Books showing colorful images of elementary particles....is this true? Do electrons or protons or any other elementary particle have colors of their own or do they have smell?

You can understand why they don't by understanding how smell and vision work

The mechanism by which we (or other animals) smell is not fully understood. But the key is that receptors in your nose are activated when certain molecules interact with them. The receptors consist of a wide variety of complex proteins with cavities of particular shapes. The cavities also have different patterns of surface structures which interact with the target molecules in a variety of ways (there might be some polar patches, patches that can hydrogen bond to another molecule and some that prefer non-polar interaction). When an appropriate target fits into the cavity, the cavity can send signals to nerves to signal that a target has been recognised. The targets can be very specific and there are a lot of different types.

Elementary particles don't (usually) exist alone but only as parts of other atoms or molecules. So they won't smell. We haven't evolved ways to detect free elementary particles. It is the 3D structure of molecules that creates smell.

The mechanism of sight is simpler. Molecules in the eye can absorb particular frequencies of light and then send a signal to the nervous system that they have detected some type of light. Light consists of photons. We see light when something emits photons, but only a narrow range of photons of about one octave. We can't see radio waves or X-rays as our eyes are not built to do detect potions of those energies. We see colour because photons of different energies can excite different receptors in the eye (human eyes have three different detectors tuned to red, green and blue light).

What creates coloured light is the either the emission of photons of the right energy or the selective absorption of light coming from another source. selective absorption happens when there is some way for the thing absorbing the light to capture the incoming photon's energy. This only happens when there is some part of the absorbing entity that can shift from one energy level to another and where that energy level matches the energy of the incoming photon. Some molecules have that property, but in many the energy levels don't match anything in visible light so they don't show colour.

Elementary particles don't have any energy levels in the appropriate range so won't be coloured. In molecules that do have energy levels absorption can happen, but this is a property of the whole molecule, not the atoms that make it up. It isn't the atoms that have colour, it's the molecule. Also atoms are about the same size as wavelengths of light, so even if you had a powerful microscope you could not see the individual components using light so the idea of them having colour makes no sense.

When presenting models of molecules, chemists often draw the atoms in different colours. But this is so they can tell them apart and in no way represents what an imaginary microscope would see. The colours are entirely arbitrary.


Does light absorption result in a momentarily weaker bond? I may have just misunderstood the topic, but light absorption by a molecule is a result of a promotion of an electron into an empty antibonding orbital; hence, does that mean that the strength of the bond is momentarily weakened by absorption of light? What processes account for the actual radiation part (release of heat) associated with the absorption of light?

http://www.chemguide.co.uk/analysis/uvvisible/bonding.html

it may happen but it is not always necessary as 1S antibonding molecular orbital is lower in energy as compared to 2S bonding MO If electron jumps from 1S ABMO to 2S BMO  then BOND ORDER willhence bond will become stronger increase σ(1s) <σ∗(1s) < σ(2s) <σ∗(2s) < π(2px) = π(2py)


How to derive the rate laws for three consecutive reactions? Given the following three consecutive unimolecular reactions $$\ce{A ->[$k_1$] I_1 ->[$k_2$] I_2 ->[$k_3$] P}$$ write the rate of change for each.

I have gotten the following:$\newcommand{\diff}{\mathrm{d}}$
\begin{align}
  \frac{\diff [A]}{\diff t}   &=-k_1[A]\\
  \frac{\diff [I_1]}{\diff t} &=k_1[A]-k_2[I_1]\\
  \frac{\diff [I_2]}{\diff t} &=-k_3[I_2]+k_2[I_1]\\
  \frac{\diff [P]}{\diff t}   &=k_3[I_2]\\
\end{align}
Is this correct? If it is, then why cant we write
\begin{align}
  \frac{\diff [A]}{\diff t}   &=-k_1[A]\\
  \frac{\diff [I_1]}{\diff t} &=k_1[A]-k_2[I_1]\\
  \frac{\diff [I_2]}{\diff t} &=k_3[I_2]+k_2[I_1]+k_1[A]\\
  \frac{\diff [P]}{\diff t}   &=k_3[I_2]\\
\end{align}

Your first set of four equations look correct. Next you need to integrate the first equation to get [A] vs. time i.e. $\ce{[A]}= \ce{A_0}\exp(-k_1t)$ where $\ce{[A_0]}$ is the value at $t=0$, and substitute [A] into the second equation. Integrate this equation (integrating factor method) and substitute it into the third and so on. It gets messy!  You can obtain $\ce{[P]}$ easily as this is the sum of all other values less the amount you start with.

The general solution for a chain of n reactions in the special case that only the first species A is present at $t=0$ has the form

$$N_n = C_1\exp(-k_1t)~+ C_2\exp(-k_2t) ~+ ...+ C_n\exp(-k_nt) $$

$$C_1= \frac{k_1k_2...k_{n-1}}{(k_2-k_1)(k_3-k_1)...(k_n-k_1)}\ce{[A_0]} $$

$$C_2= \frac{k_1k_2...k_{n-1}}{(k_1-k_2)(k_3-k_2)...(k_n-k_2)}\ce{[A_0]}$$

$$ C_3 = .....$$
etc,  for as many terms as necessary.

good luck!


Why is carbon dioxide nonpolar? I understand that polarity corresponds to an electronegativity difference and that the larger the electronegativity difference, the more polar the bond.

However, I have read that carbon dioxide is nonpolar. This doesn't make sense to me.

In carbonyl compounds, the carbonyl carbon is partially positive and thus carbonyls are polar. $\ce{CO2}$ is simply two such units bonded together, so wouldn't it be that you would have an even more positive carbon and two partially negative oxygens causing $\ce{CO2}$ to be overall polar?

On top of that, $\ce{CO2}$ dissolves to a much larger extent in water compared to $\ce{O2}$, despite both molecules supposedly being nonpolar.

so wouldn't it be that you
  would have an even more positive carbon and 2 partially negative
  oxygens

Yes, your analysis is correct to this point.  A chemist would say that the bonds in $\ce{CO2}$ are polar (or polarized) and therefor each $\ce{C=O}$ bond has a bond dipole moment.  However the molecule itself is linear and the two bond dipole moments are oriented 180 degrees with respect to one another and cancel each other out, so overall the molecule does not have a dipole moment and is non-polar.

EDIT:  There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.
$$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$
This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.


Carboxylic Acid Nomenclature According to my chemistry textbook, 2-hydroxy-1,2,3-propanetricarboxylic acid is represented as:

But 5-bromoheptanoic acid is given as:

In the former naming, propane is considered in the base chain while the COOH carbons are not. In the latter one,  the COOH carbons are considered in the base carbon chain.

On checking NCBI Pubchem, I found the following:

https://pubchem.ncbi.nlm.nih.gov/compound/16217547#section=Top
https://pubchem.ncbi.nlm.nih.gov/compound/5-Bromovaleric_acid#section=Top

Why are the carbons of carboxylic acids excluded in the former name but included in the latter?

When identical groups are present such that neither can be preferred over the other then, such nomenclature is preferred to emphasise on the equivalence of the groups such as in the one we have here. We cannot name in such a way that includes the three groups together. So, we name it that way instead.

As such, the rules of IUPAC Nomenclature are themselves arbitrary therefore, it is just something that is the way it is. For more information you might like to see https://old.iupac.org/publications/books/principles/principles_of_nomenclature.pdf, Page 34 onwards.

Edit: I almost forgot that a compound can have more than one name, even by the IUPAC nomenclature rules. As such, the names used commonly as IUPAC names are the preferred ones.


Would sodium explode in salt water? I’ve seen the video etc of sodium being put into freshwater and the reaction of flames/small explosion but I wonder if the same reaction would occur if conducted in salt water such as the ocean.

Yes, sodium metal is also going to react exothermically with salt water or any other aqueous solution as long as it comes in contact with water:

$$\ce{Na (s) + H2O -> Na+ (aq) + OH- (aq) + 0.5 H2 (g)}$$

eventually leading to explosion of hydrogen-oxygen mix forming near the water surface. Presence of sodium chloride in salt water isn't going to influence the reaction as it neither capable of shifting equilibrium to the left (gas formation), nor passivate sodium metal surface as it would do, for instance, some viscous oil.


Meaning of 1,1,1,2 in 1,1,1,2-Tetrafluoroethane & similar What is the meaning of the 1,1,1,2 part in the compound 1,1,1,2-Tetrafluoroethane, a refrigerant commonly known as R-134a, some info:
Formula: $\ce{CH_2FCF_3}$
Structure:

Reference: http://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane

The numbers indicate which carbons the substituents - the fluorine atoms - are on. The root of the molecular name is "tetrafluoro," indicating the presence of four fluorine substituents. So now where are they? That's what the numbers tell us; the "1,1,1" part indicates that there are three fluorine substituents on the first carbon, and "2" tells us there is one more fluorine substituent on the second carbon.

Which is the first carbon and which is the second carbon? The numbering scheme should be selected such that it gives the lowest number at the first point of difference.

This way, we can write "1,1,1,2" to indicate the presence of the four fluorine substituents rather than "1,2,2,2."


Predicting melting points of substances based on their molecular/atomic structure Wouldn't it be great to be able to predict the melting point of a metal based on the atomic number and not by looking in some table for the value that was measured experimentally by someone else? Or to predict the same melting point for a simple salt like NaCl or KI by applying some formula to the atomic numbers of the 2 atoms involved?

Can you help me understand why is it so difficult to build a theoretical model for this particular physical property? Were there attempts made in the scientific world for such a model?

--EDIT--

Yes, this is part of my question:
I want to understand if there is a connection between the theoretical explanations of properties like melting point and the current theoretical model of atomic structure. Can the electronic structure of $\ce{Fe 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2}$ tell us anything about it's probable melting point without measuring that point in the first place?

With regard to organic compounds, you might be interested in group contribution methods, e.g. the Joback method.

These methods divide a molecule into functional groups or similar small building blocks and estimate the melting point (and other thermodynamic properties) based on the contributions of the individual groups.


How to compute the electron configuration of an atom? How to compute the electron configuration of an (electrically neutral) atom ?

So when given the atom number how does one compute the electron configuration ?

I know the sum of electrons must equal the atomic number.

I also know the first shells fill up as $2n^2$.

I do not know how to compute quantum numbers when given the atom number.

To calculate electronic configurarion you need to know several thing first.

Electron Shell

  In chemistry and atomic physics, an electron shell, also called a principal energy level may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the "1 shell" (also called "K shell"), followed by the "2 shell" (or "L shell"), then the "3 shell" (or "M shell"), and so on farther and farther from the nucleus. The shells correspond with the principal quantum numbers (1, 2, 3, 4 ...) or are labeled alphabetically with letters used in the X-ray notation (K, L, M, …).
  Each shell can contain only a fixed number of electrons: The 1st shell can hold up to two electrons, the 2nd shell can hold up to eight (2 + 6) electrons, the 3rd shell can hold up to 18 (2 + 6 + 10), and the 4th shell can hold up to 32 (2 + 6 + 10 + 14) and so on.

Subshells

  Each shell is composed of one or more subshells, which are themselves composed of atomic orbitals. For example, the first (K) shell has one subshell, called "1s"; the second (L) shell has two subshells, called "2s" and "2p"; the third shell has "3s", "3p", and "3d"; the fourth shell has "4s", "4p", "4d" and "4f"; the fifth shell has "5s", "5p", "5d", and "5f" and can theoretically hold more but the "5f" subshell, although occupied in actinides, is not filled in any element occurring naturally.

See 'Number of electrons in each shell' section under this article.

Atomic Orbitals

  An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. Each orbital in an atom is characterized by a unique set of values of the three quantum numbers n, ℓ, and m, which correspond to the electron's energy, angular momentum, and an angular momentum vector component, respectively. Any orbital can be occupied by a maximum of two electrons, each with its own spin quantum number. The simple names s orbital, p orbital, d orbital and f orbital refer to orbitals with angular momentum quantum number ℓ = 0, 1, 2 and 3 respectively. These names, together with the value of n, are used to describe the electron configurations.

Understand electron configuration notation.

Electron configurations are written so as to clearly display the number of electrons in the atom as well as the number of electrons in each orbital. Each orbital is written in sequence, with the number of atoms in each orbital written in superscript to the right of the orbital name. The final electron configuration is a single string of orbital names and superscripts.
For example, here is a simple electron configuration: $1s^2 2s^2 2p^6$. This configuration shows that there are two electrons in the 1s orbital set, two electrons in the 2s orbital set, and six electrons in the 2p orbital set. 2 + 2 + 6 = 10 electrons total. This electron configuration is for an uncharged neon atom (neon's atomic number is 10.)

Memorize the order of the orbitals.

Note that orbital sets are numbered by electron shell, but ordered in terms of energy. For instance, a filled $4s^2$ is lower energy (or less potentially volatile) than a partially-filled or filled $3d^{10}$, so the 4s shell is listed first. Once you know the order of orbitals, you can simply fill them according to the number of electrons in the atom. *The order for filling orbitals is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s. *
An electron configuration for an atom with every orbital completely filled would be written: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6 8s^2$.

Note that the above list, if all the shells were filled, would be the electron configuration for Uuo (ununoctium), 118, the highest-numbered atom on the periodic table - so this electron configuration contains every currently known electron shell for a neutrally charged atom.

Now you can fill in the orbitals according to the number of electrons in your atom. For instance, if we want to write an electron configuration for an uncharged calcium atom, we'll begin by finding its atomic number on the periodic table. Its atomic number is 20, so we'll write a configuration for an atom with 20 electrons according to the order above.

Fill up orbitals according to the order above until you reach twenty total electrons. The 1s orbital gets two electrons, the 2s gets two, the 2p gets six, the 3s gets two, the 3p gets 6, and the 4s gets 2 (2 + 2 + 6 +2 +6 + 2 = 20.) Thus, the electron configuration for calcium is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.

Note: Energy level changes as you go up. For example, when you are about to go up to the 4th energy level, it becomes 4s first, then 3d. After the fourth energy level, you'll move onto the 5th where it follows the order once again. This only happens after the 3rd energy level.

For 'How to find quantumn numbers?', see this.


Why "initial" corrosion studies only, rather than longer time-spans? Some papers say they study the identification and quantification
of species involved in initial atmospheric corrosion. Consider the case of this article in Corrosion - The Journal of Science and Engineering, for example, it provides an exceptionally good model, but I am interested in the copper corrosion process as a whole, not just the beginning. What exactly is understood as "initial corrosion" and what's the relevance of studying such a short time-span? That paper's tests reach no longer than 300 minutes, while other studies measure time in weeks, not minutes.

Here's an attempt at an answer -

Initial corrosion may be taken to mean when the electrochemical reactions occurring are the main driving force for the material to corrode, meaning changing chemical and/or physical properties. As corrosion continues, physical factors (stresses, strains, grain defects, crystal defects, etc.) can contribute and dominate the reason why a material might fail. Studying the initial mechanism can lead to solutions that inhibit or prevent corrosion since it is currently not likely you can salvage a heavily corroded piece of metal and return it to its original service.

As for the duration variance of the studies, it could be that mechanisms change with time or that some systems are slow to corrode but still need characterization to allow for the material in question to be more accurately used. This, however, is more my opinion than actual experience.


Is there a similar characteristic of the lanthanide contraction in the other transition metal groups? Basically, does the lanthanide contraction also occur with the 4d, 6d groups as well? I am at a lack of understanding in this topic, I am just wondering what specific elements does this jump in a lower atomic radii occur. It seems to happen due to the d orbitals being occupied? I am making that judgement based on this graph.
I am hoping someone can explain, as one orbital seems to make all the difference.

Hope I interpret your question correctly:

A decrease of atomic radii occurs in every period (from left to right). Atomic radii of lower periods are in general larger (Increase from top to bottom). This simple behavior only holds true for elements that only have p-block and s-block electrons.

The first anomaly of this behavior is the d-block contraction. This can be seen in the 4p, 5p, 6p and 7p elements (Elements to the right of Ga, In, Tl). For example the atomic radius increases about 60 % from C to Si. The difference of Si and Ge with the added p-block contraction is just 20 %.

The lanthanoide contraction occurs when the f-block is added. For this reason the 4d and 5d elements are very similar in size. For example Hf (5d, 160 pm) is even smaller than Zr (4d, 159 pm).

Finally the actinoide contraction should have the same effect on 6d elements. But I'm not sure if realiable numbers for the atomic radius exist for those elements.


Why does basic peroxide react via conjugate addition rather than through direct addition? The following reaction occurs:

The $\ce{HOO^{-}}$ ion reacts as a nucleophile at carbon number three and consequently the reaction is conjugate addition. However, Why would a peroxide not react through direct addition. I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur (Baeyer-Villiger reaction: http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation). Why must it react at the alkene?

The reason that is given by Clayden et al. (chapter 23) is that because the peroxide has two adjecant lone pair on oxygen it exhibits an alpha effect, which is more or less a delocalization interaction between the lone pair orbitals. For reasons not entirely known (the basis of the alpha effect is an open problem in chemistry) this results in an increase in energy of the electrons in the resulting delocalized HOMO, making it more nucleophilic. You can imagine that this delocalized HOMO is also softer than the single localized oxygen lone pair and hence it tends to react at the soft double bond instead of the hard carbonyl site.

Excerpt from Clayden et al.


What is the nature of water? Is water neutral in nature or amphoteric in nature? I understand it as an neutral compound in complex compounds, but I've been told that it is amphoteric, too.

$\ce{H2O}$ is a both a neutral molecule and an amphoteric molecule — the two are not mutually exclusive terms. An amphoteric molecule is simply one that can act as either an acid or a base, while a neutral molecule is one in which the total number of protons is equal to the number of electrons, such that there is zero net charge. Water can act as a Brønsted acid (i.e., an $\ce{H+}$ donor), as per the following reaction:

$\ce{H2O + NH2- -> OH- + NH3}$

However it can also act as a Brønsted base (i.e., an $\ce{H+}$ acceptor):

$\ce{H2O + HCl -> H3O+ + Cl-}$

When proton transfer is involved, as in the Brønsted definition of acids and bases, the more precise term is amphiprotic. However, water can also act as both a Lewis acid (an electron pair acceptor) and Lewis base (an electron pair donor). Notably, as a Lewis base, it is not limited merely to proton abstraction, but can act as a nucleophile with any number of electrophilic molecules.


Can running water carry an electrical current indefinitely, and for long distances? Okay, not a scientist, but doing a writing project that requires a little knowledge. My question is: If there's a constant source generating enough energy into running water (like a stream, surrounded by non-conductive material) could the water carry an electrical current for an indefinite period? Basically, doing the same job as a wire.

I think the conductors in the water (minerals) may deplete over time because of chemical reactions to this process...but if it's a running stream with constantly renewed minerals, could this happen?

Edit: Clarification was requested for a couple parts. 1) Method of conduction to the water: Not that important, really. I guess a wire capable of conducting the necessary amount of power could work. 2) How fast the water flows: I'm not sure. A strong currant, I guess. But not as severe as rapids.

I assume that the fundamental problem with this idea is that due to inherent resistance of the water, heat will be generated. As a result the water will evaporate. If there is no space left for the gaseous water vapor the "pipeline" will just explode.

That is basically why conducting wires are made of copper, it mirrors a remarkably low resistance. Silver and gold would be even better but more expensive on the other hand. Consequently copper is the standard material of wires.

If constant cooling of the system would be permitted, it could basically work.
water itself does not significantly conduct electricity, but with electrolytes dissolved in the mentioned (as suggested by the author of this question) it becomes conducting.
As long as the ions do not precipitate the liquid retains the property of conductance.


What chemical properties of ethanol make it usable for drinks as compared to that of methanol? Methanol ($\ce{CH3OH}$) and ethanol ($\ce{C2H5OH}$) both are the organic compounds having an alcoholic group. The alcoholic beverages (liquors and spirits) for human consumption (albeit injurious to health) contain a certain percentage of ethanol but nothing of methanol.
What chemical properties of ethanol ($\ce{C2H5OH}$) make it usable in beverages as compared to those of methanol ($\ce{CH3OH}$)?
EDIT: I am looking for a comparison of chemical properties of ethanol and methanol with respect to suitability for drinks i.e. comparison of their mechanism upon consumption.

The problem arises from the metabolized products of methanol. Methanol oxidizes in the liver by an enzyme called alcohol dehydrogenase to formaldehyde which is further metabolized to formic acid by another enzyme called aldehyde dehydrogenase. This formic acid is the source for acute toxicity associated with methanol poisoning. Accumulation of this chemical in the blood deprives cells of oxygen by inhibiting the enzyme cytochrome c oxidase in their mitochondria, a key element of the respiratory electron transport chain.
Formic acid, together with formaldehyde, are responsible for nerve damage, blindness, and other unpleasant effects associated with methanol poisoning.

Note that ethanol is also metabolized in the same way by the same pair of enzymes to ultimately form acetic acid but human can tolerate acetic acid to an extent because it is less toxic compared to formic acid and thus can be consumed like for example, vinegar. In fact, ethanol is used as a remedy for methanol poisoning as it acts as a competitive inhibitor by more effectively binding and saturating the alcohol dehydrogenase enzyme in the liver, thus blocking the binding of methanol with the enzyme rendering it useless.
References

https://theskepticalchemist.com/methanol-toxic-ethanol/ (above reaction source)
https://metode.org/metodes-whys-and-wherefores/why-can-we-drink-ethanol-but-not-methanol.html
https://en.wikipedia.org/wiki/Methanol_toxicity


What is so special about quasicrystals? Can whatever is special about quasicrystals be understood at A-Level standard or slightly above it? The material I saw online is either too layman or too advanced for me to follow. What practical uses do they have and why does their structure lead to the properties that allow them to have these practical uses?

It's hard to know exactly what “level” you want address, as the Wikipedia entry seems a decent introduction to me. But in any case, here's what I consider “special” about quasicrystals:

From the fundamental point of view:

Though they are regular (they are formed by symmetry operations that allow them to repeat throughout space), they are not periodic (the symmetry operations are not purely translations). Although there's nothing inherently “wrong” or “impossible” about that from theoretical physics and chemistry, it goes against the classical classification of solids: crystalline or amorphous.
Though hundreds of quasicrystals have been synthesized, there is only naturally occurring quasicrystal phase (icosahedrite)!
All quasicrystals are metal alloys (most of them being binary or ternary intermetallic alloys): given that there are other types of crystalline solids, why aren't there non-metallic quasicrystals?

Regarding applications:

Quasicrystals have found, so far, few practical and industrial applications. In these applications, they are used as other alloys would, and do not have any “magic quasicrystal property”.
One specific feature they have, though, is that quasicrystals tend to be rather strong alloys and that they age well, due to the way they are synthesized


Nomenclature of common naming of alkynes vs alkenes; both -ene? Is there a difference between common naming of alkenes vs alkynes? Why is Ethylene the common name for an alkene and acetylene for an alkyne? Isn’t acet- the common form of naming ethyl groups anyway so why is ethyl used in a common name?

The official IUPAC name for ethyl with a double bond is ethene, but this is a relatively recent change.
Before the official name (as an exception) was ethylene. I’m not sure about propene/propylene but am fairly confident the official IUPAC name for propyl double bond is propene and not propylene. However, these compounds are often referred to with the unofficial name as per tradition.

For alkynes, the IUPAC name is, for ethyl, the name is ethyne.

Don’t remember to specify the location of the double/triple bond (ex: 2 - ethene)

For alkenes, the formula is $\ce{C_nH2_{n+2}}$ and for alkynes is $\ce{C_nH_{2n}}$.


How difficult is it to synthesize Novichok agents? In the light of the current UK-Russia events, how difficult is to synthesize Novichok agents (with known structures)? Is it something that can be made at any well-equipped university and/or chemical factory, or does it involve some very specialized or possibly secret knowledge?

According to Fragmentation pathways and structural characterization of
organophosphorus compounds related to the Chemical Weapons
Convention by electron ionization and electrospray ionization
tandem mass spectrometry Rapid Commun. Mass Spectrom. 2016, 30, 2585–2593:

  N-[Bis(dimethylamino)methylidene]-P-methylphosphonamidic
  fluoride (3) was synthesized by the controlled
  addition of N,N,N′,N′-tetramethylguanidine to a solution
  of methylphosphonic difluoride (1). N,N,N′,N′-Tetramethylguanidine (0.60 mmol) and triethylamine (0.65 mmol) in dichloromethane (DCM) (500 μL) were
  added slowly into the solution of methylphosphonic difluoride
  (0.40 mmol) in DCM (500 μL), while stirring at 0–5 °C. After
  30 min, the resulting precipitate was filtered off and the solution was analyzed by GC/MS.

  The corresponding O-alkyl N-[bis(dimethylamino)-methylidene]-P-methylphosphonamidates 4 were synthesized by the addition of ROH to a solution of cpd 3. Sodium hydride powder (NaH, 60% in mineral oil, 0.5 mmol) was washed with dry hexane twice. The hexane was removed and DCM (500 μL) was added and the solution was kept stirring at 0–5 °C. The appropriate alcohol (0.5 mmol) was slowly added into the suspension and the reaction solution was stirred at 0–5 °C for 30 min. Compound 3 in DCM (200 μL) was then added dropwise to the solution, while stirring at 0–5 °C for 2 h. Any precipitate was filtered off and the resulting solutions of the end product were analyzed by GC/MS. It should be noted that, due to the extreme toxicity of these materials, the separation and purification of CWC-related chemical are very difficult and therefore should be carried out only by a trained professional in an efficient fume cupboard equipped with an active charcoal filtration system.


Relationship between coordination numbers and stoichiometry In fluorite ($\ce{CaF2}$), the coordination number of $\ce{Ca}$ is $8$, and that of $\ce{F}$ is $4$. Is there any logic behind the ratio of coordination numbers being the same as the stoichiometry of the ions?

Surely there is a logic behind that, and a pretty simple one. Let's take a crystal and count all $\ce{Ca}$ atoms in it. We'll get some huge number N (possibly about as huge as $N_A$). Now how many fluorine atoms are there? Each $\ce F$ is a neighbor of some $\ce{Ca}$, so if we just count all neighbors of all $\ce{Ca}$, we won't miss anything... and will end up with 8N. But wait, each $\ce F$ is a neighbor of four $\ce{Ca}$, so it was counted 4 times! We divide by 4 to compensate for that and happily arrive at 2N, which is consistent with the formula $\ce{CaF2}$.

So yes, it is no coincidence that the ratio of coordination numbers is also the ratio of stoichiometric coefficients.


Hash and Wedge nomenclature One of the first things learnt in organic chemistry is that wedged bonds can be thought of as bonds coming out of the plane of the paper towards you, and hashed bonds can be thought of as bonds going behind the plane of the paper away from you.

What is rarely discussed, is the difference between actual 'wedged' bonds, and the 'block' variety. The diagram below from Organic Chemistry Strategy and Control by Stuart Warren goes some way to explaining that in theory at least, the two sets of bonds exist to allow us to describe absolute or relative stereochemistry.

(In practice, this is rarely followed, and individual groups/labs tend to, rightly or wrongly, favour one type over another for stylistic reasons).

What has always baffled me, however, is the direction in which the wedged bonds originate/point towards.

In most modern usage, the small end of the wedge/hash is at the chiral centre, and gets larger towards the substituent, however some (mainly older chemists, whom I'm going to give the benefit of the doubt and assume aren't just bad at using ChemDraw) chemists have the hashes differently to the wedges, as per the molecule below.

Is this something that is advisable (presumably it's meant to show some kind of 3D perspective)/are there any actual IUPAC rules/guidelines on this other than following what looks good/is conventional?

Both schools of thought were prevalent, one said that the narrow end should be considered in the plane of paper, the second said that the wide end should be considered. Perhaps the first was based on a perspective flavored approach while the second mainly relied on popular drawing schemes. Excerpts from the 2006 IUPAC Recommendations1 (more information is given there):

  The least disruptive recommendation is to reject a perspective-based approach, and rather to interpret the bonds as more-abstract graphical objects. A solid wedged bond should be interpreted as projecting above the plane of the paper from the narrow end to the wide end (with the wide end being nearer to the viewer), and a hashed wedged bond should be interpreted as projecting below the plane of the paper, also from the narrow end to the wide end (with the wide end being further away from the viewer). To summarize, the narrow end of solid wedged bond or a hashed wedged bond always should point towards the stereogenic center to which it applies.

  That said, it is possible to use the other bond types in ways that are unambiguous. In particular, any type of hashed bond is unambiguous in print when connected to an atom with no other substituents (a hydrogen atom, a chlorine atom), no other non-hydrogen substituents (a hydroxy group, an amino group), or at most one other non-hydrogen substituent (an ethyl group, an ethoxy group). In each of those cases, a viewer would have no difficulty in understanding that the hashed bond projects below the plane of the paper from the other atom to that atom. Accordingly, both the unwedged hashed bond and the hashed wedged bond with reverse directionality are acceptable when used exclusively in those cases.

However, a caveat:

  However, anyone using bonds of these types should be aware that they are not the most common bond types used to represent behind-the-plane configuration. At best, an author using these types may cause a moment of confusion as a reader figures out which convention was intended. In extreme cases, a reader (or chemical software) might actually interpret the structure incorrectly—for example, by interpreting an unwedged hashed bond in the convention where it would represent relative (and not absolute) configuration, as discussed in ST 0.8 and ST-6. Fortunately, such cases are rare. The hashed edged bond (interpreted from the narrow end to the wide end) remains preferred...

Reference

Graphical representation of stereochemical configuration (IUPAC Recommendations 2006). Pure Appl. Chem. 2006, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897. © IUPAC.


Effect of sodium hydroxide solution on copper plate I have a copper etching plate that has over 100 years of dried ink in the lines of the plate. Someone suggested flake sodium hydroxide as a possible solvent for the ink. My concern is preventing damage to the copper itself.

Suggestions? For example, what should the concentration of the solution (in water) be?

Regarding the copper metal itself, the only down side I see of using a $\ce{NaOH}$ is that it is unlikely to remove some of the oxides that would be covering any exposed copper.  The initial etching process would have mainly produced water-soluble $\ce{CuCl2}$, likely with a bit of insoluble $\ce{CuO}$.  $\ce{CuO}$, $\ce{CuCO3}$ and some $\ce{Cu(OH)2}$ would have likely formed on any unprotected copper surfaces over time.  $\ce{NaOH}$ could remove $\ce{Cu(OH)2}$ and $\ce{CuO3}$, but not $\ce{CuO}$.  It also should be unreactive toward the copper plate itself.

A common method of putting a good shine on tarnished copper is with weak organic acids like citric acid.  Lemon juice and table salt make a favorite home copper cleaning solution.  Again, this is only with regards to the cleaning oxides from the copper metal itself, not the ink residues coating it.

My bottom line advise is to do what I’ve seen in your last comment that you plan to do; test a portion of the bottom side of the plate with the $\ce{NaOH}$ solution to verify that this solution cleans the ink residues without harming the copper.  I think you can do this with confidence that the copper metal will be unharmed.  If the ink residues are removed and you want to remove any remaining oxidation to create a fresh, shiney copper plate, you could dissolve just a bit of salt into lemmon juice and wipe the plate with this solution until shiny!


How to convert CO2 ppm volume to ppm weight? I'm using a sensor that measures $\ce{CO2}$ in ppm by volume aka atmosphere.
What I'm using it for is to measure $\ce{CO2}$ concentration in water which is measured in ppm by weight.

This is what I have so far

$$\text{Ppmwco2} = \text{ppmvco2} \times 0.8317 \times 44.01 / 10000;$$

This is quicker than $\text{co2}/1000000 \times 0.8317 \times 44.01 / 1000000$. 0.8317 accounts for Henry's Law and max ppmw will be 36.6 with 10,000 ppmv.

The sensor floats on the waterline detecting any $\ce{CO2}$ bouncing out of the water. Do I even need to account for Henry's law? Is the math right?

You can't determine the concentration of CO2 in the water just from measuring the concentration in the air above the water.

The solubility of CO2 is extremely dependent upon pH. See: http://ion.chem.usu.edu/~sbialkow/Classes/3650/CO2%20Solubility/DissolvedCO2.html

There is no indication you are considering pH at all.

Solubility of CO2 also depends upon temperature.

Also, you need to consider whether or not the phases are in equilibrium.


Is electrochemically activated water a real technology? In essence what I'm asking is if the applications of Electrochemically activated water (ECA water) as a bactericidal-disinfectant is real and profitable (especially this last part). The principle at first seems very dull just take salt water through an electrolysis process and you get anolytes and catholytes, with disinfectant proprieties. Why is that only now is it viable?

Far from 'only now' becoming viable, the production of disinfectant from electrolysis of salt water (brine) is not only 'real' it is how bleach sold on supermarket shelves is produced! Industrially, this is called the  'chor-alkali process' and is well established for over 100 years. Both chlorine and sodium hypochlorite are manufactured industrially by electrolysis of brine. Where gaseous chlorine presents a significant hazard however, liquid sodium hypochlorite is offered as an alternative, however it has a limited shelf-life and degrades rapidly with temperature. Recent innovations in terms of design and manufacture of ion-exchange membranes has opened up new possibilities for small-scale on-site production of socium hypochlorite which is especially useful for remote communities.

On the question of economic 'viability' or profitability, like any question of economics, it depends upon the available technology and specific circumstances. For example, large power stations typically have the need to disinfect huge volumes of water which passes through their cooling towers. Since they literally 'generate electricity at cost', it is often more economic for such power stations to produce disinfectant on-site using eletrolysis than to truck in volumes of bleach manufactured at a chemical plant. Of course, this does not necessarily mean its 'economically viable' for you to make your own bleach at home right now, although this may change in the future.

The on-site generation of hypochlorite for disinfection may not only be useful for cooling towers on power stations but also for treating wastewater on board ships and in in remote locations. Since the chemical is produced 'on demand', there is no need for the extra waste created by packaging, shipping and storage of large volumes of hazardous chemicals.


Do self-growing hydrogels grow and strengthen infinitely? Recently, there was a publication [1] (/., YouTube) on self-growing materials using mechanoradically-generated double-network (DN) hydrogels, closely resembling self-strengthening of muscles.

  We used a series of DN gels consisting of poly(2-acrylamido-2-methylpropanesulfonic acid) sodium salt (PNaAMPS) as the brittle network
  and poly(acrylamide) (PAAm) as the stretchable network. Each network is covalently cross-linked.

  Fig. 1. Conceptual scheme of the self-growth of materials induced by mechanical training. (A) A skeletal muscle becomes bigger and stronger
  under the application of repetitive mechanical stress and a continuous nutrient supply through metabolic cycles. (B) As a result, such a tissue is
  initially weakened (purple arrows) but then recovers or even shows enhanced performance (red arrows) due to mechanical stimuli. The green outline indicates one mechanochemical cycle. (C) Our strategy to develop self-growing materials based on mechanical training of DN gels. Mechanical stress leads to breakage of the brittle network (blue), whereas the highly stretchable
  network (pink) maintains the integrity of the gel. The mechanoradicals generated at the broken ends of the brittle network strands trigger polymerization of monomers supplied from the external environment to form a new network
  (orange). The chemistry of the newly formed network can be the same or differ from that of the original brittle network, as required.

The process is reported to be repeatable and reproducible:

  With proper conditions, our DN gel exhibits persistent growth in strength and size under repetitive mechanical training. Two requirements must be satisfied: Enough monomers should be continuously externally supplied, and the newly formed
  network should serve the role of a brittle network that then breaks during the next deformation to trigger subsequent network formation.

However, with muscle growth there are limiting factors such as hormones and bones structure, and even supplying enough protein one cannot grow the mass infinitely*. Considering constant feeding with monomer, is there a theoretical/practical limit for strength and growth for these DN-hydrogels?

*not even Tetsuo Shima — sorry, couldn't resist considering the origin of this work and the anime.

References

Matsuda, T.; Kawakami, R.; Namba, R.; Nakajima, T.; Gong, J. P. Mechanoresponsive Self-Growing Hydrogels Inspired by Muscle Training. Science 2019, 363 (6426), 504–508. https://doi.org/10.1126/science.aau9533.

"Strength" is a dubious quantity. The E-modulus will rise, and as the material swells with more and more monomer diffusing in, it will at some point rupture when the brittle network cracks. Like a fruit gum that was put in water overnight.

I have no idea if the two networks actually stay separated, or if they interconnect more and more. I suspect the latter, with radicals at work. That would make the brittle part softer -> no further hardening, and the soft network more brittle -> it snaps under increasing load.

But the point is moot imho: "Self-healing" makes sense if you had a smaller damage due to light overload. Then this material here can become twice as hard, and you're fine. I don't think it makes sense to "train" sth to become fifty times as strong.


Relative strength of intermolecular forces My attempt at a solution:
First I look for hydrogen bonds:
The only compounds with hydrogen bonds are the second and fourth ones, so they have the strongest intermolecular forces. The second must have stronger intermolecular forces since it harbors three hydrogen bonds, while the fourth only has room for two.
Then I look for dipole-dipole interactions:
The first compound has the potential for a dipole-dipole moment. It would be the third strongest.
Then I look for potential for London-dispersion forces:
The third compound and The fourth experience London-dispersion forces. But I'm stuck on discerning which is stronger.
So, my working solution is: 3, 5, x, 4, x

I would say that your analysis of first, second and fourth one is correct, as H-bonds are stronger than dipole-dipole interactions. However, you made a mistake while filtering compounds having dipole-dipole interaction. The $1$ and $3$, both have dipoles!
The $\ce{C=O}$ bonds in $\ce{CO2}$ are polar and hence they possess dipoles, but unlike a water molecule, $\ce{CO2}$ shape is linear, and therefore the equal dipoles cancel each other, resulting in a non-polar molecule as seen in the following figure,

That's the reason $\ce{CO2}$ possess dipole-dipole interactions whereas $\ce{F2}$ doesn't as it has only one bond which is non-polar!
Image taken from courses.lumenlearning.com: Polar Molecules


How is purine aromatic? If we count the total number of π electrons, there are 10. But to apply Hückel's rule to this molecule, we should count only the electrons of the outer periphery and there are 8 of these. How can the molecule be aromatic?

Consider this structure of purine:

The ring on the left contains two nitrogen heteroatoms and is aromatic due to π electron conjugation. Having six π electrons, it obeys Hückel's rule.

For the ring on the right, it contains four π electrons in bonds and the lone pair on the NH group takes part in conjugation. Hence the total number of electrons taking part in conjugation is six, which makes it aromatic by Hückel's rule as well.

Since both the rings of purine are aromatic, the compound is aromatic.

For more information see: Ciesielski, A.; Krygowski, T. M.; Cyrański, M. K.; Balaban, A. T. Defining rules of aromaticity: a unified approach to the hückel, clar and randić concepts. Phys. Chem. Chem. Phys. 2011, 13 (9), 3737–3747. DOI: 10.1039/C0CP01446J.


Lewis structure of CN radical I need to draw the Lewis formula for $\ce{CN}$. However, I am not sure how to draw the single free electron.
My idea would be as follows:

$$\Large\ce{\llap{·}C#N\rlap{:}}$$

Does that make sense?

You drew the radical, which is in normal cases very unstable.
The cyanide ion has an ion pair instead of the single electron at the carbon atom.


Rule of Thumb for separatory funnel volumes Is there a general rule of thumb in terms of how one chooses the volume of separatory funnel used for an extraction? It's my understanding that it is the amount of product that determines this as opposed to the volume of the solvent it is present in. If that is correct I was wondering if someone could post what the general ranges in grams of product would be appropriate for the following sep funnel sizes:

250 mL, 500 mL, 1 L

If a general rule must be given than the volume of the separatory funnel is mostly dictated by that of the liquid to be treated.

Indeed is better to extract the product by repeating several steps each using small portions of extracting solvent. This ensures a more efficient extraction based on repartition, I.e. a more quantitative recovery of the product as well as a more convenient work-up (cost-effective, less time consuming, less waste).

I am not considering difficulties due to a slow phase separation which might prevent the use of a narrow funnel but in general use the smallest that allows an easy shaking and inter mixing. (When you get to know the actual separation you might even prefer a  specific funnel shape not just that V).

Example: 200 ml to be treated with three to four 25 or 50 ml portions : 500 ml funnel.


Do we prefer the substituent or the longest chain in IUPAC nomenclature? I've seen too many related questions already, but mine is a simple and slightly different query

Take this compound as an example:

The name written below it is generated by ChemSketch. I agree with this generated name, because according to me, "bromo" is not a functional group, but rather a substituent only. Hence, it is not necessary to include it in the parent chain. Thus, the above name has preferred the longest chain (length) over substituent inclusion.

However, my book gives the answer as 1-bromo-2-ethyl-2-methylbutane, preferring substituent inclusion over longest chain (length).

My query thus is:

  In any hydrocarbon where the substituent does not lie on the longest chain, do we prefer the longer chain, or sacrifice it for the inclusion of the substituent?

The related question and answer that were already linked in this question also cover this case. If the other answer is too long and confusing, we can limit it to the rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) that are relevant in this case:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-45.0 INTRODUCTION
  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. A parent structure is defined (P-15.1) as a parent hydride, for example, benzene, a functionalized parent hydride, for example cyclohexanol, or a functional parent compound, for example, acetic acid. Application of the criteria in this section will generate the preferred IUPAC name.

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

Therefore, for a correct name, only the longest chain is relevant for the choice of a principal chain in this case. Only if there is still a choice, substituent groups that are expressed as prefixes (e.g. ‘bromo-’) are taken into account for the preferred name according to P-45.2.1 and P-45.2.2.

Thus, the correct name for the compound that is given in the question is indeed 3-(bromomethyl)-3-methylpentane and not 1-bromo-2-ethyl-2-methylbutane since the pentane chain has the greater number of skeletal atoms.


Chemical change - decomposing single element? If I had a molecule of a single element such as atmospheric oxygen O2, and I added energy to decompose it into two separate oxygen atoms (i.e. O2 + ENERGY –– 2 O ) is this a chemical or physical change?

It is a chemical change

The aspect of this that is probably confusing you is that both the product and the result in the case of an element are still a form of the element.

But elements can exist in different chemical forms and moving between them is still a chemical process. A relevant example is the reaction occurring mostly in the upper atmosphere that converts dioxygen (O2) to ozone (O3). This is a specific example of what happens when you add energy (in this case from UV light) to an oxygen molecule in the presence of other oxygen molecules: the split-apart oxygen atoms react to form ozone (which is more stable than a free oxygen atom). This is clearly a chemical change as the molecular structure is different, though ozone is still elementary oxygen so the element has not changed.


Electron Pair Geometry in Ibuprofen (Electron Pair Geometry in Organic Compounds) I am trying to determine the electron pair geometry (and molecular geometry) of some atoms in ibuprofen. The first one I am trying to do is the double bonded oxygen. Ibuprofen is pictured below:

I counted the number of electron domains(steric number) to be 3. I counted the double bond as one and each lone pair as one. It has two lone pairs and one double bond. So this means the electron geometry should be trigonal planar.

But I am stuck on molecular geometry. There is no molecular geometry I have found in any chart that has two lone pairs and one bond. Am I making a mistake somewhere?

You are right! There is no molecular geometry in any chart that has two lone pairs and one bond, because the only possible geometry with one neighboring atom is linear.
The central atom in a chemical moiety must have at least two neighboring atoms in order to predict the geometry around this central atom (according to VSEPR model).
Let's take the case of oxygen atom of the hydroxyl group (in Ibuprofen). The oxygen atom here, as a central atom,  has the steric number: 4 ( It has two lone pairs and two simple bonds with carbon and hydrogen atoms). So, The molecular geometry around oxygen atom is bent.


Reaction of Alcohol and Metal Hydroxide Lets say there is $\ce{CH3-OH + NaH}$. The reaction leads to $\ce{CH3-O- Na+ + H2}$. However $\ce{H-}$ from the metal hydroxide is also a strong nucleophile, so why can't the reaction lead to $\ce{CH4 + OH-}$? I realize that $\ce{OH-}$ is a bad leaving group, but then again, an alkoxide isn't a good leaving group either. An alkoxide's $\mathrm{p}K_\mathrm{a}$ is even higher than $\ce{OH-}$, so its a worse leaving group.

In this simple case it is pretty obvious that a hydride ion is a much better base than it is a nucleophile. But that doesn’t answer everything; we could consider cases with much better nucleophiles ($\ce{Br-}$).

Another reason is that hydroxide is an even lousier leaving group than hydride is a nucleophile. It is literally impossible to do $\mathrm{S_N}$ displacements of uncharged oxygen atoms unless you are opening an epoxide (in which case you lose ring strain making the overall process favourable. It does not matter whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$; $\ce{RO-}$ will never leave. To remove an oxygen atom, you always need to use strong acids that can protonate it, giving the far better leaving group $\ce{ROH+-R}$ which gives the free alcohol/water when displaced. Thus, the following reaction works well due to the very strong hydroiodic acid (I tried it):

$$\ce{R-CH2-OH + HI -> R-CH2-OH2+ + I- -> R-CH2-I + H2O}$$


Is it possible to make a solution of zinc oxide in a high school chemistry lab? As part of my IB Chemistry curriculum, we are required to conduct our own experiment researching our own question. I'm in the preliminary research phase and wanted to conduct an experiment looking at the effectiveness of zinc oxide (the active ingredient in many physical sunscreens) in absorbing UV light at different pHs ranging from 4 to 7.
I've found some procedures similar to the experiment I want to conduct, but all of them seem to dissolve a commercial sunscreen in ethanol and run it through a spectrophotometer. I want to isolate the active ingredient itself, and as such, I wanted to make a solution of zinc oxide -- though the molecule is not soluble in water, might there be another way to make a solution (perhaps dissolving in 2-propanol?). Then, I would ideally change the pH of the solutions using a buffer.
Does this seem viable, or is it likely I will need to purchase a zinc oxide suspension or perhaps look at a different chemical? Thank you for your help.

Zinc oxide is not soluble. It does dissolve at very acidic or alkaline pH, but it is a chemical reaction, not true dissolution.
The role of zinc oxide ( or titanium dioxide ) is mainly to cause UV light dispersion, effectively prolonging a lot the ray paths to allow sufficient absorption.  Dissolving the sunscreen formula would create a thin suspension with very different UV properties, which would have tendency for sedimentation.
I see a sense rather in measurement of passing UV through a thin film of a cream with zinc oxide, prepared by some controlled way, e.g. controlling its mass per area.

You have to do some sunscreen solubility tests first.
The used formula would be an emulsion either "oil in water", either "water in oil". The solvent should dissolve well both.  I would perhaps dissolve sunscreen in acetone, performing filtration/centrifugation or sedimentation, whatever applicable, to get rid of eventual $\ce{ZnO}$ or $\ce{TiO2}$ particles.
Then I would evaporate the aceton and add defined $\ce{ZnO}$ amount.
Or, you can try your own sunscreen formula.


Why are HCl fumes white? I just came to know concentrated HCl forms fumes because it dissolves in water vapors present in air. But why are these fumes white, HCl solution is transparent.

Given that the fumes are scattering droplets, I would say that polar HCl acts as centre of nucleation for water vapour.

This is about the same way by which nitric acid containing small amount of nitrogen dioxide fumes white (in this case first the red-brownish oxide reacts with moisture to form the colorless acid, the latter in turn works as nucleus for water condensation and acidic droplets formation, ie white fume).

A more detailed or perhaps better answer should be under paywall:

Chang, O.-K. Why does concentrated HCl fume? J. Chem. Educ. 1985, 62 (5), 395. DOI: 10.1021/ed062p395.


Molecular orbitals and "generalized" aufbau principle for heteronuclear molecular configurations Is there any general ordering or rule for heteronuclear molecules like $\ce{AX_2}$, $\ce{AX_{n}}$, and their molecular bonding/antibonding configurations ($\sigma/\pi$ orbitals) accordingly to some order in the Molecular Orbital Theory (MOT)?

Do the molecular orbital configurations and their order depend on the scheme and the linear combination of atoms (homo and heteronuclear, di and polyatomic) or is there some "standard ordering" like the aufbau principle $\ce{1s~2s~2p~3s~3p~4s~3d~4p~5s~\dots}$ for atomic orbitals in atoms? Or are molecular configurations a totally dependent on the molecule type and geometry? That is, I want to know if there is some (general) rule for molecular orbitals (ordering and energies) just like the aufbau principle for atomic orbitals.

Do the molecular orbital configurations and their order depends on the scheme and the linear combination of atoms (homo and heteronuclear, di and poliatomic)

Not only orbitals arising depends on atom configuration, but also on atomic orbitals available (s,p,d) and their ordering depends on geometry (i.e. size) and relative energy of atomic orbitals available. Unfortunately, there is no simple rule here.

Lucky enough, there is some amount of free computational packages you can use to calculate and then visualize molecular orbitals.


Significance of Proton Tunneling It has been repeatedly emphasized that proton transfers "involving" three or four membered ring transition states are not favorable and should not be drawn as the primary mechanism for any reaction.

However, with regard to a recent discussion involving $\ce{H3CSH}$ it was noted that the rate of proton tunneling in the anionic form $\ce{H3CS-}$ was likely extremely high.

Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof)

So:

a) Is this a valid rule of thumb: avoid intramolecular proton transfers unless the transition state has 5 or 6 members in its ring?

b) What's more significant with regard to proton transfer - proton tunneling or physical proton transfer? Or is the answer an "it depends"?

a) Is this a valid rule of thumb: avoid intramolecular proton
  transfers unless the transition state has 5 or 6 members in its ring?

Yes

  b) What's more significant with regard to proton transfer - proton
  tunneling or physical proton transfer? Or is the answer an "it
  depends"?

It depends, at temperatures around absolute zero, tunneling can become significant.  At ambient temperatures or above, physical proton transfer is much more common.

As to the methyl mercaptan anionic example you cite, to the extent that the various carbon-sulfur double bond resonance structures contribute, I would expect a lowering of the barrier around carbon anion planarization - and consequently a lowering of the barrier to inversion at the carbon atom in the anion.  Therefore, it is not clear to me that tunneling of the sulfur bound proton would be preferred to carbon inversion in order to bring about the syn-anti interconversion, especially since any presumed experiments to generate the anion would not be performed near absolute zero.


What happens if I have an acid/base equillibrium and remove hydronium ions until there is 50/50 of acid and conjugate base? Suppose I had an acid HA, and it reaches equilibrium in water. Then I remove the hydronium ions. I don't know how I would do this, maybe adding some hydroxide ions. My question is can I do this and will I be able to eventually create a 50/50 mixture of acid and conjugate base? Is this a very good buffer?

You can't just remove hydronium ions because the solution needs to remain electrically neutral.

You can't just add hydroxide ions, but you could add sodium hydroxide until the HA/A- ratio was 50/50.

  Is this a very good buffer?

It is the optimum ratio of HA and A- for buffering.


In electrolysis, does it make any difference if the electrolyte is an molten, aqueous or dilute aqueous solution? I'm an IGCSE Chemistry-extended student, and my question is: in electrolysis, does it make any difference if the electrolyte is in a molten, aqueous or dilute aqueous solution form?
An example mentioned in my book:

Explain  the difference in the electrolysis products of
a) molten sodium chloride;
b) dilute aqueous solution of sodium chloride.

In molten sodium chloride, the only available ions are $\ce{Na^+}$ and $\ce{Cl^-}$. With electrolysis, the products will be sodium metal and chlorine gas.

In an aqueous solution, the sodium chloride still moves the charges from one electrode to the other, but the potential for electrolysis of water is less than $\ce{Na^+}$ or $\ce{Cl^-}$. So the products of a solution of sodium chloride is hydrogen gas and oxygen gas.


Is adsorption exothermic, and if so, why? Why is adsorption exothermic?

The explanation given in my textbook is:

  For a process to be spontaneous, the thermodynamic requirement is that, at constant temperature and pressure, $\Delta G < 0$.
  As the entropy change of adsorption is negative ($\Delta S < 0$), the enthalpy change must therefore be negative ($\Delta H < 0$).
  If $\Delta H$ were to be positive, then $\Delta G = \Delta H - T\Delta S$ would always be positive.

It basically says adsorption is exothermic because $\Delta G$ has to be negative. Isn't there any other reason?

There are two common arguments presented as for why $\Delta H < 0$:

Argument (1): Well, indeed I see nothing wrong with the argument presented by the textbook. If adsorption takes place spontaneously, then one can conclude that the change in Gibbs free energy of the process is indeed negative. Since, the entropy change associated with process is necessarily negative (if we assume the entropy of the adsorbent is necessarily greater in the gaseous or liquid state than it is in the adsorbed state), we need a sufficiently large negative value for the change in enthalpy to ensure spontaneity.

Argument (2): Now, adsorption as a phenomenon is associated with "surface energy" (not, unlike surface tension). The surface of the adsorbent molecule attracts and "attaches" adsorbates either via weak van der waals forces (physisorption) or stronger chemical interactions (chemisorption)--in either cases, the surface energy of the system is minimised due to the formation of these new attractions. Thus, one would say adsorption would be exothermic in nature.

The key difference in physisorption and chemisorption is that the electronic structures of atoms/molecules of the adsorbent and adsorbate remain largely unperturbed in the first case (i.e no chemical reaction takes place so to speak).

Anyway, these seem to be the oft cited reasons. Both these arguments are generally sound and hold true under most, but not all conditions. These arguments are often introduced in most introductory level courses, but they do have some holes.

Food for thought - what if a positive entropy change occurs? Or, the chemical interactions that take place during chemisorption are endothermic in nature? So is adsorption always exothermic, no exceptions? Well, short answer: No.

The following discussion is inspired (largely) by ref 1.

If we just look physisorption exclusively, since this is essentially a condensation process analogous to liquefaction on the absorbent.
This is so because the number of degrees of freedom of the adsorbed species is less than the number it possessed prior to adsorption, and because the entropy of the absorbent is unaltered, since there is no chemical interaction. So argument (1) can be applied.

The same need not hold true for chemisorption, since a chemical reaction is indeed taking place. In argument (2), the assumption was that whatever chemical reaction that is taking place at the surface is exothermic, however this need not be the case. If an endothermic reaction takes place then for the process to remain spontaneous the entropy change should not only be positive, but $T\Delta S$ must exceed, numerically, $\Delta H$ so that the net Gibbs free energy change as a whole is negative

Recall: $\Delta G= \Delta H-T\Delta S $

This can, indeed be realised. For instructional purposes consider the following hypothetical system: A molecule, $A_2$, is dissociatively chemisorbed on the surface of a solid, M; consider strength
of the M-A bond equal to half that of the A-A bond. This would give us a net enthalpy change of zero. If the adsorbed atoms, A, have complete two-dimensional mobility, it would then follow that a positive entropy
change, would result, corresponding to a net gain of one degree of freedom, and the associated free energy change is a) negative and numerically equal to $T\Delta S $.

Similarly, one can envisage a system where the net positive entropy change is caused not because of increased mobility of the adsorbate molecules; the entropy of the adsorbate may very well decrease, but an associated large positive entropy change of the adsorbent (consequence of the chemical reaction taking place at the surface) can make up for it (possible, again).
Again, one can expect endothermic adsorption in such a scenario.

I'll quote the conclusion from the paper:

  Summarising, it may be said that in all adsorptions entropy changes due to surface-structural changes in the adsorbent itself must be considered along with the entropy changes of the adsorbate. In physical adsorption, the contribution of the first of these two entropy changes to the total entropy change is always likely to be insignificant compared to the contribution of the second (though further investigations alone can establish this unambiguously). In chemisorption, however, there is ample evidence already available to show that the entropy change due to surface structural changes in the adsorbent contributes significantly to the total entropy of adsorption. Whenever the positive entropy change due to adsorbent surface changes exceeds numerically the negative entropy change due to loss of freedom of the adsorbate, the net entropy of adsorption is positive. Failure to recognise this has led to serious error when using equation (1) to arrive at the sign of the heat of chemisorption. This, in essence, is why endothermic chemisorption has been discounted until very recently, though its existence should be considered as normal as the existence of endothermic solution.

You can look up specific examples, which are themselves referenced within ref 1. One such example is in ref 2, which provides some detailed potential energy diagrams illustrating endothermic chemisorption.

References

(1) Thomas, J. M. The existence of endothermic adsorption. J. Chem. Educ. 1961, 38 (3), 138. DOI: 10.1021/ed038p138.

(2) Dowden, D. A.; Mackenzie, N.; Trapnell, B. M. W. 9 Hydrogen-Deuterium Exchange on the Oxides of Transition Metals. Adv. Catal. 1957, Vol. 9, 65–69. DOI: 10.1016/S0360-0564(08)60154-6.


Mechanism for formation of amide by reaction of amine with ester This is the mechanism given in my lecture handout:

I have a problem with the third species along. Surely the negatively charged oxygen would protonate before the other oxygen would. Also, why does this reaction work at all? The amine group is more basic than the "alcohol group" that must leave to form the amide so why wouldn't that protonate first and then just leave (i.e no reaction)?

Surely the negatively charged oxygen would protonate before the other
  oxygen would.

Most of the time, but not all of the time.

  The amine group is more basic than the "alcohol group" that must leave
  to form the amide so why wouldn't that protonate first and then just
  leave (i.e no reaction)?

Most of the time, but not all of the time.

Your reasoning is correct, but remember, all of these steps are equilibria.  Although what you say will happen most of the time, some of the time the reaction will take the forward route.  Presumably the reaction is run with a large excess of ammonia to "force" the reaction in the forward direction.


Letting Copper sit in NaOH for 5 min (for monolayer assembly) i made a self-assembled monolayer of dodecanethiol on copper, before that i let the copper in a solution of $\ce{NaOH}$ for 5 min, and i need to know what this step is for, anyone can help me?

Such steps are usually cleaning steps to remove deposited organic materials. $\ce{NaOH}$ is also known as lye and will dissolve basically all organics. In this case, it should leave the $\ce{Cu}$ and copper oxides alone, leaving a clean surface.

I don't have as much experience with copper monolayers, but similar cleaning steps are used in my lab, where you might treat with ozone or some other step to remove any deposited organic "gunk" from the metal surface.

In this case, I believe you don't want to use an acid, because the thiols react with the copper oxide surface, not strictly the copper itself.


Is "stainless" a bulk or a surface property of stainless steel? In particular, if I cut a block of stainless steel in half, would the newly formed faces be stainless as well?

Let's divide the steel world into two classes: 1) rusting steel and 2) stainless steel.

Rusting steel, in the presence of oxygen and moisture, will oxidize, forming hydrated iron oxides/hydroxides which have a greater volume than the original iron, and which have relatively little adhesion to the metal. They curl up and continue to expose bare metal, and so rusting iron/steel will continue to rust.

It was discovered in the early 1900s that addition of 12% or more chromium to iron would produce an alloy that oxidized very slowly, forming a very thin oxide layer that was adherent and did not continue to thicken. It resembled chromium in that respect, and the oxide was probably rich in chromium. Nickel was also found to increase the corrosion resistance of the alloy.

Now the bulk composition affects the surface oxide, but it is the surface that stains, or rusts - or doesn't. If you broke a block of stainless steel in half, in a vacuum, the fresh surfaces would be bare metal, active to many reagents. Exposure of the fresh surfaces to oxygen will oxidize them fairly rapidly (minutes, hours) to a passive state which resists many chemicals, like acids, especially oxidizing acids like nitric. (Interestingly, plain iron dissolves in dilute HNO3, but in concentrated HNO3 (>30%), the oxidizing power of the acid is so great that the surface of the iron is rendered passive: an oxide coat is produced which does not flake off and does not dissolve in the acid.)

Stainless steels are less resistant to chloride ion, which has a way to infiltrate the oxide layer and corrode the base metal. Stainless steels which have been brushed with a wire wheel made of ordinary steel will have a surface contaminated by tiny fragments of regular steel, which will rust. The oxide layer on the stainless steel body will be imperfect, and oxidation will progress thru and under the passive oxide until the whole stainless steel is corroded.

Since it is the surface of the steel which contacts the active reagent, it would seem that the surface (oxide) is the determining factor, but, of course, the bulk composition strongly affects the surface oxide layer. And the surface needs oxygen to resist further corrosion!


Bromohydration regioselectivity In which of these cases would the addition of $\ce{Br2/H2O}$ be highly regioselective?

$$\ce{CH3CH=CHCH2CH3}$$ or $$\ce{(CH3)2C=CH2}$$

since both are asymmetric alkenes, how do I answer the question?

With molecule (2), you have two possible cations from addition of Br+; one is tertiary, one is secondary. Tertiary is much more stable so you will get very little product resulting from water reacting the secondary cation = high selectivity observed.

With molecule (1), Br+ addition gives 2 possible secondary cations; nothing much to choose between them = poor selectivity, ~1:1 mixture observed.


Why does potassium permanganate turn colourless when reacted with sulfur dioxide? I learnt at school that when sulfur dioxide gas, $\ce{SO2}$, is combined with potassium permanganate, $\ce{KMnO4}$, the purple solution will become colourless.

What happens in the solution to cause this to happen?

This is a redox reaction. The permanganate ion (purple) is reduced to manganese(II) ion which is a very pale pink, or almost colourless.

$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$

while sulfur dioxide is oxidized to the sulfate ion, which is colourless.

$$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$$

So, the solution is decolourised. The overall equation is

$$\ce{2MnO4- + 5SO2 + 2H2O -> 2Mn^2+ + 5SO4^2- + 4 H+}$$


What is the bond angle of water? I have been trying to find out the bond angle of $\ce{H2O}$, but every site I visit has a different answer.

So far, I have found the following angles listed:

Site 1: 104.4º
Site 2: 107.5º OR 104.5º, depending on where you are in the article.
Site 3: 104.5º

In your comment, you say you are interested in "liquid water".

The values in your self-answer are for gas phase water, and the second source is citing the first source, so it is really just one source.  The value is determined from gas phase rotation spectroscopy.

For liquid water, the value is not as precisely known:

105.5° (calculated) and 106° (experimental) as reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.


Why burn methane to CO2 for isotopic analyis? I am wondering why mathane is burned to CO$_2$ for isotopic analysis, e.g with a mass spec?
I have looked into several text books, they all describe how to do it but not why.
Is it because of a interference on mass 16? Analysing methane it it is would give you also information about the D/H ratio. So why?

There are different types of mass spectrometers with different modes of sample introduction, analyte ionization and ion selection/focusing. You cannot generalize "with a mass spec."
In many cases of isotope ratio mass spectrometry (IRMS), compound specific isotopic analysis (CSIA), or Gas Chromatography Combustion Isotope Ratio Mass Spectrometry organic compounds (not just methane) are burned and oxidized to carbon dioxide and water and analyzed as such to determine the total ratio of carbon-13/carbon-12 and hydrogen-2/hydrogen-1. However, combustion is not the only way to introduce compounds in the mass spectrometer.
Another secondary issue is that there if often more noise at low mass ranges. Methane is a small molecule (m/z =16) and carbon dioxide is relatively large (m/z=44).


When are the terms diphosphate and pyrophosphate used? On the Wikipedia article about pyrophosphate, it states that:

  The group (pyrophosphate) is also called diphosphate or dipolyphosphate.

But for example, why is ADP apparently called adenosine diphosphate but not adenosine pyrophosphate? The same for geranyl pyrophosphate (GPP) instead of geranyl diphosphate?  In which context the terms diphosphate and pyrophosphate are used?

Great question, but unfortunately without a great answer.  Chemistry is very inconsistent in its naming of phosphates.  In addition to the terms diphosphate and pyrophosphate, the term bisphosphate is also used.

Someone will probably put in an answer that cites "official" IUPAC or IUBMB rules.  In many cases (but not all) these rules are more sensical and well thought-out than the terms in common usage in the field, but unfortunately, they are not in universal (or even wide) use.

Here is a rough breakdown of my sense of how these three terms are used:

Inorganic chemistry

Pyrophosphate indicates a phosphate anhydride.
Diphosphate indicates a salt that contains more than one (ortho)-phosphate anion.
Bisphosphate is not widely used.

Biochemistry

Pyrophosphate: used inconsistently to name an organic derivative of a phosphate anhydride.  This is what you see in geranyl pyrophosphate (GPP).  It is also used to refer to "inorganic" pyrophosphate anion, often abbreviated as PPi.
Diphosphate: an unfortunate term that is ambiguous, although usage is shifting towards synonymity with pyrophosphate.  It is often used to mean "pyrophosphate" and but sometimes still used to mean "bisphosphate".  Biochemically, the most important phosphate anhydrides are undoubtedly ATP, ADP, etc.  These are never referred to as pyrophosphates even though that term would be more accurate.  History has chosen "triphosphate" and "diphosphate" as the terms to use for adenosine-substituted phosphate anhydrides, icky though the term may be.
Older literature will also use "diphosphate" to mean "bisphosphate".  The name "fructose diphosphate was formerly in wide usage for what is now known as "fructose-1,6-bisphosphate", for example.  That compound contains no phosphate anhydrides.
Geranyl diphosphate is still a fairly widely used synonym for geranyl pyrophosphate.  You see abbreviations for this molecule of both GPP and GDP.  I prefer the name GPP, but unfortunately not everyone uses it.
bisphosphate: Now in wide usage as the preferred term for an organic molecule with multiple ortho-phosphate esters, such as fructose 1,6-bisphosphate.  Biochemists have done a relatively good job of adopting this usage in modern times.

A common confusion that stems from this nomenclature concerns the stability of organophosphate compounds. Phosphate esters, e.g. glucose-1-phosphate or glucose-6-phosphate or adenosine monophosphate, are much more stable and less reactive than phosphate anhydrides such as acetyl phosphate (a mixed anhydride of acetic acid and phosphoric acid) or the phosphoanhydride bonds in adenosine triphosphate (ATP).


How hot is elephant toothpaste? So, I remember in high school chemistry when my teacher combined hydrogen peroxide with potassium iodide and it exploded into a yellowish foam. He said it was an exothermic reaction so it's really hot. I was just wondering exactly how hot is it? Is there a way to calculate the heat from the amount of chemicals used?

The enthalpy change of the reaction can be calculated using the enthalpies of formation of the compounds in the reaction.  To find the temperature change of the reaction, divide the quantity of heat released by the heat capacity of the solution in which the reaction occurred.

This won't give the exact answer.  Some of the heat will be absorbed by the surroundings including the flask.  The reaction produces gas and some of the energy released by the reaction is used up by the work needed for the gas to be released into the atmosphere.

Rough estimate:

Let's say the teacher used a 500 ml bottle of peroxide.  The solution contains about 0.5 mol of peroxide.  The enthalpy of formation of peroxide is about −190 kJ/mol.  The enthalpy of formation of water is about −240 kJ/mol and that of oxygen is 0.  So, 0.5 mol of reaction gives about 25 kJ of energy released by the reaction.  We'll assume all of it went into the 500 ml of water.  The 500 ml of water has a heat capacity of around 2100 J/K.  So, the solution gets about 12 K hotter.


Precise estimation of air properties as a function of temperature at 1 atm Are there any tables with empirical equations to calculate properties of air as a function of a temperature at 1 atm?
For example, NIST provides such equations for a large number of gases, but I was not able to find such solution for air, where I would specifically like to obtain:

Dynamic viscosity,
Specific heat,
Prandtl number,
Thermal conductivity,
Thermal diffusivity.

It would be best, if I could put these calculation in MS Excel, because I need to change these values with respect to temperature in multiple equations.

Here are some links that look like they might be useful, there seems to be promising information on some engineering resource sites:

https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470516430.app2
This is the appendix from what appears to be an engineering textbook, it gives a chart of the information you're looking for at 1 atm and 15 C. It also walks you through the temperature dependent equations for some of the properties.
https://www.me.psu.edu/cimbala/me433/Links/Table_A_9_CC_Properties_of_Air.pdf
This is a table of the properties of air at 1 atm from -150C to 2000C. It doesn't provide equations but you could maybe enter the data into excel and run a regression to extract a an equation.
https://irc.wisc.edu/properties/ This is a handy little calculator that outputs a whole host of properties of fluids, including dry air.

The point that has been made about water vapor fluctuations is valid, but most calculations of air properties assume dry air with constant composition.


Dissolution of Pentahydrate of Copper Sulfate I'm having some difficulty understanding how the pentahydrate of copper sulfate dissolves in water.  I understand how polar molecules work, and how things dissolve in water, but I can't get my head around this in particular.

What happens to the water molecules in the pentahydrate?  Does the copper sulfate split up at all?  What makes the solution blue, if the solute is split?

When copper(II) sulfate pentahydrade dissolves in water it actually just becomes further hydrated, in what is called an aquo complex (a coordination compound containing only water molecules as ligands). According to this Wikipedia article:

  The pentahydrate ($\ce{CuSO4·5H2O}$), the most commonly encountered salt, is bright blue. It exothermically dissolves in water to give the aquo complex $\ce{[Cu(H2O)6]^2+}$, which has octahedral molecular geometry. The structure of the solid pentahydrate reveals a polymeric structure wherein copper is again octahedral but bound to four water ligands. The $\ce{Cu(II)(H2O)4}$ centers are interconnected by sulfate anions to form chains.

The aquo complex maintains the blue coloration that is present in the pentahydrate solid.


Alphabetical priority of the isopropyl group This passage is contradictory. I've boxed the section that confused me.

Why would the book tell me to alphabetize substituents, only to clearly ignore its own rule about alphabetizing substituents?

Seems to me that "isopropyl" should come before "methyl," as the letter "i" is most definitely before the letter "m" in the alphabet. And the book makes no mention about ignoring the "iso-" prefix.

So:

1) Should methyl come before isopropyl?

2) If so, why?

The iso prefix is included in the alphabetic order.

For eg:- The compound Thymol has the following IUPAC name :- 2-Isopropyl-5-methylphenol and not 5-methyl-2-Isopropylphenol.

Therefore the name of your compound should be 4-isopropyl-2,3,3-trimethylheptane.


Titrating iodine starch solution with sodium thiosulphate - Colour change I investigated two mixtures with different solvents, one with water and one with n-heptane. Both contained iodine $\ce{I2}$ as a solute. To both solutions I added a bit of starch.

As I remember this resulted in a colourchange. So the solution turned from yellowish to dark blue (if I remember correctly!).

Now according to wikipedia starch and iodine indeed form a structure which has a dark blue colour. But it only forms in the presence of $\ce{I^-}$.

This leaves me wondering, why do I remeber the solution to be dark blue, eventhough I think there was no $\ce{I^-}$ present? Could it be the solution turned dark blue only after I added some sodium thiosulfate? Because in the next step I did a titration with $\ce{Na2S2O3}$.

In this case I don't see which reaction could have produced the $\ce{I^-}$ though. I thought only $\ce{NaI}$ is produced after adding the sodium thiosulfate.

$$\ce{I_2 + 2Na_2S_2O_3 -> 2NaI + Na_2S_4O_6} \tag{1}$$

So at which point did the solution turn dark blue and where did the $\ce{I^-}$ come from, that was needed for the formation of the starch-iodine-compound? Could it be there is an intermediate step to (1) in which $\ce{I^-}$ is formed and this $\ce{I^-}$ was used to produce the dark blue starch-iodine compound?

I don't think your memory is serving you right. That is why we write everything in the notebook, especially color changes.

I think you are doing distribution experiments where iodine is distributed between aqueous layer and an organic layer. When we add indicator for titration, it is not a solid starch but starch which is boiled in water. So when you added starch $solution$ to heptane which contained iodine, I would not be surprised if the starch solution turned blue.

Remember that iodine is strong oxidizing agent as well. A very small fraction of it can easily convert into iodide. You really really need a trace of the triiodide ion to form a dark blue iodine complex.


Is it possible to boil a liquid by just mixing many immiscible liquids together? In open air, when vapour pressure reaches 1 atm, boiling takes place.

I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to $p = p^*_A + p^*_b$, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.

If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that  $p = p^*_A + p^*_B + ... > \pu{1atm}$ , will boiling results at room temperature?

I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make $p>\pu{1atm}$), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (e.g. rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.

Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil†. This is what steam distillation is all about.

As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.

So it goes.

(†) There seem to be some ambiguity as to what exactly "boiling" entails.

Will vapor bubbles form in the whole bulk of both liquids? No.
Will both liquids eventually evaporate completely, if left in a cylinder under a piston and maintained at these pressure and temperature until they reach equilibrium? Yes.

For a single pure liquid these two questions are equivalent, hence the confusion.


Why does lead(II) acetate does not ionize appreciably? As this page in this book says, lead(II) acetate is soluble, but does not ionize appreciably. It seems contradictory to me, since, to call a substance soluble means it has high $K_\mathrm{sp}$ (when compared with other substances with the same elemental ratios in their formula units). But how can the book say it doesn't ionize appreciably? Why call it soluble then?

EDIT: I forgot that it is wrong to say that "to call a substance soluble means it has high $K_\mathrm{sp}$", so I revised it accordingly.

Here's the picture of the actual paragraph for those who can't access the link:

I recognize that this question actually reflects my inability to distinguish between dissociation and dissolution, now that I have cleared it from me, I rephrased the question to reflect my actual question. I did not erase the previous version to be able to show the reference of MEL Science's and alphonse's answers.

ACTUAL QUESTION: Why does lead(II) acetate not dissociate appreciably in water, considering that the ionic forms of acetate and lead(II) ion can be hydrated well due to the presence of partial charges in the water molecule?

A source in my native language says that lgK of PbAc2 is 4.08. It seems that lead acts as a central atom with two acetates as two-dentant ligands and two molecules of water acting as ligands too.
That means that lead acetate has its lattice ruined during dissolving, but it still forms a coordination compound.
 I am not really good in coordination chemistry of lead, but I just found out that lead wants to have octahedral geometry around it, so I would assume the structure of that would be

All bonds from lead ion are not really covalent rather than coordination bonds. Mind the charges!
P. S. aq=water
P. P. S. People in the comments are saying lead is too large for acetate to be bidentate,thus I would assume acetate just forms coordination compound with lead just like a simple monodentate ligand.


Detailed description of hydroxy(phenyl)methanesulfonic acid? Hydroxy(phenyl)methanesulfonic acid, or benzaldehyde adduct is the product of reacting benzaldehyde with sodium metabisulfite ($\ce{NaS2O5}$) solution.
It is obviously a well-known and often used reaction, and yet, I'm not able to find much info on the adduct, or the reaction itself.

One thing I managed to dig out is its corresponding PubChem entry.

I want to find out the exact mechanism of the reaction with an equation, a bunch of videos of someone performing it will be nice too...

But most importantly - the hydroxy(phenyl)methanesulfonic acid: What is its color, texture and so on? What is its crystalline structure? Is it supposed to smell of benzaldehyde, how soluble it is in different solvents, how should it be stored, is it harmful or dangerous, does it decompose in some conditions, melting point, and so on.

Hydroxy(phenyl)methanesulfonate is a compound of a class commonly called "bisulfite addition products" or "bisulfite adducts", since bisulfites are commonly used for the reaction you have mentioned. But there's an equilibrium in aqueous environment between metabisulfites and bisulfites
$$\ce{S2O5^{2-} + H2O \rightleftharpoons 2HSO3^{-} }$$
in favor of the bisulfite, this is why metabisulfites can be used as well, probably.

The bisulfite anion is proposed to have two tautomers:

The form at the left has a lone electron pair at the sulfur atom, and can act as a nucleophile.

The carbonyl/aldehyde group $\ce{>C=O}$ has a partial positive charge at the carbon atom.

The reaction in now no mystery, it's a nucleophilic addition to the carbonyl group.

If I may cite a crude reaction mechanism from Clayden/Greeves/Warren's Organic Chemistry, short chapter Bisulfite addition compounds:

And from the March's Advanced Organic Chemistry:

  The reaction is reversible (by treatment of the addition product
  with either acid or base) and is useful for the purification of the starting compounds, since the addition products are soluble in water
  and many of the impurities are not.

If you want more details (kinetics, etc.), there is freely available couple of scientific articles, Sulphur isotope effects in the bisulphite addition reaction of aldehydes and ketones

I. Equilibrium effect and the structure of the addition product
II. Bond-formation effect


An electric potential is generated when a liquid moves with respect to a solid (streaming potential). Is the reverse true? Streaming potential is an electrochemical phenomenon that a that relates
electric currents with the relative movement of solid and liquid phases in contact with
each other.

So when a liquid moves past a solid a potential is generated. The equations for the values of the potential and current generated can be found in this Wikipedia article. My question is, is the reverse true? If I applied an electrical potential difference (instead of applying a pressure difference), would a pressure difference be generated?

Your assumption is correct. Electrokinetic pumps use a potential difference across an osmotic membrane, i.e. through narrow channels, the inverse of flow causing a streaming potential difference. There are other aspects of the phenomena such as electrophoresis, capillary osmosis etc.; see:

Electrokinetic Phenomena a

On the Efficiency of Electrokinetic Pumping...

and

short video demo


Are adamantane crystals covalently bonded? I'm writing a report and need one thing clarified after reading the Wikipedia article on adamantane:

'All the above methods yield adamantane in the form of polycrystalline powder. Using this powder, single crystals can be grown from the melt, solution or vapor phase (e.g. with the Bridgman–Stockbarger technique). Melt growth result in the worst crystalline quality with a mosaic spread in the X-ray reflection of about 1°. Best crystals are obtained from the liquid phase, but the growth is inpracticably slow – several months for a 5–10 mm crystal. Growth from the vapor phase is a reasonable compromise in terms of speed and quality.[2] Adamantane is sublimated in a quartz tube placed in a furnace, which is equipped with several heaters maintaining a certain temperature gradient (about 10 °C/cm for adamantane) along the tube. Crystallization starts at one end of the tube which is kept near the freezing point of adamantane. Slow cooling of the tube, while maintaining the temperature gradient, gradually shifts the melting zone (rate ~2 mm/hour) producing a single-crystal boule.[16]

Would I be write in thinking that these crystals are not randomly shaped diamond structures you would get if the adamantanes covalently fused with eachother? In other words, are the crystals formed from polycrystalline adamantanes (with different numbers of basic ten carbon adamantanes in one derivative crystal) bonding together without covalent bonds? Or perhaps without many covalent bonds? For example, one C-C bond between two crystals? What other type of bonding could be at work here? Is the macro-crystal structurally weak as a result?

Adamantane is a molecular compound, much like naphthalene, paraffin, and many, many others. Molecular compounds consist of molecules, which stay pretty much the same in the gas and solid states. A molecule has certain molecular formula. Each carbon in a molecule is linked to a certain number of hydrogens and other carbons. There are no dangling covalent bonds to be formed between molecules, let alone between crystals. Consequently, you can't just take some adamantanes and covalently fuse them; some people would discuss diamond crystal structure in such terms, but only in a metaphorical sense.

Adamantane molecules in a crystal (being non-polar molecules without any specific interactions) are held together only by the weak van der Waals forces. Because of that, yes, the crystal is not mechanically strong at all (again, think of paraffin wax).


Does 1,3-dichloroallene possess a C2 symmetry axis? Is there an $C_2$ axis of symmetry in $\ce{Cl-CH=C=CH-Cl}$?

A $C_2$ axis of symmetry means that I should be able to rotate the molecule about an axis by $180^\circ$. But I would need a composition of the following two rotations

$90^\circ$ rotation about $\ce{C=C=C}$ axis
$180^\circ$ about the perpendicular bisector of $\ce{C=C=C}$.

Would that count as a $C_2$ axis of rotation?

A $C_2$ axis is indeed here—although not the one that you are thinking of: your suggestion of a $C_4$ then $C_2$ rotation does not qualify as a $C_2$ symmetry operation. There is a true $C_2$ axis which bisects the allene and is tilted by $45^\circ$ from both the $\ce{Cl-C-Cl}$ and $\ce{H-C-H}$ planes.
One can visualize symmetry elements for small molecules using, for example, Jmol.
Launch the program GUI and do the following:

File → Get MOL
Type in an identifier, e.g. the name, 1,3-dichloropropa-1,2-diene, or CAS number, 83682-32-0.
Open console (File → Console) and enter show pointgroup to list symmetry elements and draw pointgroup to add them to the picture: