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For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)?
In a heterogeneous reaction (where the states are varied) we do not include liquids and solids in the equilibrium equation because their concentrations do not change.
E.g. Chemguide.co.uk

However, when it is a homogenous equation we DO include solids and liquids.
E.g. Chemguide.co.uk

1) Why do we include liquids and solids in the Kc equation in homogeneous equilibrium equations? I get that if we didn't there would be nothing on the right hand side of Kc=.... , but why is it physically different to the case where there are varied phases?
2) What if there was an equation that involved only liquids and solids  (if this is possible)? As this is a heterogeneous equation would we still not include solids and liquids? If so how would you write the Kc equation where there is nothing but solids and liquids?

Because experimentally solids and liquids in the first case hardly affect the rate of the reaction.

Thanks for the response. So in the case where it is say all liquids (or all solids) the rate of reaction would be effected by the liquids and solids since there is nothing else right? Does that mean if we have a reaction which includes both liquids and solids (but not gases or aqueous solutions) then we would include both the liquids and solids in the equilibrium constant, even though it is a heterogeneous equilibrium equation?

According to the website you cited, "you don't include any term for a solid in the [heterogeneous] equilibrium expression".

Thanks for pointing that out. Is there any reason why liquids would be different?

Because liquids are fluid whose concentration still affects the rate of effective collisions.

Chemguide is simplified for A-levels and therefore in this case is strictly speaking incorrect.
The equilibrium constant $K$ is defined as a product of activities. I described this in a previous answer here.
The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for $K$ without affecting the value.
In your first reaction
$$\ce{H2O(g) + C(s) <=> H2(g) + CO(g)}$$
the chunks of carbon in the reaction are necessarily pure because they don't mix with the gases.
In the second reaction (yes, I am lazy, please feel free to edit for me)
$$\ce{EtOAc(l) + H2O(l) <=> AcOH(l) + EtOH(l)}$$
none of the liquids are pure, hence their activities deviate from unity.
Just as a final example, in the dissociation of a weak acid
$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}$$
water is omitted from the expression for $K_\mathrm{a}$ because water, as the solvent, is in large excess over $\ce{HA}$ and is therefore effectively "pure".

I don't like the statement "The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for K without affecting the value." In the reaction shown for carbon and water the carbon doesn't necessarily need to be "pure." Rather I think it is better to say that the equilibrium is based on the gas phase and since the carbon is in a different phase (a solid) the equilibrium is independent of the amount of carbon present - be it 1 milligram or 1 metric ton.

@MaxW The equilibrium constant uses activities, not amounts. You are of course right that the amount of carbon doesn't affect the equilibrium but I felt that the direct link is that it doesn't affect the activity of the solid, hence doesn't affect equilibrium constant.

The point that I'm trying to make is that new students to chemistry haven't probably heard the word "activity." If K-Feldspar had any notion of what activity was this question wouldn't have been asked. I think explaining in terms of phases wouldn't introduce a new concept.

Fair enough, feel free to add your own answer - I'd +1.

@orthocresol Hi Ortho :) In case of dissociation of weak acids, isn't $\ce{K\times[H2O]=K_a}$? I mean here $\ce{[H2O]}$ isn't totally ignored, is it? Why isn't $\ce{[H2O]}$ just set to $1$ like other cases  you discussed instead of multiplying it by $K$? Thank you very much. It's a very clear answer ;-)

@user8718165 I'm sorry, but I don't entirely get what you're asking. The equilibrium constant $K$ and the dissociation constant $K_\mathrm{a}$ are both defined in terms of activities, and the activity of the solvent is 1, as mentioned above. So in no circumstances do we actually "set" a concentration to be equal to 1. The only reason why concentrations come into play is because the activities of the solutes can be approximated by their (molar) concentration. So we can write [HA], [H+], [A-] as approximations for their activities (notwithstanding units). But you should not write [H2O].

@orthocresol I got you. Sorry, I'm just learning equilibrium. I was talking about this. That's what I was saying. why isn't the activity taken in water's case? Thank you.

@user8718165 It is used in the expression for $K$, but because it's equal to 1 it doesn't really matter whether you put it in or not.

@user8718165 yes pretty much - it is not technically correct, but this is always done in introductory chemistry

@orthocresol Thank you very much for you reply. I fully understood now. I was very confused and now it's all clear. :-)


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Essential oil extraction using ethanol
I want to make extracts of Black seeds. I do not posses any professional equipment, I just want to ask if it's possible to make extracts by soaking the seeds in ethanol for a long time and then evaporating it. Will the dissolved extracts evaporate with the solvent?

How about identifying and checking the boiling point(s) of these "essential oils" and compare them to ethanol?

The primary compounds of potential theraputic interest are all far less volatile than ethanol.  Ethanol based tinctures are sold and even have   available MSDS, so at least some of the oils of interest are soluble.  In looking at the chemical structures of the predominant oils of nigella sativa I would guess that they should all be fairly soluble in ethanol.  In some ways it seems that pure acetone (available at pharmacy) might make a better solvent and is even much more volatile even than ethanol.  If you used acetone, be certain to dry it via heating and time such that there is no scent of acetone remaining.  This shouldn't be difficult.
Anyway, to answer your question: ethanol should be a good solvent and it should not be difficult to remove the majority of the ethanol from the far less volatile oils just by drying in air for sufficient time (until you don't smell alcohol).

Ethanol itself may or may not make a suitable solvent.
Use caution when experimenting with denatured ethanol for such purposes.  Denatured ethanol (alcohol) is tempting to use for home chemistry because of its low cost and availability.  However,  the alduterants used to denature the ethanol may not be volatile. https://en.m.wikipedia.org/wiki/Denatured_alcohol . Then the extract would be denatured black seed oil. FYI

I read that some US government laboratories suffer from too many hoops to jump through to order impurity free 96.5% or 99+%  alcohol for in-house use that they have decided to standardise on 90+% Everclear spirits which has water as the major adulterant.  This is a reasonably safe source of food safe (when diluted) ethanol. Random alcohol (is denatured unless otherwise indicated and) has various compounds that are either harmful or objectionable for human consumption.

source: Stack Exchange

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Bad slaked lime?
I recently purchased some slaked lime online and suspect it is not good (not fully slaked?).
When added to water around 35 °C, it heats to about 80 °C. My understanding is that slaked lime should not have this reaction and I am therefore guessing there is still some quicklime remaining in it somehow. Is this correct, or is it normal for even slaked lime to heat so dramatically?

You misunderstand what slaked lime is. Slaked lime is the common name for calcium hydroxide and the reaction to dissolve calcium hydroxide is very exothermic as you have noticed.
You also misunderstand about quicklime. Quicklime is Calcium oxide, $\ce{CaO}$. It is formed by heating calcium carbonate and driving off carbon dioxide. Quicklime is reacted with water to form calcium hydroxide so that the quicklime, which has now cooled, won't react with carbon dioxide in the atmosphere to form calcium carbonate again.

Thank you for that – I did completely misunderstand. I am using the calcium hydroxide in the preparation of a natural indigo vat and was under the impression that I should not let the starter vat get over 50c, which leaves me a bit puzzled as to how to do that when adding calcium hydroxide heats it so much (or maybe that referred to the initial water solution for dissolving the indigo?). Anyway, thank you for you explanation! Very helpful.

Please do not get confused by @MaxW regarding mixing it with water.
you would have trouble dissolving much Slaked lime $\ce{Ca(OH)2}$ in water as solubility in water is very low and the solubility decreases with increasing temperature.
1.89 g/L (0 °C)
1.73 g/L (20 °C)
0.66 g/L (100 °C)
please see https://en.wikipedia.org/wiki/Calcium_hydroxide
Seems like you have purchased Quicklime $\ce{CaO}$, not Slaked lime $\ce{Ca(OH)2}$
Reactivity drops quite rapidly with the addition of water into Quicklime $\ce{CaO}$

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Writing Ksp when the ions are on the left of the equilibrium equation?
For the equation:
$\ce{Ca^{2+} {(aq)} + ^-OOC-COO- {(aq)} <=> Ca(OOC-COO) {(s)}}$
My book gives $K_{sp} = \pu{3.0e-9 M} = [\ce{Ca^{2+}}] \times [\ce{^-OOC-COO-}]$.
However, as per Chem.Libretexts, for any  $K_c$, the right hand side substances go in the numerator and the left hand side substances go in the denominator.

"It is really important to write down the equilibrium reaction whenever you talk about an equilibrium constant. That is the only way that you can be sure that you have got the expression the right way up - with the right-hand substances on the top and the left-hand ones at the bottom."

So for $K_{sp}$, why is it not the same?
I guess intuitively for a $K_{sp}$ I would probably put the $K_{sp}$ = product of the ions. But is there any other exceptions to when the above rule from Chem.Libretexts is not followed (in regards to K values)?

By convention, the solid is on the left of the side dissolving reaction. If you do it the $K_{sp}$ has the correct form.
The value of the $K_{sp}$ given above suggests that it conforms to this convention.

Thanks for the response. How about if my exam question gives a certain Ksp value but has the equation with the solids on the right side and the aqueous products on the left side? Would I need to write 1/the given Ksp=[product][product]? They did not do that in the example question in the book.

But based on http://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant/Calculating_An_Equilibrium_Concentration_From_An_Equilibrium_Constant/Writing_Equilibrium_Constant_Expressions_involving_solids_and_liquids,  if it was  Kc value I would have to do that (if I interpreted it correctly) as it says..

"It is really important to write down the equilibrium reaction whenever you talk about an equilibrium constant. That is the only way that you can be sure that you have got the expression the right way up - with the right-hand substances on the top and the left-hand ones at the bottom."

The inverse of $\pu{3e-9 M}$ would not be a reasonable value.

For the equation:
$\ce{Ca^{2+}_{(aq)} + ^-OOC-COO-_{(aq)} <=> Ca(OOC-COO)_{(s)}}$
$K_{sp} = \pu{3.0e-9 M} = \ce{[Ca^{2+}] \cdot [\ce{^-OOC-COO-}]}$
The equation for the solubility product, Ksp, is correct.
The reason $\ce{Ca(OOC-COO)}$ is not in the equilibrium equation is because it is a solid and in another phase as indicted by the (s) subscript. The calcium oxalate is in the solid phase not the liquid phase of the solution. The gist is it doesn't matter how much solid calcium oxalate is in contact with the solution. It could be  1 milligram, or 53.6 kilograms, and the solubility constant would have the same value.

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How can elements have the same number of electrons, but different electron configurations?
$\ce{Ti}$ and $\ce{Cr^2+}$ are isoelectronic, yet their electron configurations are different, with the former's being $\ce{[Ar] 4s^2 3d^2}$ and the latter's being $\ce{[Ar] 3d^4}$.
How can that be explained?

When you fill electrons in different orbitals, you keep in mind the $n+l$ rule, but when you remove electrons you have to consider the principal quantum number. Between 3d and 4s, 4s has a lower value of $n+l$, so electrons will be first filled in it, that's why titanium metal has the atomic configuration $\ce{[Ar] 4s^2 3d^2}$.
In case of $\ce{Cr^2+}$ you have to first write configuration of $\ce{Cr}$, which is $\ce{[Ar] 4s^2 3d^4}$ and then remove two electrons from the orbital having highest value of $n$ (principal quantum number) which is 4s. after removing 2 electrons from 4s you will get configuration of $\ce{Cr^2+}$ which is $\ce{[Ar] 3d^4}$.

for n+l rule(https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0ahUKEwiXxrHq8oTPAhVFto8KHUwmCvUQFggjMAI&url=http%3A%2F%2Fwww.citycollegiate.com%2Fatomic_structureXI_EC02.htm&usg=AFQjCNFikSSLqreQWFN1VB6PtssXz8yZVQ)

This would produce the wrong result for $\ce{V+}$ which should be $\ce{[Ar] 3d^4}$ but would be $\ce{[Ar] 4s^1 3d^3}$ instead if we use your method.

The reality is more complicated than a simple rule as you present, as is stated in the first sentence of my answer.

If not completely and utterly wrong, then at least that is halfhearted and incomplete.

The reason is very complicated, but in short:
The energy gap between $\ce{3d}$ and $\ce{4s}$ is wider in $\ce{Cr^2+}$ than in $\ce{Ti}$ because $\ce{Cr^2+}$ has more protons.
When you promote an electron from $\ce{3d}$ to $\ce{4s}$, the electron gained the energy thanks to orbitals, but you also lost the energy thanks to repulsion (of electrons in the same orbital). Remember that electrons like to be in states of less energy, not more.
The energy gap between $\ce{3d}$ and $\ce{4s}$ is less in $\ce{Ti}$, so the electron, by promoting, lost more energy thanks to repulsion than it gained thanks to the orbital, so its configuration is $\ce{[Ar] 4s^2 3d^2}$.
In $\ce{Cr^2+}$, the enrgy gap widened, so the electron would gain more energy than its lost by promoting, making its configuration $\ce{[Ar] 3d^4}$.
See also chemguide.co.uk.

The general configuration $\ce{Sc}$ to $\ce{Zn}$ is $\ce{[Ar] 3d^n 4s^2}$, except for $\ce{Cr^2+}$, $\ce{[Ar] 3d^5 4s^1}$ and $\ce{Cu}$, $\ce{[Ar] 3d^{10} 4s^1}$ and which are attributed to the stability of half-filled and filled $\ce{d}$ shells respectively. In the half filled $\ce{3d}$ shell there is no spin pairing energy needed (one electron per orbital) compared to that energy needed to put two electrons into the $\ce{4s}$ orbital.
Titanium has the configuration $\ce{[Ar] 3d^2 4s^2}$ and chromium $\ce{[Ar] 3d^5 4s^1}$, thus loosing two electrons to make $\ce{Cr^2+}$ produces the configuration $\ce{[Ar] 3d^4 4s^0}$ (although $\ce{4s}$ and $\ce{3d}$ orbitals are close in first row transition elements) then the configuration is as one would expect.


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Oxidation of 1,4,5,8-tetrahydronaphthalene by meta-chloroperoxybenzoic acid
I came across this question in Himanshu Pandey. I can't understand why the m-CPBA attacks the central pi bond and not the lateral ones. The answer given is C.

As the following figure indicates, the terminal oxygen in a peracid (the oxygen involved in the epoxidation) has a partial positive charge ; therefore this oxygen is electrophilic.

(image source)
The central double bond in your compound is tetrasubsituted, the lateral bonds are only disubstituted with alkyl substituents (remember that alkyl groups are electron donating).  Consequently the central double bond is the most electron rich and will be the preferred site of attack for the electrophilic oxygen.

But why arent we giving hyperconjugation a priority here,in that case the middle bond is more stable

@DIPANJANChowdhury  While hypercojugation is important it is usually outweighed by the effect of a substituent directly attached to the carbon in question.


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Why does calcium chloride cause a greater freezing point depression in water than magnesium chloride?
So I saw this equation for the freezing point depression, when you add a solute: $\Delta T= K_\mathrm fmi$
Where,

$\Delta T =$ The freezing point depression
$K_\mathrm f =$ The freezing point depression constant (Cryoscopic constant)
$m =$ Molality
$i =$ van't hoff factor.

I read online that $\ce{CaCl2}$ would be better than $\ce{MgCl2}$, but I cannot understand why in the terms of this formula since $i$ for both of them is the same and $K_\mathrm f$ seems to be independent of the solute since I saw values of $K_f$ of water online. (without any solute given)
So, could someone answer why this happens? Because it seems to me that the $K_\mathrm f$ could be the only factor that would explain this. Specifically I would like to know if the charge or the electronegativity of the dissolved ions would make a difference.

$K_f$ is indeed independent of the solute. It is $m$ that matters. $\ce{CaCl2}$ has better solubility, so...

As quantified in Fig. 17 of Manual of Practice for An Effective Anti-Icing Program (FHWA-RD-95-202):
By weight percent, MgCl2 depresses the freezing point of water to a greater degree than CaCl2, up to about 24%.  This is consistent with the freezing point depression formula, as Mg has less mass than Ca.
Beyond 24%, CaCl2 is superior.
This is because the eutectic point of MgCl2-water is at 21.6 percent, whereas the eutectic point of CaCl2-water is at 30%.
The freezing point depression formula is only valid at reasonably low concentrations.  It is certainly not valid beyond the eutectic point.

My most recent chemistry lab experiment was determining the effectiveness of various deicers. We tested the freezing point of $\ce{NaCl}$, $\ce{KCl}$, $\ce{CaCl2}$, and $\ce{MgCl2}$. I was told by my professor that $\ce{MgCl2}$ produces an exothermic reaction. When $\ce{MgCl2}$ is dissolved in water, the temperature actually increases. This could be why the freezing point of $\ce{MgCl2}$ isn't as low as that of $\ce{CaCl2}$.

This is irrelevant. Besides, the dissolution of CaCl2 is also exothermic.


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How does the pH of fresh glacial meltwater compare to marine seawater?
How does the pH of fresh glacial meltwater (at the time of melting, so it has not travelled downstream for example) compare to marine seawater? I assume that glacial ice is more neutral since it cannot uptake as much alkalis into its structure. Any insight would be helpful.

In general, the pH of seawater can range from 7.5 to 8.4 depending on where you are in the world, this is due to temperature differences, shipping activity, pollution, and several other factors. Reefs are quite sensitive to pH and are usually found in bodies of water with a pH ranging from 8.0 - 8.4. Glacial water should be nearly neutral as most glaciers were formed thousands of years ago when pollution levels were relatively low. It is quite possible that the glacier would actually be very slightly acidic due to the carbon dioxide that would have been trapped in the ice when it formed bubbling through the water creating a very weak carbonic acid. Scientists actually measure the levels of carbon dioxide in the ice to get an idea of what the Earths atmosphere was like thousands of years ago.
Other users please feel free to correct me if I am incorrect in my answer,  hopefully you can find this useful.

Glaciers advance and retreat seasonally. If $\ce{CO2}$ in the cryosphere can correctly reflect its atmospheric concentration of thousands of years ago, than fresh snow must reflect its present atmospheric concentration. See: http://chemistry.elmhurst.edu/vchembook/191acidsnow.html


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Forming half equations
I'm having trouble finding the ionic and half equations for the following equation:
$\ce{2Sr + O2 -> 2SrO}$
From this I would have to tell which elements are being oxidised and reduced.
This is IGCSE.

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.

$\ce{SrO}$ is an ionic compound comprising $\ce{Sr^2+}$ ion and $\ce{O^2-}$ ion.
The oxidation state of $\ce{Sr}$ changed from $0$ to $+2$, and $\ce{O}$ from $0$ to $-2$.
Therefore, $\ce{Sr}$ is oxidized while $\ce{O2}$ is reduced.

The ionic equation is just that, since there is no individual ion in the equation ($\ce{SrO}$ is a solid!).

Are you really told to find a half equation for a reaction not performed underwater?

So does the big number make a difference to the ionic equation?

@AnnabelleSykes Which big number?

If you see in my question, its 2Sr.

2 is not a pretty big number imo.

And no, you can include it outright

How do I include it? I'm an IGCSE student not a university scholar.

@AnnabelleSykes The ionic equation is $\ce{2Sr + O2 -> 2SrO}$


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Work done when P,V, and T are changing simultaneously
I am given the following P-V graph for an ideal gas:

I have been asked to find the work done in the process from C to A, for which $P$ is directly proportional to $V$. This may be calculated in two ways:

Work done= Area under the line CA, which is the area of the trapezium, given by

$$w = -\frac{1}{2}(P_0+2P_0)V_0 = -\frac{3P_0V_0}{2}$$
(where the negative sign has been introduced since the work done is negative due to decrease in volume.)

Work done = $\int -P\,\mathrm{d}v$. However, in this case, since $P$ is directly proportional to $V$, how to integrate this?

How to calculate the work done in a process where all 3 variables, $P$, $V$, and $T$ are changing simultaneously and $P$ is directly proportional to $V$?

Who said that $\ce{PV}$ is the work done?

The second method $\ce{P}\Delta\ce{V}$ only works when $P$ is constant, otherwise you should use integration.

Ohh, I see. But how to integrate for this case in which I can't substitute $P$=$nRT/V$ since $P$ is directly proportional to $V$?

It does not matter if $\ce{P}$ is directly proportional to $\ce{V}$. The energy is still $\displaystyle\frac23\int\ce{P}\ \mathrm d\ce{V}$.

I'm sorry, but how did you obtain that expression?

The energy is not $\ce{PV}$, but $\dfrac23\ce{PV}$. It is derived from the last formula of this page.

Okay, but how does this help me to integrate this?

Well, $\mbox{work done}=\displaystyle\frac23\int_{\ce{V0}}^{\ce{2V0}}\ce{P}\ \mathrm d\ce{V} = \frac{2\ce{P0}}{3\ce{V0}} \int_{\ce{V0}}^{\ce{2V0}}\ce{V}\ \mathrm d\ce{V} = \frac{2\ce{P0}}{3\ce{V0}} \left(\frac{\ce{V}^2}2\right)_{\ce{V0}}^{\ce{2V0}} = \frac{2\ce{P0}}{3\ce{V0}} \left(\frac{\ce{3V0}^2}2\right)=\ce{P0V0}$

Let us continue this discussion in chat.

I don't understand how the factor of 2/3 is relevant here. The question is asking for work done not the average kinetic energy of the gas molecules.

@orthocresol: That's what I've been thinking too... :/

Your calculation of the area of the trapezoid is the same thing as getting the integral of pdV, aside from the sign  of the result (which has to be taken into account).

As far as I am aware, there is no generic formula for when all three variables are changing at once.
In this case, the easiest way is to simply use the area of a trapezium.
If you insist on integrating, then you need to know what $P$ is as a function of $V$. This derives from basic principles of integration - if you have a function $f(x)$, then $\int_a^b f(x)\,\mathrm{d}x$ gives the area under the curve of $f(x)$ from $x = a$ to $x = b$.
The line $\mathrm{C \rightarrow A}$ can be easily found since the beginning and end points are given. The gradient of the line is
$$m = \frac{2P_0 - P_0}{2V_0 - V_0} = \frac{P_0}{V_0}$$
and the the intercept is clearly 0 from the diagram (you can work it out anyway). In general the equation of a line $y = y(x)$ is given by $y - y_i = m(x - x_i)$ where $(x_i,y_i)$ is any point on the line.
That was perhaps an unnecessarily long explanation but the point is that you now know the relation
$$P = mV$$
holds (where $m$ was determined earlier). Therefore
$$\begin{align}
w &= -\int_{2V_0}^{V_0} P\,\mathrm{d}V \\
&= -\int_{2V_0}^{V_0} -mV\,\mathrm{d}V \\
&= m\left[\frac{1}{2}V^2\right]_{2V_0}^{V_0} \\
&= m\left(\frac{1}{2}V_0^2 - 2V_0^2\right) \\
&= \frac{P_0}{V_0}\left(-\frac{3}{2}V_0^2\right) \\
&= -\frac{3}{2}P_0V_0
\end{align}$$
which is the same as you found via the first method (you also forgot to put the negative sign in the right-hand side of the equation, but I assume that is just a typo).

I thought you need to multiply by $\dfrac23$ :o

@user34388: Bit why?

@user34388 why? The formula you quoted is from the kinetic theory of gases and uses the mean velocity of the gas molecules. It's not relevant to our present question of thermodynamics.

Yes, I realize that it's not relevant! I was asking @user34388, who quoted it in the first place why it is relevant for our present discussion.

@KaumudiHarikumar I wasn't talking to you.

Oh, shucks. Sorry :P


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Lewis dot structure of beryllium fluoride
I'm so confused because I'm trying to draw $\ce{BeF2}$ … and the way it's drawn (something like $\ce{F-Be-F}$) just doesn't make sense to me because Be can only hold 2 valence electrons. But this structure shows it having 4.

It doesn’t exactly have four valence electrons the way you drew it, unless you assume the oxygen of water to hold eight valence electrons.

You are correct. $\ce{BeF2}$ is octet deficient.
To solve this problem, they would form polymers by themselves, but that would be out of syllabus.
(I cannot find the polymer of $\ce{BeF2}$, so here is the polymer of $\ce{BeCl2}$:)


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Determine oxidation in organic chemistry through oxidation numbers?
I want to determine the carbons that are oxidatized in an organic reaction.
Is it an acceptable way to first figure out the oxidation number of every carbon in reactants and products and then compare the oxidation numbers (for example: $\ce{+I -> +II} =$ Oxidation; electron loss)?* Or is my attempt to work with oxidation numbers false (for example: an aldehyde and keton are on one oxidation level, but their carbon has different oxidation numbers?). Maybe I should go over the oxidation levels of functional groups given in many textbooks as tabulars (but my structures aren't that easy and have not the most common functional groups)?
*I have read in many posts here, that this is a legitimate way — but there is "my problem" with the keton and the aldehyd, as mentioned above.

Yes, determining oxidation numbers and thereby finding out what is oxidised and what is reduced is a valid method, especially for reactions that include oxidations or reductions.
Your ‘problem’ with an aldehyde and a ketone is actually not a problem at all. Ketones have an oxidation state of $\mathrm{+II}$, which means that the only thing they could be oxidised to in a two-electron (i.e. non-radical) process would be $\ce{CO2}$ — hence why they are stable towards oxidation. Aldehydes, on the other hand, have an oxidation state of $\mathrm{+I}$, which means that they can easily be oxidised (to carboxylic acids, oxidation state $\mathrm{+III}$) by simplet two-electron processes.
Likewise, the different oxidation states of primary ($\mathrm{-I}$), secondary ($\mathrm{\pm 0}$) and tertiary ($\mathrm{+I}$) alcohols can be used to help understand why it makes sense to distinguish.
However, do note that many organic reactions — while definitely involving a change of oxidation states — do not fit into the classical redox paradigm well. Throughout the mechanism of a Horner-Wadsworth-Emmons reaction, the (tertiary) phosphonate carbon changes its oxidation state from $\mathrm{-II}$ to $\pm 0$ in two distinct, one-electron steps, while the aldehyde carbon goes from $\mathrm{+I}$ to $\mathrm{-I}$ also in two distinct steps — but nobody would go as far as calling it a redox reaction.


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To find the value of x in a metal chloride MClx
A solution containing 0.1 mol of a metal chloride MClx requires 500 ml of 0.8 M AgNO3 solution for complete precipitation. The value of x is:
(a) 1
(b) 2
(c) 4
(d) 3
Please answer with an explanation.

$$\ce{MCl_x + xAgNO3 -> xAgCl + M(NO3)_x}$$
$500$ mL of $0.8$ M is $0.4$ mol, so the mole ratio between $\ce{MCl_x}$ and $\ce{AgNO3}$ used is $1:4$.
From the equation above, the mole ratio is $1:x$.
Therefore, $1:x=1:4 \implies x=4$.

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.

@Martin-マーチン Sorry, noted.


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Altitudes effect on taste
Over dinner, some vegan friends of mine discussing plans to open a 5 star organic restaurant inside the cabin of an airbus A380.
Exactly what I thought too! But they raised an interesting point: apparently the taste of certain foods changes depending on the altitude you consume it at.
I looked into it a little and discovered lots of interesting facts about Heston Blumenthal cooking salmon and whatnot for astronauts, but not must hard-chemistry behind why our sense of taste depletes the higher in the atmosphere we go.
So what causes such a taste discrepancy? Is it compounds within the food that "lose" their flavour, or it just our perception of taste that changes?

It looks like a question for Biology SE or Cooking (Seasoned Advice).

I've heard that it's a partly due to the low humidity (around 12% or so)

@Mithoron Agree, offtopic question.

@Greg: How? Somethings "Taste" - per say - is defined as the interactions  between various chemical compounds within a food-product(s) and our saliva (or rather the compounds in such) in the presence of an atmosphere of some kind. Can't get much more chemically involved than that can it? I'm asking if the process of "tasting" ( the molar/chemical interactions between product and saliva) are affected by the height  these interactions take place at, eg: due to the abundance - or lack of- oxygen.

The difference in perception of taste is altered both by the lower air humidity as well by the lowered cabin pressure while on flight, the later typically kept to what corresponds to an altitude of about 2'400 m above sea level (source).
Regarding the actual changes in taste, BBC Future once nicely shed light on context, causes and outcome; in probably more accessible words than ongoing research articles like this about it.


text:
Won't the net effect of a catalyst be zero if it creates a new path with lower activation energy?
A catalyst will provide a new path with a lower activation energy (Figure 1). Won't this mean the forward and backward reactions will both speed up (as they both have a lower activation energy path to take)?
So then what is the point of using a catalyst? Or alternatively, how does the catalyst provide a net benefit for the forward reaction if both the forward and backward reaction are sped up?
I am assuming it might have something to do with the reactants having a higher energy than the products, but I can't think up a proper formal reason that a catalyst has a net benefit for the forward reaction.

Figure 1 (source: Chemguide.co.uk)

Your realisation is correct and something chemistry teachers try to hammer into their students’ heads time and time again (and yet, the point is still often lost):
Catalysts will never change the thermodynamics of a reaction. They only ease the path of the reaction. Forward and backward reactions will be accelerated equivalently.
So what is the benefit of a catalyst? There are multiple ones.
Speed
Take for example the Haber-Bosch process to synthesise ammonia from nitrogen and hydrogen.
$$\ce{N2 + 3 H2 <=> 2 NH3}\tag{1}$$
$$\Delta_\mathrm{r} H^0_\mathrm{298~K} = -45.8~\mathrm{\frac{kJ}{mol}}$$
This reaction is exothermic and thus should, theoretically or thermodynamicly, proceed spontaneously, e.g. if you mixed nitrogen and hydrogen in the appropriate ratio and added a spark. It does not, however. Significant activation energy is required to cleave the $\ce{N#N}$ triple bond.
Typical methods to add activation energy include heating. In the Haber-Bosch process, the mixture is heated to $400$ to $500~\mathrm{^\circ C}$ to supply the required activation energy. However, since the reaction is exothermic, heating will favour the reactant side. Increasing the pressure improves the entropic term of the Gibbs free energy equation, hence why pressures of $15$ to $25~\mathrm{MPa}$ are used.
Catalysts, based on iron with different promotors, are used to accelerate the reaction. By using catalysts, one can lower the temperature required in a trade-off between speed of reaction and favouring the product side of the equilibrium. With the conditions and catalysts used, one achieves a yield of $\approx 15~\%$ of ammonia within a reasonable timeframe. Not employing a catalyst would give much lower yields at much longer timeframes — economically much less feasible.
Direct reaction path not accessable
This is mainly true for many transition-metal catalysed carbon-carbon bond formation reactions, but is also true for some inorganic processes like the disproportionation of hydrogen peroxide as in equation $(2)$.
$$\ce{2 H2O2 -> 2 H2O + O2}\tag{2}$$
Hydrogen peroxide is a reactive chemical that cannot be stored forever, but the direct disproportionation path is not typically what degrades it. However, you can add $\ce{MnO2}$ to it. Upon addition, oxygen gas vigourously bubbles out of the solution. In this case, there was a kinetic barrier impeding the direct transformation due to reactants and products having different multiplicities (oxygen gas’ ground state is a triplet, all others are singlets). The $\mathrm{d^3}$ ion manganese(IV) is a radical itself that can partake in different radical reactions, allowing the diradical oxygen to be liberated.
Selectivity
This is exceptionally true for transition-metal catalysed organic carbon-carbon bond formation reactions. Note first that the action of a catalyst is frequently depicted as a catalytic cycle: A reactant reacts with the catalyst to some intermediate species, this rearranges or reacts with other reactants/additives/solvents in a set of specific steps until finally the products are liberated and the catalytic species is regenerated.
Many such reactions require organic halides as one of the reacting species. And the first step is typically an oxidative addition as shown in equation $(3)$, where $\ce{X}$ is a halide ($\ce{Cl, Br, I}$).
$$\ce{R-X + Pd^0 -> R-Pd^{+II}-X}\tag{3}$$
Palladium typically prefers oxidatively adding to bromides or iodides and tends to leave chlorides alone. I myself have performed a reaction with near-quatitative yield in which a reactant contained both a $\ce{C-Br}$ and a $\ce{C-Cl}$ bond — selectively, only the $\ce{C-Br}$ bond took part in the palladium(0) catalysed Sonogashira reaction.
Although I did not try it myself, I am pretty sure that switching to a nickel(0) catalyst species would shift the reaction in favour of reacting with the carbon-chlorine bond rather than the carbon-bromine one.
Mildness
This is basically a reiteration of the first point albeit with different intentions. Many a time in organic synthesis, one has a rather sensitive reactant that would degrade or undergo side-reactions if subjected to standard reaction conditions, such as high pH-value or elevated temperatures. As an example, consider a transesterification as shown in equation $(4)$.
$$\ce{R-COO-Et + Me-OH <=> R-COO-Me + EtOH}\tag{4}$$
This reaction is, of course, an equilibrium and by using methanol as the solvent we can shift it to the product side. For the reaction to happen, one would need a base strong enough to deprotonate methanol, giving the methanolate anion, which can then attack the ester functionality. However, methanolate being a strong (and nucleophilic) base itself can introduce undesired side-reactions, including epimerisation of the α-carbon.
One can catalyse this reaction by using $\ce{Bu2SnO}$, which will activate the carbonyl group, making it more susceptible to a nucleophilic attack. The reaction speed is the same but the conditions are milder (no additional base required) and the number of side-reactions is those strongly limited. In particular, I noticed no epimerisation of the α-carbon in the tin(IV) catalysed method.

Does the uncatalyzed reaction pathway takes place even when the catalyst is present?

Yes, one expects both forwards and backward reactions to speed up as you suggest; there seems to be no reason why microscopic reversibility would be suspended.  The point of the catalyst will be to speed up a reaction so that it can happen on a reasonable time-scale, say hours not days or even years, and also (industrially) to use less energy to do the same reaction. The back reaction will still be slower vs. forward reaction as the activation energy to return to products is still larger and so the equilibrium could still be greatly in favour of products. If this were not the case, either the products could be physically removed (say by precipitation, or escaping as gas/vapour) or they could react with something else, as in an enzyme, then the rate of the back reaction becomes far less significant.

What if the reaction favored reactants than products? Then does it make sense to use a catalyst?

I think the fundamental point here is that a constant reduction in one constant matters a lot more than the other, e.g.:
$$e^{-1} - e^{-10} = 0.367$$
If you reduce both "barriers" by, say, $0.5$ units, i.e. push the activated state down by (say) $0.5$ units, then
$$e^{-(1 - 0.5)} - e^{-(10-0.5)} = 0.606$$
Yes, you reduced both barriers by $0.5$ units, but you still increased the rate by almost $2\times$.
In fact, a lot of the qualitative explaining away done by chemistry classes is only rigorously understandable if you start doing the math.


text:
Why is copper a better conductor than iron?
Metals are good conductors as the electrons in their outer shell are loose and can plunge out of the atom with the application of the slightest force (voltage).
Silver is the best metallic conductor, then comes copper.

Why is one metal better at conduction than the other one?
What determines the difference in conduction between 2 metals?
Why is silver a better conductor than copper, and why is copper a better conductor than iron if all of them have loose electrons?

This is very simple, I think you didn't googled https://www.quora.com/Why-are-some-materials-better-conductors-than-others https://www.researchgate.net/post/Why_is_copper_a_better_conductor_than_other_metals_say_nickel

i don't really understand the answers as I am a 10th class student...is it too advanced for me?

Rather too basic then too advanced as it doesn't really answer the problem.

I've never seen Sommerfeld's theory https://en.wikipedia.org/wiki/Free_electron_model applied to actual metallic compounds, but that should be the answer. It's in the band structure of the metal.

http://pipingmart.blogspot.com/2018/10/why-is-copper-good-conductor-of-heat.html

This is, ultimately, a question on solid state physics rather than chemistry. Further, the OP indicates that they are in high school, which kind of limits the depth of the answer that might be useful to them. However, I will try to make a simple, yet detailed answer.
As atoms are brought closer and closer together, their electron clouds overlap and interact. When a crystalline solid is formed, one abandons consideration of the atomic energy levels. Instead, one seeks out the Bloch functions to describe the electonic states - the Bloch functions explicitly consider the symmetries of the crystal. Further, they represent extended electron states, those that extend across the entire crystal. They represent the allowed (energy and momentum) combinations of electrons.
For a metal, the highest Bloch states occur as a band of states which is only partially filled. Unoccupied states lie just above states with electrons. To move an electron about, it needs just a little bit of energy, and off it can go. (Fully filled bands do not allow conduction - for any electron going one way there is one going the other way. This needs a good grounding in solid state physics to fully grasp).
OK, in general then the conductivity should just be the number of electrons near the line between occupied and un-occupied. Kind of. This line is called the Fermi surface, and things get more complicated quickly. The shape and connectivity of the Fermi surface impact conduction dramatically. Lets take a deeper look at copper and iron.
Copper is a simple noble metal with a beautiful near-ideal Fermi surface (see Ashcroft and Mermin's Solid State Physics book, or C.Y. Fong et al., Comparison of band structures and charge distributions of copper and silver). It turns out that the Fermi surface of copper is almost exactly what one would expect from the free electron sphere. Electrons can move easily in any direction, the surface is fully connected, life is good.
Iron is much more complicated. Following along from J. Callaway and C.S. Wang, Energy bands in ferromagnetic iron one finds two major differences. First, because iron atoms indeed have individual magnetic moments, the overall band structure is split into two different spin states for the electrons. So, each spin state is separated from the others, reducing how each can move (relative to copper, were such splitting does not exist). Second, and more importantly, the Fermi surface looks nothing like a free electron sphere. Instead, for iron the Fermi surface is not fully connected, comprising instead of multiple 'pockets' of electrons that do not communicate directly with the other pockets (and each spin state has a different set of pockets). To move electrons around, they have to 'jump' from one pocket to another through scattering. This makes it much harder to move the 'free' electrons in Fe around to get net conductivity.
An analogy would be the difference of crossing a stream using either (1) a nice smooth bridge (copper) vs (2) jumping from rock to rock (iron). The band structure for copper just makes it so much easier, so the resistivity is much lower.

TL;DR.
To get the "real" answer, you have to have a good understanding of the underlying quantum mechanics. I'm pretty sure you're not ready for that, yet! (and note that this topic would require several semesters of quantum mechanics, it's in the sub-discipline of solid-state physics...) So, my answer will have to be simplified.
You're taught in school (perhaps) that there are 3 types of chemical bonds: ionic, covalent, and metallic. The world isn't that simple in reality, while all chemical bonding is electromagnetic, there is no "perfectly pure" ionic, covalent or metallic bond. If you accept the idea that metallic bonding involves all of the valence electrons of the atoms in the crystal (metals generally exist in small crystals called grains or crystallites) then we can simply say that Ag, Cu, and Au all have a single valence electron (per atom) and it turns out that this electron is fairly well shielded by each atoms' inner electrons from the nuclear charge which makes it much easier for these electrons to move, compared to, say iron. One way to look at it, I suppose, is that the road is less sticky (I wonder if calling it smoother would be a better analogy?) so that these puppies can move more easily than the valence electrons of iron. So, the answer is that there are degrees of "looseness". There are also grain defects and grain boundaries to cross. And also you should always keep in mind that electrical conductivity is temperature dependent. Stating categorically that silver is "the best" electrical conductor is more fan-boy enthusiasm than objective fact. The "best" electrical conduction is with superconductors - but of course then you need to consider the amount of current they can carry before they fail so "best" takes on a magnetic field dependence..."best" isn't a very scientific word is all I'm saying...there's almost always too many qualifications involved.
If you want to get a more correct answer, go to wikipedia's articles on Nearly free electron model, Metal, Fermi surface, Electrical resistivity and conductivity but I'm afraid that they're pretty hard to understand even for someone who has had some quantum mechanics.
I personally like my analogy that the road for silver's electrons isn't as sticky, but that's not to be taken too far... You actually have to consider not only the 'resistance' due to the nuclear (attractive) charges as the electron zooms by, but also the path the electron follows (the crystal lattice, how zig-zag the road is) and the defects and boundaries is must pass through (stop and pay a toll).

Rambling argument and no paragraphs make a bad answer...


text:
Born-Oppenheimer approximation
I'm studying about formulation of Born-Oppenheimer approximation from Atkins, Molecular quantum mechanics, 5th edition. In chapter 8.1 I found the following:
I keep wondering where did this W come from in equation 8.4.? It says later in text that it is responsible for non-adiabatic effects, but I don't see how it just appeared here. If someone is willing to explain, I'd be greatful.
Also, if anyone has tips on where to find more nice and clean text about more general formulation of Born-Oppenheimer approximation, since this one is confined to particular system.
Thanks!

The question is at least related to that one and might be even a duplicate.

JFYI: Atkins' book is just an introduction to the big field, and as such, quite frequently it contains no derivations but just merely states something.

Note though that in this particular case the derivation is quite straightforward. The most important thing to watch for is that the parametric dependence of the electronic wave function on nuclear coordinates is assumed to be continuous and differentiable. So, when taking the second-order derivative of the $\psi \chi$ product with respect to nuclear coordinates in $T_N \psi \chi$ term, remember to use the product rule twice. As a result you'll get three terms: $\psi T_N \chi$ and the other two collected in $W$.

See it on sciencedirect in the Spectroscopy part

Atkins did a good job on explaining what parametric dependence of the electronic wave function (and electronic energy) on nuclear coordinates is, but failed to mention that in the derivation of the Born-Oppenheimer approximation this dependence is assumed to be continuous and differentiable and that both first and second derivatives of this quantities with respect to nuclear coordinates are in general non-zero.
In particular, for the system described in the text, we have
$$
  \frac{\partial \psi}{\partial Z_j} \neq 0 \, , \quad
  \frac{\partial^2 \psi}{\partial Z_j^2} \neq 0\, .
$$
Now, when the solution of the form (8.3) is substituted into the Schrödinger equation (8.2), the term involving $T_\mathrm{e}$ is trivial: since the nuclear wave function $\chi$ is not a function of electronic coordinates it is just a constant when differentiating with respect to them, so we get
$$
  T_\mathrm{e} (\psi \chi) = \chi T_\mathrm{e} \psi \, .
$$
But the term involving $T_\mathrm{N}$ does not trivially transform in a similar way,
$$
  T_\mathrm{N} (\psi \chi) \neq \psi T_\mathrm{N} \chi
$$
since both $\psi$ and $\chi$ depend on the nuclear coordinates. Rather, applying the product rule twice, we get
\begin{align}
  \frac{\partial^2}{\partial Z_j^2} (\psi \chi)
  &=
  \frac{\partial}{\partial Z_j} \left( \frac{\partial}{\partial Z_j} (\psi \chi) \right) \\
  &=
  \frac{\partial}{\partial Z_j} \left( \psi \frac{\partial \chi}{\partial Z_j} + \chi \frac{\partial \psi}{\partial Z_j} \right) \\
  &=
  \psi \frac{\partial^2 \chi}{\partial Z_j^2}
  +
  2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j}
  +
  \chi \frac{\partial^2 \psi}{\partial Z_j^2} \, ,
\end{align}
so that
$$
  T_\mathrm{N} (\psi \chi)
  =
  - \psi \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \frac{\partial^2 \chi}{\partial Z_j^2}
  -
  \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \left(
  2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j}
  +
  \chi \frac{\partial^2 \psi}{\partial Z_j^2} \right) \, ,
$$
where the first term is nothing but $\psi T_\mathrm{N} \chi$ and the remainings are designated as $W$.


text:
Difference between voltaic cell in chemistry and battery in physics
In chemistry's voltaic cell, the anode (negative electrode) of the battery supplies electrons to the cathode (positive electrode) of the battery. In physics, the electrons flow from negative terminal (cathode) of the battery to positive terminal (anode) of the battery.
I don't understand this because in thermionic emission, electrons flow from cathode to anode too through an electric source (which I think is a voltaic cell).

I think you've confused the flow of electrons with the flow of current. Before it was discovered that electricity is the result of the flow of negatively charged electrons it was proposed that electricity result from the flow of a positive charge. We called this current and this convention has remained to this day.
So current is effectively the flow of positive charge not the flow of electrons, which are negatively charged. So current flows from the cathode to the anode, but electrons flow from the anode to the cathode. This is true in both physics and chemistry.

Sorry but the confusion remains even on the web pages on the net. Chemistry sites tell us that the anode is the source of electrons and are attracted by the cathode. I just finished looking at the anode on a natural gas  hot water heater which is magnesium rode. No current is applied. Mg is to the left on the periodic table and will give up electrons to form compounds. It is called an anode. This is true for batteries and chemistry.
Physics, however, specifically in electronics the anode attracts electrons and the cathode gives off electrons. The cathode ray tube is a case in point. The electronic gun is the source of electrons and the front of the old television screen are the anode. The negative of the battery is the cathode and the positive is the anode. Chemistry and physics are at odds with each other.

The anode is where oxidation takes place, the cathode is where reduction takes place.
Thus, in a battery the anode supplies the electrons that enter the circuit and the cathode accepts them. But in an electrochemical cell the anode is where anions are attracted to to be oxidised by electrons supplied by the power source.
This can be confusing.

I can sympathise with your situation; for reasons unknown to me, people are taught in ways that often leaves gaps where such questions/misconceptions originate.
In general, to avoid confusion it is best to define anode and cathode in terms of reduction/oxidation. The anode is the electrode where oxidation takes place, and the cathode is the one where reduction takes place.
Also, in an electrochemical cell, electrons don't flow, but ions do.
Now, let's address the differences between galvanic and electrolytic cells.
In a galvanic cell the reaction proceeds without an external potential helping it along. The oxidation reaction which produces electrons takes place at the anode, results in a build-up of negative charge at the anode in the course of a reaction till equilibrium is reached. Thus the anode is negative.
On the other hand, reduction takes place at the cathode, and consumes electrons, and thus leads to a build-up of positive charge (or depletion of negative charge). Thus the cathode is positive.
The situation is reversed in an electrolytic cell where you do provide an external potential, and you are essential pushing the reaction in the reverse direction (say, when you charge a cell). So while oxidation still takes place at the anode, the anode is now positive and while the cathode is still the site of reduction, it is now negative.
You can read more here
In physics, circuits usually depict a current which moves from the positive electrode of a batter to the negative one. This is not representative of the flow of electrons; it is a historical artifact of sorts, because this convention was agreed upon before the discovery of electrons. Read more here


text:
How to round correctly when calculating the average and standard deviation?
Is my Chemistry book wrong on the significant figures in this exercise?
Why did they round to 0.16 instead of 0.1580 ?

Also why did they round down the means?

why the thumbs down?

Your question has little to nothing to do with significant figures, you are calculating the standard deviation, ordinary rounding rules apply. As for the down-vote I can only speculate: You used excessive markup, some people really dislike that. It also would have been better to explain, why you think it should have been one value instead of another. Give more context to your thoughts to avoid down-votes.

I figured it out.
The means are rounded to only two sig figs because its subtraction.
And significant figures follow different rules under subtraction/addition vs multiplication/division.
And its 0.16 for the same reason that the means are only two sig figs.
Because its addition, not multiplication/division.
derp, two hours down the drain

Significant figures need not be constrained to Chemistry.


text:
Polar vs. Amphipathic
How can I tell if a molecule is purely polar or amphipathic?
For example, fatty acids are amphipathic because they have a polar carboxyl group and a non polar chain of hydrocarbons. However, sometimes there are molecules that have hydrocarbon chains and side polar groups but are categorized as polar.
What is it that I should look for?
Thank you so much!

The answer is that it is a sliding scale. There is no such thing as "purely polar" or "purely nonpolar." Water is very polar and hydrocarbons are very nonpolar, but there is no clear dividing line between the two categories.
Further confusion in biological molecules such as amino acids can occur, because different individuals categorize them in different ways, considering either just the side chain or the whole molecule. For example the amino acids have the formula $\ce{H2N-CHR-COOH}$ and in the simplest case, Glycine, R=$\ce{H}$. This is clearly a very non polar side chain, but the molecule as a whole is mostly polar, as the side chain is such an insignificantly small part of the molecule. Thus authors who categorize amino acids by side chain (mainly chemists) put Glycine into the non polar category, while authors who categorize amino acids by the whole molecule (mainly biochemists and biologists) put it into the polar category. When it comes to Alanine, R=$\ce{CH3}$ all authors agree it has a non polar side chain and most seem to categorize it as a non polar molecule.
A simple search of the internet shows that there is very little agreement about the amino acid Tyrosine, R=$\ce{CH2-(C6H4)-OH}$. It is classified as polar, non polar, having a polar side chain and having a non polar side chain by different authors. Clearly it has some characteristics of both, having both a benzene ring and a hydroxyl group. The correct definition would be "moderately polar."
You might hope that it is possible to give molecules a number which describes their polarity. This can be done to some extent. Googling "solvents by polarity" will give you lists for common solvents. This is, however, complicated if the molecule of interest happens to be acidic or basic. An acidic molecule will be polar in acidic solution but will become charged (and therefore much more polar) in basic solution, due to loss of $\ce{H+}$. Similarly, a basic molecule will be polar in basic solution but will become charged (and therefore much more polar) in acidic solution due to gain of $\ce{H+}$.
An amphipathic or amphiphilic molecule has a part that is highly polar and a part that is highly non polar. These terms also generally mean not just a molecule that is moderately polar, but one in which there is some distance between the polar and non polar parts, so the molecular weight would be at least into the hundreds.
The example you give of a fatty acid is a good one. (The deprotonated, negatively charged fatty acid anion in basic solution would be an even better one.) Another example would be many cell membrane proteins, which have a non polar, hydrophobic part (that stays in the cell membrane) and polar, hydrophilic parts (that extend out into the aqueous media inside and outside the cell.)
In summary, different substances should not be pigeonholed as polar or non polar as it would require that you draw an arbitrary line, but in general the polarity of two different substances can be compared fairly objectively. And pH may make a difference.

Good thorough answer in my opinion! I made some formatting changes based on the formatting criteria discussed here. It's very well written and makes for a good primer/guideline.


text:
Can carboxylic acids like act as nucleophiles in substitution reactions?

Can carboxylic acids like formic acid act as nucleophile in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reactions?
For example can formic acid react with an alkyl/aryl halide like benzyl bromide?

I'm asking because I found a very similar question while going through some old organic chemistry test papers. However none of the book's I have read says that carboxylic acids can act as nucleophile in such cases.

Carboxylic acids can act as nucleophiles of course. Consider reactions with isobutene or thionyl chloride (although base certainly helps in this latter case). However, whether it's in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reaction mechanisms is another matter. If you beast something with enough heat it will usually react. I would not be surprised if a carboxylic acid reacted with triflouromethanesulfonic anhydride without base. It's still $\mathrm{S_N2}$ as mechanistic studies have shown the S=O bonds do not open up like acyl electrophiles.
I suspect you would need to resort to a continuous as opposed to a batch approach to push a reaction between the acid and an alkyl halide without using any base. Expose the reaction to high temperature (I'm thinking 250C and higher)  for 5 seconds or less under continuous liquid flow.


text:
Is the SN1 reaction faster with an axial or equatorial leaving group?

Why does compound 1 undergo the $\mathrm{S_N1}$ reaction faster than 2 even though both proceed via the same carbocation intermediate?

You had the two compounds the wrong way round, by the way. The cis diastereomer reacts faster. Also, please no MathJaX in titles.

@orthocresol do you know what the relative rates of solvolysis are for these two compounds?

@ron I went searching a bit. Can't find anything on the chlorides, but acetolysis of the tosylates proceeds with $k_\mathrm{ax}/k_\mathrm{eq} = 3.24$. JACS 1955, 77, 5562

In compund 2, both substituents can be placed in equatorial positions, whereas in 1 the $\ce{Cl}$ group is forced into an axial position since the bulky t-butyl group has to be placed equatorial. This makes compound 1 more unstable (higher energy), leading to a faster rate of formation of the carbocation.

I think it's highly unlikely that the cyclohexane will adopt a non-chair conformation. The chlorine will just be stuck in an axial position.

Just to add: an alternative way of looking at it is that, with compound 1, upon leaving of the $\ce{Cl-}$, there is relief of 1,3-diaxial interactions. But it is really just a different way of describing the same thing.


text:
Why does octet bring stability?
After studying chemical bonding I am stuck at a question that is fundamental to my understanding of chemistry:

What is it that makes 8 electrons in the valence shell stable?

I googled it and got answers like it is a "noble gas electron configuration". But that doesn't answer my question. What is so special in the configuration of noble gases that makes them so stable? How does octet give stability to an atom?

A simple answer would be that there are no empty, low-energy orbitals (it is good to fill those), and also no filled, high-energy orbitals (it is bad to fill those). In truth the octet rule has its limitations. It is only an empirical observation and is not always true.

@orthocresol how do you classify a low-energy orbital and a high-energy orbital. Isnt it a comparitive term  (what I can infer from ur comment is that s and p are called low-energy while d and f are called high-energy. Am I right?)

No. The energy increases in the order 1s, 2s, 2p, 3s... for the purposes of this discussion, we talk about Ne. Here we might consider 1s, 2s, 2p "low" and 3s "high" because it is in a higher shell. Therefore it is advantageous to fill 1s, 2s, 2p, but not to fill 3s. The same logic applies when discussing why fluorine forms $\ce{F-}$ and not $\ce{F^2-}$. The rationalisation of the octet rule is therefore just a generalisation of that to covalent compounds.


text:
Why do we have surface charges on a polymer sheet immerged in water ?
I am working with a conductive polymer sheet where I have dropped off a drop of water (with ions). I have plugged the polymer with a conductive wire and an other wire is on the drop. With a voltmeter I measure a voltage (open circuit) between these two points the drop and the polymer (around 300 mV)
And when I use a resistor in my circuit, I can measure an intensity. So What happens on my polymer ? Where come from the electrons crossing my resistor ?

Related: http://chemistry.stackexchange.com/questions/47811/where-does-the-charge-on-an-electrode-responsible-for-the-formation-of-a-double/47812#47812


text:
What is the correct way to illustrate a bond to something irrelevant in a structural formula?
You can probably tell I'm not really a chemist (hi from physics!), but I've suddenly found myself having to draw some structural formulae, hence the stupid question.
I've drawn a diagram of the four DNA bases:

I don't care what they're attached to, and I don't fancy drawing an entire DNA molecule, but I'd like to show where they bond to the rest of the nucleotide. I've used \cdots ($\cdots$) currently, but that doesn't feel right. I considered R, but I'm not sure if an entire DNA molecule really counts as an R group. I saw a wavy bond once, but it feels like those are confined to stereochemistry.
What's the convention in cases like this? (And what do I Google to find it next time I want to know?)

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), in the structure diagram of the parent structure, variable substituents should be indicated by an unambiguous label.

This label should be chosen so that it cannot easily be confused with other valid element symbols or fragment abbreviations. Traditionally, the letter “R” is widely used, followed by a superscripted number (R1, R2, etc.) or prime(s) (R', R", etc.) if necessary to distinguish different groups. It is not acceptable to use a letter followed by a subscripted number, since subscripted numbers are more commonly used to indicate repeat counts, and confusion between the two is inevitable. Other common symbols include X, Z, and G. (…)

The label may be used with or without an explicit list of allowed substituents (e.g. in tabular form).
The wavy line is preferred in structure diagrams of the variable substituents, in cases where the external connection points must be indicated for the variable substituents.

Therefore, if you consider nucleobases with variable substituents, you may use labels such as “R”. However, if you consider the nucleobases to be the variable substituents of another (larger) parent structure, you may use a wavy line.

Just as a side note, I think it is easier to use ChemDraw to make the graphics then import them into your LaTeX file, rather than using the chemfig package which (I find) is cumbersome. Obviously it's entirely your choice.

I don't think there is an established convention per se.
If you wrote $\ce{R}$, nearly everybody would understand it, especially if you also mention at the side that $\ce{R}$ refers to the deoxyribose sugar. In chemistry, $\ce{R}$ is used for anything and everything.
I haven't seen $\cdots$ being used for this purpose in chemistry before.
A wavy line parallel to a bond isn't used for indicating stereochemistry. It is usually used to show that a group or moiety connects to something else. For example, if I was too lazy to write this reaction out in full

I would instead write

In your case it would be acceptable as well.
Examples
With that said, I went to look up some textbooks.
Berg & Stryer, Biochemistry, 7th ed.: Nothing

Clayden et al., Organic Chemistry, 2nd ed.: R

Blackburn et al., Nucleic Acids in Chemistry and Biology, 3rd ed.: Nothing

McMurry, Organic Chemistry, 8th ed.: Wavy lines

TL;DR: You can probably get away with anything you like.

Thanks! I'm impressed by how quickly you managed to show me textbooks. ChemDraw probably would be easier, but I don't have a copy (yet) and I'm comfortable enough with LaTeX to be able to manage the relatively simple diagrams I'm working with. I'll at least change the ChemFig settings to match ChemDraw (with wider double-bond spacing and such) at some point.

Nothing is obviously the worst choice since it looks like a methyl group.

@Jan Yes, it is certainly strictly wrong, but I think it is fine as long as the context is established. Every book describes the structure of a deoxyribonucleotide before going on to talk about Watson-Crick base pairing, so there is little opportunity for misinterpretation.

That’s why I said worst choice — it takes me two looks minimum to establish the content that that is not a methyl group ;) But yeah, with context it’s not wrong ^^

@dendodge As a side note: taking into consideration your background mentioned in your profile (exposure to different operating systems, HTML), ChemDoodle represents an interesting alternative to ChemDraw.


text:
Thermochromic, fluorescent fluid
I'm a total layman in chemistry and am working on an interactive sculpture project and therefore am researching into fluids that have certain properties - long story short:
we are thinking on having a transparent sculpture where a fluid is flowing through it, a fluid that reacts visually on the heat of peoples hands on their touch - so if people touch the sculpture, the fluid changes colour or even emits light in a fluorescent way.
Sorry, I am conscious this may sound super vague and really naive - I'm just trying to reach out to check out possibilities and may come up with interesting insights.
Is there any known way of achieving such an effect? Where could I continue looking after? Any help or pointer towards a certain direction is highly appreciated. Thanks!

Not even close?

While thermochromicity already was known for me, for example as thermometer for fish tanks like here, a search based on "thermochromic fluorescent fluid" as keyword yielded this video.
Beside safety, yet indeed, it is easier to fill the sculptures with a liquid / a solution of a fluorescent dye excited by black light from the inner of the (glass) sculpture and to wrap the container with thermochromic dyes. These are even commercialized as here and here.  "Black light" is an other name for ultraviolett light, and there are even hand-size battery powered lamps found in the chemistry lab, or in civil life to check banknotes.
Disclaimer: I have no relationships to the said corporation.
Addendum:  Examples of fluorescence triggered solely by the gentle warmth of a hand are not known to me (yet).

addition: prolonged exposure to UV may cause eye burn, resulting in severe pain.

Such fluorescence has nothing to do with touch.

@permeakra Of course, extended exposure towards UV to skin and eye is harmful ("the dose makes the poison" as Paracelsus said) and I still wonder if the black light deployed in discos was regulated.  On the other hand, if the container were like an amphora and made of glass, the excitation leading  to the the fluorescence of the liquid should be done by a light source immersed into the inner of the container.  But as we know, you don't catch a sun burn while sitting on a sunny day behind a window -- because normal glass adsorbs UV-light greatly.

@MaxW  #1: To quote the OP: "[aiming for an instance where] the fluid changes colour or even emits light in a fluorescent way". From the thermometer at the outer wall of a fish tank, I extrapolate the color changes induced by touching by my hand could happen as well if such dyes were heated as solution, across a sufficiently thin glass wall. #2: And correct, I do not know yet about an example of a dye starting to fluoresce, triggered by warming with a hand worth of 37 Celsius -- were worth a new question, don't you think so, too?

I'm a chemist. I did XRF as an analytical chemist. I collected fluorescent minerals (UV excitation in dark to see visible light...). Using a UV light is not like shinning a X-ray tube around. // I stand by my NO. There is no such chemical that would behave such as the OP wanted.

@MaxW - I appreciate your scientific precision and effort of taking me word by word. But I do much more appreciate advise and guidance and a bit of explanations rather than quite simple Yes or No. I reckon this might not be 100% the right platform then, but you know, people are people.

@alberto2000 - I'm trying to avoid wasting your time looking for something that doesn't exist. I can't imagine any minor tweak to your requirements which would allow such a chemical.


text:
Naming convention for adduct ions
In order for a compound to be analysed in mass spectrometry it must first be ionised, forming an adduct ion. In many papers (e.g 1 [PDF], 2) these adduct ions are described in the form:

[M+H]+
[M+Na]+
[M-2H]2-

Where M is the analyte, the other content of the square brackets is the ion, and outside the brackets is the resulting charge. The IUPAC Gold Book also mentions this format for protonation without, as far as I can tell, describing it further.
Is there some reference actually describing this convention - ideally encompassing more complex scenarios such as dimerisation or neutral losses as well - around?

Molecular Ion peak is represented by M+ without bracket. i think this convention is used to describe the adduct is now the part of molecule

Your convention is generally correct; as more complicated adducts and ions are formed (clusters, adducts, neutral losses, etc) things start to become a free for all where the clearest terminology wins.
As of now, just go with (where A means adduct):
$$\ce{[M + nA]^{n+}}\text{ or }\ce{[M - nA]^{n-}}$$
Protonated molecules can be denoted as $\ce{[MH]+}$ (common) or $\ce{[M + H]+}$ (technical). The $\ce{[M + H]+}$ convention is more precise because if you write $\ce{[MH]-}$ for a deprotonated ion the interpretation becomes muddled (i.e is this a hydride adduct or is this a deprotonated ion?).
One important think to keep in mind that is Molecular Ions ($\ce{M^{+.}, M^{-.}}$) are not the same as $\ce{[M + H]+}$ or $\ce{[M - H]-}$ and that the terminology is not interchangable.

$\ce{M^{+.}}$ or $\ce{M^{-.}}$ retains the elemental composition of the ion and is ionized through the removal or addition of an electron.
$\ce{[M + H]+}$ and $\ce{[M - H]-}$ are protonated and deprotonated molecules, respectively. They are not 'pseudomolecular' ions as there are no such thing as pseudo ionized specties.

What I was really hoping for was a standard (IUPAC or whatever) which laid something out, but I guess that doesn't seem to exist?

Looking at the requirements from scientific journals in mass spectrometry, only the Journal of the American Society for Mass Spectrometry references to "Helpful information about nomenclature and descriptions specific to mass spectrometry can be found in Mass Spec Desk Reference, 2nd Ed., by O. David Sparkman, Global View Publications, Pittsburgh, PA, 2006." So I guess this book could be considered a standard in the field. The problem is that it seems to be out of print.

And Patrick B.'s answer seems to me as accurate as possible.


text:
Is there a chemical that when added to water makes the water flash evaporate?
Im wondering if there is a chemical that you can add to water that will make it evaporate extremely fast, either by raising the temperature of the mix or some other method.

Related: How can 30 ml of water be heated in less than 10 seconds?

In principle, the evaporation of water is a purely physical process. In order to evaporate water, the considered water sample has to be heated to its boiling temperature at the given pressure, and then further heat must be supplied, which corresponds to the enthalpy of vaporization.
For example, the specific enthalpy of liquid water at an initial temperature of $T_0=25\ \mathrm{^\circ C}$ and a pressure of $p_0=1\ \mathrm{bar}$ is $h_0=104.9\ \mathrm{kJ\ kg^{-1}}$. The boiling temperature of water at this pressure is $T_\mathrm b=99.6\ \mathrm{^\circ C}$. The specific enthalpy of liquid water at this boiling point is $h_\text{liquid}=417.5\ \mathrm{kJ\ kg^{-1}}$; the specific enthalpy of steam is $h_\text{steam}=2674.9\ \mathrm{kJ\ kg^{-1}}$, which corresponds to a specific enthalpy of vaporization of $\Delta_\text{vap}h=2257.4\ \mathrm{kJ\ kg^{-1}}$. Therefore, in order to evaporate the water, the specific enthalpy of the cold water has to be increased by
$$\begin{align}
\Delta h&=h_\text{steam}-h_0\\[6pt]
&=2674.9\ \mathrm{kJ\ kg^{-1}}-104.9\ \mathrm{kJ\ kg^{-1}}\\[6pt]
&=2570.0\ \mathrm{kJ\ kg^{-1}}
\end{align}$$
The enthalpy of the water can by increased by any kind of heat transfer. The heat may be supplied by a purely physical heat source (e.g. a sufficient amount of a red-hot substance that is dropped into the water); the heat may also be supplied by an exothermic chemical reaction or any other exothermic process (e.g. nuclear fission), provided that the transferred heat is sufficient for the evaporation of the considered amount of water.
Anyway, even if the heating process may be very fast, this is not flash evaporation. Flash evaporation occurs when a superheated liquid undergoes a quick reduction in pressure (e.g. when the liquid is passing through a small opening, or when the pressure vessel suddenly fails).
For example, water at an initial pressure of $p_0=150\ \mathrm{bar}$ and temperature of $T_0=300\ \mathrm{^\circ C}$ is liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1338.3\ \mathrm{kJ/kg}$. When the pressure vessel suddenly breaks, the pressure drops to the ambient pressure of $p=1\ \mathrm{bar}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.6\ \mathrm{^\circ C}$) at the new pressure. A new equilibrium of liquid water and steam is established at the new temperature and pressure. The corresponding specific enthalpy of the steam is $h_\text{steam}=2674.9\ \mathrm{kJ\ kg^{-1}}$; the specific enthalpy of the liquid water is $h_\text{liquid}=417.5\ \mathrm{kJ/kg}$. Since the flashing process is very quick, there is no significant heat exchange with the environment. Therefore, the vaporized fraction $x$ may be estimated according to the following enthalpy balance:
$$\begin{align}
h_0&=x\cdot h_\text{steam}+(1-x)\cdot h_\text{liquid}\\[6pt]
x&=\frac{h_0-h_\text{liquid}}{h_\text{steam}-h_\text{liquid}}\\[6pt]
&=\frac{1338.3\ \mathrm{kJ/kg}-417.5\ \mathrm{kJ/kg}}{2674.9\ \mathrm{kJ/kg}-417.5\ \mathrm{kJ/kg}}\\[6pt]
&=0.408
\end{align}$$
This means that approximately $40.8\ \%$ of the superheated water flashes to steam when the pressure vessel suddenly breaks.

Yes, the idea is to add a solvent that forms a lower boiling point azeotrope with water.
See "section 8.2 Azeotropic Drying" in The Synthetic Organic Chemist's Companion for specific suggestions such as benzene.


text:
Can CH3COOH be written as C2H4O2?
I was reading a chapter on acids and bases .It said something about acetic acid.And then I thought whether I could write CH3COOH as C2H4O2? please explain your answer.
Thanks!!

Yes you can, if you like being misunderstood. There are several $\ce{C2H4O2}$'s, but only one $\ce{CH3COOH}$. What you absolutely can't do is getting you uppercase/lowercase letters wrong. In chemistry, like in some programming languages, this is considered a major infraction.

oh sorry i will just gloss that up. thanks for the info, sir!

$\ce{C2H4O2}$ is the molecular formula of acetic acid. As such, it's a correct method to write the molecule. However, you should note that, as Ivan Neretin mentioned, molecular formulae are not necessarily unique for a given molecule.


text:
Mathematical properties of free Gibbs energy
I'm wandering, if some mathematical properties of the free Gibbs energy (also called thermodynamic potential) are well known.
For instance in chemistry, if $n_i^\alpha$ denotes the number of moles of species $i$ in phase $\alpha$ and G is the free Gibbs energy.
I know that G must be an homogeneous function of degree one in $n_i^\alpha$.
Do we have more information like continuity or coercivity properties of G ?

n is discontinuous. The least amount you can change it by is ~1 /6.022E23 of 1 mole. You'll have to be more specific about what type of (mathematical) continuity you are referring to, but it doesn't seem useful to ask about its fine grained continuity since it is a macroscopic concept.  Statistical (quantum) Thermodynamics is used both for microscopic systems and as the basis for macroscopic quantities such as Gibbs Free Energy. You need to be careful: GFE is one meaning of Thermodynamic Potential, but not the only one. I don't have any idea what you mean by G's coercivity.

Stat mech quantities are actually defined on expectation values, so they are continuous.


text:
How to judge accuracy and precision of measurement devices?
I am unsure about the accuracy and/or precision of measuring instruments.
For example, if we have a graduated cylinder, a pipette and a beaker, how and why can we assert them that they can be used with accuracy, precision, or both?

Beakers, Erlenmeyer flasks, dropping funnels and the like are not volumetric instruments. They are not precisely calibrated, and their scales serve only as approximate guides. Therefore, the beaker that is mentioned in the question is right out.
Graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes are volumetric instruments. They are calibrated; however, like any other measuring instrument, they only have a limited accuracy and precision.
The accuracy and precision do not only depend on the type of the instrument (e.g. graduated cylinder or pipette). They also depend on the dimensions, graduating divisions, and accuracy class of the instrument, which are usually internationally standardized.
In principle, volumetric instruments with a small diameter have a better precision because the same difference of the liquid level (i.e. the meniscus) corresponds to a smaller difference in volume. Therefore, typical long and thin pipettes are usually better than short and thick graduated cylinders. For example:

A standardized graduated pipette (Class A or AS) with a nominal capacity of 10 ml and an overall length of about 360 mm has a maximum permissible error of ±0.05 ml.
A standardized graduated measuring cylinder (Class A) with a nominal capacity of 10 ml and a maximum overall height of 140 mm (tall form) has a maximum permissible error of ±0.1 ml.
If you use an even larger graduated measuring cylinder (Class A) with a nominal capacity of 100 ml and fill it just to the 10 ml mark, the maximum permissible error is ±0.5 ml.

These errors represent the maximum permissible error at any point on the scale, and also the maximum permissible difference between the errors at any two points.
Since the same sources of error are, naturally, inherent both in calibration and use, a volumetric instrument should have a good precision in order to be able to be calibrated with a good accuracy.
However, note that graduated pipettes are usually calibrated ‘to deliver’ (TD, Ex), whereas graduated measuring cylinders are calibrated ‘to contain’ (TC, In). Hence, graduatted pipettes and graduated measuring cylinders may not be interchangeable at will.

There are too many measurement devices to give an answer for all types. However, assuming the scope of your question is limited to measuring fluid volume the way this can be done in a professional laboratory is simple.
You will need:
(1) high purity water
(2) a high precision balance which has been recently calibrated
(3) a type of container called a weigh boat

Tare the weigh boat. This means place the weigh boat on the balance
and (if the balance is electronic) press the tare button.  This will
cause the balance to read zero.  (With the old pivot beam style
balance another identical weigh boat could have been placed on the
other end of the balance to level it.  These pivot beam balances are
rarely used in modern lab practice as most balances are electronic.)
Using the measuring tool in question, measure a certain volume of
high purity water.  Pour the measured quantity of high purity water
into the weigh boat on the electronic balance.  The balance should
read the mass of the water.  Assuming the temperature of the lab is
close to 20°C divide the mass of the water by 0.9982 g/ml.  This is
an accurate way to tell how many ml were actually measured using the
tool.
Repeat the measurement three times.  Three true measurements can be used to determine an average and standard deviation.  The difference between what was measured and the average of the three true measurements is related to the accuracy of the measurement.  The standard deviation is related to the precision of the measurement.

If a pipet is supposed to dispense 10 mL and it actually dispenses 15, 16, and 15 mL then it has high precision (because the numbers are close.)  Its accuracy is way off because the average of 15.3 mL is half of an order of magnitude different than 10.  It could be used, though it would need to be either calibrated so the measurement is correct or the end numbers adjusted for the inaccuracy.
If a pipet is supposed to dispense 10 mL and it actually dispenses 5, 9, and 16 mL then despite being accurate, its precision is off.  Because the precision is off, this pipet is too unreliable to be used in scientific practice.
Realistically, if a pipet were supposed to dispense 10  mL and it dispenses 10.0010, 10.0000, and 9.9998 mL then it would be suitable for lab work.  Because both the accuracy and precision is high, this measurement is said to have high trueness.  Trueness -- which is both accuracy and precision -- is desirable in a good measurement.
It is always good practice to verify the calibration of any measurement before using the measurement to generate publishable data.  Recently published data in high-quality publications is a good source of information to judge if a certain set of data has acceptable trueness.

The term precision means two different things. The number 1.23456789 has more "precision" than the number 1.23. That is, there is a mathematical (and computer science) meaning that is different than the one used by physical scientists.
In physical science, precision means the property of reproducibility of a value or result. If I am precise, I get the same answer every time (or nearly so). If I am accurate, I get the correct answer. Here is one example: I have a machine which counts people going through the door. Whenever someone walks through, it adds 1.1 to its total. This machine is precise but not very accurate. I modify it and now it adds between 0.95 and 1.05 to the count each time someone walks thru. It is definitely not more precise, but may or may not be more accurate.


text:
What type of reactions are used in this synthesis?

The compound $\ce{(CH3)2CHCH2NH2}$ can be synthesised by the following route.
$$\ce{CH3CH=CH2 ->[1] CH3CHBrCH3 ->[2] (CH3)2CHCN ->[3] (CH3)2CHCH2NH2}$$
What types of reaction are used in stages 1, 2 and 3?

A 1=substitution; 2=addition; 3=reduction

B 1=substitution; 2=addition; 3=hydrolysis

C 1=addition; 2=substitution; 3=reduction

D 1=addition; 2=addition; 3=reduction

E 1=addition; 2=substitution; 3=hydrolysis

Could someone please explain to me how they worked out what reactions were used in the stages? I have been able to figure out stage 1 and 2, but I am really stuck on stage 3, I think it should be addition but this isn't one of the answers.

Well, I had an organic chemistry course more than a decade ago, but I think, in some sense, what happens on the third stage can formally be called an addition reaction. However, it is a very important type of reaction in organic chemistry which usually comes under it is own name.

If you correctly identified types of reactions 1 & 2, then you have a simple choice between reduction & hydrolysis for reaction 3. Clearly, only one of these reaction does what is expected.

Have you identified the functional groups of the concerned compounds?

C) 1-Addition, 2-Substitution, 3-reduction.
1) The hydrogen bromide is becoming an "addition" to the starting reagent.
2) The cyano functional group "substitutes" for the bromine group.
3) In this reaction the substituted cyano group is reduced to an amine.

Oh thank you! I didnt realise that this was a reaction that I needed to learn, I assumed they would have all been able to be worked out!

Most reactions can be classified in a number of different ways, just like many compounds can be classified in a number of different groups. For each transformation, your question only allows for two choices, so you either work out which one works or you just need to identify the wrong choice.

The first reaction is to be classified as either a substitution or an addition. If we look at the structures of reactant and product, we realise that all the reactant’s atoms are present in the product — but an additional $\ce{HBr}$ have been added.
That should give you more than enough clues that it has to be an addition.
The second reaction is to be classified as either an addition or a substitution. Again, just by checking the atoms present in reactant and product, we realise that we lost the bromine but gained a nitrile group.
‘Loss and gain’ is definitely not an addition but a substitution.
The third reaction is to be classified as either a hydrolysis or a reduction. It does get a bit more tricky here. To identify a reduction, you would need to do one of the following:

allocate oxidation states and determine whether they have changed
check if formally an oxygen has been removed
check if formally $\ce{H2}$ has been added.

Identifying a hydrolysis is easier. It is a reaction with water that can occur in one of two ways:

the original molecule is split into two fragments, one of which contains a hydrogen, the other a hydroxy group
the molecule has formally added $\ce{H2O}$

Just by counting carbon atoms, we realise that the molecule has not fragmented, and by identifying atom symbols we can see that water has not formally been added. Thus, it cannot be a hydrolysis, it must be a reduction. In fact, two of the three reduction identification conditions are met, but I’ll leave the proof up to you.

Note that reduction is not the only way in which you can classify reaction 3. You could also call it:

hydrogenation
hydrogen addition
reductive nitrile cleavage
saturation
and probably others.

Step 3 is definitely not reductive amination. Cleavage usually refers to sigma bonds breaking, so reductive nitrile cleavage doesn't seem right either.


text:
What happens if a proposed name for a new element cannot be abbreviated to a two-letter symbol?
Take the C-elements for example: C, Ca, Cd, Ce, Cf, Cl, Cm, Cn, Co, Cr, Cs, Cu.
Imagine some newly synthesized element is to be named after the University of Colorado: Coloradium. Its symbol can't be C, Co, Cl, Cr, Ca, Cd, Cu, or Cm, so we're forced to choose Ci.
Then another new element is synthesized, and it is to be named after Celine Dion: Celinium. Its symbol can't be C, Ce, Cl, Ci, Cn, Cu, or Cm. What would they choose for a symbol?

I'm sure someone (@Loong?) will give a detailed answer but the short version is that IUPAC oversees the process. Here is a link to "Naming of new elements (IUPAC Recommendations 2002)" which outlines the procedure and ground rules.

Yes, I saw that document. It doesn't seem to provide any input regarding cases of this kind.

@ToddMinehardt A new version (IUPAC Recommendations 2016) can be found here.

Fun fact: the document @Loong linked to does not note anything about atomic symbols except that once proposed they cannot be used again (hence $\ce{Cp}$ is not allowed, because lutetium was formerly referred to as cassiopeium.)

If they couldn't make up a suitable 2-character symbol from the name then IUPAC wouldn't approve the name. So "Col" will never be the name of an element since neither Co or Cl work work as an atomic symbols. // Now for more hypotheticals, suppose we get to the location of the former planet Krypton and find the element Kryptonite. Do we make up a new atomic symbol or use the atomic symbol that the original Kryptonians used?!?

In view of the fact that new elements are rare, and that Y, Tb, Er, and Yb could all be named after the same place Ytterby, the IUPAC Inorganic Chemistry Division will probably be able to find another suitable symbol for a new element that avoids any conflict. (“Plenty of letters left in the alphabet.“ – Picard)

@MaxW -- Kalelium?


text:
Reagents used to prepare organic ICP standards, from scratch
I am trying to figure our how I can make my own calibration standards for organic ICP analysis. I assume that I need organometallic compounds to mix with base-oil/mineral-oil to prep the standards, but I can not figure out what specific compounds are commonly used for the different metals (example: For Na, maybe Sodium isopropylcyclopentadienide?). I have looked online and through some analytical chem texts but no luck. Any source for this information would be greatly appreciated.

Well what metals other than Na? My first thought was what organic medium? Typically water is used as the solvent and the metals will be low concentration. So the viscosity of the samples and standards are the same. If you look at the signal from 1 microgram per liter of Na in ethanol, and hexane and mineral oil I'd absolutely expect different signal strengths for Na.

MaxW,
K, Fe, Al, Ca, Cu, Ni, Si, V just to name a few... the matrix is mineral oil and hence should be the calibration standards. I purchase the standards from a distribute but would rather prepare them myself.

8 elements out of 92 and that is just a few?!?

Buy the standard. The hassel you will have with water/impurities/.... is propably too much to bother with. You would need highly purified reagents, oranometalic compounds which never even saw some oxigen/water form a mile distance... Also, normally ICP does a sample preparation either by burning the sample or using strong acid. Does the counter ion of the metals really matter? (rhetoric question. there are some cases that matter, but most probably not organic).

I agree with @Julian, better to buy stock solution of different Metal ( 1000ppm) and prepare you standard solution accordingly. after drawing calibration curve, you can save this program and don't need to do calibration every time for comparison of your results.  In my opinion oganometallic compounds can,t be used for the preparation of standard solution in ICP-OES/OAS.

Running an ICPOES or ICPMS on organic solvents is not easy, you may have problems such as coke / carbon deposits in the torch.
Also I would like to point out that sodium dicyclopentadienide is very sensitive to oxygen, it decomposes with great ease on contact with air. Ask almost anyone who has made ferrocene in the lab about this question.
I would like to suggest that you consider making an emulsion in water of the organic solution using a detergent such as triton-x100. I have heard of this method for running organic samples in a normal ICPMS but I have never tried using it.
In the ideal world what I would do would be to use a wet combustion (maybe use a acid digestion bomb such as a Parr 45 ml bomb) to convert the organic sample into a aqueous sample.
Another option would be to use neutron activation anaylisis, but these days it is very hard to find a strong neutron source suitable for this type of work. You would have to seal up samples in quartz tubes. I will warn you that even with a nuclear reactor and a good gamma spectrometer this type of work is far from trival.


text:
Storage of methane in mineral oil or other liquid hydrocarbons
Is it practical (or even possible) to store gaseous alkanes such as methane or ethane in liquid non-polar solvents, such as mineral oil or vegetable oil, and how much more (or less) methane can be stored for a given pressure with solvent vs. no solvent?

After some research, I found that butane is sometimes dissolved in gasoline to make it evaporate more easily in cold weather. So it's definitely possible. However, I cannot find solubility figures (hence this not being an answer).

You seem to be putting methane and butane in the same basket, which is not entirely appropriate. Butane itself can be liquefied by applying a little more pressure; it is safe and convenient enough to be used in cigarette lighters. Methane is quite a different story.

@IvanNeretin Yes, I know that -- I've liquefied butane by applying pressure to it in a syringe, it is very easy. However, since they are both nonpolar molecules, I figured that the answer would be the same. I will change butane to ethane, though, as ethane can't be easily liquefied.

Related: How could I find the solubility of hydrocarbons such as iso- and n-Butane in liquid Methane?

Or even in water: https://en.wikipedia.org/wiki/Methane_clathrate. BTW, acetylene gas cannot be compressed much without exploding, so the gas sold for use in welding is dissolved in acetone: "A 40 litre acetylene cylinder will hold approximately 12-13kg of acetone in which we can dissolve 5500 litres of acetylene," http://wssproducts.wilhelmsen.com/mediabank/store/182/Understanding-Acetylene.pdf

"Practical" can mean different things to different people, so I am not going to comment on that.
Instead I am going to address the question, of solubility of methane in hydrocarbons, without any thought about "practicality" :)

This graph is taken from the book: Fuel and Lubricants Handbook (a simple google search lead me to it, and the relevant section appears in the preview :))
The data is reported at $50\ \mathrm{^\circ C}$ and $ 5000 \ \mathrm{psig}$ (pounds per square inch gauge, ungodly non-SI units. ca. $3.4 \times 10^7\ \mathrm{Pa}$ according to an online conversion)
Also, the following table from this publication:

I did not read it in detail, but the data was collected in the vicinity of 1 atm and between $25$ and $200\ \mathrm{^\circ C}$
FYI: Of possible interest could be this paper.

I'll dig some more, and if I find something I'll edit my answer to include it

From what I see online, the most comprehensive data can be found in: Solubility data series (methane), Volume 27-28, Pergamon Press, Oxford, 1987 authors Clever,H.L., Young, C.L., and Kertes, A.S., IUPAC. At roughly room temperature, again from what I see online, you can store about 10 mole % in n-hexane at about 400 psi. (see Hua, et al, Chinese J Chem Eng.,13(1)144-148(2005)). I have zero idea what "practical" means...it's meaningless without some context.

A partial answer, after a few experiments: Butane is soluble in vegetable oil. The specific type of vegetable oil used is olive oil. This is the process I followed:

Get some liquid butane out of a butane cylinder and place it into a test tube.
Pour the liquid butane into vegetable oil and agitate it.
Repeat until vegetable oil appears to be saturated.

The vegetable oil could then be ignited directly, and burned for about 5 minutes using about 100mL of vegetable oil. Usually, oil does not burn unless heated significantly. This indicates that the butane was, in fact, dissolved into the oil.
I have yet to try the same experiment with propane, ethane or methane.