Buffer Systems and Ion Reaction If we have a buffer solution as follows where HA is a weak acid and $\ce{A-}$ is its conjugate base, $\ce{HA + H2O<=>H3O+ + A-}$, do we ever consider $\ce{A- + H2O<=>HA + OH-}$? This is because in all of the buffer system problems I've done we only use the first reaction to find pH and other values but, we never consider the reaction of the conjugate base with water to form hydroxide ions which should increase the pH. Why is this so?

Yes the equilibrium certainly exists. In fact, there exists a third equilibrium, the self ionisation of water. Considering any of the two equilibria gives the concentrations of all the species involved, there is no need to write down the third equilibrium as the third equation will be a combination of the first two. The problem arises only when you invoke stoichiometry. For example, if $0.1\ \mathrm{mol}$ of the acid dissociated, it would certainly give $0.1\ \mathrm{mol}\ \ce{A-}$, but $0.1\ \mathrm{mol}$ would not be the amount of $[\ce{H3O+}]$ as it is part of another equilibrium. However, if we do make this approximation, ignoring the hydrolysis of $\ce{A-} $, we can justify it in some cases. Say if $K_\mathrm{a}$ is around $10^{-5}$ then the equilibrium constant for the hydrolysis would be $K_\mathrm{w}/K_\mathrm{b}$ which would be around $10^{-9}$. Thus it would not be a bad approximation to ignore the second equilibrium for finding the pH.


Strong exothermic mixing of two 'safe' liquids I have read (thanks to Stack Exchange) that mixing water and sulfuric acid is exothermic, as is NaOH and HCl.

Are there two 'safe' liquids that can be mixed and produce anywhere near the same amount of heat? By 'safe' I mean they don't come with the same heavy warnings as sulfuric acid or lye. I would like them to be safe enough to be stored in a home and that a spill would probably cause some damage but clean up could be performed by the home owner safely.

I understand that the actual mixing might not be considered safe if they do release a lot of energy.

I'm only considering liquids because they can be kept separate and pumped together.

If you need a quick source of moderate heat, why not use a single liquid? Sodium acetate trihydrate ($\ce{CH3COONa·3H2O}$), for example, releases heat energy on crystallization... and it takes very little to trigger that crystallization! Commercial heat packs contain a "clicker", which makes a sharp vibration. You can make your own heat pack fairly safely and inexpensively: sodium acetate can be found for ~US$8/450 g, and a Safety Data Sheet states, "it's  not classified for physical or health hazards under GHS."

Not only can you get a noticeable temperature rise quickly, it's reusable.

It would be more difficult to find liquids that quickly release energy that do not have physical hazards (e.g. ignition) or health hazards. Hypergolic fuels, for example, tend to be rather toxic, as well as explosive. You might get a bang out of reading John Drury Clark's Ignition!, though.


Determination of timescales using characterization methods There are 4 different timescales in material characterization (as classified in the Atkins textbook):
a. Time during which a quantum of radiation or a particle can interact with a molecule.
b. Lifetime of excited state of a molecule.
c. Minimum lifetime species must have to be seen as a distinct species.
d. Total time of experiment in which the species is observed.
Now the question is how to determine each of these? Say I have a certain compound and want to know what is the time of interaction of a photon with the molecule (part a). Is there an experimental method or instrument to determine these timescales?

The absorption of a photon en route to an electronically excited Franck-Condon state typically typically proceeds in the femtosecond range.

The lifetime of an electronically excited singlet state mostly is in the range of several nanoseconds. In some cases, however, fluorescence lifetimes in the range of several tenth of nanoseconds have been observed.

In cases where the initial excited singlet state undergoes intersystem crossing to a triplet state, much longer phosphorescence lifetimes (seconds range) are observed.

In order to detect a transient, its lifetime must be longer than the laser pulse needed to generate it.

Once you've populated a state, you may either use

time-resolved absorption spectroscopy to record it's spectrum and the decay of the absorption at a particular wavelength over time or
use time-resolved emission spectroscopy to measure the decrease of fluorescence (or phosphorescence emission) intensity at a particular wavelength over time.


Steric inhibition of resonance vs. hydrogen bonding I have to compare the acidic strength of these compounds:

Now, in A and B hydrogen bonding will be present which will help stabilize the negative charge of the conjugate base. However, the hydroxyl group is not bulky and won't cause steric inhibition of resonance, hence the +M effect of phenyl will still be present.

In C, methyl will cause steric inhibition of resonance and hence stop/reduce the +M effect of phenyl which will stabilize the conjugate base.

How do I compare the relative stabilization offered by these two effects (H-bonding and SIR)?

I think the main effect in play here is straight electron withdrawing potential of the oxygens. In my opinion this is a classical BSc question that tries to avoid the grey zones of pka eye balling.

I would invite you to review the concept of steric inhibition of resonance since I'm not entirely sure I understand your argument there.

Here's a good starting point if you do not have a text book covering the subject
https://www.quora.com/What-is-steric-inhibition-of-resonance


Kc only effected by temperature? This makes little sense to me, with a lot of contradictory statements I have learnt.
For example:
$$\ce{N2 + 3H2. <--> 2NH3}$$

($\mathrm{\Delta h = -46.2~kJ~mol^{-1}}$)

I have been told a higher kc = large amount of product being formed in the equilibrium.
So therefore why if I increase the pressure why don't I get a increase in kc as the forward reaction is favoured creating more product?

After asking around I have come to the conclusion that a reaction has many kc values depending on temperature , but then I don't see how the kc equation can work as you can get different concentrations of A B C and have constant temperature by increasing pressure.
And also the statement of higher kc = more product is also wrong.

Obviously I'm wrong but I'm just explaining where my confusion lies in my logic.
Where have I gone wrong? Or is there an explanation that will clear things up?

We want to perform the reaction you mentioned:
$$N_2+3H_2\longrightarrow 2NH_3$$

At the beginning we have certain concentrations of reagents, say $[N_2]_{in}$, $[H_2]_{in}$ and $[NH_3]_{in}$. The reason why I am assuming we have some ammonia is just a way for not dealing with infinite values of energy as it will be clear soon. Since Chemists often work at constant temperature and pressure we will follow the spontaneity of this reaction following the changes in the Gibbs Free energy $G_r(T,P)$ of reaction. A model describing the Gibbs free energy changes is the following:
$$\Delta G_{r}(T,P)=\Delta G_r^0(T,P=1bar) + RT \ln\biggl(\frac{[NH_3]^2}{[N_2][H_2]^3}\biggr)$$
The reason we have chosen not to have a zero concentration of ammonia is because in that case we would get $\Delta G_{in}=-\infty$ and this is not sensible. It is important to note that $\Delta G^0$ only depends on the temperature since the pressure has been fixed at 1 bar. Another aspect to note is that the concentrations in square brackets are the concentration at a certain point of the reaction. Now we are ready to find the answer to the question. What happens at the equilibrium? By definition the equilibrium is reached when the Gibbs free energy is minimum, thus when $\Delta G_r =0$. Taking the above equation and setting this condition, isolating the concentrations from the logarithm, we get that the concentrations of our reagents/product at the equilibrium are
$$\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}=\exp\biggl(-\frac{\Delta G_r^0}{RT}\biggr)=K_{eq}$$
Clearly the equilibrium constant depends only on the temperature and not on the pressure since in $\Delta G_r^0$ pressure has been fixed at 1 bar by definition. Changing the total pressure does not effect the amount of ammonia you get as you can see from a "mathematical" point of view. However in order to increase the product concentration we can increase the concentration of the reagents (in this case the partial pressure since we are in gas phase) since what is really fixed at the equilibrium for a given temperature is $\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}$ and not the concentrations taken singularly.


Help me identify this glassware! I'm a freshly graduated physics and math teacher moved into a small school from the late 1920s.  The equipment here is old and confusing, at the very least.  I have found many things which I have never seen before.  I would like some help identifying what I have and what I can do with them, if anything.

(Click the images to view them at full size.)

1. A load of copper pots.  For boiling something perhaps?

2. Some sort of wash bottle?  Why is the cork on there?

3. I managed to make out "unitized wash bottle" on the back.  This serves the same function as the last one, I presume.

4. It reminds me of a graduated cylinder until the frosted top and the "arms" that poke out the sides.  What is this for?

5. This reminds me of halogen bulbs.  Again, I have no idea what this is for.

6. I've taken to calling this "trumpet" glass pipe.  What is it used for?

7. This large bulb has confused me since day one.  I feel like it needs some sort of current passed through the ends.  It reminds me of a cathode ray tube in a way.  I'm completely lost on this one.

8. Here.  Have a free safety poster!  Not as good as old Carol, though.

#4 is a collection vessel of some sort.

The ground glass fitting would have connected to some other ground glass fitting. A set of the ears would be on the connecting piece of glassware also. Then springs would hold the two pieces together.


Difference between polypeptides and polyamides I can't seem to find a very good explanation of the difference between polypeptides and polyamides. So far, I know that polypeptides are a type of polyamides, but that is about it. My guess is that the term polyamides refers both to the polymers made from amino acids and those made from diamines and dicarboxylic acids, but polypeptides are only those made of amino acids, but this is just a guess...

Thank you in advance!

Your guess is correct. Polyamides are polymers where the repeating units are linked by amide bonds. Polypeptides are a specific type of polyamide where the repeating units are amino acids.


Is there any kind of reaction with two types of reactants (gases) and one product (also gas) such that the total volume after the reaction increases? I would like to think of the following equation:

$$\ce{aA(g) + bB(g) -> cC(g)}$$

where $a + b < c$ and $\ce{A},$ $\ce{B}$ and $\ce{C}$ are different gases.

Is there such chemical reaction?

Decomposition of ozone:

$\ce{O3 + O3 -> 3 O2}$

Technically, not 3 different gases, but the introduction of isotopes may be judged as in compliant (and yes, contrived).


Making a non-CN dissolved gold solution ideal for electroplating I've had success electroplating $\ce{Ni}$ and $\ce{Cu}$, but interestingly never $\ce{Al}$. My focus here is on plating with $\ce{Au}$, specifically onto an $\ce{Ni}$ surface.

In general from my own work and others' suggestions, I have found applying a voltage (direct current) of 5 V through a relatively dilute solution of relevant metal, e.g. 2.53 g $\ce{Ni(NH4)2(SO4)2*6H2O}$ dissolved in 81.9 mL 0.42 % acetic acid at ~60 °C makes a very nice, lustrous plating of $\ce{Ni}$ on a clean $\ce{Cu}$ surface after ~90 s of submersion.

I have been able to dissolve $\ce{Au}$ in warm, concentrated $\ce{HCl}$ and minimal $\ce{HNO3}$, and have heard that you can dissolve it in $\ce{HCl}$ and $\ce{H2O2}$, which makes sense because $\ce{Cu}$ does the same.

I'm looking for an ideal solution here, starting with pure metallic gold. I want to avoid any mistakes because I only have 1 g to work with. By default I'm going to try a comparable concentration of dissolved $\ce{Au}$ to what I mentioned with $\ce{Ni}$, but any advice is appreciated – especially a specific method for making an "ideal" solution.

I've found that several commercial solutions contain $\ce{(CN)4AuK}$, but I really want to avoid using cyanide.

This paper by Paul A. Kohl is immensely helpful and interesting, but I'm wondering if anyone has had any success with something simple.
It makes mention of using sulphide, thiosulfite, thallium, etc.

The problem is your amount of gold that you have. Normal chemical baths contain several grams per liter. So you will need to operate in small volumes. You can use this article - Chemical Papers
June 2012, Volume 66, Issue 6, pp 617-620
Date: 23 Dec 2011
Etching and recovery of gold from aluminum substrate in thiourea solution, that is using thiurea. Be aware of the low stability of this bath. The bath will degrade quite fast, but that is common problem of all non-cyanide gold baths.

As the article deals with a gold dissolution, I would suggest to reduce the gold back to a solid state. Use the article or this Hydrometallurgy
Volume 55, Issue 2, March 2000, Pages 137–152, Gold electrowinning from aqueous–alcoholic thiourea solutions and then do the electrodeposition.


What is peak identification? I am sorry if this question is too vague, but what is peak identification? I am working with a chemistry professor to program an autosampler and he says he would like to "program the robot for a peak identification feature that he would use on his project." Can someone please explain in detail what peak identification is?
Many thanks in advance for any help

Autosampler - is a piece of equipment that aspirates the sample from a vial and introduces it into a chromatographer or a detector. It has nothing to do with peaks or their identification - that comes later.
After the sample is injected into the detector and the measurements are taken you get back data describing physical properties of the sample. If you inject directly into the detector you can plot it as signal intensity vs. physical property, here's an example of spectra from Mass Spectrometer:

You may need to find peaks on such graph in order I don't know.. to identify how many impurities there are.
But often before introducing the sample directly to the detector chemists would run chromatography - it'd separate different components of the mixture and make them elute at different time. Now we'd measure physical properties not once - but do it at some rate. If we measure (and sum up or average) the signal at different time we may get peaks when the compound is eluted from chromatographer:

Chemists would then have to define the boundaries of the peaks and integrate their area in order to find how many substances there are in the mixture, in some cases calculate how much of each there are. Software can automate some of this work - identify the peak boundaries for you. This may be what your professor is asking for. Or maybe he wants you to write algorithms determining what substance is represented by this or that peak.
Note that this problem is very challenging and not universally solved - finding peaks can be problematic because:

different chromatograms will have different amount of high-frequency noise
there could be a baseline drift (low-frequency noise)
peaks often are skewed
compounds can elute very closely to each other, so you get merged peaks (rider peaks, shoulders, round peaks)

Due to these reasons you may not always be able to find the peaks reliably e.g.

So for autodetection you may want to go over several algorithms. And the person who uses them may have to understand and tune the parameters for it to work successfully on a particular chromatogram.


Is phosphoric acid to be considered a hazardous waste? I've found some English sites proclaiming phosphoric acid to be used for porcelain cleaning and e.g. clogged drains, even though I mainly bought it to use to clean coffee makers and water kettles.

In my country (Finland) phosphoric acid can be found from the list of hazardous waste. However, there are contradictory views regarding whether phosphoric acid can be poured in to the sewer or not. If it's a hazardous waste, then it shouldn't. But I have this English resources saying that it can.

Should I abide by the local rules saying that it's a hazardous waste or is there a reason to believe it can be used safely e.g. in clogged drains or for cleaning the toilet?

Phosphoric acid is a "food" additive because its acidity prevents the growth of microorganisms.  By itself it can cause burns, and high levels of phosphate in the diet have been linked to weaker bones (greater risk of fractures).

If you pour it down the drain, its acidity should not be the issue.  You could neutralize it with baking soda, and it will tend to become neutralized and buffered in nature on its own, but I think the primary issue is that it will yield phosphate ions which act as a fertilizer/nutrient source for algae growth and can cause algal blooms in ponds and streams that can dominate the ecosystem and choke out plant and animal diversity.

For that reason, don't pour it down the drain even though it will likely become neutralized as it acts on the clog.

If you just need to manage it, neutralize it with baking soda, and boil off the water, or let it evaporate to yield sodium phospate salt.  We don't want the sodium phosphate to enter directly into the local watershed, but you could put it on houseplants.  I don't know if phosphate fertilizer is allowed where you live, but if it is, you could put it on your lawn.  It PROBABLY would also be OK to dispose of the solid Sodium Phosphate in the trash.


Obtaining mercury from mercury(II) nitrate I was thinking about obtaining mercury from mercury(II) nitrate. Is it possible?
I've tried putting chunks of Fe into it, I've got the logic from the voltaic table (where Fe is on the left side of Hg, so I think it would get it out of the compound). Can anyone please correct me and show me the right way? I've tried searching on Google, but found nothing.

I'm just a rookie in chemistry, and was interested in it when my teacher announced about a talent show which I'll perform some cool chemistry there (I'm going to perform using fires, I've tried the colored fires and the alcohol carboy). I'm also planning to make a mirror, if anybody has some ideas please comment below.

The process you described would be more appropriately called "reduction of mercury(II) to elemental mercury".
Unfortunately, the trick with iron likely won't work (something more inert like copper would be a better choice though).

Mercury(II) oxide is weakly basic, so mercury salts in general would easily undergo hydrolysis and form basic oxosalts in aqueous solution unless acidified.
Mercury(II) nitrate quickly hydrolyses to poorly soluble yellow mercury(II) oxide upon dilution or addition of alkali:

$$\ce{Hg(NO3)2(s) + H2O(l) -> HgO(s) + 2 HNO3(aq)}\label{rxn:R1}\tag{R1}$$

which can be converted back to nitrate by adding excessive amount of nitric acid, which in turn won't leave any chance for iron not to be oxidized, so you end up with both metals in solution.

I didn't want to dive deep into the discussion of how mercury(II) salts hydrolyze and leave it simple, but after receiving critique from Maurice claiming $\ce{Hg(OH)NO3}$ to be the “real” product, I think I allow myself to add a paragraph or two.
Raman spectroscopy and x-ray scattering studies in late 1960 demonstrated that hydrolysis of mercury(II) salts produces series of polynuclear oxo-bridged species of types $\ce{Hg2OH(H2O)2^3+},$ $\ce{Hg3O(H2O)3^4+}$ or $\ce{Hg4O(OH)(H2O)3^5+}$ [1, 2].
Formation of mercury(II) hydroxide nitrate $\ce{Hg(OH)NO3}$ as a hydrolysis product was taught in 1940–1950s era, and it stuck in several textbooks published later probably because it's listed in all editions of Pauling's General chemistry up to 1988.
However, not only it's oversimplified (granted, reaction \eqref{rxn:R1} is also an oversimplification in a sense it's a boundary case), but is also an incorrect concept.

The most up-to-date summary of what is really happening when $\ce{Hg(NO3)2}$ is dissolved in water can be found in Mercury handbook [3, p. 115]:

  $\ce{Hg(NO3)2}$ solutions are only stable in the presence of a certain amount of nitric acid, which prevents hydrolysis.
  $\ce{Hg(NO3)2}$ quickly hydrolyzes in excess water and produces a precipitate of
  $\ce{Hg3O2(NO3)2 · H2O}$ or, when boiled in dilute solutions, forms mercury(II) oxide $(\ce{HgO}).$

Regarding the reduction of mercury(II), there are two pathways: $\ce{Hg^0}$ may be obtained from mercury(II) nitrate using a dry or a wet method.

Speaking of a dry method, the most straightforward way to obtain metallic mercury from nitrate is to heat $\ce{Hg(NO3)}$ in a distillation apparatus (b.p. of mercury is 357 °C).
Above $\pu{400 °C}$ it readily decomposes:

$$\ce{Hg(NO3)2(l) ->[\pu{400 °C}] Hg(g) + 2 NO2(g) + O2(g)}\label{rxn:R2}\tag{R2}$$

\eqref{rxn:R2} is a brutto-reaction; the nitrate first decomposes to the red mercury(II) oxide at lower temperatures (which, in turn, decomposes to the elements):

$$\ce{2 Hg(NO3)2(l) ->[\pu{350 °C}] 2 HgO(s) + 4 NO2(g) + O2(g)}\tag{R3}$$

Wet method suggest mild conditions and a reaction in solution.
For example, formic acid (also used in silver refining) being a strong reducing agent would cause a precipitate from ammoniacal mercury(II) nitrate solution.

Since nitrogen dioxide, mercury vapors as well as mercury salts and oxides are highly toxic, the reactions must be carried in a fume hood which makes it a poor fit for a talent show.
Considering the possible risks and your level or preparation (no offence), I'd strongly advise to take extreme caution doing mercury chemistry and avoid public demonstrations until you become more experienced.

Note: chemical reactions are adopted from [4, p. 310–312]

References

Cooney, R.; Hall, J. Raman Spectra of Mercury(II) Nitrate in Aqueous Solution and as the Crystalline Hydrate. Aust. J. Chem. 1969, 22 (2), 337. https://doi.org/10/b6t3h2.
Johansson, G.; Haugsten, K.; Rasmussen, S. E.; Svensson, S.; Koskikallio, J.; Kachi, S. An X-Ray Investigation of the Hydrolysis Products of Mercury(II) in Solution. Acta Chem. Scand. 1971, 25, 2787–2798. https://doi.org/10/bn5j2g. (PDF)
Kozin, L. F.; Hansen, S.; Kit, M. Mercury Handbook: Chemistry, Applications and Environmental Impact; RSC Publ: Cambridge, 2013. ISBN 978-1-84973-409-7.
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Calculate the pH of acetic acid for a 0.02 M solution We have a solution $\ce{CH3COOH}$ (acetic acid) with $c=0.02~\mathrm{mol/L}$  and $K_\mathrm a(\ce{CH3COOH}) = 1.8\cdot10^{-5}$. Calculate the $\mathrm{pH}$ of this solution.

All I know is that it is a weak acid and that
$$\mathrm{pH}= \mathrm{p}K_\mathrm a + \log\left(\frac{B}{A}\right).$$
I know $A$ but there is no information about $B$.

This type of problem is best solved using the ICE (Initial Change Equilibrium) method.
First, write down the chemical equation and the law of mass action for this reaction.
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$$
Second, determine the initial concentration of each of the three species. Make a little table to keep everything straight:
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\ \hline
\end{array}
At equilibrium, some of the acetic acid will be ionized. How much? We don't know right now, so let's have that be a variable; $x$. At equilibrium, the concentration of acetic acid will have decreased by $x$. How much will the concentration of $\ce{H+}$ and acetate anion increase? Since the stoichiometric coefficient of all species is 1, then those concentrations will increase by $x$. Let's add that to our table. Let's also write an expression for the equilibrium concentrations in terms of the initial concentrations and $x$.
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\
\text{C}& -x &&+x  &&+x \\
\text{E}& 0.02-x  &&x &&x \\ \hline
\end{array}
Now, plug these expressions into the law of mass action:
$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{(x)(x)}{(0.02-x)}$$
As long as $x\ne 0.02$, which is very unlikely at equilibrium, we have a solvable quadratic equation:
$$K_a=1.8\times 10^{-5}= \dfrac{(x)(x)}{(0.02-x)}$$
$$3.6\times 10^{-7}-1.8\times 10^{-5}x-x^2=0$$
Some texts will encourage you to make the simplification $0.02-x \approx 0.02$ since $x<<0.02$, however, WolframAlpha can solve the original equation (and so can you).
The equation has two roots: $x=5.91\times 10^{-4}$ and $x=-6.09\times 10^{-4}$. The second one is nonsense (we cannot have negative concentrations).
If you make the assumption that $0.02−x \approx 0.02$, then the positive root is $x=6.0 \times 10^{−4}$, not a big difference given the significant figures. If you knew your initial concentration more precisely, this assumption would be invalid.
Now, we know what $x$ is. What do we still need to know to determine pH? How is $x$ related to pH?


Preference of grignard between aldehyde and double bond? Quick question, in this reaction:

2-Butenal + CH3MgBr  -->  ??

Will the grignard attack the double-bond or the aldehyde carbon?
Please explain this preference, thanks!

The Grignard will first attack the aldehyde carbon, giving you a secondary alcohol. Since the alcohol is strongly polarized, it would be difficult for the double bond to be attacked.


Why is (NH4)2[CuCl4] square planar complex but Cs2[CuCl4] is tetrahedral? Why is $\ce{(NH4)2[CuCl4]}$ square planar complex but $\ce{Cs2[CuCl4]}$ is tetrahedral even though both have same oxidation number of copper and same ligands?

In solution, the $\ce{[CuCl4]}$2- is expected to exhibit terahedral, or nearly tetrahedral geometry.

The Jahn-Teller theorem states that degenerate orbitals cannot be unequally occupied. Molecules with unequally occupied degenerate orbitals will distort so as to render the orbitals non-degenerate, with more electrons occupying the lower energy states. Generally, this distortion leads to a reduction in symmetry. In the case of Cu (II), it is a d9 cation with a tetrahedral electronic configuration of (e)4(t2)5, so the three degenerate t2 orbitals are unequally occupied, therefore a distortion towards the square planar geometry of (eg)4(b2g)2(a1g)2(b1g)1, with no unequally occupied degenerate orbitals, is electronically favoured. Nonetheless, experimental evidence (1) indicates that this d-orbital stabilization energy is very small. Since copper is a first-row transition metal (not very big), steric effects dominate electronic effects allowing $\ce{[CuCl4]}$2- to assume a tetrahedral geometry. Most spectroscopic methods I've read involve looking at the molecule in crystal form (because that is where this effect gets interesting), but DFT studies of the $\ce{[CuCl4]}$2- hydrate in concentrated aqueous solution indicate that the anion's tetrahedral structure is more stable (2).

The issue of $\ce{(Cs)2[CuCl4]}$ vs $\ce{(NH4)2[CuCl4]}$ comes up when the complex precipitates and crystals are formed. Due to the borderline electronic effects, the $\ce{[CuCl4]}$2- anion exhibits a great degree of structural flexibility. With large cations ($\ce{Cs+}$ radius is 0.174 nm while $\ce{NH4+}$ radius is 0.143 nm) , it crystallizes as a distorted ("squashed") tetrahedron.

$\hskip1.4in$ (3)

Square-planar co-coordination occurs in such a way that you effectively get octahedra with tetragonal distortion (4). I would imagine that this phenomenon is simply a result of the smaller cation allowing the unit cell to crystallize like that. The general idea with unit cells is that the cation is never allowed to "rattle" around, so the structure will find a way to fill in the space. Given the preexisting electronic favourability of square planar geometry, $\ce{(NH4)2[CuCl4]}$ this distortion favoured by close-packing is made even more favourable.

$\hskip2.1in$ (4)

The canonical reference for $\ce{(NH4)2[CuCl4]}$ 's crystal structure is (5) if you want to take a look at the unit cell.

(1) Emeléus & Sharpe. Advances in Inorganic Chemistry and Radiochemistry, vol. 21, p. 128

(2) https://www.ncbi.nlm.nih.gov/pubmed/21462945

(3) L. Helmholz & R. F. Kruh, J. Am. Chem. Soc., 1952, 74, 1176

(4) Molecular Structure by Diffraction Methods, vol. 1, The Chemical Society, London, 1973, p. 642

(5) R. D. Willet, J. Chem. Phys., 1964, 1964, 41, 2243


How can I clean up from electrolysis? So, I was doing a little experimental electrolysis in my basement. I used 2 (presumably) galvanized steel screws as my electrodes and ran 5A through tap water with a lot of NaCl dissolved in it (to decrease resistance). It worked great: I got a lot of bubbles from the - side. Problem is, my cup started filling up with this black crud.

Now, I have a little plastic disposable cup with really salt water, 2 screws (black with what I presume is some corrosion), and a lot of black and red insoluble crud. How do I get rid of it?

Also, what it that crud (most likely)?

EDIT: I figured out that putting NaCl in an electrolytic cell means that I was actually generating chlorine gas instead of oxygen gas. That means that I have a bunch of iron and zinc chlorides in that water. It's also probably pretty basic (I saw no bubbling at that electrode). Any disposal problems?

The black one is probably $\ce{ZnFe2O4}$; insoluble compound of very weakly toxic metal ions, or $\ce{Fe3O4}$. The red one is $\ce{Fe2O3}$ In terms of disposal all above counts as non-toxic mineral materials, similar to sand.

It is also possible to have some carbon in this setup, but I doubt it.

Anyway, you have nothing too toxic in this setup, you can safely dump it into ground without any meaningful lasting environmental consequences.

The problem may arise, if the steel used was stainless steel or another iron alloy, but it doesn't seem to be the case from your description (and they are rarely galvanized anyway)


How to determine whether inversion takes place? Predict in which of the following reaction inversion of configuration takes place:

In my view, C) involves $\mathrm{S_{N} i}$ and therefore doesn't involve inversion of configuration. In D) the syn addition of $\ce{H2}$ across the double bond takes place and doesn't change the configuration. However I am not sure about A) and B).  Any ideas?

A) is $\mathrm{S_N2}$ due to the secondary $\ce{C}$ and the sufficiently good nucleophile $\ce{I-}$. So inversion takes place.
B) is $\mathrm{S_N1}$ due to the tertiary $\ce{C}$. The solvent is also more polar than in A). So the stereochemical information would be lost.


Confusion over percentage volume We've been told our lab report has to be written with "% v/v" instead of volume. I understand how to calculate it, however I'm unsure if I'm doing the right thing when it comes to using it later on. Example:

  2ul 'W', 4ul 'X', 2.2ul 'Y', 1.8ul 'Z' is mixed in an eppendorf. It is incubated for 1 hour. After incubation, 0.95ml of 'B' is addded and then 150ul of the final mixture spread on plates.

I know the first bit would be:
20% v/v 'W', 40% v/v 'X', 22% v/v 'Y', 18% v/v 'Z' is mixed in an eppendorf.

But I dont know what to do when it comes to after the incubation?

I'm generally pretty good with this stuff but I've never had to think of it this way. Would, when 0.95ml is added, I do 0.95/0.95+2ul+4ul+2.2ul+1.8ul?

My assumptions:

'ul' Refers to microlitres, be careful that the addition of B is in millilitres.
You are trying to calculate the % v/v that is being removed from the total solution at the end?

So before incubation the total volume is 2ul + 4ul + 2.2ul + 1.8 ul = 10ul (0.01ml)
You are adding in 0.95ml, totalling 0.96ml.
You are then taking 150ul (0.15ml)
This equates to (0.15/0.96)*100 %
= 15.6%


Efforts made to reduce the amount of CFC Till now I have heard $\ce{CO2}$ is the main reason of global warming. I have studied the remedies to control $\ce{CO2}$ many times through books, awarness programs, etc.

But today I learned about Chlorofluorocarbon ($\ce{CFC}$). It have 38 million times more "Global Warming Potential" than $\ce{CO2}$    according to UN.

In my book they have mention the sources of $\ce{CFC}$ like aerosols, refrigerators, air-conditioners, foams in fire extinguisher, etc.

But it does not tell what effort we should make or scientist are doing in order to reduce harmful of $\ce{CFC}$.

I have heard that $\ce{CFC}$ is replaced by 1,1,2,2-Tetrachloroethane, nothing else.

So, I want to know what else we are trying to reduce harmful effect of  $\ce{CFC}$?

After learning more about environmental chemistry I got this.

On 16 september 1987, 95 countries accepted not to use ODS(Ozone Depleting Substances)  after signing Mortal Protocol. Later other countries also accepted it.

After that United Nation have decided to celebrate 16 september as "Ozone protection day" at international level to spread awareness.


What is the mechanism of the reaction of ethyl prop-2-ynoate to ethyl (2Z)-2,3-dibromoprop-2-enoate? It's really weird for the product is 99％ Z-comformation?
Please tell me why?
This reaction can be found in scifinder.

A search of SciFinder will also produce the formation of the (E)-dibromide 3. The conditions described for the formation of the (Z)-dibromide 2 are taken from a paper by Myers1 as reported earlier by Tippett.2 Rossi3 has reported the formation of the (E)-dibromide 3 using pyridinium hydrobromide perbromide.
Given the more vigorous conditions for the formation of the (Z)-isomer 2 compared to the milder conditions for the formation of the (E)-isomer 3 suggests that the former reaction is under thermodynamic control while the latter is kinetically controlled.  There is the distinct possibility that bromine radicals may effect the isomerization via species 4 as may surreptitious acid.

1) A. G. Myers, M. M. Alauddin, M. A. M. Fuhry, P. S. Dragovich, N. S. Finney, and P. M. Harrington, Tetrahedron Letters, 1989, 30, 6997.
2) R. G. Hall and S. Trippett, Tetrahedron Letters, 1982, 23, 2603.
3) F. Bellina, A. Carpita, M. De Santis and R. Rossi, Tetrahedron Letters, 1994, 35, 6913.


Is it possible to calculate chromacity coordinates using absorbance values? I have absorbance values for blood plasma. Absorbance was measured in range of 380 – 600 nm every 5 nm ending up with absorbance values for 45 wave lengths.
I was browsing through literature, and  I came  across chromacity coordinates (XYZ) and this seems to be a very convenient way of summarizing data about color in just two data points.
I have been reading some literature about different colorimetric systems, but since my background is in genetics, it didn’t make much sense for me.
Therefore, I would like to ask you if it is possible to calculate these chromacity coordinates using my absorbance values and if so does anyone now some kind of software for this or exact formulas on how to do it?

This is a partial answer to your question.  I have used chromaticity coordinates for various scientific endeavours (air quality mainly).  I found that it is a very powerful means of determining colour differences.

A bit of searching and there is somewhat of a precedent for using chromaticity for blood plasma.  An example is the article Features of Color Characteristics of Blood Plasma of Patients with Liver Cirrhosis in the Colorimetric System XYZ as Compared with Healthy People (Dolomatov et al. 2012) - even though it may be a different context to your study, but does show that it can and has been done before.  Specifically, their method was

  by colorimetric method, based on absorption electromagnetic radiation in visible area.

There are several more examples of how it can be used.


Applied Potential and Electrochemical Cell Potential Electrolysis of molten lithium hydroxide at $\pu{750 K}$ to form lithium metal
$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.$$
Electrolysis will only proceed at an appreciable rate when the applied potential exceeds the electrochemical cell potential by $\pu{0.60 V}.$
(c) Calculate the minimum potential that should be applied.
$$ΔG^\circ = -nFE^\circ_\mathrm{cell}$$
$$E^\circ_\mathrm{cell} = \frac{ΔG^\circ}{nF} = \pu{-2.60 V}$$

Why should the difference of $\pu{0.60 V}$ be added onto the absolute value of $E^\circ_\mathrm{cell}~(+2.60)$ and not $-2.60$?
I understand  $E_\mathrm{cell}$ is the measure of the potential difference between two half-cells in an electrochemical cell, but what is applied potential in this case?  I thought electrochemical cells produce an electric potential difference (in this case $\pu{-2.60 V}),$ so why would they need an applied (positive) potential difference?

You are mixing two areas, electrode kinetics and electrode thermodynamics. If we have 3 apples and 2 oranges, this is not equal to 5 oranges or 5 apples but there are 5 fruits in total. Electrode thermodynamics tells you a "yes" or "no" answer in the sense that it is saying the reaction is spontaneous or not. Electrode kinetics tells you how fast the reaction will be.

When you wish to electrolyse something, it means that you want do a non-spontaneous reaction i.e., if you leave lithium in the presence of air and moisture, it will spontaneously form LiOH

$$\ce{4 LiOH <- 4 Li + 2 H2O + O2}.--(1)$$

Now, by forcing electric current through molten LiOH, you want to reverse this reaction and want the following to happen

$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.--(2)$$

The negative Ecell of -2.60 V only tells you that reaction (2) is not spontaneous at all. It has no other meaning whatsoever.  That is, if you wait forever, you will not see any Li spontaneously forming out of LiOH.

Now as you read in the question, they are saying that if you apply exactly +2.60 V potential difference between two electrodes the rate of reaction is very slow. Remember negative sign was just for saying that the reaction is non-spontaneous. However if you apply +2.60+0.60 = 3.2 V of potential difference between the electrodes, the rate of formation of Li from molten LiOH will be quite fast. They don't say how fast but you will not have to wait for a long time.

So +2.60 V is the apple = electrode potential
0.6 V is the orange, it is called the overpotential in formal language.

You can only add them (not subtract) and get the total voltage required to start the electrolysis at an appreciable rate.


What Would Hypochlorite Ions Prefer to React with, Hydrogen Ions or Ozone? If I had a box that had inside it a UV light an equal quantity of ozone ($\ce{O3}$), and $\ce{H3O+}$, which would be broken down because of the UV light to $\ce{O}$, and $\ce{O2}$, as well as $\ce{H2O}$, and $\ce{H+}$ because that is what the $\ce{H3O+}$ would be broken down into. If I introduced one hypochlorite Ion ($\ce{ClO-}$) or a chloride Ion ($\ce{Cl-}$), what would the chloride Ion or hypochlorite Ion react with? Would it react with the $\ce{H+}$, or the $\ce{O}$ & $\ce{O2}$?

First of all, hydronium ion, $\ce{H3O+}$ is basically a proton in aqueous solution. A proton undergoes hydrogen bond in aqueous solution to forms various forms of hydrated proton. But for simplicity, we denote $\ce{H+}$  which is considered an acid. Now, UV light dissociates hydronium ion in ozone atmosphere into oxygen, singlet oxygen, proton and water. So, if hypochlorite  salt or its corresponding acid is introduced, it might react with these. Chloride ion in form of salts is very much stable and is unlikely to react with any of the dissociated products.

So, first of all, any hypochlorite salt or hypochloric acid added will dissociate into component ion in presence of UV light. Then will react with the following dissociated products:

Singlet oxygen

$$\ce{ClO- + 2O ⇄ ClO3-}$$

(Source, considered hypochloric acid)

Proton

$$\ce{2HClO + 2 H+ + 2  e{−} ⇌ Cl2(g) + 2H2O}$$

Water is present as moisture so it is unable to attack hypochlorite ion readily. Also oxygen is present in minute amount(mainly singlet oxygen is present), so it cannot oxidize hypochlorite ion.


How strong are molecular bonds in layman's terms? Suppose you had a single sheet of graphite the size of an A4 page. What would it feel like to handle it?

Would your fingers instantly tear through it as you tried to grasp it?

Could you punch your finger through on purpose?

Could you tear it like a sheet of paper?

Could you fold it into a paper aeroplane without breaking it?

I will assume that by "a single sheet of graphite" you mean graphene.

The link in the comments to the wikipedia page on orders of magnitude of different forces, provided by @Alchimista in the comments, is helpful in getting a perspective on at least a few of your questions. Two useful reference values are the force exerted by a 1-kg weight due to gravity at sea level, ~10 N, and the force required to rupture a typical covalent bond, ~1 nN.

Perhaps the first question would seem to be the easiest to answer, but I found it is not: mechanoreception by sensors on skin is complicated. In the same way you can sense very gentle contact, you may be able to sense the graphite sheet touching your skin. It would probably be an odd feeling. Note also that the sheet would be just barely visible.

The strength of a material is described by more than one property. Perhaps the simplest property to describe is the tensile strength, the stress required to rupture the material by pulling it apart under tension. You can use estimates of the geometry of a sheet of graphite and the C-C bond enthalpy to arrive at the force per unit length of the material require to rupture it:

$$\pu{F_l(N/m)} \approx \frac{\pu{350 kJ/mol}}{(\pu{2.46 Å*1.42 Å})}$$

That's $F_l\approx\pu{16 N/m}$, so you might expect a single layer of graphite that is 1 meter long to be able to hold up a 1-kg weight suspended normal to this dimension, assuming the force is evenly distributed. If we assume the sheet is ~$\pu{2 Å}$ thick then we arrive at an estimate of the tensile strength of ~$\pu{10^11 Pa}$, making it the "strongest material ever tested."

Here I've estimated the number of bonds per unit length normal to the direction of tension as 1 per 2.46 Å based on this figure depicting the dimensions of a sheet of graphite. I also assumed the average force required to break one CC bond is $\Delta_{bond}H/r_{CC}$ and used a low estimate of $\Delta_{bond}H$. You could derive a similar value of $\Delta_{bond}H$ from the heat of sublimation of graphite.

Based on this information I would assume that it would be easy to tear the sheet, but that you would not necessarily tear the sheet by grasping it, provided you did this very carefully. On the other hand, you could certainly punch a hole with your finger. Ignoring subtle issues regarding the geometry at which forces are applied, we can use the previous estimate of the rupture force per unit length and an estimate of the radius of a finger (~1/2 cm) to arrive at the force required to punch the sheet, or $\approx\pu{0.5 N}$, or equivalently the pressure:

$$p=\frac{2F_l}{r_{finger}}$$

or $\approx\pu{7 kPa}$. A fat cat would only exert a pressure of about 300 Pa.

Could you fold it into a paper aeroplane? Apparently graphene is tough enough to be significantly bent out of shape without rupturing (Refs. 1 and 2), so I'd have to guess "yes".

References

Briggs et al., Electromechanical robustness of monolayer graphene with extreme bending. Appl. Phys. Lett. 97, 223102 (2010)
Edges of graphite and refs therein


Does 2-chloro-1,3-dinitrobenzene show resonance? I am preparing for my exam, and then I stuck up here. The answer to this question given by my teacher was

  No, this doesn't show resonance because there are two bulky groups alongside chlorine, due to this chlorine will leave the plane and as we know resonance is a planar phenomenon, it won't show resonance

My question to you is:

Why did chlorine leave the plane on the first place? And what is the whole concept behind it? Please help me with it.

My guess to this is that since these two groups are really big and bulky, and are present near Chlorine, they take up almost all the space around which causes chlorine to leave the plane.

2-chloro-1,3-dinitrobenzene is shown below:

I believe your question is referring to the lack of conjugation between the nitro groups and the aromatic ring. this is best shown in an X-ray crystal structure such as this one:

https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=722415

Due to the sterics between the nitro groups and the chloro substituent, the nitro groups cannot adopt a conformation planar with the ring. This prevents the $\pi$ system of the nitro group from achieving substantial overlap with the $\pi$ system of the ring and prevents full conjugation of the electrons. The sterics are better seen in a spacefilling view:

So, it is not the chloro group that is out of the plane, but rather the nitro groups.


Why would electrode reaction in galvanic cell stop when there's no neutrality in each cell? I know that the salt bridge produces ions to balance out the positive and negative ions in each solution, but why won't it work without a salt bride? Would it even work at the start since there's few electrons gained and lost and few ions produced?

It would work vanishingly briefly. The separation of charge that occurs when you connect two separated half cells without a salt bridge produces an electrostatic potential difference that opposes the electrochemical potential difference. The cell reaction would proceed until the two voltages balance, and it doesn't take much charge separation to hit the 1–2 volt electrochemical potential of typical cells, so this process is very fast.


Why do tosylation and mesylation of alcohols follow different mechanisms? Why do the tosylation and mesylation of an alcohol proceed via different mechanisms?

More specifically,

In tosylation, the lone pair of the alcohol attacks the sulfur in TsCl, but why does it not do so with MsCl? Surely, TsCl should be more sterically hindered due to the benzene ring.
Why is the sulfene intermediate even formed in mesylation? Can that position be deprotonated by $\ce{Et3N}$? It's clear that this mechanism cannot happen with TsCl which has no alpha protons, but I fail to see why it should happen with MsCl.

Tosylation with pyridine:

Mesylation with triethylamine:

The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.

For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.

While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.

$^{[1]}$ King, J. F.; Lam, J. Y. L.; Skonieczny, S. Journal of the American Chemical Society 1992, 114 (5), 1743–1749.


Why ionic radii of Cu2+ is less than Zn2+? Fully filled orbital has more effective nuclear charge than incompletely filled orbital. So atomic or ionic radii of elements or ions having fully filled orbitals should be less than that of elements having  incompletely filled orbitals (e.g. $\ce{F}$ and $\ce{Ne}$).

Now $\ce{Cu^2+}$ has a $\ce{d^9}$ configuration and $\ce{Zn^2+}$ has  a $\ce{d^10}$ configuration. So, $\ce{Zn^2+}$ has full filled d orbital and is thus stable. According to my assumption above, $\ce{Zn^2+}$ should have less ionic radii than $\ce{Cu^2+}$. But it is found that $\ce{Zn^2+}$ has more ionic radii than $\ce{Cu^2+}$.

Ionic radii of $\ce{Cu^2+}$ is $\ce{73}$ pm and that of $\ce{Zn^2+}$  is $\ce{74}$ pm. Why is this so?

(source)

To answer your question, it is good to see the full picture. I used data from the Wikipedia data page on ionic radii to plot the dependence of the ionic radius of dications by the number of d-electrons. The plot includes 1st d-row data for high-spin and low-spin states (these are terms from crystal field theory and are properly introduced elsewhere). It also includes data for Pd-Cd, Pt and Hg.

As we can see, a rather complex dependence is observed. However, it is clear, that the increase in ionic radius from the d8 to the d10 state is not a random fluctuation, it is observed for all three rows. To understand the nature of this dependence it is critical to understand that the values are for an octahedral ligand field (the values would be difference for tetrahedral and other coordinations). In such a field d-orbitals become unequal — three of them are lower in energy while two are higher. In high-spin complexes the difference is not enough to overcome electron pairing energy, while in low-spin ones it is. As result, for low-spin complexes we see a clear minimum at d6, corresponding to the three lower d-orbitals being fully filled, while for high-spin complexes we see changes in the trend for the 4th (first on higher orbital) 6th (all orbitals half-filled) and 9th (first paired on higher orbital) electrons. Now, let's check out the orbitals mentioned (thanks to chemwiki)

$\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ are the orbitals with higher energy. As it can be seen, they are located between the metal atom's center and the ligands. As such, electrons on said orbitals have clear effect on metal-ligand distance, so for low-spin state the observed ionic radius monotonously descreases up to 6 electrons (fully occupied lower orbitals) and then monotonously increases. For the high-spin state, the increases are also located on steps when an electron is added to higher orbital, but in this case the electrons are added on d4-d5 and d8-d10 configurations, so the increase happens in two steps. For the 2nd and 3rd rows, only few last steps on the curve are observed as most elements in these rows tend to form metal-metal bonds in low oxidation states and thus their true ionic radii are unknown.

And now, why are these dependences of limited use. Consider the graph of $\ce{M-Cl}$ distances for chlorides of the first d-row below. The data was obtained from the open crystallography database. No correlation with the plot above can be established, despite most of the chlorides except for $\ce{ZnCl2}$  having their metal in sixfold coordination. This is because said chlorides are not isostructural. Layered $\ce{MoS2}$ and $\ce{CdI2}$ structures together with chains of squares and 3D-lattices occur. Furthermore, for iron two different structures (layered and rutile-like ones) were found with, obviously, different $\ce{M-Cl}$ distances (the layered one is plotted. As a fun fact, the longer distance is observed in the higher-pressure rutile modification). An obvious conclusion for the plot is that only for isosctructural fragments we can conclude something based on electron structure of the central metal atom.


Compare the work done during dissolution of carbides 1 mole of each $\ce{CaC2}$, $\ce{Mg2C3}$, $\ce{Al4C3}$ react with excess water in separate open flask. Work done during dissolution shows the order

$\ce{CaC2}=\ce{Mg2C3}<\ce{Al4C3}$
$\ce{CaC2}=\ce{Mg2C3}=\ce{Al4C3}$
$\ce{Mg2C3}<\ce{CaC2}<\ce{Al4C3}$
$\ce{Mg2C3}<\ce{Al4C3}<\ce{CaC2}$

Calcium carbide gives ethyne on reaction with water. One mole of calcium carbide reacts with $2$ moles of water.
Magnesium carbide gives prop-1-yne on reaction with water. One mole of magnesium carbide reacts with $4$ moles of water.
Aluminum carbide gives methane on reaction with water and one mole of aluminium carbide reacts with $12$ moles of water.
Work done $=P\Delta V$. But since the amount of water is large compared to the amount of carbide, will the work done differ?
(Answer given is $1$)

The relevant equations are:

$\ce{Mg2C3(s) + 4 H2O (aq) -> C3H4 (g) + 2Mg(OH)2 (aq)}$

$\ce{Al4C3 (s) + 12H2O(aq) -> 3CH4(g) + 4Al(OH)3(aq)  }$

$\ce{CaC2(s) + 2H2O(aq) -> C2H2(g) + Ca(OH)2(aq)}$

As you can see, most number of gaseous moles are released in the 2nd reaction while equal number of gaseous moles are released in the first and third reaction, therefore maximum work is done in the first reaction under same conditions of temperature and pressure.

Note that:

In open vessel reaction, gases expand against a constant external atmospheric pressure, thereby doing work on the surroundings , thus the work done $= P_{\text{atm}}\Delta V$ is independent of the gas's pressure and only dependent on $\Delta V$

Also, recall that Avogadro's law states:

  For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

Hence, here,  $\Delta V$ only depends on the moles of the gas.


What are the other end products in sulfur oxoacid formation? I got this reaction and I can't figure out the end products
$$\ce{SO2 + S + CaO + Na2SO4 -> Na2S2O3 +  ~?}$$
I am guessing some calcium sulfur salt should form, but which one?

I am not one hundred percent certain, but I will try to give you some guidelines. There is the well known reaction of sulfur dioxide with calcium oxide.
$$\ce{SO2 + CaO -> CaSO3}$$

Under aerobic conditions, this would further react with oxygen to form sulfate, since we have excess sulfur here I think it is fair to assume it will react to form thiosulfate.
$$\ce{1/8S8 + CaSO3 -> CaS2O3}$$

Adding sodium sulfate is probably just a way to control/ buffer the pH during the reaction, I fail to see the actual relevance. But judging from the side reaction
$$\ce{SO2 + 1/8S8 (s) + 3H2O <=> S2O3^{2−} + 2 H3+O}$$
it is actually necessary to have a buffer system based on sulfate.
$$\ce{Na2SO4 + H3+O <=> 2Na+ + HSO4- + H2O}$$

I am not aware of any reaction of sulfur with sulfate.

In total I would write your equation like this:
$$\begin{multline}\ce{SO2 (g) + S (s) + CaO  (s) + Na2SO4 (s) ->[\ce{H2O}]\\ 2Na+ (aq) + Ca^{2+} (aq) + S2O3^{2-} (aq) +  SO4^{2-} (aq)}\end{multline}$$


Why does NaCl dissolve in H2O despite its strong ionic bond? If $\ce{NaCl}$ is such a strongly bonded ionic compound, why it does it disassociate so readily in $\ce{H2O}$? One would think that because of the very strong ionic bond it would not disassociate it all. Further, why do other compounds like $\ce{AgCl}$ or $\ce{HF}$ not disassociate as readily even though they also have strong bonds?

Although others have marked this as duplicate, I prefer the wording of this question as it explicitly refers to NaCl has having a strong bond (which includes the search term I used to find this website). I have been writing a manuscript in which I have been attempting to answer this question (among others). The real answer becomes a bit complicated, but I will try to give a short answer.

It is my opinion that thermodynamic data is not always properly understood. The first thing we should understand is that the same thermodynamic change in energy should result whether the reaction were to take place in the gas phase (Born-Haber cycle) or in solution. This is a principle of conservation of energy, Hess's Law. In the Born-Haber Cycle, the sum of the energies for vaporization, ionization, and lattice energies are determined. If a hypothetical solution phase Born-Haber calculation were to be performed, the same amount of energy would exchange, except for the solvation energy, which is small. That is what the poster was asking.

If you were to think about the conditions of a solution phase Born-Haber cycle, then you would take the elements and convert them directly to solvent separated ions. What this should tell you is that you have just calculated the solution phase energy of a redox reaction of sodium and chlorine to give sodium and chloride ions. This bypasses the implication of sodium chloride having a strong affinity for its ions. Yes, it would take a lot of energy to separate sodium and chloride ions in the gas phase. That is why the lattice energy is so large, but if we did the reaction in solution, we will not have a gas phase lattice energy component in our calculations. Of course this seems plausible as the solution phases of HCl and NaOH are ionic. This is consistent with our expected properties of HCl to act as an acid (chloride having a low affinity for cations) and NaOH to act as a base (sodium having a low affinity for electrons). The sum is we should not expect NaCl to have a strong affinity (not none) for the ions of the opposite charge.

By the way, I think you will also find a good correlation of heterolytic solution phase bond strength with bond length (actually, ion gap). This is different from the homolytic bond energies you may find reported in many tables, including the poster's question, "Why does NaCl dissolve in H2O despite its strong ionic bond?"

PW

The electrostatic forces between the positive cations and anions are strong. Coulomb's Law expresses the potential energy of two charges separated by a distance.

$$E_p = \frac{Q_1Q_2}{4\pi \epsilon_0 r} $$

As you may already know, water molecules stick to each other through dipole-dipole forces due to the permanent dipole that water faces due to the high electronegativity of the oxygen atom.

As ionic solids are added to water, water molecules proceed to surround each ion on the surface of the solid, forming a sphere of hydration. In the process, ions are separated from each other.

The $\delta^-$ charge on the oxygen atoms of water are attracted to cations and inversely, repels the $\delta^+$ hydrogen atoms. Thus, for cations, the oxygens of water point inward, and for anions, the hydrogens face inward respectively. The most important thing is that the ion-dipole interactions and separation of ions with little change in energy.

We can relate the potential energy of the ions to the two partial charges of a polar molecule like water:

$$ E_p \propto - \frac{|z|\mu}{r^2}$$

Z is the charge number of the ion and $\mu$ is the dipole moment of the polar molecule. Potential energy is lowered by the interaction between the solvent molecules and the ion. The $r^2$ term indicates that the interaction between ions and dipoles depends more on distance than the charges between two ions.

Thus, for hydration to occur, ion-dipole interactions must occur at the surface of the ion, and thus, ion-dipole interactions are strong for small, highly charged ions such as $\ce{Mg^{2+}}$, $\ce{Li^{2+}}$ etc.

$\ce{AgCl}$ is very slightly soluble in water and will not dissociate into it's ions. $\ce{HF}$ is a weak acid thus it does not deprotonate easily.


Why does cinnamaldehyde pass the ferric chloride test? I was given some data to determine what the unknown compound was and figured out that it was cinnamaldehyde. After reviewing the classification test given to me by my professor, I was a bit confused as to why it would pass the ferric chloride test as well as pass a Lucas test but yet show no reaction.

Can someone explain to me why that is possible?

Also, in terms of solubility, why would cinnamaldehyde be soluble in sulfuric acid only and not both sulfuric acid and $\ce{NaHCO3}?$

I think the cinnamaldehyde gives $\ce{FeCl3}$ test due the similar reason why acetic acid can give $\ce{FeCl3}$ test as $\ce{Fe^3+}$ can act as oxidizing agent.

As in case of acetic acid $\ce{Fe(CH3COO)3}$ (red colour) is formed, a similar complex may be produced.


What are the factors that affect the corrosiveness of an acid? Acids are usually corrosive but what determines this? Is it the concentration of the acid or strength of the acid ($K_a$ value)?

Corrosiveness of an acid is a more complex concept, of course the concentration of the acid and the strength of the acid are two main factors that determinate the success of the corrosion reaction. For example this is the Pourbaix diagram of iron both concentration and $K_a$ determine the pH of the solution so if this is higher then about 9 the metal is passivate and you can't have any corrosion.

I add to these another important factor:  temperature.

In some case the corrosiveness is due to a more complex interaction that is directly linked to the material you are dealing. One example is acqua regia in this case the strong corrosiveness is due to the couple effect of two different acid that are able to de-passivate the metal allowing  an enhanced corrosive effect on noble metals. However acqua regia is not the best choice for an organic material in this case Piranha Solution has a greater corrosive effect.


Thermal Decomposition of Sulfur Hexafluoride What is the thermal decomposition temperature for $\ce{SF6}?$ I've heard it decomposes into $\ce{F2}$ and $\ce{SF4}$ only to decompose further into $\ce{S2}$ and $\ce{F2},$ but I haven't found much on the topic.
I became curious since I was considering using it to flush out oxygen for some reactions I was doing, due to its density and stability (plus we have a bunch lying around), but was concerned about it decomposing into more toxic compounds if the reaction or apparatus were to get too hot or if a spark were to occur.

I do not plan to actually do this. It was a silly idea I had, and I realized soon after posting this that $\ce{SF6}$ is a bad idea for these purposes since it is a greenhouse gas.

From an IEEE paper1:

$\ce{SF6}$ begins to decompose at $\pu{300 °C}$ and the main decomposition components contain $\ce{CO2 , SO2F2 , SOF2 , SOF4 , SO2}$ and $\ce{H2S}$; [...] $\ce{H2S}$ is the special component only appears when thermal fault proceeds to some degree (above $\pu{360°C}$).

For more information about the decomposition mechanism and product yiels, see here.
References

F. Zeng et al., "Decomposition characteristics of SF6 under thermal fault for temperatures below 400°C," in IEEE Transactions on Dielectrics and Electrical Insulation, vol. 21, no. 3, pp. 995-1004, June 2014, doi: 10.1109/TDEI.2014.6832242.


Distinguishing among SN2/SN1/E1/E2 for a secondary bromide a. 2-Bromo-3-phenylbutane reacts with $\ce{NaOMe}$ in methanol. What is the mechanism?
b. This compound then reacted with $\ce{CN-}$ in DMSO. What is the mechanism?

My Attempt
a. I am debating between $\mathrm{E1}$ and $\mathrm{S_N1}$ since it is a polar protic solvent and the carbocation can be stabilized by the tertiary carbon and phenyl group. I am leaning towards $\mathrm{E1}$ since $\ce{OMe-}$ is a strong base. Am I correct?
b. If the answer to the first question was $\mathrm{E1}$, then 2-phenylbut-2-ene would be the product. I don't see how that would react with $\ce{CN-}$ in DMSO. So, would no reaction occur?

Reaction a proceeds via an $\mathrm{E2}$ mechanism. Firstly, elimination reactions are much more favorable in this molecule because of the conjugation of the resulting structure. If the reactant had been 2-bromo-2-phenylbutane, this would have proceeded via an $\mathrm{E1}$ mechanism, as this would delocalize the positive charge around the phenyl ring. There is no such delocalization in the ejection of the $\ce{Br-}$ from 3-bromo-2-phenylbutane though—it would simply be a 2º carbocation. Instead, the delocalization of the negative charge around the phenyl ring resulting form the extraction of the $\ce{H+}$ by $\ce{OMe-}$ is more stabilizing.

$\hspace{1.5cm}$

There are no possible reactions between an alkene and $\ce{CN-}$ in DMSO. If this were a haloalkene, further elimination could occur, and if it were in $\ce{H2O}$, hydrocyanation could occur.

$\hspace{4cm}$


Finding Gibbs energy at different temperatures How can I find $∆G$ at a higher/lower temperature if I am given $∆G°$ and $∆H°$ $\pu{298 K}$, and a reaction equation (solubility)?

You might know that Gibbs free energy is given as:

$$\Delta G=\Delta H-T\Delta S$$

This implies that:

$$\Delta G^\circ=\Delta H^\circ - T \Delta S^\circ$$

Which can be rewritten as:

$$\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}$$

Often it is a good approximation to assume that $\Delta H$ and $\Delta S$ are temperature independent. Now that you have found the entropy you can calculate Gibbs free energy at different temperatures using the first equation.

If you want to find the equilibrium constants, you might have seen this equation before:

$$\Delta G=-RT\ln\left( K_\text{eq} \right)$$


Why isn't 4s1 electron of Chromium the last electron? I know that to aid the symmetry of singly occupied orbitals one electron jumps from 4s orbital to 3d orbital thus giving Chromium the configuration 3d5 4s1.

But my question is, when we're evaluating the quantum numbers of the last electron, shouldn't we definitively use the farthest electron from the nucleus? Irrespective of which electron is assorted last. Because it's not like a Chromium atom is created by sequentially inserting electrons one after the other voluntarily.

In a perfectly stable Chromium atom, which has the configuration Ar(18) 3d5 4s1, the last or ultimate electron in terms of energy level, orbital and radius, should be the 4s1 electron. Why is it that most references I found tell me to consider the 3d5 electron?

3d is the highest energy, 0.04 Hartree higher than 4s according to NIST's Atomic Reference Data for Electronic Structure Calculations, Chromium


When can a molecule be considered freely rotating at room temperature? This question sparked from a long discussion in chat about the nature of $\ce{H2O2}$ and whether that molecule can be considered to rotate around the $\ce{O-O}$ axis (and hence display axial chirality) or not.

Considering two rather clear cases:
Ethane is considered to rotate freely around the $\ce{C-C}$ bond. The activation energy for rotation (equivalent to the energy difference between the staggered and eclipsed conformations) is given as $12.5\,\mathrm{\frac{kJ}{mol}}$.

Ethene on the other hand is considered to not freely rotate around the $\ce{C=C}$ bond. The energy difference between the planar and perpendicular conformations (the rotation barrier) is given at $250\,\mathrm{\frac{kJ}{mol}}$. (I am unsure of this value which I found on Yahoo answers. If I understand correctly, this should correspond with the excitation from $\unicode[Times]{x3C0}$ to $\unicode[Times]{x3C0}^*$ which I found elsewhere as $743\,\mathrm{\frac{kJ}{mol}}$.)

Somewhere between these two extreme cases must lie some kind of barrier around which rotation is severely hindered to entirely inhibited.

Attempting to calculate this myself, I remembered the Boltzmann distribution: If two states differ in energy by $\Delta E$, their relative populations $F_1$ and $F_2$ can be calculated as follows:

$$\frac{F_2}{F_1} = \mathrm{e}^{-\frac{\Delta E}{k_{\mathrm{B}} T}}$$

($F_2$ being the higher energy state and $k_\mathrm{B}$ being the Boltzmann constant.)

If I calculate relative populations from the energy differences $12.5, 19, 25$ and $250\,\mathrm{\frac{kJ}{mol}}$ for ethane, butane, $\ce{H2O2}$ and ethene, respectively (which correspond to $0.13, 0.20, 0.26$ and $2.6\,\mathrm{\frac{eV}{particle}}$) I get the following results at $300\,\mathrm{K}$:

Ethane: $6.55 \times 10^{-3}$
Butane: $4.79 \times 10^{-4}$
$\ce{H2O2}$: $4.29 \times 10^{-5}$
Ethene: $2.09 \times 10^{-44}$

Or logarithmic (ln) values of:

Ethane: $5.03$
Butane: $7.64$
$\ce{H2O2}$: $10.1$
Ethene: $101$

Again, somewhere between butane and ethene must be a threshold-ish value, above which free rotation at $300\,\mathrm{K}$ cannot be considered, but what range are we looking at; similar to what energy difference does this correspond to?

Asked the other way round, if I have an energy barrier of $25\,\mathrm{\frac{kJ}{mol}}$, what would I need to put on into the equation to calculate the temperature above which free rotation can be assumed?

I'm not a transition-state physical chemist, but I think a good approach to this problem is transition-state theory, specifically the Eyring equation:

$$k = \kappa \frac{k_b T}{h} e^{\frac{-\Delta G^{\ddagger}}{RT}}$$

This equation tries to predict the rate constant $k$ from an assumed pseudo-equilibrium between the transition-state and the starting material, which are assumed to differ in energy by $\Delta G^{\ddagger}$.  $k_b$ is Boltzmann's constant, $h$ is Planck's constant, $T$ is temperature, and $\kappa$ is the "fudge factor" or transmission coefficient, the fraction of transition states that react productively instead of converting back to starting material.

In the conformational reactions you are talking about, I think a good starting place would be to assume the eclipsed conformations are the transition state, although I don't know if this has been shown to be 100% true experimentally or computationally.   As a first approximation, let's assume there isn't a huge entropy term in $\Delta G^{\ddagger}$, so that it is well-approximated by the energy level differences you found.  If we also assume that $\kappa$ is 1.0, then we can just multiply the $e^{\frac{-\Delta E}{RT}}$ you've already calculated by $\frac{k_b T}{h} = 6.2\times10^{12}\mathrm{~s^{-1}}$ at 298 K to get (very) approximate rate constants.

Ethane: $6.55 \times 10^{-3} \rightarrow 4.1 \times 10^{10}\mathrm{~s^{-1}}$
Butane: $4.79 \times 10^{-4} \rightarrow 3.1 \times 10^{9}\mathrm{~s^{-1}}$
$\ce{H2O2}$: $4.29 \times 10^{-5} \rightarrow 2.0 \times 10^{8}\mathrm{~s^{-1}}$
Ethene: $2.09 \times 10^{-44} \rightarrow 1.3 \times 10^{-31}\mathrm{~s^{-1}}$

That shows that rotational equilibrium of everything except ethene should be very rapid, but that the half-life for ethene equilibration much greater than the age of the universe ($4 \times 10^{17} \mathrm{~s}$).

Suppose we want to find an energy difference that should lead to a rate constant that is neither very fast nor very slow, say we'd like $k\approx\mathrm{1~hr^{-1}}$.  Inverting our very very crude approximation gives:

$$\Delta E = - RT \ln{\left(\frac{k h}{\kappa k_b T}\right)}$$

Plugging in $k=\mathrm{1~hr^{-1}=2.7\times10^{-4} s^{-1}}$ gives (if I did my math right) 93 kJ per mol.  Some well-known atropoisomers do have estimated energy differences in that range.


How to calculate membrane potentials? I have recently been reading about membrane potentials in the cell and how they are derived. The textbook (Guyton and Hall Textbook of Medical Physiology, 13th Ed) gave this equation:
$$EMF=\pm\frac{61}{z}\log\frac{C_i}{C_o}$$
Where $i$ refers to the interior of the cell, $o$ refers to the exterior of the cell, $z$ is the relative magnitude of the charge of the cation/anion, $EMF$ is the potential in the cell, taking the external potential as 0. Additionally, the sign is + for anions and - for cations.
I understand that this comes from the Nernst Equation:
$$E=E^o-\frac{RT}{\nu_eF}\ln Q$$
Where $\nu_e$ is the number of electrons transferred per mole of reaction.
I understand that $E^o=0$ because at equilibrium, the concentrations of ions on both sides of the membrane are equal and I also understand where the magnitude comes from: $$61=\frac{1000RT}{\log eF}$$
Where $\frac{1000}{\log e}$ is used to change the base of the logarithm and change the units to mV and T is taken to be the body temperature of 37 C, converted to Kelvins.
However, what confuzzles me is

The sign convention: Why are cations - and anions +.
How do we derive the $-\frac{RT}{\nu_eF}\ln Q$ for transfer of ions, since unlike the Nernst Equation, which is derived for electrons, ions are being transferred instead?
How do we measure the membrane potential? According to the text, they put one electrode in the extracellular fluid and pierce another electrode (a typical Ag/AgCl electrode) into the cell, in contact with the intracellular fluid, as shown below. Since there is no transfer of electrons, how does this apparatus actually work?

Your textbook description is correct. Recall the principle of potentiometry, the potential difference is measured when there is no current flowing in the system. This is true under ideal conditions. Do you recall the pH meter? What is the electrode made of? It is made of glass, do you feel that current can flow across glass when you are measuring pH of your solution?
BTW, the working of the glass electrode puzzled electrochemists for ages and the still do not know how it works exactly. So if you are confused, welcome to the club.
You should rather ask how do high impedance voltmeters work? Such voltmeters work when there is no current flow and you are solely interested in potential difference between the electrodes.

Since there is no transfer of electrons, how does this apparatus
actually work?

This is perhaps a misconception. Potential difference only develops when electrons are transferred (only once) but the is no continuous flow of electrons (which constitutes current). If you are combing your dry hair in winter, a huge potential difference develops between your hair and the comb. You comb is still "charged" when you move it away from your hair. The electrons were transferred but only at one instant. There is no current flow between your comb and the hairs.


The appliance of LAB in amide reduction Lithium aminoborohydride is a versatile reducing agent. In the scheme below, I don't understand how does unhindered LAB afford an alcohol in the reduction of amides while a hindered LAB affords an amine. Why are two different products obtained and by what mechanisms do these reductions take place?

Very interesting question and not very obvious. As can be seen from the picture, LAB's reduce only tertiary amides. It is stated in the publication(1) about LAB reductions:

  LAB reagents do not
  reduce primary and secondary amides even in refluxing THF.

In the same publication it is provided reason for different outcomes of reductions. They say that:

  We found that the outcome of these reductions depends on
  the sterics of the amide as well as of the LAB reagent

Which means that if LAB and/or amide is sterically hindered it can't interact with nitrogen like in compound 5

In lower part of the picture is described reaction where more hindered LAB leads to amine product 9

(1)Pasumansky, L., Goralski, C. T., & Singaram, B. (2006). Lithium aminoborohydrides: powerful, selective, air-stable reducing agents. Organic process research & development, 10(5), 959-970.


Intramolecular Reaction with carbocation rearrangement I came across a question recently which is as follows,

I came up with a mechanism but I am stuck at a step as I am unable to reason out the driving force.

Step(3) has a hydride shift and its next step involves methyl shift. I want to know the driving force for step (3) or any other approach to this problem. Do let me know if I am going wrong with my approach.

Your approach is feasible. Thermal cyclizations of appropriate dienes, enones, enynes, and other related unsaturated systems are known in literature. For example, there are extensive examples of intramolecular ene-reactions in Organic Synthesis and some of them are reviewed (Ref.1 and Ref.2). To justify your first step, I'd include the abstract of reference 1, which states that:

  Thermal cyclizations of appropriate dienes, enynes and related unsaturated systems, some of them carried out on an industrial scale, demonstrate increasingly the preparative power of the intramolecular ene reaction. A variety of substituted, fused and bridged ring systems, including natural products, are thus easily accessible in a regio‐ and stereo‐selective manner. Numerous examples are discussed systematically illustrating the possibilities, limitations, and common features of this cyclization reaction and its reverse ring‐opening process.

Nonetheless, see following example given in reference 1 for thermal cyclization:

Although those are different than the problem in hand, the intramolecular cyclization has possible clear path. For example, various types of intramolecular cyclizations has been discussed in Ref.1. Now, I modified your mechanism using arrows to make it clear:

The electrophile here is the carbonyl carbon of 1, which would get more electron deficient by carbonyl oxygen’s complexation to a Lewis acid such as $\ce{AlCl3}$, or  $\ce{AlEtCl2}$, or even $\ce{AlEt2Cl}$ (Ref.2). This is evident in the intermediate I. The neucleophile, the alkene group of the molecule is in close proximity to make a bond with elecrophilic carbonyl carbon (intermediate II). The driving force here is making a less strained 5-membered ring and relatively stable $3^\circ$-cabocation. Keep in mind that all of these intermediates are reversible. The reversible hydride transfer is possible because energy different between resulting $3^\circ$-cabocations are minimal. Yet, this transfer facilitate pinacol-pinacolone type rearrangement of methide shift to give final stable product 2.

References:

Wolfgang Oppolzer, Victor Snieckus, “Intramolecular Ene Reactions in Organic Synthesis,” Angew. Chem., Int. Ed. Engl. 1978, 17(7), 476-486 (https://doi.org/10.1002/anie.197804761).
Barry B. Snider, “Lewis-acid catalyzed ene reactions,” Acc. Chem. Res. 1980, 13(11), 426-432 (https://doi.org/10.1021/ar50155a007).


Bromination of Phenylacetone Looking at the following reaction, I understand how the product forms, however, can somebody help me understand why two products are formed? Would the identical mechanism produce the two products?

Edit: This is how I understand the mechanism.

The mechanism that I have submitted in the edit of my post is correct. The reason both products are formed, is due to the the enol intermediate being planar with respect to the oxygen, allowing it to be attacked from either plane. This allows for the formation of a the so-called racemic mixture.


How to calculate the specific heat of ethanol? I am currently stuck with this question. Can somebody please help me with this?

  100 g of water was heated from $\pu{15 °C}$ to $\pu{70 °C}$ by burning $\pu{20 g}$ of ethanol. How much energy per mole does this ethanol have?

So far I have calculated that there are $\pu{0.43 mol}$ in $\pu{20 g}$ of ethanol using the formula $n = m/M$.
\begin{align}
m &= \pu{20 g}\\
M &= 12 + 3 + 12 + 2 + 16 + 1 = \pu{46 g/mol}\\
n &= 20/46 = \pu{0.43 mol}\\
\end{align}

From here, I'm not sure if I should divide the energy used to heat up the water by the amount of substance. By using the specific heat formula,

$$Q = c \cdot m \cdot \Delta{}T
= \pu{4.2 J//g \cdot K} \cdot \pu{100 g} \cdot \pu{55 K}
= \pu{23100 J}$$

$$\text{Energy per mole} = \frac{\pu{23100 J}}{\pu{0.43 mol}} = \pu{53720.93 J//mol}.$$

Is this the right answer?

Each gram of water has an specific heat of 4 J/(K.g)*. So heating $\pu{100 g}$ of water from $\pu{15 °C}$ to $\pu{70 °C}$ would take:

Heat = (70°C - 15°C) * 100g * 4 J/(K.g) = 22,000 joules = 22 kJ

$\pu{20 g}$ of ethanol would be:

Ethanol mol = (20g / 46,07 g/mol) = 0.43 mol

So this means that:

0.43 mol = 22 kJ
     mol = 22 / 0.43
     mol = 50677 J = 50.67 kJ

So ethanol has 50.67 kJ/mol vs 53.72 kJ/mol said by you. I would say that you have made fine your homework, there is a very little difference (I have used some decimal to calculate, maybe you not. (Below you can find a more accurate answer)).

* Water specific heat:  water changes it's specific heat when it changes it's heat. For example: (Wikipedia)

Water at  100 °C (steam)  2.080  J/(K.g)
Water at   25 °C          4.1813 J/(K.g)
Water at  100 °C          4.1813 J/(K.g)
Water at  −10 °C (ice)    2.05   J/(K.g)

Normaly it's used:        4.1813 J/(K.g)

A better table can be found in a french page.

Using this table we can do a better aproximation:

Water heat = (70°C - 15°C)        * 100g
Water heat = (293.03 J - 63.04 J) * 100g
Water heat =    229.99 J          * 100g
Water heat = 22,999 J = 22.999 kJ ≈ 22.9 kJ

Ethanol mol = (20g / 46.07 g/mol) = 0.43 mol
Exactly: 0,4341219882787063164749294551769 mol

0.43 mol = 22999 J
     mol = 22999 / 0.43
     mol = 52978.19 J = 52.97819 kJ ≈ 52.97 kJ
Exactly: 52978,196500000000000000000000001 J

Answer: Ethanol has 52.97 kJ/mol. A value closer to your own value.

PD: I am only a normal person who like chemistry, maybe I am wrong. For example in the theory I have no idea what means $Q = c \cdot m \cdot \Delta{}T = \pu{4.2 \frac{J}{g \cdot K}} \cdot \pu{100 g} \cdot \pu{55 K} = \pu{23100 J}$ Well, I have some ideas but I haven't learned it in the school.

Edit:

I notice that you have your own mol of ethanol (46 g/mol instead of my 46.07 g/mol) and your own specific heat of water (4.2 J/(K.g)). You have to do:

Water heat = (70°C - 15°C) * 100g * 4.2 J/(K.g)
Water heat =      55°C     * 100g * 4.2 J/(K.g)
Water heat =
Water heat = 23100 J = 23.1 kJ

Ethanol mol = (20g / 46 g/mol) = 0.43 mol
Exactly: 0.43478260869565217391304347826087 mol

0.43 mol = 23100 J
     mol = 23100 / 0.43
Using mol decimals (0.43478260869565217391304347826087)
     mol = 53130 k = 53.13 kJ
Using only 2 decimals (0.43)
     mol = 53720,93 J ≈ 53.72 kJ
     Exactly: 53720,930232558139534883720930233 J

Answer: Ethanol has 53.13 kJ/mol (With decimals) or 53.72 kJ (with 2 decimal). So yes! You have done right!


Spin-forbidden transition iof bromine In one of the lectures of my spectroscopy course, the professor claimed that there is a strong absorption band in the UV/Vis spectrum of bromine, associated with a singlet-state to triplet-state transition. He asked us to think about why this spin-forbidden transition occurs.

So far I haven't really come up with a satisfactory answer. Nevertheless, here are two of my ideas:

Broadly speaking, spin-forbidden does not necessarily mean that a transition is not observed, but rather that the intensity of the spectral line will be considerably diminished. Hence, a spin-forbidden transition could be observed in a spectrum.
An electronic transition is usually associated with a vibrational transition. Through the vibrational excitation the geometry of the molecule is changed such that the spin selection rules are "relaxed".

Am I on the right track with these ideas?

The selection rules are related to the transition dipole moment that determines the intensity of a given transition. If a given selection rule holds perfectly, all that it says is that a forbidden transition has zero intensity and an allowed transition has non-zero intensity.

If the selection rule does not hold exactly, just like the in the case of the spin selection rule for heavy atoms, a forbidden transition can be intense. Some examples: the $\text{B}^3\Pi_u-\text{X}^1\Sigma^+_g$ electronic transition in $\ce{Br2}$ and $\ce{I2}$, that are strong when compared to the same transition in $\ce{F2}$ and $\ce{Cl2}$; and the strong $^3P_1-^1S_0$ transition in $\ce{Hg}$, while the analogous one in $\ce{He}$ is very weak. As $L$ and $S$ are not good quantum numbers when there is strong spin-orbit coupling, the selection rules involving them can't be used.

The second answer you came up with makes sense, but not for spin selection rule. Consider the parity selection rule for molecules with an inversion center. As the molecule vibrates (in an asymmetric mode) it can be distorted into a geometry that does not have an inversion center, so the selection rule does not hold. In this case, this vibronic transition will be allowed.


How to decide anode and cathode form electrochemical series? To calculate cell potential one must know that from given half reactions which metal is to be anode and which is cathode.

So from standard reduction potential in electro-chemical series how to find this ?

If one element(say A) is above another element(say B) in the series then which is cathode A or B ?

If one element A is above B in the standard electrode potential series, then the one with greater oxidising power or reducing agent(ie,B) is anode. While the one with greater reducing power or oxidising agent (ie, A) is cathode.

Standard Reduction Potential Table


Autoclaving certain plastics? My professor is letting me use the autoclave in the future, but I am a novice. I asked him what materials were safe to autoclave for my notes, materials meaning glass or plastic-ware. He stated, "It depends on alot of factors, but some hard and opaque plastics are generally ok, also Teflon products."
I couldnt really ask him any more questions since it is still winter break and I guess he isnt responding to emails, so I decided to google the issue.
I found this source of an autoclave guide at: https://uwaterloo.ca/safety-office/sites/ca.safety-office/files/uploads/files/guideline-for-safe-use-of-autoclaves.pdf

Good Choices:
borosilicate glass (Pyrex) has very low thermo
expansion property and therefore resistant to breaking due to heating
polypropylene (PP) and polycarbonate (PC) are heat resistant plastics
stainless steel is a good heat conductor and thus facilitates
sterilization
Poor Choices:
polystyrene (PS), polylethylene (PE) and high density polyethylene (HDPE) do not resist heat well.
if there is a risk of material melting ensure they are placed in a
secondary container which is resistant to heat.

Would this be an accurate  summary of what plastics / glassware to use in an autoclave? I would ask about the validity of the website itself, but I assumed it was reliable because it was based off of a college.

The list looks accurate. For example, the melting point of polypropylene ranges from ~130 C to ~160 C, so it's likely safe to autoclave.  Commercially available polyethylene melts ~100 -110 C, so is unsuitable for autoclaving.

Another common plastic, not listed above, polyethylene terephthalate (PETE), used for packaging and soft-drink bottles, softens below 100 C and is also unsuited for an autoclave.

Electrical and optical items used in endoscopic procedures are often heat-sensitive and cannot be autoclaved, posing difficulties in removing contamination.

As suggested, if you're unsure but must autoclave an item, place it in an open or perforated container so as to reduce risk of damage to the sterilizer.


The mechanism of the reaction between 2-chloroethanol and hydroxyl ions in aqueous solution The mechanism of the reaction between 2-chloroethanol ($\ce{CH2ClCH2OH}$) and hydroxyl ions in aqueous solution forming ethylene oxide ($\ce{CH2CH2)O}$ probably contemplates two steps:

$\ce{CH2ClCH2OH + OH- <=> CH2ClCH2O- + H2O}$
$\ce{CH2ClCH2O- <=> (CH2CH2)O + H2O + Cl-}$

Is the mechanism valid for the following experimental reaction rate?

$\frac{-d[\ce{CH2ClCH2OH}]}{dt} = K_{exp}[\ce{CH2ClCH2OH}][\ce{OH-}]$

Yes it is. Because alcohol is a weak acid and it is reasonable to consider first step is irreversible because of the second step almost occuring simultaneously. And also because that the second step is much faster than the first.


Why Fe(CO)5 exists, but Fe(NH3)5 doesn't? Like in title: Why $\ce{Fe(CO)5}$ exists, but $\ce{Fe(NH3)5}$ doesn't?

Edit:

As Jan Dvorak pointed out, it may exist. If so: why is it so unstable?

While we cannot be sure that $\ce{Fe(NH3)5}$ does not exist, the bonding between ligand and metal offers a good explanation for why it is expected to be less stable than $\ce{Fe(CO)5}$.

In metal carbonyls like $\ce{Fe(CO)5}$, the bonding between CO and the metal consists of three single bonds, 1 $\sigma$ and 2 $\pi$ bonds. The $\sigma$ bond is formed by overlap of the carbon $sp$ hybrid orbital (which contains the lone pair) with an empty $d$ orbital of the metal. Electron density is donated from the orbital of CO into the metal orbital, so CO acts as a $\sigma$ donor. The two other $\pi$ bonds are formed by overlap of a filled metal $d$ orbital with a pair of $\pi^*$ molecular orbitals of CO. Electron density is donated from the metal orbital into the molecular orbitals of CO, so CO acts as a $\pi$ acceptor.

For the $\pi$ bonding to occur, the metal is required to have sufficient $d$ electrons, and thus preferably a lower oxidation state. This explains the relatively low oxidation state of iron (0) in $\ce{Fe(CO)5}$. The combination of one $\sigma$ and 2 $\pi$ bonds gives the M-CO bond a partial triple-bond character, while the C-O triple bond is weakened by the transfer of electron density into the $\pi^*$ orbitals (reference).

Because of its ability to serve as a $\sigma$ donor and $\pi$ acceptor simultaneously, CO is a stronger ligand than $\ce{NH3}$ which can only act as a $\sigma$ donor with its lone pair in the nitrogen $sp^3$ hybrid orbital. $\ce{NH3}$ lacks a $\pi$ system and therefore cannot act as a $\pi$ acceptor. Because $\pi$ backbonding is crucial for stabilization of a metal center in low oxidation state, ammonia is unlikely to form stable ammin complexes which are analogous to the corresponding metal carbonyls.


Writing pressure equilibrium constant expression for a reaction with a single solid product How would you write the $K_\mathrm p$ expression for the reaction: $$\ce{4 Ag(s) + O2(g) \rightarrow 2 Ag2O(s)}$$

I know normally it would be products over reactants, but there is a solid on the product side and they aren't included, so I don't know how you would write it?

The expression $\mathrm{K_p}$ is an expression of the equilibrium of partial pressures. So, it just includes the the partial pressure of the oxygen because that's the only gas.

$$\mathrm{K_P} = \frac{1}{\ce{P_{O2}}}$$

More information about the equilibrium constant $\mathrm{K_p}$ can be found here.


Use of water in IR spectroscopy I am familiar with the process of IR spectroscopy and the preparation of IR samples.

However, I am curious as to why you cannot use hygroscopic or protic solvents for IR spectroscopy. Is it just because it dissolves the cell (KBr/NaCl), or does it give an OH peak, or are there other reasons?

And are there any protocols that allow you to use water as a solvent for obtaining an IR spectrum?

I applaud this question. I have not really tried to find any solution, but I definitely recognize the trouble with water... nonsense/worthless reading. As far as why... easy; water absorbs like, all of the IR spectrum. Below is the IR spectrum of liquid water (from NIST's Chemistry Webbook).

I actually have an ATR IR Spec, which is (supposedly) meant for reading aqueous solutions...

...but it doesn't really work... it can't be used for figuring out dilutions (well maybe, give or take 20%). In other words, 50% water + sample looks a lot like 95% water + sample... which looks a lot like any sample + water. Water basically drowns the reading- so it all looks alike.

Don't get me wrong... it's super nice that I don't need to oven dry my sample or deal with KBr. Basically, I just put the sample on the top and I get a decent reading, but if the sample has a lot of water in it (like greater than 10%), it loses a lot of definition.


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

Conductance and conductivity are distinct, although related, magnitudes.

Conductance, $G$, is a measure of the total electric current, $I$, that can pass through a conducting element (such as a wire, a space through which a charged beam flows, or an electrolytic solution) when a given voltage, $V$, is applied through it:

$G = \cfrac{I}{V}$

Correspondingly, the IS unit for conductance is the siemens (S), equivalent to an ampere per volt. Since conductance is the inverse of resistance, a siemens is also the inverse of an ohm.

However, in general, conductance does not only depend on the nature of the material; it also depends on the geometry of the conducting element. For instance, the conductance of copper wires of different length and/or different section will be different. Conductivity, $\sigma$, is defined in order to compare the contribution to conductance of materials, isolating it from shape effects.

In general, for a linear electric flow across a conducting element, conductivity increases with cross-section (as more flow is able to pass through simultaneously) and decreases with length (as more resistance is found). Conductivity is thus defined as

$\sigma =  G \cfrac{l}{A}$

which, ideally, should be identical for all conductors made of the same material (in practice, there are a number of effects that can make the relationship between conductance and geometry non-linear, but for macroscopic ohmic conductors it holds fairly well). This allows direct comparison of the electric properties of materials, effect of doping, temperature, etc.

The IS units of conductivity are therefore siemens per metre (S/m).

Note that unlike conductance, which is an extensive quantity, conductivity is an intensive quantity which doesn't depend on the size of the conducting element. According to 19th century naming conventions, conductivity can also be seen as a specific conductance (in the same way specific gravity is an intensive measure for mass, or specific heat is an intensive measure for heat capacity) and indeed that name was widely used in the early days of both electrochemistry and electromagnetism. While in electromagnetism that standard has changed and currently conductivity is the almost universal name for $\sigma$, in electrochemistry that name lingered and conductivity measurements in solutions were reported as "specific conductance" for longer.

Compound the facts that conductivity measures in electrochemistry are usually made with standard-sized cells in which geometry isn't a factor, and that specific values of conductivity (such as molar conductivity or equivalent conductivity) need an additional qualifier, and these magnitudes were often called "molar conductance" or "equivalent conductance" short for "molar specific conductance" or "equivalent specific conductance". However, you'll find that these magnitudes are always reported in units of conductivity (typically $\mathrm{\mu S/cm}$ or $\mathrm{mS/cm}$).

The use of specific conductance to mean conductivity is now considered obsolete; see for instance the IUPAC Gold Book's page for conductivity.


What are polydisulfide polymers? What are polydisulfide polymers? How are these polymers applicable in energy sectors such as rechargeable batteries?

Disulfide batteries use a redox shuttle between disulfide bonds ($\ce{R-SS-R}$) and two thiolates ($\ce{R-S-}$), e.g.:

Often the counter-ions are $\ce{Li+}$ but this is hardly a requirement. Since two electrons can be transferred in the charging / discharging cycle, the theoretical energy density is very high.

One downside (e.g., in my past research) is that you're breaking the polymer bonds during discharge, creating oligomers and monomers that might be more soluble in the electrolyte than the original polymer. This might cause problems with cyclability, since your electroactive material might leave the electrode.


In this simple redox reaction, why does the Hydrogen oxidize the calcium metal? I am understanding that the Hydrogen ions oxidize the calcium metal, neutralizing itself and creating a calcium cation. My question is why does the Hydrogen oxidize the calcium? Doesn't it prefer to have an empty outer shell? Does it want more to be neutral or to have an empty outer shell?

This is a simple example of the reactivity series. More reactive metals will displace less reactive metal ions, and they switch places, with the more reactive metal becoming the ion and and the less reactive ion becoming a metal. This can be very cool in many situations: for example, copper metal can displace a silver salt to plate silver metal and produce a blue copper salt. See this video! In any case, hydrogen is consider a reference "metal" in the reactivity series, and it is less reactive than calcium, so the displacement reaction you cite occurs.


Why is the ionic product of water not divided by the concentration of water? I know the ionic product of water is $[\ce{H+}][\ce{OH-}]$, by following the equation:

$$\ce{H2O <=> H+ + OH-}$$

But actually it's just another way of writing:

$$\ce{2H2O <=> H3O+ + OH-}$$

Since, $K_\mathrm{w}=K_\mathrm{c}[\ce{H2O}]$ and

$$K_\mathrm{c}[\ce{H2O}]= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$$

Why can't the following equation be used? $$K_\mathrm{w}= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$$

The short answer is: substances in their standard form are not included in equilibria expressions, and most commonly this includes pure liquids and solids.

The long answer is: the equilibrium expression relates all species to a standard activity (this is why they are all unitless). This question and answer detail that well.


rate of reaction and RDS An article on Khan-Academy, states the following:

The overall reaction rate is determined by the rates of the steps up to (and including) the rate-determining step

Here, it says that rate is determined by steps before and including that of RDS. But from what I know the rate is determined by one single elementary reaction, which is the slowest, that is, the RDS. So why does it include steps previous to RDS?
(I would appreciate an example of this)

The interpretation of the statement depends on exactly what is meant by "overall rate."

The rate-determining step corresponds to the largest free energy difference from a transition state to any preceding stable state (reactant or intermediate).
In the example above, the biggest forward barrier is from the reactant to the second transition state, so the second step is rate-determining. However, the barrier, as identified above, is comprised of both the first and second step, so they both figure into the overall rate of the reaction, as determined by the slow step.
EDIT:
Based on comments, I think a clarification is in order. Note that the above conclusion is not true in the general case. You could make the indicated claim for the depicted reaction specifically, but not, for example, in the case below:

The second step here is still rate-determining because the transition state is involved in the largest barrier. However, the biggest barrier is from the intermediate to the second transition state. In this example, I would not say that the overall rate depends on the first step.