Is it possible to make liquid nitrogen with a bottle of CO2 gas and a freezer? I really want to make liquid nitrogen(or oxygen) without using dry ice, a specially made freezer, or other chemicals.

I have a mixture of some sponges, rotten grass, sugar and water, which makes pressure in the bottle(produces $\ce{CO2}$). Which temperatures can I expect if I freeze it and then quickly opens the bottle after 24 hours of freezing. Is it possible to make liquid nitrogen out of it?

Short answer, no.

The method you are suggesting here is to use Joule-Thompson expansion to cool the air as it escapes and hope that it liquefies. This is a valid way to make liquid nitrogen and is used in both lab-scale and industrial air liquefaction. Unfortunately, you will be many orders of magnitude off from where you need to be. Plans I've seen for liquefying air on a lab scale require starting pressures of around 3000 psi. That air is sent through a long copper tube where it is rapidly expanded through a nozzle into an insulated vessel to atmospheric pressure (14.7 psi). Even then it does not liquefy. The cooled air is vented through another tube concentric with the first tube, where it chills the pressurized gas entering the vessel. My recollection is that if you start with a full-sized cylinder of nitrogen, after about 1/2 the tank is gone, you will start to condense liquid nitrogen and when the tank is empty, you'll be left with about a half-liter of liquid nitrogen. Those numbers are probably off, but the order of magnitude is correct. (Note: if you try this based on these "instructions", you will probably get yourself killed given the pressures involved).

Nitrogen liquefies at 77 K, or -196 C. Dry ice is only −78.5 C, and my freezer at home sits at about -5 C. I think a PET Coke bottle gets pressurized to 80 psi and a soda keg is rated to 120 psi. As you can see, the temperatures, pressures, and volumes you're likely to be able to achieve at home are off by one or more orders of magnitude.

Sorry.


Estimation of change in enthalpy of a gas phase reaction using bond enthalpies Estimate ∆H for the following gas phase reaction:
$$\ce{C2H4 + 3 O2 ⟶ 2 CO2 + 2H2O}$$

I have been having trouble on this problem for a long time, and I was hoping someone would help me with it. I tried using the average bond enthalpies, but that always got me the wrong answer. Below is a table of average bond enthalpies in kJ/mol.

Use Hess' Law:  $\Delta$H = $\Sigma$ Enthalpy of Bonds Broken - $\Sigma$ Enthalpy of Bonds Formed

Reactants

Bond enthalpy of C2H4 = 602 + (413 * 4) = 2254 kJ mol-1
Bond enthalpy of O2 = 3 * 498 = 1494 kJ mol-1
$\therefore$ $\Sigma$ Enthalpy of Bonds Broken= 2254 + 1494 = 3748 kJ mol-1

Products

Bond enthalpy of CO2 = (799 * 2) * 2 = 3196 kJ mol-1
Bond enthalpy of H2O = (463 * 2) * 2 = 1852 kJ mol-1
$\therefore \Sigma$ Enthalpy of Bonds Formed = 3196 + 1852 = 5048 kJ mol-1

Overall

$\Delta$H = 3748 - 5048 = -1300 kJ mol-1

After having a quick Google search for the enthalpy change of combustion for ethene, it came up as -1411 kJ mol-1. Our calculated answer isn't far off. I'm hoping we got the same answer.


What is the difference between a polar molecule and polar bond? I don't understand the difference between a polar molecule and a polar bond. Please explain what the difference between those two is.

A polar bond is one where the charge distribution between the two atoms in the bond is unequal. A polar molecule is one where the charge distribution around the molecule is not symmetric. It results from having polar bonds and also a molecular structure where the bond polarities do not cancel. For example, water has polar bonds and its bent geometry means that there is a non-zero dipole moment. On the other hand, carbon dioxide has two polar carbon–oxygen bonds, but they are such that the polarities cancel out, so there is zero dipole moment.


Why do gases need to be cooled to liquefy/solidify? My book states:

  Gases do not liquify on compression only, although molecules come close to each other and Weak forces operate at a maximum.

Why is it that we cannot just keep on compressing  gas till it solidified or liquified? Why do we need to reduce its temperature?
Also, if that is true, what would happen if I kept compressing a gas, without siphoning out the excess energy?

What your book states is not generally true. Two counter-examples:

Ammonia ($\ce{NH3}$) will liquefy at room temperature and a pressure of approximately $\pu{10bar}$. (CRC Handbook of Chemistry and Physics 44th ed; information as cited on Wikipedia’s data page).
Butane ($\ce{C4H10}$) will liquefy at room temperature and a pressure of just over $\pu{2bar}$. (Same source as cited on Wikipedia’s data page).

Especially the second is well-known to all those who have a butane lighter of transparent material. If it is half-full, a clear distinction between the liquid and gaseous phases can be seen inside at room temperature. Naturally, the inside of the lighter must be under positive pressure for gas to flow out if a flame is required.

On the other hand, there are a great many gases that cannot be liquefied at room temperature by applying pressure. One of these is nitrogen ($\ce{N2}$). In this case, applying enough pressure will make the sample transform directly from the gaseous to a solid phase.

The difference between nitrogen and ammonia/butane is their thermodynamic data, most importantly their critical point. The critical point defines the last position of the liquid/solid separation of a phase diagram. If the temperature is higher than the critical temperature, no liquid phase will be observed under any compression. Likewise, if the pressure is higher than the critical pressure, a substance will not liquefy even under extreme cooling (it will go straight to the solid phase).

The critical points of the compounds mentioned herein are:

$$\begin{array}{lrr}\hline
\text{Compound} & \vartheta_\text{crit} [\mathrm{^\circ C}] & p_\text{crit} [bar] \\
\hline
\ce{NH3} & 132.4 & 112.8\phantom{0} \\
\ce{C4H10} & 152\phantom{.0} & 37.96\\
\ce{N2} & -146.9 & 33.90\\ \hline \end{array}$$

As you can see, the critical pressures of all compounds are well above standard pressure (hence they can be liquefied under reduced temperature). Nitrogen’s critical temperature is below room temperature, so it cannot be liquefied by compression; the others can.


Equivalencies and catalytic hydrogenation Can we actually control catalytic hydrogenation by introducing specific molar quantities of hydrogen gas? My prof insists on no. No we cannot use a certain equivalency of hydrogen gas to stop at a desired product.

Yet in the book I keep seeing examples of specific molar quantities of hydrogen gas being used ...

Using the equipment that I have and working on the scale that I tend to work on (<1 mol typically), I can't deliver only one equivalent of hydrogen gas. We can measure the amount of hydrogen consumed as the reaction proceeds. The problem you have shown could be interpreted that way, i.e. 1 mol is consumed but more may be present.

On industrial scale it may be possible to deliver a more accurate amount of hydrogen. I'll leave it to others that might have more experience on large scale to confirm or deny.

There are alternative sources of hydrogen that can more easily be delivered accurately on small scale. Either ammonium formate or cyclohexadiene can be used to generate hydrogen in the presence of palladium. The former liberates ammonia and carbon dioxide as byproducts, while the latter forms benzene as byproduct.


Boiling point of a liquid increasing with temperature? In my textbook, a statement is given, as follows:

  Boiling point of a liquid increases with increase in temperature

However, I was wondering, isn't this wrong? The boiling point of a liquid always remains the same, it doesn't change.

Hence, my question is:
Is the above statement wrong? If it is, then under what conditions does the boiling point of a liquid change?

It sounds that they actually meant:

  Boiling point of a liquid increases with increase in temperature pressure.


How to calculate the sufficient ratio of acid and corresponding base of the indicator that allows to clearly observe the endpoint of a titration? Ram can see the purple colour of an acidic indicator below pH = 5.5 only and the blue colour only above pH = 8.5. How many times that of acidic form must the concentration of the basic form at least be for him to be able to see the end point of the titration of HCl and NaOH.

Attempt at Solution

Since the indicator is acidic it is of the form $\ce{HIn}$ and dissociates as follows - $$\ce{HIn<=>H^+ + In^-}$$

where $\ce{HIn}$ is acidic form and $\ce{In^-}$ is basic form so we need to find the minimum value of $\frac{[\ce{In-}]}{[\ce{HIn}]}$. The colour change happens at ph 5.5 and 8.5

At the end of point of titration of HCl and NaOH ${pH = 7}$ which automatically falls within the pH range of indicator ( 5.5 - 8.5 ), so i don't know what to do.

Now from the Henderson Hasselbach equation we have, $${pH = pK_{In} + \log \frac{[\ce{In-}]}{[\ce{HIn}]}}$$ So I guess we put 7 in place of pH and then I again do not know what to do now.

Please help me out with this problem.

Answer in my book is 32

If you assume that the $pK_{a}$ of the indicator is 7.0 and you assume that you're going to the end point at pH 8.5 then the Henderson Hasselbach equation

    ${pH = pK_{a} + log \frac{[In^-]}{[HIn]}}$

can be rearranged to:

    $\frac{[HIn]}{[In^-]} = 10^{(pH - pK_{In})} = 31.62+ \approx 32$

However:

(1) There is no real statement in the problem that $pK_{a}$ of the indicator is 7.0 which was assumed from:

    $\dfrac{1}{2}(5.5 + 8.5) = 7.0$

There is little real chemical validity behind this assumption.

(2) There isn't a real justification of why color at pH 8.5 is any better than color at pH 5.5. So the ratio is either 32 or 1/32. Flip a coin :-(


How do I arrive at the α-trimethylsilyloxy ketone from an unmodified ketone using TMSCl, triethylamine and mCPBA? I'm confused about what is going on here. I gather that step 1 is typically supposed to protect an OH group, but there isn't any in the starting material.

In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.

In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:

γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system

By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.

If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.

The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.

Scheme 1: mechanism of the Rubottom oxidation.[1]

Reference:

[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.


Cell Representation of a galvanic cell: I know basic cell representations of galvanic cells and can identify the anode/cathode. But here's a weird representation that I encountered and couldn't understand it:

$$\ce{Pt | H2(g)}(\pu{1.0 atm})\ce{ | HX || KCl(1M) | Hg2Cl2(s) | Hg | Pt}$$

What does this actually translate into?

You will probably know that the notation for galvanic cell reads from left to right as anode || cathode, where || typically denotes a salt bridge.

Within each half cell, different phases are separated by |.

In your case, the anode is a platinum rod in an acidic solution under a hydrogen atmosphere. This is a Standard hydrogen electrode (SHE).

The cathode consists of a platinum rod which is contact with a layer of mercury and a layer of solid mercury(I) chloride. It is an aqueous cell with 1M KCl solution as the medium. This electrode is known as Saturated calomel electrode (SCE).

Both electrodes are used as reference electrodes.


Raoult's Law explained with thermodynamics/ free energy I got some confusion going on with this explanation of Raoult's  Law.

This is what my book say:

  Because the ideal solution has a higher entropy than the pure solvent, the energy of a vapor increases as its molecules expand into a larger volume, that is, if they are at a lower pressure. Thus, solutes in a solution reduce the vapor pressure of the solvent.

Also, I should mention both solutions have the same $\Delta S$ based on the same $\frac{\Delta H}{T}$.

The point is I am confused about the last part of this explanation about how molecules expand. First off, this is all in a closed container, otherwise there's no vapor pressure. If so, the volume should be the same there is no larger or smaller volume.

Basically a material in a mixture will have a lower vapor pressure than if it were pure.  We know that the vapor pressure of a liquid can be determined by:
$$\Delta G = \Delta G_v^o + RT \ln(P_x)$$
Given that $\Delta G$ is $0$ at equilibrium, we get:
$$\Delta G_v^o = \Delta H_v^o -T\Delta S_v^o = -RT\ln(P_x)$$
Being an ideal mixture $\Delta H$ is $0$ and the equilibrium vapor pressure is determined by:
$$-T\Delta S_v^o = -RT\ln(P_x)$$
$$\Delta S_v^o = R\ln(P_x)$$
Meaning that a higher difference in molar entropy ($\Delta S_g^o$) between the phases means there will be a higher equilibrium vapor pressure ($P_x$). Since gases have higher molar entropy ($S_g^o$) the process of vaporization creates an increase in entropy (positive $\Delta S$).  However since a mixture has a higher molar entropy ($S_m^o$) in the initial state than the pure material ($S_p^o$) the phase change to vapor has a lower change in molar entropy:
$$S_g^o>S_m^o>S_p^o$$
and thus:
$$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$
meaning the $\Delta S_g$ will be higher for a pure material than a mixture.

Q.E.D.


Why is sodium bisulfite needed in syn-dihydroxylation? On the reaction of syn-dihydroxylation with stoichiometric quantities of OsO4 (not a catalytic amount) the second step is $\ce{NaHSO3/H2O}$ or $\ce{Na2SO3/H2O}$.

According to my understanding, $\ce{H2O}$ is the only thing needed to hydrolyze the cyclic osmate ester resulting from the first step of adding $\ce{OsO4}$.

So, what is the use of $\ce{NaHSO3}$ or $\ce{Na2SO3}$ here?

You need to reduce remaining oxidizer (OsO4) and maintain pH of the solution. NaHSO3 acts both as the reducing agent and a buffer.


Why $\psi(r_e, r_n) = \sum \psi_e^i(r_e|r_n) \chi^i(r_n)$ in Born-Huang expansion? Given a molecular system, the time-independent Schrodinger Equation (SE) that we want to solve is
$$\hat H \psi(r_e, r_n) = E \psi(r_e, r_n)$$
$\hat H$ is hermitian, so the set of solutions $\psi(r_e, r_n)$ are orthonormal.

Under Born-Oppenheimer approximation, $\{r_e \}$ decouples with $\{ r_n \}$, thus each $\psi(r_e, r_n)$ can be represented as
$$\psi(r_e, r_n) = \psi_e(r_e|r_n) \chi(r_n)$$
in which $\hat H_e(r_n) \psi_e(r_e|r_n) = E_e(r_n) \psi_e(r_e|r_n)$.

My confusions:

In every reference I read, the eigenfunction of $\hat H$ is represented as

$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty}\psi_e^i (r_e|r_n) \chi^i(r_n)$$
in which {$\psi_e^i(r_e|r_n)$} is the set of eigenfunctions of $\hat H_e$

Why an eigenfunction of $\hat H$ can be represented in such a way? For me, this expression is just
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty}\psi^i (r_e, r_n)$$
it makes no sense that an eigenfunction can be expanded in some bases.

After reading this lecture note I have a better understanding.

The reason I'm so confused is that the form
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \psi_e^i(r_e|r_n) \chi^i(r_N)$$
seems to be a linear combination with coefficients that are all $1$. And this is very importand for the derivation of the set of entangled nucleus time-independent SEs since there is no arbitrary coefficients in the set of equations.

After I read the lecture note, here is my understanding:

For a fixed atomic configuration $R^{\alpha}$, the electronic state can be represented by a linear combination of the eigenstates of $\hat H_e(R^{\alpha})$
$$\psi(r_e| R^{\alpha}) = \sum_{i = 1}^{\infty} c_i(R^{\alpha})\psi^{i}_e(r_e|R^{\alpha})$$
although $\psi(r_e| R^{\alpha})$ is an electronic state, this is essentially a solution to $\hat H \psi = E \psi$ at a fixed atomic configuration.

Since by varying $R^{\alpha}$, the set of linear coefficients {$c_i$} will change. So the general $\psi(r_e, r_n)$ can be represented as
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \chi^i(r_N) \psi_e^i(r_e|r_n)$$
in which the set of {$\chi^i(r_n)$} is just the set {$c_i({R^{\alpha}})$}.

Essentially, the $\psi(r_e, r_n)$ that we plug back into $\hat H \psi = E \psi$ is still an electron wavefunction.


How can a decomposition reaction be exothermic? Decomposition reactions involve breaking of bonds which requires energy. Therefore, they are generally endothermic.

But how can decomposition reactions release energy? Examples are the respiration reaction or the decomposition of vegetable matter.

It just can be so.  Look for something that is favored by cooling instead of heating.
In the case of decomposition of something to pure elements that is exceedingly rare, but decomposition involving other compounds offers more possibilities.  During hot processing of steel (in the hot rolled coil, actually) we can get this reaction which is favored by cooling below 570°C:
Wustite --> Magnetite + Iron
$4 \ce{Fe_{1-x}O} \rightarrow \ce{Fe_3O_4}+(1-4x)\ce{Fe}$
$x$ typically 0.05-0.10.
See the phase diagram here noting what happens to the $\ce{Fe_{1-x}O}$ phase on cooling.


Is HF the least acidic hydrogen halide? Fluorine is the most electronegative halogen and therefore, there is larger difference in electronegativity between the atoms of $\ce{HF}$ than any other hydrogen halide, which means the positive charge on hydrogen atom is the greatest in this compound and hence a comparatively small negative charge is needed to attract it. If so, why is it classified as least acidic hydrogen halide if the $\ce{H+}$ is easy to remove?

First of all as @chipbuster says $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim.
It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to fluorine anions with hydrogen bonds, in what is called tight ion pairs. It prevents from detecting the true strength of $\ce{HF}$ with methods like acid–base titration - they show dissociation of ionic pair. Similar effect is present in ion exchange resins.
Strength of this acid is revealed in more concentrated solutions, where $\ce{HF}$ molecules replace hydronium ions, creating bifluoride anions, freeing them - it's homoassotiation mentioned by Wildcat.
More to the point - does it mean that $\ce{HF}$ isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.
In terms of $\ce{K_a}$ in diluted aqueous solutions $\ce{HF}$ is probably still weaker than $\ce{HCl}$ and the rest but getting exact value is greater problem than normally (generally values of $\ce{K_a}$ for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to this question
For concentrated solutions one needs to use Hammett $H_0$ function which gets lower than -10 in pure $\ce{HF}$ (pure sulfuric acid has -12) - it means it's very strong, although not superacidic.
See also:
The Hammett Acidity Function $H_0$ for Hydrofluoric Acid Solutions


Balanced equation for neutralisation reaction betweeen potassium carbonate and hydrochloric acid They seem all correct to me, is there any reason why one reaction would happen over the other(s)?

$$\begin{align}
\ce{K2CO3 + 2 HCl &-> 2 KCl + H2CO3}\\[6pt]
\ce{K2CO3 + 2 HCl &-> 2 KCl + H2O + CO2}\\[6pt]
\ce{K2CO3 +   HCl &-> KHCO3 + H2O}
\end{align}$$

They are all correct and they will all stay in equilibrium with one another, whereas the products of the second equation will be the most favoured, since carbonic acid is an unstable weak acid and will likely dissociate into $\ce{CO_2}$ and $\ce{H_2O}$. At the same time potassium hydrogen carbonate will get protonated easily: $$\ce{HCl + KHCO_3 \rightarrow KCl + H_2CO_3}$$ again resulting in more carbon dioxide and water.


Why is antimony oxide used as a flame retardant? What is the mechanism of action of antimony oxide as a flame retardant? Is this to do with radical absorption?

According to FLAME RETARDANT THEORY:

Antimony oxide by itself is not a fire retardant, and the halogens by themselves, mainly bromine and chlorine, are weak fire retardants. However, when they are combined they become synergistic and are the most effective and most widely used flame retardant system for plastics.

...

Two mechanisms exist in the synergistic system.
First is the "free radical capture" process that takes place in the vapor phase. On combustion at a temperature of over 600 [degrees] F, the halogen forms hydrochloric or hydrobromic acid that reacts with the antimony oxide to form antimony trichloride, antimony oxychloride, antimony tribromide, or antimony oxybromide. The flame retarding action takes place in the vapor stage above the burning material. It is thought that "free radicals propagate" the flame. But, the antimony trihalides or antimony oxyhalides act as "free radical traps", and take up free radicals. They inhibit ignition and pyrolysis in the solid, liquid, and vapor phases.
A second process occurs in a solid phase and is the "char process". The antimony oxide promotes the formation of "char" (essentially carbon) on the substrate which reduces volatile gas formation. The barrier between the substrate and the vapor phase reduces the available oxygen to the underlying substrate. The barrier effect is obtained by almost any inert additive. In plastics there is a cross linking with antimony to produce a more stable thermoset polymer. Additionally in the solid phase, the formation of SbCl3 and SbOCl acts as a dehydrating agent that increases charring. Sometimes phosphorous compounds (TCP), magnesium oxide, alumina trihydrate, molybdic oxide, zinc borate, or zinc oxide are used in combination or in place of antimony oxide to reduce costs, to increase char formation, or to reduce smoke. However, the substitution of the other retardants greatly reduces the flame retardance normally rendered by antimony oxide. Testing of the amounts of the halogen and antimony oxide in each formulation is necessary to optimize the flame retardance and lower costs. Alumina trihydrate is not synergistic with halogenated flame retardants. It functions as a flame retardant by the release of its water of hydration and cannot be used in high temperature processes. Zinc borate, molybdic oxide, zinc oxide, and magnesium oxide can be used in conjunction with antimony oxide to augment char formation and decrease smoke. Replacement of the antimony oxide to meet a smoke requirement compromises the flame retardance.

See also Flame retardant mechanism of antimony trioxide


What is the meaning of writing a catalyst name with two compounds split by a hyphen? In literature, sometimes I see catalysts written with "-" between elements/compounds. Does this mean the first element/compound is supported on the second, or is the second a promoter? (e.g. $\ce{Fe}$-$\ce{KCl}$).

If we consider palladium on carbon or alumina then it means that the palladium particles are dispersed onto the carbon or alumina. The carbon or alumina acts as a support for the catayltic particles of the palladium.

If we consider Lindlar's cataylst this can be written as Pd/CaCO3/Pb

The Pd (palladium) is the hydrogenation cataylist
The CaCO3 is the support
The Pb is a poison which is used to modify the chemistry of the palladium

Sometimes one of the things separated by a / line is either a promotor or a poison.


What's the difference between reactant, excess reactant and limited reactant in a balanced equation? In some reactions, the reactant is mentioned as being "excess". For example, "phosphorus pentoxide forms in excess oxygen" or "pentahalides form with excess halogen".

But since the balanced equation shows exactly how many moles of a reactant are necessary, what is "excess" about the reactant? And how does it contrast with "limited reactant" and just "reactant"? (Ie: What's the difference in a balanced equation between oxygen, limited oxygen and excess oxygen?)

Does each balanced equation need to explicitly say somewhere that this is specifically an excess/limited reactant, or can this be inferred from the stoichiometric coefficients of the balanced equation? Do you need at least two equations side by side to see that one is excess and the other isn't?

Additionally, taking for example the following two reactions

$$\ce{P4 + 3O2 -> P4O6}\\[1em]
\ce{P4 + 5O2 -> P4O10}$$

it seems to me that mathematically on a mole basis, sure we can say that tetraphosphorus hexoxide will form with 3 moles of O2 available (per mole of P4), but tetraphosphorus decoxide needs 5 moles of O2, however, given the huge amount of actual molecules in the reaction, how do the molecules know when there's excess? I mean, why doesn't the phosphorus just form a lot of P4O10 and then just a few molecules of P4O6 with the few leftover O2 molecules, or would this be some kind of unstable situation and the phosphorus molecules are pretty considerate and don't start forming P4O10 until every P4 has at least 3 O2 attached to it?

To sum up, my questions are:

What's the difference between "reactant", "limited reactant" and "excess reactant" in a balanced equation?
Can the fact that the reactant is excess/limited be inferred from the balanced equation?
Why is it that the same reactant (such as P4) proceeds in two different ways depending on whether the second reactant is excess/limited?

Suppose that you want to make a cake. You would first need to go your fridge to check whether you have the necessary ingredients to make one. So you go to your fridge. Luckily, you have all the ingredients: milk, eggs, flour, and frosting. So you decide to make one. You also decide to make the cake as big as possible to avoid the hassle of cleaning/preparing again. Now, what determines how big you can make the cake?

It is the quantity of ingredients of course. But what happens if you  have enough milk to make a 10 cm cake, enough eggs to make a 1000 cm cake, enough flour to make a 1000 cm cake, and enough frosting to make a 10000 cm cake? How big can you make the cake?

You would only be able to make a 10 cm cake. Because you would run out of milk. And since you no longer have milk, you cannot make anymore cake.

Lets go back to your question of excess reagents and limiting reagents. The limiting reagent is the milk in the example above. It is what determines how much of a product you will obtain. It is what limits the amount of product you get (hence the apt name limiting reagent). Everything else are excess reagents: the eggs, the flour, the frosting. They are leftovers; they are in excess.

Now question 2. You cannot tell which reactant will be in excess by just looking at the balanced  equation. Going back to the cake example, the balanced equation is sort of like a recipe for the cake. You cannot tell whether you will have eggs, flour, and frosting left over by just looking at the recipe. You also have to check your fridge. In other words, you have to know what you are starting with.

Now question 3. I think question 3 will be answered in more detail as you study more chemistry. A complete answer involves thermodynamics and statistical mechanics. Nevertheless, going back to the example of the cake: Whether or not you can make cake depends on whether you have all four ingredients. It also depends on how much of each ingredient you have. If you have no milk, you won't be able to make any cake.

But say that you want to make boiled eggs as well. You only need eggs to make boiled eggs so you would could turn all eggs leftover eggs from the cake into boiled eggs. The fact that you can make boiled eggs with the leftover eggs does not change the fact that you can make a 10 cm cake.

This is what the two equations are trying to get at. The major product, will be determined by how much of a certain reactant you have. It doesn't mean that the reaction is solely determined by what it is in excess or not.


What is the major product of the reaction given? I think the substituent of option (a) should attack the meta position as the resultant carbanion gets stabilized by an inductive effect, but no such compound is given. So where am I making a mistake?

Your reasoning is quite correct for an uncatalysed reaction.
I think you have forgotten about the hydrochloric acid you also add (in catalytic amounts). You will therefore protonate the double bond first, with the carbo cation being more stable in the ortho position. The alcohol can now act as a nucleophile and attack at that carbon. Therefore (c) is the correct answer.


Salt Water Metal etching 50 (Euro-)Cent Coin I want to etch the 50 Cent Euro Coin using this following technique that I came across on some DIY site: Immerse the coin and some disposable piece of metal in salt water such that they do not touch and attach current. Cover the non-to-be-etched parts of the coin with nail polish.

The article proposed to use a 9V battery block and used brass in their example. I understand the 50 Cent coin is made out of Nordic Gold: Is this etching method still viable, and does the 9V block deliver enough energy to at least remove the imprinting of the coin? Is there something else to consider (safety isseus)?

Unfortunately, I have no noteworthy background in chemistry. Thanks a lot in advance for any help.

ECM (electrochemical machining) is a safe (besides risk of getting electrocuted: keep your hands dry) way to carve metal. However, only electrically conductive materials can be machined. Both aluminum and copper are conductive, therefore you can carve it safely.


opperating a distillation column far from flooding I know that flooding will ruin the performance of a distillation column. There are many correlations (Eckert, kister, billet, etc.) that will tell you when a column will flood based on your fluid conditions. Most design guides suggest choosing a diameter that will lead 70-80% of the flooding condition. This makes sense as the size of the column (IE, the cost) is as small as possible while still giving you a 20-30% margin of error.

However, what if due to other constraints you want the column to be large so that it is only operating at 10% of the flooding conditions (IE, a much larger diameter than normal). Would this column still operate? Is this type of operation just avoided because it is not economically practical and a smaller column would do the same thing operating closer to flooding?

Using a much too large column means you loose a lot more heat through its walls, meaning it will take longer for the equilibrium temperature/concentration gradient in it to form, and the destillate comes not only later but also considerably slower.

And you'll have to put more power into heating the still, with following danger of decomposition, unwanted side-products, etc.

If you use an expensive column with silvered vacuum insulation, this effect might not be too large even at only 10% of the maximum capacity. I wouldn't want to judge without trying, though.

Education and experience taught me to always use the proper type and size of equipment. It's faster, cheaper, and gives higher yields. I you have a large column that costs thousands, you can sure afford an additional smaller one at a tenth of the price, can't you?


A question about the equality of heat and changes in enthalpy I am wondering how the heat release or absorbed by a chemical reaction can be equal to the change in enthalpy of the reaction at constant pressure if there are changes in the amounts and potentials of each chemical in the reaction mixture. To be more specific, at constant pressure,
$$\mathrm dH = T\,\mathrm dS + \sum_{i} {\mu_i\,\mathrm dN_i}$$
How can we just ignore the summation for something like a chemical reaction?

In the absence of non-PV work (at constant pressure), the change in enthalpy of the system is equal to the heat exchanged between system and surrounding. This follows from the first law of thermodynamics and the definition of enthalpy.
The equation you are showing is based on the definition of the Gibbs energy (the chemical potential is defined via the Gibbs energy, and the Gibbs energy and the internal energy are different by TS). Just because you define some new thermodynamic functions and express the change in enthalpy with it does not invalidate the first law of thermodynamics.

How can we just ignore the summation for something like a chemical reaction?

Here is a silly calculation to model what is going on. Let's start with a variable $a$ known to be five:
$$a = 5$$
Now we add $b$ to both sides:
$$a + b = 5 + b$$
Let's define a new variable to simplify the right hand side:
$$c = 5 + b$$
and substitute that into the equation:
$$a + b = c$$
Now we solve for a:
$$a = c - b$$
It looks like $a$ depends on $b$ and $c$. However, we know $a$ is equal to 5, so it is constant. If you substitute the definition of $c$ back into the bottom equation, you get back to $a = 5$.


Properties of f-orbitals I am not a Chemist, but I took enough undergraduate Chemistry classes to understand the basic properties of s, p and d orbitals and how the behaviour of electrons contributes to different kinds of chemical bonds and elemental/molecular properties.

I stopped before we got to actual quantum mechanics - my maths wouldn't have been good enough to stand it. It still isn't.

However, I was always fascinated as to why none of the lecturers would ever tell us about f-orbitals. When asked, they muttered dark hints that they were significantly different from lower electron shells, and that explaining why was not only a waste of time but might spoil our existing understanding.

I left university 20 years ago, but I was suddenly blindsided by remembering this question out of nowhere. And now I want to know the answer. Google searches reveal a lot of maths that I can't follow. So can someone explain:

What are some of the basic properties of elements with electrons in f-orbitals, and how do they differ from lower orbital positions?
I assume this is related to the fact that most (all?) elements with f-orbitals manifest as metals, so how and why? And why do some elements without f-orbitals display metallic properties?
Why were my tutors so unwilling to try and explain this to non-Chemists who were supposed to stop at their appreciation of lower orbitals and basic covalent bonds?

I think there are multiple reasons contributing.

First of all, the atomic orbitals of the f-block and their properties and interactions are complicated. You have a lot of electrons, and they are far away from the nucleus, on many differently oriented orbital lobes. (well, that is the visualisation at least, in reality atomic orbitals are described by quantum mechanical wavefunctions, and those get more complicated as well for the f orbitals)
Since you have a lot of electrons repelling each other, and on top of that there are relativistic effects coming into play, it is rather difficult to do quantum chemical calculations with these elements to determine their energy levels, etc.
A lot of "rules" (and approximations) valid for low atomic number elements stop working.
Overall, heavy elements have special properties, that cannot be explained by the simple rules and approximations used for light elements. As you have said there is a lot of difficult math in the background. The rules and visualisations are just there to be an  easy to grasp abstraction of the underlying physics. The side effect of this is that you need more and more exceptions as the system gets more complicated.

Second, they are usually rare and expensive elements.(actinides are also radioactive) They have no known biological role, and usually only have few significant uses. So usually one does not encounter them unless one is working in some niche field.


How to calculate concentrations and what is the concept of 'mole'? Say I have a $50~\mathrm{ml}$ solution of a liquid mixture which is $30~\%$ concentrated. Now if I take $20~\mathrm{ml}$ of that liquid out from that mix, what would be the resulting concentration of both the liquid remaining and the liquid extracted? Will the concentration of both still be $30~\%$?

Concentration of a liquid can be given by the formula C=N/V, where V is the volume of the solution and N is the number of moles present in it.

To help you understand exactly what a 'mole' is, think of it like this: when you ad a certain amount of powder to water, say sugar or a fruit drink, how many spoons/ grams of powder you add is the number of moles. (1 mole is actually 6.02214129 * 10^23 particles of the substance)

The concentration can be determined by how much water you add to the drink: If you add more water, the solution is said to be diluted, since the ratio of solute:solvent is lower. So in a drink with 30% concentration, the ratio would be 3:7, 3 moles of powder for 7 moles of water. The concentration can ONLY BE CHANGED if the amount of water changes. The ratio remains same throughout, so concentration would be 30% if you take 20ml, 10ml, or any measurement of volume.

Hope this train of thought helped your mind to get the reasoning :)


Simple explanation of dative bonding I need a simple explanation of dative bonding. I am a student that has learnt simple covalent bonding and ionic bonding, but I need help to learn dative bonding. This if for an urgent assignment, so please help. Thanks.

Alright, I will keep this extremely brief. You know the following things I presume:

Ionic bonding - formed when electrons are transferred from one atom to another, forming a cation and anion.

Covalent bonding - formed when two atoms share an electron pair

Dative covalent bonding - formed when two atoms share an electron pair, but this electron pair comes entirely from one of the atoms

For example, in a Hydrogen to Hydrogen covalent bond:

H-H

They share a pair of electrons, one electron coming from each.

However,in the carbon monoxide molecule, there are two normal covalent bonds and one dative bond. In the dative bond, both electrons are initially from the oxygen, and are then shared between the two.

Hope this helps


When a solution is diluted, the number of moles of the solute decreases. what would happen to the molarity? When a solution is diluted, the number of moles of the solute decreases, so the molarity decreases??

The number of moles of solute does not decrease on dilution.
For example suppose you add 2 moles of NaOH to 100cm3.
Now if you dilute this solution to 1000cm3 you  are still going to have 2 moles of NaOH.
But now the total volume has changed ,hence the concentration also changes.


How to calculate mass of calcium fluoride that will dissolve in sodium fluoride solution? The solubility product of calcium fluoride, $\ce{CaF2},$ is $\pu{1.46E-10 mol^3 dm^{-9}.}$
What mass of calcium fluoride will dissolve in $\pu{500 cm^3}$ of $\pu{0.10 mol dm^{-3}}$ sodium fluoride solution? (Molar mass of $\ce{CaF2}$ is $\pu{78.1 g mol^{-1}}.)$

I tried using the following method:
$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(2x)^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
x &= \pu{3.3E-4 mol dm^{-3}}
\end{align}
$$
Solubility $s$ of $\ce{CaF2}$ can be found as such:
$$s(\ce{CaF2}) = xM(\ce{CaF2}) =  \pu{3.3E-4 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{2.6E-2 g dm^{-3}}.$$
Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be
$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{2.6E-2 g dm^{-3}}\times\pu{500 cm^3} =\pu{1.3E-2 g}.$$
However, my answer differs from the given answer, that is $\pu{5.7E-7 g}.$
How should I approach this question instead?

In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect.

NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial concentration of $\ce{F^-}$ is $\pu{0.10 mol dm^{-3}}.$

$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(2x+\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
\end{align}
$$

The issue  here is that to find $x$ you need to solve a cubic equation…

However, $K_\mathrm{sp}$ of $\ce{CaF2}$ is very small and $x$, that is its solubility, must be much less than $\pu{0.10 mol dm^{-3}}!$ This allows you to make the approximation that $2x + \pu{0.10 mol dm^{-3}} ≈ \pu{0.10 mol dm^{-3}}.$

$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
x &= \pu{1.46E-8 mol dm^{-3}}
\end{align}
$$

Solubility $s$ of $\ce{CaF2}$:

$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{1.46E-8 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{1.14E-6 g dm^{-3}}.$$

Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be

$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{1.14E-6 g dm^{-3}}\times\pu{500 cm^3} =\pu{5.7E-7 g}.$$


What exactly is hydrolysis reaction in organic chemistry?What mechanism does it follow? According to my study material:

  Hydrolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) where water acts as both nucleophile and a solvent molecule.

Is this definition correct? If so how to solve the following problem (picked up from my textbook exercise which has no solution)?

  When alkyl bromides (listed here) were subjected to hydrolysis in a
  mixture of ethanol and water ($80\ \%\ \ce{C2H5OH} / 20~\%\ \ce{H2O}$)
  at $55~\mathrm{^\circ C}$ the rates of reaction showed the following order:
  $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$.

  Provide an explaination for this order of reactivity.

But my doubt is how is this order $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$ possible?

I mean I know the first compound will form a tertiary carbocation. That is okay for $\mathrm{S_N1}$ but how will the second compound undergo $\mathrm{S_N1}$ (because methyl carbocation is very unstable)? Moreover if really $\mathrm{S_N1}$ is taking place the fourth compound should form a more stable carbocation compared to the third compound. These things are confusing me. Is hydrolysis really $\mathrm{S_N1}$ or something else?

According to me the order should have been $(1)>(4)>(3)>(2)$.

Your textbook is almost correct. Here is the corrected version of its statement:

  Hydrolysis is a special type of nucleophilic substitution where water acts as both nucleophile and a solvent molecule.

Note me omitting the $\mathrm{S_N1}$ part. This is because a hydrolysis mechanism can be any nucleophilic substitution, whether $\mathrm{S_N1, S_N2, S_N}$ for heavy atoms or $\mathrm{S_NAr}$.

That also explains the given order of reactivity. tert-Butyl bromide is very fast at $\mathrm{S_N1}$ reactions. Bromomethane very rapidly undergoes $\mathrm{S_N2}$ reactions. The other two are in-between with bromoethane being just a tad better at $\mathrm{S_N2}$ reactions than 2-bromopropane is at $\mathrm{S_N1}$ reactions.

($\mathrm{S_N1}$ is preferred in 2-bromopropane as the solvent is polar protic and it can stabilize the intermediate carbocation well)


Why does opened wine develop the sour taste and smell of vinegar? An opened bottle of wine developed into a sour taste and smell of vinegar. Why?

The ethanol (alcohol content) of the wine, is oxidised, by the oxygen in the air into ethanoic acid. Vinegar is dilute acetic acid which is where it gets its smell.

The reaction happens in to stages,

The ethanol is partially oxidised to ethanal
The ethanal is further oxidised to ethanoic acid


Under which criteria HNO3 is a concentrated or a diluted acid? I would like to buy nitric acid to remove impurities of copper on an aluminum plate. I read that a 10% nitric acid will do the job. However I've also read that concentrated nitric acid is very corrosive to human skin.

So is a 10% nitric acid solution is still very corrosive to human skin. What am I risking if one drop falls on my arm?

Is there any tables that shows the reactivity of acids under various concentrations to human skin?

According to the European regulation on classification, labelling and packaging of substances and mixtures, the classification of nitric acid depends on concentration.

At a concentration of 20 % ≤ c, nitric acid is classified as skin corrosive category 1A. (For a pure substance on the basis of the results of animal testing, this category corresponds to a substance that produces destruction of skin tissue, namely, visible necrosis through the epidermis and into the dermis, in at least 1 of 3 tested animals where responses are noted
following up to 3 minutes exposure and up to 1 hour observation.)

At a concentration of 5 % ≤ c < 20 %, nitric acid is classified as skin corrosive category 1B. (This category corresponds to a substance that produces destruction of skin tissue in at least 1 of 3 tested animals where responses are described following exposure between 3 minutes and 1 hour and observations up to 14 days.)

Furthermore, at a concentration of 65 % ≤ c, nitric acid is classified as oxidising liquid category 3.

Therefore, nitric acid is labelled in accordance with the Globally Harmonized System of Classification and Labelling of Chemicals (GHS) as follows:

Concentration 65 % ≤ c

H272: May intensify fire; oxidiser
H314: Causes severe skin burns and eye damage

Concentration 20 % ≤ c < 65 %

H314: Causes severe skin burns and eye damage

Concentration 5 % ≤ c < 20 %

H314: Causes severe skin burns and eye damage


The influence of lemon juice & sugar on the ph-value If I add equal amounts of lemon juice and sugar into water, does the ph-value remain unchanged, as both cancel each other out?

This question arose in my head as I learned that pouring milk into coffee increases the ph-level of the beverage making it taste less sour (You might want to check out the following video at 5:43: https://www.youtube.com/watch?v=eTYsQUqodeE; it's in German but you will get the overall point, I'm sure).

Now, if I add lemon into my tea, it tastes more acidic, so the ph-value would be more acidic right? If I add additionally some sugar, the tea tastes not sour anymore. So, I wonder if sugar is able to make a acidic drink more alkaline as it tastes less acidic...Or am I comparing apple with pears?

(Sorry, if my question sounds a bit unscientific...)

You're right about the lemon juice and the milk but wrong about the sugar. Lemon juice is weakly acidic (hence its sour taste) and milk is slightly basic (or alkaline, the terms are used interchangeably). A sugar solution is completely neutral since sugar can't take hydrogen ions out of the water or donate them in.

If you don't have background in chemistry, then let's not talk about pH - I thought pH was confusing when I learned it because "lowering the pH" makes a solution more acidic, and "raising the pH" makes it more basic! (Ironically, the pH scale was originally devised to make the concepts of acidity and alkalinity more accessible to the general public)

Your observations about tea probably have nothing to do with acidity or alkalinity since the foods you've described are not significantly acidic or alkaline (except lemon juice). Tea is slightly bitter because of tannins in the leaves. You've evolved to think of things that taste bitter as unappetizing because they often do not have nutritional value and may even be poisonous.

Sugar (and sweet things in general), on the other hand, is very tasty because you've evolved to know that consuming sugars (carbohydrates) will give you energy. Milk also contains sugars, as well as lipids (fats) which give it its creamy taste, which you've also evolved to find appealing because they give you energy - more than simple sugars.

The creamy taste of milk or sweet taste of sugar are usually enough to cover up the bitter taste of tannins (if the tea is brewed right, that is, but that's an issue for another discussion) and make your tea tasty.


Can triethylamine/pyridine be used in the synthesis of Labetalol? As part of my homework, I had to plan the synthesis of labetalol, a blood pressure drug, from its component parts. One of the steps involved was the following SN2 reaction:

The solutions manual just shows the two being added together, but my mind harkened back to previous reactions I had learned in which triethylamine and pyridine were used when $\ce{HCl}$ was being eliminated. Would that be a reasonable thing to do in this instance, or would these amines compete with 4-phenylbutan-2-amine?

I disagree with IT Tsoi's answer.  Triethylamine absolutely will compete, and form ammonium salts.  Those salts may react with 4-phenylbutan-2-amine to lead to product, but more likely they will precipitate.  And be general gunk in the reaction flask.  Pyridine will probably be less of an issue.

In general direct alkylation of amines is a pretty ugly reaction with overalkylation products likely.  A better reaction in this vein would be to make the acetamide and alkylate that with the chloroacetophenone, and then deacetylate, or similar strategy.

If you must alyklate the amine, use Hunig's base (N,N-diisopropylethylamine) which is too sterically hindered to alyklate.  Also, add the electrophile to excess nucleophile, if the reaction is faster this will decrease polylaklyation probabilistically, and the carbonyl alpha to the amine may have enough electron withdrawing character to slow down the increase in nucleophiliciity that you get from a more substituted amine.

Further mucking up this route is the phenol with two carbonyls on the ring, that is probably quite acidic as well, and may be deprotonated with the base scavengers.  Use a phosgene protecting group to cover up both the phenol and the aramide, and this problem will be side-stepped.


How to remove fluorine from 2-[2-(3-bromophenyl)ethenyl]-5-fluoro-1,3-benzoxazole? I tried to develop a synthesis strategy for 2-[(E)-2-(3-bromophenyl)ethenyl]-1,3-benzoxazole below. I was able to find the synthesis plan shown on the picture below, however my target molecule is one without a fluorine on the benzoxazole. Hence, my questions is: would this reaction go the same way if I were to react the benzoxazole without the halogen (F)? If not, is there a way to remove it from the target molecule?
Here's a link to the synthesis plan I currently have:

And here's my target molecule:

edit:
Here's the reagents and the procedure as stated on reaxys:

With  N-benzyl-N,N,N-triethylammonium chloride; sodium hydroxide in  >dichloromethane; water
Inert atmosphere;
2.1. Synthesis
General procedure: The SBOs were prepared by the base-catalysed >condensation of the appropriate 5-halogeno-2-methylbenzoxazole with the >requisite aromatic aldehyde under phase transfer conditions. In a typical >experiment, equimolar quantities (5 mmol) of the starting materials were >dissolved in dichloromethane (20–50 ml) in the presence of >benzyltriethylammonium chloride (3 mmol) and stirred magnetically under a >nitrogen atmosphere as an aqueous solution of sodium hydroxide (50percent, >w/v, 5 ml) was added dropwise over a period of 10 min. After being stirred >for 2–36 h until analytical thin layer chromatography indicated that the >reaction was complete, the mixture was diluted with water (50 ml) and the >SBO was extracted with dichloromethane (3×20 ml), dried (MgSO4), filtered, >evaporated under reduced pressure and recrystallized from aqueous methanol >or ethanol.

How to remove fluorine from this compound?

There's no need to try that!

Just use  the commercially available 2-methyl-1,3-benzoxazole and try to react it with 3-bromobenzaldehyde in N,N-dimethylformamide under reflux for 2-4 h using self-prepared $\ce{KF-Al2O3}$ as a catalyst.

The reaction has been described (DOI) for several benzaldehydes, such as $\ce{4-H, 4-Cl, 4-Me, 4-OCH3, 4-N(CH3)2}$, with yields > 80%.


Placing the atomic mass and number of an element... Why are the atomic and the mass numbers of an element flipped in the textbooks relative to the periodic table?

In the periodic table, the largest number is indeed the atomic number. This is the number of protons. Since this defines the element, it is considered the most important number.

What you probably mean is
$^A_Z\mathrm{Element}$. Here, by convention the lower $Z$ (from the German word Zahl is the atomic number, while $A$ is the sum of protons and neutrons. This defines a particular isotope. This notation is often used in the context of NMR spectroscopy ($^{13}\ce{C}$ NMR), or when radioactive decays are described:

\[\ce{^{14}_6C -> ^{14}_7N} + e^- + \bar\nu_e\]

In summary, I think this is just a set of different conventions which show that in differents contexts either the element itself or a particular isotope and its properties are considered more important.


The distinction between dynamic & static equilibrium How could we be sure that one certain reaction in equilibrium is dynamic? Since there would be no visible change in vessel, which property of the reaction would provide a proof besides color? Which laboratory technique could be used?

My answer starts with the premise that a static equilibrium exists (although that is debatable).

  How could we be sure that one certain reaction in equilibrium is dynamic?

If your reaction is at dynamic equilibrium, changing the temperature or the concentration of one of the reactant or product species (by adding some solute, for example) should disturb the equilibrium. You would see a net reaction, i.e. change in the concentrations of species in solution or in the gas phase. The reason you did not see any changes before is that the forward and reverse reaction were occurring simultaneously at the same rate.

  Since there would be no visible change in vessel, which property of the reaction would provide a proof besides color? Which laboratory technique could be used?

Color would not help either, unless you disturb the equilibrium first. Once you disturb it, any method sensitive to changes in concentration will give you evidence, as will calorimetric methods.

  [Devils advocate question]: What about static reaction equilibrium?

If your reaction is at static equilibrium, changing the temperature or the concentration of one of the reactant or product species, in contrast, would not result in any macroscopic change. For example, if you lower the temperature such that the forward and reverse rate are practically zero, and then add some reactant or product, nothing would happen. Not sure why you would call it a reaction equilibrium though when the reaction never occurs under these conditions. In other words, I know of no reactions where forward and reverse reactions suddenly stop with reactants and products still present under conditions where the reaction proceeded to reach equilibrium.


Virial equation of state - Taylor expansion I don't get how Taylor expansion over $ \frac{1}{\overline{V}} $ of Virial equation of state is caried out: $Z = \frac{P\overline{V}}{RT} $ which yields
$$ Z = 1 + \frac{B_{2V}(T)}{\overline{V}} + \frac{B_{3V}(T)}{\overline{V^2}} + ...
$$ I thought I understood calc2 Taylor expansion of form $$f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k $$ so maybe I was in illusion. What's is $Z$ function of? At what expansion point do we expa

Here, Z is explicitly a function of temperature.

  At what expansion point do we expand?

Can you clarify what you mean? I can state that it's common to truncate this particular expansion at the second term, so we'd have:

$$\ce Z = 1 + {B_{2V}(T)\over \overline{V}}$$

The virial coefficient in this case accounts for deviation from the ideal case by accounting for attractive and repulsive forces between the particles in the system.


Is the shorthand electron configuration of Br- just [Kr]? Is the shorthand electron configuration of $\ce{Br-}$ just $\ce{[Kr]}$?

Also what about $\ce{Ca^2+}$—is it just $\ce{[Ar]}$? And, is $\ce{Na+}$ just $\ce{[Ne]}$?

You could render bromide ion that way.  But the outermost electrons in the "krypton core" are potentially reactive, valence electrons; therefore $\text{[Kr]}$ is not an accurate description of how the electrons really act.  Only the argon core acts like a noble gas, so you are better off rendering the bromide ion as $\text{[Ar]3d}^{10}\text{4s}^2\text{4p}^6$.

Cations stripped to the noble gas core, however, really are stripped to unreactive electrons, so calcium ion can accurately be rendered as $\text{[Ar]}$.


food chemistry maltodextrin vs organic maltodextrin what is the difference between 'normal' food additive maltodextrin and "organic maltodextrin"? Is it only the primary source (i.e.: corn, wheat etc) that is organic or is it a different process to create maltodextrin? And is maltodextrin the same as food additive 1405?

"Food additive 1405" refers to the process for the modification of starch; in this case the process uses enzymes to break up the starch molecules into oligosaccharides between 3 and 20 saccharides.

The difference between regular maltodextrin and 'organic maltodextrin' would be that the source was grown without synthetic pesticides and synthetic fertilizers. 'Organic maltodextrin' may also need to use natural enzymes.


Does ionization energy have anything to do with how reactive an element is? According to my textbook, fluorine has a pretty high (compared to elements like sodium) first ionization energy. But why does it have such a high requirement to ionize? After all, it actively seeks out atoms to bond with, even tearing through already existing bonds in its drive to do so.

I would expect that fluorine would have a very low ionization energy, as it is soooooooo close to having a full octet (just like sodium). Yet unlike sodium, it has a higher first ionization energy. Why is that?

(My guess is that fluorine's protons' affinity for electrons is so great that it simply "overpowers" the octet rule.)

Apparently, neon can also be ionized. Doesn't that mean it can bond with cations? But it is unusually stable, so why doesn't it?

I think there is an underlying law behind this that can explain these, please help!

EDIT: Fluorine has now been explained, but why can neon be ionized? After all, if you give it a positive charge by taking away electrons, won't it be able to bond to anions? But in reality, it doesn't. What is going on?

This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases.

We find that fluorine is at the top right of the periodic table, where the effective nuclear charge is high, and the radius of the atom is small. This means that it has a small condensed positive charge permeating through the inner shells. This causes the reactivity of fluorine to other atoms due to it's electron affinity from the unshielded nucleus. It is hard to ionize compared to sodium because of its smaller than sodium, and it's z eff is higher. These both cause the electron to be attracted to the nucleus more, causing a higher energy barrier to take it away.


Leaving Group Tendencies in Amide Hydrolysis When we do hydrolysis of amides in basic conditions, we come across a mechanism in which the $\ce{OH-}$ ion attacks the carbonyl forming an intermediate as given in the photo:

When the carbonyl gets formed again it ejects off $\ce{H2N-}$-that’s what my teacher says. But I have a problem that why this happens, why not the $\ce{OH-}$ group leaves off when the carbonyl gets formed again? After all, $\ce{OH-}$ is a better leaving group than $\ce{H2N-}$.

You can imagine following situation in this hydrolysis:
$$\ce{R-C(=O)NH2 + OH- <=>[$k_1$][$k_2$] R-C(NH2)(OH)O- \\->[$k_3$] R-C(=O)OH + H2N- ->[$k_4$] R-C(=O)O- + NH3}$$
Keep in mind that:
$$\ce{R-C(=O)NH2 + OH- ->[$k_1$] R-C(NH2)(OH)O- }$$
$$\ce{R-C(NH2)(OH)O-  ->[$k_2$] R-C(=O)NH2 + OH-}$$

In this reaction, $k_1 \gt k_2$ because of LeChateliers principle since $[\ce{OH-}]$ is in excess thus forward reaction is in favor.
Also, $k_2 \gt k_3$ because of the reason given by OP ($\ce{OH-}$ is better leaving group than $\ce{H2N-}$).
Since acid-base reaction is much faster than any other reaction, we can conclude that $k_4 \gt \gt k_3$.

Therefore, as soon as $\ce{R-C(=O)OH + H2N- }$ formed they would convert to $\ce{R-C(=O)O- + NH3}$ quickly so that $\ce{R-C(=O)OH + H2N- }$ can not able to go back to $\ce{ R-C(NH2)(OH)O-}$ intermediate. That means, when boiling with extra basic solutions, amide will be hydrolyze slowly to its corresponding carboxylate.


How to determine the sign of the enthalpy of formation? During the formation of $\ce{Al2O3}$ from 5.4 grams of $\ce{Al}$ and enough oxygen, the temperature of 2 kg of water climbs by 20 degrees. What’s the enthalpy of formation of $\ce{Al2O3}$ (per mole) ?

What I tried:

The balanced reaction:

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

5.4 grams of $\ce{Al}$ are 0.2 moles.
If 2 kg of water climbed by 20 degrees then 40 000 calories were invested during the reaction.
To get one mole of $\ce{Al2O3}$ we need 2 moles of $\ce{Al}$ , therefore the enthalpy of formation (per mole) of $\ce{Al2O3}$ is 400 000 calories. However, the correct answer is −400 000.

Can someone please explain the negative sign?

$\ce{Al2O3}$, upon the formation of its lattice, releases heat. This heats up water. The increase in the temperature is in water, not $\ce{Al2O3}$. We usually refer to the water as the surroundings and $\ce{Al2O3}$ as the thermodynamic system.

  A thermodynamic system is the content of a macroscopic volume in space, along with its walls and surroundings; it undergoes thermodynamic processes according to the principles of thermodynamics. Wikipedia

So, upon the formation of lattice (among with other, mostly, endothermic reactions) a lot of heat is released and that's what you have estimated. Since $\ce{Al2O3}$ is the system, it loses heat. Thus, $\Delta H_f$ is negative.


Why is solid phase peptide synthesis limited to 70 amino acids? I'm reading document about Solid phase protein synthesis (SPPS) from Wikipedia, and according to the document:

  SPPS is limited by yields, and typically peptides and proteins in the
  range of 70 amino acids are pushing the limits of synthetic
  accessibility. Synthetic difficulty also is sequence dependent;
  typically amyloid peptides and proteins are difficult to make. Longer
  lengths can be accessed by using native chemical ligation to couple
  two peptides together with quantitative yields.

I'm a bit not understanding why this synthesis method is limited to 70 amino acids in length. Anyone please help me to explain what is the issue?

The issue is one of yield. Each step has a certain yield percentage of yield, that the synthesizers will maximize as much as possible. However, given that it is fairly impossible to generate 100% yield all the time, the amount of desired protein you get over time will deteriorate exponentially with length.

For example, say that each step gives you 95% yield. If you have a protein that is ten amino acids long, your final step will give you .95^10 yield, which is about 60%.

However, with the same yield per step, but a 70 amino acid chain, the amount of protein you want drops to 2%.

Even if you are at 99% yield per step (a difficult feat), your final yield will be under 50% for a 70 amino acid chain.


Calculating molecular weight of an acid based on given values of mass, volume, pH and pKa This is what I thought was an easy problem but according to a given answer I am doing it wrong:

  $\pu{11 mg}$ of a monoprotic acid was dissolved in $\pu{0.5 l}$ of water. The resulting $\mathrm{pH}$ of said solution was measured to be $5$. What is the molecular weight of said monoprotic acid? $\mathrm{p}K_\mathrm{a} = 6.$ $(t = \pu{25°C})$

My solution

Given values are:

$$m = \pu{0.011 g} \qquad V = \pu{0.5 l} \qquad \mathrm{pH} = 5 \qquad \mathrm{p}K_\mathrm{a} = 6$$

Formula used to calculate molecular weight:

$$M = \frac{m}{n}$$

Formula used to calculate concentration of acid in the solution, that is, the (hopefully correct) formula for calculating the acid equilibrium constant:

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

Formula used to calculate molar concentration:

$$C = \frac{n}{V}$$

Putting them together:

$$M = \frac{m}{\frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}}\cdot V} \label{eqn:alpha}\tag{α}$$

Concentration of $\ce{H+}$ and $\ce{A-}$ based on $\mathrm{pH}$ value:

$$[\ce{H+}] = [\ce{A-}] = 10^{-5}$$

Calculating the acid equilibrium constant from $\mathrm{p}K_\mathrm{a}$:

$$K_\mathrm{a} = 10^{-6}$$

Result:

Entering the above vaules for $m,$ $V,$ $K_\mathrm{a},$ $[\ce{H+}]$ and $[\ce{A-}]$ into equation \eqref{eqn:alpha} yields the molecular weight of $\pu{220 g mol-1}$. This is answer is wrong, supposedly. Did I make any mistakes?

Ok here's what you're doing wrong.

When you write $$C = \frac{n}{V}$$

$C$ is meant to be the concentration of the monoprotic acid dissolved initially, not at equilibrium. Plugging in the equilibrium concentration is disasterous, as it is bound to be different from the actual ratio of amount of substance in moles of $\ce{HA}$ to the volume of solution in liters.

This approach will better suit the problem:

$$\ce{HA \ <=> \ H+ \ + \ A-}$$
$[\ce{HA}]_{\mathrm{initial}}=\frac{m}{M}.\frac{1}{V}$

Assuming $x$ moles/liter to be consumed at equilibrium,

$[\ce{HA}]_{\mathrm{eq}}=\frac{m}{M}.\frac{1}{V}-x$

Now $x$ moles/liter each of $\ce{H+}$ and $\ce{A-}$ would have appeared.

$[\ce{A-}]_{\mathrm{eq}}=[\ce{H+}]_{eq}=x$

$x$ happens to be $10^{-5}$, as $\mathrm{pH}$ is given to be 5.

$K_a$ would be $10^{-6}$, as you have already mentioned.

$$K_a = \ce{ \frac{[H^+]_{eq}[A^-]_{eq}}{[HA]_{eq}}=10^{-6}}$$

Plugging in, I get

$$10^{-6}=\frac{10^{-5} \times 10^{-5}}{\frac{0.011}{M \times 0.5}-10^{-5}}$$

Simplifying a little, I get

$$M=\pu{200 g/mol}$$


Acid catalyst decarboxylation of 1-carboxyl-tryptoline One of my labs had the following reaction:

I understand the Pictet Spengler mechanism. However, I am confused about the decarboxylation. I cannot come up with a way in which this decarboxylation would be acid catalyzed. Here is what I have for the Pictet mechanism:

I have been trying to figure out the acid catalyzed decarboxylation for several days now but I cannot. This is the closest I can come up with, but I'm not sure that this is correct:

Thanks!

Your suggested mechanism should be acceptable (need two more steps, protonation and deprotonation, see last two steps of my scheme below), if you have isolated the carboxylic acid product (compound $\bf{1}$ in my scheme). If compound $\bf{1}$ is not isolable, I suggest following path would be taken to decarboxylation:

Your reaction is an analog to Pictet Spengler reaction (Wikipedia). When the intermediate $\bf{I}$ is formed during the reaction as depicted in the mechanism, 1,2-hydride transfer is possible to give more stable carbocation $\bf{II}$, which is resonance stabilized $(\bf{\ce{II <-> III}})$. The intermediate $\bf{III}$ can easily undergo decarboxylation to give some what stable enamine product, $\bf{2}$. In acidic medium, this enamine $\bf{2}$ would undergo protonation to give the intermediate $\bf{IV}$, which would then deprotonated to gain aromaticity and give thermodynamically stable indole product, $\bf{3}$ (the intended product).


Ability to produce hydrogen gas on reaction with Na/K Out of a given set of compounds, I was asked to identify which would produce hydrogen gas on reaction with Na/K. One of the given compounds was:

I could not detect any acidic hydrogen in it, so concluded that it won't produce hydrogen. But the answer says the opposite.

So my questions are:

Why does the above compound produce hydrogen gas?
What can be the general set of rules to decide which hydrogen shall be acidic enough so that hydrogen gas is produced on reaction with an alkali metal?

First of all, this is not an acid-base reaction. Alkali metals are not bases, rather strong reducing agents. For example, in this site, you may find a excellent explanation. Also remember, metallic $\ce{Li}$, $\ce{Na}$, or $\ce{K}$   does not abstract acidic hydrogen from organic molecules, not necessarily. For example,  metallic $\ce{Na}$ reacts with acetylene ($\mathrm {p}K_a$ of terminal alkyne-$\ce{H}$ is $25$) to give sodium acetylide, but it does not undergo aldol condensation with acetone with more acidic $\ce{H}$, $\mathrm {p}K_a$ of which is $19.3$ (https://owl.oit.umass.edu/departments/OrganicChemistry/appendix/pKaTable.html). The reaction of acetone and metallic $\ce{Na}$ rather undergoes dimerization (e.g., see here). Another example is the reaction of triphenylmethane (the $\mathrm {p}K_a$ of $\ce{H}$ on the central carbon is $33$) and metallic $\ce{K}$ in 1,2-dimethoxyethane, which has given solutions of triphenylmethyl potassium containing various cleavage and reduction products derived from triphenylmethane (Ref.1), but authors stated that no hydrogen was evolved in the course of the preparation.

However, the given compound has potential to undergo redox reaction with alkaline metal and release $\ce{H2}$ gas as a byproduct because the resultant dianion stabilized by resonance and show hint of aromaticity (Ref.2). Tentative mechanism is illustrated in following scheme:

References:

R. O. House, V. Kramar, “The Chemistry of Carbanions. I. The Reaction of Triphenylmethane with Potassium$^1$,” J. Org. Chem. 1962, 27(12), 4146-4149 (DOI: 10.1021/jo01059a007).
G. Boche, H. Etzrodt, M. Marsch, W. Thiel, “The Dianion 1,2,3,4‐Tetraphenylcyclobutadienediide,” Angew. Chem. Int. Ed. Engl. 1982, 21(2), 133-133 (https://doi.org/10.1002/anie.198201331).


How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas? How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas? What are the conditions of that happening? The related underlying chemical reactions are hydrogen and butane combustion in oxygen, right? What is the reaction mechanism? We should avoid using any electrical appliances, electric light, electric door bells and smartphone , just anything that could produce sparks and turn off any flames, right? So I should turn off the electrical appliances immediately after noticing a leakage or should I just keep them on?

Aside of open flame and operating electronic devices, another danger hard to eliminate is static electricity, mostly due synthetic material of clothes, or rubber soles of shoes.
The safest and the most available prevention choice at emergency scenario is intense ventilation. All others are rather hope it would be enough, as it is hard to eliminate or not to forget all possible ignition sources.
Be aware that switching devices off may be more dangerous than keeping them on, as interrupting high voltage  AC circuit usually creates sparks, unless this is addressed by parallel capacitors. This is typical for devices transforming AC to  low voltage, like most low power electronics.

Coal gas is of little use these days, with major fuel components  carbon monoxide and hydrogen.
Much more common is natural gas, based on methane, or LPG, based on butane (typically 80%) and propane.
The oxidant is in all cases aerial oxygen.


What does it mean to say "the 222 space group"? I know there are supposed to be a few rotation axes, that it is non-centrosymmetric, it belongs to the orthorhombic crystal system, and it has four equivalent positions.

What do the 2s mean?

This is a crystallographer’s shorthand for a certain type of space group.

You may have heard of the 14 Bravais lattices, e.g. cubic, monoclinic, orthorhombic. These lattices all assume perfect sphere’s at the corners of the elementary cell and potentially also in certain central positions (if the lattice is not a primitive one). In actual X-ray structure determination, these perfect Bravais lattice spheres are typically not spheres. Instead, we should consider each sphere a ‘hook’ and a molecule hung to that hook. Since not all molecules are symmetric with respect to all symmetry operations a certain Bravais lattice has, the symmetry of the crystal is markedly reduced. For example, an enantiopure chiral compound cannot form crystals that contain planes of symmetry, so even if that compound were to crystallise in the orthorhombic lattice, there would be no planes in spite of them being required in simple orthorhombic crystals.

This reduction in symmetry leads to a large number of lower-symmetry sub-divisions of the Bravais lattices called space groups. Strictly speaking, the Bravais lattices are merely highly symmetrical space groups. There is a total of 219 or 230 — depending on whether you count enantiomeric space groups as two or one.

The shorthand to unambiguously label these space groups basically describes which symmetry operations are present in a certain direction. Orthorhombic is the easiest to understand: the primitive Bravais lattice has:

in $x$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

in $y$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

in $z$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

You could label this is $\mathrm{P\ \frac2m \frac2m \frac2m}$, but it is tiresome to always include the ‘2 over m’ part. Hence, it is shortened to $\mathrm{P\ mmm}$, the shorthand being ‘m replaces $\frac2m$’. If there is no plane of symmetry, then only the two-fold axis remains and that dimension is given a 2 label instead. Thus, $\mathrm{P\ 222}$ is the same lattice but there are no planes of symmetry in the actual crystal structure. The crystal could also include a screw axis, in which case the symbol $2_1$ would be used, or a glide plane in which a different letter, depending on the orientation of the glide plane, is used.


Copper (II) chloride reduction with Hydrogen gas I've been looking into making pure copper (I) chloride, but using ascorbic acid is relatively expensive for me. I'd also like to avoid having water in any part of the synthesis to minimize the contamination with copper (II) chloride. Therefore, I came up with the idea of doing the synthesis in ethanol and bubbling through H2 according to the reaction

$$\ce{2 CuCl2 (aq) + H2 (g) -> 2 CuCl (s) + 2 HCl (g)}$$

So my main question is: could this work?

There are better ways to do it.  Copper (II) can be reduced to copper (I) with a variety of reagents, especially in the presence of a soft-base halide ion such as chloride which precipitates or complexes with the copper (I) (which is a good way to prevent "overshooting" to metallic copper).  Some specific approaches can be seen here.


Why is temperature regulation an important factor for producing explosives? I was wondering why temperature regulation is an important factor for producing explosives. I know many are exothermic reactions but what would happen if you did not use an ice bath and allow the temperature to go too high (Considering the explosive substance has not already been formed in the reaction)? Would it explode? For example when mixing acetone with sulfuric acid and hydrogen peroxide to make TATP.

My advice on TATP is simple:

If you're an inexperienced amateur researcher, do the synthesis on paper - and only there!

Actually, the synthesis on TATP can be performed at temperatures above 0 °C. When Richard Wolffenstein accidently found in 1895 during his works on coniine that TATP had been formed in a mixture of 10% hydrogen peroxide and acetone, he did not use any acidic catalyst and had kept the mixture at room temperature for weeks:

  Ueber die Einwirkung von Wasserstoffsuperoxyd auf Aceton und Mesityloxyd, Ber. dtsch. chem. Ges., 1895, 28, 2265-2269 (DOI)

When he had collected the precipitate and recrystallized it from diethyl ether, it exploded, which Wolffenstein attributed to to the formation of peroxides in ether.

One of the first reports on the synthesis of TATP under acidic conditions was given by Milas and Golubovic:

  Studies in Organic Peroxides. XXVI. Organic Peroxides Derived from Acetone and Hydrogen Peroxide,
  J. Am. Chem. Soc., 1959, 81, 6461-6462 (DOI)

Btw, the experimental part of this article starts with

  These peroxides are dangerously explosive and should be handled with extreme care.

Under controlled lab conditions, the decomposition of TATP in solution seemingly has been examined:

  J.C. Oxley, J.L. Smith and H. Chen,
  Decomposition of a Multi-Peroxidic Compound: Triacetone Triperoxide (TATP), Propellants, Explosives, Pyrotechnics, 2002, 27, 209-216

and the activation energy for this first-order reaction was calculated to be around $\mathrm{150\,kJ\cdot mol^{-1}}$.

What makes TATP different from most of the other known explosives is

the molecule (obviously) does not contain nitro groups. Typical detectors for explosives thus fail here. Together with the simplicity of the synthesis, this is a real problem in law enforcement!
the explosion does not generate a lot of heat, but is an entropic explosion.

The latter was througly examined by Ehud Keinan and coworkers:

  Decomposition of Triacetone Triperoxide Is an Entropic Explosion, J. Am. Chem. Soc., 2005, 127, 1146-1159. (DOI)

If you don't want to go though all their calculations of a possible decomposition mechanism, the following hopefully is impressive enough:

  The approximate pressure increase, assuming an
  infinitely rapid conversion that occurs in the volume of 1 mol
  of solid 1 (=TATP), is of the order of 100 bar. The associated energy
  content is approximately 2500 kcal/mol. This is the origin of
  the explosive power of solid 1.

I'm aware that I probably haven't fully anwered your question, but as a take-home message I can only suggest again to not attempt any experiment with this peroxide!


How to assign oxidation number of sulfur in thiocyanate anion? Recently I was trying to balance a redox chemical equation:

$$\ce{CuSCN + KIO3 + HCl <=> CuSO4 + KCl + HCN + ICl + H2O}$$

The equation is in principle simple to solve; one element ($\ce{I}$) is reduced and two elements ($\ce{Cu}$, $\ce{S}$) are oxidized. The number of exchanged electrons in both cases must be the same. However, my problem is that I don't know how to assign the oxidation number to sulfur in a thiocyanate anion.

Later, I solved the equation using the matrix method and via obtained reaction coefficients ($4$, $7$, $14$, $4$, $7$, $4$, $7$, $5$) and concluded that the oxidation state of sulfur in thiocyanate is zero. This result is very surprising, because the oxidation number zero in heteroatomic molecules is not common.

My questions are:

How to assign the oxidation number in such cases?
How is it possible that oxidation number of an atom bound to a more eletropositive atom is zero?

In the end the oxidation state of sulfur does not really matter for balancing equations.  Rather, identifying the real oxidizing and reducing agents is.

Clearly the iodine in the iodate ion is the oxidizing agent, dropping from $+5$ to $+1$.  But copper, sulfur, and even carbon are all being oxidized or reduced from the cuprous thiocyanate.   We have multiple parts of the $\ce{CuSCN}$ participating in the electron exchange.

Why not then identify the whole $\ce{CuSCN}$ molecule as the reducing agent, thus covering all the reducing agent atoms in their proper proportions?  With this approach you see that the sum of oxidation states for this reductant goes from $0$ to $+7$ ($+2$ for $\ce{Cu}$ in the copper sulfate, $+6$ for the sulfur in the same compound, and $-1$ for $\ce{CN}$ in the hydrogen cyanide).  Then $\ce{CuSCN}$ gives $7$ electrons, iodate takes $4$ electrons, and thus you get $\ce{4 CuSCN + 7 KIO3}$, just like with the matrix method.


Hydrolysis of Sucrose over Sn1Ac or Sn2Ac In "March's Advanced Organic Chemistry" I have read about the general mechanism of the acetal hydrolysis. It names the acid-catalyzed SN1 or SN2 as a possible mechanism for a Acetal-hydrolysis.

Now I am interested to know if SN1 or SN2 is mainly going on at the Hydrolysis of Sucrose to Glucose and Fructose. So I compared the different parameters which influence SN1/SN2.

Polarity of the solvent: Water -> really polar -> SN1 preferred.
Stabilization of a possible Carbokation-Intermediat: Primary (Anomeric C!), so not really stabilized by Hyperconjunction -> SN2 preferred. (Is there Hyperconjunction also in the ether-O?)
Mesomeric stabilization There is also a mesomeric stabilization between the resulting carbocation and a Oxonium-Ion in the SN1-Mechanism.
Steric Hindrance: In my opinion, through the planar region around the anomeric C (planar ether), there is no steric hindrance -> SN2 possible.
Temperature: high temperature -> SN1 preferred.

So to conclude: Many different aspects. Because of the main points "steric hindrance" and the destabilized primary carbocation intermediate I would conclude that the hydrolysis of Sucrose goes mainly (maybe a little bit over SN1) over SN2.

Have I forgotten something important? Are my assumptions "No steric hindrance" and primary carbocation-intermediate correct?

I have attached my possible SN1-mechanism.

The mechanism supplied in the question statement is wrong according to this paper, where the evidence favored Carbon-Oxygen bond cleavage on the fructosyl-side instead of the glucosyl-side of the bridge Oxygen. As that reference states, "the initial products of the hydrolysis are a-D-glucose and a fructosyl carboxonium ion"

But don't feel bad, many textbooks have shown the same mechanism (detailed in the Discussion section of the paper). It's possible that the reaction mechanism changes with conditions, but I know of no evidence for that.

However, the same paper discusses that the mechanism was thought to depend on the type of sugar undergoing bond cleavage:

Results from a computational study show that SN1 mechanisms via a carbocation on fructose and glucose are both feasible, with the one on fructose slightly more likely.

Some other points:

1) The question statement refers to "primary" carbocations twice, and so do three of the comments. But in this reaction the 2 alternatives are a secondary or tertiary carboxoniun ion (for this terminology see Olah (1998) Onium Ions or ref 11 of the above paper).

2) It makes sense that an SN1-type mechanism would take place on the fructosyl-side of the bridge Oxygen since the resulting carboxoniun is tertiary rather than secondary.

Now, the real question is: How would you experimentally determine the mechanism? That is where the science comes in. The fact that so many textbook authors have presented this classic reaction with different mechanisms should be a big lesson to all students.  That expert authors would go to the extent of drawing detailed reaction mechanisms in a textbook that are not backed up by experimental reality, shows how mistrustful we should be of any non-experiment-based opinions. And this is for a classic, textbook example.  How much more unreliable "expert opinions" are for much more complicated things!


Does concentration remain constant in an electrochemical cell's half cells? We have from Nernst equation:

$\mathrm{E_{cell} = E^\circ - \dfrac{RT}{nF} \ln Q_c}$

But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing?  In that case, won't that lead to a decreasing emf? If it does, then what's the use of such a cell that yields decreasing emf?

But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing?  In that case, won't that lead to a decreasing emf?

Have you ever wondered how battery meters on phones, laptops and other electronics can tell how charged your battery is? The question you just asked is the answer. As energy from the battery is consumed, the voltage will drop which can be correlated to a percent charge if known.

  If it does, then what's the use of such a cell that yields decreasing emf?

These cell are still useful.  Astute engineers know this and design around it using batteries that provide more voltage than needed when charged. If I have a $\pu{3.3 V}$ circuit, I can use a $\pu{3.7 V}$ Lithium battery (note $\pu{3.7V}$ is the charged voltage) to run the device without damaging it and will continue to work until the battery declines to below $\pu{3.3V}$ (actual/rated voltage). Similarly car batteries are typically rated for $\pu{12V}$ but if you measure it with a galvanometer (voltmeter) you will find the voltage is on the order of $\pu{13-14V}$.


How is the time for a test tube in a centrifuge determined? I came across some hairy math that's used for analytical centrifugation, mostly for things like determining the sedimentation rate for liquids.

However, what I don't understand is how someone determines how long a test tube should be inside a centrifuge? Is it based off of others experimental data or is there a fixed way of doing it?

The short answer: check out equation 5 below

I will explain the whole story in terms of sedimenting particles, but you can easily read this as sedimenting droplets as well.

Necessary condition for sedimentation

If you want particles to sediment the first thing to check is whether their density is higher than the density of the liquid, if not: don't even bother. This can be easily shown on the basis of a force balance with buoyancy and gravity acting on the particles:
$$\tag{1} F_{net}=(\rho_{p}-\rho_{l})V_{p}g $$
where subscripts $p$ and $l$ stand for particle and liquid respectively.
What I should also mention here is that I neglect Brownian motion which can, for very small $V_p$ become a significant effect that avoids sedimentation even when the density of the particles is above that of the liquid.

For now, let's assume the particles are big enough to ignore Brownian motion. If we have a centrifuge it is typical to specify the centrifugal acceleration in terms of $a=Ng$ where $N$ is a multiplication factor dependent on the rotation speed of your centrifuge and the distance of the test tube to the center of rotation ($a=\frac{v^2}{r}$, see the wiki for details). The net force acting on the particles in a centrifuge is thus:
$$\tag{2} F_{net}=(\rho_{p}-\rho_{l})V_{p}Ng $$

Settling velocity

After the initial acceleration of the particles there will quickly be a force balance between the net force introduced by the centrifuge and the drag force on the particle. For a perfectly spherical particle that doesn't settle to fast this is simply the Stokes drag:
$$\tag{3}F_{drag}=6\pi\mu_l R_p v_s$$
where $\mu_l$ is the liquid viscosity, $R_p$ is the particle radius and $v_s$ is the settling velocity of the particle.

Balancing the drag with the centripetal force we obtain:
$$\tag{4} v_s=\frac{2(\rho_p-\rho_l)R_p^2Ng}{9\mu} $$.

Finally, to determine the settling time we need the liquid height in the test tube $h$ so we find $$\tag{5} t_s=\frac{h}{v_s} =  \frac{9\mu h}{2(\rho_p-\rho_l)R_p^2Ng } $$
Note that with respect to the settling height we take the worst case scenario here because we want all particles (also those close to the surface of the liquid) to settle.

Equation 5 nicely shows the influence of various parameters on the time a test tube has to stay in the centrifuge. Smaller particles means longer times, lower density difference also and obviously higher centrifugal acceleration means shorter times. Plugging in some typical values for $1 \mu$m particles in water $(\Delta\rho=100$kg/m$^3)$ and a strong centrifuge ($N=10^4$) we find that we need $O(10 s)$.

Discussion of the simplifications

Obviously there are quite some assumptions in the derivation above.

Brownian motion might not be negligible therefore either increasing the settling time, or avoiding settling altogether.
Particles need to accelerate to their settling velocity first, this will also increase the settling time a bit.
In general particles will not be spherical. If, for example, the particles are more rod-shaped then the orientation will determine whether the settling time increases or decreases with respect to the spherical particle prediction.

For many types of particles a Svedberg coefficient is tabulated in literature which will allow you to calculate the settling time without having to assume sphericity of the particles. In this document they indicate the use of Svedberg coefficients.

Clearly, these simplifications can mean that the prediction of equation 5 is of by a factor 10 (or maybe even 100 if you're unlucky) therefore in most practical situations you would use this equation as a starting point for some trial-and-error experiments. You can also use it to determine what will happen to the settling time when you e.g. run a new experiment with a different liquid or particle type.


How to make sense of the 'Relaxation Method' for measuring reaction kinetics? I get the concept that one can measure the rate of reaction by measuring the rate at which an equilibrium state reaches another equilibrium state after the system is perturbed (usually by heating). However, I do have an issue with it. Consider the reaction $$\ce{A <=> B}$$

My lecture handout suggests that the rate of change of $[\ce{A}]$ is given by:

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]+k_\text{reverse}[B]$$

However, surely it's just $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]$. The equation in the lecture handout would make sense at equilibrium because that is when $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=0$ but I think it seems to insinuate that this is always valid (without explicitly saying so). Am I right or is that equation always valid? If it is: why?

You have two simultaneous reactions,

$$\begin{array}{rcl}
\ce{A} &\xrightarrow{k_1}& \ce{B}\\[6pt]
\ce{B} &\xrightarrow{k_{-1}}& \ce{A}
\end{array}$$

and $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ equals $-k_1 [\ce{A}]$ for the first reaction alone and $k_{-1} [\ce{B}]$ for the second alone. To get the total $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ for the reactions running simultaneously, you have to add the contributions from each reaction:

$$\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_1 [\ce{A}] + k_{-1} [\ce{B}]$$

This will always be true; at equilibrium you'll have $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = 0$ and

$$K = \frac{k_1}{k_{-1}} = \frac{[\ce{B}]}{[\ce{A}]}$$

If you did it your way, this wouldn't be the case.


Why does white phosphorous exhibit chemiluminescence? My work - I only know that chemiluminescence means glowing in dark.

From here:

  The best-known and first well-documented example of deliberately
  induced, man-made chemiluminescence was the discovery of elemental
  phosphorus by a German alchemist H. Brand. The waxy white phosphorus
  is slowly oxidized at room temperature giving off faint greenish glow.
  The formation of excited reactive intermediate $\ce{(PO)2}$ and
  $\ce{HPO}$ is responsible for the green glow.

That's where the term phosphorescence come from.


Crystallization Process In my lab, my teacher asked me to not cover the beaker or crucible during the crystallization process of different solutions (e.g, FeSO4, NaCl etc). But on the internet, on Youtube, I've come across videos where they suggested that the glassware containing the solution should be covered but loosely.

I accidentally covered my solution with a watchglass and later found crystals formed all around the crucible, inner and outer both surfaces.

So, what's the reason behind it? I think we need to let the water to evaporate and leave the crucible, but why?

Usually you leave a reaction uncovered to induce crystallization, because this allows the solvent to evaporate. As solvent evaporates, the solubility of the product decreases, causing it to exit solution and begin to nucleate crystals. Once nucleation has occurred, crystals precipitate more readily.

Whether partial or totally uncovered isn't always important depending on how volatile the solvent is or how saturated the solution is. Partially covered will slow the rate, especially if the solvent isn't volatile, but it can block loose particles from falling in. On the other hand, a saturated solution doesn't need much evaporation to go below the solubility limit of your product and instigate nucleation. Whatever your case may be, a watchglass is not a sturdy seal against evaporation so it seems like your product was still able to crystallize.

Also, depending on the heat of the reaction and melting point of the product, sometimes evaporation isn't necessary at all, but it's still convenient to allow as much solvent as possible to evaporate in order to isolate dry crystals. If your solvent dries over night then you don't have to remove it the next day, so as a rule of thumb a reaction product is left uncovered unless there's something air/moisture sensitive present.


When you fill a bath, Is it different that you fill it directly or non-directly? Yesterday I was taking a shower that this thought occurred to me that when you are filling a bath, flask, etc with water in two different ways in specified time, Will the amount of water in bath be different?

I did a experiment myself in 'Crocodile Chemistry' and got these results as seen below:

First level was pouring water from a faucet directly to the bath from a specified height for 10 seconds. The result was that approximately 250 cubic centimeter water was poured into it.

Second level was pouring water from a faucet to a 'already filled beaker' and filling the bath with overflowed water of the beaker at the same height and the same time. The result was similar to the first level.

I saw all these but I'm not sure that if they're completely accurate, So I wanted to ask somebody that has experience in this manner: Do these experiments show the right results or it's different in the real world?

If the beaker is at or below the outlet of the faucet, the two values should be nearly identical.  There are very small effects, like water sticking to the outside of the beaker as it overflows or the surface of the water changing shape constantly (splashing out more at some times and less at others as a result.)  Those are insignificant for practical purposes.

If the top of the beaker is above the outlet of the faucet, that is, the faucet is submerged in the beaker, then there will be a small addition to the height to which the water must be pumped.  It's only centimeters, but the pressure increase and flow decrease is entirely measurable and would have a small but significant and predictable effect over the course of running a bath.


What are the effects of radio waves on matter? Different regions of the electromagnetic spectrum interact differently with matter:

Microwave radiation can induce molecular rotation.
Infrared radiation can induce molecular vibration.
Visible and ultraviolet light can induce electron excitation.
X-rays and gamma rays can induce ionization.

Is there any activity specific only to radio waves? I kept on thinking how radio waves are used in NMR but I don't know what to call it.

This is a bit of a stretch, but the Zeeman and Stark effects are linked to radio frequencies, in that line splitting of a visible spectral frequency (~10^15 Hz) can give rise to a beat-frequency in the RF. Your answer of NMR, though, is more applicable. The most general term for precession in a magnetic field is Larmor precession.