What would explain urine that resists freezing? A frozen urine specimen is required for an organic acids lab test, but the urine refuses to freeze after 4 days in a regular freezer turned as low as it will go. What would explain this, and how cold might it need to be to freeze?

Urine is an aqueous solution of urea, uric acid, sodium salts and whatnot. These solutions usually exhibit freezing-point depression.

As Brian has correctly pointed out in a comment, the effect isn't that strong and a quick and dirty look in the literature shows that a freezing point around -1.3 to -2.2 °C.

You might want to try a cooling bath with dry ice in isopropanol or acetone or other drastic methods to achieve shock-freezing.


Does reaction of alkali and alkaine earth metals' reaction with water an acid-base reaction? In my chemistry notes the teacher gave me the note that reaction of alkali, alkaline earh metals' reactivity with water, acids(e.g. H2S) increases down the group and bases(e.g.NH3) decreases down the group,

And also I have some doubts that

-alkali metals such as Li first react with water and produce oxides first and those oxides react with water and produce hydroxides as end product, is this true?

-and normally water has H2O molecules, (H3O+,OH- ions at small concentration). And when alkali metals react with water do they react with H2O molecule or H3O+ molecule?, If someone could give me a reaction mechanism of an alkali metal with water?

-do alkli, alkaline earth metals act as bases when reacting with water?

Any help would be most greatfull

Na and K react with the O-H bond of the water molecule. This removes the H atom that is reduced to an H atom and then to a H2 molecule. It is not an acid/base reaction. It is a redox reaction. Na or K atom is oxidized at the end. No metal oxides are produced as intermediating state.

Alkaline and alkaline earth metals act in the same way.


Stability of superoxides of alkali metals Why are superoxide ions stable only in presence of large cations such as $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$?

According to the Wikipedia page Alkali metal oxides, the superoxide of $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ form when these metals are burnt in air, and according to the Bodner Research Group page Active Metal Reactions, they

  react so rapidly with oxygen they form superoxides, in which the alkali metal reacts with $\ce{O2}$ in a 1:1 mole ratio.

Given, according to the Chem-Guide blog article Alkali metals that

  The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of these oxides.

Which, according to the blog, results in the  $\ce{Li+}$ ion (which is small, possessing a larger positive field around it) only being able to naturally combine with a small anion $\ce{O^{2-}}$ forming $\ce{Li2O}$, whereas the substantially larger $\ce{K+}$, $\ce{Rb+}$ and $\ce{Cs+}$ have weaker positive fields and can stabilise the larger superoxide anion $\ce{O2^{-}}$ forming superoxides.


Overview of material characterization methods I have been searching for quite some time now, but I feel like it is impossible to find some sort of chart where I can determine which characterization method (microscopy, spectroscopy, electrochemical, DMA etc.) is the best in which situation. There are so many methods and an overview would really help me to understand them better.
Does anyone know if a book or a site provides such an overview? I am pretty desperate, that's why I'm asking it here.

There is a collection available at Libretexts.org, by students and faculty from Franklin & Marshall College. The intent is to provide a quick overview for readers of the literature (rather than practitioners doing the experiments), so the focus is the type of result each method yields, and how to interpret it.


Can grignard reagents react with any electrophillic sites? Yeah I am confused about the above question. Like since textbooks always say how C atom bonded to Mg is very nucleophillic, I was wondering why it does not bond with carbon atom that is single bonded (not the one in carbonyl part) to one of the oxygen atom of ester? Since O atom is more electronegative than C atom, C atom is electrophillic, so shouldn’t grignard reagent also bond with that C atom instead of one double bonded to the other O atom. I always see example where GR reagent attack the carbonyl carbon atom so I was wondering why? Thanks.

To give you an idea of the nucleophilicity of the two carbons, you may have a look at $\ce{^{13}C}$ NMR data (here for example). These data tell you which carbons have less electron density (the ones which are more deshielded).

On methyl formate, you'd find $\delta_\text{COO} = 161.8$ ppm whereas $\delta_\text{OCH3} = 50.0$ ppm. This should give you a little intuition of how much the electron density is reduced on each carbon, due to the effect of the electronegative atoms. Clearly, the ester-bearing carbon is much more affected, and this accounts for the general behaviour that you encounter.

However, as Ben Norris pointed out, this is maybe a simplistic picture. In fact, it's not impossible to break preferentially the $\ce{COO - C}$ bond. This is called hydrolysis by alkyl-oxygen cleavage (see here).

To get back to the question of Grignard reagent, the problem is also explained by HSAB theory. To attack the carbon which is not bearing the ester, you would generally need a "soft" organometallic compound. I can give you an example that corresponds to what you are asking (reference here: Kawashima et al., Tetrahedron 1989, 45, 403.), but note that this is a special case. What drives the reaction is the opening of a strained ring, and the Grignard is modified by addition of copper halide:

I represented this one in particular to show the regioselectivity (breaking of bond represented in red), but the article shows plenty of examples.


Mechanism of Oxymercuration-Demercuration reaction I am trying to write a mechanism for OMDM reaction.

I could write the mechanism for oxymercuration part of reaction but I couldn't write the mechanism for remaining part (demercuration) of the reaction. Could anybody help me to proceed forward.

Here is the mechanism I wrote out for you


Predicting molecular geometry from a formula Can someone explain the following types of molecules and predict their structure and shape?

It will help my understanding if you could provide an example.

AX3E,AX5 and AX6

note that A = central atom and X = bonding atoms.

It will depend on the number of lone electron pairs as well.
For example:
$BCl_3$ will be triangular planar as it has no lone electron pairs.
$NH_3$ however will be triangular based-pyramidal because there are 3 bonded pairs of electrons and one lone electron pair.
Basically, it comes down to the number of electron pairs (both bonded and unbonded) around the central atom.

Two electron pairs - Linear

Three electron pairs - Triangular planar

Four electron pairs - Tetrahedral

Five electron pairs - Triangular based bi-pyramidal (it's like triangular planar, but with one on top and one below, or two triangular pyramids with their base stuck together)

Six electron pairs - Octahedral (Square based bi-pyramidal)

Those geometries include the lone electron pairs. For the molecular geometry, you will then need to exclude the electron pair.
E.g. $H_2O$ : central atom O. O has 4 electron pairs: 2 bonded, 2 lone. The shape, including lone electron pairs is Tetrahedral. Ignore the lone pairs now, you will be left with a bent molecular geometry.

Blue is oxygen;
Yellow is hydrogen;
White/Grey are lone/unbonded electron pairs.
I hope this makes sense. :)


How much does butter shrink when frozen? I just bought a butter dish which is a millimeter or two too small (the problems are that the corners are rounded which they forgot to mention before I ordered it and asked for the dimensions) to fit a piece of butter as it is sold in the supermarket. Since I don't want to return it my idea was to freeze the butter before I put it in this butter dish but for different reasons I can't make a "field test" just now (will update the question when I get the opportunity to do that). Instead I have to fall back to on theoretical answers.

How much does butter shrink when it's cooled from approximately 4°C (39°F) to -18°C (-0.4°F)?

Typically butter has about 1% salt, 82% fat and the rest is water.

Given that fat (consisting mainly of assorted triglycerides of varying saturation) is the major constituent by a wide margin, it's probably reasonable to compare the change in density w/r/t temperature of butter to the change measured in various vegetable oils, since they have roughly comparable chemical composition (and because extensive data on them actually exists). I found a study in which the densities of various vegetable oils were measured over a range of temperatures. On average, the increase in density resulting from cooling from 110.0°C down to 23.9°C was roughly 6-7%. The data points graphing density against temperature arrange very linearly, and the regression lines generated fit them exceedingly well. I see no reason to suppose that the trend wouldn't continue as temperature decreases further. Assuming the comparison is reasonably valid, you could extrapolate that the increase in density you'd observe over the temperature range you specified should be on the order of approximately 2% at best.

As for the amount of shrinkage, strictly considering volume, the ratio of densities (assuming the 2% approximation is reasonable), would be:

$\frac{D_2}{D_1} = 1.02$

That is, the density after cooling, $D_2$, is 2% greater than $D_1$. That implies, holding mass constant:

$\frac{V_1}{1.02} = V_2$

Hence, given that volume is in units of cubic length, you can get a rough idea of how much the length of each side/edge should shrink by taking the cube root of both sides of the equation (obviously, this is only completely accurate for a perfect cube, but it emphasizes that the shrinkage in each dimension is smaller than the shrinkage in volume):

$\sqrt[3]{V_2} \approx 0.993\sqrt[3]{V_1}$

Unless your stick of butter is really gargantuan, a decrease of 0.7% on an edge is not going to account for 1-2mm.

Edit: Additionally, I've read that, in general, the magnitude of relative thermal expansion/contraction that liquids experience is usually greater than that experienced by solids, meaning that liquid density varies more over a given temperature range than solid density. Of course, I'll add the caveat: that's a very broad statement that admits of various exceptions. Nevertheless, this suggests that the estimate of 2% I made above by comparison to data on liquid vegetable oils is, if anything, more likely to be too large than too small.


What exactly causes equilibrium to shift when pressure is decreased? Take the reaction:

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$

If pressure is decreased, equilibrium shifts to the left because it has more moles. I understand the basics of why this is, i.e. due to Le Chatelier's principle, but is there an actual explanation as to why the system does this (e.g. using collision theory)? What's actually happening to cause the shift?

My chemistry teacher says she doesn't know of an explanation.

This happens because the equilibrium constant $K_p$ is independent of the pressure, although it does depend on temperature, which we will assume to be constant. The equilibrium constant can be written as

$$\displaystyle K_p =\frac{p_A^2}{p_Np_H^3}$$

where for simplicity I have abbreviated the chemical names. The partial pressure $p$ can be written in terms of the degree of dissociation and total pressure $P_T$, i.e. $p=aP_T$ where $a$ is a mole fraction term in the degree of dissociation. As partial pressure is proportional to the degree of dissociation and total pressure, for example, $p_A =a_A^2  P_T^2$ then the equilibrium constant

$$K_p=\frac{a_A^2}{a_Na_H^3} \frac{1}{P_T^2}$$

where the terms in $a$ we have not yet evaluated and contain the degree of dissociation $\alpha$. However, you can see that as $K_p$ is constant if $P_T$ decreases then $1/P_T^2$ increases and so  $\displaystyle \frac{a_A^2}{a_Na_H^3} $ must decrease. This means that the degree of dissociation decreases and so the equilibrium 'moves to the left'.

Working out the partial pressure for this reaction is a bit messy.

Let $\alpha $ be the degree of dissociation then $(1-\alpha)$ of nitrogen reacts , $3(1-\alpha)$ hydrogen and $2\alpha$ ammonia is produced. The total of these is $6-2\alpha$ and so the mole fraction of ammonia is, for example, $a_A=\displaystyle \frac{2\alpha}{6-2\alpha}$ and the partial pressure $p_A=\displaystyle \frac{2\alpha}{6-2\alpha}P_T$. You can make a similar expression for the other species and substitute to get the mole fraction term $\displaystyle \frac{a_A^2}{a_Na_H^3}=\frac{16\alpha^2(\alpha-3)^2}{27(\alpha-1)^4}$. This expression increases as $\alpha$ increases. As you see the expression is messy but the main point is that as the pressure decreases the mole fraction term must decrease which it does by reducing $\alpha$.


Why do the left- and right-handed enantiomers of a chiral molecule react differently? So I was researching about thalidomide when a thought struck to me. Why does the left enantiomer react differently than the right? They have the same elements and structure.

With other chiral molecules, the transition states are necessarily diastereomeric. Those transition states thus have different energies, which implies that rates will be different. Also, the products with other chiral reactants are diastereomeric, so equilibrium mixtures will be different.


What are the differences between monotropic and enantiotropic polymorphic transformations? I'm studying polymorphism and came across the terms "enantiotropic" and "monotropic" transformations. The first is described as reversible, whereas the latter, irreversible due to metastability. But what does "reversible" and "irreversible" means practically? What exactly is metastability? What makes these two types of polymorphism distinguishable between each other?

For instance, it's used as an example of monotropic polymorphism the transformation of graphite to diamond. However, the same source claims the reverse operation is possible. As for enantiotropic polymorphism, it used the example of silica polymorphs (cristobalite and alpha-quartz) which also can be transformed from one another. If both are reversible, why are they different?

In enantiotropic there exist an transition temperature between two crystalline state in equilibrium but reverese is true for monotropic


Bond rotation energy comparison of aromatic compounds Which of the above compounds has a lower bond rotation energy?

Here is my progress: We basically need to find which compound has more single bond character. Now both compounds will break the double bond, and give the electron pair so as to make both cycles aromatic. But what do I do after that?

Cleavage of the double bond will lead to both compounds having two rings that are aromatic, however the amount of resonance energy(hence stability) will not be the same.

Both systems will have a Cyclopentadienyl anion, but one will have a Cyclopropenium cation
while the other has a Tropylium cation . As the Tropylium cation has more resonance structures, it will have a higher resonance energy and hence greater stability than the Cyclopropenium cation. So (b) has a lower bond rotation energy.


What does a molecules color have to do with its bond/orbital energies? For example, elemental iodine is deep violet. Its sigma bond or perhaps the lone pairs are capable of absorbing all visible light frequencies except violet which is why we see it as that color.
Lithium, on the other hand, is red which means its valence electrons must be reflecting red frequencies.
So, does that mean the valence/bonding electrons of $\ce{I2}$ are higher energy than the lithium valence electrons?
In other words, is there a relationship between an orbitals energy and the type of light it can reflect?

You could perhaps compare $\ce{I2, Sn^{II}I2}$, and $\ce{Sn^{IV}I4}$, and which are violet, yellow, orange (in decreasing order of visible light energy).

Perhaps it has to do with symmetry? $\ce{I2}$ is $D_{\infty\mathrm{h}}$, stannous iodide is $C_{2\mathrm{v}}$, and stannic iodide is $T_\mathrm{d}$.

Rather than give answers to your specific question I try to explain what happens in general as there is some confusion in your question. You should really consult a book on spectroscopy, of which there are many, for a thorough explanation.

In a molecule the arrangement of electron in orbitals is called a state, hence ground state, first exited state etc. When all the electrons a molecule has fill up all the orbitals with no gaps left, this is the ground state. They will fill up half of all the orbitals, two to an orbital.  When a photon is absorbed and it has sufficient energy (usually in the visible or UV part of the spectrum) an electron is promoted from one of the bonding orbital to a anti-bonding orbital which has higher energy. This arrangement of electrons is called an excited state. This transition between states is what gives molecules and atoms their colour.

In iodine the difference between the lowest excited state and the ground state occurs in the visible part of the spectrum. (for simplicity, I'm ignoring any vibrational transitions accompanying the electronic excitation). Benzene has a larger energy gap between ground and excited state than I$_2$ and so absorbs only in the UV and looks transparent to us.

The same basic idea applies to all atoms and molecules. Where the molecules or atoms absorb is a function of the exact electronic energy difference between ground state and excited state of the particular species. An understanding of the nature of the bonding and so energetics allows us to predict trends in absorption spectra, but exact values can only be obtained from quantum calculations.


Total magnetic moment of atom Whenever I read about coordination compounds in my textbooks, I always find a discussion about spin-only magnetic moment which is given by $\sqrt{n(n+2)}\cdot\mu_\mathrm{B}$, where $n$ is the number of unpaired electrons and $\mu_\mathrm{B}$ the Bohr magneton.

I'd like to know how to calculate the total magnetic moment of a given atom (spin only + orbital). For example let's take $\ce{Ni^2+}$. The spin only magnetic moment comes out to be $\sqrt{8}$ BM. Now how do I add the orbital magnetic moment? I've seen a bunch of expressions for orbital magnetic moment in terms on quantum numbers $n$ and $l$, but I am unable to understand which one to add.

Firstly I recommend you to understand the concept of "Magnetic Susceptibility". My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition.

Consider Ni2+, its electronic structure is 4s03d8. This is a d8 ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals.

Also you should know gyromagnetic ratio which is the ratio of magnetic dipole moment to angular momentum of the system. (We got a system where quantum mechanics may be obeyed).

So, considering orbitals and electrons that fill these orbitals we got two magnetic moment here. Spin magnetic moment(sum of it is S) and angular magnetic moment(sum of it is L).
I don't know properly write all calculations with codes. So I attached an image.


"Gasoline" in Other Languages I speak Russian and the term used for gasoline is "бензин", pronounced like benzene. I've also noticed that in Italian, it can be called "benzina". What's the significance behind this, if any? I think gasoline contains benzene, but that seems like a very minor part of the overall composition.

I can't speak about the etymology of the word, since I don't speak any of the languages in question (plus, it it is purely concerned with that, I believe this question is off topic here).

However, if you are interested in the contents of gasoline, I can tell you a little something about them.

Gasoline (or petrol if you're not from North America) is typically a mixture of hydrocarbons with 4 to 12 carbon atoms per molecule. It is a comprises of alkanes, cycloalkanes, alkenes, or in industry parlance parrafin, naphthenes and olefins.

Moreover, as you suspected, it does contain aromatics in general and benzene in particular, and most countries set limits on their content in the fuel. For example, in the EU the benzene content should be under $1\%$ (v/v) for automative grade gasoline$^{[1]}$ Benzene, Toluene, and Xylene (BTX) are important petrochemicals, and valuable chemical feedstocks, and most are often recovered during catalytic reforming.

EU: Fuels: Diesel and Gasoline


For which pair of species is the difference in radii the greatest? For which pair of species is the difference in radii the
  greatest?

  (A) $\ce{Li}$ and $\ce{F}$

  (B) $\ce{Li+}$ and $\ce{F^-}$

  (C) $\ce{Li+}$ and $\ce{O^2-}$

  (D) $\ce{O^2-}$ and $\ce{F^-}$

I am pretty sure it is not (D) as $\ce{O}$ and $\ce{F}$ are quite close to each other on the periodic table and should have atomic radii pretty much the same.

I also know that gaining an electron increases the atomic radius whereas losing one decreases atomic radius. Also, as you move to the right of the periodic table atomic radius decreases.

I think that the answer could be (A) because $\ce{Li}$ is larger than $\ce{F}$, and the other choices result in $\ce{Li}$ becoming smaller and $\ce{F}$ becoming larger which decreases the difference in radii, but this is not the correct answer.

The answer given was C. How can this be explained?

Your argument that $r_\ce{Li} > r_\ce{F}$ is valid only for the neutral atoms.Consider the lithium ion. Its electronic configuration is exactly similar to helium but the nuclear charge is greater by one unit, and hence is even smaller than the helium atom. On gaining an electron, fluorine now has the electronic configuration of neon with lesser nuclear charge and hence is larger than a neon atom. Similar, oxide ion, with even lesser nuclear charge is the largest. The decrease in radius of lithium on losing one electron (reduction by one entire shell), more than compensates for the difference in the radii of neutral lithium and fluorine. Finally, the order appears to be:-

$$\ce{Li+ < He < F < Li < F- < O^{2-} }$$


Confusion in naming a carboxylic acid with a cyclohexyl group For the below compound  .

According to me the answer should be 1-carboxymethylcyclohexane.
But the answer given as 2-cyclohexylethan-1-oic acid
Why the name given by me is wrong .
Whereas the name given to

Is cyclohexane-1-carboxylic acid
Why it is not 1-cyclohexylmethan-1-oic acid

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the senior parent structure has the maximum number of substituents corresponding to the principal characteristic group.

  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).

  (…)

If there is still a choice (for example, if the ring and the chain each have a substituent corresponding to the principal characteristic group), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

The compound given in the question contains a characteristic group $(\ce{-COOH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-COOH}$ substituent corresponds to the principal characteristic group.

Since the principal characteristic group is a substituent of the chain, and the ring does not have any substituents corresponding to the principal characteristic group, the senior parent structure of the compound given in the question is the chain and not the ring. The rules P-44.1.2.2 and P-52.2.8 for the selection between a ring and a chain as senior parent structure are not relevant in this case.

Simple carboxylic acids are named substitutively using the suffix ‘oic acid’ as follows.

  P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

Therefore, the systematic name for the unsubstituted carboxylic acid structure that is given in the question is ethanoic acid (not ‘ethan-1-oic acid’ since no locants are used to denote the positions of the carboxylic acid groups in a hydrocarbon chain). However, the preferred IUPAC name (PIN) of ethanoic acid is the retained name acetic acid according to rule P-65.1.1.1.

The acetic acid chain is substituted using the usual principles of substitutive nomenclature.

However, according to rule P-14.3.4.6, locants are omitted where there is only one kind of substitutable hydrogen.

  P-14.3.4.6 All locants are omitted for parent compounds when all substitutable hydrogen atoms have the same locant.

Therefore, the preferred IUPAC name name for the complete compound given in the question is cyclohexylacetic acid since the locant ‘2’ of ‘2-cyclohexylacetic acid’ is omitted.


How to obtain barium hydroxide I am trying to perform the demonstration of a strongly endothermic reaction reacting barium hydroxide with ammonium nitrate.  I have been scouring the web for ways to get barium hydroxide, but none have showed any promise.  The only way I have seen to obtain it is by heating the nitrate to decomposition and reacting the oxide with water.  I would rather avoid this as I have no way to deal with the nitrogen dioxide gas that would result.

You can buy barium hydroxide. See the listing from Flinn Scientific although Flinn is by no means the only seller. However, Flinn's safety information is written at a very accessible level.


Formic Acid vs Formica - Which came first? Source: Google(Oxford)

So, we know that 'Formic Acid' was derived in formic acid in the late 18th century from the Latin word 'formica'. From Wikipedia:

  Formica is a genus of ants of the family Formicidae, commonly known as wood ants, mound ants, thatching ants, and field ants. Formica is the type genus of the Formicidae, and of the subfamily Formicinae.

I think I've read in some junior classes that the biological name 'Formica' was given to ant because of the secretion of 'Formic Acid' by it which is the cause of the irritation at the stung area.

So, I want to know - Is formic acid derived from 'Formica' (ants) or are ants' (biological name) derived from the presence of 'formic acid' in them?

The name 'formic acid' has been derived from the name 'formica' as it is derived literally from ants. Wikipedia clearly states:

  The word "formic" comes from the Latin word for ant, formica,
  referring to its early isolation by the distillation of ant bodies.

If you type 'formic acid etymology' in Google, it will give two links that proves the above statement:

  1791 (in formic acid), literally "from ants," coined from Latin
  formica "ant". The acid first was obtained in a fairly pure form in 1749 by German chemist Andreas Sigismund Marggraf (1709-1782), who
  prepared it by distilling red ants.(etymoline)

  As early as the fifteenth century, some alchemists and naturalists
  were aware that ant hills gave off an acidic vapor. In 1671, English
  naturalist John Ray became the first person to describe the isolation
  of formic acid by the distillation of large numbers of ants. These
  insects secrete the substance for attack and defense purposes. Thus
  the name "formic acid" was coined from the Latin word for ant,
  formica.(newworldencyclopedia)


Spontaneity of an exothermic chemical reaction When the process is exothermic (ΔHsystem<0), and the entropy of the system increases (ΔSsystem>0), the sign of ΔGsystem is negative at all temperatures. Thus, the process is always spontaneous.

It is easy to prove this using the formula of Gibbs free energy, but how can we prove it by logical reasoning?

I was thinking that in an exothermic process, the system loses heat. Shouldn't "the loss in heat" normally cause more ordering of the system and eventually a decrease in entropy? Might be a stupid question, but what am I missing here?

Not all reaction is like this but, exothermic reactions are usually spontaneous because molecules always want to be more stable. So by reacting to another molecule and releasing energy, they can lower their own energy thus making them more stable and unreactive.


Why are lactate and lactic acid used synonymously in biochemistry? In science there's always a misuse of the two terms. I know that lactate is the conjugate base of lactic acid.

Why are they used like is the same thing? For exemple: in lactic acid bacteria (LAB) fermentation it's often said that anaerobic glycolysis produces lactic acid, but if you actually look at the reaction it produces lactate.

Is this relevant? Because lactate is a negative compound while lactic acid is a neutral one.

If you were to produce any lactic acid in the body, it would pretty much immediately get deprotonated, so by right it's always going to be lactate. The $\mathrm pK_\mathrm{a}$ of a typical carboxylic acid is at most 5, and at physiological pH levels (~ 7, give or take) the ratio of $\ce{HA}:\ce{A-}$ is on the order of $1:100$. (As getafix mentioned, lactic acid is somewhat more acidic than usual, because of the electron-withdrawing hydroxyl group, so the ratio would be expected to be even larger. But these are just meant to be very rough numbers, for illustration.)

You could say that it's sloppiness on the part of the biologists, but past a certain point you don't really care whether it's $\ce{HA}$ or $\ce{A-}$, and that's why the distinction isn't really enforced. From a biochemical perspective the more important thing is the oxidation state as well as other byproducts (ATP, carbon dioxide, etc.) Likewise the citric acid cycle is called that despite citrate being the actual compound in the cycle.

And perhaps the most egregious case of sloppiness: DNA stands for deoxyribonucleic acid, but in the body the phosphate groups are deprotonated, so chemically speaking it really should be called the conjugate base of DNA: "deoxyribonucleate". But again, nobody really cares - as long as we know that it has a negative charge, we can call it by a strictly incorrect name and it's fine.

There are probably tons of physicists out there who hate us chemists for using non-rigorous terminology, too!


Why decarbonate soda for spectroscopy experiment? I did an experiment with a nanodrop photometer where I had to find the amount of caffeine and benzoic acid in soda. During the experiment, we were to decarbonate the soda by heating it. I've done a similar experiment before where I had to add salt to decarbonate soda. I've also read similar procedures where soda was decarbonated by shaking.

My question is this: Why should the soda be decarbonated?

My thought is perhaps the air pockets in the soda (due to $\ce{CO2}$) could cause discrepancies in absorbance due to changes in the refractive index.

I also thought that perhaps $\ce{CO2}$ absorbs in the same region as caffeine and/or benzoic acid, but NIST does not appear to have UV/Visible spectrum data for $\ce{CO2}$, whereas it UV/Visible spectrum data for caffeine. Please pardon my naiveté here, but I'm then assuming $\ce{CO2}$ thus does not absorb in UV/Vis?

The primary reason is that when soda warms up, the $\ce{CO2}$ would escape slowly as bubbles. These bubbles would significantly scatter light affecting the absorption spectra. Usually, the effect of such scattering is quite pronounced and the absorption bands of individual molecules will be (might be) buried below the broadened feature in the lower wavelength region.

In colloidal and suspension systems light is scattered by individual particles. These particles are larger in size than the wavelength of light and such scattering phenomena is termed as Tyndal scattering or Tyndall effect.  For such a process, the scattering coefficient has $wavenumber^{4}$ dependence which means that for lower wavelengths the scattering increases significantly. Thus when the sample has particles then absorption spectra is skewed and we see an increasing absorption peak when going to a lower wavelength. (You can yourself try this with milk which is a suspension).

About the absorption spectrum of $\ce{CO2}$:

$\ce{CO2}$ being a simple molecule is expected to have electronic absorption bands below 300nm. It does have electronic absorption between 200 to 270 nm. It is more difficult to find on the internet since usual absorption of $\ce{CO2}$ is referred to the vibrational absorption observed in the in the infrared region. Hence, 'UV absorption of $\ce{CO2}$' is a better search term.

For your reference: T. Joutsenoja, A. D'Anna, A. D'Alessio, and M. I. Nazzaro, "Ultraviolet Absorption Spectra of Carbon Dioxide and Oxygen at Elevated Temperatures," Appl. Spectrosc. 55, 130-135 (2001)


Reaction of diacetyl with acid What happens when diacetyl, that is, butan-2,3-dione is treated with acid?

I guess there would be acid catalysed aldol formation.
But what would the final product be? I can see too many possibilities. Please help me out here.

Thanks!

AKM: In your response to MM the self-condensation of diacetyl 1 occurs stepwise and not with the double enol of diacetyl 1 (biacetyl; butan-2,3-dione). Although your question relates to the acid-catalyzed condensation of diacetyl, historical condensations of diacetyl were conducted under alkaline conditions. The products formed from the base or acid-catalyzed are numerous and dependent on reaction conditions.
Quinone 2 was prepared under basic conditions by von Pechmann.1 Diels and coworkers2 isolated from an alkaline reaction a dimer, C8H12O4, which was assigned lactonic structure 3 with no spectroscopic evidence.3 The structures are color-coded to display the four carbon moiety of diacetyl. Clearly, lactone 3 shows a branched red carbon framework. An undefined "benzylic acid rearrangement" was suggested without details. The structure was later revised to diketone 4 based upon spectroscopic evidence and a more reasonable normal carbon chain.4
Trimer 5, C12H18O6, has also been isolated under alkaline conditions and the stereochemistry confirmed by NMR and IR.5
An acid-catalyzed condensation of diacetyl in dilute aqueous acetic acid (70oC, 13 days) leads to C10 structure 10.6 The structure was confirmed spectroscopically and by periodate cleavage with loss of acetic acid with concommitant formation of diketone 11. Structure 10 does not form a hemiketal as does structure 5 owing to the resonance stabilizing effect of the α, β-unsaturated ketone. The mechanism proposed is described in the sequence of steps 4 --> 10.
As can be seen from the examples provided --- and that's not all of them --- the self-condensation of diacetyl is not straightforward.

1) Pechmann, H. v., Berichte, 1888, 21, 1417; Otte, R. and Pechmann, H. v., 1889, 22, 2115.
2) Diels, O., Blanchard, W. M., Heyden, H. v. d., Berichte, 1914, 47, 2355.
3) Machell, G., J. Chem. Soc., 1960, 683.
4) Birch, A. J. and Moye, C. J., J. Chem. Soc., 1957, 412.
5) Cresswell, R. M., Smith, W. R. D. and Wood, H. C. S., J. Chem. Soc., 1961, 4882.
6) Shapiro, R., Hachman, J., Wahl, R., J. Org. Chem., 1966, 31, 2710.


Equilibrium Concentrations of Products/Reactants Does increasing the concentration of reactants mean the equilibrium concentration of the products will be greater (than the concentration of the reactants?)

Adding a reactant increases the concentration of products at equilibrium, but it doesn't necessarily mean that the product concentrations will be larger than the reactant concentrations at equilibrium.

Suppose you have a reaction $\ce{A <=> B}$. Then
$$K = \frac{[\ce{B}]}{[\ce{A}]}$$
where $[\ce{A}]$ and $[\ce{B}]$ are the concentrations at equilibrium and $K$ is a constant. The ratio $[\ce{B}]/[\ce{A}]$ is constant at equilibrium, so if you add some $\ce{A}$ to the system, some $\ce{A}$ must react and become $\ce{B}$ to maintain the ratio. The equilibrium concentration of the products will increase.

But that increase in product concentration won't necessarily mean that the equilibrium product concentration becomes higher than that for the reactants. If $K$ is bigger than one, the equilibrium concentration of the products will be greater than that for the reactants; if it's less than one, the equilibrium concentration of the reactants is bigger.


Raman scattering and "electron and proton center" of molecule I am currently studying the textbook Infrared and Raman Spectroscopy, 2nd edition, by Peter Larkin. In a section entitled The Raman Scattering Process, the author says the following:

Light scattering phenomena may be classically described in terms of EM radiation produced by oscillating dipoles induced in the molecule by the EM fields of the incident radiation. The light-scattered photons include mostly the dominant Rayleigh along with the very minor amount of Raman scattered light. The induced dipole moment occurs as a result of the molecular polarizability $\alpha$ where the polarizability is the deformability of the electron cloud about the molecule by an external electric field. Fig. 2.9 shows the response of a nonpolar diatomic placed in an oscillating electric field.

Here we represent the static electric field by the plates of a charged capacitor. The negatively charged plate attracts the nuclei, while the positively charged plate attracts the least tightly bound outer electrons resulting in an induced dipole moment. This induced dipole moment is an off-resonance interaction mediated by an oscillating electric field.

Can someone please better explain what the "electron and proton center" in figure 2.9 is supposed to be, and how it relates to Raman scattering? Is the "electron and proton center" of the diatomic molecule only at that single point that is pointed to by the arrow in figure 2.9? If so, then why? Or is each of the three points with the identical symbols an "electron and proton center"? Again, if so, then why?
I would greatly appreciate it if someone would please take the time to clarify this.
EDIT:
What does it even mean for a diatomic molecule to be "homogeneous"? I understand the concept of "homogeneous" mixtures, but not a "homogeneous" molecule.

I guess for homogeneous read homo-nuclear. He is trying to explain how the electric field of the radiation induces a dipole by shoving the electrons around, hence reference to polarisability: this relates to the standard description. This induced oscillating field then radiates, hence scattering. The electron and proton centre is when the field is zero, i.e. whatever that distribution is in the molecule in no external field


Reaction of alkyne with sodium in ethanol What is the product obtained on the reaction of but-2-yne with $\ce{Na}$ in ethanol?

I know that $\ce{Na}$ in liquid $\ce{NH3}$ converts an alkyne to an alkene by anti addition, and $\ce{Na}$ in ethanol is the principle reagent in Bouveault Blanc Reduction, but what about this reaction.

The answer given in my book is a trans-alkene. Is this a standard reaction? Also does it depend if it is a terminal alkyne or not? (I guess yes, because the terminal alkyne will try to deprotonate)

the reduction of alkynes using Na, proceeds through the anion radical, which can obtain trans conformation for the stability. Now considering the usage of NH3 and ethanol are proton donors. So, there is no effect on trans conformation altering the proton donors.


Do molecules change shape when they become vibrationally excited? In quantum mechanics we learn the the classical picture of a vibrational excitation is wrong. The atoms will 'vibrate' only while the molecule is in a coherent superposition of two vibrational states, and will find itself in a stationary state once this coherence is broken.

Let's say I have a carbon monoxide in vacuum, and I excite it from v = 0 to v = 1. In order to describe the electron distribution of the molecule, I would expect the electron distribution to be given by the sum of the electron distributions for the CO at each bond distance, weighted by the nuclear wave function.

So... An excited CO in vacuum would look similar to a stationary molecule of C-o-o, where the little o's indicate that the electron density is smaller than that of a regular oxygen. Is this interpretation correct?

Yes they change 'shape'. A diatomic only  in the sense that the bond extends and contracts, but for  a triatomic or polyatomic molecule there are bond stretches and bends. The number of different ways these vibrations change the 'static' shape of a molecule, such as may be measured by x-ray diffraction, depends on the symmetry and this is described by the point group. Only certain combinations of bond stretches and bends are allowed and these are called 'normal modes'.

Some other points.

A molecule will vibrate even when it has zero point energy, i.e. when $v=0$; by the Heisenberg principle this must happen. If it were stationary then it would have infinite energy because its position is determined exactly. The electron density, if it were to be measured would thus be an average of that at the zero point energy. Calculations of this made at zero potential energy would thus be slightly different. Slightly because the bond extension is only a few percent of the bond length at this energy, and also most nuclear wavefunction density is located at the minimum energy.

We assume that the Born-Oppenheimer condition applies, briefly that electrons being far lighter than nuclei instantaneously adjust to the slow nuclear motion as a molecule vibrates. This allows us to draw the potential energy profile, for example a Harmonic or Morse potential. When we draw the familiar wavefunctions for vibrational motion these are the nuclear wavefunctions at each energy and are the stationary solutions to the Schroedinger equation and we interpret these as meaning that the molecule has a probability of being at each position given by the square of the wavefunction.

In a long time spectroscopic experiment (e.g. using a spectrophotometer) only the spectrum is measured, i.e. energy gaps between levels.

In experiments using a femtosecond laser pulse$^*$ it is possible to excite several vibrational wavefunctions almost instantaneously, i.e. in a time shorter than a vibrational period, in this case a 'wavepacket'$^{**}$ is produced and this 'particle' moves left and right inside the potential. This motion has been observed by using a second femtosecond laser pulse as a probe. As the energies of the vibrational levels are not all the same the wavepacket, which starts off in phase, will become out of phase then 'in phase' again and so goes on repeatedly recurring until it is disturbed, say by a collision with another molecule. (In these time resolved experiments Fourier transforming the data gives the spectrum.)

$^*$ As a femtosecond pulse is short it has a wide spectral width and can therefore excite several energy levels.

$^{**}$ A wavepacket is the direct summation of the vibrational wavefunctions multiplied by a time dependent term related to the vibrational frequencies. Thus the wavepacket evolves in time. The wavepacket recurs at regular time intervals due to the time dependent terms coming into phase just as they were at time zero. This is a property of a collection of oscillators; children on swings would show the same effect.


How can a Zn-Cu galvanic cell have a voltage of 1v in stead of 1.1v Let s say a galvanic cell based on the reaction of zinc and copper half-cells. How is it possible to get the voltage at 1V in stead of 1,1V?

My guess: maybe by modifying the salt bridge in some way as this influences the charges in some way, but I am really not sure.

You can get a smaller potential if you alter the concentration of one or both ion solutions away from ideal 1M concentration. This behavior is described by the Nernst equation:

$$E_{cell}=E_{cell}^{o} - \frac{RT}{zF}lnQ$$

where $E_{cell}$ is the actual potential in the cell, $E_{cell}^o$ is the potential of the cell under ideal conditions, R is the ideal gas constant, T is the temperature, F is Faraday's constant, z is the number of electrons in the reduction oxidation reaction and Q is the reaction quotient. By altering Q to be greater than 1, the actual voltage of the cell dips below the ideal value.


Carbohydrates: What does "typical nature" mean here? I am a 12th grader in India. I have understood basic theory on biomolecules. I know about carbohydates (and structures of important saccharides) and $\alpha$-amino acids, proteins, and peptides.

This is a section given in my book:

I understand what is the meaning of "D"-glucose, and also how to identify the functional group. But, notice the last column in this table. I don't understand what they mean by the "typical nature" in each case.

I searched the net but couldn't find anything. Is my book using obsolete/unknown terminology for an existing concept? What is it? Please tell. Thank you!

References: Page 1, Page 2, Page 3

Looks like your book's come up with a "localized" version of the D-L notation.

$\ce{D-L}$ notation in brief: A $\ce{D}$-sugar has the $\ce{-OH}$ group (on the stereogenic carbon atom on the parent chain, furthest away from the aldehyde group) facing right in the Fischer representation.
Likewise an $\ce{L}$-sugar has the $\ce{-OH}$ group facing left.

What your book's attempting to do is extend the $\ce{D-L}$ classification to individual carbon atoms in the parent chain.

The "3rd($\ce{L}$)" bit is supposed to mean, "3rd carbon atom has an $\ce{L}$-type configuration".

The "2,3($\ce{L}$)" bit is supposed to mean, "2nd and 3rd carbon atoms have an $\ce{L}$-type configuration".

You can figure out the rest ;-)

This appears to be some sort of memory aid to help you remember the configuration of the various kinds of 6-carbon sugars. The "typical nature" term was, in all likelihood, concocted by the authors (perhaps they thought it sounds better than "mnemonic"?)


Oxygen and sulfur bonding Why does oxygen form a double bond in $\ce{O2}$ but sulfur, also from group 16, forms single bonds in $\ce{S8}$?

This is essentially the same question as to why carbon concatenates through up to four bonds with binding energies low enough to be broken again yet stable enough to last given typical terrestrial environmental conditions. Great for entities like us.
On the other hand, the binding energies of Silicon, Germanium, from the same group IV, for instance, are higher. The binding energies are directly related to the bond length/bond distance, which is great for determining the energies through crystallography, along with proper model fitting.
Sulfur is one of few elements amenable to catenation, - forming long molecules with each other.
So whichever configuration releases the greatest enthalpy is typically the most stable one, but depends upon the initial energies thus giving rise to allotropes - elemental structures of different molecular configurations. The bound system typically has lower potential energy.
If a double bond lowers the overall energy of the bound system more it will typically form over a single bonded one.
Update: I found this (see also this table of dissociation energies):
Update 2: I'd like to add that from a quantum mechanical perspective, which is a viewpoint much closer to nature, the bonds arise from 3D space where electron densities are highest - which means most energetically favorable.
Since for whatever weird reason, nature is really quantized, and I cannot stress that "really" part enough, one gets standing waves for the electron equations around a given nucleus - rather than a continuum - as is the case of a sphere, like say a falling waterdrop to add a few exterior forces into the mix as well. The classical and the quantum object have only in common the "energetically favorable" state.
Somehwat simplified then, those standing waves become "orbitals" when doing the calculations for 3D space. The orbitals change in shape with increasing interference with other electrons and higher potential energies due to larger nuclei , meaning higher positive proton counts.
The energies involved result in very different packing i.e. different lattices. A necessity for life is breaking bonds or "bond accessibility". There simply isn't "enough energy" in today's protein-scaffold based catalyists on earth, to use anything else than the constituents of the periodic table with few protons in case of homogenous bonds (i.e. atoms of the same species like say Oxygen O2).
It does not have to be this way. We from earth are biased by our own biochemistry we observe. Every human generation breaks down the biases of the previous ones and encounters new ones.
Bond energies

1 S8: 1808 kJ/mol
1 O8: 1168 kJ/mol
4 S2: 1408 kJ/mol
4 O2: 1976 kJ/mol


How to measure the amount of calcium in a supplement? In chemistry lab, my group was assigned to determine the actual amount of calcium in a supplement.  We did this successfully using HCl to dissolve the tablet, vacuum filtration, and then the addition of sodium carbonate.  Next week, it has been suggested that we attempt it again using a different method.  Thus, we are seeking the most interesting way, albeit using elementary tools and chemicals, that one might do this.

For the determination of calcium (at least) two methods are conceivable:

gravimetric analysis
Sulfate is a rather bad choice here. As compared to the carbonate, the sulfate has a much higher solubility. A better choice would be the oxalate with a solubility around $6\,\mathrm{mg\,l^{-1}}$.
photometry
Calcium forms a purple complex with the Arsenazo III dye that can be detected at wavelengths of 660 and 700 nm.
Alternatively, you can use o-cresolphthalein and detect the complex at 545 nm.
The photometric methods are typically used in automated high-throughput clinical analysis of calcium in serum, but given that a photometer isn't available in the students' lab, Karl's comment on EDTA is exactly what you might be looking for. Complexometric titration is the keyword you might want to look for. Here, an indicator that forms a coloured complex with $\ce{Ca}$ is replaced by EDTA. The liberated indicator then shows a different colour.


TKN: Why are they called Missouri tablets? When analyzing for Total Kjeldahl Nitrogen (TKN), you can make a copper sulfate digestion solution in the lab or purchase a variety of pre-prepared digesting tablets for example on VWR. My question is, why are some formulations called Missouri tablets?
As far as I can tell, the state of Missouri does not specifically require this formulation, and I haven't seen any digestion tablets named after any other states. Were they perhaps invented in Missouri?

As suggested in the comments, the predecessor of the method that uses "Missouri" tablets was developed at the University of Missouri (Columbia, Missouri, USA). I believe the first report was from 1972 and named the method as the "Missouri Automated Nitrogen Method (MANM)". They also refer to it as the "Missouri Technicon" method because it uses a Technicon AutoAnalyzer instrument. Many variations of this method have been described, and the specific formulation of Missouri tablets is for one of them.

An automated spectrophotometric method, utilizing Technicon AutoAnalyzer modules, has been developed to determine total nitrogen in fertilizers containing only ammoniacal, nitrate, and urea nitrogen. This colorimetric system employs the Berthelot ammoniaphenate-hypochlorite reaction. A homogeneous chromous/titanous reduction system for the automatic reduction of nitrates was interfaced with the digestion unit and color development manifold, resulting in a total nitrogen system for fertilizers with an effective analysis rate of $\pu{20-25 samples/hr}$. A new sample retrieval system removes digested samples from the helix. The totally automated system was optimized to obtain the highest sensitivity for nitrogen. Data were compared for $\pu{458 samples}$ analyzed by the automated method and the comprehensive nitrogen method (CNM), 2.053–2.054. The average difference between the 2 methods (MANM – CNM) was +0.04% nitrogen. The average relative difference was 1.12%. The average value for $\pu{140 samples}$ of $\ce{KNO3}$ primary standard analyzed by the MANM was 13.84% nitrogen (theoretical content 13.85%). The results of these standards ranged from 13.58 to 14.14% nitrogen and the standard deviation was 0.11%. The relative standard deviation was 0.80%. Of fertilizer samples received for analysis, less than 1% contained insoluble organic nitrogenous material which was unsuitable for analysis by the MANM. These few samples were not included in the statistical evaluation. The MANM is rapid, accurate, and precise and is a convenient and dependable automated method for the determination of total nitrogen in fertilizers containing only ammoniacal, nitrate, and urea nitrogen.
 Source: Charles W. Gehrke, John S. Killingley, Larry L. Wall, Sr. (1972) Missouri Automated Nitrogen Method (MANM) for Fertilizers. Journal of Association of Official Analytical Chemists 55:467-480. https://doi.org/10.1093/jaoac/55.3.467

A rapid automated method for total protein nitrogen has been developed, using the Technicon AutoAnalyzer with block digestor. Ammonia is determined by the automated ammoniasalicylate reaction at a rate of $\pu{40 samples/hr}$. A number of variations in digestion parameters have been evaluated. Hydrogen peroxide was used as a digestion accelerator. A salt-acid ratio of 1:1 and a block temperature of $\pu{425 ^\circ C}$ were chosen. The catalysts evaluated included $\ce{HgO, CuSO4, SeO2,}$ and $\ce{TiO2/CuSO4}$. Maximum nitrogen recovery in the shortest time (30 mill) was achieved with HgO as the catalyst. The results from multiple analyses of 10 experimental samples of different refractoriness with the block digestor method, using $\ce{HgO}$ or $\ce{CuSO4}$ as a catalyst, compared well with the results by the official AOAC Kjeldahl method. The accuracy, precision, economy, and saving of space offered by the Missouri-Technicon block digestor method make this an attractive alternative to classical Kjeldahl analysis for large numbers of total protein determinations in samples of different refractoriness. The AutoAnalyzer cartridge manifold is simple and reproducible in its performance, and a large dilution of the sample is made on dialysis which eliminates the matrix effects from the sample digests. At least $\pu{250 samples}$ can be analyzed in $\pu{9 hr}$ with 1 instrumentation setup.
Source: Larry L. Wall, Sr., Charles W. Gehrke (1975) An Automated Total Protein Nitrogen Method. Journal of Association of Official Analytical Chemists 58:1221-1226. https://doi.org/10.1093/jaoac/58.6.1221


If dark surfaces absorb/store more light/energy than light surfaces, then does dark skin, on average, store energy from light most efficiently? If darker surfaces absorb/store more light/energy per square inch than light surfaces, then does dark skin, on average, store energy more efficiently than light skin? (edited).

Melanin is a chemical compound(s) found in the skin and is responsible for skin pigmentation.  The darker the skin tone, the more melanin is present.  In light-toned-skin, melanin is produced as part of the tanning process when the skin is exposed to UV radiation.  Melanin is a very effective filter for UV radiation.  The more melanin present in the skin, the less light is available for light-catalyzed chemical processes that occur in the skin.

As Dissenter noted in his comment, light energy is not stored per se in the human body.  Rather light catalyzes certain reactions in the skin and the products of these reactions can be stored (at least for a while) for use in the body.

Vitamin D production is an example of an extremely important light-mediated process that begins in the skin.  Vitamin D is thought to play a role in a large number of regulatory processes throughout the body ranging from bone health to cancer.  Because melanin is such an effective filter of sunlight, non-white populations are at an increased risk of vitamin D deficiency.


Aldol condensation in acidic medium Attempt:

I did aldol condensation in acidic medium followed by $\ce{E1cB}$ to get 1,3-diphenylbut-2-enone but the answer given is:

I am unable to devise a mechanism to reach the above product from the reactant. Could someone please suggest the mechanism?

I have drawn a three step mechanism as opposed to the concerted step reaction proposed in the other answer.

This mechanism has the added benefit of not leaving any room for formation of any side products as there is only one source carbon of acidic hydrogens in each compound.


Why don't electrons pair up in an oxygen radical? In an oxygen radical (produced by the break down of O3 for instance), there are 2 unpaired electrons in the outermost subshell which make the molecule unstable. Why don't these electrons pair up which would make the atom more stable (I think)?

Pairing up of electrons in this case is unfavourable due to increased inter-electron repulsion; the average spatial separation of the electrons is higher in the triplet state where the two electrons are unpaired. Regular oxygen molecules also exist in a triplet state and have unpaired valence electrons; they are said to be biradicals.


Confirming the presence of 4-bromoacetanilide during bromination of acetanilide In my lab, acetanilide was added to bromine in aqueous acetic acid, and discolouration of the reaction mixture was obtained, indicating that bromine was consumed.

After some discussion, it was concluded that an electrophilic aromatic substitution might be taking place. This was backed up by a procedure we found online. What was left to do was to confirm the presence of the product, 4-bromoacetanilide.

The challenge is that the reaction mixture contains unreacted acetanilide as well as 4-bromoacetanilide, acetic acid, and possibly some polyhalogenated byproducts. Any conditions that could result in amide hydrolysis, e.g. heating with base, should also be avoided.
No residue was observed in the reaction mixture.

So, what chemical property can be used to determine the presence of 4-bromoacetanilide in the reaction mixture?

I advise against fishing in the mixture in this case. Try to isolate the product, purify it and determine its properties.

Have a look at the melting points of the starting material and the possible monobromination products:

acetanilide 114 °C
2-bromoacetanilide 96.5–100.5 °C
3-bromoacetanilide 81–85 °C
4-bromoacetanilide 165–169 °C

This promising! You can of course use further spectroscopical methods (NMR), but I'd start with the melting point.

You mentioned possible complications for the workup, but I disagree.

The reaction is typically performed in glacial acetic acid with a slight excess of bromine to make sure that there's (almost) no starting material left.

For workup, you pour the whole mixture into a beaker with water that has been placed in a bowl with ice. If you want/need to destroy an excess of bromine, add some bisulfite solution until the aqueous phase is colourless.

There should be some crystals. Filter them off, and recrystallize the material from ethanol.


Improving copper contact and preventing oxidation I have a model railway in which I use pure copper on the tracks and in the engine to power the train with electricity.

Is there any solution to prevent the copper surfaces from corrosion, and maybe improve the conductivity?

Though I have not tried them DeoxIT products claim to do what you want, available as liquid and spray.

If anyone has used these, please add a comment as to effectiveness.


DFT calculation of solids with different periods I would like to simulate a layer of graphene on a solid substrate, say Cu or quartz. I'm interested in their binding energy and other physical or chemical properties. However, their periods do not match each other, so if I want to use some standard DFT package like VASP or Quantum Expresso, I cannot use a single period for the whole supercell. I had some experience with molecules absorbed on some specific surface, which is an easier case because the period of the supercell is just that of the substrate.

Does anyone have experience with this situation? Thanks.

I have found a useful reference here

Basically the idea is to avoid the mismatch by best fitting their primitive cell unit vectors.


Why is covalent bonding stronger than electrostatic attraction? I guess if we look at this problem from a "conjugate" perspective then the conjugate of a covalent bond is two elements with electrons lying around. On the other hand, with two electrostatically attracted things, the lack of a bond leaves two separate, charged particles.

But that still doesn't answer the question; why are electrons less stable than ions? I guess in the case of an ion though there are still electrons but also a nucleus the stabilize the electrons.

Ionic bonds can easily be as strong as covalent ones.

First off, let's be clear that almost everything has some ionic and some covalent character. Moreover, it's not an either-or situation. Some bonds have both strong ionic and strong covalent character.

For now, I'll stick to simple cases.

Let's consider $\ce{LiF}$, which seems a prototypical ionic bond. The interatomic distance is $\pu{156 pm}$ according to NIST. That gives a potential energy of $\pu{903 kJ/mol}$. Now, that assumes a full electron is transferred from $\ce{Li+}$ to $\ce{F-}$, which isn't the case (more like $0.8 - 0.9$). This simple calculation also ignores Born repulsion which is similar to the short-range Van der Waals repulsion between atoms.

A more complete analysis J. Chem. Educ. 2000, 77 (8), p 1076 gives a much better agreement. The real bond dissociation energy is $\pu{577 kJ/mol}$ (Source, PDF via the Internet Archive). A simple spherical ion model from the paper gives the bond enthalpy as $\pu{608 kJ/mol}$.

  calculations based on a simple ionic model reproduce the bond energies of $\ce{LiF(g)}$ and $\pu{NaF(g)}$ to within $5\%$.

If we look up covalent bond dissociation energies, the triple bond in dinitrogen is $\pu{945 kJ/mol}$. Ok, that's perhaps stronger, but $\ce{LiF}$ only has a "single" charge and clearly some ionic compounds have greater charge transfer.

More "prototypical" covalent bonds include $\ce{C-H}$ ($\pu{410 kJ/mol}$) and $\ce{O=O}$ ($\pu{497 kJ/mol}$), both less than $\ce{LiF}$.

Now, as I said earlier, it's not the case of a bond being only covalent or only ionic.

To quote from R. Gillespie J. Chem. Educ. 2001, 78 (12), 1688:

  The large atomic charges in $\ce{BF3}$ and $\ce{SiF4}$ are important to understanding the great strength of these bonds—which are the two strongest known single bonds, with bond dissociation enthalpies of $613$ and $\pu{565 kJ/mol}$, respectively. In comparison, the $\ce{C–C}$ bond in ethane has a bond dissociation enthalpy of only $\pu{345 kJ/mol}$. The great strength of the $\ce{BF}$ and $\ce{SiF}$ bonds arises from both the relatively large amount of shared density and the large atomic charges. So it is not strictly correct to say that these bonds are “predominately ionic”; if we are forced to use these terms we have to say that they are both strongly covalent and strongly ionic.

Finally, you can of course treat both covalent and ionic bonds using MO theory. For example, an approximate picture of $\ce{LiF}$ under MO theory might be something like this:

We readily see $\ce{LiF}$ has very high ionic character. For all intents and purposes, the "bonding" $\sigma$ orbital is localized on the $\ce{F}$ atom, there's very little energy difference between the $\pi$ and $\sigma$ here.


How to calculate partition coefficients? Below is a transcript of this doubtful question:

  A weak monobasic organic acid $\ce{HA}$ is soluble in both water
  and $\ce{CHCl3}$.  $\pu{500.0 cm3}$ of a solution of $\ce{HA}$ in $\ce{CHCl3}$, with a concentration of
  $\pu{0.057 mol dm-3}$,
  is shaken well with $\pu{500.0 cm3}$ of water and allowed
  to attain equilibrium at $\pu{27°C}$.  An aqueous layer and
  $\ce{CHCl3}$ layer then separate out; the pH of the aqueous layer is
  found to be 3.21 under these conditions.

  The dissociation constant of $\ce{HA}$ in water at $\pu{27°C}$
  is $\pu{1E-5 mol dm-3}$.

  Calculate the partition coefficient at $\pu{27°C}$
  for the partitioning of $\ce{HA}$ between water and $\ce{CHCl3}$.
  In a second experiment, a further $\pu{500.0 cm3}$ portion of the
  same $\ce{HA}$ solution in $\ce{CHCl3}$ in which the concentration
  of $\ce{HA}$ is $\pu{0.057 mol dm-3}$, is shaken well with
  $\pu{500.0 cm3}$ of a $\pu{0.027 mol dm-3}$ aqueous $\ce{NaOH}$
  solution and allowed to reach equilibrium at $\pu{27°C}$.
  Calculate the $\mathrm{pH}$ of the aqueous layer under these conditions.
  State the assumptions you make, if any, in the above calculations.

My solution for Question 1.

Since pH is given, $\ce{[H^+](aq)}$ is determined by

\begin{align}
\mathrm{pH} &=-\log[\ce{H^+~(aq)}]\\
[\ce{H^+}~(\ce{aq})] &= 10^{-3.21}\\
    &= 6.16 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}\\
\end{align}

By appropriately applying $K_\mathrm{a}$, the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer is $0.038~\mathrm{mol\,  dm^{-3}}$.
Next, $[\ce{HA}~(\ce{aq})]$ present in the organic layer can be determined by subtracting the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer
from the initial concentration of $[\ce{HA}~({\ce{aq}})]$, which results in $0.019~\mathrm{mol\,  dm^{-3}}$.
Finally, I got $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.019~\mathrm{mol\,  dm^{-3}}} = 2$.

I hope that this is correct, but  I am not quite sure.

My incomplete solution for Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ is

$$\ce{HA}~(\ce{aq}) + \ce{NaOH}~(\ce{aq})  \longrightarrow \ce{NaA}~(\ce{aq}) + \ce{H2O}~(\ce{l})$$

The number of moles of $\ce{NaOH}$ present is $0.0135~\mathrm{mol}$.

I am confused with the concentration of $\ce{HA}$ that I needed to take for this calculation. Initially I have found the concentration of $\ce{HA}$ present in organic layer as $0.019~\mathrm{mol~dm^{-3}}$. Do I need to use that here or $0.057~\mathrm{mol~dm^{-3}}$?

What would be the assumptions that I need to assume to take out this calculation?

Equilibrium are really dependent on the chemical activities of the species not their concentrations. So an overriding assumption for both parts is that the activities of the chemical species is the same as the concentration. That is a reasonable for these solutions. As a rule of thumb, for solutions with concentrations more than 0.1 molar the assumption is dicey.

I think you're very close to the right solution for part 1, but I'd state it a bit differently.

Since pH is given, $\ce{[H^+]}$ in the aqueous phase is determined by

    $[\ce{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.21} = 6.2 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}$

      The mantissa of the pH only has two significant figures, so should the $\ce{[H^+]}$

We know that in the aqueous layer the charges have to balance so $\ce{[H^+] = [A^-] + [OH^-]}$, but $\ce{[A^-] >> [OH^-]}$ so we can assume that $\ce{[H^+] = [A^-]}$
Using $\ce{[H^+] = [A^-]}$ the $K_\mathrm{a}$ equation can be solved for the $\ce{[HA]}$ present in aqueous layer which gives $0.038~\mathrm{mol\,  dm^{-3}}$.

$$\ce{[HA]} = \dfrac{\ce{[H^+][A^-]}}{1\times10^{-5}} = \dfrac{(6.2\times10^{-4})^2}{1\times10^{-5}} = 0.038$$

Next, the $\ce{[A-]}$ in the aqueous phase is insignificant compared to the $\ce{[HA]}$ in the aqueous phase, so the moles of $\ce{HA}$ present in the organic layer can be determined by subtracting the moles of $\ce{HA}$ present in aqueous layer (0.038*0.5=0.019)from the initial moles of $\ce{HA}$ (0.057*0.50 = 0.029), which results in a final concentration of $0.020~\mathrm{mol\,  dm^{-3}}$ of $\ce{HA}$ in the organic layer.
Finally, $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.020~\mathrm{mol\,  dm^{-3}}} = 1.9$

    The $\ce{[HA]}$ in both layers is known to two significant figures so the $K_D$ value should have two significant figures too.

For Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ in aqueous solution is, as you noted,

$$\ce{HA + NaOH -> Na^+ + A^- + H2O}$$

Assume none of $\ce{NaA}$ migrates into the organic phase.
Assume that no $\ce{NaA}$ forms in the aquaeous phase either, which is to say that both the $\ce{Na^+}$ and $\ce{A^-}$ ions are completely solvated.
Since $\ce{[Na^+]}= 0.027$ and the volume is 0.5000 $\mathrm{dm}^{-3}$, the number of moles of $\ce{NaOH}$ present is $0.0135$.
Started with 0.0295 moles of HA total, so the solution will be acidic.
Since the charges must balance $\ce{[Na^+] + [H^+] = [A^-] + [OH^-] }$, but $\ce{[A^-] >> [OH^-] }$ and $\ce{[Na^+] >> [H^+]}$ so $\ce{[Na^+] \approx [A^-] }$ which means that there are $0.0135$ moles of $\ce{[A^-]}$ in the aqueous solution, and 0.016 moles of $\ce{HA}$ split between the aqueous phase and the organic phase.
From above $0.016 = \ce{0.5[HA]_{aq} + 0.5[HA]_{org}}$ and using $K_D$ we can determine $\ce{[HA]_{aq} = 0.021}$ and $\ce{[HA]_{org} = 0.011}$
We can use the $K_a$ to solve for $\ce{[H^+]}$

\begin{align}
1\times10^{-5} &= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\
\ce{[H^+]} &= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\
\mathrm{pH} &= 5.11
\end{align}


How to read TINKER's force field parameters I'm going to parameterize AMBER force field for the simulation of some small molecules. I have found the parameter sets provided by TINKER at here.

I'm stuck here because I couldn't find any explanations about the parameter values. They should be the parameters that appear in the formula of AMBER force field. However, I have no idea which numbers mean $k_b$, $l_0$, $k_a$, $θ_0$, or the rest parameters.

Okay I will take a stab at this.

Consider the following bond stretching parameters in the TINKER FF file:

      ##################################
      ##                              ##
      ##  Bond Stretching Parameters  ##
      ##                              ##
      ##################################

bond          1    1          310.00     1.5260

This tells you that this is a bond parameter between atom 1 and atom 1.  The index of atoms is given in the FF file as

   #####################################################
   ##                                                 ##
   ##  TINKER Atom Class Numbers to Amber Atom Types  ##
   ##                                                 ##
   ##    1  CT      11  CN      21  OW      31  HO    ##
   ##    2  C       12  CK      22  OH      32  HS    ##
   ##    3  CA      13  CQ      23  OS      33  HA    ##
   ##    4  CM      14  N       24  O       34  HC    ##
   ##    5  CC      15  NA      25  O2      35  H1    ##
   ##    6  CV      16  NB      26  S       36  H2    ##
   ##    7  CW      17  NC      27  SH      37  H3    ##
   ##    8  CR      18  N*      28  P       38  HP    ##
   ##    9  CB      19  N2      29  H       39  H4    ##
   ##   10  C*      20  N3      30  HW      40  H5    ##
   ##                                                 ##
   #####################################################

So clearly this is a bond parameter between CT and CT.  What do these atom notations stand for?  They are outlined in Table 1 of this paper.  The value 310.00 is $K_r$ (in kcal mol$^{-1}$)  and 1.5260 is $r_{\mathrm{eq}}$ (in Ang).

For angle bends:

      ################################
      ##                            ##
      ##  Angle Bending Parameters  ##
      ##                            ##
      ################################

angle         1    1    1      40.00     109.50

angle denotes what you'd expect.  The last two parameters, 40.00 ($K_{\theta}$ in kcal (mol radian$^2$)$^{-1}$) and 109.50 ($\theta_{\mathrm{eq}}$ in deg.) are for CT-CT-CT (atom1-atom1-atom1).

For torsional angles:

      ############################
      ##                        ##
      ##  Torsional Parameters  ##
      ##                        ##
      ############################

torsion       1    1    1    1            0.156    0.0  3

This is clearly a torsion between CT-CT-CT-CT.  Here is where it gets a little tricky.  If anyone finds this to be erroneous, please let me know and I will correct it.  The parameter, 0.156 is simply determined by the following formula:

$$\frac{\frac{V_n}{2}}{\mathrm{no.~of~paths}}$$

where $V_n/2$ is the magnitude of torsion in kcal mol$^{-1}$.  In Table 14 of the previously cited paper, for Torsional Parameters you find for X-CT-CT-X:

no. of paths = 9
$V_n/2$ = 1.40

Therefore, $1.40/9$ $\approx$ 0.156 which is exactly what is given in the TINKER FF file.  I've done this with a few other torsional parameters and it seems to add up. I believe footnote $f$ in Table 14 explains this.

The 0.0 is simply $\gamma$ (phase offset in deg.) and 3 is $n$ (the periodicity of the torsion).

Functional Form of Amber94 FF

It may be useful (and convenient) to spell out the functional form as presented in the cited paper here.

$$E_{\mathrm{tot}} = \sum_{\mathrm{bonds}} K_r(r-r_{\mathrm{eq}})^2 +$$
$$~~~~~~~~~~~~\sum_{\mathrm{angles}} K_{\theta}(\theta - \theta_{\mathrm{eq}} ) ^2 + $$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{\mathrm{dihedrals}} \frac{V_n}{2}[1+\cos(n\phi - \gamma)] + $$
$$~~~~~~~~~~~~~~~~~~~~~~~\sum_{i<j} \left[ \frac{A_{ij}}{R_{ij}^2} - \frac{B_{ij}}{R_{ij}^6} +
\frac{q_i q_j}{\epsilon R_{ij}} \right]$$

Reference: Cornell and co-workers. J. Am. Chem. Soc. 1995, 117, 5179.


Mathematical proof of ΔQ=ΔH at constant external pressure Consider a system with an ideal gas. A process is being carried out in which the gas is being compressed at a constant external pressure ($P_{ext}$). For this process ΔQ should be equal to ΔH, but this is where I am stuck.

Work done by the system will be: $$ΔW= -P_{ext}ΔV$$ and
$$ΔH=ΔU+Δ(P_{sys}V)$$Now we can write: $$ΔQ=ΔU-ΔW = ΔH-Δ(P_{sys}V)-(-P_{ext}ΔV)$$ In this expression $Δ(P_{sys}V)$ and ($P_{ext}$ΔV) are not equal as the process is irreversible and the so the external pressure will remain constant throughout the process but the internal pressure varies. I know that at the initial and final stage the external and internal pressure will be equal as the system will be then at equilibrium with the surrounding but the magnitude of the internal pressure will be equal to the constant external pressure($P_{ext}$) only at the final stage. So how will ΔQ be equal to ΔH when the process is carried out at constant external pressure?

In an irreversible process, the total force per unit area exerted by the gas on the piston face is comprised of the local thermodynamic gas pressure at the piston face (as determined by the local gas density and temperature at the piston face) plus a viscous stress related to the rate of deformation of the gas in close proximity to the piston. So the thermodynamic pressure is not uniform throughout the gas and there are viscous stresses present.  By comparison, in a reversible process, the pressure is uniform throughout the gas and the viscous stresses are negligible.  In both cases, if the piston is massless, the force per unit area exerted by the gas on its side of the piston is exactly equal to $p_{ext}$, the force per unit area on the outside of the piston, and the work will be the same, as calculated from the force on either side of the piston.

If a frictionless piston has mass and the cylinder is horizontal, the piston will oscillate about its final position until the viscous stresses damp out its motion (i.e., its kinetic energy dissipates).  In the end, the final gas pressure will then be equal to $p_{ext}$, and, from a transient force and energy balance on the piston, one can show that the total work done by the gas will still integrate to $p_{ext}\Delta V$.

If the compression takes place irreveribly by, at time zero, suddenly increasing the external force per unit area from the initial gas pressure $p_0$ to the constant external force per unit area $p_{ext}$, the change in enthalpy of the gas from its initial state to its final state will not be equal to Q.  We will have $$W=p_{ext}(V_f-V_0)$$ and $$\Delta U = Q-W=Q-p_{ext}(V_f-V_o)$$so$$Q=\Delta U+p_{ext}(V_f-V_o)$$while$$\Delta H=\Delta U+(p_{ext}V_f-p_0V_0)$$


How to calculate the pH of a sodium bicarbonate-sodium hydroxide buffer solution? This website claims that if you add $\pu{50 mL}$ of $\pu{0.05 M}$ sodium bicarbonate, and $\pu{5 mL}$ of $\pu{0.1 M}$ sodium hydroxide (and dilute to $\pu{100 mL}$), you should create a solution with $\mathrm{pH }= 9.6$.

I'm struggling a bit getting this result.

My attempt:

The sodium bicarbonate reaction would be:

$$\ce{NaHCO3 + H2O -> H2CO3 + OH- + Na+}$$

We know the $\mathrm{p}K_\mathrm a$ of $\ce{H2CO3}$ is 6.4, so the $\mathrm{p}K_\mathrm{b}$ of $\ce{HCO3-}$ is:

$$14 - 6.4 = 7.6$$

Furthermore, $\ce{pOH} = \mathrm{p}K_\mathrm b + \log\frac{[\ce{B+}]}{[\ce{BOH}]} = 7.6 + \log\frac{[\ce{H2CO3}]}{[\ce{HCO3-}]}$

However, how do we find $[\ce{H2CO3}]$ ?

I'm not entirely sure this is the right approach, but none of the other resources I've used to determine this have helped.

Your idea is good, but you have thought of the wrong reaction.

You have written that $\ce{HCO_3^-}$ hydrolyses to give $\ce{H_2CO_3}$ and $\ce{ OH^-}$, but think carefully when you add base, $\ce{OH^-}$ is in excess in this medium. So, shouldn't your proposed reaction proceed in backward direction instead? Moreover, your reaction is actually not going to occur.

When you are adding strong base, $\ce{HCO_3^-}$ acts an acid to give $\ce{CO_3^2-}$ like:

$$\ce{NaHCO_3 + NaOH -> Na_2CO_3 +H_2O}$$
 or, more simply
$$\ce{HCO_3^- + OH^- -> CO_3^2- + H_2O}$$

Initially, number of millimoles of $\ce{HCO_3^-}$ were $=50\times0.05 = 2.5$ and number of millimoles of base added $=5\times0.1 = 0.5$

So, $\ce{CO_3^2-}$ produced will be also 0.5 millimoles as the base added was a limiting reagent, and $\ce{HCO_3^-}$ left $= 2.5-0.5 = 2$ millimoles.

As, you can see now, the solution acts an acid buffer as, $\ce{HCO_3^-}$ is a weak acid and $\ce{CO_3^2-}$ is the salt after reacting with a strong base. According to Henderson-Haselbach equation,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{[\text{salt}]}/[\text{acid}]$$
$\mathrm{p}K_\mathrm{a}$ of $\ce{HCO_3^-}=\mathrm{p}K_{\mathrm{a2}}$ of $\ce{H_2CO_3^}=10.3$ . Now [$\ce{HCO_3^-}$] = $\pu{0.02 M}$, and [$\ce{CO_3^2-}$] = $\pu{0.005 M}$. Now if you calculate $$\mathrm{pH}= 10.3 +\log\frac{0.5}2 = 10.3- \log(4) = 9.69$$ So, the final $\ce{pH}$ of the medium has come out to be $= 9.69$.

The difference in digits after decimal may be due to some given $\mathrm{p}K_\mathrm{a2}$ which is experimental and you should do calculation according to that.


Why does DFT-D finds different optimized geometries for the same structure when using different input geometry files? When I change c lattice parameter and optimize the structure with DFT-D, every time I get a significantly different geometry with different energy? Why is this happening with DFT-D?

Simply put, a geometry optimization finds a local minimum. Many such minima may exist for a given system and it is up to the user to ensure that a reasonable minimum is used for further calculations. Techniques may include using many handmade starting geometries, using a random mutation algorithm or starting from known, experimental structures.


What is A in the following equation This equation describes the total current in bulk solution in a battery caused by migration in the electrolyte.

Then the source says what A is: In the bulk solution (away from the electrode), concentration gradients are generally small, and the total current is carried mainly by migration. All charged species contribute. For species j in the bulk region of a linear mass transfer system having a cross sectional area, $A$...

$$i = \sum_j i_j = \frac{FA\,\Delta E}{l} \sum_j |z_j| u_j C_j$$

I am confused if $A$ is the electrode area or the solution cross sectional area.

I will describe what the rest of the variables are

The $i$ is the current in the bulk solution.
F is Faraday's Constant
∆E is the change in potential (i.e. Voltage measurement of the battery)
l is the distance of how far the electrodes are separated (or at least that is what I think)
z is the charge of species j
u is mobility of species j
C is the concentration of species j

$A $ is the cross-sectional area in solution. The  "bulk" is the relatively homogeneous "mess" of charged particles and solvent -- $A $ is not the area of the electrode.

What this equation tells you in terms of $A $, is that, the larger the area through which some number of charges are passing in a given moment, the greater the possible current, because more area means more room for more charges that can pass through in a moment. If you squeeze the area you start to make a bottleneck wherein less charged particles can pass in a given time. (This is why copper wires with larger diameter (or, smaller US gauge) have higher current capacity).

This makes sense, right? Current is measured as charge per unit time. The implication there, is that it is motion of charge through some point (1D) or cross sectional area (2D, $A $ here) per unit time.


How does density of alkali metals increase down the group? Size of alkali metals increases down the group, so volume also shows increment, and since volume is inversely proportional to density, how does density also increase down the group?

$Density=mass/volume$

1) If mass is increasing and volume is decreasing, then density (mass/volume) will increase.

2)If mass is decreasing and volume is increasing simultaneously, then the density (mass/volume) will decrease.

*3)If both mass and volume are increasing, then we need to check which one of them is increasing at a faster rate(since both are contradictory factors)

a) If mass is increasing at a faster rate than volume, then density will increase.

b) If volume is increasing at a faster rate than mass (i.e denominator in mass/volume is increasing making the overall fraction smaller), then density decreases.

You can apply same logic when both volume and mass are decreasing simultaneously.

Generally, we see that in alkali metals the rate of increase of mass is greater than rate of increase of volume, therefore the density increases down the group.

Also, note that there is no definite reason for why mass is increasing at a faster rate than volume. So asking "why" does it happen isn't a great question.


Calculating partial pressures from mixture of gases A gas is enclosed in a container at $35~\mathrm{m^3\,mol^{-1}}$ and has the following composition at $49\,\mathrm{^\circ C}$
$\ce{N_2} \ 2\,\% $
$\ce{CH_4} \ 79 \,\%$
$\ce{C_2H_6} \ 19 \,\%$
Assuming an ideal gas, calculate the partial pressure of each component (in kPa).

My working so far:
I chose $100 \,\mathrm{mol}$ because I've been given percentages and it's a nice number to work with, and I multiplied this with $35 \,\mathrm{m^3\, mol^{-1}}$ to get $3500 \,\mathrm{m^3}$
I used $P=\frac{nRT}{V}$ to calculate an overall pressure but got $76\, \mathrm{Pa}$ which seems way too low. Have I done anything wrong?

First calculate the total pressure from the ideal gas law, explicitly keeping track of units to avoid confusion and mistakes:

$$\text{Ideal Gas Law:}~~~~{P=\frac{n}{v}*R*T}$$

where:
$\frac{n}{v}=\pu{\frac{1}{35}\frac{mol}{m^3}}$
$R=\pu{8.31\frac{m^3~Pa}{mol~K}}$
$T=\pu{322K}$

Then plug the values in to the ideal gas equation to get the total pressure:

$$P_{total}=\pu{\frac{1}{35}\frac{mol}{m^3}*8.31\frac{m^3~Pa}{mol~K}*322~K}$$

$${P_{total} =\pu{76.5Pa}}$$

Then just multiply the total pressure by the fraction of each gas as stated in the problem:

${P_{\ce{N2}} = \pu{76.5Pa * 0.02 = 1.5 Pa}}$
${P_{\ce{CH4}} = \pu{76.5Pa * 0.79 = 60 Pa}}$
${P_{\ce{C2H6}} = \pu{76.5Pa * 0.19 = 15 Pa}}$


Are the breaking of bonds simultaneously endothermic and exothermic? I understand that energy is NEEDED to BREAK bonds. But don't chemical bonds already have some form of potential energy in them already? So wouldn't it make sense that even though energy is required to break them, their potential energy is also RELEASED once they are broken?

In other words, shouldn't the bond breaking process be considered simultaneously endothermic AND exothermic?

Side note: I'm studying grade 12 level chemistry right now, so I don't have too much knowledge on the nitty gritty details of this concept. That being said, It would be greatly appreciated if anyone could explain this in very simple terms.

Bond breaking does NOT release energy!!

In the course of a chemical reaction, you typically break some bonds and form some new bonds. The bond breaking is always endothermic. The formation of new bonds is exothermic, so depending on whether the old bonds or the new bonds were stronger, the reaction overall can be either endothermic or exothermic.

When somebody says chemical bonds “contain potential energy” it is a HIGHLY misleading statement. You justifiably interpreted this as: when you break bonds, they release the potential energy that was stored within. This is, unfortunately, wrong. That phrase should honestly be banned, but somehow it has always persisted.

What that phrase really means is more nuanced. It means that some chemical bonds can be broken in a reaction which subsequently produces more energy via bond formation.

Bond breaking is always endothermic!

For further reading see: Is Bond Formation "Strictly" Exothermic?


Equation to find specific gravity (or density) from a solution of sugar and water Is there an equation or simple way to calculate the final density of a solution made from x grams of sugar and y liters of water?

And would it work in reverse? if we have y liters of water at z density or specific gravity, can we know how much sugar is in the solution?

I'd like to run an experiment for my nephews, something like This.

However, lacking any background in chemistry, the precise math is tough to nail down. It would be great to have some deeper understanding of how this works...

Here an empirical approach:

1) A simple way to calculate density: mass/volume

mass = x g of sugar + (y L of water*1000 mL/L*1g/mL)

1 g/mL = water density

Then measure the total volume (total mL) of the mix:

density = (x g of sugar + (y L of water * 1000 mL/L * 1g/mL))/(total mL)

2) For the reverse calculation, you can determine density as a function of sugar content by preparing solutions with different sugar concentrations and use this to interpolate your specific density value to a sugar concentration.

You can also use preexisting tables or graphs as Felipe suggested


Pi molecular orbitals of polyenes As an organic chemist, I'm comfortable deriving the pi molecular orbitals of linearly conjugate systems to give the following result:

In a qualitative sense, these molecular orbitals are easily arrived at. For a polyene of n-atoms, the lowest energy combination will have 0 nodes (all in-phase) and the highest energy combination will have n-1 nodes (in, out, in, out...).

Is there a way of arriving at the same result using a more rigorous, formal molecular orbital approach?

These pi-type MOs are most commonly derived using Hückel MO theory. I've never been a fan of Wikipedia's technical articles, so for further reading I'd suggest using the QMUL resources online, which are very comprehensive. $\require{begingroup}\begingroup \newcommand{ket}[1]{\left|#1\right>} \newcommand{bra}[1]{\left<#1\right|} \newcommand{braket}[1]{\left< #1 \right>}$

Theory

Simple Hückel theory makes some key assumptions about the forms of the MOs as well as some of the integrals involved in a typical quantum chemical calculation:

pi-type MOs are linear combinations of p-orbitals, and no other orbitals contribute (so-called "sigma-pi separation")
the value of $\braket{a|b}$ (where $\ket{a},\ket{b}$ are p-type AOs on atoms $a$ and $b$) is equal to $\delta_{ab}$ ($1$ if $a = b$, $0$ otherwise)
the value of $\braket{a|H|b}$ is

$$H_{ab} = \braket{a|H|b} = \begin{cases}\alpha \text{ if } a = b \\ \beta \text{ if atom }a\text{ is bonded to atom } b \\ 0 \text{ otherwise}\end{cases}$$

With these simplifications the secular equations

$$\mathbf{H}\mathbf{c} = E\mathbf{S}\mathbf{c}$$

can be readily solved (the derivation of the secular equations will not be presented here). Here $\mathbf{H}$ refers to the Hamiltonian matrix and $\mathbf{S}$ refers to the overlap matrix. These are defined as the matrices with elements $\mathbf{H}_{ij} = \braket{i|H|j}$ and $\mathbf{S}_{ij} = \braket{i|j}$ respectively. In simple Hückel theory the matrix $\mathbf{S}$ can be neglected as it is simply equal to the identity matrix.

$\mathbf{c}$ is a column vector and one column vector represents one MO. The elements of $\mathbf{c}$ represent the contribution of each AO to the MO. This will be clearer with an example (to follow).

In general, for $n$ atoms there will be $n$ vectors $\mathbf{c}$ that will satisfy the secular equations, and $n$ corresponding values of the energy $E$. Thus, for the allyl cation (for example), there will be three column vectors $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3$ and four associated energies $E_1, E_2, E_3$.

The allyl cation

Using the atom numbering scheme as shown above, and the assumptions mentioned in the previous section, we find that the Hamiltonian matrix $\mathbf{H}$ is:

$$\mathbf{H} =
\begin{pmatrix}
\alpha & \beta & 0 \\
\beta & \alpha & \beta \\
0 & \beta & \alpha
\end{pmatrix}$$

For example the entry $\mathbf{H}_{12} = \beta$ because atoms 1 and 2 are bonded to each other, and the entry $\mathbf{H}_{13} = 0$ because atoms 1 and 3 are not bonded to each other.

The secular equations then reduce to the eigenvalue problem

$$\begin{pmatrix}
\alpha & \beta & 0 \\
\beta & \alpha & \beta \\
0 & \beta & \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix} = E\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix}$$

Obtaining the eigenvalues and eigenvectors of a matrix is a process outlined in many maths books. The crux is that the "secular determinant" must equal zero:

$$\begin{vmatrix}
\alpha - E & \beta & 0 \\
\beta & \alpha - E & \beta \\
0 & \beta & \alpha - E
\end{vmatrix} = 0$$

from which one obtains the eigenvalues (here I chose an example where it's easy to do by hand, but in general you can use software like MATLAB). In increasing energy (note that $\beta < 0$), they are

$$\begin{align}E_1 &= \alpha + \sqrt{2}\beta & E_2 &= \alpha & E_3 &= \alpha - \sqrt{2}\beta \end{align}$$

and the corresponding eigenvectors

$$\begin{align}
\mathbf{c}_1 &= \begin{pmatrix} 1/2 \\ 1/\sqrt{2} \\ 1/2 \end{pmatrix}
&
\mathbf{c}_2 &= \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}
&
\mathbf{c}_3 &= \begin{pmatrix} 1/2 \\ -1/\sqrt{2} \\ 1/2 \end{pmatrix}
\end{align}$$

This means that the lowest-energy pi-type MO, $\mathbf{c}_1$, is

$$\psi_1 = \frac{1}{2}\mathrm{p}_1 + \frac{1}{\sqrt{2}}\mathrm{p}_2 + \frac{1}{2}\mathrm{p}_3$$

where $\mathrm{p}_i$ is the 2p atomic orbital on carbon $i$. Note that the coefficients all have the same sign, indicating that the p orbitals are all in phase with each other. However, the coefficient for carbon 2 (the middle carbon) is larger than the others. Hence, a sketch of this MO should technically show that the middle p-orbital makes a larger contribution than the outer two p-orbitals. This is what TAR86 meant by their comment on your question:

  Hückel MO theory will also give you different absolute values for the linear combination coefficients, whereas your drawing suggests the same coefficients throughout.

Anyway, likewise for the LUMO we have

$$\psi_2 = \frac{1}{\sqrt{2}}\mathrm{p}_1 - \frac{1}{\sqrt{2}}\mathrm{p}_3$$

Note that 1) the p orbitals have different phases (different sign of coefficient), and 2) the p orbital on carbon 2 does not contribute to the MO - this leads to the node observed in the second MO. This is why the allyl cation is electrophilic on C-1 and C-3, but not on C-2: the LUMO has zero coefficient on C-2.

All in all, considering that the allyl cation only has two electrons in the pi system, this lets us construct the MO diagram


tap water and distilled water for electrolysis my water electrolyzer which I've built using this pdf, produces less gas than I expected. In that file the author insists on using distilled water instead of tap water and I have used tap water. would it be the reason?

Why is it so important to use distilled water since it has high resistance for electricity?

This article tells you to use distilled water because the salt in tap water will damage the booster. The author also tell you to add NaOH or KOH to the distilled water. Did you add these salts?


Why No Change in Polarizability in Asymmetric Stretch? In order for a molecule to be Raman active, there must be a change in the polarizability, meaning that there must be change in the size, shape or orientation of the electron cloud that surrounds the molecule. And this change said to occur in symmetric stretching, but not asymmetric stretching.
Please explain using plain english (without any rules or tables) why is that in asymmetric stretching like the following, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule? Specially considering each side independent has its own electron cloud.

Please explain using plain english (without any rules or tables) why is that in asymmetric stretching like the following, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule?

There is a change. The size and shape of the electron cloud does change for the individual bonds affected by the asymmetric stretch. Polarizability relates to the ease with which electrons can be moved from their equilibrium position.  Electrons in weaker (longer) bonds are more easily displaced than electrons in stronger (shorter) bonds.
From your diagram, I believe that you are discussing the case of a linear $\ce{AB2}$ molecule. Picture the asymmetric stretch in our $\ce{AB2}$ molecule with the two $\ce{B}$ atoms remaining stationary and the $\ce{A}$ atom oscillating back and forth. Simultaneously one $\ce{A-B}$ bond becomes longer and the other shorter.  One becomes more polarizable, the other less polarizable.  In the linear $\ce{AB2}$case these changes in polarizability are equal and opposite, hence for the molecule overall they cancel out and there is no change in overall polarizability (for example, see here).

In order for a molecule to be Raman active, there must be a change in the polarizability

Yes, but it must be a change in polarizability for the entire molecule, not just an individual bond. Further, keep in mind the  Principle of Mutual Exclusion. It states that, for centrosymmetric molecules (molecules with a center of symmetry, your linear $\ce{AB2}$ is an example), vibrations that are IR active are Raman inactive, and vice versa.
The asymmetric stretch in such a molecule changes the dipole moment in the molecule from zero to non-zero; hence the asymmetric stretch is IR active. Therefor by the Principle of Mutual Exclusion, the asymmetric stretch must be Raman inactive.  Even though the asymmetric stretch changes the polarizabilty of some bonds in the molecule, the polarizability of the whole molecule does not change (as explained above) and the asymmetric stretch is Raman inactive as predicted.
The links I posted in the comments show how these concepts can be applied in the linear $\ce{AB2}$ examples of $\ce{CO2}$ and $\ce{XeF2}$.