Where does the residue accumulate in the boiling point composition curve? My teacher was teaching me the boiling point composition curve for fractional distillation and in the class notes he gave the following diagram (I have given below). It says that the residue starts accumulating at the point on the vaprous line just above the liquidus line (above $χ_{A},χ_{B}$). Note that $χ_{A},χ_{B}$ don't represent the coordinates, they represent the liquid composition of the two components $A$ and $B$. Similarly, $y_{A}, y_{B}$ represent the vapour composition of the vapour in equilibrium with the liquid with composition $χ_{A},χ_{B}$. But why does residue accumulate at that point? My teacher literally gave no explanation as to why. I am so confused. Please can someone explain. Please help.

Let's start with the mixture $x_A$, which is made of 85% A + 15 % B. When heating this liquid, the first bubble vaporized appeared at a temperature just a little above $T'_A$, as given by the horizontal line. And the composition of the vapor is nearly pure $A$, with $1$% $B$. As a consequence, the liquid is loosing some $A$. its volume decreases, but as it does not loose $B$, its composition changes from $15$% $B$ to maybe $20$% $B$. As it has more B, it is necessary to heat the liquid to a higher temperature. The composition of the liquid is a point following the lower curve upwards to the right hand side of the graph. Meanwhile the composition of the vapor slightly changes and contains a little bit of $B$.
When the volume of the liquid has decreased by more than one half, the remaining liquid contains more $B$ than $A$. Its composition is $20$% $A$ and $80$% $B$. The boiling point of this mixture has increased : it is now about half way between the boiling temperatures of the pure $A$ and $B$ stuff. And the vapor gettin out of this liquid can be obtained in drawing a horizontal line from the point $0.8$ on the lower curve. This line cuts the upper line $y$ at a point $y_A$ = $85$%, and $y_B$ = $15$%. It is possible to continue heating the rest of the liquid. The amount of liquid is smaller and smaller, and the vapor has a composition following the upper $y$ curve.
Now, if you have a hot vapor containing the same proportion ($15$% B , $85$% A), and if you coli it down, the first drop of liquid happening at the beginning of the condensation has a composition corresponding the th horizontal line passing at $y_A, y_B$. So this liquid droplet contains $80$% $B$ and $20$% $A$.


What's the best way to interconvert between molarity and percentage? Let's say that I want to convert concentration of an acid from % to molarity.
One way I found to do this is by using the following formula:
percentage = (molarity * molar mass) / 10.

Another way is to use an equivalency chart like this one (see bottom of page): http://www.ochemonline.com/Acid_and_Base_Concentrations

For example, to convert 38% HCl to molarity I calculated:
38% = (molarity * 36.46) / 10 and molarity = 10.42 = 10 M.
However, the equivalency chart says that 38% is actually 12.4 M.

The same discrepancy happened when I tried to convert 68% nitric acid molarity. Using the equation, I got molarity to equal 11 M, however, according to another source 68% nitric acid is closer to 15 M.

So my questions are:

1) Is there something I'm doing wrong that I get such a huge discrepancy?

2) What's the best way to figure out molarity from % and vice versa?

Thanks to LDC3's comments, I understand that the best way to convert from % to Molarity is as follows:

1) multiply density by volume at hand to get weight

2) solve ratio for x, where x is grams of solute (in volume at hand)

(given% / 100%) = (x / weight from step 1)

3) multiply x by molar mass to get moles (in volume at hand)

4) if needed, multiply resulting moles/volume in hand to get Molarity

Example: Let's say I have 100 mL of 38% HCl.

1) 1.19 g/ml * 100 mL = 119 g

2) (38%/100%) = (x/119) where x = 45.22 g

3) 45.22 g * 1mol/36.46g = 1.240 moles

4) 1.240 mol/100 ml = 12.40 mol / 1000 mL = 12.4 M

Now, the result matches the equivalency chart!


Finding molar mass of an unknown acid "In the titration of a solid acid, an endpoint is reached after 22.0mL
  of 0.120 M $\ce{NaOH}$ has been added. Assuming that each acid particle
  contains two acidic hydrogens and both are neutralized, what is the
  molar mass of the unknown acid?"

I really don't know how to solve this. I've figured out how to calculate the moles of acid were in the sample but don't see how to find the molar mass from that...any ideas?

Your data is not enough to calculate molecular weight. You may have missed the mass of the solid acid you have weighed.
Let's assume you have weighed $a$ gram of acid. Then:
$Molecular Weight = \frac{Weight}{Moles_{acid}} = \frac{Weight}{2 \times Moles_{NaOH}} =  \frac{Weight}{2 \times Molarity_{NaOH} \times Volume_{NaOH}} = \frac{a}{2 \times 0.120 \times \frac{22.0}{1000}}$


Finding required drug concentration given Kdiss The enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9 mol/l). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7 mol/l).
Compute the total concentration of [D1]tot that is needed to bind 90% of the HIV-RT at the given concentration [E]tot. You do NOT have to consider competition betwwen the drugs D1 and D2! They are administered separately.

I understand how to calculate the concentration when [E]eq = [ED]eq when 50% of [D]eq is bound to HIV-RT because Kdiss = [D]eq and it makes for a simple calculation because [E]eq and [ED]eq cancel out when calculating Kdiss, but what changes mathematically need to be made to account for the change in ratio (90% rather than 50%)? Any help would be appreciated.

Chemical reaction is

$$\ce{D1 + E <=> ED1}$$

Initial concentrations: $[\ce{D1}]_\textrm{tot}; \quad [\ce{E}]_\textrm{tot}; \quad 0$.

If $X$ percent of $E$ "reacts", then stoichiometry tells us that:

Concentrations at equilibrium: $[\ce{D1}]_\textrm{tot} (1- X) ;\quad [\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X;  \quad [\ce{D1}]_\textrm{tot} X$.

The equilibrium expression is

$$K_\textrm{diss} = \frac{ [E][D_1]}{[ED_1]}$$

Therefore

$$K_\textrm{diss} = \frac{ [\ce{D1}]_\textrm{tot} (1- X) ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)}{[\ce{D1}]_\textrm{tot} X}$$

Which we can simplify to

$$K_\textrm{diss} = \frac{(1- X)}{X} ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)$$

So you can solve this for $X = 0.90$, $[\ce{E}]_\textrm{tot} = 2.00 \ \mu M$ and $K = 1.00 \ nM$.


What does pH of NH3 solution refers to? I was introduced with an exercise which asks to find $K_\mathrm{b}$ of $\ce{NH3}$ when the $\mathrm{pH}$ of $\ce{NH3}$ solution with concentration $M_1$. Well, in order to solve it, I used the fact that $\mathrm{pOH} = 14 - \mathrm{pH}$ and the dissociation $\ce{NH3 + H2O <=> NH4+ + OH-}$ to compute $\ce{[OH-]}$ and then $K_\mathrm{b}$. However I don't fully understand what does $\mathrm{pH}$ mean, since in this dissociation there aren't any $\ce{H3O+}$ molecules. I think it relates to the conjugate dissociation of the $\ce{NH4+}$, i.e.: $\ce{NH4+ + H2O <=> NH3 + H3O+}$, so the $\mathrm{pH}$ of the $\ce{NH3}$ solution is in fact related to the concentration of $\ce{H3O+}$ which is formed from the $\ce{NH4+}$ dissociation. Is it indeed the case? I would really appreciate any explanation.

As Ivan Neretin said in the comments, $\ce{H3O+}$ is already there, via the autodissociation of water:

$$\ce{2H2O <=> H3O+ + OH-}\tag{1}$$

Now water reacts with the ammonia that you add, according to the equilibrium reaction you provided:

$$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{2}$$

This increases the hydroxide concentration further, so the first reaction is out of equilibrium, and consumes some hydronium and hydroxide by going in the net reverse direction, increasing the pH.

Alternatively, you can combine the two reactions to show that adding ammonia will lower the hydronium concentration:

$$\ce{NH3 + H3O+ <=> NH4+ + H2O}\tag{3}$$

All three reaction and the reverse reactions will occur simultaneously.

  However I don't fully understand what does pH mean, since in this dissociation there aren't any H3O+ molecules.

For basic solutions, the hydronium ion is a minor species. For acidic solutions, the hydroxide ion is a minor species. Sometimes in calculations, we get away with ignoring a minor species for a while, but they are still there.


How accurately does (pKa1 + pKa2) / 2 estimate the pH of an amphoteric salt? This question came up when answering Find the pH and amphiprotic salt is added to water at room temperature. If we dissolve $\ce{NaHA}$ in water, the species $\ce{HA-}$ can act as acid or base:

$$
\begin{align}
\ce{H2A &<=> AH- + H+}\\
\ce{HA- &<=> A^2- + H+}
\end{align}
$$

The $\mathrm{pH}$ of the solution can be estimated as

$$\mathrm{pH} = \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2}$$

My question is how accurate this estimate is, depending

on concentration of the amphiprotic salt (as the concentration approaches zero, the $\mathrm{pH}$ should approach 7),
on the average of the $\mathrm{p}K_\mathrm{a}$ values (the closer these are to neutral, the smaller the difference between $[\ce{H2A}]$ and $[\ce{A^2-}]$), and
on the difference between the $\mathrm{p}K_\mathrm{a}$ values (the bigger the difference, the lower the percentage of $\ce{NaHA}$ undergoing acid or base reactions).

So depending on these three variables, how accurate is the estimate? My guess is that high concentration, average of $\mathrm{p}K_\mathrm{a}$ near neutral and difference between $\mathrm{p}K_\mathrm{a}$ low would give good estimates. When any of these parameters a very different from those "ideal" conditions, the estimate probably gets worse.

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a  good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$,  then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$.

As the consequence, they have  just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values.

We have then a mix of 2 inferior buffers.

If e.g.the $\mathrm{p}K_\mathrm{a}$ difference = 4, the buffer capacity is roughly 25 times smaller, compared to $\mathrm pH=\mathrm pK_\mathrm {a1}$ or $\mathrm pH=\mathrm pK_\mathrm {a2}$.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )


Is it correct to say that H2SO4 is an acid in this reaction? Is it correct to say that $\ce{H2SO_4}$ acts as an acid in the following reaction?

$\ce{2Ag + H2SO_4} \rightarrow \ce{Ag_2SO_4 + 2H_2O + SO_2}$

I know it acts as an oxidizing agent, but is it correct to say it shows acidic properties?

Do you get the same reaction with sodium sulfate?

I do not think so, at least under "normal" conditions.

When the sulfuric acid acts as an oxidizing agent giving off $\ce{SO_2}$, the excess oxygen comes off as oxide ions which must be somehow put into more stable form.  The protons from the sulphuric acid do that by turning them to water.

Getting the reaction to go requires the sulfuric acid to act as both an acid and an oxidizing agent.


Boiling and Melting point of Lithium (or any other element that does not occure in free state) The melting and boiling point of Lithium as compared to other members of the Alkali group metals is high. But, Lithium does not occur in the free state.

So, what does it mean that it has high melting and boiling point? How are its BP and MP measured if it isn't in the free state?

Your premise here is wrong, pure metal Lithium exists and can be handled without much difficulty.

However, materials which are extremely reactive can always be kept in an inert gas when you want to measure their physical properties. Of course, there are compounds which can don't need an excuse to explode (and  will explode in a vacuum/inert gas as well), in which case it is harder to find out BP/MPs (one can put them at low pressure and then extrapolate, but that's not as accurate).


Rate of catalytic hydrogenation of alkenes What does the rate of catalytic hydrogenation of alkenes depend upon? What'd be the increasing order of rate towards catalytic hydrogenation of the following alkenes, A (2-methylpropene), B (cis-but-2-ene), and C (trans-but-2-ene)?

I suppose this has something to do with sterics and orientation - as catalytic hydrogenation is a surface phenomenon (occurs on surface of metal catalyst with adsorbed hydrogen), but I'm not sure about the above order in particular (I don't think a clear comparison can be made without looking at data, as the alkenes are quite similar), and the factors on which rate of hydrogenation of hydrocarbons depends in general.

Catalytic hydrogenation is more or less a surface phenomenon so sterics play the deciding role in the rate of hydrogenation.

From Wikipedia:

Here's a newman projection of the three Alkenes.

 A = cis-dimethylbutene, B = trans-dimethylbutene, C = 2-methylpropene

From here it is easy to see that the cis-dimethylbutene will have the least hindered approach towards the surface of the catalyst as both the methyl groups can be made to face away from the surface.

The rates of hydrogenation in the case of trans-dimethylbutene and 2-methylpropene will be decided by which one of them is more strained.

2-methylpropene is more strained than trans-dimethylbutene.

So, we can now establish the order.

$$\text{A > C > B}$$

References

Catalytic Hydrogenation:A Core Technology In Synthesis; Ryoji Noyori
Wikipedia article on Hydrogenation


Relationship between inhibition constant Ki and interaction energy? Inhibition of an enzyme $E$ by an inhibitor $I$:

$$
E + I \rightleftharpoons EI
$$

The inhibition constant $K_i$ is defined as follows:

$$
K_i = \frac{[E][I]}{EI}
$$

where $[E]$ is the enzyme concentration, $[I]$ is the inhibitor concentration, and $[EI]$ is the enzyme-inhibitor-complex concentration. Inhibition is strong when $K_i$ is low; that is, when the equilibrium above is shifted toward the product.

Can the strength of inhibition be approximated by the relative interaction energy: $E_{Interaction} = E_{EI complex} - E_{Enzyme} - E_{Inhibitor}$? This seems like a logical and valid hypothesis to me, but I cannot find any sources confirming that this is justified. If the enzyme-inhibitor complex is very stable, I would think that the equilibrium above naturally is shifted toward the product, given a relatively fast kinetics. In the case of a commercial drug inhibiting the HIV-1 protease, for example, where it is safe to assume fast kinetics, I would model the strength of inhibition as the relative interaction energy: stronger interaction $\Rightarrow$ lower $K_i$ value.

The context is a simple theoretical modelling of a protein-ligand interaction using density functional theory.

One complication of enzymatic inhibition constants is that there a multiple mechanisms to inhibition, and the type of inhibition can change how the inhibition constant relates to binding energy.

On one side of the spectrum you have competitive inhibition, which means that the $EI$ complex is functionally inactive, and cannot bind substrate. (The easiest way of getting competitive inhibition of an enzyme is to have the inhibitor bind in the active site, competing with the substrate binding. But that's not the only way. You can also have allosteric inhibitors which exhibit "competitive inhibition", if they bind the apo enzyme and by doing so prevent substrate binding.) This is the sort of inhibition you're assuming in your question, but isn't the only type of inhibition.

On the other side of the spectrum, you have uncompetitive inhibition. (N.B. uncompetitive - noncompetitive inhibition is something different.) This means that the inhibitor requires the substrate to be bound to the enzyme prior to binding itself, and it's the $ESI$ complex which is functionally inactive.

And then you have a whole range of variation where you have a mix of effects (binding to both the apo and substrate-bound forms, with various affinities), of which noncompetitive inhibition is a particular type.

You're correct in that for pure competitive inhibition, the $K_i$ is equal to the $K_D$ of the inhibitor. The problem arises when you have one of those mixed effect or uncompetitive inhibitors, where catalytic steps and substrate concentrations get folded (partially) into the $K_i$.

Personally, I would recommend Enzyme Kinetics and Mechanism by Paul F. Cook and W.W. Cleland, which goes into great details about reaction mechanism, kinetics and inhibition constants. It should include derivations and relations for all the relevant scenarios.


Is it possible to identify the source of a protein hydrolysate I have a powder which only declares to be "Protein Hydrolysate"
Meaning that the all the protein has been broken down.

We've tried to identify the source of the powder with DNA research. But no usable DNA (parts) where found.

Is there a way to identify the source of this powder? (ie. chicken, pork, cow, shark, dolphins or unicorns)

For clarification, we failed to find any DNA in the powder. so we're looking for alternative ways to analyse it.

DNA is not a protein, as @MARamezani states. See http://www.answers.com/Q/Is_DNA_a_protein.
There is a prevalence (or lack) of some amino acids in certain sources. For example, maize ("corn" in America) is deficient in tryptophan and lysine, so if your sample is short on those two, it is more likely that it is from corn. A list of some foods and amino acid content is at http://www.ljhs.sandi.net/faculty/aquesnell/Biology%20Notes/Chapter%204/4.8-4.13/Essential%20Amino%20Acids.pdf [there appear to be some errors, though]. According to that source, yeast may be deficient in phenylalanine, so that is another indicator.

You would need an analysis of the relative abundance of each amino acid, some reference showing their abundance in various sources, and additional analysis to calculate relative abundances in a mixture, e.g. corn supplemented with yeast hydrolyzate.


What are B and n in this expression for charge repulsion? In a study of the creation of ionic bonds of in this case $\ce{Na}^+$ and $\ce{Cl}^-$ into $\ce{NaCl}$, I have come across this equation for the overall energy of the system:

$$E=-\frac{e^2}{4\pi\epsilon_0 R}+\frac{B}{R^n}+1.4 \,\mathrm{eV}$$

$e$: electron charge. $\epsilon_0$: vacuum permittivity. $R$: the distance between them.

The first term is the attraction energy between the two equal but oppositely charged ions $\ce{Na}^+$ and $\ce{Cl}^-$.
The second term is the repulsion between the like charges within the shells. When the distance $R$ is very small, this term has large (opposite) effect.
The last term is the difference in free creation energy for the two free reactions ($\ce{Na \rightarrow Na^+ + e^-}$ requires $+5.1\,\mathrm{eV}$ and $\ce{Cl + e^- \rightarrow Cl^-}$ gives $-3.7\,\mathrm{eV}$)

My question is two-fold and about the repulsion energy term: What is $B$? And what is $n$? I have been told that the exponent is around $n \approx 10$, but is this a general constant or material specific?

There isn't an exact form for the Pauli Repulsion term, but it is often approximated as a $\frac{1}{r^n}$ term for simplicity.  For example, in the Lennard-Jones expression for Van Der Waals interactions, $$E(r) = \epsilon \times[(\frac{\sigma}{r})^6-(\frac{\sigma}{r})^{12}]$$, the $\frac{1}{r^6}$ term is the London force term (which in reality, does go as $\frac{1}{r^6}$), and the $\frac{1}{r^{12}}$ is the Pauli repulsion term (just an approximation, but in this approximation, $n=12$).
In the OPLS force field (which uses a Lennard-Jones expression for Van Der Waals interactions), the parameters for $\ce{Na+}$ and $\ce{Cl-}$ are:

$\ce{Na+}$: $\sigma = 3.500 A$, $\epsilon = 0.0660\ \ce{kcal/mol}$

$\ce{Cl-}$: $\sigma = 4.180 A$, $\epsilon = 0.11779\ \ce{kcal/mol}$.

OPLS is parameterized for a geometric mixing rule, meaning that for an interaction between the two, the $\epsilon$ and $\sigma$ parameters are equal to the geometric mean of the parameters for each atom.  Thus, for an $\ce{Na+ - Cl-}$ interaction:

$\sigma = 3.8249 A, \epsilon = 0.08817\ \ce{kcal/mol}$

Now, the Pauli part of the Lennard Jones expression is $\epsilon \times (\frac {\sigma}{r})^{12}$, so $B$ in this case is $\epsilon \times \sigma^{12} = 864554\ \ce{kcal A^12 /mol}$

However, there are other force fields parameterized differently, and some of these approximate the Pauli repulsion using a different value of $n$ (and some like $MM3$ don't even assume a $\frac{1}{r^n}$ dependency).  So since $n$ is assumed (and since different force fields are parameterized differently), there are other possible values for $B$ and $n$ to use in the approximation.


What are the products in this SN1 reaction? I am asked to identify the organic/inorganic products of an SN1 reaction between 3-iodopentane and ethanol. I have identified correctly that the substrate is the 3-iodopentane, the leaving group is the iodide, and that the nucleophile is the ethanol. My problem is that I do not understand what the products will be, and I am unsure why.

I decided that the products should be 3-pentanol and ethyl iodide (or iodoethane) because iodide is a good leaving group, and the oxygen takes two electrons and leaves ethane to bond with the pentane group where the iodide left off. Now it makes sense that the ethane and iodide react to create ethyl iodide.

Unfortunately, I am wrong in this case, and I just can not put my finger on why. What is the error in my thinking and what is the concept that I am missing?

I realize that my major organic product will be the ether compound which I am having trouble naming, and the other compounds are a protonated ethanol plus the iodide. The most confusing part now is the protonated ethanol and the iodide, why not just create the ethyl iodide?

OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$

So it's SN1, so the $\ce{I}$ is going to leave spontaneously and give you the carbocation: $\ce{(CH3CH2)2CH+}$

The oxygen is in the ethanol so that will attack the carbocation giving the oxonium ion: $\ce{(CH3CH2)2CH-O+} ~ \ce{H-CH2CH3}$

Which will then be deprotanated by something acting as a base. There's probably some water around, but I suppose saying protonated ethanol and iodide is fine. This gives the final product of: $\ce{(CH3CH2)2CH-O-CH2CH3}$

For ethyl iodide to form, one of two things would need to happen:

SN1 mechanism: $\ce{OH-}$ would have to spontaneously leave forming a primary carbocation. It is SUPER DISFAVORED, so much so that it straight up won't happen.
SN2 mechanism: $\ce{I-}$ would have to act as a nucleophile AND $\ce{OH-}$ would have to leave. Iodide is an OK nucleophile, but OH- is a pretty bad leaving group. Both things considered this pathway is unlikely.

This was a pretty weak question by your (I am assuming) professor. This reaction would be horribly slow, and probably would barely go. Now, if they used sodium ethoxide instead of ethanol you get a nice Williamson ether synthesis and form the same ether and NaI, much more favored thermodynamically.

I just realized you probably meant why doesn't $\ce{I-}$ perform an SN2 on the protonated ethanol and have water leave. There are a couple of halfway decent chemical explanations for this:

Assuming there is any water around (there is pretty much always water around in the real world (especially with ethanol), unless remarkably heroic measures are taken to avoid it), the protonated ethanol will quickly react with it and form hydronium, ethanol and $\ce{I-}$. (Protonated ethanol has a $\mathrm pK_\mathrm a$ of about -2.4 and hydronium has a $\mathrm pK_\mathrm a$ of about -1.7.)
Assuming there is not water around, you are right it will probably react and form small amounts of ethyl iodide, however this will generate water which will stop this from continuing for the above reason.
You could also write the mechanism where the iodide deprotonates to form the ether (and avoid this whole fuss). BUT $\ce{HI}$ has a $\mathrm pK_\mathrm a\approx10$ where as oxonium proton $\mathrm pK_\mathrm a$s are usually between -7 and -2, so this would NOT be an accurate way to write the mechanism (in my opinion), but some professors get a little hand wavy here and allow it.

I know those are kind of sketchy workaround reasons, but honestly it won't be formed in any real quantity. There is probably another (probably better) argument from a thermodynamic stand point, but I don't know the bond energies off hand.


Effect of rain on water and soil samples contaminated by petroleum I am about to perform research on the impact of petroleum fire disaster on the soil and water sources within the region of disaster, using heavy metals and total petroleum hydrocarbons as impact markers.

My supervisor suggested I collect samples during raining season and dry season and compare.

My question is, would there really be any significant difference between the  two samples in the rainy and dry season?

It has much sense since the rain will modify the concentrations of both the heavy metals and the petroleum.
The initial quantities will be redistributed.
One part will penetrate deeper in the ground, another will be moved due to water runoff and what is left will be measured in your samples.

If possible, you should take the first samples as soon as possible and the the next ones towards the end of next season.
It is also important to get the value of the total precipitation during the rainy season.
This will serve to estimate how long it will take to eliminate the effects of the disaster.


Why are synthetic pH indicators used over natural indicators? Synthetic indicators seem to be exclusively used when determining the pH of a substance with an indicator (with the exception of that school experiment where you boil cabbage to demonstrate natural pH indicators) or in acid-base titrations etc. over natural indicators. Why is this the case when naturally occurring indicator can be easily prepared from accessible and sustainable methods, such as boiling cabbage?

Intuition seems to suggest that perhaps synthetic indicators are much more accurate and reliable, being specifically designed for the task, is this the case? Or are there other reasons why natural indicators are regularly used as a cost-effective, environmentally friendly and easily prepared alternative to synthetic pH indicators?

What are the advantages/disadvantages of using synthetic indicators over natural ones (or vice versa)?

In acid-base titrations, synthetic indicators are exclusively used to find accurate end-point determinations because they always have a highly defined color change at certain pHs. For example, phenolphthalein ($\mathrm{p}K_\mathrm{a} = 9.7$ at $\pu{25 ^{\circ}C}$) is colorless in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 8.2$), but it is pink in basic conditions when the pH of the solution goes above 8.2 (precisely $8.2 \lt \mathrm{pH} \lt 12.0$), which makes it ideal for strong acid-strong base titrations (Wikipedia).  On the other hand, bromothymol blue ($\mathrm{p}K_\mathrm{a} = 7.1$ at $\pu{25 ^{\circ}C}$) is yellow in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 6.0$), but it is blue in basic conditions when the pH of the solution goes above 7.6 ($7.6 \lt \mathrm{pH}$). However, it shows greenish-blue color in a neutral solution (precisely $6.0 \lt \mathrm{pH} \lt 7.6$; see the picture below).

It also change color from yellow to pink in high acidic conditions such in concentrated hydrochloric acid ($0 \gt \mathrm{pH}$; see most left test tube in the picture below), thus it is ideal to use as a universal indicator (Wikipedia). These two examples would clearly show you, as you correctly put it, synthetic indicators are much more accurate and reliable since they are specifically designed for the task.

Sure, natural indicators could be easily prepared from easily attainable row materials, and hence, could be regularly used as a cost-effective and environmentally friendly alternative to synthetic pH indicators, but I regret to say that it can be used only in high school science activities to fascinate science-loving kids. Since those natural sources contain several color changing substances (e.g., flavanones, flavones, flavonols, and anthocyanidins), the color change at the end-points is not sharp, perhaps, due to the difficulty of determining $\mathrm{p}K_\mathrm{a}$ of relevant indicator(s). For example, although boiled red cabbage water can be used as a universal indicator in high school projects and show the end point of an acid-base titration (see the following figure), it cannot be used in a day-to-day analytical lab to do accurate calculations.

Also, another reason for not using natural indicators is that most of these dyes tend to decompose in higher pH values.  The best example is most common natural color indicator, anthocyanidins. The stability of anthocyanidins is dependent on pH. At a low pH (acidic conditions), colored anthocyanidins are present, whereas at a higher pH (basic conditions) the colorless chalcones forms are present (see scheme below) (Wikipedia).

Most importantly, aside from the above, flavonols including anthocyanidins have been studied extensively in drug development due to their plethora of biological activities (Ref.1). Thus they have been targetted in much more important tasks than a usage as pH indicators.

Reference:

M. Rahnasto-Rilla, J. Tyni, M. Huovinen, E. Jarho, T. Kulikowicz, S. Ravichandran, V. A. Bohr, L. Ferrucci, M. Lahtela-Kakkonen, R. Moaddel, “Natural polyphenols as sirtuin 6 modulators,” Scientific Reports 2018, 8, 4163 (11 pages) (DOI:10.1038/s41598-018-22388-5).


What are the products for the redox reaction HCl + SnCl₂ + KMnO₄? In the redox reaction, where $\ce{HCl}$ is the excess reactant do these reactions produce following half reactions?
\begin{aligned}
(1)&&\ce{SnCl2 &-> Sn^{4+} + 2e-}\\
(2)&&\ce{8H+ + KMnO4 + 5e- &-> Mn^{2+} + 4H2O}
\end{aligned}

Where does the $\ce{HCl}$ come into play here, or do you consider it at the end when you have a final equation?

This is a redox reaction where $\ce{KMnO4}$ was added to $\ce{HCl}$ and $\ce{SnCl2}$ (hydrochloric acid was simply used as an excess reagent)

When using $\ce{MnO4-}$ as an oxidant, pH is crucial.

In acidic medium: $\ce{Mn(VII) -> Mn(II)}$

$$\ce{8 H3O+ + MnO4^{-} + 5 e- -> Mn^{2+} + 12 H2O}$$

Under neutral conditions: $\ce{Mn(VII) -> Mn(IV)}$

$$\ce{2 H2O + MnO4^{−} + 3 e− -> MnO2 + 4 OH^{−}}$$

In strongly alkaline medium: $\ce{Mn(VII) -> Mn(VI)}$

$$\ce{MnO4^{−} + e− -> MnO4^{2−}}$$

As far as an excess of $\ce{HCl}$ is concerned, one might take into account that $\ce{Sn(II)}$ and $\ce{Sn(IV)}$ possibly exist as their anionic chloro complexes
$\ce{[Sn(Cl)4]^{2-}}$ and $\ce{[Sn(Cl)6]^{2-}}$, respectively.


What is the difference between the chemical potential μ and the molar Gibbs free energy Gm? I'm not too sure if I understand this correctly:
Is the $G_\mathrm{m}$, the Gibbs free energy of the entire system, divided by the amount of substance making up the system, while $\mu$ is the Gibbs free energy of a component of the system?

If so, what does ${\partial G_\mathrm{m}}/{\partial n}=\mu$ signify? As in, what's the difference between this and $\mathrm{d}G/\mathrm{d}n=\mu$?

The chemical potential of a species is the partial derivative of the Gibbs free energy of the mixture of components with respect to the number of moles of that particular species, holding the temperature, pressure, and number of moles of all the other species constant.  For a single component system, this is reduces to the total free energy divided by the number of moles.


Basic Water Dissolving Salt Questions I have a basic question about how water dissolves salt. In the Khan Academy explanation, it says that the H and Cl atoms attract each other while the O and Na atoms attract each other.

My question: Does this weaken the bond between Na and Cl within a single NaCl molecule? Or does it weaken the bonds between Na of one molecule and Cl of another salt molecule? Or does dissolution weaken both?

There are no $\ce{NaCl}$ molecules in the solid salt nor in solution.

In solid salt, there are $\ce{Na+}$ ions, surrounded by 6 $\ce{Cl-}$ ions, while each $\ce{Cl-}$ is similarly surrounded by 6 $\ce{Na+}$ ions. Here is the Wikipedia picture of the structure.

During the dissolution, water molecules reach for a particular ion, oriented toward it by the opposite partial charge of their electric dipole.

$\ce{Na+}$ ions are torn away and wrapped by $\ce{O}$ side of $\ce{H2O}$ molecules, which have a partial negative charge.

$\ce{Cl-}$ ions are torn away and wrapped by $\ce{2H}$ side of $\ce{H2O}$ molecules, which have a partial positive charge.

Ions end hydrated in solution like $[\ce{Na(OH2)}_n]^+$ and $[\ce{Cl(H2O)}_m]^-$


Reaction of hydrogen peroxide with sodium carbonate In my book the following acid base reaction is written:

$$\ce{H2O2 + Na2CO3 -> Na2O2 + H2O + CO2}$$

But I found that the $\mathrm{p}K_\mathrm{a}$ value of $\ce{H2CO3}$ is less than the one for $\ce{H2O2},$ which means $\ce{H2CO3}$ is more acidic than $\ce{H2O2}.$

As $\ce{H2CO3}$ is getting formed as the product (and then decomposing to give $\ce{CO2}$), shouldn't the reverse reaction be more favourable due to greater acidity of $\ce{H2CO3}$ and the above reaction not happen significantly?

Is $\ce{H2CO3}$ maybe getting decomposed as soon as it forms allowing this reaction to occur?

The impossibility of a reaction producing simultaneously $\ce{Na_2O_2, H_2O_2}$ and $\ce{CO_2}$ can be proved by the following arguments, drawn from the Merck Index, under the reference "$8728$. Sodium Peroxide".

1) $\ce{Na_2O_2}$ is produced by burning metallic sodium in a current of air, from which carbon dioxide has been removed.

2) $\ce{Na_2O_2}$ reacts so easily with $\ce{CO_2}$ that it is used for rendering air charged with $\ce{CO_2}$ respirable as in submarines and diving bells.

3) $\ce{Na_2O_2}$ absorbs water from the air forming sodium hydroxyde and hydrogen peroxide, the latter quickly decomposed into Oxygen and water.

Ref.: Maryadele O'Neil, Ann Smith, et al., The Merck Index, An Encyclopedia of Chemicals, 13th Ed. 2001, Merck & Co Inc., Whitehouse Station, NJ, USA.


What is the correct pH value of weak and strong alkalis? In the book titled fundamental chemistry for Cambridge O level, it states at page 149,

  Ammonia solution with a pH of 11.1 is a weak alkali and that's why it shows much lower conductivity of 0.50 US/cm.

However at p. 147 it's written that

  a solution with a pH of 10 is a strong alkali.

So either the correlation of pH with conductivity is incorrect  or the statement that ammonia solution with a pH of 11.1 is a weak Alkali is wrong.   What are the pH values for a weak and strong Alkali?

Is there anything wrong with the two diametrically opposite sentences/ statements given in the book?

There might be confusion as below:

strong/weak "Alkali"(1) - or rather an alkaline solution, which is related to the $\mathrm{pH}$ value. The conductivity and the strength increases with $\ce{pH}$, but conductivity can be obviously affected by presence of salts.
There is not a strict threshold between strong or weak alkaline $\ce{pH}$. It is often considered relatively and/or it depends on context. As a guide the threshold can be between 10-12, context dependent.

strong/weak "Alkali"(2) - or rather a base ( as an opposite to an acid ), which is related to the base dissociation constant. IF a base has too low dissociation constant, so there is significant fraction of not dissociated base ( like ammonia ), the base is considered a weak base. There is ongoing equilibrium $\ce{ BOH <=> B+ + OH-}$, resp. $\ce{ B + H2O <=> BH+ + OH-}$

E.g.
$\ce{NaOH}$ solution with $\mathrm{pH} = 10$ is weak "alkali"(1), but strong "alkali"(2).

$\ce{NH3}$ solution with $\mathrm{pH} = 11$ is stronger "alkali"(1) then above, but weak "alkali"(2).

Note: For historical reasons, "alkali" is reserved to hydroxides of alkalic metals, so it should not be used for ammonia. As the general term, a base is preferred, with alkalic/alkaline ( along with basic ) as adjective being accepted.


Resonance of 4-isopropylbenzenethiol Can someone draw what the missing structure is? I can't find it.

The missing resonance structure is

If you replace "SH" with "OH" in the 4 structures you've drawn (your 5th structure is the same as the one pictured on the exam) and the one I've drawn, you'll have all of the analogous resonance structures for phenol.

EDIT : response to OP's comment

In my view, it doesn't really matter how you order the structures, but your suggestion (between the 3rd and 4th structures from the left) seems logical.  I've redrawn the figure to show the electron movement.


Determining boiling point on a created temperature scale 1.Assume that you construct a thermometer using gallium as the fluid instead of mercury, and that you define the melting point of gallium as 0 °G and the boiling point of gallium as 1000 °G. What is the melting
  point of sodium chloride (801 °C) on the gallium scale?

First we determine the ratio of $^{\circ}\mathrm G/^{\circ}\mathrm C$ by dividing the melting points. This states that $0.4599\ ^{\circ}\mathrm G=1\ ^{\circ}\mathrm C$.

$$\dfrac{1000\ ^{\circ}\mathrm G-0\ ^{\circ}\mathrm G}{2204\ ^{\circ}\mathrm C-29.78\ ^{\circ}\mathrm C}=\dfrac{1000\ ^{\circ}\mathrm G}{2174.22\ ^{\circ}\mathrm C}=0.4599\ ^{\circ}\mathrm G/^{\circ}\mathrm C$$

To calculate the boiling point, this is done: $$T=0.4599\ ^{\circ}\mathrm G/^{\circ}\mathrm C\times\left(801\ ^{\circ}\mathrm C-29.78\ ^{\circ}\mathrm C\right)=355\ ^{\circ}\mathrm G$$

My question is, why are we subtracting by $29.78\ ^{\circ}\mathrm C$?

Unlike an absolute temperature scale like degrees Kelvin, Celsius and your hypothetical °G are relative scales, where the zero points are set to the melting points of water and gallium, respectively. In your first step, you've accounted for the difference in the size of one degree Celsius and one °G, but not the difference in the zero points.

Instead of sodium chloride, think about what the melting point of gallium is on each scale. You've defined it as 0 °G, but you know it's not 0 °C. What do you have to do to correct for this to get a value in Celsius?

Incidentally, this is also why the conversion between Celsius and Fahrenheit is not just multiplying by a constant.


Calculated 13C NMR Shifts of brominated Carbons Since I calculate NMR spectra for a while by simply using one of recommendations by CHEmical SHIft REpository:

mPW1PW91/6-311+G(2d,p)-SCRF//B3LYP/6-31+G(d,p)

without thinking to much how it actually works*, there is a common misprediction that I'd like to understand.

It seems to be that the shift/shielding of the brominated carbons is always way off the measured shift. Compared to all other "normal" substituted carbons (C, H, O, N...) it is also far from being predicted good.

Luckily for me, as I don't want to show my own molecules :D, the Repository gives an example (Aplydactone) with a brominated carbon that shows exactly what I mean:

The overall $^1$H Mean Absolute Deviation (MAD) is about 0.08 ppm and for the $^{13}$C shifts the MAD is about 1.5 ppm but without the brominated carbon (right side of picture). The predicted shift has a failure of 12.25 ppm (measured 65.5 ppm vs. calculated 77.75 ppm) which is about ten times the MAD of all other carbon shifts.

Is there a common explanation for that? Is it simply that there were no brominated samples within the test set or is there a much deeper reason? Like Bromine must be treated with a different/super special basis set or sth similar.

This comes right now into my mind ... As there might already be some influences of the relativistic effect, could that be a reason?

Please tell me of your deeper insight into the calculation of NMR shieldings that could yield to this misprediction that I either understand how to correct those errors or that I at least have an explanation.

I did no further research throughout the internet yet, as I have not much time right now. But as this question keeps in my mind, maybe someone of you has done the research and would share it with me.

* I know that it uses a regression function which was calculated through comparing lots of experimental vs. predicted shifts for a bunch of small organic substances. What I mean is more the quantum theoretical aspect.

The authors note in their review article* that such effects are expected as "heavy atom effects" due to electron-correlation effects and spin-orbit coupling: (references stripped for clarity)

  This being said, one way to improve the computed chemical shifts of heavy-atom-bound carbon atoms is to utilize a method that better captures electron correlation effects, such as MP2 or higher-level methods along with correlation-consistent basis sets (see section 4.3). This will likely result in significant improvement for carbon atoms attached to sulfur or phosphorus as well as to halogens, although the improvement is expected to be greater for sulfur or phosphorus. Further improvement requires the use of methods that determine relativistic spin–orbit corrections.
  [...]

  Another option for improving chemical shift calculations with heavy-atom errors is to utilize a linear regression method, as discussed in section 3.6. Due to the systematic nature of heavy-atom effects, they are amenable to improvement via this approach, provided that the linear regression data consists of nuclei similar to the nucleus of interest. That is, a linear correlation correction to carbon nuclei attached to a given number of bromine atoms, for example, should be derived from empirical data for other carbon nuclei attached to the same number and type of halogen atoms. In fact, this approach has been demonstrated specifically for such cases.

The latter section cites two relevant articles:

"Theoretical investigation on $\ce{^1H}$ and $\ce{^13C}$ NMR chemical shifts of small alkanes and chloroalkanes" J. Chem. Phys. (2006) 125, 144309.
"A hybrid quantum mechanical and empirical model for the prediction of isotropic $\ce{^13C}$ shielding constants of organic molecules" Phys. Chem. Chem. Phys., 2002, 4, 5498-5507.

Both articles suggest ways to use DFT methods with empirical corrections that work fairly well.

* Michael W. Lodewyk, Matthew R. Siebert, and Dean J. Tantillo, Chem. Rev., 2012, 112 (3), 1839–1862.


How do amino acids look like at the isoelectric point? At the isoelectric point amino-acids are neutral to the outside. It is however not known to my knowledge if the molecules are indeed in a zwitterionic form or just in its "normal" totally uncharged form, i.e. the dissociation constant for this reaction:

is not known. Does anyone know how they actually look like? What form is more probable, how realistic is the zwitterionic form?

Edit: I did a test calculation for cysteine with DFT, using a PBE functional and implicit water and found that the zwitterionic form is about 7 kcal/mol more stable, which would rather strongly indicate that the zwitterionic form is indeed dominating a lot and you would not at all find the structure on the left as the bars from wikipedia indicate (supposing that I found low lying minima in both calculations and not only local ones). Are there some other studies proving this?

Edit2: Ok the zwitterionic form is definitely more stable, that one can somehow reason from chemical knowledge or pKa's. But how much more stable, can one somehow get a dissociation constant for the reaction above?

Edit3: Since there is much confusion about my question. I want a dissociation constant for the reaction (COOH)(NH2) -> (COO-)(NH3+). Until now I think only the dissociation constants for (COOH)(NH3+) -> (COO-)(NH3+) and (COO-)(NH3+) -> (COO-)(NH2) are known if I am right and I cannot tell how these would help to calculate the constant I want...

Edit4: To clear up my question even a bit more, here is a diagram:

The red arrow and question mark is what I want to know about. The pks1 and pks2 are known, that was clear to me, but how to get any information about the equilibrium in the middle?

By Henderson relation we have

$$\mathrm{pH=pI=p}K_\mathrm{a}+\log\left(\frac{[\ce{A-}]}{[\ce{AH}]}\right)$$

So if $\mathrm{pH - p}K_\mathrm{a} < 0$ then $[\ce{A-}] < [\ce{AH}]$

And if $\mathrm{pH - p}K_\mathrm{a} > 0$ then $[\ce{A-}] > [\ce{AH}]$

If you want then to know the exact proportion you need to know the concentration of your amino-acid.


Why is Sodium Transported Across the Cation Exchange Membrane in the Chlor Alkali Process? In the Chlor Alkali Process, a sodium ion is transported across a cation exchange membrane to create NaOH in the Anode Chamber:

The Half reactions are:
2Cl− → Cl2 + 2e−
and
2H2O + 2e− → H2 + 2OH−
and the full reaction is:
2NaCl + 2H2O → Cl2 + H2 + 2NaOH
Why does the sodium ion get transported across the membrane? Sodium isn't involved in the anode/cathode reactions, so there's no concentration gradient driving it. I understand that the Anode reaction produces OH-, and that there is now a charge imbalance so that the sodium ion crosses in order to maintain electroneutrality. But how can this be described mathematically? I don't see how the standard mass conservation/ Nerset-Plank equations can describe this motion.

There is a rather interesting explanation as to why only sodium ion can pass through a cation exchange membrane and chloride ion is repelled by it.
Let us recall some properties
a) Ion exchangers are very good electrical conductors. This is well studied in the 1940s. The current is carried, not by electrons but by mobile ions in ion exchange resins. In a cation exchanger, current is carried by mobile cations.
b) Imagine in the cation exchanger, it consists of a sulfonated polymer. Let us call that R-SO3H, where H is mobile and R-SO3(-) unit is immobile.
c) Ignore the flow of the chlor-alkali cell for the time being.
d) The punch line is that the bulk solution has to be electrically neutral all the time.
Now consider an electrolytic cell which consists of NaCl solution. There is a partition of a cation exchange membrane in between. On the left there is an anode where chloride ions are being oxidized and the right, water is being reduced to hydrogen.
Right hand side:
When water is being reduced, neutral hydrogen gas is leaving the solution, leaving behind negatively charged hydroxide ions. There is a charge imbalance in the bulk which cannot exist. What has to happen? Na(+) ions must cross the cation exchanger from the left hand side to ensure there is no bulk charge imbalance.
Now chloride cannot pass through the ion exchange membrane because the R-SO3(-) has a negative charge. All it can do is to repel the negatively charged chloride ion. You can read more about Donnan effect and Donnan potential.
Left hand side:
You are losing negatively charged chloride ion as neutral chlorine gas. There is excess positive Na(+) ions there. The positive ions have to cross the membrane in order to maintain electrical neutrality in the bulk and eventually join the free hydroxide ions.
As a combined effect, the right hand side becomes a solution of NaOH (contaminated with NaCl) and the concentration of NaCl on the left hand side depletes.


Looking for a radio-transparent epoxy/resin I have much hobby experience with 2-part resins, but this time I'm looking to
make a repair part for the inside of a microwave oven (everything I've tried so far heats up, but the original part does not).  I need the result to be hard enough (shore durometer D ~ 75) to rotate the glass turntable tray within the oven, withstand ~ 250 degrees Fahrenheit (it will not be in contact with food) and be radio-transparent at the frequency of 2.4 GHz.

I've reached out to over a dozen epoxy/resin manufacturers, but am having trouble finding such a product.  Thank you in advance.

You might not be able to find a home-usable thermosetting resins such as epoxy or RTV silicone rubber that meets your needs: low dielectric loss (dissipation factor), moderate hardness and stability at microwave cooking temperatures.

Bakelite (urea-formaldehyde thermosetting resin) might be usable, as it's hard and temperature resistant, though its dissipation factor is ~0.005, a tiny bit higher than ideal. However, one-off molding of Bakelite is non trivial, requiring pressurized and temperature controlled molds.

A table of electrical properties of some plastics shows polystyrene, polypropylene and some fluorinated plastics such as polytetrafluoroethylene with dissipation factors <0.001, but these are less hard than Bakelite.

Glass-filled plastics such as glass-epoxy laminate (printed circuit board, PCB, material) and RO4000® hydrocarbon ceramic laminates might be better choices. However, instead of casting a part, you'd need to machine the part from one or more PCB's laminated together. This might actually be easier, for a one-off part, than casting.


Unusual Titration Curve Explanation From Costanzo's Physiology book in Acid-Base Physiology chapter, this titration curve of a weak acid ($\ce{HA}$), in the context of discussing buffers is provided:

What confused me is that in this diagram, the $\mathrm{pH}$ is on the $x$-axis rather than the $y$-axis. Also, why does the peek of the curve has ($\ce{HA}$) only and the trough has ($\ce{A-}$) only?

Acid (Brønsted–Lowry Acid) is a substance that gives away its $\ce{H+}$, but the environment should be able to accept that $\ce{H+}$. After it accepted it - the acid (which is now its conjugate base) can accept it back. There are 2 extremes:

The environment already has plenty of $\ce{H+}$ already. So even if acid gives its $\ce{H+}$ to the environment, it immediately receives $\ce{H+}$ back. Hence the acid always has its $\ce{H}$ and you see $\ce{HA}$ on the picture.
The environment is basic - it can take lots of $\ce{H+}$ and won't give them back to the acid. Hence acid stops being acid - it gives away $\ce{H+}$ and becomes $\ce{A-}$.

$\mathrm{pH}$ is a measure of how many $\ce{H+}$ there are in the solution. Low $\mathrm{pH}$ means there's plenty of $\ce{H+}$ (our 1st case), high $\mathrm{pH}$ means there's a "shortage" of $\ce{H+}$ (2nd case).
In the middle (where $\ce{H+}$ equals acid's $\mathrm{p}K_\mathrm{a}$ which in your case is 6.5) you'll have a case when 50% of $\ce{H+}$ is taken by the environment and the other 50% of acid molecules are $\ce{HA}$.
When you titrate you do exactly that - you change the $\mathrm{pH}$ of your environment which then changes how many molecules are ionized. And then at some point your indicator changes the color which means you reached some $\mathrm{pH}$ you were looking for.
Titration curves that I've seen have the volume of solvent added (your graph probably shouldn't be called Titration Curve?) - not $\mathrm{pH}$. But these are related since if you're adding a basic titrant to acidic solution you're changing its $\mathrm{pH}$.


How to apply Faraday's law to calculate the charge of metal ions? Electrolysis of a molten ionic compound with a current of $\pu{0.50 A}$ for $\pu{30 min}$ yielded $\pu{0.700 g}$ of a metallic element at the cathode. If the element has a relative atomic mass of $150.0$, calculate the charge on the metal ions.

My attempt at solving:
$$n = \frac{m}{M} = \frac{0.7}{150} = \pu{4.67e-3}\\
Q = Ft = 0.5 \times (30 \times 60) = \pu{900 C}$$

The main problem I'm having here is that, how do I relate this unknown metallic ionic compound to the amount of electrons gained in this reaction?

After I obtain that can I just use $$Q = n\ce{e-} \times F?$$

I then solved for $$F = Q/n = \pu{1.92E5 F}$$. I was experimenting with some stuff and divided this by $\pu{96500 C}$ which got me $1.99$. The correct answer was $\pu{2 F}$.

Is this just coincidence because I don't see the connection between dividing by $\pu{1 F} = \pu{96500 C}$ when I already solved for $F$?

When you're confused by a problem like this it is definitely time to use dimensional analysis.

For the amount of electrons $\text{mol}_e$

\begin{align}
\text{The ampere} &= \text{Amp} = 1.00 \frac{\text{coulomb}}{\text{second}} \approx 6.242\times10^{18}~\frac{\text{electrons}}{\text{second}}\\
\text{Farady} &= 26.801 \frac{\text{Amp}\times\text{hours}}{\text{mol}_e}\\
30~\text{minutes} &= \frac{30 \text{minutes}}{60~\text{minutes/hour}} = 0.500~\text{hours}\\
\text{mol}_e &= \frac{0.50~\text{Amp} \times 0.500~\text{hours}}{26.801~\text{Amp}\times\text{hours}\text{/mol}_e} = 9.328\times10^{-3}~\text{mol}_e
\end{align}

For the amount of substance of metal $\text{m}_m$

$$\text{mol}_m = \frac{0.700~\text{g}}{150~\text{g/mol}_m} = 4.667\times10^{-3}~\text{mol}_m$$

Charge per atom = $z$

$$z = \frac{\text{mol}_e}{\text{mol}_m}= \frac{9.328\times10^{-3}}{4.667\times10^{-3}} = 1.998 = 2.0 \frac~{\text{electrons}}{\text{atom}}$$


Possibility of SET mechanism of Wurtz reaction I was studying the Wurtz reaction, and I wanted to figure out the mechanism. When I searched the web, I came across the single electron transfer (SET) mechanism explanation. How can the alkyl group and the halogen form free radicals when the halogen is clearly much more electronegative then the alkyl group?

How can the alkyl group and the halogen form free radicals when the halogen is clearly much more electronegative then the alkyl group?

To answer your first question, it is important to note that the halogen does not form free radicals in this mechanism. Instead, the driving force is the net formation of sodium halides, as illustrated in the mechanism I drew below. In the preceding sections, I will try to compare the Wurtz mechanism to a perhaps more familiar one, the Grignard mechanism. This should help elucidate some general aspects of organometallic mechanisms.

If you are familiar with the Grignard reaction, the following comparison may help clear up the involvement of radical species in the mechanism. The Wurtz reaction is in fact very similar to the Grignard reaction. Both involve the 'insertion' of a metal species in between the carbon-halogen bond. In the case of the Grignard reaction, this metal is magnesium. This allows for the formation of a simple alkyl magnesium halide; however, with the Wurtz reaction, the addition of a sodium atom in between the carbon-halogen bond results in the formation of neutral sodium halide and a carbon radical. This carbon radical will rapidly react with another equivalent of sodium metal to form a carbanionic species, which will in turn displace another halogen in an Sn2 reaction.

The Wurtz reaction is not a super useful reaction due to the formation of this reactive radical. Most chemists would opt to use the much gentler Grignard reagent. Below, I have drawn out the mechanism of the Wertz reaction of 1-bromo-3-chlorocyclobutane to form bicyclo[0.1.1]butane. In this, you can see that a carbon radical is formed. This radical must come in contact with another equivalent of sodium before it is neutralized to form a carbanionic species. For this reason, the radical will also have plenty of time to react with other parts of the molecule in unwanted ways. The Grignard reaction, on the other hand, results in the formation of a radical in the 'reduction' step; however, this is neutralized very quickly in the recombination step and will thus not react unnecessarily with other parts of the molecule. I have included a diagram of the Grignard reagent formation below.

Mechanism of the General Wertz Reaction

Mechanism for the Formation of a Grignard Reagent


Action of HCl on PbS I was wondering if it would be possible to use $\ce{HCl}$ acid to dissolve $\ce{PbS}$.

I know that I can dissolve it in $\ce{HNO_3}$ since, the nitric acid oxidises the sulphide ion to elemental sulphur, and some sulphides like $\ce{ZnS}$ do dissolve in $\ce{HCl}$. But what about $\ce{PbS}$?

What is the extent of $\ce{pH}$ sensitivity of lead sulphide precipitate.

Hydrogen sulfide is a weak acid:
$$\begin{align}
\mathrm{p}K_{\text{a}1} &\approx 7\\
\mathrm{p}K_{\text{a}2} &\approx 13\\
\end{align}$$

Therefore, the concentration of sulfide $c_{\ce{S^2-}}$ is very low in acidic solutions:

$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{c_{\ce{H2S}}}{c_{\ce{H+}}^2}$$

Besides, hydrogen sulfide is a gas. Its solubility in water is approximately $c_{\ce{H2S}}\approx 0.1\ \mathrm{mol\ l^{-1}}$.

For example, in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$, the concentration of sulfide may be estimated as

$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{0.1\ \mathrm{mol\ l^{-1}}}{\left(1\ \mathrm{mol\ l^{-1}}\right)^2}=10^{-21}\ \mathrm{mol\ l^{-1}}$$

On the other hand, the solubility product of lead(II) sulfide is extremely low:

$$K_\text{sp} = c_{\ce{Pb^2+}} \cdot c_{\ce{S^2-}} \approx 10^{-28}\ \mathrm{mol^2\ l^{-2}}$$

Thus, the solubility of lead(II) sulfide (i.e. the maximum concentration of $\ce{Pb^2+}$) in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$ is

$$c_{\ce{Pb^2+}} = \frac{K_\text{sp}}{c_{\ce{S^2-}}} \approx \frac{10^{-28}\ \mathrm{mol^2\ l^{-2}}}{10^{-21}\ \mathrm{mol\ l^{-1}}} = 10^{-7}\ \mathrm{mol\ l^{-1}}$$

Therefore, lead(II) sulfide precipitates from acidic solutions and will not significantly dissolve in dilute hydrochloric acid.

However, lead(II) sulfide can be dissolved in hot hydrochloric acid with $c_{\ce{HCl}}=8\ \mathrm{mol\ l^{-1}}$ because of the formation of anionic complexes $\ce{[PbCl3]^-}$ and $\ce{[PbCl4]^2-}$.


Unusual color from leaking battery. What happened? I was removing old batteries from a camera the other day. One of the batteries had leaked something very bright blue. I have seen batteries leak before and typically the colors look a bit like a corroded penny - pale green to white. This time it was a beautiful bright blue; what was it? I initially thought this could have been the metal making the connection was corroding as opposed to the color of the contents of the battery. However, I am not so sure about that.

Note: I believe they were double or triple A batteries and not LiION.

It is, as you suspected, likely a copper corrosion product. Copper compounds are usually blue or green, and are used as pigments. Malachite, http://en.wikipedia.org/wiki/Malachite, is one of the likely corrosion products of an alkaline battery and copper terminals.

BTW, you may need to rinse the battery compartment and use a file to remove the corrosion from the contacts to make connection.


Why does my blood taste like rust? I thought it is just me, but when I searched it on Google, it revealed that there are many people who experience this:
Why is it that whenever I taste my own blood, I always think it tastes like rust?  I have never eaten or tasted rust, so how can I relate something to it?

What you are calling "taste" is actually produced by the olfactory sense – "smell."  True taste, which requires contact of a substance with your tongue, is limited to sensations of sweetness, sourness, saltiness, bitterness, and umami.  The chemistry of taste is certainly quite interesting, but anything beyond those "flavors" is produced by smell (via far more complex physiological processes).  You have certainly smelled rust (or, more accurately, various oxidation states of iron that exist in blood, likely from wet or weathered steel).  So you are, in fact, correctly associating the smell of iron oxides from two different sources: steel and hemoglobin.


Why is phenyl shift preferred over methyl shift in cumene hydroperoxide reaction? In the cumene process, which converts cumene to phenol and acetone, there is a step where cumene hydroperoxide undergoes acid-catalyzed rearrangement:

(Image source: Wikipedia)

Why does the phenyl shift happen preferentially? The methyl group is smaller and should have a higher migratory aptitude.

Despite the mechanism shown in your question, which is commonly used to depict a phenyl shift, phenyl shifts occur by a slightly different mechanism.

Phenyl shifts do not usually happen by the normal 1,2-sigmatropic "hop". Instead, the phenyl group shifts by more of a "walk". It's not really a sigmatropic rearrangement. First, one of the $\pi$-bonds in the phenyl group can attack the electrophilic center. This leads to formation of a spirobicyclic intermediate where the cation is delocalized throughout the rest of the ring. Then, the three-member ring breaks open at the expense of a carbon-carbon bond to rearomatize the phenyl ring and form the new resonance stabilized cation.


Using Arrhenius equation to calculate the ratio of two reaction rates For two reactions,

  $$\begin{align}
\ce{A ->[$k_1$] B} \tag{1} \\
\ce{A ->[$k_2$] C} \tag{2}
\end{align}$$

  you have determined that the activation energies are $\pu{71.9 kJ/mol}$ for $(1)$ and $\pu{142.8 kJ/mol}$ for $(2)$. If the rate constants are equal at a temperature of $\pu{321 K}$, at what temperature will $k_1/k_2$ be equal to $2$?

I know I have to use the Arrhenius equation

$$k=A\exp\left(-\frac{E_\mathrm A}{RT}\right),$$

and I've tried solving for $T$ after filling in my data into the formula as ratios, ie $k_1/k_2$, $E_\mathrm{A}^{(1)}/E_\mathrm{A}^{(2)}$, etc, but I can't seem to get a coherent answer. Any guidance on figuring this out would be much appreciated.

To a first approximation, the activation energy of a single given reaction is a constant at all temperatures (see reasons here, although more precisely, there is some temperature variation). However, the activation energies for both reactions need not be the same.

If you consider both reactions at $\pu{321 K}$,

$$\begin{align}
\ln k_1 &= \ln k_2 \tag{1}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{(\pu{321 K})R} &= \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{(\pu{321 K})R} \tag{2}\\[4pt]
\ln A_1 - \ln A_2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{(\pu{321 K})R} \tag{3} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{321 K})(\pu{8.31446 J K-1 mol-1})} \tag{3} \\[4pt]
&= -26.56 \tag{4}
\end{align}$$

Now, denote the desired temperature at which $k_1  = 2k_2$ as $T^*$. At this new temperature, we have

$$\begin{align}
\ln k_1 &= \ln 2 + \ln k_2 \tag{5}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{T^*R} &= \ln 2 + \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{T^*R} \tag{6}\\[4pt]
\ln A_1 - \ln A_2 - \ln 2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{T^*R} \tag{7}
\end{align}$$

Here we can reuse the previously calculated value for $\ln A_1 - \ln A_2$, and we can also plug in the same activation energies:

$$\begin{align}
T^* &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{R(\ln A_1 - \ln A_2 - \ln 2)} \tag{8} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{8.31446 J K-1 mol-1})(-26.56 - 0.693)} \tag{9} \\[4pt]
&= \pu{313 K}. \tag{10}
\end{align}$$


Moist Chlorine & Hydrogen Psychrometric or Mollier Chart In our electrolysis of brine, $\ce{Cl2}$ and $\ce{H2}$ are produced as gaseous products along with aqueous $\ce{NaOH}$. In this scenario I require what amount of maximum moisture can $\ce{Cl2}$ and $\ce{H2}$ hold up at specific temperature & pressure (precisely 85 celsius & 1.2bar absolute). Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.

I have searched hard but without proper psychrometric data am unable to even estimate preliminary values. Link to data or other alternatives are highly welcome.

regards,

I have searched hard but without proper psychrometric data am unable to even estimate preliminary values.

You have to google for 'vapor pressure' of water and ideal gas model. In cases ideal gas model is close enough to true, the total pressure is divided into partial pressures, attributed to different gases, and the partial pressures are proportional to amount of molecules in the gas mixture. So, in the case the equilibrium partial pressure of water vapors above the liquid can be found in any table of vapor pressures, the rest of the pressure would be other gases. In the case of vapor above saline water, the equilibrium pressure is reduced by the fact that the salt is present. The exact reduction is guarded by Raoult's Law and is linearly proportional to the concentration of the salt.

  Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.

To dehumidify these gases effectively, if this is really required, consider bubbling them through layer of concentrated $\ce{H2SO4}$, it is a quite common and easy way to dehumidify gases.


Nucleophile in a reaction How can we determine which atom is a nucleophile in a compound in a reaction?

I know we see resonance, but I am little confused between electronegativity, formal charge, solvent, and size.

Here are some rules i remember from organic chemistry in Collage.
When trying to distinguish between nucleophile and base,

Look for resonance
If atom is involved in resonsance, this makes ato/group more nucleophilic,
if electrons are more localized its more basic.
Eg. A carboxylate is a nucleophile, the -ve charge is delocalized (i.e - charge is  spread out across carbonyl and oxyanion. Compared to nitrogen on an amine,  electron are restricted.

Sterics,
more sterics less nucleophilic.
Smaller radius atoms, valence electrons are more localized in space, thus more basic. Big radius atoms, poor localization relative to small, thus more nucleophilic.
More electronegastive, less basic
Electronegativity is a measure of the nuclear positive charge effect on near -ve charges.
The more electronegative the less basic. Because the atom doesnt want to give up its electrons. Why? Because it has high electronegativity
Conduction effects
If there are electronegative atoms within 3 bonds (taking into account electronegativities, and proximity) more nucleophilic less basic.

Nitrogens are usually bases, oxygens nucleophiles. I swear it was always like that. Halogens are weak nucleophiles.

Will edit for legibility when i get to computer


Bent's rule for Sulfur trioxide and Sulfur dioxide How is Bent's rule applied on $\ce{SO3}$ and $\ce{SO2}$?

Bond length of $\ce{SO3}$ < $\ce{SO2}$. Is this a direct consequence of applying the Bent's rule?

Here are structures of dominant resonance structure (only one of 3 and 2 differencing only by rotation) for sulfur tri- and dioxide. Despite common misconception, d-orbitals apaprantely play very insignificant role in chemistry of p-block elements, so the structures include formal charges.

As can be seen, the average order of sulfur-oxigen bond lowers with move from di- to trioxide from $1.5$ to $1.(3)$. However, average formal charges grows from $+1$ to $+2$ on sulfur and $-0.5$ to $-0.(6)$,  on oxygen, thus greatly increasing ionic bonding (roughly from $1*0.5 = 0.5$ to $2*0.(6)=1.(3)$) Generally, ionic bonding greatly outperforms covalent bonding, and here charges grown considerably so bond shortening may be attributed to that. The point, however, is, that in this particular structure more than one effect is in play, and their effects outperform whatever small effects are considered when Bent's rule is applied.


"Editing" atom nucleus First of all. I am really bad at chemistry and all that. So this might or more likely will sound at least primitive... But this question has been bothering me for far too long.

So, I was reading about an atom. How all different kinds of it differ by the number of protons in its nucleus. And that gave me this utopian thought: maybe in far (like, really far) future we will have some sort of super "printers" which could take anything as their "ink" and somehow turn that anything (a brick for example) into manageable atoms.

Then it probably should identify each atom, by counting its protons and edit it by "deleting", "adding" or separating needed number of the protons and neutrons.

So my question is... Do you think this is theoretically possible?
Are there any known ways of changing the number of protons?

*Edit: as it turns out, there is a talk going on about this kind of utopia: https://www.youtube.com/watch?v=RzgVWpa4fzU

Big thanks for your answers! I wasn't expecting to receive any answers :)*

In part, your vision already exists and uses electrons as "ink". In a process kown as electron capture, a proton "swallows" an electron of sufficient energy and is thus converted to a neutron.

$$\ce{^{A}_{Z}X + e- -> ^{A}_{Z-1}Y + \nu_{e}}$$

As a consequence, the atomic mass number A remains the same, but the atomic number Z is decreased by one: Element X is converted to element Y.

A typical example would be the conversion of aluminium to magnesium:

$$\ce{^{26}_{13}Al + e- -> ^{26}_{12}Mg + \nu_{e}}$$


Net ionic equation with only aqueous ions When you turn a balanced equation into a net ionic equation, you ignore all the ions that are spectator ions (aqueous on one side as well as the other). You only write the solid and the aqueous ions that form that solid.

So, what happens when you try to write a net ionic equation if ALL the ions are aqueous? Would you write down every ion in the reactants/products, or write nothing?

If you have a balanced equation like the following:

$$\ce{NaCl_{(aq)} + KBr_{(aq)} -> NaBr_{(aq)} + KCl_{(aq)}}$$

and the solubility products for all the species are big enough that there's essentially no solid, there's really no distinction between the left and right sides of the equation. It's a balanced equation, but nothing has changed. We started with $\ce{Na+_{(aq)} + K+_{(aq)} + Br^{-}_{(aq)} + Cl^{-}_{(aq)}}$ in solution and we still have that on the right side. So there's really not a net ionic equation because there's no net change.


Getting the molality of a solution given molarity and density If a solution of $\ce{HNO3}$ has a molarity of 16 and a density of
  $1.42\ \rm{g/mL}$, what is the molality of the solution?

Well, a molarity of 16 means that there are 16 moles of solute for every liter of solution.

Based on density, if we have 1000 milliliters, we have 1420 grams of solution.

To calculate the molality I need:

$$\frac{16}{\text{kilograms of solvent}}$$

Since we have 16 moles, and the molar mass of $\ce{HNO3}$ is 63, we know that we have 1008 grams of solute, meaning there are 412 grams of solvent.

So I calculate the molality to be 0.038.

This answer is wrong because it is not among the options. What did I do wrong?

If a solution of $\ce{HNO3}$ has a molarity of 16 and a density of
    $1.42\rm{g/mL}$, what is the molality of the solution?

Molarity is an old word for amount concentration and it is defined by the IUPAC Goldbook as

  Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by  B. The common unit is mole per cubic decimetre ($\mathrm{mol\,dm^{−3}}$) or mole per litre ($\mathrm{mol\,L^{−1}}$) sometimes denoted by $\mathrm{M}$.

In mathematical terms $$c(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{V(\ce{HNO3(aq)}}.$$

   In this case the concentration is presumably given as $c(\ce{HNO3(aq)})=16~\mathrm{mol\,L^{-1}}$

The molality, $b$, is defined by the IUPAC Goldbook as

  Amount of entities of a solute divided by the mass of the solvent.

In mathematical terms this means
$$b(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{m(\ce{H2O})}.$$

The (mass) density, $\rho$, is defined by the IUPAC Goldbook as

  Mass of a sample or body divided by its volume.

In mathematical terms this means
$$\rho(\ce{HNO3(aq)})=\frac{m(\ce{HNO3(aq)})}{V(\ce{HNO3(aq)})}.$$

   In this case the density is given as $\rho(\ce{HNO3(aq)})=1.42~\mathrm{g\,mL^{-1}}.$

Now you can go ahead and derive the formula for conversion all by yourself, which I recommend, but you can also check if someone has already done that. I chose to be lazy today, so I checked Wikipedia:
\begin{align}
c &= \frac{\rho b}{1 + bM} &
b &= \frac{c}{\rho - cM}
\end{align}

Now you only have to figure out the molar mass of nitric acid, and you can calculate the result.

   I assume $M(\ce{HNO3})=63~\mathrm{g\,mol^{-1}}$.

Solution:

   You should find $b(\ce{HNO3(aq)})=38.8~\mathrm{mol\,kg^{-1}}$.

So in principle you have done quite good. If you would have used proper notation and included all the units in every step, you would maybe have noticed the mistake all on your own.

(Hover over the hidden fields to reveal their content.)


Relation between activation energy and temperature sensitivity In my reaction engineering textbook (Levenspiel), it is stated that reactions with high activation energies are very temperature sensitive, while reactions with low activation energies are relatively temperature-insensitive. This principle was then applied to a homework problem of ours where for two reaction pathways in competition, at high temperature, the reaction rate of the higher activation energy pathway would dominate, with the opposite being true at low temperatures.

Could someone explain why the high activation pathway dominates at high temperatures? In my mind, I still expect the rate of the low activation energy pathway to dominate due to path of least resistance principle.

In chemicals reactions that have more than one product, being one of them the kinetic product and the other the thermodynamic product, the temperature dependency is a relevant factor. For example, the kinetic vs. thermodynamic control can manipulate the regiochemistry of the alkylation of enolates. In other words, you have a competition between reactivity and stability.

Reactions are usefully to describe it in terms of potential energy diagrams, $\Delta G$ vs. reaction coordinate ($\xi$), where a one-step reaction have a global maximum corresponding to a transition state leading to the product and the free energy of this transition state, relative to the reactant, determine only the rate of reaction. The enthalpy part of $\Delta G$ deal with the stability of the molecule or intermediate but this doesn't give information about the rate of reaction.

Absolute rate theory (or transition state theory), by Eyring, have an expression for the rate constant $k$:

$$k=\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}e^{-\Delta H^{\ddagger}/RT}$$

where $\kappa$ is the transmission coefficient, $k_{B}$ the Boltzmann's constant, $h$ the Planck's constant, $R$ is the gas constant and $T$ is the absolute temperature. Here, the term $\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}$ varies slightly with temperature compared to the enthalpic term. Arrhenius equation also reflects the temperature dependence of reactions:
$$k=Ae^{-E_{a}/RT}$$
For remark, both relationships are temperature dependent.

So, if you have a reaction under kinetic control, the ratio of the products is determined by the relative energies of the transitions states. Whereas under thermodynamic control, the ratio of the products is determined by the relative energies of the products. This leads to the question if there are any relationship between $\Delta G$ or $\Delta G^{\ddagger}$, where you might search about Bell-Evans-Polanyi formula, Hammond's postulate and the Marcus equation.

Suppose you have a reaction
$$\text{A}\leftarrow\text{R}\rightarrow\text{B}$$
with the energy profile:

The product A have a lower $E_{a}$ or $\Delta G^{\ddagger}$ (kinetic product), whereas B have a higher $E_{a}$ but a lower $\Delta G$. The potential energy profile generally is temperature independent, i.e. the barrier and the heat of reaction are determined by the molecular structure. When you raise the temperature, the kinetic energy of the molecules is raised and this helps the molecules to pass the potential energy barrier more easily (you can check it by a Maxwell-Boltzmann distribution for a molecule changing the temperature and fixing an energy corresponding to $E_{a}$), being obvious that if the barrier is smaller will be more easy to pass.

If the temperature is high enough to make the molecules pass over all the barriers on the energy profile, the molecules will choose the pathway with the most stable product. All of this is determined by whether the products are or aren't able to equilibrate under the reaction conditions (search for the principle of microscopic reversibility). So, at lowers temperature you will disfavor the equilibrium and only get the kinetic product.


Formula of Manganese (IV) Ion When determining the formula of manganese (IV) oxide, since it’s made up of $\ce{Mn^4+}$ and $\ce{O^2-}$ ions, why is the formula $\ce{MnO2}$ and not $\ce{Mn2O4}$?

Any help would be greatly appreciated

For ionic compounds there is really no such thing as a "molecule" of the compound. In the solid state the ions exist in a 3D crystalline structure and there is no unique way to associate a particular $\ce{Mn^{4+}}$ ion with a particular pair of $\ce{O^{2-}}$ ions. Therefore the simplest integer ratio is used as the "molecular formula."


How would I measure the dissolved carbon dioxide content of a liquid (water to start with) I would like to measure the dissolved carbon dioxide content of a sample of water.  I was thinking that conductivity might change with carbon dioxide level, and also pH.  I would like to have the measurement sample electronically, so current, voltage, conductivity / impedance (resistance, capacitance, inductance) would all be fine.  My chemistry goes up to A-level; I am an engineer.

This paper looks helpful: http://www.sciencedirect.com/science/article/pii/S0144860910000683

Quoting from the abstract:

  Many methods are available for the measurement of dissolved carbon dioxide in an aqueous environment. Standard titration is the typical field method for measuring dissolved CO2 in aquaculture systems. However, titrimetric determination of dissolved CO2 in marine water aquaculture systems is unsuitable because of the high dissolved solids, silicates, and other dissolved minerals that interfere with the determination. Other methods used to measure dissolved carbon dioxide in an aquaculture water included use of a wetted CO2 probe analyzer, standard nomographic methods, and calculation by direct measurements of the water's pH, temperature, and alkalinity. The determination of dissolved CO2 in saltwater based on partial pressure measurements and non-dispersive infra-red (NDIR) techniques with a CO2 gas analyzer are widely employed for oceanic surveys of surface ocean CO2 flux and are similar to the techniques employed with the head space unit (HSU) in this study. (...)


Why is nitration of veratrole specific in these positions? Why do we not need $\ce{H2SO4}$ as an acid catalyst for the di-nitration of veratrole?

Why does the 2nd nitro group go to positions 4 or 5, following the directing effects of the methoxy-group? Doesn't the first nitro group, that has been substituted onto the compound, have any effect on its position?

When there are activating and deactivating groups on an aromatic ring, the direction of the electrophilic aromatic substitution is generally controlled by the activating groups.
Think that activating groups improve the substitution in certain positions by increasing its electron density. On the other hand, deactivating groups favor one position over another, not by activating it, but deactivating it less than the other positions. So, the directing effect of a deactivating group is masked by the directing effect of the activating ones, which is stronger.

You most likely don't need the sulphuric acid due to the presence of two activating groups. In contrast, as an example, last nitration step of trinitrotoluene is so disfavored that it has to be carried out in boiling sulphuric acid.


Plotting a calibration curve for Gas-Liquid Chromatography Question:

The given graph of abundance of different samples of butanol with respect to time is-

Textbook solution:

But why aren't the points like this:
$(20,5);(40,10);(60,15);(a,7)$
after which, to plot the curve we can use $y=mx+c$ and then find the value of $a$?

If I remember correctly, the amount of your compound isn't proportional to the peak height but to the peak area.

Back in the days, when plotters were just plotters (=no numerical integration of the peaks), people used to cut out the curves and weighted the paper pieces on a balance to determine the peak area.


Why does adding salt to water help distillation? I've been looking at instructions on how to get lanolin from fleece and a lot of the instructions mention that adding salt to the water before distillation will increase the lanolin yield. Why is this?

I would think that the distillation would separate the water but not the salt leaving the lanolin contaminated with salt.

Good question. We frequently "salt out" semi-soluble organic compounds from water because the ionic nature of the salt is more compatible with the water and less compatible with the (less ionic) organic compound. This is used, for instance, to drive sodium stearate (soap) from water. Adding salt could help drive more lanolin out of the water, or what is probably the situation, making an emulsion of water and lanolin less stable so that you can get more lanolin and less "grunge". The lanolin is then purified by filtering, which would remove salt crystals that formed as the water evaporated. And filtering would remove other dirt.

One description of obtaining lanolin suggests driving off all the water, then purifying by adding olive oil and water to remove dirt. But then it raises the question: why not remove the lanolin before all the water is evaporated? The salt would remain in the water. https://permies.com/t/65837/fiber-arts/extract-lanolin-wool

It's also possible that the water pulled out by the wool extracts a great deal of the salt, leaving the lanolin with only a small portion of the original salt and water (depending on how much water you use in the first place).


How do I calculate the cell voltage using the following half-reactions and solubility products? The concentrations of $\ce{NaCl}$ and $\ce{KI}$ are at $\pu{0.120 M}$ and $\pu{0.090 M}$, respectively, in the following electrochemical cell:

$$\ce{Cu_{(s)}|CuI_{(s)}|I^{–}_{(aq)}||Cl^{–}_{(aq)}| AgCl_{(s)}|Ag_{(s)}}$$

How do I calculate the cell voltage using the following half-reactions:

\begin{align}
\ce{Cu+ + e- &<=> Cu(s) & E^\circ &= \pu{0.518 V}} \\
\ce{Ag+ + e- &<=> Ag(s) & E^\circ &= \pu{0.7993 V}}
\end{align}

The solubility products $(K_{sp})$ for $\ce{AgCl}$ and $\ce{CuI}$ are $\pu{1.8e–10}$ and $\pu{1.0e-12}$, respectively.

I think you need the Nernst equation. This is something we were just introduced to today in my class, so I’m not so sure about the notation you’re using in your question at the top, but you should be able to correct my answer to fit your needs if you understand the equation I’m about to implement.

$$E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{zF}\ln Q$$

Where $E^\circ_\text{cell}$ = what the $E_\text{cell}$ normally is, under STP with the solutes at a concentration of $1\ \mathrm{mol\ dm^{−3}}$.

$$E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}$$

Since, the reduction potential of copper is lower, it will be the reducing agent in this reaction.

$$E^\circ_\text{cell} = -0.7993\ \mathrm{V} + 0.518\ \mathrm{V}$$

$$E^\circ_\text{cell} = -0.281\ \mathrm{V}$$

Where $R$ is the familiar gas law constant:

$$R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$$

$T$ is the absolute temperature (it’s always assumed to be $298.15\ \mathrm{K}$):

$$T = 298.15\ \mathrm{K}$$

$z$ is the number of moles of electrons transferred in the half-reaction:

$$z = 1$$

And $F$ is Faraday’s constant, which is measured in coulombs per mole:

$$F = 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}$$

And $Q$ is the also familiar reaction quotient, but as I discovered, it takes some work to calculate, and you have to use the quadratic equation.

$$Q=\frac{[\ce{red}]}{[\ce{ox}]}$$
$$Q=\frac{[\ce{Cu+}]}{[\ce{Ag+}]}$$

$$K_{\text{sp},\ce{CuI}}=10^{-12}=[\ce{Cu+}][\ce{I-}]$$
$$1.0 × 10 ^{-12} = [X][0.090 + X]$$
Do the quadratic formula
$$X = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$
$$[\ce{Cu^+}] = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$
$$K_{\text{sp},\ce{AgCl}} = 1.8 \times 10^{-10}\ \mathrm{mol\ l^{-1}} = [\ce{Ag+}][\ce{Cl-}]$$
$$1.8 \times 10^{-10} = [X][0.120 + X]$$
Do the quadratic formula
$$X = 1.50 \times 10 ^{-9}\ \mathrm{mol\ l^{-1}}$$
$$[\ce{Ag+}]=1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}$$

$$Q=\frac{1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}}{1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}}$$

$$Q=0.00740$$

So, plugging this all into the Nernst equation, the answer should be:

$$E_\text{cell}=-0.281\ \mathrm{V}-\frac{8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 298.15\ \mathrm{K}}{1 \times 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}}\ln(0.00740)$$

$$E_\text{cell}=-0.155\ \mathrm{V}$$

I should add that since $R$ is a constant, $F$ is a constant, and $T$ is usually something like $298.15\ \mathrm{K}$ or just not measured in elementary chemistry, the equation can just be

$$E_\text{cell}=E^\circ_\text{cell}-\frac{0.0257}{z}\ln Q$$

Which will give the same answer:

$$E_\text{cell}=-0.281\ \mathrm{V}-0.0257\ \mathrm{V}\ln(0.00740)$$
$$E_\text{cell}=-0.155\ \mathrm{V}$$


Selenium oxydichloride bond angle According to me , the $\ce{Cl—Se—Cl}$ bond angle should be greater than $\ce{Cl —Se—O}$
bond angle because since oxygen is more electronegative than $\ce{Cl}$ , the orbital directed towards $\ce{O}$ will have more p character hence, the bond angle should be less than $\ce{Cl—Se—Cl}$ bond angle.

But in my book its written that the $\ce{Cl—Se—O}$ bond angle is greater. Why?

My suspicion is that the info in the book relies on single-crystal x-ray diffraction experiment (1), where geometry of selenium oxydichloride cannot be determined solely from MO principles. So, here is the structure of dioxane-seleninyl-dichloride:

As far as you can see, Se1 atom has three more neighbors: O1 from another $\ce{[SeOCl2]}$ unit, and O2 + O3 from dioxanes. The bond angles correspond to those you discovered in the book, $\angle \ce{Cl1-Se-Cl2} = 95.6(3)$, and $\angle \ce{Cl1-Se-O} = 101.6(7)$:

The reason for that is given in the paper as well:

  APPLICATION of the Gillespie-Nyholm theory to the molecules of
  seleninyl dichloride, $\ce{SeOCl2}$, ... leads to the same prediction
  for each:  pyramidal geometry with the fourth position of a
  tetrahedron occupied by a lone pair. However, more detailed study of
  many non-metal compounds containing lone  pairs has shown that the
  Gillespie-Nyholm approach gives an incomplete description of the
  bonding. The central atoms very frequently form additional bonds, of
  greater length than the primary covalent links. In the case of
  pyramidal molecules, it has been found that there are often three of
  these secondary bonds, completing a distorted octahedron around the
  central atom. Striking conformation of this has come from complexes
  (or solvates) of $\ce{SeOCl2}$. All show $\ce{Se-O}$ and $\ce{Se-Cl}$
  secondary bonds, with distorted octahedral geometry, although in some
  cases the third secondary bond opposite the $\ce{Se-O}$ bond may be
  significantly longer than the other two secondary bonds. Furthermore,
  there are rare examples where a secondary bond has been replaced by
  two longer contacts, and it may be inferred from these results that
  the steric importance of the lone pair of electrons on $\ce{SeOCl2}$,
  varies. In those cases where an octahedron can be observed, the angles
  $\ce{X-Se-Y}$ ($\ce{X, Y}$ = $\ce{O}$ or $\ce{Cl}$) have been found to
  lie in the range 142 -- 173, and there appears to be no clear way of
  deciding which atoms, chlorine or oxygen, are going to form the
  secondary bonds. The present structure of $\ce{SeOCl2 * C4H8O}$, is,
  however, the first in which $\ce{SeOCl2}$, is involved in three
  secondary bonds to oxygen with the $\ce{SeOCl2}$, being classed as
  amphoteric.

Original

Long story short, your assumption would probably be correct for the "idealized" molecule in gas phase, whereas in reality there always will be a distortion as $\ce{As}$ strives to complete unsaturated coordination environment.

(1) Alcock, N. W.; Sawyer, J. F. J. Chem. Soc., Dalton Trans. 1980, No. 1, 115–120. DOI 10.1039/DT9800000115


Can solutions of polar covalent compounds conduct electricity? I learned in class that solutions of polar covalent compounds are weakly conductive, while ionic solutions are strongly conductive.

But I'm getting different answers online. According to this lecture, "They do not conduct electricity in the liquid state, or when soluble in water, do not conduct electricity in aqueous solution."

I saw in lab, however, that a dextrose (D-glucose) solution weakly conducted electricity, and so did that of levulose (D-fructose).

So can solutions of polar covalent compounds conduct electricity? If so, how? Since it's covalent, it's not like anything can dissociate to form ions, right?

Consider for example this lab where polar covalent bonds are said to be strong, weak or non-electrolytes : ELECTROLYTES AND NONELECTROLYTES CHEM 151

The part of the molecule which dissociates has strongly ionic character but the molecule itself is described as being polar covalent. Not not all bonds in the molecular have the same level of covalent character, but most are covalent and it is polar.

There are also other ways for molecules to allow conduction, conjugated systems can do it by a very different process see for example the wiki on Conductive Polymers


Aerobic respiration to produce water Could the aerobic respiration reaction occur outside the body to produce water? If it is possible how would it occur

If you look at the net reaction, it is just combustion of glucose:

$$
\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}
$$

So you could just burn a sample of glucose, and it would produce water.


When we add soap in low concentrations to water it reduces the surface tension. Why? What is the purpose of this? I understand how soaps lower the surface tension of water but I don't know what the purpose of this is. How does this help with cleaning, removing stains etc?

The lowering of surface tension is a side effect, though it does relate to the reason why we use soaps when cleaning things.

Many of the things we want to clean are not naturally soluble in water (e.g. things like vegetable oil). The structure of soaps and detergents usually consist of molecules with two parts: one is water-loving and one is oil-loving. When soaps are added to mixtures of oily things and water, they align so the oil-loving parts of the molecules surround the oil and present only a water loving surface to the bulk water. Thus the oil is made much more soluble than it would be otherwise (so it is possible to remove it from your clothes, hands or dishes despite most of the liquid being water).

In reality the situation is quite complicated but the general principles of how soaps and detergents work is captured by this basic view.

The effect on surface tension is related to this. The soap molecules tend to cluster at the surface of the water (to get their oil-loving ends away from the water environment). This breaks up the surface structure of water molecules that normally create strong surface tension. But this effect is basically the same effect that enables soaps to solubilise normally insoluble things in bulk water.


How does simulation software work at a fundamental level? There are lots of software packages like Schrödinger Maestro or Argus Lab, which are used for simulations of docking.

I have always been fascinated as to how such software works. What is the fundamental principle working "under the hood" of this kind of sofware?

Let's say I wanted to make a simple piece of software which, when I input the atoms and atom number (e.g., C1H4) can be used for finding different energy levels (or other properties) in a molecule such as methane ($\small\ce{CH4}$).

Where does one have to start in attempting to make such a package, given a considerable knowledge of programming? What theory is applicable?

Well, that's a rather diffuse question. From what I have seen, the software packages you mention do Molecular Mechanics (MM), which is a treatment of molecules with classical physics. For each atom in a given molecule, the forces acting upon it due to chemical bonding with other atoms is calculated using a set of empirically determined parameters (a force field), and then an estimate of the potential energy is provided. But given that this question is tagged with "Computational Chemistry", I also want to mention quantum chemistry, i.e. the application of quantum mechanics to atoms and molecules. This field is both far more interesting and far more difficult than the classical MM treatment.

A first shot would be the Wikipedia article about Computational Chemistry which already points you to other relevant topics, i.e. the article about Molecular Mechanics, which is quite concise about the topic (although I don't know how exhaustive it is in this regard, since I don't deal with MM in my research). For quantum chemistry, the necessary knowledge is more spread-out; relevant Wikipedia articles would be Eigenproblems, the Schrödinger equation and Ab initio methods. I don't know any literature about MM, but if you want to look deeper into quantum chemistry, I would recommend the following two books (and lots and lots of time and patience for reading them):

Attila Szabo, Neil Ostlund: "Modern Quantum Chemistry". Provides a good mathematical introduction and thorough treatments of the Hartree-Fock and Moller-Plesset methods. It's cheap, also.
Ira Levine: "Quantum Chemistry". Covers a wider range of methods, but is in my opinion a bit more difficult to consume than Szabo/Ostlund.


Greater bond distance for silicon diamond structure than for carbon diamond structure? In the silicon diamond structure the bond distance is about two times the bond distance of the carbon diamond structure. Is this caused only by the greater dimension of $\mathrm{Si}$ atoms with respect to $\mathrm{C}$ atoms? Or are there other reasons for a greater bond distance in the case of silicon?

Yes, both structures belong to the same diamond cubic type, and the interatomic distances for diamond and silicon (you can use "Measurements" tool from the JSmol menu)

$$
\begin{align}
d(\ce{C-C}) &= \pu{1.54 Å} \\
d(\ce{Si-Si}) &= \pu{2.35 Å} \\
\frac{d(\ce{Si-Si})}{d(\ce{C-C})} &\approx 1.53
\end{align}
$$

are in a good agreement with the corresponding covalent radii

$$
\begin{align}
r_\mathrm{cov}(\ce{C}) &= \pu{0.76 Å} \\
r_\mathrm{cov}(\ce{Si}) &= \pu{1.11 Å} \\
\frac{r_\mathrm{cov}(\ce{Si})}{r_\mathrm{cov}(\ce{C})} &\approx 1.46
\end{align}
$$

which would set the silicon atoms further apart from each other.


Can HI/P reduction be used to reduce nitroalkanes and nitroarenes to alkyl amines and aryl amines? Can $\ce{HI/P}$ reduction be used to reduce nitroalkanes and
  nitroarenes to alkyl amines and aryl amines ?

I learnt from my textbook that heating the nitroalkanes and nitroarenes with $\ce{H2}/\ce{Pt}$ is sufficient for the above conversion. But I am not sure whether $\ce{HI}$ and red $\ce{P}$ reduction will give the same products. Also, I could not find related information about this on the Internet.

I've never seen the $\ce{HI/P}$ method being used for the reduction of aliphatic or aromatic nitro compounds.

  I learnt from my textbook that heating the nitroalkanes and nitroarenes with $\ce{H2/Pt}$ is sufficient for the above conversion.

I've reduced a lot of nitroarenes to their corresponding amines by catalytic hydrogenation under standard conditions in ethanol (or methanol) over $\ce{Pd/C}$. In the cases I examined, heating wasn't necessary.


Optical Isomerism in Coordination Compounds of type MA₂B₂(CC) I was asked the following question:

How many optical isomers are there for $\ce{[Cr(NH3)2Cl2(en)]^+}$ ?

First of all I figured out the following four geometrical isomers of the above complex:

Out of which, (iii) and (iv) are optically inactive, due to a plane of symmetry (shown by faint line) passing through them as shown below:

For (iii), you can suppose: green as $\ce{en}$, orange as $\ce{NH3}$ and red as $\ce{Cl}$. Similarly, for (iv), you can suppose: green as $\ce{en}$, orange as $\ce{Cl}$ and red as $\ce{NH3}$.
Therefore only (i) and (ii) will show optical isomerism, which I've confirmed myself.

But, the following four isomers was provided as answer, in which I don't think that the second configuration of complex will show any optical activity due to plane of symmetry (as discussed above for the (iv) geometrical isomer).

What I am doing wrong?

I came to the same conclusion as you, total 4 stereoisomers (2 of them create one enantiomeric pair).

In the 3rd party solution, they depict the same isomer at bottom, i.e. only 3. The missing one is that with two $\ce{NH3}$ ligands in the same plane with $en$, and $\ce{Cl-}$ ligands in straight line (trans), perpendicular to that plane (not that horizontal, which is depicted)

Here is my simplified schematic representation of the octahedral complex with ethylenediamine ligand bound by back-pointing bonds in horizontal plane, hidden; the rest is vertical axis and two front-pointing horizontal bonds (seesaw geometry, not to be confused with Fischer projection for tetrahedrals!) Dotted line is mirror plane, equal sign means identity (by 180° rotation around observer-to-model-line axis).

has B ligands in the (en)-defined plane, A ligands perpendicular (axis)
(it's their bottom left isomer, idenitical with bottom right, for A=$\ce{NH3}$, B=$\ce{Cl}$; note that they have en oriented vertically, everywhere, with one end in dotted horizontal plane, other one in upper vertical position, I have it somewhat simpler, rotated for en to be horizontal);
has the opposite, A's and B's swapped (and it's not mirror image to 1.)
(it's their missing isomer);

If A and B are in the plane then there are two possibilities how to relatively place A and B in the perpendicular axis (resulting in mirror images),

(it's their top right);
(it's their top left).