Does copper(III) nitrate exist? I was balancing this reaction when I found an unusual compound had been formed.
$$\ce{Cu(s) + HNO3(aq)-> Cu^{3+}(aq) + NO(g)}$$

My attempt:

Balancing oxidation number:
$$\ce{Cu -> Cu^{3+} +3e^-}$$
$$\ce{3e^- + NO_3- -> NO}$$
Balancing hydrogen and oxygen atoms:
$$\ce{3e^- + 4H^+ + NO3- -> NO + 2H2O}$$
Adding $\ce{3NO3-}$ on both sides to form compounds:
$$\ce{3e^- + 4H+ + 4NO3- -> NO + 2H2O + 3NO3-}$$
Adding both half reactions
$$\ce{Cu + 4H+ + 4NO3- -> NO + 2H2O + Cu^{3+} + 3NO3-}$$
$$\ce{Cu + 4HNO3 -> NO + 2H_2O + Cu(NO3)3}$$

So does this compound $\ce{Cu(NO3)3}$ actually exist? or have I done a mistake.

Copper(III) nitrate cannot be obtained from aqueous nitric acid, and likely doesn't exist.
Reaction carried under strongly oxidative conditions between $\ce{Cu(NO3)2}$ and fuming $\ce{HNO3}$ yields in nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]},$ [1, 2] sometimes written as adduct $\ce{Cu(NO3)2 · N2O4},$ which is contradictory to the crystal structure consisting of copper(II) centers octahedrally coordinated by nitrate groups and the voids in cross-linking mesh of octahedra are occupied by nitrosyl ions [3]:

Figure 1. Crystal structure of nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]}$ (ICSD-422033). Color code: $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.
According to Wiberg's Anorganische Chemie [4, p. 1707], oxidation of $\ce{Cu^2+}$ to $\ce{Cu^3+}$ is possible only with strong oxidation agents $(E^\circ(\ce{Cu^2+/CuO+}) = \pu{+1.8 V}),$ and copper(III) is highly unstable in water and has a tendency to oxidize the medium $(E^\circ(\ce{H2O/O2}) = \pu{+1.23 V}).$
The reduction potential or $\ce{Cu^3+}$ can, however, be reduced down to $\pu{+0.45 V}$ by coordinating it with appropriate strongly donating ligands (which nitrate really isn't).
Examples of such complexes are:

$\ce{K3CuF6}$ (oxidant: fluorine $\ce{F2}$; $\ce{CuF3}$ is unknown);
$\ce{K5[Cu(IO6H)2]},$ $\ce{K5[Cu(TeO6H2)2]},$ $\ce{Na5[Cu(TeO6H2)2]}$ (oxidant: peroxodisulfate $\ce{S2O8^2-}$ in strongly alkaline medium);
$\ce{Na[Cu(OH)4]},$ $\ce{KCuO2},$ $\ce{Ba(CuO2)2}$ (oxidant: hypobromite $\ce{BrO-}$/oxygen $\ce{O2}$; $\ce{Cu2O3}$ is unknown).
Cu(III) complex of 2-ethoxy-5,10,15,20-tetrapentafluorophenyl-3,7-diaza-21,22-dicarbaporphyrin (Cu(III)-$\ce{N2CP}$, a doubly N-confused porphyrin) [5]
A variety of Cu(III) peptide complexes [6, pp. 213–223]

References

Znamenkov, K. O.; Morozov, I. V.; Troyanov, S. I. Synthesis and Crystal Structure of Copper(II) Nitrato Complexes $\ce{NO[Cu(NO3)3]},$ $\ce{Na2[Cu(NO3)4]},$ and $\ce{Ag2[Cu(NO3)4]}.$ Russian Journal of Inorganic Chemistry 2004, 49 (2), 172–179. (in Russian)
Gagelmann, S.; Rieß, K.; Wickleder, M. S. Metal Oxidation with $\ce{N2O5}$: The Nitrosylium Nitrates $\ce{(NO)Cu(NO3)3},$ $\ce{(NO)2[Zn(NO3)4]}$ and $\ce{(NO)6[Ni4(NO3)12](NO3)2·(HNO3)}.$ Eur. J. Inorg. Chem. 2011, 2011 (33), 5160–5166. https://doi.org/10.1002/ejic.201100565.
Blackwell, L. J.; King, T. J.; Morris, A. Crystal Structure of the Copper Nitrate–Dinitrogen Tetroxide Adduct $\ce{Cu(NO3)2 · N2O4}.$ J. Chem. Soc., Chem. Commun. 1973, No. 17, 644–644. https://doi.org/10.1039/C39730000644.
Wiberg, E.; Wiberg, N.; Holleman, A. F. Anorganische Chemie. Nebengruppenelemente, Lanthanoide, Actinoide, Transactinoide, 103. Auflage.; De Gruyter: Berlin; Boston, 2017; Vol. 2. ISBN 978-3-11-049573-7. (in German)
Furuta, H.; Maeda, H.; Osuka, A. Doubly N-Confused Porphyrin:  A New Complexing Agent Capable of Stabilizing Higher Oxidation States. J. Am. Chem. Soc. 2000, 122 (5), 803–807. https://doi.org/10.1021/ja992679g.
Karlin, K. D.; Tyeklár, Z. Bioinorganic Chemistry of Copper; Springer Netherlands: Dordrecht, 1993.


Does dilithium exist? Lithium has two electrons in $\mathrm{1s}$, filling the first energy level; and one valence electron in $\mathrm{2s}$. Could two $\ce{Li}$ atoms form a bond to create $\ce{Li2}$; like $\ce{O2}$ or $\ce{C2}$?

The dilithium dimer exists- but it isn’t particularly common, and we usually only see it in the gas phase.

Essentially, the 2s electrons interact and form a bonding orbital. (s-p mixing is particularly pronounced here, but I presume you don’t know MO Theory.)

Due to the diffuse nature of lithium’s atomic orbitals, the bonding is pretty weak (Wikipedia gives a binding energy of $102\text{ kJ/mol}$, about one fourth that for a hydrogen molecule); and the inter nuclear distance between the 2 lithium atoms is actually quite large.


When is the product a compound and when is it 2 ions in these reactions? When finding the half equation and overall redox equation for the following:

a) $\ce{Fe^2+(aq) + Cl2(g) <=>}$
b) $\ce{N2(g) + H2(g)  <=>}$

The solution from the textbook is given below. I do not understand why the products (in the overall reaction) for a) are 2 individual ions (and not $\ce{FeCl3}$ whilst the product for b) is a compound (and not $\ce{2N3- + 6H+}$).

One difference I noted was, when the electrons are balanced for a) it gives $\ce{2Fe3+}$ and $\ce{2Cl-}$ , which cannot go together unless its $\ce{2Fe3+}$ and $\ce{6Cl-}$, whilst for b), when the electrons are balanced we have $\ce{6H+}$ and $\ce{N2}$ which combine as is to form $\ce{2NH3}$. Is this the reason?

The solution given is:

a)
Oxidation: $\ce{2Fe^2+(s) -> 2Fe^3+ + 2e- }$
Reduction: $\ce{Cl2(g) + 2e- -> 2Cl- }$
Overall: $\ce{2Fe^2+ (aq) + Cl2(g) <=> 2Fe^3+ + 2Cl- }$

b)
Oxidation: $\ce{3H2 -> 6H+ + 6e-}$
Reduction: $\ce{N2 + 6e-  -> 2N^3-}$
Overall: $\ce{N2(g) + 3H2(g)  <=> 2NH3}$

While FeCl3 is a possible product, the reaction in A is aqueous. Wikipedia shows that is readily dissolves in water (92 g/100 mL @ 20C). The products exist as ions, and are written as such.

I would suspect you're not supposed to have solubility constants memorized though, so in general if it is in water they will exist as ions.

B is a gaseous reaction and the product is a compound, not an ion set like A.


Find the minimum temperature at which the reaction will become spontaneous This question came in my entrance exam and I think 273 K should be the answer because the reaction is already spontaneous. I just want to know if I've attempted it correctly.

For a certain reaction $ΔG^\circ = \pu{-45 kJ mol-1}$ and $ΔH^\circ = \pu{-90 kJ mol-1}$ at $\pu{0 °C}$. What is the minimum temperature at which the reaction will become spontaneous, assuming that $ΔH^\circ$ and $ΔS^\circ$ are independent of temperature?
(a) $\pu{273 K}$
(b) $\pu{298 K}$
(c) $\pu{546 K}$
(d) $\pu{596 K}$

You are only partially correct: negative Gibbs free energy at $\pu{273 K}$ is a sign of a spontaneous reaction.
However, the question asks when the reaction will become spontaneous and not to select the minimal temperature value from the multiple choices list which corresponds to the spontaneous process.

So, you were supposed to find the equilibrium temperature, e.g. when $ΔG = 0$.
You have correctly started with the equation for the free Gibbs energy:

$$ΔG = ΔH - TΔS = 0  \quad\implies\quad T_\mathrm{eq} = \frac{ΔH}{ΔS}$$

Unknown entropy change $ΔS$ can be determined from the Gibbs free energy $ΔG_1$ at initial temperature $T_1$:

$$ΔS = \frac{ΔH - ΔG_1}{T_1}$$

Finally,

$$T_\mathrm{eq} = \frac{ΔH}{ΔS} = \frac{ΔH}{ΔH - ΔG_1}T_1 = \frac{\pu{-90 kJ mol-1}}{\pu{-90 kJ mol-1}-(\pu{-45 kJ mol-1})}\cdot\pu{273 K} = \pu{546 K}$$


Why do puffed rice stick to wet hands? My mom gave puffed rice to eat, I had quite wet hands. I saw that, each puffed rice pieces started hanging to the hand, when I brought my hand in contact with them. I thought about electrostatics initially to be the cause. Actually, the electrostatics usually deals with rough surfaces. Here due to hand being wet, it has got smooth surface. Then, what might be reason? I started thinking, suddenly I thought it to be because of hydrophillic nature of kadale puri (puffed rice). Is it really because of hydrophillic nature of puffed rice?

This thesis, (Texture and Hydration of Expanded Rice, Clive Norton, September 1998), has a vast amount of information on rice, including puffed rice.

Rice is mostly starch, which when heated is gelatinised. This causes "the breaking of hydrogen bonds between poly-α-1,4-glucan chains, leading to loss of crystallinity" allowing those hydrogen bonding sites to absorb more water. The loss of crstallinity also leads to air pockets making puffed rice sponge like. The starches become "interspersed with gas pockets" which are filled by water through capillary action.

So, puffing makes the starch more hydrophillic, exposing hydrogen bonding sites, and the air pockets act as capillaries. The puffing also increases the ratio of surface area to weight. These factors cause the rice to be liftable with a "water chain".


Periodic trend in difference of energy between the s and p orbitals Why does the difference of energy between the 2s and 2p orbitals of the second period elements increase with increasing atomic number? Does this difference increases by moving down a group, e.g. is the difference between 3s and 3p bigger than 2p and 2p and why?

  Atomic Orbital Ionization Energies, eV
  $$
\begin{array}{lccccc}
\hline
\text{Atom} & \mathrm{1s} & \mathrm{2s} & \mathrm{2p} & \mathrm{3s} & \mathrm{3p} \\
\hline
\ce{H} & 13.6 & & & & \\
\ce{He} & 24.5 & & & & \\
\ce{Li} & & 5.45 & & & \\
\ce{Be} & & 9.30 & & & \\
\ce{B} & & 14.0 & 8.30 & & \\
\ce{C} & & 19.5 & 10.7 & & \\
\ce{N} & & 25.5 & 13.1 & & \\
\ce{O} & & 32.3 & 15.9 & & \\
\ce{F} & & 46.4 & 18.7 & & \\
\ce{Ne} & & 48.5 & 21.5 & & \\
\ce{Na} & & & & 5.21 & \\
\ce{Mg} & & & & 7.68 & \\
\ce{Al} & & & & 11.3 & 5.95 \\
\ce{Si} & & & & 15.0 & 7.81 \\
\ce{P} & & & & 18.7 & 10.2 \\
\ce{S} & & & & 20.7 & 11.7 \\
\ce{Cl} & & & & 25.3 & 13.8 \\
\ce{Ar} & & & & 29.2 & 15.9 \\
\hline
\end{array}
$$

Data source: www.colby.edu

In naive terms I think this could be explained as follows:

$2s$ orbital is lower in energy than $2p$ orbitals because it is closer to the nucleus, and thus, experiences a bigger attractive force to the positive nucleus. The so-called effective nuclear charge is bigger for an electron on $2s$ orbital than on one of $2p$ orbitals since in the later case there will be an additional shielding from electrons on $2s$ orbital.
As the nuclear charge increases, both $2s$ orbital and $2p$ orbitals become lower in energy due to the increased attractive force to the nucleus, but the $2s$ orbital is affected to a larger degree for the same reason: it is closer to the nucleus and experiences a bigger gain in attractive force to it.

Thus, the energy gap between $2s$ orbital and $2p$ orbitals becomes larger as the atomic number increases.

That is all what they can tell you in school. But, of course, the situation is more complicated, and i don't think it can be addressed without doing electronic structure calculations. Yes, effective nuclear charge experienced by an electron is often expressed as $Z_{\mathrm{eff}}= Z - s$, where $Z$ is the real nuclear charge and $s$ is the so-called screening constant; $s$ is different for electrons at orbitals corresponding to different values of the principal quantum number $n$ and there are some empirical Slater's rules which provide numerical values of $s$ for different values of $n$. But these picture is "not quite right".

First, as I already pointed out and as one indeed can see from the Wikipedia page linked above, $s_{2s} = s_{2p}$ (same value of $n=2$). But surely if one calculate these screening constant by solving the corresponding Schrödinger equation he/she will indeed found that $s_{2s} \neq s_{2p}$. But even without solving the Schrödinger equation the naive argument given above in (1) favour this inequality. So, $s$ is not just a function of $n$, but also of $l$.

Secondly, screening constants in Slater's rules does not change their values as atomic number increases which is also wrong. Loosely speaking, the more nuclear charge electron can screen, the more will it screen. Below is the related graph from the article with screening constant calculated at the Hartree-Fock level linked above.

So, $s$ is also a function of $Z$, and besides for different $n$ and $l$ it depends on $Z$ in different ways. The "real" picture (in quotes since orbitals aren't real anyway) as we can see is much more complicated, and, honestly speaking, I do not even believe that there is a trend you are looking an explanation for.


Internal Energy Definition, Why is dU zero when dT is zero? According to my Equation sheet, $$\delta U=\left(\frac{\delta U}{\delta T}\right)_V\,\delta T+\left(\frac{\delta U}{\delta V}\right)_T\,\delta V=C_V\,\delta T-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$$ Where $\delta U$ is change in internal energy.

So one of the homework solutions said that for a system with a constant Temperature, $\delta T=0$ $\Delta U=0$. And I remember one of the TAs saying that $\Delta U=C_V\,\delta T$ and since $\delta T=0$ the value goes to zero, which makes sense. But she didn't even include the second term: $-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$.

I was trying to make sense of the derivative but I'm really having trouble with the second term. The small letter $U$ outside the parenthesis is supposed to mean constant $U$.

Why am I allowed to ignore the second term?

The equation for enthalpy is very similar, and I have the same problem:

$$\delta H=\left(\frac{\delta H}{\delta T}\right)_p\,\delta T+\left(\frac {\delta H}{\delta H}\right)_T\,\delta H=C_p\,\delta T-C_p\left(\frac{\delta T}{\delta p}\right)_H\,\delta p$$

The second term in each of your two equations can be expressed solely in terms of the equation of state for the solid, liquid, or gas:$$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
In the case of an ideal gas, the right hand sides of both these equations are equal to zero.


Why are the given compounds in a homologous series? Why are the given compounds in one homologous series?

$$\ce{CH3-OH}$$
  $$\ce{CH3-CH2-OH}$$
  $$\ce{CH3-CH2-CH2-OH}$$
  $$\ce{CH3-CH2-CH2-CH2-OH}$$

I know that homologous series can be represented by a common general formula and have the same chain length. But I could not answer this question. Can anyone help me?

Because they can all be described by the general formula $\ce{CH3-(CH2)_n-OH}$. The only varying parameter in this case is the number of middle $-\ce{CH2}- $ groups.

So while the the number of middle $-\ce{CH2}- $ groups may vary $\left(n = 0,1,2,\ldots\right)$ they all fit in this general formula.

The formula $\ce{CH3-(CH2)_n-OH}$ tells you that:

The chain starts with a $\ce{CH3}-$ group
The chain ends with an $-\ce{OH}$ group
Contains a variable number, $n$, of $-\ce{CH2}-$ groups in between.


Hyphenation of chemical names When a word has to be split broken into multiple lines, it is almost always hyphenated. Chemical names can be considered words, or at least they consist of words. So the hyphenation should apply to them as well. However, hyphens often are integral parts of systematic chemical names. Therefore confusion might arise, whether such a hyphen emerged from hyphenation or if it is "mandatory". Are there any official rules for the (possible) hyphenation of the chemical names? I could not find any, yet. I am sure some journals could have it mentioned in their style manuals, but I am primarily interested whether this topic is mentioned by IUPAC.

In fact, I am undecided whether to represent inserted optional breaks in long chemical names (that have to fit into some fixed-width infobox) HTML(5) representation as &shy; or <wbr>, which however means hyphenation, and non-hyphenation, respectively.

IUPAC project 2014-003-2-800 addresses this issue. However there seems to be no officially released recommendation to date.


Equivalence point vs inflection point in the titration curve of a weak acid Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions coming directly from the base is equal to the amount of hydronium ions due to the acid, here we have the equivalence point.

When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.

In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?

In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point.

The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount equivalent to the originally present weak acid. At this point,  $[\ce{H3O+}]<[\ce{OH-}]$, so $\mathrm{pH} \gt 7$.

  When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.

For the approximative experimental chemical approach, it is true.

But mathematically, even this is not exactly true, as the curve is slightly asymmetric. When there is the same molar amount of hydroxide as was the acid, the solution volume is 3 times bigger. And hydroxide concentration 3 times lower.

It would be symmetric, if we constructed the graph for the constant total volume and complementing strong acid and hydroxide solution volumes, as shown in the formula below.

$$\mathrm{pH}=f\left(\frac{V_\mathrm{acid}}{V_\mathrm{acid} +  V_\mathrm{hydroxide}}\right)$$

And the concentrations of both should be the same.

  In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?

For weak acids, the coincidence is mathematically even less true, as the curve shape before and after equivalence differs.

But for the chemical experimental point of view, the difference is still negligible, as the equivalence/inflection coincidence error is much less then experimental errors.

Both inflection and equivalence points ( approximately identical ) lay in the alkalic region.

  In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

For weak acid/strong base ( and vice versa ) titrations, the neutral $\mathrm{pH}$ does not play any special role, so this point on the curve is insignificant.

Neither is there at the neutral point any inflection point, unless by chance $\mathrm{p}K_\mathrm{a}=7$. In such a case, the inflection point  at the maximal acid/salt buffering capacity at the half equivalence would be at $\mathrm{p}K_\mathrm{a}=7$.

For further reference, see e.g  Titration_of_a_Weak_Acid_with_a_Strong_Base


H0 value for fluoroantimonic acid I am confused about $\mathrm{H_0}$ value of fluoroantimonic acid. Wikipedia says $\mathrm{-23<H_0<-21}$. Others say around $-31$. There's a huge gap. I won't worry for $3-4$ fluctuations but worry for $10$.

Can anyone provide confirmed data?

Superacids consisting of a mixture are going to have a certain range for the Hammett acidity function $\mathrm{H_0}$. For example, $\mathrm{H_0}$ for both Magic Acid $\ce{HSO3F-SbF5}$ and fluoroantimonic acid $\ce{HF-SbF5}$ relies on concentrations defined by the equilibrium of the acid's formation, which, in turn, is affected by how much $\ce{SbF5}$ has been added [1]:

  $$
\begin{align}
\ce{2 HF + 2 SbF5 &<=> H2F+ + Sb2F11-}\\
\ce{2 HSO3F + 2 SbF5 &<=> H2SO3F+ + Sb2F10(SO3F)-}
\end{align}
$$

  The acidity function of $\ce{HSO3F}$ increases from $-15.6$ to $-21.0$ on addition of $25$ mole percent $\ce{SbF5}$ ... as shown in Fig. 1. Extrapolation of the $\ce{HSO3F-SbF5}$ curve in Fig. 1 would lead to an $\mathrm{H_0}$ value of about $-25$ for Magic Acid. Fluoroantimonic acid is even stronger. As shown in Fig. 1, with $4$ mole percent $\ce{SbF5}$ the $\mathrm{H_0}$ value for $\ce{HF-SbF5}$ is already $-21.0$, a thousand times stronger than the value for fluorosulfuric acid with the same $\ce{SbF5}$ concentration. At present it is difficult to estimate the acidity of $1:1$ $\ce{HF-SbF5}$, but a value of $-28$ can be predicted ... on the basis of isomerization kinetics data.

  Fig. 1. Relative acidities of $\ce{HF}$ and $\ce{HSO3F}$ on addition of $\ce{SbF5}$.

That is why most textbooks (e.g. [2]) list $\mathrm{H_0}$ from $-21$ to $-25$ for Magic Acid and $\mathrm{H_0}$ from $-21$ to $-28$ for fluoroantimonic acid.

References

Olah, G. A.; Prakash, G. K. S.; Sommer, J. Superacids. Science 1979, 206 (4414), 13–20. https://doi.org/10.1126/science.206.4414.13.
Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, Fifth edition.; Pearson: Boston, 2014. ISBN 978-0-321-81105-9.


Importance of Overall Reaction Order All my texts make it a point that in a differential rate law, the sum of the exponents on the reactant concentrations add up to the overall reaction order. However, I have yet to see any use in this. So you have two reactants, each undergoing a first-order reaction, and the overall reaction is second-order. So what? We analyze reactions for being first-order, second-order, and zero-order (at my level anyway), but that hasn't to do with the overall reaction order—just the reaction order pertaining to each individual reactant.

So my question is simple: what is the point of the overall reaction order? Why is it important?

This answer is mostly a compilation of the excellent comments by Spontification and Chet Miller.

Differential rate law

The overall reaction order tells you what happens to the rate when you change all concentrations by the same factor at the same time. For example, if you decrease all concentrations by a factor 2 and the overall order is 1, the reaction will be slower by a factor of 2. If the order is 2, slower by a factor of $2^2 = 4$, if the order is 3, slower by a factor of $2^3 = 8$ and so on (overall reaction orders higher than 3 are unusual for elementary reactions). Dilution by a factor 2 would be one way of lowering concentrations; however, not all species will decrease by a factor of 2 (solvent, hydronium concentration in a buffer, any other species buffered by a fast equilibrium).

Integrated rate law

Integrated rate laws describe how the rate changes (decreases) over time. All species concentrations that are part of the differential rate law will change over time unless they are catalysts. If one species concentration is very high compared to others (e.g. the solvent) then its change will be negligible. So the integrated rate law depends on more than the differential rate law.

However, if all of the species in the differential rate law are reactants, and these reactants are present at stoichiometric ratios, the overall reaction order is directly related to the integrated rate law (e.g. overall second order reaction gives second-order integrated rate law). If not, you can get, for example, a pseudo-first order reaction even though the overall order of reaction is two.


Lewis structure for SO4 (2+) This was a question on a review sheet for Lewis structures.

Now I could draw $\ce{SO4^2-}$ with no issues, but I can't make sense out of $\ce{SO4^2+}$ because I can't even make bonds to all $\ce{O}$ without exceeding the number of bonding electrons.

This is on a high school AP review so it is possible that there are exceptions to the normal rules, but I can't see how any of them apply. I could just chop electrons off the oxygens and leave it depleted, but I don't imagine that is a stable form of a cation which would exist?

A quick check doesn't turn up anything so is this a real thing or was it just a typo?

How about
$\ce{O=O+-S+=O+-O-}$ ? I don't know if it exists but it is a closed shell species and has a delocalized pi bonded structure that might give it some stability.


Which sodium citrate is used in silver nanoparticles production? In my studies, I have a group project to realise: to make some $\ce{Ag}$ nanoparticles using Turkevich method (boiling $\ce{AgNO3}$ with sodium citrate in water). In most of articles authors write about using sodium citrate solution, but when I searched for this compound on Sigma-Aldrich, they offer me a wide range of compounds named "sodium citrate". Mono-, di-, tri- sodium citrate.

Searching for a more specific description, I found this article:

http://www.sciencedirect.com/science/article/pii/S1878535214003645

Here they wrote citrate's formula. It is tri-sodium citrate. But Sigma-Aldrich offers it in hydrated versions. Which version should I use? Is hydration even relevant when I'm going to boil it in water?

You can use hydrated tri-sodium citrate, it's readily soluble in water. As for its chemical reactivity, it's identical to tri-sodium citrate.
You should take into consideration the increase in the molecular weight in the hydrated form due to the presence of water molecules.


Is this really a comparison of the absolute energies in the reaction energy profile for ester formation? In Clayden et al., Organic Chemistry, p. 332, the authors compare the reaction energy profile for ester formation from a) acyl chloride and b) anhydride:

They then write:

  The energy levels of the starting materials, the transition state, and the intermediate are all lower in the anhydride reaction than in the chloride reaction.

Are they really comparing absolute energies between different species?

These authors are comparing relative energies. Look at what they set as their reference... the products. They could have easily made "starting materials" as their 0.0 reference point and stated that the products of the acyl chloride reaction is lower in energy relative to the anhydride reaction mechanism.


Why are halogens base sensitive groups (Wolff-Kishner Reduction)? So while studying about Wolff-Kishner reduction reaction to form alkanes from carbonyl groups. I read that it is not suitable for base sensitive groups as the environment is highly basic and as an example the following reaction was given:

Now the desired product that should've formed is

But this doesn't happen as -Cl is a base sensitive group and so reacts with KOH. But how can a halogen be base sensitive when it already  has 7 e- and is highly electronegative? How is it possible for it to react with OH-?
Also, I don't understand the reaction mechanism of this reaction so can someone please explain how this takes place?

You've probably realised already that a possible side reaction is the SN2 nucleophilic substitution of Cl by HO-, which would look like this:

Evidently, this is not the product we actually get. Hydroxide ions are considered 'hard' nucleophiles, as they have a high negative charge density, and they tend to react best with 'hard' electrophiles with high positive charge density. The electrophilic carbon here is reasonably 'soft' whilst the proton adjacent to it is considered 'hard' due to its small size (you can learn more about this here). As a result, hydroxide ions like to act as bases instead of nucleophiles, and the mechanism favoured is not the SN2 shown above, but in fact the competing E2 elimination shown below.

This proton may not look particularly acidic, but formation of a new alkene alongside loss of a reasonably stable chloride leaving group drives the deprotonation.
Another contributing factor is that whilst the substitution has a reasonably small change in entropy, the elimination results in a greater number of species than in the reactants, so has a large positive entropy. This is relevant because Wolff-Kishner reduction is usually done at a reasonably high temperature, and entropy becomes a greater influence on the spontaneity of a mechanism as the temperature increases. This further favours E2 over SN2.
The chloride is considered a base-sensitive group for the above reasons, and this second pathway is inevitable alongside a Wolff-Kishner reduction in a highly basic solution.


Why does the mixing of sigma 2s and 2p orbitals lower the energy of the sigma 2s orbital and raise the energy of the sigma 2p orbital? My general chemistry textbook (General Chemistry Principles and Modern Applications, Tenth Edition) says that for bonding in diatomic molecules with Z ≤ 7, orbital mixing occurs between the $\sigma_{2s}$ and $\sigma_{2p}$ molecular orbitals because they are close enough in energy. As a result, the energy of the $\sigma_{2s}$ molecular orbital is lowered and the energy of the $\sigma_{2p}$ molecular orbital is raised.

My question is why: if the $\sigma_{2s}$ orbital mixes with the $\sigma_{2p}$ orbital, wouldn't the modified $\sigma_{2s}$ orbital be higher in energy, as a result of mixing with an orbital slightly higher in energy?

Likewise, wouldn't the modified $\sigma_{2p}$ orbital be lower in energy as a result of mixing with an orbital slightly lower in energy? Why is it the other way around?

This is a frequent source of confusion in my experience - you'll see there are at least two other very similar questions here already.

Let's take the first point: why do the $\sigma_{2s}$ orbitals get lower in energy when "mixing" with the $\sigma_{2p}$?

I'm going to ignore "mixing" and use the word "interact" instead. If I consider the 1s orbitals in $\ce{H2}$, I take two 1s atomic orbitals and they (of course) interact. So one orbital goes down in energy, and one orbital goes up in energy.

This is a common theme - if there's any degree of overlap between two orbitals, one will become more stable and one will become less stable. In the diagram, we claim "bonding" and "anti bonding" but in more complex molecules, these terms become hard to identify (e.g. benzene).

The theme remains: if orbitals interact (i.e., have the same symmetry), some will go down in energy, some will go up in energy.

That's the answer to your first question. Since the $\sigma_{2s}$ orbitals do interact with $\sigma_{2p}$ the 2s will go down in energy.

Your second question is about the $\sigma_2p$ going up in energy.

I could answer with a formula from quantum mechanics, but let's argue qualitatively. Here's a diagram for two interacting orbitals:

This is much like our case with the 2s and 2p orbitals. We see that the result is a system that's most likely more stable - the orbital lower in energy is even lower because of the interaction.

If we consider the reverse case - that the 2s goes up in energy, and the 2p goes down in energy, we would have a system that's less stable. In theory, this could happen - if the interaction is negative. In other words, the system would be more stable without the interaction than with.

In short, the "mixing" or interaction between 2s and 2p orbitals occurs because the species becomes more stable.


Colored Flames Experiment I am going to be doing a gender reveal for my sister's pregnancy. To do this, we wanted to light a huge fire pit and throw in a chemical to change the fire into the color (red/pink for girl, blue/violet for boy).

From my research, I found that strontium chloride is the best option for red/pink and copper I chloride is the best for blue. I found that Iron Filings make great gold sparkles, as well as magnesium shavings make for a great silver spark.

It is my goal to have my sister and brother in law throw a satchel full of chemical-enriched wax pucks into the fire pit fire and have the fire first glow many different colors, a rainbow of colors until the other colors die down and the pink or blue for the gender remain.

To do this, I was going to create many wax discs with many different colored chemicals and methanol (to create a more vivid color) by first melting wax, adding methanol and the chemical, mixing and setting to dry. The pink or blue one would be the LARGEST wax puck/disc and would, hence, color the longest after the rest of the colors died down.

I was going to add all the discs and the iron filings into a satchel with the largest disc (the gender color in it's own satchel so it remains last), and then when they throw it in, it will create this spectacle.

My question is, does this sound like a viable option and experiment?
Is using methanol in the wax an unnecessary step?
Is it ok to mix methanol with wax? Will it still harden?
How can I insure that the blues, yellows, oranges, greens will die down and the gender color (lets say RED in this case/girl), remains?
How long will it take the color to shine through? How can I make sure this is done within the course of a minute or so?
How much of a chemical should I use to ensure it is seen in a large fire pit (about 3-4 feet in height)?

Any further advice? Tips? Tricks?

Colorful flame
To create a colorful flame you could just buy some packages of Colorful flame which is not toxic.
To buy it on amazon click here.

Gender part
For the gender part you could treat some sawdust with aqeous Copper(I)-chloride ($\ce{CuCl}$). Let that sawdust dry and throw it in the fire (after some time passes so they can be amused by the "Colorful fire"). It'll create a blue colour which would be great if its a boy.
However, if its a girl you could do the same thing, but instead of the Copper(I)-chloride you would use Strontium(II)-chloride ($\ce{SrCl_2}$) or Strontium(II)-nitrate ($\ce{Sr(NO_3)_2}$).
I don't know how long it'll  last but it depends on how much you put in the fire. (More stuff lasts longer)

Precautions

Be careful! You could be burned or something could catch on fire.

Definitely don't breathe in the fumes from the fire.

Don't use methanol because its a poison if ingested.

And the general advice: Wear safety gear!

Sources linked here and here.


Conversion of potassium dichromate solution to a less toxic solution I have a solution of potassium dichromate ($\ce{K2Cr2O7}$) and I want to convert it to a less toxic solution. What would be the best way to do this?

I was thinking of using the dissolved potassium dichromate as an oxidizer in the oxidation of alcohols to organic acids, to eventually produce the chromium(+III) ion (what is less toxic)

Or are there any other, easier methods?

That looks like one way to reduce the dichromate ion which is known to be toxic. MSDS sheets indicate that potassium dichromate has category 1B carcinogenicity.

Another way I believe is to just start from an aldehyde and convert it into a carboxylic acid:

$$\ce{3RCHO + 2Cr_2O_7^- + BH^+ -> 3RCOOH + 2Cr^{3+} + 4H_2O} $$

Alternatively, one can put dichromate in acidic solution with iron:

$$\ce{Cr2O7^{2-} + 6Fe^{2+} + 14 H^+ -> 2 Cr^{3+} + 6 Fe^{3+} + 7H_2O }$$

Dichromate oxidizes $ \ce{Fe^{2+}}$ ions and produces $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ ions.


Why is it important to minimize impurities when doping a semiconductor? I was reading "Electronic Devices and Circuits" by Boylestad, and in there the author talks about how the semiconductor industry seeks to add as little impurities as possible when manufacturing semiconductors.

What I don't understand is why it is so important to keep the material's purity? What characteristics are improved by doing that? If the objective of doping is improve the conductivity of the material (in a controlled manner), wouldn't reducing the ammount of doping material decrease that efficiency?

Semiconductors are based on impurities.  They just call it doping.  The addition of group 3 (Boron group) or group 5 (Nitrogen group) impurities create a hole or an extra electrno, respectively.  These form charge carriers which are necessary for a semiconductor to be used in a circuit.  However, the dopants need to be added in highly controlled and precise amounts to get the necessary mobility.  The silicon also needs to be highly pure for regular electronic response, for regular electronic properties over arbitrary distances.  Another important consideration is that the dopant allows some degree of tuning of the electrical conductivity of the semiconductor.

Also, highly pure crystalline silicon  fractures nicely along the crystal planes, meaning it can be cut with high precision.  Since most mass-produced electronic processors are made on 12- or 16-inch silicon wafers with dozens of devices manufactured at once, and initially all physically connected, the ability to precisely excise each tiny circuit is critical for high through put and yield in manufacturing.


What's so special about chelation? There is another closely related post here , and I've also read the referenced wiki article including applications of chelation, but I still don't see what's so special about chelation. I understand the structure of the bond - the 'capturing' of a metal cation by an organic agent, but I still fail to see why this chemical reaction might be so special relative to other possible organic chemical reactions with a metal cation.

Why is chelation so special?

Chelation is an entropy driven reaction, i.e. complexes with polydentate ligands are more stable.

Consider what happens when you dissolve copper sulfate in water:
\begin{align}
\ce{CuSO4 + 6H2O <=> [Cu(H2O)]^2+ + SO4^2- (aq)}\\
\end{align}

This complex forms and with it comes some kind of ordering of the water molecules around it. Dissolution is often favoured because of the solvation energy. Now consider a bidentate ligand like ethylene diamine, $\ce{H2N-(CH2)2-NH2}$, short $\ce{en}$. Adding this ligand to a copper sulfate solution will result in a ligand exchange:
\begin{align}
\ce{[Cu(H2O)]^2+ + 3en <=> [Cu(en)3]^2+ + 6H2O}\\
\end{align}

In this reaction you increase the number of particles from four on the left side to seven on the right side. Taking into account that you also disturb the order of the surround water molecules, this reaction is very much favoured by entropy. In an attempt of zeroth order approximation, one could assume that entropy roughly doubles. Since the Gibbs energy is inversely correlated to the entropy, this means, that the reaction will be spontaneous. This is usually referred to as the chelate effect.$%edit$


What is difference between density of the equilibrium and vapour density of the mixture While solving question on calculating degree of dissociation,  I came across this question:

At 627 degrees Celsius and 1 atm $\ce{SO_3}$  is partially dissociated into $\ce{SO2}$ and $\ce{O2}$ by the reaction;
$$\ce{SO_3 <=>  SO2 + 1/2O2}$$
the density of the equilibrium mixture is 0.925 gram per liter what is the degree of dissociation?"

In  answer, they calculate the the molecular mass of mixture at equilibrium(Mmix) by using density of equilibrium which comes 63.348 and we know molecular mass of $\ce{SO3}$ is 80 , vapor density of $\ce{SO3}$ will be 80 by 2 is equal to 40 ,vapour density of mixture is equal to molecular mass of mixture by 2 is equal to 68.348 by 2 is equal to 34.174 and from here we can calculate degree of dissociation by using the formula(x=D-d/(n-1)d ) but my doubt is what is the difference between vapor density of mixture and density of the equilibrium mixture?

Let's start from $m = 0.925 g˙\ce{SO3}$ that will be heated at $900 K$. This is the amount of $\ce{SO3}$ that will occupy a volume of $1.00 L$ after heating it at $900 K$ at $1 atm$. and after partial decomposition. In moles it is also  $n_o$ moles $\ce{SO3}$,  $$\ce{ $n_o = \frac{m}{80 g/mol} = \frac{0.925 g}{80 g/mol} = 0.01156 mol$}$$ At $900 K$,  a fraction $\alpha$ of these moles are decomposed into $\alpha$ moles $\ce{SO2}$ plus $\alpha/2 $ moles $\ce{O2}$. Then the total number of moles at equilibrium will be : $$\ce{$n$ = $n_o$˙($1$ - $\alpha$ + $\alpha$ +    \alpha/2}) = n_o(1 + \alpha/2) = 0.01156 mol (1 + \alpha /2)$$ The total volume is : $$\ce{$V = nRT/p= 0.01156 mol$(1 + \alpha/2)$RT/p = 0.001 m^3$}$$  Simple rearrangement of this formula gives : $$\ce{1 + \alpha /2  = \frac{0.001 $m^3·p$}{0.01156 $mol·RT$} = \frac{0.001 $m^3$·101325 $Pa$}{0.01156 $mol$·8.314 $J mol^{-1}K^{-1}·900 K$} = 1.172 }$$
The degree of dissociation $\alpha$ is equal to $\alpha = 2 ·0.172 = 34.4$ %


Conceptual confusion about NMR I am quoting Clayden et al 2nd edition:

The sample is irridiated with a short pulse of radio-frequency energy. This disturbs the equilibrium balance between the two energy levels: some nuclei absorb the energy and are promoted to a higher energy level.

When the pulse finishes, the radiation given out as the nuclei fall back down to the lower energy level is detected using what is basically a sophisticated radio receiver.

Now, my confusion is about the equilibrium of nuclei in higher & lower energy level , after the introduction of radio wave.
Are the nuclei in higher level steady i.e. not radiating and coming back to the lower level constantly?
Or, are they radiating and constantly re-energized by the radio wave till the radio wave is on(still keeping a equilibrium constant between nuclei in lower and higher energy level)?

The nuclei absorb the rf radiation and the spin population in the upper level increases at the expense of that in the lower level. There is now a chance that the upper levels will radiate away their extra energy and so return to equilibrium. However, in an nmr experiment the excitation pulses used in most if not all nmr machines are very short, just a few microseconds (and so spectrally wide) but the natural radiative lifetime of the excited levels is some seconds. Thus it is possible to excite the nuclear spins and build up a relatively large population before any significant decay occurs.

(This is one of the reasons why nmr works so well considering the tiny population difference that is produced because the energy levels are so close in energy and so all levels are significantly thermally populated.  The long lifetime of upper levels means that it is possible to coherently build up a significant macroscopic magnetisation in the sample and this what is detected.)


How might ammonia be created in this mechanochemical reaction? Can mechanical agitation catalyze reactions?  It appears so (thanks to Mithoron for the pointer!).

We observed the following: We were using 10-micron hBN powder to impact-plate copper bearings (e.g., by running them together in a vibratory tumbler, and other similar processes).  Even though the mechanical processes ran in a standard atmosphere and did not exceed 60C we consistently observed that ammonia was released.  As a control:  I have run the same copper with all sorts of other media through the same mechanical processes and never detected ammonia.

Our conclusion is that the BN is decomposing during the mechanical processing.  But hBN is very thermally and chemically stable: hBN doesn't react with Cu below 1000C!

So is it plausible that mechanical abrasion at this low a temperature would break the B-N bond?  If so, what subsequent reactions (again, in a standard atmosphere) might cause the free N to form $\ce{NH3}$ in detectable amounts, instead of just forming $\ce{N2}%edit$?

Fine hexagonal boron nitride powder is hygroscopic, so this implies there might be some water in your system. According to Reaction of Hexagonal Boron Nitride Nano-crystals under Mild
Hydrothermal Conditions "The reaction between water and hBN nano-crystals starts at very low pressure and temperature, i.e. 220 C and 1.0 MPa [~10 atmospheres]."

Those conditions are likely to be met in a ball-mill or vibratory tumbler, at least momentarily at the points of impact (in a similar way, cavitation can heat water to luminescence!).

A simple test would be to preheat the tumbler and its contents to 150 or 200 C for an hour or two, then seal it airtight before running the tumbling cycle. If there's no ammonia odor, then it indicates that adsorbed water in the hBN had been reacting with it, perhaps catalyzed by the copper bearings, as well as by impact heating and pressure.


Calibration Curve Error Propagation I am using ED-XRF for geochemical analysis for a geology dissertation. The ED-XRF provides a ppm reading for elements and a 2sigma error, for example 472 +- 5.11ppm Sr.

An example calibration curve made for Sr comprising 37 standards has a linear regression of y=1.26x + 32.1 (R^2 0.9991), which I can easily apply to correct my unknown analyses. I need to plot my corrected readings with their error bars, so how is the 2sigma error associated with this unknown calibrated?

Does this error value also simply get calibrated with the curve equation, or just the slope value, or is this error propagation far more complex? I have searched for a solution and keep finding an equation for uncertainty propagation with calibration curves, but this only calculates the uncertainty caused by the curve, and has no relation to the uncertainty of each analysis? So is the total error the original 2sigma error provided by the machine + calibration curve uncertainty?

Thanks

I will try to offer a detailed way to propagate errors here, but before beginning it's worth asking what your audience is likely to expect "error bars" to mean.  In your field are error bars usually determined simply by replicate analyses of the same sample?  Or replicate samples of the same experiment or field site?  If so, it's probably best just to stick with what your audience is likely to expect.  That said...

The theoretical best way to fit your calibration curve would be with weighted linear least-squares algorithms.  In that algorithm, in addition to the {x, y} data, which in your case is $\{\ce{Sr_{known}, Sr_{meas}}\}$, you also supply weights corresponding to the uncertainty in the y values.  The weight for each y data point would be $\frac{1}{\sigma^2}$, which you could calculate from the $2\sigma$ uncertainty given by your instrumentation for that data point.  In practice, if you did this, I doubt that the parameters of your calibration would change very much at all from what you did previously.  But formally the weighted variant of least-squares is required if the uncertainty in y depends on y.  See wikipedia for more.
After doing the (weighted) least squares calibration regression, you can find the uncertainty in the estimation using the calibration equation.  $\sigma_{est} = \sqrt{\frac{\sum{\left(y_{pred}-y_{meas} \right)^2}}{N}}$.  In that equation $y_{pred}$ is the calibration-predicted Sr concentration and $y_{meas}$ is the measured Sr concentration, and for you $N$, the number of data points, is 37.  This uncertainty is one part of the uncertainty you'll eventually report, and reflects the uncertainty in your calibration curve.  Based on the R-squared you reported, my guess is that this uncertainty will be low.
After all that (!), you need to estimate the other major source of uncertainty, the uncertainty in the measurement of your unknowns ($\sigma_{meas}$).  There are several OK methods for doing this, the best of which would be replicate analysis of several samples.  It might be OK to use the $1\sigma$ value calculated from the $2\sigma$ uncertainty reported by your instrument too.
Finally, combine the uncertainties with the standard equation for combining multiple sources of uncertainty $\sigma_{total} = \sqrt{(\sigma_{meas})^2 + (\sigma_{est})^2}$.


What are the chemical process responsible for the warping of wood? Background:

According to this Wikipedia article:

  Wood warping costs the wood industry in the U.S. millions of dollars per year. Straight wood boards that leave a cutting facility sometimes arrive at the store yard warped. This little-understood process is finally being looked at in a serious way. Although wood warping has been studied for years, the warping control model for manufacturing composite wood hasn't been updated for about 40 years.

Factors contributing to the warping of wood include plant species, temperature and temperature changes, humidity and humidity changes and wet-dry cycles.

Discussion:

I'm not interested in which species are more prone to warping, how to prevent warping by treatments, etc. Nor am I interested in the biological processes of living or freshly cut wood. I'm interested in the processes happening over time on a molecular level that lead to macro-scale warping of wood. Water, in particular, seems to play a major role in the warping of wood, implying that possibly the lignin and cellulose matrix of wood is disturbed by the formation of hydrogen bonds. But if this is the case, or whatever the case, why would the microscale effects not cancel out rather than causing macro scale warping in a particular direction? This may be more of a physical than a chemical question, however.

Question:

What are the chemical process responsible for the warping of processed (as opposed to living or freshly cut) wood?

Edit 1:
A link to an interesting and relevant video was given in a comment, in which several types of wood were placed in a chamber which was evacuated of air, then pressurized with ammonia gas. The video also gave a link to a patent for this process. The following is an excerpt from the patent:

  The rigidity of wood is the result of crosslinking between adjacent cellulose chains by water molecules which hydrogen bond between sites on adjacent cellulose chains. Anhydrous ammonia is extremely reactive with water, and it is believed that anhydrous ammonia scavengers water from wood, breaking the crosslinking between cellulose chains, and permitting the cellulose chains to slide, which renders the wood very flexible so that it can readily be bent or twisted. When the ammonia volatilizes from the treated wood, water vapor reestablishes crosslinking and rigidity.

The reason for this edit is to suggest that the disruption of hydrogen bonding between adjacent cellulose chains, when the wood is dried then rehydrated, for example, may also play a role in "natural" warping. I'm going to look into this more and I hope it gives somebody else a clue to a good answer.

Warping (be it a bow, crook, cup, or twist); as well as loosening of knots, checking, splitting, and collapse; of wood is caused by uneven shrinkage.

To explain uneven shrinkage chemically, consider wood as 50% cellulose.  Cellulose, which is a polymer consisting of linked glucose monomers, is arranged into fibrils.  Fibrils are part of larger structures which make up the cell walls of wood fiber.  Binding these cell walls together is lignin, which is a phenylpropanol polymer.  Cellulose and lignin (along with hemicellulose) are the major structural compounds in wood.

Cellulose in wood is a tangle of rigid crystalline cellulose and flexible amorphous cellulose.  While the structure of crystalline cellulose makes it water insoluble, amorphous cellulose has sites like its base monomer glucose which are capable of absorbing water. When water is absorbed both the shape and volume of the amorphous cellulose change.  While amorphous cellulose changes its shape, the crystalline cellulose (as well as lignin) retains its shape and volume.

Cellulose may also change shape and volume due to other factors such as temperature fluctuation and biological degradation.  Because (1) wood is not uniform; (2) moisture, temperature, and biological changes are rarely uniform throughout the bulk of a piece of wood; and (3) most wood is exposed to some degree of these changes over time; uneven shrinkage occurs.  Because of uneven shrinkage in wood, it warps.


Can 99% isopropyl alcohol be stored in a clear container? I'm planning to get hold of some 99% isopropyl alcohol (aka rubbing alcohol) for the purpose of cleaning electronics, and am going to transfer it from the plastic bottle it comes in to a smaller, clear dropper bottle or similar. However, I vaguely remember being told that isopropyl alcohol shouldn't be stored in a clear container - possibly something to do with speedier evaporation, but I'm really not sure.

Is this true? Are there any downsides to storing 99% isopropanol in a clear container? Will storing the solution in direct sunlight change this at all?

I would not expect that isopropanol would evaporate appreciably from a properly sealed container. Rather the main purpose of storing isopropanol out of light is to prevent peroxide formation.  Isopropanol can form peroxides and when distilled they can concentrate and explode. A clear container allows more light to pass and thus more peroxides to form.


Implication of negative order on kinetics In the reaction

$$\ce{2 O3 -> 3 O2}$$

the order of reaction that I have been taught is

$$\ce{O_3 \underset{$k_{-1}$}{\overset{$k_{1}$}{<=>}} O_2 + O}$$
$$\ce{O_3 + O \overset{$k_{2}$}{->} 2O_2}$$

$$\mathrm{rate} = k_{2}[\ce{O3}][\ce{O}]$$

$$ k_{1}[\ce{O3}]= k_{-1}[\ce{O2}][\ce{O}]$$

So $$\mathrm{rate} = \frac{k_{1}k_{2}}{k_{-1}} [\ce{O3}]^2[\ce{O2}]^{-1}$$

What does this $[\ce{O2}]^{-1}$ part mean? For example, initially when there is no $\ce{O2}$, the rate should be $\infty$; how is that possible? Or is it normal and I should accept it?

What does this $[\ce{O2}]^{-1}$ part mean?

It means the higher the concentration of oxygen, the lower the rate. The two-step mechanism you show relies on oxygen atoms (radicals) to react with ozone in the second step. Dioxygen acts as a radical scavenger (first reaction, reverse direction). So if dioxygen concentrations are high, the reaction will not proceed via this mechanism.

When ozone is produced as part of smog in the lower atmosphere (i.e. the air we breathe), this mechanism is not expected to help lower the ozone concentration. The equilibrium of the first reaction would be too far on the side of the reactants.

In the higher layers of the atmosphere, however, this sequence is more feasible. Concentrations of all species are extremely low, so bimolecular reactions are slow as well. In fact, in the image linked below, concentrations are given as molecule per mL (as apposed to mol / L) and rate a similarly expressed as molecules reacting per time unit.

https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry_files/image018.jpg

What actually happens in the ozone layer is a bit more complicated than the two step process you show, see for example https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry.htm

  For example, initially when there is no O2, the rate should be ∞; how is that possible? Or is it normal and I should accept it?

See the excellent comment by Nicolau Saker Neto. Another way to say this is "the molecules don't know the concentration of species, but they still do their thing without worrying about division by zero".


Is a spectrophotometric method of measuring the rate of a complex reaction valid? A few months back I designed and carried out an experiment to measure the rate of the reaction between ferric chloride and copper.The dependent variable was the rate of the reaction; the independent variable was the the concentration of ferric chloride.

I searched up the reaction and got it from chemiday. There, it stated it as the following:

$$\ce{2FeCl3 + Cu -> 2FeCl2 + CuCl2}$$

So the method I used to measure the rate of this reaction was to use a spectrometer to find a rising absorbance peak for one of the reactants, which I selected to be: $\ce{CuCl2}$.

Then I got some anhydrous $\ce{CuCl2}$ and made a standard curve (Absorbance vs Concentration) for it, so that I could use that to convert the absorbance at it's $\lambda_\mathrm{max}$ to a concentration value.

I carried out the experiment taking the readings at that $\lambda_\mathrm{max}$ at set time intervals, and I reached the conclusion that the relationship between the concentration of ferric chloride and the rate of the reaction between it and copper was a linear one, at least for the concentrations I tested ($\pu{0.1 .. 0.6 M}$).

But now I'm doing more reading to write my report and I found out that this was a complex reaction, meaning that there were several steps to the reaction. After searching up some papers, the consensus seems to be:

\begin{align}
\ce{FeCl3 + Cu &-> FeCl2 + CuCl}\tag{1}\\
\ce{FeCl3 + CuCl &-> FeCl2 + CuCl2}\tag{2}\\
\ce{CuCl2 + Cu &-> 2CuCl}\tag{3}
\end{align}

As seen from above, it isn't as simple as the original single reaction that I mentioned earlier. First, $\ce{CuCl}$ is produced, then it becomes a reactant in the second reaction to produce $\ce{CuCl2}$, and lastly $\ce{CuCl2}$ becomes a reactant to produce $\ce{CuCl}$.

I'm seriously questioning my method after finding out that a series of reactions takes place, not a just single one. Although the data that I got and the conclusion that I reached matched nicely with other papers that I've read, I'm not sure if the explanation behing my method is correct.

Any input would be greatly appreciated. Thank you.

This is a perfectly valid method of determining rates of chemical reactions. It is obviously limited to those reactions where either the product or the initial reactant can be monitored easily using spectroscopy as well as reactions which are slow enough to get an absorbance measurement at set intervals. I have used this method multiple times in undergraduate chemistry labs and it works, and in principle it does not matter if there are intermediates.

The intermediates should not matter because the rate of a reaction is defined as the rate at which product is formed. That is, the rate of a first-order reaction is,
$$
\text{rate}=k_{tot}[A]
$$
If the reaction involves only one elementary step then the prefactor $k_{tot}=k_1$, the rate constant of the one step (assuming no backwards reaction). Note that $[A]$ can be the concentration of either a product or reactant, but you might need a stoichiometric coefficient in there depending on the reaction. On the other hand, if there are intermediate steps, the prefactor can become arbitrarily messy. Maybe there's an equilibrium between two intermediates so you have a term in there $k_2/k_{-2}$ and whatever else you can imagine. This doesn't matter though because your experiment measures the change in concentration at either end, and the slope is then the effective rate constant for the reaction which has all of the intermediate stuff lumped together.

In fact, this method you use is probably the most direct means of measuring a rate in the sense that a rate is defined as a change in concentration over time. You are measuring a change in absorbance over time, so there is only step (and one assumption) in going from your data straight to the reaction rate. That assumption is simply the validity of Beer's law, which is usually not a problem for a large range of concentrations.

As a sanity check, look at this website which has a description of the same experimental setup and what appears to be real data. Note that they are measuring change in absorbance of a reactant while you say you are measuring the change in absorbance of a product, $\ce{CuCl2}$


Why do we need to add hydrogen to amorphous silicon? I was recently studying about Staebler-Wronski effect observed in amorphous silicon (a-Si). I realised it was due to imperfections in the lattice structure and breaking of Si-Si bond to Si-H bonds.

So if hydrogen atoms will be less the effect will be minimum. Why do we add hydrogen atoms to a-Si?

Is is just to passivate the unsaturated Si atoms at the edges, or am I missing something?

Well, one either adds it on purpose: In order to reduce the defect volume - by passivating the dangling bonds. Unsaturated silicon atoms in a-Si is a serial killer of charge carriers. But then the issue is the SWE - you have to handle it somehow. Options are few but research is being done. The common way of handling it right now is to be honest about it. This pv cell will lose some efficiency the first months, but then stabilize. Not a good selling point, but that is where it is at.

Or one adds it accidentally / coincidentally by using a deposition technique or raw material that contains hydrogen. ($\ce{SiH4}$ or $\ce{SiH_xCl_{4-x}}$). But you can usually thermocycle these problems away, as long as your substrate can handle it. If you did it, accidentally, on a plastic surface - it is usually referred to as "scrap" and recycled...


How to compare viscosity of alcohol to water? Are there any simple at-home experiments to test the relative viscosity of alcohol to water?
If they are the same or practically the same, are there any household liquids that are less viscous than water? If so, how can I demonstrate this to first-graders?

A common experiment would be to have two concentric cylinders. The outer one can rotate almost freely (with some slight and constant retardation/breaking applied), while the inner one is driven by a motor, like so. Between them is a layer of the liquid that you wish to determine the viscosity of. The speed of the outer cylinder is a measure for the viscosity.

However, since you want to explain this to first-graders, you might wanna start with something more obvious than the difference in viscosity between ethanol and water. How about the difference in viscosity between water and honey? As a demonstration experiment, you could use a funnel in which you pour an equal amount of either liquid, and then measure the time it takes until it has passed through the funnel. This should be more tangible and easier to understand. You could also prepare beakers with water, honey and maybe peanut butter, and have your students run a spatula through all of them. That way they can experience the difference in force that is necessary to move the spatula first hand.


Constitutional isomers of cis-1,2-dibromocyclopentane The following is a problem from 'Fundamentals of Organic Chemistry' by McMurry-

  Draw two constitutional isomers of cis-1,2-dibromocyclopentane.

Constitutional isomers must have different connections. But the question specifies everything including the spatial arrangement. I don't get how there can be constitutional isomers for the above-mentioned compound.

If I understood you correctly, your objection is to the relative cis- designation in the name of the compound. As far as I can tell, the use of any stereochemical designation within the context of that question is either deliberately misleading, incorrect, or simply pointless. Per the definition of constitutional isomerism, the only criteria that the two molecules in your answer must satisfy is that they have (a) the same molecular formula, and (b) different structural connectivity. Stereochemistry, whether relative or absolute, is totally irrelevant. Note the wording in the IUPAC Gold Book entry for constitution (emphasis mine):

  The description of the identity and connectivity (and corresponding bond multiplicities) of the atoms in a molecular entity (omitting any distinction arising from their spatial arrangement).

Hence, any molecule with the formula $\ce{C5H8Br2}$ that differs in connectivity from a cyclopentane with two vicinal bromine atoms should be a valid answer. Two such isomers are given by simply permuting the positions of the bromines (i.e., 1,1-dibromocyclopentane and 1,3-dibromocyclopentane). Of course, there are numerous other possible answers.


Can a long polymer chain interact with itself via van der Waals forces? It is known that van der Waals interaction is an intermolecular forces. So is it possible that a long polymer chain interacting with itself via non-bonded interactions (such as van der Waals forces / electrostatic forces / hydrogen bonding)?

You have a possible answer to your question in proteins, an example which includes some long polymer chains.

Intramolecular interactions - while not necessarily the driving force for formation of a collapsed protein globule (usually argued to be due to the hydrophobic effect, requiring intermolecular interactions) - are the basis for higher order structure in proteins and even some forms of RNA. I won't mention DNA since most commonly this forms a bimolecular double helix.

All of the interactions you mention may play a role in stabilizing a particular protein structure or "fold". A good and basic example of protein structure acquisition due to intramolecular interactions in proteins is polyalanine, which acquires alpha-helical secondary structure stabilized by intramolecular hydrogen bonds, typically between amide groups separated by 2 amino acids along the chain. Other types of structure stabilized by intramolecular interactions (H-bonds, usually) in polypeptides include other types of helices, beta sheets and various turns.


Why does salt dissolved in water taste like salt? If it's just Na+ and Cl- ions When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$. It stays in this form until the water evaporates and then the ions go back to normal $\ce{NaCl}$. So why does water with salt in it still taste like salt? I am asking because if the molecule $\ce{NaCl}$ is broken down into $\ce{Na+}$ and $\ce{Cl-}$ then how does it have the characteristics of $\ce{NaCl}$?

When you taste salt, you're not pushing crystalline $\ce{NaCl}$ into your taste buds. It dissolves in your saliva and dissociates. When one tastes salt, the saltiness taste receptors respond specifically to the sodium cation. That type of taste receptor is a cation channel. This is why lithium and potassium cations also taste salty (though they also stimulate other receptors which make them taste somewhat different).

There seem to be at least two types of receptors that respond to saltiness. One responds almost specifically to sodium at low concentrations, but at higher concentrations, the other type responds to many cations. See here


Why does the mixture's temperature change while phase transition? Consider a mixture made up of alcohol and water at room temperature. If I freeze this mixture, why does it change its temperature during the phase change?

I don't understand the difference between the case when you have a pure substance (only water or only alcohol) and that one. How could it be different (I mean, in molecular thinking...)?

The concentration of water decreases as the mixture cools, so one is left with fairly pure ice, freezing at ~273 K, and pure(r) ethanol, freezing ~159 K (well, the eutectic mixture, about 93% ethanol). Since the ratio of the components are changing as it freezes, there is no distinct freezing point (unless you start with the eutectic mixture). See the phase diagram for ethanol-water mixture. Instead, the slush mix has a range of solidus and liquidus temperature. BTW, this is a method to make apple jack or brandy by freezing out water from cider or wine.

The same is noticed with solder, where it crystallizes gradually and dully, except for the sudden freezing of a eutectic mix (~63% Sn, ~37% Pb), making a shiny joint.


Molecular term symbols for excited state oxygen What are the possible term symbols for excited state oxygen with configuration $(1\pi_u)^3$,$(1\pi^*_g)^3$ (all other occupied orbitals are closed shell)? Based on the possible values of S and $\lambda$ it seems that $^1\Delta_u$, $^3\Delta_u$, $^1\Sigma_u$ and $^3\Sigma_u$ are possible. However, I am not sure in this case how to determine the +/- labels for the $\Sigma$ term symbols. For ground state oxygen, triplet states pair with - and singlet states with + to give overall antisymmetric wavefunctions. However, in this case, two different orbitals are involved, so does this rule no longer apply?

What are the possible term symbols for excited state oxygen with configuration $(1\pi_u)^3(1\pi^*_g)^3$ (all other occupied orbitals are closed shell)? Based on the possible values of $S$ and $\lambda$ it seems that $^1\Delta_u$, $^3\Delta_u$, $^1\Sigma_u$ and $^3\Sigma_u$ are possible.

Term symbols for $(1\pi_u)^3(1\pi_g^*)^3$ - i.e. 1 hole + 1 hole - is equivalent to finding terms for $(1\pi_u)^1(1\pi_g^*)^1$, i.e. 1 electron + 1 electron. I will continue answering the question for the latter case as this is easier to handle.

The fastest way is to use group theory. $O_2$ molecule belongs to $D_\infty^h$ point group. We look at the character table for this group (e.g. here) and perform a direct product of irreducible representations corresponding to our electrons, i.e.:

$\Pi_u \otimes \Pi_g = \Sigma_u^+ \oplus \Sigma_u^- \oplus \Delta_u$

Note this only pertains to spatial symmetry, i.e. orbital angular momentum. It tells us nothing about possible spin states. However, since the two electrons are in different orbitals, all spin combinations are allowed in all cases. The resulting terms are:

TLDR: $^3\Sigma_u^+,$$^3\Sigma_u^-,$$^3\Delta_u,$$^1\Sigma_u^+,$$^1\Sigma_u^-,$$^1\Delta_u$

Your answer is thus correct, the missing piece being the +/- signs of the $\Sigma$ terms. Note that in answering we relied heavily on Pauli exclusion principle not playing a role - since the two electrons occupy different orbitals. Were that not the case, some of the terms would disappear (terms of e.g. $(1\pi_u)^1(2\pi_u)^1$ are $^3\Sigma_g^+,$$^3\Sigma_g^-,$$^3\Delta_g,$$^1\Sigma_g^+,$$^1\Sigma_g^-,$$^1\Delta_g$, while terms of $(1\pi_u)^2$ are a non-trivial subset, specifically $^3\Sigma_g^-,$$^1\Sigma_g^+,$$^1\Delta_g$).

  However, I am not sure in this case how to determine the +/- labels for the $\Sigma$ term symbols.

The following derivations draw heavily from this pdf, read for further details.

We refer to definition of +/- signs as describing the symmetry with respect to reflection in plane that contains both atoms, let's use an arbitrary choice of xz-plane. Wavefunctions symmetric when reflected in this plane are "+" terms, anti-symmetric are "-" terms. However, imagining reflections is not practical. Therefore, we introduce operator notation. The "+" term corresponds to a wave function with eigenvalue of +1 (with respect to $\hat{\sigma}_{xz}$ operator); "-" term corresponds to eigenvalue of -1.

We note:

$$\hat{\sigma}_{xz} \pi_x = \pi_x$$
$$\hat{\sigma}_{xz} \pi_y = - \pi_y$$
$$\pi_1 = -N(\pi_x + i\pi_y)$$
$$\pi_{-1} = N(\pi_x - i\pi_y)$$

, where N is some boring normalization constant. Following up:
$$\hat{\sigma}_{xz} \pi_1 = - \pi_{-1}$$
$$\hat{\sigma}_{xz} \pi_{-1} = - \pi_1$$

We need combinations of configurations (Slater determinants = SD) that are eigenfunctions of $\hat{\sigma}_{xz}$.
E.g.:

$$^1\Sigma^+_u = N(|SD_1\rangle - |SD_2\rangle + |SD_3\rangle - |SD_4\rangle) =$$$$= N(|\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1}{\pi}_{g,-1}\rangle + |\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1}{\pi}_{g,1}\rangle) $$
Let's apply the $\hat{\sigma}_{xz}$ operator:
$$\hat{\sigma}_{xz}{ }^1\Sigma^+_u = \hat{\sigma}_{xz} (N(|\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1}{\pi}_{g,-1}\rangle + |\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1},{\pi}_{g,1}\rangle)) = $$
$$ = N(|-\pi_{u,-1},-\bar{\pi}_{g,1}\rangle - |-\bar{\pi}_{u,-1}{-\pi}_{g,1}\rangle + |-\pi_{u,1},-\bar{\pi}_{g,-1}\rangle - |-\bar{\pi}_{u,1},{-\pi}_{g,-1}\rangle) =$$
$$ = N(|\pi_{u,-1},\bar{\pi}_{g,1}\rangle - |\bar{\pi}_{u,-1}{\pi}_{g,1}\rangle + |\pi_{u,1},\bar{\pi}_{g,-1}\rangle - |\bar{\pi}_{u,1},{\pi}_{g,-1}\rangle) = $$
$$= N(|SD_3\rangle - |SD_4\rangle + |SD_1\rangle - |SD_2\rangle) = { }^1\Sigma^+_u$$

In going from first to second line we apply the operator, i.e. flip "1" subscript to "-1" and vice versa; in addition we multiply each orbital by "-1". Since we have two orbitals this multiplication disappears: ($-1*-1=1$), which is what we do in going from second to third line. In the last line we see that $\hat{\sigma}_{xz}|SD_1\rangle = |SD_3\rangle$;$\hat{\sigma}_{xz}|SD_2\rangle = |SD_4\rangle$ etc.

We could have used a WF in the form e.g.: $WF = |SD_1\rangle + |SD_3\rangle$. Such a WF is indeed an eigenfunction of $\hat{\sigma}_{xz}$. But it is not an eigenfunction of $\hat{S}^2$.

Similarly, a corresponding "-" term is:
$$^1\Sigma^-_u = N(|SD_1\rangle - |SD_2\rangle - |SD_3\rangle + |SD_4\rangle) =$$
$$\hat{\sigma}_{xz}^1\Sigma^-_u = N(|SD_3\rangle - |SD_4\rangle - |SD_1\rangle + |SD_2\rangle) =$$
$$ = -1 * N(- |SD_3\rangle + |SD_4\rangle + |SD_1\rangle - |SD_2\rangle) = -1 * { }^1\Sigma^-_u$$

The four SDs we use here are all eigenfunctions of $\hat{L}_z$ and $\hat{S}_z$ with $M_l = 0$ (i.e. one $\pi_1$ + one $\pi_{-1}$ orbital) and $M_s = 0$ (i.e. one alpha and one beta orbital). There are two more combinations of these SD and these are part of the $^3\Sigma^-_u$ and $^3\Sigma^+_u$ states.

Lastly, non-$\Sigma$ terms have both a "+" and "-" component that flips to each other upon reflection, e.g.: $\hat{\sigma}_{xz} \Delta^+ = \Delta^-$. But since $\Delta^+$ and $\Delta^-$ are different SDs (different configurations), we don't use the label in these cases - they are reserved for $\Sigma$ terms.

The take-away message here is that "+/-" label relates to some kind of symmetry. Similarly to "g/u" label, it only is relevant in some special molecules ("g/u" in molecules with inversion center; "+/-" in diatomic molecules). We use group theory where possible or a very technical procedure when not.

  For ground state oxygen, triplet states pair with - and singlet states with + to give overall antisymmetric wavefunctions. However, in this case, two different orbitals are involved, so does this rule no longer apply?

There is some confusion here. As mentioned above, for ground state ($(1\pi_u)^2$) the terms are indeed $^3\Sigma_g^-$ and $^1\Sigma_g^+$. But there is also the state $^1\Delta_g$ that has no "+/-" label, yet it is still anti-symmetric. The "+/-" label has nothing to do with anti-symmetry. Anti-symmetry is ensured by use of Slater determinants. The "rule" you mentioned is a result applicable to $\pi^2$ (and other doubly degenerate 2 electron) configurations, but is not related to anti-symmetry and gives no predictions for other configurations.


What surface treatment can turn nickel black? Questions:

The detailed surface treatment to turn nickel plated steel black.
The reaction mechanism of the process.

Methods:

(preferred) Anodizing, oxidizing : The only black nickel oxide is $\ce{Ni2O3}$ but I'm not sure if it's suitable for surface protection. I do not know the color or other Ni(II) or Ni(III) compound. There are nickel blackening solution for sale, but I don't know their ingredients.

(not preferred) Electroplating onto the existing nickel: US military have a standard to electroplate black nickel (MIL-P-18317). However this process is quite a fuss.

The article provided by Karl has answered the question with electroplating method, which is not preferred but this is the only answer I have.

The method is based on MIL-P-18317. The electroplating solution consists of $\ce{NiSO4, Ni(NH4)2(SO4)2, ZnSO4}$ and $\ce{NaSCN}$

The black nickle coating are $\ce{NiS}$ and $\ce{ZnS}$.
$\ce{NiS}$ and $\ce{ZnS}$ are deposited by $\ce{M^2+ + H2S -> MS + 2 H+}$.
$\ce{H2S}$ is generated by $\ce{SCN- + 3H+ + 2e- -> H2S + HCN}$

note: Is reducing thiocyanate a standard method to generate sulfide in electroplating ? I know soluble sulfide salts smells horrible, but it is worth the trouble ? not mentioning the cyanide byproduct, in acidic solution.

Metal $\ce{Ni}$ and $\ce{Zn}$ are also deposited $\ce{M(NH3)2^2+ + 2e- -> M + 2NH3}$.
The role of $\ce{NH4+}$ is to generate $\ce{M(NH3)2^2+}$ prior to metal deposition.


What is the significance of an organic compound's M peak? Background

On the topic of identifying an organic compounds using peaks generated in Mass Spectrometry, a rule of thumb expressed in the UK educational textbook, CGP, is that that "the M peak is the one with the second highest mass/charge ratio".

Question

What is the significance of this M peak anyway?

"M" stands for molecular ion.  It is often represented as $\ce{M, M^{+} or M^{+~.}}$. In the mass spec experiment, an electron is knocked out of some of the sample molecules.  The peak at M/q (q=1) represents the molecular weight of the entire molecule, less an electron, before it fragments to smaller ions.  From precise measurement of the molecular ion (e.g. 82.34567890) you can determine the exact molecular formula of your sample compound.  This can help greatly with sample identification.


Basic thermodynamics question about work I am getting into thermodynamics, and I have a basic fundamental question about the definition of the change in internal energy $\mathrm{d}U$:

$\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V + \sum\mu_i\,\mathrm{d}n_i$

Firstly, I do not understand why it's not the case that $\mathrm{d}W$ is defined as $P\,\mathrm{d}V + V\,\mathrm{d}P$, but only as $P\,\mathrm{d}V$. I can imagine a situation where the volume is fixed, but number of particles increases, thus increasing the pressure. I would think that this increase in pressure would also increase the internal energy.

Secondly, in the same manner I also don't understand why $\mathrm{d}Q$ is defined as $T\,\mathrm{d}S$ and not as $T\,\mathrm{d}S + S\,\mathrm{d}T$. I would think that a change in temperature would surely change the internal energy.

Well, the answer is tricky. But before the really tricky part, you may want to take a step back and consider a general statement of the 1st law,

$$dU = \sum_i \left(\frac{\partial U}{\partial q_i} \right)_j dq_i$$

where $q_i$ is an extensive coordinate of the system. Note this is consistent with our standard description of the relation between generalized forces and energy (the derivatives wrt coordinates are forces).

In the case of pV work, $q_i = V$ and

$$ \left(\frac{\partial U}{\partial V} \right)_j = -p $$

The partial derivative in question is evaluated holding all other extensive variables (including number of particles) constant. If there is a change in chemical composition of the system we can define a chemical force (although referred to as the chemical potential of substance i) as

$$ \left(\frac{\partial U}{\partial n_i} \right)_j   = \mu_i  $$

and an infinitesimally small amount of chemical work as

$$ \left(\frac{\partial U}{\partial n_i} \right)_j dn_i  = \mu_i dn_i $$

As noted in the comments, the formulation combining the first and second laws of thermodynamics presented in the OP is a particular representation of the first law for the case of reversible pV work, a more general formulation of the 1st law being

$$\begin{align} dU &= dq + dw\end{align}$$

Note that while $dU$ is exact, the differentials on the right-hand-side of the previous equation may be inexact. Distinguishing between pV and other types of work for a reversible process (for which the differentials $dw$ and $dq$ become exact) we can write

$$\begin{align} dU &= dq_{rev} + dw_{pV,rev} + dw_{nonpV,rev}   \\ &= dq_{rev} + dw_{pV,rev} + \sum_i \mu_i dn_i   \\ &= TdS + dw_{pV,rev} + \sum_i \mu_i dn_i \end{align}$$

The second equality follows from the first using the 2nd law. Then since $dw=-p_{ext}dV$ in the case of pV work  and since $p_{ext} = p_{sys}=p$ when the work is carried out reversibly, you arrive at the expression in the OP.

  I can imagine a situation where the volume is fixed, but number of particles increases, thus increasing the pressure. I would think that this increase in pressure would also increase the internal energy.

You have to be careful how you define the process that alters the number of particles. Do you have an open system that allows the particles in from the surroundings?

Assume you transfer an ideal gas at constant T from the surroundings into the system, which consists of a vessel at lower pressure with a small orifice which we seal at some point. Since the gas is ideal there is no change in energy with pressure at constant T. There is no mechanical work done, since there is no volume change, but the change in the number of particles clearly amounts to a transfer of energy. This is accounted for in the term $\sum_i \mu_i dn_i $ of $dU$, provided the entropy is constant.  Changing the number of particles amounts to "chemical" work.

  Secondly, in the same manner I also don't understand why dQ is defined as TdS and not as TdS+SdT. I would think that a change in temperature would surely change the internal energy.

A change in temperature can change the internal energy, but that does not require a change in the definition of heat or entropy. The equation $dQ=TdS$ holds only for a reversible transfer of heat, it is not general.

Now to the tricky part.

You can write

$$dU = -pdV-Vdp +SdT + TdS + \sum_i \mu_i dn_i + \sum_i n_i d\mu_i  $$

since

$$U = -PV + TS + \sum_i n_i \mu_i$$

The thing is you can show that

$$ 0 = -Vdp +SdT + \sum_i n_i d\mu_i  $$

This is known as the Gibbs-Duhem relation.


Does standard electrode potential of a reaction depend on stoichiometric coefficients? Take a reaction like this

$$\ce{Fe^3+ + 3e- -> Fe}$$

now will the standard electrode potential change if we take 2 moles of $\ce{Fe^3+}$ and 6 moles of electrons?

$$\ce{2Fe^3+ + 6e- -> 2Fe}$$

If you think of the Nernst equation, you'll see that: $$\Delta G° = -nFE°$$

Since $n$ corresponds to the number of electrons transferred, the ratio $\frac{\Delta G°}{n}$ remains constant, because any change in the stoichiometric coefficients will multiply both terms by the same value.

Therefore, we see that the potential is an intensive property, not depending on the dimensions of the system under study.


Why does adding an acid speed up electrolysis of water? I've read in several places that when electrolyzing water into $\ce{O2}$ and $\ce{H2}$, that adding a salt or an acid will speed up the process. Why is this?

Primarily due to

  Pure water conducts an electric current very poorly and, for this reason, is difficult to electrolyze.

(Source: Water Electrolysis)

and that salt and acids are able to act as an electrolyte catalyst as most salt and acids dissolve (hence dissociate) in water, specifically, according to the Wikipedia page,

  If a water-soluble electrolyte is added, the conductivity of the water rises considerably. The electrolyte disassociates into cations and anions; the anions rush towards the anode and neutralize the buildup of positively charged $\ce{H+}$ there; similarly, the cations rush towards the cathode and neutralize the buildup of negatively charged $\ce{OH-}$ there. This allows the continued flow of electricity.

A diagram of the electrolysis of water with dissolved salt/acid ions is shown below:

Image source

Further information can be found in the article A Review of Water Electrolysis (Zoulias et al).


How do/did chemists observer/infer the structure of organic molecules and reactions? As I understand organic chemistry there is no simple generalization that allows us to predict reaction outcomes- rather we transform molecules using a sequence of known reactions.

This has me wondering how these known reactions came to be known. My first guess would be that the structure was empirically observed before and after the reaction.  If this is the case what instrumentation would be used to observe this?

If not, how else were these reactions determined and if they were not all developed by trial and error, what is the name of the branch of chemistry that deals with hypothesizing them?

Various crystallographic techniques exist, the lab technique I have used most is x-ray crystallography

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1186895/pdf/mp53000008.pdf

Currently I specialize in computational chemistry, which can estimate (with very high accuracy) the geometry of a molecule which gives the lowest energy (aka the equilibrium geoemetry) through a process called geoemetry optimization

https://www.shodor.org/chemviz/optimization/students/background.html


Boiling sodium hydroxide in stainless steel cup: Solution turning to a blue color I boiled highly concentrated sodium hydroxide in a stainless steel cup. This created a blackish layer on the bottom of the cup and turned the colour of the sodium hydroxide solution to blueish.
Am I right to assume that there was some oxidation happening at the surface of steel? Are any oxides of metals, present in common stainless steel, known to have a blue colour when dissolved in an aqueous solution?

Do you say stainless steel? Stainless steel is an alloy of $\ce{Ni, Cr, Fe}$ with other trace elements, and owes its apparent resistance to corrosion to a protective, adherent, coating of mixed chrome, nickel, and iron oxides. A large amount is probably $\ce{Cr^3+}$, which is amphoteric and will dissolve in hydroxide solution. Once the protective coating is breached chromium will react in base similarly to aluminum. The potential to $\ce{[Cr(OH)4]^-}$ is about $\pu{+1.2 V}$. Stainless steel can be much more reactive than pure iron if the protective layer is continually disrupted.


Is it reasonable to use natural orbitals of an unrestricted HF calculation as a basis to start a CASSCF calculation for a radical Let's say we want to run a CASSCF calculation of a radical, for example the cation of a neutral closed-shell molecule (therefore, we calculate an open-shell radical cation). MOLCAS, as well as probably every other program package out there, needs starting orbitals when performing a CASSCF calculation. For a neutral species, these are normally assumed to be the regular old Hartree-Fock orbitals. Usually, when dealing with a cation, it seems that the Hartree-Fock-SCF calculation is performed for the neutral species or the di-cationic species to yield the necessary HF starting-orbitals. They are in these cases the result of closed-shell calculations. MOLCAS also permits the use of a UHF calculation as the basis for a subsequent CASSCF calculation. Then, the natural orbitals of the unrestricted calculation are used.

Is the usage of the natural orbitals of an unrestricted HF calculation reasonable as a basis for a subsequent CASSCF calculation? If not, why? When is it a good idea / when is it a bad idea? What are the differences in using natural UHF orbitals in contrast to the SCF orbitals of the neutral or di-cationic species? What are the pitfalls?

Edit: I am aware (or at least it is my understanding) that conceptually it is no problem to perform the CASSCF using the natural UHF orbitals, as it could for example also be performed using only the alpha orbitals of the UHF calculation. My question is therefore aimed at finding out in which cases it is generally a good or bad idea to use the natural UHF orbitals.

The best starting orbitals for a CASSCF calculation are optimised orbitals of another CASSCF (or RASSCF) calculation. That sounds a bit ridiculous, but this is probably the best way to figure out the active space.

At first you are probably choosing something quite small, like CAS(2,2) to CAS(4,4). For these calculations it barely matters what kind of starting orbitals you are choosing. The change will be negligible and probably depending heavily on the system you are using.

The advantage of calculating the cations by dismissing the radicals is often a good choice since you can run a very fast RHF calculation and the orbitals are symmetric, which is what you need for CAS.
Natural orbitals of an UHF calculation are also a quite sensible choice, because they already allow for fractional occupation and should also be symmetric. The only downside I can think of is, that this kind of start orbital generation takes a lot longer.
Similarly you can choose to run a ROHF calculation, if you are uncomfortable with too much cationic charge.
As you have said you can choose also an UHF calculation by throwing away one set. I would advise caution here. Sometimes - probably especially when CAS is necessary - spin contamination in UHF is significant. Therefore one set might not give you anything close and you could end up choosing the wrong CAS.

I am not aware of any major pitfalls except for the just mentioned choosing of a completely wrong active space.

In any case it is a good idea to give the CASSCF calculation a few cycles and see how the active space is developing. Sometimes you need to adjust the active space long before the calculation converges and then you will have already a much better guess than with your initial starting orbitals.

I personally use the following routine

R(O)HF [x+] -> CAS(x, ⌈½x⌉ +1)
            -> RAS(x+2n, ⌈½x⌉ +n +1)
            -> CAS(x+2n, ⌈½x⌉ +n +1)
            -> RAS(...)
            -> CAS(...) -> ...

to narrow down the active space and include all orbitals I want and need. In most cases I am not converging the intermediate calculations, except for the minimal CAS.
As you can see the procedure that follows after your start orbitals is much™ more time consuming, so it really does not matter.


How to add a color to vapors? I want to add a color to vapors, like the vapor that comes out of glycerine when heated.

Is there a chemical compound I can put into glycerine that makes the vapor colorful? Or is there a chemical compound over which I can pass the vapors and they get colored?

By means of physics, just vaporise the glycerine and shine light on the vapours -- like it is done for water curtains.  Similar results may be achieved with fog machines (here).  The advantage: your light source defines when and what kind of colour is visible behind a dark background / in a darker environment.  To quote from a lecture by French chemist Hervé This: "Egg white appears only white because of the white light normally used in the kitchen, as soon we use red light it were egg red."

(picture source)

If, on the other hand, you want to use chemical means, mix glycerine, water and appropriate food colours, and evaporate it in a fume machine. A how-to like this may be a good starter.


Why does Chloride attack CrO? In certain applications some sorts of stainless steel (V2A, 1.4301) corrode when exposed to $\small{\ce{Cl^-}}$ in water, or at least it is strongly advised to use other steels.

The conditions that were explained to me as being especially bad were: redox potential >0, acidic environment.

Under these conditions, why do I need $\small{\ce{Cl^-}}$ to corrode steel? I would think that any electrolyte would have a bad effect.

Edit to add 26.7:
Most stainless steel contains chrome, a layer of $\ce{CrO}$ is what's protecting the steel. Most articles mention that $\ce{Cl^-}$ damages this layer. More resistant steels use $\ce{Ni}$ (and are more expensive because of that). So, to narrow down the question, how does $\ce{Cl^-}$ attack the passive layer?

This is an area of corrosion science. Chloride ions do not aid general corrosion  of stainless but rather the special processes called crevice corrosion and pitting corrosion. General corrosion is of large areas. Crevice corrosion is like　a hole in the metal and local corrosion effects occur. Pitting corrosion is corrosive fluid forms its own hole and similar local corrosion effects occur. The local corrosion effects for chloride crevice and pitting corrosion corrosion are complex and dependant on the fluid but a major one is the increase in acidity from  ferric ions (which normal complex with water to be acid) as the chloride ions displace a water molecule and the new complex is more acidic. This acidity damages the complex chromium iron oxide layers. Another effect is that the chloride penetrates the oxide layer and opens it up to allow corrosive species mainly oxygen to penetrate to base metal.

Nickel gives some benefit to corrosion protection of  stainless steels by substituting for iron in the iron-chromium-oxide but that's not main reason its included but rather allows the stainless steel to be austenitic- basically more ductile and easier to weld.
Molybdenum is main element that is added to improve pitting/corrosion resistance. 304 is the most common basic number stainless steel 16-18% Chromium and 8-10% Nickel  (they add letters to give more information) . 316 is second as it contain  16-18% Chromium, 8-10% Nickel and 2-3% Molybdenum- greatly reduces pitting corrosion- conditions need to be more severe to have pitting and crevice corrosion.


Mass of Ice during Phase Change If 18g of ice at a temperature of -6 ℃ is placed in 250 g of water at 20 ℃, we know that there is enough energy supplied by the water to completely melt the ice and transform it into liquid. My question is: during the phase change of the ice, when the temperature of the ice has reached 0 ℃, is mass from the ice transferred to the surrounding water? In other words, is the mass of the ice changing during phase change or is it constant? By the mass of ice, I mean mass that is still solid.

When the temperature of the ice has reached 0°C, additional heat transferred from the liquid water to the ice will not increase the temperature of the ice above 0°C. Instead the transferred heat is consumed as heat of fusion. This lets ice melt, which means that there is a phase change from solid to liquid, which means that a certain mass of the ice is transferred to the liquid water. The mass of the ice consequently decreases during the phase change.


Which is the central atom in Lewis Dot Structures? What is the correct Lewis structure for hypochlorous acid, a compound containing chlorine, hydrogen and oxygen?

As shown in the following video, the central atom in Lewis Dot Structures is always the least electronegative atom of all the elements that make up the atom.
https://www.khanacademy.org/science/chemistry/chemical-bonds/copy-of-dot-structures/v/drawing-dot-structures

This way, you would expect chlorine to be the central atom. However, it is shown that oxygen is the central atom.

The ambiguity exists for molecules like $\ce{OCl_2}$ as well. What's the issue? Is this not a very consistent rule of thumb?

The video is wrong if it says "the central atom ... is always the least electronegative atom".

There are chlorine oxides with Cl central such as perchlorate, chlorate and $\ce{ClO2}$, even though $\ce{Cl2O}$ has O central.

There are $\ce{SeCSe}$, $\ce{CAt4}$, $\ce{NI3}$ and $\ce{Li6C}$ each with the more electronegative element central.


Non-Scanning Mass Spectrometry Does anyone do mass spectrometry with a magnetic sector to separate ions by mass and an array of detectors in a line to measure the ions? If not, why don't they?

Something like this this setup for measuring the spectrum of light?

Yes, this is called distance-of-flight mass spectrometry. It offers some advantages for large (e.g. biological) molecules and because of its high speed for identifying multiple targets, in chemical and biological warfare.


What is the meaning of "superionic"? I see the term "superionic" applied to high pressure water in articles like Giant planets may host superionic water, but I don't understand what the term really means.

How is "superionic" distinguished from simply "ionic"?

Why does the article refer to hydrogen "atoms", which would imply that they are neutral, as opposed to protons or hydronium or some other actual ion?

Superionic water is water under extreme heat and temperature with the properties of both solids and liquids.
It is where the water only exists as a mixture of $\ce{H+}$ and $\ce{OH-}$ ions. At high pressures, the oxygen crystallises, leaving the hydrogen ions free - thus making superionic water highly conductive.
It's crystal structure is a face-centred cubic lattice, and it is thought that Uranus and Neptune hold a layer of superionic water.


Can we really beat the speed of light? This is a rather physics question but I wanted to get an insight from a chemistry perspective. I heard from some that you can beat the speed or go faster than it but is that really true. Physicists as we all know look at the theoretical possibilities but what is your insight from a chemistry-ish view.

Here and here for more information.

If we can then how much more faster is it? I heard one view into this that goes along the lines of:

If you distorted space and moved it then you would be light-years away from where you started. This seemed very dumb at first but as I thought about it is this anyway possible.

There are a number of things that exceed the speed of light:

Particles in a medium such as water can enter faster than light [in that medium] but are slowed and release energy as Cherenkov radiation.
The phase velocity of a wave can exceed that of light [in that medium].
An arbitrary point, such as the point on a pair of scissors where blades,
meet, can move faster than light.
Most puzzling to me: entangled particles. Einstein called this "spooky action at a distance". Although each particle moves at C or less, information might be shared between them in an indefinitely short time; see superluminal communication. Entanglement does apply to chemistry, as do other odd quantum effects; e.g. the detection of single photons with silver halide photography as they form an interference pattern in the double-slit experiment.


How to predict ease of oxidation or reduction for a given oxidising or reducing agent? I have never understood which agent will be able to oxidize or reduce which type of reactants and yield which type of products will be formed. For example I want to know about DIBAL-H and how is it able to reduce some groups but not carboxylic acid. Here is what all I want to know:

What are the criteria for judging which agent will oxidize or reduce?
What will be the condition in which it will be able to oxidize or reduce a given molecule.
What type of products will be formed?

Note: I am in high school and thus need an explanation which I will be able to understand.

This goes to the heart of organic chemistry and most of it is just about learning the properties of the most common reagents, then understanding where to find the detailed information. A good Organic chemistry text book is the starting point, then there are lists to help, for examples list of oxidising agents, list of reducing agents or entire series of books on Organic chemistry reagents e.g. Fieser & Fieser which a good library should have. For individual transformations databases such as Reaxys or Scifinder are searchable by structure and can find the conditions under which a transformation has previously been done. If the transformation has not previously been reported then you have to use your judgement as to which conditions to try, that is where the real art of synthesis comes in. It is important to note that solvent and temperature play a huge role in such transformations. Reactions that proceed in one solvent may give a different outcome in another.

Dibal-H does reduce carboxylic acids, but require multiple equivalents and elevated temperature or additional reagents such as TMSiCl paper here