic chain to form additional bonds by dispersion forces.Also dehydration of carboxylic acids with phosphorus pentoxide (P2O5) yields acid anhydrides,because it strong dehydrated anhydride ;P2O5 itself reacts with water molecules to form phosphoric acid. That enables it to remove free water molecules by a simple hydrolysis reaction,


text:
Why are fluoride ions (and ions in general) reactive?
Please bear with me—-I have a pretty meager background in chemistry but am trying to learn some!
My question is about ion reactivity—-specifically of the reactivity of $\ce{F-}$.
I often hear that the main reason the “weak” acid $\ce{HF}$ is so dangerous is because of the fluoride ion rather than the $\ce{H+}$. The impression I’ve gotten is that fluoride is the real hit man in this situation, with the $\ce{H+}$ only really serving to corrode barriers to entry.
A search for “fluoride ion reactive” yields the following sentence: (source)

Unlike other hydrogen halide acids, the anion (fluoride ion) is quite reactive, and can form fairly insoluble salts with alkaline earth metals such as calcium and magnesium.

However, I think I’m fundamentally failing to grasp something. Why is this insoluble salt effect specific to fluoride? It’s obvious why fluorine is reactive, but why is the octet-satisfied fluoride reactive as well? And how does fluoride reactivity compare with the other halides?
I have also heard people say that fluoride is a bad nucleophile, which makes sense given how tightly it holds on to its electrons. So I can’t quite wrap my brain around why fluoride would want to react with anything at all. Wouldn’t that just force it to share its extra electron with another atom? (Is there such a thing as “electronegativity” but for ions rather than atoms?)
After thinking about it, I realized my confusion extends to all ions in general. Why would “fully dressed” ions with full octets do anything but sit around on the couch eating Cheetoes?
There’s definitely something stupidly fundamental that I’m missing — hopefully someone can point me in the right direction!

Fluoride activity, or more properly danger, comes from the fact that formation of some salts of fluoride is energetically highly favorable for aforementioned metals.
Calcium for instance, rather than in calcium carbonate, is more stable in $\ce{CaF2}$ form. Then, consequently, it will etch your bones eventually if there is a dermal contact with $\ce{HF}$. Other acids like sulfuric acid make a superficial effect and with a copious amount of water you mostly get rid of it. $\ce{HF},$ on the contrary, is a weak acid and slowly seeps through your epidermis to your bone.
$\ce{HF}$ is also reactive with glass, through silicon. It may also etch the beaker or any other container through time so you don't want to store $\ce{HF}$ for a long time in glass containers, actually same with bases to a lesser degree.
$\ce{F}$ can even react with noble gases, you should just look at the last product and compare the overall stabilities of the current form and the final form. Fluoride simply more stable in $\ce{CaF2}$ solid than in solution.


text:
Does the wavelength of light absorbed by a complex depend upon extent of splitting or stability?
Background: we were doing Crystal Field Theory in school when this question popped up. Specifically we were talking about stability obtained due to the chelate effect and if the extra stability in a complex would affect the wavelength of light it absorbed.
The exact question is:

Arrange in decreasing order of wavelength of light absorbed:
  $\ce{[Fe(ox)3]}$, $\ce{[Fe(CN)6]^{3-}}$ and $\ce{[Fe(H2O)6]^{3+}}$

If we considered wavelength depending on stability, the answer differed than if we consider splitting only.

The more the ligand splits the $t_{2g}$ and $e_g$ levels (often called $10 D_q$ or $\Delta_0$) the shorter the wavelength of light absorbed.  The size of $D_0$ is related to the position in the (experimentally determined) spectro-chemical series.  Look up the order of ligands in this series and you have your answer.

The spectrochemical series we have done does not account for Chelsie effect.  My question is if the effect will change the wavelength of light absorbed compared to the wavelength it should based on the spectrochemical series

I think the stability achieved due to chelating effect is already accounted for in the spectrochemical series, so that series can be used as a correct basis for deciding the wavelength absorbed by the complex.


text:
Is there a polar liquid with a high electrolysis potential?
I'm trying to use $\ce{Na2SO4}$ for intercalation of sulfate ions in van der Waals stacked materials (as anodes). With water as the solvent, there is some indication of hydroxyl radical forming due to the applied potential being larger than the standard potential of water electrolysis (1.23 V).
I am aware that hydroxyl scavengers such as ascorbic acid can be used to, well, scavenge hydroxyls. Yet I'm wondering if there is a solvent for $\ce{Na2SO4}$ that would only decompose at a higher potential difference.

Either you won't have electrolysis at all, or you will have products of electrolysis, there is no third option. What kind of products would you want?

Thanks for your input Ivan. I've changed the details of the question to more accurately describe what I'm trying to achieve.

So I guess you don't care much about sodium? Maybe you should change it to tetraethylammonium? That would make the salt better soluble in various organics.

I will definetly look into using lipophilic salts, thank you!