text:
Improving efficiency of absorption refrigerator (ammonia+hydrogen)
Apologies if this is not the right forum for this question, I checked the Physics side and there's no questions on this, whereas there's at least one other question here about absorption refrigeration.
I have a very standard camping fridge which runs on the absorption cycle, I'm making a leap of informed guesswork and assuming it's ammonia + hydrogen based.
I'm investigating ways the efficiency or performance of the system could be improved without major surgery but I'm neither a chemist nor physicist (nor surgeon).
Here's the system as explained by the manufacturer:

And here's the actual unit, minus some lagging round the boiler:

So you can see it's pretty basic, they haven't changed the design in decades.
My main train of thought, based on the bleeding obvious, is that either more heat could be added at the boiler end, or more cooling (basically, a fan) could be added to the condenser.
However, I don't know enough about how the coolant behaves to know if this would be of any benefit - adding heat could just ruin everything, cooling beyond a certain point may make no difference, etc. etc.
Additionally, I don't know if things could be done at other points in the cycle (cooling or heating other areas such as the absorber being the obvious easy wins) to improve matters.
I would also like to add that this is mostly driven by my curiosity for tinkering with things, I realise that the real-world answer to a better fridge is to buy a compressor type one - but they're expensive (seriously, compressor camping fridges are more than a full-size household fridge+freezer) and I won't learn anything that way ;)
EDIT: Thanks for the responses, ultimately the thing was never going to be viable running on 12v so we bought a compressor-type which is about 10x more efficient (in electricity) by comparison.

Your title refers to efficiency, but your post discusses increasing heat flow.  While I suspect you're looking for more cooling, I find the efficiency idea more intriguing.  :-)

I've come here for basically the same reason, however I have some suggestions that I've done and I believe they have helped.
And my fridge is basically the same as your, just a bigger model, same manufacturer.
Firstly I should suggest messing with the process (heat etc) would not be a good idea as I would assume the manufacturer would have optimized the design years ago, and I have heard over heating the boiler can cause catastrophic and irrepairable damage .
My fridge has a freezer box (shoebox size, quite useless) within the fridge and a heat sink next to it for the fridge part. I took the door off the freezer and cut the base out of the box, that made a huge difference, infact with it set on high on a cool day of 20c or less items in the bottom of the fridge will freeze.
But as the camping car is used mostly in summer and it's hot that doesn't happen then.
If you need a freezer use the engel.
The fridge is mounted permanently into the camping car with a door on the outside to access the back of the fridge, it has vents top and bottom, on the inside of the bottom door vents I've installed 2 mini computer 12v power fans, they are very week but do set up a positive upward circulation of air over the absorber/condenser. As these use so little electricity they can be connected into your auxiliary battery without fear of draining it.
I think the added circulation helps on hot days.
And finally if you can't avoid parking it with the rear of the fridge on the sun exposed wall try a shade cloth over the area.
Hope that helps.

Good tips for those sticking with these. I gave in and bought a compressor-type Waeco/Dometic (CRX50) as cutting large vents in the vehicle / running on gas was not a viable option. The Waeco uses <1Ah of power on average compared to ~7Ah for the 3-way.

Adding heat is risky and could lead to overpressure or damage.  I'd also expect it to boil more of the water off with the ammonia, which is probably not ideal. Not sure about efficiency per se, but you can definitely improve the effectiveness with fans.  It's a heat pump that maintains a given temperature delta not between outside and inside exactly but between the condenser and evaporator.  So, forced air across the condenser can definitely help (as long as you're not creating excess cooling at the boiler).  So can forced air across the evaporator heat exchanger fins on the inside.
In practice, it's also worth looking at the installation instructions on the fridge to see if the installer followed them.  They're pretty sensitive to the exact placement of vents, baffles, and insulation.

I worked as a production manager for a company that manufactured similar units. On these absorption units, the inside temperature is directly related to the ambient temperature. The warmer it’s outside, the warmer inside. Good ventilation over the condenser to get rid of heat does help and will make a difference. As long as you don’t disturb the natural convection that takes place in the boiler’s flue. The chest type units work a bit better as all the cold air inside does not get lost every time the door is opened and closed. There are a few other points to consider though. The gasses (hydrogen and evaporated ammonia) and liquids (condensed ammonia and water) are circulated with heat, pressure and gravity. Making sure that the unit is level is important, and an out of level unit may not function optimally. I also noted the removed casing and insulation around the boiler. When installing the insulation (usually with a type if mineral wool), it is important to achieve good compaction and density for optimal insulation. Loosely packed insulation reduces the efficiency significantly. Also, inside the flue there should be a spiral twisted plate suspended on wire. Make sure it's present as it swirls the heat from exactly the correct height above the heat source. Then lastly, make sure your gas burner has a blue flame, not a bigger flame and not a yellow flame, but a blue flame. Hope this can useful.


text:
Why does benzene have a D6h point group?
I understand the fact that the molecule has a principal axis of rotation of C6 however when following the point gorup flow chart why is the answer to the second question NO when it contains two C3 axis? Hence the answer should be a YES right? (I understand the molecule has C2 axis as well)

Only different axes count.

Also, you chart seems to miss the groups $T,;T_h,;O,;\text{ and }I$.

As @IvanNeretin stated - There's rather 1 C6 axis, than 2 C3 axis

Does it have two or more $C_2$ axes with $n>2$?

This question is asking if there is more than one $C_{n>2}$ axis not if there is more than one $C_{n>2}$ symmetry operation.
Yes, benzene has some $C_3$ operations. However, as pointed out in the comments, the $C_3$ axis lies along the $C_6$ axis - both run through the center of the ring perpendicular to the plane of the molecule. This axis is also a $C_2$ axis. There is only one $C_{n>2}$ axis - it just happens to be both a $C_6$ and a $C_2$ axis. All other rotation axes in benzene are $C_2$.

While reading https://www.nature.com/articles/srep25670 I double checked the definition of $D_{6h}$ and this question came up in a search. While it's not related to benzene per se the following question in Math SE is about graphene-like xene materials and probably requires an understanding of symmetries to address. Any thoughts or recommendations? Determining if a coincident point in a pair of rotated hexagonal lattices is closest to the origin

source: Stack Exchange

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text:
Where does the pH come from in the equation of Standard Hydrogen Electrode?
$$\mathrm{E=-{2.303RT \over F}pH - {RT \over 2F}\ln {p_{H_2}/p^0}}$$
This is a equation on Wikipedia page of Standard Hydrogen Electrode, but I have no idea where the $pH$ come from.
Since $\mathrm{a_{H^+}=f_{H^+} C_{H^+} /C_0}$, how does the $\mathrm{f_{H^+}}$ disappear?

There are a few logarithm rules that you need to apply here.
\begin{align}
E &= \mathrm{RT \over F}\ln {a_{H^+} \over (p_{H_2}/p^0)^{1/2}}\\
  &= \mathrm{RT \over F}\ln a_{H^+} - \mathrm{RT \over F} \ln \left({p_{H_2} \over p^0}\right)^{1/2}\\
  &= \mathrm{\ln(10) RT \over F} \lg a_{H^+} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)\\
  &= -\mathrm{2.303 RT \over F} \mathrm{pH} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)
\end{align}

You need a few logarithm rules:

$\log\left({a \over b}\right) = \log(a) - \log(b)$
$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$
$\log_b(x^r) = r \log_b(x)$

I tried editing but there's a 6 character minimum. You've got a small typo in the derivation. Next to last line from bottom, lg(CH+) should be log(CH+)

I simply don't understand why $f_{H^+} \approx 1$, since the concentration we are dealing is not low.

You don't need $f_{\ce{H+}} \approx 1$, because the definition of $\mathrm{pH}$ is not $\mathrm{pH} = - \log (c_{\ce{H+}})$ but rather  $\mathrm{pH} = - \log (a_{\ce{H+}})$. The definition based on the concentration is just an often used approximation which assumes $f_{\ce{H+}} \approx 1$ and is only valid in cases where this assumption is valid.

@SeanDoris: Thank you, but this is correct. "...  $\log_{10}(x)$ should be written $\lg (x)$ and $\log_e(x)$ should be $\ln (x)$., so it is correct.

@Philipp: Thank you, you are right ... changed it.

@pH13 Fair enough. I haven't seen that convention (and I personally don't care for it now that I have seen it), but by all means keep using it if you think it's clearer

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Why hydrofluoric acid (HF) has two Raman bands?
Hydrofluoric acid ($\ce{HF}$) is a diatomic molecule with only one vibrational coordinate. Therefore I would expect only one Raman band, however there are two. Why?
(source)

How is it even possible that HF is vibrating at 600 wavenumbers? It's gas phase vibrational mode is up around 4400? I wonder if these are the modes of vibration between HF-HF dimer (the higher one) and HF-H2O dimer. I just imagine those would be lower frequency. Also, both peaks, particularly an HF-HF dimer should increase when HF is added which is what the percentages are I believe.

I've been doing some digging and I think I have a possible answer so I'm gonna post it and if I'm misinterpreting information, I'm open to being told so!
As I said in my comment, this vibration has to be coming from some dimer which is formed because if one of these modes were the $\ce{HF}$ fundamental it would be up around $4000\ cm^{-1}$.
So, I found this paper which uses FTIR to analyze $\ce{H2O/HF}$ complexes. They find two librational modes (which I believe is just a solvated rotational/rocking mode basically) which they identify as coming from the $\ce{H2O--HF}$ dimer. They assign these two frequencies at $635.5\ cm^{-1}$ and $753.1\ cm^{-1}$. Those two values are extremely similar to the frequencies we're seeing in the spectra you posted which is why I think we might be on to something here. Unfortunately I don't have access to the paper so I can't go into any more detail really.
In answer to your question of how we could even be seeing multiple peaks in an $\ce{HF}$ spectrum, the answer is that we're not looking at just $\ce{HF}$ here!
We know that both $\ce{H2O}$ and $\ce{HF}$ are very strong hydrogen bonding molecules, so it's no surprise that we find peaks which are associated with the various vibrational modes of the dimers present. And, once we have a dimer, we expect to find all sorts of rotational and vibrational modes.
So, to conclude, I'm fairly certain that the modes we're seeing in this spectra are two (and I think the only two because my digging around says this dimer has $C_s$ symmetry) librational modes of the $\ce{H2O--HF}$ dimer, each rotating about a different axis of the water.

Source:
Andrews, L., & Johnson, G. L. (1983). FTIR spectra of water–hydrogen fluoride complexes in solid argon. Evidence for inversion doubling in the HF librational modes of H2O––HF. The Journal of chemical physics, 79(8), 3670-3677.

(1) aren't IR and Raman wavenumbers of the same vibration supposed to be quite different (while you say that the IR wavenumbers are very similar to the Raman ones in the charts). (2) both IR and Raman originate in intramolecular transitions so how can an intermolecular interaction be expressed in the spectra?

Not necessarily. The paper I cite has the water and HF in a solution of really really cold Argon so we could expect that the vibrations will be something like a regular solution. The only thing that makes IR and Raman quite different is that certain modes are active for one and not active for the other and Raman is usually in solution. Being in a liquid generally changes the frequency of vibrations. Also, you can think of a dimer as being a new molecule that's just held together by a weak bond. Then it's no surprise to see new Vibrational modes show up.

Also, to convince yourself that Raman can resemble IR, by simply looking at the frequency which is identified as coming from water, we can see that this is water's bending mode but is shifted down about 200 wavenumbers due to hydrogen bonding and is also wider because there are many waters in varying hydrogen bound geometries. This same shift does not apply to the HF/H2O dimer because it only exists when there's hydrogen bonding anyways! There might be slight shifts associated with a third water complexing in, but this is probably a small effect.


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Software for creating the QSAR model?
There is the .csv file with a large amount of descriptors and known experimental activity.
Which software can find and describe model for prediction. Most important - numerical representation of model - which descriptor is taken with coefficients.
I am looking for any software, not necessary free.

There are multiple packages capable of designing QSAR models.
At a basic level, QSAR is about building a statistical model fitting some property/activity based on a set of multiple descriptors. You can use any multivariate statistical package to do that, including R (as indicated in the other answer) or Python.
If you're new, I would highly suggest KNIME an open "platform" for connecting multiple software packages for analyzing data. It's heavily used for QSAR.
There are several examples for KNIME QSAR, e.g. using CheS-Mapper - a quick search turns up several.
A critical concern, though is validation, either through separate training and test sets, cross-validation / bootstrapping methods, or other techniques. It's easy to design models that fit noise.

Package ‘rcdk’ https://cran.r-project.org/web/packages/rcdk/index.html start here ...


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What is the product of the reaction of 2-methyl-5-phenyl-2-pentene with sulfuric acid?
The reaction is 2-methyl-5-phenyl-2-pentene with $\ce{H2SO4 }$

I have tried multiple attempts at this question but am still getting it wrong. I have no idea why any of these answers are wrong.
I know that the chain off the aromatic ring is weak aromatic activator and an ortho-para director. I figured para would be better than ortho because of the size of the chain but obviously both answers were incorrect. I then figured that $\ce{H2SO4}$ is just a strong acid so its basically $\ce{H3O+}$ but adding $\ce{H}$ and $\ce{OH}$ across the double bond in the chain did not work either.

I think your top answer is the closest. If $\ce{H^+}$ from $\ce{H2SO4}$ adds first to the double bond, what adds to the other side? You have $\ce{HSO3^-}$ adding to the other side. Something is missing. Of course, it could be a cyclization due to Friedel-Crafts, the intramolecular verision of the top left reaction here.

Actually, the more I think about it, the more sure I am that you will have Friedel-Crafts-type cyclization occur.

Yes, you're right!!! I don't know how I didn't see the possibility of a Friedel- Crafts reaction, but thank you so much!! The final product was a benzene ring attached to a cyclohexane with two methyl groups coming off the bottom carbon. Once again thank you so much! I will update the question with the correct answer :)

No problem. By the way, for just an alkene, make sure you're adding $\ce{HSO4^-}$, not $\ce{HSO3^-}$. You would form an organosulfate, not an organosulfonate (or an organosulfuric acid, not an organosulfonic acid, really).

With the help of user SendersReagent in comments, the answer to the question was a simple intramolecular Friedel–Crafts reaction:

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What does dl- mean in nomenclature
I came across the compound dl-stilbene dibromide, and I am confused by exactly what this means. I thought it might have meant a racemic mixture but there is only one chemical structure given.

That picture is showing relative stereochemistry. Essentially, it is just saying that, in the shown confirmation, the hydrogens will be on one side and the bromines will be on the other, but half of the mixture will have the bromines coming out and the other half will have the bromines going back.
If the S or R labels had been shown in the molecule (or they had used just a d- or l- prefix) the picture would have been showing absolute stereochemistry. By saying dl-stilbene dibromide, they are saying that it is a racemic mixture of the compound shown and its enantiomer, but not any diastereomers.

Here's a snippet with surroundings from the book (Experimental Organic Chemistry – A Miniscale and Microscale Approach), for the context:

Should we accept the name “stilbene dibromide” (which is almost as bad as would be e.g. “ethylene hydrochloride” for chloroethane (ethyl chloride)), there are some other problems.
49 is defined in the text as a diastereomer, but the reaction scheme it's named  “dl-Stilbene dibromide”, denoting a racemic mixture (i.e. of both enantiomers), therefore should apply to entire 49 structure including the “(+ enantiomer)” text. (Preferred notation is “and enantiomer” text, or including structure of the other enantiomer, without ‘+’ sign, but that's for the standalone context, here we have a reaction scheme.)
It's true that sometimes the constant-width bold/hashed bond is used to denote relative stereochemistry, together and in contrast with wedge bold/hashed bonds for absolute stereochemistry. This mixed convention is not supported by IUPAC, even it's described as “obscure” in Graphical representation of stereochemical configuration (2006):

ST-0.8 Mixtures of diastereoisomers
There are a variety of obscure conventions for depicting arbitrary mixtures of diastereoisomers. The
  most common (but still obscure) of these conventions says that bold and hashed wedges should be used
  for absolute configuration, unwedged bold and unwedged hashed lines represent relative configuration
  and  racemic  character,  while hollow  wedges  and  dashed  lines  represent  relative  configuration (…)

However this is not the case for this book, according to the following example depicting a single enantiomeric product:

It should be also emphasised that dl- stereodescriptor usage is strongly discouraged ((E)- and meso- are OK), the recommended one for racemates is rac-.

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What type of analytical chemistry (such as spectroscopy) is suitable for high resolution air analysis
I want to find some type of analytical chemistry method, such as spectroscopy, that is capable of analysis the chemical composition of air. If it can be used for high resolution analysis would be preferable.

It really depends on what you want to monitor. Do you want to analyze for radioactive radon in a basement or are trying to monitor air pollution?!?

Headspace GC-MS is the way to go. It's used to detect vapors coming off a solution. For example, it is used in the fragrance industry to test the time-dependent development of a perfume, since the lighter molecules will evaporate first and form a higher concentration in the headspace initially.
Unfortunately, for detection of noble gases in an extremely low concentration this is difficult. The technology has just been created and involves concentrating the rare gases via adsorption over high surface area carbon before analysis. Each gas adsorbs in a specific temperature range that is different.
From an article on the International Noble Gas Experiment, a modern method to detect trace amounts of Xenon from nuclear explosions:

“Xenon is isolated from the air by a charcoal-containing purification device resulting in higher concentrations of the gas in its stable and unstable, i.e. radioactive, forms.  The isolated xenon is then measured for its level of radioactivity which helps to quantify the radioactive xenon contained in the sample.” Source

I know that a company called Air Products liquifies air before distilling Nitrogen and Oxygen, and then purifies Xenon for medical use. An in-depth process is described here.

Awesome, thanks very much. I have been looking into it and it looks  to be what I need. Thanks again.

Do you think that this would also be able to detect esters carried in the air too, such as from food?

That is exactly the kind of thing headspace GCMS is for.

You can use gas chromatography attached to a mass spectrometer. I've used headspace GC-MS to analyze the vapor above a sample. If you're looking to analyze oxygen/nitrogen/etc, this may not work (it might, but I'm not positive), but to test the vapors in the air, it should work.


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Calculating KC Given Equilibrium Concentrations
Ex. N2O4(g) 2NO2(g)
• If you place 0.0350 mol N2O4 in 1 L flask at
equilibrium, what is KC?
• [N2O4]eq = 0.0292 M
• [NO2]eq = 0.0116 M
i want to know from where this coming
• [N2O4]eq = 0.0292 M
• [NO2]eq = 0.0116 M

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Those are the concentrations of $\mathbf{N_2O_4}$ and $\mathbf{NO_2}$ at equilibrium. They are just given to you as part of the question for you to use in calculating $K_c$.


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What is the difference between enthalpy (H) and heat (q)?
It seems that in some cases it's the same (that is, $\Delta Q_p=\Delta H$)  yet in others they differ.
For example,
\begin{align}
\ce{2H2(g) + O2(g) &->2H2O(l)} & \Delta H &= -572~\mathrm{kJ}
\end{align}
The answer for one mole of $\ce{H2O}$ produced is $\mathrm{-572\times 0.5~kJ}$.
Is this under the assumption of constant pressure? Therefore it's $\Delta Q_p=\Delta H$?

Enthalpy and heat are entirely different things.
Enthalpy is a function of state. If you know the state of a system, you know its enthalpy. If you know the starting and ending states of a process, you can find the enthalpy change.
Heat, on the other hand, is an inexact differential. Knowing the initial and final states of a process is not enough information to tell you the heat transfer. Instead, the heat transfer depends on the particular path taken between the states.
For a simple example of why this is important, consider a heat engine. The process is a cycle, meaning the initial and final states are the same, so a cycle of a heat engine has zero enthalpy change. However, the entire point is that the heat engine converts heat into work, so the heat exchange during a cycle is not zero.
That is why the heat exchanged is equal to the enthalpy change only under constant pressure. We need to give extra information (such as that the pressure is constant) to know the path taken during the process in order to say anything meaningful about heat exchange.

Enthalpy is defined as $$H= E + PV$$
Using First Law, we get $$\mathrm d H =\delta Q - \left(\underbrace{\sum Y\,\mathrm dX + W'}_{\delta W}\right) + \mathrm d(PV)$$
where \begin{align} \sum Y\,\mathrm dX &= \,\,\textrm{configuration work}\\  W'& = \,\,\textrm{dissipative work}\end{align}
When the process is reversible, then the dissipative work $W'$ is zero.
We can further write the expression for constant pressure as \begin{align}\mathrm dH & = \delta Q - \underbrace{A'}_\textrm{non-compression work} - P\,\mathrm dV + P\,\mathrm dV\\ &= \delta Q- A'\end{align}
Thus, basically $$\mathrm dH = \delta Q - A'$$
When $A' = 0,$ then only $\mathrm dH = \delta Q$ at constant pressure.

Further look:
$\bullet$ Thermodynamics, Kinetic Theory, and Statistical Thermodynamics by Sears, Salingar.

It seems that in some cases it's the same (that is, $\Delta Q_p=\Delta H$)  yet in others they differ.

Heat is an energy transfer, whereas enthalpy is a kind of energy content of a sample. When we want to know the change in enthalpy (e.g. before and after a reaction), we use the "$\Delta$" prefix to indicate that. For heat, which already refers to a transfer of energy, the "$\Delta$" prefix is not used. So it should say $$Q_p=\Delta H$$

For example,
\begin{align}
\ce{2H2(g) + O2(g) &->2H2O(l)} & \Delta H &= -572~\mathrm{kJ}
\end{align}

If you want to describe the enthalpy of reaction without knowing the amount that reacts, it is best to use the molar enthalpy with units $\pu{kJ/mol}$. This depends on how you write the reaction, e.g.
$$\ce{2H2(g) + O2(g) ->2H2O(l)}\ \ \   \Delta_r H = \pu{-572 kJ/mol}$$
or
$$\ce{H2(g) + 1/2 O2(g) -> H2O(l)}\ \ \   \Delta H_f(\ce{H2O}) = \pu{-286 kJ/mol}$$
Unfortunately, there is no generally accepted and followed convention for distinguishing enthalpy change and molar enthalpy change, but the unit tells us which one is meant.

The answer for one mole of $\ce{H2O}$ produced is $\mathrm{-572\times 0.5~kJ}$.
Is this under the assumption of constant pressure? Therefore it's $Q_p=\Delta H$?

Here, $\Delta H$ symbolizes the extensive quantity enthalpy (i.e. not molar enthalpy) so the equation is dimensionally correct. There is one more assumption we need - no work other than pressure-volume work. So if you run this reaction in a fuel cell to do electrical work, less heat will be exchanged.

Strictly speaking, that $\Delta H$ assumes that the initial state consists of 2 moles of pure $\ce{H2}$ and 1 mole of pure $\ce{O2}$, each at 298 K and 1 bar, and the final state consists of  2 moles of pure liquid water at the same temperature and pressure.  Since enthalpy is a state function independent of path, this enthalpy change is independent of any and all processes used for transitioning the system from state 1 to state 2.
Since, at this pressure, $\ce{H2}$ and $\ce{O2}$ behave essentially as ideal gases (and their mixing is ideal), the gases can be mixed first and then allowed to react at constant pressure.  In this case, the heat added to the system with be equal to the stated $\Delta H$ (in this case negative).


text:
Why are critical micelle concentrations so large?
Dissolving a non-polar molecule in water incurs a free energy cost of about $10 kJ/(mol \cdot nm^2)$ (source: http://bionumbers.hms.harvard.edu/bionumber.aspx?&id=101826)
The entropy gained from dissolving a solute is $k_B \ln(c_0/c)$ where $c$ is the number density of solutes in the water and $c_0$ is the number density of solutes when separated out from the water. Thus, I would expect the critical micelle concentration to be approximated by
$$c_{crit} \approx c_0 \exp\left(-\frac{10 kJ / (mol \cdot nm^2) SA}{k_B T}\right)$$
with $SA$ the surface area of the surfactant.
For SDS, using the information at https://en.wikipedia.org/wiki/Sodium_dodecyl_sulfate, I estimated that
$$c_0 \approx 1/(.5 nm^3)$$
$$SA \approx 4 nm^2$$
and
$$k_B T \approx \frac{1}{40} eV$$
This gives me
$$c_{crit} \approx 2*10^{-7} mol \cdot L^{-1}$$
but the measured value is nearly $10^{-2} mol \cdot L^{-1}$.
Where are the missing five orders of magnitude coming from? Are there, perhaps, additional degrees of freedom gained when SDS dissolves besides just the location of the molecule?


text:
How to figure out which substance has the highest osmotic pressure?
In preparing for an exam and I'm trying to solve the following question. I'm clueless. What should I do?

Which of the following water solutions has the highest osmotic pressure?
\begin{align}
\tag1 \pu{0.1 mol/kg} &~ \ce{NaCl}\\
\tag2 \pu{0.1 mol/kg} &~ \ce{KF}\\
\tag3 \pu{0.1 mol/kg} &~ \text{sucrose}\\
\tag4 \pu{0.1 mol/kg} &~ \ce{MgCl2}
\end{align}

$\ce{MgCl2}$ has the highest osmotic pressure because it breaks up into three ions in water, as opposed to all the others, which only break up into two.


text:
How to calculate the rate constant for the reaction of mercury chloride and oxalate?
If for the reaction
$$\ce{2HgCl2 (aq) + C2O4^2- (aq) -> 2Cl^- (aq) + 2CO2 (g) + Hg2Cl2 (s)}$$
The initial concentration of $\ce{HgCl2}$ is $0.0836~\mathrm{M}$ (first order), of $\ce{C2O4^2-}$ is $0.202~\mathrm{M}$ (second order), and the initial rate of formation of $\ce{Cl-}$ is $0.52\times10^{-4}~\mathrm{M\,s^{-1}}$, then to find $k$ I would do:
$$\frac{1}{2}(0.52 \times 10^{-4}~\mathrm{M\,s^{-1}})=k(0.0836~\mathrm{M})(0.202~\mathrm{M})^2$$
And isolate for $k$. I included the one half in front of the rate of chloride ion formation because the coefficient in front of the chloride ion is 2 in the equation.
Is my method correct? The solution key states that I should set up my equation without the one half. Thus, its answer was twice as large as the one I got. If the answer key is correct, please explain why.

Maybe the initial rate already includes the factor of 1/2

@AdityaDev That would be very confusing for a student doing this question on a test. As far as what I can tell, it only states the rate at which chloride ions are formed. That suggests to me it has not included the factor of 1/2.


text:
Reaction of aluminum hydroxide and nitric acid
I've been focusing a lot on aluminum hydroxide recently, or specifically breaking it down into its base components. I know that when aluminum hydroxide dissolves in nitric acid it forms aluminum nitrate ($\ce{Al(NO3)3}$), which decomposes when it boils into aluminum and nitrogen dioxide. The end goal of this process is to recover the aluminum from the aluminum hydroxide, but that can't be done until the aluminum hydroxide dissolves into the nitric acid. I don't know if aluminum hydroxide readily reacts with nitric acid or not. Can aluminum hydroxide react with nitric acid readily? Or does it need to be catalyzed, or should I use a different reaction to isolate the aluminum?

At the right ratios, you should end up at same compound.

I have heard that a passive layer of oxide forms on aluminum instead of forming a nitrate

@AdityaDev that works in air, but acids degrade this coating, I believe.

I'm fairly certain that the atmospheric oxide layer is stripped away in nitric acid

Aluminium metal is hard to produce. Industrial processes consume vast quantities of electricity to electrolyse molten aluminium salts. If there were easy ways to do this, they would be used industrially. There is a big incentive to find better ways but yet none have appeared. That suggests your proposed method of aluminium recovery won't work.

Aluminum oxide could be the product of fusing Al(NO3)3 at high enough temperature. Aluminum metal would not be a product, in water or in fused salt.
2.Aluminum metal does not readily dissolve in nitric acid because of formation of passive oxide/hydroxide film. The nuclear power industry had need of large amounts of aluminum nitrate, and it was made by dissolving aluminum metal in hydrochloric acid (very fast) and treating the solution with nitric acid, boiling off the HCl.

Aluminum metal is often prepared for welding by removing the oxide film with dilute hydrofluoric acid. The aluminum surface is bare metal for, oh, maybe a microsecond, before oxidizing back to the oxide. But the aluminum oxide film will be very thin for a few hours and easy to weld. The oxide film grows thicker in air as it ages.

Industrial preparation of Al using electrolysis of molten $\ce{AlF3/AlO2}$ mixture at $\pu{1500 ^\circ C}$ is done not because they like using high temperatures, but because there are no other good ways to make Al. Hydrolysis of aqueous solutions of aluminum salts will not yield metallic aluminum.
Can $\ce{Al(OH)3}$ react with $\ce{HNO3}$? It really depends. Fresh $\ce{Al(OH)3}$ will dissolve in dilute $\ce{HNO3}$. Old $\ce{Al(OH)3}$ ages to $\ce{Al(OH)3}$ and is much less reactive.
Electrolysis of water solution of aluminum salt will not yield metallic aluminum. Hydrolysis of $\ce{AlCl3}$ yields $\ce{H2 + O2/Cl2}$. $\ce{Al(NO3)3}$ will be reduced by forming hydrogen, so you well get a a mix of products: $\ce{NO2}$, $\ce{NO}$, $\ce{H2}$, but not aluminum metal. If this were easy, industry wouldn't bother with hydrolysis of molten $\ce{AlF3/Al2O3}$.

What you mean by:-1."Old Al(OH)3 ages to Al(OH)3 and is much less reactive" 2."Would it be this easy plants wouldn't bother with hydrolysis of molten AlF3/Al2O3"

Please explain how old aluminium hydroxide ages to aluminium hydroxide — the same compound.


text:
Which of the following pair of molecules have identical bond dissociation energies (approx)?

Which of the following pair of molecules have identical bond dissociation energies (approx)?

$\ce{H2}$, $\ce{F2}$
$\ce{F2}$, $\ce{I2}$
$\ce{N2}$, $\ce{CO}$
$\ce{HF}$, $\ce{O2}$

I think it's either (2) or (3). (Single correct type).
Bot $\ce{F2}$ and $\ce{I2}$ bonds are weak because of inter electronic repulsions and due to large sizes respectively.
In case (3), both molecules have bond order 3.

Your large sizes argument is weak. I is much bigger than F.  F2 bond length: 142pm.  I2 bond length:267pm.  Both 2 and 3 are very close in bond dissociation energies.

The answer here has nothing to do with how the molecules relate to each other, e.g. similar bond order or length. This is actually a question that highlights an unexpected break in a trend.
The bond enthalpies for the halogens are:
$$
\begin{array}{|c|c|}\hline
\text{Halogen}&\text{Bond energy (kJ/mol)}\\\hline
\ce{F-F}&\pu{156}\\\hline
\ce{Cl-Cl}&\pu{243}\\\hline
\ce{Br-Br}&\pu{193}\\\hline
\ce{I-I}&\pu{151}\\\hline
\end{array}
$$
Plotting these data we would expect the bond enthalpy of $\ce{F2}$ to be the strongest of them all, something around $\pu{300 kJ/mol}$, but in fact it much weaker than for chlorine.  Indeed it is so weak, it as almost as weak as for iodine.
The reason for this is that fluorine is so small that it breaks the trend.  The atomic radius is so small that the electrostatic repulsion between the nuclei is significant, and the non-bonding electrons surrounding each atom also repel each other due to the small amount of space available.  These forces counteract the strong attraction between the nuclei and the bonding pair of electrons and weakens the bond compared to what might be expected.


text:
Cyclobutylamine with sodium nitrite in dilute HCl
What are the products formed in the reaction of cyclobutylamine with $\ce{NaNO2}$ and $\ce{HCl}$?
I thought the answer was only cyclobutanol. But the answer given says the product contains cyclopropyl carbinol as well. How does the second product form?
This is basically a diazotization reaction which results in  the conversion of $\ce{- NH2}$ to $\ce{- N2^+Cl-}$ which is a good leaving group. Then water molecules will attack the carbon containing the leaving group. Is the reaction $\rm{S_N1}$ or $\rm{S_N2}$?

The water is attacking the carbon next to the one that has the diazonium cation on it, this leads to a bond migration and subsequent loss of nitrogen.  As to whether the reaction occurs in the order described is another matter.

It is an SN1 reaction, see this earlier answer.

@ron why does the nucleophilic attack a carbon that does not contain a leaving group?

The initially formed cyclobutyl cation quickly equilibrates with the cyclopropyl carbinyl cation which has positive charge on the necessary carbon.

Should I mark this question as duplicate @ron ?

I think your question covers the same general ground, so yes, mark it as a duplicate.


text:
Electrophilic substitution reactions
Why is the para position preferred more than ortho in the electrophilic substitution reactions of haloarenes?

It is true already in the case of halogenation, nitration, sulphonation, fiedel crafts alkylation and fiedel crafts acylation.

Klaus Warzecha is right of course.Well,if you aren't still satisfied then you could think of steric hinderance (which is less in para isomer than ortho).But as Klaus Warzecha mentioned,it is completely dependent on reaction condition.

But in my ncert textbook it's written that para position is preferred....

yup i am also studying from ncert book itself.then let me ask why do we get salicylic acid in kolbe and riemmer an tieman reaction?both give ortho product instead of para.(its salycilaldehyde in the latter one.doesn't matter its still ortho product only)

Okay. I'm not doubting your answer but what's the reason behind it and is what's written in the text book wrong?

in para isomer steric hinderance is less.whereas in ortho its more because the two functional or substituent groups are very close to each other in ortho isomer.our textbook talks about the reaction under normal condition only.the textbook isn't wrong,neither Klaus Warzecha's answer.

Is it really? Always?
In an older article (J. Chem. Educ., 1988, 65, 367), the authors described an experiment that demonstrates the outcome of the alkylation of chlorobenzene (1) with 2-chloropropane (2) under kinetic control.
They first added 25 mL of 1 and 2 mL of nitromethane to 1.2 g anhydrous aluminium chloride, and stirred until the aluminium chloride was fully dissolved. To the cooled mixture (3-5 °C), a solution of 3 mL 2 in 20 mL of 1 was dropwise added within 10 min, while keeping the temperature of the reaction mixture below 10 °C. The reaction was stirred for another hour, allowed to come to room temperature.
After workup, gas chromatographic analysis showed that the o/m/p ratio for the resulting isopropyl chlorobenzenes was 63/5/32.
To cut the long story short, under the conditions describe above, the ortho isomer prevailed.

This says the ortho and para positions are equally reactive, and the ortho dominates because there are two potential reaction sides leading to the same product.

@Lighthart Yes, two for the price of one :)


text:
Sn1 mechanism and concentration of neucleophile
Why is low concentration of neucleophile  preferred in Sn1 mechanism?

You already asked this here.


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Meaning of Flammability Limits and reaction in Air of Ammonia
I have read on Wikipedia that the flammable limits of gaseous ammonia are between 15–25% and I have also read that the autoignition temperature is $651\rm~^\circ C$.  And that the combustion of ammonia in air is $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$  My understanding of this is that with a 20% air to $\ce{NH3}$ that I could hold a flame.  Outside of that range, I cannot hold a flame.
My question is if I am at 30% air and above $651\rm~^\circ C$ will Ammonia still combust/react with Air or will it stay $\ce{NH3}$?
My follow up question would be that when if I sprayed ammonia into a heated chamber (above $650\rm~^\circ C$) would flames/the exothermic reaction only happen when the ammonia vapor and the air reach that 15–25% range as the vapor propagates into the air and not react outside that range?

Well you're mixing concepts which is leading to your confusion.
In the section named "Combustion" the Wikipedia text correctly states (per reference) that ammonia burns with difficulty when mixed with 16-25% of air. As you noted, the combustion of ammonia in air is: $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$

In the "Properties" section of the same Wikipedia article there is a subheading named "Solvent properties" which has the weird statement: "Ammonia does not burn readily or sustain combustion, except under narrow fuel-to-air mixtures of $15–25\%$ air." The statement is weird both because of the placement of the sentence as well as the wording. It should be that the fuel-to-air mixture is $16-25\%$ of ammonia, not $15–25\%$ air.

The gist is that you'd mix gaseous ammonia and air in range of $16-25\%$ and blow it through something like a Bunsen burner, which once lit, would allow the gas mixture to continue to burn. But if you just blew such a mixture through a burner, then the gas mixture would not autoignite at room temperature.
Now as to the autoignition temperature of $651^\circ \rm{C}$, ammonia the Wikipedia section "Laboratory use of anhydrous ammonia (gas or liquid)" notes that in the range of $16-25\%$ air that an explosive mixture can result. So at or above $651^\circ \rm{C}$ and outside the limits of $16-25\%$ air, the reaction would still take place but you wouldn't get an explosion. The reaction would proceed until most of the ammonia or the oxygen is used up.

With an explosion you can't really account for the temperature in all points of the explosion so there isn't really any well defined equilibrium.
Without an explosion there would some well defined equilibrium constant for the reaction which would be temperature dependent. (We're assuming that the container of gas is heated or cooled to maintain whatever the desired equilibrium temperature is.) Thus the kinetic energy of the gas molecules at the given temperature is providing the activation energy necessary for the reaction to occur. The equilbrium would be given by the equation of the partial pressures: $$\ce{K_{eq} = \dfrac{{p_{\ce{N^2}}}^2 {p_{\ce{H2O}}}^6}{{p_{\ce{NH3}}}^4{p_{\ce{O2}}}^3}}$$

@ MaxW is there a way to calculate the time it would take for this reaction to take?  I do not know what an equilibrium constant is or how it is used.  I assume it is for calculating the time of equations.  I also assume the equation would be dependent on pressure as well as temperature.

How fast the reaction occurs at a fixed temperature (assuming no explosion...) is a matter of chemical kinetics which is another level of complexity separate from the equilibrium. // If the gas were to start to burn, then as the gas burns it gets hotter which causes it to burn faster. The next result is an explosion.

Chemical reaction would proceed in an air/ammonia mix, even outside the concentration range at which the reaction is self-sustaining, particularly if there were a catalyst to help. It is just that at the specified concentration range, when it reaches $651^{\circ}\mathrm{C}$ at one point in the mixture, the reaction proceeds fast enough to heat the rest of the mixture to that ignition temperature.
This is similar to an Otto cycle spark-ignition engine: with the correct mixture, the flame front propagates from the spark plug throughout the cylinder.
In the Ostwald process, a $1:10$ ration of $\ce{NH3}$ and air can sustain the reaction at the surface of a catalyst.
In a similar fashion, hot copper can catalyze the oxidation of acetone, methanol or ethanol, providing a surface where the reactants can combine fast enough to keep the surface hot.


text:
Concentration of a species outside buffer region (polyprotic acid titration)
I'm wondering if you titrate a polyprotic acid, say, $\ce{H3PO4}$, with a strong base, and you successfully pass the first equivalence point and are maybe at the $\mathrm{pH}=\mathrm{pK}_{a2}$ point now, how would you find the concentration of $\ce{H3PO4}$?
Is there none left? I don't have a specific homework question, I'm just wondering.
Attempt at a solution:
I thought maybe you would find the equilibrium concentration of $\ce{H2PO4-}$, find the $\mathrm{pH}$, and use the $\mathrm{pH}$ to find $\ce{OH-}$. Then, you could write the equilibrium, $$\ce{H2PO4- + H2O <=> OH- + H3PO4},$$ and without making an ICE table, substitute your equilibrium values into the $\mathrm K_a$ expression.

At the one side, you can always calculate the percentages of the acid or it's ions in solution:
$$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{a,j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{a,j}}$$
with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons.

But on the other hand there are some special points during titrations. One of those points did you choose, which is where $\mathrm{pH} = \mathrm{pK}_{a,i}$.
For $\ce{H3PO4}$ there exist three different acidity constants, which describe the three possible dissociation steps. For the first step, the reaction is as you described: $$\ce{H3PO4 + H2O \xrightleftharpoons{k_{a,1}} H2PO4- + H3O+}$$
A little work on it gives you the Henderson-Hasselbalch-equation, which usually is good approximation to buffer regions: $$\mathrm{pH} = \mathrm{pK}_a - \lg\left(\frac{[HA]}{[A^-]}\right)$$
If you enter the same values for pH and pKa you have to solve $$-\lg(x)=0$$ which is true for $x=1$. So you know that you have a 1:1 ratio at this point.
This means for phosphoric acid:
At $\mathrm{pH=pK}_{a1}$, you've got exactly $50~\%~\ce{H3PO4}$ and $50~\%~\ce{H2PO4-}$ (blue, yellow). The same procedure can be applied to the other dissociation steps. Therefor at $\mathrm{pH=pK}_{a2}$ you've got $50~\%~\ce{H2PO4-}$ and $50~\%~\ce{HPO4^2-}$ (yellow, green) and at the last possible $\mathrm{pH=pK}_{a3}$ there are $50~\%~\ce{HPO4^2-}$ and $50~\%~\ce{PO4^3-}$ (green, red).

This can be applied to most acid-base dissociation steps but you have to pay attention for close successive $\mathrm{pK}_a$ values, as if the gap gets too small, the 1:1 ratio doesn't hold.

When you have some solution of $\ce{H3PO4}$, or its various anions, then you'll always have some of all the species ($\ce{H3PO4}$, $\ce{H2PO4^-}$, $\ce{HPO4^{2-}}$ and $\ce{PO4^{3-}}$)present. The relative proportion changes with pH of course.
If you titrate $\ce{H3PO4}$ with a strong base like $\ce{NaOH}$ then there will be three equivalence points. If the $\ce{H3PO4}$ is 1.00 molar, then the first equivalence point will be when 0.33 moles of $\ce{NaOH}$ have been added. The second equivalence point will be when 0.66 moles of $\ce{NaOH}$ have been added. The third equivalence point will be when 1.00 moles of $\ce{NaOH}$ have been added. So the equivalence point is based on stoichiometry and not really the concentration of a particular species of ion.
Of course using the values for pKa1, pKa2, and pKa3 you can calculate the pH at which the three equivalence points take place.
The other point here is the "sharpness" of the equivalence points. If the pka values are close together there isn't a large pH change at the equivalence point. This means that it is hard to detect the precise equivalence point. So essentially you'd like a large pH change for a small volume of titrant change. If the pKa values are close together then you get the opposite, a relatively small pH change for a large volume of titrant.


text:
Determining structure using HNMR and IR
I have a homework problem where I'm to draw the structure of a compound of molecular formula $\ce{C10H12O2}$ using the following IR data and $^1$H NMR spectrum:
IR absorption at 1718 cm$^{-1}$
NMR Data:
Triplet (3H) at ~1.4 ppm
Singlet (3H) at ~2.4 ppm
Quartet (2H) at ~4.3 ppm
Doublet (2H) at ~7.2 ppm
Doublet (2H) at ~7.9 ppm

The two doublets lend to a disubstituted benzene. The IR absorption lends to a C=O.
I came up with this structure:

My thoughts were that the singlet owing to $\ce{CH3}$ would be shifted to the left of the other $\ce{CH3}$ due to the C=O.
However, I'm incorrect. This is the correct answer:

What makes this answer more correct than mine (other than the fact that it's an ester and we happen to be studying esters)? Thank you for any help you can offer.

The only argument I could make for why yours is wrong is that it has a conjugated ketone which will often shift that carbonyl absorption down to the high 1600's while the unconjugated ester absorbs around 1740, so when it's conjugated we'd expect something around 1720 as is given.

I agree with @jheindel's analysis. There's no way I would make a definitive conclusion based solely on this data though. 13C NMR would be really helpful.

One slight NMR-based "issue" is that the ether oxygen would act as an electron-donating group, so the oxygens ortho to that position would likely have lower chemical shifts, closer to 6.5 ppm. Maybe not all the way to 6.5 ppm, but I would expect lower than 7 ppm.

Okay, I looked up the spectra for both. 4'-ethoxyacetophenone has a doublet at 6.85ish ppm, and ethyl 4-methylbenzoate has a doublet around 7.25 ppm.

Just been creating an answer and notice you have edited your peak list table. This makes a world of difference 4.3ppm vs 4.0ppm and 7.2ppm vs 7.0ppm. Are you working off real spectra, or have you been given a peak list?

It's an overly simplified mockup. I'll include the image.

Ok. Given your original 1H shifts, your answer would be the more likely candidate. Given these new shifts, the correct answer is the standout candidate.

What about IR spectrum?

The only information given was the 1718. No image or anything specific about the peak.

Your answer is reasonable, even if not correct.

A methylene group bounded to -O(O)CR has a chemical shift of about 4ppm. A methylene group bonded to -OR group has a significantly smaller shift (about 2ppm).

I think this is a great question/ example. You have used your reasoning and understanding to come to a logical conclusion. If I was marking this in an exam I would give full marks as your answer is well thought out and reasoned. It shows you understand the problem and how to approach the problem. You got the wrong answer? So what!

Okay, following up on my comment, I found the spectrum for the "correct" answer:

As you can see in this IR the absorption of the carbonyl peak is shifted down to 1718 as the problem statement claims.
Now, consider this IR spectrum for a conjugated aldehyde which is similar to a conjugated ketone:

In this spectrum, we see that the IR absorption for an aldehyde (normally around 1715) is shifted down below 1700.
Everything else in your answer seems right though. You always need to be careful to consider conjugation when looking at the peak of a carbonyl stretch.
If you want to look around at HNMR/CNMR/IR for basically any reasonable organic molecule, you should check out the SDBS Database. I actually found this to be a useful study tool and just kind of interesting to look around in.

So with the spectrum finally in hand, we can finally resolve these two structures. This is not a trivial case, but serves to show the importance of accurately reporting chemical shifts.
Initially, your reported chemical shifts supported your proposed structure 4'-ethoxyacetophenone, and I was at a real loss trying to ignore NMR data and justify IR data. However, now I am happy to do what I normally do, and ignore the IR data and explain by NMR.
The general approach to solving problems like this is to correctly do what you have done, and come up with a possible structure and then test the data against it. Whilst it is nice to find authentic spectra of the two possibilities and pick from that, that really doesn't provide a learning pathway to predict more novel unknowns.
There are three pieces of evidence that support the correct structure over your answer: two based on chemical shift, and one based on peak appearance.
Using simple additive tables (See Pretsch, or Silverstein etc), there are two standout chemical shifts. First the methylene protons. For your answer,  4'-ethoxyacetophenone, 4.0ppm is the expected shift, while for ethyl p-toluate, the correct answer, this comes in at around 4.3-4.4ppm. 0.3 ppm is a very significant difference for such small molecules.
The second big difference is in the upfield aromatic proton group. I calculate about 7.0ppm for the protons ortho to your ethoxy group in 4'-Ethoxyacetophenone and around 7.2ppm for the same group in Ethyl p-toluate.
There is very little difference in the expected shifts for the methyl protons. A close inspection of the data would likely show long range coupling between the methyl group and the upfield aromatic protons (7.2ppm) for ethyl p-toluate, although would possibly be within line width and so only appear as broadening. In the aromatics, this manifests in the upfield protons appearing lower in intensity than the signals at 8ppm. This is indeed the case, and further supports the correct answer. For your proposed answer, both sets of aromatic signals would be the same intensity. Often very subtle differences in linewidth can tell an awful lot about bonding pathways. Also, given a real spectrum, data massaging would enhance this coupling in the methyl group and make it more obvious.
Of course, you would also run a 13C{1H} and show instantly that you had an ester group and not a ketone.
I'll be honest and say I couldn't really be confident in using IR to distinguish these. Someone else may give that one a shot though.

‘However, now I am happy to do what I normally do, and ignore the IR data’ — stopped reading and upvoted there 8D

Look at the standard reference for solvent impurity shifts. Acetone in chloroform is $\delta\ 2.17$. Toluene is $\delta\ 2.36$. I think your structure is perfectly reasonable. In practice, you would try to derivatize your compound. LAH reduction gives very different products that should be easy to identify via GC-MS.


text:
Deuterium removal from water
What percentage of deuterium might be removed from water after a typical electrolysis procedure? From a post on this site I read that, "In the first stage, $\ce{NaOH}$ solution (initially $0.5\rm{M}$) is subject to electrolysis, until only $\frac{1}{32}$ of the original volume remains. This increased the original concentration of deuterium by a factor of $12$."
This seems to be, for the electrolyzed hydrogen, an approximate $38\%$ decrease from the original concentration. If we assume the original concentration was $150\rm{ppm}$ then after the first pass, we might expect to have $93\rm{ppm}$ in the deuterium reduced hydrogen. If my estimate is valid, may I expect that each successive pass would reduce the deuterium content by a similar percentage?

Looks like so. Also, the procedure is likely optimized for getting deuterium; if your goal is opposite, you might want to stop at an earlier point.

You are quoting from my answer here.
Looking at the source reference Inexpensive Equipment for the Preparation
and Concentration of Pure D2O or of
Deuterium-Rich Water. Description of
Construction and Operation of the Former
Ohio State University Heavy Water Plant The Ohio Journal of Science. v41 n5 (September, 1941), 357-365:

With NaOH electrolyte we obtained, with Armco iron, an electrolytic
  separation factor[footnote 8] of a little more than 8, whose magnitude was
  practically independent of current density and of the deuterium concentration
  in the electrolyte. This is the true separation factor applicable
  to the dry gas. The practical separation factor is somewhat less than
  this (about 6 or 7) due to water vapor carried out with the electrolytic
  gases.

So, like Ivan Neretin is commenting, it is not optimal to electrolyze most of the first stage solution if the goal is obtaining deuterium depleted hydrogen.  The initial hydrogen coming from natural abundance water will be deuterium depleted by a factor of "about 6 or 7".  Since natural abundance water is essentially free, it would be best to only electrolyze a small fraction of a large amount of water.  Then you will deplete deuterium down to about 25 ppm in one stage.
After the first stage, if you attempt a second stage, burning the water to make deuterium depleted water, you will then have to add electrolyte, which, if it is NaOH, would introduce more deuterium.  So if you used a hydrogen-free electrolyte it would be better.  By the second stage, the starting material is valuable, and you would want to electrolyze a significant fraction of it, but not 31/32.


text:
E1 elimination and alkyne formation
Are E1 elimination reactions possible from a trans dihalide alkane (like 1,2-dibromoethane) all the way to the alkyne (like acetylene)? Or is only E2 possible from the alkene to alkyne? What is the rate of reaction?
My thoughts are that E1 is a possible reaction to form the alkyne, with E2 taking place if there are any anti peri-planar hydrogens and a strong base. So, another question might be that which of these has the faster rate of reaction - an E1 reaction or an E2 reaction with anti periplanar hydrogens that are only available through conformational energies and torsional strain?

What is the proposed E1 intermediate?  Do you have any idea if this is a high-energy intermediate?

Yes, in your example, it is possible to form acetylene. E1 mechanism does not depend on whether the leaving group is anti to $\ce{H}$ or not.
All vicinal halides will not give alkynes, as E1 mechanism involves a carbocation intermediate which can rearrange to the most stable position. Example: In 4-methyl-2,3-dibromopentane, the product will be a conjugated alkene not alkyne.
Rate of E1 elimination depends on the stability of the carbocation whereas rate of E2 mechanism does not. Since in your example, the carbocation formed is a 1 degree, it is not very stable so the rate will be higher for E2 mechanism.


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Ranking gases by value of the van der Waals equation-of-state constant 'a'?
How can I find out which gas has the highest value of the van der Waals constant $a$?
Example set of gases in ascending order of $a$: $\ce{NO2}$, $\ce{CO2}$, $\ce{H2}$, $\ce{H2O}$ and $\ce{CCl4}$.

Taken from your comment to Max' answer, I think what you are really asking is how you can determine the value of a for any gas without looking it up. I think that would be a valid question. However, now your question was treated as if it was a homework question, because as it is posed, you simply have to look up the values. Hence it is not a good fit for our format. I recommend [edit]ing your question, so that it can be reopened and properly answered.

You look up the values of "a" in a table of constants for the van der Waals equation.
\begin{array}{lrr}
   \text{gas}                  &
    a / \mathrm{L^2bar\,mol^{-2}} &
    b / \mathrm{L\,mol^{-1}}      \\\hline
   \text{Hydrogen}             &    0.2476  &         0.02661 \\
   \text{Carbon dioxide}       &    3.640  &         0.04267 \\
   \text{Nitrogen dioxide}     &    5.354   &         0.04424 \\
   \text{Water}             &    5.536  &         0.03049 \\
   \text{Carbon tetrachloride} &  19.7483  &         0.1281  \\
\end{array}

Yes, but if I do not have this table which criterion should ultilizar to determine which has the greatest 'a'? type polarity?

So I find some function of the critical temperature that gives the constant "a." How do you get the critical temperature? You look it up in a table. What is the difference?!? https://en.wikipedia.org/wiki/Van_der_Waals_equation#Reduced_form


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Carbocation Stability

A homework question I was given asked me to identify the most stable carbocation between these 4. I was able to narrow it down to either the trifluoromethylcyclohexane carbocation or the methoxycyclohexane carbocation, but I’m not sure which between the two of them.
My gut feeling is that the methoxycyclohexane is more stable because of its contributing resonance structures, but I’m not sure if that trumps the hyperconjugation that the fluorine atoms provide. Any insight would be appreciated!

The fact that the methoxycyclohexane carbocation is resonance stabilized greatly increases its stability over the the trifluoromethylcyclohexane carbocation.
The same power of oxygen’s ability to donate electrons can be seen in its effects on benzene rings. Both $\ce{-OH}$ and $\ce{-OR}$ groups are strongly activating groups, while $\ce{-CF3}$ is a ring deactivating group, meaning it actually pulls electrons from the benzene ring. This further shows that the methoxycyclohexane carbocation is the most stable of the 4.


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Which method can be used to determine the pKa of an acid-base indicator like bromothymol blue?
I have a question about how one would determine the pKa of an acid-base indicator such as bromothymol blue.
I know that spectrophotometric analysis is one option, but is it possible to perform an acid-base titration to find the pKa (7.1)? How would one set up this experiment?
I was thinking I would use a pH meter/titrant curve and treat the indicator like a weak base, but what titrant would I use? Would I use a strong acid like HCl?

Yes, HCl would work with a pH meter. If you know the mass of base (indicator) you use, you can as 0.5 equivalents of a strong acid such as HCl and determine the pH. At that point, since the weak base would be half protonated, the pH would be equal to the pKa, based on the Henderson-Hasselbalch equation. Or you could just so the whole titration and find the pKa halfway to the equivalence point.
In this case, you have to make sure you start with the basic form of the indicator, however. Not sure what form it is supplied in.


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How do people know HCN smells like almonds?
I was told by my chemistry teacher that $\ce{HCN}$ smells like almonds. She then went on to tell a story about how some of her students tried to play a prank on her by pouring almond extract down the drain to make her think that they had inadvertently created $\ce{HCN}$ gas. She said that she knew that it wasn't $\ce{HCN}$ because if she had smelled the almond scent, then she would have already been dead.
I never asked her, but how do people know $\ce{HCN}$ smells like almonds if they would die before they knew what it smells like?

That's an exaggeration. Some compounds would probably kill you before you have the chance to feel their smell, but $\ce{HCN}$ would not.

I have a compound in my synthesis right now that smells like almonds. I instinctively am weary of the smell, despite having never (knowingly) smelled HCN. In addition to what everyone else has said, it should be noted that bitter almond oil (apparently) actually contains HCN. I knew amygdalin was a component, but I found out a while back that, supposedly, there is free HCN in the oil (likely a decomposition product of amygdalin). See "side effects" section here.

@DGS Very interesting! Reminds me in the question by a writer on amygdalin a while ago: http://chemistry.stackexchange.com/questions/24528/how-do-i-extract-cyanide-from-apple-seeds

As to the comment by your teacher, the smell of the almond extract was probably quite pronounced, so perhaps her comment was along the lines of, "that much HCN would have produced symptoms already, so it isn't HCN".

https://www.youtube.com/watch?v=WYagO-nup6c Nile Red's cyanide video

The odour threshold for hydrogen cyanide $(\ce{HCN})$ is in fact quite a bit lower than the lethal toxicity threshold. Data for $\ce{HCN}$ can be found in many places, but here and here are a couple of good references. That subset of the human population that can detect bitter almonds do so at a threshold of $0.58$ to $\pu{5 ppm}$. The lethal exposure dose is upwards of $\pu{135 ppm}$. That's a whole $\pu{100 ppm}$ range in which to detect and report the fragrant properties.

You did beat me by the length of a shower and a shave :D  Further data on inhalation effects can be found in the SCOEL Recommendation 115. One should however keep in mind that a number of people can't smell HCN at all.

I wonder who was the first to volunteer for this experiment. "100ppm and you won't die, I swear, just sniff this jar and tell me what you smell!"

For all we know HCN at 100ppm and above may smell substantially different than the dilute one at 5ppm.

One way to look it at this is that our olfactory system wouldn't have been much of an evolutionary advantage if it didn't warn us about poisons before they killed us.

@Crashworks True! But also why we can make use of sentinel species like the canary in the coalmine :-)

Missing a major point: inhaling 100 ppm for a short time has absolutely no ill effects, as you need to accumulate a certain amount (about 10 mg) of cyanide in your blood for it to kill you. One breath would have to be highly concentrated to kill you - on the order of 10,000 ppm. See my answer for details.

Recognize that a whiff of most toxins, even in high concentration, will probably not kill you. You need a sufficient concentration in your blood - which means you have to actually get a certain number of HCN molecules to penetrate across the mucosa of the lung and into the blood stream. Typical breathing volume is about 500 mL (tidal volume), about 1/40th of a mole, so a single normal breath of 100 ppm HCN would contain 2.5 µmol of HCN.
According to the CDC, humans can tolerate inhalation of 50 ppm HCN for half an hour "without immediate or delayed effects", while 100 ppm for more than half an hour may be fatal.
The Mayo clinic reports that blood cyanide levels over 2 µg / ml are toxic; if you have 5 liters of blood, that corresponds to 10 mg. The molar mass of HCN is 27, meaning that 10 mg is 370 µmol. At 2.5 µmol / breath, you should be able to take 148 breaths; at 6 breaths per minute, that would be about 25 minutes.
All these numbers are quite consistent, and it tells you that you can tolerate a whiff of cyanide without ill effects. But it's a good idea, once you smell it, to open the windows, turn on the extractor fans, and get yourself to some fresh air.

Gatterman reports (Org. Synth. 1927, 7, 50, as a footnote) that people who smoke regularly have enhanced sensitivity to the smell of cyanide gas, and he recommend smoking while preparing it!
Organic Synthesis Collective Volume 1 1941 314-315
Just opening the NaCN container, most regular (and former regular) smokers can smell the trace amount of HCN formed from the water vapor in the air.
I suspect your teacher was exaggerating slightly.

Even if cyanide gas isn't a reactive fuel for a cigarette butt, surely other gases in the lab are -- that doesn't sound like sound advice to me :P

It isn't sound advice, the recommendation is from 1927.  Old-fashioned, outdated and dangerous.  I wasn't smoking when I was making hydrogen cyanide.

It may not be sound advice, but it is true that smokers smell cyanide at much lower levels. It is probably one of the few positive benefits of smoking.

Hydrogen cyanide $(\ce{HCN})$ is variously described as smelling of bitter almonds, marzipan, ratafia, or peach kernels. While some people can smell $\ce{HCN}$ at very low concentrations, many people cannot perceive the odour at all. The odour threshold is about $1{-}6\ \mathrm{mg/m^3}$ for people who are actually sensitive to the odour of $\ce{HCN}$.
Inhalation of $\ce{HCN}$ at low concentrations above the odour threshold is not necessarily lethal. Toxicity of inhaled cyanides is strongly dependent upon concentration and exposure time. By way of comparison, guideline figures taken from Marrs, T. C.; Maynard, R. L.; Sidell, F. R. Chemical warfare agents: toxicology and treatment; John Wiley & Sons, 1996; p 204 are shown in the following table:
$$\textbf{Inhalation toxicity to humans}\\
\begin{array}{lll}
\hline
\text{Time} & \mathrm{LC_{50}} & \mathrm{LCt_{50}} \\
\text{in}\ \mathrm{min} & \text{in}\ \mathrm{mg\ m^{-3}} & \text{in}\ \mathrm{mg\ m^{-3}\ min} \\
\hline
\hphantom{0}0.25 & 2\,400 & \hphantom{0\,}660 \\
\hphantom{0}1 & 1\,000 & 1\,000 \\
10 & \hphantom{0\,}200 & 2\,000 \\
15 & \hphantom{0\,}133 & 4\,000 \\
\hline
\end{array}$$
Hence, for example, it is possible to smell $\ce{HCN}$ for a period of $1\ \mathrm{min}$ at a concentration of $100\ \mathrm{mg/m^3}$, which is well above the odour threshold $(1{-}6\ \mathrm{mg/m^3})$ but well below the lethal concentration for this period $(1\,000\ \mathrm{mg/m^3})$. However, after an extended period, inhalation of $\ce{HCN}$ at this concentration will probably become lethal.
Even at acute exposure to high concentrations of $\ce{HCN}$, it is possible to smell $\ce{HCN}$ before the toxic effects occur. This may be illustrated by a description given in Vedder, E. B. The Medical Aspects of Chemical Warfare; Williams and Wilkins, 1925; p 187:

In an atmosphere containing a lethal concentration an odour of bitter almonds is noticed. This is followed by a sensation of constriction of the throat, giddiness, confusion and indistinct sight. The head feels as though the temples were gripped in a vice, and there may be pain in the back of the neck, pain in the chest, with palpitation and laboured respiration. Unconsciousness occurs and the man drops. From this moment if the subject remains in the atmosphere of hydrocyanic acid for more than two or three minutes death almost always ensues, after a brief period of convulsions followed by failure of respiration.

However, the figures given above are extremely uncertain. Nevertheless, note that $\ce{HCN}$ does not obey Haber’s rule $(c \cdot t = k)$. One important reason for the dependency of the toxicity upon concentration is the existence of various pathways for detoxication. Detoxication explains the ability to withstand very low concentrations of cyanide indefinetely. However, it is unlikely that detoxication plays any significant role in acute cyanide poisoning.
A significant contribution to the uncertainty of acute inhalation toxicity is caused by the variable breathing rate. Standard reference values taken from ICRP 66 (1994) are shown in the following table:
$$\textbf{Breathing rates for adult males}\\
\begin{array}{lll}
\hline
\text{Acitivity} & \text{Breathing rate} \\
 & \text{in}\ \mathrm{m^3\ h^{-1}} \\
\hline
\text{Resting (sleeping)} & 0.45 \\
\text{Sitting awake} & 0.54 \\
\text{Light exercise} & 1.5 \\
\text{Heavy exercise} & 3.0 \\
\hline
\end{array}$$
However, such values may not be applicable to acute cyanide poisoning because of the respiratory stimulation caused by $\ce{HCN}$.

how do people know HCN smells like almonds

Bit late to the party, but I'm missing a crucial part in the existing answers. So here are my 2ct: HCN does not smell like [bitter] almonds.
Personally, I find the description "HCN smells like almonds" very confusing: what we typically associate with the smell of almonds is benzaldehyde rather than HCN.
I believe that confusion comes from crime stories probably written by people who knew that bitter almonds "contain" cyanide (and that [some] people can smell cyanide) and also knowing the smell of bitter almonds (after all, they are/were used for their arome) - but not knowing/realizing that the major arome component is not the cyanide.
Bitter almonds contain amygdalin, a cyanogenic glucosid. Amygdalin can be hydrolysed into gentiobiose (disaccharid of 2 x glucose) and mandelonitrile, the nitrile of mandelic acid (Mandel = German for almond) which in turn is hydrolyzed into HCN and benzaldehyde. This leaves us with 2 volatile compounds, benzaldehyde and HCN.

Benzaldehyde has a very strong smell (odor threshold 0,2 mg/m3 ≈ 0.04 ppm).
In addition, also the HCN can be smelled (not by everyone, btw. but by many people, and with varying detection limits - this has been studied including the genetics in detail in the 1950s/60s, odor threshold 0.2-5 ppm)
Bitter almods are used as spice for baking, but keep in mind that HCN is not very stable (e.g. you cannot keep small amounts of KCN or NaCN very long at ambient air: they will be oxidized into cyanate). The HCN basically doesn't make it into the baked cookies.
What we associate with "almond" is the benzaldehyde.
Also the artificial "almond arome" you can buy (at least here in Germany) is benzaldehyde without HCN. (An oil exctracted from bitter almonds or other prunus kernels may contain or form HCN, though)

Personal experimental anecdata:

I believe I'm a reasonably good smeller of HCN (I did solve that part of a fellow student's anion analysis in our first semester that way by sniffing, and comparing to the smell over KCN salt)
Still, mashing bitter almonds in water to me leads to an overwhelming benzaldehyde smell that masks the HCN smell.
I do smell the HCN, though, if the bitter almonds are mashed in a (non-smelling) acidic solution (e.g. dilute sulfuric acid): that way, HCN smell comes first before benzaldehyde takes over.

So the chemistry teacher could (should!?) have known that the joke smell was benzaldehyde.
(And that the risk of expensive natural bitter almond oil being used instead of cheap benzaldehyde was minute - not to say non-existent given that the students were doing well ;-))


text:
Line spectrum of boron
I'm having trouble with this question from a first year undergraduate chemistry paper. The question states

How many lines will appear in the line spectrum of the Boron atom if the valence electron in the ground state is excited to the 3p orbital?

If in another experiment, the electron of $\ce{B^4+}$ in $n=6$ is excited to $n=7$, how many lines will appear in this line spectrum?

Boron has an electron configuration of $\rm{1s^22s^22p^1}$, so this means it would have to jump from $2\rm{p}$ through $3\rm{s}$ to $3\rm{p}$. I don't understand how I would work out the number of lines that would be seen in the emission spectrum though.
Update:
The memo gives an answer of 6 possible lines - I think the reasoning behind this is that for the valence electron in the ground state (which is 2p) it could jump (depending on how much energy is given to the system) from 2p -> 3p or 2p->3s or 3s->3p but because we don't know the spin on the electron, whether positive or negative we say there are 6 possible lines (3 lines for each possible electron state). Not sure if that reasoning makes sense or not?

What does 'the 5b atom' mean here? Can you clarify?

Sorry, mistake on my part. It should just be B, Boron atom. Have corrected.

Indeed, the first part doesn't make much sense. The 3p <- 2p transition is Laporte forbidden. I assume the intention of the question is to ask about spin-orbit coupling. I think you might be better off asking one of your tutors.

I will certainly ask my tutor. I have a feeling you are right with regards to the question relating to spin-orbit coupling. The memo gives an answer of 6 possible lines - I think the reasoning behind this is that for the valence electron in the ground state (which is 2p) it could jump (depending on how much energy is given to the system) from 2p -> 3p or 2p->3s or 3s->3p but because we don't know the spin on the electron, whether positive or negative we say there are 6 possible lines (3 lines for each possible electron state). Not sure if that reasoning makes sense or not?

The 2p-3p transition is forbidden https://en.wikipedia.org/wiki/Laporte_rule Also spin-orbit coupling is not as simple as that. Search for "Russell-Saunders coupling"


text:
Uses of oxidizing and reducing flames

Why do we use oxidizing and reducing flame in different analytical tests?

For example, in the charcoal block reduction test, we use reducing flame. Why? How does this affect the salt to be reduced into metal?

From Wikipedia: "the oxidizing flame is usually undesirable for welding and soldering, since, as its name suggests, it oxidizes the metal's surface."
" The reducing flame is also called the carburizing flame, since it tends to introduce carbon into the molten metal." Here is the link: https://en.m.wikipedia.org/wiki/Oxidizing_and_reducing_flames

I need detailed explanation as to why we are using it

As the name suggests the oxidizing flame is used to oxidize probes, while the reducing flame is used to reduce probes. Hereby oxidation is done by oxygen, while reduction is done by reducing agents in the flame, such as carbon monoxide or hydrogen.
In the charcoal block reduction test with a blowpipe one often will oxidize the probe in the first step to get rid of nonmetal elements in the probe, and then will use the reduction flame to reduce oxidic metal compounds in the melt to the elemental metal. Volatile metals will hereby be precipitated at the colder regions of the charcoal as oxides. Metal and oxide precipitate can then be further analyzed.


text:
excitation upon emission from 2 LEDs
I came across a problem:
I have 2 LEDs with different emission spectra (blue and green).
I have 1 molecule with a known excitation spectrum (channelrhodopsin2 protein).
I turn on 1 LED at a time.

I want to approximately estimate the ratio of the (level of excitation generated upon green over blue emission) in the full spectrum.
I initially thought it was the dot-product between the emission and excitation functions. But my colleague told me that I have to add the lambda and he offered equation 3 from this website
:
$$J = \int_o^{\infty} F_D(\lambda)\epsilon_A(\lambda) \lambda^4 d\lambda$$

FD(λ) is the normalized emission spectrum of the donor
$\epsilon_{A}$ standards for the molar absorption coefficient of the acceptor
λ is the wavelength.

I'm not sure how this equation applies to the problem at hand, in special I don't know what should the exponent from lambda be.
I have measured the power of both LEDs, and I don't care about the real level of excitation, just the ratio between the excitation generated from the 2 LEDs.

re: "I don't care about the real level of excitation, just the ratio...". You can't get the ratio without assuming something about the power. Is the peak intensity of green and blue LEDs assumed to be the same, or is the area under the peak assumed to be the same, or is the electrical power assumed to be the same?!? After that decision it is just a matter of doing the integral.

We measured the power at the peak intensities for both LEDs. We assume the peak is the same, not the integral.

We measured the power at the wavelength of peak intensity for both LEDs. We assume the peak intensity is the same, not the integral. During the experiment, we applied the same power (at the peak intensity) on the blue and on the green LEDs.

@MaxW, what integral are you talking about? does it involve multiplying $\lambda$ or not?


text:
Why can't two Silicon tetraoxide tetrahedrons share an edge or a face?
Our teacher told us about structural types of silicate minerals (neosilicate, sorosilicate, cyclosilicates ...) saying that those tetrahedrons can only share one corner (oxygen), but he didn't give us any explanation, could anyone help me with this?

That would be too much of angular strain.

What if the angles facing each other, symmetrical in respect to the potential shared edge,  were equal?

They are equal; so what?

In that case, both angles would be 109°, which does not qualify as "abnormal" for a tetrahedron, i don't see source of the strain here.

I suppose you are talking about the $\ce{O-Si-O}$ angles. OK, but what about $\ce{Si-O-Si}$?

I found in the following page that the optimal angle for a bridging oxygen is 140°, and that it can vary from 120 to 180, when i visualize the two tetrahedrons, i would say that the angle must too sharp in the case of an edge or face sharing, causing the said strain, i'm i right?

Here is the link : https://books.google.dz/books?id=yMGzmOqYescC&pg=PA142&lpg=PA142&dq=bridging+angle+si-o-si+angle&source=bl&ots=whvDBEnuMt&sig=SWFRYFEV8MZ11G4dVWLRyhCw6QU&hl=en&sa=X&ved=0ahUKEwjA9anSm6DLAhVDqQ4KHfGlBD8Q6AEIIzAB#v=onepage&q=bridging%20angle%20si-o-si%20angle&f=false

That's right. If you have a 4-membered cycle, some of the angles must be too sharp.

In solid-state chemistry, we know a certain set of rules called the Pauling rules. One of them states that sharing faces or edges is unfavorable. Imagine your silicate-tetrahedra to share faces. That would equal the closest distance between two highly charged silicon cations. They both have a charge of +4, so there is really some repulsion. Same goes for the edges which is also quite close.
If you increase the overall size of the tetrahedron however by exchanging oxide by sulfide you will end up in the silicon(IV)sulfide structure which is essentially long chains of tetrahedra that share edges. Here by introducing a much larger anion the distance between the two silicon centers is increased and edges can be shared.


text:
Galvanic cell of copper and zinc
When we dip zinc metal in zinc sulfate solution, does zinc sulfate ionize into Zn and sulfate ions? Zinc ions come into the solution as Zn$^{+2}$ so why is the metal not completely dissociated?
Why is zinc sulfate used as an electrolytic solution?

Basically, this is because we want to know the concentration on $\ce{Zn^2+}$ to plug it into the Nernst equation. The galvanic cell would work just as well with pretty much any other electrolyte.


text:
Ammonia Reaction with O2
I asked a similar question earlier that lead me to ask this question.
NH3 and O2 react in the following way $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$  From my understanding Ammonia will explode(react quickly/release a lot of heat quickly is how I understand the term explode) when there is 15%-25% ammonia vapor in Air.
If I injected ammonia vapor into the air (above autoignition temperature of 651°C so that it's fuel to air ratio was more like 30% ammonia to air.  Would it follow this reaction slowly until it reached the 25% mark then explode/react quickly until it reached the 15% ammonia balance air, and then react slowly from 15% ammonia to lower?
Here is the link to my similar question. Meaning of Flammability Limits and reaction in Air of Ammonia

Are you certain that the oxidation of ammonia under these conditions does not yield nitric oxide instead of nitrogen?

I am not certain.

No.  Because remember, the oxygen is being consumed too.  If we assume it's reacting via that reaction (realistically, you're going to get a complex mix of products) then we start with something like: 30% NH3 56% N2 14% O2 (the mix is 70% air, which is an 80/20 mix of N2 and O2).  For every 4% reduction in Ammonia, you're getting a 3% decrease in oxygen.  So O2/NH3 reactions will be rarer (slower), and slow down heat generation.  Also, the amount of inert gas in the vessel will be increasing, which also need to be heated up, so less of the enthalpy of reaction will go into increasing the temperature.
That said, above the autoignition temperature, it's going to be a pretty fast reaction anyway...  What we can be sure of is it will stop reacting at about 10% NH3 when all the oxygen is gone.


text:
Why are potential electrolytes poor conductors in the pure liquid state?
In my textbook, "In the pure liquid state, a true electrolyte is a good conductor of electricity. In contrast, a potential electrolyte is a poor conductor in the pure liquid state."
 definitions
"A true electrolyte consists of ions in the pure state. Most salts are true electrolytes. A crystal of NaCl, CuSO4, or MgS consits of positive and negative ions."
"A potential electrolyte consists of uncharged molecules in the pure state, but when dissolved in a solvent, it reacts with the solvent to some extent to yield ions"

You've pretty much answered your own question. If a potential electrolyte (the term sounds strange to me, but whatever) in the pure state consists of uncharged molecules, then how would it conduct electricity?

I read it more that it was uncharged(not an ionic crystal) but when you put it in water, it becomes an electrolyte. Compared to true electrolytes, NaCl which has ionic structure and is electrolyte.

Geez, you heard about covalent compounds dissociating after being mixed with water, didn't you? If they aren't dissociated earlier, they don't conduct. Provide yourself example

@Mithoron How about bubbling CO2 through water? You'd end up with an acidic solution, which will be more conductive than pure water, by adding a neutral compound. I have to say though, I don't really understand the question either.

@SeanDoris Well, OP's asking about liquids so rather acetic acid. And answer would be probably: because they don't autodissociate strongly.


text:
What type of isomers are cis-but-2-ene and 2-methylpropene?
One of my homework problems is asking whether cis-but-2-ene and methylpropene are cis-trans isomers or constitutional isomers. But I'm not sure about whether cis-trans isomers include those which the two functional groups exist on the same carbon in the double bond. So does it?
I've searched on Google and looked in my textbook, but they doesn't seem to have anything.

(2Z)-But-2-ene and 2-methylprop-1-ene are constitutional isomers as they have different connectivities, i.e. the sequence in which the atoms are bonded to each other is not the same. For example, the carbons highlighted here are different: in (2Z)-but-2-ene it is only bonded to two other carbons (through a single bond and a double bond), but in 2-methylprop-1-ene it is bonded to three other carbons:

Cis/trans isomers are a subset of stereoisomers, which need to have the same constitution, or molecular connectivity. The two isomers of but-2-ene satisfy this:

Note how both highlighted carbons are now bonded to the same partners, but with a different spatial arrangement.


text:
What are the steps to balance this equation? confused
$$\ce{C_4HO + O_2  ->  CO_2  + 2H_2O}$$
the book tells me that the balanced equation is:
$$\ce{2C_4HO + 13O_2  ->  8CO_2  + 10H_2O}$$
but when I try to balance it to get the answer I get something totally different. I want to know why, and also why is what Im doing wrong?
I get :
$$\ce{4C_4HO + 15O_2  ->  16CO_2  + 2H_2O}$$
Both equations balance so how do you know which one is correct?
Thanks for any help

What is $\ce{C4HO}$? Is it supposed to be methanol?

@orthocresol Exactly! I agree that it's probably $\ce{CH3OH}$.

it says its butane, C4H1O

oh wait...i think  this is all due to formatting error..

If you start from $\ce{C4H10}$, you'll probably agree that you need 2 oxygen atoms for every carbon and one oxygen atom for each two hydrogen atoms. Remember that oxygen comes as $\ce{O2}$. Count the oxygen atoms that you will need. If the number is odd, multiply the number of butane molecules by two.

The equation in your textbook is imbalanced.
On careful inspection, one finds oxygen imbalanced.
On the reactants' side we find 28 atoms of O. On the products' side we find only 26 atoms of O.
Your balancing seems correct.
Although the compound $$\ce{C_4HO}$$ seems ridiculous. So I'm not sure if the reaction is correct.


text:
Does aluminium (III) react with copper?
We all know know that Aluminium react with cupric producing Copper and Aluminium (III), but does Aluminium (III) react with Copper?

Do you knot that is better not to mix aluminum and copper wires? http://homeguides.sfgate.com/problems-aluminum-copper-wire-electricity-86313.html


text:
What is this SMILES notation: C[Si](C)C |^1:1|
When converting InCHI to canonical SMILES using Indigo, I will occasionally get an output SMILES string like this:
C[Si](C)C |^1:1|

What does the "|^1:1|" part mean? I can't find this explained anywhere in either SMILES grammar documentation, nor Indigo documentation, although other chemical software seem to accept it as a valid SMILES string. Any ideas on the meaning of this annotation?

This is part of an extended SMILES syntax used by ChemAxon. Please have a look at the docs

The SMILES syntax in question is neither mentioned in the Daylight Theory Manual, nor in the OpenSMILES specification.
It is not part of the canonical SMILES syntax, but an extension, developed  by the ChemAxon company.
According to their documentation, extensions are tagged by leading and trailing | (pipe) symbols. ^1:, followed by an atom index, indicates a

monovalent radical center:


text:
Why is the oxidation state of carbon in carbon monoxide +2? Why does its dipole moment point from oxygen to carbon?
Shouldn't it be +3 since 3 of its electrons are being pulled away by the electronegative oxygen?
Also, shouldn't the dipole moment point from the carbon to the oxygen since most of the electron density in this molecule is surrounding the more electronegative oxygen?

It's +2 because the 3th electron pair is the non-bonding electron pair of oxygen.

http://chemistry.stackexchange.com/questions/30797/how-can-the-dipole-moment-of-carbon-monoxide-be-rationalised-by-molecular-orbita http://chemistry.stackexchange.com/questions/32161/can-i-predict-the-preferred-resonance-structure-of-carbon-monoxide-from-its-mole


text:
How to use the enthalpy of formation to assess the energy stored in a molecule?
The standard enthalpy of formation for glucose is -2801 kJ/mol, while the standard enthalpy of formation for carbon dioxide is −393.5 kJ/mol.
If one simply examine the numbers attached to the molecules, one may have the impression that far more energy is released from the formation of glucose and thus come to understand glucose as a molecule with a low "level of energy" but this is clearly not the case as glucose is in fact combustible in most circumstance (produce carbon dioxide in the process, spontaneously)
This impression is probably inappropriate as glucose is a far larger molecule than CO2, and thus more changes is done to a greater amount of single-element reactants, but is there a better way to appreciate the amount of energy stored in a kind of molecule based on the enthalpy of combustion?

To calculate enthalpy of combustion you need full equation of combustion.

Sorry, I believe you have misunderstood my question. I want to know if there is a way to assess the amount of energy stored in a molecule. The fact that glucose can combust to form carbon dioxide shows that there are more energy in the glucose molecule before it is converted to carbon dioxide and water. NOw that you have mentioned it, I think may be asking for an invalid comparison, maybe the energy stored in a molecule can not be calculated and compared with other species of molecules.


text:
Farnesyl Diphosphate: Chemical commercial synthesis
Are there any practiced commercial routes for the synthesis of "farnesyl diphosphate"? I'm especially interested in options that would scale to industrial scale processing.
To add some context, there's a bunch of secondary metabolites that branch out from FPP. Assuming one can immobilize the requisite enzyme the next question is whether the substrate FPP could be somehow manufactured at scale using a chemical synthetic route.

Farnesol is commercially available as a distillate of essential oil or synthetically derived from Nerolidol. These chemicals are commonly used in the fragrance industry, as such efficient synthetic processes should exist.
Here's a potential synthesis route for Farnesol (1) from the starting material Nerolidol (2). Farnesol is then phosphorylated into Farnesyl diphosphate.

Nerolidol (2) is treated with anhydrous acetic acid under basic conditions to produce Farnesol (1). From The Total Synthesis of Natural Products
edited by John ApSimon, page 201. Also referenced in US1663817A.
If we assume that isopentenyl pyrophosphate is similar to farnesyl diphosphate (1 isoprene unit vs. 3 isoprene units), then this technique "An Improved Synthesis Of Isopentenyl Pyrophosphate" can theoretically be used.

The double phosphorylation has stated yields from 32%-51%. Let's assume the yield of Nerolidol to Farnesol is 70%. Given the price of Neroidiol is 39 USD/kg†, the least efficient price would be about 174 USD per kilogram. The price of Farnesol is 129 USD/kg†, and Geraniol is 19 USD/kg†. (If one would like to synthesize Geraniol into Nerolidiol, steps are given on page 201 above.)
This is pure conjecture, but should most likely work given that Farnesol is similar to both Geraniol and Isopentenol. Also the reason I included the step from Nerolidol to Farnesol is because Farnesol is too expensive compared to the price of Nerolidol despite being one synthetic step away.
†Prices are quoted from Vigon International, a North American reseller.

Great answer! This solves a lot of my problems. Thanks very much!

Glad to see it helped! There was a small vial of Farnesol that I was smelling as I composed your answer :)

Diphosphates (and sulfates) of isoprenoid alcohohols are rather sensitive to hydrolysis. (Back in the days, we gave up further work on sulfates and performed our experiments on biomimetic cyclizations using the acetates until I developed something better.)
To my knowledge, the ammonium salt of (E, E)-farnesyldiphosphate is commercially available on a milligram scale and I wouldn't be astonished if it is only synthesized upon customers' demand.
The synthesis of the diphosphates and their hydrolysis is described in two articles published 1967 in German in Tetrahedron by Rittersdorf and Cramer.

F. Cramer, W. Rittersdorf, Die Hydrolyse von Phosphaten und Pyrophosphaten einiger Monoterpenalkohole: Modellreaktionen zur Biosynthese der Monoterpene, Tetrahedron, 1967, 23, 3015-3022 (DOI)
W. Rittersdorf, F. Cramer, Die Hydrolyse von 2,3-Dihydroterpenylphosphaten und -pyrophosphaten, Tetrahedron, 1967, 23, 3023-3028 (DOI)

Thanks! To add some context, there's a bunch of secondary metabolites that branch out from FPP. Assuming one can immobilize the requisite enzyme the next question is whether the substrate FPP could be somehow manufactured at scale using a chemical synthetic route.

@curious_cat The procedure described for the smaller homologue of FPP might be applied. I wasn't aware that there's still a lot of research on this. Cationic biomimetic cyclizations of isoprenoid polyalkenes were a huge topic back in the eighties, when van Tamelen , Johnson, Corey and Nishizawa published lots of articles on this. However, they hardly used the pyrophosphates, but prepared other starting materials. I never worked on the cationic cyclisations myself, but on photoinduced radical ring closures. Btw, do you care to accept my answer as marked?

The reason to prefer FPP is because there's a enzyme available for converting FPP to the desired product. But if there's no good chemical route to FPP one must start with another substrate higher up the chain.


text:
Effect of Steric Environment on Selectivity of Protection by Sisyl Chlorides

(sorry about the poor diagram. I don't know how to add in the arrows and stuff)

Draw the product when the following compound is reacted with sisyl chloride in the presense of DMF catalyst at $\pu{-10^\circ C}$?

My Attempt
I know that sisyl methyl is a silicon protecting group that readily forms $\ce{Si-O}$ bonds. However I am not entirely sure how the different steric environments will effect which oxygen is protected.
I am guessing that the primary alcohol group will get protected since it is a primary alcohol, hence there will be less steric hindrance for the bulky sisyl chloride molecule to attack the primary alcohol group compared to the secondary alcohol group.
I am not sure if there are other factors that effect which oxygen will get protected. For example, I am not sure the effect of the double bond on the secondary alcohol (I don't think that there is an effect since it is too far away).

This has some influence too: https://en.wikipedia.org/wiki/Allylic_strain