Does a freezing point of a substance always increase as its boiling point increases? I am studying for an introductory to physical sciences test and came upon a chart describing substances found in petroleum and the distillation process.

Each freezing point increases as each boiling point increases.  Each of these properties also increases with the rise in density.

Is this just true for substances in petroleum (methane, ethane, propane, butane, pentane...) or is this consistent throughout all substances?

Update:

(Pure substances)

Yes and no.

Yes in the sense that boiling point and freezing point usually correlate. If there is a line of homologous compounds, than this properties rise in general, as both are increased with increase in number of dispersion interactions involved. However, it is likely to find several compounds with odd behavior, especially around beginning of the line.

As a counterexample. Toluene has higher boiling point than benzene, but much lower freezing point. This is because toluene has less symmetrical form and consequently has harder time packing into strictly ordered lattice.

Another counterexample. Gallium melts around 30 celsium, but boils well above 2000 celsium. However, Tallium melts around 300 celsium, but boils around 1700.

However, for similar compounds the values usually correlate in the series. Say, for alkanes both values steadily grow with increase of number of carbons... give or take occasional peaks due to details in crystal packing.

However, if you consider compounds of different families, it is not predictable without looking into structures of the compounds.


Is Fluorine more toxic than Chlorine? Fluorine is more reactive than Chlorine. But does it mean that Fluorine can cause more damage to living tissues? If so, why wasn't Fluorine used in WW1 instead of Chlorine?

Fluorine is much more reactive than chlorine and would certainly cause more damage to living tissues. You can even check out https://www.youtube.com/watch?v=vtWp45Eewtw for some fun demonstrations of its oxidizing power too!
Likewise, compared to chlorine gas, I'd assume fluorine munitions would be significantly harder to manufacture, store, and handle for any side attempting to wield them in combat. (And, as Waylander mentioned in the comments, especially for "WW1-era technology" too!)
Also, due to its great reactivity, fluorine gas will readily react even with the atmospheric humidity to form hydrogen fluoride: $$\ce{2F2(g) + 2H2O(g) -> 4HF(g) + O2(g)}$$ which, while still toxic, is much lighter than chlorine gas and so may not be as good at staying "close to the ground" and/or "flowing down" into an enemy's trenches as intended. Or even worse -- potentially thus dispersing even back to your own troops and damaging them as well!


When titrating a weak acid should only undissociated acid molecules be considered? Let's say we do the titration of a weak acid (acetic acid in this case)  and a strong base (sodium hydroxide) represented by the following equation:
$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$

My question is that, since $\ce{CH3COOH}$ is partially dissociated, does $\ce{NaOH}$ react  only with the partially dissociated part or the complete acetic acid solution?

Acetic acid and acetate are connected by the dissociation equilibrium $(1)$:

$$\ce{CH3COOH + H2O <<=> H3O+ + CH3COO-}\tag{1}$$

So you basically have two acidic particles that can react with hydroxide: hydronium or acetic acid. Hydronium will actually react much faster on a molecular level even though its concentration is significantly lower as it is much more mobile in solution. However, any hydronium consumed is immediately reformed by acetic acid molecules reestablishing the equilibrium.

Therefore, on a very minute level hydroxide reacts with neither acetic acid nor acetate but with hydronium which is produced in the dissociation of acetic acid. In practice, that still means that hydroxide will be added until $n(\ce{CH3COOH})+n(\ce{CH3COO-})=n(\ce{OH-})$. From this point onwards, there will be more hydroxide than acidic particles causing the $\mathrm{pH}$ value to rise significantly, the indicator to change colour and the student to realise that they reached (and passed) the equivalence point.


Reviving a 6 Volt lead-sulfuric acid battery I have a 6 Volt sulfuric acid battery used to power a golf cart that has all the acid drained out, and the usual solution was to add more water to the battery however I am afraid that would not work this time since there is no acid left. Could I solve it by adding more acid from other batteries? or do I need to be accurate in the amount of acid added? or could I just solve it by adding water? If you could also give me an explanation as to how these batteries work and why adding water has any benefit I would appreciate it.

Depending on how long the battery has been sitting empty even adding electrolyte might not help.

Your best bet is to fill the battery with distilled water and put it on a charge, charge it for a couple of days if possible.

Get yourself a hydrometer (used for checking the specific gravity of your electrolyte) and check the water's gravities. After you charge the battery for awhile check them again. If the gravities start to rise, and the liquid is reasonable clear stop charging the battery and let it sit 12-24 hours. (If however the water turns cloudy/grey the battery is scrap get a new one).

After the battery has sat and cooled down, empty the water into a container, and neutralize with soda ash, or baking soda. BE CAREFUL even though what you put it was water, what will come out might/will be corrosive. Dispose of properly.

Now go to your local auto parts store and buy a box of battery electrolyte its usually sold as 1.28 spg, fill the battery with electrolyte re-charge until your gravities come up, and voila, use just re-conditioned a battery.

Be sure to wear rubber gloves / goggles or face shield and any other appropriate safety gear. Remember that when a battery charges it releases hydrogen sulfide gas which is explosive so no open flames sparks etc....

Of course the battery may only last your a couple of months or another year or so, and it may not even come back at all. But I have used this procedure with great success to revive old batteries....and squeeze a bit more life out of them.

Jeff
Industrial battery specialist for 30 years.


Can powdered beta-tin be made from raising the temperature of alpha-tin? When white tin ($\beta$-tin) is cooled to a temperature below $13.2\ \mathrm{^\circ C}$, it creates the allotrope of gray ($\alpha$-tin), a gray, amorphous powder.

My question is that once you have the powdered gray tin, doe just raising the temperature above the point of stability turn it back to white tin, but in a powdered form?

Is there a change in appearance of the gray powder when that happens, or is it mostly unnoticeable?

Does the conversion to gray tin change the melting point?

$\alpha$Sn and $\beta$Sn are the two solid allotropes of Sn.  As you not, below 13C the stable phase is $\alpha$Sn, which has a diamond cubic crystal structure (like diamond, Si, and Ge) and is a semi-metal.  Above 13C, the thermodynamically stable phase is $\beta$Sn, a body-centered tetragonal crystal. So, cycling back and forth around 13C varies the thermodynamically stable phase back and forth, with the only question being the kinetics of the transformation.  The video in the comment by @JasonPatterson shows that the transformation indeed takes place reasonably quickly (unlike diamond to graphite for carbon).

As for melting, if you rapidly heated $\alpha$Sn and avoided the phase transformation, you would find that the melting temperature would be lower. Using the SGTE thermodynamic data (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)), one finds that the $\alpha$Sn -> liquid phase transition would occur at about 430.7K, ~75K lower than the standard $\beta$Sn -> liquid melting point at 505.06K.


Reaction mechanism for the conversion of acyl hydrazide to acyl azide How is an an acyl hydrazide converted into an acyl azide in the presence of $\ce{HNO2}$? According to me, the $\ce{-NH2}$ group of hydrazide will get converted into $\ce{-N2+}$ through $\ce{HNO2}$. However, I am unable to proceed after that.

You are correct acyl hydrazide can be converted into acyl azide by adding $\ce{HNO2}$ also you wrote that the $\ce{NH2}$ group of hydrazide will be diazotisated making a $\ce{N2+}$ group hence I am assuming that you know the mechanism, after diazotisation the adjacent $\ce{N}$ donate the lone pair to the $\ce{N2+}$ group making the final product as acyl azide.

The mechanism scheme is in the following picture :


Calculating conductivity of a solution given molar conductivities and concentration From my understanding the conductivity of a solution ($K$) is given by the sum of the conductivities of ions in solution:

$$K = c(K_{m,1}+K_{m,2})$$

Where $c$ is the concentration, and $k_{m1}$ and $k_{m2}$ are the respective molar conductivities of the two ions. For example, with a solution of $\pu{0.1 mol L-1}$ $\ce{NaOH}$:

\begin{align}
K &= 0.1(K_{m,\ce{Na+}} + K_{m,\ce{OH-}})\\
K_{m,\ce{Na+}} &= \pu{50 S cm2 mol-1}\\
K_{m,\ce{OH-}} &= \pu{197.9 S cm2 mol-1}\\
K &= 0.1(247.5)\\
K &= \pu{24.75 S m^-1} (?)
\end{align}

However I think the units are throwing me off here as my calculations aren't making much sense - I was led to believe the conductivity was much lower.

Could you please offer some help with these calculations?

Just be explicit with all the units, including for the molar concentration, and be sure they cancel out properly.  You have $\pu{0.1 mol L-1}$ for the solution concentration, but your $K$ values are in $\pu{S cm2 mol-1}$.

So convert your molar concentration to cubic centimeters (there are $\pu{1e3 cm3}$ in one liter) and you get $\pu{1e-4 mol cm-3}$.

Then when you multiply both values and your units cancel nicely to give you

$$\pu{1e-4 mol cm-3} \times \pu{247.5 S cm2 mol-1} = \pu{2.475e-2 S cm-1}$$


How dangerous is superglue? A long time ago, when I used to make plastic models with glue, I never wore white gloves when gluing together the pieces, so I often got a lot of superglue on my hands. The superglue, which was very viscous and liquid, dried on my hands and later created small flakes in the process of coming off my skin.

At the time the back of the superglue container were listed instructions that read “in case of contact with skin, rinse with warm water until the glue is completely removed” or something similar.

My questions are

Is it okay to have ever had glue on my skin? Will it not create a lasting problem?
Is it okay that the flakes could have been spread on the bathroom towel when I wiped my hands, on the bedsheets, my clothes, the surface of the desk, the floor, and in other areas in the home?

I'm worried that the dust could have remained or been absorbed through the mouth, eyes or even nose (inhaled)?

*by glue I mean the type of glue that can be found on this link:

https://www.amazon.com/Tamiya-87038-Extra-Cement-Plastic/dp/B01F1S4N8U

You should not regard this answer as medical advice. If concerned about your health consult a physician.

The glue you link to (TAMIYA  EXTRA-THIN  CEMENT, made by IKEGAMI PAINT INDUSTRY COMPANY), according to a material safety data sheet (MSDS) I was able to retrieve online, appears to be a 1:1 mixture of acetone and butyl acetate. Such a mixture works in a similar way to dihloromethane as a solvent to weld together plastic parts. Quoting the wikipedia,

  Acetone is believed to exhibit only slight toxicity in normal use, and there is no strong evidence of chronic health effects if basic precautions are followed.

OSHA classifies acetone under the following health codes ("principal effect(s) of exposure to each substance [...] used in determining if a violation of an air contaminant standard is serious"):

  HE16     Irritation-Eyes, Nose, Throat, Skin---Mild

OSHA classifies butyl acetate as follows:

  HE8    Nervous System Disturbances---Narcosis

  HE15   Irritation-Eyes, Nose, Throat, Skin---Moderate

For completeness, I include a screenshot from the beginning of the MSDS I found for the adhesive. For more on the GHS classification (divisions or categories) you can refer to the method employed in classifying substances.


Cause of imaginary part of Warburg element in EIS In electrochemical impedance spectroscopy experiments (EIS) the diffusion of electroactive species at flat, macroscopic electrodes is commonly described by the Warburg element with the impedance expression

$$Z_W = \frac{σ}{\sqrt{w}} - i\frac{σ}{\sqrt{w}}$$

This means that the values have a phase angle of $-45^\circ$.

What is an intuitive way to understand the fact, that the Nernst-diffusion layer contributes both a real and imaginary part to the impedance of the system?

I find that it is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. This would explain the real part. I am unsure however, where the imaginary part, or energy-storing component, fits into this picture.

It is very hard to find any information in this regard, as practically all literature treats this just as a result of the mathematical derivation of the faradaic impedance, but does not go into any kind of intuitive explanation.

I am sorry, this is not an answer but a comment, too long to be edited in the comments section.

I don't agree with "straightforward" in the sentence :
" It is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. "

This would be an oversimplified view which implicitly considers that the phenomena involved are only on the energy dissipative kind (commonly modelized with resistors elements in the equivalent circuits). This cannot be so in case of Warburg impedance. Necessarily the phenomena involved includes electrical energy storage elements which are stocking energy during a part of the period and destocking during the other part of the period of the alternative current (commonly modelized with capacitor elements). One cannot forget the part of the alternative current due to the phenomena of energy storage kind. If we forget that, there is no phase shift and no complex impedance, thus no Warburg element.

In an intuitive way, but discutable because alternative models can be proposed, if we think that the more the diffusion layer grows, the shallower the concentration gradient gets also the shallower the electrical gradient of potential gets. So, the energy storage phenomena also decreases and the "out of phase" part of alternative current decreases as well. This is a loosely way for an intuitive approach.

The case of Warburg element (which is a particular case of CPE : http://www.consultrsr.net/resources/eis/cpe1.htm) can be related to dissipative energy phenomena and storage energy phenomena $\textbf{homogeneously}$ distributed. A possible equivalent circuit can be made of repeated identical pattern of resistor-capacitor. This is a particular case of network for example of the kind represented page 10, figure 7(a) in the paper : https://fr.scribd.com/doc/71923015/The-Phasance-Concept , only with the part noted $P_\varphi$, without $R_p$. Of course different interpretations and equivalent circuits can be proposed, depending on the kind of electrochemical system. This is briefly discussed page 26 in the above referenced paper.


Why are hydrocarbon flames yellow or blue? A candle gives a yellow flame while kerosene gives a blue flame on burning.
Now, saturated hydrocarbons give a blue flame but unsaturated ones give a yellow flame. I want to know why the colour of the flame is yellow and blue respectively?

It depends on the amount of oxygen available and on the ratio of hydrogen to carbon in the fuel.

Hydrogen burns with an almost invisible flame no matter how much oxygen is present, but solid carbon burns with a yellowish (black-body) flame. If the fuel combusts in a multi-step process where all reactants are gaseous, you get a bluish flame. For example, the combustion of methane produces methanol, then formaldehyde, formic acid, carbon monoxide etc.

The combustion of acetylene at lower temperature, with little oxygen, however, starts with cracking of the C-C triple bond, releasing carbon particles that have a luminous flame.

So the greater the hydrogen:carbon ratio, the easier it is to get a blue flame.


NMR chemical shift range of different elements A typical $\ce{^1H}$ NMR runs from approximately 0 to 10 ppm, give or take a bit. $\ce{^13C}$ NMR runs from 0 to 200. And $\ce{^59Co}$ NMR runs from -5000 to 15000 ppm!

There seems to be some correlation with atomic number, although three points hardly make a trend. Is there an explanation for this? I'm not interested in the exact numbers, but rather the relative orders of magnitude.

The are a number of important factors that contribute to the shielding of a nucleus. Chemical shifts arise due to differences in the local magnetic field in the different environments within a molecule, driven by overall electron density about that nucleus. That is to say, the effective magnetic field experienced by a nucleus is a function of the shielding constant, σ. There are a number of components of σ.

$$B_\mathrm{eff} = B_\mathrm{o}(1- \sigma)$$

σdia is the local diamagnetic contribution
σpara is the local paramagnetic contribution
σm is the neighbour anisotropy effect
σrc is the ring current contribution
σef is the electric field effect
σef is the solvent effect

The two most important factors for the chemical shift range are the diamagnetic and paramagnetic contribution. All other contributions are quite small and operate within the effects of these fist two.

The diamagnetic contribution comes predominantly from electrons in the s-orbital; for most nuclei this is almost constant for all arrangements. Its effect diminishes with increasing distance from the nucleus, which is why s-orbitals play a more important role than p-orbitals. Only for those nuclei that have their outer orbitals as s-orbitals ($\ce{^1H, ^2H, ^7Li, ^{23}Na}$ for example) will the diamagnetic contribution be the dominant contributing factor, and this effect is generally small, providing only modest chemical shift ranges of up to tens of ppm.

The paramagnetic contribution is the dominating factor for all nuclei that have electrons in outer shell p, d and f orbitals. Paramagnetic shielding comes from the imbalance of electron density of the valence electrons in the outer orbitals. The way I get my brain around this is by thinking that the larger the non-spherical orbitals, the harder it is to get uniform distribution. The effect of paramagnetic contribution is large; up to many thousands of ppm.

To address your question, there is a general relationship that describes the expected chemical shift range being proportional to the inverse cube of the valence shell radius and also to the populations of the p and d orbitals. So, you are correct in your observation of chemical shift range being basically proportional to atomic number.

A useful reference for you is:
W. von Philipsborn, Chem. Soc. Rev. 1999, 28, 95. DOI: 10.1039/A706424A


Reaction of primary nitroalkane with nitrous acid I was reading the reaction of primary nitroalkane with nitrous acid. I read this reaction

Another mechanism which I saw:

I am now confused  whether the first mechanism occur or the second one.I tried to search it on google but didn't get anything which could clear this doubt .Please help me regarding this.This I need to know because the reaction written in my book after 2nd reaction is-

If 1st reaction is right then what will occur after adding $\ce{NaOH}$ in the product. Also when I read Victor Mayer test the 1st reaction was written there. Due to this I am in a dilemma. Please explain.

As the comments say, it's both.  The two products can tautomerize in the same way as a ketone and it's tautomeric enol.  In fact, this is basically a keto-enol tautomerization, disguised by occurring around a nitrogen atom instead of carbon.


How to interpret a singlet term symbol for an s¹p¹ configuration? I am interested in the $^1P_0$ singlet state of $\ce{Hg}$ ($\ce{5d^10 6s^1 6p^1}$).

I interpret this to be the two following electron configurations below, because for an $S = 0$ state the two electrons have to have different spins.

Are these two both a valid state? Do they both have the same energy? How would I tell if it's only one of them?

The two states you have drawn correspond to the (antisymmetrised) wavefunctions

$$\begin{align}
\Psi_i &= \mathrm{s}(1)\alpha(1) \cdot \mathrm{p}(2)\beta(2) \\
\Psi_j &= \mathrm{s}(1)\beta(1) \cdot \mathrm{p}(2)\alpha(2)
\end{align}$$

Neither of the two states above are valid, as they do not obey the requirement for indistinguishability. For fermions such as electrons (spin-1/2), this is achieved by making the wavefunction antisymmetric upon permutation of any two electron labels. However, neither $\Psi_i$ nor $\Psi_j$ obey this:

$$\begin{align}
\mathcal{P}(\Psi_i) &= \mathrm{s}(2)\alpha(2) \cdot \mathrm{p}(1)\beta(1)  \neq -\Psi_i \\
\mathcal{P}(\Psi_j) &= \mathrm{s}(2)\beta(2) \cdot \mathrm{p}(1)\alpha(1)  \neq -\Psi_j
\end{align}$$

The appropriate wavefunction is a linear combination of these states (and a couple more):

$$\Psi_\text{singlet} = \frac{1}{\sqrt 2}[\mathrm{s}(1)\mathrm{p}(2) + \mathrm{p}(1)\mathrm{s}(2)] \cdot \frac{1}{\sqrt 2}[\alpha(1)\beta(2) - \beta(1)\alpha(2)]$$

If you expand these you can see the constituent wavefunctions

$$\begin{align}
\Psi_\text{singlet} &= \frac{1}{2} [\mathrm{s}(1)\alpha(1) \cdot \mathrm{p}(2)\beta(2) \\
&\qquad + \mathrm{p}(1)\alpha(1) \cdot \mathrm{s}(2)\beta(2) \\
&\qquad - \mathrm{s}(1)\beta(1) \cdot \mathrm{p}(2)\alpha(2) \\
&\qquad - \mathrm{p}(1)\beta(1) \cdot \mathrm{s}(2)\alpha(2)] \\
\end{align}$$

You need four "simple" wavefunctions in total because you have to account for the spatial indistinguishability of the two electrons (hence need a linear combination of two spatial wavefunctions) as well as the spin indistinguishability (linear combination of two spin wavefunctions).


Dependency of rate of diffusion on mean free path and velocity of molecules? What is the dependency of rate of diffusion ( Graham's law) on mean free path, molecular size , average velocity, collision frequency and pressure? I have searched everywhere on internet but just found that rate of diffusion is inversely proportional to square root of density or molecular mass.

My reference is Concepts in Thermal Physics, Blundell and Blundell, Ch. 7, 2nd ed. Everything here is from there.

Consider a gas trapped in a rigid box. We begin by evaluating the molecular flux incident on one part of one wall of the box. Let there be $n$ particles per unit volume, so $n$ is the particle density of the gas in the box. Let $A$ be the area of a small patch of the wall, and let $\theta$ be the angle between the particle's direction of approach and the normal to the wall. Let $f(v)$ denote the speed distribution function of the particles, which we assume to be a Maxwell-Boltzmann distribution.

Consider all the molecules with speed between $v$ and $v+\text{d}v$ approaching the wall at an angle between $\theta$ and $\theta+\text{d}\theta$, which we write in shorthand as ($v,\theta$). There are $n$ particles per unit volume, and a fraction $f(v)\,\text{d}v$ of them are moving within the desired speed range. A further fraction $\sin\theta\,\text{d}\theta/2$ of them are moving in the desired direction, so the particle density for ($v$, $\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2.$$ (To derive the fraction $\sin\theta\,\text{d}\theta/2$, we consider a unit sphere with surface area $4\pi$. The surface area of the differential annulus is $2\pi\sin\theta$ (circumference) $\times\;\text{d}\theta$ (differential width), so the fraction of surface area occupied by the annulus is as stated.)

Now consider how many particles traveling at ($v,\theta$) hit the patch of wall in a time $\text{d}t$. These particles must be contained within a volume $$A\,\times\,v\,\text{d}t\cos\theta$$ (distance particle travels normal to the wall in a time $\text{d}t$), for all particles traveling at ($v,\theta$) must impinge on the wall within $\text{d}t$. Actually this area is slightly enlarged by a width $v\sin\theta\,\text{d}t$ uniformly, but this contribution to the area is of order $\text{d}t^2$ and vanishes in the small $\text{d}t$ limit.

All in all, then, the number of molecular collisions in $\text{d}t$ on a patch of wall of area $A$ at ($v,\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2\,\times\,Av\,\text{d}t\cos\theta.$$

The flux $\Phi$ is the rate of molecular collisions per unit area, so we divide by $A$ and $\text{d}t$, and integrate over all $v$ and $\theta$. This yields $$\Phi = \int_0^\infty\text{d}v\int_0^{\pi/2}\text{d}\theta\,nvf(v)\sin\theta/2\cos\theta = \frac{1}{4}n\langle v\rangle = \frac{p}{\sqrt{2\pi mk_BT}},$$ assuming an ideal gas and given the average speed of gases within the Maxwell-Boltzmann distribution.

By definition of the molecular flux, the rate of effusion is simply $\Phi A'$ for $A'$ the cross-sectional area of the hole out of which gas effuses, and the resulting dependencies can be read off from the equation for $\Phi$.

Interesting notes.

Due to the extra factor of $v$ in the integral for the molecular flux, the distribution of speeds of the effusing particles is $v^3e^{-\beta\epsilon}$ , rather than the standard $v^2e^{-\beta\epsilon}$. This extra factor of $v$ comes about because speedier molecules hit the walls of the box faster than do slower ones.
The hole through which particles effuse must be small, and the length scale that governs smallness is the mean free path. Make the hole too large, and the box eventually reaches equilibrium with the outside world. Consider a system of two boxes placed side by side, and a small hole between the neighboring walls of the boxes. For holes small enough that effusive behavior is observed, at equilibrium the molecular fluxes must balance and hence $$\frac{p_1}{\sqrt{T_1}} = \frac{p_2}{\sqrt{T_2}}.$$ On the other hand, if the holes are too large, then we simply have regular thermodynamic equilibrium, with $$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$$


Synthesis of Polypropylene glycol Polypropylene glycol is produced by ring-opening polymerization of
propylene oxide. The initiator is an alcohol and the catalyst a base,
usually potassium hydroxide. When the initiator is ethylene glycol or
water the polymer is linear. With a multifunctional initiator like
glycerine, pentaerythritol or sorbitol the polymer branches out.
Source: https://en.wikipedia.org/wiki/Polypropylene_glycol
Polypropylene glycols ... are formed by action of base on propylene oxide in the presence of 1,2-propanediol as an initiator.

I don't understand the meaning of an initiator here. My mechanism is the first the OH- adds to the propylene oxide giving $\ce{HOCH2CH2O-}$ which again adds to another molecule of the oxide and the chain continue and the polymer is formed.
My mechanism requires no role of the "initiator". I want to know what is the requirement of the diol in initiating the polymerisation, whether my mechanism is correct and if wrong, what is the actual mechanism?

If you just use hydroxide, then the initiator is water.


Yellow precipitate in the presence of ammonium molybdate? We had analytical chemistry experiments to determine acidic radicals from a mixture of two salts. After the chloroform-chlorine water test for distinction between halides $(\ce{salt\ solution + H2SO4 + CHCl3 + Cl2_{(aq)}})$ when I emptied the test-tube in the sink, all of it suddenly turned yellow.

What reaction could have occurred to cause this?
The sink mostly contained diluted $\ce{H2SO4}$ and $\ce{HNO3}$ along with small quantities of $\ce{AgNO3, BaCl2, NH4OH}$, lead acetate and ammonium molybdate.

All of these would mostly be in small quantities since I had rinsed the sink with water before emptying the test-tube.

[...] diluted $\ce{H2SO4}$ and $\ce{HNO3}$ [...] $\ce{NH4OH}$ [...] and ammonium molybdate.

The yellow precipitate indicates that phosphate was present too. Under these conditions a yellow ammonium molybdatophosphate is formed:

$\ce{H2PO4^- + 22 H+ + 12 MoO4^2- +3 NH4+ -> \underset{yellow}{(NH4)3[P(Mo3O10)4]} +12H2O}$


Finding the mole fraction At $25\ \mathrm{^\circ C}$ the vapor in equilibrium with a solution containing carbon disulfide and acetonitrile has a total pressure of $263\ \mathrm{torr}$ and is $85.5\ \mathrm{mol}\text{-}\%$ carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At $25\ \mathrm{^\circ C}$ the vapor pressure of carbon disulfide is $375\ \mathrm{torr}$. Assume that the solution and the vapor exhibit ideal behavior.

I was gonna use Raoult's law but like how is the molar percentage different from the molar fraction, isnt it the molar fraction times 100? also I checked the solution manual and it had the molar fraction sign but with an L or V above it, and theres no such thing in the boook, can anyone help or explain what is that?

As you said, the molar percentage is just the molar fraction multiplied by 100. As to the L and V above the molar fractions, they refer to the molar fraction of the component in the liquid and vapor phase, respectively. They are needed since (in general) the molar fraction of a component in a mixture are different in the liquid and vapor phase.

So, from the data in your problem, you know that the molar fraction of carbon disulfide in the gas phase is 0.855, and let's call it $x_{CS_2}^V$. The total pressure is P=263 torr, and since the vapor exhibits ideal behavior, the partial pressure of CS2 is: $P_{CS_2} =x_{CS_2}^V P$. Using Raoult's law (since the liquid exhibits ideal behavior), this partial pressure must be equal to the molar fraction of CS2 in the liquid times the vapor pressure of pure CS2: $P_{CS_2} =x_{CS_2}^L P_{CS_2}^o$. And I am sure you can follow from here.


How can I isolate glycolic acid from sugarcane juice? This beginning of a bachelor thesis uses ethylene glycol as a catalyst in the isolation of glycolic acid. I can't find the rest of the article (it ends in the middle of the description) to find out how they did it.
I'm wondering if someone could please help me I want to learn how to do this as an amateur dermatologist. Currently studying pre-med, small knowledge of chemistry.

Ethylene Glycol was not used as a catalyst! Kataoka described in 2001 the production of Glycolic acid (GA) from ethylene glycol using glycol-oxidizing microorganisms isolated from soil, which is what the author of this thesis is trying to say (the writing is unclear though). It will be difficult to reproduce this in your home since you would need to cultivate those microorganisms inside your own house, chemical oxidation would also be a problem if you mean to use it for medical purposes (the oxidation of ethylene glycol releases formaldehyde, which is toxic and probably carcinogenic), besides the reaction needs potassium dichromate to happen which is also hazardous.

Fruits produce a small amount of glycolic acid and you can try to extract it from banana or sugarcane pulp by an acid-base workup, though I believe you won't get much pure GA using this method (specially if you don't have a deep knowledge of organic chemistry).

Glycolic acid is very easy to find online and it will be a lot cheaper to buy it from ebay or other source then to synthesize it if your goal is to use it for its skin effects.

Source: Glycolic Acid Production Using Ethylene Glycol-Oxidizing Microorganisms
Michihiko KATAOKA, Mié SASAKI, Aklani-Rose G.D. HIDALGO, Michiko NAKANO & Sakayu SHIMIZU
Bioscience, Biotechnology, and Biochemistry Vol. 65 , Iss. 10,2001


Chromatography: Why does the insoluble compound get pulled along? In paper chromatography, we see that a paper with an ink spot on it get dissolved in the liquid and travels up with water due to capillary action. Then the compounds with more affinity to the paper get deposited on the paper. My question is why do the molecules with more affinity to the paper move with the liquid in the first place and then ultimately settle down after some time?

Chromatography only works when the affinity for the paper vs. affinity for the water is in a dynamic equilibrium.  Suppose there are three compounds in the ink, called $\ce{I}$, $\ce{N}$, and $\ce{K}$.  Lets call the  spots on the paper fiber that bind ink molecules as $\ce{P}$.  Then let's write the dyanmic equilibria this way:

$$\ce{I(aq) + P <=>> PI} $$
$$\ce{N(aq) + P <=> PN} $$
$$\ce{K(aq) + P <<=> PK} $$

This means that any given $\ce{I}$ molecule will be absorbing to paper, then becoming unstuck from the paper, then absorbing to a (probably different) site on the paper, then desorbing again and returning to aqueous solution.  The same is true of $\ce{N}$ and $\ce{K}$.

There are two further facts that we need to realize to understand the chromatography.  The first is that the equilibria are balanced at different points.  This is why the size of the arrows in the reactions I drew above are different.  As drawn, component $\ce{I}$ has stronger affinity for the paper and a lesser affinity for water, as indicated by the longer rightward top arrow and the shorter bottom left arrow.  The reverse is true for component $\ce{K}$, and $\ce{N}$ is somewhere in between.

The second fact is that the water is moving due to the capillarity you mentioned. So when the compounds are in aqueous solution, they will flow with the water.  When they are stuck to the paper, they will not move.  Since all of the molecules are very rapidly transitioning between paper-bound and water-"bound" states, they will all move some of the time, and be still another part of the time.  The key is that the different components spend, on average, different fractions of time moving vs. being still.  This makes their average speed different.

As for the question of why they eventually come to rest, this is because the capillarity-driven flow eventually runs out of steam (i.e. gravitational forces compete with the capillary forces after the fluid rises far enough), and the water itself stops flowing very much.  Thus the bands of separated molecules eventually settle into a relatively fixed position.

So to answer your title question, truly insoluble compounds would not be separated by chromatography.  Compounds need to be soluble to at least a degree in order to be separated.


How to differentiate magnetite from metallic iron? Both metallic iron and magnetite, $\ce{Fe3O4}$, are magnetic. Now, iron is grey in color, whereas hematite is black, so that they can be distinguished visually.
Now, I have a powdery mixture of iron/magnetite and lighter grains so that the mixture appears grey, making visual differentiating a bit tricky. I assumed it was iron, but someone else claimed it could also be magnetite, and now I'm not sure.

Is there a really easy way to be sure what I am dealing with? I don't want to do an XRD just for this.

These are the methods that come to mind:

LIBS
XRF
Streak color. When scratching on unglazed porcelain, magnetite leaves charcoal-black traces, whereas iron leaves shiny steel-gray streaks. Having a powdery mixture this test can be tricky, unless there are reasonably big shards left. Alternatively, one can try to assemble a bulky piece by gluing the powder together with silicate adhesive or cyanoacrylate (as paracetamol suggested in the comments, "rubbing some of the mysterious powder between two shards of porcelain" can also do).
Compare density of your sample with $\rho(\ce{Fe}) = \pu{7.874 g cm-3}$, $\rho(\ce{Fe3O4}) = \pu{4.9 .. 5.2 g cm-3}$ and, assuming this mechanical mixture consists of iron and magnetite exclusively, calculate the percentage of each phase.

Magnetite in its mineral form often contains $\ce{(Fe,Mg)O}$, $\ce{TiO2}$ and isomorphous impurities of $\ce{Mn}$, rarely $\ce{Cr}$, $\ce{V}$; therefore physical methods of analysis (1, 2) are more preferable. Due to the same reason I'm not suggesting treatment with $\ce{HCl}$ and other methods based on chemical transformations as primary for this analysis.


Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride In my inorganic class, we carried out the synthesis of chlorotribenzyltin, $\ce{SnCl(CH2C6H5)3}$.

The synthesis begins with a reflux I'll describe below:

In a hood, place 2.0 g (17 mmol) of 325 mesh tin powder, 4.0 mL of $\ce{H2O}$ and 6.0 mL (6.6 g, 52 mmol) of benzyl chloride in a 50 mL round-bottom flask. After the reaction mixture has been heated to reflux for 3 hours*, remove the heat and cool the flask in ice. The lower liquid phase should solidify to form a white solid.
*The white solids that appeared shortly after refluxing commenced should have largely disappeared, although small amounts of white or gray solids on the sides of the flask are usually still present. The reaction mixture at this point should consist of two liquid phases, plus a small amount of unreacted tin.

So the question the textbook asks is

Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride.

I have a number of ideas as to what this could be:

The chloride bonded to the tin forming some kind of salt like
$$\ce{SnCl2}$$

The reaction did some kind of partial reaction creating something like
$$\ce{SnCl2R2}$$

The reaction went too far and created
$$\ce{SnR4}$$ and $$\ce{SnCl4}$$ and keeping them in reflux redistributed the ligands  based on stoichiometric amounts like
$$\ce{SnCl4 +3SnR4->4SnClR3}$$

This last one seems most likely based on supporting evidence from the text:

$\ce{SnCl4}$ reactions with Grignard reagents to give organotin compounds:
$$\ce{SnCl4 +4RMgX->SnR4 +4MgXCl}$$ ...
(it goes on to say)...the product, chlorotribenzyltin $\ce{SnCl(CH2C6H5)3}$, is a colorless solid that is stable in air for several days. In some respects, the reaction is analogous to the preparation of Grignard reagents, $\ce{RMgX}$ by treatment of magnesium metal with organic halides.

Is my reasoning for choice 3 solid? Is there something I'm not thinking of?

Turned in this report and my professor commented with his thoughts
$$\ce{Sn + 2RCl \rightarrow R2SnCl2 \rightarrow R2Sn }\text{ (white solid)} \ce{+ SnCl2 \rightarrow  R3SnCl}$$
but I didn't get the question wrong. So unless if someone contributes another answer, I'll accept this as the answer.


Looking for a safe exothermic reaction Is there an exothermic reaction which produces 1 to 2 minutes of heat at around 175 degrees Fahrenheit, no toxic gasses and is considered to be more or less safe?

Crystallisation of supersaturated sodium acetate is a good one, as seen in my answer here.

Another is the reaction of water with calcium oxide, which is commonly seen in self-heating coffee and soup cans. This reaction can get very hot, though.


Does the term "Alkaline" necessarily indicate the presence of an actual alkali? A month back, @CowperKettle had asked this at The Periodic Table chat room:

  Does "alkaline" imply the presence of an actual alkali?

Now I've come across the term "alkaline" on numerous occasions, and pretty much all examples that I know of (in Chemistry) actually do involve the presence of an alkali (often NaOH or KOH).

Wikipedia defines an alkali as:

  In chemistry, an alkali is a basic, ionic salt of an alkali metal or alkaline earth metal.

Further, it states:

  An alkali also can be defined as a base that dissolves in water.

Now when something is a "base", it could mean a lot of things (An Arrhenius base? A Lewis base? A Bronstedt base?), which doesn't really make the Wikipedia definition any clearer.

Take for example, this case:

  ...mucosal cells of the small intestine secrete bicarbonate ions into the intestinal lumen to render it alkaline...

I know that was from a "Biology perspective", but that raises the following questions:

1) Is there a universally accepted (I guess that would be IUPAC) definition for an alkali? (Since the Wikipedia definition seems a bit oversimplified)

2) Is it "acceptable" to refer to, say, as suspension of a (Lewis) base in water as "alkaline"? And what about that excerpt from my Bio. text book? Is that use of "alkaline" acceptable as well?

Most every English language dictionary definition of alkaline that I've found defines it as the adjective form of alkali and includes something like "having a pH greater than 7".  As pointed out in the comments there is no IUPAC definition, so it's going to be difficult to come up with an authoritative scientific answer.  Here are some interesting items I've found on the word's origin, usage in different contexts, etc.:

Online Enemology Dictionary:
alkali: late 14c., "soda ash," from Medieval Latin alkali, from Arabic al-qaliy "the ashes, burnt ashes" (of saltwort, which abounds in soda due to growing in alkaline soils), from qala "to roast in a pan." Later extended to similar substances, natural or manufactured. The modern chemistry sense is from 1813.

Cambridge Dictionary :
alkali: chemistry - a substance that has the opposite effect or chemical behavior of an acid. adj. alkaline

Biology Online:
alkaline: (Science: chemistry) Having the reactions of an alkali. Origin: L. Alkalinus Relating to or containing an alkali; having a ph greater than 7; alkaline soils derived from chalk or limestone.A substance that has a ph higher than 7.

Merriam-Webster:
alkaline: :  of, relating to, containing, or having the properties of an alkali or alkali metal :  basic; especially, of a solution :  having a pH of more than 7

The almighty Wikipedia, fairly representative of a half-dozen other online definitions:
In chemistry, an alkali (/ˈælkəlaɪ/; from Arabic: al-qaly القلي, القالي , “ashes of the saltwort”) is a basic, ionic salt of an alkali metal or alkaline earth metal chemical element. An alkali also can be defined as a base that dissolves in water. A solution of a soluble base has a pH greater than 7.0. The adjective alkaline is commonly, and alkalescent less often, used in English as a synonym for basic, especially for bases soluble in water. This broad use of the term is likely to have come about because alkalis were the first bases known to obey the Arrhenius definition of a base, and they are still among the most common bases.

I think in the end its usage is simply tied to the jargon and vocabulary of the context in which it is used. And with so many definitions, most of the ways that we hear it used are probably at least partially correct.  And as is so often the case, I think the Wikipedia definition probably comes as close to the best definition as any.


When reacting a strong acid with a weak base, would all of the strong acid react with the weak base If I do the following reaction:
$$\ce{Na3PO4 + 3HNO3 -> 3NaNO3 + H3PO4}$$
Does nitric acid completely react with the Trisodium phosphate to form a weak acid and a salt, or does it form a small amount of the weak acid or nothing? If it does form the weak acid, will this reaction create heat?
I have only really dealt with neutralization reaction so I don't really understand what will happen, any help would be greatly appreciated.

This would be a simpler question to answer for a salt of a monobasic acid, like sodium acetate. But, for the specific reaction you ask about:

First, we must consider the conjugate acid of trisodium phosphate: phosphoric acid, which will be formed when nitric acid is added. It has three $\mathrm{p}K_\mathrm{a}$ values for the three decreasingly acidic hydrogens: $2.15,$ $7.20,$ and $12.32.$ These values are measured in dilute water solutions, so they do not apply exactly to a reaction between pure nitric acid and pure trisodium phosphate. But these values do indicate that, by combining pure trisodium phosphate with three equivalents of a strong acid like nitric acid $(\mathrm{p}K_\mathrm{a} < 0)$ you would effectively have a complete conversion to phosphoric acid and sodium nitrate. And, yes, the reaction will be exothermic.

That was the simple answer. But if the reaction is carried out in water solution, which would help to keep the reaction from overheating and allow control of the rate of combining the reactants, we note that the first $\mathrm{p}K_\mathrm{a}$ of phosphoric acid is only $2.15$ units higher than that of nitric acid in water (effectively, hydronium ion with a $\mathrm{p}K_\mathrm{a}$ of $0)$ — in other words, phosphoric acid is $10^{-2.15}$ or about $1\%$ as strong an acid as hydronium ion.

So, after the reaction is complete, we would have about $99\%$ of the phosphate in the form of phosphoric acid and about $1\%$ in the form of the dihydrogen phosphate anion. There would also be small amounts present of monohydrogen phosphate dianion and phosphate trianion, whose  concentrations can calculated as $10^{-\mathrm{p}K_\mathrm{a}}$ for the other two $\mathrm{p}K_\mathrm{a}$ values. So, technically the reaction will be about $99\%$ "complete".

Meanwhile, keep in mind that the sodium ions are "observer" species that do not participate in the acid-base reaction.


Solubility of Organic Compound confusion My teacher wrote the following on the board today:

  A solute will only dissolve if the energy required to break the bonds
  in the solute and solvent is less than the energy released from the
  bonds formed between the solute and solvent.

I am aware that "like dissolves like" but not entirely sure what the sentence above means.

From the bond enthalpy chapter I learnt that, in an exothermic reaction the bonds formed are stronger due to lower in energy and thus more stable. Although I am not sure whether this is related to solubility, it seems to me that they are contradictory.

So can someone explain to my as to why bonds formed has less energy released than the broken bonds?

There are two regions of bond energy to look at here:

The bonds IN a molecule: a carbon-hydrogen bond has about $\pu{100
    kcal/mol}$ of stabilization. An oxygen-hydrogen bond is a little
stronger, about $\pu{110 kcal/mol}$. Stronger means HIGHER in
stabilization energy; but we draw diagrams where the stronger bonds are
designated by lines that are LOWER on the diagram. We hope no confusion
results!
The bonds BETWEEN molecules of a pure material or a mixture. These
(much weaker) bonds between molecules of a substance can be
estimated by heats of fusion or vaporization. The heat/enthalpy of
fusion of $\ce{H2O}$ is $\pu{1.435 kcal/mol}$, and for $\ce{CH4}$, $\pu{0.223
kcal/mol}$. Hydrogen bonding (not the regular bond to hydrogen, but the polar
bonding between molecules) falls in this region of energy and is
frequently encountered when examining solutions.

The bonding between water molecules is strong enough to resist separation by methane molecules because the gain could be no more than the stabilization in methane ($\pu{0.223 kcal/mol}$), but the loss would be as much as $\pu{1.435 kcal/mol}$. The first sentence in the question refers to this region of bond energies, not to the much higher energy region (~ $\pu{100 kcal/mol}$) that would be extensively covered in a chapter on bond enthalpy.


Why is coulomb integral is called local while the corresponding exchange integral is called non-local In the book "A chemist's guide to DFT" page 27-29 they said that coulomb integral is local interaction term while exchange integral is non-local. I can visualize, to some extent, that repulsion between two orbital/wave function at two different position (say x1 and x2) is called coulomb integral and that is why it's local. But I am having problem with the exchange term. It is also local, I feel, just the difference is two orbital can interchange their position between them. Can anyone please shed light on these concepts?

The definitions of local and non-local operators are given at p. 12 of the book. Basically, an operator $\hat{A}$ is called local if the value of $\hat{A} f(x)$ at some point $x=x'$ depends only on the value of $f(x)$ at the very same point $x=x'$ and not on the values of $f(x)$ at any other points. Otherwise an operator is called non-local.

For instance, any $j$-th Coulomb operator in Hartree-Fock theory is local since the value of $\hat{J}_{j} \psi_{i}(q)$ at $q=q'$ depends only on the value of $\psi_{i}(q)$ at $q=q'$, while any $j$-th exchange operator is non-local since the value of $\hat{K}_{j} \psi_{i}(q)$ at $q=q'$ depends on the values of $ψ_{i}(q)$ at all points $q$.


Why does pH of a buffer solution change according to Henderson–Hasselbalch equation? If I add a little amount of acid or base to a buffer solution, the pH of the buffer won't change. That's what the diagramatic mechanism tells us.

But according to Henderson–Hasselbalch equation, the $\mathrm{pH}$ changes!

For example, a buffer solution of $\ce{CH3COOH}$ and $\ce{CH3COONa}$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)$$

Now I add a little amount of acid $(\pu{0.01M},\;\pu{5 mL}).$ It disturbs $\ce{CH3COONa}$. Reaction:

$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$

It is noticed here that a small amount of $\ce{CH3COOH}$ is produced and a little amount of $\ce{CH3COONa}$ is spared. So simply $[\ce{CH3COOH}]$ and $[\ce{CH3COONa}]$ go high and low, respectively. As a result, the ratio between $\ce{CH3COONa}$ and $\ce{CH3COOH}$ decreases. As the $\mathrm{pH}$ depends on the ratio, the $\mathrm{pH}$ changes as well.

Suppose you have a $\pu{1.0 L}$ of acetate buffer solution made by $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COONa}$ solution adding to $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COOH}$ solution. Thus, $[\ce{CH3COOH}] = [\ce{CH3COONa}] = \pu{0.5 M}$. Therefore, according to Henderson–hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.75$ ($\mathrm{p}K_\mathrm{a} = 4.75$ for acetic acid).

Now you add a little amount of acid ($\pu{0.01M}, \; \pu{5 mL}$). It disturbs the equilibrium and change both $[\ce{CH3COONa}]$ and $[\ce{CH3COOH}]$, according to the reaction:

$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$

Now, the amount of $[\ce{CH3COONa}]$ is reduced by $\pu{0.01M} \times \pu{0.005 L} = \pu{0.00005 mol}$ and, similarly, $[\ce{CH3COOH}]$ is increased by $\pu{0.00005 mol}$.

Hence, new $[\ce{CH3COONa}]$ is $\pu{0.49995 mol/L}$ and $[\ce{CH3COOH}]$ is $\pu{0.50005 mol/L}$ (neglecting the volume change for convenience). Thus, new $\mathrm{pH}$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)= 4.75 + \log \left(\frac{\pu{0.49995 mol/L}}{\pu{0.50005 mol/L}}\right) = 4.7499 \approx 4.75$$

You see, it is a very minimal change. It is true that if you add more acid or base, change could be significant. However, all buffers have buffer capacity (usually, $\mathrm{p}K_\mathrm{a} \pm 1$). Nobody ever said $\mathrm{pH}$ of buffer is constant. It fluctuate withing the capacity. If you add too much acid or base, its $\mathrm{pH}$ would drastically change.


why does phosphorus makes 5 bonds with oxygen in P4O10 whereas 3 bonds in P4O6? how phosphorus forms 2 types of bonds with oxygen i.e. 5 in $\ce{P4O10}$ and 3 in $\ce{P4O6}$ when only three are needed to complete its octet?

Phosphorus has electronic configuration of [Ne]3S2 3P3 As it is a 3rd period element it also have a vacant 3D orbital.

In case of P4O6 only the 3P3 electrons take part. Resulting in formation of 3 bonds with O. But in case of P4O10 one of the electron of 3S is excited to vacant 3D orbital. Now P has 5 unpaired electron and it can form 5 bonds. Therefore P forms 5 in P4O10.

Now, about the octet. Not every compound follow the octet rule. Just to understand The whole chemistry depends upon STABILITY i.e lesser energy of molecule. Therefore, if by forming more bonds (and does not obeying Octet rule) the molecule is gaining more stability, it will do so.


What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays? When carbon-14 decays, the decay products are nitrogen-14 and an electron (and an electron antineutrino, but that's chemically irrelevant*):

$$\ce{^14_6C -> ^14_7N + e- + \overline{v_e}}$$

Let's assume that the carbon atom in question is part of a carbon dioxide molecule in the atmosphere. What would happen to the molecule when the atom decays into nitrogen? Will it be converted into a $\ce{NO2}$ molecule, or will it split apart? Will the electron created in the decay have sufficient energy to escape the molecule and form a positive ion?

Here's a somewhat related question dealing with the formation of radioactive carbon dioxide.

* Of course, not all the energy from the defect will be transferred to the beta particle's kinetic energy, so this is in fact relevant for the rate. See Loong's answer for details.

An article by Snell and Pleasanton, 'The Atomic and Molecular Consequenses of Radioactive Decay', (J. Phys. Chem., 62 (11), pp 1377–1382, $1958$) supports Ben Norris's comment.

  It is clear ... that $\ce{^{14}CO2}$ remains predominantly bound as $\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\ce{^{14}CO2 -> NO2^+}$ dissociation yielding $\ce{NO+}$, $\ce{O+}$ and $\ce{N+}$ follows [in], respectively, $8.4$, $5.9$, and $3.6$% of the decays.

A table summarising the results is given.

$$\begin{array}{|c|c|}
\hline
\mathbf{Ion} & \mathbf{\%\  abundance} \\
\hline
\ce{NO2+} & 81.4(16) \\
\ce{NO+} & 8.4(4) \\
\ce{O+} & 5.9(6)  \\
\ce{N+} & 3.6(4) \\
\ce{NO2^{2+}} & 0.40(06)\\
\hline
\end{array}$$


Could chromatography induce a conformational change? We'd been studying chromatography in our instrumental analysis course and I was wondering if chromatography could induce a conformational change.

For example, you have a straight chain with a hydrophobic side and hydrophilic side; when relaxed, its conformation is such that it has a hydrophilic outer surface and a hydrophobic core.

If you run it through a normal phase column, could the stationary phase cause a change in conformation to, let's say, a hydrophilic core and hydrophobic outer surface?

Also, if this could happen, then it would initially go through the column at a rate similar to the mobile phase and then slow way down, yes? If so, what would be the implication? Would the peaks be broadened?

While I do not have numbers at hand to support it, I think it is possible, chromatography on silica indeed may influence, may restrain the conformational flexibilty of molecules in the mobile phase.

Normal phase chromatography depends on adsorption.  A complete rotation around a $\sigma$-bond, even in absence of other molecules around, needs some energy to pass from one stagggered to an other staggered conformtion.

(figure source)

If there were now -- instead of methyl groups separated by three bonds, as depicted in the diagram -- two groups with a high affinity to the stationary phase, this said energetical barrier would increase.  Assuming lowered temperature and affinity of these groups "just intense enough" (e.g., deportontated sulfonic acids, not per-alkylated ammonium groups), the propability to find one molecule in such $\ce{s-cis}$ conformation will increase.  I just see a potentially relevent publication Langmuir, 2004, 20, 10639–10647 about proteins decoiling on silica.  I imagine, amphiphlic structures like reverse micelles may be equally influenced and will "swap their inner to the outer" as if they were (again) exposed to aqueous systems when facing silica.

Yet I am still hesitant to state if this may by become a constraint, leading to a situation where once conformation is looked, as in the instance of atropisomers.


What is primary valency This is from an adapted version of Concise Inorganic Chemistry by J.D Lee.The part on primary valency is confusing me.It mentions that primary valency is the oxidation of the central metal (a number) ,on the other hand I have read that primary valencies are non directional (this description suggests primary valencies are a physical thing).In compounds like $\ce{[Co(H2O)4Cl2]Cl}$ the primary valency is 3 but 2 of the chloride are directional.Can a ligand be both primary and secondary valency simultaneously?

I am sorry to see that they are still teaching these concepts of the 19th century as facts. The concept of primary valence and secondary valence is a translation of Werner's terminology (Nobel) as Hauptvalenz and Nebenvalenz. Haupt means something "main" and neben means something "extra, supplementary". What is valence in old school terminology? It is the combining power of an element.

Suppose you have $\ce{Co^{3+}}$, how many counter anions do you expect? $\ce{Cl^-}$? $\ce{CoCl3}$. This is the primary valence of 3. Standard electrostatics, that is, the number of positive charges must equal negative charges. Charged particles cannot stay alone under normal daily conditions.

However, Werner found that compounds of definite composition can also be found such as $\ce{[Co(NH3)6]Cl3}$. But cobalt could only take three chlorides with it? How come there is additional ammonia, that is a part of the molecular "Komplex".   He invoked the idea of Nebenvalenz, i.e. supplementary valence to explain the additional ammonia molecules built into this molecule. Indeed this was a complex idea for him and he named it as such Komplexverbindung (complex compound).

  I have read that primary valencies are non directional (this
  description suggests primary valencies are a physical thing).

Delete this from your mind. This does not mean anything today.


Adding non stoichiometric amount of NH3 to copper solution Consider the following reaction: $$\ce{Cu(H2O)_n^2+ + 4NH3 -> Cu(NH3)4^2+}$$
The amount of ammonium hydroxide added to your penny solution was not controlled (i.e. you did not add
“stoichiometric” amounts). Discuss what would happen to its absorbance value, and explain why, if the amount
added was too low and too high. (Assume the mark on the volumetric flask was not overshot in either scenario.)

I was hoping to get a second opinion on what happens to the absorption in the case where you add too much $\ce{NH3}$ and the case where you add too little.
I think that if I had added not enough $\ce{NH3}$ that some of the copper would not turn into $\ce{Cu(NH3)4^2+}$ and thus the absorption reading would be lower.
As for adding too much... I think it would be okay to add too much. I don't think it would affect the absorption.

Both your conclusions are absolutely correct.

I'll point out that you'll never get "all" the $\ce{Cu^{2+}}$ to the amine complex. You're trying to get greater than 99% of the $\ce{Cu^{2+}}$ to the amine complex. Because 99% and 99.9% and 99.99% would have virtually the same absorption reading.

To give even more detail, in mildly acid solution the copper will be in a octahedral complex $\ce{Cu(H2O)6^{2+}}$ which is a light blue. As you add ammonia at first $\ce{Cu(OH)2}$ forms, which is then dissolved as the various ammine complexes form.

$\ce{Cu(H2O)_6^2+ + NH3 <--> Cu(NH3)(H2O)5^2+}$
$\ce{Cu(NH3)(H2O)5^2+ + NH3 <--> Cu(NH3)2(H2O)4^2+}$
$\ce{Cu(NH3)2(H2O)4^2+ + NH3 <--> Cu(NH3)3(H2O)3^2+}$
$\ce{Cu(NH3)3(H2O)3^2+ + NH3 <--> Cu(NH3)4(H2O)2^2+}$

The tetraammine complex, $\ce{Cu(NH3)4(H2O)2^2+}$, is a much darker blue than the corresponding complex with water molecules alone. Thus colorimetry is much more sensitive measuring the tetraammine complex that the water complex.

The tetraammine complex has shorter Cu-N bonds than the C-O bonds. Thus the four ammine groups are one plane and the two water molecules at the apexes of the square bipyramidal complex.

Hopefully now the point about not converting all the copper to the tetraammine complex is clear. The various ammine complexes are in equilibrium with each other. Given that $\ce{[NH3] >> [Cu^{2+}]}$  then one can imagine the overall reaction as

$\ce{Cu(H2O)_6^2+ + 4NH3 <--> Cu(NH3)4(H2O)2^{2+}}$

In extremely concentrated solutions of ammonia the final two ammine groups will attach.

$\ce{Cu(NH3)4(H2O)2^{2+} + NH3 <--> Cu(NH3)5(H2O)^2+}$
$\ce{Cu(NH3)5(H2O)^{2+} + NH3 <--> Cu(NH3)6^2+}$


Theoretical yields and determining the number of moles in a reaction I'm working on a question which asks me to do two things. I've balanced the related equation as:

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

Now, if I have $44$ grams of $\ce{CO2}$ and $36$ grams of $\ce{H2O}$, how many moles of glucose will I have? The molecular mass of $\ce{CO2}$ is $44\mathrm{~g/mole}$, and that of $\ce{H2O}$ is $18\mathrm{~g/mole}$. Do I need to determine how many moles of each element are in the certain amount of grams for each compound, and then use that to determine how many moles are in the glucose?

Secondly, for the same amounts of $\ce{CO2}$ and $\ce{H2O}$, I need to figure the theoretical yield. Am I correct in saying since for ever 6 molecules of $\ce{H2O}$ and $\ce{CO2}$ I get 1 molecule of glucose, I would do

$$36\text{ grams of } \ce{H2O} \times \left( \frac{1~\mathrm{mol}~\ce{H2O}}{18\mathrm{g}~\ce{H2O}} \right) \times \left(\frac{1~\mathrm{mol}~\ce{C6H12O6}}{6~\mathrm{mol}~\ce{H2O}}\right)$$

Then take the same for $\ce{CO2}$? I'm unsure of what to do with the values after that.

You are correct in your thinking.

First, knowing the mass of the compounds, use the molar mass relationship for each (g/mol) to convert them all to moles. This is because when it comes to reactions, compounds react on a mole or number basis.
This lets you know the "starting amounts" of $\ce{H_2O}$ and $\ce{CO_2}$. Now, our forum has a policy against providing direct answers to homework, so I'll give examples. If you hypothetically had, say, 12 moles of both water and carbon dioxide, then using the numbers in the balanced chemical equation, they can fully react to make 2 moles of glucose. Thus, 2 moles is your theoretical yield. However, in another case, if you had 9 moles of water and 14 moles of carbon dioxide, the $\ce{H_2O}$ limits the amount of glucose you can produce, and it is the limiting reactant. (14 moles of $\ce{CO_2}$ would require 14 moles of $\ce{H_2O}$. There will be some $\ce{CO_2}$ left over because you are limited by only 9 moles of $\ce{H_2O}$). In this hypothetical case, the theoretical yield would be 1.5 moles of glucose. (Your problem may or may not involve limiting/excess reactants).

And if you're wondering, it's called "theoretical" yield because if you were carrying out this reaction in real life, some inefficiencies in combustion would yield less than the calculated amount. Hope this helps.


What does the prefix "dihydro" in the systematic name of luminol refer to? I'm doing a research project on the fascinating chemiluminescent molecule luminol, and figured it would be a good idea to start breaking down the systematic name of the chemical and linking the terms to the structure. The systematic name of luminol is 5-amino-2,3-dihydrophthalazine-1,4-dione, and I get it all apart from the "dihydro" - any help? Thank you very much in advance! :)

On the left is 5-amino-2,3-dihydrophthalazine-1,4-dione and on the right is the molecule which the name is based on, phthalazine. The 2,3-dihydro refers to the fact that at the 2 and 3 positions (the two nitrogens) hydrogens have been added (hydro is the prefix for hydrogen).


What makes some metals melt at higher temperature? I'm looking at the melting temperature of metallic elements, and notice that the metals with high melting temperature are all grouped in some lower-left corner of the $\mathrm{d}$-block. If I take for example the periodic table with physical state indicated at $\pu{2165 K}$:

I see that (apart from boron and carbon) the only elements still solid at that temperature form a rather well-defined block around tungsten (which melts at $\pu{3695 K}$). So what makes this group of metals melt at such high temperature?

Some factors were hinted, but let me put them in an order of importance and mention some more:

metals generally have a high melting point, because metallic interatomic bonding by delocalized electrons ($\ce{Li}$ having only a few electrons for this "electron sea") between core atoms is pretty effective in those pure element solids compared to alternative bonding types (ionic $\pu{6-20 eV/atom}$ bond energy, covalent 1-7, metallic 1-5, van-der-Waals much lower). Also, ionic lattices like $\ce{NaCl}$ have a higher lattice and bonding energy, they have weak interatomic long-range bonding, unlike most metals. They break apart or are easily solvable, metals are malleable but don't break, the electron sea is the reason for their welding ability.
the crystal structure and mass play an inferior role among your filtered elements (just look up the crystal structure of those elements), as metallic bonding is not directional unlike covalent bonding (orbital symmetry). Metals often have half filled $\mathrm{s}$ and $\mathrm{p}$ bands (stronger delocalized than $\mathrm{d}$ and $\mathrm{f}$) at the Fermi-edge (meaning high conductivity) and therefore many delocalised electrons which can move into unoccupied energy states yielding the biggest electron sea with half or less fill bands.
noble metals like $\ce{Au,Ag}$ have a full $\mathrm{d}$ orbital, therefore low reactivity/electronegativity and are often used as contact materials (high conductivity because of "very fluid" electron sea consisting only of $\mathrm{s}$-orbital electrons. Unlike tungsten with half or less occupied $\mathrm{d}$-orbitals they show no interatomic $\mathrm{d-d}$ bonding by delocalized $\mathrm{d}$-electrons, and more importantly, a half filled $\mathrm{d}$-orbital contributes 5 electrons to the energy band, while a $\mathrm{s}$ only 1, $\mathrm{p}$ only 3, the electron sea is bigger among the $\mathrm{d}$-group.
The "packaging" of core atoms in the lattice (interatomic distance) among the high $Z$ atoms (compared to e.g. $\ce{Li}$) is denser (more protons, stronger attraction of shell electrons, smaller interatomic radius), means stronger interatomic bonding transmitted by the electron sea:

You can see here that in each series ($\ce{Li,\ Na,\ K}$) the melting points rise to a maximum and then decrease with increasing atomic number (lacking unoccupied energy states for delocalized $\mathrm{d}$-electrons), bigger electron sea being here a stronger factor than a bit more dense packaging.

Boron as a semi-metal shows metallic and covalent bonding, Carbon strong directional covalent bonding and is able to build a network of bonds unlike other non-metal elements showing covalent intramolecular bonding, e.g., in diatomic molecules but not strong intermolecular bonding in macromolecules because of lacking unpaired electrons.

So there are some bigger trends for melting points explaining the high melting points of $\mathrm{d}$-metals, but also some minor exceptions to the rule like $\ce{Mn}$.


Why does phenolphthalein form in this reaction? This reaction (http://orgsyn.org/demo.aspx?prep=CV1P0476) gives quinizarin as the major product in about 70% yield.

However, when the same reaction is conducted with phenol insteas of p-chlorophenol, the product we get is not 1-hydroxyantraquinone (as would be expected if we only knew the above reaction), but phenolphthalein.

Why does phenolphtalein form instead of 1-hydroxyantraquinone? Would 1-hydroxyanthraquinone form if we added only 1 mol of phenol instead of 2 in the phenolphthalein synthesis?

In the Org. Syn. link you provided phthalic anhydride 1 and p-chlorophenol 2 in 95% H2SO4 and boric acid at 200oC undergo condensation and cyclization. Phthalic anhydride 1 under acid catalysis reacts with p-chlorophenol 2 at the ortho position as shown below to form ketoacid 3. Chlorine is a weak o,p-director and boric acid, I believe, is complexing with the phenolic hydroxyl and the ketone to reduce bond rotation and aid in cyclization to form anthroquinone 4. There is an aqueous work-up at this point followed by treatment with hot aqueous KOH which forms quinizarin 5 by nucleophilic aromatic substitution which is aided by the adjacent keto group.
In the reaction of phthalic anhydride 1 with phenol 7 in the presence of sulfuric acid, (presumably no boric acid), cyclization is repressed because of a lack of electron density at the phenol meta-position in 6. In addition, protonation of the ketone in 6 gives extended conjugation with the phenolic hydroxyl. Preferrable addition of phenol to the ketone of 6 leads to 8 and on to phenolphthalein 9.


Can we classify all the d-block elements as transition metals? I thought that properties of d-block elements are transitional between those of s-block and p-block elements, and that is the reason for calling them transition metal.

My textbook says that not all d block elements are called transition metals.

My doubts:

What are these non-transition metals that are in d-block?
Why are these specific elements not called transition metals?

The idea that the properties of d-block elements are transitional between those of s-block and p-block elements, and that is the reason for calling them transition element is absolutely correct, but the thing is transition metals are defined is slightly different manner.

Definition of a transition metal/Criteria for an element to be a transition metal

  A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals

Appreciate that the general electronic configuration for d-block elements are slightly different for those of transition metals.

EC of d-block elements: $\ce{(n-1)d^{1-10}ns^{1-2}}$
EC of transition metal/ion : $\ce{(n-1)d^{1-9}ns^{1-2}}$

Some examples of elements which are in d-block but not a transition metal

Scandium has the electronic structure $\ce{[Ar] 3d^{1} 4s^2}$. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The $\ce{Sc^{+3}}$ ion has no d electrons and so does not meet the definition.
Zinc has the electronic structure $\ce{[Ar] 3d^{10} 4s^2}$. When it forms ions, it always loses the two $\ce{4s}$ electrons to give a 2+ ion with the electronic structure $\ce{[Ar] 3d^{10}}$. The zinc ion has full d levels and does not meet the definition either.

There's a small thing that one must take into account that some elements are capable of forming multiple metal ions in those cases we must consider more common ion (the most stable one)

The pefect example for this case is $\ce{Cu}$ ($\ce{[Ar] 3d^{10} 4s^1}$), $\ce{Cu}$ is capable of forming two ions :

$\ce{Cu^{+1}}$ ($\ce{[Ar] 3d^{10}}$) (not the most common ion of $\ce{Cu}$)

$\ce{Cu^{+2}}$ ($\ce{[Ar] 3d^{9}}$)  (most common ion formed by $\ce{Cu}$)

Hence, copper is definitely a transition metal because the $\ce{Cu^{+2}}$ ion has an incomplete d subshell.

Using this idea we can also say  $\ce{Zn}$, $\ce{Cd}$ and $\ce{Hg}$ are not considered as transition elements.

How are these metals different from rest of the d block elements?

$\ce{Zn}$, $\ce{Cd}$ and $\ce{Hg}$ are not hard metals (as compared to other d block elements), which can be attributed to the fact that they have no unpaired electrons which makes their metallic bonding weak.


How to explain the structure of HNO3 with a Molecular Orbital diagram? Take an example, $\ce{HNO3}$

  The total valence of $\ce{HNO3}$ must be 24. By using formula $\pi =6\mathrm{n}+2-\mathrm{V}$ where $\pi$ is number of $\pi$ electrons, $\mathrm{n}$ is number of atoms excluding all hydrogen atom, and $\mathrm{V}$ is total valence, the $\pi$ is $2$. It means that there would be one double bond in the structure of $\ce{HNO3}$ and the remaining bonds other than $\pi$ bond are $\sigma$ bond.

I understand the double bond has 1 $\pi$ electron and 1 $\sigma$ electron. It means 10 electrons are used for the bonds and 14 electrons are remaining.

I know it would be impossible to violate octet rule because the calculation of $\pi$ already stated.

I have trouble how to spread the 7 lone pairs. Why does the $\ce{O}$ in the single bond $\ce{N-O}$ have 3 lone pairs, while the $\ce{O}$ in the double bond has 2 lone pairs? I don't expect the answer like "Just spread the lone pairs to somewhere".
I'm worried about the odd number of unshared electrons.

I know the $-$ means rich of electron and $+$ means lack of electron. To check the formal charge in each atom is easy, but how to calculate the formal charge, if I don't know where to put the $-$ and $+$.

I think this question is not a duplicate of The Lewis structure of HNO3. I already saw that question , but I'd like to ask here again, because I need the answer for my misunderstanding, even though the answer would be general structure of $\ce{HNO3}$.

To be precise, I just don't know where I need to put unshared electrons. It's like I'm missing the bridge. I know resonance, as arrow-pushing, and mechanisms are required in organic chemistry.

You have 24 valence electrons. Counting the electrons: 6 of these will go into the $σ$ framework and there are 14 electrons in non-bonding, either $sp^3$ or $sp^2$ hybrids, on the oxygens. This leaves 4 electrons for the $π$ system.

We can think of the Nitrogen and Oxygen in the double bond being as both being $sp^2$ hybridised, with vacant $p$ orbitals orthogonal to the plane of the molecule. These 4 remaining electrons will then go into the $π$ system formed by the overlap of these 2 orbitals, which overlap to form 2 molecular orbitals, the $1π$ and the $2π$ MOs.

We expect the $1π$ to have greater contribution from the Nitrogen due to the lower orbital energy of the $p$ in N compared to O, and the $2π$ to have greater contribution from the O instead of the N.

A full MO calculation of the molecule would also take into account the non bonding $sp^3$ electrons on the oxygen, of which one lobe will be able to interact slightly with the $ p$ orbitals, giving a delocalised system over the Nitrogen double and single bond, exactly as we see drawing resonance structure for the molecule.


Why does ATP inhibit glycogen synthase? Why is ATP an inhibitor? I mean, if there are high levels of ATP, it means that the cell has adequate amounts of energy. Therefore, if it doesn't need the energy, then glucose needs not go through glycolysis to produce that energy and is more practical to be stored as glycogen. So why is ATP an inhibitor of glycogen synthase?

Firstly, I will put across main points so that it will be easy to understand this complex control mechanisms of the enzymes involved in both processes:

Glycogen synthase control

Control of glycogen metabolism is effected via reciprocal regulation of glycogen phosphorylase and glycogen synthase. Thus, activation of glycogen phosphorylase is tightly linked to inhibition of glycogen synthase, and vice versa.

Both glycogen synthesis and breakdown are exergonic under the same physiological conditions. If both pathways operate simultaneously, however this is deemed to be wasteful hydrolysis of UTP.

Glycogen phosphorylase and glycogen synthase therefore must be under stringent control such that glycogen is either synthesized or utilized according to cellular needs.
Regulation involves both allosteric control and covalent modification, with the latter being under hormonal control.

  Glycogen synthase also exists in two distinct forms that can be
  interconverted by the action of specific enzymes: active,
  dephosphorylated glycogen synthase I (glucose- 6-P–independent) and
  less active, phosphorylated glycogen synthase D (glucose-6-
  P–dependent). Glycogen synthase, is activated by G6P.When there is
  high demand for ATP (low [ATP], low [G6P], and high [AMP]), glycogen
  phosphorylase is stimulated and glycogen synthase is inhibited, so
  flux through this pathway favours glycogen breakdown.

Why is this so?

ATP also binds to the allosteric effector site, but in the T state, so that it inhibits rather than promotes the T-R conformational shift.
Having said that here is an illustration of control of glycogen phosphorylase activity;

  The enzyme may assume the enzymatically inactive T conformation or the
  catalytically active R form. The conformation of phosphorylase b is
  allosterically controlled by effectors such as AMP, ATP, and G6P and
  is mostly in the T state under physiological conditions. In contrast,
  the modified form of the enzyme, phosphorylase a, is largely
  unresponsive to these effectors and is mostly in the R state unless
  there is a high level of glucose.

Exactly this situation occurs in glycogen metabolism through the opposition of the glycogen phosphorylase and glycogen synthase reactions.

Another point to note is that glycogen synthase is regulated by covalent modification through complex cyclic cascade. For more information, please see here: Glycogen Metabolism

If you need more clarification don't hesitate to put in comments. Hope it helps

References

Biochemistry (Grisham)
Biochemistry (Voet and Voet)
Harper’s Illustrated Biochemistry


Naming compounds with an OX group How are compounds which have a bond between an oxygen atom and a halogen atom named according to IUPAC rules? For example this compound:

I looked in many sites, but couldn't find how such compounds are named.
Also what is their seniority compared to alcohols, esters, ketones, and carboxylic acids? Or are they always a prefix?

The preferred IUPAC name (PIN) of the compound given in the question is butan-2-yl hypofluorite. It is named as ester of hypofluorous acid. In the seniority order of classes, esters are ranked between anhydrides and acid halides.
The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-67.1.3.2 Esters of mononuclear noncarbon oxoacids
Esters of mononuclear noncarbon acids are named in the same way as esters of organic acids (…). Alkyl groups, aryl groups, etc. are cited as separate words, in alphanumerical order when more than one, and followed by the name of the appropriate anion. Partial (acid) esters of polybasic acids are named by citing alkyl groups, aryl groups, etc. as separate words, in alphanumeric order if more than one, followed by the word ‘hydrogen’ (with the appropriate multiplying prefix, as necessary) also cited as a separate word, and the name of the appropriate anion. Salts of partial (acid esters) are named by citing the name of the cation before the name of the organic group; remaining acids groups are denoted by the word ‘hydrogen’ as described above. Structural specificity for esters of chalcogen analogues of mononuclear noncarbon oxoacids is provided by the appropriate italic element symbols O, S, Se, and Te, prefixed to the name of the group, as needed.


Reuse 20 litre plastic drums for drinking water? Would you consider it safe to clean and reuse a 20 litre plastic drum that contained 36% Hydrochloric Acid for drinking water? If so how would you recommend cleaning it?

Regardless of what it takes to remove any traces of hydrochloric acid from the drum, the container is probably not certified as food-grade. So in addition to any risks from the acid itself (including things like heavy metals, which are common contaminants in mineral acids), the plastic is not certified to be free of toxic substances left over from manufacturing or that the plastic will not degrade to release such substances over its lifetime.


Explain this boiling point order please I read it from a book but don't know the reason. Please explain...

Boiling point orders are -

Ethanamide > Ethanoic Anhydride > Ethanoic Acid

Ethanamide has a higher boiling point than ethanoic acid since amide has two hydogen atoms it can form more Hydrogen bonds than ethanoic acid and their molar masses are 45gmol-1 and 60gmol-1 which makes the difference brought by the London forces somewhat negligible .
you may suppose Ethanoic anhydride should be having a lower boiling point than ethanoic acid and ethanamide since it only has London forces and dipole dipole attractions.But ehanoic anhydride(molar mass-106gmol-1) has a higher molar mass than Ethanoic acid(molar mass-60gmol-1) making the London forces in ehatnoic anhydride more significant.
Hence we have
ethanamide(222 °C)>Ethanoic anhydride(140°C)>Ethanoic acid(118°C).


Molecular Orbitals of Hydrogen Fluoride (HF) This source states that the three s-orbitals of hydrogen and fluorine interact to form three new molecular orbitals, while other sources say that the 2s orbital is non-bonding.

Which one is more correct? Also, if they indeed form three new molecular orbitals, how do they look like?

Bearing in mind the energy and symmetry conditions, in the case of $\ce{HF}$ one can construct MOs using the $1s$ AO on H and the $2s$ and $2p$ AOs on F.

In general, the contribution to MOs is determined by the coefficients in the linear combination.

Here, one observes that the $1s$ electrons are almost completely localized on the $\ce{F}$ atom. Also, $1 \pi$ electrons are completely localised on the $\ce{F}$ atom because the $2p_x$ and $2p_y$ orbitals on F have a zero net overlap with the $1s$ orbital on $\ce{H}$.

Electrons in MOs localized on a single atom are referred to as nonbonding electrons.

Also, I would note that the $3 \sigma$  MO has less bonding character and the $4 \sigma^*$ MO has less anti-bonding character.

Note that the total bond order is approximately one because the $3 \sigma$ MO is largely localized on the F atom, the $3 \sigma$ MO is not totally bonding, and the $1\pi$ MOs are completely localized on the F atom.

On account of fluorine being a more electronegative atom, in the $2 \sigma$ bonding orbital the electron density is much greater on the more electronegative fluorine than on the hydrogen. However, in the anti-bonding $4 \sigma^*$ orbital, this polarity is reversed.

Caveat: The preceding paragraph may support your intuition and may be right in a few simple cases, but I wouldn't rely on it too heavily.

Given below is a diagram showing $2 \sigma$ , $3 \sigma$ and $1 \pi$ MOs in HF


Number of hyperconjugation structures Is the number of hyperconjugation structures for a hydrocarbon with number of 'alpha' hydrogens 'a' equal to 'a' or 'a+1'?

Let's use the n-propyl carbocation as an example.  It has two hydrogens alpha to the cationic carbon and two resonance structures involving hyperconjugation of those alpha hydrogens can be drawn.


Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid? Why does the methyl group at the ortho-position stabilize the carboxylate ion? I recognize that there will be a steric clash between the carboxyl group and the methyl group but I don't understand why this would help increase acidity. If there was any resonance occurring then this steric clash would DESTABILIZE the anion and make it LESS acidic given that the carboxylate group would not be in the same plane as the ring and thus delocalization, which would stabilize negative charge, would be impossible. Clearly there is no conjugation with the aromatic ring in "benzoic acid based compounds" from the carboxylate group. So why does having a methyl group at the ortho-position increase acidity?

Having one methyl group ortho to the carboxylate group will not prevent the carboxylate group from becoming reasonably planar with the aromatic ring.  If you had a second methyl group in the other ortho position, then planarity would be a problem.

A methyl group attached to a benzene ring is electron releasing in both an inductive and a resonance sense.

Inductively, the electrons in an $\ce{sp^3}$ hybridized methyl group prefer to shift towards the lower energy (more s-character in the orbital) $\ce{sp^2}$ aromatic carbon.
The methyl group is also electron releasing in a resonance sense due to carbon-hydrogen hyperconjugation (see figure below).

The resonance structure pictured below on the left is one of many that we can draw for the o-toluic carboxylate anion.  Note that it places a positive charge on the methyl bearing ortho carbon. Both inductively and through resonance (see the hyperconjugated resonance structure below on the right) the methyl group will stabilize this resonance structure.  It is this stabilization provided by the methyl group that makes o-toluic acid a stronger acid (pKa=3.91) than benzoic acid (pKa=4.19).

Note: As pointed out by both @Mithoron and @GaurangTandon, this answer does not explain why p-toluic acid is a weaker acid than o-toluic acid or benzoic acid. The answers in this earlier question better answer this question.


Why does the rate-determining step determine the overall rate of reaction? I can appreciate that the rate determining step affects the rate of reaction. What I take issue with however is the exact wording given in the definition of the RDS.  From here:

  The rate determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds.

The last part about the RDS determining the rate baffles me. The RDS would affect the rate by slowing down a succeeding step, but in what way does it determine it?

Think of a chemical reaction as a series of processes that must happen in order. If reaction $3$ is the slowest (rate determining) step of some overall reaction, then the intermediates formed in reaction $2$ just accumulate and wait for their turn to participate in the reaction. Depending on the order and kinetics of the reaction, the rate of reaction $3$ can be increased by this build-up of reactants, but there will always be an excess waiting to participate in this slow step. Once the reactants get past step $3$, they breeze on through the remaining faster steps of the reaction. Since reactions $4$ and above are fast compared to $3$, it makes little difference how fast they are.

Note that it is somewhat of an approximation to say that the overall rate is equal to that of the rate determining step. As mentioned in the example above, depending on the order and kinetics of the reaction, a buildup of intermediates from reaction $2$ could increase the rate of $3$, thus the rate of $2$ becomes somewhat relevant. Likewise, it's possible that a buildup of products from reaction $3$ could inhibit its rate depending on the rate of reaction step $4$. Wikipedia's definition of the rate-determining step of a reaction includes this mention that it is an approximation:

  In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step. For a given reaction mechanism, the prediction of the corresponding rate equation (for comparison with the experimental rate law) is often simplified by using this approximation of the rate determining step.


Sintered vs perforated plate Büchner funnels I just bought a sintered-glass style Büchner funnel for some vacuum filtration tasks. There's two issues I'm having with it:

Filter material, especially activated charcoal, ends up stuck in the glass disc. I did a little bit of research trying to find out how to clean it out and the only thing I found (for the carbon, at least), was piranha solution, but I have neither the equipment nor the experience to use it safely so I'm SOL on cleaning. Fortunately the presence of carbon isn't a huge deal for my uses, though.
For paper filtering; the sintered disc goes all the way to the edge of the funnel, so unless I've got filter paper exactly the size of the funnel (which I don't), I'm worried that the 0.5-1.0mm gap I usually end up with around the edge of the paper gives fluid a path to bypass the paper. If I err on the side of larger paper it's still a poor fit with large channels around the edge.

It seems to me that the perforated-disc style of Büchner funnel would solve both of these problems; since the funnel itself can't trap filter material and also the holes don't go all the way to the edge of the disc:

(Incidentally, the pictured funnel on the left is exactly the one I own.)
My question is, given the two apparent disadvantages I've observed with the sintered-glass style funnel, what are the pros and cons of each type? More specifically, what could I do with a sintered-glass funnel that I couldn't do (as easily) with a perforated-disc funnel?
The context here is "kitchen chemistry"; I'm not a chemist I just do a few DIY projects here and there and all my knowledge comes from YouTube and Google searches. The only lab equipment I own is this funnel, a flask, and a vacuum pump. The only relevant PPE I own is a respirator, various gloves, and some chemical safety goggles.

The thing on the left is not usually a called a Büchner funnel, but a frit. Frits are generally meant to be used for stuff that can be redissolved afterwards, because they tend to clogg permanently, if very small particles get stuck inside.
You do not normally put a filter paper on it, because that would leak on the edge. The filter grade should be etched onto the glass somewhere. https://en.wikipedia.org/wiki/Fritted_glass Sometimes a frit can be used to hold another, powdery filtering substance, if that has a coarse enough grain size.
Advantage of the frit is the well defined pore size, you can heat it for drying, it can be mounted in a glass tube with two ground glass connectors (for evacuating), it can be used with concentrated and/or oxidising acids or other solutions, ...
Advantage of the Büchner funnel is that it does not get clogged with insoluble stuff, ease of use, cheap, available in large diameters, very mechanically solid, tolerates alkalic solutions (glass frits do too, but only briefly), ...


Deviation of Ytterbium from general trend What is the reason for deviation of ytterbium from the general trend of lanthanides in terms of hardness, melting point, density etc.?

To compliment Jon Custer's comment, we can likely trace this deviation to the electron configuration of ytterbium. Note that ytterbium is not the only deviation among the lanthanides - Europium deviate as well:

Why is that? Note the electron configurations of europium and ytterbium:

Europium - $\ce{[Xe]} 4f^7 6s^2$
Ytterbium - $\ce{[Xe]} 4f^{14} 6s^2$

In europium, the $4f$ subshell is half-filled, an arrangement that is particularly desirable from a quantum mechanics standpoint. In ytterbium the $4f$ subshell is completely filled, another 'happy' arrangement. These configurations lead to changes in bonding behavior, for example noting that most lanthanides have the hexagonal close pack or the double hexagonal close pack crystal structures. The only ones that are different are samarium, which is rhombohedral, europium which is body-centered cubic, and ytterbium with is face-centered cubic.

The link between electron configuration and physical properties exists elsewhere in the periodic table, but the trends are not as clear. However, transition elements with filled or half-filled $d$ subshells are likely to have lower melting points than their neighbors.


Free energy of an ideal gas I want to calculate the Gibbs free energy for an ideal monoatomic gas, i.e. $$U=\frac32nRT.$$
Free energy is defined as $G(T,p)=H-TS$ and using the definition of enthalpy $$G(T,p)=U+pV-TS=U+nRT-TS.$$

To calculate the entropy I use the total differential of entropy given by
$$\mathrm{d}S=\left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T
  +\left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p.$$

Using the Maxwell relation
$$\left(\frac{\partial S}{\partial p}\right)_T =
-\left(\frac{\partial V}{\partial T}\right)_p,$$
and for an ideal gas with $V=\frac{nRT}{p}$ we hence get
$$\left(\frac{\partial S}{\partial p}\right)_T=-\frac{nR}{p}.$$
Further
$$\left(\frac{\partial S}{\partial T}\right)_p =
\frac{c_p}{T}=\frac{\frac52nR}{T},$$
 so all in all
$$\mathrm{d}S=\frac{\frac52nR}{T}\mathrm{d}T-\frac{nR}{p}\mathrm{d}p.$$

How do I calculate $S$ from here, because integrating would leave me with $\Delta S$ which I cannot use?

The point of reference should be standard conditions ($1~\mathrm{atm}$, $1~\mathrm{mol}$, $298~\mathrm{K}$, ...)

Let $H^0$ be the heat of formation of the gas at $298~\mathrm{K}$ and $1\ \mathrm{atm}$. and let $G^0$ be the free energy of formation of the gas at these same conditions.  The enthalpy of the gas at temperature $T$ (any pressure) is then:
$$H(T)=H^0+C_p(T-298)$$  First let's get the free energy at temperature T and 1 atm.  At constant pressure, the change in free energy with respect to temperature can be calculated from:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
If we integrate this equation by parts, we obtain:

$$\frac{G(T,1~\mathrm{atm})}{T}-\frac{G^0}{298}=\frac{H(T)}{T}-\frac{H^0}{298}-\int_{298}^T{\frac{C_p}{T}dT}$$
Some algebraic manipulation of this equation yields:
$$G(T,1~\mathrm{atm})=G^0+\frac{(G^0-H^0)}{298}(T-298)+C_p(T-298)-TC_p\ln(T/298)$$
If we define $S^0=\frac{(H^0-G_0)}{298}$, this equation becomes:
$$G(T,1~\mathrm{atm})=G^0-S^0(T-298)+C_p(T-298)-TC_p\ln(T/298)$$
Adding the free energy change between $1~\mathrm{atm}$ and pressure $P$ at temperature $T$ yields:
$$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[C_p\ln(T/298)-R\ln(P/P_0)]\tag{1}$$
where $P_0=1~\mathrm{atm}.$  The term in brackets in this equation is the same as $S(T,P)-S^0$.  If we substitute this into the previous equation, we obtain:
$$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[S(T,P)-S^0]$$
If we combine terms in this relationship, we obtain:
$$G(T,P)-G^0=C_p(T-298)-(TS(T,P)-298S^0)$$
The term $C_p(T-298)$ is $\Delta H$, and the term $(TS(T,P)-298S^0)$ is $\Delta (TS)$.  So, as expected, the equation is the same as $$G(T,P)-G^0=\Delta H-\Delta (TS)$$
where
$$S=S^0+C_p\ln(T/298)-R\ln(P/P_0)]\tag{2}$$
Either Eqn. 1 or Eqn. 2 can be applied, depending on preference.