Is pure water very corrosive? In other site, someone claimed that pure water is very corrosive.

Water distilled for some laboratory uses (like the Z-pinch reactor) is distilled to 0.00000009% purity, and is caustic enough to burn through most things.
Water, purified enough, cannot be distributed through stainless steel pipes. Instead, they use glass pipes to transport it within the plants where it's made.

For sure, caustic is not the correct word, since water is neutral.
But, is it that corrosive? Or is it even less than water with salt?

That quote had false statements all over it.

Type I ultra pure water (Milli-Q water, or others) are fairly common among all labs.  They are not corrosive to stainless steel.

As @chipbuster said, purified water is never distributed through stainless steel partly because concern of contamination.  Metal reacts not only with oxygen, but also many pollutants in the air.  Overtime on metal surface will build up a layer that has all kind of compounds in them.  When water flows through, ions get picked up.

However, the most major reason would be that glass is better in all respect.  The reason glasswares are commonly used in the lab have little to do with pure water.   Glasswares are relatively stable across wide range of pH and very hard to be oxidized or reduced.   Plus, glasswares are cheap, extremely easy to manufacture, and transparent, so you can see through.  You will really only need metal if extreme pressure or temperature is required.

As a matter of fact, it is also unwise to use glassware to distribute ultrapure water--you will get ion contaminations from glass as well.  Purified water are actually almost always stored in plastic containers and distributed through plastic tubes.

As stated by others, the original post if full of falsehoods.  There are also falsehoods in the replies.  I have a lot of experience with ultrapure water, my company makes very large ultrapure and bench top labwater systems.

1) UPW (ultrapure water) is not caustic.  Conversely, in it's purest form it because acidic when exposed to atmosphere because it readily absorbs CO2, which converts to carbonic acid.  However, by default, UPW can only have a pH of 7 (again when not exposed to air).  So proper pH measurement can only be done inline or with proper buffers. We actually wrote a paper on this topic because researchers were always complaining that their water was acidic.

2)  UPW will corrode iron, brass, poor quality steel.  However, it will not corrode stainless steel as long as the welds are proper, even then hard to make SS conducive to corrosion by UPW.  Example, resistivity meters that we use are made from SS (need to be to conduct electricity across 2 metal probes to get a reading).

3) UPW will not corrode Pyrex as stated above.  After all, HPLC and LCMS grade water (not going to define so look it up) is placed in glass containers. Distilled water is even made in glass stills for laboratory use.  Having said this, it will leach some silica and boron, so they actually should use special glass.  But it won't corrode it.  I had a bottle of LCMS grade water sitting in the lab for years.  Also, exotic plastics don't need to be used, such as PFA, PVDF, Halar, etc. The possible exception is in a fab of a semiconductor plant.  We use polypropylene, PEX (cross-linked polyethylene), and even PVC (but not ideal because in some cases the glue can leach TOC's - total organic/oxidizeable carbon that are undesireable, for example in a waste water treatment plant.).

4) the comment that "pure water is less corrosive than water with impurities", is not completely true.  Put a brass or iron bolt in distilled water and see what happens compared to tap water.  Also, power plants don't use UPW because tap water is more corrosive.  All steam powered plants (coal, gas, nuclear, etc.) use UPW, because residuals will form on the turbine blades and throw them off balance and destroy them.  Also, impurities from the water will leave residue on the evaporators in the cooling towers and render them useless.  One can't compare the corrosiveness of tap water and UPW.  I've seen very corrosive tap water (classic example is the city water issue in Flint, MI.  The water is corroding the pipes and leach lead into the water because it contains chlorates).  UPW can be corrosive to other metals because it forms carbonic acid when exposed to air. etc......

By definition, ultrapure water is defined as ~17.0-18Meg-ohm or type I water.  Meaning does not contain dissolved ions to conduct electricity.  In theory, if you throw your TV into a tub full of UPW it won't short out.  If swimming in a lake of UPW and lightening hits the water you won't be electrocuted, because it contains no ions to conduct electricity.  The theoretical limit for UPW is ~18.26 Meg-ohm.

DO NOT DRINK UPW, it will corrode metal fillings, leach calcium from your bones, eat the enamel in your teeth, even can cause your cells to explode (due to osmosis).  Fortunately, before you can do to much damage you likely will feel sick to the stomach.  The irony is that the purer the water is the worst it is for you to drink.  Though, you can denature UPW by mixing it with Tang, powdered drink mix, tea, coffee, etc.  It will make very strong tea and coffee.  Even though distilled water is not considered UPW (by definition, it is a low grade Type II), it isn't good for you to drink because it has no minerals.  Also tastes insipid.   While on the subject, home RO or 'purified water' you buy in the store, is considered as ASTM Type IV, but in our industry we call it Type III.

Yes, UPW is one of the strongest solvents known to mankind.  But that doesn't mean it will eat through everything.  It just means it's always trying to reach equilibrium by leaching out ions, VOC's and CO2.


Why do Ionic Bonds exist? This is a legitamate question.  If one atom wants to lose an electron and another wants to gain an electron why don't they just transfer the electrons and just part ways?

If a Na+ ion and Cl- ion exist as NaCl,(ionic) it doesn't mean that, that particular Na donated electron to Cl. Na may lose electron somewhere else and Cl may gain electron somewhwere else.
The main point of existence as Na+Cl-, ie NaCl is the electrostatic attraction between them.

Ionic bond is not a bond in real. It’s just a force attraction between two ions placed closely. It’s something similar to the van der Waals’, but it has a lot of physical significance. Many have this query and I have seen a lot troubling on ionic bond. So no transfer or such thing takes place. Only attraction to hold them tight. This is all about ionic bond.


How do you name a bicyclo compound with a fused benzene ring? What is the IUPAC rule for naming a bicyclo compound with a fused benzene ring? Are you required to state the number of double bonds by adding something like 2,4,6-ene to the name or do benzene rings have a specific name ?
For example:

What will this compound be named? Will it be bicyclo[4.3.0]non 1,2,4 triene or is there some specific rule for fused benzene rings, as I stated before?

Beside systematic nomenclature about organic structures, there are other systems in use.  One of them is the by Hantzsch and Widman which departs from a number of stem structures.  Some of these archetypical motifs are named systematically, others (still) retain their traditional name which does say little  to you about the structure (e.g., pyridine) until you learn them.  One (incomplete) compilation of these names, for example, is this one.
This being said, you either look up, or over time (i.e., training) recognize that the parent structure of your molecule is indene:

except that the double bond in the five-membered ring of the structure you showed is completely saturated.
You remember reactions on benzene, and recall about the importance to denote the sites in organic structures.  The approach to use numbers equally exists for larger compounds, too; a few positions are labeled.  While there are cases where you may shorten 1H-indene to indene, this no longer is the case to describe unambiguously the compound shown by you, which is 2,3-dihydro-1H-indene.


Kb values of N,N-dimethylaniline derivatives I need the $K_\mathrm{b}$ values for ortho-, meta- and para- positions of the following compounds:

I've looked and couldn't find them anywhere.

chemicalize.com is  a cheminformatics website which is the kind of resource you are looking for.

You can use the search tool to find your compounds. The input is the structure itself, which you can draw in-situ.

Here is the $\mathrm{p}K_\mathrm{a}$ data I retrieved from the aforementioned site. Note that these are calculated / estimated, not measured $\mathrm{p}K_\mathrm{a}$ values.

Table 1. $\mathrm{p}K_\mathrm{a}$ values (of the conjugate acid)

$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & 3.71 & 5.19 & -0.98 & 1.44 & 2.66 \\
\textit{meta-} & 3.89 & 5.15 & 1.73 & 4.77 & 3.45 \\
\textit{para-} & 5.55 & 5.46 & 0.65 & 4.68 & 3.65 \\
\hline
\end{array}
$$

Since the question is asking for $K_\mathrm{b}$ values, I converted them too.

Table 2. $K_\mathrm{b}$ values

$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & \pu{5.13e-11} & \pu{1.55e-9} & \pu{1.05e-15} & \pu{2.75e-13} & \pu{4.57e-12} \\
\textit{meta-} & \pu{7.76e-11} & \pu{1.41e-9} & \pu{5.37e-13} & \pu{5.89e-10} & \pu{2.82e-11} \\
\textit{para-} & \pu{3.54e-9} & \pu{2.88e-9} & \pu{4.47e-14} & \pu{4.79e-10} & \pu{4.47e-11} \\
\hline
\end{array}
$$


Why does aqueous NaCl conduct electricity? Could someone please explain why an aqueous solution of NaCl conducts electricity? I've Googled this up but couldn't get a satisfactory answer. This has an answer on Physics.SE but that's too complicated. My textbook doesn't answer my question.

Original Question:
Why an aqueous solution of $\ce{NaCl}$ conducts electricity
Because $\ce{NaCl}$ is an electrolyte. This means it yields ions in solution.
Simply put, solid $\ce{NaCl}$ consists of $\ce{Na+}$ cations and $\ce{Cl-}$  anions bound together in a rigid crystal lattice. When it melts or is dissolved in water, the crystal lattice breaks. The ions are now able to move around. Similar to charged particles in a metal conductor (in this case electrons), in liquid form or aqueous solution the ions are the charged particles that can move, allowing the solution to conduct electricity.
Follow-up question:
Why $\ce{NaCl}$ dissociates in water
The (rather oversimplified) answer is that dissociation of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
The water molecules (dipoles) are attracted to the ions  and cause the crystal lattice to destabilize and ions to dissociate. The charged ions in solution are surrounded and stabilized by the water molecules (dipoles).
Note: Some ions migrate as ion pairs, but for a strong electrolyte most or all the ions will be dissociated and surrounded by water molecules.

As you seem dissatisfied with the answers received so far, I assume you want to fully understand the mechanism involved in dissociation. Unfortunately, this is rather complex.
There is an excellent  paper by Ballard & Dellago [1] that explains their work on the subject, but you will probably need a bit more physical chemistry / thermodynamics knowledge to understand it fully.
Reference

Ballard, AJ & Dellago, C "Toward the Mechanism of Ionic Dissociation in Water," *J. Phys. Chem. *B 2012, 116, 45, 13490–13497
Publication Date:October 19, 2012
https://doi.org/10.1021/jp309300b

$\ce{NaCl}$ is strongly ionic compound. It gets completely ionized and dissociates in $\ce{Na+}$ and $\ce{Cl-}$. Both $\ce{Na+}$ and $\ce{Cl-}$ are surrounded by water molecules, $\ce{Na+}$ is surrounded with $\ce{O}$ of $\ce{H2O}$ facing towards $\ce{Na+}$, similarly $\ce{Cl-}$ is surrounded by $\ce{H2O}$ molecules with $\ce{H}$ facing towards $\ce{Cl-}$. Presence of positive and negative ions helps in conduction of electricity.
When electrodes are inserted and current is passed through these electrodes, movement of ions in opposite direction creates current.
After passing current for a long time (especially DC current), the concentration of $\ce{Cl-}$ ions will decrease. The concentration of $\ce{OH-}$ will increase gradually and there will be competition between $\ce{Cl-}$ and $\ce{OH-}$.
Electrolysis of $\ce{NaCl}$
Anode: $\ce{2 Cl- -> Cl2 + 2 e-}\quad E=-1.36\ \mathrm V$
Cathode: $\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$
Electrolysis of $\ce{NaOH}$
Anode: $\ce{4 OH- -> O2 + 2 H2O + 2 e-}\quad E=-0.40\ \mathrm V$
Cathode: $\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$
After Sufficient long time there will be competition between $\ce{Cl-}$ and $\ce{OH-}$ to get oxidized on Anode, whereas there is only one possible reaction on cathode. Reaction on Anode depends upon the concentration and reduction potential of $\ce{Cl-}$ and $\ce{OH-}$. Actually on cathode there is also possibility of reduction of $\ce{Na+}$ but reduction potential is $-2.7\ \mathrm V$ which is difficult than $\ce{H2O}$ so only one reaction takes place on cathode.


Convert .car file PBC to basis vectors? I have a .car file that contains dimensions (periodic boundary conditions) for the unit-cell of a hexagonal crystal material I'm working with:

PBC   27.1979   27.1979   15.4999   90.0000   90.0000  120.0000 (P1)

That's the a, b, c lengths and $\alpha$, $\beta$, $\gamma$ angles.

I need to convert this information into basis vectors in cartesian space, e.g.

basis1 27.1979 ?       0.0
 basis2 0.0     27.1979 0.0
 basis3 0.0     0.0     15.4999

This may be better suited as a math question but it's computational-chemistry specific.

I found the way to convert the hexagonal parameters:

I have a,b,c and $\alpha$, $\beta$, $\gamma$. The solution in basis vectors is:

basis1 = (a, 0, 0)
basis2 = (xy, yx, 0)
basis3 = (xz, yz, c)

where
\begin{align}
xy &= b\cdot\cos(\gamma)\\
yx &= b\cdot\sin(\gamma)\\
xz &= c\cdot\cos(\beta)\\
yz &= \frac{b\cdot c\cdot \cos(\alpha) - xy \cdot xz}{\sqrt{b^2 - xy^2}}\\
\end{align}

Thus my solution was:

basis1 = (27.1979,  0,       0     )
basis2 = (-13.5990, 23.5541, 0     )
basis3 = (0,        0,       15.499)


What is the hybridization state of E (= Sn, C) in ECl₂? What is the hybridization state of tin in tin(II) chloride, $\ce{SnCl2}$, and is the explanation similar for dichlorocarbene, $\ce{CCl2}$?

For $\ce{SnCl2}$, I read in my book that it is sp² but I am not able to explain it. I am not sure about the carbene.

I might have a misunderstanding about hybridisation, I think that the energy required for hybridization has to be compensated by the energy released in forming bond between hybridized orbitals.

Hybridisation is a mathematical concept, you never gain or lose energy due to it. Descriptions of the wave function involving hybrid orbitals or not are equally as valid. The linear combination of orbitals that result in the equilibrium geometry can however be interpreted as hybrid orbitals. Sometimes this may lead to a better understanding of the bonding situation.

From the above, it is obvious that hydbridisation is always a result from the molecular structure, never a cause for it. Hence we can deduce from the known bond angles the following:

Tin(II) chloride has a ∠Cl-Sn-Cl of about 95° (see Wikipedia). We know that p-orbitals are 90° towards each other. Therefore the tin will have more or less unhybrididised orbitals. See also this previous answer of me and this answer of Ben Norris.
Dichlorocarbene on the other hand has a ∠Cl-C-Cl of about 109° from both experimental and calculated data (Phys. Chem. Chem. Phys. 2002, 4 (14), 3282–3288). We know that the angles between sp³ hybrid orbitals are tetrahedral angles of about 109.5°. We can therefore deduce that the carbon employs sp³ hybrid orbitals in first order approximation.

The differences in the molecular structure and the utilisation of the s-orbitals for bonding stem from what is termed the inert pair effect, there is an explanation available in the this Q&A: What is the inert pair effect?

1.The answer of question one is this that as Sn contain 4 electrons in its valence shell and will have to combine with two Cl atoms to form SnCl2 by sharing one electron with each Cl atom so it need two orbital which have one electron each then two free electrons will remain since there are maximum chances of these 2 free electrons to be present in the same orbital and be called as a lone pair hence Sn will undergo sp2 hybridization forming 3 sp2 hybridized orbitals and SnCl2 will have bent structure in which central atom Sn will be surrounded by 2 bond and 1 lone pair☺


Stability of Carbanions For comparing stability of carbanion, we use the fact that "negative charge is stable on more electronegative atom." So, The stability order is, {electronegative order: $\mathrm{sp \gt sp^2 \gt  sp^3}$} and therefore, $\ce{HC#C^- \gt H2C=C^-H \gt CH3-C^-H2}$.
My doubt is that, we generally say, carbanion is stable when there is less electron density on $\ce{C^-}$.
In $\ce{HC#C^-}$, $\ce{C}$ is $\mathrm{sp}$ hybridized that means it is more electronegative, so it will pull electrons more as compared to other two. So, electron density order will be reverse of above order. And, $\ce{HC#C^-}$ will be least stable.
By, my thought, answer is completely opposite.
Please help me to get out of this doubt and please correct me if I said something wrong.

$\mathrm{sp}$ hybridized carbanions are more stable as the lone pair of electrons is placed in an orbital consisting of high s-character. In general, electrons are more stable in s orbitals rather than the p orbitals of the same shell. This can be understood in a simple way, by thinking that the s orbitals are closer to the nucleus, and can also be verified by the n+l rule.
Now, when we say $\mathrm {sp}$ hybridized species are more electronegative than $sp^2$ hybridized species, we mean that the $\mathrm{sp}$ orbital are more accomplished in holding an electron within them when compared to their $\mathrm{sp}^2$ counterparts. As electrons are more stable in an $\mathrm{sp}$ orbital, electrons are pulled more towards them rather than the $\mathrm{sp}^2$ orbitals, which makes us say that $\mathrm{sp}$ orbitals are more electronegative than $\mathrm{sp}^2$ orbitals.

The stability works on the following principle:

More spread out the electron density, more stable is the compound.

As you can see, sp hybridized carbon atom pulls the electron density to a larger extent, thereby, effectively spreading out the charge.
Can you also think of why this is the reverse for cations?
In the case of carbocations, the sp hybridized carbon atom pulls electron density causing even more positive density on the carbocation.


What is the enol form of 4-methylcyclohex-2-en-1-one? What is the enol form of this compound?

I have made an attempt, but it doesn't look quite right.

Which hydrogen should be taken off (left or right hand side of the carbonyl)? And I don't think the ring is aromatic so I don't know how it is stabilized.

There are two possible enol forms of this ketone. The first one is:

This is the one you drew, which shows the α hydrogen being abstracted. It is perfectly valid.

The other one is:

Note how the alkene next to the carbonyl group "extends" the range of the carbonyl group. This allows the γ hydrogen (i.e. the one three carbons away) to be removed, and then you can "push" electrons into the carbonyl group through the alkene group. In general, this is a property of α,β-unsaturated ketones (i.e. ketones which have a double bond between the α and β carbons).

Note that in this enol form, the methyl substituent is now in the same plane as the C=C next to it. This means that, if you were to reverse the tautomerisation to get back the keto form, there's an equal chance of the hydrogen coming in from above the plane and from below the plane, and racemisation would occur. In practice, there is a rapid equilibrium between keto and enol forms, and this means that if you were to dissolve this compound in any kind of acid (or base), the racemisation would occur pretty quickly.

Here, I have drawn the mechanisms for acid-catalysed tautomerisation. I assume you are now able to draw the mechanisms for the reaction in base (which in general forms an enolate instead of an enol).

At this stage, you might wonder why we cannot abstract the other α hydrogen. There's a couple of reasons. Firstly, it's attached to a $\mathrm{sp^2}$ carbon, and the C-H bond is therefore somewhat stronger. Secondly, even if you could take off that proton and form the "enol", you would have two double bonds next to each other. That's a huge problem because those double bonds want to be linear, but don't quite have the space to do so since there's a ring:

Regarding the stability of the two legitimate enol forms, I would not say that they are any more stable than the keto form. In general, the keto form is more stable. There are cases in which the enol form is relatively stabilised, most often when there is aromaticity or extra conjugation in the enol form (other factors such as solvent will also influence the position of equilibrium):

However, that is not the case here. The starting keto form is as conjugated as any enol you can form.

yes your answer is right. this is more stable as it is a conjugated diene. but, initially the H+ attacks on the lone pair of =O then the oxygen takes up the pie bond to neutralize the formal + charge on it.


What is the order of acidic strength among phenol, catechol, resorcinol and hydroquinone? This question is from a book that gives the answer as

$$
\text{hydroquinone} < \text{resorcinol} < \text{phenol} < \text{catechol}
$$

but does not provide any explanation for the answer. If I go by the stability of the conjugate base due to resonance, phenol seems to have the most number of resonance intermediates and seems to be most stable. Considerations about number of resonance intermediates of conjugate base gives an order that is nowhere close to that provided as an answers.

Please provide an explanation as to the provided answer or, if the answer is wrong, please supply the correct order with explanation.

A quick search on says that resorcinol is more acidic that catechol, followed by phenol.

Hydroquinone - strong +M effect and weak -I effect

Phenol - the normal resonance you'd expect in all phenols

Resorcinol - -I effect

Catechol - +M effect, stronger -I effect and hydrogen bonding

Taking these factors into account, it is easy to see why hydroquinone is least acidic. Hydrogen bonding and the vicinity of the $\ce{-OH}$ to the anion (for the stronger -I effect) make cactechol more aciic than the phenol despite its +M contribution. Resorcinol has only -I effect and no counteracting electron-donating effects, which makes it the most acidic.

Catechol is less acidic than resorcinol and quinol because in case of Catechol present hydrogen bonding and do not give easily H+ ions.. In case of resorcinol - I work but in case of Quinol work +M and - I but +M is dominant hence the acidity order of.     resorcinol) Quinol) Catechol.


The violation of a certain rule or principle in the following electronic configuration I'm to find the violation of a certain rule or principle in the following electronic configuration:

$$\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$$

$n+l$ rule is not violated because: for $\ce{1s^2}$, $n+l=1+0=1$ so it is filled first, for $\ce{2s^2}$, $n+l=2+0=2$ so it is filled second, and so on.
Aufbau Principle is not violated because: the orbitals with minimum energy are filled first with electrons and $\ce{1s < 2s < 2p}$ which is the sequence of increasing energy orbital judged by the $n+l$ rule.

I'm quoting here a paragraph from my chemistry textbook of the chapter Atomic Structure:

  The orbitals given by a particular value of $l$ if $n$ is same, have the same energy and such orbitals are called degenerate. Like $l=1$ gives three $p$ orbitals specified as $\ce{p_x, p_y, p_z}$ and the p orbitals are said to be three fold degenerate. These orbitals are filled according to Hund's Rule.

The numerical value of the principle quantum number $n$ for $\ce{2s}$ and $\ce{2p}$ is the same; however, they don't have a particular value of the azimuthal quantum number $l$. For $\ce{2s}$, $l=0$, and for $\ce{2p}$, $l=1$, so $\ce{2s}$ and $\ce{2p}$ are not degenerate. So none of the three principles or rules whether Aufbau's, $n+l$ or Hund's is violated in this electronic configuration.

However, my friend says that there is a violation of the Hund's Rule in this electronic configuration, and according to him the right configuration is: $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$, which in turn violates the Aufbau Principle, because the orbitals of lower energy level have to filled first completely and $\ce{2s < 2p}$.

  So, which one of the electronic configurations is right, $\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$ or $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$?

Note: The required electronic configuration is for the electrons in their ground state.

If we talk only about electronic configurations and rules violation then you are correct. The 2nd one (excited state) is referred for hybridization that is sp2. An excited state  configuration is meant only when the particular atom gets ready or excited for forming a bond. Like boron does it with 3 fluorine atoms to form $\ce{BF3}$. So, if we talk about electronic config and rule violations we must go for ground states exceptional cases are $\ce{Cu}$ , $\ce{Cr}$ and $\ce{Mo}$.

It is violating Hunds Rule..
 " If degenerated orbitals are available then electrons want to live separately with same spin rather than mutually in same orbital with opposite spin."


Electrophilic addition to alkenes Consider the addition of HBr (in absence of any radical initiators or environment that promotes radical mechanistic pathway), to 3-bromocyclopentene.

The first step is the attack of $\ce{H^+}$
On the exposed π electron cloud. But my query is whether it forms a classical carbocation like this one to oppose the negative inductive effects of the halogen atom:

Or does it proceed with the formation of bromonium ion like this one to partially complete its octet using lone pairs on Br.

These two would lead to different products but I'm interested in the major one.
The first one would lead to 1,3-dibromocyclohexane, while the second one would lead to 1,2-dibromocyclohexane.

Since it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom


Can you approximate the change in entropy of a reaction? I found a problem in which I am given a hypothetical reaction in which from 2 liquids (reactants) I get 4 moles of gas and 1 mole of a solid substance (products):

$$\ce{A (l) + B (l)  -> 4C (g)  + D (s)}$$

I am asked about the change in entropy (an approximation of it), so I am curious if there is some kind of approximation for the change in entropy of a reaction based on the change in the amount of substance of gaseous compounds (since I know that the highest influence on entropy is that of gases).

The molar standard entropy of formation ($S_{f}^{\ominus}$) for many common substances is documented in tables. You can approximate the overall change in entropy of the reaction ($\Delta S^{\ominus}_{reaction}$) with with Hess's law:

$\Delta S^{\ominus}_{reaction} = \sum S^{\ominus}_{f,molar,products} - \sum S^{\ominus}_{f,molar,reactants}$

You notice that there is no $\Delta$ in front of $S^{\ominus}_{f,molar}$, since entropy is an absolute physical quantity. Every system has an absolute entropy.

Don't forget to multiply every $S^{\ominus}_{f,molar}$ with the number of moles that you have. This is only an approximation because the molar entropy of a substance depends on Temperature as well, and the reaction likely doesn't take place at standard temperature.

As a sidenote please be aware that the molar entropy of formation for elements is not zero, although the molar enthalpy of formation is!

In such a case, you may simply count the number of gaseous molecules on each side of the arrow. In your reaction there are no gas molecules in the reactants, and 4 in the products. The number of moles of gaseous substances increases. So the qualitative entropy increases.


What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation? We know that the Gibbs free energy is related to the equilibrium constant by the following equation:
$$\Delta_\mathrm{r}G^\circ=-RT\ln K$$
We also know the Van't Hoff equation:
$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

$\left(R = 8.314\ \mathrm{J~mol^{-1}~K^{-1}}\right)$

My question is: what kind of $K$ do we use for these equations? Concentration quotient ($K_c$) or pressure quotient ($K_p$)?

The first equation actually contains the definition of the standard equilibrium constant:
$$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$
With this definition the equilibrium constant is dimensionless.

Under standard conditions the van't Hoff equation is
$$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$
and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant.
$$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) =
  \frac{{\ - \Delta H^\circ }}{R}
  \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$

Now the ordinary equilibrium constant may be defined in various forms:
$$K_x = \prod_B x_B^{\nu_B}.$$

Probably one of the best representations for the standard equilibrium constant involves relative activities, for an arbitrary reaction,
$$\ce{\nu_{A}A + \nu_{B}B -> \nu_{C}C + \nu_{D}D},$$
this resolves in
$$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$

The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = \pu{1 mol/L}$.
At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as
$$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$
See also a very detailed answer of Philipp.

The partial pressure is connected to the activity via
$$a(\ce{A}) =
\frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$
with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=\pu{1 bar}$ or traditional use of $p^\circ=\pu{1 atm}$.
For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since
\begin{align}
\lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &=1, &
p(\ce{A}) &= y_{\ce{A}}\cdot{}p.
\end{align}

Of course concentrations and partial pressures are connected via the ideal gas
\begin{aligned}
pV\ &=nRT\\
p\ &\propto \text{const.} \cdot c,
\end{aligned}
and therefore it is valid to write:
$$K_c\propto \text{const.} \cdot K_p.$$
It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of $(\mathcal{R}T)^{\sum{}\nu}$.

While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.

Strictly speaking these equations should contain some correction of activity (concentration quotients) or non-ideal behaviour of gases (pressure quotients).
If you don't need to deal with those and you can treat your system as ideal, the concentration of a gas and pressure are equivalent. Be careful with the units, though!


not understanding buffering region vs equivalence point "During the titration of a weak polyprotic acid with a strong base, the first stoichiometric equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.  Prior to the equivalence point, a buffering region forms in which both the weak acid and its conjugate base are present and the pH is determined by the pKa and the ratio of the acid and base species concentrations.  When the concentrations of the acid and base species are equal, pH = pKa."

The bolded statements seem like they're saying the same thing. What is the difference between the buffering region and the equivalence point?

Isn't the buffering region when [HA] = [A-], so that in the HH equation, pH = pka? The first bolded statement makes it sound like the first equivalence point is also when [HA] = [A-].

This is a matter of definitions which you must learn. Let's say that the titrant is NaOH and that we have a monoprotic acid.

Equivalence point - The equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.

Notice that this doesn't infer anything about the pH of the solution.

Now for a polyprotic acid there are multiple equivalence points. The first equivalence point occurs when the moles of added base are equal to the moles of the initial acid species. The second occurs when twice as many moles of added base are added as compared to the initial acid species. The third three times as much base as acid and so on.

Again notice that this doesn't infer anything about the pH of the solution at the multiple equivalence points.

When the concentrations of the acid and base species are equal, pH = pKa."

This is a poor statement. Let's let the acid be $\ce{HA}$ so the conjugate base is $\ce{A^-}$. The statement would have been much better expressed as:

When the concentrations of the acid and its conjugate base are equal, pH = pKa."

Now to wrap this up, imagine doing this experiment. We start off with $\ce{HA}$ in solution. As we add more base we get to the pH where $\ce{[HA] = [A^-]}$ and where (1) there is maximum buffering capacity, and (2) the pH = pKa of the acid. We now add more base and we finally get to the equivalence point.

PS - For titrations you should also know the endpoint which is where the chemist gets some sort of signal to stop the titration.

I started reading about chemistry 2 months ago, so take this with a grain of salt. I also learned this in norwegian, so terms migt be different..

As far as i know, the equivalence point is when there is no longer any of the reactant, because it has all reacted with the added base/acid. Whereas the buffering/halfway titration point is when, as they say, there are equal amounts of H- as HA, making the solution a buffer.

Did i understand your question right?


Reaction of NaOD/D2O with semicarbazone I managed to solve the question till D but I'm stuck at the last reaction.
I tried to find its solution and the source suggest it as

Upto (D) you are correct From there it's :

This is what I reffered frm book and conclude d..


What are the currently most accurate sources for lists of the values for elemental properties I have a hopefully simple question. I am a tutor working with a student that wants to know where to find the most recent accurate measurements for elemental properties. He's only in Chemistry I, so he's not looking for every property out there calculated multiple ways. I think he (and I) just like looking at the best sources. So we're looking for the following:

Latent heat of fusion
Latent heat of vaporization
Melting point
Boiling point
Density
Specific heat
Ionization energy
Atomic radius (metallic and covalent)
Atomic mass
Electronegativity
Commonality and uncommonality of charges (like oxygen is commonly -2 but is also -1 in peroxide)

I can point him to wikipedia or https://ptable.com/#Properties/Series for some of the information, but I myself don't know where the most legitimate sources for these values can be found. My CRC handbook (97th edition) lists ionization energies, but for atomic weight it doesn't list the weight of the most common isotope of elements like Ac. Also, it's a few years old.
My question is, what are the current best sources of accurate values for each of these properties?

You should teach your student not to use numerical values with as many significant figures as possible. Knowing how to make sense to decimals is part of the scientific culture. The distance between two cities cannot be given to the least micrometer : it has no meaning. The weight of an object has no use if it is given such as $123,45678910111213 g$. The least figures which can be given for sure is the milligram (here 6). Put on a balance, such an object will display fluctuating next figures. Already the figure $7 (0.1 mg$) will fluctuate at the least draft. To be sure of the next decimals you must put your balance in an insulated and closed room without nobody entering during the whole day, in order to avoid drafts. And even, the least dust grain falling from the air will change one or two of the following figures. So please do learn to use numbers with maybe $2, 3, 4$ or $5$ significant figures. But not with ten or twenty significant figures. It is not the way science works. So why looking for the newest published values which have one or two more significant figures ? The given weight $123.45678910111213 g$ is not better than $123.456 g$. It only gives an irrational feeling of superiority. That's all. It has no meaning for usual applications. Nobody will believe you if you say that you have calculated a weight of $123.4567891011121314 g$, or measured a temperature of $12.3456789 °C$.


Why can't Pd/C and H2 reduce both the alkene and carbonyl portions of α,β-unsaturated carbonyls? Why is it that the major product of the reduction of chalcones the ketone and not the monoalcohol? In other words, Why isn't the major product a benzyl alcohol?

From what I understand, catalytic hydrogenation can be used to reduce carbonyls as well as alkenes.

My TA told me that nucleophilic hydrides are preferred for reducing carbonyls. Why wouldn't hydrogenation work as well? Does it have to do with resonance involving the carbonyl since in a chalcone, the carbonyl is adjacent to an aromatic ring and is also conjugated with the alkene? Could the reason that catalytic hydrogenation can't effectively touch the α,β-unsaturated carbonyl be the same reason catalytic hydrogenation can't effectively reduce carboxylic acids, esters, and amides — all of which are also resonance stabilized?

My first thought had to do with heats of hydrogenation and how resonance-stabilization found in carboxylic acids and its derivatives reduce the heat of hydrogenation. However, later I found a resource online implying that all pi bonds - even the delocalized, resonance-stabilized ones found in benzene — could be reduced through catalytic hydrogenation with enough time.

In addition, it was noted elsewhere that chalcones could be completely reduced to the benzylic alcohol, although not with ease.

This leads me to believe that there is an activation energy barrier that's impeding the hydrogenation of certain substrates within the confines of a 3 hour undergraduate lab period … am I on the right path? Sterics, perhaps? It was noted that the more highly substituted an alkene is, the slower it is reduced because of the difficulty in getting a highly substituted alkene to approach the catalyst's surface with the $\ce{M-H}$ bonds in an appropriate manner.

You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers ($\ce{\Delta \mathrm{G^{‡}}}$) that prevents the carbonyl reduction from occurring at the same rate as the alkene reduction. This means to produce the alcohol faster, we must manipulate the kinetics of the reaction.

To understand why the carbonyl-reduction transition state is higher in energy, we should consider the differences between the carbonyl and alkene bonds. The bonds are much more polarized, and as such the carbonyl $\pi$ bond is considerably stronger than alkene ($93\ \mathrm{kcal\ mol^{-1}}$ vs. $63\ \mathrm{kcal\ mol^{-1}}$)$^{\mathrm{[1]}}$. This means that at low temperatures, only a small fraction of the molecules have enough energy for the carbonyls to associate onto the $\ce{Pd/C}$ surface. Catalytic hydrogenation of carbonyl compounds (aldehydes, ketones, and especially esters) requires high temperatures and pressures to increase the presence of both the substrate and hydrogen on the catalyst.

$^{\mathrm{[1]}}$ Fox, M. A.; Whitesell, J. K. Organic chemistry; Jones and Bartlett: Sudbury, MA, 1997.

the answer to your question lies in the catalyst used!!

if we use only Pd then complete reduction will take place .

but if we want to reduce the reactant partially i.e. reduce only one double bond ; we use Pd alongwith C(carbon) because "C" acts as a "poison" and influences the reduction reaction to complete at partial stage.

hope this helps!!!


How to show that when the equilibrium constant is larger than one, some real-life equilibria are still reactant favoured? My instructor assigned us with this question:

  When real examples of chemical equilibrium are used, the conclusions about reactant- or product-favoured versus the value of $K$ must be modified (e.g. real equilibria can have $K > 1$ but still be reactant-favoured). Discuss this idea, using examples of real chemical equilibria and assumed values of equilibrium concentration.

I was taught that when $K > 1$, it will be product favoured, assuming it’s already at equilibrium ($Q = K$).

How would I go about answering this question? Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

The terms reactant-favored and product-favored
As discussed in the comments and in YB609's answer, these are not technical terms with precise definitions.
First possible definition
One straight-forward definition would be to call reactant-favored reactions those for which $K$ is smaller than one, and product-favored those for which $K$ is larger than one. You would have to specify the reaction (with phase information), the temperature and the standard state you are using. With this definition, you would be correct to say that such a reaction, if initiated at standard state, would go in the net forward direction to reach equilibrium if $K$ is greater one (and therefor, according to this definition, it is product-favored). Of course, if you take this definition, the question does not make any sense.
Second possible definition You could make a definition based on whether there is "more" reactant or product at equilibrium. This is problematic for multiple reasons. First, you could have a lot of one product, little of another product, and something in between for the reactant. So is there more reactant or more product? Second, if you use amounts of substance (moles) as the criteria for "more" and have a heterogeneous reaction, the relative amounts depend on the volumes (compare oxygen in equilibrium between hemoglobin and the dioxygen in a test tube vs in the entire atmosphere).
Third possible definition If you look at the concentrations to define "more", it still is confusing under some circumstances, even for a homogeneous reaction (everything in the same phase). If the sum of coefficients of reactants is not equal to that of products, diluting the system will disturb an equilibrium, so reactant: product ratios will depend on the dilution of the system.
Answering the question

How would I go about answering this question?

You would start by asking for the definition of reactant-favored and product-favored.

Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

Once the terms are clearly defined, it would be sufficient to look at equilibrium states (and you need to know the definition of the standard state, otherwise the value of $K$ has no meaning).

Actually you should specify correctly what exactly product favoured and reactant favoured mean according to you. Some text use product favoured or reactant favoured term to notate the direction of reaction and some text use product favoured or reactant favoured term to notate the domination of concentration.

If product favoured term is used to notate direction of reaction:
If $K < 1$ that means reactant concentration is greater than product concentration. Now at equilibrium concentration of reactant can be greater than that of product or vice-versa. Say for a reaction $\ce{A -> B}$, we get $K = \frac{\ce{[B]}}{\ce{[A]}}$ and let's say this $K$ is less than 1. Now this ratio will always remain constant. Now if you add some more reactant, then it will get consumed and yield product so as to keep $K = \frac{\ce{[B]}}{\ce{[A]}}$ constant. Now adding reactant made the reaction move towards right i.e product favoured (i.e reaction moves towards the product) but still $K < 1$ is being maintained.

If product favoured is used to notate domination of concentration in the reaction i.e. if by the term product favoured you mean to say at equilibrium concentration dominates product concentration, then according to this notation:
Again taking the example of $\ce{A -> B}$, we get $K = \frac{\ce{[B]}}{\ce{[A]}}$ and let's say this K is less than 1. That means at equilibrium reactant concentration is greater than product concentration. So according to the notation we are using now, we should say it to be reactant favoured. Now $K < 1$ will always be reactant favoured and can never be product favour if you are using the term reactant favoured or product favour to notate domination of concentration in the reaction.

If you are using the terms product favoured and reactant favoured to notate the direction of reaction, then saying K < 1 and saying product being favoured makes sense. According to your professor I think product favoured and reactant favoured mean the direction of reaction. So, it depends how you use those terms.


Why is wavenumber used in IR spectroscopy rather than wavelength? In IR spectroscopy, the $x$-axis is used to represent wavenumber, in $\mathrm{cm^{-1}}$. Why is wavenumber, equal to $1/\lambda$, used in place of wavelength, which is simply $\lambda$?

Sources I’ve already found explain why it was chosen rather than energy of waves, but the conversion from wavelength to wavenumber is never explained. Below are two relations from Wikipedia, which explain how it can be used in equations, but in all of these cases, $\lambda$ seems to be a choice that’s easier to work with.

What explanations are there, if anything other than “historical reasons”, for why $1/\lambda$ is favored over $\lambda$?

A spectroscopic wavenumber $\tilde\nu$ can be converted into energy per photon $E$ via Planck’s relation:

$$E = hc\tilde\nu$$

It can also be converted into wavelength of light via

$$\lambda = \frac{1}{n\tilde\nu}$$

where $n$ is the refractive index of the medium.

Not only in IR spectroscopy. Wavenumber is unit of energy and therefore you can directly deduce the difference of energy between states.

In addition, humans like to think in acceptably small numbers (0.01 - 10,000). Wavenumber allows this for IR and conveniently supplements the eV unit in small energy separations range. Admittedly, the conversion factor of 8,065.73 won't win beauty contest.

Reciprocal length (cm-1) or wavenumber (cm-1) used in vibrational spectroscopy is actually an intuitive concept in reciprocal space (used heavily in diffraction world). From literature: "Reciprocal length is used as a measure of energy. The frequency of a photon yields a certain photon energy, according to the Planck-Einstein relation. Therefore, as reciprocal length is a measure of frequency, it can also be used as a measure of energy. For example, the reciprocal centimetre, cm−1, is an energy unit equaling the energy of a photon with 1 cm wavelength. That energy amounts to approximately 1.24×10−4 eV or 1.986×10−23 J.

The higher the number of inverse length units, the higher the energy.


How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution? Below was question 34 in the USNCO 2017 exam:

  If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?

  (A) Roughly 50%
  (B) Roughly 75%
  (C) Roughly 95%
  (D) Over 99%

My solution is as follows:

Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K_\mathrm{sp}$.
Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$
Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$
Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation:
$$
(0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$
Find percentage of calcium hydroxide precipitated:
$$\frac{0.019}{0.02} \times 100\% = 95\%,$$
hence (C)

I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).

Is there a simpler method?

A less analytic aproach:

Initial concentrations:

$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$

$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.

Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.
The concentrations that remain in solution are:

$\ce{[Ca^{2+}}]=0.0126(1-p)$

$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$

Since the solubility product remains the same:

$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$

Now, since there are only four scenarios, try them out substituting every value of $p$ in  the equation and take the closest one to $8\cdot10^{-6}$:

\begin{array}{|c|c|}\hline
    p&K_\mathrm{sp}\\\hline
    0.50&\pu{7.99E-05}\\
    0.75&\pu{3.56E-05}\\
    0.95&\pu{6.46E-06}\\
    0.99&\pu{1.27E-06}\\\hline
\end{array}

The closest one is for roughly 95% precipitation, without solving any equation.

Note what the question actually asks you, "roughly". Approximation is an important tool in ionic equilibrium calculations.

Here, you have $0.0126M \text{ }\ce{Ca}^{+2}$ and $0.0252M\text{ } \ce{OH-}$ in the original solution. By addition of $0.1M \text{ }\ce{NaOH}$, which is a strong electrolyte, you'll not only add $0.1M\text{ } \ce{OH-}$ to the solution, but also cause $Q$ to exceed $K_{sp}$, hence, the salt will be precipitated out.

Now, the concentration of our ions before precipitation is $0.0126M \text{ }\ce{Ca}^{+2}$ and  $0.1252M\text{ } \ce{OH-}$. Let $y$ be the fraction of existing ions that are not precipated out. So, we'll be left with $0.0126 \cdot y M \text{ }\ce{Ca}^{+2}$ and $0.1252 \cdot (2y-1) M\text{ } \ce{OH-}$ ions. Equating their ionic product to the salt's $K_{sp}$, we get:

$$0.0126 \cdot y\cdot(0.1252\cdot (2y-1))^2=8\times10^{-6}$$

$$y\cdot(2y-1)^2\approx0.0405$$
$$y\cdot(1+4y^2-4y)\approx0.0405$$

Magic trick$^\text{TM}$: Neglect $y^3$ at this step. If $y$ comes out be $\le5\%$, we'll assume this neglection to be correct.

$$4y^2-y+0.0405\approx0$$
$$y\approx0.050838$$

That means nearly $(100-5)\% = 95\%$ of the salt was precipitated out. Hence, your answer. Without a graphing calculator.

PS: If you actually solve the original equation, you'll get $y=0.05$. Hence, our approximate answer is very reasonably close.

How did I get $(2y-1)$ as fraction of final concentration of $\ce{OH-}$ ions?

Fairly easy. Observe that $y$ is the fraction of concentration of ions not precipitated out, while $x$ is the fraction of concentration of ions precipitated out. Then $x+y=1$. So, $1-x=y=\text{ concentration of }\ce{Ca}^{+2}$ ions. And $1-2x=1-2\cdot(1-y)=2y-1=\text{ concentration of }\ce{OH-}$ ions.

But why did I take $y$ anyway? What was wrong with the original $x$?

Our aim was to reduce this problem into an equation which can be solved without a calculator, i.e., by approximation. Taking $y$ ensured that in the end we could neglect higher powers of $y$ (note that we cannot neglect higher powers of $x$) and solve the question easily.

Crux of my answer: if while solving an ionic equilibrium question you're stuck in a tough calculation/unsolveable equation, consider modifying your approach to allow the use of neglection.

Hope it helps!

Ok, I admit, it wasn't any magic trick :P But just a clever observation that if $x\le 0.05$, then $x^3\le0.000125$, which is too small to affect our final result significantly.


What form of zinc's found naturally in vegetables & meat? What kind of zinc naturally occurs in plants? Zinc oxide, zinc sulphide, or something else?

The most likely form is $\ce {Zn^{2+}}$ bound inside a protein, or maybe chelated to some amino acids. It is not very likely that there is zinc oxide, or zinc sulfide, or zinc sulfate in organic matter.

Different vegetables contain different types of Zinc compounds, so we can't say specifically which type of zinc compound is there in vegetables.

Zinc Oxide is the most used supplement in the U.S.

Zinc carbonate is also present, which our body cannot absorb.

Also they can contain  zinc picolinate, zinc gluconate or zinc citrate (zinc picolinate is better than zinc gluconate.), zinc glycinate, etc.

Reference: Wikipedia


What is the difference between physical and chemical bonds? If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds chemical and physical?

From the freedictionary.com, chemical bond:

  Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.

What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these "several forces" excactly?

In the answer I would like to see the lists of chemical and physical bonds.

In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is the sharing of electrons between two or more nuclei. In contrast, other interactions are often said to be intermolecular (which is somewhat more specific than the term “physical”.

In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).

Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In The Nature of the Chemical Bond (1960), he gives the following definition:

  The Chemical Bond Defined.—We shall say that there is a chemical bond between two atoms or groups of atoms in case that the forces acting between them are such as to lead to the formation of an aggregate with sufficient stability to make it convenient for the chemist to consider it as an independent molecular species.

A bond is what links atoms into molecules, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…
The complete opposite: consider all interactions as chemical bonds, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like this one:

  A chemical bond is the physical phenomenon of chemical species being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons and is a phenomenon that is fully described by the laws of quantum electrodynamics. In general, strong chemical bonds are found in molecules, crystals or in solid metal and they organize the atoms in ordered structures. Weak chemical bonds are classically explained to be effects of polarity between molecules which contain strong polar bonds. Some very weak bond-like interactions also result from induced polarity London forces between the electron clouds of atoms, or molecules. Such forces allow the liquification and solidification of inert gases. At the very lowest strengths of such interactions, there is no good operational definition of what constitutes a proper definitional "bond".

This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).
You can classify interactions by energy: decide that chemical bonds are those interactions that have an energy higher than a certain threshold, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.
Finally, what I believe is the most common description is to look at the nature of the interaction, and classify it following a certain convention. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.
I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the New Oxford American Dictionary as well, which I quite like:

  chemical bond a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.

Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is sharing (or transfer) of electrons. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).

Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…

Bond means force of attraction. Matter is made up of molecules and and atoms. The physical force of attraction which holds atoms and molecules in a matter is called physical bond. Vander waal's forces, coulombic forces are physical forces. The bond which is made due to the behaviour of atom to become stable is called chemical bond. The physical bond is due to property of particles called mass,electric charge(positive or negative).


What decides whether a reaction releases light or heat I understand the reason that a chemical reaction would create photons (physics is more my strong point), but why would it create heat instead? My only guess is that the light released somehow causes the molecules around it to gain kinetic energy through some process and thus create heat, but I don't know how this would occur. What decides whether a given reaction produces photons or heat?
I'm a chemistry noob so try to explain using noob-friendly terms. Also, I don't know what tags suit so please put appropriate tags if my tags aren't.

The amount of energy released by a reaction is what most often creates light. Burning, the energetic reaction usually caused by oxidation of something, creates light because the amount of heat released is large enough to heat the reactants or products to temperatures where the radiation emitted contains a lot of visible light (coals in a fire glow red because they are hot enough that black-body radiation has a significant component in the visible).

Most exothermic chemical reactions don't generate this amount of heat and, although they release heat, you don't see it. Hand warming pads, for example, often contain supercooled sodium acetate which can be made to crystallise by popping a small metal insert. The exothermic reaction (crystallisation) releases heat to warm your hands, but not enough to see visible light.

A very small class of chemical reactions produce light directly as the reaction leaves molecules in an excited state which decays with the emission of visible light. Glow sticks are an example of this.

Overall, though, most reactions produce heat because the products have lower energy than the reactants and the released energy has to go somewhere. And it usually just ends up as heat. No photons are usually involved as the energy passes directly into the vibrational and translational energy of the products. So the excess energy in the reactants' bonds ends up as vibrational or translational energy in the products.

Let me put this into a physics friendly form.

When you say heat, heat is not actually an entity of absolute definition like a photon. heat is transference of energy between two bodies. so when you say when heat and when photon is actually like saying, when energy, when photon.
it doesn't make sense, since photon has energy.
I assume you mean when it does transfer energy in form of light and when it doesn't.

Every chemical reaction has an enthalpy based on how different is the products from reactants. when the reactants has more energy than products, the reaction results heat to transfer into environment (exothermic) when the vice versa, heat transfers from environment to reaction (endothermic) which makes our energy balance solely right. in exothermic case, heat gets absorbed by compounds in environment. if the temperature rises from critical limit that makes electrons in atoms emit light, then photons emit and when not, we just have raise in temperature.

I hope I was simplistic enough.


Calculation of molarity using dilution equation Using the formula $C_1V_1 = C_2V_2$ I am supposed to calculate the molarity of dilutions.

The original solution is $\pu{0.100 M}$ at $\pu{10 mL}$. I diluted it by mixing a $\pu{7.50 mL}$ stock solution with $\pu{2.50 mL}$ water, and made another one by mixing a $\pu{5 mL}$ stock solution with $\pu{5 mL}$ water.

Intuitively, I should get a higher molarity for my first dilution since it is higher in concentration, but when I calculate using
$$0.1(10)=C_2(7.5)$$
for the first dilution and
$$0.1(10)=C_2(5)$$
for the second, the second dilution's molarity is higher than the first. I know I am doing something wrong but I just can't figure out what.

You're mixing up the terms and formulae. Instead of rote learning the formulae, remember that the moles of the solute will remain the same in a solution - before and after dilution.

With that in mind, let's inspect your $\pu{7.5ml}$ stock solution. Before dilution, it had $\pu{7.5ml}\times\pu{0.1M}$ moles of the solute. After dilution, it has an unknown molarity. Let that molarity be $x ~\pu{M}$. Then, the new moles are $\pu{10ml}\times x~\pu{M}$. Equating and solving, you get $x=0.075$.

If you do that again for the second solution (I give it you as a homework), you'll get $x=0.05$. This easily verifies your own intuition that you should get "a higher molarity for the first dilution since it is higher in concentration".

It should be $7.5\cdot0.1=(7.5+2.5)\cdot C$ then find the value of C which is the concentration of 1st solution....for the 2nd one similarly, $5\cdot 0.1=10\cdot C$ then similarly calculate concentration....


Does rearrangement occur in Friedel Crafts alkylation or not? In the reaction of alkylation of aromatic systems with the help of Friedel Crafts alkylation - the reagent used is alkyl halide in presence of appropriate Lewis acid.

When is rearrangement observed?

If the alkyl halide is primary, the carbocation is usually not stable - as a result, a free carbocation shouldn't be formed and the attack should occur from the acid base complex itself (this is what my instructor said). However, in cases where the alkyl halide is tertiary, or where the carbocation is stable enough to be formed - rearrangement (if possible) is observed and the reaction proceeds through a free carbocation intermediate which acts as electrophile.

So in the former case, where a full fledged carbocation isn't formed, is rearrangement possible?

For example, is rearrangement possible in alkylation of benzene with n-propyl chloride in presence of aluminium chloride? Clearly, the primary carbocation is not stable - and may arrange to secondary carbocation (only if it is formed in first place)

So what really happens, and when in general and in particular can we expect rearrangements in Friedel Crafts alkylation? Also, when not to expect rearrangements? (Of course, when the carbocation isn't formed, but when is that?)

It'd be great if you could quote some credible sources along with your answer, if you wish to add certain specific reactions etc.

P.S.

Which Lewis acids allow for rearrangement to occur? As far as I know, only in presence of stronger Lewis acids does rearrangement occur - but how strong?
Does temperature also play a role for rearrangement to occur?

When you conduct the alkylation of benzene with n-propyl chloride, the product you obtain is isopropylbenzene, so a rearrangement occurred by virtue of the fact that the primary carbocation that forms will undergo a 1,2-hydride shift to form a secondary one. Most alkylation ( aromatic ring and alkyl halides ) are conducted in the presence of aluminium chloride which is one of the strongest Lewis acids. The principle is that the alkyl halide reacts with the aluminium chloride to form a salt: a carbocation and a complex ion of tetrachloroaluminate. You can have methyl chloride react in this way because even if it forms an unstable carbocation, it will readily react with the aromatic ring ( which acts as a Lewis base ). You can have methyl carbocation in solution such as in alkylation reactions, but you can’t isolate a salt containing methyl carbocation in solid state. If you had methyl chloride and aluminium chloride in solid state ( in contact with each other, they would not react to form carbocations.


pKaH of pyridine versus other imines Why is pyridine so much more acidic than other similar imines (5.5 versus ~9, see picture)? Wouldn't delocalization of the positive charge into the aromatic system stabilize it and hence decrease acidity? Also the lone pair of pyridine is not delocalized into the perpendicular $\pi$-system and hence not any less available than the one of the projected imine.

(Source: Organic Chemistry by Clayden et al. p. 1149. )

Why is pyridine so much more acidic than other similar imines?

When we discuss the acidity of a compound, we are really discussing where the equilibrium between the acid and its conjugate base lies.  Therefore, when we compare acidities we are really comparing equilibria, the relative stabilities of the reactant (acid) and product (conjugate base).

The following diagram illustrates the equilibrium between the iminium ion and the corresponding imine, it's conjugate base.

We can't draw any good resonance structures for the imine, but we can draw 2 resonance structures to describe the iminium ion, A and B in the figure. In the pyridinium ion - pyridine case we can draw 2 resonance structures for pyridine (just like in benzene), but we can't draw any resonance structures for the pyridinium ion that delocalize the positive charge without destroying the aromaticity of the ring.

So, in the iminium ion - imine case

we can stabilize the iminium ion through resonance - this stabilizes the iminium ion and shifts the equilibrium to the left making the iminium ion less acidic.

Whereas in the pyridinium ion - pyridine case

we can't draw good resonance structures that stabilize the positive charge in the pyridinium ion - this makes pyridinium ion more acidic.

Both factors operate in the same direction and tend to make the pyridinium ion ($\mathrm{p}K_\mathrm{a}=5.5$) more acidic than the iminium ion ($\mathrm{p}K_\mathrm{a}$ $\sim$ 9).

One interpretation of acidity is the ability to stabilize a negative charge (since the electron is generally kept by the acid, and the proton harbors the positive charge, if we go by bronsted lowry)

Thus, the negative charge is delocalized and thus stabilized. Its impact is not felt as greatly on one atom, and therefore it is more stable, not seeking a proton as much.

Also, the negative charge is better for nitrogen, an electronegative atom which can better stabilize it (but this doesn't explain the difference. I think the main difference is resonance)

This website may help:http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/heterocyclen/arom_gessaet_heterocyclen.vlu/Page/vsc/en/ch/12/oc/heterocyclen/sechsringaromat/sechsring_aromat.vscml.html


Why do the pennies in the pre/post 1982 penny experiment sometimes dissolve completely? USA pennies were nearly pure copper pre-1982.  Post 1982 they are copper clad zinc.  A common experiment in US high schools is to notch pennies with a file and add them to an HCl solution.  The old pennies remain untouched and the zinc of the new pennies dissolves, leaving the copper cladding.

Sometimes the copper cladding also dissolves and sometimes a considerable amount of the old penny dissolves.

Why does the copper dissolve?

I propose that the copper in the penny is reacting with atmospheric oxygen, forming copper oxide that dissolves in the acidic solution.

Another proposal is that the formation of a complex ion between copper and chloride ion drives the equilibrium, allowing the copper to react in HCl.  This is analogous to how aqua regia dissolves gold.

Another proposal is that the copper oxide layer on the penny reacts with copper metal in a comproportionation to form copper (I) ion that then reacts with oxygen to make copper (II) and dissolves?

Which is right?  Why does the copper dissolve?

Your conclusion about formation of the Copper Oxide is correct, but not about the reaction with atmospheric oxygen.

The origin of the Oxygen is from the hydrochloric acid solution. The solution does not mean is pure HCl , it means is a dissolve in water.  There is a redox reaction with water

I leave to you to find the redox reaction.


Aldolisation reaction and chalcone formation I realized a reaction of mixed aldolisation between acetone and 2-nitrobenzaldehyde, using a solution of 10% NaOH in ethanol.

I obtained a mixture of products at the end of the reaction. The problem is when I added the ethyl acetate I saw a blue liquid mixed with brown solid.  After a night of this reaction I found a lot of white crystals mixed with the blue liquid.

Can someone explain the presence of this blue liquid with the brown solid, knowing that the crystals are soluble in ethanol?

Well, actually you've made a reaction called the "Baeyer–Drewson indigo synthesis" (original article here), which was route developed in the nineteen century to obtain the indigo dye, nowadays replaced by other methods. I'm not writing the mechanism for this reaction because it is quite long and well-documented elsewhere.

Your blue compound is therefore the indigo dye represented here. As for the brown compound, it's difficult to say without more data (NMR spectra for instance), but the synthesis can give many colored intermediates or by-products such as isatin (bright red).

I hope this helps.

In the aldol reaction, we would expect to come up with the following product:

However, this reaction is bound to happen poorly. This is due to the presence of the nitro group, which has so much electron-withdrawing power that you can consider it to be twice as powerful as a carbonyl group (although the nitro group is not as powerful here due to its attachment to the benzene ring). Hence the nitro group then subsequently attacks the ketone in many intramolecular reactions, an example is shown here:

Since there are so many aromatic compounds that are fully conjugated forming, these should all contribute to the blue colour that you are getting.

I believe the brown solid is the result of some polymerization of the ketone with either the aldol or the nitro group.


Comparing Acidic Strength of H3PO4 and H3AsO4 The $\mathrm{p}K_\mathrm{a}$ values for $\ce{H3AsO4}$ and $\ce{H3PO4}$ are $2.30$ and $2.12$ which indicate that $\ce{H3PO4}$ is a stronger acid compared to $\ce{H3AsO4}$. I looked up their electronegativity values to find that they are almost the same for both arsenic and phosphorus. I feel $\ce{H3AsO4}$ should be the stronger acid because arsenic atom is larger in size and the charge is spread out in much more volume which should make the conjugate base more stable. What am I missing out?

The stability of an acid is judged by the strength of it's conjugate base. If the conjugate base is stable, it implies that it is a weak base and hence the corresponding acid is stronger.
Now consider $\ce{H3PO4}$. The structure has one $\ce{P=O}$ bond and $\ce{3 P-OH}$ bonds.

The conjugate base would be one Hydrogen short from an $\ce{OH}$ group. Similarly with $\ce{H3AsO4}$. However, the conjugation (resonance) will be better in the case of $\ce{P=O}$ bond, since it is an orbital overlap of oxygen's $\mathrm{2p}$ and phosphorus' $\mathrm{3p}$ as against arsenic's $\mathrm{4p}$. $\mathrm{2p}$ and $\mathrm{3p}$ are of comparable size and account for a better overlap and better resonance, distribution of electronic charge and hence a more stable conjugate base.
Consequently, $\ce{H3PO4}$ is better than $\ce{H3AsO4}$ as an acid.

H3PO4 has more EN atoms than H3AsO3 first of all. So it wins that first (checkmark more ability to stabalize a negative charge therefore more stable and that means stronger acid.
P + 4 oxygens > As +3 oxygens. Secondly, there are more terminal oxygens on the H3PO4 as well. All of the oxygen atoms on the H3AsO3 are bonded to the 3 H atoms.
IF they had an even number of electonegative atoms bonded to the central atom and equal (or even closely similar) number of terminal (not bonded to H) oxygens... then we can go ahead and compare their size since they are in the same column on the periodic table. When you have binary acids bonded to one other anion... go ahead and jump to electonegativity comparisons ( for comparisons in the same period) or size of anion comparisons (if in the same column or group.) first. With oxoacids like this first compare which one has more electronegative atoms .... then compare their electronegatively/ size.


Compound Polarity Lookup I know from the literature that the following compounds have decreasing polarity: Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds of similar structure it is manageable to compare bond types, groups etc etc by which to order their polarity, but what about compounds with very different structures, groups and bonds, especially when you want to rapidly order 10's-100's of compounds.
My questions:

Is there such a thing as an absolute measure of "polarity" for a compound by which other compounds can be compared/ordered?

Depending on 2) is there a resource whereby you can lookup your compound to obtain its polarity?

I am asking this question from a Chromatography perspective. I would like to know the order of separation for a group of compounds and how to know what "polarity" of mobile phase to start my experimentations with.

Polarity is an umbrella term which is popular among organic chemists. It is as vague as saying the weather is good today. "Good weather" may include, bright light, sunny, temperature, right humidity and so on.  Now read your own statement

Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds
of similar structure it manageable to compare bond types, groups etc
etc

Since you correctly use etc. etc. ...yes polarity includes et cetra et cetra. Its usage is discouraged by IUPAC.
The correct term is dipolarity which is a measure of the dipole moment (you can look them up in tables). Even that dipole moment is of no good because you have plenty of conformers and each conformer has a different dipole moment (at least in the gas phase).


Optional determination techniques in Atomic Absorption Spectroscopy I have tried to find information about this case but I couldn't find anything specific.

The key point of instrumental methods of analysis, is for the student to be able to know which one of the possible methods to use for a specific determination. As for me, that is not an easy thing to do because every method has its pros and cons. For the moment I am analysing AAS (Atomic Absorption Spectroscopy), and I simply cannot tell:

What's the reason that $\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ is preferred to be determined only by the hydrid method? $\ce{Pb}$ for example can be determined also with the other methods in AAS

Does anyone have any tip, how to choose between methods if it's required for any element?

I replied with a similar question earlier about the mercury determination.

Remember this is "atomic" spectroscopy so it is expected that the "sample" is being totally ripped apart. So one "best technique" per element.

$\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ are analyzed by the graphite tube technique rather than in a flame because the technique provides more sensitivity.

$\ce{Hg}$ is an odd one since you use a cold vapor technique.

Of course depending on the particular sample matrix what you might need to change.

You are right, every method has its pros and cons and if you were to choose one, the choice depends on what are you trying to analyse.
The important things to keep in mind are:

With AAS you will rarely be looking for every single element in the sample, but only the ones you expect to find or the ones you're specifically searching for

For example, you have an organic product and during its synthesis you used the metal catalyst (Pt is a common one). You want to make sure the catalyst is not present in your product. You will then want to search only for that particular metal. Based on what your analyte is, you choose on a method.

The other important thing to consider is how sensitive your technique needs to be. You don't need ultra high sensitivity in all analyses and this will depend on your research (again, what are you trying to find with your measurement)

For example, if some product that's sent to you for analysis (e.g
 toys, pharmaceuticals) is regulated to have <5 ppb lead concentration, it is enough to use a technique which has sensitivity on ppb level. The sample may still contain traces of lead on ppt level and there are techniques which are that sensitive, but they are in this case unnecessary

I hope this answer is satisfactory to you because I don't believe any analyst can give you a singular answer. You may use different kinds of techniques for the same type of analyte.


What molecules can be eliminated in condensation polymerization? Is there any specific rule to what kind of molecules can be eliminated in condensation polymerization? From Wikipedia I found that "small molecules" like $\ce{H2O}$ can be eliminated. Is there a more accurate criteria to what these "small moecules" are?

As an abstract answer, to me it would be whatever is commercially viable to manufacture and sell. Commercial polymerisation processes eliminate ethylene glycol (in the manufacture of PET) which is probably recycled,  methanol in transesterification reactions, ethene or propene in metathesis type,  phenol (recycled) in the use of diphenylcarbonate with bisphenol A, etc. In home applications,  acetoxysilicone sealant gives off acetic acid. So, anecdotally, it seems "simple" molecules of MW around 100 or less might meet your criteria.

yes , only small molecules such as H2O or CO are eliminated , as such it is no rule . if large and bulky molecules are extracted as waste, it will no longer be called an elimination product but a side-product. i hope you understand now !


Questions about van't Hoff factor in colligative property I learned from many websites that $i$ of strong acids and bases and some weak bases could be determined by how many ions are dissociated. But, I realized that I need to calculate the $n$ in $i=1+(n-1)a$ ($a$ is the degree of dissociation), not the $i$ in my problems.

Is there any rule to calculate the $n$ for strong acids and bases and weak acids and bases? Do I also need to check solubility table to solve colligative property problems that using van't Hoff factor?

In general, 'n' is the number of species the compound dissociates into, and such formulas are applicable only when the acid itself is monobasic, i.e.only one dissociaton is involved.For sulphuric acid,this is not applicable as there are two dissociations with two dissociation constants.The best way to find the van't hoff factor is to find the ratio of the final number of particles to the initial number of particles.

You must know what van't hota factor is right. Its the ratio of observed colligative property and calculated or theoritical colligative property. And there are few more formulas like this.

n is the no. Of mole after dissociation or association what ever according to the problem.

Generally this formula is used to find degree of dissociation or association i.e, alpha.


Does mercury have an NFPA 704 classification? I've searched on a lot of websites if elemental mercury (Hg) has an NFPA classification but I can't find any.

Does mercury have an NFPA classification?

According to Sigma-Aldrich

Some symbols:

GHS06: Acute toxicity (oral, dermal, inhalation), categories 1,2,3
GHS08: Respiratory sensitization, category 1; Germ cell mutagenicity, categories 1A,1B,2; Carcinogenicity, categories 1A,1B,2; Reproductive toxicity, categories 1A,1B,2; Specific Target Organ Toxicity – Single exposure, categories 1,2; Specific Target Organ Toxicity – Repeated exposure, categories 1,2; Aspiration Hazard, category 1
GHS09: Hazardous to the aquatic environment; Acute hazard, category1; Chronic hazard, categories 1,2

And according to ScienceLab:

Health: 3
Fire: 0
Reactivity: 0

I believe the ScienceLab information is the 'NFPA 704 classification'.

EDIT****

An email exchange between NIST and myself asking for clarifiation:

My Email (Pre-filled form on the website, I don't have the original):

Something along the lines of "Hello, I see NIST and a few other manufactures
of have different NFPA classifications (I provided links to the SDS's),
could you explain the difference between your ratings and theirs?"

Their response:

Good afternoon,
Our NFPA rating Health = 2 Fire = 0 Reactivity = 0 matches other manufacturers.
You can check Aldrich website for 99.99 Mercury
http://www.sigmaaldrich.com/catalog/product/sigald/215457?lang=en&region=US

Thank you,
Office of Reference Materials

Other manufacturers (Note, plural) apparently now equals 1 example

according the Federal Final Rule of hazard Communication Mercury has a NFPA rating of 0 because it does not burn.


What is difference between aluminium mercury alloy and gallium aluminium alloy? Aluminum mercury alloy is used as a chemical reagent to reduce compounds, such as the reduction of imines to amines,in this case mercury destroy aluminium oxide layer which protect aluminium to react with water.

Alloy of aluminium and gallium which similar to above alloy make the aluminium more reactive by preventing it from forming an oxide layer.

In both cases mercury and gallium plays a same role of destroying oxide layer on aluminium to react with water but why can't gallium aluminium alloy reduce imines to amines like mercury aluminium alloy ?

We can rationalize the differences in the reactivity as follows.  Gallium has a higher ionization energy compared to mercury.  During a redox reaction the metal is oxidized and it loses an electron, therefore how "easy" it is for a metal to lose its electon is iimportant. Since gallium is harder to inonze than mercurey, it implies that it is "easier" for mercury to lose its electron compared to Gallium, making mercury a better stronger reducing agent.

I may come back and add some quantitative information later, but i'm going to go eat now. I think looking in a table of standard reduction potentials would be reasonable.

I cannot think of any reason that it can't. Mercury and gallium should both be acceptable mediators for the electron transfer. In fact, elemental Gallium may even be preferred over salts of Gallium unlike Mercury, where only Mercuric salts can be used (maybe due to mercury having such a low melting point?). Elemental Gallium might actually have an added advantage of having the perfect melting point to mediate electron transfer.

It would actually be quite ironic if all these chemists are taking elemental mercury and making salts of it just to give it the properties of elemental Gallium!

And I'm not even touching on the toxicological superiority of gallium!


Acidity/basicity of oxides across the periodic table Recently while self-studying my chemistry book, it dawned on me that metals form basic oxides and nonmetals forma acidic oxides. Why is this?

It is one of the characteristics of the metals as for electrical conductivity, heat conductivity, mechanical properties. All these basically originate from the outer electrons to be "loosely" bound to the nucleus, when compared to non metals. Bonds in metal oxides are ionic, with positive charge on the metal. The same is for their hydroxides, i.e. their compounds containing an OH group. This leads to their basic dissociation in water  giving the stable hydroxide ion.

Contrary non metals have tightly bound electrons and are prone to fill their outer shell (Octet rule). On reaction with a non metal as oxygen the situation is therefore a bit more complicated depending on what specific element the non metal is. Non metals oxides are essentially molecular, although the covalent bond is polar to various extents. However, their compounds containing an OH group will retain electrons on the fragment consisting of the two non metals (of which, one is oxygen of course) leading to acid reaction, i.e. H+

(Note that HClO would be better wrote as ClOH).

It should be clear that things are going as ionisation potential - electron affinity - atom size (distance of the outer electrons from the nucleus). For a better explanation I would need more time, it is basically a big chapter of introductory general chemistry, which goes under the name of periodic law and properties of the elements, or something like that.


Is there an easy way to detect methanol in spirits (which may have been adulterated)? Is there an easy way to detect methanol in spirits, which, you suspect, may have been adulterated?

Well that depens on your possibilities.

a) Destillation / Boiling Point (easy, unsafe)

b) Refractive index (depends on impurities)

c) GC (best, very expensive)

d) 1H NMR


Why do we use helium in balloons? While I was looking at the periodic table today, I realised that there were gases that were much lighter than helium such as hydrogen. If hydrogen is lighter than helium, why do we insist on using helium in balloons?

As other answers have noted, the only gas lighter than helium is hydrogen, which has some flammability issues that make it more difficult to handle safely than helium.

Also, in practice, hydrogen is not significantly "lighter" than helium.  While the molecular mass (and thus, per the ideal gas law, the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the difference between the density of the gas inside the balloon and the air outside.

The density of air at STP is about $\rho_{\ce{air}}=\pu{1.2754 kg m-3}$ , while the densities of hydrogen and helium gas are $\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$ and $\rho_{\ce{He}}=\pu{0.1786 kg m-3}$  respectively.  The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences  $\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$  and $\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$.  Thus, helium is only about $7.5\%$ less buoyant in air than hydrogen.

Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant.  For example, if you wished to go ballooning on Jupiter, which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply sink, and even a pure hydrogen balloon (at ambient temperature) would not lift much weight.  Of course, you could always just fill the balloon with ambient Jovian air and heat it up to produce a hot hydrogen balloon (not to be confused with a Rozière balloon, which are used on Earth and have separate chambers for hot air and hydrogen / helium).

Ps. A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about $\pu{2 Da}$, while a helium atom has two protons and two neutrons, for a total mass of about $\pu{4 Da}$.

Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about  $\pu{32 Da}$ (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to  $\pu{28 Da}$ (one proton and one neutron per atom less than oxygen).  Thus, the average molecular mass of air should be between $28$ and  $\pu{32 Da}$; in fact, since air is about three quarters nitrogen, it's about  $\pu{29 Da}$, and so the buoyancies of hydrogen and helium in air are proportional to $29 - 2 = 27$ and $29 - 4 = 25$ respectively.  Thus, hydrogen should be about $\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$ more buoyant than helium, or, in other words, helium should be about $\frac{2}{27} \approx 7.5\%$ less buoyant than hydrogen.

Pps. To summarize some of the comments below, there are other possible lifting gases as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices.

For example, methane (molecular mass  $\approx \pu{16 Da}$) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas.  However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's more dangerous by others (total energy content per volume).  In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium.

A much less flammable choice would be water vapor which, with a molecular mass of  $\approx \pu{18 Da}$, is only slightly less buoyant than methane at the same temperature and pressure.  The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all.  This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster — whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses.

Despite these drawbacks, hot steam balloons have certainly been suggested, studied and tried in the past — alas, not always particularly successfully (although, apparently, there have been much more successful attempts as well).  There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope.  So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas.

Other potential lifting gases, with molecular mass similar to methane and water, include ammonia and neon.  Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium.

Ammonia, on the other hand, while much less flammable than methane, is rather toxic and corrosive (not to mention really stinky, which, given its other properties, is probably a good thing).  I don't think I'd like to fly in an ammonia balloon, but apparently, some people do!  It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form.

Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second.  If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical.  Then again, so would hydrogen — its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.

From my thinking helium is a stable gas and is a noble gas whereas hydrogen being the lightest gas but is flammable that's why helium is filled in weather balloons.


Stoichiometry and POAC Contradiction Lets say we have the reaction

$$\ce{H2 + O2 -> H2O}$$

And we are given 10 moles of $\ce{O2}$ and are asked that
How many moles of hydrogen gas do we need to complete the reaction ?

Stoichiometry gives the answer as 10 moles but by applying Principle Of Atom Conservation,  we get

$$\text{Moles of }\ce{H2}  =2\cdot\text{moles of }\ce{O2}$$

Why is the POAC giving wrong results?

You applied the POAC incorrectly . Even your stoichiometric answer is incorrect.

First Balance the reaction

$$\ce{2H2 + O2    -> 2H2O}$$

We get that one mole of $\ce{O2}$ requires 2 moles of $\ce{H2}$.
Thus 10 moles $\ce{O2}$ need 20 moles of $\ce{H2}$.


Reduction of molality of methanol How many grams of water would you add to $\pu{1.00 kg}$ of $\pu{1.44m}$ $\ce{CH3OH (aq)}$ to reduce the molality to $\pu{1.00 m}$ $\ce{CH3OH}$?

I have tried $M_1V_1=M_2V_2$. I have tried converting it to kg of $\ce{CH3OH}$ and subtracting it from the $\pu{1.00kg}$ to get kg of $\ce{H2O}$. All of which have worked to no avail.

If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.

To prepare this solution, we add the mass of  $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.

The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.

Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol.

The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$

The objective of the problem is to prepare a solution of molality $1 \ce{m}$,

i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.

It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56
\ce{g}$ of water.

Our solution already contains $955.95\ce{g}$ of water.

So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.

Firstly, 1.44 m of CH3OH yields a solution of 1.046 L, (where 1.44 m means (1.44*32=46.08) grams of CH3OH in 1 Kg of H2o, so solution weight will be 1.046 Kg) but we have a 1 Kg solution, so we say:

1.046 Kg solution contains 46.08 g of CH3OH

1.000 Kg solution contains    x  g of CH3OH

x=CH3OH grams in your solution = 46.08×1000/1046.08 = 44.05 g

and water weight in your solution is 1000−44.05=955.95 g

now, to prepare the new solution (1 m) we say:

1 kg of water contains 32 g of CH3OH

y Kg of water contains 44.05 g of CH3OH

y= weight of water required for new solution =44.05×1000/32=1376.56g

then 1376.56 - 955.95 = 420.6 g  is water that should be added to the 1.44 m solution to have the 1 m solution .

Note: 1 Kg almost equals 1 L (for water)


O-carboxylation in the Kolbe-Schmitt reaction In the Kolbe-Schmitt reaction, we know that the pi electrons of the benzene ring attack the carbon in carbon dioxide, but why doesn't the reaction occur on oxygen instead (as shown in the diagram below)?

The resulting product would be a carbonate hemiester monoanion, specifically monophenyl carbonate.  Similar species have been observed in carbonated alcoholic beverages.  (Instead of phenol, the alcohol involved is ethanol so the carbonate hemiester is monoethyl carbonate.)  There is an equilibrium between bicarbonate and the hemiester:

$\ce{HCO3- + ROH <=> RCO3- + H2O}$

According to the paper I linked the equilibrium constant is about 2.0, i.e. $K_{eq} = \frac{[\ce{H2O}][\ce{RCO3-}]}{[\ce{HCO3-}][\ce{ROH}]} \approx 2$.  Under aqueous conditions, the concentration of water is very large and the concentration of bicarbonate is much lower, so usually the equilibrium concentration of the carbonate hemiesters is very low, on the order of a few mM.

In contrast, under the conditions of the Kolbe-Schmitt reaction, the carbon-carbon bond formation is effectively irreversible.  There is no equilibrium, and the desired product can be obtained in high yield.

The phenoxide doesn't attack on oxygen because the oxygen on the carbon dioxide is electron withdrawing leaving an electron deficient region on the carbon atom.