text:
Synthesis of 2-Phenylcyclopentanone from cyclopentanol
My best try was to use aldol condensation after the reduction of the substrate(cyclopentanol) but that gives me a product with a methylphenyl substituent(2-Phenylcyclopentanone), nor a phenyl only.

This can be done in quite a few ways. It would be helpful if you told us what level of organic chemistry you were in to provide context for what types of reactions you have likely learned about.

@zhe now I'm studying reactions in alcohols, I already know about Sn1,Sn2,E1 and E2 reactions as like any reactions involving reductions and oxidation of functional groups. Actually, I'm learning nucleofilic substitutions. If you give one or more ways of proceeding that I don't fully understand, I will study that.

Step 1 dehydrate cyclopentanol to cyclopentene with conc Sulfuric acid
Step 2 epoxidise cyclopentene with m- Chloroperbenzoic acid
Step 3 open the epoxide with phenyl grignard to give 2-phenyl cyclopentanol
Step 4 oxidise to required product

1) oxidise using your favourite method (pyridinium dichromate; metal catalyst and oxygen; hypochlorous acid etc.) to cyclopentanone.
2) Pd catalysis using a bidentate ligand e.g. DPPF, with a hindered base, coupling the cyclopentanone to PhBr, to give the alpha-arylated product.
See JACS 1997, 119, p.12382 by John Hartwig.  This method works with isopropyl ketones also


text:
Migratory aptitude of substituents for carbonation rearrangement in Pinnacol Pinnacolone rearrangment
Generally, the observed migratory aptitude for carbocation rearrangement is (as given in various books)  :
$$\ce{H- > Ph- >CH_3-}$$
But the fact is that the carbocation at the benzylic carbon is more stable.
This is counter-intuitive, so what is a reasonable explanation for this?

@JaideepKhare You aren't aware of the \ce notation for chemical equations. Do check out https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here before suggesting wrong edits

@PrittBalagopal Thanks, I wasn't knowing about that notation of chemical equations., (I am mostly active on Math SE.)I'll take care next time.

https://chemistry.stackexchange.com/questions/31599/migratory-preference-in-a-rearrangement-involving-carbocations

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sort: votes


text:
How can I calculate the percentage of urea in animal feed?
I'am having some doubts in the calculation  that concerns the urea percentage in animal feed using the method AOAC $967.07$

How can I calculate the urea percentage using the following data:

The formula is: $\text{Urea %} = \displaystyle\frac{1.0 \times A (\text{test solution}) \times 100}{A \text{ (standard solution})\cdot \text{mg (test portion aliquot)}} $
If ($A$) absorbance of test solution = $0.187$ and its correspondence concentration from calibration curve = $\pu{0.41 mg/5ml}$.
If ($A$) absorbance of standard solution = $0.502$ and its correspondence concentration from calibration curve = $\pu{1.09 mg/5ml}$
Sample is prepared by weighing $\pu{1g}$ in $\pu{500 ml}$.
What can be the urea percentage value?

I edited the question to make it more readable, did I write the formula correctly?

Thank you , but the formula must be divided by the mg(test portion aliquot) and not multiplied

Oh I'm sorry, I'm editing it again.

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text:
Is the t-butyl carbocation more stable than the benzyl carbocation?
Various authors have different views regarding stability order of the benzyl and t-butyl carbocations.
$$\ce{PhCH2+ ; (CH3)3C+}$$
In my opinion, resonance effect dominates, so the benzylic carbocation should be more stable. But in the other case, both the inductive and hyperconjugation effects are present, which stabilize the intermediate carbocation.
What should is the correct stability order? Especially when you consider them in SN1 reactions, and what their effect on their rate is.

Phenyl group have sp2 hybridized carbon and methyl group have sp3 hybridized carbon which is less electronegative than sp2 hybridized carbon.

Thus 1 phenyl group shows 1 +R and 1 -I effect and 3 methyl groups show 3 +H and 3 +I effects

Thus, electron donating nature of 3 methyl groups >> 1 phenyl group

And the correct order is (CH3)3C+ > (C6H5)C+

Anyone else think that solvent effects can potentially matter here?

@Zhe Yes, and the counter ion.

What do you mean by "stable"?

I am using a very simplistic quantum chemical approach of the following isodesmic reaction:

I have used Gaussian 16 Rev. A.03 and the DF-B97D3/def2-TZVPP level of theory. The summaries of the calculations are included below.
On this level of theory the depicted reaction has an energy change of $\Delta G = \pu{- 37.1 kJ mol-1}.$ Therefore one could assume that the 2-methylpropan-2-ylium cation is more stable than the phenylmethylium cation. These values were estimated at $T = \pu{298.15 K}$ and $p = \pu{1 atm}.$

One certainly can do more calculations, but it is, however, a start.

Calculation summaries

INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch2_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -270.5777462860 hartree
zero-point corr.             (ZPE):            +0.114824 hartree/particle
thermal corr.                  (U):            +0.120571 hartree/particle
ther. corr. enthalpy           (H):            +0.121515 hartree/particle
ther. corr. Gibbs en.          (G):            +0.085614 hartree/particle
entropy (total)            (S tot):              +75.561 cal/(mol K)
heat capacity (total)       (Cv t):              +22.578 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch3_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -271.4854244270 hartree
zero-point corr.             (ZPE):            +0.125049 hartree/particle
thermal corr.                  (U):            +0.131428 hartree/particle
ther. corr. enthalpy           (H):            +0.132372 hartree/particle
ther. corr. Gibbs en.          (G):            +0.093359 hartree/particle
entropy (total)            (S tot):              +82.110 cal/(mol K)
heat capacity (total)       (Cv t):              +23.831 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h10_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -158.4201368210 hartree
zero-point corr.             (ZPE):            +0.128849 hartree/particle
thermal corr.                  (U):            +0.134606 hartree/particle
ther. corr. enthalpy           (H):            +0.135550 hartree/particle
ther. corr. Gibbs en.          (G):            +0.101200 hartree/particle
entropy (total)            (S tot):              +72.297 cal/(mol K)
heat capacity (total)       (Cv t):              +20.521 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h9_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -157.5167928240 hartree
zero-point corr.             (ZPE):            +0.114033 hartree/particle
thermal corr.                  (U):            +0.120557 hartree/particle
ther. corr. enthalpy           (H):            +0.121501 hartree/particle
ther. corr. Gibbs en.          (G):            +0.083653 hartree/particle
entropy (total)            (S tot):              +79.659 cal/(mol K)
heat capacity (total)       (Cv t):              +20.965 cal/(mol K)

(Compiled with tools-for-g09.bash, which somewhat surprisingly works for g16.)
Optimised geometries

14
phch2_q1/b97d3tzvpp.freq.xyz
C          0.08663       -0.19347        0.09384
C          0.05691       -0.11475        1.46208
H          1.02702       -0.19311       -0.45239
H         -0.82919       -0.25883       -0.48899
C         -1.20323       -0.11721        2.16356
C         -1.22180       -0.03857        3.53614
C         -0.00341        0.04387        4.24119
C          1.24453        0.04873        3.58467
C          1.28552       -0.02918        2.21254
H         -2.12409       -0.18167        1.59151
H         -2.16067       -0.03932        4.07940
H         -0.02695        0.10566        5.32621
H          2.15902        0.11360        4.16439
H          2.23039       -0.02773        1.67720

15
phch3_q0/b97d3tzvpp.freq.xyz
C          0.00007        0.00000       -0.04694
C          0.00682       -0.00000        1.46143
H          1.01682       -0.00000       -0.44951
H         -0.51968       -0.88199       -0.43936
H         -0.51968        0.88199       -0.43936
C         -1.19805       -0.00000        2.17854
C         -1.20438       -0.00000        3.57251
C          0.00015        0.00000        4.27958
C          1.20614       -0.00000        3.57927
C          1.20606       -0.00000        2.18275
H         -2.14093        0.00000        1.63572
H         -2.14998        0.00000        4.10835
H         -0.00257        0.00000        5.36594
H          2.14940        0.00000        4.11910
H          2.15082        0.00000        1.64397

14
t-c4h10_q0/b97d3tzvpp.freq.xyz
C          0.00369       -0.00664        0.00284
C         -0.01764        0.03020        1.53527
C          1.44139        0.03012       -0.52791
C         -0.74698       -1.23338       -0.52790
H         -1.04360        0.03406        1.92005
H          0.49077       -0.85020        1.94859
H          0.49182        0.92075        1.92008
H          1.46219        0.03417       -1.62347
H          1.97408        0.92055       -0.17569
H          2.00042       -0.85041       -0.18649
H         -0.76068       -1.24961       -1.62346
H         -0.26410       -2.15775       -0.18616
H         -1.78452       -1.24933       -0.17586
H         -0.51485        0.89146       -0.36388

13
t-c4h9_q1/b97d3tzvpp.freq.xyz
C         -0.00184       -0.00056        0.03904
C          0.01681        1.45949        0.05366
H         -0.46367        1.80346       -0.88071
H          1.00698        1.90257        0.15089
H         -0.66770        1.83073        0.83269
C          1.25179       -0.74789        0.02785
H          1.76765       -0.48991       -0.91708
H          1.93393       -0.35706        0.79738
H          1.13840       -1.82916        0.09679
C         -1.27605       -0.71190        0.03180
H         -1.24998       -1.55839       -0.66820
H         -1.35257       -1.20345        1.02233
H         -2.15046       -0.07921       -0.11824

(g09.chk2xyz does not work for g16.)

Can you elaborate the results of the calculations? I'm finding it hard to understand the output data.

@ApoorvPotnis Not really. The interpretation of the calculation results is the actual answer. The rest only gives the details for each molecule calculated, which are necessary to reproduce the reaction enthalpy. This is just additional data, if someone decides to run this as a practise calculation, so that they can compare their calculated data. And it's a measur of good manners to include these details.

Is the trimethyl carbocation more stable than the benzylic carbocation?

There are a number of approaches we can take to try and answer this question.  We'll start by first comparing solvolysis rate data to see which carbocation is more stable in solution, and then we can look at thermochemical data to see how the carbocation stabilities compare in the gas phase.
Solution Stability
By comparing the rates at which two compounds solvolyze we can infer which compound leads to the more stable carbocation. For example, allyl chloride solvolyzes ~8.5 times faster than i-propyl chloride (1) in agreement with the idea that the allyl carbocation is more stable than the 2-propyl carbocation. Of course, reactions must be run under conditions to minimize any non-solvolytic pathways ($\mathrm{S_{N}2}$) and insure that an $\mathrm{S_{N}1}$ mechanism is operating.
Jones further reports (1) that at $\pu{45 ^\circ C}$ in 50% $\ce{EtOH}$, t-butyl chloride solvolyzes almost 20,000 times faster than i-propyl chloride.  This is in accord with the expectation that a tertiary carbocation is more stable than a secondary carbocation.
\begin{array}{|c|c|c|c|} \hline
\ce{R-X} & \mathrm{k_{rel}} \\ \hline
\ce{iPr-Cl} & 1 \\ \hline
\ce{tBu-Cl} & 1.76 \times 10^4 \\ \hline
\end{array}
Later in the book (2), we see that benzyl chloride solvolyzes 145 times faster than i-propyl chloride. at first glance this seems to tell us that (using isopropyl chloride as a common reference point) t-butyl chloride solvolyzes ~120 times $\left(\frac{1.76 \times 10^4}{145}\right)$ faster than benzyl chloride.
However this solvolysis is run at slightly lower temperature ($\pu{25 ^\circ C}$) and in pure $\ce{EtOH}$.
\begin{array}{|c|c|c|c|} \hline
\ce{R-X} & \mathrm{k_{rel}} \\ \hline
\ce{iPr-Cl} & 1 \\ \hline
\ce{PhCH2-Cl} & 145 \\ \hline
\end{array}
If we were to raise the reaction temperature to $\pu{50 ^\circ C}$ (supply more thermal energy to the reaction), this would tend to decrease the difference in relative rates.  Similarly, since the dielectric constant of water is greater than the dielectric constant of ethanol, and since a higher dielectric constant facilitates ionization, if we were to rerun the second set of reactions in water-ethanol, both reaction rates would be enhanced and the difference in relative rates would decrease. So both the reaction temperature and solvent dielectric effects operate in the same direction; if we were to rerun this second set of reactions under conditions identical to the first set of reactions we would expect the relative rate to be something less than 145. If the relative rate for the second set of reactions is really 100 than t-butyl chloride would solvolyze ~176 times $\left(\frac{1.76 \times 10^4}{100}\right)$ faster than benzyl chloride. If instead of 100, the relative rate for the second set of reactions is really only 10, then we would estimate that t-butyl chloride solvolyzes ~1,760 times $\left(\frac{1.76 \times 10^4}{10}\right)$ faster than benzyl chloride.
In any case, the t-butyl chloride solvolyzes faster than benzyl chloride, suggesting that the t-butyl carbocation is slightly more stable than the benzyl carbocation in solution.
Gas-Phase Stabilty
Let's examine the following gas phase reactions.
$$\ce{t-Bu-H -> t-Bu^{+} + H^{-}}$$
$$\ce{PhCH2-H -> PhCH2^{+} + H^{-}}$$
We are looking for the energy difference between them, so when we subtract them the $\ce{H^{-}}$ term cancels out. NIST provides the standard heat of formation of gaseous isobutane as ~$\pu{ -32 kcal/mol}$, while that for gaseous toluene is ~$\pu{ 12 kcal/mol}$. Using this information along with the thermochemical data provided in this answer by user55119 for the corresponding ions leads to an estimated difference in stability of ~$\pu{ 6 kcal/mol}$ (~$\pu{13 kcal/mol}$ if we use $\pu{162 kcal/mol}$ as the heat of formation of the t-butyl cation; see user55119's comment below) favoring the t-butyl carbocation.  The same general result as we found above in solution, now the magnitude is larger since there is no solvent to stabilize the ions in the gas-phase.
References

Maitland Jones, Jr, In Organic Chemistry; Third edition, W. W. Norton & Co.: New York, NY, 2005, p. 585 (ISBN: 978-0-393-92408-4).
Maitland Jones, Jr, In Organic Chemistry; Third edition, W. W. Norton & Co.: New York, NY, 2005, p. 658 (ISBN: 978-0-393-92408-4).

In you last paragraph, did you mean isobutane (-32 kcal/mol). That would bring the difference down to a ~6 kcal/mol difference. But fear not, t-butyl cation has also been reported as +162 kcal/mol which gets you back to a difference of ~13 kcal/mol. BTW: The gas phase equilibrium between PhCH2Cl + t-butyl cation and PhCH2+ and t-butyl chloride has been studied. It slightly favors t-butyl cation. Nice analysis though, ron.

@user55119  Yes, I meant isobutane. Thanks for catching that!

Gas phase measurements give:
\begin{align}
\Delta_\mathrm{f} H^\circ (\ce{PhCH2+}) &= \pu{+219 kcal mol-1} \\
\Delta_\mathrm{f} H^\circ (\ce{t-C4H9+}) &= \pu{+169 kcal mol-1}
\end{align}
The tertiary-butyl cation is seemingly more stable than the benzyl cation in the gas phase. These data do not speak to the condensed phase [1].
Addendum (12/14/2017): The latest data for the gas phase heats of formation of these two cations is here along with other radicals and cations. I thank G. B. Ellison, University of Colorado, for these data.
\begin{array}{llcclc}
\hline
& \text{Radical} & \Delta_\mathrm{f} H_{298} (\pu{kcal mol-1}) & \qquad & \text{Cation} & \Delta_\mathrm{f} H_{298} (\pu{kcal mol-1}) \\
\hline
\text{methyl} & \ce{CH3} & 35.06 \pm 0.07 & & \ce{CH3+} & 261.9 \pm 0.1 \\
\text{$t$-butyl} & \ce{C(CH3)3} & 11.9 \pm 0.2 & & \color{red}{\ce{C(CH3)3+}} & \color{red}{166.4 \pm 0.7} \\
\text{benzyl} & \ce{C6H5CH2} & 50.5 \pm 0.2 & & \color{red}{\ce{C6H5CH2+}} & \color{red}{217.6 \pm 0.2} \\
\text{tropyl} & \ce{C7H7} & 66.5 \pm 0.3 & & \ce{C7H7+} & 210.0 \pm 0.3 \\
\hline
\end{array}
Reference

Jo Anne A. Jackson, S. G. Lias, P. Ausloos, J. Am. Chem. Soc., 1977, 99 (23), pp. 7515–7521. DOI: 10.1021/ja00465a020.

Aside from the stability difference that we seek, won't the different numbers of carbon and hydrogen atoms in the two compounds also influence the heat of formation?  For example, the heat of formation of hexane is ~ -200 kJ/m while octane is ~ -250 kJ/m.  Is octane really 50 kJ/m more stable?

Benzylic is more stable because of the priority given to resonance. Like when inductive effect, hyperconjugation and resonance occur together, resonance is given preference then hyperconjugation and then inductive effect. In this case it's resonance.
Moreover the explanation of aromaticity also favours this answer.

That is not the correct conclusion.

The idea is not to dismiss hyperconjugation just because it does not fit well with valence bond structures.  It is the valence bond structures we render that must be adjusted to reflect nature, not the other way around!

From my knowledge, the benzylic carbocation should be more stable. This is due to the fact that it is aromatic as well. Aromaticity further enhances stability.

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text:
In a mixture of NaOH and Na2CO3, why doesn't phenolpthalein change its color until Na2CO3 is half neutralised?
Let us consider a mixture of $\ce{NaOH}$ and $\ce{Na2CO3}$. Based on what I have understood, double titration works in the below manner :

Phenolpthalein turns pink when added to the mixture. As $\ce{HCl}$ is added, the pH of the mixture keeps reducing and phenolpthalein becomes colorless when the mixture becomes of a pH less than $8.3$ ie. at a stage when $\ce{NaOH}$ is completely neutralised and $\ce{Na2CO3}$ is half neutralized. Then, we add methyl orange which makes the solution yellow. As we add more $\ce{HCl}$, the pH reduces and hence, methyl orange turns red when the neutralization of $\ce{Na2CO3}$ is complete.

My questions are:
a) When phenolpthalein is added, I have read elsewhere that $\ce{NaOH}$ reacts with $\ce{HCl}$ first. In that case, phenolpthalein should make the solution colorless as $\ce{NaOH}$ neutralizes with $\ce{HCl}$ to form $\ce{NaCl}$ which has a pH of $7$ (not between $8.3$ and $10.5$). Why doesn't phenolpthalein change its color until $\ce{Na2CO3}$ is also half neutralized? Is it because the reactions take place simultaneously?
b) When methyl orange is added, $\ce{NaHCO3}$ is neutralized to $\ce{NaCl}$. But pH range of methyl orange is 3-4.5, but the solution obtained is neutral. How does methyl orange change its color, then?I think it is because of the $\ce{H2CO3}$ ($\ce{H2O + CO2}$) that is formed along with $\ce{NaCl}$, which makes the solution slightly acidic. Is that a correct explanation?

Using $\ce{ insert chemical here }$ will save you time and keystrokes. Not to mention it looks neater.

Phenolphthalein changes color only above a $\text{pH}$ of about 11 or 12. So it shows a change only after quite alot of base is added.

a) That's right, $\ce{NaOH}$ reacts first. But you are wrong in thinking that $\ce{NaCl}$ has a pH of 7. In a way, $\ce{NaCl}$ has "no pH at all". It just sits there and does nothing. It is other component(s) of the solution that determine its pH. At that point, the other component is $\ce{Na2CO3}$, which is alkaline due to hydrolysis. As the titration progresses, it turns to $\ce{NaHCO3}$, hence the change in pH.
b) Well, $\ce{H2CO3}$ would indeed make the solution slightly acidic, but not enough so to trigger the color change in methyl orange. It is $\ce{HCl}$ that does the job. After the neutralization is completed, one tiny extra drop is enough.


text:
pH level in human body
For my alkaline diet, I need to maintain a pH that's above 7, I need to achieve the objective like this:
I have eaten 300 g of grapefruit (pH = 13), and eaten 70 g of chocolate (pH = 5,5). What pH level have got my body from both these products?

please read at least 1st paragraph of https://en.wikipedia.org/wiki/Alkaline_diet

please note grapefruit is acidic, pH between 3 and 4, according to publicly available info.

unfortunately this image stating grapefruit pH 13 is wrong, see https://scholar.google.com/scholar?q=grapefruit-pH&hl=en

I wonder why don't you believe your chemical sensors in the tongue, that tell you, by sour taste, that grapefruit is acidic. Then, probably your bad or good feelings are caused by something else than acidity / pH.

Human blood maintains a pH of about 7.3 very consistently. For example if the pH of your blood drops to even 7.1 you're either dead or looking at a ventilator. Maintaining a blood pH not reasonable or practicle, your body already does it automatically.

Additionally, the body has several of mechanisms to maintain homeostasis in the blood and extracellular fluid. So what you consume does not have the final effect on pH of your body as this is tightly regulated around the 7.35-7.45 pH range.
The most important way that the pH of the blood is kept relatively constant is by buffers dissolved in the blood. The most important buffer system in the body is the carbonic acid-bicarbonate buffer:
$$\ce{2H2O(l) + CO2(g) <=>  H2CO3(aq) + H2O(l) <=>  H3O+(aq) + HCO3-(aq)}$$
Other organs help enhance the homeostatic function of the buffers. The kidneys help remove excess chemicals from the blood,  It is the kidneys that ultimately remove (from the body) H+ ions and other components of the pH buffers that build up in excess.
The lungs also play a big role to help control the pH of the blood. The increased-breathing response  helps to counteract acidosis (the pH-lowering effects)  by removing CO2, a component of the principal pH buffer in the blood vice versa.

This question has no answer, from two main reasons.
Chemistry does not work that way
It's impossible to calculate resulting pH of solution obtained by mixing known volume or mass or amount of substance of solution A with pHA and known amount of solution B with pHB without knowing composition of substances in those solutions, and their acido-basic properties (pKa).
Biology does not work that way
Your body has balance systems to preserve proper internal parameters (homeostasis). If you are in hot environment, you begin to sweat, etc. If you are in cold environment, you start to shiver, etc.
In you stomach is strongly acidic environment, because of hydrochloric acid produced by "proton pumps". In other parts of your digestive system, duodenum, the acid is intentionally neutralized and pH is kept slightly alkaline.
Another, maybe more important issues of your question have been discussed just below it.

Blood pH is very tightly controlled to be 7.35-7.45. Any deviation from this is going to result in serious medical issues. The alkaline diet you refer to is dangerous unscientific nonsense

@Waylander I fully agree, but you might've commented on the wrong post!

Your body's pH will stay the same! Thankfully our bodies are able to regulate our pH levels or else we would be in major trouble as bodily functions only work within set pH levels.


text:
What are appropriate techniques to ensure homogeneous solidification of an agar-based aqueous solution?
I am making a T1 weighted phantom for an MRI project. I'm trying to see if I can calculate how much a persons head heats up during an MRI scan. It consists of distilled boiling water, $\pu{7 g/L}$ agar, $\pu{10 g/L}$ $\ce{NaCl}$ and $\pu{1 g/L}$ $\ce{CuSO4}$. As the phantom cools it will get a jelly like texture. What I would like is for the phantom to be as homogeneous as reasonably possible when it solidifies. What do you think is the best way to do so?
I was thinking of having the mixture on a vibrating plate, or maybe a centrifuge or even just inverting the mixture at regular intervals. Any ideas would be greatly appreciated.

Neat. I think this question might not be well-received since you aren't asking for something specific for this problem, while simultaneously not asking the real question, which is how to make a uniformly-disperse solid colloid.

Thank you for that! Would you have any advice on making a uniformly dispersed solid colloid by any chance? :D Or should I make a new topic?

If it doesn't matter how quickly the phantom solidifies, you could use any aggressive mixing technique and then do a rapid cooling to lock the suspension in place.

Let me suggest two avenues :
Vertical set up à la potter's wheel: At school we once did something related to your aim: using an old $\pu{75 rpm}$ record player, obviously no longer needed, with a beaker containing previously molten bees wax in the centre.  Under constant rotation (rather than the revolution by a centrifuge), the surface of the wax formed a nice meniscus which frooze by the slow cooling. While I am not aware if the alread-gel particles would keep nicely floating in your suspension, until all is solidified, it may be worth to giv it a try.  Centering of the whole ensemble is important (unbalance and spill), especially if the sample to be generated should really become as large as a human head.  Scaling up appears possible (larger beaker / bucket, and potter's wheel).
Horizontal set up: If you were interested in a smalller sample, a roller mixer would be an alternative:

(screen photo)
A wide mouth bottle (like the still cylindrical $\pu{1 L}$ polymer centrifuge bottles used in biochem) would be filled by immersion with liquid and -- still immersed in solution -- closed with the lid to ensure absence of air bubbles in the bottle.  After external cleaning of the bottle, they would be layed along the rolls.  Subsequently, you would turn on and crank up the rolling and hopefully the gel would form sooner than the liquid dropping out from the bottles.  (One may orient the bottles with the lids facing you, and place a bucket underneath them to collect potential spill.)  You may find such a roller mixer in biochem groups cultivating their own cell cultures.

As the gel solidifies, couldn't the horizontal setup have issues with radial mixing (meaning from the long axis of the cylinder out to the circumference)?

Thanks a million for the detailed answer! They look like some really interesting methods to try!

@J.Ari Depending on the model of the mixers, some keep the bottles like on a conveyor belt (like in a market), inducing axial rotation only.  Other models include optional or even adjustable precession  motion which offers mixing in more than one direction -- as in the example above video, where the bottles contain air bubbles. (Except air bubbles could be trapped into the gel, i) rendering the phantom likely mechanically less similar to human tissue.  And more importantly, ii) changing the $t_1$ / relaxation times (since in this regard (f)MRT often is closer to 1H than 13C NMR spectroscopy).

@JamesBlackwell Maybe i) a brief degassing (e.g. some pump thaw cycles while using an ultrasound bath) of the freshly prepared solution prior to the addition of agar-agar and ii) admixing agar-agar without trapping much air would be beneficial to keep the solution, and eventually the forming gel, "air free".  I'm curious about the outcome, too.

Hi Buttonwood, I think a small amount of ethanol can be used as a degassing agent and it shouldn't affect the properties of it too much. I was also wondering about an autostirrer? I've seen some papers where they use one at 120RPM and allow it too cool very slowly. Mix while heated for 30min and then 30min as it cools. What do you think?

@JamesBlackwell If the gel still forms "well enough" in the presence of EtOH, fine. Without practical simultaneous experience of agar-agar gels and ethanol, pump thaws was my first idea.  In analogy to crystal growth, I expect the agitation may influence how well the gel nuclei are kept suspended, if they growth into more larger / smaller gel-centres (coarse / fine gel). You want to balance with "not too much disturbance" / catching air (if stirred in an open beaker, if there is a significant vortex -- depending on rpm, stirrer size/shape, progress of gel formation). It is an experiment.

@Buttonwood I don't have much experience of it either. My supervisor tried it before and got ok results with a basic setup of a kettle and stirring haha, but wants to refine the method now to make it more accurate. I've read papers where a mechanical stirrer is used at 120 rpm for an extended period of time. I'll be starting small scale experiments next week and I'll keep you posted on how things turn out!


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Lubricating distillation glassware with coconut oil or olive?
For the joints
Is there parts that don't need to be lubricated ?
How often should I lubricate if I plan on distilling water for a long time.

Often joints are greased only if either a) you distill under vacuum, b) you heat above say 130 or $\pu{150 °C}$, c) you distill something agressive (against glass), like strongly basic or acidic material, or something that over time may polymerise, d) you need to pertain flexibility of the joints during the distillation, like rotating around an axis -- for example you collect several fractions.  If so, you need something that is not liquid at the temperature range of application; otherwise, it potentially may drop into the inner of the distillation.  Hence there are several greases foreseen for such an application.
Otherwise (taking into account your profile name, too), especially if you aim to isolate otherwise volatile materials by steam distillation, you may seal the joints with sleaves of PFTE / Teflon.  Nothing will drop into your distillation; but after the concluded distillation, start to dissamble carefully the still warm glass ware.

(picture source)


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Repairing acetone damage and discoloration to plastic
I used acetone on the back of my monitor to get rid of old stickers from LED strips. However it severely damaged my monitor. Anyway to undo this or repair it?

What do you mean by "severely damaged" If it's only plastic damaged then it's like non-issue, but if electronics got damaged then it's not the site for it?

Its the plastic, was wondering if I could reverse my mistake, but seems that i have to respray it

The short answer is no.

Possible duplicate of Does acetone eat through Thermoplastic polyurethane and Polycarbonate

Acetone is quite good at dissolving polymers. Do not use acetone on plastic before checking out whether that plastic can withstand acetone.
Let it dry for a couple hours. Try turning it on again. If it does turn on, good. If not, you're pretty much screwed.
However from what I can see from the picture you didn't really damage anything vital. You can try to polish the rough surface again, but I doubt you would get any good results.
If it does turn on, I would cover up the ugly parts with black tape or something and be happy I don't have to buy a new screen.

The monitor works fine indeed, was hoping to fix the ugly back now, i used nail polish remover, i read that the white stains could be some oils, it's weird that it is only white on the sides while i used it in the middle as well.

@Wybrem you probably have a white material under your black material. I would refrain from doing any more abrasive or dissolving work and count my blesses to be honest.

Yee seems to be that, I'm going to respray the monitor as I can't live with this, thanks for the help :D

@Wybrem if you feel your question has been answered you should check that answer as "accepted" so others know the question has been answered and doesn't need attention anymore.

To understand what to do you have to understand what the plastic is and what has happened. typically cases for electronics are a fiberglass-ABS composite resin. This is a cheap mixture that has a high mechanical resilience to fracture and reduces combustibility.
When the plastic is cast at the factory it is forced into a mold and rapidly cooled. This process causes the polymer molecules to retain a high degree of internal stress.  The part however is smooth because of the surface of the mold.  When you use acetone on ABS, you soften is and allow these polymer molecules to partially relax the stress resulting in a rough surface as the residual stress pulls molecules together creating a rough surface (this is a bit simplified). Also you have dissolved and wiped away the dyes that make the plastic black at the surface.
To restore the finish you can attempt to mechanically buff the surface to make it smooth again but the underlying fiber glass may be an issue. Using nail polish may be another way but this is more to make it look less bad than good again.

I think ethanol might be a better option in the future! That's some pretty amazing dissolving power though. As Fl.pf has suggested, black tape would help cover up the worst of the damage.

Black permanent marker might look marginally better than black tape.

If you want to see "amazing dissolving power" try a similar experiment, but substitute acetone for dicloromethane.... many a mechanical pencil has been lost in my lab, melted by my messy ways

Thanks guys, I'ma repaint the monitor. I removed the rest of the sticky stuff with WD-40.. This is what it looks like now https://puu.sh/vYpcx/5c17f73cf8.png

@Wybrem It is similar like with a suit, a good pair of trousers, a fine dress, a couch: try first at a spot where if something goes wrong won't hurt (this) much.  (At least for three first you may get a little patch to try out safely.)  For polymers, you often see some marks (like PE, PP, for example) that you find in guiding resistance charts like http://www.plasticsintl.com/plastics_chemical_resistence_chart.html again.

With WD-40 you have a good chance of crazing the surface even more. Use isoprop or ethanol.

I was taking stickers off a casetter player and I thought nail polish remover would do a better job than alcohol in taking off the residue (stupid I know now). After like 3 coats of headlight restorer the plastic went back to its clear state and i’m so happy that it worked out. I have no idea how it would do on other plastic or what residue headlight restorer would leave but I used it on clear plastic. It’s worth a try though.

I have read elsewhere that toothpaste or Brasso might do the job. But have not tried it. I like the sound of automobile headlight restorer better.

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


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Solving a kinetic rate law
Here is a problem I just don't see how to get around... The following equation describes the oxidation by $\ce{O_2}$ of a phenolic compound, catalysed by an enzyme $\ce{E}$.
$\ce{4R–OH + O_2 ->[r] 4R–O. +H_2O}$
$r=k\frac{\ce{[R-OH]}}{\ce{K_M +[R-OH]}}[\ce{E}][\ce{O_2}]$
Rate $r$ is the speed at which $[\ce{R–OH}]$ is consumed. It tells us that our reaction follows a Michaelis-Menten kinetic law with respect to our substrate $[\ce{R–OH}]$ and first order kinetics with respect to oxygen.
First, we are asked to find a way by which the rate equation could be simplified by approximation.
We are told that initially,  $[\ce{R–OH}] = 10 mol/m^3$ and that we want a $95\%$ conversion of that reactant. We also know that $K_M =0.04 mol/m^3$. Therefore $[\ce{R–OH}]$  can vary between being $250$ times and $12.5$ times greater than $K_M$, so we could approximate the problem by getting rid of $K_M$ which would leave us with the following rate:
$r=k[\ce{E}][\ce{O_2}]$. Since $[\ce{E}]$ is constant, we could write $r=k'[\ce{O_2}]$. (This sort of trick has been used before in my chemical engineering class.)
The next question is the one I am struggling with.
We are asked to use our new kinetic rate law to write down the material balance equation expressing how $[\ce{R–OH}]$ would changes with time in a batch reactor (perfectly stirred, no input or output), then solve for $[\ce{R–OH}]$.
So I would be inclined to write $\frac{d[\ce{ROH}]}{dt}=-r=-k'[\ce{O_2}]$ but then I don't see how this can be solved since we have two different variables.
From the stoichiometry we can see $\frac{d[\ce{ROH}]}{dt}=4\frac{d[\ce{O_2}]}{dt}$ therefore $\frac{d[\ce{O_2}]}{dt} = -\frac{1}{4}k'[\ce{O_2}]$ and this can be integrated to give us $[\ce{O_2}]=[\ce{O_2}]_0e^{-k't/4}$. But how can the rate law be solved for $[\ce{R–OH}]$ ?
Any insight into this problem would be hugely appreciated.

Hello, and Welcome to StackExchange. Looks like a pretty good question for a first post. You can write chemical equations using \ce. For example, $$\ce{NH4OH <=>[k_1][k_2] NH4+ + OH-}$$ would give: $$\ce{NH4OH <=>[k_1][k_2] NH4+ + OH-}$$ Good luck!

In your simplification of the rxn. equation, you are assuming that the change in [R-OH] negligibly impacts the rate, so your mass balance would reflect that with the new rate law. Have you tried writing an expression for the concentration of $[\ce{O_2}]$ in terms of R-OH, knowing that overall mass is conserved?

Is there a reason why you cannot relate the concentration of oxygen with the concentration of the alcohol?


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In the reversible reactions of acyl substitution, how do backward reactions happen by being against the forward reaction drives?
Consider this mechanism for acyl substitution by nucleophilic addition-elimination:

To start with, the nucleophile having a lone pair of electrons attacks carbon (because of oxygen being more electronegative?) and forms the conformation/arrangement of atoms as shown in the image.
But (there is an equilibrium sign) doesn't this indicate that the atoms are going back to the initial state against their original tendency which was responsible for the formation of the product in the first place? For example, in second step, what would make a nucleophile attached to a positive carbon site, leave to form back the reactants of step one?
Which force would make nucleophile again leave the carbon?
Edit (25/5/17): Above reaction mechanism seems to be supported by "isotope labeling experiments". See Nucleophilic acyl substitution -- Wikipedia.
Edit (26/5/17): For more clarification:
(These comments were present below the deleted answer of Berry Holmes)
Are the intermediates given in any mechanism experimentally verified by any of the techniques (though they may be having short life span?)? I will search for it, if you know it, it will be helpful.
I have found one such experiment: http://pubs.acs.org/doi/abs/10.1021/ac60221a003?journalCode=ancham ; I am still seeing the reasoning behind every action of atoms in the reaction using even your answer, I will say you after completion of that process.
Let us consider only nucleophillic addition, i.e. first two steps; nucleophile in the second step is the only product, there may be energy for it due to temperature, this kinetic energy can be used to collide with either of the reactants in the first step. If it collides with nuclephile containing compound :Nu-H, Nu+ in second step may form bond with :Nu-H from by being attached with carbon of second step, this seems to not form the first reactants back again.
If we consider another possibility of collision of second reactant with the acyl compound, the attack of Nu+ of second step goes to oxygen of acyl compound, here I am now not seeing any easy way of the initial products being formed again. Even if energy profile diagram of the above reaction has less energy of activation for the back reaction, which could be easily attainable from the initial given energy for forward reaction, I am not seeing any way of this energy being used to form back the reactants.
There seems to be one possibility for the compound in second step to produce the reactants back. The negative charge in second step of O- might produce tendency for its lone pair to form double bond with carbon at the center, this needs to make -(Nu+-H) group leave the central carbon. We need to now explore on how -(Nu+-H) group is a better leaving group than other groups, and even on the time taken by oxygen's lone pair to form double bond, to know on whether this happens before H of -(Nu+-H) attaches to L.

Image Source: Solomons, Fryhle and Snyder's "Organic Chemistry" [International Student Version (India) 11th edition (Page No. 784)].

Does anyone know who gave acyl substitution mechanism? I am searching for it, if you know it and can say it, it will be helpful. Knowing the person might help in reading his reasoning for the above mechanism.

I am also searching on how reactions in mechanism are tested as either reversible or not. If you know it and can say it, it will be helpful.

You can award the bounty now if you wish, by clicking on the +50 button below the up/downvote buttons. I cannot assign a bounty for you.

@orthocresol: Thank you for replying. It seems that I can only award it after 7 hours, as indicated when I try to award it. I will do it later itself.

You seem to be confusing a number of different concepts here. Let's try address a few of them.
1) As to the original question, addition-elimination reactions are reversible (indeed all reactions are reversible in some sense, via the principle of microscopic reversibility, it's just a matter of scale), depending on the partners being exchanged. If, for example, you were to add water to an acyl chloride (that is, "L" in your figure is Cl and "Nu" in your figure is OH), the reaction is, for all intents an purposes, irreversible. The resulting Cl is such a bad nucleophile that it's much happier just being a $\ce{Cl-}$.
If, instead, water was being added to a methyl ester (that is, "L" is OMe) the reaction would be reversible (also incredibly slow without a catalyst, but that's a separate issue).
The best way to look at this is to look at the tetrahedral intermediate. It is unstable and it will collapse back down to a carbonyl. It must expel one of the groups bound to the carbonyl carbon in order to do that. If there is an obvious best choice (that is, if there is one path that lowers the total system energy much more) for leaving group (like $\ce{Cl-}$) then that group will always be the one expelled and the reaction really only goes in the direction in which the leaving group has left. If there isn't an obvious choice (like between OH and OMe, where the energy of the system is similar regardless of which one is bound and which is free), either one could be expelled and the reaction is reversible.
It's important to keep in mind that chemicals don't have an agenda. They don't always go in the direction of your reaction arrows. In reality, chemicals try to react in all possible ways but only those ways that are accessible according to the physical realities of thermodynamics are successful. If only one product is thermodynamically favorable, the reaction will appear irreversible. If many products are thermodynamically favorable (and the reaction barriers are low enough), the reaction will appear reversible.
2) Yes, the carbonyl carbon is the most electrophilic site because the more electronegative oxygen is drawing electron density away from that carbon.
3) In addition to evidence from isotope labelling studies, as you mentioned, there is considerable evidence for this mechanism in the form of molecular beam studies, in which a microscopic jet of nucleophiles is shot at different angles at the acyl group.
I'm not sure I understand what you are saying after the paragraph with the ACS paper, but hopefully the above answers your questions.

Thank you for the answer. I might need to learn QM and particle physics to know exact and precise reasoning on each action of atoms in the reaction. I have now stopped thinking on this.

I will come back to this later, after reading physics.


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In a hydrogen fuel cell stack, is pure oxygen required?
I have been researching making a hydrogen fuel cell to produce electricity. However, since pure oxygen is expensive, is pure oxygen needed for the dry fuel cell stack or can the oxygen in the air be used.

You will have to clean the air supply, dry the air supply, and have to deal with the low $\ce{O_2}$ concentration of normal air.

Clean air, without sulfur compounds or other chemicals that "poison" the catalyst, should be sufficient.


text:
What is the role of NaCl in the functionalization of gold nanoparticles?
In the functionalization of gold nanoparticles (GNPs) through the sulfur-metal bond, this article (among others) states that $\ce{NaCl}$ is slowly added in the GNPs solution after addition of thiol compound.
What is the role of $\ce{NaCl}$ in the functionalization step?

$\ce{NaCl}$ acts as a charge suppressor so that the negatively charged gold nanoparticle and a negatively charged ligand (with thiol) are not longer charge repelled and can experience a chemical bond. $\ce{NaCl}$ is easily broken into $\ce{Na+}$ and $\ce{Cl-}$ so that the $\ce{Na+}$ can pacify the negative charge on the surface of the gold nanoparticle to become less negative and approach zero charge. Other higher charged salts such as $\ce{Mg^2+}$ impart even greater pacification.
Here are some good references:

J Liu, Y Lu. Preparation of aptamer-linked gold nanoparticle purple aggregates for colorimetric sensing of analytes. Nature protocols, 2006. nature.com

B Du, Z Li, Y Cheng. Homogeneous immunoassay based on aggregation of antibody-functionalized gold nanoparticles coupled with light scattering detection. Talanta, 2008. Elsevier. From this last ref:

in $\pu{10 mM}$ phosphate-buffered solution (pH 7.0) containing $\pu{1.0 mmol L−1} ~\ce{NaCl}$ at room …


text:
How can I emulsify Beeswax with Aloe Vera Gel?
I was looking thru ingredients for non-silicone based makeup primers and two ingredients always seem to be used: aloe vera gel and some type of wax, like jojoba wax, plus some type of oil to thin out the consistency like sweet almond oil or apricot kernel oil.
I thought it'd be interesting to see if it'd be possible to see if I could make my own, not least the companies that make them charge an arm and a leg for them.
My question is: what procedure and chemicals would you use to emulsify aloe gel with beeswax to make a cosmetic?
I vaguely remember that borax is used to make cold cream, maybe that'd come in useful?

Aloe vera gel is water based, whereas beeswax is, well, wax-based. As they say, "oil and water don't mix" -- at least, not on their own.
So, the trick is to make them willing to mix. To do this, you need what's called an emulsifier. These are chemicals (natural or synthetic) that have both oily and watery parts to them, and thus are able to make oil and water play nice(r) with one another.  Soap is one well-known, everyday example of an emulsifier; in particular, it's an example of a surfactant.
Anyways, you need an emulsifier. If you're looking to stick with natural products, probably the most readily available option is lecithin, commonly sourced from soy but also available derived from other plants such as sunflower.  While I don't have detailed recommendations, I would melt the beeswax together with lecithin and the aloe vera gel over very low heat, mixing thoroughly, then allowing the blend to cool.  You'll have to experiment until you find a ratio among these that gives the consistency you want in the final product.

A great way to help is adding corn starch, which is an overall benefit. I blended various carrier oils, aloe vera, corn starch, cocoa butter (small percentage), and shea butter (small percentage). My product took a few times of trial and error! After 4 different times of adding ingredients, it finally blended well.


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When recovering Copper from Copper Chloride via Aluminum, why can't we use Sodium instead to make Sodium Chloride?
About a week ago in my Middle School Science Class, we used an aluminum washer and some copper chloride, which we had gotten from creating circuit boards, in a closed system to recover the copper (if it was possible). We resulted in getting some copper flakes, but as a consequence the copper was replaced by aluminum in the chloride solution. So technically it is still a waste material that can't be used in everyday life, despite having recovered the copper.
So I got thinking: If Salt, or Sodium Chloride, is Sodium and Chlorine, and is much more useful than Aluminum Chloride or Copper Chloride since we need it in our everyday diets, what prevents us from using it to recover copper from chlorine?
I asked my teacher what she thought about the subject, and she explained that apparently sodium is very reactive when not bonded to anything and so is hard to come by naturally without extracting it from something else. I gathered that Aluminum, as reactive as it is, is apparently more common to find, or cheaper to manufacture, than sodium is. This was an unsure answer, as the costs and waste produced when manufacturing aluminum vs. manufacturing sodium were not looked up or explained.
However, in an earlier lesson we had learned that manufacturing aluminum produces a lot of hazardous waste, and so does not have a great impact on the environment to make. I do not know about sodium and its costs and waste production.
With that information provided, and a consideration of the environment at hand, would it be better to consistently use sodium in the recovery of copper from copper chloride than aluminum? Consider all waste produced from manufacturing and all costs to create when answering, please.
Thank you.

Welcome to Chemistry! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

Go ahead, try it. All you have to do is obtain some metallic sodium, then when you add it to the water avoid side reactions with the water that will consume the sodium ... and possibly ignite. Can you use magnesium, maybe?

As your teacher mentioned, sodium is a very reactive material. She mentioned that it is hard to obtain it naturally as a result. However, this isn't quite the reason as to why sodium isn't used for recovery.
If we were to only consider the cost in energy, aluminum would be a poor choice.
The main process for production of aluminum metal requires a current of 400 kiloamperes and temperatures of 950 C. In contrast, the main process for sodium metal production requires a current of 30 kiloamperes and temperatures of 600 C.
The main reason would be because sodium is too reactive to be used. The sodium metal would react with the water and form sodium hydroxide. This reaction would be too fast and violent that you wouldn't be able to recover copper.
So your teacher was right in pointing out that sodium is too reactive.

What if the water was evaporated away so there was just solid CuCl? Water will evaporate at a much lower point than the melting point of Copper Chloride, and I would think it would be possible to boil it off. Would there be a problem in using sodium then?

Check that - The CuCl was brought to its melting point in addition, to make it a liquid.

@PyramLinum I am not sure. However, at that point there's too much waste.  I think that if your overall goal was to reclaim copper, you would be better off doing an electrolysis on the copper chloride: http://www.docbrown.info/page01/ExIndChem/electrochemistry06.htm
That way, you obtain chlorine, which has industrial uses, and the desired copper.

Alright, I guess that works. Thank you for helping me understand this.


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White mucous substance forming around iron paper clip during electrolysis
I'm using a graphite cathode and an iron paper-clip anode, in a saltwater solution, basically to produce rust.
I've done this before with no complications, but this time I noticed a white, mucous material forming around the paperclip, quite late into the process.
I had eaten about half-way through the clip when I started noticing the cloudy white stuff. That suggests that whatever is producing it, is at the core of the clip.
If I try to move it, the mucus immediately detaches and sinks to the bottom of the jar, where it stays above the iron hydroxide.
Everything else seems to be in order, colour-wise.
Another thing is, if I shake the mucus off, the paperclip starts off clear, before turning black again, as it should be.
Does anyone have a clue of what this might be?
I know zinc oxide is white, but I'm pretty sure zinc and iron don't mix.
I've turned the power off, just in case it's something I -really- don't want to handle without protection.

Can you please add a picture? That can help us find out better. By the way, welcome to StackExchange!

Where did you get your water? If it's from the tap, then the question gets more interesting! Please describe the water, container, and as @PrittBalagopal mentions, if you can reproduce the experiment again, a photo of the stuff in question. What happens if you pour the water off slowly, does the stuff stay lumpy, or collapse into a very thin layer?

Apparently there's a character limit and I'm too lazy to make two posts, so here's a text image


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Dehydrobromination of 3-bromo-3-cyclopentylhexane and number of possible alkene products
I faced a question like this:

The total number of alkene products possible from the dehydrobromination of 3-bromo-3-cyclopentylhexane is:

The answer is given 5. But I only found 3.

3-cyclopentylhex-2-ene
3-cyclopentylhex-3-ene
(hexan-3-ylidene)cyclopentane

I can't find the other 2. Any hint?

Stereoisomers are counted apparently. Both 1 and 2 have (E)- and (Z)-isomers.

I agree with Orthocresol, do consider stereoisomers,sometimes the question doesn't mention it.

The structures (i) and (ii) have two isomers each,namely E and Z isomers.
For the structure (iii) there is a plane of symmetry (marked in the diagram) so it is not a geometrical isomer.
Thus Number of isomers=2 (of structure i) +2 (of structure        ii) + 1(of structure iii)= 5 isomers

Umm, plane of symmetry should not be considered here, as we are looking for geometrical isomerism. Instead you can say that two groups attached to the double bond are identical.

In fact, the molecule does not have a plane of symmetry there, since an ethyl residue (chain left-hand side) and a propyl residue (chain, right-hand side) are not identical. What you should be discussing are the identical substituents on the other side of the double bond.


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Reduction of hemiaminal by triethylsilane / boron trifluoride
I was trying to understand how the following reaction occurs (taken from a synthesis of homopipecolic acid[1]):

I thought that triethylsilane, $\ce{Et3SiH}$, would reduce the ester to an aldehyde. However, it turns out that the hemiaminal carbon has also been reduced with cleavage of the $\ce{C-N}$ bond. How does this happen?
References:

Chiou, W.; Chen, G.; Kao, C.; Gao, Y. Syntheses of (−)-pelletierine and (−)-homopipecolic acid. Org. Biomol. Chem. 2012, 10 (13), 2518. DOI: 10.1039/C2OB06984A.

My organic chemistry is rusty by now, but the full mechanism for the reduction itself is likely something along these lines:

In a "typical" lactone / ester, the alkyl oxygen can't be lost so easily, but in this case there's a nitrogen lone pair that can assist in that. The resulting iminium ion is reduced by the silane. The acid at the end will remove the remaining boron trifluoride.

Are you sure that carbmate nitrogen is available to assist this reaction?

Given the role of BF3, the carbamate doesn't have that much choice but to play along. Not that rusty. ;)

This is what I came up with.
The starting material first reacts with the Lewis acid - $\ce{BF_3}$-etherate. There are two sites which can attack the acid - the carbonyl of the ester, or the carbonyl of the amide in the protecting group. Considering the first step to be kinetically controlled (which it generally is), the sterically less hindered carbonyl of the ester group will react.
Now, the lone pair of the nitrogen plays its part, breaking the $\ce{C-O}$ bond, relieving the positive charge on the oxygen.
The iminium formed will be converted to the corresponding amine by the action of the triethylsilane.
Hydrolysis will lead to the carboxylic acid.
You can see the stereochemistry in the images.
Justification: Intramolecular rearrangement is faster than intermolecular encounters.


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Solubility curve for uranium hexafluoride (in any solvent)
I am doing a project and need a solubility curve or at least solubility data for uranium hexafluoride in any substance. It is impossible for water because $\ce{UF6}$ reacts violently with $\ce{H2O}$.
Is there any resource I could use for free to find this information?

You can use mhchem ($\ce{H2O}$) to write chemical formulas, it will save you a few keystrokes and it looks neat.

Ok. Thanks, I usually go on math SE or physics SE so I am not used to chem SE format.

No worries, this is a short help article for formatting on Chem.SE: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

"normal MathJax" still works too, but the extra \ce turns on the chemistry-specific stuff

I don't have the time to look through it, but did you already try DOI:  10.2172/4025868. Have you tried other databases yet? Please add them to avoid doubling the effort.

So far I have found that it is soluble in:

Bromine Pentafluoride
Chlorotrifluoroethylene Liquid Polymer, Dichlorotetrafluoroethane, Perfluorotributylamine, Perfluoroisopropyl ether, 1,2-Dichlorohexafluoropropane, hydrogen fluoride and more.
chloroform,  carbon  tetrachloride  and  fluorocarbon  solvents;
liquid chlorine and bromine; dissolves in nitrobenzene to form a dark red solu-
tion that fumes in air

I think the resource given by Martin (2) is the best. Resource 1 gives a solubility curve, but resource 3 doesn't. However, I think there may be data floating around on compounds mentioned by 3.

Wikipedia states that reacts violently with water* but is soluble in chloroform($\ce{CCl4}$), liquid chlorine/bromine and
dissolves in nitrobenzene.
This paper discuss solubility of uranium hexafluoride in some low temperature solvents.

Solubility of $\ce{UF6}$ has been studied in liquefied gases like phosgene,
  nitrous dioxide and sulfur dioxide below 273K. The solubilities were
  compared using real and ideal gas theory and found out that phosgene
  provide better solubility than sulfur dioxide. Solubility of $\ce{UF6}$ has
  also been studied on halocarbons and fluorinated halocarbons** all below
  273K.

** Perfluoroether is widely used as a solvent for $\ce{UF6}$. There are many papers which  discuss this topic and can be found on Google. The following is an abstract of one of the papers.

The polyperfluoroethers are compatible with uranium hexafluoride
  ($\ce{UF6}$) and are suitable for use in diffusion pumps and in
  mechanical vacuum pumps which rely on oil as both the lubricant and
  the seal. The $\ce{UF6}$ is soluble in all fluids with which it is
  compatible. Because a number of vacuum pumps in the BOP facilities of
  the GCEP plant employ these perfluoroether oils as the working fluid
  and have oil chambers which are large, questions have been raised as
  to the relationships governing the solubility of $\ce{UF6}$ in these
  materials and the maximum quantities of $\ce{UF6}$ which could be
  dissolved in these oils under credible accident conditions. This
  report summarizes these solubility relations and the interaction of
  the  $\ce{UF6}$ solubility and the pumping capability of this type of
  vacuum pump. It will be shown that, whereas the solubility of
  $\ce{UF6}$ in Fomblin Y25 fluoroether fluid under a
  $\ce{UF6}$ pressure of 760 torr and at the pump operating temperature
  of 1600F is about 500 g of $\ce{UF6}$ per liter of oil, the system
  controls are such as to isolate the system from the pumps before the
  quantity of $\ce{UF6}$ dissolved in the perfluoroether exceeds about
  10 g of $\ce{UF6}$ per liter of oil.

*


text:
Equilibriums and Overshooting
If we have some reaction $\ce{A(aq) + B(aq) -> AB(aq)}$  and right now we have less $\ce{A}$ and $\ce{B}$ then we would have at equilibrium, is there ever a time where the amount of $\ce{A}$ and $\ce{B}$ would overshoot the equilibrium of $\ce{A}$ and $\ce{B}$ if the reaction proceeded without any new sources of energy or reactants. In other words, will the concentrations of $\ce{A}$ and $\ce{B}$ always increase towards their equilibrium concentrations but never actually reach those points or will they sometimes overshoot that concentration a large amount (not just an amount that can be ascribed to chance) but then fall back towards the equilibrium concentration.

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https://chemistry.stackexchange.com/questions/15511/can-a-multi-species-system-oscillate-around-equilibrium?rq=1

Not really. Overshooting doesn't happen, rather the concentrations gradually approaches the equilibrium concentration as an asymptote.
You'll soon see what I mean.
As an example, for simplicity, lets consider:
$$\ce{P <=>[k_1][k_2] Q}$$
I will assume here that the forward and backward reactions are first order, so we get the rate to be:
$$\text{Rate}_{\text{forward}}=k_1\ce{[P]}$$
and
$$\text{Rate}_{\text{backward}}=k_2\ce{[Q]}$$
As a result the net rate would be:
$$\frac{d\ce{[P]}}{dt}=-k_1\ce{[P]}+k_2\ce{[Q]}$$
Let's take the initial concentration of $\ce{P}$ to be $\ce{[P]_0}$, and the initial concentration of $\ce{Q}$ to be zero.
Suppose at any time $t$, the concentration of the reactant is $\ce{[P]}$. Thus, $\ce{[Q]} = \ce{[P]_0} - \ce{[P]}$. Now, we get the rate law to be:
$$\frac{d\ce{[P]}}{dt} = k_2\ce{[P]_0} - (k_1 + k_2)\ce{[P]}$$
Integrating this from $\ce{[P]_0}$ to $\ce{[P]}$:
$$\int_{\ce{[P]_0}}^{\ce{[P]}}\frac{d\ce{[P]}}{k_2{[P]_0}-(k_1 + k_2)\ce{[P]}}=\int_0^tdt$$
Solving the integral, we get:
$$\frac{-1}{(k_1 + k_2)}\log_e{\left(\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}\right)}=t$$
$$\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}=e^{-(k_1 + k_2)t}$$
Finally, on simplifying it, we get our integrated rate equation as:
$$\ce{[P]}=\ce{[P]_0}\frac{k_2 + k_1e^{-(k_1+k_2)t}}{k_1+k_2}$$
Let's plot this on a graph to understand what this is actually saying:

(purple-$\ce{[P]}$ and orange-$\ce{[Q]}$)
You can clearly observe that the concentrations never overshoot the asymptotes (black dotted line). The asymptotes denote the equilibrium concentrations, so it's evident that the reaction tends toward the equilibrium concentration.

Then again, there is a number of clock reactions with oscillations which kinda may be interpreted as "overshooting".

In normal experiments vast numbers of molecules are involved, $10^{18}$ would not be untypical, and as a result the kinetic behaviour observed as equilibrium is approached is rather like that of a highly over-damped  mechanical system.  If, however, molecules could be observed one by one at each of many small time intervals then considering a very small sample it is possible that overshoot could be observed. As an example consider a cis-trans  isomerisation $\ce{A <=>}B $ that is initiated by a very short light pulse of perhaps femtosecond duration. Consider this only as possible  mechanism with which to start a reaction so that we might study it.
The molecule when excited into its excited state has a certain finite lifetime, its reciprocal is the rate constant. If we observe just a single molecule we cannot know exactly when it will react but only that if we repeatedly measure the times it remains in the excited state and generate a histogram of these times that an exponential decay is produced from which a lifetime can be measured. Now if we observe just a few molecules, say 10, it is possible, albeit with a small probability,  that all of them will react within a small time, say, 10% of the lifetime, thus we have a situation that in such a small time interval after the reaction starts that all the molecules are now products. The 'equilibrium' at this time is now product 10, reactants 0.  The product also has a lifetime with which it transforms back into reactants in just the same manner. And so depending on pure chance the s
'snap-shot' ratio of reactant to products can vary widely.
Of course the chance of all molecule reacting in the same short time-span is small, and if we observed the same molecules many times over the 'normal' equilibrium constant would be observed. The same is true if we observed very many molecules that reacted in the same short time interval as did our initial 10.  Thus this variability is not observed in practice because by observing over longer times and/or larger numbers of molecules the average properties are obtained.


text:
Number of F-Br-F angle of 90 degree in BrF5?
I faced a question like this:

The number of $90$ degree $\ce{F-Br-F}$ angle in $\ce{BrF5}$ according to VSEPR theory is:

The answer is given $0$ or $8$. I know $\ce{BrF5}$ has $\pu{sp3d2}$ hybridization. But there is no $90$ degree $\ce{F-Br-F}$ angle in this geometry due to the lone pair electron in $\ce{Br}$. But I find no other alternative structure to present $8$ $\ce{F-Br-F}$ angle of $90$ degree.
Any hints?

Please format your questions using mhchem. You've written over 38 posts, so by now you should be aware how to use proper formatting. Please respect other people's time by taking the trouble to make the post as readable as it can be.

There is no such thing as sp³d² hybridisation, that is the main and principle shortcoming of VSEPR. The answer, however, is https://en.wikipedia.org/wiki/Square_pyramidal_molecular_geometry.

Have a look at it's structure: https://i.stack.imgur.com/qDpYw.png

The starting structure would be an octahedron with 5 of the 6 positions occupied by an F atom and the last being the lone pair. Since the F is quite small, and VSEPR is "all about" lone pair repulsion, you can assume the lp would push the four equatorial F atoms up away from it (assuming it is in the south polar position) and so their angles would be less than 90° (both their angles between two adjacent equatorial Fs  and their angles to the (north) polar F). According to various internet sources (of questionable reliability) the angles are nearly 90° anyways: 86° and 85° respectively. The key thing would be to realize this is square pyramidal rather than a triangular bipyramid. I think it's arguable whether or not VSEPR would or would not "predict" the bond angles to within a couple of degrees, and that's the difference we're talking about here (between 85° and 90°), so I'd characterize this as a very poor homework "first principles" question. I think it would be in the "too close to call" category.


text:
Covalent and Ionic Bonds
Is sodium nitrate a covalent or ionic compound? This is a test question. I put the answer ionic compound but my teacher says that statement is false because it is only partially ionic as there is covalent bond between nitrogen and oxygen in the nitrate ion. Is that true?

Life is full of inadequate test questions. Bear with it. Sodium nitrate is a completely ionic, and at the same time a covalent compound, like a great deal of other salts.

So what will the answer be?

The answer will be whatever your teacher wants it to be, as is customary with inadequate questions.

Yup, stupid test question.

Is the test's language English? There might be subtle differences in terminology that may apply if you're translating - you might want to include the original text if that's the case. Though I have to admit I can't imagine how you could possibly give a better answer than "neither" or "both" - I certainly wouldn't say that it's fine to call it "covalent" while it's not fine to call it "ionic".

Related: Is pyrite (FeS₂) an ionic or a covalent compound?

And: https://chemistry.stackexchange.com/questions/40729/is-khf2-an-ionic-compound-or-a-covalent-compound

Yup the test was in English

Bonds are usually used to describe how electrons are shared. "Partially ionic bonds" are polar covalent bonds. Which is not the same as saying that a compound has different types of bonds in the whole compound. I think that your teacher gave you a bad question and a bad answer. If the question had asked whether all the bonds in sodium nitrate are ionic, then it would be a different story.
However, it is true that the bonds between the nitrogen and oxygen are covalent. Nevertheless, stating that sodium nitrate is ionic shouldn't be wrong.

If you are asked on a test "Is sodium nitrate ionic or covalent?", say "ionic" and if given the opportunity, justify your answer with electrical conductivity of molten sodium nitrate and aqueous sodium nitrate.
Chemistry: The Molecular Science :

Polyatomic Ions
  In addition to atomic ions such as Na+ and Cl-, ionic compounds often contain groups of atoms with a net electrical charge. Sodium nitrate is an example of an ionic substance that contains such a group.

The answer your teacher seems to have been looking for was to specify that the the sodium-nitrate bond is ionic and nitrate's nitrogen-oxygen bond is covalent.
This is under the assumption that the question was not multiple choice but was short answer. You may be able to contest the marking, but I wouldn't expect a change.

Yup, No changes. I argued but she was very rigid holding on to her answer. But the point is I got to know the true answer and explanation.


text:
Expression for an expanded configuration interaction wave function
I am trying to write down an expanded expression for the configuration interaction wave function. We know that the wave function $\Psi$ is a linear combination (LC) of configuration state functions (CSF), each CSF is a LC of Slater determinants, each determinant is an antisymmetrized product of molecular orbitals (MO's), and each MO is a LC of basis functions (BF). Without saying anything about the antisymmetrizer operator $\mathcal{A} $, are the equations below in general correct? What would the expanded wave function $\Psi$ look like? I have trouble putting the expression for the MO's into the expression for the determinants.
\begin{align}
\Psi &= \sum\limits_{i} c_i ( \text{csf} )_i \\
\text{csf} &= \sum\limits_{j} d_j ( \text{det} )_j \\
\text{det} &= \mathcal{A} \prod\limits_k (\text{mo})_k \\
\text{mo} &= \sum\limits_{l} e_l (\text{bf})_l  \\
\end{align}

What about simply doing $\Psi=\sum_i c_i \mathcal{A} \prod_k (\text{mo})_k$ ?
Your equations are fine, but 2 remarks: 1. CSF: Unless you want to go into details about how spin is treated, you can skip the CSF step. It is just a way to couple certain CI coefficients together.  $\Psi=\sum_i d_j (\text{det})_j$  is true as well. 2. indices: Since you usually have more than one MO/Determinant (or wave functions in case of excited states) it is common write these equations with one more index, e.g. $\text{mo}j=\sum_i c{ij}(\text{bf}_i$


text:
The preparation of 0.005 M Iodine Solution
The dissolution of iodine crystals with potassium iodide is very slow. To prepare a 0.005 M Iodine solution will require you keep the vessel until all crystals dissolve. With the low concentration and the volatile nature of Iodine, the solution might lose its concentration at the time of complete dissolution. Can the solution be heated to speed up the dissolution?

Since $\ce{I2}$ is a non-polar covalent molecule, it does not ionize in water. It cannot also be soluble in the polar water.
$\ce{KI}$ which is a polar, ionic compound, will ionize and dissolve in water. When $\ce{KI}$ dissolves in water, it ionizes to $\ce{K+}$ and $\ce{I-}$.
The $\ce{I−}$ will react with $\ce{I2}$ to form the complex ion $\ce{I3-}$. $\ce{I3-}$ being negatively charged will dissolve in water.


text:
How come noble gases have full outer shells?
Argon has $18$ electrons overall and $8$ in the outer shell. But the $3^{rd}$ shell is supposed to have $18$ electrons.
Please use A-level terms so that I can understand.

Possible duplicate of What is meant by complete outer shell? Why do the noble gases have zero valency?

What you're referring to is the $\pu{KLMN}$ configuration, so the electronic configuration which is on your mind for argon is: $2, 8, 8$. So far so good, the third shell can hold $18$ electrons, no doubt, but there's another side to this coin.
Every shell is made up of subshells. Nah, it's not that weird; basically each of $\pu{K,L,M,N}$ is divided into $s, p, d, f \ldots$ Hey, writing $\pu{K,L,M,N}$ is getting cubersome (I'm really lazy) why don't we switch to simply $1,2,3,4 \ldots$? Alright, sounds good. Now we'll try to look at how many subshells are there in a shell.
Here's a formula for that: $n^2$. So basically the first shell ($n=1$) has $(1)^2=1$ a single subshell. And every subshell can hold $2$ electrons. So you can have $2$ electrons in your first shell. So we've reached at $2, 8, 18, \ldots$ Another similar formula, and some more calculations, I was finally able to figure out that, these shell actually look like this:

Each square $\square$ box can hold $2$ electrons. Now when you fill up argon's $\pu{K}$ shell (that's $1s$) and $\pu{L}$ shell (that's $2s$ and $2p$ together) and then finally you fill up $3s$ and $3p$, you've already filled $18$ electrons.
Think of electrons as food and argon as a gentleman. When you ask him: "Hey argon, there are more electrons!", he replies: "Oh sorry, my $\pu{M}$ shell feels full, can you please move it to over to $4s$? I'll fill up my $3d$ after a while."


text:
Will octahedral splitting occur in tribromidotricyanidocobaltate(III)?
Octahedral splitting occurs strong field ligands are coordinated with the central atom. What if we have both types of ligands attached to the central atom as in $\ce{[Co(CN)3(Br)3]^{3-}}$, knowing that $\ce{CN^-}$ is a strong field ligand while $\ce{Br^-}$ is a weak field ligand.

In reality it would perhaps boil down to experimental reasoning, but if it's an exam asking you this, I'd like you to remember a rule of thumb as you'd call it:

If the complex has a coordination number of $6$, with $3$ strong field ligands and $3$ weak field ligands, then consider all the ligands to be strong field ligands.
If the complex has a coordination number of $4$, with $2$ strong field ligands and $2$ weak field ligands, then consider all the ligands to be weak field ligands.


text:
Common tangent construction

I'm still rather confused as to how the common tangent construction ensure the chemical potentials are equal in the two phases?
(Also, if the one-phase regions and the two-phase metastable regions both concave upwards, why do the one-phase regions not undergo phase separation upon nucleation?)

Notation. I will use uppercase letters to represent extensive quantities and lowercase letters intensive molar quantities. Let $G$ be the solution Gibbs free energy, $G_i$ be the Gibbs free energy of pure species $i$, and $\Delta G_\text{mix} := G - \sum_iG_i$ the extensive Gibbs free energy of mixing.
First, it's important to understand what the common-tangent construction actually is. Concretely, consider a solution of species $A$ and $B$. The solution may be of a single phase, or of two phases $\alpha$ and $\beta$; this scenario corresponds to that shown in the provided image. Let's say that $\alpha$ is 20 mol% $A$ and $\beta$ is 60 mol% $A$.
The common-tangent construction represents $\Delta g_\text{mix}$ in the two-phase system, whereas the green curve in your image represents $\Delta g_\text{mix}$ in the one-phase system. Since the Gibbs free energy is minimized at equilibrium, whether or not the system actually has one or two phases will depend on whichever system has the lower energy, which is easily compared visually.

With regards to your second question, it is worth noting that a two-phase system cannot dominate over the entire range of $x_A$. Indeed, consider a solution of 15 mol% $A$. Such a system cannot exist in the two phases $\alpha$ and $\beta$, as it has a lower mole fraction of $A$ than either phase. That $\Delta g_\text{mix}$ be concave up for some region of the one-phase system indicates that a two-phase system will exist over some $x_A$, but not all $x_A$.

We turn now to your first question: how does the common-tangent construction ensure that both phases have equal chemical potential?
A physical answer. The common-tangent construction minimizes $\Delta g_\text{mix}$ over some range of $x_A$, and the equilibrated system over this range must consist of the two phases $\alpha$ and $\beta$, and necessarily two phases at equilibrium have the same chemical potential.
A mathematical answer. Recall that the chemical potential of species $i$ in a mixture is given by $$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{\{n_{j\neq i}\}} = \left(\frac{\partial\Delta G_\text{mix}}{\partial n_i}\right)_{\{n_{j\neq i}\}} + g_i = \left(\frac{\partial\Delta g_\text{mix}}{\partial x_i}\right)_{\{n_{j\neq i}\}} + g_i.$$ We will apply this equation to species $A$ in both phases $\alpha$ and $\beta$. By construction the first term on the right is equal for both phases $\alpha$ and $\beta$, and by definition the second term is equal for both species. Hence $\mu_A^\alpha = \mu_A^\beta$.

Thank you so much! Would that mean since $\Delta g_{mix}$ is minimized over some range $x_A$, by construction $(\frac{\partial \Delta g_{mix}}{\partial x_A})=0$ ?

@Chloe, no, that's not the case: that would mean that the tangent line intersects the green curve exactly at its minima. $\partial\Delta g_\text{mix}/\partial x_A$ is exactly the slope of the tangent line.
I was trying to say that (given that some part of the green curve is concave up) there would be some region over which the common-tangent construction would lead to a lower Gibbs free energy of mixing than the green curve.

Several answers (full disclosure - mine) are available that shed at least some light. So read them first.
https://physics.stackexchange.com/questions/196686/what-causes-a-eutectic-point
Plot Binary Eutectic from Temperature of Melting and Enthalpy
OK. Now to your confusion (I think). What does a straight line across a plot of free energy vs composition actually mean? Well, mathematically that straight line represents the weighted average of the pure free energies, $x_{A}G_{A} + x_{B}G_{B}$ (perhaps more obvious using $x_{B} = 1-x_{A}$). So, $G_{A}$ is the intercept at $x_{A}=1$ and $G_{B}$ is the intercept at $x_{B} = 1$. If the free energies are a fixed value, the chemical potentials are a fixed value. So, the tie-line that is a common tangent to two free energy curves is the solution where the free energies (chemical potentials) are equal in both phases.
As for why the one-phase region does not chemically separate, the question is why should it? There is no second composition that has a common tangent, so there is not way to make the chemical potentials the same in two phases while in a one-phase region.


text:
An equilibrium reaction in the presence of other fast reactions -- is it necessary also to consider the individual rate constants?
When mathematically describing an equilibrium reaction, we generally treat it as though it is enough just to specify the equilibrium constant, as it relates the concentrations of products and reactants.
If this equilibrium reaction is occurring in a highly reactive environment, however, where the reactants and the products are constantly changing due to other reactions, it seems to me that the equilibrium reaction can no longer be described by equilibrium constant only. Instead, the forward and/or reverse reaction coefficients are needed (in addition to the equilibrium constant) to describe the relationship between the reactants and the products. Is this  correct? If not, why not?

Pretty much OK.

Hello and welcome to Chemistry.SE. Feel free to visit the [help] if you have any questions about how to best use this site. Best of luck!

Note that "time-dependent equilibrium reaction" is oxymoronic, as equilibrium assumes time invariance.

@a-cyclohexane-molecule Good point. Title revised...better?

You're correct. If other reactions are occurring that are fast on the time scale of the equilibrium reactions, then you must treat the forward and reverse reactions separately, including by using their separate rate constants.
"Equilibrium" is a mathematical model that never actually perfectly matches reality.  In order for most systems to actually be at "perfect equilibrium" we would have to let them sit and react for an infinite amount of time. Of course, practically speaking, that's not necessary: equilibrium is reached to within measurement precision in quite reasonable amounts of time for many reactions. What's necessary in order for the concept of 'equilibrium' to be useful in analyzing a reaction system is for the time scale of those reactions to be "very short" (or, equivalently, for the rate of those reactions to be "very fast") when compared to all other reactions of interest in the system.
In other words, when we write the following system of reactions:
$$
\ce{A + B <=>[K_1] C + D} \\ ~ \\
\ce{C + E ->[k_2] P}
$$
what we're really writing is:
$$
\ce{A + B \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} C + D} \\ ~ \\
\ce{C + E ->[k_2] P} \\ ~ \\
\begin{align}
k_1\ce{[A][B]} &\gg k_2\ce{[C][E]} \\
k_{-1}\ce{[C][D]} &\gg k_2\ce{[C][E]} \\
K_1 &= {k_1 \over k_{-1}}
\end{align}
$$
The relative values of the forward and reverse rate constants can vary, with their ratio described by $K_1$, but at all points through the reacting process of interest, both the forward and reverse equilibrium reactions are assumed to be much faster than the final, product-forming reaction. If this assumption doesn't hold, then we can't treat the first reaction as "just" an equilibrium reaction.


text:
pH of ammonium acetate solution
I'm trying to calculate the pH of a 1M $\ce{NH_4CH_3COO}$. I know that I'll have these reactions:
$\ce{NH_4CH_3COO <=> NH_4^+ + CH_3COO^-}$
$\ce{NH_4^+ <=> NH_3 + H^+}$
$\ce{CH_3COOH <=> CH_3COO^- + H^+}$
I know the $K_a$s of the last two, so I'm able to calculate the $K$ of the first one (it's $\frac {K_{a1}}{K_{a2}}$), which gives me these equations:
$\frac {K_{a1}}{K_{a2}} = \frac {(x-y)(x-z)} {1 - x}$
${K_{a1}} = \frac {y(y+z)} {x-y}$
${K_{a2}} = \frac {z(y+z)} {x-z}$
But I have only 2 independent equations (the first one is just the ratio of the second and third) and three variables so I'm unable to solve for $[\ce{H^+}]$, which is $y+z$...
What do I do?

It would be good to know what are your x, y, and z. Also, you don't really have K for the first reaction, nor do you need one.

You're missing the conservation of matter constraint. The total amount of ammonium, ammonia, acetate, acetic acid is equal to the amount you started with.

Are you sure the concentration is 1 M? If so, I think your problem is even more complicated.
At this high concentration you should perhaps also consider the activity coefficients of all the protolytes involved in order to make a reasonable calculation.

@Bive I think you would just assume (possibly incorrectly) that concentration effects are not significant.

Ok, I will follow the assumption proposed by @Zhe above (possible incorrect as he states, but please do not be confused by that).
In order to solve this problem we need two acidity constants:  $\mathrm{p}K_{\mathrm{a}(\ce{NH4+})} = 9.25$ and
$\mathrm{p}K_{\mathrm{a}(\ce{CH3COOH})} = 4.76$.
First  we state the proton balance (the amount of protons taken up have to be equal to the amount of protons given off in the system): Initially we have $\ce{H2O}$ and $\ce{CH3COONH4}$.
Proton balance:
$[\ce{H3O+}] + [\ce{CH3COOH}] = [\ce{OH-}] + [\ce{NH3}]$
At $\mathrm{pH} = 7, [\ce{H3O+}] = [\ce{OH-}] = \pu{10^{-7} M}$:
At this $\mathrm{pH}$ the proton balance can be simplified as $[\ce{CH3COOH}] = [\ce{NH3}]$.
The simplified proton balance will be true only at a $\mathrm{pH}$ that is exactly in the middle of the two $\mathrm{p}K_{\mathrm{a}}$-values. We get $\mathrm{pH} = (4.76 + 9.25)/2 = 7.005 \approx 7$ (only one significant figure is given, since you have stated the concentration as $\pu{1 M}$).

Simple answer
The salt ammonium acetate composed of the anion acetate ion (conjugate base of weak acetic acid)and of the cation ammonium ion(conjugate acid of a weak base ammonia), both cation and anion hydrolyzed  in water equally ${(k_a=k_b)}$, so the solution is neutral
 $${[H3O^+]=[OH^-]=10^{-7}and\ pH=7}$$
I will give a more theoretical answer to this question using equilibrium constant and deriving formula:
Four equilibria are possible in the ammonium acetate solution; the auto ionization of water, the reaction of the cation and anion with water, and their reaction with each other :
$$
\begin{array}{ll}
\ce{NH_4^+ + H2O <=> H3O+ + NH_3} &\quad\left(K_{a(NH_4^+)} = \frac{K_w}{K_{b(NH_3)}}=\frac{10^{-14}}{10^{-4.74}}=10^{-9.26}\right)\\
\ce{CH_3COO^- + H2O <=> OH^- + CH_3COOH} &\quad\left(K_{b(CH_3COO^-)} =\frac{K_w}{K_{a(CH_3COOH)}}=\frac{10^{-14}}{10^{-4.74}}= 10^{-9.26}\right)\\
\ce{H_3O^+  + OH^-  <=> 2H_2O }&\quad\left(\frac{1}{K_w}\right)\\
\ce{NH_4^+ + CH_3COO^-  <=> CH_3COOH  + NH_3} &\quad\left({K_{eq}}=\right)\\
\end{array}
$$
The last equation is the sum  of the first three equations, the value of $K_{eq}$ of the  last equation is therefore
$$K_{eq}=\frac{10^{-9.26}\times10^{-9.26}}{K_w}=3\times10^{-5}$$
Because $K_{eq}$ is several orders of magnitude greater than $ K_{a(NH_4^+)}\ or\ K_{b(CH_3COO^-)} $ , it  is valid to neglect the other equilibria and considering only the reaction between the ammonium and acetate ions .Also, the products of this reaction will tend to suppress the extent of the first and second equilibria,  reducing their importance even more than the relative values of the equilibrium  constants would indicate.
From the stoichiometry of ammonium acetate :
$$\ce{[NH_4^+]=[CH_3COO^-]\  and\   [NH_3]=[CH_3COOH]} $$
Then
$$K_{eq}=\frac{[CH_3COOH][NH_3]}{[NH_4^+][CH_3COO^-]}=\frac{[CH_3COOH]^2}{[CH_3COO^-]^2} =\frac{K_w}{K_{a(CH_3COOH)}K_{b(NH_3)}}$$
From the acetic acid dissociation  equilibrium :
$$\frac{[CH_3COOH]}{[CH_3COO^-]}=\frac{[H_3O^+]}{K_{a(CH_3COOH)}}$$
Rewriting the expression for $K_{eq}$ ,
$$K_{eq}=\frac{[CH_3COOH]^2}{[CH_3COO^-]^2}=\frac{[H_3O^+]^2}{K_{a(CH_3COOH)}^2} =\frac{K_w }{K_{a(CH_3COOH)}K_{b(NH_3)}}$$
Which yields the formula
$${[H_3O^+]}=\sqrt{\frac{K_wK_{a(CH_3COOH)}}{K_{b(NH_3)}}}={\sqrt{\frac{10^{-14}\times10^{-4.74}}{ 10^{-4.74}}}  =\ 10^{-7}}$$
$pH=7$


text:
Hash and Wedge nomenclature
One of the first things learnt in organic chemistry is that wedged bonds can be thought of as bonds coming out of the plane of the paper towards you, and hashed bonds can be thought of as bonds going behind the plane of the paper away from you.
What is rarely discussed, is the difference between actual 'wedged' bonds, and the 'block' variety. The diagram below from Organic Chemistry Strategy and Control by Stuart Warren goes some way to explaining that in theory at least, the two sets of bonds exist to allow us to describe absolute or relative stereochemistry.
(In practice, this is rarely followed, and individual groups/labs tend to, rightly or wrongly, favour one type over another for stylistic reasons).

What has always baffled me, however, is the direction in which the wedged bonds originate/point towards.
In most modern usage, the small end of the wedge/hash is at the chiral centre, and gets larger towards the substituent, however some (mainly older chemists, whom I'm going to give the benefit of the doubt and assume aren't just bad at using ChemDraw) chemists have the hashes differently to the wedges, as per the molecule below.
Is this something that is advisable (presumably it's meant to show some kind of 3D perspective)/are there any actual IUPAC rules/guidelines on this other than following what looks good/is conventional?

My understanding of the direction of the wedge vs the hash is akin to looking at an object moving toward and away from the horizon respectively. An object approaching you takes up an increasing portion of your vision, and an object moving away   shrinks to a point.

Both schools of thought were prevalent, one said that the narrow end should be considered in the plane of paper, the second said that the wide end should be considered. Perhaps the first was based on a perspective flavored approach while the second mainly relied on popular drawing schemes. Excerpts from the 2006 IUPAC Recommendations1 (more information is given there):

The least disruptive recommendation is to reject a perspective-based approach, and rather to interpret the bonds as more-abstract graphical objects. A solid wedged bond should be interpreted as projecting above the plane of the paper from the narrow end to the wide end (with the wide end being nearer to the viewer), and a hashed wedged bond should be interpreted as projecting below the plane of the paper, also from the narrow end to the wide end (with the wide end being further away from the viewer). To summarize, the narrow end of solid wedged bond or a hashed wedged bond always should point towards the stereogenic center to which it applies.

That said, it is possible to use the other bond types in ways that are unambiguous. In particular, any type of hashed bond is unambiguous in print when connected to an atom with no other substituents (a hydrogen atom, a chlorine atom), no other non-hydrogen substituents (a hydroxy group, an amino group), or at most one other non-hydrogen substituent (an ethyl group, an ethoxy group). In each of those cases, a viewer would have no difficulty in understanding that the hashed bond projects below the plane of the paper from the other atom to that atom. Accordingly, both the unwedged hashed bond and the hashed wedged bond with reverse directionality are acceptable when used exclusively in those cases.

However, a caveat:

However, anyone using bonds of these types should be aware that they are not the most common bond types used to represent behind-the-plane configuration. At best, an author using these types may cause a moment of confusion as a reader figures out which convention was intended. In extreme cases, a reader (or chemical software) might actually interpret the structure incorrectly—for example, by interpreting an unwedged hashed bond in the convention where it would represent relative (and not absolute) configuration, as discussed in ST 0.8 and ST-6. Fortunately, such cases are rare. The hashed edged bond (interpreted from the narrow end to the wide end) remains preferred...

Reference

Graphical representation of stereochemical configuration (IUPAC Recommendations 2006). Pure Appl. Chem. 2006, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897. © IUPAC.

In addition, the recommendations document mentioned by Barry Holmes suggest preferred way to represent relative configuration / racemic mixtures, so the 1st formula in your 1st image representing (±)-syn-27 would be depicted better

or


text:
HOMO-LUMO and perturbation theory
I have some confusion about perturbation theory, and just how it fits into the hard-hard/soft-soft interactions within organic reactions. I have come across a formula for perturbation that is proportional to an overlap constant squared but is inversely proportional to the energy difference of the orbitals in question. This suggests that for orbitals of poor energy match, there is least perturbation, and vice versa for good orbital energy matching. How does this translate into a hard/hard and soft/soft interpretation? I am not looking for a quantitative answer because my chemistry is not advanced enough!

Could you add the equation or a reference to your question?

Maybe related: https://chemistry.stackexchange.com/questions/59281/hard-and-soft-acid-and-bases-in-terms-of-homo-and-lumo

What you are referring to as perturbation theory sounds like a very crude approximation to the 'charge-transfer' or CT intermolecular interaction.  In quantum chemistry lingo, the CT interaction is the mixing of occupied orbitals on one molecule/fragment with the virtual orbitals on another fragment (this is the 2-body CT interaction).  This permits the transfer of electron density from one fragment to another (charge-transfer).
Instead of redoing the HF (variational) calculations with the new orbitals included, perturbation theory is instead used to calculate the nth order correction as a result of occ(a)+vir(b) mixing.  Differencing the HOMO on fragment A with the LUMO on fragment is just a way to quickly approximate the more complicated (and expensive) perturbation theory expressions for the CT iteraction.
So this seems to make sense in the context of HSAB theory and Lewis acids where acid/base exchange electron pairs and thus participate in a strong charge-transfer interaction.
*EDIT:  I'd like to add that the orbital energy of the LUMO orbital (in fact, all virtual orbitals) is sensitive to the basis-set size. When a large enough basis-set is used, the LUMO can always be forced to near 0.00.  Orbital energies, at the end of the day, are not nature.