Why transition metal chloride complex have lower reduction potential than metal aqua ions?

The  anion  $\ce{AuCl4−}$ is  the  most  known  complex  of  gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we are dealing with:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &<=> Au(s)} &\quad E^\circ_\mathrm{cathode} \tag1\\
\ce{[AuCl4]-(aq) + 3 e- &<=> Au(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} \tag2
\end{align}
$$
If we substract the equation $(2)$  from the equation $(1)$, we get:
$$\ce{Au^3+(aq) +  4 Cl- (aq) <=> [AuCl4]-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag3$$
The equation $(3)$ is the reaction for the formation of $\ce{AuCl4−}$ with formation constant $K_f$. If you applied the Nernst equation for the equation $(3)$, then at $\pu{25 ^\circ C}$:
$$E^\circ_\mathrm{cell} = \frac{RT}{nF} \ln Q = \frac{RT}{nF} \ln K_f = \frac{0.0592}{n} \log K_f \tag4$$
This last solution of the equation $(4)$ (where $n = 3$) is because, when at equilibrium $Q = K$. If we know the $K_f$, we could have find the $E^\circ_\mathrm{cell}$, and hence $E^\circ_\mathrm{anode}$ (which is for $\ce{AuCl4−}$). Unfortunately, I couldn't find $K_f$ for $\ce{AuCl4−}$ ion formation. Thus, I'd show how big $K_f$ for $\ce{AuCl4−}$ by applying the given values of $E^\circ_\mathrm{cathode} = \pu{+1.52 V}$ and $E^\circ_\mathrm{anode} = \pu{+0.93 V}$ on the equation $(4)$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f \ \Rightarrow \log K_f = \frac{n}{0.0592} E^\circ_\mathrm{cell} = \frac{3}{0.0592} (1.52 - 0.93) = 29.9$$
$$\therefore \ K_f = 10^{29.9} = 7.91 \times 10^{29}$$
It is also in similar way you can find the $E^\circ$ for redox half-reaction of $\ce{PtCl4^2−}$ since we know its formation constant ($K_f = 1.0 \times 10^{16}$):
$$
\begin{align}
\ce{Pt^2+(aq) + 2 e- &<=> Pt(s)} &\quad E^\circ_\mathrm{cathode} = \pu{1.18 V} \tag5\\
\ce{[PtCl4]^2-(aq) + 2 e- &<=> Pt(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} = ? \tag6
\end{align}
$$
If we substract the equation $(6)$  from the equation $(5)$, we get:
$$\ce{Pt^2+(aq) +  4 Cl- (aq) <=> [PtCl4]^2-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag7$$
From equation $(4)$ where $n = 2$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f = \frac{0.0592}{2} \log( 1.0 \times 10^{16})  = 0.474$$
Since, $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.474 V}$, and $E^\circ_\mathrm{cathode} = \pu{1.18 V}$,
$$E^\circ_\mathrm{anode} = 1\pu{1.18 V} - \pu{0.474 V} = \pu{0.706 V}$$
$$\therefore \ \ce{[PtCl4]^2-(aq) + 2 e- <=> Pt(s) + 4 Cl- (aq)} \quad E^\circ = \pu{0.706 V}\text{ (calculated)} $$
This value is in good agreement with the experimental value, which is given as $\pu{0.755 V}$ in electrpchemical series of CRC Handbook of Chemistry and Physics.
Source of formation constant of platinum complex ion


In a galvanic cell why does using a more reactive metal increase the voltage?

It is a good question because textbooks do not discuss it. Scientific explanations are of two types, how and why. The "why" part of the explanation is very difficult to answer. Often we don't know the answer at a fundamental level. A simple example is that of what is universe is made? What is the origin of mass or charge? In the same way the electrode potential can also be understood at various levels. The most advanced discussion of electrode potentials can be done in terms of Fermi levels, but this is not my area of expertise. The macroscopic level discussion is that one can consider the electrode potential as a (thermodynamic) tendency of a substance to gain electrons. If you look up modern electrode reduction tables, all entries are listed as reduction process (all electrochemists agreed to do that after a long academic fight to write it this way). Hydrogen has been given an arbitrary value of 0.0000 V for the sake of comparison. You always need a reference in electrochemistry so electrochemists decided to use H2 electrode as a reference.

a) If you look at the table below, look at lithium entry: It shows that Li+ has very low (or negative electrode potential) tendency to gain electrons. Read it the other way (right to left), that Li metal has a very high tendency to become a lithium ion.
b) Look at the silver entry: It shows that Ag+ has a high tendency to gain electrons (positive electrode potential). Read it the other way (right to left) that Ag has a very low tendency to become Ag+.

If I make a galvanic cell consisting of two beakers connected by a salt bridge, having a lithium electrode in 1 M Li+ solution, and a silver electrode dipped in 1 M Ag+ ions. Connect the leads to an voltmeter.

Now Ag+ has a tendency to accept electrons and become Ag and lithium metal has the tendency to lose electrons and become Li+, what do you think happen? Electrons would feel a "pressure" or better "potential difference" causing them flow from higher potential to a lower potential, which is given by

Ecell=  Ecathode - Eanode= 0.7996 - (-3.040) = + 3.8396 V, this should be your voltmeter readings. Why did we choose Ag+ /Ag as a cathode? Because Ag+ ions have a higher tendency to reduce than lithium ions.

In the same way you can choose Li and Be cell. What would be the cell voltage?

For example, electrode potential tables look like:

Li+(aq) + e−  ⇌  Li(s)      –3.040

Be2+(aq) + 2e−  ⇌  Be(s)    –1.99

Al3+(aq) + 3e−  ⇌  Al(s)    –1.676

2H+(aq) + 2e−  ⇌  H2(g)  0.00

Ag+(aq) + e−  ⇌  Ag(s)       0.7996

Ce4+(aq)+e−⇌Ce3+(aq)        +1.44

PbO2(s) + HSO4−(aq) + 3H+(aq) + 2e− ⇌ PbSO4(s) + 2H2O(l)    +1.690

H2O2(aq) + 2H+(aq) + 2e− ⇌ 2H2O(l) +1.763

F2(g) + 2e−⇌ 2F−(aq)         +2.87


The number of optical isomers of 3,4-dichlorobutan-2-ol

The carbon marked with the asterisk cannot change its configuration, for it to be counted as a stereocentre, it must be able to. hence 2 isomers only


900 mCi is a lot of activity, the problem is that it is impossible to know how bad it will be without knowing which radionuclide is present.

If it is a tritium light with 900 mCi in it, then the external threat is close to zero. Just do not break open the glass vials in a tritium light.

If it is 900 mCi of radium-226 or cobalt-60 then I would suggest to stay a very long way from the object and call the fire brigade or police. They will not have the ability to resolve the problem. What they will do is to seal off the area and then send for expert advice and an emergency team from a nuclear site or some other place where people have experience of working with large amounts of radiaoctivity.

Even a pure beta emitter at 900 mCi is a lot of activity if you are dealing with a high energy beta emitter such as Sr-90 / Y-90.


The Tollen's test is safe, but silver amine solutions should be prepared as needed just before use.  The formation of explosive silver nitride is a slow process and can be prevented by treating the solution with mineral acid after use.


I am guessing that this is the compound you are interested in:

Sigma labels this compound as p-Terphenyl-4,4′′-dicarboxylic acid (or 1,4-Bis(4-carboxyphenyl)benzene).

In your case, the p,p' indicates that the 2 carboxylic acids are in the para positions on 2 of the phenyl rings.


Yes, mechanism of this reaction involves internal attack.


Why enthalpy is defined at a constant pressure?

How are absolute enthalpy and enthalpy change defined?

Enthalpy $H$ is defined as in your question. The most basic definition of enthalpy change when going from state 1 to state 2 is the enthalpy of state 2 minus the enthalpy of state 1:
$$\Delta H = H_2 - H_1$$

  What is the clear difference between enthalpy and heat?

Enthalpy is a state function and heat is not. Heat is energy transferred from a hotter system to a cooler system (not to be confused with thermal energy, that part of internal energy due to the kinetic energy of particles randomly moving, i.e. what makes a sample have a certain temperature).

A simple experiment to experience the difference: If you rub your hands together, they warm up. If you wait a while until they are back at the original temperature, they will be in the same state (no change in enthalpy). At the same time, there was a heat exchange with the environment, increasing its temperature a bit.

In essence, some mechanical energy was converted into thermal energy. There was some heat transfer $q > 0$, but the enthalpy did not change ($\Delta H = 0$) in this thermodynamic cycle. So enthalpy change and heat are different concepts, and there are cases where their values are different.

  Is it the constancy of pressure what makes the enthalpy a state function?

No, it is because of how enthalpy is defined. The internal energy is a function of state, and P and V are a description of the state. This means that by the definition of enthalpy, the same state will have the same enthalpy. (If you want to know why internal energy is a function of state: it is related to the first law of thermodynamics, but more fundamentally - it is what all measurements so far have shown).

There is a reason why looking at enthalpy instead of internal energy is often simpler, see below, related to how much of what we experience is at constant pressure (usually atmospheric pressure).

  My teacher told me in very simple words that when heat goes inside a system, it is called enthalpy. Is this correct?

This is a shortcut for the following: If you run a process in a way that the only forms of energy transferred between the system and the surroundings are heat and pV-work, the change in enthalpy will be equal to the heat transferred. (In my example above, there was mechanical work; that would break the equality of $\Delta H$ and $q$, as would electrical work done by or on the system, or electromagnetic radiation going in or out.) Also, the system has to be closed (no matter entering or leaving the system).

pV-work is a term for the work of expanding the system's volume by $\Delta V$ into the surrounding. When this happens, you work against the pressure of the surrounding. Enthalpy is defined in a clever way so that it does not change due to pV-work at constant pressure:

$$ H = U + PV$$
$$ $$
$$ \Delta H = H_2 - H_1$$
$$ = U_2 + PV_2 - (U_1 + PV_1)$$
$$ = \Delta U + P\Delta V$$

It turns out the change in internal energy due to pV-work (at constant pressure) is $-P\Delta V$ (it takes work to push against the surrounding, done by the system, so the system loses energy), which cancels out against the $P\Delta V$ term in the calculation above. Thus there is no change in enthalpy due to pV-work at constant pressure.

  Would constant pressure make the volume constant in a thermodynamic system?

No. Take an ideal gas (in a closed system) whose temperature rises for some reason. If you keep the pressure constant (by having the gas inside a floppy mylar baloon), the volume will increase. Enthalpy is a useful state function for systems at constant pressure because it does not change just because the volume changes.

  Why is this answer so long and complicated (I added this question)?

At my college, enthalpy is introduced as a "fact" in our first chemistry course. A semester later, we learn about first and second law of thermodynamics, and we discuss the tools of thermodynamics a year or two later. So it is not taught from first principles, and you have to learn some stuff without a full picture of the why and how. What I summarized here would be taught in three different courses at my college (along with other material, of course). If you have not been introduced to all of these topics yet, this answer will contain some concepts unfamiliar to you.


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

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

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

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


Can we interpret the "extent of reaction" as the number of reactions that happened?

If I understand this correctly you want to measure $\ce A$ to infer the extent of reaction. I think you should have started with mathmatically defining the extend of reaction:

$$\xi(t) = \left[\frac{n_i(\ce A) -n_t(\ce A)}{n_i(\ce A)}\right]\tag 1$$

Where $n_t$ is the concentration at a given time $t$. Here $n_i(\ce A) -n_t(\ce A)$ gives the consumed concentration of $\ce A$ after time $t$ which is normalized by the initial concentration to yield  an extent of conversion.

  Assume $\mathrm{2\ a}$: If the stoichometric constants are normalized $({\alpha} + {\beta}= 1)$ to be taken as the initial concentrations then we can substitute: $\alpha = n_i(\ce A)$ for example $$\ce{0.33A + 0.67B -> 0.5C + 0.5D}$$
  $$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right]\tag {2 a}$$
  otherwise:
  $$\left(\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1\right) \implies \frac{\alpha}{\alpha + \beta} = n_i(\ce A)\tag {2 b}$$

Which we can rearrange to :

$$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right] \implies \alpha-n_t(\ce A) = \alpha\ \xi(t) \tag{3} $$

Subtracting $\alpha$ and then negating gives:

$$n_t(\ce A) = \alpha - \alpha\ \xi(t) = n_i(\ce A) - \alpha\ \xi(t) \tag{4 a}$$
Which would make you correct, or in the case of assumption $\mathrm{2\ b}$:
$$n_t(\ce A) = n_i(\ce A) - \frac{\alpha}{\alpha + \beta}\ \xi(t) \tag{4 b}$$
Which would make you almost correct.


What solutes will give the largest reduction in volume when dissolved in water

(1) Your calculations are correct.

(2) $\ce{CaCl2}$ has a density of 2.16 kg/L

           kg        vol(L)
  CaCl2    18.00     8.38
  H2O      41.62    41.62
           -----    -----
           59.62    50.00 L

But when dissolved the solution has a density of 1.300 kg/L, so the total volume is 45.86 L. (Data about $\ce{CaCl2}$ solutions.)


What is the timescale of disulfide bond formation?

I agree that the actual redox process is very fast. The rate limiting issue with proteins and other macromolecues involves the reactive groups finding each other. Thus, from the time of translation to the time of disulfide formation, the peptide chains need to fold properly. Perhaps other subunits need to be assembled and attached. Once everything is in the right place, the disulfide bond should form quickly provided there is an appropriate oxidant present.

You could measure the rate of the entire process by attaching reporters to positions on the protein - say a fluorophore and a quencher. When the process is complete, the quencher and the fluorophore are close in space, and fluorescence stops. This approach will measure the overall rate of the entire process. I am not sure if you can separate out the different components, e.g. specifically the S-S bond forming. I can't point to a specific paper where this is strategy is used, since I am away from my usual computer that likes my journal subscriptions. Hopefully I will be able to update my answer.


Carnauba wax is a high-melting wax that should be suitable. In the Wikipedia article cited,
"Carnauba wax can produce a glossy finish... It was commonly used in its purest form as a coating on speedboat hulls in the early 1960s to enhance speed and aid in handling in salt water environments. It is also the main ingredient in surfboard wax, combined with coconut oil."


How to re-suspend silver from the dried out silver paste?

According to SPI Supplies:

  If the Dotite Silver Paint dries out, unlike some other silver paints, it can be re-suspended with the addition of SPI #05004 Silver Paint Thinner followed by an ultrasonic bath treatment.

Generally, ketones are used as diluents, also for conductive paints (acetone, methyl ethyl ketone (MEK), methyl isobutyl ketone (MIBK)).


Is chromate a suitable indicator for the titration of Ag⁺ with Cl⁻?

You got the solubility part reversed.
The solubility of $\ce{AgCl}$ is lower than the solubility of $\ce{Ag2CrO4}:$

$$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}$$

$$s(\ce{Ag2CrO4}) = \sqrt[3]{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt[3]{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$

Therefore, if the $\ce{AgNO3}$ solution is gradually added to the solution containing the both $\ce{Cl-}$ and $\ce{CrO4^2-}$ ions, then initially the formation of a sparingly soluble $\ce{AgCl}$ salt occurs.
After the $\ce{Cl-}$ ions are almost completely isolated in the form of $\ce{AgCl},$ the $\ce{Ag2CrO4}$ precipitation starts to occur, signifying the equivalence point is reached.

The same reasoning can also be applied to the titration of even less soluble silver bromide $\ce{AgBr}$ with $K_\mathrm{sp}(\ce{AgBr}) = \pu{5.3E-13}$ (try it yourself).

Note, however, that while using Mohr's method it's imperative to titrate halide salts solutions with $\ce{AgNO3}$ and not vice versa.
Otherwise the precipitation condition

$$c(\ce{Ag+}) · c(\ce{Cl-}) > K_\mathrm{sp}(\ce{AgCl})$$

will be overridden by

$$c(\ce{Ag+})^2 · c(\ce{CrO4^2-}) > K_\mathrm{sp}(\ce{Ag2CrO4})$$

due to high concentration of silver ions in solution, favoring silver chromate precipitation (note squared term $c(\ce{Ag+})^2$) and thus shifting the equivalence point.


Why does benzene have a D6h point group?

Does it have two or more $C_2$ axes with $n>2$?

This question is asking if there is more than one $C_{n>2}$ axis not if there is more than one $C_{n>2}$ symmetry operation.

Yes, benzene has some $C_3$ operations. However, as pointed out in the comments, the $C_3$ axis lies along the $C_6$ axis - both run through the center of the ring perpendicular to the plane of the molecule. This axis is also a $C_2$ axis. There is only one $C_{n>2}$ axis - it just happens to be both a $C_6$ and a $C_2$ axis. All other rotation axes in benzene are $C_2$.


In principle, the acetalisation reaction is indeed reversible. Typical conditions to reverse it would be to use $\ce{TsOH}$ or $\ce{PPTS}$ and water.

However, both full acetalisation and liberation of the aldehyde require acidic conditions. Hemiacetalisation — the prime mechanism of mutarotation — proceeds without acidic catalysis, i.e. the inherent (low) acidity of water is sufficient. And to observe mutarotation, you typically dissolve the pure sample in a pure solvent (with no acid added).

If you dissolve e.g. methyl α-ᴅ-glucopyranoside in water and add catalytic amounts of acetic acid, you will observe mutarotation but not if you simply dissolve the O,O-acetal in water.


Identifying the oxygen atoms that undergo reduction

The oxygens in the dichromate ion on the right have -2 oxidation number, the chromium being +6.
http://ch302.cm.utexas.edu/worksheets/basic-redox-wkst-KEY.pdf
https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-the-dichromate-ion-cr2o7-2

As you seem to have deduced, the peroxide ions are -1 so all oxygens, as shown, have been reduced.


Certainly water and ethanol and all mixtures thereof will completely evaporate without boiling given enough time.

Water and ethanol form a postive azeotrope that is 95.6% ethanol and 4.4% water and boils at 78.2 degrees C.

If you boil a mixture of water-ethanol that has less than 95.6% ethanol, the concentration of water in the liquid phase will increase, but it will only asymptotically approach pure water as all the liquid boils away.  There will not be a time when all the ethanol is gone and you are left with a finite amount of pure water.

If you boil a mixture of water-ethanol that has more than 95.6% ethanol, the concentration of ethanol is the liquid phase will increase.


You should be careful with simple associations such as "metal + non-metal = ionic bond". These tend to throw out the idea of understanding the chemistry involved in favour of rote memorization. Note for example that mixing caesium metal with gold will produce a salt instead of an alloy, caesium auride ($\ce{Cs^+ Au^{-}}$). Mixing barium metal and platinum can also produce salts, though their structures are somewhat more complex. One can also argue that there is significant ionic character in solid xenon difluoride, even though both atoms are non-metals.

The idea of using electronegativity to determine covalent/ionic character is also meant as a helpful guide, not as a strict rule with black-and-white limits. Firstly, all bonds have both ionic and covalent character; both concepts are an oversimplification, and in reality it is more correct to say that a bond has a certain contribution from each type of bonding. This means there is a smooth transition from compounds with mostly ionic character and those with mostly covalent character. Also, the inequalities you mention rely on Pauling electronegativities. Electronegativity is surprisingly still a hotly debated topic, as we continue to search more general, more fundamental and more precise ways of defining it. Pauling electronegativities are based on empirical thermodynamic data regarding bond energies after applying a certain equation that was "picked", not derived from scratch. The values are particularly poorly defined for transition elements, such as the $\ce{Cu}$ in your problem. You get some not-so-easy to explain situations, like $\ce{HF}$ as a gas that is a borderline ionic compound.

Finally, in light of these comments, the answer to your question is that bonding in $\ce{CuCl_2}$ (I'm pretty sure that's what you actually meant to write) has intermediate characteristics between a purely ionic and a polar covalent bond, with similar contributions (though pinpointing which is highest sounds like an exercise in futility). A good way to study it more in depth is to analyze Fajans' rules. After a little self-calibration, you can get a good feel for the degree of ionicity and covalency of a compound. Some further but less certain evidence (lots of caveats!) for the intermediate character of $\ce{CuCl_2}$ can be found by looking at the substances' melting and boiling points ($\pu{498°C}$ and $\pu{993°C}$ [decomposition], respectively, according to Wikipedia). They are both quite high compared to substances with polar covalent bonds (dimethylformamide boils at around $\pu{150°C}$), but rather low compared to substances with very ionic bonds ($\ce{NaCl}$ boils over $\pu{1400°C}$).


When boiling water, how is evaporation rate affected by the surface area of the container?

Boiling water

When you boil water on a stove, the temperature on the bottom of the pot is higher than the boiling point. Steam bubbles are generated, and they help to heat the water at the top of the pot (convection also helps, and conduction of the pot itself). Once you have a rolling boil, the rate is determined by how quickly you can supply more heat (almost all of it now goes into the phase transition). If you use the same burner, and mostly heat the pot (so no tiny pot on a big burner), at that stage the rate of boiling off should be the same.

Evaporating water

When the temperature of the water is below boiling, the mechanism is different. Rather than steam displacing the air above the water, single water molecules break away from the liquid and mix with the air. Evaporation rates depend on the partial pressure of water just above the surface (this influences condensation, i.e. the reverse reaction), the surface area, and the vapor pressure of water in the liquid state. Anything that lowers the partial pressure of water (such as "wind" bringing in dry air), increases the surface area (different pot geometry) or increases the vapor pressure (temperature) will increase net evaporation rates.

Beaker vs Erlenmeyer

Tall vs. wide pot changes the contact area with the stove top and the area of the liquid:gas interface at the same time. If you compare a beaker with an Erlenmeyer (same bottom area), it should show similar boiling rates (maybe a big mess with overboiling for the Erlenmeyer) but very different evaporation rates.


Why iron(II) spectrometry is not measured in distilled water and it requires a complexing agent like o-Phen?

I'm sure OP get the answer by reading all comments and the answer elsewhere. However, I'd like to point out few things on this spectrophotometric determination of $\ce{Fe^2+}$ in drinking water (it is not distilled water):

Iron can exist in natural waters where it originates from the dissolution of minerals. Even though most iron containing minerals are not very soluble in water as evident by the solubility products of iron-containing salts. However, a trace amount of iron can still dissolve in water. The dissolve iron is predominantly found in the form of $\ce{Fe^3+}$, because dissolved oxygen in water oxidizes $\ce{Fe^2+ -> Fe^3+}$.
Based on above fact, water cant be green, but probably yellow. But, it can be still determined by UV-vis spectroscopy, only if you have a lot of $\ce{Fe^3+}$ ions in your unknown drinking water sample to do so (see Ivan Neretin's comment and Maurice's answer), which is not possible.
On the other hand, 3 molecules of 1,10-phenanthroline (ortho-phenanthroline, a bidentate) chelate with one ion of $\ce{Fe^2+}$ to give highly intense orange-red $\ce{[Fe(o-phen)3]^2+}$ complex ion in water (Ref.1). The equilibrium constant for this reaction $\left(\ce{Fe^2+ + 3 o-phen -> [Fe(o-phen)3]^2+}\right)$ is $2.5 \times 10^6$ at $\pu{25 ^\circ C}$  in the $\mathrm{pH}$ range between  $\mathrm{pH} = 3$ and $\mathrm{pH} = 9$. Thus, a $\mathrm{pH}$ of about $3.5$ (using acetate buffer) is recommended to prevent precipitation of iron salts.
The molar absorption coefficient (ε) of this ferrous complex, $\ce{[Fe(o-phen)3]^2+}$ is $\pu{11,100 dm3 mol-1 cm-1}$ at the wavelength of maximum absorbance, $\lambda_{max} = \pu{508 nm}$. Thus, spectrophotometric absorbance of this complex can be easily measured in $\pu{10^{-4} M}$ to $\pu{10^{-5} M}$ or even lower levels at $\pu{508 nm}$, which is compatible with $[\ce{Fe^3+}]$ levels in drinking water (usually well water).
However, there is a problem: 1,10-Phenanthroline complexes quantitatively only with $\ce{Fe^2+}$ ions. But that can be solve using a reducing agent such as hydroxyl amine ($\ce{HCl}$ salt, which is soluble in water):
$$\ce{4Fe^3+ + 2HO-NH2.HCl -> 4Fe^2+ + N2O + 6H+ + 2Cl- + H2O}.$$ The reducing agent should be used in excess to prevent $\ce{Fe^2+}$ oxidizes back to $\ce{Fe^3+}$ by dissolve oxygen.

References:

R. Belcher, "The application of chelate compounds in analytical chemistry," Pure Appl. Chem. 1973, 34(1), 13-28 (DOI: http://dx.doi.org/10.1351/pac197334010013).


Which solvent is good for using SU-8 particle in Optical tweezers?

I've done some experiments recently with SU-8 and polistirene and figured it out, that SU-8 solvent melts polistirene. Just wanna say, that this won't work out well.

But to use something instead of water: why not using alcohol (ETOH) or isopropyl. I use the last as much as possible since it's not leaving any stains on stuff I do.


why are lp-lp repulsions greater than bp-lp and bp-bp?

So, from Wikipedia article on VSEPR theory we read:

  The overall geometry is further refined by distinguishing between
  bonding and nonbonding electron pairs. The bonding electron pair
  shared in a sigma bond with an adjacent atom lies further from the
  central atom than a nonbonding (lone) pair of that atom, which is held
  close to its positively charged nucleus. VSEPR theory therefore views
  repulsion by the lone pair to be greater than the repulsion by a
  bonding pair. As such, when a molecule has 2 interactions with
  different degrees of repulsion, VSEPR theory predicts the structure
  where lone pairs occupy positions that allow them to experience less
  repulsion. Lone pair-lone pair (lp-lp) repulsions are considered
  stronger than lone pair-bonding pair (lp-bp) repulsions, which in turn
  are considered stronger than bonding pair-bonding pair (bp-bp)
  repulsions, distinctions that then guide decisions about overall
  geometry when 2 or more non-equivalent positions are possible.

As I said in my comment, the key point is given in the second sentence: the bonding pair shared in a bond lies further from the atom than a nonbonding pair of that atom which is held close to its positively charged nucleus. Consequently, lone pairs are closer to each other than any other combination of pairs (lone pair-bonding pair and bonding pair-bonding pair), and thus, repell each other more strongly.


How to check which compound will show optical isomerism?

In the first image due to the absence of any chiral carbon or steriogenic centre it does not show optical activity.
In the second image there is a plane of symmetry thus it ceases to be optically active(the plane of symmetry is perpendicular to the substituted butane ring).finally Alenes are optically active as the successive π orbitals lie on a different plane.You can definitely get the structure of alene from a reference book and the due to successive π bonds they lie on a different plane (the π orbitals).thus they are optically active.
Thus in compounds where there are no chiral centre check centre of symmetry,axis of symmetry and plane of symmetry as a CHIRAL carbon is essentially an unsymmetrical molecule.Hope you understood:).


Why are bonding interactions so much stronger than interactions of other types?

Bonding interactions effectively share electrons. In a gedankenexperiment, you can think of two isolated species with many unpaired electrons coming together, bonding and forming a species with less unpaired electrons. The exact nature of bonds is slightly different, involves quantum mechanics and orbitals, but the end of the story is that the electrons are in a less energetic state after bonding than before it. The energy gained by this process is rather significant, typically hundreds of kilojoules per mole.

These bonds that are created typically generate a molecule or some type of network solid (ionic, metallic or covalent). If you have generated a molecule, then you can rather easily separate a single molecule from all the surrounding molecules by physical methods such as high vacuum. It is not possible, however, to break the bonds by any simple, low-energy physical methods. (UV irradiation will do, but that is high-energy.)

The other interactions, London dispersion forces, hydrogen bonds, etc. are interactions between molecules. You already have two species that by themselves are in a rather stable, i.e. low energetic, state and you bring these two together. The energy that can still be gained is very minor compared to that obtained by bonding. This is also why this type of interaction is often called secondary interaction.

A second reason is the distance. As you note, most of these interactions can be explained electromagnetically. However, the distances between molecules are typically significantly larger than the distances between atoms of a molecule. Since electromagnetical interactions decline exponentially with rising distance, these forces are a lot weaker at higher distances.


Is it possible to prevent saponification by modifying free fatty acids?

In principle You can replace the Sodium ion with any other cation. But all You get is another kind of soap: Potassium gives a smoother substance, Calcium/magnesium gives the soap-stone you see in washing machines, etc.

Another possibility is to esterify the fatty acid with an alcohol. This is widely used in the industry to produce non-ionic surfactants, and these do not react with cations. But off course, then it really isn’t a fatty acid anymore.


How does Aloe Vera leave a cool feeling on your skin? Why does it evaporate fast?

The leaf of aloe vera is composed of three layers: 1) An inner clear gel that contains 99% water and rest is made of glucomannans, amino acids, lipids, sterols and vitamins. 2) The middle layer of latex which is the bitter yellow sap and contains anthraquinones and glycosides. 3) The outer thick layer of 15–20 cells called as rind which has protective function and synthesizes carbohydrates and proteins. So overally , it consists of about 99% water and remaining are vitamins, enzymes, minerals, sugars, lignin, saponins,etc.

So what makes inner layer gelly ?? Well, the innermost part of the leaf is a clear, soft, moist and slippery tissue that consists of large thin-walled parenchyma cells in which water is held in the form of a viscous mucilage consiting of cellulose, hemicellulos and acetylated mannans.

As for you question, as most of it consists of water(99%) the cool feeling might has been felt due to its evaporation but some of its remaining constituents are absorbed in the skin. Mucopolysaccharides present in it help in binding moisture into the skin.Its one of the important medical applications is that it could potentially be used as a
penetration enhancement agent for the transdermal delivery of drugs and intestinal absorption of coadministerd drug.link

Saponins that are the soapy substances are found in the gel of aloe vera and have cleansing and antiseptic properties link so you can say that hand sanitizer and aloe vera are somewhat connected.


Reaction of Grignard reagents with esters

Reaction of a Grignard reagent with an ester is a standard method for producing tertiary alcohols where at least two of the substituents (the "$\small\ce{R_2}$" group attached to the Grignard) are the same.  Judicious choice of the starting ester allows for the preparation of a tertiary alcohol where all 3 substituents are the same. Alternately, you could react 3 equivalents of the Grignard with a cheap dialkyl carbonate (a "double" ester, if you will) to obtain a tertiary alcohol where all 3 groups are the same.

Typically a slight excess of Grignard reagent is used and the tertiary alcohol is produced after acid work-up.  If an excess of Grignard is not used then a mixture (ugh!) of ester, intermediate ketone and alcohol may result.


The time-dependent pH value in a fermentation

The pH is not corrected mathematically, it is stabilized chemically by addition of NaOH when the pH sensors get out of range. Your pH is controlled to be constant over the fermentation period (+/- a little bit).

The acid produced by the fermentation is neutralized by addition of NaOH to keep the pH constant. What you can calculate from the data you get is the rate of acid production. The moles of acid produced will equal the number of moles of NaOH added, to a good approximation. The pKa of H2PO4- is 7.2 and the pKa of acetic acid (a typical carboxylic acid) is 4.76, so the acid produced will be "strong" - i.e., fairly acidic in your medium, which appears to be aimed for about pH = 7 - and the NaOH is strongly basic, so they will balance out (approximately).

Interestingly, sodium acetate has a pH range of 8.5 - 9.9 in real life, according to SDS. Calculation is so much more pleasant, where you can calculate pH to three decimals and feel good about it (until you test it in real life and have to calibrate your meter, etc, etc.)


Mechanism for basic hydrolysis of α-chloronitrile to ketone?

Your analysis is slightly incorrect, actually. If you do a $\mathrm{S_N2}$ on the chloride, hydrolyse the nitrile, and decarboxylate, you end up with the wrong oxidation state:

Instead, if you actually get to the cyanohydrin, you could (in theory) simply lose $\ce{CN-}$ to form the ketone:

However, a $\mathrm{S_N2}$ reaction at that centre is going to be difficult, to say the least.

The reaction has been studied by Shiner et al.,1 who determined that the primary site of attack is likely to be the nitrile. They wrote:

  No mechanistic studies of the hydrolysis step have previously been reported, although pathways involving nucleophilic displacement of chloride or cyanide, affording intermediate cyano- or chlorohydrins, have been proposed. A priori we regarded both $\mathrm{S_N2}$ and $\mathrm{S_N1}$ displacements as improbable in these systems.

The immediate product of nitrile hydrolysis is the amide, and the α-chloroamide intermediates were actually isolated:

The mechanism after this involves a cyclisation of the amide onto the $\ce{C-Cl}$. The exact steps are not entirely clear. However, the following was proposed:

Reference

(1) Shiner, C. S.; Fisher, A. M.; Yacoby, F. Intermediacy of α-chloro amides in the basic hydrolysis of α-chloro nitriles to ketones. Tetrahedron Lett. 1983, 24 (51), 5687–5690. DOI: 10.1016/S0040-4039(00)94173-X.


If an atom loses electrons, it develops a positive charge and becomes a cation. Likewise, if an atoms gains electrons, it develops a negative charge and becomes an anion.

"Charges only EXIST on IONS."

Covalent bonding rose the oxidation number.
We are aware that in a Covalent Bond sharing of electron takes place.
The more electronegative atom will take more bonding electrons and it will get a negative oxidation number (It depicts Anion).
Similarly less electronegative atom mimics a Cation.

In other words you can say Oxidation Numbers are fictional ionic charges


What reaction happens when carboxylic acids are treated with calcium hydroxide and distilled?

Acetic acid (as an example of a typical carboxylic acid) reacts with calcium hydroxide to form calcium acetate. This is a simple acid-base reaction which forms a salt and water:

$$\ce{2CH3COOH + Ca(OH)2 -> (CH3COO)2Ca + 2H2O}$$

Upon heating, calcium acetate undergoes decomposition to acetone and calcium carbonate. The acetone can be distilled off from the mixture, leaving behind solid calcium carbonate:

$$\ce{(CH3COO)2Ca ->[\Delta] CaCO3 + 2(CH3)2C=O}$$

At even higher temperatures calcium carbonate itself can be decomposed to calcium oxide and carbon dioxide:

$$\ce{CaCO3 ->[\Delta] CaO + CO2}$$

References

https://www.tut.fi/ms/muo/vert/3_organic_chemistry/carboxylic_reactions.htm
http://www.docbrown.info/page04/OilProducts10.htm


Can chemicals be identified by their molar mass?

It depends.

This was pretty much the way that the elements were discovered.

For inorganic chemicals in general the problem is really one of specificity. So I want to work with a chemical sample between 1 gram and 20 grams and I have an analysis good to 1%. That only yields about 300 statistically significantly different numbers $1.01^{300} = 20$. So for various inorganic salts this wouldn't work too well.

To take this into another field think of geology. Hardness and specific gravity are simple tests used to help identify minerals. Neither is very good by itself, but combined they sort out the minerals into a lot more bins.


Regioselectivity in Diels–Alder reaction between 2-nitrobuta-1,3-diene and acrylonitrile

You pointed out correctly that $-CN$ has a great $-I$ effect but it won't bring a $+\delta$ on adjacent carbon atom, reason being it's instability. Apply electromeric effects on propanenitrile.

If you need help then just comment below :-)


Does the temperature during this reaction remain constant?

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


What could cause a peak to split into 2 resonances as temperature increases in NMR spectroscopy

Since I do not know which compound it is, my guess is that you have two different  $\ce{H^1}$ that have almost the same chemical shift at room temperature. The chemical shifts change with temperature as for instance in sucrose:

Here  are represented mainly the  $\ce{H^1}$  of the $\ce{-OH}$ but the $\ce{C-H}$ (the peak of 1) shifts also.

In your case,  one is almost fixed, while the other one moves to higher $\delta$  and, therefore, they split into two.

More info on the sucrose spectrum here.


Though the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly.  The Wikipedia article for cyanoacrylates shows the polymerization more clearly than I can easily explain in words.  The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool provides a large number of sites for the glue to cure.

When I read your question I was curious; I had never heard of this before.  I searched the internet and was only able to find one or two examples of people actually trying this successfully, and both of them looked dubious.  I've spilled superglue on cotton shirts and pants any number of times and while it cures almost instantly, it has never caught my shirt on fire.

I decided to try it myself, and I soaked a cotton ball in superglue to see what would happen.  The glue cured extremely rapidly, producing very irritating (colorless) fumes, and the mass of glue and cotton got warm, but it was nowhere near hot enough to ignite - my tap runs hotter than the glue got.  There may be a particular type of glue that cures significantly more exothermically, but I'm inclined to think that this is an internet hoax, and Popular Science agrees.


Mechanism of rearrangement reaction found in synthesis of retinol

When trying to propose a mechanism, especially one where the bond changes are not obvious, I like to start by numbering the atoms of the product and trying to map them to the reactants. Here, I've taken the numbering given by ChemDraw for the product and translated it to the reactant.

Carbons 1-3 are pretty logical. In the product, carbon 4 is alkyl with 2 hydrogens and between to other carbons. In the starting material, the carbon connected to C3 is a carbonyl with zero hydrogens. It seems unlikely to me that this would be C4.

So I look for another, more logical mapping. For example, carbons 7 and 9 are methyl groups in the product, and there are two methyl groups in the starting material, which I label as 7 and 9. In both structures, they are bridged by a single carbon, which I call 6.

In the product C5, is part of an alkene and so is the carbon connected to C6 in the reactant, so that seems logical. And the next carbon in the starting material has two hydrogens, so that seems a better mapping as C4. The last carbon of the reactant must be lost as carbon dioxide, so I've labeled it C10.

With a hypothetical atom mapping, now I can identify bonds broken and formed. C10 and its adjacent oxygens are being lost as carbon dioxide, which makes it easy to identify the bonds broken as C3-C10 and C6-O, with a shift of the pi-bond from C4-C5 to C5-C6.

Without being aware of [3,3]-sigmatropic rearrangements, or the Claisen rearrangement specifically, you might be stuck. The Claisen rearrangement is a cyclic movement of electrons converting an allyl vinyl ether into a gamma,delta-unsaturated carbonyl compound. This is driven by the formation of a carbonyl at the expense of an alkene.

The given starting material isn't quite an allyl vinyl ether, but as a 1,3-dicarbonyl, it is equilibrium with one by tautomerization. From the tautomer shown, the Claisen rearrangement is possible, which forms the C3-C4 bond we noted earlier and shifts the pi bond into the desired position. Also note that the resulting intermediate is the one shown in the solution manual. Decarboxylation through a similar cyclic movement of electrons gives the tautomer of the product along with carbon dioxide.


It is always easier to buy ferrofluid than to make it. Some of the steps don't work well unless the conditions are just right and it is hard to get them right on your first attempt. Luckily there are plenty of suppliers (this UK supplier is top of my Google search for "ferrofluid sales" and they also supply lots of cool magnets).

Ferrofluid is a suspension of small magnetic particles (often magnetite, the magnetic oxide of iron) in a fluid. To avoid the heavy and magnetic particles clumping together or precipitating, they need to be coated in a surfactant. And this only works if the particles are the right size (very small) and you have the right mix of surfactant that works in the carrier fluid (the fluid is usually mineral oil or vegetable oil; alcohols will be a poor choice).

Laserprinter toner isn't usually a mix of suspended magnetic particles: only specialty toner inks are (the ones used for machine readable printing). If you can get the right sort, it will save a lot of effort, but the wrong toner won't work at all and will just make a mess.

There are quite a few recipes on places such as Youtube, but some of them don't work well or don't explain some of the important details. So, if you don't want to spend a lot of time experimenting to get the synthesis right, just buy the ferrofluid.


Dilute the feed solution tenfold, etc.  Inelegant.  Try solution engineering,

http://www1.lsbu.ac.uk/water/kosmos.html
Chaotropes and kosmotropes.

http://en.wikipedia.org/wiki/Theta_solvent
Add a little "poor" solvent like ethanol or acetone to ball the polymer, disrupting hydrogen bond crosslinking and reducing chain entanglement.

You want a novel solution not obvious to one skilled in the art.  That qualifies it for patent protection.  A large fraction of the hard answers you need already exist.  They were done by other people doing other things.  Accumulate bone piles of knowledge.  The process for creating ceramic dislocation-pinned nickel superalloys works for Plexiglass, which "dissolves" 30 wt-% copper phthalocyanine as a dye masterbatch.  So it can't be done, so what?  Do it the "other" way.


Le Châtelier’s Principle and Addition of Water

Yes adding more water will affect the position of the equilibrium simply because you are adding more volume to the solution by adding more water. So adding volume is inversely proportional to pressure where the system needs to reduce the amount of water. Then the equilibrium will shift to the side with increased amount of pressure.


As a result of the -I effect of the trifluoromethyl group, triflate is a better nucleofuge than mesylate and tosylate.
As compared to tosylate and nosylate, triflate is rather small. This might be an advantage in reactions at "crowded" centres.
Triflates do not undergo side reactions, such as hydrogen abstraction reactions, etc. A triflate, once introduced, can easily survive several steps in a synthesis.


Failed synthesis of aspirin due to excess sulfuric acid

But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

The short answer is "side reactions".  Without more information about your exact experimental conditions, or even better, data about the obtained yields of the desired product and of byproducts, it isn't possible to explain conclusively what went wrong, but here are some ideas.

Acetyl sulfate formation.  In the comments you you implied that you used anhydrous sulfuric acid.  Under those conditions, acetic anhydride can undergo solvolysis to yield acetyl sulfate and water.  $$\ce{CH3COOOCCH3 + 2H2SO4 <=> 2 CH3COOSO3H + H2O} $$  Acetyl sulfate might be a less active acetyl transfer reagent than acetic anhydride, or the water formed may hydrolyze some of the acetic anhydride or even the formed acetyl sulfate.
Dehydration/sulfation of the starting material instead of acetyl transfer.  If you really did use anhydrous sulfuric acid, it is a powerful dehydrating reagent.  It could be possible that you are sulfating the alcohol or even the acid group (or both?) of the substrate, forming either o-sulfoxybenzoic acid (sulfation at $\ce{-OH}$ group) or sulfosalicylic acid (sulfation at $\ce{-COOH}$ group).

The heat of mixing of concentrated sulfuric acid with water is about $-74~\mathrm{kJ~mol^{-1}}$, judging from Figure 34b of a NIST treatise on the subject.  The heat of reaction for hydrolysis of acetic anhydride is only $-41~\mathrm{kJ~mol^{-1}}$.  Thus under anhydrous conditions, I think sulfated products will be more thermodynamically favored than acetylated products.  Sulfated products will be far more soluble in water or sulfuric acid than would be acetylated products.

I would think that using more water in the reaction would help.  The reason is that the heat of mixing of sulfuric acid is a very strong function of water concentration, but the heat of hydrolysis of acetic anhydride is not.  So raising the concentration of water a little bit will disfavor sulfation significantly by lowering the heat of mixing of sulfuric acid, but will leave the heat of hydrolysis of acetic anhydride the same.  It's worth noting that this argument is purely thermodynamic, not kinetic.  But I don't think sulfation reactions are kinetically hindered under ambient conditions, so although it isn't always true, I think thermodynamic arguments are valid in this case.


Ring Strain and C=O Stretching Frequency

Why does C=O stretching frequency increase with ring strain?

The carbonyl carbon prefers to be (more or less) $\ce{sp^2}$ hybridized.  Imagine taking the two other bonds attached to the carbonyl carbon and bending them towards each other so that the R-C-R angle is forced from 120 to 90 degrees (the same thing as incorporating these bonds into a smaller and smaller ring).  As we do this, the hybridization of the orbitals the carbonyl carbon uses to form these bonds changes from $\ce{sp^2}$ to $\ce{p}$.  As these bonds use more p-character, more s-character remains for use in the C-O sigma bond.  This would strengthen the CO bond and raise the energy required to stretch it (increase its ir stretching frequency).

An alternate explanation can be made based on resonance structures.  Look back at the acylium ion resonance structure used in our earlier acid chloride discussion (resonance structure IV).  As we strain the carbonyl by placing it in a small ring, the bonds between the carbonyl carbon and the adjacent carbons become weaker.  This means that resonance structure IV, with a strong $\ce{C#O}$ triple bond, becomes more important.  Here, in the present case, we would say that resonance structure IV involves carbon-carbon hyperconjugation.

  Why does conjugation decrease C=O ring strain?

Do you mean, "why does conjugation to the carbonyl decrease the ir stretch frequency"?  Look at the resonance structure on the right side below for the conjugated ketone, cyclohexenone.  It's inclusion leads to a weakening of the carbonyl pi bond.


Does tolerance to alcohol affect blood/breath alcohol levels?

Typical roadside 'breath-testing' measures metabolites produced in the metabolism of alcohol (primarily, they measure ethanal, oxidised ethanol). The level measured level of intoxication therefore definitely differs from person to person, and even day-to-day for everyone. Some people have higher metabolic rates, and will metabolise the alochol faster. Some people have greater amounts of alcohol dehydrogenase (enzyme that breaks down alcohol) than others. It follows, that a larger person will likely be able to consume more alcohol compared to a smaller person (compare the concentration of 1 g of salt in a small glass, compared to a large glass)

However, there isn't really any fudge room for arguing against the tests. And most police-forces will back up breath testing with far more rigorous blood tests.


Why does sp hybridised orbital have more directional character than p orbital?

Ignore everything else in image , this image is just for shape comparison.

As you can see in image that so orbital is a lobe with unsymmetrical shape this leads to its more directionality.

P orbital possess equal Electron Density on both sides of the orbital , but in case of sp orbital electron density isn't equal and that's the reason why sp is more directional than p .

That's also the reason for their different Electronegativity.


Why must both the critical temperature and pressure be exceeded to achieve the supercritical phase?

The critical point is a phase transition of phase transitions.  It's a different beast than your everyday run-of-the-mill phase transition like water boiling to vapor or ice melting to water.

The diagram you included is very misleading in its use of color.  The "yellow" supercritical regime doesn't really exist.  Beyond the critical point, gas is indistinguishable from liquid.  So instead of having a special yellow region, the supercritical region should be some gradient of green (liquid) and pink/orange (gas).  The vapor pressure curve between the triple point and critical point is the curve along which liquid is in equilibrium with gas, i.e. those phases have the same molar free energy $G$.  However, they obviously differ in many other properties, including molar enthalpy $H$, molar entropy $S$, molar volume $V$, etc.  Moving along a trajectory in $PT$ space that crosses the vapor pressure curve results in a discontinuity of those values.  Thus, such a phase transition is a first-order phase transition, as first-order derivatives of $G$ (such as $S=\left (\frac{\partial G}{\partial T}\right )_{P}$ and $V=\left (\frac{\partial G}{\partial P}\right )_{T}$) are discontinuous through the transition.

As you move along the vapor pressure curve closer to the critical point, the size of the discontinuous jump in $S$, $H$, $V$, etc. gets smaller and smaller.  Finally, at the critical point, the discontinuity disappears.  So the existence of a first-order phase transition goes away, and no phase transition at all takes its place.

Say you were to start with liquid $\ce{CO2}$ at a $T$ slightly below and $P$ slightly above the critical point, and move through $PT$ space in a counterclockwise cyclic orbit around the critical point.  As you lowered $P$, there would be a distinct first-order phase transition to gas -- the $\ce{CO2}$ would boil -- but then as you reached the bottom of the circle and started increasing $P$ and $T$, moving past the critical point, there would be no phase transition at all as you went to the yellow region in your chart.  There would similarly be no phase transition as you reached the top of the circle and moved back into "green" territory.  You'd get liquid carbon dioxide again, and when you go again through the circular path through $PT$ space, it would boil -- a distinct first-order phase transition -- again.  But it would never condense, in the sense of a sudden change from gas to liquid.  The change back to liquid $\ce{CO2}$ would be gradual, with no discontinuities.

So to specifically answer your questions:

  what is the physical explanation for why it's necessary for both the critical temperature and critical pressure both to be exceeded simultaneously in order for this phase transformation to occur?

The pseudo-fallacy in the question is using the phrase "phase transformation" to describe critical phenomena.  Passing the critical point is not a first-order phase transition like boiling water under normal conditions.  It's a much higher-order phase transition.  Here's a good chapter with more info.

  The fact that increasing pressure on a supercritical fluid can eventually drive it to a solid phase would seem to imply that it's not quite that simple. See, for example, the phase diagram of carbon dioxide.

Think in terms of free energy landscapes.  Solid, liquid, and gas phases are at the bottom of potential wells or pits in this $G$ surface.  Past the critical point, the liquid "well" and the gas "well" are no longer distinct minima, but are the same.  However that doesn't change the fact that if you move "up" in the free energy landscape by increasing $P$, you will eventually find the still-distinct pit (free energy minimum) for solid.


Redox Reaction Equation with elusive oxygen

First of all you need to write equations with real ions and molecules that do exist in solution. Your oxidation agent here is $\ce{BrO3-}$. In neutral media you can use $\ce{H+}$, $\ce{OH-}$ and $\ce{H2O}$. Your half-reaction can be made from this scheme:

$$\ce{BrO3- -> Br-}$$

There is no oxygen in the right part so we need to add $\ce{H2O}$ there, we need 3 molecules.

$$\ce{BrO3- -> Br- + 3H2O}$$

Now there are no H atoms in the left part and 6 H atoms in the right part. Fix this by adding 6 protons, $\ce{H+}$, in the left part:

$$\ce{BrO3- + 6H+ -> Br- + 3H2O}$$

Charge of left part is +5 and charge of right part is -1. Now equalize charges in left and right parts by adding 6 electrons to the left part.

$$\ce{BrO3- + 6H+ + 6e- -> Br- + 3H2O}$$

Half-reaction is ready.


Pressure is defined as force exerted per unit area. Therefore, the diameter of the mercury column is taken into account when we calculate the area of the column.  A mercury column with a large diameter will exert a greater force than a column of the same height, but with a smaller diameter - but the force divided by the area will be the same.


Clarification in a particular application of Le Chatelier principle

For starters, I wouldn't classify your equilibrium as a buffer. A buffer requires excess weak acid, and it also requires either a strong base or the salt of the weak acid (which you don't have). The equilibrium you stated is the dissociation of CH3COOH into its conjugate base and a conjugate acid.

Say you had the following buffer:

$$CH_{3}COOH + NaOH ⇌ CH_{3}COONa + H_{2}O$$

The equilibrium of this would be:

$$CH_{3}COOH ⇌ CH_{3}COO^{-} + H^{+}$$

And the ionic equation would be:

$$H^{+} + OH^{-} → H_{2}O$$

As per definition, a buffer is a system that minimises pH change on small additions of acid or alkali. How? All down to equilibrium...

If you were to increase $[H^{+}]$:

In terms of the equilibrium, the added $H^{+}$ would react with the base, $CH_{3}COO^{-}$, to form
$CH_{3}COOH$. This is Le Chatelier's principle in the fact that increasing $[H^{+}]$ shifts the equilibrium to the left.

If you were to increase $[OH^{-}]$:

In terms of the ionic equation, the added $OH^{-}$ would react with the $H^{+}$ to form $H_{2}O$. This is Le Chatelier's principle in the fact that increasing $[OH^{-}]$ shifts the equilibrium to the right.

In both of these, it's clear that the addition of $H^{+}$ or $OH^{-}$ would just be an acid-base reaction. The products formed minimise the $pH$ of the system due to their relative strengths compared with the added acids/alkalis.

The $pH$ does change though, but just not as much as it would if it wasn't in the buffer system. A buffer just keeps the equilibrium central.


What would the name and formula be for Mo+O+Se?

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

More formally:

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

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


Reaction of aluminium trichloride with water

Aluminum chloride is very soluble in water, according to wikipedia 44-49g/100ml water depending on the temperature, so you shouldn't expect any preciptation. The reaction is very aggresive and exothermic.
Assuming you dont have any buffers or bases in solution when you add $AlCl_3$, the reaction will lower the pH. At a pH lower than 4-ish, $Al^{3+}$ is the most stable cation. Thus, the reaction
$$\ce{AlCl3 + 3H2O -> Al(OH)3 + 3HCl}$$
is pretty accurate. Aluminum hydroxides are decreasingly stable at low pH, as you can see below. Since HCl is formed upon dissolution of $AlCl_3$ you can safely assume that $Al^{3+}$ is the only Al-species present.

!1


The two $\ce{CH2}$ groups are, as you mentioned, in chemically different environments. Thus, each one will couple separately. But within each of the two groups, there are two protons. So you will not get a doublet of doublets, but a triplet of triplets.

Note that for each $\ce{CH2}$ group the hydrogens are chemically equivalent but not magnetically. Therefore, if you include a chiral environment, the pattern will further complexify itself into a higher-order signal that cannot easily be analysed. But since you’re probably measuring in chloroform you need not worry — there is no chiral information present and the couplings will be degenerate.

If I had to label a spin system here, I would probably choose $\ce{A3BB'CC'MM'X3}$.

By the way, it can be assumed that both coupling constants to the two different $\ce{CH2}$ groups be reasonably similar that the overall pattern looks like a quintet without actually being one.


Stability of cyclobutyl methyl carbocation

George Olah and coworkers have examined the cyclobutyldicyclopropylmethyl and the cyclobutylmethyl cation.

(Yes, this is an awfully short answer and I can't really provide more information on the topic.)


Is there a characteristic that characterizes all, and only, metals?

The Universe is full of continuous and smoothly-transitioning properties, yet which we often stubbornly try to assign a discrete yes/no answer in order to make the world easier to cope with. Metallicity is one of them. The Van Arkel-Ketelaar triangle is a diagram which attempts to show the transition between covalent, ionic and metallic types of bonding. Here is one populated by a variety of compounds:

The closer a substance is to one of the vertices, the more that vertex is important in describing the nature of bonding in the material. As you can see, the entire triangle is populated quite uniformly. There are certainly many more materials which would fit in the gaps between the materials shown, and even simply varying the ambient conditions (pressure, temperature, etc), a same substance can be put in different regions of the triangle.

Given this behaviour, it should be clear that any attempt to rigorously define a line dividing substances between metals and non-metals will be fraught with difficulty and likely not useful in the first place. No matter where the line is drawn, there will be some compounds that will look strange for "barely being a metal", while others will be strange for "barely being a non-metal".


Mechanism of protein precipitation with TCA

The paper "Trichloroacetic acid-induced protein precipitation involves the reversible association of a stable partially structured intermediate" (Rajalingam et al. 2009 Protein Science 18(5), 980–993.) may be relevant.

In the introduction, they mention the conventional explanation is that "TCA forces protein to precipitate by sequestering the protein-bound water" - that is, it may force protein precipitation in a related fashion to chaotropic agents or antichaotropic agents like ammonium sulfate.

Their experiments indicate a more complex explanation. They hypothesize  that the negatively charged TCA disrupts electrostatic interactions in the protein, resulting in a partially folded state which is prone to aggregation. The fact that TCA precipitation is less effective for intrinsically disordered proteins lends support to this hypothesis (as they don't have an aggregation-prone partially unfolded state).

The acidic nature of the TCA (the pH effect) helps with the rate of precipitation, but does not appear to be necessary, as the sodium salt of TCA is able to precipitate proteins in neutral solutions, although slower than acidic TCA.


There are a number of ways of looking at Ewald's summation, but one is that instead of trying to solve the original problem where the potential is due to a set of point charges you, instead, solve the system for a set of smeared charges and add a correction. Now there are limitations on what are good ways to do the smearing, but there are still quite a lot of possible choices. The most common is to replace the original delta function charge with a Gaussian, the width of which is determined by alpha in your question. This leads to the error function and exponentials you are used to. However it's not the only possible choice. In equation 22 in the paper you refer to a different way of smearing the charges has been chosen, a linear ramp centred on the particle. This is described in equation 21. As the smearing is different the individual terms in the Ewald summation have different forms when compared to Gaussian smearing, but it's no less valid a way of proceeding.

If you want to learn more about this look into Fourier Transforms and the Convolution theorem. A bit of reading on that and the Ewald Summation in all its forms becomes pretty straightforward.


Can I get ride of dichlorobenzene in honey?

Just theoretically, if you were to heat the honey (primarily fructose and glucose) to 373 K, the boiling point of water, it would speed up the evaporation of the dichlorobenzene (vapor pressure ~8,000 Pa). That might be enough to get rid of the odor, if kept at that temperature for a few days.

That said, the flavor of the honey would likely be destroyed, as well. Even though pasteurization of honey is not normally recommended, that only requires heating to 145° F (336 K) for 30 minutes, and it still changes the flavor.

So toss the honey, or use it as an embalming agent (see also National Geographic). The dichlorobenzene "moth flakes" should make this honey even more effective!


Why isoelectronic species give similar spectral lines?

You are completely forgetting the nuclear charge of isoelectronic series. Their valence shell configuration is the same but the electrons experience a different attraction in each. As you stated

$\ce{Mg^{2+}}$ and $\ce{Al^{3+}}$ give series of similar spectral lines
(this was the question of my exam i.e.NEET)

Yes the spectrum pattern is similar but not identical. Atomic spectrum and that of ionized atoms is like a fingerprint, otherwise identifying the elements in the stars will become difficult and perhaps impossible. There is certainly some overlap in atomic spectrum but not identical as your question title suggests. A quick search of Google shows that this non-sensical question is quite popular in local exams. It is wrong. Isoelectronic species will not have an identical matching spectrum. Even the wavelengths are different.
Search this title in Google Scholar to see the wavelengths of isoelectronic series, where it available free of charge. None match with neutral sodium spectrum.
Wavelengths and levels of the Na I isoelectronic sequence from K IX through Mn XV
Journal of the Optical Society of America Vol. 66, Issue 9, pp. 899-904 (1976)


Why are methanol flames less visible than other flames?

Here's a link to a YouTube video showing a dish of methanol burning next to a dish of ethanol.  The methanol flame is more difficult to see.  Methanol was also widely used as a fuel in the auto racing circuit, until fires that resulted from crashes or fuel filling accidents resulted in "invisible fires" that were difficult to see during daylight, resulting in severe burns to crews and drivers ("you can't fight the fire, if you can't see it", this is one reason why it was replaced by ethanol as a fuel for Indy racing).

Flame color (visibility) is dependent upon the elemental composition of the material being heated and the temperature to which it is heated.  This is because heat will excite electrons in the element(s) to an excited state; when these electrons return to the ground state visible light (flame test) will often be emitted

If we look at the carbon content of methanol and ethanol we find the following weight percents:

methanol - 38% carbon by weight

ethanol - 52% carbon by weight

If we compare the heats of combustion for methanol and ethanol we find:

methanol $\ce{~~}$ -726 kJ/mol (exothermic)

ethanol $\ce{~~}$ -1368 kJ/mol (much more exothermic)

There's just not much carbon there to burn in methanol!  Consequently burning methanol doesn't get that hot and its flame will not be as colored as materials that burn more exothermically and result in higher temperatures.


Strength of intramolecular vs intermolecular hydrogen bonds

It is mainly due to kinetic motion in molecules

The kinetic energy of molecules follows the Boltzmann distribution. Therefore, at room temperature molecules are constantly moving.

Intermolecular interaction, such as intermolecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.

Intramolecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.


Both of these compounds are coordination complexes. The label complex is preferred by many chemists because despite the appearance of some of these compounds as simple ionic compounds, they often display much more complex bonding motifs.

For example in $\ce{Pd(PPh3)2Cl2}$, there are two neutral ligands, $\ce{PPh3}$ that interact with the metal center through dative bonds. There is also evidence that the $\ce{Pd-Cl}$ bond is more covalent than ionic, since palladium forms a complex ion $\ce{PdCl4^-}$ with chlorine.

Palladium acetate man seem like a more clear cut case until you look at the crystal structure. Each acetate anion bridges two palladium, centers in a polymeric chain:

  It is worth noting that IUPAC does not recommend using the term complex to describe these compounds given that complex can refer to a number of other, more loosely bound, species. IUPAC prefers coordination entity, though I have never heard a practicing chemist or seen a textbook use that phrase.


Yes, they are possible; several have even been detected.  They all have the general formula $\ce{H2O_{n}}$ and belong to the class of hydrogen polyoxides.

Of course, the $n=1$ member of the series is water $(\ce{H2O})$, and the $n=2$ member is hydrogen peroxide $(\ce{H2O2})$.

The $n=3$ member is trioxidane $(\ce{H2O3})$, and it has the following structure:

It can be prepared in a number of ways as outlined in the linked article, such as the reaction of ozone and hydrogen peroxide. Trioxidane is also an active antimicrobial ingredient. Trioxidane has a half-life around 16 minutes in organic solvents, but only milliseconds in water.

Hydrogen tetroxide ($n=4$, $\ce{H2O4}$) is also a known compound as is hydrogen pentoxide ($n=5$, $\ce{H2O5}$).  Hydrogen pentoxide is formed as a byproduct in the preparation of trioxidane.

This abstract of a computational study provides bond dissociation energies for the hydrogen polyoxides. Starting from water, they steadily decrease (the $\ce{O-O}$ bonds become weaker) to $\ce{H2O6}$ and then start to increase again.

Abstracts from other computational studies dealing with the expected structures and stabilities of these molecules can be found here and here.


Why does this reaction produce an alcohol in one case and an alkene in the other?

Please also see Why substitution and elimination reactions are favored respectively at lower and higher temperature?
The reaction conditions are the clue you are looking for.

The first thing to recognize is that in any situation where there is competition between substitution and elimination, you will always get a mixture of products.  The question is, which one will dominate.

The effect of temperature is subtle, and sometimes misleading. In practice, organic chemists heat many (maybe even most) reactions, in order to get them to occur in a reasonable timeframe.  When we are teaching about the effect of temperature on the competition between substitution and elimination, we often indicate heating ($\Delta$) is present for the elimination and not the substitution.  You should take this to mean that higher temperature favours elimination.

Now, subtleties aside, let's look at why heating favours elimination.

One difference between the two situations is the relative changes in entropy.  We can see that in the case of elimination, the overall entropy of the system is increased after the reaction (there are more different particles) while in the case of substitution, the entropy change is negligible.

We can consider the competition between the two reaction pathways like this:

$\Delta G_{S_N2} = \Delta H_{S_N2} - T\Delta S_{S_N2}$

and

$\Delta G_{E2} = \Delta H_{E2} - T\Delta S_{E2}$

At low temperature, the enthalpy term ($\Delta H$) dominates, while at higher temperature, the entropy term ($\Delta S$) dominates.

Therefore, the increase in temperature favours the reaction with the larger entropy contribution.


Can I use pure zirconium as an anode in electrolysis?

No, you cannot use zirconium as an anode for electrolytic rust removal.
Zirconium is a member of the family of 'valve metals' (an archaic term), which form a strongly passivating oxide film when exposed to air and/or water.  From here:

Owing to their low electrochemical potential the group IVB and VB valve metals Ti, Zr, Hf, V, Nb and Ta readily react with water or oxygen to form a dense, protecting passive layer.

You're right that these metals, including Zr, tend to be exceptionally chemically resistant (same source):

Because of these protecting oxide films, the valve metals show an exceptional resistance towards corrosion in many aggressive environments. This explains why valve metals are widely used in the construction of chemical apparatus.

However, a feature of these metals is that the resistance of the passivating oxide film increases significantly when an anodic bias is applied to them (same source):

Valve metal oxides are formed conventionally electrochemically ... .

The Wikipedia article on anodizing concurs (emphasis added):

Anodic films are most commonly applied to protect aluminium alloys, although processes also exist for titanium, zinc, magnesium, niobium, zirconium, hafnium, and tantalum.

Thus, were you to construct an electrochemical cell with a zirconium anode and a rusty steel cathode, I expect you would get a brief current transient as the Zr-oxide passive film grew a bit, but then it would tail off to essentially zero.  While I haven't worked specifically with zirconium, the above is the observed behavior of all of the valve metals I have experience with: tantalum, niobium, titanium, and aluminum. I would be quite surprised if zirconium behaved any differently.


The hydrogen ion in acidified potassium manganate (VII)

I think you have known that the permanganate(VII) ion is a strong oxidizing agent because manganese is in the highest +7 oxidation state.

But in different pH conditions, $\ce{MnO_4^-}$ Will be reduced to different product:

In a acidic solution, it is reduced to the colourless manganese(II) ion, $\ce{Mn^{2+}}$.
$\ce{8H^+ + MnO_4^{−} + 5e^{−} → Mn^{2+} + 4H_2O}$
In a strongly basic solution, it is reduced to the green manganate ion, $\ce{MnO_4^{2−}}$.
$\ce{MnO_4^{−} + e^{−} → MnO_4^{2−}}$
In a neutral solution, it is reduced to the brown manganese dioxide $\ce{MnO_2}$.
$\ce{2H_2O + MnO_4^{−} + 3e^{−} → MnO_2 + 4OH^{−}}$

But the more understandable way to say, I think, is that $\ce{MnO_4^-}$ can be reduced to different oxidation states; however, it is an idea about chemical balance. If there is $\ce{H^+}$, the first reaction dominates (since we put in more reactant). While if there is $\ce{OH^-}$, neither reaction 1 nor 3 happens (1: lack of reactant; 3: too many product), so the third dominates.

Hope this helps.

Addition: in those different reactions, there are also different tendency to take place and make $\ce{MnO_4^-}$ reduced. In fact, the first reaction (happening with acidic solution) have a much larger tendency to take place. Thus, if $\ce{H^+}$ presents, $\ce{MnO_4^-}$ will have a larger tendency to oxidate other compounds. Thus some reactions require $\ce{MnO_4^-}$ with $\ce{H^+}$ instead of only $\ce{MnO_4^-}$.


$h\,\nu$ is photon's energy; $h\,\nu _0$ is the potential energy of the electron. All energy excess is converted in kinetic energy of the eletron - that's the left side of the equation.


In sucrose molecule the  glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose.

source :

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_25%3A_Biomolecules%3A_Carbohydrates/25.08_Disaccharides

As seen, in either alpha d-glucose or beta d-fructo furanose may react to give sucrose.

Let us examin.

Alpha D-gluco Pyranose gives an intermediate 1 and 2 by protonation at anomeric carbon.

Similarly  Beta D-fructo furanose gives an intermediate 3 and 4 by protonation at anomeric carbon.

If you examine , structure 3 is more stable then structure 4.

The negative inductive effect of CH2OH in fructofuranose destablizes this intermediate 4.

Hence structure 2 , an intermediate of aplha D-gluco pyranose (more stable) attacks beta D-fructo furanose togive sucrose.

The above diagrams are author drawn in CHEM DRAW


Is Kp and Kc only effected by temperature?

According to LeChatelier's principle on increasing temperature in an exothermic reaction "which is at it's equlibrium initially" the reaction proceeds in backward direction and in endothermic reaction it proceeds in forward direction, this can be thought as in exothermic reactions the heat should be provided to the products to yield the reactants and in endothermic the heat should be supplied to reactants to yield the products!
And the value of equilibrium constants depend only upon temperature and the pressure only effect the equilibrium conditions and not the value of Kp because partial are so adjusted to provide the same value of Kp at the same temperature.


The paper specifically addresses this issue and ascribes it to the tendency for boronic acids to exist as solvent and concentration dependent oligomers:

  The dependence of the intensity ratio for these signals and the
  chemical shifts in the 1H NMR spectra on the concentration of the
  samples under study as well as on the nature of the solvents used
  indicates that anhydrides of boronic acids with different degrees of
  oligomerization can exist along with monomeric forms of these
  derivatives (ArB(OH)2) in solutions of substituted arylboronic acids.
  The conclusions about the existence of oligomeric forms of boronic
  acids and the absence of byproducts in the compounds isolated are
  supported by the results published earlier.

This type of equilibrium can be expressed like:

Higher order oligomers are also possible.


Mechanism of oxidative dearomatisation with hypervalent iodine

The conventional mechanism is as follows:

The phenol displaces one acetate group on iodine – this makes the iodine itself act as what is essentially a fancy leaving group. Nucleophilic attack at either the ortho or para position allows the loss of iodobenzene and another molecule of acetic acid. What I can't explain right now is why it chooses to go para rather than ortho; it probably depends on the exact substitution pattern as I have seen examples of both. A good reference for this is Zhdankin, Hypervalent Iodine Chemistry (2013).


Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride

Turned in this report and my professor commented with his thoughts
$$\ce{Sn + 2RCl \rightarrow R2SnCl2 \rightarrow R2Sn }\text{ (white solid)} \ce{+ SnCl2 \rightarrow  R3SnCl}$$
but I didn't get the question wrong. So unless if someone contributes another answer, I'll accept this as the answer.


Find amount of benzoic acid in dissolution

$\pu{0.037L}$ of water dissolves $0.037\times \pu{1.7g}$ of benzoic acid at $\pu{0^\circ C}$ = $\pu{0.0629g}$.

Therefore $1 - 0.0629 = \pu{0.9371g}$ benzoic acid remains undissolved


Hydrolysis products of chlorides of group 15

Nitrogen trichloride

Nitrogen trichloride is quite unstable and hydrolyse easily:

$$\ce{NCl3 + 3 H2O → NH3 + 3 HOCl~~~~~~~~(1)}$$

$$\ce{NH3 + H2O -> NH4OH~~~~~~~(2)}$$

$$\ce{NCl3 + 4H2O -> NH4OH + HOCl~~~~~~~(1) +(2)}$$

Phosphorus trichloride

$$\ce{PCl3 + 3H2O → H3PO3 + 3HCl}$$

Arsenic trichloride

$$\ce{AsCl3 + 3 H2O → As(OH)3 + 3 HCl~~~~~~~~(1)}$$

This is the first and main reaction of hydrolysis of arsenic chloride to form arsenous acid. But other species (arsenite ions) like $\ce{[AsO(OH)2]−}$ and $\ce{[AsO2(OH)]^2−}$ also exist in solution and are the conjugated bases.

$$\ce{AsCl3 + 4H2O ⇄ H[As(OH)4] + 3HCl~~~~~~~(2)}$$

$$\ce{AsCl3 + 2H2O ⇄ HAsO2 + 3HCl~~~~~~~~~(3)}$$

And at last aresnic trichloride fully hydrolyse to form arsenic(III) oxide

$$\ce{2AsCl3 + 3H2O → As2O3 + 6HCl~~~~~~~~~~(4)}$$

Antimony trichloride

$$\ce{SbCl3 + H2O → SbOCl + 2HCl~~~~~~~~(1)}$$

With more water it forms $\ce{Sb4O5Cl2}$ which on heating to 460° forms $\ce{Sb8O11Cl2}$.

$$\ce{SbCl3 + H2O ⇄ Sb(OH)Cl2 + HCl~~~~~~~(2)}$$

$$\ce{4SbCl3 + 5H2O ->[50 C] Sb4O5Cl2 + 10HCl~~~~~~~(3)}$$

Final reaction to form antimony trioxide.

$$\ce{2SbCl3 + 3H2O ->[\Delta] Sb2O3 + 6HCl~~~~~~~~~~(4)}$$

Note that speculation on the composition of antimony oxychloride has been raised. Some say it is a mixture of antimony trioxide and antimony trichloride.

Bismuth tricloride

$$\ce{BiCl3 + H2O -> BiOCl + 2HCl~~~~~~~~(1)}$$

Main reaction to form bismuth oxychloride. Further hydrolysis yield and intermediate monohydrate $\ce{BiCl3.H2O}$.

$$\ce{BiCl3 + H2O(vap) ->[50 C] BiCl3.H2O~~~~~~~~(2)}$$

$$\ce{$$BiCl3 + 2H2O → Bi(OH)2Cl + 2HCl~~~~~~~~~(3)}$$

Explaination

Now, you may been be wondering that hydrolysis of nitrogen and phosphorus yield one product but that of arsenic, antimony and bismuth yield 3 or 4 products. This is because of presence of vacant d or f orbitals where electrons can reside easily. For this reason, $\ce{Sb^3+, Bi^3+}$ solvate in aqueous solution to form antimonyl or bismuthyl ($\ce{SbO+, BiO+}$) and other species like $\ce{[Bi6O4(OH)4]^6+, [Bi6(OH)12]^6+ ,  [Bi(H2O)9]^3+}$ and similar antimony ions. Other factors also play role like relativistic effects, lanthanoid contraction, inert pair effect etc.


Symantics of parts per (million, billion, etc)

The notation "parts per" is just a dimensionless unit to describe small amounts.
In chemistry it's used for concentrations of dilute liquids.. whereas in physics it can be used for defining change with respect to another quantity and in Six sigma it's used to define "defects per million opportunities".

In chemistry however, $ppm$ is mostly used as concentrations of dilute solutions in water and sometimes used interchangeable with $mg/l$. The notion of "one particle per million particles of solvent" can be used for solid-solid solutions or initial understanding of the term.


Why is hydrobromic acid a stronger acid than hypobromous acid?

The structure of $\ce{HBr}$ and $\ce{HBrO}$ is different, and also the place from where the proton leaves the molecule is also diferent.So, we have to take account of this and also the stability of the conjugate base formed after deprotonation.In $\ce{HBrO, }$ the proton is attached to the $\ce{O}$ atom of the molecule.The molecule can lose proton as, $$\ce{H -O -Br <=> H^+ + (O -Br)^-}$$ The negative charge here mainly resides on the Oxygen, which makes $3$ lone pairs on $\ce{O}$ atom, adjacent to that, there is also electron rich $\ce{Br}$ atom, which also has $3$ lone pairs on it, so there are inter electronic repulsions. For the conjugate base to get stabilisation, the weak $\ce{-I}$ effect of $\ce{Br}$ atom is not at all sufficient. Instead, if the lone pairs can somehow delocalise that will contribute better to stability of the conjugate base. But the lone pair resides on the $\ce{2p}$ orbital of Oxygen, which can't at all effectively overlap with filled$\ce{4p}$ orbital of $\ce{Br}$, and  $\ce{4d}$ orbitals are hugely high in energy. So, due to high electronic repulsion, no chances of delocalisation, and very weak $-I$ effect of $\ce{Br}$ can't make the conjugate base stable at all. So, $\ce{HOBr}$ has high $\ce{pK_a}$.But if you come to $\ce{HBr}$, the bond between $\ce{H}$ and $\ce{Br}$ is very weak. More over if you dissolve it in a sloution, the weak bond is easily broken, and the individual cations and anions can get extensively solvated which makes the  whole process of dissociation thermodynamically very much favourable.That's why, the $\ce{pK_a}$ of $\ce{HBr}$ is very low which means it can dissociate  very easily in a solvent.


Nomenclature of Dihydroxyacetone Phosphate

Because it is the phosphoester of dihydroxyacetone. One of the -OH groups was used in the esterification reaction.

See Here


How can decreasing in temperature indicate an endothermic reaction?

Let’s say that you have a beaker of water. You measure the water’s temperature using a thermometer, and you determine that it is 25 °C. Now you add two reactants to the beaker, and mix them together well until they are completely reacted. If the reaction between these two reactants is endothermic, then the chemicals will absorb heat from the water. If you were to now remeasure the water’s temperature, you might find that it was something like 23 °C. This is because the heat that was absorbed by the chemicals was lost by the water.


Are the words "apparatus" and "equipment" synonyms?

Within a lab they are synonymous. But if you consider in general, apparatus refer to non-machines. Equipment on the other hand can be divided into a couple of sections:

Non-mechanical which can be things like kitchen equipment, measuring cups, knives (all hand tools)
Mechanical machines which are things like sewing machines, mechanical diggers.
Powered machines that are powered by either gas, steam or electricity.


How to calculate the concentration H3O+ in a solution with pH=6.99?

If you take a sample of pure water, there will be few hydroxide and hydronium ions. Of course, they can combine to form water and yes they do combine but there will be few water molecules which break/combine to form the ions again. Hence, there exists a dynamic equilibrium between concentration of ions and water molecules.

$\textrm{pH}$ by definition is the negative logarithm of hydronium ion concentration.

$$\textrm{pH} = -\log [\ce{H^+}] = -\log [\ce{H3O^+}]$$

You can obtain the concentration of H+ ions by substituting the value of pH in the following formula,

$$[\ce{H3O^+}] = 10^{\mathrm{-pH}}.$$

Your attempt 2 is flawed because your assumption that all the ions combine to form water molecules is incorrect. There will always be some concentrations of the ions and all of them needn't combine to produce water molecules. Your attempt 1 is correct.

It appears like you are not aware of the concept of equilibrium and self ionization of water, I have picked few good materials which you might(should) want to refer to,

Chemical Equilibrium

Self Ionization of water

The concept of chemical equilibrium is very important and you will come across it frequently in chemistry, so you must learn it. Also, self-ionization of water along with chemical equilibrium are central concepts for learning acids and bases.


is it possible to store formic acid longer if frozen?

Formic acid is thermodynamically unstable to decomposition to water and carbon monoxide: HCOOH ==> CO + H2O. I assume this is what you are referring to.

A common use is acidification of silage to prevent smells. The fact that it decomposes makes it more environmentally friendly than alternatives. I was not aware that it was used in bee keeping to control varroa parasitic mites.

Lowering the temperature would reduce the decomposition rate from both a thermodynamic and kinetic point of view.

The MSDS I have looked at indicate that the bottle should be vented periodically to release the pressure built up from decomposition. However this one also says that formic acid should not be stored below its freezing point of 8C. http://ionics.ca/wp-content/uploads/MSDS-Formic-Acid.pdf Whether this is just for convenience of use or for reasons or safety I do not know.

To me, the build up of pressure seems far more of an issue than any loss of strength in the acid. In order to lose 50% of its strength, a 1L bottle of formic acid (1200g/46=26mol) would have to lose 13mol (about 312L) of carbon monoxide. That's a lot of venting! I would think a bottle should have a shelf life of 2 or 3 years.

I did my university chemical engineering project on a formic acid plant. It is made as follows:

1. CO + Methanol (20 bar, 80C, potassium methoxide catalyst) == Methyl Formate
2. Methyl Formate + H2O (acidic autocatylisis)== Formic acid + Methanol

From this you might conclude that formic acid is stable to decomposition at 20 bar, but that is not the case, because excess reactants and careful use of distillation are used to push both reactions in the desired direction.

That was a long time ago, but I will see if I can find any relevant data in the decomposition reaction. However you would probably be best off just following the manufacturer's recommendations.

Oh, and keep it in the original glass bottle. Many metals catalyse the decomposition.


Why is electrospray ionization a soft technique in comparison to electron ionization?

The title of the Nobel Lecture of the person who invented electrospray ionization was "Electrospray Wings for Molecular Elephants". By that he meant that this technique is good for very large molecules and it does not break them. The interesting part is nobody fully understands how the big molecules get a charge. There are two models and who can explain it better than John Fenn, the inventor. Have a look at his lecture: https://www.nobelprize.org/prizes/chemistry/2002/fenn/lecture/

Coming to the electron impact ionization, it is a physical contact (don't take it literally) of a very high energy electron beam with the molecule, the energy of this process is sufficient enough to knock out an electron and break other bonds too in the gas phase. This breakage is understood in terms of molecule's cross section. Note that in ESI, no free electrons are used to ionize molecules. The ionization process is a mix of solution phase chemistry (i.e., what ions were present, pH, etc) and the molecule's own properties. Not all can be ionized by ESI. Certain ions will suppress electrospray phenomenon like trifluoroacetic acid.


Chemistry behind lip balm coloring based on skin pH

CI 45410 (A.K.A. Phloxine B or acid red 92) is a pH dependent dye which is clear below about pH=4.1 and pink at pH>4. The formulation is mildly acidic and is neutralised by saliva upon application to become pink.

(Thereby "uniquely creating the perfect shade of pink lips just for you.")


I'd suggest you do it with graphite electrodes.

Because:

Its  more inert than copper
If copper is used, it readily reacts and more copper will be deposited at the cathode compared to silver
the yield of silver will be greatly reduced


Calculating energy level from wavelength

All you need are two equations and two constants (speed of light and Planck constant):
\begin{eqnarray}
\mathrm{E} & = & h\nu\\
h & = & \mathrm{6.62606957 \cdot 10^{−34}   J\cdot s}\\
\mathrm{c} & = & \lambda\nu\\
\mathrm{c} & = & 299792458\ \mathrm{m\cdot s}^{-1}\\
\end{eqnarray}

Using these data, the energy of a photon with $\lambda = 350\ \mathrm{nm}$ calculates to:

$$\mathrm{E} = \frac{6.62606957 \cdot 10^{-34} \cdot 2.99792458 \cdot 10^8}{350 \cdot 10^{-9}} \cdot \frac{\mathrm{J \cdot s \cdot m}}{\mathrm{s \cdot m}} = 0.05676 \cdot 10^{-17} \mathrm{J} = 5.676 \cdot 10^{-19} \mathrm{J}$$

This is the energy difference to the higher level.


Primary carbocation in addition of HBr to ethylene?

There is quite an evidence that carbocations are involved in this process, and that no concerted pathways exist for this transformation.

While the case of ethyne may be difficult to analyse, lets consider 3-methylbut-1-ene:

If you consider the regular Markovnikov's rule to predict the product, you would probably get this:

While in actuality, the major product is actually this:

You can't really explain why that happens without accepting the existence of carbocations and the following mechanism:

This shows that carbocations are involved in the addition of $\ce{HBr}$ to an alkene.

Note: Since this question is about carbocations, I'd like to mention George A. Olah, a reputed chemist who studied carbocations by preparing them in superacid environments, where they were stabilized and studied using spectroscopid techniques. It was his research that led to the conclusion that carbocations have a planar structure. Have a look at his wikipedia page for more information.


How does the radial distribution function of Vanadium differ from that of Calcium and how does this affect the ionic electron configurations?

I think @NicolauSakerNeto mentioned a number of excellent points in his linked answer. Frankly, I think he deserves a bounty.

There are several effects occurring, which give different electron configurations to the transition metals. A very nice review article is available: "The Full Story of the Electron Configurations of the Transition Elements" J. Chem. Educ., 2010, 87 (4), pp 444–448

"d orbital collapse": Because of the difference in angular-momentum "centrifugal force" in quantum mechanics, incomplete shielding of core electrons, and nuclear attraction, the $Z_{eff}$ - effective nuclear charge - for 3d and 4s electrons are different. As a rule of thumb, electrons are typically lost from the 4s first,
electrostatic repulsion in 4s vs. 3d. It's interesting you mention V, since $\ce{V^0}$ is $[3d^3 4s^2]$ and $\ce{V+}$ is $[3d^4]$. The usual explanation is that while 3d is closer to the nucleus and energetically favorable, once you start filling the 3d electrons, the electrostatic repulsion is high, and for the neutral atoms, it's favorable to have one or two 4s electrons, which are more diffuse.

There are other things going on, particularly later in the periodic table, where relativistic and spin-orbit coupling effects occur. Also, many times the electron configuration can change in a free atom or ion compared to a complex.


Find the structural compound from carbon and proton NMR spectrum

Interesting Problem and quite a nailbiter.
I will try to sort some of my ideas, even thought this won't be a full answer.

This looks like a question from an exam. Did you have any reactions during study with hydrochloric acid and propanoyl chloride (acid chloride of propionic acid). So maybe look into different reactions of acid chlorides.
As @MethewMahindaratne mentioned, it is hard without integration but for carbon we dont need integration. We have 7 Carbons, some hydrogen, pretty sure some oxygen and for sure some nitrogen. Why? Because only nitrogen is able to produce odd numbered masses. If we have CH compound, there will be always an even number of hydrogens and carbon has a mass of 12  so overall its even. When we create double bonds or add oxygen, the amount of hydrogen is decreased by 2 so it remains uneven. Only with nitrogen, we can add 1 hydrogen per nitrogen, therefore we have 1 or 3 or 5 etc. nitrogens in it.
The proton g is also quite specific. This high shift in proton NMR usally comes from a carboxylic group. So we have $C_7O_2H_xN_y$.The idea of the carboxylic group is also supported by the C atom at 175 which is a really high shift.
With a molecular mass of 201 g/mol this leaves quite some but not endless options. We can take of 7 x 12 (for the seven carbons) and 2 x 16 (for oxygen carboxylic group) and 1 (for hydrogen carboxylic group) leaving us with a mass of 84 split along oxygen, hydrogen and nitrogen.
This is a lot. On the bottom line it is mentioned, that W consists of certain things. halogenides which usally make compounds heavy are not included there.
Appart from the carboxylic group we are left with 6 carbons which could give the idea of a benzene. But I think it is not benzene because we could only attach $NH_2$ or $OH$ groups onto it and even with 5 $NH_2$ groups we get not enought mass. In addition, aromatic C atoms have high shifts usally 120-150. But we only have 3 others (next to our carbon from the carboxylic group) so not enought for a six membered ring.
But also not enought for just one alkene which I would prefer. Alkene shift ist around 110-140 and fits well.
Now that is all I could provide without crazy speculations.

I can throw in some suggestions but they are realy wild and you need to look in deeper.

The big peek a looks a bit like some alkohols I worked with so maybe a hydroxylic group?
Maybe Carbon Atom u contains 2 or even 3 peaks? (the line looks thicker than the others)= that would contribute to the idea of a substituted benzene ring and provide more carbons so we could add other groups.
Could $H_f$ be two dupletts instead of a quartett so that could come from a double bond?
With integrals you could look at the singuletts and determine wheather it is an amine (singulett with an integrall of 2)
Maybe my whole carbon acid idea is wrong and it is an ester or something.
If we are allowed to include hallogens, then it is pretty sure a benzene with a carboxylic group maybe hydroxylic and a hallogen atom.
If you have more information, maybe you can use the sdbs data base https://sdbs.db.aist.go.jp
I was not able to find an entry that fits.

Hope this brings the project forward.I am very happy to hear, what the compound was when you get a result.