What is the condensation point of water? The boiling point of water is $\pu{100^\circ C}$. However, we find that even at room temperature, water can evaporate into gas.
Therefore, water vapor can exist at temperatures of, say, $\pu{50^\circ C}$.

At what temperature, can you say for certain that all water vapor will turn into liquid?
In other words, what is the condensation point of water?

You're right - water vapor can exist at temperatures such as $\pu{50^\circ C}$. This is a phenomenon known as evaporative cooling, where molecules of water with higher kinetic energies tend to "release" themselves from the system, and as a result, less and less water molecules are held within that system.

Temperature is a sort of proxy for kinetic energy, and vice versa. Reference, also, the Maxwell-Boltzmann distribution, which reveals that at $\pu{100^\circ C}$, for example, not all molecules in the system possess the kinetic energy that they would at $\pu{100^\circ C}$, but rather the average exists at that temperature.

To answer your question, never in a real scenario.


Why thermodynamic state variables of a sytem are only defined at thermodynamic equilibrium and not during any other instant? As you have read the title so my question is why are thermodynamic variables only meaningful when the thermodynamic system is in thermodynamic equilibrium and is always meaningless in any other case.

Variables such as T and p are not well-defined when a system is fluctuating. Thermodynamic state variables are defined for systems that are static or quasi-static (in the latter case, transitioning very slowly - and therefore reversibly - through a series of equilibrium points). The main reason is that macroscopic fluctuations are absent in true equilibrium states. Each phase in such states is homogeneous. The argument is thus somewhat circular: thermodynamic states are homogeneous states that have fixed properties. The only way to guarantee fixed properties is by being at equilibrium.

Note that it is possible to obtain thermodynamic parameters by imposing restraints that trap a system in a local free energy minimum, and thereby prevent the system from transitioning to a global free energy minimum.


Why does ammonia have sp hybridization? Apparently it has $sp^3$ hybridization, but I don't understand why. Ammonia ($\ce{NH3}$) seems to me to not require $sp$ hybridization because all of its bond lengths are already equal. It has 3 hydrogens bonded to the $p$ orbitals. Why can't the lone electron pair in nitrogen's $2s$ shell stay put where it is?

Well, I can think of these reasons that can justify the hybridization in ammonia molecule.

Bond angle: If the molecule did not have hybrid orbits, and instead had unhybridized p-orbitals taking part in the bond formation, then the bond angle between the orbitals would be 90 degrees. And as for the real situation the bond angle is nearly 107 which makes the molecule more stable, decreasing the bond pair-bond pair and bond pair-lone pair repulsion.
Energy: Recalling the definition of hybridization, it is the mixing of the atomic orbitals having slightly different energies to form new orbitals that have equal energies. This stabilizes the molecule. As for ammonia molecule due to hybridization the energies of the lone pair of electrons and the bond pair of electrons becomes almost equal, thereby increasing the stability of the molecule.
Geometry: Thinking about the spatial arrangement of the atoms in the molecule of ammonia if there is no hybridization in the molecule then the size of the orbital containing lone pair of electrons would be different from that of the orbitals containing bond pair of electrons. Also if hybridization would not be taking place, explaining the geometry of the molecule, that is it's trigonal pyramidal shape, would not be possible.


Acid value of linseed oil For this we weigh $\pu{1.015 g}$ of linseed oil which is dissolved in alcohol. 3 drops of phenolphthalein (pH indicator) are added and a solution of potassium hydroxide of concentration $C = \pu{9.5e-3 mol L-1}$ is poured. The indicator turns to $V = \pu{7.4 mL}$

I calculate like that:
$$n_\mathrm{acid} = n_\mathrm{OH} = C_\ce{OH}\cdot V_\ce{OH} = 9.5\cdot 10^{-3} \cdot 7.4\cdot 10^{-3} = \pu{1.444e-4 mol}$$

$$\pu{1.444e-4 mol} → \pu{1.015 g}$$
$$\pu{1.423e-4 mol} ← \pu{1 g}$$
$$\implies I_a = m_\ce{KOH} = 1.423\cdot 10^{-4} \cdot 56 = \pu{7.967 mg}$$

But the teacher responds with $I_a = 3.8$. How to solve?

Unfortunately, the proportion you are using is not correct. Acid value $\mathrm{AN}$ determination procedure is not much different from an ordinary titration with indicator with the only difference that the mass of titrant is normalized to the mass of oil $m_\mathrm{oil}$:

$$\mathrm{AN} = \frac{T\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}$$

where $ΔV_\mathrm{eq}$ is equivalent volume (ideally you have to use a difference with an external standard ("spiking solution"), but you probably didn't do that); $T$ is a so-called titer, or basically mass concentration of the titrant:

$$T = CM$$

so that

$$
\begin{align}
\mathrm{AN} &= \frac{C(\ce{KOH})\cdot M(\ce{KOH})\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}\\
  &= \frac{\pu{9.5e-3 mol L-1}\cdot\pu{56.106 g mol-1}\cdot\pu{7.4e-3 L}}{\pu{1.015 g}}\\
  &= \pu{3.89 mg g-1}
\end{align}
$$

Also, a couple of side notes:

Always carry units across all steps and never omit them.
I recommend not to do intermediate calculations if you don't have to. Instead, first find an algebraic equation and only then plug in all the numbers.
Mind significant figures. If the mass of oil is given with 4 significant figures, it's a good idea to use a molecular mass of the titrant with at least 4 sigfigs too.


How to set up a home laboratory? I have spent the last few years self-studying chemistry and I want to try doing practical experiments with chemistry in a safe manner. I don't plan on doing anything dangerous but would still like to keep precautionary measures. I allotted myself a medium-sized (8ft by 6ft) space in the garage to carry out any experiments. What equipment would you recommend me to keep for precautionary measures as well as carrying out chemistry experiments?

You will be better off if you join a real lab. Researchers can help you get started in a safe manner. Chemistry experiments killed many professionals. Here is a list of things you would need.

A chemistry lab needs at least this:
Hood with 650 cubic feet per minute air flow. You CANNOT make it with a small pump.
Water.
Chemistry waste disposal. Flushing it down the drain is not a good approach.
Freezer and fridge for chemicals only.
Fire extinguisher.
Rotovap. Basic one costs 5000 dollars.
Source of vacuum.
Glassware.
Emergency shower.
Analytical instruments. NMR costs 100 000 dollars to buy, plus you will need to refill liquid helium once in a while. You can also ask chemistry department to run you samples for 10-20 dollars per piece.

Lab rules strictly prohibit working alone. In case of emergency someone should be around to save you.

Bottom line: you will poison yourself without a hood and set garage on fire. Find a university professor and ask to join their lab. If you show decent knowledge of chemistry they might agree to help you.


Why does hydration of alkene takes place in presence of an acid such as H3PO4 or H2SO4? Why acid is used? Can’t the hydration be done without acid? If not, why?

The acid is used as a catalyst. As can be seen from the mechanism below, the electron-rich double bond first attacks the positively-charged acid. The positively-charged hydronium ion is important because it is rather electrophilic and allows the nucleophilic pi bond to attack it, breaking the pi bond. This serves to initiate the reaction. Without an acid catalyst, the neutral water molecule would not be electrophilic enough for the pi bond to want to attack it, thus the reaction would not proceed.

(from http://slideplayer.com/slide/11423361/)


Substituent effects on acidity It is said that 4-methylbenzoic acid will be a weaker acid than benzoic acid. 4-nitrobenzoic acid will be a stronger acid than benzoic acid.

I don't understand why the methyl group on 4-methylbenzoic acid would affect the acidity. I know that it has an inductive effect (hyperconjugation) but I thought induction only had an effect over a one to two carbon distance. The carboxylic acid is separated from the methyl group by four carbons.

(1) It is true that the inductive effect is weak across several bonds, BUT conjugated bonds (i.e. alternating simple and multiple bonds) propagate this effect much farther, even more for mesomeric effects. You can think of "conjugated bonds" like a tunnel for electrons.

(2) In your case, the electronic effects will stabilize the salt more/less than the acid. As they are in equilibrium, the salt is more/less favored than its acid counterpart.

I think that the most important thing here is that the more electric charges are "diluted" (i.e distributed among more atoms), the more stable a molecule is. This is an effect of entropy, which is too often poorly described as disorder.


Predicting bond-strength of metal carbonyls The metal carbonyls (and similar organometallic compounds) involve a combination of sigma bond, a pi bond and backbonding. The bond strengths under consideration are the metal-carbon bond and the carbon oxygen bond. Here is the original question:-

  Predict the order of $\ce{C-O}$ bond strength in the following:-
  I) $[\ce{Mn(CO)_6}]^+$
  II) $[\ce{Cr(CO)_6}]$
  III) $[\ce{V(CO)_6}]^-$

All of these are isoelectronic, and hence whatever the difference of bond strengths, must arise from the difference in the resultant nuclear force of attraction. Since $\ce{Mn+}$ has the highest charge density, I would assume that the $\ce{M-C}$ bond strength of the manganese compound is the strongest (vanadium being the weakest) and therefore, the $\ce{C-O}$ bond strength must follow the opposite order, as the strengthening of the metal carbonyl bond should weaken the carbon oxygen bond.

But the answer given is III<II<I, exactly the opposite of my prediction. Why is it so? A relevant factor which I didn't consider is the back bonding, but I am unsure of how to integrate that into my prediction. Also, how would we make a prediction if the given compounds were not isoelectronic?

As the π-antibonding orbital of CO is filled by electrons from the metal, they weaken the C-O bond compared with free CO.  The metal-carbon bond is strengthened in kind. That's what the books says.


What old chemistry textbooks are good until present day? I have found old chemistry textbooks like Walter Moore's Physical Chemistry and Morrison&Boyd's Organic Chemistry to be really good source of information, as good as actual reference textbooks.

What other books do you recommend, regardless the area, that still useful nowadays despite its age?

The reason I ask this is because the language used by these authors seems to be different from today, and I find it to be a pleasant reading.

Charles E. Mortimer's Chemistry book was very good for our General Chemistry I & II courses. It has detailed Molecular Orbital diagrams, coordination chemistry information, sufficient examples of electrochemical cells and also a good text of nuclear chemistry. It does not have too many colorful images and pages so it is easier to focus on the knowledge.

In general, old books are only for general chemistry and history of chemistry research etc. When one starts to delve into things, it is dangerous not to use the most up-to-date textbook.


How to round significant figures correctly? Calculate how many grams of p-nitrophenol are required to prepare $\pu{250 mL}$ of a $\pu{11 mM}$ solution. (Answer to 3 significant figures.)

I worked it out as

\begin{align}
\frac{250}{1000}\,\mathrm{L} \times (\pu{11E-3 M}) \times \pu{139.11 g//mol}
  &= \pu{2.75E-3 mol} \times \pu{139.11 g//mol} \\
  &= \pu{0.3825525 g}
\end{align}

Is that $\pu{0.383 g}$ or $\pu{0.382 g}$ when approximated to 3 significant figures?
I think it is $\pu{0.383 g}$, but that's only going by the general rule of "up" approximating a five.

In rounding with significant figures, you want to avoid undue precision, yes, but it's also important to minimize error. It's this error minimization which is the primary goal of all "round to nearest" approaches.

So in short:

\begin{align}
0.383 - 0.3825525 &= 0.0004475\\
0.3825525 - 0.382 &= 0.0005525
\end{align}

$0.0004475$ is less than $0.0005525$, so you should round up to $0.383$, as that's the nearest value, and will minimize any error you introduce by rounding.

Even & Odd?

So what's this about the even/odd rule? Is there any validity there?

This technique is called "Banker's rounding" and is also attempting to minimize error.

When a number falls exactly in the middle of the two values to be rounded to (that is, when the digit to round is "5" exactly), each of the two difference is equally far away. So to a first order, both will result in an equal amount of error.

But picking a consistent direction (e.g. "always round up") introduces a subtle bias in the sums over the long run. This is because the errors are biased.

.     |  Always  |  Up   | Banker's | Banker's |
Value | Round Up | Error | Rounding | Error |
------+----------+-------+---------+------+
2.0   |     2    |  0.0  |    2    |  0.0  |
2.1   |     2    | -0.1  |    2    | -0.1  |
2.2   |     2    | -0.2  |    2    | -0.2  |
2.3   |     2    | -0.3  |    2    | -0.3  |
2.4   |     2    | -0.4  |    2    | -0.4  |
2.5   |     3    | +0.5  |    2    | -0.5  |   <<<<<
2.6   |     3    | +0.4  |    3    | +0.4  |
2.7   |     3    | +0.3  |    3    | +0.3  |
2.8   |     3    | +0.2  |    3    | +0.2  |
2.9   |     3    | +0.1  |    3    | +0.1  |
3.0   |     3    |  0.0  |    3    |  0.0  |
3.1   |     3    | -0.1  |    3    | -0.1  |
3.2   |     3    | -0.2  |    3    | -0.2  |
3.3   |     3    | -0.3  |    3    | -0.3  |
3.4   |     3    | -0.4  |    3    | -0.4  |
3.5   |     4    | +0.5  |    4    | +0.5  |   <<<<<
3.6   |     4    | +0.4  |    4    | +0.4  |
3.7   |     4    | +0.3  |    4    | +0.3  |
3.8   |     4    | +0.2  |    4    | +0.2  |
3.9   |     4    | +0.1  |    4    | +0.1  |
4.0   |     4    |  0.0  |    4    |  0.0  |
------+----------+-------+---------+-------+
63.0  |    64    | +1.0  |    63   |  0.0  |   - Totals

With "always round up" you get a consistent $+0.5$ bias per decade, and this is consistent even if you extend the range to more decimal places. You can balance the errors for $0.00...01$ to $0.49...99$ with those for $0.99...99$ to $0.50...01$, but there's that dead center $0.50...00$ (aka $0.5$) which is the odd man out - consistent rounding up (or down) will introduce a bias because of this entry.

So what are you to do? Well, introduce a rule which rounds that entry up half the time and rounds down the other half. Bankers' rounding is the most common such rule. This rule says that if the digits to be rounded off are $5$ exactly, then round such that the last digit of the number after rounding is even. (Note that this is equivalent to a rule to round up if the digit to the left of the $5$ is odd, but round down if the digit to the left is even - two different ways of saying the same thing.)

0.255 -> 0.26 ... because 6 is even (or round up because 5 is odd)
0.265 -> 0.26 ... because 6 is even (or round down because 6 is even)
0.275 -> 0.28 ... because 8 is even (or round up because 7 is odd)
0.285 -> 0.28 ... because 8 is even (or round down because 8 is even)

But again, this is the rule only if the number to be rounded falls exactly between the two possibilities. If the number is closer to one or the other, then round to the nearest.

0.2453   -> 0.25 (Round to nearest)
0.2553   -> 0.26 (Round to nearest)
0.2651   -> 0.27 (Round to nearest)
0.275000 -> 0.28 (Falls exactly between 0.27 and 0.28, apply even/odd rules)
0.275378 -> 0.28 (Round to nearest)
0.285000 -> 0.28 (Falls exactly between 0.28 and 0.29, apply even/odd rules)
0.285023 -> 0.29 (Round to nearest)

Note also that you should only round once, and don't cascade your rounding:

0.14555 -> 0.1    (Round to nearest)
0.14555 -> 0.1456 -> 0.146 -> 0.15 -> 0.2  (WRONG!)


Equilibrium calculation gives negative molarity There is a reaction: $$\ce {ClF3 <=> ClF + F2}$$ with $\ce {K_\mathrm{c} = 2.48 x 10^{-3}}$ Initially, $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, and $\ce {[F2] = 0.02 M} $. $\ce {F2}$ is added until the concentration of $\ce {F2} $ is $\ce {0.1 M} $.
  What is the concentration of these substances after equilibrium?

My attempt:
$\ce {ClF3 <=> ClF + F2} $

Initial concentration: $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, $\ce {[F2] = 0.1 M} $
Concentration that react: $\ce {[ClF3] = +x M} $, $\ce {[ClF] = -x M} $, $\ce {[F2] = -x M}$
Concentration after equilibrium: $\ce {[ClF3] = 0.25+x M} $, $\ce {[ClF] = 0.031-x M} $, $\ce {[F2] = 0.1-x M} $
So, $\ce {Kc = [ClF].[F2].[ClF3]^{-1}} $

I applied the concentration after equilibrium in $\ce {Kc} $, so I got:
$\ce {2.48.10^{-3} = [0.1-x].[0.031-x].[0.25+x]^{-1}} $

After calculate the equation above, I got $\ce {x = 0.1111 M}$.

Then my question is: if $\ce {x = 0.1111 M} $, so $\ce {[ClF] = -0.0801 M} $ and $\ce {[F2] = -0.0111 M} $. The concentration is negative. Is it possible? Are my steps incorrect or the answer is correct?

The equation $2.48\times 10^{-3} = \frac{(0.1-x)(0.031-x)}{0.25+x}$ is correct, but it simplifies to a second degree polynomial, so it has two solutions, not one. You just picked the incorrect root (negative concentrations have no physical meaning). Can you find the right answer now?


Is phenolphthalein used in laminar wood construction panels? Recently someone is claiming that the laminar wood panels used for building our new primary school contains a "carcinogenic and dangerous compound", namely phenolphthalein.

I couldn't find any evidence of it being used as that on the net.
Could you confirm my findings?

Thank you

Could it be that he meant phenol-formaldehyde resin?

These resins are often used to bond wood chips. They contain unreacted formaldehyde and the resin decomposes slowly, giving off more formaldehyde, which is volatile and very toxic. This is a known health concern. Many companies offer indoor air testing for formaldehyde, you can even buy dubious tests to do it yourself.

My bet would be on such a resin.

There are indeed some resins that use phenolphtalein:

Phenolphtalein might be cancerogenic, even though it was used as a laxative

  Phenolphthalein was commonly used as a laxative for most of the 20th century. The use of phenolphthalein in laxatives has decreased since 1997 when the United States Food and Drug Administration (FDA) proposed to withdraw its classification as an over-the-counter drug (21 CFR, Part 310). Phenolphthalein has been previously evaluated in 2-year carcinogenicity studies by the National Toxicology Program (1996). The major route of human exposure to phenolphthalein is via ingestion, dermal contact, and inhalation of contaminated air originating from process units manufacturing the compound.

In any case, phenolphtalein is a powder that is not very volatile and would not pose a big risk.
Commonly the formaldehyde resin is used for particleboard, I have never heard about the phenolphthalein resin being used for this purpose (but this is far from my area of work)

Of course, there is a simple test. Phenolphtalein changes color in a basic solution and looks like this

Dissolve the material in ethanol and add NaOH (drain cleaner), phenolphtalein will turn fuchsia.


Does a mixture of gases contain equal volume to that of the container they're in? Why? In my textbook, the derivation of Dalton's Law of Partial Pressure is done by keeping the volume of a mixture of gas in a container constant which is equal to the volume of container itself.

How can this be so? Let's say we have 100 moles (fairly large) of Nitrogen and 1 mole of hydrogen gas taken in a container of a certain volume. Then is the volume still considered equal to that of the container in this case? Because this is what is done when deriving Dalton's law of partial pressure?

Also, if we have 100 moles of a  gas and another just a single molecule then what happens?

Well, the word "container" means a fixed volume. Let's look at your room for example (don't take into account any openings, like doors or windows, just the walls). It has some amount of gas in it at this moment. Now, if someone decided to "push in" let's say 100 extra moles of O2 gas, then that would not change the volume of your room. The volume of the containter remains constant, no matter how much gas we put into in.

This is also true for each component of the mixture. The oxygen in the original air of your room had the volume of your whole room, and even when we "push in" that 100 extra moles, it still has the same volume - because your room did not change. What can change then? Obviously, the pressure or the temperature can change.


Why shouldn't Uranium come in contact with water? In the documentary about Chernobyl, it was mentioned that Uranium should not  come in contact with water, otherwise an explosion occurs. What is the reason for that? What kind of reaction makes it explode?

Interesting question. The documentary was almost certainly referring to the extremely hot lava generated by the nuclear meltdown at Chernobyl coming into contact with water. This lava is made of a substance colloquially known as 'corium' and is a combination of molten fuel rods, moderator, reactor walls and whatever else is melted by the incredible temperature of the reactor failure. This is as far as I know why the failure mode is called a 'meltdown', because the corium melts and pools in the bottom of the reactor vessel. In a molten pool, the fissile material may achieve uncontrolled criticality and just get hotter and hotter.

If the temperature is high enough the corium may melt through the bottom of the reactor. This is really bad, because at least hypothetically the lava may melt down to the water table and cause a steam explosion. This is quite a bit like what happens if you try to extinguish an oil fire by throwing water on it (something you should never do) - the water violently boils and expands and sprays burning oil everywhere. However in this case the burning oil is radioactive lava.

This is not a chemical reaction, per se, but rather a case of extremly vigorous heating of water and associated violent expansion.

Additional

If you remember the Fukushima disaster, at least one of the buildings was destroyed in a hydrogen explosion. As I understand it, in the event of a meltdown, the fissile material can get hot enough to effect the thermolysis of water. (The wikipedia article of the event seems to confirm this) Inside a sealed building, the water will develop a headspace filled with hydrogen and oxygen gas. This mixture is very explosive.


Solubility of urea in water I'm hoping that this isn't too obvious of a question (it's been awhile since I've taken chemistry), but I was looking at the solubility table on Wikipedia, which gives the solubility as grams per 100 mL. My understanding is that 100 mL of water weighs 100 grams.

So here's what I was having trouble wrapping my head around: it lists the solubility as 108 (at least at 20 °C). If I understand that right, that would mean that I can absorb 108 grams of urea in 108 grams of urea in only 100 grams of water, right? Can someone explain why it's possible to have more urea than water in this case?

Well, that is just the way that it is...

You should also look at this a bit differently. Urea has a molecular weight of 60 and water has a molecular weight of 18.

$\dfrac{108}{60} = 1.8$ moles urea

$\dfrac{100}{18} = 5.6$ moles water

$\dfrac{5.6}{1.8} = 3.1$ molecules of water for every molecule of urea


Is there an easy way to test for the amount of iron (Fe) in water? I have a rusty metal rainwater tank, and I'd like to find the amount of iron in the water. (The water I've collected is perfectly clear.)

How would chemists in the 1800's have gone about this? What chemicals/apparatus would I need to purchase?

So far, I've soaked a tissue in the water and am letting it air dry to see if any rust discoloration becomes apparent.

The following info from the WHO might be helpful:

  Iron (as Fe2+) concentrations of 40 µg/litre can be detected by taste
  in distilled water. In a mineralized spring water with a total
  dissolved solids content of 500 mg/litre, the taste threshold value
  was 0.12 mg/litre. In well-water, iron concentrations below 0.3
  mg/litre were characterized as unnoticeable, whereas levels of 0.3–3
  mg/litre were found acceptable (E. Dahi, personal communication,
  1991).

  In drinking-water supplies, iron(II) salts are unstable and are
  precipitated as insoluble iron(III) hydroxide, which settles out as a
  rust-coloured silt. Anaerobic groundwaters may contain iron(II) at
  concentrations of up to several milligrams per litre without
  discoloration or turbidity in the water when directly pumped from a
  well, although turbidity and colour may develop in piped systems at
  iron levels above 0.05–0.1 mg/litre. Staining of laundry and plumbing
  may occur at concentrations above 0.3 mg/litre (4).

  Iron also promotes undesirable bacterial growth ("iron bacteria")
  within a waterworks and distribution system, resulting in the
  deposition of a slimy coating on the piping (4).
  http://www.who.int/water_sanitation_health/dwq/chemicals/iron.pdf

In a rainwater tank in contact with air, dissolved iron is probably oxidized to iron(III) ($\ce{Fe^3+}$).

A simple and sensitive test for $\ce{Fe^3+}$ in water uses thiocyanate ions ($\ce{SCN-}$, also known as rhodanide), which form the blood-red coloured complexes $\ce{[Fe(SCN)(H2O)5]^2+}$, $\ce{[Fe(SCN)2(H2O)4]+}$, and $\ce{[Fe(SCN)3(H2O)3]}$.

$$\ce{[Fe(H2O)6]^3+ + SCN- <=> [Fe(SCN)(H2O)5]^2+ + H2O}$$

Take a precise volume of the water (e.g. $10\ \mathrm{ml}$), acidify with dilute sulfuric acid (e.g. $2\ \mathrm{ml}$), add a thiocyanate solution (e.g. $5\ \mathrm{ml}$ $\ce{KSCN}$ or $\ce{NH4SCN}$, $c =0.5{–}2\ \mathrm{mol/l}$), and fill up with distilled water to a precise volume (e.g. $25\ \mathrm{ml}$).

An orange to red colour should appear.

The colour of the solution can be compared to that of standard solutions that contain known concentrations of $\ce{Fe^3+}$, e.g. prepared from iron(III) nitrate ($\ce{Fe(NO3)3.9H2O}$).

In a laboratory, the intensity of the red colour could be measured using a photometer.

Similar simple test kits are commercially available, e.g. for aquariums, garden ponds, or other water or soil samples.


Can we prepare aryl halides by heating with halogen acids? My book says, we can't prepare aryl halides ($\ce{RX}$, where $\ce{R}$ is aryl group and $\ce{X}$ is any halogen) by treating phenol with halogen acids ($\ce{HX}$, where $\ce{X}$ is halogen). It is said to be, because of partial double bond character of carbon-oxygen bond in phenols ($\ce{C6H5OH}$), which is difficult to break (I don't know how there is partial double bond character). What I thought was, if we assume there is no partial double bond character between $\ce{C}$ and $\ce{O}$, we can heat the reactants (i.e. phenol and halogen acid), so that $\ce{OH}$ bond eliminates from the phenol (which has the least bond dissociation energy than $\ce{CH}$), and $\ce{X}$ can get dissociated from the $\ce{HX}$, then it can add up to $\ce{R}$ (aryl group) to form aryl halide. So, can we prepare aryl halides by heating with halogen acids?

The partial double bond character of the C-O bond in phenol becomes obvious from the resonance structures. Contributions from the structures with $\ce{C=O}$ double bond explain why the character of the C-O bond is in between single and double bond, and also show that oxygen lone pairs are less available for protonation to form $\ce{Ph-OH2+}$.

Protonated phenols are therefore fairly acidic, with a $pK_a$ around -6. $\ce{HBr}$ and $\ce{HI}$ have significantly lower $pK_a$ values (-9 and -10, source), so they would indeed be able to protonate phenol. However, $\ce{H2O}$ is unlikely to be eliminated from this compound due to the instability of the resulting phenyl cation. Its positive charge cannot be delocalized into the aromatic system as it resides in an empty $sp^2$ hybrid orbital which is perpendicular to the delocalized $\pi$ bonds (source). Formation of the phenyl cation by elimination therefore requires much better leaving groups like $\ce{N2}$, which can be eliminated from aryl diazonium cations.

Substitution of the protonated OH group by halides with an $S_N2$-like mechanism is hindered by Coulomb repulsion between the $\pi$ electron cloud of the benzene ring and the lone pairs of the nucleophile. This reaction only works with aromatic compounds in which the electron density in the $\pi$ system is decreased by multiple electron-withdrawing substituents, like nitro groups.


Stoichiometry Chemistry Lab: Copper (II) Sulphate Pentahydrate and Iron Single Displacement Reaction? In my grade 11 chemistry class, we performed a lab in which we added 1.00g of iron to 6.00g of copper (II) sulphate which was dissolved in water. Using stoichiometry, we predicted that if the iron was ferrous iron, 1.14g of copper would be produced and if the iron was ferric iron, 1.71g of copper would be produced. We waited 24 hours for the copper to dry and then we weighed the copper. The copper had a total mass of 1.24g. We then asked our teacher how this could be as we could not have ended up with more copper than 1.14g (as she already told us that the copper was ferrous). She told us that we need to make sure we rinse the copper really well.

So we decided to do a second trial. We were extra careful in our measurements and we made sure to rinse the copper well with distilled water. We waited 24 hours for the copper to dry and weighed the copper. The copper had a total mass of 1.21g. This was an improvement, but we also understood that we cannot have a percentage yield of over 100%, while ours was 106%.

What could have caused the copper to have a mass larger than predicted? Also, what was the reasoning behind my chemistry teacher advising us to rinse the copper before we let it dry?

Note: I can provide photos and calculations, as well as any other information if it is needed. Thank you!

Your results are actually very good.  The reason for thoroughly washing your copper is to remove any residual water-soluble un-reacted reagents.  If the copper were simply dried by sitting in room-air, the bulk of your extra 6% could still be water.  That's only 70 mg or 70 ul of water.  They way you produced your copper would likely result in a high-surface area product which can be tricky to dry.  Ideally, you would do multiple rinses in hot DI water then dry overnight at 105 C (really 70+C should do it).  If your yield goes down just a couple percent or so, you can still verify the excess water theory by repeating the drying.  If your yield goes down again then you are just having trouble getting all the water out.  Make sure you cool the copper, in a desiccator, before weighing.

If you don't have lab time for this here's what I'd try: take you copper home and put it an oven on low, about 200F or so, for several hours.  While still hot, place in the smallest reasonably air-tight glass jar you can find an get it back to the balance ASAP!  Good luck!  Oh, and NEVER touch the copper with your sweaty, oily fingers ;-)


Why can't sodium borohydride reduce carboxylic acid and ester groups? Sodium borohydride ($\ce{NaBH_4}$) is a reducing agent used in various organic reaction to reduce aldehydes, ketones and acyl chlorides.
Why can't sodium borohydride reduce carboxylic acid groups and esters? Why can it reduce acyl chlorides but not esters, both which are derivatives of the carboxylic acid group?
I read somewhere that it had to do with the oxidizing property of sodium borohydride, but I'm not sure if it is correct or how it explains the same.

The reason for this is because OH and OR groups are much more electron-donating. Typically, when you have an aldehyde or ketone, the electronegative electron is slightly more negative and thus the carbonyl carbon is more positive. However, with an OH or OR group attached (i.e in carboxylic acid and esters), the group can push electrons towards the carbon since they are electron-donating groups. This makes the carbonyl carbon less positive and less open to nucleophilic attack.


Openbabel: Generate 3D depiction? The process of creating 2D depictions is explained in detail here. This way, it is possible to create a 2D image via running e.g.

obabel  -O 2d_depiction.png

However, how do I generate a ball and stick 3d depiction (as image) instead?

Open Babel does not generate "ball and stick 3D depictions" directly, because it's not a molecular visualization program.

That's why we started Avogadro.

There are a few formats supported in Open Babel to write out 3D depictions, e.g.:

POV-Ray
STL - for 3D printing
Point Cloud - outlining the Van der Waals surface

But you can make a pseudo-3D "ball and stick" depiction in the SVG format:

obabel [file] -O file.svg -xS


What is the name of the crinkly plastic used in clear parts of envelopes? I've always wondered what it is, as it makes a lot of noise on being deformed. No other common plastic that I've come across has quite the same properties.

It usually is BOPS: Biaxially oriented polystyrene.


How many critical points? Why? Critical points are defined (by wikipedia) as the "end-point of the phase equilibrium curve".

I would say this diagram therefore has three critical points. But this is unnatural. I doubt its true.

Other curves are different from the ones with critical points because "they can run to infinity". But I think thats ridiculous because its physically impossible.

I don't know if I'm interpreting PDs correctly but I think that what defines that you are on a phase in these diagrams is that if you travel along a horizontal line you meet with a phase equilibrium curve.

If you don't then you were either on a single phase the whole time or on a sort of "critical phase".

I don't like it though. For example if you travel along a horizontal line that wont intersect the L-V curve, how would you know when the LV critical phase is present? Regions are not clearly defined here, except for monoclinic.

Whats wrong with my ideas?

How many critical points does this diagram really have and why?

Thanks.

This phase diagram has one critical point - it is where the line separating the liquid and gas comes to an end - it is now one fluid. So, you can go from liquid to gas (or vice versa) by swinging T,P around the critical point without undergoing a first order phase transition.

The points with arrows pointing to them are triple points, which are also special points, but different from the critical point. Movement away from a triple point means that one (or more) of the phases have to undergo a first order phase transition to the stable phase(s).

Thermodynamically, these are very different things.


Can buffer solutions be used to control the pH of ethanol? For my experiment, I want to investigate the effect of differing pH on the solubilities of a sparingly soluble drug. However, the solubility is too low for me to conduct a reliable study on the solubility (0.003 mg/ml in water). In ethanol, the solubility increases to 10 mg/ml. Can I simply add buffer solutions and acids into ethanol to change its pH as I would do with water? Are there alternatives to adjusting the pH of ethanol?

Although pH scale is rather strictly used for aqueous solutions, the concept of pH in organic or non-aqueous solutions is not trivial. It is called apparent pH or operational pH when organic solvents are present. There is a practical problem here as well, do you think your buffers will dissolve in ethanol? Once you figure this out, you can read a section in  "pH measurements in Nonaqueous Solvents" by Laitinen & Harris, “Chemical Analysis ─ An Advanced Text and Reference". It is a very useful book.


Mixing naphthalene, formaldehyde and sulfuric acid (Marquis reagent + naphthalene) Excuse me if this a dumb question but it has haunted me for years. Back in my old school days I mixed  naphthalene, formaldehyde and sulfuric acid in a mad attempt to make $\mbox{2-Methoxynaphthalene}$. I can't remember if I heated the mixture or not.

It turned black instantly and my teacher at school couldn't tell me what had happened, even less why. Did I produce nano-carbon? Just soot? Or something else?

Edit:
way back then, I did try the reaction without naphthalene, and also without formaldehyde, with no (noticeable) result. My teacher also repeated the experiment with the considerably more pure chemicals available at the school lab.

Edit2: as of the comment by Loong, what I did was using Marquis reagent on naphthalene. What I found on the net about this was very little and poor. I couldn't even find anything on the color of $\mbox{polymethylnaphthene}$. (Instinct tells me it should be colorless.)

Marquis Reagent (1) is a spot test for alkaloids, first reported in 1896. The original agent was a mixture of 2 drops of 40% formaldehyde and 3 milliliters of concentrated sulfuric acid and was used to detect certain (e.g., opium) alkaloids and distinguish between them. The signature of the alkaloid is both the initial color produced and the sequence of color changes with time. Morphine reacts with Marquis' Reagent to give a purple to violet color. Two molecules of morphine and two of formaldehyde are proposed to condense in the presence of sulfuric acid to the dimeric product which is protonated to the oxoniumcarbenium salt.
The key reaction is protonation of formaldehyde, which attacks an aromatic ring, ortho to the activating phenol group, producing a hydroxymethyl derivative which is immediately protonated and dehydroxylated to give a benzylic cation which attacks another aromatic ring. And it happens again at ortho positions to close a new ring.

Now the ingredients in your old school days reaction (naphthalene, sulfuric acid and formaldehyde) are used to make what was a common concrete admixture. Kao Soap (2), who made Mighty 150 (“Naphthalenesulonic acid-formaldehyde condensate sodium salt”) offered it to my company. We were making a similar material for paint dispersion and promoted our material for concrete strengthening by water reduction.
The condensate is a polymer and is made by first sulfonating napthalene in the beta position with concentrated sulfuric acid. (The alpha sulfonate is the first product, but quickly rearranges to the more stable beta isomer.) Then formalin, 40%, is added, slowly. The formaldehyde protonates, attacks the alpha position of the (still activated) unsulfonated ring (the sulfonated ring is deactivated), forms a hydroxy methyl aryl group, is immediately protonated, dehydrated and attacks another naphthalene sulfonate in the alpha position of the unsulfonated ring.
$H_2SO_4$ + $H_2CO$ --> $H_2COH^+$ +$HSO_4^-$
$H_2COH^+$ + $HO_3SNap$ --> $HO_3SNapCH_2OH$ + $H^+$ -->
$HO_3SNapCH_2^+$
$HO_3SNapCH_2^+$ + $HO_3SNap$ --> $HO_3SNapCH_2NapSO_3H$
This methylene group is the first link of a polymer chain. The other alpha positions (one on each unsulfonated ring) are still open and will be attacked if there is enough formaldehyde. Controlling the amount of formaldehyde added is a method of controlling ultimate molecular weight of the polymer, which affects its dispersing power because of the ability of longer polymer chains to adsorb more strongly and repel other particles without deabsorption of the polymer.
Controlling the rate of formaldehyde addition is key to preventing runaway polymerization. The addition of formaldehyde at near room temperature lowers the temperature of the 500-1000 gallon batch; however, the batch temperature must be raised to cause polymerization, which is exothermic - sulfuric acid generates heat as it hydrates - and the batch viscosity is already very high, so the heat cannot be quickly removed.
There is a possibility of a runaway reaction in your mixture: the naphthalene is not even sulfonated, so both rings are activated (relative to benzene). If you did get 2-methoxynaphthalene as an intermediate, the sulfuric acid would protonate the methoxy oxygen, generating a benzylic cation, and water and heat of hydration, and a polymer with chains coming out of up to 4 alpha positions per naphthalene. Now the typical naphthalene sulfonate condensates (the acids and the sodium salts) are brown (light to dark), but I could imagine a polymer coming out with a black color.
This approach to 2-methoxynaphthalene seems destined to fail, because the protonation of 2-methoxynaphthalene is easier than  protonation of formaldehyde. But organic chemists frequently generate tars.
Ref.1 https://en.wikipedia.org/wiki/Marquis_reagent
Ref.2 https://chemical.kao.com/en/products/B0012344_en/?region=gl


How to dialkylate a cyclohexene, and convert a carbonyl to an alkene? I'm trying to design a synthesis of D from C. All I can think of is using $\ce{Me2CuLi}$ to add the methyl group to the alkene. I thought maybe then if isobutyl bromide was in the reaction mixture when $\ce{Me2CuLi}$ pushed the negative charge onto the oxygen and it came back down that could do an $\mathrm{S_N2}$ with isobutyl bromide, but I've never seen that before. Then for converting the carbonyl to an alkene I have no idea.

You are right about using the Gilman in a 1,4-Michael addition. The enolate created from this can then be used in the nucleophilic attack of a haloalkane. Then all that's left to do is a Wittig reaction, swapping the carbonyl oxygen for a carbon. The full stepwise synthesis would be:


Why is 5,6-dimethylidenecyclohexa-1,3-diene non-aromatic? The compound 5,6-dimethylidenecyclohexa-1,3-diene is supposedly non-aromatic:

But I have no idea why this is so. The compound is probably planar. All the atoms on the ring are $\ce{sp^2}$ hybridized. So, the p-orbitals are in conjugation. There are 4-$\ce{\pi}$ electrons, as I suppose the external pi bonds do not participate in resonance. So, it should be antiaromatic. Why is it non-aromatic?

I am a school student, so an detailed description would be helpful.

In this question, conjugation of p-orbitals in the system is misrepresented. It should be noted that, a π-system to be an aromatic or anti-aromatic, there should be a closed loop(s) of p-orbitals (conjugated), which is planner. What makes it an aromatic or anti-aromatic whether it follows Hückel's
rule of (4n+2) π-electrons or not.

Aromatic (most stable): If a planer, closed loop(s) of p-orbital system has (4n+2) π-electrons (in the closed system), it is aromatic. Example:, benzene (a closed p-orbital system with six π-electrons) and naphthalene (a closed p-orbital system with ten π-electrons).

anti-Aromatic (least stable): If a planer, closed loop(s) of p-orbital system has (4n) π-electrons (in the closed system), it is anti-aromatic. Example:, cyclobutadiene (a closed p-orbital system with four π-electrons) and cyclopentadiene carbocation (a closed p-orbital system with four π-electrons).

non-Aromatic (stability depends on the system): Every other systems are non-aromatic. Examples: 5,6-dimethylidenecyclohexa-1,3-diene and heptafulvene (In both cases, each p-orbital system is not closed).

See following diagram:


crystalline structure of Si wafers used for MEMS When they say that we have a {1 0 0} single crystal wafer, I know that the surface normal to the wafer is oriented in the <1 0 0> family direction, but if I draw a line on the wafer, parallel to my primary flat, does it indicate a crystallographic direction family too? Like <1 0 0> or <1 0 1>. I mean it's obvious that it cannot be <1 1 1> due to the FCC lattice of Si(I am making the assumption that the thin plane of the primary flat is at exact right angles to the wafer surface, if that angle is allowed to be 54 degrees, it can also be a <1 1 1> ), but it can surely be directed at none of the crystallographic directions at all, so the question is, if I draw a line on the wafer surface parallel to the primary flat, will it be directed towards the <1 0 0> or the <1 0 1> family, or does the manufacturing process just slice an arbitrary flat out of the Si ingot before slicing it into wafers?

The flats on Si wafers are not arbitrary.  The largest, or primary flat, is perpendicular to the <110> orientation in the plane of the wafer. If it exists, the secondary flat helps identify the wafer orientation and doping type (and can easily vary based on wafer maker and the company ordering the wafers). All modern wafers (anything 200mm or bigger, almost all 150mm) are (100) and boron doped, so the secondary flat isn't used.

The primary flat has a specific orientation since the chips need to be aligned along the <110> crystallographic orientations so the wafer can be diced up.  This usually uses a diamond scribe and careful breaking rather than sawing, because the saw is wider than a scribe line (so fewer chips/wafer) and dirtier .

So, you wafer has <001> pointing up, and <110> in plane towards the flat.


Why is the less substituted alkene favoured in Hofmann elimination? In the reaction, $\ce{OH-}$ group attacks a hydrogen ion in an E2 reaction, forming a transition state in which simultaneously $\ce{NCH3}$ is removed. In this reaction, the $\ce{OH-}$ should attack, like in other case of E2 so as to form the Zaitsev product, but due to "steric hindrance" the less substituted alkene is formed. But surely the $\ce{OH-}$'s attack is not affected by $\ce{CH3}$ groups? It can be affected by the $\ce{NCH3}$ group - but whichever way it attacks, it's going at the beta carbon - so is at the same distance from that group in all cases. How can steric hindrance affect its attack? Why is the Hofmann product (less substituted product) favoured?

There are 3 main types of elimination reactions, the E1, E2 and E1cb.  Here is a generalized picture of the transition state for the E2 elimination reaction.

Breaking of the $\ce{C-H}$ bond and the $\ce{C-LeavingGroup}$ bond are roughly equally advanced, as is formation of the double bond. Because the 2 bonds are broken to roughly the same degree, it can be said that the E2 reaction is in the "middle" of the possible elimination mechanisms.

If we change this pattern and make the $\ce{C-H}$ bond breaking more advanced than the $\ce{C-LeavingGroup}$ bond breaking (we can do this by making the leaving group a poor leaving group [i.e. groups like ammonium, $\ce{-NR3^+}$, sulfonium, $\ce{-SR2^+}$, etc.]), then we shift the mechanism towards the E1cb side of the elimination spectrum.

In the E1cb case, because breaking of the $\ce{C-H}$ bond is advanced, partial negative charge develops on the carbon where the proton is leaving from.  Now primary carbanions are more stable than secondary carbanions, which, in turn, are more stable than tertiary carbanions (electron releasing alkyl groups will destabilize a negative charge).  Therefore, when we have a choice of which proton to remove, we remove the proton on the least substituted carbon as this carbon will better stabilize the developing negative charge.  As the LeavingGroup departs and the reaction completes, this will produce the olefin with the least substituted double bond, the Hofmann elimination product.

Sometimes it is argued that base size is responsible for elimination patterns and that when a large base is used the proton on the least hindered carbon is abstracted.  This is true but the effect is relatively small.  Changing the base from methoxide to t-butoxide only increases the amount of Hoffmann product by a modest 10-20% in simple haloalkanes.

So while sterics play a role, it is usually a small role with the electronic factors discussed above being more important.

For completeness, let me briefly mention the E1 elimination which is on the "other side" of the elimination spectrum of mechanisms. In this case, breaking of the $\ce{C-LeavingGroup}$ bond is advanced and partial positive charge develops on the carbon where the LeavingGroup is departing from.

Elimination of a proton from this intermediate can often proceed in two directions.  Since more highly substituted double bonds are more stable than less substituted double bonds and since the transition state for proton elimination involves partial double bond character, the transition state  leading to the more stable double bond (e.g. the more highly substituted double bond) will be lower in energy.  Hence, the product with the more highly substituted double bond (Zaitsev product) will form in E1 reactions.

Edit: Response to OP's comment

  I meant the pyrolytic elimination of quaternary ammonium hydroxide,
  which follows E2 mechanism

I wasn't saying that this reaction (the thermal Hofmann elimination) goes through an E1cb mechanism, rather I said "we shift the mechanism towards the E1cb side of the elimination spectrum."  In other words, it is an E2 reaction but $\ce{C-H}$ bond breaking is somewhat more advanced than $\ce{C-LeavingGroup}$ bond breaking. Therefore the primary $\ce{C-H}$ bond is more likely to be involved than the secondary or tertiary $\ce{C-H}$ bond, for the reasons outlined above.

Here's another approach.  Instead of electronic factors, let's look at steric factors - in this reaction they both lead to the same conclusion.

E2 eliminations prefer a trans or antiperiplanar geometry.  Below are Newman projections of the 2 conformations with this geometrical arrangement of the substituents.

Conformation A is set up to eliminate towards the least substituted carbon, the methyl group (Hofmann), while conformation B depicts elimination towards the more highly substituted carbon (Zaitsev).  The trialkyammonium group is extremely large and generates an unfavorcbale steric interaction in conformation B.  This makes conformation B higher in energy, hence elimination from conformation A to produce the less-substituted double bond will be preferred.


What does it actually mean by stability of an atom There are several rules related to stability of an atom, like octet rule, half filled or fully filled d orbital etc. But none of them (at my best knowledge) does not explain why an atom is stable when it obeys those rules. So, is there any reason at all or are they just based on experimental results?

First you need to understand what stability is. Normally, octet rule is based on noble gas configuration. Since, noble gases hardly react with other atoms, and they have 8 electrons as their shell configuration, thus we consider them "stable".

So, taking noble gas 8 electrons shell configuration, we tend to generalise all other atoms/elements to have 8 electron filled orbit to gain stability.

You must understand that scientist create models to explain natural phenomena. Rules are  designed to explain already existing phenomena.  Stability truly means "less energy" or inert behaviour where the atom neither gains nor lose electrons to other atoms/molecules. The lesser the energy of the atom/system, more the stability.

Thus your doubt on "does not explain why an atom is stable when it obeys those rules" is very vague. Think of inert gases, their configuration, their inert nature and you will understand.


Replacement For RF sputtering In my experimental work, I'm following an article. It said that:

  A 30 nm ZnO seed layer was deposited on the glass substrates by RF sputtering of a ZnO target at room temperature.

But I haven't any tools for "RF sputtering". Is there any replacement for this technique?

If you need a high quality thin film of $\ce{ZnO}$, especially on insulating substrates, you're mostly limited to sputtering techniques, PVD (including things like pulsed laser deposition), CVD (plasma-enhanced CVD, spray pyrolysis, and atomic layer deposition are common). Unfortunately, all of these require fairly sophisticated instruments. More accessible methods include sol-gel synthesis and electrodeposition.

The sol-gel synthesis of $\ce{ZnO}$ only requires a spin coater and an oven that can do 600 °C. It's suitable for coating large substrates, but if you really need a layer that's only 30 nm thick, it may not be the best technique.

$\ce{ZnO}$ can be electrodeposited onto conductive substrates from aqueous solutions. This requires a potentiostat and controlling thickness and orientation can be tricky. One possible method: $$\ce{Zn^2+ + NO3- + 2e- -> ZnO v + NO2-}$$

I should add that if you don't have access to any of these tools, there are certainly commercial suppliers of $\ce{ZnO}$ coated glass substrates.


How does X-Ray Crystallography determine the components (as opposed to the structure) of a molecule? Thanks for some online resources (i.e. here) I have an understanding of what X-Ray Crystallography does.
By blasting X-Rays at a crystal and recording the scattering patterns on a plate behind the crystal, it becomes possible to determine the shape of the molecule in the crystal.

Where I'm confused though is in diagrams of protein molecules mapped out by x-ray crystallography, scientists are, incredibly, able to map out extremely complex structures. But what I don't understand is, how are they able to determine which atoms, or other molecules, they are seeing in the structure?
How do they know which part of the structure is which amino acid? Which part is hydrogen? Which part is nitrogen? Etc. I don't see any way of determining this from a diffraction pattern on a plate.

That is because the solution to the structure isn't done just via the plate. The signal you see on the plate is what is called "structure factors" of the experiment. They are associated with the fourier transform of the electron density. There are computational ways of obtaining the electron densiy once you have the structure factors. When you do that, you arrive at something like this:

Then, once you have something like that, at least for proteins, you need to find the best possible geometry that fits the density field. So if you succeed you might end up with something like this:

The trick here is that the molecular biologists and physicists who work with proteins already know something of what they are looking for. They don't try to fit a haphazard collection of atoms to the electron density map, they start with a polypeptide chain, often one for which the sequence has already been experimentally determined so they know a lot about the composition of the molecule well before any structure resolution takes place.
This is the tl;dr version of the thing. In reality there are a number of things I have omitted. For instance, it isn't always necessary to obtain the electron density in order to resolve the structure. There are ways to theoretically compute the structure factors from a test geometry such that a structure is considered resolved if the theoretical and the actual form factors coincide up to a certain tolerance.
The structure is also resolved in steps. Often when you crystallize something, a lot of elluents and additives are added to the mix in order to make the crystallization possible, particularly for proteins, which have very fragile and sensitive crystallization behaviors. So researchers first try to resolve the geometry of the main protein chain, after that is done there will be residuals which can be adjusted by trying to find molecules of the additives and elluents in the unit cell. That is how we end up with sulfates, waters and other small molecules and analytes floating amidst the protein structure for the majority of the PDB structures resolved via X-Ray crystallography.
That said, for very small molecules and an electron density with a very high resolution, it is possible to distinguish one atom from the other by the shell structure around them. But, again, composition is often an information that the researcher has a priori, not something the diffraction is expected to provide.
P.S. If you have time, you can check the Yale Course of Professor McBride Freshman Organic Chemistry I. It is totally free an in the internet. In it, he does a very pedagogical and awe-inspiring introduction to the ins and outs of X-Ray diffraction (and other experimental techniques) that go well beyond the scope of this answer. He does not focus on proteins, but on the basics, small organic molecules, but it does wonders to clear the fog. The strictly relevant classes are classes 5 and 6, but I desperately urge you to go through everything that came before instead of just jumping in.


Lower Heating value calculation from chemcial combustion equation of Butane QUESTION
    Butene, C4H8 is burned in an engine with a fuel-rich air-fuel ratio. Dry  analysis of the exhaust gives the following volume percents:                   $\ce{CO2} = 14.95%$, $\ce{O2} = 0%$, $\ce{CO} = 0%$, $\ce{H2} = 0%$, $\ce{C4H8} = 0.75%$, with the rest being $\ce{N2}$. Higher heating value of this fuel is 46.9 MJ/kg. Write the balanced chemical equation for one mole of this fuel at these conditions. Calculate the air-fuel ratio, equivalence ratio, lower heating value of the fuel, and energy released when one kg of this fuel is burned in the engine with a combustion efficiency of 98%.

My attempt at solution:
I write the stoichiometric equation like so:
$$ \ce{C4H8 + \frac{1}{\phi}(4+(8/4))(O2 +3.773N_2)\rightarrow n_1 C4H8 +n_2 CO2 +n_3 H2O}$$
where, $\phi$ is the equivalence ratio(a engineering term used to describe the ratio of $(F/A)_{\text{actual}}$ to $(F/A)_{\text{stoichiometric}}$.

So,now I have 3 unknowns, $n_1$, $n_2$ and $\phi$.

I unfortunately get $n_2=0$ after doing the mole balance of carbons,hydrogen and nitrogen.

I understand that the percentages given by volume are the mass fractions which should later form a separate equation. But, I still get $n_2=0$. Where have I gone wrong?(Should I still consider putting in $\ce{O2}$ and $\ce{CO}$ into the above equation even though their %by volume are zero?)

I see three flaws in your reasoning.

First, butane is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of butene. I will keep using your formula, but you should specify whether you mean butane or $\ce{C4H8}$.

Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables.

$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$

Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:

$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}$$

To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.

$$\ce{C4H8 + (6-6n_1)O2 -> n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$

Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.


Ethanol for wood polishing We use ordinary spirit to polish wood, and it contains ethanol. What makes ethanol a good polishing agent for wood? What effect does it causes on wood so that it shines?

Ethanol is an organic solvent. I would guess that it makes wood shine by:

Cleaning the surface, dissolving away organic compounds
Maybe slightly dissolving (?) a thin layer of cellulose at the surface of the wood, removing irregularities and making it more malleable, hence easier to polish

There may be more specific interactions between wood and ethanol, but I don't know about them.


Can isotopes of a given element be represented by different symbols? Can isotopes of any given element be represented using a completely different chemical symbol? What's the IUPAC's take on this?

Sure, ordinarily you would add the isotope's mass as a superscript to the element's symbol to differentiate it from other isotopes: For example, carbon-12 ($\ce{^{12}C}$) and carbon-14 ($\ce{^{14}C}$); however the base-symbol $\ce{C}$, for carbon, doesn't change.

But the isotopes of hydrogen don't seem to follow this strictly. Often, I see deuterium ($\ce{^{2}H}$) and tritium ($\ce{^{3}H}$) represented by $\ce{D}$ (I see this one in organic chem textbooks a lot) and $\ce{T}$ respectively. Does this "convention" fit in with IUPAC norms? If so, can isotopes of other elements be represented differently as well?

IR-3.3.1 Isotopes of an element
The isotopes of an element all bear the same name (but see Section IR-3.3.2) and are
designated by mass numbers (see Section IR-3.2). For example, the atom of atomic number
8 and mass number 18 is named oxygen-18 and has the symbol $\ce{^{18}_{}O}$.
IR-3.3.2 Isotopes of hydrogen
Hydrogen is an exception to the rule in Section IR-3.3.1 in that the three isotopes $\ce{^{1}_{}H}$, $\ce{^{2}_{}H}$ and $\ce{^{3}_{}H}$ can have the alternative names protium, deuterium and tritium, respectively. The symbols D and T may be used for deuterium and tritium but $\ce{^{2}_{}H}$ and $\ce{^{3}_{}H}$ are preferred because
D and T can disturb the alphabetical ordering in formulae (see Section IR-4.5). The
combination of a muon and an electron behaves like a light isotope of hydrogen and is
named muonium, symbol $\ce{Mu}$.⁵
These names give rise to the names proton, deuteron, triton and muon for the cations $\ce{^{1}_{}H+}$, $\ce{^{2}_{}H+}$, $\ce{^{3}_{}H+}$ and $\ce{Mu+}$, respectively. Because the name proton is often used in contradictory senses, i.e. for isotopically pure $\ce{^{1}_{}H+}$ ions on the one hand, and for the naturally occurring undifferentiated isotope mixture on the other, it is recommended that the undifferentiated mixture be designated generally by the name hydron, derived from hydrogen.

Source:
N.G. Connelly, T. Damhus, R.M. Hartshorn, A.T. Hutton (eds) (2005). Nomenclature of Inorganic Chemistry (PDF). RSC–IUPAC. ISBN 0-85404-438-8.

Addendum:
The small subscript ⁵ present in the source is a reference to Names for Muonium and Hydrogen Atoms and Their Ions, W.H. Koppenol, Pure Appl. Chem., 73, 377–379 (2001) which can be viewed over at:
https://www.iupac.org/publications/pac/pdf/2001/pdf/7302x0377.pdf

$\cdots$ A particle consisting of a positive muon and an
electron ($\pu{\mu^+ e^–}$) is named “muonium” and has the symbol $\ce{Mu}$. Examples: “muonium
chloride,” $\ce{MuCl}$, is the equivalent of deuterium chloride $\cdots$


Oxidation States of D Block Elements These might be basic question but they are quite confusing to me?

If $\ce{Sc+}$ doesn't exists, why does $\ce{Ti^{2+}}$ exist?

on the similar account why does this trend increases at first( till Mn) and then decreases?
My book just concludes that the general trend is due to increase in $Z_{\mathrm{eff}}$ and decrease in atomic radius.

A little insight will be really  appreciated.

Actually an organometallic scandium(I) compound is known, from P.L. Arnold at al. (1996).  As with organometallic compounds generally, the metal is covalently bonded, and the use of $3d$ orbitals as valence orbitals (there are 14 shared valence electrons associated with each scandium atom) in this and similar compounds favors including scandium as a transition metal.

In predominantly ionic settings, however, it is true that scandium(I) would be unlikely.  Because scandium is only at the beginning of the transition metals it has relatively low ionization energies for its valence electrons, and this low energy cost tends to be overcome by increasing lattice or solvation energy associated with higher ion charges.  With ionically bonded scandium the balance is all in favor of more ionization and more electrostatic attraction, whereas with titanium the slightly higher ionization energies and the fact that a noble gas configuration requires four rather than three positive charges makes lower charge ions more accessible.


Is the colour of leaves due to d-d transitions in chlorophyll? While reading a chapter on transition metals, I came to know that d-Block elements have colour compounds due to d-d transitions.  A question came to my mind as to whether there is any d-d kind of transitions in plants as they also show characteristics colours.

I know the green colours of leaves is due to presence of chlorophyll and also there are coloured pigments in plants.

I may be wrong in my thought, but I really want to clarify my doubt.

Let's rephrase your hypothesis and test it:

  The green colour of leaves results from a d-d transition of the central atom of metal complexes known as chlorophylls.

Green leaves contain two types of chlorophyll, Chl a and Chl b, both have $\ce{Mg}$ as the central atom and show two distinct bands in the UV-VIS absorption spectrum.

The stronger ($\epsilon \approx 100000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) band in the 400 nm (blue) region is known as the Soret band, the weaker  $\epsilon \approx 30000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) in the 660 nm (red) region is known as the Q band.

Now imagine that Chl a is treated with weak $\ce{HCl}$. In the resulting product, known as Pheophytin a, the central atom ($\ce{Mg^{2+}}$) is replaced with two protons.

In the UV-VIS absorption spectrum, however, we would still observe the two bands. They are slightly shifted and show somewhat different extinction coefficients, but the pattern is fundamentally the same.

We may conclude that our initial hypothesis is wrong. If it is not the absorption of the central atom in our complex, the ligand must be responsible.

Our new working hypothesis therefore is:

The colour of green leaves results from the absorption of the polycyclic pheophytin ligand and its extended $\pi$ system.

Let's simplify the Pheophytin a molecule: Removal of the E ring and replacing all the fancy substituents with ethyl groups leads to octaethylporphyrin as the core structure.

In the UV-VIS spectrum, we would still be able to see the strong Soret band, which corresponds to a $\pi,\pi^*$ transition from the ground state to the second excited singlet state ($S_0 \rightarrow S_2$).


How to optimise battery usage on modern electronics? What is the best way to use an external power bank with a modern smart phone or similar device if the goal is to get as much usage out of the combination?

Premises:

The devices has a battery of 2 Ah and the power bank's capacity is 10 Ah
As I understand it, a rule of thumb is that the thermal efficiency η when charging a battery is about 50 %. That is, charging the power bank from (1) requires 20 Ah.

With 1 and 2 in mind, the power bank could charge the phone from 0 to 100 % 2.5 times.

Therefore, my question is: wouldn't it be better to keep the power bank attached to the phone all the time to keep the phone at 100 % as long as possible rather than using it to charge the phone when the battery is half empty or so? My idea is that keeping the power bank attached all the time would make η ≈ 100 % since the phone's battery would just "pass by" the current from the power bank. Am I right?

There is always some internal resistance in a battery. All other things being kept constant, the larger the battery, the less internal resistance, so less energy is wasted inside a larger battery due to internal heating and energy loss.

This is particularly important in high-drain devices such as phones and laptops, where batteries are discharged in just a few hours. By having the power bank in parallel, the load is shared and batteries are being drained over a longer time.

For example, a once-cell penlight with an incandescent lamp might burn for an hour. Two cells, one replacing the other, would then last two hours. But two cells in parallel might last 2.2 hours, because of the lower internal resistance and decreased energy waste in the cells.


The property F in an equation for chemical potential While studying the lamba sensor that works on yttria-strabilised-zirconium (YSZ) I came across the following equation:

$$p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$

What is $F$ in this equation?

Some more info

The "wall" of the lambda sensor is as shown in the picture with the voltage $V_{ref}$ measured across:

The equation apparently is derived directly from the chemical equilibrium (chemical potentials equals to zero):

$$\mu_{O_2}+4\mu_e=2\mu_{O^{--}}$$

They come to the following where they end at the final formula:

$$V_{ref}=\frac{RT}{4F}\ln\left(\frac{p_{O_2}^{gas}}{p_{O_2}^{ref}}\right) \implies p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$

I understand the partial pressures. The gas constant $R$ and temperature $T$ are also clear. If I am correct, $V_{ref}$ is the voltage across the material of the sensor wall, as in the picture, so this is also clear.

The equation is the final result from which the oxygen amount in the gas can be found, since the partial pressure of the oxygen in the gas can be calculated. My question is: What is $F$ in the equation?

$F$ is the Faraday constant: $9.64853399(24)\times10^4~\mathrm{C~mol^{-1}}$.

A quick dimensional analysis confirms that this makes sense.

$$\frac{\mathrm{J~K^{-1}~mol^{-1}~K}}{\mathrm{C~mol^{-1}}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V}$$


What is the geometry of an alkyl radical? I encountered this question which asks how many possible radicals are formed when $\ce{CH3CH2C(CH3)3}$ is monosubstituted by $\ce{Br2}$. The answer given is 3 while I think it  should be 4, reasoning that there is a "chiral" carbon if the radical is formed on carbon 3. While arguing with my tutor two questions came to my mind:

What is the geometry of the $\ce{CH3\dot{\ce{C}}HC(CH3)3}$ radical?
If the radical is trigonal pyramidal, does it transition quickly between the two possible states? If so are they considered two radicals?

My findings so far:

This answer on SE hints that the exact shape depends on the substituents on the alkyl radical.

Modern Physical Organic Chemistry states that all other (non-methyl) localized radicals are not planar.

Organic Reaction Mechanisms states that the geometry of free radicals is still controversial.

First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.

In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar.  On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons.  Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.

This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center.  In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so.  Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.

So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers.  Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.


Good chemistry experiments/projects Im not sure if this is the right place to ask, but what are some good chemistry experiments or projects that could be conducted in grade 12?

Here's what I have found so far:

Preparation of aspirin
Synthesis of dyes
Analysing the percentage of metals present in coins

In my opinion, these projects are not something that grade 12 students should be doing, I'm looking for something more complex, more in touch with research being conducted in universities.

Edit: Please refrain from suggesting topics that involve spectroscopy or crystallography or any other equipment unlikely to be found in a high school lab.

What about browsing across

the section Laboratory Experiments in the Journal of Chemical Education?  This month (march issue), for example with "Unboiling an Egg: An Introduction to Circular Dichroism and Protein Refolding" by Hoben et al., (DOI:10.1021/acs.jchemed.6b00319), or "Demonstrating the Effect of Surfactant on Water Retention of Waxy Leaf Surfaces" by Chiu et al., (DOI: 10.1021/acs.jchemed.6b00546)
more broad than the journal, visiting educational ressources the American Chemical Society offers
as an example from the other side of the pond, a collection provided by the Royal Society of Chemistry (you may refine your search by age of the audience)
albeit less chemistry is not the barycentre, bring science home in the Scientific American

and then, choosing while taking into account what resources are at your disposition.


Why does water dissociate to H3O+ and OH- rather than H+ and OH-? Why does water dissociate to
$\ce{H_{3}O^{+} + OH^{-}}$
instead of $\ce{H^{+} + OH^{-}}$?

This question came to surface when I was learning about acids and bases, and learned this definition:

$\mathrm{pH=}-\log_{10}[\ce H^{+}]$

Which after looking at wikipedia, looks like a simplified definition, but it got me thinking - how can water have a pH of around seven if it dissociates into $\ce{H_{3}O^{+}}$ instead of $\ce{H^{+}}$? Why does it do this instead of the expected dissociation?

You have to analyze the reaction mechanism behind the self-ionization of water to understand better what happen. In fact is not strictly self but a matter of "couple-ionization": you know the water atoms have different electronegativity, the more electronegative oxygen attracts the less electronegative hydrogen of another water molecule forming an hydrogen bonding.

From there the electron from Hydrogen forms a lone pair with the oxygen of the same molecule forming $\ce{OH^-}$. Hydron ($\ce{H^+}$) is in fact a proton, a very reactive specie so is not so simple found it uncoupled and moreover the hydrogen is suppose to be bonded with an hydrogen bonding with the other water molecule so is more accurate use hydronium ($\ce{H_3O^+}$) instead of hydron...

However you don't have to worry about it for your pH calculation:

$\mathrm{pH=}-\log_{10}[\ce {H^{+}}]= -\log_{10}[\ce {H_3O^{+}}]$

Because, in fact, you have the same ratio between basic species ($\ce{OH^-}$) and acid species( $\ce{H_3O^+}$ or $\ce{H^+}$ as you prefer) in both cases!


Acidic tendency of copper(II) ion vs. calcium(II) ion In my Chemistry textbook, a table is shown wherein ions originating in various salts are categorised according to their acidic or basic tendencies. Among others, $\ce{Cu^2+}$ and $\ce{Ca^2+}$ are present in the table; however, the copper ion is classified as being acidic, whilst the calcium ion is classified as being neutral.

I don't really see how they differ, apart from the amount of electrons they have in non-valence shells and the subshell that holds their valence electrons. Either way, since this is in the context of Brønsted–Lowry and not Lewis, I have no clue how the textbook got to its conclusion; does it have to do with copper parttaking in any redox reactions, or perhaps a certain stability calcium has?

Cu2+ has an incomplete 3d orbital (4s0 3d9) and can therefore be included in the chemistry of transition metals. As it can be coordinated (form dative covalent bonds) with ligands to create complex ions, for example H20 in hexaaquacopper(ii), it acts as a Lewis acid.

But in terms of Brønsted–Lowry theory, perhaps the textbook is referring to the fact that when transition metals are dissolved in water they can donate an H+ ion to another H2O molecule to form H3O+. This can be observed with Fe3+:

[Fe(H20)6]3+ + H2O -> [Fe(H20)5(OH)]2+ + H3O+

The complex ion is acting as an acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a base by accepting the hydrogen ion.


What is the meaning of electrode potential? When we say that the electrode potential for $\ce{Co}^2$$^+$/$\ce{Co}$ is -0.28 V and $\ce{Cu}^2$$^+$/$\ce{Cu}$ +0.337 V, what is the physical meaning?

Can you conceptually explain when two electrons react with $\ce{Cu}^2$ or $\ce{Co}^2$to form the metal, what happens and what is the source of these different potentials?

The potential difference between two electrodes represents the voltage that is generated between the two electrodes that will allow the transfer of current/electrons. Physically it's basically a measure of how strongly one electrode attract electrons over the other which leads to the flow of current.


Concentrations of gases at equilibrium The equilibrium constant for the reaction:
  $$\ce{2NO(g) <=> N2(g) + O2(g)}$$
  is $2.60 \cdot 10^{-3}$ at $\pu{1000^\circ C}$. If $\pu{0.820 mol}$ of $\ce{NO}$ and $\pu{0.223 mol}$ each of $\ce{N2}$ and $\ce{O2}$ are mixed at $\pu{1100^\circ C}$, what are the concentrations of $\ce{NO}$, $\ce{N2}$ and $\ce{O2}$ at equilibrium?

I'm really stuck, because with these types of questions usually the concentration of at least one reactant or product at equilibrium is provided, so I know the direction of the equilibrium shift.

Usually I allow the variable $x$ to equal the change in concentration of each product/reactant, where I subtract or add different ratios of it from the initial concentrations of the reactants and products (depending on the direction of equilibrium shift).

However, since I do not know the direction of equilibrium shift, how am I supposed to know whether to subtract or add $x$ to the reactants/products?

In cases like this calculate the reaction quotient, Q (https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Equilibria/Chemical_Equilibria/The_Reaction_Quotient) using the initial concentrations, and compare the value with the equilibrium constant, K:

Q=K implies that the system is at equilibrium
Q > K means that there is an excess of products so the system will evolve towards the equilibrium consuming products (which means that you should subtract x from the products)
Q < K just the other way around, the system will shift to the right to reach equilibrium (subtract x from the reactants)


1H NMR of p-methoxy phenol, 32 possible lines? This is a follow-up question, emerging from answering this one.

The Silverstein book 1 aims to rise awareness that not every 1H NMR signal that on first inspection may look like a doublet actually is a doublet.  Triggered by counting possible number of lines of signals for p-methoxy phenol

-- which would yield just 16 (two ddd) I came across this part of the book about p-chloro nitrobenzene:

(loc. cit., p. 175)

Question: For the sake of understanding this part I leave out a potential simplification of the 1H NMR spectrum if some of the centres of the signals of the aryl H's accidentally may coincide, or some parts of the multiplets overlap.  For a moment, I assume each aryl H's multiplet may be recorded without disturbance of the signals by any other aryl H.

Is my understanding of this section correct, in theory there were up to 32 (thirty-two) lines observable, as each of the aryl H's couples with the other three H's relatively in ortho, meta, para position, hence four ddd?

On the next page, the book actually displays an experimental 1H NMR spectrum of p-chloro nitrobenzene

and assuming the more intense peaks could be a partial overlap of two lines (not all of them are smooth), I "count" only 16 lines. As if -- a speculation follows -- obviously the p-coupling, as the one with the lowest coupling constant, was not-at-all resolved in this acquisition.

1 Spectroscopic Identification of Organic Compounds, Wiley, 6th edition, 1998, ISBN 0-471-13457-0

I think your theory is correct, but my tests were inconclusive. You can try your own hand at simulating the spectrum, for instance with Hans Reich's program DNMR. It resulted in the following:

This is the downfield half of the spectrum. I did not attempt to replicate the experimental spectrum exactly, it's only to illustrate. I used the following parameters:

$\sigma_A = \pu{7.5 ppm}$
$\sigma_X = \pu{8.19 ppm}$
$J_{ortho} = \pu{8 Hz}$
$J_{meta} = \pu{3 Hz}$
$J_{para} = \pu{0.5 Hz}$

On to your question. If you increase the J sufficiently, while retaining symmetry among the J, you get...

... 20 peaks! (only one set of multiplets is shown, so the total number of peajs is twice the number visible above) . I continued tinkering with the values, and breaking the symmetry among the couplings led to ...

... 26 peaks!

Presumably breaking the symmetry in the shifts will increase the number of peaks further, but surprisingly this had just the opposite effect.

But this is not the most illuminating approach. Better would be better to dig into the QM.


Proper definitions of resistivity of conductivity Is resistivity the same as specific electrical resistance?
And conductivity is same as specific conductance when I am dealing with electrical properties ?

The specific modifier simply means that stated property per a given quantity, usually mass, usually a gram. In other words resistance or conductance per gram of resisting or conducting material.  But the fundamental definition of the properties for the different word forms of resistance and conductance are not changed by this wording.


Coarse-grained force field for ion–water interactions I want to simulate, at a very coarse grained level, the interaction of water molecules with metal ions, specifically Calcium and Potassium. Is there a standard force field for this. I was told to use to the Martini potential but it doesn't seem to have metal ions...

Forcefields for alkali ions in water have been developed for decades now, so the literature is rich with examples… and yet there still are significant developments (see this very new paper for only one such example).

If you want to have a forcefield for alkali ions that mix well with the TIP3P water model, you better look at what existing studies on such systems have done… this is an example, but you can find many others with searching through the literature.


Deriving the equations for two-electron integrals in Hartree-Fock theory In Szabo and Ostlund's Modern Quantum Chemistry, the matrix elements for the hamiltonian of a two-electron system, with an operator $\hat O$, are written on pages 64-66 as

$$\langle \Psi_0 | \hat O | \Psi_0 \rangle=\int dx_1 dx_2 [2^{-\frac{1}{2}} (\chi_1(x_1)\chi_2(x_2)-\chi_2(x_1)\chi_1(x_2))]^* \cdot \hat O [2^{-\frac{1}{2}} (\chi_1(x_1)\chi_2(x_2)-\chi_2(x_1)\chi_1(x_2))]$$

for a spin orbital $\chi$ of electron x defined as the product of a spacial orbital $\psi(r)$ and spin $\alpha(\omega)$ or $\beta(\omega)$, e.g., $\chi(x)=\psi(r)\alpha(\omega)$.

This is multiplied out into

$$\langle \Psi_0 | \hat O | \Psi_0 \rangle=\frac{1}{2} \int dx_1 dx_2 \{\chi_1^*(x_1)\chi_2^*(x_2)\hat O \chi_1(x_1)\chi_2(x_2)+\chi_2^*(x_1)\chi_1^*(x_2)\hat O \chi_2(x_1)\chi_1(x_2)-\chi_1^*(x_1)\chi_2^*(x_2)\hat O \chi_1(x_1)\chi_2(x_2)-\chi_2^*(x_1)\chi_1^*(x_2)\hat O \chi_2(x_1)\chi_1(x_2)\}$$

This expanded equation is what I will base my question on.

For the core-hamiltonian (one-electron integrals),

$$\hat O=h(r_1)+h(r_2)$$

and for the electron repulsion (two-electron integrals),

$$\hat O=\frac{1}{r_{12}}$$

For the core hamiltonian, you can integrate out the electron which is not being acted upon by the operator to reduce the expanded equation to a one-electron integral.

For the electron repulsion, on page 66 the two-electron integral is simplified by stating the first two terms of the expanded equation are equal. This means:

$$\int \chi_1^*(x_1)\chi_2^*(x_2)\frac{1}{r_{12}} \chi_1(x_1)\chi_2(x_2)= \int \chi_2^*(x_1)\chi_1^*(x_2)\frac{1}{r_{12}} \chi_2(x_1)\chi_1(x_2)$$

This is what I do not follow. How is this derived? I can permute the electrons to achieve the same form, but that would introduce a negative sign.

Furthermore, for the equality to be true, does it not imply one of two scenarios: that either the spatial functions are degenerate, i.e.,

$$\chi_1^*(x_1)=\chi_2^*(x_1)$$
$$\chi_2^*(x_2)=\chi_1^*(x_2)$$
$$\chi_1(x_1)=\chi_2(x_1)$$
$$\chi_2(x_2)=\chi_1(x_2)$$

or otherwise that the electrons have the same spin and coordinate such that $x_1=x_2$

$$\chi_1^*(x_1)=\chi_1^*(x_2)$$
$$\chi_2^*(x_1)=\chi_2^*(x_2)$$
$$\chi_1(x_1)=\chi_1(x_2)$$
$$\chi_2(x_1)=\chi_2(x_2)$$

neither of which would make sense.

I believe your statement

  I can permute the electrons to achieve the same form, but that would introduce a negative sign.

is mistaken. The exchange integrals arise from permuting one pair of integration variable instances, not two. The latter is the case necessary here. Szabo and Ostlund state on the same page:

  Since $r_{12} = r_{21}$, we can interchange the dummy variables of integration in the second term [...].

where the second term appears to be you are troubled with.


Periodic cleavage of carbohydrates Carbohydrate like glucose undergoes periodic cleavage by $\ce{HIO4}$ and forms respective products are five moles of formic acid and one mole of formaldehyde.

Here is my question , what is mechanism and what are the products by fructose with the same? Do we have any specific rules for that?

Yes, there are specific rules regarding the products of periodic acid cleavage of 1,2-diols (Malaprade reaction). As to the mechanism, it is illustrated here. It is similar to the mechanism for KMnO4 oxidation of 1,2-diols. The structures of the products of 1,2-cleavage of carbohydrates is a diagnostic tool used as an aid in determining their structures. The oxidative cleavage of ethylene glycol (1, Fig. 1) affords two equivalents of formaldehyde. Primary alcohols give formaldehyde. Cleavage of the first single bond in glycerol 2 necessarily affords formaldehyde and α-hydroxyacetaldehyde. Aldehydes are readily hydrated in water. As far as periodic acid is concerned, the hydrate is a 1,2-diol. Cleavage of the second C-C bond produces not only another equivalent of formaldehyde but also an equivalent of formic acid. Secondary alcohols produce formic acid via aldehydes. (continued)

From the foregoing information, D-glyceraldehyde (3, Fig. 2) yields formaldehyde from the primary alcohol and 2 equivalents of formic acid, one from the secondary alcohol and one from the aldehyde group. Cleavage of one C-C bond of 1,3-dihydroxyacetone (4) yields formaldehyde and glycolic acid, which in turn provides formaldehyde and CO2. Ketones, and ultimately carboxyl groups, produce CO2. The oxidative transformations are summarized in the red box in Fig. 2.

The expected products from periodic acid cleavage of D-glucose, D-mannitol and D-fructose are predicted below based upon the formulated rules.


What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table? What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table?  I noticed when I look at a periodic table of elements, the elements of the F-block between group 3 and 4 doesn’t have a group number.  Are they just refer as Lanthanide,and Actinide or Inner Transition Elements,  But why doesn't they have a group number, it least it appear that way.

Link to a wide version of Periodic table of Elements so you can see what I am talking about.
http://en.wikipedia.org/wiki/Template:Periodic_table_%28wide_large%29

I suppose IUPAC hasn't bothered with group numbers for the lanthanides/actinides quite yet because there would only be two elements for each group, and because there are few vertical similarities among the f-block elements. The entire row of lanthanides behaves very similarly due to the 4f orbitals being anomalously compact and not being available for bonding or ionization. The first half of the actinides is quite different to the lanthanides, as the 5f orbitals there can actually participate in chemical interactions, though the second half becomes more similar as the 5f orbitals are once again strongly trapped to the atom, this time by the very high effective nuclear charge.


Do all sigma bonds are stronger than pi bonds? Many textbooks refer that sigma bonds are stronger than pi bonds . But each individual bond ( between different atoms) have different bond energy. So how we know for sure that every possible sigma bond is stronger than every possible pi bond ?

Every possible sigma bond is not stronger than every possible pi bond.

From this table at Wikipedia, we can see that the bond dissociation energy for the F-F bond in $\ce{F2}$ is 157 kj/mol. Meanwhile, the bond dissociation energy for the C-C bond in ethane $\ce{H3C-CH3}$ is on the order of 360 kJ/mol. The bond dissociation energy of the C=C bond in ethene $\ce{H2C=CH2}$ is approximately 710 kJ/mol. The note for this compounds suggests that the pi bond is around 270 kJ/mol (65 kcal/mol), leaving the sigma bond at 440 kJ/mol. Thus, the carbon-carbon pi bond is stronger than the fluorine-fluorine sigma bond.

A general statement of "sigma bonds are stronger than pi bonds" must thus be interpreted in the "all else equal" mode. The relative strength of the carbon-carbon pi bond is less than the relative strength of the carbon-carbon sigma bond. However, just as the strength of the carbon-carbon sigma bond has no bearing on the strength of the fluorine-fluorine sigma bond, the comparison of pi bonds to sigma bonds involving different elements is not helpful. The relative strength of the bonds depends as much or more on the atoms in the bond as it does the type of orbital overlap used to form the bond.


Energetics of Bond formation Back to chemistry after 2 months and got stuck in the beginning! That's why I don't like it. Anyways, come to the question please.
Following are the words from my textbook (Don't ask the name ;-)

$$\ce {Na -> Na^+ + e^-}\quad\Delta{H}=+495 \frac{KJ}{mol}$$
$$\ce {Cl + e^- -> Cl^-}\quad\Delta{H}=-348 \frac{KJ}{mol}$$
$$\ce {Na^+ + Cl^- -> NaCl}\quad\Delta{H}=-786 \frac{KJ}{mol}$$
We conclude that it is essential for the formation of ions and ionic bond  that the sum of energies released in second and third reactions must be larger than what is required in the first step.
The great attractive forces (bonds) reduce the energy of the system greatly as it emits 786 KJ of energy.

a. Why should it happen? i.e. sum must be greater et cetera.
b. How does attractive forces tend to reduce the energy?

Ok your question is a little ambiguous but I'll try.

a. Why should it happen? - If you look at Na, it has a lone electron in its outer shell and that essentially unbalances the atom, so it prefers to lose it and become an ion, this is almost spontaneous and hence requires less energy. When $\ce{Na+}$ joins up with $\ce{Cl-}$ they neutralise the electrostatic charge introduced by them becoming ions, releasing more energy. More energy because they have gone from an energy state A (medium energy) to B (higher energy) to become an ion. Then when they come together they both, each atom, falls to a lower energy state C than state A even and hence the change (NB) from B to C is greater than the change from A to B. In your book the change from my A to B is the change of an electron ( first two equations in your book) and then my change from B to C is the Na + Cl part.

b. How do attractive forces tend to reduce energy? - This is an important one to note the point of the text book, which is pretty badly written if you ask me, is referring to the energy of the system and not the energy of attractive forces. When it says system it means the newly bonded together $\ce{NaCl}$. The energy is reduced because of the neutralisation mentioned above. Positive and negative prefer to be neutral and hence, are attracted to one another. But before that, you must realise that being charged or an 'ion' is less stable and therefore in a state of higher energy than a neutral system. So it is a matter of stability as well.

If you want quantum explanations please say but I get the feeling you'd prefer the simpler option.

I hope this helps, even if only a bit.


Can carbon subnitride be used as a fuel with similar properties to gasoline? I am trying to figure out if Dicyanoacetylene, aka carbon subnitride, aka $\ce{C4N2}$, can have a similar reaction with air as gasoline, or if it only creates a lot of heat but not much expansion?  As I understand it, the thing that makes gasoline useful is that when it combusts, the product has a much lower density than the reactants, causing a huge expansion to reduce pressure (which drives pistons in an engine).  Does $\ce{C4N2}$ have a similar effect, or does it have only comparatively negligible expansion compared to gasoline and simply generates more heat? Or might it produce a similar expansion or more than gasoline?

No
You wouldn't want to burn this thing anywhere, it burns like, 9010 degrees... Fahrenheit. It's nearly the temperature of the surface of the sun (around 900 degrees off). Not much can contain it. So unless you had some kind of super-metal, than you are out of luck trying to use it as fuel.


How can I change the bond type (single, double) of an existing bond in Avogadro? I have a molecule extracted from the PDB. When I load it into Avogadro, it has some misplaced bond types. I want to correct this, by setting the bond type (single, double) for these bonds. How can I do this?

As discussed in the Avogadro website about the draw tool you can change the bond order:

By clicking on the bond - the bond order will go from single, to double, to triple, back to single.
By setting the bond order and "drawing over" the bond. The new setting (e.g., double) will replace the existing bond.


How to calculate the magnitude of magnetic field of an orbital? I read that according to QM  the electron in the ground state orbital produces no magnetic field, if that is true, can you explain why so and in what states there is a magnetic field and how that is produced? Why the difference?

Anyway, I am interested in learning how the classical model calculated it, since I read different values in different papers: for example hyperphysics says its value is .4 T, elesewhere I found .5 and .6, and If I use the classical formula I get .125 T

Does 0.4 Tesla refers to the exact value of the magnitude of the magnetic field od 1S state?

Hyperphysics means that the energy difference between the 2p J=3/2 state and the 2p J=1/2 state is as if the magnetic moment of the electron were aligned with and against a magnetic field of 0.4T.

The energy of the two states, relative to the ground state, is (according to NIST)

2p J=3/2 : 10.19885142904 eV
2p J=1/2 : 10.19880606470 eV

The difference is 0.00004538204 eV or $7.271004 \times 10^{-24}$ J

The electron magnetic moment is $−9.28476462 \times 10^{-24}$ J/T

So the energies are as if they are raised and lowered by being in a field of 0.3915556 T.

For the ground state, 1s, there is no orbital angular momentum.  The 1s state is only split by interaction with the proton.  This is sometimes referred to as hyperfine structure. The energy difference is $5.87433 \times 10^{-6}$ eV.