What is the correct value of the Avogadro constant? And how was it derived? I found different values of Avogadro constant in different places. So what is the correct value?

$\pu{6.0221367*10^{23}}$ or
$\pu{6.02214129*10^{23}}$ or
$\pu{6.0221415*10^{23}}$ or anything else?

Whenever you're looking for accurate fundamental physical constants, CODATA recommended values are the way to go. As of May 20, 2019, Avogadro's constant is now truly an exact value, with infinite significant digits. Behold!

$$6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$

It is unlikely this value will change within our lifetimes, which is exciting in its own way!

For historical reasons, I'll leave my previous answer below:

As of 2015, the latest data for the Avogadro constant is from 2014. According to CODATA, the most accurate value is:

$$6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,074\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2014)}$$

The relative uncertainty in the measurement is thus only 12 parts per billion!

Interestingly, the Avogadro constant may be redefined in the near future to be an exact value, that is, a constant with zero uncertainty by definition, much like the speed of light. This would come as a consequence of redefining the SI kilogram as a function of the number of atoms in an ultrapure monoisotopic $\ce{^{28}Si}$ monocrystalline sphere engineered to extreme precision. A great video on this can be found in the Veritasium YouTube channel.

All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry.

Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, CODATA released updated values for the physical constants, so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was:

$$6.022\,141\,29\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,27\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2010)}$$

This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science!


Why is it not dangerous to dissolve NaCl? $$\ce{H2O~(l) + NaCl~(s) ->[\Delta] Na+~(aq) + Cl^{-}~(aq)}$$

When table salt is placed in water, it dissolves due to the polarity of water molecules. When solvation takes place, negatively polar sides of water molecules attach to $\ce{Na+}$ ions, and positively polar sides of water molecules attach to $\ce{Cl-}$ ions. The ions are then carried away by water molecules through diffusion. Since the sodium is now isolated from the chlorine, what keeps it from violently exploding with water in the following chemical reaction?

$$\ce{2Na + 2H2O -> 2 NaOH + H2}$$

Also, since chlorine is poisonous, wouldn't that make drinking salted water extremely lethal?

[A really-good answer to this question should include quantitative evidence, but here is my hand-waving version...]

Solvated sodium ions ($\ce{Na+}$) aren't all that reactive.  Surrounded by a lot of polar water molecules, they're actually quite stable.  What you're thinking of is solid sodium (Na) which will "react violently with water" until it's all dissolved - dissolved that is, into stable sodium ions.

Same kind of logic applies to chlorine.  Solvated chloride ions ($\ce{Cl-}$) are stable and not nearly as poisonous as chlorine gas. (Note:  everything is poisonous at some level, i.e. "the poison is in the dose".  In saying chlorine gas is more poison, that really just means it is a more 'potent' poison:  it will kill at lower doses compared to chloride ions.)

Another way to think of it:
Solid sodium and chlorine gas are at high energy states.  Since they can reach much lower energy states by ionzing in the presence  with water, the WILL break down and ionize in the presence of water


Why is the gram no longer a function of the atomic mass number? According to Wikipedia,

  ... the definition of the gram is not (as of 2011) mathematically tied to that of the atomic mass unit

Why did we change that?

Unless I misunderstand the question, the gram was never defined that way.

For historical reasons the SI unit for mass is an oddity:

The base unit is not the gram, but the kilogram. This is the only case where the base unit has a prefix. The gram is defined as $\frac{1}{1000}$ of a kilogram.
The kilogram is the only remaining SI unit that is based on a man-made object, the International Prototype of the Kilogram.

At its 25th meeting in 2014, the General Conference on Weights and Measures (CGPM) has (again) declared its intention to abandon this atificial standard and link

  the definitions of the kilogram, ampere, kelvin, and mole to exact numerical values of the
  Planck constant $h$, elementary charge $e$, Boltzmann constant k, and Avogadro constant $N_A$ […]

(Source)

One of the possible approaches for the redefinition of the kilogram would be a side-effect of the redefinition of the Avogadro constant with in the Avogadro project.


Why is the graph of fluorescence vs. wavelength continuous? If the emission of a photon from an excited state to the ground state occurs with a precise energy, then why is it that wavelengths that differ from lambda max have values on a fluorescence vs. wavelength graph?

There are two main reasons for what you're describing.

The first is the fundamental properties of the molecules involved, as mentioned in the comments. Even for isolated molecules, the transition is not always between the same vibrational states of the excited and ground state. Since the vibrational states are quantized, we would expect a series of very closely spaced emission lines, each representing a different combination of vibrational states.

If the molecules are in a solution or in liquid or solid phase, then the interactions of the emitting molecule with the molecules around it can also cause small changes in the energy of the emission as the electronic structure of the molecule is perturbed.

The general term for these phenomena is "line broadening," and most reference books about spectroscopy will have a discussion of the various mechanisms that lead to it.

The second reason for it is the error in measurements. Most spectrometers cannot filter wavelengths perfectly, so, for example, light of wavelength 590 nm will contribute to a signal at 589.5 and 591.5 nm, less as you get farther away. Sometimes the slit width is an adjustable parameter, but even at the narrowest setting, most instruments will not give extremely high resolution. As a result, a series of closely spaced lines will appear instead as a smooth curve.


How to rationalise the protonation state of amino acids I am learning about amino acids in various $\ce{pH}$ environments and how their groups are protonated or deprotonated in these environments.
I don't need to mention a specific acid for my question, I will just use the generic form of one:
$$\ce{H2N-CH(R)-COOH}$$

The $\ce{H2N}$ group has a $\mathrm{p}K_a$ of $8$, I believe, and the acid has a $\mathrm{p}K_a$ of $3.1$.

If I drop these into a solution with $\ce{pH} = 12$, the $\ce{COOH}$ will donate it's hydrogen ion, but the $\ce{H2N}$ will not. It will only add another hydrogen ion if it is basic compared to the solution to become $\ce{H3N+}$. Why is it, that it won't donate its hydrogen ion like $\ce{COOH}$, when it is acidic compared to the solution, to become $\ce{HN-}$?

Let's assume you drop the amino acid in water at $\ce{pH} = 7$.
You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$
Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.

You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$
This means $\ce{NH3+ -> NH2}$.

You could define another $\ce{p}K_a$ for $\ce{NH2 -> NH^{-}}$.
The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35.
You may look at this site: How to use a $\mathrm{p}K_a$ table
You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.


Calculating the heat of reaction for Mg metal and HCl Adding $\pu{2.00g}$ $\ce{Mg}$ metal to $\pu{95.0 mL}$ of $\mathrm{1.00 M}$ $\ce{HCl}$ in a coffee-cup calorimeter leads to an increase of $\pu{9.2 ^\circ C}$. If the molar heat capacity of $\pu{1.00 M }$ $\ce{HCl}$ is the same as that for water $(\pu{C_p = 75.3 J mol-1 K-1)}$, what is the heat of reaction?

Two relevant equations are: $${\Delta H_{(rxn)} = -q}\\
{q = n \times C_p \times \Delta \mathrm T}$$

The answer given is $\pu{-44.4 kJ mol-1}$. I wonder if I'm not way off the mark here by using the wrong equations, because the answer I got was $\pu{122.83 J mol-1 \; ( 0.1773 mol \times 75.3 J \times 9.2)}$.

note: The heat of reaction = $\Delta H_{(rxn)}$

The equations you listed can be used to partially solve the problem.  However, with the equations you used, you have made one error which was in your choice of "moles."  Your moles value must match up with your $\ce{C_{p}}$, so you must find the moles of Water in $\pu{95.00mL}$!  Alternatively, you can use $4.184 \pu{\frac{J}{g\ ^\circ C}}$ and $\pu{1 \frac{g}{mL}}$ and $\pu{95.00mL}$ of water along with the equation $\mathrm{Q = n \times C_p \times \Delta T}$.

On the other hand, the premise of your solution is close but wrong, which is obvious if you examine the units:

  $\pu{0.1773 mol} \times 75.3 \frac{\pu J}{\pu{mol K}} \times \pu{9.2 K} = \pu{122.8 J}$

The units for Heat of Reaction are $\pu{\frac{J}{mol}}$, not $\pu{J}$!

To finish off your solution, you must divide by the moles of reaction that occurred (given by the limiting reactant).  $\pu{-44.4 kJ}$ is the value calculated based on the moles of Mg reacted.  However, I suspect that the given answer is wrong because the balanced equation does not yield a 1:1 reaction ratio between $\ce{Mg}$ and $\ce{HCl}$, which means that $\ce{HCl}$ is the limiting reactant, which would make the Heat of Reaction Larger.


Which gas makes Orange juice bottles swell? I forgot a bottle of 1L Orange Juice (100% orange juice with pulp) in the fridge for about a week, the bottle was previously open so air got inside as well.

After a week the bottle got inflated and when I opened it I could hear the gas coming out. No odours, orange juice still the same colour (not sure if the same taste, I threw it away).

Which gas was produced and how did it happen? The air that got in seems to have played a part on it (these bottles stay in the supermarket closed for a long time without getting inflated).

I can't imagine the gas was anything except $\ce{CO2}$ formed by the action of microorganisms contaminating the orange juice.  They degrade the organic components of the juice to power their metabolisms, and produce carbon dioxide as a byproduct.

A common contaminant of pasteurized fruit juice is a genus of bacteria called Alicyclobacillus, but this paper says that Alicyclobacillus bacteria do not cause bottle swelling and do cause a medicinal smell.

By searching Google for ["bottle swelling" "orange juice"], I found an English-language abstract of a thesis from China which implies that the culprit could be a yeast.


What is the product formed after nitrosation of ketones (HNO2 + ketone)? On C-nitrosation of ketones we get oximes. After this, the oxime formed is hydrolyzed ($\ce{H3O+}$ is added). What will be formed in the end? Does it follow Beckmann rearrangement or what?

Hydrolysis of the ketone oxime gives the 1,2-diketone as reported in this Synlett reference:

  Treatment of alpha-methylene ketones with excess sodium nitrite and
  aqueous HCl in THF at reduced temperature provieds an effective tool
  for the preparation of a variety of 1,2-diketones


Acids that donate just one proton What is the general name for acids that can only lose one hydrogen ion? (i.e. $\ce{HCl}$)
What is the general name for acids that can lose $n$ hydrogen ions? (i.e. $\ce{H_3PO_4}$, which can lose 3)

The answers to these questions are probably mentioned a million times on the internet, but I could not find them.

An acid that can donate one hydrogen ion is called monoprotic (i.e., $\ce{HCl}$). An acid that can donate two is called diprotic (i.e., $\ce{H_2SO_4}$). An acid that can donate 3 is called triprotic ($\ce{H_3PO_4}$). I don't know of any which can donate four.


Positive or Negative Anode/Cathode in Electrolytic/Galvanic Cell In a galvanic (voltaic) cell, the anode is considered negative and the cathode is considered positive. This seems reasonable as the anode is the source of electrons and cathode is where the electrons flow.

However, in an electrolytic cell, the anode is taken to be positive while the cathode is now negative. However, the reaction is still similar, whereby electrons from the anode flow to the positive terminal of the battery, and electrons from the battery flow to the cathode.

So why does the sign of the cathode and anode switch when considering an electrolytic cell?

The anode is the electrode where the oxidation reaction

\begin{align}
 \ce{Red -> Ox + e-}
\end{align}

takes place while the cathode is the electrode where the reduction reaction

\begin{align}
 \ce{Ox + e- -> Red}
\end{align}

takes place. That's how cathode and anode are defined.

Galvanic cell

Now, in a galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative.

At the cathode, on the other hand, you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive.

Electrolytic cell

In an electrolytic cell, you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are "pushed out" of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak. So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode.

At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode.

A tale of electrons and waterfalls

Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a waterfall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole. The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into.

Ok, what happens at the electrolysis reaction? At the cathode, you have the high altitude situation. So the electrons flow to the "edge of their waterfall". They want to "fall down" because behind them the river is pushing towards the edge exerting some kind of "pressure". But where can they fall down to? The other electrode is separated from them by the solution and usually a diaphragm. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the "pushed out" electron immediately. And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket.

Now the anode: At the anode, you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little "water" that is added to it doesn't change the river much. It stays the same, unaltered so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance.


Which is more stable: Ortho or Para Hydrogen? It is known that Para-Hydrogen has opposite spin and Ortho-Hydrogen has same spin.
Therefore, the energy of Para-Hydrogen is lower than that of Ortho-Hydrogen.
Therefore, the stability order should be: Para-Hydrogen > Ortho-Hydrogen.
Is this true or is the stability order the opposite of what I predicted?. What is the reason for this stability order?

At room temperature, about $75$% of $\ce{H2}$ is $\ce{o-H2}$ and $25$% is $\ce{p-H2}$. As the temperature drops, the relative amounts of $\ce{p-H2}$ increases. The two forms are in a temperature-dependent equilibrium $\ce{o-H2<=> p-H2}$, $\Delta H = \pu{-1.66 kJ/mol}$.
Ref. M. Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, New York $1994$, 1202 p. ISBN 0-89925-4547-8.


Why only two atoms share an electron and not three? In a covalent bond between two atoms, an electron from one of the either atom is shared by overlapping of their orbitals.

So,
Why can't three atoms share an electron and overlap their orbitals?

The normal case for a covalent bond is indeed a 2-centre 2-electron bond.

There are however cases, where 3 centres (= atoms) either share 2 or 4 electrons.

In organic chemistry, the most famous case for an electron-deficient 3-centre 2-electron bond is the 2-norbornyl cation.
In inorganic chemistry, boranes with their banana bonds are the role models.

Hypervalent molecules with 3-centre 4-electron bonds, like $\ce{SF6}$ or the bifluoride anion $\ce{HF2-}$ have been discussed here too.


Finding rate of reaction from a concentration vs. time graph A tangent drawn on “conc. of product vs. time” curve of a 1st order reaction makes an angle 30 degrees with the $y$-axis at 20 min. Find the rate after 20 min.

I first tried to find the equation like this:
$$k=\frac{2.303}{t}\log\frac{a}{a-x}$$
Let $\frac{k}{2.303}$ be $c$ to make calculations less messy. Here, $a$ is the initial concentration of reactant.
$$\frac{a}{a-x}=10^{ct}$$
$$x=a\left(1-10^{-ct}\right)$$
$$\frac{\mathrm dx}{\mathrm dt}=a\left(c\cdot\ln10\cdot10^{-ct}\right)=ak10^{\frac{kt}{2.303}}$$
Now, the tangent makes an angle of 30 degrees with concentration axis (or $y$-axis in this case). So the angle made with time axis (or $x$-axis) is 60 degrees. Hence $\frac{\mathrm dx}{\mathrm dt}=\tan\theta=\sqrt{3}$. But the value of initial concentration is not given for finding rate constant. Should I go for a different approach or am I making any conceptual mistake?

Let the reactant be $S$ and the product be $P$.  Let $p$ be the dimenionless product concentration, $\frac{P}{S_0}$.  First order kinetics:

$$\frac{P}{S_0} = p = 1- e^{-k t}$$

Slope of first-order kinetics:

$$\frac{dp}{dt} = k e ^{-kt}$$

Slopes and angles.  One confusing thing about this problem is the definition of $\theta$ is different than what you usually see in math class.  It's the angle with respect to the $y$ axis, not the $x$ axis.  As you note, if the $y$ angle is 30 degrees, then the $x$ angle is 60 degrees, meaning the slope is $\tan~60^{\circ} = \sqrt{3}$.
Units problem.  However there is a big units problem in this question.  The slope is $\sqrt{3}$ what?  Gigamolar per nanosecond?  Zeptomolar per eon?  The angle of the tangent curve will change depending on the units used to plot the data.  If you use nanoseconds as the time unit, the angle of the tangent curve will be very flat, but conversely if eons is the time unit, the the angle will be very sharp.  The question does not explicitly say what units are used in the plot.  And the expression for the derivative of a first-order reaction shows that the slope should have units of concentration per time, but $\sqrt{3}$ is dimensionless.  So there is not really enough information to solve the problem.
Dicey assumptions about units.  Let's assume that the time unit is minutes.  Let's further assume that the $y$ axis plots a dimensionless concentration, i.e., $p$. Then the derivative (slope) is $-k e^{-k t}$ and apparently has a value of $\sqrt{3}~\mathbf{{min}^{-1}}$.

The question is asking for the rate, so you are done.  But there are a lot of weird assumptions in this problem.


Comprehensive resource on diazo coupling? I am starting to do research on diazo coupling reactions and am looking for a book or literature review that is a comprehensive exploration of everything diazo coupling.

I found one book which is pretty amazing: https://ia802302.us.archive.org/8/items/aromaticdiazocom031270mbp/aromaticdiazocom031270mbp.pdf

It has a whole chapter devoted just to diazo coupling with various classes of compounds like phenols, amines, phenolic ethers, hydrocarbons, etc (which is exactly what I am looking for).  BUT, it's from the 1930's! It is quite challenging to read because it mostly describes reactions rather than showing them pictorially. Also, some of the theory is a bit dated!

My question: Does anyone know of a more modern resource?

If you prefer books as secondary reference, the The chemistry of diazonium and diazo groups in the Patai series (back 1978) may be one source.  More recently, the two books about diazochemistry by Zollinger, 1994 here and here.

Maybe not as comprehensive as the two sources above, but offering an overview are Merino's critical review about diazocoupling reactions in Chem. Soc. Rev. (doi 10.1039/C0CS00183J) in 2011, and Oger's overview about their application in flow chemistry in Organic Chemistry Frontiers (doi 10.1039/C5QO00037H) in 2015.


What would the name and formula be for Mo+O+Se? I do cross-stitch and want to spell the word "moose" (for a project focusing on the animal) with squares from the periodic table and present a compound formula and name that encapsulates all three elements, Mo+O+Se, in that order. I'm not looking for strictly realistic chemistry, just a formula and a name that would work IF the compound had actually existed. The number of atoms of each element is up to you. Can you help me with a formula and a name for Molybdenum + Oxygen + Selenium?

Well, molybdenum oxysulfide is apparently a real compound with the formula $\ce{MoOS2}$ or $\ce{MoO2S}$.  Since selenium is a chalcogen and has (some) chemistry that is analogous to sulfur, I think molybdenum oxyselenide would be an appropriate choice.

More formally:

Molybdenum(IV) oxide is a compound with the formula $\ce{MoO2}$.
Molybdenum(IV) selenide, also called molybdenum diselenide, has a formula of $\ce{MoSe2}$
Molybdenum sulfide selenide exists and has the approximate formula $\ce{MoSSe}$.

Thus, perhaps molybdenum(IV) oxide selenide would be another choice.


Why is it necessary to measure the NMR frequency relative to a standard such as TMS? The horizontal axis of an NMR spectrum is "chemical shift". A standard environment such as tetramethylsilane is chosen as the zero of chemical shift. The frequency at which resonance occurs for other environments is then measured relative to the frequency at which resonance of the standard environment occurs e.g. a particular environment might resonate at a frequency 1ppm lower than standard environment.

Why can't we measure the frequencies at which resonance occurs absolutely, not relative to a standard environmnent? Then we would have a horizontal axis of frequency instead of chemical shift.

The use of an standard makes it possible to compare spectra. The frequency itself is dependent on the used magnetic field, while the the chemical shift is independent of the magnetic field and thus comparable using different spectrometers.


What is the reaction mechanism of a selective catalytic reaction with urea? What is the reaction mechanism of the following reaction:

$$\ce{4NO + 2(NH2)2CO + O2 -> 4N2 + 4H2O + 2CO2}$$

I expect that the catalyst is vanadium oxide and titanium oxide.
Where can I find a detailed description of the reaction mechanism?

After researching this topic, I've found the mechanism to be:


Titration to determine the amount of hydrogen peroxide I have a mixture of sodium carbonate, sodium bicarbonate, ammonium carbonate, copper (II) sulfate, hydrogen peroxide, and sulfuric acid. I'm titrating potassium permanganate against this mixture to find out the amount of hydrogen peroxide in it. Will anything else in this mixture react with potassium permanganate? If so, is there another better way of determining the amount of hydrogen peroxide in it? Thank you.

I would not use potassium permanganate with sulfuric acid present:

You'll throw off your result due to the competing reaction..and risk a fire depending on the concentration of sulfuric acid.

If you measure the amount of $\ce{O2}$ produced by the decomposition of $\ce{H2O2}$ you can back-calculate how much $\ce{H2O2}$ was present in the sample. I think (with a decent certainty) that none of the species you list here should produce gas besides the peroxide.

That said I would use a catalyst (like $\ce{MnO2}$) for decomposition instead of a potentially reacting species (like potassium permanganate). You will likely get side reactions that consume the catalyst but they don't produce gas. So simply use excess $\ce{MnO2}$.

Measure the volume of $\ce{O2}$ produced (i.e. wait until the gas volume no longer increases) and use the ideal gas law (as a fair approximation) to get moles of $\ce{O2}$ produced. Once you have that, you can stoichiometrically determine moles of $\ce{H2O2}$ in your sample of known volume, thus concentration of $\ce{H2O2}$ (recalling that $2\ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2}$).

I've done this before with $\ce{MnO2}$ from batteries, washed thoroughly with water. I capture the gas produced in an upside-down grad-cylinder in a tub/beaker of water, e.g.:

P.S. if you manage to / happen to get a pure $\ce{H2O2}$ sample, you can very accurately get concentration (in %) by the following:

$C = \frac{d - 0.9946}{0.00456}$ where $d$ is the density in g/mL at 20deg. C.

That's a standardized function I derived from known and experimental densities of $\ce{H2O2}$


Is thin layer chromatography suitable to characterize distillate samples? Would I be correct if I state: Thin layer chromatography (TLC) is a separation method for non-volatile mixtures. Because distillate samples are pure, this would not be appropriate.

I'm unsure and would like some insight into this.

Your statement is unnecessarily complicated and not easy to follow. You can use TLC as a quick way to assess the purity of a sample. It doesn't matter if your sample is a distillate or not. It only has to be non volatile as you said and soluble in some solvent or a mixture of solvents that can (the solvent/s) make it move on the TLC plate.

I assume you want to use distillation to separate a mixture and TLC in order to assess the purity of your fractions. Obviously TLC becomes an unsuitable method for low b.p. distillates


Nomenclature of complexes: Cyano vs Cyanido ; Chloro vs Chlorido What is the correct nomenclature for $\ce{CN^-}$ and $\ce{Cl^-}$ ligands in coordination compounds?

Wikipedia says the coding name(?) as cyanido and chlorido while this suggests cyano and chloro.

The IUPAC Red Book is the official reference for recommended inorganic nomenclature. The full change to the additive nomenclature system was made in the 2005 update:

  Some new recommendations represent breaks with tradition, in the interest of increased clarity and consistency. For example, the application of the ending ‘ido’ to all anionic ligands with ‘ide’ names in additive nomenclature (e.g. chlorido and cyanido instead of chloro and cyano, and hydrido throughout, i.e. no exception in boron nomenclature) is part of a general move to a more systematic approach.

This is discussed in more detail in section IR-7.1.3 on page 112. Basically, anionic ligand names used as prefixes get changed from '-ide', '-ate', and '-ite' to '-ido', '-ato', and 'ito', respectively. Neutral and cationic ligands are usually used unchanged except for things like aqua, ammine, carbonyl, etc.


Deposition of saturated solution and affect of evaporated water 20g of water is evaporated from 200g of saturated solution of KNO3(aq) at 60°C by heating and the solution was cooled to 20°C. How many grams of KNO3 were deposited? The solubility of KNO3 in 100g of water is 32 at 20°C and 109 at 60°C.
I have a doubt on "20g water is evaporated from 200g KNO3" statement. I'm assuming that as 20g evaporated that means the initial saturated solution has mass of 220g. This is at 60°C. I also assume the solubility in g/L.
Now, at 60°C the max amount of solution that can dissolve is 109. Thus, if we get 220g this leaves $220-109=111g$ of solvent that is not dissolved.
Then, if the saturated solution is cooled down to 20°C, the solubility will decrease to 32g/L while the amount of precipitation we have remains the same. This leaves $111-32=79 grams$ of KNO3 precipitated. Therefore, the final amount of KNO3 deposited is 79g.
Did I treat the amount of water evaporated in a correct way? I'm not sure whether the water is affecting the total concentration or 'overall' volume of the saturated solution.

Unfortunately, your calculations are a little bit wrong because of the poor choice of the total initial mass. It is $200 g$, and not $220~ g$. So here are the right calculations.
As $100 g$ water dissolves $109g$ $\ce{KNO3}$ at $60$°C, producing a total mass of $209g$, you may deduce that $200g$ of your initial solution contains $(200/209)·109g = 104.3 g$ $\ce{KNO3}$ and $(200/209)·100g = 95.7g$ water.
Now you loose $20 g$ water by evaporation. The amount of $\ce{KNO3}$ is still $104.3 g$. Remaining water is $95.7g - 20g = 75.7g$ water. At the final temperature $20$°C, $100g$ water can only dissolve $32g$ $\ce{KNO3}$. As a consequence, $75.7$ g water can dissolve $(75.7/100)·32g = 24.2 g$ $\ce{KNO3}$. The rest is deposited in the flask. It is $104.3 g - 24.2 g = 80.1 g$.
Please note that the mass has a unit, the gram. It is incorrect to say "the solubility of $\ce{KNO3}$ in $100g$ of water is $32$ at $20°$C and $109$ at $60°$C". You should have said that "the solubility of $\ce{KNO3}$ in $100$g of water is $32 g$ at $20$°C and $109$ g at $60$°C


Solvent Effects in Friedel–Crafts Reaction While reading about the factors affecting the product of Friedel–Crafts acylation with acetyl chloride, I came across a reaction involving napthalene in two different solvents: in nitrobenzene substitution took place at the beta position; in carbon disulphide substitution took place at the alpha position.

What is the reason for this?

On Friedel-Crafts-acylation reactions, two types of Friedel-Crafts-acylation mechanisms, namely an ion pair mechanism and a dipolar ion mechanism, have been proposed (Ref.1).  Normal acylations are presumed to proceed via the ion Pair mechanism, which seems to be more important in sterically hindered reactions:

The product complexes with aluminum chloride at the end. This complex liberates the final product after addition of water in the final step as shown in following diagram:

The Friedel-Crafts acetylation of naphthalene has been extensively reportedand frequently studied as of today (Ref.1). Basically, two isomeric ketones, 1-acetylnaphthalene and 2-acetylnaphthalene, can be prepared by the reaction of unsubstituted naphthalene with acetyl chloride and aluminum chloride. A variety of solvents have been used for this acylation, and the solvent effect on naphthalene acylation has been studied intensively (Ref.1). The product ratio of 1-acetylnaphthalene to 2-acetylnaphthalene is found to be solvent-dependent. For example, when the reaction was performed in non-polar solvents such as $\ce{CH2Cl2}$, $\ce{CH2ClCH2Cl}$, $\ce{CHCl3}$, or $\ce{CS2}$, the 1-acetylnaphthalene formation is preferred and when reaction is done in polar solvents such as nitrobenzene or $\ce{CH3NO2}$, the reaction has proceeded to give 2-acetylnaphthalene,  exclusively. This phenomenon has been explained considering reversibility of the reaction:

  The aluminum chloride-acylaromatic complex is insoluble in non-polar solvents carbon disulfide, but very soluble in polar solvent such as nitrobenzene. The 1-acetylnaphthalene-$\ce{AlCl3}$ complex can form rapidly (kinetic product) and it would precipitate from the solution, preventing further reaction. On the other hand, the 1-acetylnaphthalene-$\ce{AlCl3}$ is soluble in the polar solvents such as nitrobenzene. Thus, the reaction proceeds further and 1-acetylnaphthalene would deacetylate, and the slower 2-acylation reaction may occur to give more stable thermodynamic product (Ref.2).

REFERENCES:

P. H. Gore, “The Friedel-Crafts Acylation Reaction and its Application to Polycyclic Aromatic Hydrocarbons,” Chem. Rev. 1955, 55(2), 229-281 (https://doi.org/10.1021/cr50002a001).
Gary Warren Keen, "The acetylation of naphthalene and the formation of (chlorovinyl)naphthalenes therein," The Ph.D. Dissertation, Oklahoma State University, Stillwater, OK, 1976.
Wenpeng Li, Haibo Jin, Suohe Yang, Xiaoyan Guo, Guangxiang He, Rongyue Zhang, “An environmentally friendly acylation reaction of 2-methylnaphthalene in solvent-free condition in a micro-channel reactor,” Green Processing and Synthesis 2019, 8, 474–479 ( https://doi.org/10.1515/gps-2019-0015).


Why is Enthalpy of Compound at Some Temperature Equal to its Enthalpy of Formation at the Same Temperature? This is a question I don't understand in thermo. Enthalpy of compound would be sum of U + PV at some temperature, but since we can't calculate absolute value of H, we calculate it relatively to some reference. Enthalpy of formation is heat needed to be exchanged between system and surroundings at some pressure to keep the system at the same temperature when compound is formed from elements in the most stable state. Why are these two quantities equal?

They are not equal.  The enthalpy of a compound at temperature T is relative to the pure elements at the reference temperature $T^0$.  So it is equal to the standard heat of formation at the reference temperature plus $C_p(T-T^0)$.


Which solution has the highest total molarity of ions and why? (a) 0.008 M $\ce{Ba(OH)2}$
(b) 0.010 M $\ce{KI}$

The answer is "a" but I do not understand why. The book says that "a" is a strong electrolyte with a total ion concentration of 0.024. How did they calculate the concentration from the information that they gave? And why is it a strong electrolyte based on that calculation. I thought strong electrolytes were (solutes) ions that dissolved completely in a solvent/solution?

You are right in thinking that a strong electrolyte is something that dissociates completely in solution.$\ce{Ba(OH)2}$ and $\ce{KI}$ then, being a strong electrolytes, will effectively dissociate into $\ce{Ba^2+}$+ $\ce{2OH-}$ and $\ce{K+}$+ $\ce{I-}$, respectively.

Going into your example, it's a just a matter of multiplication. We can assume that the volume of the solution is 1L for simplicity. Since you begin with 0.008 M $\ce{Ba(OH)2}$, we can say that you begin with 0.008 mol of $\ce{Ba(OH)2}$. In solution this becomes 0.008 mol of $\ce{Ba+}$ and 2(0.008) mol of $\ce{OH-}$, or 0.024 mol of ions in total. Likewise for KI, you are left with 0.010 mol of $\ce{K+}$ and 0.010 mol of $\ce{I-}$, or 0.020 mol of ions in total.

Converting these values back to molarity (by dividing by 1), we get that the concetration of ions in the $\ce{Ba(OH)2}$ solution is 0.024 M, whereas it is only 0.020 M in the $\ce{KI}$ solution.


Carbon/silicon pentacovalent intermediate during substitution During SN2 reaction with a silicon compound, a pentacovalent intermediate is formed, while there is no such intermediate formed with carbon compounds (only a pentacovalent transition state). Why is it so? I need an MO explanation (no d orbitals involved).

Bonds involving carbon are typically shorter than bonds involving silicon.  Therefore a pentacoordinated carbon species would be sterically more crowded around the central carbon atom than its pentacoordinated silicon analogue would be. This crowding would raise the energy of pentacoordinate carbon compared to pentacoordinate silicon; which is consistent with the observation that the SN2 reaction with carbon produces a higher energy pentacoordinate transition state, while in the case of silicon a more stable pentacoordinate intermediate can be observed.
\begin{array}{lcc}
\hline
\ce{Bond} & d(\ce{X=C})/\pu{pm} & d(\ce{X=Si})/\pu{pm} \\
\hline
\ce{X-O} & 143 & 163 \\
\ce{X-Cl} & 177 & 202 \\
\ce{X-Br} & 194 & 215 \\
\ce{X-I} & 214 & 243 \\
\hline
\end{array}
Recasting this into MO terms, we would say that both the carbon and silicon pentacoordinate species involve a trigonal bipyramid structure where the two ligands (the "attacking" and "leaving" groups in the carbon case) occupy axial positions in the trigonal bipyramid arrangement.

(source)
The figure above makes it easy to see that we have a hypercoordinated species involving a 3-center 4-electron bond. Since a silicon p-orbital is larger than a carbon p-orbital, the 4 electrons are spread out over a larger volume in the silicon case, reducing electron-electron repulsion.  This makes the pentacoordinate silicon case more stable than pentacoordinate carbon.


Are molecules with bond orders less than one stable? Are molecules with bond orders less than one stable? My professor commented that they were "barely alive" but what does he mean, scientifically?

I know that molecules with fractional bond orders greater than one can exist - i.e. nitrogen oxide has a bond order of 2.5 (through MO calculations) and it exists just fine.

So are molecules with fractional bond orders below one just very unstable?

Molecules possessing a bond of order below 1 can be perfectly stable, in the sense that their resulting molecular structure lies in an energetic potential well. Strictly speaking it is enough that at $T=\pu{0 K}$ and in the absence of any interactions with matter or fields, the molecule will not spontaneously disassemble. However, there is no need to go quite so far to protect such a molecule from decomposing; there are examples of species which are chemically important in regular laboratory conditions.

All other things being equal, it is true that species with bond orders below 1 are relatively unstable. This is mainly because the fractional bond is comparatively weak (requiring comparatively little activation energy to break, i.e. a smaller $E_\mathrm{a}$), and because in most cases the molecule can react with other substances in such a way as to form products with all covalent bonds of bond order 1 or higher (increasing the exergonicity of most reactions, i.e. a more negative $\Delta_\mathrm{r}G$).

Since the kinetic barrier for reaction is comparatively low, and the thermodynamic drive for reaction is comparatively high, species with bond order below 1 tend to need some extra protection against the world to stay in one piece. For example, diborane and trimethylaluminium are compounds possessing bonds of order 0.5, and while being indefinitely stable when pure, they spontaneously ignite in air from exposure to oxygen and moisture. As rightly pointed out in the comments, noble gas compounds require fractional bond orders, yet several compounds can be obtained and stored (especially xenon compounds), though they tend to be sensitive to moisture and heating. Boron is also responsible for a very interesting class of compounds in which many boron atoms attach to  each other in cage-like structures with very complicated bonding, in which fractional order bonds are involved. Some of the larger and more symmetric structures can be relatively stable, especially with adequate substituents.

In space, there's really not much of anything around, so you might expect to find some molecules with fractional bond orders floating about. Indeed, one can find the trihydrogen cation, which is actually one of the most common ions in the Universe!


How would the "-COOCl" functional group be named? I had this question that has been bouncing around for over 2 years.

Basically, I imagined this molecule:

This molecule, as far as my former (and current) knowledge of nomenclature goes, couldn't be named by me. At first, when I was in about 11th grade, I used to call it ethanoyl chloride, but that's $\ce{CH3COCl}$, not $\ce{CH3COOCl}$. I then thought of chlorine ethanoate, but that's ridiculous too, since chlorine can accomodate the negative charge way better than what ethanoate can.

I looked up a while, and I found out that the $\ce{-OCl}$ group is called a hypochlorite group, that gave rise to my third guess ethanoyl hypochlorite. However, I am not very sure about that either, since ethanoic acid isn't called ethanoyl hydroxide, and methylmethanoate isn't called methanoyl methoxide.

So, what would the "-COOCl" group be called?


Calculating Mixing Enthalpy in GROMACS What is the most reliable way to calculate mixing enthalpy of binary solvent mixture using GROMACS?

[EDIT]

I have simulated two-component systems of water and organic solvent with GROMACS. In a research paper from my field of study, authors calculate 'enthalpy of mixing' with the software in order to compare it with experimental mixing enthalplies. Under the Experimental Section of the paper, authors show that they do this according to this term:

A comment added in the text says the following:

"The internal energy, U, of the liquid mixture was obtained directly from the potential energy (see Tables S1
and S2) during the simulation, and yi is the molar fraction of component i."

I am interested in calculating same thermodynamical properties for my systems. I am aware that I can do that with g_energy function in GROMACS, or at least some parts of it (Pressure, Volume, Potential Energy).

My question is:
Firstly, is this an appropriate way of calculating enthalpy of mixing in GROMACS via this equation?
Secondly, how can one specify with the g_energy function which of the liquid's U(or potential energy), P and V is calculated if I initiate the function with .edr files, i.e. how can I choose to calculate i'th component's properties?

Reference for the image and quote:

Aguilera-Segura, S. M.; Di Renzo, F.; Mineva, T. Structures, Intermolecular Interactions, and Chemical Hardness of Binary Water-Organic Solvents: A Molecular Dynamics Study. J. Mol. Model. 2018, 24 (10), 292. https://doi.org/10.1007/s00894-018-3817-2.

GROMACS won't be able to calculate the enthalpy of mixing directly but using the energy function in GROMACS you will be able to get energy, pressure, and volume of the simulations.

Going through the paper that you posted it looks like the ran multiple simulation boxes for the different components and mixtures. They did this because you will not be able to get U,P, and V for individual components in GROMACS. So in order to calculate the enthalpy of mixing for a 2 component mixture they ran 3 simulations: one of the mixture, and one of each component by themselves.

Once you have the simulation boxes for the different components and mixtures it should be pretty straight forward to calculate the enthalpy of mixing.

Just a side note: I am always a little nervous calculating thermodynamic expressions using pressure from GROMACS because the pressure will fluctuate so wildly. So if you're going to make sure you're careful about your equilibration run and your NPT equilibration run as well.

Below are the GROMACS manual for energy which you would use to get U,P, and V from your simulation. The next is a tutorial for an NPT run from Virginia Tech and their tutorials are really well written and should help as a starting point and finally the last is a chat between a user and GROMACS developer discussing pressure fluctuations in NPT.

http://manual.gromacs.org/documentation/current/onlinehelp/gmx-energy.html

http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin/gmx-tutorials/lysozyme/07_equil2.html

https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2011-May/061370.html


Hexagonal unit cell vs crystal system Is the hexagonal unit cell (resulting from hexagonal closed packing) same as the unit cell of hexagonal crystal system?

The unit cell of hexagonal crystal system is described by the following sides and angles: $a=b\neq c$, $\alpha=\beta=90^\circ$, $\gamma=120^\circ$,
where $a,b,c$ are the sides of unit cell and $\alpha,\beta,\gamma$ are the angles.

This is the unit cell of hcp

In the unit cell of hcp there are 17 atoms whereas in unit cell of hexagonal crystal system there are 8. Is this true?

I this, you are totally confuse with meaning hexagonal unit cell and hexagonal crystal system. First of all, I make you to clear about these two terms.

Hexagonal unit cell is single entity of hexagonal crystal system. It means there are a no. of repetitive hexagonal unit cells inside a hexagonal crystal system.

Yes, the diagram given in the question is of hcp i.e. hexagonal unit cell.

But, your next statement : In the unit cell of hcp there are 17 atoms whereas in unit cell of hexagonal crystal system there are 8.Is this true?

is incorrect. One thing I want to clear you that as per saying there are 17 atoms in unit cell of hcp is wrong. We can make the statement after correction is : To make a unit cell of hcp, we need 17 atoms. It means that there are 17 atoms present in unit cell hcp but to make a unit cell of hcp we require 17 atoms.

Now, we calculate the no. of atoms present inside the unit cell of hcp or we can say this as no. of effective atoms of unit cell of hcp.

Calculation of no. of effective atoms of unit cell of hcp :

No. of effective atoms of unit cell of hcp = 3 (there are a total of three atoms present inside the unit cell of hcp) * 1 (as all three atoms are in main unit cell of hcp, no part of atom is outside the unit cell of hcp) + 2 (there are two atoms present a centre of each base of hexagonal unit cell) * 1/2 ( half part of each atom is inside the main hexagonal unit cell) + 12 (there are a total of 12 atoms present at 12 corners of each base of hexagonal unit cell) * 1/6 ( one-sixth part of each atom is inside the main hexagonal unit cell) = 6 (not 8)


Why is the hydrolysis of ATP to AMP more energetic than the hydrolysis of ATP to ADP? I am confused as to why the hydrolysis of ATP to AMP and pyrophosphate is more energetic than hydrolysis to ADP and inorganic phosphate, especially since pyrophosphates are more unstable than inorganic phosphates.
My only guess was that the pyrophosphate undergoes further hydrolysis to inorganic phosphates due to its instability and this drives the reaction forward via Le Chatelier's principle, but I think my logic is flawed. I would like to know the factors which cause this difference of -2.6 kcal/mol between the two reactions.

Dr. Larry Moran (Professor Emeritus in the Department of Biochemistry at the University of Toronto) has posted an nice article about the Free Energy of ATP Hydrolysis. Accordingly, some important concepts in biochemistry may be widely misunderstood and/or not well described in most textbooks. One of them is the free energy of ATP hydrolysis:

The traditional standard Gibbs free energy changes do not apply in biochemical system because there are $\ce{Mg^2+}$ ions, ionic strength, and $\mathrm{pH}$ involve in the transformation. The traditional standard Gibbs free energy changes are at $\pu{25 ^\circ C}$ $(\pu{298 K})$ and $\mathrm{pH} \ 7$. For reactions like ATP hydrolysis, we want a new "standard" that includes free $\ce{Mg^2+}$ ion concentration of $\pu{10^{-3} M}$ and an ionic strength $(I)$ of $\pu{0.25 M}$ (Ref.1).
The transformed Gibbs energy (denoted as $G'$) is defined by the following Legendre transform of the Gibbs energy $G$:
$$G' = G - n'(\ce{H+})\mu(\ce{H+}) - n'(\ce{Mg^2+})\mu(\ce{Mg^2+}) \tag1$$
where $\mu(\ce{H+})$ and $\mu(\ce{Mg^2+})$ are the specified chemical potential of $\ce{H+}$ and  $\ce{Mg^2+}$ at equilibrium, respectively, and $n'(\ce{H+})$ and $n'(\ce{Mg^2+})$ are the total amount of these ions (free and bound) in the system. Accordingly, following are the calculated new values for specific reactions:
$$
\begin{array}{l|c}
\text{Biochemical reaction} & \Delta G'_\text{hydrolysis} \ (\pu{kJ mol-1})  \\
\hline
\ce{ATP + H2O -> ADP + P_$i$ + H+} & -32  \\
\ce{ATP + H2O -> AMP + PP_$i$ + H+} & -45  \\
\ce{AMP + H2O -> Adinosine + P_$i$ + H+} & -13  \\
\ce{PP_$i$ + H2O -> 2P_$i$} & -29  \\
\hline
\end{array}
$$

References:

Robert A. Alberty, Robert N. Goldberg, “Standard thermodynamic formation properties for the adenosine 5'-triphosphate series,” Biochemistry 1992, 31(43), 10610–10615 (https://doi.org/10.1021/bi00158a025).


How determine the number of significant figures of a feet/inch system? How many significant figures are there in $5~\mathrm{ft}\ 10~\mathrm{in}$.
I know that $5~\mathrm{ft}\ 10~\mathrm{in}= 5.8\bar{3}~\mathrm{ft}=70~\mathrm{in}$, so I am not sure exactly how much significant digits are in this number.

Significant figures is a rule of thumb, to avoid giving overly precise results when the input doesn't warrant it.

In actual scientific calculations - where keeping track of the precision matters - significant figures aren't used. Instead, explicit error ranges (confidence intervals) are used. Notation of these varies, but often takes the form of a plus-minus format (14.5±0.3 m). The concept of significant figures uses implicit error ranges, conventionally set to one half of the last significant digit.

When error ranges are explicitly given, the error of a calculation result is found with formal propagation of uncertainty, rather than the rule-of-thumb significant figure rules. Sometimes this is based on a model of the distribution of error around the central value (for example, assuming that the error is normally distributed around the central value with a given standard deviation), but most commonly the error is taken as the minimum and maximum of an acceptable range of values around the main measurement. Under this scheme, the propagation of uncertainty often takes the form of the "crank three times" method, whereby you run the calculation once for the central value, once for the minimum value, and once for the maximum value.

So if you were doing a calculation like $\pu{14.5\pm0.3 m} - \pu{1.5\pm0.2 m}$, you would calculate once for the central value ($\pu{14.5 m - 1.5 m = 13 m}$), once to give you the smallest value you could get with the range ($\pu{14.2 m - 1.7 m = 12.5 m}$), and once to give the largest value your could get with the range ($\pu{14.8 m - 1.3 m = 14.5 m}$). You'd then convert this into a range, normally rounding both values such that the error value only has 1-2 significant figures: $\pu{13\pm0.5 m}$

It turns out that the rules for significant figure calculations are just a quick-and-dirty rule of thumb of doing more-or-less the same thing. I'd encourage you to try it out with a number of examples. If you assume that the error in each quantity is one half of the last significant digit, and you do the crank three times error propagation, you should come up with a result that has an error that is about one half the last significant digit of the answer by the significant figures approach - give or take a digit.

So really what you want to do is to convert the measurement of $\pu{5 ft 10 in}$. into one which has an explicit error associated with it. For this measurement, I'm guessing you'd have an error of $\pu{\pm0.5 in}$. - I say this because rounding to the nearest 10 inches seems rather silly to me. (Feel free to work the calculations below with different assumption, say with ±5 in., if you disagree.) So you have a measurement of $\pu{5 ft 10 in. \pm 0.5 in}$ in other words $\pu{5 ft} (\text{exact}) + 10\pm\pu{0.5 in}. = \pu{5 ft} (\text{exact}) \times \pu{12 in./ft }(\text{exact}) + \pu{10\pm0.5 in.} = \pu{70\pm0.5 in.}$ Note that is the same as $70$. in with two significant digits.

Since we have explicit error bars, we can use the crank three times method to convert it to feet: $70~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.8\bar{3}~\mathrm{ft}$ and $69.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.791\bar{6}~\mathrm{ft}$ and $70.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.875~\mathrm{ft}$. This is $5.8\bar{3}±0.041\bar{6}~\mathrm{ft}$. Convention holds that we round the error to one or two significant figures, and also round the main value to the same decimal place: $5.83±0.04~\mathrm{ft}$. -- This is basically what you'd have from $\pu{5.8 ft}$ with two significant digits.

The upshot is that you'd never use $\pu{5 ft 10 in.}$ in a calculation. You'll want to convert it to a single consistent unit first. When you do that, keep in mind the actual error associated with the measurement, and use that error as a gauge at how precise the converted quantity is.

(Note that if this is a question on a test or on a homework, ask your teacher how they think you should get the "correct" answer. When dealing with rule-of-thumb things like significant figures, intelligent people can vary in their interpretations. This means an answer valid in one context can be marked as wrong on the answer key because it doesn't match the formalism the teacher is using.)


Differences between H-H-O and H-O-H I'm not an expert.

In simple words how do they differ:

$\ce{H-H-O}$ and $\ce{H-O-H}$ molecules?

$\ce{H_{2}O}$ is the water molecule, two hydrogen atoms attached to a central oxygen atom, $\mathrm{C}_{2v}$ symmetry, thermodynamically minimal structure of these atoms, Adam's ale, etc.

HHO is a poorly defined term often bandied around by 'water powered car'... enthusiasts. I'm not sure that it's supposed to represent a molecule so much as a state of hydrogen and oxygen that conventional chemistry has no concept of.

Here's a post on skeptics.SE about it
And here's an acerbic article on rationalwiki that mentions 'HHO'

I think we're supposed to believe that it's a gas of hydrogen and hydroxyl radicals, or of atomic hydrogen and oxygen, or protons and hydroxide ions, or something, that is metastable at room temperature and that also happens to violate the first law of thermodynamics when produced and combusted. Why this glaringly obvious anomaly in quantum electrodynamics and thermodynamics has never been observed anywhere in nature before is anyone's guess (probably part of the conspiracy).

These guys should write up a few papers, get published in Nature, collect the Nobel chemistry, physics and peace prizes...


Conventional heating compared to microwave heating of chemical reactions In order to try to increase yield of my cross-metathesis, it was suggested to carry out the reaction in the microwave (a lab microwave, not a kitchen microwave).

The rationale for this was that microwaves accelerate the reaction, but thinking on how a microwave actually works, it's basically just heating in the same way as I could achieve with a PEG bath.

When I searched the internet, there are lots of reviews and even books published about microwaves in organic synthesis, but I can't find a coherent argument for what microwaves are doing that makes reactions 'magically work'.

The radiation in theory simply causes the molecules to move creating heat, as far as I know, it isn't like photochemistry where something mechanistically different can occur.

Your question about the 'magic' microwave effect is one that I think most chemists who've used a microwave have asked. Anecdotally, some reactions do often just work better when carried out in a microwave. The whole debate is quite nicely summarised in this quote from a (fascinating...) book on the use of microwaves in chemical synthesis:

  Since the early days of microwave synthesis, the observed rate accelerations and sometimes altered product distributions compared to oil-bath experiments have led to speculation on the existence of so-called “specific” or “non-thermal” microwave effects. Historically, such effects were claimed when the outcome of a synthesis performed under microwave conditions was different from that of the convention- ally heated counterpart at the same apparent temperature. Reviewing the present literature, it appears that today most scientists agree that in the majority of cases the reason for the observed rate enhancements is a purely thermal/kinetic effect, i.e., a consequence of the high reaction temperatures that can rapidly be attained when irradiating polar materials in a microwave field

  Source: Microwaves in Organic and Medicinal Chemistry

Microwave radiation:

Source: Microwaves in Organic and Medicinal Chemistry

From the table above, its clear that microwave radiation is not strong enough to cause any actual chemical change itself (other than via the temperature it creates), for microwave radiation used by common household and lab microwaves (2450 MHz, which is important as it avoids interfering with communication equipment), the energy is sufficiently low that it couldn't in theory even over-come a hydrogen bond.

Thermal/pressure effects:

There are several reasons why the use of a microwave may be beneficial compared to a standard reflux setup in a PEG bath (or any other conventional heating method):

the microwave is able to quickly reach a given temperature, with only a short induction period. The result of this is that the reaction reaches the desired temperature quickly, rather than slowly warming from room temperature. This means that the reaction gets going immediately, which can be especially beneficial in cases where the catalyst/reagents being used aren't terribly stable (your metathesis is one such example where the catalyst can quickly degrade in solution).
the microwave heats the solution directly and uniformly, compared to heating conveniently where the glass is being heated and in turn the glass heats the solution.

Source: Biotage. Left image = microwave heating, right image = oil bath, both taken after 1 min of heating

the vials used in microwave reactions are sealed (they're closed systems), which means in addition to temperature, they're able to carry out reactions under increased pressure, allowing the reaction to be heated above the boiling point of the solvent being used.

Source: Biotage

Overall, the easiest way to show all of this is by looking at the read-out from a microwave reaction, showing how quickly the desired temperature can be reached and how this remains constant, along with the pressure, throughout the entire reaction run:

Non-thermal effects:

Non-thermal effects, or 'specific microwave effects' are defined as accelerations of chemical transformations in a microwave field that cannot be achieved or duplicated by conventional heating, but essentially are still thermal effects. These have been considered by many groups and indeed a wikipedia page exists discussing them, however they're difficult to show/prove experimentally.

For solution phase organic synthesis, its generally accepted that there is no specific microwave effect. Only in solid phase systems (ionic lattices for example) is there anything to suggest something unique going on that cannot be achieved using standard heating methods alone.


Can ethanol be both a Lewis base and a Lewis acid? I'm asked to determine if ethanol is a Lewis acid, Lewis base, or both.

Since $\ce{O}$ has two lone pairs of electrons I would think that it could act as a base, since they could easily be donated. If it was a Lewis acid, it should have an empty orbital, and I can't find such one. All of the carbons are bonded and oxygen's orbitals are full. Still my book tells me it can act as both.

Is the textbook wrong or could one say that the 1s orbital to $\ce{H}$ (bonded to $\ce{O}$) acts as "empty", since $\ce{O}$ is more electro-negative than $\ce{H}$?

Lewis acidity and basicity require some amount of context because they can't function in those roles without at least a putative base or acid, respectively. Certainly the oxygen here has a lone pair so it could be basic, and the $\sigma^{*}$ orbital of the O-H bond (on the hydrogen side) can accept an electron pair. What matters more is the context of what acids and bases are present to interact with these moieties. There is no reason why these pieces can't separately fulfill their roles independent of each other.


Apple Cider vs. Cleaning Vinegar My grandparents stumbled on some cheap  Heinz Apple Cider Vinegar while cleaning their house.

  Ingredients: Made from the juice of apples diluted with water to a uniform pickling and table strength of 5% (50 grains) acidity.

Barring the origin and 1% difference in acidity, how does it differ from Heinz® All Natural Cleaning Vinegar? Is Apple Cider Vinegar equally as potent for cleaning clothes and the house?

  Ingredients:
  Made from select sun-ripened grain, diluted with water to a special cleaning strength of 6% acidity.

Beverly Leestma purports that the brand beneath at Walmart didn't indicate "the acidity but did state it was “Not For Food Consumption." Some vinegars (like this Industrial Strength Cleaning Vinegar) are still 5% but aren’t refined or purified to be food quality."

Assuming that the industrial strength cleaning vinegar's acidity is also 5%, then why can't it be consumed?
Why must raw vinegar be refined or purified for human consumption?
Is Beverly Leestma correct in her wariness beneath?

  Part of me thinks Heinz is just looking to get in on the popularity of homemakers who want natural cleaning.  I’m not convinced that the 1% of extra acidity warrants the higher price tag and that maybe they’re just trying to cash in on a trend at our expense.

The acetic acid in "Cleaning vinegar" may come from an industrial process. One way to produce acetic acid industrially, is carbonylation of methanol. And methanol is toxic!
So unless the product has been analyzed for residues of methanol and approved for food consumption, it is probably unwise to use it for consumption.


If I drink water without sipping, is there a chance of saliva diffusing into my bottle? I always prefer to drink water without sipping, like this:

I do this so that my saliva doesn't contaminate the water contents, which would make the water impossible to share with my friends.
But lately, I have been thinking: Can the saliva diffuse through the water stream that pours into my mouth? Saliva is mainly composed of water, some salts, and amylase. Amylase is a rather bulky protein, so I don't expect it to diffuse so quickly, but I'm a little concerned about the electrolytes. Ions like $\ce{Na+, K+, Ca^2+, Mg^2+, Cl-, HCO3-, PO4^3-}$ are quite small and might diffuse into the bottle.
I don't have access to the values of the ionic speeds, so I'd like to ask a chemist for confirmation whether these ions can diffuse into the bottle.
To make the question a bit more informative, lets assume I keep the bottle about $\pu{3cm}$ away from my mouth, so the water would be flowing at a speed of approximately $\pu{77cm s^-1}$.

This question reminds me of a germaphobic classmate I had back in Middle-school. He would always stay clear of public lavatories/urinals because he was under the impression that urinating at one would result in bacteria from the urinal climbing up the stream of urine and into his urethra...needless to say, he was a very lonely boy.

The situation you described ("Drinking without sipping") is a modification (albeit, a more reasonable one) of the "Bacteria-climbing-up-a-stream-of-urine" theory.

I'm not a chemist myself, and like you, I am faced with... ahem, "issues", while searching for speeds of common ions in water (at a given temperature, and in the absence of an external electric field). But I feel this question can still be satisfactorily approached even without these values.

From the wording of your question, I suppose you assumed that potassium, sodium, magnesium and other ions would diffuse from a region of higher concentration (the saliva in your mouth) to a region of lower concentration of those ions (the stream of water entering your mouth). Now this would seem perfectly plausible if it weren't for a few things:

1) You're dealing with flowing water

This isn't a simple case of spitting into a glass of water and waiting for the contents of your saliva to (passively) diffuse into the water (and eventually form a dilute saliva solution). By drinking it the way you do, the passive diffusion of electrolytes in your saliva won't be able to match the speed of the water flowing into your mouth, much less exceed it in order to enter the bottle.

2) All the electrolytes in your mouth are trapped in a viscous protein-soup

Human saliva contains a variety of proteins (among other things). Common experience tells us that saliva is (a little more) viscous than water, and doesn't "instantly" dissolve in it (<----- Coming from a guy who had to spit in test tubes for his Biology lab assignments). Now not only do the electrolytes have to actively diffuse upstream to get into the bottle, but they also have to "free" themselves from the saliva first.

3) What do you mean by "saliva"?

This is the philosophically inclined conclusion to my answer. You're worried that the entry of electrolytes from your mouth into the bottle (won't happen) counts as contaminating it with your saliva. At first glance, even if the electrolytes seem to be able to make it to the bottle, you're pretty sure that larger molecules (such as proteins) won't be able to do the same.

If you don't count the proteins, epithelial cells and bacteria that normally constitute human saliva...how would this make saliva any different from a simple laboratory-prepared cocktail of sodium, potassium, magnesium, calcium, etc? Besides, ordinary (un-branded) bottled water would already contain these electrolytes; so by that logic, the water's already contaminated with your "saliva" even before you opened it!


Forming AgCl from electroless Ag We are looking to design flexible substrates with electroless plated silver on them (for use in sensor applications).  We need convert the surface of the silver to silver-chloride (AgCl) for conduction to hydrogel/hydrocolloid.  What are simple ways to do this.  The material will be on rolls and we need a simple production method that can be applied to the rolls.  Thank you.

I would recommend this patent which generates AgCl via the electrolytic chlorination of sheet silver, to quote:

  The present invention relates to a method for chlorination of silver materials in the forms of wire, thick film paste and metallic sheet/disc. The method is environmentally friendly since the process does not produce hazardous chemicals or flammable gas. The electrochemical setup includes position the electrode containing silver materials in the anode and a platinum wire at the cathode. A DC voltage supply between 1 to 5 volts oxidizes the silver surface which receives chloride from imidazolium chloride solution to produce silver chloride on silver wire, film or disc. The imidazolium chloride reagent can be in aqueous solution or polar organic solvents or naturally exists in ionic liquid form. The imidazolium radical cation decomposes to produce a stabilized organic radical and an imidazole molecule. The stabilized radical recombines to produce volatile organic compound that can be recovered by simple distillation. The chlorination reagent can be regenerated...

In place of platinum wire perhaps a diamond electrode.

Also, some pertinent comment from atomistry, to quote:

  The rose-red to purple-brown mixtures of silver subchloride and silver chloride are called by Carey Lea " photochlorides." They are formed by the interaction of colloidal silver and a suspension of silver chloride; by the action of organic reducers and of light on silver chloride; and by the electrolytic chlorination of sheet silver. The photochlorides are characterized by their great sensitiveness to light, blue light producing a blue coloration, and red light a red coloration. This action is reversible, the blue coloration being transformed into red by the action of red light, and so on. This phenomenon is inapplicable to the production of colour-photographs, for white light causes a darkening in colour.


Does 3,6-di-X-cyclohexa-1,4-diene have stereo centers? The following question has multiple correct answers:

The book out of which the question is taken  shows that a, b and d are correct answers.

According to me c should also be correct, as it shows cis-trans isomerism, both X could go to the front (wedge bond), or one could go to the front (wedge) and one to the back (dash bond).  After all geometrical isomerism is also a sterioismersim.
I came to this conclusion considering the definition given on Wikipedia:

  A stereocenter or stereogenic center is any point in a molecule, though not necessarily an atom, bearing groups, such that an interchanging of any two groups leads to a stereoisomer.

Good point. The (c) compound has two tetrahedral stereogenic carbon atoms, with four “substituents”

$\ce{-H}$,
$\ce{-X}$,
$\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$,
and the same one, $\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$ but from the other side, thus with the opposite configuration, or its mirror image.

(‘*’ denotes (the other) stereogenic carbon atom)

They are called pseudo-asymmetric carbon atoms.

A different relative configuration at these two pseudochiral centers leads to these two ‘cis’ and ‘trans’ isomers.

The definition of stereogenic center (for more complete one, see e.g. this answer) is fulfilled.

(Maybe this topic is too early, for this textbook stage, and the author did not notice this complication.)


Does the translational part of internal energy mean the kinetic energy of the gas? Internal energy of an ideal gas consists of energy due to translational, rotational , vibrational etc.

This line from WP says:

  In thermodynamics, the internal energy of a system is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields

At the same time it says:

  The ideal gas is a gas of particles considered as point objects ... Such systems are approximated by the monatomicgases, helium and the other noble gases. Here the kinetic energy consists only of the translationalenergy of the individual atoms. Monatomic particles do not rotate or vibrate, and are not electronically excited to higher energies except at very high temperatures.
  Therefore, internal energy changes in an ideal gas may be described solely by changes in its kinetic energy. Kinetic energy is simply the internal energy of the perfect gas and depends entirely on its pressure, volume and thermodynamic temperature.

So my question is:

Does the translational internal energy of the diatomic or polyatomic molecules also refer to their kinetic energies?

Thanks to @IvanNeretin for pointing me in the right direction.

Imagine a box full of an ideal gas. Let the box and gas together be of a mass $m$.

Now imagine that the box is moving with a velocity $v$. Thus the system will have a kinetic energy $\frac{1}{2}mv^2$. Now this energy is of the system as a whole and is therefore not counted as internal energy.

Similarly imagine the box placed at height $h$. It will have a potential energy $mgh$. Again this will not be counted as internal energy.

When talking about diatomic or triatomic gases, we will say that that the total kinetic energy is the sum of all the possible ways in which the particle can move in space. This includes motion like translational motion, rotational motion and vibrational motion. The number of dimensions of such motion are called the degrees of freedom($f$) of the molecule and the total kinetic energy of the system is given by

$$U = \frac{f}{2}nRT + \phi$$

where the first term represents the total kinetic energy and the last one represents all the other energies associated with the gas molecules

For example for a diatomic gas, $f=5$ because it can move in $3$ dimensions and rotate in $2$. So $$U = \frac{5}{2}nRT + \phi$$

For gases in high temperature, the degree of freedom would increase by due to the vibrational motion of the gas particles.

Wikipedia


How do P-H bonds in phosphorus acids impart reducing properties? My textbook states:

The oxoacids of Phosphorous which contain $\ce{P-H}$ bond have strong reducing properties. Thus, hypophosphorous acid is a good reducing agent as it contains two $\ce{P-H}$ bonds and reduces, for example, $\ce{AgNO3}$ to metallic silver.
These $\ce{P-H}$ bonds do not ionize to give $\ce{H+}$ and do not play any role in basicity.

On the basis of what's given in the text; how do these $\ce{P-H}$ bonds account for the reducing properties of phosphorus acids?

The $\ce{P-H}$ bond imparts an oxidation state of -1 per bond to the acid. Upon oxidation, these are converted into $\ce{P-OH}$ bonds, which imparts a +1 oxidation state per bond. You can clearly see this in the reaction:
$$\ce{H3PO2 + 4AgNO3 + 2H2O -> 4Ag + H3PO4 + 4HNO3}$$
There are two $\ce{P-H}$ bonds in $\ce{H3PO2}$, both of which are converted into $\ce{P-OH}$ bonds in $\ce{H3PO4}$.

Moreover the $\ce{P-H}$ bond has an enthalpy of 322kJ, while the $\ce{P-O}$ bond has an enthalpy of 335kJ/mol. This slight increase in bond energies drives the reaction in the forward direction. Resonances may also be a factor that contributes as well.


Nucleophilicity in polar aprotic solvent (DMSO) The strength of a nucleophile in polar aprotic solvent is proportional to its basicity. Lone pair on O- in the molecules will donate an e- pair to an atom. I'm not sure why the alkene ring in 2 makes it a worse nucleophile compared to 1. Benzene ring is more stable than cyclohexane but the double bond is not the nucleophile in the molecule. Is it due to steric hindrance?

let me start with nucleophilicity of halides. In protic polar solvent, $\mathrm{F^-}$ is more solvated (hydrogen bonding) than $\mathrm{I^-}$ and hence less reactive.

In polar aprotic solvent $\mathrm{F^-}$ solvation is less and therefore more reactive.

Such behavior may not be applicable here, since phenoxide is resonance stabilized while other compound is not. Therefore phenoxide $\ce{2}$  is less nucleophilic then cyclohexanolate $\ce{1}$.


How to create a 3 M solution of ammonia from 25% ammonia We have ammonia $25\%$. We also know $\rho_{\text{ammonia}} = \pu{0.86 \frac{g}{cm^3}}$, $\pu{25 ^\circ C}$. We want to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.

This means that there are some calculations we need to do in order to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.

I try to calculate the required mass by the following method: $\text{m}=\text{C}\cdot \text{V}\cdot M \cdot 0.25$. Is this method correct?

$\text{Produce a solution of }3\;\text{M}\;0.5\;\pu{dm^3}\;\ce{NH3}$

$\\\text{Ammonia}\;25\%\;\text{(mass-percent)}\\\rho_{ammonia}=0.82\pu{\frac{g}{cm^3}}$

Suppose 100ml

$$0.82\times 100\times 0.25=20.5\;\pu{g}\;\ce{NH3}\text{ per 100 ml of solution}$$

$$n_{ammonia}=\frac mM=\left(\frac{20.5}{17.031}\right)\pu{moles}$$
$$C_{ammonia}=\frac{n_{ammonia}}{100\times10^{-3}}\approx 12.037 \;\pu{M}.$$

To decide $V_1$, which is needed in order to prepare the solution of $0.5 \;\pu{dm^3}$ $3\;\pu{M}$ ammonia,

$$V_1=\frac{C_2V_2}{C_1}\approx0.1246 \pu{dm^3}=124.6\; \pu{ml}.$$


What does a square mean in a chemical formula? We find in the book Minerals and their Localities by Jan H Bernard and Jaroslav Hyrsl   several chemical formulas containing a square. What does this square means? An example: Franciscanite $\ce{Mn3 (V,\square)(O,OH)3 (SiO4)}$

Thanks for your answer for  I have been  looking for answer for a long time.

That square refers to a vacancy (or framework vacancy) in the crystal structure. Here's one reference from Advances in Electronic Ceramics. Ceramic Engineering and Science Proceedings. Volume 28, Issue 8. Ed. C. Randall ${\it et\;al.}$

In short, a framework vacancy is a feature of some inorganic materials in which a "guest atom" (or more than one) can fit into the structure or denotes a structure that is missing interstitial or packing atoms (as outlined here).

Here's the info on your mineral where you can see that the square is instead denoted by square brackets [].


Denominator in rate law? What the possible rate laws are for the decomposition of nitramide by the following process:

  $$
\begin{align}
\ce{O2NNH2(aq) &<=>[$k_1$][$k_{-1}$] O2NNH-(aq) + H+(aq)} &\quad &\text{(fast equilibrium)} \tag{1}\\
\ce{O2NNH-(aq) &<=>[$k_2$] N2O(g) + OH-(aq)} &\quad &\text{(slow)} \tag{2}\\
\ce{H+(aq) + OH-(aq)&<=>[$k_3$] H2O(l)} &\quad &\text{(fast)} \tag{3}\\
\end{align}
$$

  (A) $\text{rate} = k[\ce{O2NNH-}]$
  (B) $\text{rate} = k[\ce{O2NNH2}]/[\ce{H+}]$
  (C) $\text{rate} = k[\ce{O2NNH2}][\ce{H+}]$
  (D) Both A and B

I selected A, but the answer is actually D. The answer key gives the following information:

  At the first glance, you may think that Option A is the best answer due to the information available in the question stem in which details the way a rate of reaction is written. However, option B also deserves some attention. It is important to understand that $K_\mathrm{eq}$ is the equilibrium constant for step 1, as this is the fast equilibrium step

  $$K_\mathrm{eq} = \frac{[\ce{O2NNH-}][\ce{H+}]}{[\ce{O2NNH2}]}$$

  Rearranging for $[\ce{O2NNH-}]$, and substituting it into the rate law gives an equally valid expression which leads to Option B also being correct. The best answer is therefore Option D.

I am confused about how this can be so.

Firstly, why can we just sub the equilibrium equation in like this, and what happens to the Ksp term, it seems to just disappear.

Secondly, doesn't this mean that the overall order of the reaction is inconsistent between option A and B (i.e. in A the overall order is 1, but in B it is 1-1=0).

Thirdly, conceptually I am confused about why the concentration of the product should change the rate of the forward reaction, this does not really make sense to me, why should the reaction proceed less quickly just because there is more product around? Further, if it is the case that the product changes the speed of the forward reaction, then why don't all rate equations feature the product?

I am left very confused, perhaps someone can help me out?

The $\ce{O2NNH^-}$ is formed very rapidly but reacts slowly in the second step. The rate limiting step is therefore loss of $\ce{O2NNH^-}$ in which the back reaction with $\ce{N2O}$ is considered (in the question) to be so slow as to be insignificant when writing the rate expression. Thus the rate involves just $\ce{O2NNH^-}$, however, as the fist reaction is a fast equilibrium substituting for $\ce{O2NNH^-}$ with this equilibrium is valid, provided the $k's$  are marked as being different, which they are not, and in the second reaction $k$ should be $k\times K_{eq_1} $.


What chemicals are going to be used to make the smoke black or white? For the papal election, white and black smoke will be used to signal success or failure in the voting process.

From what I understand, the ballots are burnt and chemicals are added to create the color.

But what chemicals are used to get black or white smoke?

I think Titanium Tetrachloride would be a good option for white smoke (I've heard it's used quite often in smoke bombs). But it releases quite a lot of $HCl$ gas and so it might not be very safe.
A quick Google search comes up with a lot of answers to your question though. This site even answers your question in the same context (The elections going on at the Vatican):
http://thehappyscientist.com/science-experiment/black-and-white-smoke
Apparently you can make white smoke using Magnesium powder as well. (The white colour is due to fine particles of Magnesium Oxide $MgO$ )
It seems the Vatican has experimented using many sources before (Military flares, Smoke bombs and so on) though apparently they didn't turn out too well...
EDIT
The Wikipedia article on Coloured smoke gives another possibility. You can add Titanium Dioxide ($TiO_2$) to a mixture of Potassium Chlorate ($KClO_3$) and Lactose for white smoke. $KClO_3$ is a powerful oxidizing agent which reacts with Lactose (which is a reducing sugar) to create an explosion. $TiO_2$ is safe and is used as a pigment in most white paints and as a food colouring as well.
EDIT2
The official recipes have been revealed by the Vatican press office. See this story on the New York Times website.
The previous edit was partially correct. They use a mixture of $KClO_3$, lactose and pine rosin for white smoke. For black smoke, they use Potassium perchlorate ($KClO_4$), anthracene (an aromatic compound made of three fused benzene rings) and sulfur.
The process (quoted from the site):
"The chemicals are electrically ignited in a special stove first used for the conclave of 2005, the statement said. The stove sits in the Sistine Chapel next to an older stove in which the ballots are burned; the colored smoke and the smoke from the ballots mix and travel up a long copper flue to the chapel roof, where the smoke is visible from St. Peter’s Square. A resistance wire is used to preheat the flue so it draws properly, and the flue has a fan as a backup to ensure that no smoke enters the chapel."


Electrolysis of water in 0.1 M hydrochloric acid For water electrolysis in acidic media ($\ce{TiO2}$ anode and $\ce{Pt}$ cathode) the half reactions for water splitting and $\ce{O2/H2}$ evolution are as follows:

Anode: $\ce{H2O + 4 H+ -> 4 H+ + O2}~~(E = \pu{-1.23 V}\text{ vs NHE})$
Cathode: $\ce{4H+ + 4 e- -> 2 H2} \quad\quad~~(E = \pu{0.00 V}\text{ vs NHE)}$

If I use $\pu{0.1 M}$ $\ce{HCl}$ as my acidic electrolyte, what reaction will occur at the anode (the standard oxidation potential for $\ce{2Cl-}$ to $\ce{Cl2}$ is $\pu{-1.36 V}$ vs NHE). In $\pu{0.1 M}$ $\ce{HCl}$ electrolyte, there should be $\ce{H_2}$ evolution at the cathode. Will the anode maintain $\ce{O2}$ or $\ce{Cl2}$ evolution (or maybe both?)

Possible reactions :

At Cathode:

$\ce{2H2O{(l)} + 2e- -> H2{(g)} + 2OH-{(aq)}}$

$\ce{2H+{(aq)} + 2e- -> H2{(g)}}$

At anode:

$\ce{2H2O_{(l)} -> O2_{(g)} + 4H+_{(aq)} + 4e-}$

$\ce{2Cl-{(aq)} -> Cl2{(g)} + 2e-}$

General rules of electrolysis of aqueous solution using inert electrodes :-

Reduction potential of $\ce{H2O}$ > reduction potential of cations of 1, 2 and 13 groups of periodic table. So reduction of $\ce{H2O}$ takes place at cathode.
Reduction potential of $\ce{H2O}$ < reduction potential of other cations (than cations of 1, 2 and 13 group). So reduction of $\ce{H2O}$ does not occur at cathode.
Oxidation potential of $\ce{H2O}$ > oxidation potential of fluoride, sulphite, sulphate, nitrate, etc. So oxidation of $\ce{H2O}$ occurs at anode.
Oxidation potential of $\ce{H2O}$ < oxidation potential of chloride, bromide and iodide, etc. So oxidation of these ions takes place.

Keeping these rules in mind, at cathode reduction of hydrogen ions will take place and hence $\ce{H2}$ gas will be evolved. At anode, oxidation of chloride ions will take place and chlorine gas will be evolved and $\ce{O2}$ won't be evolved.

Although the oxidation potential of $\ce{H2O}$ > oxidation potential of chloride ion, chloride ion gets oxidized because of over potential of $\ce{O2}$ i.e. extra potential is required to initiate a reaction at a required rate. And formation of $\ce{O2}$ from $\ce{H2O}$ is kinetically very slow.

Therefore oxidation of chloride ion takes place.


How does temperature affect buffer capacity? I am currently leaning towards an experiment on how certain factors affect the capacity of a phosphate buffer, and am wondering if temperature in any way affects the capacity of the buffer. I know changing the temp will change the Ka of the acid, but am not sure if that changes capacity (how much it resists pH change) or just changes the pH range in which the buffer effective e.g. from 5-8 pH to 4-7 pH.

If you follow this derivation of buffer capacity, the last equation shows you that buffer capacity is a function of $K_a$, so a change in the Ka will affect the buffer capacity. Also, since you know that $K_a$ is affected by temperature, that should lead you to realize that the actual chemical environment is different at different temperatures due to a different concentration of acid and weak base pair. The magnitude of effect maybe small, but your experiment can prove that one way or the other!


Do different oxidation states of the same element have the same emission spectrum? I looked at the flames of copper (I) chloride and of copper (II) chloride through a spectroscope and they looked the same. The flame colour is the same too. But since they have different oxidation states, and therefore different electron configurations, should their emission spectra be slightly different?

Would this logic also apply, for example, to elemental sodium in a discharge tube, vs sodium chloride burning in a flame? Would their emission spectra differ?

Interesting question. Keep in mind that the elemental emission spectrum in a flame or plasma and even a discharge does not remember its history in solution or a solid phase.

The punchline is that the emission spectrum is dependent on the elements gas phase chemistry in the flame/plasma/discharge. Prof. Ed has explained you the example of sodium. Let us say we have the following

(a) a block of element sodium
(b) a block of sodium chloride crystal
(c) solution of sodium chloride in water

If you introduce (a), (b) and (c) in the flame, the flame will be colored yellow in each case, which means that the emission is coming from a common emitter. That emitter is a elemental sodium atom excited by high temperature in the gas phase. Thus atomic emission spectrum is a fingerprint of the element.

You may ask that you introduced Na(+) in the flame in the case of b and c. Flames can easily reduce an ion to the elemental state.

How to see the sodium ion spectrum: As we just said, the emission spectrum is independent of this original state. You can only cause ionization by increasing the temperature. This is the way to see the spectrum of an ion. If we were using a high temperature flame/electrical discharge, we would start seeing Na(+) spectrum along with elemental sodium spectrum.

Coming to your particular example: You introduced copper (I) and copper (II) into a flame and they all colored it beautiful blue-green. The reason is that if the flame temperature is low, compounds cannot fully dissociate into atoms (not enough energy to break the bonds). In such cases, very simple diatomic or triatomic molecules are formed in the flame which emit their characteristic colors. In the case of copper, CuCl is formed in the flame. CuOH may be formed as well. Whether you introduced Cu(I) or Cu(II), as a chloride, it does not remember its solution phase or solid phase history.

If you were indeed using a high temperature flame, you will never ever such a blue green coloration, because this time, the emission is from Cu atoms (in UV). Hope that clarifies your confusion.


Baeyer-Villiger Oxidation restricted to Ketones According to Wikipedia, Baeyer Villiger oxidation occurs at the electron deficient oxygen site, with elimination of the carboxylate ion, and migration of an R group. But it is restricted to ketones and lactones. Why are aldehydes not included in it? Moreover, since migratory amplitude of hydrogen is lower than alkyl or aryl, shouldn't aldehydes be more prone to participate in this reaction?

May be Baeyer Villiger oxidation is not for aldehydes in the past. But that has changed recently (Ref.2). The abstract of this reference states that:

A conceptually distinct, modern strategy for Baeyer−Villiger oxidation (BVO) was developed. Our novel method involves initial hydration of water to carbonyl compounds, followed by ligand exchange of hypervalent aryl-$\lambda^3$-bromane on bromane(III) with the resulting hydrate, yielding a new type of activated Criegee intermediate. The intermediate undergoes BV rearrangement and produces an ester via facile reductive elimination of an aryl-$\lambda^3$-bromanyl group, because of the hypernucleofugality. The novel strategy makes it possible to induce selectively the BV rearrangement of straight chain primary aliphatic as well as aromatic aldehydes, which is missing in the classical BVO: for instance, octanal and benzaldehyde afforded rearranged formate esters with high selectivity (> 95%) under our conditions, while the attempted classical BVO produced only carboxylic acids. This firmly establishes the powerful nature of new methodology for BVO.

The suggested mechanism and comparison it with classical Baeyer−Villiger oxidation is shown in following scheme $(\bf{A})$:

The part $\bf{B}$ of the scheme depicted computational calculation of energy diagram for acetaldehyde oxidation.
In addition, similar oxidation of acetal (an aldehyde derivative) to ester have been known for years. For example, conversion of cyclic acetals to hydroxy esters by mCPBA oxidation is published in 2002 (Ref.2).

Reference:

Masahito Ochiai, Akira Yoshimura, Kazunori Miyamoto, Satoko Hayashi, Waro Nakanishi, "Hypervalent $\lambda^3$-Bromane Strategy for Baeyer−Villiger Oxidation: Selective Transformation of Primary Aliphatic and Aromatic Aldehydes to Formates, Which is Missing in the Classical Baeyer−Villiger Oxidation," J. Am. Chem. Soc. 2010, 132(27), 9236–9239 (https://doi.org/10.1021/ja104330g).
Jin Yeon Kim, Hakjune Rhee, Misoo Kim, "Conversion of Cyclic Acetals to Hydroxy Esters by MCPBA Oxidation," Journal of the Korean Chemical Society 2002, 46(5), 479-483 (DOI: 10.5012/jkcs.2002.46.5.479).


What is the formula for migration rate of DNA in agarose gel electrophoresis? Let's consider a typical agarose gel electrophoresis experiment. An appropriate quantity (not so much that it is overloaded, but not so little that it cannot be visualized) of linear DNA pieces with identical sequence and chemical structure is loaded onto the gel, and constant voltage is applied for a given time. I suspect that there is a function:

$x = f(V, t, l, c, k)$

Where the variables are:

$x$, cm, distance from well
$V$, Volt, potential difference across the gel
$t$, minutes, time over which the voltage is applied
$l$, bp, length of DNA fragments
$c$, gr/100 ml, density of the gel given in grams of agarose per 100 ml of buffer
$k$, a constant which is empirically determined for a given running buffer and gel tank

Basically, I want a formula into which I can plug in the size of my DNA and a few other parameters, and figure out how long I have to run it to get decent separation (the output variable is x because I intend to plot $x$ over $t$ for a given planned electrophoresis).

Is such a formula possible? What is it?

I have previously worked with gel documentation software which fits linear, polynomial and exponential functions to the DNA (I couldn't figure out exactly what kind of fitting the software was doing). However, after training a best-fit model on my ladder (with ~12 bands) and applying it to my samples of known size, I found all of these fits to be very inaccurate (5-10% error). Therefore, I suspect the equation governing the migration rate of DNA is in fact not a simple exponential function (as is often taught in college) but more complicated. Unfortunately, I lack the background in physical chemistry to derive the equation myself.

If you have a good argument for why the relation should be a simple exponent, eg. $x = e ^ {\alpha \cdot l}$, that is also an acceptable answer.

I guess my answer is very old, but here it is for newcomers searching the internet for this formula:

$$D = a - b \log(m)$$

where

$D$: Distance traveled for the restricted fragment.
$a$ and $b$: Constants depend on the electrophoresis experiment.
$\log(m)$: The logarithm of the molecular weight ($m$)of the fragment.

Knowing $a,b$ and the distance traveled in your gel, you can easily determine the molecular weight of your fragment and its size by the following formula:

$$m = e^ \left(\frac{a-D}{b}\right)$$


Why are branched alkanes used as fuels instead of straight chain alkanes? My textbook states that branched alkanes are thermodynamically more stable than straight chain alkanes. Of course, iso-octane has an enthalpy of formation of $\Delta H_\mathrm f$= $\pu{-259.3.40 kJ mol-1}$ and n-octane has an enthalpy of formation of $\Delta H_\mathrm f$= $\pu{-249.73 kJ mol-1}$ which indicates that iso-octane is relatively more stable than n-octane at room temperature. When they are used as fuels, both produces energy and the enthaply of combustion of iso-octane would be less than enthalpy of combustion of n-octane. So, n-octane should be more efficient fuel than iso-octane. But in petrol, iso-octane has more composition than n-octane. Why?

The main reason for the high quantity of iso-octane is to prevent Engine Knocking. The Wikipedia page for Octane explains this quite well:

The octane rating was originally determined by mixing fuels from only heptane and 2,2,4-trimethylpentane (a highly branched octane), and assigning anti-knock ratings of zero for pure heptane and 100 for pure 2,2,4-trimethylpentane. The anti-knock rating of this mixture would be the same as the percentage of the latter in the mix. Different isomers of octane can contribute to a lower or higher octane rating. For example, n-octane (the straight chain of 8 carbon atoms with no branching) has a -20 (negative) Research Octane Rating, whereas pure 2,2,4-trimethylpentane has an RON rating of 100. Some fuels have an octane rating higher than 100, notably those containing methanol or ethanol.

A fuel made of only octane would have such a high enthalpy of combustion, that it would combust spontaneously in the cylinders, causing knocking, as seen by the RON number of -20 for it. Thus, iso-octane content is high because it reduces knocking and prevents engine wear.


What would a (gas) chromatogram look like if two compounds have the same retention times? In gas chromatography, the number of peaks represents the number of compounds in an unknown sample, as retention times for each component differ.
But what would happen if two components have the same exact retention time? Would you see two separate peaks or one?

] Complete overlap (= identical retention) is a regular feature of experimental chromatography with very different compounds and complex mixtures. It just happens when the selectivity happens to be equal to 1.00. You can have one, two, or three different compounds hidden under a single peak and this usually happens in early eluting peaks in all modes of chromatography of a complex mixture.
Peak overlap is undesirable, but it happens all the time. All you need to do is to change the method.
People have spent a portion of their life in something called as peak deconvolution to find out what is hidden in a single peak. Note how the area adds up.