There's a technicality in that the 2013 IUPAC recommendations (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book)) have slightly awkward preferred IUPAC names for simple amines like these. The preferred IUPAC name for your molecule is "N,N-dimethylethanamine" (as described by P-62.2.2.1 on p 671, but note that there is an erratum for the triethylamine example shown there).

Putting that aside, though, what you're conceptually missing is that the methyls and the ethyls in your proposed name are (implicitly) not on equal footing. The methyls are names of substituents, but ethyl is not a substituent - it is the parent chain (since it is longer than methyl). So, $\ce{CH3CH2NH2}$ is named "ethylamine", and $\ce{CH3CH2N(CH3)2}$ is named as a N,N-disubstituted form of "ethylamine", i.e. "N,N-dimethylethylamine". In common usage the "N,N" is often dropped, leaving "dimethylethylamine". I should emphasise that this is not in line with the IUPAC recommendations, but people can and will use it, and this is why you will see it on the Internet. Returning to the preferred IUPAC name in the previous paragraph, we can see that the use of "ethanamine" rather than "ethylamine" makes the difference between methyl and ethyl a lot clearer.

In fact, according to the recommendations, it is also permissible to name tertiary amines in the manner (alkyl1)(alkyl2)(alkyl3)amine (although it does not lead to the preferred name). If we were to adopt this method, then the name would be ethyl(dimethyl)amine, because now there is no notion of a parent chain and all three substituents are merely substituents - which should be listed in alphabetical order ignoring prefixes like "di". The parentheses around "dimethyl" are required (but not around "ethyl") because of a separate rule (P-62.2.2.2, p 672 - see also the discussion in the comments).

To sum up:

N,N-dimethylethanamine: Preferred IUPAC name
ethyl(dimethyl)amine: "Alternative" IUPAC name; permissible but not preferred
dimethylethylamine: Doesn't comply with 2013 IUPAC recommendations, but the logic behind the name is similar to that used in generating the preferred IUPAC name.


How to minimise solution concentration changes due to evaporation?

What if I stored some water in exactly the same conditions together with my sample, would I be able to approximate water loss in my sample from the water loss in this reference container, or does the difference in the concentration of the solute affect the rate of evaporation too much?

That would be a gross approximation, not suitable for a standard solution. Your solution and pure solvent do have different vapor pressures according to Raoult's law.

  What temperature?

It depends on solvent, solute, and their interaction. I'd check two main things: i) solubility at the storage temeprature, which must be as close as possible to the solubility at the working temperature; ii) vapor pressure of the solvent, that gives you a rough indication of the storage temperature. You could take a look at the temperature dependence of the solvent vapor pressure and solute solubility - or, if seriously in doubt, even try simulating the thermodynamics of your solution using a software.

  What container should I choose?

Volumetric flasks are usually the best choice. They minimize evaporation compared to other glassware (bottles, beaker, etc.) and, most important, they have the most accurately placed graduation mark. Use the appropriate rubber cap and close accurately with laboratory film. PTFE band between cap and flask cone might also help preventing losses.

At times, a slight overpressure above the liquid can be desirable. In this case, a gas valve (or a quality glass faucet) will be placed on top of the flask and nitrogen will be flushed inside. The flask can be stored capped by the sealed valve. Also consider the opportunity of degassing your solution by placing sieves on the bottom of the flask or by bubbling nitrogen through the liquid. Finally, consider the need for a dehydrating agent in the case of a moisture-sensitive solution.

But, most of all, don't keep standard solutions for two years but replace them every ~6 months.


How does strong acids conduct electricity in water?

In water at 25 ºC $[H+]=[OH-]=10^{-7}M$, so a concentration of $[H+]=10^{-6}M$ is very low. With a strong acid like HCl you have to dissolve only around 35 micrograms in a liter of water to get this concentration.
Sea water on the other hand has a high conductivity and, in average, 35 grams per liter of salts dissolved in it. If you dissolve 35 grams of HCl en 1 liter of water you have $[H+]=1M$ and surely you can expect a high conductivity.
These are gross numbers, but I think they help to get the idea.
Conductivity of a solution depends on the free ions present on the mixture. With a strong acid you can get a high concentration of ions using a little quantity of acid. With a weak acid you need much more.


How does temperature affect buffer capacity?

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!


Deriving an expression for dU as a function of V and T

I realize now what you're looking for:
Consider $\mathrm{d}S(V,T)$ as

$$\mathrm{d}S(V,T) = \left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V +  \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T$$ and place into the first law:

$$
\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V = T\left[\left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V +  \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T\right] - p\mathrm{d}V
$$

Combine like terms of $\mathrm{d}V$:

$$
\mathrm{d}U = T \left(\frac{\partial S}{\partial T}\right)_{V} \mathrm{d}T + \left[T \left(\frac{\partial S}{\partial V}\right)_{T} - p\right]\mathrm{d}V
$$

and you have $\mathrm{d}U$ a function of $V$ and $T$.

It turns out that $\displaystyle T \left(\frac{\partial S}{\partial T}\right)_{V} = \left(\frac{\partial U}{\partial T}\right)_{V} = \mathrm{C}_V$ and $\displaystyle \left(\frac{\partial S}{\partial V}\right)_{T} = \left(\frac{\partial p}{\partial T}\right)_{V}$ by a second-derivative Maxwell relations of $A$ (below) with respect to $T$ and $V$, and a "prettier" version is:

$$
\mathrm{d}U = \mathrm{C}_V\mathrm{d}T + \left[T \left(\frac{\partial p}{\partial T}\right)_{V} - p\right]\mathrm{d}V
$$

(In the special case of an ideal gas, $pV = nRT$ and the term in brackets is zero - just write $p = nRT/V$ and evaluate.)

To derive $\mathrm{d}A$:
What you're looking for is a Legendre transform. You know

$$
\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V
$$
from the First Law of thermodynamics.

Hence the natural variables for $U$ are $S$ and $V$, not $T$ and $V$ (which is desired). So instead of $U(S,V)$ you want $?(T,V)$ where $?$ happens to be the Helmholtz free energy $A$, sometimes denoted $F$ by physicists.

To do the transform, you take your original function - the exchange variable times its differential form, i.e.

$$ A(T,V) = U(S,V) - S\left(\frac{\partial U}{\partial S}\right)_{V}$$

You know also that $$\left(\frac{\partial U}{\partial S}\right)_{V} = T$$ because

$$\mathrm{d}U(S,V) = \left(\frac{\partial U}{\partial S}\right)_{V}\mathrm{d}S +  \left(\frac{\partial U}{\partial V}\right)_{S}\mathrm{d}V = T\mathrm{d}S - p\mathrm{d}V$$
by total derivative formula for a function, in this case $U$.

So you have $A = U - ST$ or as commonly written, $A = U - TS$ so the differential form of $A$ is $\mathrm{d}A = \mathrm{d}U - T\mathrm{d}S - S\mathrm{d}T = T\mathrm{d}S - p\mathrm{d}V -T\mathrm{d}S - S\mathrm{d}T$, thus $\mathrm{d}A = -p\mathrm{d}V - S\mathrm{d}T.$ That is a function of $V$ and $T$.


What are the best chemical liquids to clean metallic surfaces?

Assuming clean means anti-pathogenic, I would use Hydrogen Peroxide "oxyclean" since it can "clean" metals by removing pathogens. $\ce{H2O2}$ is marketed in various strengths. 1% dilution can kill up to 99.99% bacteria in 30 seconds.

Any butyl based cleaning product is a very good degreaser but it may harm rubber products.

Check here and here for more.


At what pressure will hydrogen start to liquefy at room temperature?

$\ce{H2}$ cannot be liquified at room temperature, whatever the pressure. Generally speaking, all gases can only be liquified when the temperature is under its critical value.


What is the oxidation state of nitrogen in nitrate ion and what is its structure?

You are correct that the oxidation state is +5. The second period elements, particularly carbon, nitrogen and oxygen, do show variable oxidation states regularly.

The $\ce{NO3-}$ ion cannot be depicted using a single Lewis structure. However, its resonance structures are shown below:

What this means is that the actual structure of the nitrate ion is somewhere in between the three structures shown below. It is sometimes also depicted like this to show the symmetrical nature of the ion and the distribution of charges across it.

http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
http://en.wikipedia.org/wiki/Nitrate#Structure


Why do the trends in reactivity not apply for francium?

Francium is radioactive. All isotopes of francium are radioactive. The most stable isotope of francium has a half life of only 22 minutes. From the Wikipedia article I linked:

  Bulk francium has never been viewed. ... However preparing such a sample is impossible, since the extreme heat of decay (the half-life of its longest-lived isotope is only 22 minutes) would immediately vaporize any viewable quantity of the element.

This extreme radioactivity makes it hard to determine the chemical properties of francium. Some are known. For example, francium's ionization energy is 392.8 kJ/mol (cesium is 375.7 kJ/mol). Since francium has a higher ionization potential than cesium, which it does because of relativistic effects, francium should be less active than cesium. However, since observable quantities are difficult (nearly impossible) to come by. It is hard to say for sure.


Is it possible to calculate atomic radius with electron configuration?

There's no measuring the radius of a single atom, mainly because electrons are around the nuclei: We can't define a radius for any single atom. Take a loom at the uncertainty principle:

  Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the standard deviation of position $\sigma _x$ and the standard deviation of momentum $\sigma _p$ was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928: $$\sigma _x \sigma _p \geq \frac{\hbar}{2}$$ Wikipedia

Thus, if you want to measure an atom's radius, you have to measure the distance between two nuclei, and then divide it by two. This results in different types of radius defined for atoms. The most common are covalent radius, ionic radius, and Van der Waals radius. Radii trends are what that are mostly studied in undergrad chemistry, rather than the how of measuring them with advanced techniques.

Chemguide is pretty informative when it gets to teaching the trends:

chemguide: atomic radius


Is a triplet state or singlet state more stable?

The key issue here that we are comparing excited states. So let's begin by defining Singlet and Triplet Excited States.
Singlet and Triplet Excited States:  A singlet or a triplet can form when one electron is excited to a higher energy level.  In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired).  In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in the illustrative Figure below. Singlet, doublet and triplet is derived using the equation for multiplicity, $2S+1$, where $S$ is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up ($s = +1/2$) or spin down ($s = -1/2$).  If we were to calculated the S for the excited singlet state, the equation would be $2[(+1/2) + (-1/2)]+1 = 2(0)+1 = 1$, therefore making the center orbital in the figure a singlet state.  If the spin multiplicity for the excited triplet state was calculated, we obtain $2[(+1/2) + (+1/2)]+1 = 2(1)+1 =3$, which gives a triplet state as expected.

Illustrative Figure: Spin in the ground and excited states The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. ***For this reason, the lifetime of the triplet state is longer than the singlet state by approximately 104 seconds fold difference.***The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.


How to calculate the concentrations of conjugate species present in the solution of diprotic acid?


Chemical bond-covalent-valence electrons OH-

In the photo the orange electrons are non bond electrons.
And the black electrons are bond electrons.

There are 2 electrons took part in creating chemical bond in $\ce{OH}$.
The first one is from oxygen.
The second is from hydrogen.

This kind of bond is called "Covalent bond".
For more about Covalent bond :
https://en.wikipedia.org/wiki/Covalent_bond

The minus is  related to  the black and white electrons that are free (in Oxygen).

The white electron is come from other element.


How does a group 15 oxide react with water?

The reaction between the two substance you have mentioned above in synthesis
it is an acid-base reaction .
and about the other question you asked above i can say that we have reactions where two ionic substances take part in oxidation reduction reactions instead of double replacement.


Why does a substance with an endothermic heat of solution dissolve?

Dissolution happens in three steps.

Solute-solute attractions must be broken (consumes energy, endotherm), solvent-solvent attractions must be broken (also endotherm), and finally solute-solvent attractions form. This results in a lower energy state and is exotherm.

Dissolution will be endotherm if it takes more energy to break the mentioned attractions, than is released in the last step. You are asking then, why is this reaction spontaneous? In other words, why does it occur? To find out we need chemical thermodynamics. $$\Delta G=\Delta H - T\Delta S$$

$\Delta G$ = Change in Gibb's free energy for a reaction. $\Delta G<0$ = reaction is spontaneous. Otherwise it is NOT spontaneous (ie. will not occur on its own).

$\Delta H$ = Change in enthalpy. $\Delta H <0$ = reaction is exotherm. If larger than 0, the reaction is endotherm. Already now we can see that an endotherm reaction will make $\Delta G$ for the reaction more positive, and thus the reaction is less likely to be spontaneous.

$T$ = temperature in K.

$\Delta S$ = Change in entropy. $\Delta S > 0$ = reaction leads to a state of higher entropy. Entropy describes disorder in a system. Dissolution of a solute in a solvent will always lead to a state of higher disorder, since we go from having all the solute concentrated in a cluster (more ordered) to being spread evenly throughout the solution (more disordered). We can see from the equation that if the reaction leads to higher entropy, then higher temperatures increase likeliness of $\Delta G < 0$, and the reaction will be spontaneous. On the other hand, if we're going to a more ordered state, lower temperatures increase likelihood for a spontaneous reaction.

So how do we apply this to dissolution?
You know that the dissolution reaction is endotherm. That means that $\Delta H > 0$ for this particular reaction. Yet we know that the reaction is spontaneous, since you report that it does indeed occur! This means $\Delta G < 0$. The only way to make that happen is for $\Delta S$ to be positive. So the reaction must go to a more disordered state. Since it's a dissolution, we know that it does. The solute goes from being concentrated in a cluster (ordered) to being spread throughout the solvent (disordered). Pay attention to where temperature is in the equation. It is quite easy to see, that for a dissolution reaction where  $\Delta S > 0$, you can increase the chance of the reaction being spontaneous by increasing temperature!

Bottom line
Your dissolution reaction is endotherm, but because dissolution reactions lead to states of higher entropy, your particular reaction is spontaneous (it occurs) at your specific temperature nonetheless! If it wasn't, heating the system could make it so.


Does adding lemon juice to water chemically alter its nutrients and thereby negating any of its health benefits?

Does water really chemically alter the nutrients in lemon juice rendering a lemon drink of no health benefit whatsoever other than the obvious water base?

No, of course not. Otherwise, humanity would have succumbed to scurvy a long time ago.

I encourage you and your friend to visit PubMed where you can peruse hundreds of refereed abstracts of studies that illustrate that nutrients from citrus fruits are bioavailable to humans and other species and necessarily require administration with water.

Here is a one-pager on lemon juice from the Cleveland Clinic, which is probably a reliable enough source in this case if you want to skip the more technical items at PubMed.


Why does neighbouring group participation happen?

There may be some confusion over what neighbouring group participation (NGP, also sometimes referred to as anchimeric assistance) is. Carey puts it quite succinctly:

  When a molecule that can react by nucleophilic substitution also contains a substituent group that can act as a nucleophile, it is often observed that the rate and stereochemistry of the nucleophilic substitution are strongly affected. The involvement of nearby nucleophilic substituents in a substitution process is called neighboring- group participation.

  Source: Advanced Organic Chemistry (Part A). Francis Carey.

Rather than a discrete carbocation forming, which is then stabilised, NGP usually (though not exclusively) occurs via two consecutive SN2 processes, with overall retention of stereochemistry.

The reason why the NGP happens, rather than an external nucleophile just doing the substitution is generally due to the fact that intramolecular reactions happen at a faster rate than intermolecular ones, since this removes the necessity of encounter between nucleophile and electrophile (the two are essentially tethered together). The initial internal attack forms a ring (usually 3 or 5 membered for reasons that should be familiar to you) which may then be opened by the external nucleophile.

A classical example is of a sulfide displacing a halide, with subsequent opening of the reactive 3-membered intermediate:

  Source: Organic Chemistry. Clayden/Warren.


I found something interesting in Acta Crystallogr. 1952, 5 (2), 292–292, which is attached below.

As someone rightly pointed out in the comments, we'll know this only if we know the crystal structure of the compound in question.

What's interesting to note is that hydrogen bonding, and the arrangement of atoms in space under influence of the same varies from compound to compound, and is dependent on the crystal structure of the same.


My text seems to say that the high(est) energy orbitals are not valence orbitals - everything else says otherwise. What's true?

The problem is the words used to describe energy. I am not native english speaker so I am not the best person to answer your question but I'll try.

As smaller $n$, smaller energy. Problems arises when the words higher lower are used to compare numeric values (as those of the energy). They must say greater or smaller.

Without accurate words all of us intuitively agree in that:
"2 is higher that 1", and we should say that 2 is greater than 1 (2>1)

When negative numbers are involved it is not so clear. For some authors "-2 is higher than -1" and "-2 is higher than 1", for others "-2 is lower than -1" and "-2 is lower than 1". In the later cases the correct word is "smaller" (-2<-1).

In this context energies are negatives because of the definition of the reference states. So the energy of an $1s$ orbital is smaller than the energy of a $2s$ orbital.

I have some discrepancies about the meaning and usage of orbital energies but I adopted the common usage to avoid extra confusions.


Help on alternative terpene names and naming schemes

Chemical nomenclature is officially defined by IUPAC (International Union of Pure and Applied Chemistry). You can find most IUPAC names for these terpenic compounds by searching their popular names on Wikipedia, they are quite long and confusing terms though, and can only be interpreted if you have a certain knowledge in organic/inorganic chemistry.

What you are probably more interested in are databases with hundreds of thousands of compounds with unique identifying numbers that are unequivocally assigned to a specific substance. Two of the largest are CAS Index (https://en.wikipedia.org/wiki/CAS_Registry_Number) and ChemSpider (https://en.wikipedia.org/wiki/ChemSpider). I believe working with numbers will facilitate your work a lot, so I would suggest that you look for one of these databases (but be careful, each database uses a different numbering system so you should pick one and stick to it).


Are there bonds which share single electron?

There exists an interesting molecule dicarbon with bond order two and two pi bonds instead of the usual one sigma and one pi. This is the case when 2s-2p mixing is allowed in molecular orbital theory. However when the aforementioned condition is not allowed, dicarbon exists as a molecule with one sigma and two pi bonds of, interestingly, one electron each. So in a way, one electron bonds do exist.


How many moles NH4Cl must be added to NH3 to create buffer with pH=9?

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


What is a benzene ring minus a hydrogen atom called?

You could be referring to "aryl".  "Phenyl" is specific to $\ce{C_6H_5}$, "aryl" is generic to monovalent benzenoid  aromatic ring groups.


What is the purpose of ammonium chloride in anti hydrogenation of alkynes?

Products from reaction step (1) are the trans-alkene and 2 equivalents of the amide $\ce{LiNHEt}$, a strong base. Step (2) is the workup of the reaction mixture with aqueous $\ce{NH4Cl}$ solution, which serves several purposes. First, it quenches the lithium amide and unreacted lithium metal.

$$\ce{LiNHEt + NH4Cl ->~ EtNH2 + NH3 + LiCl}$$
$$\ce{Li + H2O ->~ LiOH (aq) + \frac{1}{2} H2\uparrow}$$

Secondly, the trans-alkene is usually less water-soluble than ethyl amine and the lithium and ammonium salts, and can be separated from the workup mixture by precipitation or extraction with an organic solvent.


What's the reaction when bleaching hair with lemon juice (citric acid)?

Hair is mainly keratin, the same protein found in skin and
  fingernails. The natural color of hair depends on the ratio and
  quantities of two other proteins, eumelanin, and phaeomelanin.
  Eumelanin is responsible for brown to black hair shades while
  phaeomelanin is responsible for golden blond, ginger, and red colors.
  The absence of either type of melanin produces white/gray hair.

I think the citric acid (in lemon) acts as a bleaching agent. The "bleach" reacts with the melanin in hair, removing the colour in an irreversible chemical reaction. The bleach oxidises the melanin molecule.

The illustration of the molecules perhaps might give a better picture on which groups are oxidised in the process.

The melanin is still present, but the oxidised molecule is colourless. However, bleached hair tends to have a pale yellow tint. It behaves as an oxidising agent at least according to my findings.

Acknowledgements

https://www.thoughtco.com/salon-hair-color-chemistry-602183
http://www.keratin.com/as/as002.shtml


If a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral (achiral).

The all hydrocarbon [2]catenane will be achiral. The molecule contains a plane of symmetry; picture a plane that contains one of the rings and bisects the other ring. In the following image, you can see that the plane of the screen contains the ring on the right and bisects the ring on the left.

You might say that if we replaced the circles with cyclohexane rings, because of their chair-like shape, the plane of symmetry is lost.  That is true, but you can flatten one of the cyclohexane rings and the plane of symmetry is restored.  Remember, if any thermally accessible conformation of a molecule is achiral, then the molecule is achiral.

Sometimes it helps to build a model or use a Newman projection to help with the stereochemical analysis.  Here is such a projection for your chloro-[2]catenane.  Note that it still possesses a plane of symmetry (a plane perpendicular to the screen).

If we go further and add another chlorine atom, but now to the back ring, then the molecule becomes chiral — just like 1,3-dichloroallene.

Regarding:

  The compounds attached seem to be enantiomers. I don't see how by rotation they could be super-imposable.

Look at the bottom figure in my second drawing.  I've taken your interlocking 5- and 6-membered catenane and rotated the 6-membered ring a bit further along (a conformational isomer; remember, as I mentioned above, if a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral [achiral]).  The plane of the screen now contains the 5-membered ring and it bisects the cyclohexane ring.  The plane of the screen also contains the methyl group.  The plane of the screen is a symmetry element of this molecule; therefore this molecule is achiral.


thinking about osmotic pressure when liquid is replaced by gas

In both the case of the osmotic pressure of a dilute solution and the case of the pressure exerted by an ideal gas, the solute or gas may be described as composed of non-interacting (ideal) particles, and the mathematical expressions (equations of state) describing the two situations are very similar (in one case $p=cRT$, in the other $\pi = cRT$)$^\dagger$. However it might be less confusing if equivalent to state that in both cases the equations describe similar relationships between the work required to change the volume of the system and the accompanying change in the concentration of gas or solute. In both cases work can be done by the system through an expansion, but in one case the expansion results from pressure exerted by the gas, while in the other it results from pressure exerted by the solvent. In the case of osmotic pressure, since the chemical potential of the solvent is coupled to that of the solute (as described by the Gibbs-Duhem relation) it is possible to relate the osmotic pressure to the solute concentration  (in the limit of an ideal solution as described by Henry's Law).

$^\dagger$As I commented, the equations are analogous but in my opinion, "The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature." is a misstatement. The entire solution, if evaporated, would not exert the same pressure. It is more subtle than that. It is the solute that would exert an equivalent pressure if a gas.


Why is the Chlorine missing an electron in Chlorine Monoxide?

In fact the unpaired electron belongs neither to Cl nor to O, but to a molecular orbital spread over both atoms [but not equally: see below].  The Lewis structure is simply not a complete representation of what is going on.  (And for what it's worth wikipedia gives the same Lewis structure you drew: http://en.wikipedia.org/wiki/Chlorine_monoxide)

Edit: Actually it seems the highest occupied molecular orbital (where the unpaired electron is) is mostly localized on the oxygen atom.

MO Diagram (taken from here):

That is to say, your structure is more correct than the one in the answer key.  However the justification involves molecular orbital arguments outside the scope of a general chemistry course.

Another edit: I too like your argument about formal charges and would accept it as a justification that the unpaired electron belongs to O rather than to Cl.


What is the reason for strong acids/bases dissociating in water?

The easy answer:

It comes down to the strength of the bonds. Or that covalent bond strength is directly related to the bond dissociation energy and thus dissociation.

A little Physical Chemistry:
(The following is how Pauling explained this.)

In physical chemistry, they have something called bond dissociation ($D_o$).

From Atkins (9th Ed 10.4): "bond dissociation energy, $D_o$, the energy required to separate the atoms to infinity or by the well depth $D_e$." - What is meant here by well is a potential well – for now just think of that as the 2D box talked about in physics.

Now lets shift gears. You may be familiar with bond enthalpy (sometimes just called bond energies). A common formula used in general chemistry is $\Delta H_\mathrm{rxn} = n(\text{bonds broken}) - n(\text{bonds formed})$

It turns out that $\text{bonds broken}$ or $\text{bonds formed}$ is actually referring to the bond energy or bond enthalpy.

It actually works out that there is a direct relationship to bond dissociation energy and bond enthalpy; which makes sense. It is elegantly related by:

$\Delta H^o_\mathrm{bond} = D_o + \frac32 RT $

I guess the natural question to ask now is why do some bonds have a greater enthalpy of formation?

This was not well understood for a long time. In reality there is a lot of factors that can come into play. The way that Pauling initially came up with dissociation energy, $D_o$, in The Nature of the Chemical Bond followed this protocol:

Spectroscopy, (I did this with FTIR for example, but there are several choices) to get a molecular potential energy curve.
Extract the well depth, $D_e$
Find the dissociation Energy using $D_o = D_e - \frac12 h\nu$

In modern thermo:

Because of the relationship with enthalpy, we can measure enthalpy with calorimetry and back calculate.

But the best description uses statistical thermodynamics … Which I will not get into.


You actually are correct, the value is greater than 100g because when you burn a hydrocarbon, it is actually an exothermic redox reaction with oxygen.  Meaning that the carbon in the compound is combining with atmospheric oxygen.  So even if you use 100g of malonic acid, you could end up with more than 100g of CO2.  Your answer is correct.


Identification of Sir Raman's handheld scope

Sir Martyn Poliakoff did the video about the spectroscope on his Periodic Videos channel on youtube.

To my knowledge, he presented the original instrument but did not mention where to obtain a replica.


Why does molten plastic (PLA) stick to metals (AL)?

Aluminum reacts with oxygen in the air to produce aluminum oxide, which forms a fine coating on any exposed aluminum surface protecting against further oxidation.

$$\ce{ 4Al + 3O2 -> 2Al2O3}$$

Aluminum oxide is polar, amphoteric, and hygroscopic: it absorbs moisture from the atmosphere. The surface of aluminum oxide is likely coated with $\ce{Al-OH}$ groups. The polar aluminum oxide and aluminum hydroxide surface is attracted to the polar ester functional groups in PLA.

Aluminum surfaces can be rendered nonpolar by silanization. This article may be behind a paywall for you. I apologize. I could not find a good open access reference. The Wikipedia article on silanization is pretty terrible. The surface is rendered nonpolar, which may decrease the interaction with PLA.


As was mentioned in the comments, with reference to the Myers $\mathrm{p}K_{\mathrm{a}}$ table, the $\mathrm{p}K_{\mathrm{a}}$ values of a protonated ester and a protonated ketone are largely similar (in the region of $-7$ to $-6$).

Therefore, it will likely depend on external factors, which group is protonated first. Are there electron withdrawing or electron donating groups in the vicinity of either? Can the proton on either maybe intramolecularly hydrogen bond to something?

All these effects will also slightly modify the $\mathrm{p}K_{\mathrm{a}}$ of the functional group you’re observing. The one with the higher $\mathrm{p}K_{\mathrm{a}}$ will be protonated first. Thus, it is impossible to give a general answer here, it depends on which molecule you have and which groups are where.

Side fact: $\ce{HCl}$ has almost the same $\mathrm{p}K_{\mathrm{a}}$, so the protonation will be very close to equilibrated between all three components.


Why is fermentation of cellulose to produce biofuel and nutrients so difficult?

Your question is a little bit all over the place, but I believe I can answer it anyway. First, though, allow me to point out that your sum formula for cellulose is wrong. While glucose is indeed a single $\ce{C6}$ compound with the exhaustive sum formula of $\ce{C6H12O6}$, cellulose is, in fact, a poorly defined polymer consisting of multiple $\ce{C6H10O5}$ subunits, linked together both linearly to form long chains but also cross-linked to form extensive networks. Therefore, the better sum formula would be $\ce{(C6H10O5)_nH2O}$ or $\ce{C_{6n}H_{10n+2}O_{5n+1}}$ for very large values of $n$.

There are two reasons why glucose, its dimer maltose and its polymers amylose, amylopectin and glycogen are much more easily digested by all organisms than cellbiose (the glucose dimer that can be found in cellulose) and cellulose itself.

The first one is the type of linkage: the first set of compounds all contain α$\ce{{1}\bond{->}4}$ connections while cellbiose and cellulose contain β$\ce{{1}\bond{->}4}$ connections. This seemingly minor difference means that while the starch subunits form spirals, the cellulose subunits actually form fibres and all the enzymes designed to break up spiral subunits cannot work with fibres.

The second is the number and type of cross-linkages. Each organism has a set of enzymes that can break up the cross-links present in amylopectin or glycogen (or both) but these again are different in shape from those in cellulose owing to the subunits’ β configuration in the latter.

Enzymes to break down amylose, amylopectin and glycogen are pretty much present in every multicellular organism since these compounds are used to store energy in an available form — which would not make sense if the organism could not break the storage form back down to monomers. However, cellulose was not ‘intended’ to be energy storage and thus there was no evolutionary pressure to evolve enzymes that could digest it. Therefore, only very few species that found and populated this ecologic niche are able to digest cellulose and even less are able to digest the additional compounds present in wood that makes it hard.

Now on to the process design part of your question. In general, if we have the enzymes we should be able to degrade wood to useable monomers much like whichever organisms do that in whichever environment they live. Not only the symbionts of termites are of interest but also wood-digesting fungi, naturally. However, the key difficulty here is acquiring the enzymes required in sufficient amounts to make the process economic. I am only aware of very few enzymes that can be fermented and isolated in sufficient quantities and purity to make for a possible economic use; among them a generic lipase. That is one significant hurdle process chemists must overcome.

However, there’s another bit. It may work if we have access to the enzyme but it doesn’t have to. A number of times the most significant issue for chemical engineers is the scale-up problem: reactions work fine on a small scale but once you try to scale them up to higher quantities for more throughput they break down in one way or another. I am sure without knowing the exact titles that there are shelves of books concerning the scale-up problem. To the best of my knowledge (caveat: I am not an industrial chemist or chemical engineer), it is often down to trial and error which processes work well and which don’t; and what works well for one type of product does not work at all for others.

Needless to say that one has to have access to a significant amount of funds to bridge the research gap until one actually is able to make money so it doesn’t surprise me (although I am unhappy) that the company you mention went bankrupt.


How do I know that an acid and conjugate base are present in equimolar amounts?

I'm really confused here!

No, you are not! You only (temporarily) mixed up two things, both of with you have already understood separately.

$$\ce{HA + H2O <=> H3O+ + A-}$$

1. Position of the equilibrium

This is described by the equilibrium constant $K$, which is the quotient of the rate constants for the forward and the back reaction: $K = \frac{k_{forward}}{k_{back}}$. For strong acids, of which you know that they completely dissociate, the equilibrium is on the product side.

2. Equimolarity

There's nothing out of nothing! If one molecule of $\ce{HA}$ dissociates is water, there exactly one molecule $\ce{H3O+}$ and exactly one molecule of $\ce{A-}$ generated. Not more, not less.

Q: How could another molecule of $\ce{A-}$ appear out of nothing?

A: It can't and it doesn't happen - not even in the LEGO universe.

It is just a proton transfer reaction from $\ce{HA}$ to $\ce{H2O}$; the overall number of atoms does not change!

If is however possible that some $\ce{A-}$ may come from a different source, just by adding a salt $\ce{MA}$ of the conjugate base (think sodium acetylsalicylate) to the solution of your acid. jerepierre has mentioned this in his comment! But for the moment, this is another story.


Why do halogens have odd numbers as oxidation number?

Because of the way we construct oxidation numbers.  Put an atom into a molecule with all electrons paired and all the bonds heteronuclear, and by assigning all the valence electrons to the more electronegative atom of each bond we render even number on each atom.  Combine that with an odd number required for a neutral atom and you count an odd difference, thus an odd oxidation number.

Note that I have assumed all electrons to be paired and all bonds as heteronuclear.  But, it is not always the case.  There could be homonuclear bonds as in the halogen element itself ($\ce{X_2}$), with an oxidation number = $0$.  Or there could be unpaired electrons on the halogen so we count an odd rather than even number of valence electrons per halogen atom.  $\ce{ClO}$ (chlorine oxidation number $+2$) and $\ce{ClO_2}$ ($+4$) are such examples.


Melting point of alkaline earth metal carbonates

Melting and decomposition are different things, as you guessed in your question.

Melting:

Something can melt, and then re-solidify, undergoing a physical but not chemical change.

A simple example would be ice, but it is also true for many chemicals. For example tert-butanol is a solid at room temperature (at least in a well air-conditioned lab), but melts readily at slightly elevated temperatures without any degradation. Dimethylsulfoxide is another example of a chemical which often solidifies in a cool lab, but can be melted back down with only the heat of your hand.

Decomposition:

Decomposition is a different thing, whereby the material undergoes a chemical change (usually in addition to physical change). This is often seen in laboratory settings when trying to measure the melting point of molecules with high melting points, such that at the temperatures needed to melt the solid, side reactions are often observed, destroying the material. For this reason, in the literature, you will see melting points reported as  mp. (decomp.), i.e. the temperature at which your molecule decomposed rather than melted.


Looking for a non-toxic substance that evaporates very quickly

I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it.

Nope ;-)

Your concerns are justified and I wouldn't expose the audience to anything but water either.

you can

use water from all above

and let the audience get wet
and hand out rain coats

use water and only let it rain to the left and right of a marked path or
apply a more sophisticated technique, as used by the random international art studio in their rain room. Here, it rains thoughout a whole room, but the visitors are tracked by cameras and whereever they move, the water is selectively switched off.


Evaluate mass of salt needed to add to a buffer solution knowing only pH

I think there is all the data you need. $\mathrm{pH}$ of a buffer formed by a weak acid $\ce{HA}$ and its potassium salt $\ce{KA}$ can be found as

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{KA})}{C(\ce{HA})}}$$

On the other hand

$$C(\ce{KA}) = \frac{m(\ce{KA})}{M(\ce{KA})V}$$

where $m$ and $M$ are mass and molecular mass of $\ce{KA}$; $V$ is volume. In general, first equation can be rewritten as

$$
\begin{align}
\mathrm{pH}_i &= \mathrm{p}K_\mathrm{a} + \log{C_i(\ce{KA})} - \log{C(\ce{HA})} \\
&= \mathrm{p}K_\mathrm{a} + \log{m_i(\ce{KA})} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})}
\end{align}
$$

Rearranging:

$$\mathrm{pH}_i - \log{m_i(\ce{KA})} = \mathrm{p}K_\mathrm{a} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} = \mathrm{const}$$

so that now we can equate conditions for both solutions and find the mass:

$$\mathrm{pH}_1 - \log{m_1(\ce{KA})} = \mathrm{pH}_2 - \log{m_2(\ce{KA})}$$
$$\log{\frac{m_2(\ce{KA})}{m_1(\ce{KA})}} = \mathrm{pH}_2 - \mathrm{pH}_1$$
$$m_2(\ce{KA}) = m_1(\ce{KA})\cdot10^{\mathrm{pH}_2 - \mathrm{pH}_1} = \pu{2.40 g}\cdot10^{5.0-4.8}\approx\pu{3.80 g}$$

So, in order to achieve $\mathrm{pH} = 5.0$, one has to add $\pu{3.80 g} - \pu{2.40 g} = \pu{1.40 g}$ of potassium salt.


I think this Wikipedia article answers your question quite directly:

  "This important property [rubber toughening] is entirely due to the
  small rubbery domains that aid in arresting crack propagation, thereby
  increasing toughness/energy needed to break the material."

The same article gives an example using the materials sited in your post:

  "Well-known rubber toughened polymers include high impact polystyrene
  (HIPS), rubber-toughened nylon, and acrylonitrile butadiene styrene
  (ABS). In HIPS, polybutadiene domains are dispersed in a polystyrene
  matrix; in rubber-toughened nylon, typically cis-polyisoprene is
  dispersed in a Nylon matrix; in ABS, polybutadiene domains are again
  used, but for both polystyrene and acrylonitrile."


Silanol activity is perhaps an informal term which is quite frequently used in the chromatographic literature to explain the poor band shape of primary amines. Other amines show this too but to a lesser extent. That is why you will not find a formal textbook definition of silanol activity. One might also see the term silanophilic interactions used occasionally. If you look at the surface of silica gel, you will see that a silanol group, is like an alcohol analogue, except that silanol is slightly a stronger acid.
If we disperse silica particles in water, and let it stay there for sometime, the pH of water is < 7 (around 3–4, if I remember correctly). So the point to keep in mind is that silica is acidic due to its ionizable silanols.
There are different types of silanols as drawn in this paper: Sunseri, J.; Cooper, W. T.; Dorsey, J. G. Reducing residual silanol interactions in reversed-phase liquid chromatography: thermal treatment of silica before derivatization. J. Chromatogr. A 2003, 1011 (1-2), 23–29. DOI: 10.1016/S0021-9673(03)01070-7.

Now if you are doing chromatography of relatively strong organic bases, the silanol may be deprotonated by the amine (acid–base). The locally ionized silanol Si–O− groups can now act as a cation exchanger, which is a pretty strong interaction. The reason for seeing the streaking or tailing, as it is called in the formal literature, arises from those dual modes of retention. One mode could be regular partitioning or adsorption and the other
one is this additional cation-exchange behavior.
There are dozens of chromatographic tests for assessing the silanol activity of a given silica gel. Not all silica gels were created equal!
As your question

would that imply that what I'm seeing on the TLC is no longer
pyridine, but rather $\ce{C5H5NH+}$

Remember that ionization is an equilibrium process and hence dynamic. You are seeing the combined effect of both, pyridine and protonated pyridine.


Unexplained peak cluster in proton NMR trace?

You are correct. The chemical shift of methyl protons 1 and 4 are close to one another.  You still see the two expected triplets, they just have similar chemical shifts.  That's not too surprising since they are both methyl groups attached to a methylene group.  We could tell which triplet belongs to which methyl carbon by measuring and comparing the 2 sets of CH2-CH3 coupling constants.


Why caffeine is easily oxidized to uric acid?

I doubt caffeine can be so easily oxidized to uric acid. I'd propose a way by demethylation via cytochrome P450 to the corresponding xanthines and oxidation via xanthine oxidase (but these are enzymatic reactions).

From a synthetic perspective, one will start with guanine (-> guanine deaminase followed by oxidation; or oxidative deamination of Guanine followed by another oxidation)

It's not safe to conclude that caffeine increase uric acid blood level.

Long story short:
Coffee consumption seems to lower uric acid blood levels.
Decaf coffee consumption also seems to lower uric acid blood levels.
Tea consumption seems to have no effect on uric acid blood levels.

  Serum uric acid level tended to decrease with increasing coffee intake
  (Figure 1). After adjusting for age and sex, serum uric acid level in
  individuals with coffee intake 6 cups daily was lower than in those
  with no use by
  0.43 mg/dl (95% CI 0.22, 0.65; P for trend  0.001). After adjusting for other covariates, the differences were attenuated but remained
  significant (P for trend  0.001) (Table 2).

"Coffee, Tea, and Caffeine Consumption and Serum Uric Acid Level: The Third National Health and Nutrition Examination Survey", Choi HK, Curhan G., 2007
DOI 10.1002/art.22762

It has to be noted this was the first study I found on the fly and there may be other results out there


steam distillation of 4-bromoaniline , why render it basic prior

In principle, it is difficult to distill salts (ionic compounds). Therefore you first make sure that you have a free base in the solution. This is guaranteed in alkaline solution. For the steam distillation the compounds should be immiscible, which is another reason for having a free base. Let’s compare boiling points and solubility for aniline and its salt:

Aniline b.p. 184 °C, solubility in water: 36 g/L
Anilinium chloride b.p. 245 °C, solubility in water: 1070 g/L


Experiment to show that air contains about 20% oxygen

One problem with using combustion reactions to determine the percent of oxygen in air is the additional products, namely $\ce{CO2}$ and water vapor.  If performing a water displacement reaction, these gases add to the uncertainty and imprecision of this method.

Water displacement by a combustion reaction can, however, be used if an appropriate combustion reaction is chosen.  A more reasonable experiment to determine the percent oxygen in air is to use the oxidation of iron $$\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$$ The benefits of this reaction are (a) only one product is formed and (b) the product is a solid so it will not influence the water displacement in a significant way.

This experiment has been reported in the Journal of Chemical Education and a brief search of the web found an adaptation of this experiment if you don't have access to the journal.  Briefly, one measures the volume of a test tube, adds a known amount of steel wool into the tube and inverts the tube in a beaker of water.  The oxidation of the iron is catalyzed by first rinsing it in a dilute acid (vinegar) solution.  Knowing the density of iron one can calculate the volume of the test tube occupied by the solid and then determine the volume of air before and after the reaction is completed.

There are a number of possible errors that can be introduced into this experiment, but unlike the combustion of a hydrocarbon, these errors are much more manageable for an undergraduate or high school chemistry experiment.  Most significantly, the reaction occurs near room temperature so the temperature change of the gas can be ignored.


Nucleophilic acyl substitution leaving group

Indeed, there is no reason why the protonated alcohol should not be eliminated upon the lone pair rebound. It probably happens more often than not. However, it does not lead to an observable product: it just goes back to the reactants. Thus, we can write an equilibrium arrow on the first step.

$$\ce{R-C(=O)-Cl + R'-OH <=> R-C(-O^-)(-Cl)-OH+-R'}\tag{1}$$

Once the chloride has been kicked out, there is no way back as the ester is a much weaker electrophile than the acid chloride (and chloride a weaker nucleophile than the alcohol). Thus, the reason why we only care about the lone pair rebound eliminating chloride is because only that is productive.

$$\ce{R-C(-O^-)(-Cl)-OH+-R' -> R-C(=O)-OH+-R' + Cl-}\tag{2}$$

These arrows are also given in your source like that; if everything up to a single step is given as equilibria while that one step is given as non-reversible, that should be sufficient driving force to let a reaction happen.

At which point in time the proton is eliminated is somewhat up to debate, of little relevance for the overall reaction and dependent on what is in solution. If just a pure acid chloride and a pure alcohol are mixed the only potential bases that could accept the proton of the tetrahedral intermediate are other acid chloride molecules or other alcohols — or the tetrahedral intermediate’s $\ce{O-}$ atom. All of these are very weak bases and I see no preference for the proton to go anywhere. After we have eliminated chloride, that is added to the set of potential bases. It is also not a strong base; maybe even a weaker base than some of the other options. The big difference is that it is volatile and can diffuse out of the reaction mixture. Thus, with no other bases added, I would actually prefer this sequence of steps.

Once you add a base, I would also prefer deprotonation before chloride is eliminated.


For the inorganic chemistry I recommend Holleman-Wiberg's text, also exists in english. It is reasonably readable and very thick and contains a lot.
Holleman-Wiberg's Inorganic Chemistry. If you prefer, the original is in German and updated very often.

For the organic chemistry, besides being thick also very nicely readable,  Organic Chemistry by Clayden, Greeves and Warren

Besides that, you would probably need some nice general chemistry book, but I have no good personal experience in this area.


The Maillard reaction is also called the non-enzymatic browning reaction. The reaction occurs when there are reducing sugars and amino acids present. The reaction is controlled by solvent choice, time and temperature considerations.
When browning butter, certainly, the Maillard reaction is occurring. There may also be other reactions happening such as Strecker degradation and caramelization.
As noted, browned butter is made by essentially heating butter until the water boils off at 100C and then the temperature ramps upward. So, what is causing the browning once the browned particulates are strained out? Some are chemicals that are formed from the reaction.
The color comes about due to production of melanoidins. As the Maillard reaction produces a range of chemical compounds, some of the components may have a lot of color on their own, such as furfural and its derivatives.
Professor Chi Tang Ho at Rutgers University has done extension work on Maillard reactions in model systems. Heath & Reineccius in their book Flavor Chemistry and Technology write about the Maillard reaction in more general terms that includes more on color production as part of the process.


One of the reasons to take Carbon-12 could be that it's hard to get your hands on one mole of pure hydrogen (and to weigh it). Carbon-12 is a (room temperature) solid which is relatively easy to produce, keep pure, store and weigh.


Why does potassium form peroxides but sodium does not?

If you look at the standard Gibbs free energies for the alkali metal peroxides, actually, lithium should form it most easily. The standard Gibbs free energies of formation are -564.8, -430.1, -418.4, -351.5, and -359.8 kJ/mol for $\rm Li_2O_2, Na_2O_2, K_2O_2, Rb_2O_2,$ and $\rm Cs_2O_2$, respectively.

If we're looking at reaction of an alkali metal with oxygen to form explosive compounds, perhaps we should look at superoxides. The superoxides are powerful oxidizing agents, and they can explode when mixed with water, acids, organics, or powdered graphite.

At room temperature, oxygen readily reacts with K, Rb, and Cs to form superoxides ($\rm MO_2$), while Li and Na don't. There is a direct correlation between superoxide stability and electropositivity of the metal concerned (source: Cotton & Wilkinson's Advanced Inorganic Chemistry). $\rm NaO_2$ can be obtained by high pressure and temperature reaction of $\rm Na_2O_2$ with $\rm O_2$, and $\rm LiO_2$ won't form at room temperature.


What is the degree of unsaturation of oxalate anion?

The Degree of Unsaturation can be considered as the number of Pairs of Hydrogen Atoms that could be added to the structure. One can simplify the structure to consist of two carbon-oxygen double bonds and two carbon-oxygen single bonds in one resonance contributor. Note that the single bonded oxygens have a negative charge and thus could accept a hydrogen each. This accounts for 1 degree of unsaturation. There are also two double bonds, and a pair of hydrogens could be added across these also. This gives two degrees of unsaturation. Hence there are 3 degrees of unsaturation in total (as the book says).

In order to calculate the degree of unsaturation for any compound, one can use the formula:
((2* Number of Carbons) + 2 - (Number of Hydrogens and halides) + (number of nitrogens)) / 2.
In this case, the formula gives 3, which is the degree of unsaturation.


How do I calculate the temperature of hot air in kelvin?

You already know the temperature of the hot air in Celsius: 100 degrees

Using the equation Kelvin = Celsius + (273.15 degrees), you would get 100 degrees + 273.15 degrees, or 373.15 degrees.

I am assuming that this is a preliminary step that is required in order to find the temperature to plug into Charles' law (which must be in Kelvin), in which case you can use the formula I mentioned above for your other calculations as well.


"Specifically, I'm curious if there are any programs which come close to Gaussian in breadth"

Not really, as far as I can tell. Gaussian is a massive work by many, many people over many, many years, and the reason it's still one of the most popular packages, as far as I can tell, is that it does most things (notably excluding non-GTOs), even if it's not necessarily the fastest or the best at the things it does in every case.

"but also take advantage of things such as GPU processing."

Double no. As LordStryker says above, GPU programming is pretty new, and most of the resources I've seen for it are C-based, while a significant chunk of scientific development still takes place in Fortran. I suspect library-based encapsulation of GPU-running numerical routines may improve this over time (I mean, by introducing people to the idea of getting speedup by using the GPUs more easily), but I wouldn't hold my breath. It'll be interesting to see what hybrid chips with multiple types of core, or the Xeon Phis will be able to do, though.


Find the pH of a sodium chloride solution using the extended Debye-Huckel equation

Let's first look at ionic strength:

Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.

$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$

What is the strength of 0.02 M NaCl solution?

$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$

With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient.
$$\textrm{activity} = a_i = \gamma_i[i]$$

Let's look at a typical equilibrium expression:

$$\ce{A + B <=> C + D} $$

$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$

The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:

$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$

where $\alpha$ is the size of the hydrated ion in pm.

The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$

To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.

$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$

Take the negative log and you will get pH.

$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$

In context with the problem:

$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$

$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$

$$\gamma_{\ce{H+}} = 1.1243166 $$

$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$

Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.

For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.


Which elements tend to form the most phases?

Much empirical data is needed to either produce or validate phase diagrams.  Any phase diagram would likely need to be of practical consequence to justify spending the energy and resources to produce it.

An example of a binary mixture with practical consequence would be a metal oxide formed during corrosion. The many forms of iron and oxide are well studied since iron is commonly used to construct tools.

In 2015 a group claimed to use the computer code CALYPSO (Crystal structure AnaLYsis by Particle Swarm Optimization) to discover a new phase of beryllium.  Beryllium is used in aerospace and the construction of nuclear reactors, so the expenditure of effort was justifiable.

So to answer the  sub-questions:

Is this a function of only one or both of the elements involved?
Practically .. No. In the case of iron oxide in an aqueous solution, multiple iron oxide phases can form at the same temperature and pressure if the pH of the solution or the redox potential value of the solution were changed.

Has this tendency to form more phases (or less) been quantified for the various elements?  - Yes, phases of practical consequence have been quantified at great effort; however, discoveries of new phases are likely to continue.

It is possible to estimate the existence of new phases using computer algorithms; however, these phases must be experimentally validated.  A computer algorithm could be constructed around pseudomorphs; however, if a new phase were discovered it would need practical validation by experimentation.

"Which elements form the most phases?" is an impossible question to give and absolute answer as we are limited to certain pressures and temperatures with which to experiment. The way experimentation is done is likely to bias any verifiable answer toward more practical materials.


A IChO-2014 problem involving reaction mechanism

A to B is a classic hydroboration of the least hindered double bond followed by oxidation of the trialkyl borane.


How to determine the most stable carbanion?

Wikipedia says:

The stability and reactivity of a carbanion is determined by several factors. These include
1)The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;
2)Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;
3)The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.

You can see that point 3) is not relevant here. We do not have a conjugated system, nor is there any chance for resonance.
Keeping in mind point 1), the $\ce{N^+}$ has a $\ce{-I}$ effect, but inductive effect reduces over distance. The negative charges in options a), b) and d) are either in the meta or para position.
Also, remember that $\ce{+I}$ effect destabilises the carbanion.
In c), the carbanion is primary, but in a), it is secondary. Primary carbanions are more stable due to lesser $\ce{+I}$ effect.
So your answer is c).


You may be familiar with the equilibrium that exists between a gem-diol and the corresponding carbonyl compound, as shown in the figure below.  The carbonyl double bond is very strong, so in most cases the equilibrium lies far to the carbonyl side.

The same type of equilibrium exists in the case of bis- and tris-amino compounds where all of the amino groups are attached to the same carbon atom.  In the case of such a tris-amino compound (see figure) the imino analogue of a carbonyl is generated, but since a third amino group is present, another equilibrium between the imino compound and a nitrile also exists.  Just like in the carbonyl case, the nitrile triple bond is very strong.  So when a tris-amino compound is generated an equilibrium with a very stable nitrile will be set up and the equilibrium will (usually) lie far to the nitrile side.  Most tris-amino compounds prefer to exist as nitriles.  If the amino group is fully alkylated (all of the amino hydrogens are replaced by alkyl groups), then this equilibrium cannot occur and such alkylated tris-amino compounds should be isolable.


What type of reaction is this? (Modified organic nucleoside synthesis)

My thoughts are along the lines of @Andrew and @Waylander. Step F doesn't tell us much, @orthocresol! I prefer a cyclization, 1 $\rightarrow$ 2a, followed by hydrolysis of ester 2a to alcohol 2b. Alternatively, N-O bond homolysis with light would probably involve a cage mechanism. .


Sodium Chlorite is a white powder, which dissolves in water.

Chlorine Dioxide is a gas. This gas also dissolves in water to some degree and gives a yellow color.

Sodium Chlorite is used to make Chlorine Dioxide in the laboratory.

I strongly doubt, that someone has sent you Chlorine Dioxide, as it is to dangerous to store and to send.


SN2 vs SN1 in Acidic Environments for Secondary Alkyl Halides

Note that the starting material is drawn as a chiral molecule.  Solvolysis of this starting material will generate a symmetric, achiral, allylic carbocation.  Since a stabilized allylic carbocation is generated, the mechanism likely proceeds through an $\ce{S_{N}1}$ mechanism.

Both ends (there is positive charge delocalized at both ends of an allylic carbocation - draw the resonance structures and convince yourself)  and both faces (top and bottom) of the allylic carbocation can capture the nucleophilic solvent molecule to restore neutrality as shown below. This results in the formation of two diastereomers of the 3,5-disubstituted cyclohexene - the cis and tran isomers. For each diastereomer, both enantiomers are also produced, as they must be, having passed through an achiral intermediate.


Is adsorption exothermic, and if so, why?

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

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

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

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

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

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

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

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

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

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

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

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

I'll quote the conclusion from the paper:

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

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

References

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

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


What is the reason for the different solubility of silver halides in ammonia?

Silver chloride is very thermodynamically stable in ammonia solution. The $K$ for the dissolution of silver chloride in ammonia solution is only $\pu{2.9E-3}.$ suggesting that the product — silver ammonia complex ion — is much less thermodynamically stable than the reactant, silver chloride.

$$\ce{AgCl + 2 NH3 <=> [Ag(NH3)2]+ + Cl-}$$

$$
\begin{align}
K &= K_\mathrm{sp}(\ce{AgCl})\times K_\mathrm{f}(\ce{[Ag(NH3)2]+})\\
  & = \pu{1.8E-10}\times\pu{1.6E7}\\
  & = \pu{2.9E-3}
\end{align}
$$

However, silver chloride will dissolve in an excess of concentrated ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride has a minuscule $K_\mathrm{sp}$ value, some free silver ions are present in solution. Addition of concentrated ammonia solution effectively consumes this free silver ion as ammonia, a good σ-electron pair donor ligand, combines with the silver ion.

Ions themselves are not precipitates, as all bulk matter must be approximately electrically neutral, and so the silver chloride dissolves when hit with excess, concentrated ammonia.

Also, note that the cyanide ion is also a good complexing agent with respect to silver. The cyanide ion has a surfeit of electrons on the carbon, and carbon is even less able to stabilize negative charge than nitrogen, making the cyanide ion reactive as a complexing agent. Hitting silver chloride and other silver salts with a source of cyanide ion will similarly cause the silver salt to dissolve into ions suspended in solution.


What are the oxidizing and reducing agents in the reaction?

You are correct, the Pb(s) is oxidized, which makes it the reducing agent. However, there is no change in the oxidation state of the H or the SO4−, so it is not the oxidizing agent. Rather the PbO2(s) is the oxidizing agent, as the Pb goes from +4 in the PbO2(s) to +2 in the PbSO4(s).


Solution , Solubility and Colligative Property?

Let's say we have some ice.

Beautiful structure. Ice stays solid at some temperatures because of its neat, tight lattice. Ice stacks this over and over until there's no more ice left to stack.

If I were to put a lot of ions between this stack, the stack might bend or even break; it isn't as reinforcing as it once was. The stack isn't as durable anymore, so we would expect it to melt a lot easier - there's too much structural stress for it to be a solid.

The freezing point of solutions depends, at least in this case, on how much dissociation occurs. If we can push more ions into solution, it's going to have a harder time making a sturdy lattice - a harder time freezing.

$\ce{C6H6O6 -> C6H6O6}$ It doesn't dissociate at all.

I'm not sure how much urea dissociates, but I assume it would be like glucose.

$\ce{NaCl -> Na+ + Cl-}$ It dissociates into 2 ions.

$\ce{K2SO4 -> K+ + K+ + SO4^{2-}}$ It dissociates into 3 ions.

$\ce{K2SO4}$ dissociates the most, so it's freezing point is lowest. Hope this helps!


Copper is used to avoid lithium-halogen exchange. Additional information on that reaction can be found in this discussion. Organolithiums will often, instead of acting as a nucleophile, form a new organolithium: $$\ce{RLi + R'X <=> RX + R'Li}$$ Using $\ce{Cu(I)}$ allows the formation of an organocuprate, which acts as a nucleophile, allowing alkylation rather than metallation: $$\ce{2RLi + CuX -> LiR2Cu + LiX}$$ $$\ce{LiR2Cu + R'X -> R'R + RCu + LiX}$$ I'm not sure if silver could be used instead; I'm not seeing any reference for it, but I may just not be searching the right terms. The Wikipedia page on organosilver chemistry isn't terribly extensive, but I'm guessing stoichiometric use of silver, while not terribly expensive as transition metals go, would certainly be more so than analogous use of copper.

Update:

It seems that analogous chemistry cannot be performed with silver or gold. Lithium diorganoargentates ($\ce{LiR2Ag}$) are apparently too unstable. Lithium diorganoaurate ($\ce{LiR2Au}$) complexes are apparently either not good nucleophiles, or the triorganogold(III) ($\ce{R2AuR'}$) intermediate complexes are too stable to perform similar chemistry. This is per this 2005 paper. A fair amount of gold chemistry has come out since then, so I'm not sure how much of this has changed since.

Second update (though I should have answered this to begin with):

Also, I'd like to add that this is not an $\text{S}_\text{N}2$ reaction. The organocuprate undergoes oxidative addition into the $\ce{R'X}$ bond. After that, it loses the $\ce{X^-}$ ligand to form $\ce{LiX}$ and $\ce{R2CuR'}$, then undergoes reductive elimination to form $\ce{R'R + CuR}$.


Grignards react with molecular oxygen to form hydroperoxides. According to this paper  here the yield for alkyl hydroperoxides can be as high as 90%, yields from aryl Grignards are lower (10-20%).


Delta G in an equilbrium reaction(forward and backward)

I feel Koushal has not seen the difference between $\Delta G$ and $\Delta G°$, because $G$ of all reactants and products change during the reaction. $G$ and $G°$ are very different concepts.

$G°$ is the Gibbs energy of $1$ mole of any reactant and of any product in the pure state at 25°C and 1 atm. $G°$ is a constant of the substance, independent of the equation chosen, and independent of the concentrations.  $\Delta G°$ is obtained by calculating the sum of the Gibbs energies of the products in the pure state, and substracting the sum of the Gibbs energies of the reactants in the pure state, before the initial mixture is done.

During the reaction, the $G$ values are changing all the time, because they are proportional to the concentration of the reactant (or of the products) . The $G°$ values are never changing during the course of a reaction.

As soon as you mix the initial reactants, the Gibbs energy of each reactant decreases from $G°$ to a lower value $G$ < $G°$ even if no chemical reaction occurs. And the Gibbs energy of each reactant continues to decrease due to the chemical reaction. Simultaneously the Gibbs energy of the products are moving in the opposite direction. But they are is very low at the beginning of the chemical reaction. The reaction is finished when the sum of the Gibbs energies of the reactants is equal to the sum of the Gibbs energies of the products.


Global reactivity parameters from open shell DFT calculations using Koopman's theorem

You (probably) cannot apply Koopmans' theorem. (Please also note that it is named after Tjalling Charles Koopmans, the s at the end is part of the name.)
Open shell systems are usually hard to describe and unrestricted density functional approximations might not even describe the system correct qualitatively. The problem usually comes down to the multireference character of the partially occupied orbitals. In these cases you need a multireference wave function to correctly describe the ground state, e.g CASSCF methods. In a first order approximation, unrestricted methods can probably yield reasonable results, because (in a practical sense) it is using more than one determinante. However, when you use these methods, there will be fractional occupancy of multiple orbitals, which make the approximations of Koopmans' theorem break down. I would suggest a thorough literature review on the best practises for such systems. In short, HOMO and LUMO are not defined in these systems.


Alkali metal hydrides can typically be viewed as labile hydride anions. The extent of lability can correlate with the electronegativity of the metal. Hence, less electronegative metals in alkali metal hydrides yield more labile hydride anions and more "reducing" nature.


Sample: How to calculate mass of nanopaticles with given density and Vol %

Defining the following

Density of solution: $\rho = \pu{1.18 g/cc}$
Density of water: $\rho_w$
Density of nanoparticles: $\rho_{np}$
% (v/v) n.p.s: $p_v=3.9$%
% (w/w) n.p.s: $p_{wt}$

Then solve for $\rho_{np}$ in the following equation (you'll need to look up missing value of water density):

$$\rho = \frac{p_v}{100}\rho_{np} + \frac{100-p_v}{100}\rho_{h2o}$$

With $\rho_{np}$ you can compute $p_{wt}$

$$p_{wt} = 100\frac{p_{v}\rho_{np}}{p_{v}\rho_{np}+(100-p_{v})\rho_{w}}$$

As noted in the comments, the resulting density of the particles ignores the effect of the added cationic surfactant. Also, the solvent density is assumed that of water, free (not particle bound) detergent or other additives assumed to be negligible.


Why oxidation potential of an electrode equals negative reduction potential of the same electrode?

There exist two methods for teaching electrochemistry, and of course two schools of teachers. The two methods are equivalent, but they are exactly the opposite of one another. I will take an example to explain it correctly : Let us speak of the Daniell cell Zn/Cu, which uses reduction potentials equal to -$0.76 $ V for Zn and +$0.34$ V for Cu, if the ions concentrations are 1 molar. In this case, the overall voltage is $1.10$ V.
The Daniell cell is based on the two half-reactions :  $$\ce{Zn^{2+} + 2e^- -> Zn}$$  $$\ce{Cu^{2+} +2e^--> Cu}$$In the Zn/Cu cell, Zn is oxidized and acts as emitter of electron, which is in the opposite direction of the above equation, so the equation explaining what is going on in a Daniell cell contains the equation of Zn reversed. This becomes :$$\ce{Zn->  Zn^{2+} + 2e^-}$$  $$\ce{Cu^{2+} + 2e- ->Cu}$$When you want to describe the overall equation that goes in this cell, you add these two half-equations, remove the 2$\ e^-$ and this gives : $$\ce{Zn + Cu^{2+}->  Zn^{2+} + Cu}$$  We are here at the point where the two approaches differ. Strangely enough, one uses an addition, and the second a subtraction. Let me speak of the first method, which is used in your textbook.

First method. To obtain this overall equation, you had to add the two half-equations. To follow the same logic, you may simultaneously add the corresponding reduction potentials. But the reduction potentials of Zn and Cu are respectively $-0.76$V, and + $0.34$ V. And if you add them, you do not get $1.10$ V. So the chemists of this school have decided that when a half- equation is used in the opposite sense, with the electrons at the right-hand side, in an oxidation sense, like Zn here, the sign of the potential of this cell should also be opposite. $\ E_{anode} = - E_{cathode}$. Here : $E(Zn)_{anode} = - E(Zn)_{cathode)}$ = +$0.76$ V. With this assumption, you add $+ 0.76$ V plus $+0.34$ V and you obtain $1.10$ V, which is perfect.

Second method. Here you use the same half-reactions as described previously. But the values of the potential remains the same if the electrode works as a reducing agent or as an oxidant. It is always - $0.76$ V pour Zn and + $0.34$ V for Cu. You never change its sign. But remembering from the electricity course, that a measured voltage is a difference of potential, the voltage of the cell may be obtained by subtracting the redox potential of the cathode minus the potential of the anode, that is $E_{Cu} - E_{Zn}$. The result of this subtraction is: $+0.34$V -($0.76$ V) = $1.10$ V, which is perfect. This remains true even though you have obtained the overall equation with an addition of two half-reactions.

As you see, both approaches are equivalent. Have you understood my development ?


Whenever you just mix two substances (dissolve in each other), you lower the melting point and raise the boiling point. That's a fundamental principle. If it was the other way round, the substances would not mix, or at least not mix well.

(Azeotropes of compounds with very different polarity (water/ethanol) are the exeption confirming that rule, but the acchievable raise in Bp. won't be large. And I doubt you find one you can make a candle from.)

That sounds bad, but: Paraffin wax already is a mixture, so the desired effect might be reached by fractionating it, instead of adding sth. else.

You might try to slowly crystallise a molten sample of paraffin wax, and pour off the residual liquid when the sample is about half solid. Now the residual solid should have a higher melting point than the original wax.

(Note: This is purely my educated guessing, no idea if the effect is large enough or if some secondary kinetic effect might completely preclude it.)


What you probably missed was that the order you mention is for the SAME GROUP elements. For example, down the group basicity of halide ions is-

F->Cl->Br->I-

But the the nucleophilicity order in polar protic solvents is-

I->Br->Cl->F-

This reverse in order is due to higher charge density of smaller halide ions and thus better interaction with polar protic solvents.

But along a period the charge density is similar and thus the interaction between solvent and nucleophile is almost similar. Thus, along a period the order follows basicity order.


Things to consider:

Sodium is merely a spectator ion. It will not take part in any reactions, so the initial amount of sodium you weigh in will remain constant. We can thus ignore it.
The concentrations of $\ce{H3PO4}$ and $\ce{PO4^3-}$ will be sufficiently small to entirely ignore them.

This reduces your system to the following relevant chemical equation:

$$\ce{H2PO4- + H2O <=> HPO4^2- + H3O+}$$

With the unknown concentrations of $\ce{H2PO4-}$, $\ce{HPO4^2-}$ and $\ce{H3O+}$. For these three we only need three equations and luckily we have them. Remember that we know:

the final phosphate concentration we want;
the final $\mathrm{pH}$ we want; and
the $\mathrm{p}K_\mathrm{a}$ value of our acidic species.

This gives us three equations that I am going to hide behind a spoiler tag:

   $[\ce{H2PO4-}] + [\ce{HPO4^2-}] = 0.1~\mathrm{M}$
 $10^{-7.198} = K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.4} = [\ce{H3O+}]$

We can now put the third equation into the second resulting in exactly the Henderson-Hasselbalch equation, because this exactly is how that equation was derived. The maths of one possible way to approach this problem is shown in the following spoiler tag.

   $10^{-7.198} = \frac{10^{-7.4} \times [\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.198 + 7.4} = \frac{[{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $[\ce{H2PO4-}] \times 10^{0.202} = [\ce{HPO4^2-}]$ ← Here! You have the Henderson-Hasselbalch right here! (In a different form but equivalent.)

 $[\ce{HPO4^2-}] + [\ce{H2PO4-}] = 0.1~\mathrm{M}$
 $[\ce{HPO4^2-}] = 0.1~\mathrm{M} - [\ce{H2PO4-}]$

 $[\ce{H2PO4-}] \times 10^{0.202} = 0.1~\mathrm{M} - [\ce{H2PO4-}]$
 $[\ce{H2PO4-}] \left (10^{0.202} + 1 \right ) = 0.1~\mathrm{M}$
 $[\ce{H2PO4-}] = \frac{0.1~\mathrm{M}}{10^{0.202} + 1} = 0.0385~\mathrm{M}$

The last steps to calculate the required concentration of $\ce{Na2HPO4}$ including calculating the respective masses of the water-free salts should be trivial and can be found in another castle answer.

I can only speculate why you are not allowed to directly use the Henderson-Hasselbalch equation since mathematically, every optimal approach will simplify itself to exactly that equation. Maybe the point of the excersize is to derive that equation; only your prof will know.


What proportion of a combustion engine's output energy comes from the air?

In combustion the "energy" is really the heat that is released. This heat causes the expansion that moves the piston upwards.

However, the fuel and the oxygen did not enter the engine with energy, and then leave without it. The fuel and the oxygen did not leave at all, different molecules did. The heat that was released was due to the reaction that changed the molecules into different molecules, it was not "brought" and handed off by the molecules. I don't think you can divvy this energy up, since it would not have happened without either the fuel or the oxygen. 0% of the reaction would have happened without the oxygen, but that doesn't mean the oxygen was responsible for 100% of the heat generated.

If the oxygen had entered at one temperature, and left at another, you could calculate the heat the oxygen released or absorbed by its change in enthalpy. However, this does not happen in combustion, the oxygen that is responsible for the reaction does not leave as oxygen.

In fact, if anything, the uncombusted oxygen absorbs heat that could otherwise have been used in the expansion of the piston and this quantity would be measurable. So the only thing I think you can quantify is the amount of work not done because air is used (same goes for all the nitrogen that goes in cold and leaves hot). If pure oxygen was used rather than air, engines would get more expansion work since less heat would be absorbed by  nitrogen.


Which chiral carbon do L and D in L- or D-glucose refer to?

I think your confusion arises from trying to locate a particular carbon atom, which is D or L. This is not the case.
Points to remember:
The reference for D and L notation is glyceraldehye molecule not a particular carbon atom.
D and L are not equivalent to R and S which are individually applicable to each chiral center in a molecule.
Similarly, D and L cannot tell you the direction of rotation of light before hand.
Capital “D” and “L” notations have no connection with small d and small l, which tell you the direction of rotation of light.


Decrease in temperature of an aqueous salt solution decreases conductivity

Decrease in temperature has two effects, both attributing to lower electrolytic conductivity:

decreases the mobility of the charge carriers (e.g. $\ce{H3O+}$ and $\ce{OH-}$ for pure water);
suppresses auto-ionization of water (higher $\mathrm{p}K_\mathrm{w}$), reducing the total number of charge carriers.


How is more than one pinacolone is possible?

The problem is what is a reaction which gives "only one product", the problem is that very few reactions give a 100 % yield of a substance. For almost all real life reactions there are a small fraction of molecules which react he wrong way.

I think that when more than 99 % of molecules react the "right" way that those which react the "wrong" way can be ignored normally. Unless the wrong product has some very harmful effect on the product when it is used for a purpose.

What I think will happen is that the carbocation formed by protonation of the hydroxyl bonded to the carbon bearing the two phenyl groups, followed by the loss of a water molecule will have a much lower energy than the other carbocation. Thus it will be more easy to form.

But there is a problem, for a rearrangement of a carbocation to occur normally we need to rearrange to a lower energy carbocation. By lowering the energy of the carbocation we can lose the thermodynamic drive for rearrangement. In the ideal world we would consider the rate of rearrangement of the two different carbocations and their formation. I think that the reactions forming the carbocation will be reversable while the pinacol rearrangement will be irreversable.

But in some cases such as many homogenious asymmetric rhodium hydrogenations of prochiral alkenes the lower energy alkene rhodium complex which is present at a far higher concentration than the higher energy isomer, the lower energy isomer has far too little reactivity. As a result while the majority of the complex might be in the R form the tiny trace of the complex in the S form is responsible for the vast majority of the hydrogenation reactivity.


First, many plastics are somewhat porous. Since plastics are permeable to water vapor and oxygen, coffee and other food products are sometimes packed in two-layer bags of polyethylene (for strength) laminated to aluminum foil (to reduce permeability).

Second, lipids and other non-polar substances can dissolve in and and migrate through common plastics used in food containers, such as polyethylene and polypropylene, as you and @Alchimista note. This is why carotene, lycopene and other chemicals in tomatoes eventually stain poly containers permanently: they've dissolved into the plastic.

In a similar way, mercury dissolves into gold (e.g. jewelry) in a matter of seconds, leaving a silvery stain. That stain then disappears in a week or two as the mercury migrates throughout the object.

The reverse phenomena also occurs, where substances in plastic packaging can migrate out of the packaging.

As the Bard might have put it, "Will all the water in the ocean wash this food from my cans?"


For any salt dissolving in water, its not compulsory that enthalpy change needs to be negative. Yes, if it is negative then its better dissolving for sure, But it doesn't mean the enthalpy change has to be compulsorily negative for salt to dissolve in water.

Enthalpy change when we dissolve Nacl in water is positive. This is because breaking of ionic bonds in the crystal lattice of Nacl requires more energy than that of provided by hydration of Nacl. Yes, hydration energy for Nacl is less than the energy required to break the Nacl apart. Hence the extra energy which can add up with the hydration energy to break the bonds of Nacl apart comes from the heat of the water. So the extra amount of energy which hydration could not provide comes from the heat of the water. Hence water looses heat, becomes slightly cold and also hydration energy (exothermic process) is less than the required energy to break the bonds, hence the net enthalpy change is positive.

So till now we learnt that for Nacl hydration enthalpy is not greater than enthalpy required to break the bonds and hence the extra energy required to satisfy the bond breaking comes form the heat of the water, and as hydration enthalpy is less than required energy to break the bond of Nacl, the net enthalpy change is positive, hence endothermic process.

Now the question arises how comes then Nacl dissolve in water?
As already said its not compulsory that enthalpy change needs to be negative for a salt to dissolve in water. Remember my friend, dissolving depends upon ΔG. ΔG is the most important factor which decides weather the reaction will happen or not. If ΔG is negative and ΔH is positive, salt will for sure dissolve because dissolving of salt (or any reaction) depends on ΔG. So if ΔG is negative for the dissolving of salt then salt would dissolve. Infact for Nacl ΔG is negative. Lets see how.

ΔG= ΔH −TΔS.

In dissolving Nacl in water ΔS is very much positive because is dissolving Nacl, Na and Cl gets apart, increasing the disorderedness in the system, hence ΔS is highly positive. So practically TΔS term is higher than ΔH term and hence we end up having ΔG negative.

Hence as ΔG is negative, Nacl dissolve. Remember dissolving of Nacl depends on ΔG. Negative ΔG makes Nacl dissolve in water.


Calculate the pressure of the gas samples as indicated by the manometer

In part (A), ${\rm P}_{\rm gas} > {\rm P}_{\rm atm}$, as evidenced by the column of mercury being "pushed up" in the manometer on the side of ${\rm P}_{\rm atm}$ - in this case, you'll need to add (the amount it's "pushed up") to atmospheric pressure (i.e., 764 mm Hg). You state that you tried to subtract 40.0 from 764.0: Look at the graphic again (I hold it's 7 cm - or 70 mm - Hg) and try adding that number to 764 mm Hg instead of 40 mm Hg.

For part (B), ${\rm P}_{\rm atm} > {\rm P}_{\rm gas}$, so reverse the process above. I believe what you state is your answer to part (A) is actually the correct answer to part (B).


Will Ammonia react with Gasoline mixed with Ethanol?

A fuel that is mostly ammonia is an altogether different fuel formulation that would not reap the benefits of the oxygenated fuels program in the US for gasoline. In other words, there would be no reason to mix in ethanol or any of the other approved oxygenated species used in gasoline (the specification requires 1.5% - 3.5% oxygen by weight during the winter (state dependent) and is reported as ethanol, regardless of the oxygenated species actually used).

According to this US Environmental Protection Agency document:

  Oxygenates are fuel additives that contain oxygen, usually in the form of alcohol or ether. Oxygenates can enhance fuel combustion and thereby reduce exhaust emissions. Some oxygenates also boost gasoline octane. The Clean Air Act requires use of oxygenated gasoline in areas where winter time carbon monoxide levels exceed federal air quality standards. Without oxygenated gasoline, carbon monoxide emissions from gasoline-fueled vehicles tend to increase in cold weather. Winter oxygenated gasoline programs are implemented by the states.

The octane levels of ammonia fuels are already very high at around 120. There would seem to be little benefit in the form of carbon monoxide emission reductions by adding oxygenated hydrocarbons to a fuel that is 70% ammonia.

So, the bottom line is that there is no reason to expect ethanol and ammonia to be in the same fuel formulation. Furthermore, if there were ethanol in the ammonia based fuel, there is no reason to expect the two to react in any significant way.


Nitrocellulose (flash paper) synthesis without Sulfuric Acid?

I do not think it will be possible to form nitrate esters of cellulose without sulfuric acid present, I am sure that you need the strong bronsted acid to make the reaction work. But I think that there is another issue here, your safety / health / the law on energetic materials.

I hate to have to say this, but I do not think it is safe to make energetic materials using nitric acid at home. You should also consider for a moment the question of "is it legal" to make the energetic material at home.

Keep in mind years ago some "home chemists" would make explosives / fireworks and it was regarded as "boys will be boys". But nowadays the "boys in blue" (Police) are very keen to chase after people who make explosives / things which go bang etc.

I think that the distillation of sulfuric acid at home does not sound like a good idea, I think that you would need to work in a good fume hood. Most school labs would be unlikely to have the equipment required to do it.

With all due respect I think you should choose another chemical project to pass the time while you are on "lockdown".


Second Peak in X-Ray Diffraction Pattern of bcc Metal

The reflection with $n=2,\;(h,k,l)=(1,0,0)$ is the same as $n=1,(h,k,l)=(2,0,0)$. Yes, I said "same", not "similar"; they are one thing, there are no "both". For this reason, you may just as well abandon using $n$ altogether.

As for why the peak is so small compared to others you've seen, let me respectfully suggest that maybe you haven't seen all that many. Indeed, there is no law that all peaks on the diffraction pattern must be about the same size. On the contrary, they are essentially proportional to the squares of Fourier coefficients of electron density, and as such, can be anything from 0 to the maximum. The early textbook exercises usually employ some mercifully simple cubic or tetragonal structures with few well-resolved and well-developed peaks. Not all compounds are like that. Look at the graphite diffraction pattern, for instance: you'll basically see a lonely (002) peak, large as life, and only after the most careful inspection you'll start noticing the rest of them amidst the background noise.

As for the (100) peak, it couldn't be there, since it fell victim to the systematic absences of the bcc lattice. Ditto for (111) and (210). Only those with even $h+k+l$ survived, and that's precisely what you see on the picture.