How to determine a Reynolds number for mixture of 2 liquids I'm experimenting with synthesis in microreactors and I'm using 2 reactants. I'd like to estimate the Reynolds number of the flow containing two miscible fluids. I have the values of their properties (density, viscosity, flowrate).

Do I just average both property values for both liquids or is there a more intricate relation?

Thank you very much in advance for your solutions and suggestions!

The Reynolds Number is the ratio of inertial to viscous forces and is calculated according to this equation: $$Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density, $\mu$ is the dynamic viscosity, $v$ is the fluid velocity and $L$ is a characteristic length that depends on the flow being modeled. When estimating the Reynolds number of a mixture, the term $v L$ would not change but we need to make the assumption that the term $\frac{\rho}{\mu}$ of the mixture can be derived from the properties of the pure components.  This assumption will be valid if

The components have similar structure
The components are non-polar
The components are not associated
The difference in the kinematic viscosities is small (~15 cP)

Some of these conditions could be relaxed if one species is in significant excess.

Densities are additive so long as the volume of the mixture equals the sum of the volumes of the pure solutions.  (Search for volume weighted densities for more information.)  Viscosities of mixtures is a much harder thing to estimate.  One method is the Grunberg-Nissan equation: $$Log\eta_m=x_1Log\eta_1+x_2Log\eta_2+x_1x_2d$$ Where $\eta_i$ is the dynamic viscosity of component $i$, $x_i$ is the mole fraction of component $i$ and $d$ is an interaction coefficient.  If you assume $d=0$ then you have the Arrhenius equation for the viscosity of a mixture.

So, at the end of the day, if you state you are working under the assumptions that volume weighted densities equal mass weighted densities and use the Arrhenius equation to determine the viscosity of the mixture then you can average the weighted properties of the pure components to get an estimate of the mixture.

Just a word of caution under these assumptions you should take the mass weighted average of the densities but the mole weighted average of the viscosities.

I relied on this reference for much of this answer.  It will require a subscription to the service to access the PDF, but some of it can be found on google books.


Volume in decimetres cubed dimensional analysis In chemistry class, we were learning about concentration, which is the number of moles of a substance per decimeter cubed of water, expressed as $\mathrm{mol} \cdot \mathrm{dm}^{-3}$.

However, I was confused when my teacher wrote decimeters cubed as $\mathrm{dm}^{-3}$ instead of what I presumed would be $\mathrm{dm}^3$. I asked why and he didn't know. Why do chemists write it like this?

This touches a very powerful analytical technique called unit analysis (aka dimensional analysis).

Moles per decimetre cubed is $\dfrac{\text{moles}}{\text{dm}^3} = \text{moles}\cdot \text{dm}^{-3} \ne \text{moles}\cdot \text{dm}^3$

So the -3 indicates that the unit is in the denominator not the numerator.

PS - I learned chemistry when there were only four elements - earth, wind, water and air. I think of liters not $\text{dm}^3$.


aldehydes are easily oxidised but ketones are not Aldehydes are easily oxidized to carboxylic acid but ketones are difficult to oxidise. Most books say that the hydrogen directly bonded to the $\ce{C=O}$ in the aldehyde is what aids the oxidation process. However, can someone please tell me why this is so.

In the oxidation of aldehyde, the hydrogen attached to carbonyl compound is replaced by an $\ce{-OH}$ group resulting in the formation of carboxylic compound. In case of ketones, bulkier alkyl groups ($\ce{-R}$) should be replaced by an $\ce{-OH}$ group to form carboxylic compound.

We know that hydrogen abstraction is much easier than that of bulkier alkyl groups. Likewise, after the oxidation of the aldehyde, the carbon chain containing the $\ce{C=O}$ group remains same but due to loss of $\ce{-R}$ group during oxidation, the length of the carbon chain decreases in ketones.


Why do we use the word phase and state of matter so interchangeably? I would be happy if someone explains me why the word "phase" and "state of matter" is used so interchangeably.Lets say we have HBr and another compound in the system both gases.Then am I to take two phases on a single phase when I plug in the Gibbs formula in Phase rule?

From the wikipedia page for phase (matter)

a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform [...] The term phase is sometimes used as a synonym for state of matter, but there can be several immiscible phases of the same state of matter.
A simple description is that a phase is a region of material that is chemically uniform, physically distinct, and (often) mechanically separable. In a system consisting of ice and water in a glass jar, the ice cubes are one phase, the water is a second phase, and the humid air is a third phase over the ice and water. The glass of the jar is another separate phase.

In the above system you can see that the ice and the glass jar are the same state of matter, but are two distinct phases. It's the same with oil and water. So, people who are being precise don't use the two terms interchangeably.


Equivalent weight of the anhydrous acid The cupric salt of a monobasic acid contains $3$ molecules of water of hydration per atom of $\ce{Cu}$. $\pu{1 g}$ of hydrated salt yielding on strong heating $\pu{0.3306 g}$ of $\ce{CuO}$. What is the equivalent weight of anhydrous acid?

My approach:
Let the formula of the acid be $\ce{HE}$ (where $\ce{E}$ is some element). So, the formula of the salt is $\ce{CuE2*3H2O}$

Don't know how to proceed from here. Any help would be appreciated.

Let the anhydrous acid be $\ce{HA}$ and assume its Molar Mass to be $X$. There is no a priori reason to assume $\ce{A}$ is a mere element.

The copper salt in question is $\ce{CuA2.3H2O}$.

Its Molar Mass is:

$$63.55+2\times(X-1)+3\times 18=115.55+2X$$

Assume reasonably and without losses, that:

$$\ce{CuA2.3H2O(s) -> CuO(s)}$$

So that $1\ \mathrm{mol}$ of the hydrated salt decomposes to $1\ \mathrm{mol}$ of cupric oxide.

The Molar Mass of $\ce{CuO}$ is $79.545$, so $0.3306\ \mathrm{g}$ of it is:

$$\frac{0.3306}{79.545}=4.156\times 10^{-3}\ \mathrm {mol}$$

So now we know that $1\ \mathrm{g}$ of the hydrated salt is $4.156\times 10^{-3}\ \mathrm {mol}$, so its Molar Mass is:

$$\frac{1}{4.156\times 10^{-3}}=240.6\ \mathrm{g/mol}$$

And with the above:

$$115.55+2X=240.6$$

Solve for $X$.

As the acid is monobasic, its Molar Mass is also its Equivalent Mass.


How to get the efficiency of a heat engine which undergoes an elliptical cycle? An ideal diatomic gas undergoes an elliptic cyclic process characterized by the following points in a $PV$ diagram:
$$(3/2P_1, V1)$$
$$(2P_1, (V1+V2)/2)$$
$$(3/2P_1, V2)$$
$$(P_1, (V1+V2)/2)$$

A rough sketch:

This system is used as a heat engine (converting the added heat into mechanical work).
Evaluate the efficiency of this engine

We know that the efficiency is defined as the benefit/cost ratio:
$$e = \frac{W}{Q_h}$$
Let's focus first on the work done by the engine; taking into account the quasistatic approximation, $W=PV$. Then:
$$W = (P_2 - P_1)(V_2 - V_1)$$
Note that from the given points we can guess that $P_2 = 2P_1$. Then:
$$W = P_1(V_2 - V_1)$$
Now let's focus on $Q_h$. I have the following issue here: none of the 4 steps of the cycle has either P or V constants. This means that the strategy of using:
$$Q = nc\Delta T$$
Won't work because you cannot use neither $c_p$ nor $c_p$.
However, when we deal with a rectangular cycle:

It would be really easy to derive expressions for both $Q_a$ and $Q_b$ and then the efficiency for the system would be obtained. That is because in each step $Q_h$ is added, either $P$ or $V$ are constant (and thus $Q = nc\Delta T$ works).
What to do with the elliptical cycle to get $Q_h$?
EDIT
My bad, the work done by the working substance is the area under the PV graph. So as Chet Miller pointed out, the work is:
$$W = \pi (P_2 - P_1)(V_2 - V_1)$$
I have been trying to solve the heat equation so that we get the two angles.
So what I did was deriving $P$ and $V$ wrt the angle:
$$dP = (P_{max} -P_0)\cos \theta d\theta$$
$$dV = (V_{max} -V_0)\sin \theta d\theta$$
And plugging it into 3:
$$0=[-(C_v+R)P(V_{max} -V_0)\sin \theta+C_VV(P_{max} -P_0)\cos \theta]$$
This above equation is satisfied by two $\theta$ angles. But how to solve it?

Let $(V_0,P_0)$ represent the coordinates of the center of the ellipse, and let $P_\mathrm{max}$ and $V_\mathrm{max}$ represent the maximum pressure and maximum volume, respectively over the cycle.  Then, the work is the area of the ellipse, which is given by:  $$W=\pi(P_\mathrm{max}-P_0)(V_\mathrm{max}-V_0)$$This is, $\pi$ times the product of the semi-major and semi-minor axes.  This differs from the result which you gave.

The shape of the ellipse can be represented parametrically in terms of the angle $\theta$ around the cycle, assuming $\theta$ is measured clockwise from the point $(V_0-(V_0-V_\mathrm{max}),P_0)$:
$$P=P_0+(P_\mathrm{max}-P_0)\sin{\theta}\tag{1a}$$
$$V=V_0-(V_\mathrm{max}-V_0)\cos{\theta}\tag{1b}$$
Application of the first law of thermodynamics to the working fluid over a differential portion of the cycle gives us:
$$dU=nC_vdT=dQ-PdV\tag{2}$$But from the ideal gas law, $$nRdT=d(PV)$$Substitution of this into Eqn. 2 yields:
$$\frac{C_v}{R}d(PV)=dQ-PdV$$
So the differential heat added during an arbitrary portion of the cycle is given by:$$dQ=\frac{1}{R}[(C_v+R)PdV+C_VVdP]\tag{3}$$So the differential heat is zero when dQ = 0, or, equivalently, when $d\ln{P}+ d\ln{V^{\gamma}}=0$, or, equivalently when $PV^{\gamma}=\mathrm{Const}$.  This is the equation for an isentropic line tangent to the ellipse.

There are two angles $\theta$ at which adiabats are tangent to the ellipse.  These two angles can be obtained by substituting Eqns. 1 into Eqn. 3 with dQ = 0.  Once these angles have been determined, the total positive heat added Q during the cycle can then be obtained by integrating Eqn. 3 with respect to $\theta$ between the two angles.


Dealkalization of water I want to remove Alkalinity from tap water by using an acid. The starting point is about 370pm alkalinity, and I am wondering to reduce this to about 50ppm, could I use a weak acid like citric acid? Or do I need to use a strong acid, such as HCl?

Roughly speaking, any acid is equally effective if its pKa lies below the pH 7 you want for a neutral solution.  It does not have to be HCl at all, in fact citric acid works for all three of its commonly observed deprotonations (pKa = 6.4 for the third deprotonation and of course lower for the first two; reference).  So you may use citric acid assuming three equivalents per mole and you should get where you want to go.

As a matter of taste I prefer citrate in my tea versus chloride, anyway.


The toxicity of carbon monoxide in terms of its interaction with haemoglobin On the basis of its interaction with haemoglobin, explain the toxicity of carbon monoxide.

I understand that ligand substitution is involved and carboxyhaemoglobin is formed meaning that the haem group in the haemoglobin molecule cannot form a ligand with the oxygen molecule, but i would like to know more in depth in terms of its chemistry, why this is the case.

Carbon Monoxide and dioxygen both reversibly bind to the iron atom in hemoglobin.

The two molecules compete for the same binding site.

So if both $\ce{O2}$ and $\ce{CO}$ are present, some of each will bind, but for a given amount of $\ce{O2}$, less $\ce{O2}$ will bind if $\ce{CO}$ is present. The affinity of the Fe for $\ce{CO}$ is much greater than for $\ce{O2}$, so a lower concentration of $\ce{CO}$ results in a given saturation level of hemoglobin compare to $\ce{O2}$.

The same would be true for cyanide which is isoelectronic with carbon monoxide.

A more complete explanation needs to go into the cooperation between the four hemoglobin subunits.  $\ce{CO}$ bound to two of the four subunits causes $\ce{O2}$ to being too tightly bound in the other subunits.  That the oxygen can not be released from the hemoglobin, is another aspect of toxicity.


Is converting SMARTS to SMILES a "lossless" operation? The following three different SMARTS strings represent the same structure - at least when depicting it (e.g. at Uni Hamburg Smarts viewer).

SMARTS:
[#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8]
[#6]-1-[#6]=[#6]-[#6](-[#6]-[#6]-1-[#6])=[#8]
[#6]-1-[#6](-[#6]=[#6]-[#6]-[#6]-1-[#6])=[#8]

Also, the chemical structure represented in SMILES leads to only one string, at least only one I can think of.

SMILES:
CC1CC=CC(=O)C1

Now, I do know the differences between SMARTS and SMILES, and I do know what the symbols in SMARTS mean. Still, I tried several softwares to find any differences in the use between SMARTS and SMILES representation for that specific case, and couldn't find any.

So I wonder if I could use SMILES in that case, or if there's any danger of missing information, for example when searching for substructures?

SMARTS is deliberately designed to be a superset of SMILES. That is, any valid SMILES depiction should also be a valid SMARTS query, one that will retrieve the very structure that the SMILES string depicts.

However, as a query language, SMARTS can be more general than SMILES is. For example, CC as a SMILES string depicts a single compound: ethane. As a SMARTS query, though, CC will match ethane, but will also match propane, acetic acid, cyclohexane, vancomycin, etc.

There's also SMARTS strings which are not valid SMILES strings. You list several in your question: [#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8] is not a valid SMILES representation, unless your SMILES parser is being particularly generous. Even if it is, [#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8,#7] is also a valid SMARTS query which should match your molecule, but not even a generous SMILES parser is likely to accept it.

Despite being similar, SMARTS and SMILES are intended for fundamentally different things - the purpose of SMILES is to represent particular compounds, whereas SMARTS represents a query against a range of possible molecules, or an abstract description of a set of possible molecules. As such, they're not inter-convertible, even if the strings are literally identical.

For your particular case, yes, CC1CC=CC(=O)C1 is both a valid SMILES and a valid SMARTS, but as a SMARTS query, it represents not just 5-methyl-2-cyclohexen-1-one, but also 5-propyl-2-cyclohexen-1-one and 3-hydroxy-5-butyl-6-amino-2-cyclohexen-1-one, as well as many others, all of which contain that substructure. The SMARTS viewer you link doesn't depict this explicitly, because it's implicit in the use of SMARTS that it's a substructure pattern for a broader class of compounds.


What would make guanidinium ions and urea more effective denaturants of proteins in some cases, but not in others? Some proteins do not unfold, or only partially unfold due to chaotropic agents such as guanidinium ions and urea. What are some reasons for this?

Chaotropic agents increase the solubility of nonpolar substances in
  water. Consequently, their effectiveness as denaturing agents stems
  from their ability to disrupt hydrophobic interactions, although the
  manner in which they do so is not well understood. Conversely, those
  substances  that stabilise proteins strengthen hydrophobic forces,
  thus increasing the tendency of water to expel proteins. This accounts
  for the correlation between the abilities of an ion to stabilize
  proteins and to salt them out.
  1

Among other well established chaotropic agents, guanidinium ion ($\ce{Gu+}$) and the nonionic urea, which, in concentrations in the range 5 to 10 M, are the most commonly used protein denaturants. The effect of the various ions on proteins is largely cumulative: $\ce{GuSCN}$ is a much more potent denaturant than the often used $\ce{GuCl}$, whereas $\ce{Gu2SO4}$ stabilizes protein structures.

There are factors which have been investigated to explain why these two chaotropes appear to be effective in different cases (i.e they tend to be effective in some cases but not others). In one experimental investigation, the mechanistic basis of thermodynamic destabilisation due to urea and guanidinium was tested in order to find possibility that they exert their denaturing effect through peptide group hydrogen bonding (done experimentally by measurement of acid and base catalysed hydrogen exchange (HX) in a small peptide model)[2]

Urea

It has been established that urea increases the acid catalysed rate and decreases the base catalysed rate.

The acid catalysed HX rate in urea solution is higher than in the absence of urea. This can be expected when it is understood that acid catalysed HX proceeds dominantly through transient protonation of the peptide carbonyl group and deprotonation of the peptide NH by water or another base like urea and thus the proposed explanation is that urea forms an H-bond to the peptide carbonyl.[2]
In the case of base catalysed HX rate, which strongly decreases upon urea, the peptide NH group transfers its proton to a transiently H-bonded hydroxide. If urea is H-bonded to the peptide N-H, hydroxide can’t access the proton and HX is blocked.[2]

Guanidinium

With gaunidinium, as explained above, the effect of the various ions on proteins varies by some margin as they exert differently on the ion. It doesn’t selectively H-bonds to the peptide group (base catalysis) because guanidinium cannot H-bond to the peptide NH. Instead studies seem to suggest that guanidinium tends to engage in transient interactions with itself.[2]
Guanidium is often found to be approximately twice as efficient as urea in denaturing proteins but this varies with protein target. Helical peptide studies show that gaunidinium can be up to four times more efficient than urea when planar amino acids are major contributors to the helical stability. In contrast, guanidinium is barely more efficient if stabilisation is due to salt bridges. Thus protein targets are major contributing factor to the varying efficiency of guanidinium in denaturing proteins.[2]

Tl;dr

From above discussion, the varying effeciences depend on things like:

acid/base environment
nature of ions (the case of guanidinium)
protein target (e.g planar amino acids in helical peptides)

References

Voet and Voet Biochemistry. Section 7-1. Primary Structure Determination of Proteins.
Urea, but not guanidinium, destabilises proteins by forming hydrogen bonds to the peptide group. Woon Ki Lim 2595-2600 vol 106 no 8.


Canonical structures of (1E)-prop-1-en-1-yl acetate I have to find all of the possible canonical structures of this molecule

And here's what I did

Are they correct and if yes, did I miss any possibility?

May it look like this? In those two possibles ways?


Why is methoxide a strong base? Why is methoxide strong base? In case of halogens, I get why one is weak base, and one is strong base, but in the case of ones with oxygen, this seems to be hard to figure out.

Methoxide ($\ce{CH3O-}$) is the conjugate base of methanol. Methanol is very weak acid (e.g. its dissociation constant is very small), so its conjugated base is very strong.


What is K2O-nSiO2-xH2O? In need of $\ce{K2SiO3}$ for my cucumber plants, I purchased a bag of potassium silicate. The molecular formula is $\ce{K2O-nSiO2-xH2O}$. Normally, my bags of fertilizer are easy to figure out. For example, $\ce{Ca(NO3)2}$, or calcium nitrate, has a molecular weight of 164.0878 g/mol. The purity is also 95%. Knowing the purity and the molecular weight, I normally can find out how many ppm of Ca or N I want to use in my hydroponic nutrient tank. With $\ce{K2O-nSiO2-xH2O}$, I don't understand $n$ and $x$. I've googled this only to be stumped. I just don't get it.

The bag of fertilizer ($\ce{K2O-nSiO2-xH2O}$) has the following information listed on it:

Wt% $\ce{K2O}$ 26.5~29.5,
Wt% $\ce{SiO2}$ 55~60,
Weight ratio ($\ce{SiO2/K2O}$) 2.0~2.2,
Mole ratio ($\ce{SiO2/K2O}$) 3.2~3.4,
Bulk Density g/L 400~600,
solid % 82~89,
particle size 270+/-50 and
purity 77%

I need to know the molecular weight, so I can figure out how many ppm of $\ce{Si}$ and $\ce{K}$ to add to the solution.

You're dealing with a mixture of similar hydrated salts instead of a single salt (as in the example of $\ce{Ca(NO3)2}$). The notation $\ce{K2O-nSiO2-xH2O}$ means that there are different crystals with varying proportions of silicon dioxide and water (hydration). i.e. $n$ could be $1$, or $2$, $3$, etc. and $x$ likewise, for any given molecular "speck". A hydrated salt can be de-hydrated with heat (thus changing relative weights of elements), so be careful where you keep the fertilizer, because it will change the math.

This means, to measure a reasonable concentration, you need to consider the reported bulk average quantity of each component. Based on the percent by weight you reported, you can approximate the material as being (by mass -- that is wt%)

$28\%~\ce{K2O}$
$57.5\%~\ce{SiO2}$
(Thus) $14.5\%~\text{water}$

Say you want to know the $\pu{ppm}$ of $\ce{K+}$ for a solution of $\pu{1 g}$ of this fertilizer in a gallon ($\pu{3.78 L}$) of $\ce{H2O}$. Using the good approximation $\pu{1 mg/L} = \pu{1 ppm}$, We have

$$\begin{multline}
\pu{1 g}
\times \frac{\pu{0.28 g}~\ce{K2O}}{\pu{1 g fertilizer}}
\times \frac{\pu{1 mol}~\ce{K2O}}{\pu{94.2 g}~\ce{K2O}}\\
\times \frac{\pu{2 mol}~\ce{K+}}{\pu{1 mol}~\ce{K2O}}
\times \frac{\pu{39.0983 g}~\ce{K+}}{\pu{1 mol}~\ce{K+}}
\times \frac{\pu{1000 mg}}{\pu{1 g}}
\times \frac{\pu{1 gal}}{\pu{3.78 L}}\\
 = \pu{61.5 ppm}~\ce{K+}
\end{multline}$$

Here I guessed a concentration that may be relevant for you, but you can see the way to move forward. I would note, though, that since purity is reported pretty low ($77\%$), you should multiply the above result by $0.77$. Hard to say what the impurities are though, if they are mostly other potassium salts, the concentration could be higher. At any rate it's safe to say the above example is $\approx \pu{45-65 ppm}~\ce{K+}$.

You can also think of the compound like this, based on component masses:

$$\text{Molar mass} = 0.28 \times M(\ce{K2O}) + 0.575 \times M(\ce{SiO2}) + 0.145 \times M(\ce{H2O}) = \pu{63.532 g/mol},$$ and the average molar formula is $\ce{K2O - 3.2SiO2 - 2.7H2O}$.

Finally, I would recall that dissolution of $\ce{K2O}$ in general produces a basic solution, so if cucumbers are fond of acid (I don't know), you may consider adding a little acid ($\ce{HNO3}$, perhaps) to the solution to avoid perturbing the greens. But, you're the farmer!


Are vitamins in tablet are more efficient than the vitamins in their natural source? In medical shop I have seen that there are loads of vitamins tablets available but I was wondering like does vitamins in tablet are more efficient than the vitamins which are available in their natural source like in fruits ?

For some vitamins, such as Vitamin C, which are always found in the form of a unique chemical substance, there is no chemical difference between the vitamin in a tablet--synthetic or otherwise--and the same vitamin as found in a foodstuff. Still, how well you absorb a vitamin might be affected by what else you consume at the time - for example a given amount of Vitamin C might be absorbed more or less efficiently as an ingredient in orange juice than as a pill, even though it is the same substance in both cases. But these differences might not necessarily be significant.

Other vitamins, such as Vitamin A or Vitamin D, are found in multiple closely related chemical forms, any of which the body can use to fulfill its need for the vitamin. In that case, the body may absorb or utilize the vitamin more efficiently when it is in one form relative to another. Fats in the diet can further influence your absorption of a fatty vitamin like Vitamin D.

Minerals, such as iron, are notorious for being made easier or harder to absorb depending on their chemical form and how they are bound to other components of the food. So, for example, iron in greens is known to be slower/more inefficient for the body to extract than iron in liver.


Which of the following compounds would result in the reaction being more active in the reverse direction? $\ce{Fe(OH)2 (s) <=> Fe^{2+} + 2OH-}$
The reaction will be more active in the reverse direction if .......... is added
a) $\ce{KOH}$
b) $\ce{Na2S}$
c) $\ce{Fe(OH)2}$
d) $\ce{Fe(NO3)2}$

My thoughts:
Well, adding more $\ce{KOH}$ will increase the $\ce{OH-}$ ions so the reaction will go in the reverse direction to get back to equilibrium. Adding $\ce{Fe(NO3)2}$ Will increase $\ce{Fe^{2+}}$ ions and the reaction will go in the reverse direction too.
However, the book answers states that only (d) $\ce{Fe(NO3)2}$. I searched google but all the examples involved having only choice which works.
Note: Refer me to sites if necessary. I am ready to read.

In fact, adding more ion hydroxide won't displace the equilibrium towards the reverse reaction. This is due to the fact that ion complex $\ce{[Fe(OH)4]^{-2}}$, which is stable in presence of excess of ion hydroxide, is formed by successive addition of ion hydroxide on ferrous hydroxide.
If you want to study this more in depth, I invite you to read this webpage
It's explained at the end of the page under "numerical problems".


How to calculate the number of prochiral carbon atoms in a complex organic molecule? Calculate the number of prochiral carbon in the following compound.

The Wikipedia page on prochirality states:

  In stereochemistry, prochiral molecules are those that can be
  converted from achiral to chiral in a single step. An achiral species
  which can be converted to a chiral in two steps is called
  proprochiral.

My approach:

I first considered the sulfur atom to be prochiral as it satisfies the above condition, but it was pointed out, that this is incorrect as the S atom is already chiral.

My doubt:

But should even the double bond containing $\ce{sp^2}$ carbon atoms be considered prochiral?

According to what Martin postend in the comments I conclude that ("all" carbon atoms-4) in the compound are prochiral. I subtracted 4 chiral carbon atoms in the compounds. Tell me if there is still any mistake!
However, I am feeling that a carbon containing two hydrogen cannot be considered prochiral. The Wikipedia article states

  If two identical substituents are attached to a sp3-hybridized atom, the descriptors pro-R and pro-S are used to distinguish between the two.

But H atoms are not considered substituents in organic chemistry, see substituent .

I consider the concept of prochirality as a property of an achiral compound (or centre), that can be turned into a chiral compound (or centre) in one step (proprochiral: in two steps) by means of substituting or addition. Hence all carbon centres with three different ligands (substituents or hydrogen) can be considered prochiral.

Prochirality is unfortunately a grown concept, which was used in different ways throughout the years and can be confusing at times. I find the definition in the IUPAC gold book quite enlightening and much better than the Wikipedia page that spawned off of it.

  prochirality
  Also contains definition of: proprochirality

  This term is used in different, sometimes contradictory ways; four are listed below.

  The geometric property of an achiral object (or spatial arrangement of points or atoms) which is capable of becoming chiral in a single desymmetrization step. An achiral molecular entity, or a part of it considered on its own, is thus called prochiral if it can be made chiral by the replacement of an existing atom (or achiral group) by a different one. An achiral object which is capable of becoming chiral in two desymmetrization steps is sometimes described as proprochiral. For example the proprochiral $\ce{CH3CO2H}$ becomes prochiral as $\ce{CH2DCO2H}$ and chiral as $\ce{CHDTCO2H}$.
  The term prochirality also applies to an achiral molecule or entity which contains a trigonal system and which can be made chiral by the addition to the trigonal system of a new atom or achiral group. For example addition of hydrogen to one of the enantiotopic faces of the prochiral ketone $\ce{CH3CH2COCH3}$ gives one of the enantiomers of the chiral alcohol $\ce{CH3CH2CHOHCH3}$; the addition of $\ce{CN−}$ to one of the diastereotopic faces of the chiral aldehyde shown below converts it into one of the diastereoisomers of the cyanohydrin. The two faces of the trigonal system may be described as Re and Si.

  The term prochiral also applies to a tetrahedral atom of an achiral or chiral molecule which is bonded to two stereoheterotopic groups. For example, the prochiral molecule $\ce{CH3CH2OH}$ can be converted into the chiral molecule $\ce{CH3CHDOH}$ by the isotopic replacement of one of the two enantiotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The prochiral molecule $\ce{HO2CCH2CHOHCH2CO2H}$ can be converted into a chiral product by esterification of one of the two enantiotopic $\ce{–CH2CO2H}$ groups. The carbon atom of the $\ce{CHOH}$ group is called prochiral. The chiral molecule $\ce{CH3CHOHCH2CH3}$ can be converted into one of the diastereoisomers of $\ce{CH3CHOHCHDCH3}$ by the isotopic replacement of one of the two diastereotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The stereoheterotopic groups in these cases may be described as pro-R or pro-S. Reference to the two stereoheterotopic groups themselves as prochiral, although common, is strongly discouraged.
  See: chirality centre
  The term prochirality is also applied to the enantiotopic faces of a trigonal system.

The first definition applies especially to tetrahedral carbon atoms.
You can also work off of another definition which might be a bit easier to comprehend.

  prochirality centre
  An atom of a molecule which becomes a chirality centre by replacing one of the two stereoheterotopic ligands attached to it by a different ligand, e.g. C-1 of ethanol; C-3 of butan-2-ol.

    stereoheterotopic
    Either enantiotopic or diastereotopic. (In contrast the term constitutionally heterotopic has been used to describe identical groups located in constitutionally non-equivalent positions).

As a rule of thumb you can always substitute a $\ce{-CH2\bond{-}}$ group with a $\ce{-CHD-}$ group and ask your self if it is a chiral centre now.
I personally find it easier to identify stereoheterotopic ligands as this is also important for spectroscopy (esp. NMR).
This is somewhat also mirrored in the third definition.

The second (and fourth) definition applies to doubly bonded carbon atoms especially. If there are three different ligands, then you can assign a Re- and Si-face of that centre. Hence I would also consider these carbons prochiral.

In your case that means that all carbons (except for the ones that are already chiral) are prochiral.

Marked in pink are all chiral elements (the chirality of the nitrogen depends on the inversion barrier).
Marked in dark blue are all carbon centres who contain diastereotopic protons and hence can be considered prochiral according to the first definition. This is also evident from the $\ce{^1H-NMR}$ estimation of ChemDraw.
Marked in green are all carbon centres where you can add a nucleophile to create a stereo centre and are hence prochiral according to definition two.
Marked in orange are all carbons who are part of an aromatic system. Strictly speaking they are also trigonal carbons and technically you can assign a Re/Si to them. Accordingly they fall within the second definition and can be considered prochiral centres.


Why can enolizable aldehydes undergo Cannizzaro reactions? Why does 2-methylpropanal undergo a Cannizzaro reaction even though it has an $\alpha$-hydrogen?

There is never an either–or in chemistry; it is typically a both instead. In the case of a hydroxide ion approaching an aldehyde with an α-hydrogen, it can both deprotonate the hydrogen creating the corresponding enolate and attack the carbonyl nucleophilicly to generate the required intermediate for a Cannizzaro reaction.

Most of the time, if there are two (or more) possible pathways and neither of them is intrinsically less hindered due to a strong reason, both pathways will be taken.


Change in Entropy and Enthalpy found over a temperature range and asked to compute the change in Gibb's Energy I was given the following question by my tutor:

"The variation of enthalpy ($\Delta H$) and entropy ($\Delta S$) in a reaction carried out at
constant pressure in the temperature range between 300 ºC and 350 ºC, are 1700 kcal and
15kcal/K respectively. Calculate the range of temperatures in which this reaction is
spontaneous."

But my book on thermodynamics says that the change in Gibb's energy for a process can only be calculated if we start and end at the same temperature (the temperature can vary during the process as heat is added or lost from the system to the surroundings, but the end points must have constant temperature and pressure.) In this question the changes in entropy and enthalpy are found over a range of temperatures where the final temperature is not the same as the initial temperature, but the pressure is always constant.

Is my tutor's question invalid, is my book wrong in saying that the temperature has to be the same at the initial and final values, or am I thinking about this wrong?

This should address your confusion about constant T:

$$G = H - TS \Rightarrow dG = dH -TdS - SdT$$

$$ \Rightarrow \text{(at constant T) } dG = dH -TdS \Rightarrow \Delta G = \Delta H - T \Delta S$$

So, yes, you need $\Delta T = 0$ for $\Delta G = \Delta H - T \Delta S$ to hold.

However, this holds at any T.

E.g., suppose you do the reaction at $300 K$. $\Delta T = 0$, so $\Delta G = \Delta H - T \Delta S$ holds.

But suppose you repeat reaction at $350 K$. It's still the case that $\Delta T = 0$, so $\Delta G = \Delta H - T \Delta S$ still holds.

As far as the question itself goes, I agree with Chet Miller that it appears to be badly worded.  It seems what your tutor meant to say was:

"Between 300 ºC and 350 ºC, assume that $\Delta H$ and $\Delta S$ are constant, with values of 1700 kcal and 15kcal/K, respectively. Calculate the range of temperatures over which $\Delta G <=0$, i.e., the range the textbook labels as 'spontaneous'".


Why can't Carbon Dioxide be liquid at 1 atm? Wikipedia says that Carbon Dioxide cannot be liquid under atmospheric pressure.

I thought the phase of an element was related to both pressure and temperature.

If that is so, then should you not be able to set the temperature conditions such that you could have a cup of liquid Carbon Dioxide?

This is the phase diagram from a better source:

The pressure axis is the vertical one at the left. One bar (which equals 0.986923 atmospheres) is near the bottom, as shown on the diagram. So carbon dioxide is a gas at all temperatures above −78.5 °C when the pressure is one bar. Solid carbon dioxide (dry ice) sublimes at one bar (about one atmosphere), i.e., goes directly from solid to gas.

If you want liquid carbon dioxide, you need a temperature, pressure pair in the liquid region, e.g., −40.0 °C and 20 bar.


Azeotropic mixtures and Raoult's Law So I find this everywhere:
"Mixtures showing large deviations from Rault's law form azetropic mixtures at specific compositions"
My questions are:
(a)Why is it so?Why deviation from Raoult's law is needed at all.
Wikipedia simply skips over the answer stating its a mathematical consequence.
(Quoting wikipedia:
When the deviation is great enough to cause a maximum or minimum in the vapor pressure versus composition function, it is a mathematical consequence that at that point, the vapor will have the same composition as the liquid, resulting in an azeotrope.)
(b) why can't ideal mixtures form azeotropes?
P.S: (a)A more intuitive answer would be appreciated although maths may be included if necessary.
(b)I did my research but nowhere the reason was to be found. It's presented as a very obvious consequence everywhere.

To understand why non - ideal mixtures form azeotropes, we need to understand what is ideal mixture. If you have a liquid mixture of two components A and B, ideal mixture is a mixture in which heterogeneous interactions (A - B)  are the same as the average homogeneous (A - A and B - B). This means when you mix components A and B, they don't "feel" like their enviroment has changed at all and thus no component has higher nor lower tendency to escape the liquid phase and go into vapor phase. When interactions are largely different, in most cases heterogeneous interactions are less favourable than homogeneous (molecules of A in most cases like to be around molecules of A rather than B) than component will have higher tendency to escape from liquid phase to vapor since in vapor interactions are much less pronounced due to much bigger average distance of the molecules. If this scenario happens you got yourself a lower boiling azeotrope since due to higher tendency of molecules to escape liquid phase, vapor pressure is higher than in ideal case (Raoult's Law). In some cases heterogeneous interactions can be very favourable and this is usually due to some specific interactions which mostly originate from hydrogen bonding, if so molecules will tend to stay in liquid and you got yourself higher boiling azeotrope. Example of such azeotrope is HNO3 and water. If you have any questions, fire.


For two compounds that are symmetrical, why does the more compact one have a higher melting point? In this post, I refer to two molecules B (Di-tert-butyl ether) and C (Dibutyl ether).

Wikipedia: Di-tert-butyl ether (B)
b.p.  107.2 C
m.p. - 61   C

ChemSpider: Dibutyl ether (C)
b.p.  142 C
m.p. - 96 C

I understand that Compound C has a higher boiling point than compound B, because of its greater surface area and therefore stronger intermolecular VDWs.

I don't understand, however, why Compound B has a higher melting point than compound C. A lecturer told me that this is due to compound B being more compact and symmetrical. I think that I understand the concept that some structures pack into a crystalline lattice better than others, such that the structure of the packing influences the strength of intermolecular forces within the solid, therefore affecting the melting point.

However, it seems to me that both B and C above are equally symmetrical in at least one conformation. Beyond that, C has conformations of greater surface area than B, which should further strengthen its intermolecular bonds in a solid state. I would expect the melting point of C to be higher than B; yet this is not the case.

I feel like the answer may lie in the fact that C has many more conformers than B. Even so, I can't seem to draw a link anywhere. Shouldn't the solid formed take the most stable crystalline structure, much like how water tends to just one of many crystal structures?

The terms used by your teacher are quite in line with at least a qualitative explanation.

I really suspect that there are various ways to express compactness and sort of "sphericity" of molecules. I am not a specialist so forgive me for being imprecise on this.

However, if we forget a rigorous classification in symmetry group, a potato is more spherical symmetrical than a banana.

In such a sense the ramified isomer is definitely more  symmetric and "round" than its linear counterpart.

This is even more evident if all conformers are taken into account: in the linear molecule a chain end can even turn toward the center while the opposite being extended.

Now this does not necessarily explain the observed behaviour as for the more spherical molecule can pack easily but at expense of VdW forces that depend on molecular surface.

However, it relates to the fact that the linear isomer has a number of conformers much higher than the ramified one (rotation about 4 bonds versus 1 for each side of the ether)*** and results in the first having a bigger conformational entropic term.

Although the crystals will generally contain one isomer, the crystal formation take place at expense of the intramolecular forces. As such, when keeping entalpy terms comparable, bigger is the entropy term then less stronger is the binding.

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

For comparable enthalpies and very different conformational entropy terms, than the melting occurs at lower T for the entropy rich molecule, i.e. that one for wich more conformations become accessible.

This obviously is not the case for the liquid to gas transition, as the conformational S term is similar in both phases and negligible faced to the gas entropic content.

***Conformers due to the rotation around the bonds in the terbutyl groups do not really alter the shape of the molecules (connection with the compactness mentioned above). More rigorously at each side we should have (calling x the number of conformers around a bond) that the number of conformations is circa $x^4$ and about $3x^2$ for the linear and the branched molecules, respectively.

Additional note: conformational entropic terms are crucial to many phenomena in polymers and proteins (solubility, crystallization, protein folding/unfolding, etc).
A treatment of this requires statistics that I have forgotten. So take my numerical estimated about conformers with a grain of salt.  I am satisfied if I did convey the message.

Add. Note 2. Another useful example is, for instance, a stiff and insoluble chain such as polythiophene.

It is insoluble as for electron delocalisation makes it flat and rigid. Once side substituted with alkyl chains, the entropic term brought in by the flexible chain is the main responsible for the acquired solubility.


What is the most stable conformation (Newman Projection) of CF3CH3? Is a totally eclipsed Newman projection the most stable conformation for a 2 carbon compound, with one carbon bonded to more electronegative atoms (making them partially negative), and the other carbon with something less electronegative (making it partially positive)?

For example, let's imagine $\ce{CF3CH3}$. If they are eclipsed, the fluorine side is partially negative and the hydrogen side is partially positive, so wouldn't they attract?

Or do the electron clouds always repel each other, no matter what atom is used (making a staggered conformation the most stable, not eclipsed)?

For $\ce{H3C-CF3}$, the stable conformation is staggered. I would expect this to be case for any similar, freely rotating molecule (the exception being conformations imposed by rings etc.), with the reason being the repulsion between the $\ce{C-F}$ and $\ce{C-H}$ bonds.

To confirm, I have optimized the molecular geometry on different levels of quantum chemistry, starting from a conformer halfway between staggered and eclipsed. In all cases, I have observed convergence to the staggered conformer. The levels are

HF/def2-TZVP
PBE-D3/def2-TZVP
B3LYP-D3/def2-TZVP
PW6B95-D3/def2-TZVP
PW6B95-D3/def2-QZVP

In response to comment: I have performed PW6B95-D3/def2-TZVP optimizations for propene, acetaldehyde, and propanaldehyde (the latter starting from two different conformers). The result is that in all cases, the double bond is eclipsed by a single bond (and two non-equivalent conformers exist for propanaldehyde).


What are the correct resonance structures of nitrous oxide? There was a question in one of my exam papers to draw the resonance structures for $\ce{N2O}$.

These are the ones I drew, but they were not awarded marks:

The ones in the marking scheme are as follows:

Can anyone please explain why mine aren't correct?

Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost $\ce{N}$ is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central $\ce{N}$ is participating in 5 bonds, which $\ce{N}$ cannot do. The maximum number of bonds that $\ce{N}$ can participate in is 3 (covalent bonding - like $\ce{NH3}$) or 4 (coordinate bonding - like $\ce{NH4+}$).

As long as you satisfy the octet rule (or at least as best as you can - here are some exceptions) and you wish to check the stability of a molecule that you have drawn, simply use formula for formal charge. This is given by the equation:

$$\mathrm{FC} = \mathrm{V} - (\mathrm{N_B} + \frac{\mathrm{B}}{2})$$

Wher $\mathrm{FC}$ is the formal charge, $\mathrm{V}$ is the number of valence electrons that the atom under consideration usually has, $\mathrm{N_B}$ is the number of non-bonded electrons, and $\mathrm{B}$ is the number of electrons shared in a covalent bond on the atom. If a molecule is stable, the sum of the formal charges of each substituent atom should be 0. If you were to test this on each of the resonance structures that the test provides as answers, you would see that this holds true.


Molecular explanation of the Hofmeister Series The Hofmeister Series is a useful metric in understanding how to precipate proteins. However, wikipedia and many other source indicate that the order of the series is rather empirical. I'm curious about the variety of explanations of the mechanism of the series as well as the application of the series to other non-protein macromolecules.

The molecular origin of the Hofmeister series is not well understood and still a very active area of research. For example, see the upcoming Faraday discussion conference. The syllabus for the conference says:

  Although it is now clear that the Hofmeister series is intimately connected with ion hydration in homogeneous and heterogeneous environments and with ion pairing, the molecular origin of these effects has been poorly understood. Biochemists and physical chemists have been typically using the term Hofmeister series to put a label on ion specific behaviour in various environments, rather than to reach a molecular level understanding and, consequently, an ability to predict a particular effect of a specific salt ion.

(So, this is not an answer, but rather too long for a comment…)


Boiling point of oxygen I am studying chemistry from the book titled "Fundamental Chemistry for Cambridge O Level" written by Rosemarie Gallagher and Paul Ingram.
On page 17, it states that

  oxygen boils at −219 °C and freezes at −183 °C.

Isn't this incorrect? Should it not be the other way round?

According to NIST Chemistry Webbook, for oxygen:

$$T_\mathrm{boil} = \pu{90.2\pm0.2 K}$$
$$T_\mathrm{fus} = \pu{54.8\pm0.2 K}$$
$$T_\mathrm{triple} = \pu{54.33\pm0.6 K}$$

The $T_\mathrm{boil}$ and $T_\mathrm{fus}$ are from Ref.1 while Brower and Thodos reported $T_\mathrm{triple}$ as $\pu{54.363 K}$ at pressure of $\pu{1.125 mmHg}$ (Ref.2). The uncertainties were given by Thermodynamics Research Center, NIST Boulder Laboratories, CO.

Accordingly, $T_\mathrm{boil} = \pu{-183 ^\circ C}$, $T_\mathrm{fus} = \pu{-218.4 ^\circ C}$, and $T_\mathrm{triple} = \pu{-218.8 ^\circ C}$. Therefore, as you correctly suspected, the values in the textbook is a misprint.

References:

Alex G. Streng, “Miscibility and compatibility of some liquefied and solidified gases at low temperatures,” J. Chem. Eng. Data 1971, 16(3), 357-359 (https://doi.org/10.1021/je60050a024).
Gerald T. Brower, George. Thodos, “Vapor pressures of liquid oxygen between the triple point and critical point,” J. Chem. Eng. Data 1968, 13(2), 262-264 (https://doi.org/10.1021/je60037a038).


What does a higher half cell potential mean? I was taught in class that a positive large half cell potential generally meant that the reaction was more favorable.  What exactly does this mean?  Doesn't a reaction with potential 4V need a greater input of energy to occur? Meaning that a lower electron cell potential would occur first and is therefore more favorable?

Also how why does Zn(II) +2e- ---> Zn favor the left side of the reaction?  Shouldn't Zinc naturally tend towards being neutral?

That's not quite correct.

Half-reaction potentials are only really meaningful when you complete the redox pair, that is, you must have an oxidation and reduction reaction. The sum of the potentials is the potential for the full redox reaction at standard concentration and state. In that sense, the more positive either of the potentials is, the more favorable the overall reaction will be.

There is no preference for a neutral species in general. If you put sodium in water, the electron is quickly liberated and the end products of hydrogen and sodium ion are much more favored than the reactants of water and sodium.


Writing Topology file for Ligand for Gromacs Is there any tutorial for how to manually write the topology file for any Ligand file(PDB for mol2)? There are so many sites which provides it like ProDrug or ATB but no where the actual process is written. Can anyone explain the process?

PS: I am actually not sure this question belong to which SO.

First off, you should not be manually writing topology files whenever you can avoid it - it's far too prone to making silly errors. That's why there are so many programs out there that attempt to automate the process.

If however, you just want to understand HOW the topology file is structured, I recommend reading the GROMACS manual (chapter 5). These tutorials also do a good job of describing some of the key aspects of the topology files.


1,1-diformyl propanal The correct IUPAC name of the following compound is:

  (A) 1,1-diformyl propanal
  (B) 3-formyl butanedial
  (C) 2-formyl butanedial
  (D) 1,1,3-ethane tricarbaldehyde

The structure of 1,1-diformyl propanal has been given. And the IUPAC name has been written. But I can't get the same IUPAC name because I'm a beginner. Also I don't know how to get structure here in the question, so I didn't do it.
Help me with the IUPAC naming of the compound with step-by-step explanation . Please.

The proper and straightforward way to solve nomenclature problems is to construct the systematic name for the given structure according to the current IUPAC recommendations. However, for such multiple-choice questions, it may be easier to check the given answers and try to draw a structure for each proposed name as follows – especially when the particular nomenclature rules that are considered in the exercise are not exactly known.

The name 1,1-diformyl propanal (A) can be ruled out since the atom ‘1’ (i.e. the $\ce{C}$ atom of the aldehyde group) of propanal has only one $\ce{H}$ atom that could be substituted; therefore it is impossible to place two formyl groups at atom ‘1’. Furthermore, after substitution of the $\ce{H}$ atom of the aldehyde group, the compound would no longer be an aldehyde.

The name 1,1,3-ethane tricarbaldehyde (D) does not make sense since ethane has only two $\ce{C}$ atoms; therefore it is impossible to place a carbaldehyde group at atom ‘3’ of ethane.

It is possible to draw structures for the names 3-formyl butanedial (B) and 2-formyl butanedial (C) based on the structure of butanedial:

Note that the resulting structures for 3-formyl butanedial and 2-formyl butanedial are actually identical, and both structures correspond to the structure that is given in the question. The correct choice is 2-formyl butanedial (C) rather than 3-formyl butanedial (B) since the locant ‘2’ is lower than ‘3’. (Note that the correct spelling of the name is actually 2-formylbutanedial, i.e. without space.)

Anyway, even the name 2-formylbutanedial is actually not in accordance with current IUPAC recommendations. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), aldehydes can be systematically named in various ways:

  P-66.6.1 Systematic names of aldehydes

  Aldehydes are systematically named in three ways:

  (1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;

  (2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;

  (3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

  P-66.6.1.1.2 The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.

Therefore, the preferred IUPAC name for the compound given in the question is ethane-1,1,2-tricarbaldehyde.


What forms the emulsion in chocolate? Some days ago, before buying a chocolate I read the ingredients and saw that it contains a natural emulsifier. A Google search didn't reveal anything about of what the emulsion formed is, do you have any idea?

The general composition of chocolate:

  The main components in milk chocolate is cocoa butter, sugar, and milk
  powder.

Then, a concrete example:

  Hershey’s chocolate is made of sugar, chocolate, cocoa butter, cocoa
  processed with alkali, milk fat, lactose (milk), soy lecithin, PGPR
  (emulsifier), vanillin, artificial flavor, and milk.

This is taken from The Chemistry of Chocolate on ChemistryIsLife.com

I continue wondering what the emulsion is formed by.

Chocolate consists of tiny clumps of cocoa solids, sugar, and milk solids dispersed in cocoa butter.  Because the solids aren't soluble in the butter, they tend to clump up, which reduces the quality of the texture of the chocolate, it feels less "smooth."  When you add an emulsifier, the hydrophilic end tends to stick to the relatively more polar surface of the solids (the proteins in the cocoa solids, for example, or the carbohydrates) and then the hydrophobic end interacts favorably with the cocoa butter.  The coating of emulsifier tends to help the clumps stay small and well dispersed in the fat.  That makes it easier and faster to process the chocolate, and it also lets the chocolate last longer before the fat and chocolate solids start to phase separate on a macroscopic scale.


What does it mean when it is said that an sp³ orbital has 25% s character? My textbook frequently mentions:

  $\mathrm{sp^3}$ hybrid orbital has 25% $\mathrm{s}$-character and 75% $\mathrm{p}$-character.

What are these "characters"? And how do these characters influence the bond and the hybrid orbital?

Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four equivalent orbitals. Hence $\mathrm{sp^3}$ is a contraction of $\mathrm{s^{\frac{1}{4}}p^{\frac{3}{4}}}$. Therefore you can see, that the orbital has $25\,\%$ $\mathrm{s}$ character and $75\,\%$ $\mathrm{p}$ character.
Generally, the concept of hybridisation has to be treated as a concept only, it is nothing which can be physically observed. It is a mathematical model to understand geometries better. Please also note, that hybridisation is a function of the geometry, not the other way around. Having said that, it is fairly obvious, that I am not a big fan of the concept. I would not go as far as Alexander Grushow* and preach to abandon the whole thing, but I advise caution in the use of this concept to not obtain wrong conclusions.

  Despite significant experimental evidence and theoretical advances to indicate that hybrid atomic orbitals do not exist and do not appropriately describe molecular bonding, their description still permeates chemical education at many levels, and the model still finds its way into modern chemical literature.

* “Is It Time To Retire the Hybrid Atomic Orbital?” Alexander Grushow, J. Chem. Educ., 2011, 88 (7), 860–862.


Why does the conductivity of a LiOH solution decrease when boric acid is added? If I start with pure water and add $\ce{LiOH}$ the conductivity of solution increases. However, if I then dissolve boric acid in the same solution the conductivity falls. I would have thought adding more cations or anions to a system would always increase conductivity, but in this case it does not. Why is this?

TL;DR: The addition of boric acid effectively switches the conducting anionic solute from hydroxide to borate, which has a lesser contribution to the conductivity due to its larger size and corresponding lower diffusivity. Most solute systems will not behave this way.

In sufficiently dilute solution, the contribution $\kappa_i$ of each dissolved ion to the overall conductivity is proportional to the concentration, diffusivity, and squared charge of that ion:

$$
\kappa_i \propto z_i^2D_iC_i
$$

And, the total conductivity is just the sum of these contributions:

$$
\kappa = \sum_i{\kappa_i}
$$

In this case, you're starting with $\ce{LiOH}$, which dissociates  more or less completely into aqueous $\ce{Li^+}$ and $\ce{OH^-}$, and the conductivity rises as you add more of it. Once you start adding boric acid, though, you're no longer just dissociating ions into the solution.

Boric acid is a funny creature.  Even though it looks like it might be a triprotic Arrhenius acid if we were to write it as $\ce{H3BO3}$, in reality it's a monoprotic Brønsted acid that either exchanges a hydroxide ion with the water solvent:

$$
\ce{B(OH)3 + 2H2O <=> B(OH)4^- + H3O^+}
$$

or, at sufficiently high $\mathrm{pH}$ $(\mathrm pK_\mathrm a = 9.2)$, it just takes up a hydroxide ion from solution:

$$
\ce{B(OH)3 + OH^- <=> B(OH)4^-}
$$

So, when added to a solution of $\ce{LiOH}$, boric acid just substitutes dissolved $\ce{OH^-}$ for dissolved $\ce{B(OH)4^-}$, with no net addition of anions to the solution!  Thus, the $C_i$ contribution to $\kappa$ doesn't really change much. As well, both $\ce{OH-}$ and $\ce{B(OH)4^-}$ are monovalent anions, so there's no effect on any of the $z_i^2$ terms either.

The overall decrease in $\kappa$ comes from the change in diffusivity that results from the swap of hydroxide for borate.  The diffusivity of an anion is roughly proportional to its ionic volume—bigger ions experience more "drag" from the surrounding solvent molecules, so they diffuse (move around) more slowly. I couldn't find an ionic radius for borate on a quick search, but just from their compositions, it seems sensible to assume that the borate anion is considerably larger than the hydroxide anion, and thus its diffusivity should be correspondingly less.


Structure of Benzyne I have seen in a lot of pictures that benzyne looks like:

But aren't $\ce{C_6}$ and $\ce{C_5}$ $-$ $\ce{sp}$ hybridized?

And hence bond angle should be $180^{\circ}$ or at least near about it?

Something like :

Or,

?

Yes, but.

Yes, in the conventional low-level models, one would consider the two carbons in benzyne you mentioned as sp-hybridised. And that does mean that their orbitals seem to be pointing the wrong way.

However, a better picture would be to use an orbital which is a lot closer to sp²-hybridisation, and an even better picture would calculate orbitals computationally (I can’t do that, but Orthocresol can).

The thing is, all these pictures show, that the triple bond in benzyne is rather different from those in linear molecules or large cycloalkynes such as cyclodecyne ($\ce{C10H16}$). It contains a large angular strain, not unlike very small saturated rings such as cyclopropane. The strain causes the triple bond to be rather weak and easily cleaved/attacked — which paves the way for all reactions in which benzyne is an intermediate.

Larger cycloalkynes with eight or nine carbons have been isolated, but for benzyne only in situ studies have been made according to my quick check on SciFinder two seconds ago. In layman’s terms, you can think of the angle strain by saying ‘I am a triple bond, I want a $180°$ angle, but all I can have is $120°$ so I’m rather unhappy!’ A straight-up $180°$ angle is not possible, however, because of the other two atoms in the benzyne ring.

Kukolich, McCarthy and Thaddeus published two papers (one together with Tanjaroon) on microwave-spectrum measurements to determine the structure and compare it to in silico structures. Their results give a $126°$ angle along $\ce{C^1-C^6-C^5}$ (your numbering scheme). This is more obtuse than in benzene ($120°$) but still far away from $180°$. However, the major constraint is the angle $\ce{C^6-C^1-C^2}$, which is compressed to a mere $111°$.[1,2]

Before the objection is raised and as ron pointed out in the comments: The triple bond is indeed a bond and not a diradical as can be derived from its behaviour. E.g, it concertedly and stereospecificly adds to dienes.

References
[1] S. G. Kukolich, C. Tanjaroon, M. C. McCarthy, P. Thaddeus, J. Chem. Phys. 2003, 119, 4353. DOI: 10.1063/1.1593015
[2] S. G. Kukolich, C. Tanjaroon, M. C. McCarthy, P. Thaddeus, J. Phys. Chem. A 2004, 108, 2645. DOI: 10.1021/jp031344p


Quadrupole MS: mass as a function of voltage Every time I see an article about Quadrupole MS (e.g. here), they give formulas like:

$$ a_x = \frac {8eU} {m r_0^2 \Omega^2} $$ and $$q_x = - \frac {4eV} {m r_0^2 \Omega^2}$$

I suppose the instrument changes voltage and keeps the rest constant, but I failed to find any article that would show direct formulas on how to calculate m/z from that. I could rearrange it like this:

$$ \frac {m} {e} = \frac {8U} {m r_0^2 \Omega^2 a_x} $$ But this doesn't really help because $a_x$ itself is unknown and is derived from mass.

PS: this question is about Quadrupole MS though it looks to me that Ion Trap MS has very similar concepts and math.

These parameters arise from non-dimensionalization of the Mathieu equations for ion motion in exogenous, periodically varying electric fields (i.e. oscillating potentials).  You can't calculate $m/z$ directly from $a$ and $q$.  For a fixed $U$ and $V$, you can regard $a$ and $q$ as non-dimensional parameters that correspond to $m/z$.  (Conversely, at fixed $m/z$ you can regard $a$ and $q$ as non-dimensional versions of the electrical potentials $U$ and $V$.)  The reason that those parameters don't tell you everything is that you still have to solve the Mathieu equations.

An analogy might be to the Reynolds number for characterizing fluid flows.  It, like $a$ and $q$, is a non-dimensional number that depends on properties of the molecules under study and on the system geometry.  Knowing just the value of the Reynolds number is useful for understanding some of the general characteristics of a flow, but if you want to compute fluid motion, i.e. the velocity profile in a pipe, you still have to solve the Navier-Stokes equation.

For quadrupole mass analyzers, only certain regions of $a$-$q$ space result in stable ion trajectories (i.e. where ions don't collide with the surfaces of the four rods in the quadrupole).  This map shows which values for $a$ and $q$ result in stable solutions to the Mathieu equations, and thus is called the Mathieu stability diagram.  The same document you linked to shows an example of such a diagram:

Thus, to compute $m/z$, we need to impose the criterion that $a$ and $q$ are in a stable region of the stability diagram, more specifically in the region "A" shown in the diagram.  By rough eyeballing, the "middle" of this diagram corresponds to about $q = 0.75$ and $a = 0.07$ or so.  You can use either of those values to estimate $m/z$, but a more precise estimate, including line-widths etc., would require knowing both.


How do I build a thermochromic thermometer? My Chem 152 lab is requiring us to build a thermochromic thermometer utilizing methods from previous labs such as absorbance spec and temp.

We've been taught how to calculate Kc values successfully as well as delta H and delta S from these methods, but I'm unsure of how to combine everything into creating a thermochromic thermometer.

We're given $\ce{CoCl2(EtOH)2}$ and $\ce{CoCl2(iPrOH)2}$ to react with $\ce{MeOH}$. We had to complete a pre-lab assignment by calculating $K_c$ values at $275\ \mathrm{K}$ using $\Delta H$ and $\Delta S$ values from a reaction of $\ce{CoCl2(EtOH)2/MeOH}$ and a reaction of $\ce{CoCl2(iPrOH)2/MeOH}$. I was able to determine whether each reaction was reactant or product favored based on their $K_c$ values, but I'm completely stumped on how to use all of this information in creating a thermochromic thermometer.

We calculate the $K_c$ values. Ratios that give $K_c$ greater than 1 are product favored and will turn from blue to pink at low temps. The opposite is true for the ratios that give $K_c$ values less than 1. These are reactant favored, and will turn from pink to blue at the low temp. First, look to see what is chaging color most drastically, meaning the reaction is proceding to a greater extent ($K_c$ value a lot greater than or less than 1), then look at the cost. If it's too expensive, a thermometer that is less effective but still OK, may be better.

Overall, think of a thermometer that you stick on the side of a fishtank (sort of).


What is the purpose of ammonium chloride in anti hydrogenation of alkynes? In my organic chemistry textbook, lithium is used to create a free radical of the alkynes, to then allow the anti addition of hydrogen to get an alkene.

"The Dissolving Metal Reduction of Alkynes" is a reaction that then uses $\ce{NH4Cl}$. Why put $\ce{NH4Cl}$?

Products from reaction step (1) are the trans-alkene and 2 equivalents of the amide $\ce{LiNHEt}$, a strong base. Step (2) is the workup of the reaction mixture with aqueous $\ce{NH4Cl}$ solution, which serves several purposes. First, it quenches the lithium amide and unreacted lithium metal.

$$\ce{LiNHEt + NH4Cl ->~ EtNH2 + NH3 + LiCl}$$
$$\ce{Li + H2O ->~ LiOH (aq) + \frac{1}{2} H2\uparrow}$$

Secondly, the trans-alkene is usually less water-soluble than ethyl amine and the lithium and ammonium salts, and can be separated from the workup mixture by precipitation or extraction with an organic solvent.


Ethyl Acetate: how to store I'm a 3D printing hobbyist and a few months ago I bought some Ethyl Acetate after learning online that it might have a similar effect on PLA as acetone has on ABS. From the tests I made, I was either sold a bad batch or the reaction speed is next to 0 thus making the ethyl acetate useless to smooth out PLA. I have thus placed it in a drawer to test it more at a later time. The liquid was kept inside its original plastic bottle that was then placed inside a glass jar to ensure that none of the vapors got out. Today I opened the drawer again to have a look at it and I noticed a very strong odor consistent with that of the ethyl acetate. Also the label on the plastic bottle separated as if the glue has been dissolved away. As ethyl acetate is highly inflammable I'd like to ask what would be a correct way to store it without leaking the vapors.

In laboratories, ethyl acetate is usually stored in special ventilated cupboards that are furthermore typically designed to withstand fire. This is because you will typically store not just one but many flammable solvents/reagents in that type of cupboard.

At home, you typically do not have that possibility. However, the basic advice remains the same: store in the original container (if possible) and in a well-ventilated area away from spark sources. The original bottle should, in theory, be airtight enough — remember that the ethyl acetate must have somehow found its way to you in that bottle without having lost too much content.

I would strongly advise against storing in a closed drawer. As you noticed, a few months were enough for enough solvent to evaporate but stay in the vicinity for you to smell it. The amount of solvent you lose per unit of time is typically very small and if it can diffuse away it does not do much harm. But the closed drawer prevented it from efficiently diffusing away and allowed it to concentrate.


Vibrational Spectrums of HI, HBr, HF, and HCl Do you all know of any large graphs for the vibrational spectrums of HI, HBr, HF, and HCl? I need graphs that are large enough to see the spacing between the peaks in the spectra of HCl, HBr, and HI.

Alternately, if you know of any raw data sets for any of the above, that would work. Thanks in advance.

The fine structure in the diatomic vibrational spectra is based on various population of the rotational levels.

The best demonstration of isotopic effects in vibrational spectra is to switch hydrogen for deuterium.

The frequency of vibration (from Hooke's Law adapted to a harmonic oscillator) is given by:

$$\nu = \frac{1}{2\pi}\sqrt{\dfrac{k}{\mu}}$$

where $k$ is the force constant of the oscillator and $\mu$ is the reduced mass:

$$\mu=\dfrac{M_1M_2}{M_1+ M_2}$$

Switching $\ce{HCl}$ to $\ce{DCl}$ has a much larger effect than the two chlorine isotopes (35 and 37):

$\ce{H^{35}Cl}, \ \mu\approx\frac{35}{36}=0.972$
$\ce{H^{37}Cl}, \ \mu\approx\frac{37}{38}=0.974$
$\ce{D^{35}Cl}, \ \mu\approx\frac{70}{37}=1.892$
$\ce{D^{37}Cl}, \ \mu\approx\frac{74}{39}=1.898$

The difference between $\ce{^{35}Cl}$ and $\ce{^{37}Cl}$ is in the third decimal place.

Here is a nice Google Image search for the vibrational spectra of HCl and DCl


Non - aromatic nature of quinones I have read that quinone systems are conjucated but not anti-aromatic. I had read the related questions:

Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene?
Is a quinone ring aromatic?.

In one of those answers, I read a point which said that it was non aromatic:

  [...] and you must be able to draw double bonds - internal to the ring
  - from each ring carbon atom. [...]

I don't get what this means as I had never heard of such a thing in Hückel's rule. I did read elsewhere that the system must be completely conjugated which is true in this case but due to cross conjugation the electrons are not delocalised throughout the system? I don't get why cross conjugation affects the delocalisation of the electrons this much? What is the proper explanation for this?

The reason you must be able to draw double bonds within the ring is the $4n+2$ rule applies rigorously only when the conjugated atoms are all in such a ring. If you are drawing double bonds to adjacent atoms off the ring then you don't have all the conjugated pi electrons in the ring anymore, and then there is no longer a $4n+2$ rule for aromaticity or a $4n$ rule for antiaromaticity.

In some systems it is possible to render contributing structures that have all the conjugated electrons in the ring and meet the $4n+2$ rule, in which case you can assert that this contribution is enhanced by aromaticity and the properties of the compound affected accordingly.  For example, the aromatic contribution to cyclopropenones leads to enhanced polarity of the carbonyl group and increased basic character of the oxygen atom, compared with ordinary ketones.

Illustration source: https://commons.m.wikimedia.org/wiki/File:C3H2OResonance.png.  Drawn by WP user Smokefoot.  Link to License

Quinones, however, cannot be given such a structure without leaving the highly electronegative oxygen short of an octet, so they lack any significant aromatic contribution.


Why are there two Hydrogen atoms on some periodic tables? Most periodic tables only feature one Hydrogen atom, on the top of the first group. But some, like the one I was given, also show Hydrogen in the 7th group, to left of Helium.

Why are there two Hydrogen atoms? What's the difference between the two, and why do we work with the left one more often than the right?

The two hydrogens are the same, but some periodic tables show hydrogen in both places to emphasize that hydrogen isn't really a member of the first group or the seventh group.

Hydrogen is a diatomic gas in it's elemental state, which is different from the other group one metals (and similar to the group seven elements).  At the same time, hydrogen usually loses its electron, similar to sodium and the other group one metals.


Why do alkanes have higher boiling point than their ether counterparts? Based on my understanding of inter-molecular forces, I expect dipole-dipole interactions to be significantly stronger than van der Waal's interactions. Hence, I expect ethers (which obviously have dipole-dipole interactions) to have higher boiling points (as boiling point $\propto $ inter-molecular forces) than alkanes (assuming approximately the same molecular mass). But, the experimental data suggests otherwise.

$$
\begin{array}{|c|c|c|c|}
\hline
\textbf{Compound}&\textbf{Boiling point / °C} &\textbf{Compound}&\textbf{Boiling point / °C} \\ \hline
\textit{n}\text{-pentane}  & 36.1   & \text{ethoxyethane}  & 34.6    \\ \hline
\textit{n}\text{-heptane}  & 98.42  & \text{1-propoxypropane} & 90 \\ \hline
\textit{n}\text{-nonane}   & 151    & \text{1-butoxybutane}  & 141   \\ \hline
\textit{n}\text{-undecane} & 196    & \text{1-pentoxypentane}& 184   \\ \hline
\end{array}
$$

As evident, experimental data suggests that alkanes are on par with ethers when it comes to inter-molecular forces, infact the former are slightly higher.

I tried to reason this discrepancy as follows. Ethers have weak dipole, so their dipole-dipole interactions will also be weak (though stronger than their van der Waal's interactions). Alkanes on, the other hand, have van der Waal's forces only but I suspect that size is playing a dominant role here. Ethers have a v-shaped bent structure, and therefore, have reduced surface area as compared to alkanes. Therefore, I believe that alkanes lead over ether in this case and this compensates the weak dipole of ethers. Though this is only a hypothesis from my part.

But there is another problem:

\begin{array}{|c|c|c|c|}
\hline
\textbf{Compound}&\textbf{Boiling point / °C} &\textbf{Compound}&\textbf{Boiling point / °C} \\ \hline
\text{propane} & -42 & \text{methoxymethane} & -24 \\ \hline
\end{array}

Now at this point I am out of ideas. From $\ce{Et-O-Et}$ to $\ce{Pe-O-Pe}$ we had ethers leading over alkanes which felt odd but when we compare di methyl ether and propane, everything just seems fine with ether having the dominating lead (as expected initially). What is real cause? Am I missing some important concept here? Or is there any other factor that I am not considering?

I suspect London forces dominate in the longer hydrocarbons, whereas they are disrupted by the ether bond. For shorter hydrocarbons, the dipole of the O-C bond is greater than the London forces, but this is reversed as the # of carbons increases.


Bromine monochloride equilibrium The value of $K_p$ for the reaction

$$\ce{Br2(l) + Cl2(g) <=> 2BrCl(g)}$$

is $\pu{1 atm}$. At equilibrium in a closed container partial pressure of $\ce{BrCl}$ gas is $\pu{0.1 atm}$ and at this temperature the vapour pressure of $\ce{Br2(l)}$ is also $\pu{0.1 atm}$. Then what minimum moles of $\ce{Br2(l)}$ to be added to $\pu{1 mol}$ of $\ce{Cl2}$ initially to get the equilibrium mixture from above?

In this question should I include the vapour pressure of $\ce{Br2}$ in calculating $K_p$ or not?

Yes

As the reaction is written above, you may incorrectly conclude the bromine is in a liquid state throughout and hence is omitted from the equilbrium constants.

However as the question reminds us bromine is highly volatile and significant amounts of gas will accumulate above the liquid, hence most of the reactions will be between gas molecules of bromine and chlorine rather than gas and liquid.

The vapour pressure tells us the partial pressure of these gas.  Liquid bromine will evaporate until the sealed vessel contains bromine vapour with a partial pressure of 0.1 atm.

This means it is actually the equilibrium constant for the all three gases:
$$K_p=\frac{[\ce{BrCl}]^2}{[\ce{Br2}][\ce{Cl2}]}$$
where $[\ce{A}]$ is the partial pressure of $\ce{A}$.  As the vapour pressure of $\ce{Br2}$ is known to be 0.1 atm, that is the partial pressure of bromine you use in your equilibrium constant.

Hence the minimum amount of bromine to add to establish any equilibrium with the liquid is the point where partial pressure of bromine is 0.1 atm and all the bromine is in the gas phase.  Any more bromine added will condense out back into liquid.


Clarification on the Clausius inequality In Atkins' Physical Chemistry the criteria for spontaneity using Gibbs energy was calculated using the Clausius inequality:

$$\mathrm dS ≥ \frac{\mathrm dq}{T},$$

and at constant pressure, $\mathrm dq = \mathrm dH.$ Thus, $\mathrm dH - T\,\mathrm dS$ is negative.

So, if we define a function $G = H - TS$ at constant pressure and temperature, we get

$$\mathrm dG = \mathrm dH - T\,\mathrm dS ≤ 0,$$

which we use as a criteria for spontaneity.

My issue with this argument is that it implies that the Clausius inequality is for spontaneous processes only, otherwise we would end up with $\mathrm dG ≤ 0$ for all processes.

However, the derivation of the inequality simply used the fact that

$$|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$$

when the system does work.

So, where is it implied that the derivation is for spontaneous processes only?

Denbigh, in his book the Principles of Chemical Equilibrium, considers the case where "(a)the only heat transferred to the system is from a reservoir which remains at the constant temperature T; (b) the initial and final temperatures of the system are equal, and are equal to the temperature T of the reservoir; (c) apart from the system, the only other body with has undergone a change of volume at the end of the process is at a constant pressure p (e.g. a surrounding fluid such as the atmosphere); (d) the initial and final pressures are equal, and are equal to p."

For this situation, from the first law, we have
$$\Delta U=q-w$$ so that, from the definition of Gibbs free energy, it follows that:  $$\Delta G=q-w+p\Delta V-T\Delta S$$
Applying the Clausius inequality to this situation yields:
$$\Delta S=\frac{q}{T}+\sigma$$where $\sigma\geq0$ is the entropy generated, with the = sign applying if the process is reversible and the > sign applying if the process is irreversible or spontaneous.  If we combine the previous two equations, we obtain:$$w=-\Delta G+p\Delta V-T\sigma$$ So, for any process path between two specified end states of a system under these pressure and temperature constraints, the maximum work over and above p-V work is $-\Delta G$, and all other irreversible cases will involve less work (over and above $p\Delta V$).


What is the meaning of the term 'reserve acidity'? In $\ce{CH3COOH/CH3COONa}$ buffer, the reserve acidity of the solution is due to the $\ce{CH3COO-}$ ions.

What is the meaning of the term of 'reserve acidity' in the above statement?

Reserve acidity, more commonly known as Buffer capacity, is the amount of acid/base (in gram equivalents) that must be added to a buffer solution to change it's pH by one unit.

In your question, the $\ce{CH3COO-}$ is able to react with any Hydrogen ions that are added to the solution, decreasing the $[\ce{CH3COO-}]$ while increasing the $[\ce{CH3COOH}]$. It takes care of any excess acidity, hence, it is called as the reserve acidity. Similarly, the $\ce{CH3COOH}$ should be called the reserve basicity, as it can reach with any base to prevent much pH change.

For a mathematical approach, we use the Henderson-Hasselbach equation for calculating the final pH.


Can separated and purified substances be obtained from a chromatogram? I am studying chemistry, and I am learning about chromatography.

But I am confused about how chromatography is used to ‘separate and purify’ substances.

I have learnt about chromatography paper, which is used to check if a substance if pure or not. Spots of the substances to be tested are place on the chromatography paper, which is then dipped in solvent. If there are several components, the spot separates into several spots (1 per component) as they travel up the paper.

Is it possible for separated and purified components obtained from the separate spots on the chromatogram, as usable substances, or do chromatographs only show the purity of substances?

Generally, all chromatography is done on silica gel. It's the exact same procedure for TLC (thin layer chromatography), except that instead of paper you have a thin layer of glass covered with silica. What can be done in that case is scraping the spots, then extracting the product with organic solvents and filtration. There exists plates made for this, which are thicker than just for analysis.

Of course this is not viable for large amounts of products. In that case, we use a column of silica (see the image below, taken from https://bitesizebio.com/29947/basics-chromatography-column/).

This time, gravity makes the solvent go through the column and come out at the other end. Compressed air is often used to make this process go much faster.

You will however think "how do you reveal the products IN the column?" As the products aren't always colored, you can't really tell when what you want is coming out. Therefore, everything is collected in tubes of the appropriate size and revealed afterwards. If everything goes right, there is no overlap between your products and you get everything out of the column. If there is an overlap, you have to either do the column again for the mixed fractions or just discard those and keep the tubes containing only the pure product you want.

This technique is used pretty much all the time, as it can separate anything (as long as it doesn't degrade on silica) and can be adjusted by changing which solvent or mix of solvents you use.


Relative atmospheric lifetimes of carbon monoxide and methane In atmospheric chemistry, the lifetime of most substances is dictated by their reactions with the hydroxyl ($\ce{OH}$) radical. The rates of these reactions are listed by NIST as about $\mathrm{1.5e^{-13} cm^{3}\  mol^{-1}\ s^{1}}$ for carbon monoxide ($\ce{CO}$) and $\mathrm{6.7e^{-15}\ cm^{3}\ mole^{-1}\ s^{1}}$ for methane ($\ce{CH4}$). With atmospheric $\ce{OH}$ concentrations of about $\mathrm{1.2e^6\ mole\ cm^{-3}}$, this leads to an atmospheric lifetime of about 2 months for $\ce{CO}$ and 4 years for $\ce{CH4}$. This is indeed what's observed, so the relative reaction rates are definitely experimentally true.

My question is: what's the theoretical explanation for the $\ce{CO}$ reaction being faster? I would expect the $\ce{C-O}$ triple bond to be much more stable than any $\ce{C-H}$ bond. In atmospheric chemistry texts I've seen, the reaction goes like

$$\ce{CO + OH -> CO2 + H*}$$
$$\ce{H* + O2 -> HO2}$$

I've also come across notes from a pure chemistry class that mentions a different reaction with an intermediate:

$$\ce{CO + OH -> OCOH*}$$
$$\ce{OCOH* + O2 -> CO2 + HO2}$$

I can't get my head around how the intermediate would/should increase the reaction rate, though, or how something like electronic configuration could lead to $\ce{CO}$ being more susceptible to $\ce{OH}$ than $\ce{CH4}$.

The dominant oxidation (via hydroxyl radical) pathways for CO don't have any intermediate species as unstable as the methyl radical, which is step one for methane oxidation via the hydroxyl radical.


5-methylhexa-1,3-diene reaction with HBr What will be the major product of the reaction?
I guess that rearrangement will take place and give 5-bromo-5-methylhex-2-ene as a product, but I'm completely not sure

In this reaction if peroxide is not present then reaction follows markovnikov addition in which carbonation is formed as intermediate. By stability due to resonance and 2° carbonation
5-methylhexa-1-ene cation with C+ at 3 is formed. Hence 3-bromo-5-methylhexa-1-ene is formed as product.


How to determine the value for n in the Gibbs free energy and redox potential equation? I am having a problem with this equation for redox potentials

$$\Delta G = -nFE_\mathrm{cell}$$

In this equation I never am totally sure about what the value of $n$ should be, for example for the reaction shown below, would the $n$ be 2 electrons or 1 electron? Personally, I think it should be 2 electrons because that is the number of mol of electron under the simplest whole number ratio.

\begin{align}
\ce{cyt $c$(Fe^3+) + e- &-> cyt $c$(Fe^2+)}\\
\ce{NADH &-> NAD+ + H+ + 2e- }\\
\end{align}

I think this is a different matter than in the already answered question "Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?" Because in that Nernst equation the change in $n$ would be balanced by the change in the equilibrium constant. I assume that you are insinuating that there is one of the variables here which will correct for the increase in $n$. I don't see which would, as $E_\mathrm{cell}$ should be independent of quantity and $F$ is a constant.

The accepted answer is not correct, the number $n$ is not equal to 2 simply because 2 is the lowest common multiple of 2 and 1. Furthermore $n$ is not the "number of moles of electrons", which would have units of $\pu{mol}$; $n$ is in fact dimensionless.

In fact the number $n$ can be anything you like, and it depends on how you combine the two half-equations.

To use a different (not very realistic, but I don't care) example consider the half-reactions (data from Wikipedia)

$$\begin{align}
\ce{Na+(aq) + e- &-> Na(s)} & E_1 &= \pu{-2.71 V} \tag{1} \\
\ce{Br2 (l) + 2e- &-> 2Br-(aq)} & E_2 &= \pu{+1.07 V} \tag{2} \end{align}$$

We have $E_\mathrm{cell} = E_2 - E_1 = \pu{+3.78 V}$. To construct a full reaction for this cell the "usual" procedure taught is to take (1), multiply it by 2, and subtract it from (2).

$$\ce{Br2(l) + 2Na(s) -> 2Na+(aq) + 2Br-(aq)} \tag{3}$$

In this case, we have $n = 2$ as there are two electrons being transferred in this reaction, from bromine to sodium. Equivalently, you can think of (3) as being the sum of two half-reactions, each featuring the gain or loss of two electrons:

$$\begin{align}
\ce{2Na(s) &-> 2Na+(aq) + 2e-} \tag{4} \\
\ce{Br2(l) + 2e- &-> 2Br-(aq)} \tag{5}
\end{align}$$

Note that the stoichiometric coefficient of $\ce{e-}$ - the number in front of the electrons - is 2, for both half-equations.

However, you could easily construct a different full reaction, simply differing from (3) by a multiplicative factor. There's no reason why you cannot, for example, divide (2) by 2, then subtract (1) to get:

$$\ce{1/2 Br2(l) + Na(s) -> Na+(aq) + Br-(aq)} \tag{3*}$$

and here you should only have $n = 1$, because there is only one electron being transferred here, from bromine to sodium. Again you can decompose this into its half-reactions:

$$\begin{align}
\ce{Na(s) &-> Na+(aq) + e-} \tag{4*} \\
\ce{1/2Br2(l) + e- &-> Br-(aq)} \tag{5*}
\end{align}$$

and the stoichiometric coefficient of $\ce{e-}$ here is simply 1.

The bottom line is that

$E_\mathrm{cell}$ doesn't change with the stoichiometric coefficients.
$n$ does, and hence $\Delta G = -nFE_\mathrm{cell}$ does. Your choice of the full cell reaction determines the value of $n$ you work with.
The best definition of $n$ is the stoichiometric coefficient of the electrons in the half-reactions into which the cell reaction can be divided (Atkins & de Paula, Physical Chemistry).


Electrophilic substitution in meta-nitrophenol Where will electrophilic substitution take place in meta-nitrophenol (3-nitrophenol) ? The two mesomeric effects seem to compete against each other.

Electrophilic substitution can happen at only that place where electrons are present which can be donated/shared with some other electron-deficient species. In the resonance structures of m-nitrophenol shown above , there is a negative charge on hydrogen or carbon(because charge keeps on shifting between carbon and Hydrogen) and they are not that electronegative to handle the negative charge . So they want to stabilize themselves by giving off their electrons to electron deficient species by letting electrophilic substitution happen. You can see lone pair (negative charge ) of OH group  is not delocalised by NO2 group if it is present in meta position . Although NO2 group is still in Conjugation with benzene ring but for OH group, NO2 can show nothing but only -Inductive Effect.
So , now after making changes in my answer , In my opinion , electrophilic substitution will happen on left side of the molecule at ortho position. On the right side at ortho position , electron density is less as compared to the left side because NO2 is showing -I effect making conditions less suitable for electrophilic substitution to happen.


How to find the molar mass of an iron compound based on its mass and mole fraction in the compound? I'm trying to solve this question:

  We have a molecule composed of $3$ iron atoms, and $4$ atoms of another element. We are given the following information: it has $\pu{2.36 g}$ of iron for $\pu{3.26 g}$ of molecule.

I want to find the molar mass of the compound, I have tried so far:
$$ m = \pu{3.26 g} = \pu{0.00326 kg}$$
Since it has $3$ atoms of $\ce{Fe}$ and $4$ atoms of an unknown substance, therefore:
$$3 + 4 = 7~\text{atoms},\\
  \pu{1 mol} = \pu{6.022* 10^23 atoms}\\
\frac{7}{\pu{6.022* 10^23}} = \pu{1.16 * 10^-23}$$

As we know: $M = m / n,$ I tried to divide $0.00326$  by $\pu{1.16 * 10^-23}$ and I obtained $\pu{2.79429 * 10^19}$, but the correct answer is $\pu{231.43 g/mol}$.

What I have done wrong?

Let $\ce Z$ denote the unknown element.
The total amount of iron atom in the compound is,

$$n(\ce{Fe})=\frac{\pu{2.36g}}{\pu{55.845 g\cdot mol^{-1}}}=\pu{0.0423 mol},$$

The molecule comprises of 3 iron atoms and 4 other atoms, so the amount of the unknown atom is,

$$n(\ce Z)=\pu{0.0423 mol}\times \frac43=\pu{0.0564 mol},$$

The molar mass of $\ce Z$ is,

$$M(\ce Z)=\frac{\pu{3.26g}-\pu{2.36g}}{\pu{0.0564 mol}}=\pu{15.97 g\cdot mol^{-1}}$$

So $\ce Z$ is $\ce{O}$. The molar mass is
$$M(\ce{Fe3O4})=\pu{55.845 g\cdot mol^{-1}}\times 3 + \pu{15.97 g\cdot mol^{-1}}\times 4=\pu{231.43 g}.$$


Vapour Pressure in Closed Containers When the vapor pressure exerted by the gas equals that exerted by the liquid, do their densities equal?

V, you are mixing terms a little.

Vapor pressure is a quantity indirectly describing the amount of a substance must be in the gas phase for the substance to be in equilibrium between the liquid and gaseous states.  If the partial pressure of the gas (say water vapor) is equal to the vapor pressure of the liquid (must be the same substance, so water) than the liquid will have no net flow either from the gas phase to the liquid phase or vice versa.

There is no need for the substances to have the same density. This is only true for super-critical fluids.  Super-critical fluids exist once you have raised the temperature above a critical point after which there is no phase transition between liquid or gas.


How can one determine if all reagents will react in the reaction? Is it correct that all the reagents above the arrow will participate to form an intermediate product, to which the reagents below the arrow are then added, and they react to form the final product?
Thank you in advance.

That is correct, the two numbered reagent and solvent pairs refer to two reactions, that will be done after one another.
There is an intermediate product that can be isolated, or at least identified.


Oxidation of malonic acid Question
  How many grams $\ce{CO2}$ will result when burning $\pu{100 g}$ malonic acid?
  The formula for malonic acid is $\ce{C3H4O4}$ so 3 $\ce{CO2}$ molecules should result from oxidating one malonic acid molecule.
  $$\ce{C3H4O4 + 2O2 -> 3CO2 + 2H2O}$$

My attempt

$$44.01\cdot\frac{100}{104.0615}\cdot3 = \pu{126.88 g}$$

with $\pu{44.01 g}$ being the mass of a mol $\ce{CO2}$ and $\pu{104.0615 g}$ for a mole of malonic acid.

However, this doesn't seem to be right. Can anyone hint me for the correct solution?

You actually are correct, the value is greater than 100g because when you burn a hydrocarbon, it is actually an exothermic redox reaction with oxygen.  Meaning that the carbon in the compound is combining with atmospheric oxygen.  So even if you use 100g of malonic acid, you could end up with more than 100g of CO2.  Your answer is correct.


Why is % yield in a purification table measured using activity instead of total protein? I'm trying to understand why % yield is measured with activity instead of total protein in mg. According to my biochemistry notes, activity is "1.0 units of enzyme activity = 1.0 μmol of substrate catalyzed to product under optimum conditions."

Then, yield is defined as "% of total starting protein/enzyme in final purified protein fraction."

With this definition of yield, it would make sense to say that you get 30% yield in the second step since you have 3,000 mg / 10,000 mg of the original protein amount. The definition of yield in my notes mentions nothing of activity.

Because total protein includes other proteins than the one desired.

"% yield" is expressing how much of the enzyme desired was actually recovered.


Resonance contributors and IR stretching frequency I am told that a carboxylic ester would have a higher carbonyl stretching wave number than a carboxylic amide due to there being more double bond character in the carbonyl group of the ester.

This makes sense prima facie; the resonance structures suggest that there is somewhat more double bond character in the carbonyl of the ester. And nitrogen and oxygen are nearly identical in mass.

Is this actually true and applicable?

Yes, it is true and applicable. In fact, the resonance structure you've drawn with the double bond between the carbonyl carbon and the amide nitrogen is so significant that one can actually observe restricted rotation about this bond on the nmr timescale.  That is to say, if a methyl group was attached to the amide nitrogen, you can see distinct resonances for the syn and anti isomers.  The rotational barrier about the C-N bond is around 20 kcal/mol, depending upon the substituents, etc.  In the case of the ester, for the reason you provided, the barrier to rotation about the carbonyl carbon - ester oxygen bond is much lower, typically in the range of 10-12 kcal/mol.

So in the case of the amide, the second resonance structure is more important than it is for the ester.  Consequently, the amide carbonyl bond is weaker than the carbonyl bond is for the ester.


Why do some substances ionize instead of dissolving in water? My question can also be understood as

  what make something electrolyte or nonelectrolyte?

I think this question stems mainly from my confusion with the definitions of ionizing and dissolving.

Dissolving of a ionic substance is due to ion-dipole attraction be/wn water molecules and ionic formula unit. If water molecule has enough attraction to break down the ionic compound, the the substance will dissolve.

Ionizing. Uh, all I know about it is...I don't really know. I think only ionic compound will ionize because it is composed of ions. So once it dissolves it basically ionize, right?

Dissolution is simply the mixing of two phases to produce a single homogeneous phase. Ionization is a process by which a substance with no net charge is converted into one or more ions. It includes the mundane example of dissolving an ionic compound in water that dissociates into its constituent ions as well as things like forming ions by knocking electrons off with an electron beam. They're both surprisingly complex topics, but I think the question can be answered conceptually by considering a few compounds: $\ce{NaCl}$ , acetic acid ($\ce{CH3COOH}$), and glucose:

(source)

All dissolve readily in water, but only two of them are electrolytes. As you note, $\ce{NaCl}$ is composed of ions and breaks apart into $\ce{Na+}$ and $\ce{Cl-}$ in solution. An electrolyte is any compound that ionizes when dissolved in a given solution, so $\ce{NaCl}$ is an electrolyte in water. Moreover, because there's is no way for a salt to be dissolved without ionizing in water (i.e. $\ce{NaCl_{(aq)}}$ isn't possible), $\ce{NaCl}$ is known as a strong electrolyte as when it dissolves, it ionizes completely.

$\ce{CH3COOH}$ also dissolves readily in water, but isn't an ionic compound. Because it's a weak acid, it exists in this equilibrium in solution:
$$\ce{CH3COOH_{(aq)} + H2O <=> CH3COO^{-}_{(aq)} + H3O+}$$

Because it can lose a proton to form ions, it does ionize in solution and is thus an electrolyte, but because it's a weak acid, the reaction doesn't go to completion and some $\ce{CH3COOH_{(aq)}}$ exists in solution making acetic acid a weak electrolyte as it doesn't completely ionize in solution.

Finally, glucose readily dissolves in acid, but none of its protons are readily lost so it doesn't ionize at all meaning that it isn't an electrolyte in water. Without getting too much into the thermodynamics of dissolution, the reason glucose is so soluble is because the many hydroxyl groups allow it to hydrogen bond extensively with water, where the equivalent compound without them, hexanal, is only slightly soluble due to the polar carbonyl group, and hexane is essentially insoluble.