Temperature and moles of gas given volume, pressure, enthalpy and constant pressure molar heat capacity I've been assigned this question as homework and have been attempting it for hours but cannot seem to get anywhere. Any help would be greatly appreciated!

An airbag of volume $40\ \mathrm L$ is inflated adiabatically through the following reaction:
$$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$
One mole of $\ce{NaN3}$ decomposed produces an enthalpy of $-23.1\ \mathrm{kJ}$. $60\ \%$ of this energy is deposited in the gaseous product; assume the decomposition is instantaneous. $C_{\mathrm m,p}(\ce{N2})=29.12\ \ \mathrm{J\ mol^{-1}\ K^{-1}}$. When fully inflated, the pressure of the $\ce{N2}$ gas inside the airbag is $2.5\ \mathrm{atm}$.
Estimate the mass of $\ce{NaN3}$ needed to operate the airbag and state any assumptions or approximations being made.

So far, I'm only sure that I've done the following correctly:
$H=U+PV=q+w+PV$ where $q=0$ since the process is adiabatic, so $H=w+PV$
For the expansion of $\ce{N2}$ gas:
$H=0.6n(23.1\ \mathrm{kJ})=n(13.86\ \mathrm{kJ})$ where $n$ is moles of $\ce{NaN3}$
$n(13.86\ \mathrm{kJ})=U+PV$ so $U=n(13.86\ \mathrm{kJ})-(2.5\ \mathrm{atm})(40\ \mathrm L)=n(13.86\ \mathrm{kJ}-RT)$
$q=0$, $P=2.5\ \mathrm{atm}$, $V=40\ \mathrm L$ and $C_{\mathrm m,p}(\ce{N2})= 29.12\ \mathrm{J\ mol^{-1}\ K^{-1}}$
I'm not sure what else to do from here. I'd really appreciate any advice or if anyone could point me in the right direction. Thank you!

Hint: A general outline of what I think is the idea behind this problem/ This is a quick back of the envelope estimate:
$$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$
The reaction given above takes place, and releases:

$$\Delta H = -23.1\ \mathrm{kJ} $$

60% of this heat is transferred to gaseous nitrogen, and since the process is adiabatic no additional heat enters/leaves the system. All of this energy goes into expansion work

$$Q = \frac{60}{100} \times 23.1 \ \mathrm{kJ} = 13.86\ \mathrm{kJ} $$

Temperature of $\ce{N2(g)}$ (assuming initially it is at $298\ \mathrm K$) can be found using (first, assuming one mole of azide decomposes releasing 3/2 moles of nitrogen)

$$Q = nC_{\mathrm m,p} \Delta T$$

$$T_{\text{final}} = \frac{Q}{nC_{\mathrm m,p}} + T_{\text{initial}} = 615.3\ \mathrm{K}$$

Volume occupied by 3/2 moles of nitrogen at the above temperature and given pressure is

$$V = \frac{nRT}{P} = 30.3\ \mathrm{L}$$

You need another $9.7 \ \mathrm{L}$ of gas to completely fill the airbag, I am sure you can complete the calculation. I'll come back and post a full, detailed answer later.

Complete Answer

So, we need a volume of $ 40 \ \mathrm{L}$, at an unspecified temperature.

$$\frac{PV}{n_{\ce{N2}}R} = T_\text{final}$$

Additionally, $$T_{\text{final}} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$

I am assuming since the process is adiabatic no additionally heat enters/leaves the system. The way I see it, the energy generated during this decomposition would heat the nitrogen produced to some (as of yet unknown temperature), and would be used for doing expansion work. I assume everything is initially at $298 \ \mathrm{K}$

$$\frac{PV}{n_{\ce{N2}}R} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$

From the reaction stoichiometry, $n_{\ce{N2}} = \frac{2}{3}n_{\ce{NaN3}}$

Thus, the expression we are interested in is

$$\frac{PV}{(2/3)n_{\ce{NaN3}}R} = \frac{Q}{(2/3)n_{\ce{NaN3}}C_{\mathrm m,p}} + T_{\text{initial}}$$

Now, it is just a matter of plugging in appropriate values, and solving for $n$

I used,

$ P = 2.5 \ \mathrm{atm}$
$ V = 40 \  \mathrm{L}$
$ R = 0.0821 \ \mathrm{L \ atm \ mol^{-1} \ K^{-1}}$
$ Q = 0.6 \times 23.1 \times 1000 \times n_{\ce{NaN3}} \ \mathrm{kJ}$
$C_{\mathrm m,p} = 29.12 \ \mathrm{J\ mol^{−1}\ K^{−1}} $

Answer given below:

   Solving all of this I get no. of mol of azide $\approx 1.33401 \ \mathrm{mol}$ and, no. of mol of nitrogen produced $ \approx 2.001015 \ \mathrm{mol}$. Additionally, $T_{\text{final}} = 615.307 \ \mathrm{K}$ Plugging all of this back into the ideal gas equation and solving for volume at the given pressure gives me $40.4\ \mathrm{L}$


Definition of Lattice Energy From previous studies, I recalled in my lecture theatre today that lattice energy was the energy released when an ionic compounds is formed from individual atoms in the gaseous state. I thought this idea would follow through throughout my studies at university.

However, my lecturer this afternoon defined lattice energy as the following:

  • For an ionic solid the lattice energy is defined as the energy required to
  completely separate a mole of solid ionic compound into its gaseous ions

I thought this was my definition but in the reverse order.

Which one of us is correct, the lecturer or myself?

There is not necessarily a set convention: https://en.wikipedia.org/wiki/Lattice_energy#Lattice_energy_vs_lattice_enthalpy

However, the sign must match the convention used. For example, the values in the table you provided are all positive which implies you are adding energy to break the lattice.

The definition from you prior experience would require negative values.


Is there a way to counteract the problems with weighing hot objects? I am currently performing an experiment in which it would be greatly beneficial to be able to accurately measure the mass of a hot object without waiting for it to cool. The current plan is simply to place the hot object in a crucible with a lid and attempt to measure the mass quickly, before the crucible has time to heat up. The thought is that the convection currents around the object won't matter as much if they are contained within the crucible.

I will be testing this method in the next few days for accuracy and reproducibility, but would appreciate any other ideas to mitigate the problems with weighing hot objects, or insights on how this idea could be better.

Depending on geometry, mass and temperature (hot is pretty vague), you might be able to use an inertial balance.

If the device used to hold the sample is rigid, you could calibrate its resonant frequency with and without the sample and with known weights.

See this high school lab and a video of the procedure.


How to synthesize 3‐cyclopentylpropanal from (chloromethyl)cyclopentane? I am not sure on how to start, but I think of these two possible ways:

Reacting the starting material with Grignard reagent and then oxidise the product with PCC to produce the aldehyde;
Hydrolysis the starting material to produce alcohol and then react the alcohol with PCC to form aldehyde.

But these two ways didn't give the desired product.

3 Steps:
Step 1 - react the starting material with the preformed anion of ethyl cyanoacetate (NaOEt/EtOH should do this comfortably)
Step 2 - mild base hydrolysis of the product (NaOH/EtOH) followed by acidification to remove the ethyl ester group.
Step 3 - Dibal-H reduction of the nitrile to aldehyde 1.


Metal and reactions with different temperatures of water I knew that in warm water magnesium forms $\ce{Mg(OH)2}$ and $\ce{H}$ gas, and in steam it forms $\ce{MgO}$ and $\ce{H}$ gas (this is also the case for metals lower down the electrochemical table). However I came to realize that for those higher than magnesium all form metal hydroxide when react with water.
What is the mechanism for such a reaction?

You have two reactions, the metal reacting with water and the hydroxide decomposing:

$\ce{Mg + 2 H_2O \rightarrow Mg(OH)_2 + 2 H_2}$

$\ce{Mg(OH)_2 \rightarrow MgO + H_2O}$

The second reaction, for magnesium, sets in around 350°C.  Below that you can get the hydroxide, above it any hydroxide decomposes and you have the oxide.

Heavier alkaline earth metals, and all alkali metals, have hydroxides stable to higher temperatures, making it more difficult to "direct" the reaction to oxide formation.  In contrast, most transition metals lack low-temperature reactivity with water and their hydroxides decompose easily, so they can give only the oxide.


Why does my metal phone ring stand becomes iridescent, and turns back to normal after a while? I got this metal phone ring stand from online. When I was meddling with it, I notice the ring started to produce iridescence. (See photo-only the centre portion, the moveable ring is not affected) I believe the mild heat from my hand caused this to happen. I left it alone on a table overnight, and the next morning, the iridescence disappeared. Te ring is back to normal.

Whenever I applied mild heat again (I placed it at the back of my vacuum cleaner for 5seconds), the iridescent will appear, but will then disappear again after I leave it overnight.

The iridescent consists of greenish, pinkish and purplish hues. It doesn’t seem to be forming at the surface, but rather it seems to be forming in the interior.

The seller claims that the ring is made up of zinc alloy, but it is magnetic. Some guesses of the material that might be in the product: zinc alloy, nickel, steel, chrome plating (the centre portion is very reflective like a mirror) Please correct me if my guesses are wrong.

I am just very perturbed by this phenomenon occurring in such simple object and am curious to find out what causes this!!

Thank you!!

This is an interference phenomenon, more specifically interference of light between two thin layers with an air gap. Recall the beautiful soap bubble as an example. Whatever coating your have on this object gets very very slightly separated from the surface upon mild heating. This generates an air gap between two layers.

Iridescence is a different phenomenon, which is angle dependent in the sense that you see different colors at different angles.

If you have over heated an empty stainless steel frying pan by mistake, that also forms a interference pattern (purple tinge). This is because of an oxide coating on the surface but it is not reversible. You have to scrub it away.


Courses for Molecular Modeling? I'm probably doing overkill trying to learn a bit of everything, but here's my story. I have a B.S. in Math with a minor in Microbiology, and I'm currently taking Computer Science courses online to meet the pre-requisites for a Master's in CS where I'll focus on machine learning. I want my thesis to be on the Molecular Modeling of a cellular process.

After receiving my Master's, because I want more research experience (and it will count as job experience for indistry), I want to work as a paid research assistant to a laboratory where I help code machine learning algorithms to aid biological modeling.

Let's say I do this for 3-4 years and I have permission to take up to three courses each fall/spring as a non-degree seeking student (realistic?). Would taking courses such as:

Physics: Modern Physics, Stachistical Mechanics, Thermodynamics, Quantum Mechanics, Computational Physics, Biophysics.

And

Chemistry: Physical Chemistry, Analytic Chemistry, Intermediate/advanced Organic Chemistry, Reaction Kinetics, Intro to drug discovery.

Be beneficial to me?

Sorry if this is too opinion based or off topic. However, what will probably benefit me the most when I get to a research assistantship, is that I'll speak to faculty and they'll  direct me on what courses to take. I'd still like your input though before the post is removed.

Someday I was wondering as you are doing but  I think that now my confusion has lessened considerably after talking with experts in the field and enrolling in a master program of pharmaceutical chemistry and drug design.

Different kinds of simulations can be called modelling of cellular processes. The first are the rules of systems biology and the second are the rules of chemistry or more fundamentally, quantum chemistry.

An example of systems biology is the simulation of the heart muscle fiber. a single cell, contractility as a function of the different ions concentrations, K, Ca, Cl, Na, etc.

The rules of quantum mechanics when applied to systems of atoms and molecules yields quantum chemistry. Quantum chemistry can calculate the tendency of a certain chemical reaction a function of the electron density and energy in certain regions. Molecular Mechanics and Molecular dynamics are more convenient when simulating bigger systems, larger number of atoms and molecules. With large number of molecules statistical mechanics must be taken into account to realistically simulate the relative population of molecules that have a certain energy.

In you question, I don't think that analytical chemistry has any thing to with computer simulation of cellular processes. From the the disciplines you outlined I thank that you are more interested in Quantum chemistry/Molecular mechanics/Molecular dynamics than that of systems biology. In my opinion you have to focus on: Physical chemistry, quantum chemistry, statistical thermodynamics,  Drug discovery and design. Also, In order to understand molecular dynamics you must have a background in motion laws of newton.

Occasionally you will need to gain a background in some specific points but, however, mastering the basics of the disciplines I mentioned will give you a comfortable platform from which you can move to your target easily.

Regards.


What should be the oxidation state of oxygen in HOF (hypofluorous acid)? What should be the oxidation state of oxygen in $\ce{HOF}$ (hypofluorous acid)?

Sources on the internet have confused me. Most state its oxidation state to be 0, while the others state that it is -2 (considering the oxidation state of fluorine to be +1, which is a bit surprising for me as it's the most electronegative element in the periodic table)

Source stating the oxidation state to be -2:

  J. Chem. Educ. 1972, 49 (4), 299:
  The two possibilities brought forth in the Note are (1) assign hydrogen +1, oxygen -2, and fluorine +1 or (2) assign H +1, oxy- gen 0, and fluorine -1. On the grounds that $\ce{HOF}$ is a powerful oxidizing agent, the conclusion reached in the Note is that (1) should be favored.

Other sources stating the oxidation state to be 0:

Yahoo Answers!
Answers.com

Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States.

When you apply the official pre-2016 definition (via the Internet Archive) from the IUPAC gold book, then you have to assign $\text{OS}(\ce{H}) = +1$ as it is not a compound with a metal, $\text{OS}(\ce{O}) = -2$ as it is no peroxide compound, and leaving the disturbing $\text{OS}(\ce{F})= \color\red{+1}$.

Going with the on electronegativity based alternative description, you will still assign $\text{OS}(\ce{H}) = +1$ as it has the lowest electronegativity. Then you assign $\text{OS}(\ce{F})= -1$, because of the highest electronegativity. This leaves a oxidation state of $\text{OS}(\ce{O}) = 0$.

This assignment is matched by the 2016 definition, the summary version of which is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The definition refers to the use of Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail). Consequently the heteronuclear approximation yields:

The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with Multiwfn 3.4.1 (a newer version is available). The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used.

\begin{array}{lrrr}\hline
\text{Method}    & \ce{H} & \ce{O} & \ce{F} \\\hline
\text{Hirshfeld} & +0.18  & -0.10  & -0.09  \\
\text{ADCH}      & +0.38  & -0.29  & -0.09  \\
\text{VDD}       & +0.18  & -0.10  & -0.09  \\
\text{Mulliken}  & +0.20  & -0.06  & -0.15  \\
\text{Löwdin}    & +0.09  & +0.01  & -0.10  \\
\text{Becke}     & +0.38  & -0.21  & -0.17  \\
\text{ADC Becke} & +0.38  & -0.29  & -0.09  \\
\text{CM5}       & +0.35  & -0.26  & -0.08  \\\hline
\text{QTAIM}     & +0.61  & -0.44  & -0.17  \\\hline
\text{NBO}       & +0.45  & -0.32  & -0.12  \\\hline
\end{array}


A good textbook to balance chemical equations? Is there any good textbook which after reading it I would be able to balance almost any chemical equation in inorganic chemistry?

This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically.

Consider the reaction $$\ce{C4H10 + O2-> CO2 + H2O}$$

To start, you would put a variable in front of each molecule here.
That would become:  $$\ce{aC4H10 + bO2 -> cCO2 + dH2O}$$

Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c
Repeating this process for the other elements, I get:
10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side).

After this step, you have your three equations.
4a=c
5a=d (This is simplified from 10a=2d)
2b=2c+d

Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1.
From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2.
Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields:
a=2 b=13 c=8 and d=10
Now put those values exactly where you originally put the letters.
It becomes $$\ce{ (2)C4H10 + (13)O2  ->  (8)CO2 + (10)H2O}$$

So the steps are:
Assign variables to each molecule in the reaction.
Write out those variables as equations corresponding to the amount of certain elements in each molecule.
Make one variable equal 1.
Solve for the rest of the variables.
Multiply by the denominators of any fractions to leave whole numbers.
Those numbers are your answers. You just plug them in :)

If you want any clarification, or another example done, just comment and I'll be happy to.

Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome.


Which of the following show hyperconjugation None of the given compounds have any alpha hydrogen to show hyperconjugation, but the given answer is c.

I guess you've only considered one resonance form of compound C. If you consider the other major resonances, there are alpha C-H bonds that can participate in hyperconjugative effects:


What's the longest C=C bond? There's been ample work and media coverage on how to make a long C-C single bond. To my knowledge, the longest stable C-C bond is currently 1.704 Å, prepared by Schreiner Nature 477, 308–311 in 2011.

All around us, there's considerable plastics, made from polymerizing alkenes like polyethylene. Typical bond lengths for alkenes are ~1.33 Å, with bond dissociation energies around 611 kJ/mol, vs ~347 kJ/mol for C-C.[ref] Clearly, longer C=C bonds could decrease the energy required to polymerize the alkane.

Using either chemical databases, like the Crystallographic Open Database, or computational methods, what's the longest C=C bond you can find?

Please include in your answer:

The molecular structure, with the C=C bond either obvious or highlighted (e.g., if it's a conjugated bond)
The method used to determine the bond length (e.g., DFT geometry optimization using program X, found in database Y, etc.)
Proof that the bond is really a double bond (e.g., clear evidence of substituents, planarity, etc.)
Any appropriate literature citations


The role of temperature in Nernst equation When measuring a non-standard voltaic cell’s electrode potential, regardless the standard conditions, as long as the concentration of product and reactant the same, a non-standard cell’s potential will be exactly equal to a standard cell’s potential.

$E= E^\circ -\frac{RT}{nF} \ln(Q)$

Q = 1 if the concentration of product equals to the concentration of reactant:

$E= E^\circ -\frac{RT}{nF} \ln(1)$

$E=E^\circ - 0$

In another word, no matter how temperature varies, as long as the concentration of both solution the same, temperature will have no effect on the voltage.

Is the understanding above valid? If it is, then my experimental result contradicted the Nernst equation. When I increase the temperature while having two half cell with the exact same concentration, the voltage increased as well. What is a possible cause for such result? (Assuming my voltemeter is funcotional one)

The standard cell potential is temperature-dependent itself, just like the Gibbs free energy. The easiest way to see this is to write the Gibbs free energy in terms of enthalpy, entropy and temperature:

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

This shows you that $\Delta G^\circ$ is strongly temperature dependent (even if $ \Delta H^\circ$ and $\Delta S^\circ$ were independent of temperature, which they are not).

To relate this to the cell potential, remember that $\Delta G^\circ$ is a measure of the maximal work a system can do, and that the electrical work is electrical potential times charge moved across that potential.

$$\Delta G^\circ = w_\text{electrical} = -E^\circ n F$$

So $E^\circ$ is temperature dependent as well.


Is the dimethylglyoximate anion ambidentate or not? In class, my professor said that dimethylglyoximate is a special ambidemtate ligand that can donate lone pairs via both nitrogen (which it normally does), and oxygen (very rarely).

On googling whether it is ambidentate or not, I didn't find anything. Can anyone cite an example where dmgH is donating lone pairs of electrons via oxygen and not nitrogen?

If this doesn't exist, can someone confirm that dmgH is not ambidentate?

I'm not sure if this example qualifies as an example for dimethylglyoximate acting as an ambidentate ligand, because it, for me, is more like bidentate. Nonetheless, I'm presenting this example here only because both nitrogen and oxygen are donating lone pairs within the same molecule. See trinuclear copper complex below (ref: doi.org/10.1107/S160053680301331X):


Identifying phycobiliprotein based on wavelength I have the absorption spectrum of a specific phycobiliprotein. Its maximum wavelength is 620.6 (nm),this is the only data I have. How can I know what type of phycobiliprotein it is?

Its maximum wavelength is 620.6 (nm)

I'm afraid that this is pretty much a link-only answer, but based on   Spectroscopic Properties of C-Phycocyanin and of Its $\alpha$ and $\beta$ Subunits, published in J. Biol. Chem., 1973, 248, 5679-5685 (PDF) there's a good chance that the phycobiliprotein is C-Phycocyanin.


Terminology of redox reactions: which species are the oxidizing and reducing agents? I've got a question regarding 'notation' for redox reactions. Suppose you are asked to identify the oxidizing and reducing agents for a simple reaction like this one:

$$\ce{Cl2(aq)}+\ce{2KI(aq)} \longrightarrow \ce{2KCl(aq)}+\ce{I2(aq)}$$

Now are you supposed to say $\ce{Cl}$ is the oxidizing agent or $\ce{Cl2}$? Also, is $\ce{KI}$ the reducing agent or $\ce{I}$ or $\ce{I^-}$?

By the way, if you were instead asked to identify which species got oxidized, would you again say that $\ce{KI}$ is oxidized or $\ce{I}$ or $\ce{I^-}$?

Ignoring spectator ions,

$\ce{Cl2 +2 I- ->2Cl- +I2}$

So, $\ce{Cl2}$ is the oxidant and $\ce{I-}$ is the reductant.


How can it be proved that (S)-bis-[2-(t-butylsulfinyl)phenyl] diselenide is a chiral non-racemic product? I read in paper that:

  The aim of this part of the work was to synthesise new aromatic diselenides with chiral phosphorous or sulfoxide centres in coordinative distance to the selenium as precursors for new chiral selenenylating reagents.

How can it be proved that (S)-bis-[2-(t-butylsulfinyl)phenyl]diselenide is a chiral non-racemic product?

Compounds with the general structure $\ce{R-SO-R'}$, i.e. containing a sulfinyl $\ce{-SO-{}}$ functional group attached to two carbon atoms, are called generically ‘sulfoxides’. Sulfoxides may be considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom).

When the two organic residues $\ce{R}$ and $\ce{R'}$ are dissimilar, the sulfur atom is a stereogenic center.

The compound bis[2-(tert-butylsulfinyl)phenyl]diselenide contains two chiral sulfinyl groups and thus two stereogenic centers.

Therefore, three stereoisomers exist:

bis[2-((R)-tert-butylsulfinyl)phenyl]diselenide,

bis[2-((S)-tert-butylsulfinyl)phenyl]diselenide,

and the meso compound [2-((R)-tert-butylsulfinyl)phenyl][2-((S)-tert-butylsulfinyl)phenyl]diselenide


When an iodide/iodate solution is acidified with H2SO4 instead of thiosulfate, why should it be titrated immediately? My class is currently performing a reaction to determine the oxidizing power (i.e. the amount of hypochlorite) of household bleach. The general process is as follows:

Standardize thiosulfate to be used as a titrant of $\ce{I2}$ solution.

An excess of I- ions are allowed to react with a known amount of $\ce{KIO3}$ in acidic solution.
The iodine formed from this reaction is titrated with thiosulfate using starch as an indicator of triiodide anion.
The relation between moles $\ce{KIO3}$ and thiosulfate is used to determine molarity of thiosulfate.

Analysis of oxidizing capacity of liquid bleach

Excess iodide ion solution added to bleach.
Iodide ions are oxidized to I2 after this solution is acidified simultaneously with the reduction of hypochlorite.
The iodine that is formed is then titrated with the above standardized thiosulfate solution.
The indicator (starch) is not added until the dark-brownish color of the iodine has changed to pale yellow. When it is added, the solution color changes to blue black.
The end point of titration is indicated when the solution becomes colorless.
(Note that if the starch is added too soon in the titration, the formation of the blue-black complex is not easily reversed, making the end point very slow and difficult to detect.

Pertinent reactions are given below

Reactions Pertaining to Titration

$\ce{HOCl + 2I- + H+ -> I2 + Cl- + H2O}$

$\ce{2S2O3^--  + I2 -> 2I- + S4O6^--}$

$\ce{HOCl + 2S2O3^-- + H+ -> Cl- + S4O6^-- + H2O}$ [the addition of the above two reactions]

$\ce{I3- + starch <-> starch*I3-(complex)}$  [blue-black in solution]

Reactions pertaining to the standardization of thiosulfate

$\ce{IO3- + 5I- +6H+ -> 3I2 + 3H2O}$

$\ce{3I2 + S2O3^-- -> 6I- + 3S4O6^--}$

The question

We were told that, during the standardization process, if the iodate-iodide solution is acidified with $\ce{H2SO4}$, it should be titrated immediately. My question is why? Using thiosulfate, this is not the case, correct? Why does this change when sulfuric acid is used in its stead?

After acidification with $\ce{H2SO4}$, starch starts to hydrolyze to its component sugars. Lugol's iodine does not react with those sugars to form the blue complex.

Our amylose is gone, today,
Its trace we'll see no more,
For what we thought was $\ce{H2O}$
Was $\ce{H2SO4}$.


Using Nanodrop for UV/Vis analysis of compounds at limited amounts I am no chemist, but a biologist who knows some biochemistry. I have some natural extracts at limited quantity, which I have been evaluating using mainly GC-MS. These extracts are rich in liquid alkaloids.

I was wondering whether a trivial "Nanodrop" could help me in some analyses  -- e.g. estimating concentration of some main compound --, via UV/Vis absorbance scans. A typical nanodrop can scan for absorbance within the range 200-800 nm.

Usually molecular biologists are the only ones routinely making use of Nanodrop, strictly for DNA/RNA purity and quantitative estimations.

Has anyone here ever tried making use of a Nanodrop micro-spectrophotometre for general analytical purposes with other compounds?

[UPDATE]

I cannot add an answer as others for some reason closed it as "too broad". It isn't, see below.

After completing my tests, I have decided to come back to answer my own question.

A Nanodrop machine has served my purpose quite efficiently, as far as I can judge. I used have used it to evaluate the degree of purity of a natural alkaloid extract from insects, using synthetic alkaloid analogues as controls. See our published paper discussing the method here, along with raw data.

We have noticed no alterations nor cross-contaminations to other DNA/RNA samples running in parallel, highlighting on the fact that the equipment was properly cleaned between each use (with 100% ethanol & distilled water).

Therefore, yes: I recommend using the Nanodrop as a cheap & quick method to scan alternative biological/chemical samples, provided the chemicals and solvents involved are compatible with the pedestal according with the manual.

Hope this discussion encourage others to dare into multidisciplinary research!

In short, i would say that you cant use Nanodrop for such work. Here is why:

From a quick google search it appears that a nanodrop is a UV-Vis spectrometer that can measure very small amounts of liquid. Looks like that  bio-sciences people (I am not a biologist or biochemist) use it to quantify nucleic acids by knowing the approximate molar absorptivity for DNA, RNA etc. at a specific wavelenth, measuring the sample's absorption at that wavelength, and applying the Beer-Lambert law assuming that the only compounds that absorb in the wavelenght that measurement takes place in their sample are the nucleic acids (which i guess is a fair assumption for such samples).

However, I am not sure from your question if you want to measure nucleic acid content in a sample also containing alkaloids as part of the sample matrix or you want to quantify the various alkaloids in your sample.

I think you must mean the latter since you mention GC-MS in which case the nanodrop seems to be a poor choice: UV-Vis spectroscopy in general is a technique with very low resolution in terms of compound separation due to the broadness of the peaks in the absorption spectra. If you had only a few alkaloids in your mixture (which i dont think you do) maybe you could use nanodrop or a normal UV-Vis spectrometer followed by a lot of calibration and absorption band decomvolution in order to identify and quantify each component. If on the other hand you have a complex mixture (which is usually the case with natural extracts) any kind of UV-Vis spectroscopy will fall short. Even NMR spectroscopy which has a much higher resolution would struggle.

This is why GC-MS seems to be the best way for such work: the GC separates the mixture in simpler mixtures (or in the pure compounds if your sample is not too complicated) and then the MS identifies and quantifies the compounds present in each fraction. So i think that you will either need to find more sample or further optimize your GC-MS method to increase sensitivity.

I hope this helps.


Is sodium ion an acid? As sodium ion has a 1+ charge and it can form the ionic compound NaOH with OH1-, hence it is a lone pair accepter (is this correct?). Therefore sodium ion should be an acid?

According to the bronsted lowry definition of an acid, the sodium ion is not an acid. This is because the bronsted lowry definition of an acid states that an acid is a proton donor. The sodium ion is not a proton donor and so according to this definition it is not classed as an acid.

I hope this answers your question :)


Does enthalpy of dissolution change with temperature? Dissolution can be either endothermic or exothermic. At higher temperature dissolution is much faster. However, supplying heat also leads to an increase in the internal energy of the system. Ultimately, this affects the enthalphy of product and reactant. I want to know whether an increase in temperature affects the enthalphy of dissolution?

I want to know whether an increase in temperature affects the enthalphy of dissolution?

It certainly can. You can treat the temperature dependence of dissolution with a formalism analogous to that applied to chemical reactions. The key parameter is the difference in heat capacity of the solid and dissolved solute, since the enthalpy of solubilization can be written as

$$\Delta H(T) = \Delta H(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^{T} \Delta C_p\,\mathrm{d}T$$

where $\Delta C_p = C_{p,\mathrm{solute}} - C_{p,\mathrm{s}}$. The $\Delta C_p$ term accounts for the different temperature sensitivities of the enthalpy (which for condensed phases approximates the energy) of solid and solute. If the heat capacity of the solute is greater than that of the solid, the enthalpy of dissolution will increase (become more positive) with an increase in $T.$

Note that if we associate a solubility constant $K_\mathrm{sp}$ with the dissolution process

$$\ce{A(solid) <=> A(solution)}$$

then we can write

$$\frac{\mathrm{d}\log K_\mathrm{sp}}{\mathrm{d}T}=\frac{\Delta H^\circ}{RT^2}$$

as suggested in a comment. Here both $K_\mathrm{sp}$ and $\Delta H^\circ$ are functions of $T.$


Why does chromium(III) have the highest tendency to form complexes, comparatively, among Mn(III), Cr(III), V(III) and Ti(III)? I was looking at some quick revision notes for d&f block elements and it had this random fact: Out of $\ce{Mn^3+}$, $\ce{Cr^3+}$,  $\ce{V^3+}$, and $\ce{Ti^3+}$, chromium readily forms complexes. I did not understand which properties of chromium make it advantageous for it to form complexes, was it due to its size? Its electronic configuration? Were there any other properties at play?

The premise of the statement is skewed. All four of those have the same (rather strong) tendency to form complexes since naked metal cations in aquaeous solution simply do not exist. However, a point may be made that the complexes of chromium(III) are the least labile (or most inert) ones of the series.

Th reason for this lies in the typical octahedral arrangement observed in transition metal aqua complexes. This arrangement causes splitting of the $\mathrm d$ orbitals (which are degenerate in ground state) into two sets of $\mathrm d$ orbitals: $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In very basic crystal field theory, this can be understood as the $\mathrm{e_g}$ orbitals being those pointing towards the ligands approaching along the coordinate axes and bearing a point-shaped negative charge; thus, these orbitals are destabilised (raised in energy). This affects the orbitals $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2 - y^2}$. On the contrary, the other three $\mathrm{d}$ orbitals — $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ — are considered stabilised since the entire transformation should preserve the total energy. (This concept is not physico-chemically accurate in the way it is presented here but close enough.)

The energy difference between the two energy levels is defined as (the relative value of) $\pu{10Dq}$. Again, using the ‘conservation of total energy’ principle, the higher-up orbitals are defined as having an energy of $\pu{+6Dq}$ each while the stabilised orbitals are defined as having $\pu{-4Dq}$ each. A ligand field stabilisation enthalpy or LFSE can be calculated by taking the electrons in each orbital and adding up the corresponding $\pu{Dq}$ value to give a total energy; the higher this is, the more stable the configuration is relatively.

All complexes in question are high-spin complexes, meaning that even though the $\mathrm d$ orbitals are no longer degenerate they are still filled according to Hund’s rule. Since the order of the elements in the periodic table is $\ce{Ti, V, Cr, Mn}$ and since each is in the same oxidation state ($\mathrm{+III}$), we can instantly see that they differ by the number of $\mathrm d$ electrons only: they have the configuration $\mathrm{d^1, d^2, d^3}$ and $\mathrm{d^4}$, respectively.

If we add between one and four electrons into the five $\mathrm d$ orbitals according to Hund’s rule and then calculate the LFSE, we will get the following result:

\begin{array}{cr}\hline
\text{conf.} & \text{LFSE}\\\hline
\mathrm{d^1} & \pu{-4Dq}\\
\mathrm{d^2} & \pu{-8Dq}\\
\color{red}{\mathrm{d^3}} & \color{red}{\pu{-12Dq}}\\
\mathrm{d^4} & \pu{-6Dq}\\
\hline\end{array}

The highlighted value is the configuration with the lowest energy, i.e. the one which is most stabilised during complex formation. Therefore, chromium(III) complexes are the most inert complexes of the series.


What is the benefit of using Am-241 in smoke detectors even though it is a radioactive element? I see the labels denoting Am-241 in my smoke detectors. I am having a hard time understand why Am-241 is used in smoke detectors even though it is a radioactive element. It seems to me like a lot of trouble to use a radioisotope in consumer products and notably extra effort given that Am-241 is a synthetic isotope. Why would Am-241 be used in the manufacture of smoke detectors?

The only mitigating factor I can see is that it has a relatively long half life.

The usage in ionizing smoke detectors requires a radioactive isotope to work.  In addition to a sufficient half-life to make a smoke detector with a suitable service life, Am-241 emits mostly alpha radiation (which is used for the ionizing mechanism) with relatively little of the more hazardous and useless gamma radiation.  See the Applications section in  https://en.m.wikipedia.org/wiki/Americium.


Ammonia (NH3) and nitric acid (HNO3) I understand that by adding Nitrogen to my water my Lucerne plants will grow better. Firtsly, is this correct? Secondly, if so, I further understand that I must add/mix Ammonia (NH3) and nitric acid (HNO3) to form nitrogen which is added to the water, is this correct? However I understand that these two (Ammonia (NH3) and nitric acid (HNO3)) is highly corrosive, therefore would this corrode my irrigation system or damage my lucerne crops, rather than increase the growth? Help will be appreciated!

Ammonia and nitric acid can be considered corrosive materials (though the real question would be "corrosive towards which material?"). The thing here is that ammonia and nitric acid are corrosive for strictly opposite reasons: ammonia is a base and nitric acid is an acid. So in a mixture of 1 equivalent(1) of ammonia and 1 equivalent(1) of nitric acid, both effects will cancel and the resulting mixture is no longer corrosive.

You do not create nitrogen (with such a mixture) for your lucerne, but you provide it with a form of nitrogen(2) which can be transformed by bacteria before being absorbed by your plant. Though, you do not want to mix ammonia and nitric acid because it is dangerous. As Mithoron said, you would better buy the product.

The chemistry and biology of soils are extensively (?) studied nowadays because it appears that monoculture exhausts soils and an agroecological approach can drastically outperform the monocultures we employ in western world.

(1) This requires a bit of calculations.

(2) Use of the term "nitrogen" can be very misleading as it can describe the "nitrogen gas" which composes 80% of our atmosphere (but is quite inert) or any source of nitrogen atoms for a plant.


Determining geometry of trisubstituted alkene The way I know to tell whether an alkene is E or Z -is to look at the coupling constant of the protons across the doble bond. Large value (16 Hz) indicates an (E)-alkene where as smaller values (11 Hz) usually indicate a (Z)-alkene.

In my project I've been doing a cross metathesis between two alkenes which gives me a tri-substituted alkene, and I cannot observe the coupling constant across this double bond. The reaction gives a mixture of products which I think are the (E)- and (Z)-isomers. I'm able to separate them on HPLC but they do not crystallise.

Is there another method of determining this? A lab member suggested I could derivatise the alkene but I don't think this would provide any more useful information.

As @Zhe points out, its not possibly to definitively answer your questions without knowing the structure of the olefin, as its important what is around the olefin, as well as just how many protons are attacked to the olefin.

If you read many of the original Grubbs' papers (and indeed any papers using a Grubbs' metathesis to make tri-substituted alkenes), the common method to determine the geometry is to use the nuclear Overhauser effect, or nOe for short.

The nOe identifies interactions through space (as opposed to standard NMR where we're looking at the interactions through bonds. By recording the nOe, one can essentially 'measure' how close two protons are, which in this case would allow you to distinguish E and Z.

Practically speaking, there are several different NMR experiments that one can run to observe the nOe.

The 1D nOe (1D-NOESY) difference experiment involves irradiation of a particular signal (for instance the one alkene proton), to allow measurement of what other protons are proximal in space. This experiment has the advantage of being quick to run, but in order to gain useful information, you have to run several of them (irradiate the alkene proton, then run a different 1D-NOESY in order to make sure that the proton that the alkene proton is 'seeing' also 'sees' the alkene proton
The 2D nOe (2D-NOESY) experiment shows all nOe enhancements in a molecule, this is incredibly useful and often used in structure elucidation, but can be slow to run, especially if the amount of sample available is limited.


Phase diagram analysis The figure is a portion of the titanium copper phase diagram for which only single-phase regions are labeled. Specify all temperature composition points at which eutectics, eutectoids, peritectics and congruent phase transformations occur.

I think i've found the eutectic and congruent point. (aprox 960 and 980 celsius) but how do I find the eutectoid and peritectic point?

Sorry for blurry image!

Assuming that the fields marked by x are solid and y is liquid, these are the points:

a: eutectic point. Cooling will turn the single liquid into two solids. The point is between the composition of the two solids involved.
b: eutectoid point. Cooling will turn the single solid into two solids.
c: peritectic point. Cooling from immediately to the left will turn the solid into a liquid and a solid. The point is outside the composition of the two solids involved.
d: congruent point: A solid will melt at this point to a liquid of the same composition.
e: incongruent melting points. A solid will melt to a liquid and a solid. Note that the left one is intimately associated with the peritectic point.

There are no peritectoid points in this diagram. A peritectoid point is similar to a peritectic point, but with all phases solid.


Why does phosphoric acid mask the colour of iron(III) complex in water? In the redox titration of iron(III) with permanganate or dichromate, we use phosphoric(V) acid to "mask" the color of iron(III) because it interferes with the end point color change.

What's the chemistry behind this?

When phosphoric acid is added to an aqueous solution of iron (III), the ligands in the yellow complexes $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$ ( $\ce{[Fe(H2O)6]^3+}$ are weakly colored but fairly acidic), are successively substituted by $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. This yields colorless hydrogenphosphatoiron(III) complexes, like $\ce{[Fe(H2PO4)(H2O)5]^2+}$ and the water-free $\ce{Fe(H2PO4)3}$ complex, depending on the phosphoric acid concentration (source). In less acidic solutions, where $\ce{PO4^3-}$ is present, the colorless bis(phosphato)ferrate(III) $\ce{[Fe(PO4)2]^3-}$ is formed (source).


Is it harmful to mix Hydrogen Peroxide with oven cleaning spray? My mother recently stumbled upon this mix when she was trying to remove stains from some laundry. When just the peroxide didn't work, she sprayed on some Easy Off, and the stain disappeared. Now she uses this method regularly, and I'm worried that it might be toxic or something. I just want to make sure she isn't poisoning herself.

Chemically, we are talking about mixing hydrogen peroxide with sodium hydroxide. They will react to form sodium hydroperoxide, but the latter is not as good an oxidizing agent as hydrogen peroxide, and a weaker base than sodium hydroxide. Probably the added base helps to dissolve some components of the stain. I don't see this as risky, especially if it is on a small scale (spraying laundry), and uses 3% peroxide, rather than mixing them in bulk. Note: With hydrogen peroxide, you have to watch out because some metal ions catalyze its decomposition to oxygen and water.


XAS spectroscopy - Differences in magnitude of 2p SOC and 3d SOC? Within L-edge x-ray absorption spectroscopy, the spin-orbit coupling within the 2p shell results in the splitting of the L edge into L3 and L2 edges.As given in the following paper http://dx.doi.org/10.1063/1.4896373, the 2p SOC is 8ev whereas the 3d SOC is 0.05eV. Could some please explain why the 3d SOC constant is much smaller than that for 2p SOC?

The energy of a spin orbit level for a '1 electron' or 'hydrogenic' atom such as sodium is given by

$$E_{so}=\frac{AZ^4}{n^3(l(l+1)(l+1/2))}\frac{j(j+1-l(l+1)-s(s+1)}{2}$$

where $A$ is a collection of constants and $j,\,l,\,s$ quantum numbers derived from the term symbols for the levels involved.

The second fraction in the equation is the angular momentum term, the first due to $\langle1/r^3 \rangle$ where $r$ is the distance from the nucleus.

For a D electron you might have $j$ quantum numbers $j$=5/2 and 3/2 with $l$=2 and $s$=1/2 and for a P,  $j$=3/2 and 1/2 with $l$=1 and $s$=1/2 thus the separation of the D level pair is less than that of the P.

Most good textbooks will have a derivation of this equation and other corrections such as due to relativity etc.


Is (2R,3S)-butane-2,3-diol chiral? Is the attached compound a chiral or achiral overall? My lecturer said that this is an achiral compound because it has a superposable mirror image but I don't get that. So I tried to figure out the configuration of both stereogenic centers because what I understood is when you have an even number of chiral centers and if they have opposite configuration, say (R,S) or (S,R), they will cancel out each other so that specific rotation is zero and two chiral centers are overall achiral, which gives you a meso compound.

The staggered, conformation of (2R,3S)-butane-2,3-diol (1a) that you have shown does not have mirror planes but rather a center of symmetry, which is the reason that this conformation does not rotate plane polarized light. There are two other stable, staggered conformations of the diol, namely 1b and 1c. They are a racemic pair, each of which is chiral. But like any racemate, it is optically inactive. Each of these staggered conformations has also been represented as Newman projections with the S-carbon in the front (1a', 1b' and 1c''). Rotation of 1c'' by 180o about a vertical axis passing through the center of the C2, C3 bond leads to structure 1c', which is the mirror image of structure 1b'. A time tested device for ascertaining whether or not a compound is meso is to pair like groups in an eclipsed conformation  (1a''). While this technique serves its purpose and conformation 1a'' has a plane of symmetry, its population in solution is negligible as is the racemic pair of eclipsed conformations (not shown).


Whether for all zero order reactions rate is independent of concentration of reactants For example,for the reaction $\ce{A + B \longrightarrow C}$,If rate law expression is:
$$\mathrm{Rate=k \times [A]^{(0.5)}\times [B]^{(-0.5)}}$$
It is a zero order reaction but whether it is independent of concentrations of reactants?

I think the issue is not that you can't define overall order, but that it isn't a meaningful quantity. This gets at the point that Zhe was making that while we can define the total order of a reaction as the sum of the exponents, it doesn't tell us anything about how the reaction occurs.

Your question gives a very good example of why it isn't a useful quantity: you might expect a zeroth order reaction to have no dependence on the concentration of reactants, but your reaction clearly depends on the concentration of $\ce{A} \text{ and } \ce{B}$, so the total reaction order isn't actually telling us anything about the kinetics of the reaction.

If the order of a particular reactant is zero however, you can say that its concentration won't affect the rate of reaction (within certain limits, as no reaction can be entirely independent of the concentration of an involved reactant).


Why do we see complementary colors in UV/Vis Spectroscopy? From what I read, a compound that absorbs visible light will produce a complementary color (using the color wheel), that can be seen. Why is this the case? Is it because, for example, if a molecule absorbs red, for instance, all the other colors will reflect and combine (?) to give off the color green that we see? I thought if we excite an electron to its excited state, with it absorbing a certain wavelength, when it drops back down to its ground state, it will give off a photon of that same wavelength (and same color). Also, when a molecule absorbs UV radiation and reflects all of the visible wavelengths, I read that it will appear white or colorless (isn't black the absence of all colors?), is this just two ways of saying the substance appears white or does that mean that it can be white or clear?

There are several points

  From what I read, a compound that absorbs visible light will produce a
  complementary color (using the color wheel), that can be seen. Why is
  this the case?

This has to do with the physiology of the eye and how it has been designed to precept the colors in the visible range. As you stated, if we remove green from the visible spectrum, the transmitted light appears reddish or purplish to our eye. May be a cat or dog sees it differently. Color is a human perception. This complementary chart is just an approximation. If we take potassium permanganate solution, and if you look at what colors does it allow to pass through, you will see that it absorbs mostly in the green. You can easily do such experiments with a cardboard box and a diffraction grating made from CDs and a white LED.

  I thought if we excite an electron to its excited state, with it
  absorbing a certain wavelength, when it drops back down to its ground
  state, it will give off a photon of that same wavelength (and same
  color).

This does not happen in majority of the cases i.e, in solutions and solids. The phenomenon you are referring to is called resonance fluorescence and it only happens to atoms in the gas phase. You shine 589 nm to Na atoms in gas phase, and you will see emission of 589 nm from Na atoms. This is accomplished in a flame. See Atomic Fluorescence Spectroscopy.

Normally, once light is absorbed, the absorbing material does not emit light. Instead, the absorbed energy is dissipated as heat.

  Also, when a molecule absorbs UV radiation and reflects all of the
  visible wavelengths, I read that it will appear white or colorless
  (isn't black the absence of all colors?), is this just two ways of
  saying the substance appears white or does that mean that it can be
  white or clear?

Clear and white are two different things. Something clear like water does not absorb much in the visible range, perhaps a very very small portion from the red region. Therefore when you have huge amount of water you start to see the effects of absorbing red portion of the light. Do you recall that ocean water looks bluish? Even liquid oxygen is light blue!

You can say that water transmits most of the visible light. Majority of the substances will also absorb UV. How deep UV do you want? Most of the substance will start absorbing deep UV. Even air does, otherwise we would have been fried by the Sun.

A white piece of plastic or a cotton cloth, is reflecting most of the visible light, hence  it appears white to us. If you look the same piece in red light, it will look red, and in blue light it will appear blue and so on.


Reactions of phloroglucinol Does phloroglucinol respond to the

iodoform test?
cerric ammonium nitrate test?
2,4-DNP test?

I couldn't find any good sources that answer any of these.

I thought of this:

Tautomerism can "change" phloroglucinol to a 1,3,5-triketone,  which should respond to the iodoform test.
It is a phenol, so it should respond to the cerric ammonium nitrate test.
Again, the keto form should respond positively to the dinitrophenylhydrazine test.

In aromatic compounds, the enol form is more stable than the keto form due to the aromatic character. In normal conditions, aromatic compounds exist in enol form but in basic conditions, it tautomerises to keto form.  Phloroglucinol predominantly exist in enol form under normal condition.

iodoform test

Compounds containing ketomethyl group($\ce{-COCH3}$) will give iodoform test. e.g.- propanone, acetophenone etc. Thus phloroglucinol will not give iodoform test because it does not contain ketomethyl group.

ceric ammonium nitrate

Generally, CAN test is done for alcohols which give pink or red colour. But for phenols and phenolic compounds, it gives brown or black test. So, this test helps to differentiate phenols from alcohols. See the detailed question on Chem.SE

2,4 DNP test

Phloroglucinol predominantly exist in enol form under normal condition and thus not give  2,4 DNP test. However, in basic condition (pH- 11-13) it forms 3,5-dihydroxy-2,5-cyclohexadienone dianion,  the deprotonated intermediate in the interconversion of the two forms, referred to as the enolate anion which might give the 2,4 -DNP test. See here

One final note

Don't take tautomerism into account while experimentally determining the functional group in organic compounds . If you consider tautomerism, there will be two functional group for the compound and you will get confused as of which functional group should be considered. For ex- propanone tautomerises to form prop-1-en-2-ol which contains an alcoholic group and then you will get confused which test should be used- Ester or CAN test(for alcohol) or 2,4 DNP test (for ketone).


How to convert SPC files to ASCII without original spectroscopy software? IR and Raman spectroscopy are typically saved in *.SPC files (chemical/x-galactic-spc), which are binary files. Collected data is save to SPC files by default, but can be exported to txt (ASCII) files.

I now have original SPC files and the original spectroscopy software is not available. Is it possible to somehow read those files with some other (generic?) software?

If you don't mind Python on the command line, there's an spc converter on Github.

While the trial version of the GRAMS Suite 9.2 is available with a 30 days license, it seemingly doesn't work with any data but the samples provided with it and therefore isn't suitable for your purpose :(


How would 2-(chloromethyl)-1-methylpyrrolidine recyclize into 3-chloro-1-methylpiperidine with only water? My Attempt

I always get thrown off by mechanisms with just water as a solvent. I've attached my work as far as I could get but I'm not sure where to go from there. Thank you!

Forget about the water present - its just a solvent.
The best nucleophile present is the N-lone pair; if you think about it in 3-D you will see that it can do a nucleophile attack on the CH2-Cl to displace Cl- and form the aziridinium cation. Cl- can then attack this at either carbon of the [3] ring, one attack merely reforms starting material, the other gives the 3-Cl-piperidine product shown.


Chelation-controlled aldol: why is the α substituent pseudo-axial? In this Mukaiyama aldol reaction, taken from a 1999 paper by D.A. Evans,1 when $\ce{TiCl4}$ was used as the Lewis acid catalyst, the aldehyde 4a formed a chelated intermediate whereas 4b did not (due to the bulk of the TBS group). This was demonstrated by the stereochemistry of the product. Aldehyde 4a produced a syn:anti ratio of 3:97, whereas aldehyde 4b led to a ratio of 93:7:

The chelated intermediate which aldehyde 4a goes through is a standard six-membered ring with a half-chair conformation. Perplexingly, the α-methyl group has to be placed pseudo-axial instead of pseudo-equatorial (or else the syn aldol product will be obtained):

Ordinarily, substituents would adopt a pseudo-equatorial position in the dominant conformer. Why is this an exception?

Reference

Evans, D. A.; Allison, B. D.; Yang, M. G. Chelate-controlled carbonyl addition reactions. The exceptional chelating ability of dimethylaluminum chloride and methylaluminum dichloride. Tetrahedron Lett. 1999, 40 (24), 4457–4460. DOI: 10.1016/S0040-4039(99)00739-X.

Having the methyl group psuedo-axial means that when the nucleophile approaches, the transitiation state is chair-like. The energy penalty for having the methyl group initially in the less favourable position is compensated for the low energy TS.

The ring-flipped version, in which the methyl group is psuedo-equatorial, would open via a high energy twisted TS (the opposite of the argument presented above, i.e. the low energy conformation of the starting material doesn't compensate for the high energy of the twisted TS).

Looking at the Evans' paper specifically, the TS he draws has an addition isopropyl group on the beta-carbon, which is equatorial (this would further stabilise the starting material, since the isopropyl would be far higher in energy if axial compared to the methyl).


Reaction of ethylene glycol with acidified potassium permangnate One would expect the reaction of ethylene glycol with acidified potassium permangnate to result in oxidation of ethylene glycol to oxalic acid. But a Q&A guide book tells me it is oxidized to formic acid. Which is wrong: the book or my reasoning?

I think the final oxidation product should be carbon dioxide ($\ce{CO_2)}$ if acidified potassium permanganate is present in excess amount.

Firstly, ethylene glycol will be oxidised to oxalic acid. But oxalic acid can still be oxidised. In fact, it is the concept of titration of oxalates with potassium permanganate ($\ce{KMnO_4}$). Under acidic conditions, oxalic acid ($\ce{H_2C_2O_4}$) will be oxidised to $\ce{CO_2}$ by $\ce{KMnO_4}$ as:

$$\ce{2MnO_4^- + 5C_2O_4^{2-} + 16H^+-> 2Mn^2+ + 10CO_2(aq)  + 8H_2O}$$

Thus the final product should be $\ce{CO_2}$

Also, note that, formic acid is an oxidisable acid because it gives silver mirror in reaction with Tollen's reagent $(\ce{[Ag(NH_3)_2]OH})$, and thus, formic acid can not be the end product, as it is still oxidisable and hence it will definitely react with acidfied $\ce{KMnO_4}$ if both are present in  the reaction medium.


Thermodynamics - Series Reaction - Estimating mols of reactants based on Free Energy of Rxn Question

  Three reactions occur simulteneously (assume ideal) at $\pu{T = 0^\circ C}$.

  $$\ce{A<=>B<=>C<=>D}$$

  a) With $\Delta_r G^\circ_\text{AB} = \pu{400 J/mol}$, $\Delta_r G^\circ_\text{BC} = \pu{-100 J/mol}$, and $\Delta_r G^\circ_\text{CD} = \pu{-200 J/mol}$. If you start with 10 moles of A, calculate how many moles of B, C, and D you have when the system reaches equilibrium. Show all your work. (Hint: mass can neither be created nor destoyed)

  b) Order the equilibrium concentrations from largest to smallest. Explain how you might be able to intuitively guess this ranking.

I know that given deltaG_RxN, one can calculate Keq, and use Keq to find the moles at equilibrium of reactants and products. Since the products at the end of one reaction became the reactants of the second, I fed in the mol amount of the first equation into the second equation.

Confused because in discussion - we went over how to do part B. and I don't think my math lines up.

positive delta G_RXN = more reactants than products, and vice versa
thus in the first rxn there should be more [A] than [B] @ equilibrium

in the second rxn there should be more [C] than [B]

and the third rxn there should be more [D] than [C].

what did i do wrong math-wise?


Pyranose v/s Furanose form of Glucose and Fructose According to my textbook (Class 12 NCERT Chemistry):

Fructose also exists in two cyclic forms which are obtained
by the addition of —OH at C5 to the C = O group. The ring, thus formed
is a five membered ring and is named as furanose with analogy to the
compound furan.

I understood why the pyranose structure for glucose is more stable than the furanose structure ( The pyranose form has the perfect 60° dihedral angle between two non-ring atoms which minimises this type of strain Source: Stability of furanose vs. pyranose form of glucose?). But I do not understand why the same cannot be applied to the fructose structure.
So, to summarize : Why is fructose present in furanose form and not in pyranose form?

You are correct that the same reasoning can be applied to fructose, and, as with glucose, the beta-pyranose form is most stable.
It is common in biochemistry textbooks, however, to draw fructose as a furanose. The reason for this is that fructose 6-phosphate is a common metabolite, and the phosphate group prevents the 6'-hydroxyl group from participating in ring closure. As a result, the free 5'-hydroxyl attacks the carbonyl, and a furanose is formed. Similarly, fructan biopolymers often contain 1,6 linkages that also prevent the pyranose formation.


What assumptions underlie Hess' law and and its analogues? Let me first briefly describe the laws as I understand them.

Hess' law states that the enthalpy of reaction can be obtained by taking the sum of the enthalpies of formation for the compounds involved in the reaction, weighted by their stoichiometric coefficient. This is often shortened to $\Delta_\mathrm{r} H = \sum \Delta_\mathrm{fm} H_\mathrm{products} - \sum \Delta_\mathrm{fm} H_\mathrm{reactants}$. A similar law exists for entropy. To get the enthalpy of a compound, one usually looks up the tabulated standard state enthalpy, and then integrates the compound's heat capacity $C_p$ to get the enthalpy at the temperature of interest. A similar procedure is used for computing the entropy. Although I have not seen it done, I guess there is nothing in the way of using such laws in a partial reaction (i.e. equilibrium): by summing the enthalpies of the various compounds at equilibrium and subtracting the enthalpies of the compounds you started with, you should get the "enthalpy of equilibrium".

What assumptions underlie these laws? For Hess' law, I would think "enthalpy of mixing is zero" is one of them, but I've yet to see it stated.

An admittedly wage bonus question: I am here talking about $H$ and $S$, but what about similar laws (if they exist) for other potentials (Gibbs energy $G$, Helmholtz energy $A$ and internal energy $U$)?

The typically unstated assumption in applying Hess's law is that you are going from an initial state of pure reactants (say, in separate containers) to a final  state of pure products (also in separate containers).  In the case of ideal gas reactions, the enthalpy of mixing is zero and enthalpy is independent of pressure.  So, for ideal gas situations, Hess's law can be extended and applied directly to starting and ending states in which the reactants and products are mixed (and even simultaneously present) and the partial pressures are all different from the standard pressure of 1 atm.

Of course, since U, S, and G are functions of state, the strict version of the law, applying only to pure reactants and products, still holds.  However, for ideal gas mixtures, partial molar entropy and partial molar free energy are functions of mole fraction, so this needs to be taken into account for mixtures.


Balancing complex redox reaction of gold treated with sodium cyanide How would you balance this reaction:
$$\ce{Au(s) + NaCN(aq) + O2(g) + H2O(l)-> Na[Au(CN)2](aq) + NaOH(aq)}?$$
Here’s what I’m came up with. First we can balance any atoms besides hydrogen and oxygen in the reaction (like $\ce{Na}$ and $\ce{CN}):$
$$\ce{Au(s) + 2 NaCN(aq) + O2(g) + H2O(l) -> Na[Au(CN)2](aq) + NaOH(aq)}$$
Then we can write out the half reactions:
$$
\begin{align}
\text{Anode:} &\quad &\ce{Au(s) &-> Au+} \\
\text{Cathode:} &\quad &\ce{O2(g) &-> H2O}
\end{align}
$$
Since the oxygen is going from an oxidation state of $0$ to $-2$ in the cathode half reaction, I figured you can rewrite $\ce{OH-}$ as $\ce{H2O}$ on the right hand side of the equation.
For the half reactions, I figured we can eliminate the ions whose oxidation number don’t change but I’m not sure whether you should include $\ce{NaOH}$ in the anode or cathode half reaction.

In principle, the oxidation of gold is the half-reaction $$\ce{Au ->  Au^+ + e^-}$$But as this oxidation is done in a cyanide medium, $\ce{Au^+}$ is included in a complex ion : $$\ce{Au  +  2 CN^- -> Au(CN)_2^- + e^-}$$ The second half-equation is $$\ce{2H2O + O2 + 4 e^- -> 4 OH^-}$$ So the final equation describing the dissolution of gold in an aerated cyanide solution is the sum of the two previous equations, after multiplying the first by 4 : $$\ce{4 Au + 8 CN^- + 2 H2O + O2 -> 4 Au(CN)2^- + 4 OH^-}$$ And if you prefer equations with neutral substances, you may write it this way $$\ce{4 Au + 8 NaCN + 2 H2O + O2 -> 4 NaAu(CN)2 + 4 NaOH}$$


Is the electron configuration redundant? For example the electron configuration for oxygen is 1s2 2s2 2p4. But if an element can never have an element in the p orbital unless it has 2 full s orbitals*, then why can't we write 2p4 as the electron configuration for oxygen? Is it not implied that it has 1s2 2s2? I would think so because the noble gas configuration works on a similar principle.

*My teacher told me this.

The configuration you describe is the ground state configuration, the configuration of oxygen in which all electrons are in their lowest allowed energy states.

Substances can also be their excited states, where electrons are not in the lowest allowed energy states. Excited states happen when atoms or molecules absorb energy, usually electromagnetic radiation (but not always).

An excited state for oxygen may involve one of the $2s$ electrons being promoted to one of the remaining $2p$ spots:

$$\begin{aligned}
\ce{O} + E &\ce{-> O}^* \\
1s^2 2s^2 2p^4 &\ce{-> }1s^2 2s^1 2p^5
\end{aligned}$$

This configuration still describes an oxygen atom, but now the $2s$ subshell is partially empty.

The other reason you can't always assume that the lower subshells are always filled is that there are some exceptions to the aufbau principle. Some elements do not follow it, especially the lanthanides and actinides.

Chromium: $\ce{Cr} - \ce{[Ar]}4s^1 3d^5$ ($s$ is not filled so that $d$ can be half-full)
Copper: $\ce{Cu} - \ce{[Ar]}4s^1 3d^{10}$ ($s$ is not filled so that $d$ can be full)
Palladium: $\ce{Pd} - \ce{[Kr]}4d^{10}$ ($s$ is empty so that $d$ can be full)
Cerium: $\ce{Ce} - \ce{[Xe]} 6s^2 5d^1 4f^1$ ($f$-block elements are weird)


Why did my deep eutectic solvent solidify? I'm finally getting around to beginning to test this deep eutectic electrolyte.  I mixed choline chloride with urea in a 1:2 mole ratio in a glass container.  It formed a sludge.  I sealed the container and then came back two days later to begin electrolysis and the mixture had formed a hard solid!

The whole time the mixture was at room temperature, and except for the mixing stage it was sealed.

So what am I missing?  Is this a sign my reagents are bad?  Or do these eutectics have a limited mixed life?

Update: Following Beerhunter's suggestion I ran the reagents in a blender until they formed a very liquid (low viscosity), homogenous mixture.  But one day later the DES is crystallizing again at room temperature:

You have to keep mixing the components until the liquid forms. If you have a sludge, you will have mixed something, but not everything intimately, so you won't have the correct component ratio. The DES is not going to form spontaneously,  so leaving it will not have helped.
We have had DES's remain as liquids for 6 months and counting.


Gaseous State(Dalton's Law) Why does the mixture of gases $\ce{NH3}$ & $\ce{HCl}$ not obey Dalton's Law of partial pressure at room temperature?
Also, what other mixtures don't obey Dalton's Law of partial pressure

As defined in an excerpt from this Wikipedia entry:

  "In chemistry and physics, Dalton's law (also called Dalton's law of
  partial pressures) states that in a mixture of non-reacting gases, the
  total pressure exerted is equal to the sum of the partial pressures of
  the individual gases."

Where this assumption falls apart for your question is that $\ce{NH3}$ and $\ce{HCl}$ readily react with each other, both in the gas and aqueous phases.  This alone renders Dalton's Law mute.


How to determine the equivalent weight of carbon dioxide correctly? I just started learning about equivalent weight and encountered this solved example.

  Problem Statement:

  $4.215~\mathrm{g}$ of a metallic carbonate was heated in a hard glass tube and the $\ce{CO2}$ evolved was found to measure $1336~\mathrm{mL}$ at $27~^\circ\mathrm{C}$ and $700~\mathrm{mmHg}$ of pressure. What is the equivalent weight of the metal?

My attempt:

First of all lets find the volume occupied by $\ce{CO2}$ at NTP.

We know,
$$\frac{P_1 V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Now, as it is given that
$$\begin{align}
&P_1 = 700~\mathrm{mmHg}\\
&V_1 = 1336~\mathrm{mL} = 1.336~\mathrm{L}\\
&T_1 = 27~^\circ\mathrm{C} = 300~\mathrm{K}\\
&P_2 = 760~\mathrm{mmHg}\\
&T_2 = 273~\mathrm{K}
\end{align}$$

Hence, $$V_2=\frac{P_1\times V_1\times T_2}{T_1\times P_2}=\frac{700\times 1.336\times 273}{300\times 760}=1.12~\mathrm{L}$$

Also, according to the law of equivalence, we get

$$\text{equivalents of metal carbonate} = \text{equivalents of }\ce{CO2}$$

Now comes the part where I am really stuck, i.e. how to determine the volume occupied by $1$ equivalent of $\ce{CO2}$.

I started with finding the equivalent weight of $\ce{CO2}$ by considering the reaction of carbonation of water which produces $\ce{H2CO3}$
$$\ce{CO2 + H2O -> H2CO3}$$
As we know the equivalent weight of $\ce{H2CO3}$ is $\frac{M}{2}$ because it has a basicity of $2$.

Hence, according to the law of equivalence,

$$\text{Equivalent of }\ce{CO2} = \text{Equivalent of }\ce{H2CO3}$$
which gives us the equivalent weight of $\ce{CO2}$ to be $22~\mathrm{g/equivalent}$. Hence, the equivalent volume of $\ce{CO2}$ at NTP is $11.2~\mathrm{L}$

Now consider this another method.

The equivalent weight of $\ce{C}$ in $\ce{CO2}$ is $3~\mathrm{g/equivalent}$ as $1 \ce{C}$ combines with $4$ equivalents of $\ce{O}$ to form $\ce{CO2}$, hence we get the equivalent weight of $\ce{CO2}$ as $3+2\times8=19\text{ g/equivalent}$ which is super weird.

My deal with the question:

What am I doing wrong in the second method for finding the equivalent weight of $\ce{CO2}$ and if you wanna know why I added the equivalent weights of the elements was because I had read that for finding the equivalent weight of an electrolyte we add the equivalent weights of the cations and anions. Hence, I tested this method of finding the equivalent weight on $\ce{SO4^2-}$.

Here is my attempt of finding the equivalent weight of $\ce{SO4^2-}$ using the method for electrolytes. We know that equivalent weight of $\ce{S}$ is $\frac{32}{2}=16$ and that of $\ce{O}$ is $8$, so we have

$$\text{Eq. } \ce{SO4^2-} = \text{Eq. }\ce{S} + 4 \times \text{Eq. }\ce{O} = 16 + 32 = 48~\mathrm{g/equivalent}$$

For those who are attempting to solve the problem:-

The solution provided by the book is as follows:-

   Volume of $\ce{CO2}$ at NTP=$\dfrac{1336\times273}{300}\times\dfrac{700}{760}=1120~\mathrm{mL}$
 Suppose the equivalent weight of the metal is $\ce{E}$.
So, Equivalent Weight of metal Carbonate $=\ce{E}+30$
Because, Eq. wt. of ${\ce{CO3}}^{2-}=\frac{60}{2}=30$
Now,
 $$\begin{align}\text{equivalents of metal carbonate} &= \text{equivalents of }\ce{CO2}\\\frac{4.215}{E+30}&=\frac{1120\text{(vol. at NTP)}}{11200\text{(vol. of 1 eq. at NTP)}}\\\text{So, } \text{E}&=12.15~\mathrm{g/equivalent}\end{align}$$

Metallic Carbonate means $\ce{M^_x(CO3)_y}$ .

For example there are $\ce{Li^_2(CO3)}$, $\ce{Ca(CO3)}$, and $\ce{Al^_2(CO3)_3}$

and $\ce{M^_x(CO3)_y -> M^_xO_y + yCO2}$

$PV=nRT$

n= 0.0500 moles carbon dioxide evolved

mass of the evolved carbon dioxide is 0.0500 moles $\times$ 44.0 g/mol = 2.20 grams

mass of the remaining metal oxide is 4.215 g - 2.20 g = 2.015  grams.

Of the 2.015 grams of the remaining metal oxide, 0.0500 moles $\times$ 16.00 g/mol = 0.800 grams are oxide, while the mass of metal is 1.215 grams.

There are 24.3 grams of metal per mole of carbon dioxide evolved.

Because the ratio x:y is unknown, it is unknown whether 1.215 grams corresponds to 0.05 moles, 0.1 moles or some other number of moles.

However, since there is one mole of carbon dioxide evolved per carbonate ion, and carbonate has a charge of -2, the molar mass of metal per charge of the metal ion will be 24.3/2 = 12.15 grams.  This is what is meant by equivalent weight of the metal in the question.


What is the major product formed on addition of Hydrogen Iodide to Pent-2-ene? Orginal question:

(1) One of the carbocations formed (CC on the right) is only slightly more stable than the other (both are 2°).
(2) Iodine ion ($\ce{I-}$) is a slightly bigger ion and it faces more steric hinderance while attaching to one of the carbocation (CC on the right).
These two factors are contradictory hence how do i compare them and arrive at a major product?

I would say the reason that the iodine ends up in the 2-position is because at the 2 and 3 positions the charge of the carbo-cation is on a secondary position and has no thermodynamic driving force. The adjacent secondary hydrides may shift without much energy loss leaving the major product to be kinetically determined. Since there are two positions for an addition to the 2-position carbon and one for the 3-position carbon, your 2-position addition has a statistical advantage. Combine this with the fact that two adjacent ethyl groups would be more stericly hindered to attack than a methyl and propyl group and it makes reasonable sense that the addition occurs on the 2-position.


How can I identify and neutralize the chemical leaching out of new nitrile gloves? I have a few cases of brand-new, nitrile gloves of a type and vendor that used to be good.  But apparently, the vendor has moved manufacturing to a different country and outsourced customer service to Hell. I can't return the gloves.

The gloves all have a chemical smell, resembling a burnt transistor with a hint of burnt rubber. If worn for a while, they irritate the skin.

Since this model of glove used to be outstanding, and the cases were not cheap, I'm hoping there's some way to salvage the gloves.  I've tried repeated washings in mild detergent, or just airing them out for 4 weeks. No change.

What chemical(s) is/are this likely to be?  How can I tell?
Is there a way to neutralize this/these compound(s) without otherwise destroying the gloves?

Pack a glove into a glass container, let it sit for a few hours or days, get the container to your institute's mass spectroscopy and gas chromatography departments, and have them have a go at the gas phase in the container. I think that might be your best bet at finding out what is leaking out of these gloves. An IR or UV/Vis of the gas phase might also yield results, as Spießbürger suggested above.

Actually, if you got an analytics department at your university/institute, why not go to them, they might be able to have someone do a B.Sc. or M.Sc. thesis on that. Even if you end up trashing the gloves, at least then it won't have been a complete waste.


Inductive effect of phenyl ring I completely understand how phenyl group 'releases' or 'donates' electron through resonance, but how is that a phenyl group shows a weak -I effect, despite having electrons literally smeared on itself?

As the $\mathrm{s}$ characteristic of an atom increases, so too does its electronegativity. Proof of this can be seen when comparing the $\mathrm{p}K_\mathrm{a}$'s of 2-carbon hydrocarbons. The $\mathrm{p}K_\mathrm{a}$'s of $\ce{CH3 -CH3}$, $\ce{CH2 =CH2}$, and $\ce{CH ≡CH}$ are $51$, $44$, and $25$ respectively.

When considering phenyl group for what it is, a ring of six $\mathrm{sp^2}$-hybridized carbons, it is easy to see why it shows a strong electron withdrawing effect, and a weak electron donating effect. The electronegativity of the carbons make benzene a nucleophilic structure that does not easily give up its electron cloud.


If i keep adding sulfuric acid into water, will it keep dissolving forever? If i keep adding sulfuric acid into water, will it keep dissolving  forever?

If you remove the last two or so percent of water from concentrated $\ce{H_2SO_4}$, it starts fuming, giving off SO3 until it again contains enough water to be stable.

$\ce{H_2SO_4 <-> SO_3 + H_2O}$

Pure sulfuric acid does not exist, so there is no clear yes/no answer to your question. Also $\ce{H_2SO_4}$ (like all other mineral acids) does not just mix with water, but strongly reacts. So it's clear there can be no miscibility gap like with water/ether or similar pairings.


How can a conductivity meter be used to determine the equivalence point? I know how a conductivity meter works. Power is supplied to electrodes and these electrodes change charge. The ions in the solution will be attracted to the oppositely charged electrode. The charge they carry at the electrodes is measured.

The equivalence point may be located graphically by plotting the change
in conductance as a function of the volume of titrant added.

I'm not sure this is an accurate enough answer. Is there something I'm leaving out or that could improve the answer?

I’m assuming you’re talking about acid/base titrations.

In these, you typically start with an acid (maybe strong) and slowly add a strong base until you reach the equivalence point (classical, using indicator) or are sufficiently past it (for plotting, using a conductivity meter). Of course, switching acid and base is possible but for the ease of discussion I will consider only acids to be titrated with bases.

Say for example your mixture included hydrochloric acid and acetic acid. Hydrochloric acid (strong) can be considered fully dissociated, acetic acid (weak) mainly undissociated in the initial solution. Thus, your solution contains significant concentrations of:
$$\ce{H3O+\qquad Cl- \qquad CH3COOH}\tag{1}$$
Now you start adding a base such as sodium hydroxide. The most significant macroscopic observation is that the concentration of hydronium ions $\ce{H3O+}$ decreases while additional sodium ions are added. At half-way towards having titrated away all hydrochloric acid, you would have the following species in solution:
$$\ce{\underset{less}{H3O+}\qquad Cl- \qquad Na+ \qquad CH3COOH}\tag{2}$$
At the equivalence point of hydrochloric acid, we can consider the significant portion of hydronium to be consumed; remember that only a small percentage of acetic acid is dissociated, meaning most of it is not. Thus, we have the species:
$$\ce{Cl- \qquad Na+ \qquad CH3COOH}\tag{3}$$
Henceforth, any additional hydroxide added will, in a macroscopic and sighlty simplified view, only serve to deprotonate the acetic acid to acetate. Thus, half-way towards the equivalence point of acetic acid, we can observe the following species in significant concentrations:
$$\ce{Cl- \qquad Na+ \qquad \underset{less}{CH3COOH} \qquad CH3COO-}\tag{4}$$
Finally, we will reach the second equivalence point at which all acetic acid has been consumed to acetate, meaning the solution features:
$$\ce{Cl- \qquad Na+ \qquad CH3COO-}\tag{5}$$
Hitherto, we have not observed any hydroxide because it was immediately quenched. Now, however, all species that would be able to quench hydroxide have been consumed. Thus, any additional sodium hydroxide will remain sodium hydroxide, giving us the following species:
$$\ce{Cl- \qquad Na+ \qquad OH- \qquad CH3COO-}\tag{6}$$

Now that the types of species in each solution has been established, which influence the conductivity and how? Water by itself does not conduct electricity but water with dissolved ions does. Therefore, we would immediately expect the solution’s conductivity to rise from point (3) to point (5) since a non-ion is converted to an ion (and sodium ions are constantly being added). This is indeed observed: a slow but steady rise in conductivity from (3) to (5).

The key, however, is to realise what happens from (1) to (3) (which also influences our understanding of (6)). Here, it may seem that hydronium ions are constantly being replaced by sodium ions, meaning that the total ion count should remain constant. Indeed, it approximately does. But hydronium (and hydroxide) is a much stronger conductor than any other ion.

Typical ions would conduct electricity by being drawn towards the anode or cathode depending on their charge. This is also, in general, what happens to hydronium (or hydroxide) ions. However, because they are defined merely by an excess (or missing) proton added to (or substracted from) a water molecule, they have the ability to move much more rapidly following the mechanism below:

$$\begin{multline}\ce{cathode^-\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-OH2+} \\\ce{-> cathode-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}OH2}\end{multline}\tag{7}$$

Because this proton-exchange mechanism (the Grotthuss mechanism) allows both hydronium and hydroxide travel speeds much more rapidly than diffusion control, these two ions will induce a much higher conductivity if present. Therefore, our overall graph will feature:

a high-conductivity starting poing
a strong, linear decrease in conductivity while hydrochloric acid is consumed
a global minimum
a slow, linear rise (as discussed before)
a turning point
a strong, linear increase in conductivity while hydroxide is added.

To find points 3 and 5, the best method is to determine sufficiently many data points of the linear sequences 2, 4 and 6 and extend the straight lines. The crossing points represent the equivalence points 3 and 5.

What I have outlined here with two acids also, naturally, works for a single acid; depending on the nature of the acid you will either not see 1/2 or not see 4.


Predicting the Product of a Substitution/Elimination Reaction I am asked to explain why neither $\ce{E2}$ or $\ce{S_{N}}2$ products are formed when 3-chloro-2,2,4,4-tetramethylpentane is treated with NaOH.

I realize that since the reagent is a strong nucleophile & a strong base, it can only participate in either an $\ce{E2}$ reaction or in a combination of the $\ce{E2}$ and $\ce{{S_{N}}2}$ reactions, depending on the substrate. I think the substrate is secondary, so $\ce{E2}$ should predominate and $\ce{S_{N}2}$ should give a minor product.

The textbook solution says that since the substrate is tertiary, $\ce{{S_{N}}2}$ will not occur at an appreciable rate, and there are no β carbons, $\ce{E2}$ cannot occur.

My misunderstanding seems to stem from an inability to depict the molecule as is expected, so I am hoping someone can clarify whether or not my drawing and line of reasoning are correct.

Remember that in a $S_{N}2$ reaction, the centre bearing the $\ce{Cl}$ atom would have to be attacked by $\ce{OH-}$ from the back side. This, however, is prevented by the by two bulky tert.-butyl substituents

As far as the alternative $E2$ pathway is concerned: There are no $\alpha$ hydrogen atoms that could be abstracted by $\ce{OH-}$.

I've uploaded a CML file of the structure to pastebin. I suggest to get yourself a copy of Avogadro and inspect the geometry on your own PC.


Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? Is it (a) para-bromonitrobenzene and sodium methoxide or (b) sodium para-nitrophenoxide and bromomethane?

Or are they both appropriate? (And is one more appropriate than the other?)

One book (NCERT Chemistry Class 12 [India]) and at least one source Free NCERT solution of this question mention the latter as the appropriate. [However, I doubt the credibility of this source, as most likely, it is made to show how the NCERT book is correct (for CBSE Boards' exam) which, though is credible, is still a human-made book, which may have mistakes/errors.] NCERT only gives the answer as (b) without explanation. Another book (abc of Chemistry [India]) mentions "chemically, both are equally probable" as -nitro group activates the bromine site on benzene for substitution. And IMO, this book is more credible.

Any expert opinion?

In principle, both reactions will work.

(a) is a nucleophilic aromatic substitution, in which is $\ce{Br-}$ is displaced by $\ce{CH3O-}$

(b) is a methylation of a phenoxide

I'd probably go for (a) for two reasons:

The nitro group in para position decreases the charge density on the oxygen atom of the phenol. As a result, this anion will be less reactive than an unsubstituted phenoxide.
Methyl bromide with a boiling point of just 4 °C is difficult to handle and a poor choice for a methylation reagent. I'd rather try dimethyl sulfate if I had to.


Does air dissolve in the R-113 (i.e. Freon 113)? Can anyone tell me whether or not air or N$_2$ dissolves easily in the R-113 (liquid solvent) under the saturation pressure of R-113? If yes, what is the solubility roughly? From this link one can only find its solubility in water.
It will be better if you could provide a link. Thank you in advance!

PS: In Table V of this data provided by @Jon Custer, I found no solubility data for air (or N_$2$) dissolved in R-113.

While data for air solubility in R-113 is difficult to find, Henry's law constants $k_\text{H}$ are available for $\ce{N2}$ in a series of related refrigerants (Ref. 1) comprised of variously halogenated ethanes. However, these are predicted, not experimental, Henry's law constants: the vapor-liquid equilibrium of binary mixtures containing nitrogen was modeled with an equation-of-state based on a perturbed–Chain (PC) modification of the Statistical Associating Fluid Theory (SAFT). I back-calculated the $k_\text{H}$ from the published temperature coefficients. Assuming a $\ce{N2}$ partial pressure $\pu{p_{\ce{N2}}= 0.78 atm}$ then leads to the following $\ce{N2}$ solubilities (as mole fractions) in the refrigerants in the temperature range $\pu{170-280 K}$:

Evidently the solubilities are generally below $\pu{0.02 mol \%}$, which I believe would qualify as "not high" but not negligible ($\chi=1\times 10^{-4}$ corresponds to $\pu{\approx 1 mM}$ solution concentration).

After more careful examination, I found that Vinš et al. cite a publication by Huang et al. (Ref. 2) which compares simulation and experimental (Ref. 3) Henry's law constants for $\ce{N2}$ in R-113, the value at 298 K being $\pu{k_H= 52.5 MPa}$. This is larger than the values above, which are in the range $\pu{1-15 MPa}$. Since $p_\ce{N2} = k_\text{H} \chi_\ce{N2}$, the larger $k_\text{H}$ for R-113 means nitrogen is even less soluble in this refrigerant.

Note of course the analysis here is restricted to nitrogen, but you can proceed in the same way to analyze other gases for which $k_\text{H}$ are reported.

References

Vinš, V., Hruby, J., Solubility of nitrogen in one-component refrigerants: Prediction by PC-SAFT EoS and a correlation of Henry’s law constants. International Journal of Refrigeration, 34, 2109-2117 (2011).
Huang, Y.L., Miroshnichenko, S., Hasse, H., Vrabec, J., Henry's Law Constant from Molecular Simulation: A Systematic Study of 95 Systems. Int. J. Thermophys. 30, 1791–1810 (2009).
H. Hiraoka, J. H. Hildebrand, J. Phys. Chem. 68, 213 (1964).


Unknown notation relating to stoichiometry I have never seen the highlighted symbol before. What is the name of the symbol itself, and what does it mean in this context?

It is probably supposed to be the 'corresponds to' sign[ref, p 105]; in this case simply representing that for every two moles of $\ce{NaOH}$ one mole of $\ce{H2SO4}$ is converted by reaction stoichiometry.

This sign is difficult to typeset in LaTeX which is why I presume the authors used something else instead. Creating the sign is even more annoying in MathJax. The closest I [got at first] were

$$\overset{\lower.5em\wedge}{=}\ \text{or}\ \hat{=}$$

which are $\overset{\lower.5em\wedge}{=}$ and $\hat{=}$, respectively.

Edit. After some testing, this option $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ looks best in my opinion.

$$\ce{2 mol NaOH $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ 1 mol H2SO4}$$

full code: $$\ce{2 mol NaOH $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ 1 mol H2SO4}$$

Just the bare $\widehat{=}$ leads to

$$\ce{2 mol NaOH \widehat{=} 1 mol H2SO4}.$$

ref: E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. (2007). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. (2008 IUPAC & RSC reprint). ISBN: 978-0-85404-433-7. [free copy]

Here is a picture of how it looks in the 'Green Book':


Why does the Heisenberg uncertainty principle work in an atom? Where is my logic wrong?

An electron can only assume quantized energies in an atom.

If an electron is localized in space in a hydrogen atom, its radius is known. If its radius is known, its electrostatic potential energy is known. Then, if the electron assumes quantized energies, its kinetic energies are quantized by $E_\mathrm{total} - E_\mathrm{potential}$ for all quantized possible total energies of the electron. This gives us quantized kinetic energy and thus momentum, giving us quantized frequencies. If you sum all of the waves of those frequencies, will it give the localized position wave-function? They can't right due to the fact that they're quantized.

This is probably very wrong; I just don't understand how.

Electrons don't have a well-defined orbital radius even in a hydrogen atom

An electron "orbiting" in a hydrogen atom does not have a well-defined position or even just a well-defined radius just because it has a well-defined energy.

The orbital is more like a 3D probability density function covering a cloudy space of a defined shape (spherical in the case of the lowest energy orbital but not of a single radius and extending with a finite probability from the nucleus to a vaguely defined boundary we often regard as the radius of the atom).

Different orbitals have different, increasingly complex, shapes defined by solutions of the Schrodinger equation. Heisenberg's uncertainly relationship limits our ability to define the location and momentum of the electron at the same time. The fuzzy probability clouds defined by Schrodinger's solutions are the best we can do.


Why is water transferred from a beaker of pure water to a beaker of sugar solution when placed in the a closed container? I know that there is something to do with the vapour pressure of pure water being higher than the vapour pressure of the sugar solution, but I still can’t really explain what happens on a molecular level. Could someone help?

Water molecules in gas phase collide and are absorbed by both solutions at the same rate, assuming the same humidity over the solutions.
OTOH, as sugar solution has lower molar fraction of water than water itself,  its rate of water evaporation is lower than pure water evaporation.
Therefore, there is the net transfer of water molecules from the water beaker to the sugar solution beaker.
It can be formulated also by other way: The saturated vapor pressure over sugar solution is lower than over pure water.  Therefore vapour saturated over water is oversaturated over sugar solution. And vice versa - vapor saturated over sugar solution is undersaturated over water.


What are the most important resonance structures of 5-aminocycloheaxa-2,4-dienone? Show, by means of curly arrows and electron shifts, the resonance contributors to the structure given below. Suggest which one would not be a major contributor and state why.

This is a question from my exam. I do not understand where I went wrong. My effort is placed at the end (along with a request).

Also, my TA removed 1 point out of 7 for a reason. Can anyone decipher for what? There's the writing in red but it is illegible to me. I don't understand to what he refers by 8 and 6 t or +, or whatever he wrote.

The handwriting of your TA probably stems from ancient egyptian script, I had a hard time reading it. In the end I agree with jerepierre, interpreting it as "6E" or "8E" short for 6/8 electrons. The last statement then translates to "but all have 8E."
Before continuing to answer your question, I want to make a statement about resonance structures in general. Resonance structures are used to express a complicated bonding situation in terms of simpler, Lewis type structures. None of these structures exist individually, only a mixture of all possible resonance structures at the same time exists. I am not a fan of declaring single resonance structure as having more or less impact on the overall structure. Of course one can go ahead and calculate the total wave function in terms of resonance structures, generally known as valence bond theory, but this is overkill for such simple molecules. Then you get a mathematical representation of the electronic structure which you can decompose into contributions of the individual resonance structures.
A simpler approach is to analyse the molecule with Natural Bond Orbital (NBO) Analysis. I have done this for you, but only as an educational effort. Maybe after understanding this concept it is easier to grasp why certain structures are make a smaller contribution.
Let me go ahead and explain a little NBO to you.  The electronic structure of a molecule is in general highly delocalised. All orbitals are stretched over the whole molecule. Without changing the electron density, it is possible to localise (transform) these orbitals to represent two centre interactions, that can be interpreted as bonds. This gives an approximate picture in terms of a Lewis structure. There will be electron density which cannot be assigned to a bond. The smaller this density is, the better is the fit as a Lewis structure.
For this educational approach I picked the 5 resonance structures you have posted plus an additional ionic form. (I would have detracted a point for the last structure.) There are many more ionic forms, but these will not have big effects, e.g. cleavage of a carbon carbon bond.

The following is based on NBO analyses from DF-BP86/def2-SVP.
\begin{array}{crr}\hline
\text{Structure} & \text{% Lewis} & \text{% non-Lewis}\\\hline
\mathbf{1}   & 96.4  &  3.6  \\
\mathbf{2}   & 96.8  &  3.2  \\
\mathbf{3}   & 96.2  &  3.8  \\
\mathbf{4}   & 95.8  &  4.2  \\
\mathbf{5}   & 96.3  &  3.7  \\
\mathbf{6}   & 95.2  &  4.7  \\\hline
\end{array}
The first statement that should be made is that this molecule is quite well described by Lewis resonance structures. However, delocalised bonding would describe the molecule much better. All structures are reasonably close to each other, as one would expect for a very delocalised system.
The ionic structure has the smallest impact, which is feasible because nitrogen-hydrogen bonds are better described as covalent. However, it proves that there can (in the right medium) be a proton shift from nitrogen to oxygen, and is therefore worth consideration.
The structure with the second largest error (% non-Lewis) is $\mathbf{4}$. A positively charged carbon would be by far more electronegative than nitrogen. And many imine-type structures are known, which makes $\mathbf{5}$ more important. (In this sense I would disagree with your explanation, which could also justify the detraction of a point.)
In general: Carbocation structures that are not in close proximity to (much) more electronegative elements should be considered as having a lesser influence on the electronic structure of a molecule.


Why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane? So any data sheet will tell you;

Group 1: methane boils at: -164 deg C
         methanol boils at: 65 deg C

Group 2: decane boils at: 174 deg C
         1-decanol boils at 229 deg C

that being said; why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane ?

As well is there anything else other than difference in non-polarity and IMF's that contribute to differences between molecules in individual groups (i.e. between methane and methanol; between decane and 1-decanol)

The difference in boiling point, between methanol and methane on one hand and between 1-decanol and decane on the other hand, is primary due to the presence of hydrogen bonds (HB) in methanol and 1-decanol and not in methane and decane. The density of these intermolecular forces (HB) is higher in methanol than in 1-decanol (In 1-decanol, for every ten atoms of carbon there's one possibility to have HB. While in methanol for every atom of carbon there's one  possibility to have HB.