How is ATP generated by cellular respiration? I learned in a high school level Biology class that the chemical equation for cellular respiration is

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

When I looked up the chemical formula for ATP I found that it is

$$\ce{C10H16N5O13P3}$$

Knowing that, that must mean that the first equation becomes

$$\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O + C10H16N5O13P3}$$

One of the products contains nitrogen and phosphorus, but none of the reactants contain those two elements, but there should be in order to produce a product that contains nitrogen and phosphorus. With my high school knowledge, I cannot figure this out. Where do the nitrogen and the phosphorus atoms come from?

One note at the beginning: please do not be scared! You can scroll right down to the bottom of this answer to read the conclusion which in itself should sufficiently answer the question. The bits in between are merely elaboration and need not be known in full detail.

The reaction one typically receives in high school biology classes to explain cellular respiration is already a simplification on many levels. It is presented as $(0)$, but in reality that is not one single reaction. Whether or not $\ce{ATP}$ has already been put in is irrelevant.

$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}\tag{0}$$

In fact, this reaction is a series of steps, each performed by a single, specific enzyme. The first reaction, that basically immediately happens when glucose enters an animal cell is phosphorylation as shown in $(1)$.

$$\ce{\underset{{glucose}}{C6H12O6} + \underset{{ATP}}{C10H12N5O13P3^4-} ->[enzyme] \underset{{glucose-6-phosphate}}{C6H11O9P^2-} + \underset{{ADP}}{C10H12N5O10P2^3-} + H+}\tag{1}$$

I have written $\ce{ATP}$ and will continue to do so for all products in the form they are in at physiological $\mathrm{pH}$

Glucose-6-phosphate is isomerised to fructose-6-phosphate, which cannot be distinguished by sum formulae alone, so I’m leaving this step out. Another enzyme comes along which phosphorylates fructose-6-phosphate again to give fructose-1,6-bisphosphate; see $(2)$. So far, all we have done is use up $\ce{ATP}$ at no benefit for the cell.

$$\begin{multline}\ce{\underset{{fructose-6-phosphate}}{C6H11O9P^2-} + \underset{{ATP}}{C10H12N5O13P3^4- ->[enzyme] }} \\ \ce{\underset{{fructose-1,6-bisphosphate}}{C6H10O12P2^4-} + \underset{{ADP}}{C10H12N5O10P2^3-} + H+}\end{multline}\tag{2}$$

In a retro-aldol reaction, fructose-1,6-bisphosphate is split into glycerinaldehyde-3-phosphate (commonly abbreviated GAP) and dihydroxyacetonephosphate (commonly abbreviated DHAP). These once again are structural isomers that bear the same sum formula and they are interconverted. Hence I am pretending in equation $(3)$ that only GAP is a product. This assumption is close enough to valid in sum formulae.

$$\ce{\underset{{fructose-1,6-bisphosphate}}{C6H10O12P2^4-} ->[enzyme] 2 \underset{{GAP}}{C3H6O5P^2-}}\tag{3}$$

We have now reached the phase of glycolysis (the consumption of glucose) called the pay-off phase. Previously, as you may have noticed, all we did was consume ATP. It is now almost time to get some back. But first, we need to prepare GAP for its prominent role in reaction $(4)$:

$$\begin{multline}\ce{2 \underset{{GAP}}{C3H5O6P^2-} + 2 HPO4^2- + 2 \underset{NAD+}{C21H26N7O14P2-}} \\ \ce{->[enzyme] 2 \underset{{1,3-bisphosphoglycerate}}{C3H4O10P2^4-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{4}$$

It may seem confusing that I labelled an anion as $\ce{NAD+}$ (its common name). This is because there are two phosphate diesters that can each be protonated once. Fully protonated, $\ce{NAD+}$ is indeed a cation — only not at physiological $\mathrm{pH}$.

In the previous step, we synthesised a phosphoric carboxylic mixed acid anhydride. These are very energetic and consequently its hydrolysis is now used in reaction $(5)$ to synthesise $\ce{ATP}$. We are breaking even.

$$\begin{multline}\ce{2 \underset{{1,3-bisphosphoglycerate}}{C3H4O10P2^4-} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} ->[enzyme]} \\ \ce{2 \underset{{3-phosphoglycerate}}{C3H4O7P^3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{5}$$

After this step, we again have an isomerisation that cannot be shown in sum formulae. The phosphate group is moved from position 3 to position 2 giving 2-phosphoglycerate. This can then be dehydrated as shown in $(6)$ to give phosphoenolpyruvate: a very important cellular intermediate. We have also generated our first two water molecules.

$$\ce{2 \underset{{2-phosphoglycerate}}{C3H4O7P^3-} ->[enzyme] 2 \underset{{phosphoenolpyruvate}}{C3H2O6P^3-} + 2 H2O}\tag{6}$$

Finally, a second set of two $\ce{ATP}$ molecules is formed by dephosphorylating phosphoenolpyruvate to give free pyruvate as shown in reaction $(7)$.

$$\begin{multline}\ce{2 \underset{{phosphoenolpyruvate}}{C3H2O6P^3-} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 H+} \\ \ce{->[enzyme] 2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{7}$$

We can draw an intermediate line here, since the process typically known as glycolysis has reached its close. We can sum it up in reaction $\text{(1–7)}$. We still have not generated any $\ce{CO2}$ and only $\ce{2 H2O}$. We also already have $\ce{2 ATP}$ per molecule of glucose and 2 mysterious molecules called $\ce{NADH}$.

$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 HPO4- + 2 \underset{NAD+}{C21H26N7O14P2-} ->[steps]} \\ \ce{2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+ + 2 \underset{{ATP}}{C10H12N5O13P3^4-} + 2 H2O}\end{multline}\tag{1–7}$$

The next step is still very easy to explain. It is called oxidative decarboxylation and adds a cofactor known as coenzyme A of $\ce{CoA}$ covalently to what was pyruvate before the reaction. See equation $(8)$.

$$\begin{multline}\ce{2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{NAD+}{C21H26N7O14P2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} ->[enzyme]} \\ \ce{2 \underset{{acetyl-CoA}}{C23H35N7O17P3S^3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 CO2}\end{multline}\tag{8}$$

Acetyl-CoA now enters into a magic known as the Krebs cycle or citric acid cycle. As the name implies, this is constantly going around. At the beginning, the two carbon atoms of acetyl-CoA are added to oxaloacetate. After the ten cycle steps, we will have lost two molecules of $\ce{CO2}$, effectively reforming the oxaloacetate substrate. And once again some of the steps are merely isomerisations which do not show up in some formulae. Are you ready? Here goes. Reaction $(9)$ has already been described in the text above:

$$\begin{multline}\ce{2 \underset{{acetyl-CoA}}{C23H35N7O17P3S^3-} + 2 \underset{{oxaloacetate}}{C4H2O5^2-} + 2 H2O} \\ \ce{->[enzyme] 2 \underset{{citrate}}{C6H5O7^3-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-}}\end{multline}\tag{9}$$

After isomerisation of citrate to give isocitrate, this is oxidised to give oxalosuccinate as shown in $(10)$.

$$\begin{multline}\ce{2 \underset{{isocitrate}}{C6H5O7^3-} + 2 \underset{NAD+}{C21H26N7O14P2-} ->[enzyme]} \\ \ce{2 \underset{{oxalosuccinate}}{C6H3O7^3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{10}$$

For each oxalosuccinate, one molecule of $\ce{CO2}$ can be removed to give α-ketoglutarate as shown in $(11)$.

$$\ce{2 \underset{{oxalosuccinate}}{C6H3O7^3-} + 2 H+ ->[enzyme] 2 \underset{\alpha{-ketoglutarate}}{C5H4O5^2-} + 2 CO2}\tag{11}$$

α-ketoglutarate is then once again attacked by coenzyme A in the same manner as in step $(9)$. This can happen because both pyruvate and α-ketoglutarate have the same structural feature. Refer to equation $(12)$.

$$\begin{multline}\ce{2 \underset{\alpha{-ketoglutarate}}{C5H4O5^2-} + 2 \underset{NAD+}{C21H26N7O14P2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} ->[enzyme]} \\ \ce{2 \underset{{succinyl-CoA}}{C25H36N7O19P3S^4-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 CO2}\end{multline}\tag{12}$$

We have now lost all $\ce{CO2}$. What we need to do is regenerate the oxaloacetate from the beginning of the Krebs cycle. The first step of this is to generate succinate from succinate-CoA by hydrolysis as given in $(13)$.

$$\begin{multline}\ce{2 \underset{{succinyl-CoA}}{C25H36N7O19P3S^4-} + 2 H2O + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 HPO4^2-  ->[enzyme]} \\ \ce{2 \underset{{succinate}}{C4H4O4^2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{13}$$

Next, we oxidise the saturated succinate molecule to an unsaturated fumarate. This is typically perfomed by yet another cofactor called flavin-adenine-dinucleotide or $\ce{FAD}$; see reaction $(14)$.

$$\ce{2 \underset{{succinate}}{C4H4O4^2-} + 2 \underset{FAD}{C27H31N9O15P2^2-} ->[enzyme] 2 \underset{{fumarate}}{C4H2O4^2-} + 2 \underset{FADH2}{C27H33N9O15P2^2-}}\tag{14}$$

Water is added to the double bond to give malate (reaction $(15)$):

$$\ce{2 \underset{{fumarate}}{C4H2O4^2-} + 2 H2O ->[enzyme] 2 \underset{{malate}}{C4H4O5^2-}}\tag{15}$$

And malate is finally oxidised by using $\ce{NAD+}$ as shown in $(16)$ to give the starting material of the Krebs cycle, oxaloacetate.

$$\begin{multline}\ce{2 \underset{{malate}}{C4H4O5^2-} + 2 \underset{NAD+}{C21H26N7O14P2-} ->[enzyme]} \\ \ce{2 \underset{{oxaloacetate}}{C4H2O5^2-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{16}$$

Technically, we could end it here. Let’s look again at the final products of reactions $(1–16)$. Thankfully, a lot cancels out.

$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 4 \underset{{ADP}}{C10H12N5O10P2^3-} + 4 HPO4^2- + 10 \underset{NAD+}{C21H26N7O14P2-} +} \\ \ce{2 \underset{FAD}{C27H31N9O15P2^2-} ->[steps] 6 CO2 + 6 H+ +} \\ \ce{4 \underset{{ATP}}{C10H12N5O13P3^4-} + 10 \underset{NADH}{C21H27N7O14P2^2-} + 2 \underset{FADH2}{C27H33N9O15P2^2-}}\end{multline}\tag{1–16}$$

We’re still not there yet. We have generated a lot of $\ce{NADH}$ and $\ce{FADH2}$ which somehow needs to be transformed back into $\ce{NAD+}$ and $\ce{FAD}$ for the next iteration to be performed — remember that cellular energy is, as you noted in your question, typically equated to $\ce{ATP}$. And we still haven’t used oxygen at all and are not producing enough water.

The next set of reactions to bring us closer to our goal is the electron transport chain. By a series of reactions that is not easily described (but basically involve electrons being moved in redox processes from one compound to another), oxygen molecules acquire four electrons and four protons each to become water. The electron transport chain that performs this will, at the same time, use the energy liberated by the generation of water to set up a proton potential: protons ($\ce{H+}$) get pumped from one side of an intracellular membrane to another.

This proton potential of course is a driving force since there is a lack of protons on the other side. Protons may travel back through special trans-membrane proteins. These proteins can be imagined like a watermill and start rotating as protons pass through. This rotation allows them to literally slap a phosphate onto an $\ce{ADP}$ molecule to give $\ce{ATP}$.

As a theoretical maximum, the regeneration of $\ce{NAD+}$ from $\ce{NADH}$ gives around 2.5 equivalents of $\ce{ATP}$ while $\ce{FADH2}$ can only be used to generate around 1.5 equivalents of $\ce{ATP}$ upon the reverse reaction to $\ce{FAD}$. These numbers are, however, generalised, since some energy is also lost in transport and other processes.

Therefore, our theoretical maximum yield of $\ce{ATP}$ is 25 from $\ce{NADH}$, a further 3 from $\ce{FADH2}$ and a further 4 that are directly produced in the reaction sequences. Each of these $\ce{ATP}$s is generated by the reaction of a phosphate anion onto an $\ce{ADP}$ molecule, technically involving the loss of water or an equivalent.

As you may see, writing up a complete reaction for the entire cellular respiration process is very difficult. We can, however, arrive approximately there by a few basic principles:

there are up to $\ce{32 ATP}$ generated over the course of the digestion of one molecule of glucose
each $\ce{ATP}$ is generally synthesised by some form of $\ce{HPO4- + ADP}$ — therefore, the phosphorus part is balanced.
the exact process is not necessarily important; a lot cancels out and the end result counts.

This in total may lead us to question why exactly the reaction taught at schools is often the one presented in equation $(0)$. Well, we can see that exactly six molecules of carbon dioxide are formed in the overall process per molecule of glucose. This makes sense, as it means each glucose carbon atom effectively leaves the body as $\ce{CO2}$, the carbon compound with the lowest remaining energy. The number of oxygen and water molecules cannot, unfortunately, be accurately established for the single process. However, schools want to teach the concept of a balanced equation. And if all carbon leaves the body as $\ce{CO2}$, then all hydrogen must eventually leave it as $\ce{H2O}$. To accomplish that, a certain set of oxygen molecules is needed — exactly 6.

The reaction $(0)$ thus never actually happens in our body. That is a good thing. You can see what it looks like by heating suger to very high temperatures and then adding a match — actually don’t. It’s the same combustion as any other combustion. Part of it can be adequately summarised of many partial reactions, part of it is just there so the summary fulfills the required stoichiometry. And even the $\ce{32 ATP}$ are only an approximation based on a lot of averaging.

What we can definitely say, however, is that each of the $\ce{32 ATP}$s is produced by the reaction of phosphate and $\ce{ADP}$. And therefore, at least consistent with your high-school level, the reaction should better be written as:

$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 6 O2 + 32 HPO4- + 32 \underset{ADP}{C10H12N5O10P2^3−}} \\ \ce{-> 6 CO2 + 6 H2O + 32 \underset{ATP}{C10H12N5O13P3^4−}}\end{multline}\tag{full}$$


How can the stereochemistry of the product of an E1cb reaction be determined? Say if this is your reagent.

I guess similarly to how you'd determine the stereochemistry of the product of an E1?

Since your example compound has a chirality center, let's start by picking one enantiomer. We'll do the other enantiomer later. In the presence of a suitable base, this compound can be deprotonated to form the enolate anion. The enolate group itself can be E or Z, but for this purpose it does not matter.

In the second step of the mechanism, the electrons in the new pi bond from the enolate pi electrons shift toward the beta carbon. The new pi bond will be formed from the enolate p orbital on the alpha carbon and the C-Cl sigma-antibonding orbital on the beta carbon.

For this to work, the C-Cl must be oriented perpendicular to the plane of the pi bond. We can draw Newman projections for our specific enantiomer of the enolate anion. There are two conformations that put the chlorine in the correct orientation. The second one below has less steric hindrance with the larger group (ethyl) near the H atom and the smaller group (methyl) oriented toward the enolate carbon.

The major product forms from this conformation, with the methyl group cis to the enolate/carbonyl carbon.

Now, the other enantiomer forms the same major product, but I would like you to convince yourself of that.


Why does like dissolve like? Polar solvents love polar solutes to be dissolved in it and non polar with non polar. Often said as like dissolves like.

Okay, polar loving polar can be understood with help of the facts: same polar nature, same kind of interactions etc.

But how will you explain the application of this rule with organic or non-polar solvents?

Solomons’ and Fryle’s Organic Chemistry says that there are some unfeasible entropy changes when polar solutes are added in non polar solvents and vice versa.

Why so?

The reason behind this is the hydrophobic effect. Everyone has seen it if they pour a spoonful of vegetable oil into a pot of water, e.g. to cook pasta. As long as nothing is disturbing the vegetable oil, it will collect itself together in one big bubble rather than form many small bubbles.

Polar solvents will always be arranged in a way that positively polarised areas are close to negatively polarised areas in the neighbouring solvent molecule. Hydrogen bonds — especially in water — are nothing but the extension of this concept to even more polarisation. Polar molecules, as you stated, will fit together with this scheme well. Unpolar compounds mixed in do not. As soon as you add an unpolar compound to a polar solvent, you are creating a type of artificial boundary and only the areas of the solvent molecules that are neither particularly positive or negative will be happy to be in the near vicinity. That means that you have a much higher ordering of the solvent molecules where they hit an unpolar compound, because one direction is basically doomed to be neither positive nor negative.

This ‘solvent wall’ will form no matter whether the unpolar island is large or small. However, it will also have almost the same thickness, no matter how big the contained part is. Therefore, if multiple unpolar molecules clump together, the overall number of polar molecules constrained in that wall is lower — an entropic gain. Thus, unpolar compounds tend to either not dissolve or precipitate out of polar solvent solutions.

This also works in the opposite direction. A polar molecule will prefer, for energetic reasons, to have a polar neighbour. However, an unpolar solvent cannot provide that polar environment. A neighbouring undissolved polar molecule, however, can. Here again, the polar compounds rather stay clustered together as it allows the central molecules to be more disordered while only a small layer on the border must take care to interact as well as possible with the unpolar solvent.

Note that this answer assumes some kind of black/white dichotomy. In reality, most compounds are somewhere on a scale from absolutely unpolar to very polar and are able to adapt to a wide range of organic solvents. Conversely, some organic solvents have the reputation of dissolving almost anything organic; most notably dichloromethane. However, the solubilities may greatly vary.


Why is H2O a weaker acid than acetylacetone? Shouldn't a hydrogen connected to an oxygen be more acidic than a hydrogen connected to a carbon atom? I recently started studying chemistry after a long pause in my education so pardon me if it is a simple question.
In the organic chemistry course I am taking, the professor told us that the first factor defining the stability of the conjugate base (and thereby the acidity of the compound) is the atom that is connected to the acidic hydrogen. By this definition, the hydrogen in $\ce{H2O}$ should be more acidic than the hydrogen in acetylacetone as the first one is connected to oxygen but the second one is connected to carbon (as oxygen is more electronegative than carbon).
My professor mentioned other factors such as the existence of resonance structures, inductive polarity, and the effect of side chains, but she mentioned these factors come next and only if both hydrogens in our comparison are bound to the same element.

Your instructor is right, up to a certain point. There are factors that are more important (the atom to which it is bound) and factors that are less important (resonance, etc.), and it's good to be able to order these relative factors. The devil is in the details, though. If your resonance stabilisation is strong enough (and in this case it is) then it can overcome the more important factor.
Before starting, let me point out that CH's are less acidic than OH's, but this isn't actually to do with electronegativity. If it was about electronegativity, then OH's should be more acidic than SH's as O is more electronegative than S, but the opposite is true. Same applies to HF and HCl: F is more electronegative but HF is actually a weak acid. It has more to do with the X–H single bond strength than anything, as a weaker bond is easier to break (homolytically). The X–H heterolytic cleavage, relevant for an acid–base reaction, is sort of like a homolytic cleavage, plus an electron transfer from H to X. The electronegativity comes into play in the electron transfer step, but it's just not so important a consideration, compared to the bond strength which describes the homolytic cleavage.

I'm giving pKa's to the nearest 5(ish) in this discussion. The exact numbers don't matter too much, we're more interested in broader trends.
If you compare water and methane, their pKa's are ~15 and ~50 respectively, which is a difference of about 35 units. That's roughly how much the adjacent-atom factor counts for, when comparing C and O.
If you then compare acetone and methane, pKa's are ~20 and ~50 respectively, which is a difference of about 30 units. Not quite as big as the previous one, which was 35.
Acetylacetone, though, has two carbonyl groups, that's twice as many as in acetone. Now, resonance stabilisation doesn't scale linearly in that fashion and hence the additional stabilisation over acetone is only ~10 units. However, it's still enough to bring the pKa down to ~10, which is more acidic than water.


How is rotenone a polyphenol? How is it a polyphenol if it doesn't have a hydroxyl group?

If it isn't which subclass would you put it in, and which functional groups does it contain?

How is rotenone a polyphenol?

It is not. Wikipedia says: Polyphenols are $\ldots$ characterized by the presence of large multiples of phenol structural units.

And if you're to pull up an image which gives you the structural formula of rotenone you get (Image source: https://upload.wikimedia.org/wikipedi...)

Do you see any phenols in this structure?

  If it isn't which subclass would you put it in

According to this article I've found:

  Article source

    In the late $1920$’s, the chemical structure of rotenone was determined. Rotenone is a member of a class of compounds of related molecular structure referred to as isoflavones.

And finally,

  which functional groups does it contain?

With a keen eye, if you look at the structure carefully, you'll at least be able to spot a ketone.


Selection of parent chain (IUPAC nomenclature) In the given compound, there are (apparently) two possibilities of choosing the parent (main) chain for its nomenclature in accordance with IUPAC rules:

(A) being the one containing 4 carbon atoms and the triple bond in the parent chain;

(B) being the one containing 6 carbon atoms and the hydroxyl group as a substituent in the parent chain.

Which one shall I choose while naming this compound? My book claims it to be (A), but I do not understand. (B) is clearly the longer chain and contains the hydroxyl group, which has a higher priority in nomenclature as compared to the triple bond. Where am I going wrong?

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

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-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
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.
P-45.2.3 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 to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. Apparently, you have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the aldehyde group ($\ce{-CHO}$).
The next rule (P-44.3.2) is looking at the longest chain. Therefore, the correct answer is B; the principal chain contains six carbon atoms and the hydroxyl group. Thus, the complete name is 2-ethynyl-5-hydroxyhexanal.

The next rule (P-44.4.1.1) would be looking at the multiple bond, but it is not needed since we already found the principal chain.
Note, however, that the IUPAC recommendations on this matter have been changed. In the current version, the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed.

(…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

If your book is still using old nomenclature rules, it would expect the other answer A to be correct.


Does the temperature during this reaction remain constant? Considering all gases to be ideal, the enthapy change for the reaction
$$\ce{I2 (g) + H2 (g) -> 2 HI (g)}$$
is $\pu{−106.78 kJ}$ at $\pu{350 K}$.
To find the change in internal energy the formula we use is
$$\Delta H = \Delta U + (\Delta n)RT \, .$$

To derive this equation we must have assumed the temperature during the reaction to be constant in the expression
$$\Delta H = \Delta U + \Delta(nRT).$$
I know that $\Delta U$ depends on temperature and the amount of substance, but if we assumed the temperature to be constant then $\Delta U$ should be $0$,
as the amount of substances of reactants and products are also the same, but it has a non-zero value. Is this possible?

The internal energy of an ideal gas mixture is determined not only by the temperature, but also by the chemical composition of the mixture, and, in the case of a chemical reaction, the composition changes.  For a real gas, the internal energy also becomes pressure-dependent at higher pressures.


Nail polish chemistry question When I read my ingredients on my nail polish remover, it has water as its first ingredient.  Why does nail polish have water, yet the warnings still say its flammable?

Mixtures of water with flammable liquids can themselves be flammable, largely due to the fact that the flammable liquids generally are more volatile than water, (ie they evaporate more easily.) As a result, the air above a solution of nail polish will have a lot of molecules of the flammable liquid in it. (That's why you can smell them easily). If you have a flame near the liquid, those molecules in the air above it can combust. The heat generated will evaporate more liquid, which will fuel the fire. Some of the heat is lost to vaporizing water, which doesn't burn, but if the fraction of the total liquid that is flammable is high enough, the combustion continues.

In nail polish, the flammable liquid is usually acetone or ethyl acetate.


How should you name the solid-solution compound ScY(SeO4)3? How should you name the compound $\ce{ScY(SeO4)3}$? There are 2 cations, and I know that $\ce{SeO4}$ is a polyatomic ion with $-2$ charge. However, I am unsure how to proceed.

This is similar to a double salt, but because of the chemical similarities between scandium and yttrium, $\ce{ScY(SeO4)3}$ is a solid solution in this case. And the answers is you just name the cations in alphabetical order, then the anions like any normal salt. Thus the name is scandium yttrium selenate.

If the ratio between cations had a large difference such as $\ce{Sc_{1.9}Y_{0.1}(SeO4)3}$ It might be called yttrium-doped scandium selenate, but could still correctly be called just scandium yttrium selenate too.


Total number of “hyperconjugable” hydrogens in 1-(3-ethyl-4-methylphenyl)-2-methylprop-1-ylium How do I find the total number of “hyperconjugable” structures for 1-(3-ethyl-4-methylphenyl)-2-methylprop-1-ylium?

I read some sources stating that the total number of “hyperconjugable” structures is equal to the total number of α-hydrogen atoms. But which carbon is taken as the alpha carbon in a compound having no functional groups, just carbons?

There is one obvious $\alpha$-position, which is the tertiary carbon atom adjacent to the carbon carrying positive charge. This contributes one hypercongujation structure. It looks like there are no more apparent structures, but the methyl group that is para to the cation can also participate in hyperconjugation

There are three $\ce{H}$-atoms on the methyl group, which give us three more structures. Thus, there are a total of 4 hyperconjugable hydrogens in this structure.
Interesting to note is that the ethyl group at the meta position does not participate in hyperconjugation. Can you find a reason why?


Can we recycle garbage with the principles of mass spectrometry? 'Mass Spectrometry (MS) is an analytical technique that sorts ions based on their mass (or "weight"). Mass spectrometry is used for many chemical analyses, ranging from the analysis of a complex mixture of petroleum, to the products of genetic engineering.'

Is it possible to atomise some garbage by applying a very high voltage and a very high current across it, thereby obtaining high-temperature sparks, and then apply the principles of mass spectrometry, to separate out the individual elements for use in manufacturing?

Yes it is possible, but is very expensive and would be orders of magnitude more costly than what people are willing to pay for recycled materials.  Let me give two data points to explain why it is expensive.

Some commercial material is manufactured by mass spectrometry.  The starting material isn't garbage, but reasonably chemically purified metals of various types.  The product(s) are isotopically separated metals.  Iron 57 ($\ce{^{57}Fe}$) is an isotope that is naturally present in terrestrial iron supplies at an abundance of about 2%.  It is useful to researchers that study iron materials using various spectroscopic techniques.  Pure iron 57 costs hundreds of dollars per milligram.
Mass spectrometry requires very high voltage, a very high vacuum, and large amounts of purified gases; additionally, the throughput of material separated by mass is very low.  I think that for commercial isotope purification, magnetic sector-type instruments are used.  I'm not too familiar with them, so let me go through some rough numbers for a different type of mass spectrometry: time-of-flight: Several thousand volts are applied to packets of ions which leads to their separation. Total ion currents are as high as say $10^9$ particles per second, which corresponds to about 0.1 nanoamperes or $10^{-15}$ moles per second.  So to purify one mole of iron would require $10^{15}$ seconds, or several million years of continuous operation of very expensive machinery!  Even if other types of mass spectrometry do better than time-of-flight by three orders of magnitude, it will still be cost-prohibitive.$%edit$


How much energy is needed to break a water molecule? How do you calculate how much energy is needed to break a water molecule into its constituent atoms?

note
This is not homework. I just want to know what is required to get hydrogen out of water.

The answer is given by the standard enthalpy of formation, $\Delta H_\mathrm f^\circ$ for water:

$$\begin{alignat}{2}
\ce{H2 + 1/2O2 &-> H2O(l)}\quad&\Delta H_\mathrm f^\circ&=-285.8\ \mathrm{kJ/mol}\\[6pt]
\ce{H2 + 1/2O2 &-> H2O(g)}\quad&\Delta H_\mathrm f^\circ&=-241.8\ \mathrm{kJ/mol}
\end{alignat}$$

The energy required for the following reaction, your point of interest:
$\ce{H2O -> H2 + 1/2O2}$ is simply a positive value for either one of the above $\Delta H_\mathrm f^\circ$ values depending on the state of the reactant. It means that per mole water splitting requires at least $241.8\ \mathrm{kJ}$ of energy.

Any method(thermal,electrolysis etc) followed, this much of energy will be always required. When using electrolysis, it may be convenient to express it in $\mathrm{eV}$.

Regarding calculation, I have to mention that $\Delta H_\mathrm f^\circ$s are experimental values. To know how they are calculated, here is a nice example of $\ce{NaCl}$.


How to find the molecular formula for a hydrocarbon compound from known combustion products? What is the molecular formula of a compound containing only carbon and hydrogen if combustion of $1.05~\mathrm{g}$ of the compound produces $3.30~\mathrm{g}~\ce{CO2}$ and $1.35~\mathrm{g}~\ce{H2O}$ and its molar mass is about $70~\mathrm{g}$?

Here is my work:

$$
\begin{array}{cccccc}
                  & \ce{C_{a}H_{b}} & \ce{->} & \ce{CO2} & + & \ce{H2O} \\
\text{masses (g)} & 1.05            &         & 3.30     &   & 1.35
\end{array}
$$

\begin{align*}
\ce{CO2} &\rightarrow \ce{C} \\
44~\mathrm{g} &\rightarrow 12~\mathrm{g} \\
3.30~\mathrm{g} &\rightarrow x
\end{align*}

$$
x = 0.9~\mathrm{g},~\text{moles of C} = \frac{0.9}{12} = 0.075
$$

\begin{align*}
\ce{H2O} &\rightarrow \ce{2H} \\
18~\mathrm{g} &\rightarrow 2~\mathrm{g} \\
1.35~\mathrm{g} &\rightarrow y
\end{align*}

$$
y = 0.15~\mathrm{g},~\text{moles of H} = \frac{0.15}{1} = 0.15
$$

$$
\text{empirical formula}~\ce{C_{0.075/0.075}H_{0.15/0.075} -> CH2}
$$

$$
\frac{70}{14} = 5
$$

$$
\text{molecular formula is}~\ce{C5H10}
$$

$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -> $a$CO2 + $\frac{b}{2}$ H2O}$$

Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$:
$$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$
So the empirical formula is $\ce{CH2}$
Now for actual formula $\frac{70}{14}=5$
Yes the formula is $\ce{C5H10}$.


Lowering Freezing Point of Amines What is most efficient method of lowering freezing point of amines without greatly effecting properties? Method must involve incorporating an additive at small percentage.

Add another amine. It is better to choose an amine with just few more carbons.


How to determine the calorimeter constant from its heat capacity A bomb calorimeter contains $600\;\mathrm{mL}$ of water. The calorimeter is
   calibrated electrically. The heat capacity of the calorimeter is
   $785\;\mathrm{J\,K^{-1}}$. The calorimeter constant would be closest to:

  A. $3.29\;\mathrm{kJ\,ºC^{-1}}$

  B. $4.18\;\mathrm{kJ\,ºC^{-1}}$

  C. $4.97\;\mathrm{kJ\,ºC^{-1}}$

  D. $789\;\mathrm{kJ\,ºC^{-1}}$

My (rather mindless) attempt is as follows:
$$
E=mC_PT\to E/T=mC_P\to C_{\mathrm{cal}}=mC_P=(600)(8.314)(10^{-3})=4.9884\;\mathrm{kJ\,ºC^{-1}}
$$
The closest answer to my result seems to be C ($4.97\;\mathrm{kJ\,ºC^{-1}}$), however I know I'm wrong.

To give a precise answer, the following assumptions are necessary and must be clear:

bomb calorimeter works at constant volume ($V=const$);
both water and calorimeter itself are at thermodynamic equilibrium before the experiment and during the measurement, in particular their temperatures $T_w$ and $T_c$ are equal before the experiment and during the measurement;
the system is compound by calorimeter itself plus water;
the system is an isolated one;
pressure is 1 bar.

Initially the system is at temperature $T_1$. Let's imagine that an object at $T_o>T_1$ is put inside the chamber of the calorimeter. Temperature of the system increases and, once reached thermodynamic equilibrium, it stops at a precise value $T_2$.

Since $V=const$, heat transferred from object to system is:
\begin{equation}
Q_V=\Delta U=\Delta U_{calorimeter}+\Delta U_{water}=(mc_V\Delta T)_c+(mc_V\Delta T)_w
\end{equation}
where $\Delta T_c=\Delta T_w=T_2-T_1$.

We know that heat capacity at constant volume is defined as:
\begin{equation}
C_V=\left(\frac{\partial U}{\partial T}\right)_V\approx \left(\frac{\Delta U}{\Delta T}\right)_V
\end{equation}
So, reshaping the first equation, we obtain:
\begin{equation}
C_V=\frac{\Delta U}{\Delta T}=(mc_V)_c+(mc_V)_w=(C_V)_c+(\rho Vc_V)_w
\end{equation}
Adding the following data:

$\rho_w=1000\;kg/m^3$;
$(c_V(300\;K,1\;bar))_w\approx 4.134\;J/(kg\;K)$ (source: Perry's Chemical Engineers' Handbook)

and carrying out the conversion: $V=600\;mL=6\times10^{-4}\;m^3$, we obtain finally:
\begin{equation}
C_V=787\;J/K=0.787\;kJ/K
\end{equation}
So right answer is A.


How does propylene glycol/glycerin form formaldehyde/acetaldehyde when vaporized? So apparently, e-cigarettes produce formaldehyde/acetaldehyde because they vaporize propylene glycol/glycerin. What is the chemical reaction through which propylene glycol/glycerin forms formaldehyde/acetaldehyde? Why does propylene glycol form more carbonylated products than glycerol?

I think the reaction is essentially an incomplete combustion reaction.

In a vaporizer, the idea is to get the propylene glycol, glycerine, and and "flavor" molecules into the vapor phase (or into aerosol droplets) without chemically degrading them.

However, the way this is accomplished is with very high-temperature resistor coils - the idea is that if you have a small amount of heat in a confined space, and the liquid is distributed throughout an absorbent (like cotton, for example), then the heat transfer will happen very quickly and a large amount of liquid will be vaporized.

The problem is that when you have a very high temperature heat source, and there is lots of oxygen in comparison to the liquid, there is a good chance that at least some of the liquid will reach the combustion temperature instead of just vaporizing.

What happens is that the hydrocarbons in the propylene glycol and glycerin molecules are partially oxidized - oxygen reacts with them to "steal" electrons. If the combustion was complete, you would wind up with just carbon dioxide and water. When combustion is incomplete, you can get any number of compounds, two of which happen to be formaldehyde and acetaldehyde. Carbon monoxide is another one I would be concerned about.

To see how this works, take a look at proplyene glycol (all pictures are from wikipedia):

and glycerine:

Notice how they both have three carbon atoms in the "backbone" and either one or none $\ce{O-H}$ groups (called hydroxyls or alcohols) attached to each one.

Oxygen can react with any one of those carbons and will try to form $\ce{CO2}$ and water, the most thermodynamically stable products that are possible:

If there is plenty of oxygen, it will. However, if there is not enough, you will get some intermediate species - formaldehyde or acetaldehyde, for example.

Let's look at formaldehyde:

In this case, oxygen has reacted with the the other two carbons on the backbone, leaving one with a hydroxyl. The hydrogen on the oxygen goes somewhere (either water or one of the other pieces of the original molecule) and the leftover electrons end up in the oxygen-carbon bond, forming a double bond. This molecule is more thermodynamically stable than proplyene glycol or glycerol, but less stable than carbon dioxide.

Now let's look at acetaldehyde:

Notice the similarity - it's almost the same, there is just a $\ce{CH3}$ (methyl) group in place of one of the hydrogens. You could get this by reacting the first carbon on propylene glycol with oxygen, and then removing the hydrogen from the hydroxyl on the second carbon. This is again more stable than propylene glycol, but less stable than carbon dioxide - it is also an intermediate.

  Why does propylene glycol form more carbonylated products than glycerol?

This is a guess - but I think it likely has to do with the reaction kinetics. The reason I think this is true is that all of the intermediate products are less thermodynamically stable than carbon dioxide and water. Therefore, if a propylene glycol (PG) gives more of a given intermediate product than glycerine does, that product is "easier" to make for PG. This would happen if the activation energy for forming it is lower, or if the activation energy to go from the intermediate to carbon dioxide and water is higher compared to the intermediates formed by glycerine.

Note: Many people in other answers are pointing out that there are other mechanisms that could lead to these products, and that combustion is not likely at these temperatures. I think that given the mechanics of heat transfer we can't just rule it out, but I don't have any evidence for that. My answer is meant to show you how the reactant molecules contain the same structures as the products - I wouldn't put much faith in my mechanism (or anybody else's) without some real evidence.


How many moles NH4Cl must be added to NH3 to create buffer with pH=9? How many moles of $\ce{NH4Cl}$ must be added to $\pu{2.0 L}$ of $\pu{0.10 M}$ $\ce{NH3}$ to form a buffer with $\mathrm{pH}=9$? Assume the addition does not change the volume of the solution significantly.

The solution given is as follows:

The equilibrium between $\ce{NH3}$ and $\ce{NH4+}$ is given by-
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
We know the $K_\mathrm{b}$ value and $\ce{[OH-]}$ from the $\mathrm{pH}$ and $\ce{[NH3]}$ is given, thus we can solve for $\ce{[NH4+]}$. The solution given says that the final answer is given by this concentration multiplied by the total volume $\pu{2 L}$.

I don't follow the logic in the solution. The $\ce{[NH4+]}$ we found was the total concentration of $\ce{NH4+}$ in the solution after adding the salt. But before adding $\ce{NH4Cl}$, there was already some $\ce{NH4+}$ in the solution (we don't know how much).
Thus to find the number of moles $\ce{NH4Cl}$ we need to add, shouldn't we subtract the number of moles $\ce{NH4+}$ already in the solution from the answer we got in the textbook solution?

$\ce{NH3}$ is such a weak base ($K_{\mathrm{b}}=1.8\times 10^{-5}$) that the initial concentration of $\ce{NH4+}$ can be considered negligible. You can consider the initial concentration of $\ce{NH4+}$, but the answer you will get is the same to a considerable number of significant figures.
By considering the equilibrium of $\ce{NH3}$ dissociating to form $\ce{NH4+}$ and $\ce{OH-}$ it is easy to find the initial concentration of $\ce{NH4+}$.
$$K_{\mathrm{b}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$
Since \begin{align}[\ce{NH4+}] &= [\ce{OH-}] \tag{1}\\
K_{\mathrm{b}}\cdot [\ce{NH3}] &= [\ce{NH4+}]^{2} \tag{2}
\end{align}
Plugging in values gives:
$$[\ce{NH4+}] = 1.34\times 10^{-3} (3\text{s.f.})$$
Which equates to $\pu{2.68\times 10^{-3} mol}$ in the $\pu{2 L}$ solution.
If I'm right this should be orders of magnitude smaller than the textbook answer, and therefore negligible.


How does copper reduce dilute nitric acid to nitric oxide and concentrated nitric acid to nitrogen dioxide? Here are the 2 equations:

\begin{align}
\ce{3Cu + 2NO3- + 8H+ &<=> 3Cu^2+ + 2NO + 4H2O} & E &= 0.62~\mathrm{V}\\
\ce{Cu + 2NO3- + 4H+ &<=> Cu^2+ + 2NO2 + 2H2O}& E &= 0.45~\mathrm{V}\\
\end{align}

The Nernst equation tells us that as we increase the concentration of nitric acid the first equation becomes less spontaneous.

My question is: how? What is the mechanism of the two reactions that decides between the first and second reactions depending upon how much nitric acid is available?

Also, any qualitative explanation for what the Nernst equation shows would be appreciated. Why does having more protons available end up in fewer protons being consumed by the reaction?

It was pointed out that the second equation above is improperly balanced.  Here's the correct equation:

$$\ce{Cu + 4H+ + 2NO3- <=> Cu^2+ + 2NO2 + 2H2O}$$

So, as pointed out in the comment by Neto, the more concentrated acid does consume more protons per reaction.

Does anyone have any insight on the mechanism of the reaction and how it changes with nitric acid concentration?

If we examine the Nernst equation: $E = E^\circ-\frac{RT}{zF}\ln Q$, the logarithmic term is what changes with concentration. The reaction quotient $Q$, gets smaller as the ratio of reactants to products increases, meaning that the log term will decrease (and become negative below $Q = 1$), so both equations will become more favourable as the ratio of reactants to products increases.

The correctly balanced second reaction is $$\ce{Cu + 2NO3- + 4H+ <=> Cu^2+ + 2NO2 + 2H2O}.$$

If we look at just the logarithmic term for each reaction, assuming the products part of the reaction quotient is unity and that the proton activity is equal to the nitrate activity:

\begin{gather}
\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{10}}\tag{cyan}\\
\frac{25.7~\mathrm{mV}}{2}\ln\frac{1}{[\ce{NO3-}]^6}=\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{18}}\tag{red}
\end{gather}

y axis: log term (mV), x axis: $[\ce{NO3-}]$

As expected, when multiplied to normalize the number of electrons transferred, the reaction that consumes more reagents becomes more favourable more quickly as the reagent concentration is increased and less favourable more quickly as the concentration is decreased.

Conceptually, the reason for this is that the reactions travel down a slope towards the equilibrium point (see figure) and the reaction that consumes more of the reagents to produce a given amount of the products involved in the equilibrium reaches equilibrium with fewer reactions. Because ∆G is simply the partial derivative of the Gibbs energy with respect to reaction extent, the reaction that reaches equilibrium with less reaction extent has a steeper slope.


Which compound results in the following products when mixed with AgNO3 and Chlorinated water I'm currently working on attempting to identify a particular substance of which I know the cation is Na, however I am unsure as to what the corresponding anion is based off of these characteristics.

The salt is soluble, and when drops of AgNO3 are added to the dissolved substance the liquid turns a dark brown and a nearly black precipitate forms on the top. The liquid remains relatively transparent however. In addition, I observed that when chlorinated water was added to the solution, the solution slowly turns white with a light tint of yellow, like a very light creamy color, it releases some pungent odor however I am not familiar enough with scents to determine whether or not it is an ammonia or sulphur odor. Of the sodium compounds, the following are options:

Sodium bisulfate
Sodium borate
Sodium citrate
Sodium chloride
Sodium fluoride
Sodium nitrate
Sodium nitrite
Sodium oxalate
Sodium sulfate
Sodium sulfite
Sodium thiosulfate

As far as I'm aware the reaction with chlorinated water is just a reaction with Cl

The black precipitate can either be

metallic silver, formed by reduction of the $\ce{Ag+}$ ions (most likely)
an insoluble silver salt

Possible candidates for the sodium salt:

╔══════════════════╦═══╦═══╗
║      SALT        ║ 1 ║ 2 ║
╠══════════════════╬═══╬═══╣
║Sodium bisulfate  ║   ║   ║
║Sodium borate     ║   ║ ? ║
║Sodium citrate    ║   ║   ║
║Sodium chloride   ║   ║   ║
║Sodium fluoride   ║   ║   ║
║Sodium nitrate    ║   ║   ║
║Sodium nitrite    ║   ║   ║
║Sodium oxalate    ║   ║   ║
║Sodium sulfate    ║   ║   ║
║Sodium sulfite    ║ ✔ ║   ║
║Sodium thiosulfate║ ✔ ║   ║
╚══════════════════╩═══╩═══╝

The reaction with chlorinated water tends to show it is the first case (maybe with sodium thiosulfate).

Regarding the pungent odor, $\ce{H2S}$ has a strong rotten-egg odor.

A specific test for sodium thiosulfate is adding an acid (such as $\ce{HCl}$ to the unknown salt solution). A yellowish-white sulfur colloid should slowly appear.

A specific test for sodium sulfite is the decoloration of fuschine (should be available in a microbiology lab): add a buffer at pH = 7 to 1 mL of your unknown salt solution, then 1 mL of fuschine. After a few seconds, it should turn transparent.

Can you conduct those test to get more insight on this? If the first one is positive, no need to do the second one.


Can we do combustion caused by limited supply of air in a lab? How? For example Ethyne in combustion by limited supply of air

$$     \ce{ 2C2H2 + 3O2 ->[\Delta] 2CO2 + 2H2O + 2C } $$ vs.

$$     \ce{ 2C2H2 + 5O2 -> 4CO2 + 2H2O + Energy } $$

What are the limits of air availability to make these combustions?
How to do these experiments in lab?

Hmm, I see a couple of problems in here:
Energy in the reaction is misunderstood:
In general, breaking bonds is endothermic, and making them is exothermic. In any combustion reaction we have bonds broken ($\ce{O2}$'s for example) and bonds created (forming $\ce{H2O}$, in this instance). The overall change in the energy defines a reaction to be endo- or exothermic; and in the case of combustion (any type) it's exothermic. If you want to write "heat" and "energy" in both sides of the reactions you face, it's going to look a little awkward. Note that in the biochemical reactions in which you see terms like "energy" and such in only the side of reactants or the products (e.g.: ATP energy releases) it's because they want to put emphasis on that particular intake or release of energy.
The equation you've provided for incomplete combustion is incomplete:
I don't see any balanced equations in there, do you? And, I can't get the reason for why you've written two $\ce{CO2}$s.
However, your second equation is about the complete combustion of ethyne, and is balanced and correct.
As for your answer, take a look at what wikipedia says:

The incomplete (partial) combustion of a hydrocarbon with oxygen produces a gas mixture containing mainly $\ce{CO2, CO, H2O}$, and $\ce{H2}$.

Depending on how much your burning sample will be out of reach of air, the mixture will vary in weight percent of the products of the reaction.
In a lab, you normally will get incomplete combustion reactions, for reasons:

Your sample is hardly pure hydrocarbon.
The air in the lab isn't in wild currents and overall, is "stale" in simple words.
The hydrocarbon doesn't get enough "requirements" for combustion and ignites incompletely.
etc.

Source

Source
A good sign (or alarm) of the incomplete combustion is the orange hue of fire, especially if it's accompanied by a visible "smoke". If the experiment requires complete combustion, we'll make use of burners such as Bunsen burner.
So, let's summarize: The incomplete combustion is the normal thing happening in the burning of ethyne. You usually do some modifications if you're to have a complete combustion reaction. When the examiner claims "with contact with adequate supplies of air", s\he only means that a complete combustion takes place, though this is rare in real life. Plus, ethyne is not a very good hydrocarbon to start with, because as CDC mentions, it has a potentially negative effect on the nervous system, causing headaches, asphyxia etc., as well as painful frostbites.


How does a limiting reactant relate to the concentration of ions in solution? Given the question

  A sample of $\pu{1.50 g}$ of lead(II) nitrate is mixed with $\pu{125 mL}$ of $\pu{0.100 M}$ sodium sulfate solution. What is the limiting reactant in the reaction? Calculate the concentrations of all ions that remain in the solution after reaction.

I first found the chemical equation for this problem to be:

$$
\ce{Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq)}
$$

I found the limiting reactant to be $\ce{Pb(NO3)2}$ because there are fewer moles of that substance than the $\ce{Na2SO4}$, with there being  $0.00453 \,\mathrm{mol}\, \ce{Pb(NO3)2}$.

I then used the limiting reactant to find the concentration of each ion. I multiplied 0.00453 mol by the number of moles of each ion in the equation, and put that over 0.125 L to get the molarity. However, the correct answers are

  no $\ce{Pb^{2+}}$ ions are left in solution
  $[\ce{SO4^{2-}}] = 0.0638 \,\mathrm{M}$
  $[\ce{Na^+}] = 0.200 \,\mathrm{M}$
  $[\ce{NO3^-}] = 0.0725 \,\mathrm{M}$

I only got the concentration of $\ce{NO3^-}$ correct. What have I done wrong? I think my mistake has something to do with using the limiting reactant in the equation. Also, I understand why there is no $\ce{Pb^{2+}}$ in the solution, because it is in solid form, but doesn't that mean the $\ce{SO4^{2-}}$ should also produce no ions since it is bonded to the $\ce{Pb^{2+}}$?

Of course, the concentration of $\ce{Na^+}$ is $0.2\ce{M}$, because it is a spectator ion. It doesn't participate in the  main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$.
The concentration of $\ce{Na^+}$ is $0.1 \times 2=0.2 \ce{M}$ as there are two ions of sodium in sodium sulfate.

The number of mols  of ion nitrate, as a spectator ion, is $2\times1.5/331.2= 0.00906 \ce{mol}$.
The concentration of ion nitrate is  $ 0.00906/0.125=0.0725 \ce{M}$.

Let's go back to the main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$: The limiting ion is $\ce{Pb^2+}$ as the number of moles of lead nitrate  $1.5/331.2=0.00453 \ce{mol}$ is smaller than the number of moles of sodium sulfate  $0.0125 \ce{mol}$.
So, after the reaction, there  are no  $\ce{Pb^2+}$ ions left in solution.

As for the remaining $\ce{SO4^2-}$; its concentration is:  $(0.0125-0.00453)/0.125=0.06376 \ce{M}$


What reaction happens between glucose and ammonium chloride when drying? I have the following observations:

When I put a solution of glucose (80 g/ℓ) and ammonium chloride (20 g/ℓ) in a drying cabinet at 37°C (maybe a bit warmer) a brown substance forms when all (or most) of the water has evaporated.
When I add water to the reaction product, the substance dissolves within minutes (under shaking), but the resulting solution is brownish.
From this I conclude that a reaction happened and the brown colour isn’t just the colour of some salt crystal.
If I perform the same procedure on glucose solution or ammonium chloride solution alone, nothing similar happens, which indicates that it’s not a reaction with the air or the container. PS: If I redissolve the product of drying ammonium chloride alone, the result is acidic (pH of 5 or lower for 1 g/ℓ).
No browning happens if I replace ammonium chloride with sodium chloride, which indicates that chloride is not responsible for the reaction.
The same thing doesn’t happen in the presence of water.
Even when the solution has already been dried to a highly viscous sludge, I still don’t observe any browning.
It only happens in final stage of drying.

My questions are:
What reaction is this, why does it only happen when drying, and is there some way to prevent this, e.g., with a lower temperature?
I ideally want to get a dried residue containing glucose, ammonium, and chloride in my container.
Just filling in the powders doesn’t help, because I cannot do this efficiently and precisely.
What I found so far is this:

Typical browning reactions of sugar (Maillard reaction, caramelisation) require much higher temperatures or additional ingredients.
According to the paper Reactions of Free Sugars with Aqueous Ammonia, some reactions between sugar and ammonia happen at this temperature, but with a low yield and in the presence of water.

When a sugar solution turns brown (or black), it's generally a sign of polymerization of the sugar molecules triggered by dehydration reactions that create more reactive molecules. A classic example of this is caramelization, in which sucrose is dehydrated by heating and forms many types of furans and pyrans and the oligomers caramelan, caramelen and caramelin. The latter three all contribute to the brown color.
Although caramelization typically requires heat, a similar effect can also be achieved by use of an acid catalyst. A simple example is treatment of sucrose with concentrated sulfuric acid, which very quickly results in a solid black mess.
In your case, you've started with a nearly neutral solution of glucose in ammonium chloride. The solution is nearly neutral because it is a 1:1 mix of a moderate base (ammonia) with a strong acid (HCl). As you evaporate off the water, however, some ammonia evaporates as well, resulting in a slight excess of HCl and a lowering of pH. The combination of mild heat and acid is likely causing a caramelization-type reaction at a lower temp than is usually required. The ammonia might also enhance the color via a Maillard-type reaction.
A good review of carmelization is: Sengar, G.; Sharma, H. K. Food caramels: a review. J. Food Sci. Technol. 2014, 51 (9), 1686–1696. DOI: 10.1007/s13197-012-0633-z


Is the differential, dp, exact or not? Consider the differential $$\mathrm dp=\frac RV\,\mathrm dT+\left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)\,\mathrm dV$$ (where $a$ is a constant value)
(a) Determine whether the above differential, i.e. $\mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!

I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.

Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.

Let me rewrite the differential as

$$\mathrm{d}p = A(V)\,\mathrm{d}T + B(T,V)\,\mathrm{d}V$$

where

$$A(V) = \frac{R}{V}$$

and

$$ B(T,V) = \left( \frac{2a}{V^3} - \frac{RT}{V^2} \right) $$

A differential is exact if

$$\left( \frac{\partial A}{\partial V} \right)_T = \left( \frac{\partial B}{\partial T} \right)_V $$

Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $\mathrm{d}p$ is exact!

Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to

$$\left( \frac{\partial A}{\partial V} \right)_T = -\frac{R}{V^2}$$

$$\left( \frac{\partial B}{\partial T} \right)_V = -\frac{R}{V^2}$$

If the two partial derivatives are the same, the differential is exact. I will let you be the judge.


How to find if the unpaired electron is placed in hybrid orbital or pure orbital in odd electron species? Odd electron species contain an unpaired electron which can either be placed in hybrid orbital or in pure orbital.

Example: In the case of $\ce{.CF3}$ radical, the unpaired electron is placed in $\mathrm{sp^3}$ hybrid orbital of carbon.

In the case of $\ce{.C(CH3)3}$ radical, the odd electron is placed in pure p orbital of carbon.

How do we determine whether the unpaired electron is placed in hybrid orbiral or pure orbital? What factors govern the same?

When discussing hybrid orbitals, they are justified by the experimentally observed or calculated/simulated molecular geometry. For instance, the $\ce{^.C(CH3)3}$ radical was found to be planar by ab initio calculations (http://www.tandfonline.com/doi/abs/10.1080/00268977600102711). Thus, the molecular shape is best approximated by $\ce{sp^2}$ hybridization and the unpaired electron is most likely to be found in a $\ce{p}$ orbital.

As a simple heuristic under the arguably outdated VSEPR theory, electronegative atoms want to be the furthest away from each other and other electrons as possible due to electric interactions. When considering the $\ce{^.CF3}$ radical, the fluorine atoms are highly electronegative and want to be the furthest they can be from the unpaired electron that holds negative charge. A planar $(\ce{sp^2})$ conformation would make the distance between the unpaired electron and each fluorine atom $90^\circ$. While a pyramidal $(\ce{sp^3})$ conformation would separate the two groups by $109^\circ$, which is much more favorable according to VSEPR theory. The later is what is observed under calculations (http://www.chem.mun.ca/courseinfo/c4420/handouts/handout-03a.pdf) with a $\ce{F-C-F}$ bond angle of $112^\circ$.

So, when considering hybridization problems, you must remember that hybridization fixes geometry. Thus if any plausible argument for a hybridization must be made, there should be experimental or computed evidence of the geometry. One cannot simply look at a molecule and know what the geometry could be with absolute certainty, but we can make a good guess based on simple heuristics, like those of VSEPR, molecular orbital diagrams, previous experience/similar molecules, etc.


Addition of inert gas at "constant volume"? My textbook states that

  "The addition of inert gases at constant volume does not affect the equilibrium state of the reactants and products contained in that volume."

Although I am not questioning the truth behind this statement, I am very confused by it.

Let us consider a homogeneous gaseous equilibrium system contained within a closed cylinder. The cylinder has a very small hole on its side, through which an inert gas can be pumped in. This is how I visualize the addition of an inert gas to the system. Now, upon introducing a given amount of inert gas into the system, how does the volume remain constant in any way?

While deriving Van der Waals equation for real gases, I learned that the term $V$ in the ideal gas equation stands for the empty space that is available to the gas molecules for movement. Therefore, when we introduce an inert gas into the system, aren't we essentially decreasing this amount of space that is available to the reactant and product molecules for movement?

I am very confused. Please share your insights for it would be tremendously helpful for me. Thanks ever so much in advance :) Regards.

You should not :)

"does not affect the equilibrium state" means that the thermodynamical constant is still the same. If you add an inert gas $\ce{G}$ in your system it will not react with other chemical coumponds.

For example at the beginning you have $\ce{A} \rightleftharpoons \ce{B}$ and after you add you inert gas you have $\ce{A + G} \rightleftharpoons \ce{B + G}$. So it not afftects the value of $\ce{K_{eq}}$ because $\ce{K_{eq}}$ depends only of $\ce{T}$ the temperature.

If you use a solid container the volume will not change but the pressure will.

Also we have $$\ce{\mu_i(T,P_i)=\mu_i^°(T)}+\ce{RT\ln\frac{P_i}{P^°}}$$

Then $$\left(\frac{\partial\Delta_r\ce{G}}{\partial n_{inert}}\right)_{\ce{T,V}}=0$$

If you prefer calculus.


Cerium (CeO) behavior in Glass Polishing? I am basically working in semiconductor industry and I specifically taking care of CMP (Chemical Mechanical Polishing). We did polish glass (borosilicate type of glass) and we are using slurry (which is cerium-based) as the chemical compound. Our polishing time was quite long (~40 min) and without chilling the slurry, it could deteriorate the polishing pad, material removal etc. I have decided to chill the slurry (CeO) at temp ~20 °C prior supply to my polishing machine.

How does this Ce chemically and mechanically impact my CMP performance? What is the agglomeration index of this Ce with the changes of temperature?

Any good article to share?

My guess is that for the reason why your slurry deteriorates without chilling you should look the other way. Maybe the other slurry components are to blame? I do not see how ceria could impact the slurry performance in your case. Ceria (and yes, it's $\ce{CeO2}$ not $\ce{CeO}$) and some other oxides like zirconia are used in milling and polishing chiefly because of their physical hardness and chemical inertness. In other word, $\ce{CeO2}$ should probably be regarded as purely abrasive component. You need extreme conditions to chemically alter ceria, such as concentrated acids and a lot of prolonged heating, and the chemical interaction between ceria and the other solid materials is highly unlikely unless the temperatures are so high that the whole system starts to glow.


What is a typical monolayer formation time for liquid phase adsorbate on a solid surface? Let's say a solid is submerged in a solution of species with functional groups and a self-assembled monolayer is forming. What are the typical time scales for the formation of a continuous monolayer? (seconds, minutes, days..?)

 (source)

For an adsorbate in gaseous phase and high vacuum, a rough approximation on Wikipedia is ~1 sec.

Depends on concentration and other parameters. For thiols on gold, for example, a typical concentration is 1-5 mM, and formation time is 24-48 hours; "Longer assembly times tend to result in better monolayer packing." (ref)


Why is the Dulong–Petit law not applicable to the elements Be, B, C, or Si? According to the Dulong–Petit law, the molar heat capacity of a solid element is approximately $3R$ (where $R$ is the gas constant $\pu{8.314 J K-1 mol-1}$.)

This is useful for calculating the atomic mass of an element when its specific heat capacity (i.e. heat capacity per unit mass) is given.

However, the Dulong–Petit law is said to not apply to the solid elements Be, B, C, and Si. Why is this so?

I refer to the same link and add a summary:

The deviation is quantum mechanical in nature. The Dulong–Petit Law is exact only if all vibrational modes are fully activated, in which case equipartition theory can be used. As every atom in a solid can be considered to be a three-dimensional harmonic oscillator, the contribution to the heat capacity is $3k_\mathrm{B}$ for one atom, or $3R$ for one mole.

However, this applies exactly only as the temperature goes to infinity. For finite temperatures we can still come close if the thermal energy is high enough versus the energy required to excite the vibrational modes:

$$k_\mathrm{B}T \gg \hbar\omega$$

For most elements – except the elements listed in the question – room temperature is high enough to come close to full excitation. For the elements in question, the vibrational frequency $\omega$ is large, particularly because of the strong bonds and relatively small masses ($\omega \propto \sqrt{k/m}$.)


Would an (NH3)2+ molecule be trigonal planar like BH3 rather than trigonal pyramidal? I've been learning about using MO theory to explain why $\ce{BH3}$ and $\ce{NH3}$ have different geometries and by following the line of reasoning used to rationalise the differences in geometry I came to wonder whether ionisation or the addition of electrons to a molecule can change its geometry.

Eg: $\ce{NH3}$ is trigonal pyramidal but an $\ce{NH3^2+}$ ion would have the same electron configuration as $\ce{BH3}$, which is trigonal planar; does this mean $\ce{NH3^2+}$ would be trigonal planar?

Edit: My bad when I first posted this I wrote $\ce{NH3^2-}$ but meant $\ce{NH3^2+}$, this was just an example, my main objective was to find out whether changing the geometry through ionisation was possible.

Not only is changing shape due to ionization possible, it has been observed.  Not with ammonia as suggested in the question, but with cyclooctatetraene, which is nonplanar in its neutral state but switches to a planar, aromatic ring structure when reduced to its di-anion.


C14 Nuclear Diamond to AA Battery Apples-to-Apples Engery output I was reading an article about nuclear diamonds and they made an apples-to-oranges comparison of a 1 gram C14 nuclear diamond to a single AA battery. Here is the part I'm referencing:

  "An alkaline AA battery weighs about 20 grams, has an energy density
  storage rating of 700 Joules/gram, and [uses] up this energy if
  operated continuously for about 24 hours." Meanwhile, "A diamond
  beta-battery containing 1 gram of C14 will deliver 15 Joules per day,
  and will continue to produce this level of output for 5,730 years — so
  its total energy storage rating is 2.7 TeraJ."

I'm far from a chemist or any other applicable type of scientist otherwise I'd take a crack at the answer myself. The question simply is this: according to the data provided how many Joules per gram would a C14 nuclear diamond produce if it weighed 20 grams (to compare it to the quoted AA battery that weighs 20 grams)?

To visualize the desired apples-to-apples answer:

Conventional 20 gram battery: 700 Joules/gram, for 24 hours.

Nuclear ?? gram battery: ??? Joules/gram, for 5,730 years.

Let me address this part of your question first:

  "[...] how many Joules per gram would a C14 nuclear diamond produce if it weighed 20 grams"

I think it's important to understand that the bolded/italicized phrase is redundant, because your are discussing (as did the article) the energy output per gram.  So the per-gram output of a 20 gram battery is the same as that of a 1 gram battery.  The 20 gram battery would have a total output of 14000 Joules in it's lifetime, where the one gram battery would have a total output of 700 Joules.  The fact that the article, and your question, normalize to a per-gram basis is how they've maintained an apples-to-apples comparison.

The semantics get tricky when they talk about the mass of the diamond battery, and by my thinking this is where they may fail to give an apples-to-apples comparison.  The diamond would not be pure $\ce{C^{14}}$, and at a minimum would contain a thin protective coating of non-radioactive $\ce{C^{12}}$ (protecting those on the outside from the radiation on the inside).  So when they say "A diamond beta-battery containing 1 gram of $\ce{C^{14}}$", it is not made clear what the total mass of diamond would be (e.g. we aren't given the ratio of $\ce{C^{14}}$ to the $\ce{C^{12}}$ that would also be present).

That said, I think that the apples-to apples answer to your question, based on the information given in the article, is as follows:

Conventional 20 gram battery: 14000 Joules/day, for 1 day.
Nuclear 20 gram $\ce{C^{14}}$ battery: 300 Joules/day, for 5,730 years.

or, equivalently:

Conventional 1 gram battery: 700 Joules/day, for 1 day.
Nuclear 1 gram $\ce{C^{14}}$ battery: 15 Joules/day, for 5,730 years.


Is the ionic solution of a half-cell part of the electrode in a electrochemical cell? A simple electrochemical cell consists of two half-cells, that is: a metal solid submerged in a solution. Additionally, there is salt bridge connecting the half-cells.

Ex: $\ce{Cu(s)}$ in $\ce{CuSO4(aq)}$ and $\ce{Zn(s)}$ in $\ce{ZnSO4(aq)}$

According to my text-book (Chemistry: Collins 3rd Edition, pg.506), the 'electrode' in a electrochemical cell is the whole half-cell.

"In work on electrochemical cells, however, the term electrode is extended to include what is known as a half-cell. So it refers not only to the conductor, but also to the conducting solution in which it is placed."

From other sources, I get the idea that the electrode is the metal solid (ex. $\ce{Zn}$), while the $\ce{ZnSO4}$ is an electrolyte (a solvent that seperates the cations and anions). Even when they are talking in the context of an electrochemical cell.

However, would not the saltbridge be considered the electrolyte as that is the compound that actually neutralizes the half-cells, through a seperation of cations and anions that travel to respective half-cells. While the redox reactions is actually happening between the $\ce{Zn(s)}$ and the $\ce{Cu^2+(aq)}$, which includes the metal solid and the ionic solution. As they are both conducting electrons.

Or are both the ionic solution and the saltbridge made of an electrolyte. Both consists of polar solvents, that seperate cations and anions. Therefore the electrode would be a metal solid in an electrolyte.

I guess the essence of my question, comes to down how to define electrode and electrolyte in the context of an electrochemical cell.

Your question is about definitions.

I haven't read the Collins book, but, in general, the electrode is the metallic conductor that connects a circuit to the "non-metallic" part of the circuit. The electrolyte is this non-metallic part - dissociated ions that conduct electricity. Note electrolyte has broader uses than electrochemistry.

The term electrode can be used in reference to the whole half-cell since it's the site at which the reaction is taking place. At other parts, all we have is storage of reactant and product for the reversible reaction.
Consider the Standard Hydrogen Electrode(SHE), the whole half cell is referred to as an "electrode", though this is more convention than technical.

-

I suspect the Collins book is for UK A-levels; they should use the clear general definitions;
the electrode is the solid metal connecting circuit to non-metallic part, & the electrolyte is the solution of ions:

Here are some additional brief notes on contextual half cell setups:

SolidMetal/MetalSolution half-cell: Metal Electrode in metal ion solution.
All AqueousIons half-cell: Platinum electrode in solution.
Gaseous/AqueousIons half-cell: Glass/Platinum Deliver tube (for gas) bubble into solution.

The salt bridge is an electrolyte; A-level questions on electrode, electrolyte and salt bridge definitions aren't asked too often, consult your spec & exam papers. Importantly, solutions need to be $1dm^3$,  at $298K$ and $100kPa$, & salt bridge is typically $KNO_3$


Stereochemistry: R and S notation How do we number the carbon atoms in symmetrical molecules while giving R and S notation. For instance, in 2,3-Dibromobutane, there is an ambiguity over which is carbon C2 and carbon C3. Will we call it (2R,3S) or (2S,3R)?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants
  are assigned to them in the following decreasing order of seniority:

  (…)

  (j) When there is a choice for lower locants related to the presence of stereogenic centers or stereoisomers, the lower locant is assigned to CIP stereodescriptors Z, R, M, and r (pseudoasymmetry) that are preferred to E, S, P, and s, respectively, which are preferred to the non-CIP stereodescriptors cis, trans, or r (reference), c, and t (…).

  (…)

Therefore, the chirality center R receives the lowest locant and thus a systematic name is (2R,3S)-2,3-dibromobutane and not (2S,3R)‑2,3‑dibromobutane.

However, your drawing shows (2R,3R)-2,3-dibromobutane.


Are there any gases more dense than liquids? Inspired by Is there any type of liquid, other than mercury, that PTFE teflon is known to "float" to the surface in?, are there gases that are more dense than liquids? (So bubbles would sink down instead of bubbling up?)

It depends on the conditions. Let's start decomposing your question in two related questions:

Denser gasses at SATP?

There are indeed some gasses that are quite dense. Sulfur hexafluoride has a density of 6.17 g/L  while tungsten hexafluoride of 12.4 g/L. But usually, they are not so dense compared to the density of liquids.

Lightest liquid at SATP?

The density of liquids largely depends on the atomic mass of the compounds. Hydrogen that is the compound with the lowest atomic mass has a density of 70.85 g/L, which is probably the lowest density you can find.

So at room temperature is not possible but if you increase the pressure the density of the gasses will increase while the density of the liquid won't increase much because liquids are not appreciably compressible.  It is hence theoretically possible to achieve a gas with a greater density compared to that one of a liquid, and also the coexistence of the two phases should be possible in certain conditions but the pressure required won't allow you to observe it.

Bubbles sinking down?

That won't happen also if there was a denser gas compared to a liquid. Bubble formation is quite a different process in this case inverting the gravity would be better! ;-)


Why do different alcohols have different effects on the brain? I've been trying to understand why different alcohols have different effects on the brain.

Methanol is obviously poisonous, apparently because it's converted to formic acid. Because the body seems to cut organic substances in two's, any odd-numbered alcohol would seem to yield methanol, and therefore be poisonous. But why is 1-butanol (according to certain sites) more toxic than ethanol? Does it have a different effect on the receptors in the brain? And if so, why? Same with other simple even-numbered alcohols: why are their effects different from ethanol?

As the chain length of the alcohol increases, it becomes more hydrophobic. This would make it easier for them to insert into membranes and hydrophobic pockets of proteins. This could interfere with the protein function. Hydrophobic molecules also tend to be harder to eliminate than shorter molecules and can accumulate in fatty tissues, including the brain. Longer chains can also be metabolized into a greater number of products simply because there is more to metabolize. And sometimes seemingly innocuous compounds can be metabolized into very toxic substances, such as n-Hexane, which can be converted to a molecule with an epoxide on each end. The epoxides can cross-link proteins, particularly in neuron axons.


How does the ambient temperature affect the rate of redox reactions in a salt lake? Many years ago, I used to work as a gold exploration geologist.  One of the main locations was a salt lake in Western Australia detailed in the document WMC Resources Exploration Successes inN Lake Terrains - Applications of Element Dispersion, Kambalda W.A (Carey and Dusci, 1993).

In the article, and in my research at the time, it is stated that

  gold disperses laterally along iron redox fronts that tend to occur higher in the regolith profile

Now the area is arid (up to 50C or 122F in summer), but was not always like this, fossil records suggest far milder conditions occurred in the past.

My question is, how does the ambient temperature affect the rate of redox reaction in a salt lake, such as in the example provided?

The salt lakes that occur in Western Australia are touted by the article Geochemical characteristics of naturally acid and alkaline saline lakes
in southern Western Australia (Bowen and Benison, 2009) are unique environments, in that they possess

  an extreme and rare
  range of natural hydrogeochemical conditions. Many WA lake
  waters and groundwaters are hypersaline and acidic.

The article discusses observations of effects due to temperature, especially cycles of dry and wet.  The high ambient temperatures would cause increased evaporation, and hence increase local evapoconcentrations of ions, resulting in a lowering of the local pH, storm/inundation events have the opposite effect.  This leads to further observations by the authors that found that

  Field observations of lake pH during storm events
  illustrated that fluids that have been freshened due to meteoric input
  are able to regain acidity rapidly. Part of the acidity may be locally
  sustained by ongoing $\ce{Fe}$ cycling and redox reactions that
  generate $\ce{H+}$ ions. The amount of $\ce{Fe}$ increases significantly in the
  more acid waters, and authigenic $\ce{Fe} mineralization (oxides
  and sulfates, e.g., hematite and jarosite) is abundant throughout
  the acid lake sediments.


Relationship between molecularity and the reaction order Should the molecularity be equal or less than the reaction order?
Consider the reaction:
$$\ce{H2(g) + I2(g) → 2 HI(g)}$$
If we consider the mechanism of this multi-step reaction, the molecularity is 3 (termolecular reaction):
$$
\begin{align}
\ce{I2(g) &<=> 2I(g)} &\quad&\text{(equilibrium reaction)} \\
\ce{2I(g) + H2(g) &-> 2HI(g)} &\quad&\text{(rate-determining step)}
\end{align}
$$
The rate equation for the reaction is
$$\mathrm{rate} = k[\ce{H2(g)}][\ce{I2(g)}].$$
The order of the reaction is 2 here. The molecularity is greater than the reaction order. Is it possible?
Or is the actual mechanism for the above reaction is different?

Molecularity applies to single elementary reaction steps and is seldom higher than 2.
The reaction order is the "effective molecularity" of the whole set of linked reactions, if it had been a single step reaction. It need not to be integer and has just indirect relation to molecularities of elementary steps.
The reaction order is either determined experimentally, either derived from the reaction schema via determined conditions for steady states of intermediate products.


How do non-polar substances dissolve in non-polar solvents? The case of polar solvents is clear to me - we get an attraction between opposite charges. However, how do non-polar substances dissolve in non-polar solvents? How could it be explained on a molecular level?

The electron density distribution in molecules (including nonpolar ones) is not static. Therefor, as a function of time, the electron density is not uniform. Occasionally, randomly, the electron density in a molecule will shift to produce a spontaneous dipole: part of the molecule now has more electron density ($\delta^-$) and part of the molecule now has less electron density ($\delta^+$). This spontaneous dipole is transient. The electron density will shift back to negate it and then shift again to create a new spontaneous dipole. In a vacuum, this behavior would be a curiosity.

In the presence of other molecules, these spontaneous dipoles have a propagating effect. If molecule A develops a spontaneous dipole, then the electron density in neighboring molecule B will by respond by developing a spontaneous (but opposite) dipole. The $\delta^+$ region of B will be close to the $\delta^-$ region of A. This new dipole in molecule B is an induced dipole. This induced dipole in B will then induce a new induced dipole in another neighboring molecule C. By this time the transient dipole in A is already fading, perhaps to be replaced by a new induced dipole from another molecule.

These random, transient, but continuously propagating dipoles have attractive forces associated with them. These forces are named London dispersion forces after the physicist who proposed them. The magnitude of these forces scales with increasing surface area of the molecule. Thus, larger molecules will overall have stronger London dispersion forces than smaller molecules. Linear rod-like molecules will have stronger forces than spherical molecules (spheres having the smallest surface area to volume ratio of the 3-dimensional solids).

More complex attractive forces also arise from nonpolar molecules. Benzene (and other aromatic molecules) can pi stack, which relies on the strong permanent electric quadrupole. Benzene also has a (weak but) permanent magnetic dipole due to its ring current, so some component of the attraction between benzene molecules may be magnetic and not coulombic. Or, the two attractions may be one and the same, as benzene's electric quadrupole may result from its having a magnetic dipole. As his part of the discussion is devolving into physics, I will end here.


Synthesize acyl chloride from aldehyde I know it is usually easier to go from an acyl chloride to an aldehyde,

but is there a way to synthesize acyl chlorides from an aldehyde?

Edit:

Upon some textbook reading (Wiley: Organic Chemistry 11th Ed, Solomons), I realized that aldehydes could also be oxidized to carboxylic acids with chromic acid in aqueous acetone (the Jones Oxidation) and then you could react the carboxylic acid with thionyl chloride or phosphorus trichloride to yield the acyl chloride.

Another method was to convert the aldehyde into a cyanohydrin and through hydrolysis the -CN group could be then converted into $\ce{RCOOH}$. Then react $\ce{RCOOH}$ with $\ce{SOCl2}$ to get the acyl chloride.

There are many ways to affect the transformation, as you have realised, and many of which Loong has detailed below in his post.

This answer gives one method which I would be confident of working practically, in the lab, using common reagents. A full search of Reaxys of SciFinder would reveal conditions/methods more suitable for a specific aldehyde.

A Pinnick oxidation will convert an aldehyde into its corresponding carboxylic acid, which may then be treated with a reagent such as oxalyl chloride to give the desired acyl chloride.


How to unify all the subunits in a PDB file? A PDB may contain a TER token. For example, hemoglobin has 4 subunits separated by a TER identifier.

If we take hemoglobin's biological assembly pdb file we see:

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11           N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86           N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99           O
TER    1070      ARG A 141
ATOM   1071  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58           N
ATOM   1072  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83           C
ATOM   1073  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79           C

Is there an easy way to merge all the subunits into a single VMD frame?

If you don't care about residue and atom numbering, which you probably do, then sed '/^TER/d' < 1A3N.pdb > 1A3N_combined.pdb will do. Since atom entries must be reordered, it looks like MDAnalysis will strip the TER entries automatically:

#!/usr/bin/env python2

import MDAnalysis

u = MDAnalysis.Universe('1A3N.pdb')

with MDAnalysis.Writer('1A3N_combined.pdb') as writer:
    writer.write(u)

So,

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11           N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86           N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99           O
TER    1070      ARG A 141
ATOM   1071  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58           N
ATOM   1072  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83           C
ATOM   1073  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79           C

becomes

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11      A    N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86      A    N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99      A    O
ATOM   1070  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58      B    N
ATOM   1071  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83      B    C
ATOM   1072  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79      B    C

where there are now chain IDs in the last column. See how this doesn't renumber residues? Modify the script; resids is a property, so the in-place object mutation works properly:

#!/usr/bin/env python2

import numpy as np
import MDAnalysis

u = MDAnalysis.Universe('1A3N.pdb')
u.residues.resids = np.arange(1, 1 + len(u.residues.resids))

with MDAnalysis.Writer('1A3N_combined.pdb') as writer:
    writer.write(u)

Finally:

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11      A    N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86      A    N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99      A    O
ATOM   1070  N   HIS B 142       3.670 -13.643  19.447  1.00 38.58      B    N
ATOM   1071  CA  HIS B 142       2.695 -14.734  19.744  1.00 32.83      B    C
ATOM   1072  C   HIS B 142       1.379 -14.140  20.199  1.00 30.79      B    C


Is there a chemical which fluoresces only after coming in contact with water? I require an easily available chemical compound which in its pure form, does not  fluoresce under UV light. However upon coming in contact with water (at room temperature) it fluoresces under UV light.

Many fluorescent dyes are not fluorescent in their solid states. Fluorescein and similar dyes (rhodamines, eosines) for example. They need to be dissolved in order to be fluorescent (so just wetting them with a bit of water won't work either).


What is the significance of the "Standard Temperature" in Standard Enthalpies of Formation tables? In my data book, there's a list of common compounds and their molar enthalpies of formation -- at 298.15 K. What's the meaning of this given temperature value (is it the final temperature of the compound in question?) and how does it affect the molar enthalpy of formation of a substance?

What, for example, does it mean when the molar enthalpy of formation of H2O(g) at 298.15 K is given as -241.8 kJ/mol, even though water should be in a liquid state at 298.15 K? And what would the molar enthalpy of formation of H2O(g) be at, for instance, 500 K?

Most enthalpies (or themodynamic observables whatever their type) are not constant across the temperature range — not even across the range of one phase. So the enthalpy of vapourisation of water is different a $273\,\mathrm{K}$ from what it is at $295\,\mathrm{K}$ or $350\,\mathrm{K}$.

Of course, standard conditions have been defined. However, the definition of standard conditions does not explicitly include temperature! Likely because some wanted $273\,\mathrm{K}$, others wanted $295\,\mathrm{K}$. So the temperature at which a certain entropy is valid must always be specified.

Some books put a ‘standard temperature’ at the top and call that their specification.


What is viscosity in terms of chemistry? In physics, viscosity is the force applied in a unit area times time or basically Pascal times second. We could say that it is a measure of its resistance to gradual deformation by shear stress or tensile stress.

But what does it mean in the chemistry world?
I believe it is related to the distance between molecules in a solution. When molecules are farther apart then viscosity is lower than when molecules are closer.

Could you tell me if my mature concept about viscosity is on the right track or far too off? Could you also define viscosity in terms of chemistry?

More or less similar…I will try to put brief concepts of viscosity in terms of pharmaceutics which is somewhat another branch of chemistry, where it is very much applicable in drug design.

Viscosity is an expression of the resistance to flow of a system under an applied stress. The more viscous a liquid is, the greater is the applied force required to make it flow at a particular rate.

  Viscosity is defined in terms of the force required to move one plane
  surface past another under specified conditions when the space between
  is filled by the liquid in question. More simply, it can be considered
  as a relative property, with water as the reference material.

Viscosity implies that the liquid flows even under the smallest stress and does not return to its original shape or form once the stress is removed.

Viscous deformation, i.e. viscous flow, occurs in accordance with Newton’s law,

$\ce{σ=ηỳ}$

  where the applied stress σ results in flow with a velocity gradient ỳ
  or rate of shear. The proportionality constant η is termed viscosity,
  while its reciprocal is called fluidity.

Viscosity has also been
 described as the internal friction in the fluid as it corresponds to the resistance of the fluid to the relative motion of adjacent layers
 of liquid.

The viscosity of simple liquids (i.e, pure liquids consisting of small molecules and solutions where solute and solvent are small molecules) depends only on composition, temperature, and pressure. It increases moderately with increasing pressure and markedly with decreasing temperature.

There is quite a lot to discuss on viscosity in depth but these are just shallow concepts to understanding the term.

References

Martin’s Physical Pharmacy and Pharmaceutical Sciences : Rheology
Remington: Essentials of Pharmaceutics: Rheology
Ansel’s Pharmaceutical Dosage Forms and Drug Delivery Systems : Special Solutions and Suspensions
Aulton ’s Pharmaceutics The Design and Manufacture of Medicines 4th edViscosity Rheology, and flow of fluids


How do I calculate the mass percent in the original mixture? I have the question:

  "A mixture of $\ce{KClO3}$ and $\ce{KCl}$ with a mass of 0.950g was heated to produce $\ce{O2}$. After heating, the mass of residue was 0.700g. Assuming all the $\ce{KClO3}$ decomposed to $\ce{KCl}$ and $\ce{O2}$, calculate the mass percent of $\ce{KClO3}$ in the original mixture."

I took the amount of $\ce{O2}$, converted it to moles $\ce{O2}$, then divided it to moles of just O, then divived by 3, bcause for every 3 O (for the 3 in the $\ce{KClO3}$) there is a mole of $\ce{KClO3}$ so I had moles of $\ce{KClO3}$. Then I converted that to grams of $\ce{KClO3}$, but thaat ended up as more then the original mixture's mass.

Am I even thinking about this the right way?

Looks close. Your math looks sound, but the reasoning is messy. $\ce{O2}$ is the only oxygen here.

$$\ce{2KClO3 -> 2KCl + 3O2}$$

You might have skipped writing out that reaction — it helps a lot.

If $0.700~\mathrm{g}$ of residue remained, that means $0.950\ \mathrm{g} - 0.700\ \mathrm{g} = 0.250\ \mathrm{g}$ of $\ce{O2}$ left the container (which means that $0.250\ \mathrm{g}$ of $\ce{O2}$ was made!)

We can calculate the number of moles, similar to what you did.

$0.250\ \mathrm{g} \times \frac{1\ \mathrm{mol}}{32\ \mathrm{g}} = 0.00781\ \mathrm{mol}\ \ce{O2}$

However, we have no reason to divide by 2 — $\ce{O2}$ is the only way (well, at least for this purpose) oxygen can exist in nature. Treat it as "oxygen".

Let's apply a mole ratio, now. There were $\ce{2 KClO3}$ molecules used for every $\ce{3 O2}$ molecules we got.

$0.00781\ \mathrm{mol}\ \ce{O2} \times \frac{2~\mathrm{mol}\ \ce{KClO3}}{3~\mathrm{mol}~\ce{O2}} = 0.00521\ \mathrm{mol}\ \ce{KClO3}$ which we had at the beginning. Note that we can do this because it says all of it reacted completely. If only $95~\%$ of it reacted, we would multiply by $0.95$, for example.

Find the grams of $\ce{KClO3}$ expended in reaction.

$0.00521\ \mathrm{mol} \cdot \frac{122.55\ \mathrm{g}}{1\ \mathrm{mol}} = x~\mathrm{g}\ \ce{KClO3}$

Your percent is pretty easy from there.

$\frac{x\ \mathrm{g}}{0.950\ \mathrm{g}} \times 100~\% = y~\%$

Where I'm sure you can figure out $x$ and $y$. (Remember to find $x$ to 3 significant digits).

Leave a comment if I wasn't clear on something!


Testing for ammonia by smelling it? My chemistry teacher taught us that one of the ways to see if we had produced ammonia gas in a reaction was if it released a pungent smell (Even BBC supported that here! http://www.bbc.co.uk/schools/gcsebitesize/science/triple_edexcel/qualitative_analysis/qualitative_analysis/revision/3/ ). However, as far as I knew, Ammonia was poisonous. Is it okay to breathe in small amounts of Ammonia, or is this a hazardous practice that can potentially harm us?

This question What chemical compounds used in qualitative chemistry can be distinguished by smell? did say that Ammonia could harm you, but it doesn't say how much is needed to cause the damage, which is a big part of my question.

Thanks!

First, using the smell test for chemistry is dangerous - don't do it!

Second, for this an similar questions, consult a Safety Data Sheet (ex-MSDS) for exposure limits.

The copy that I looked at indicates:

TWA of 25ppm (from the ACGIH) or 50ppm (OSHA) -- these are based on a 8 hour/day, 40 hour/week exposure.

STEL of 35ppm (short term exposure limit). Exceeding the STEL may result in 'upper respiratory tract irritation, eye damage'.

Now, the yield from your lab bench chemistry experiment is unlikely to be very high, and as you may have noticed ammonia is very noticeable (although of the several SDSs that I consulted, none list an odor thershold, surprisingly).

But, the bottom line is, do not inhale your chemistry experiments.