Find mass of CO₂ and heat released per minute in a combustion reaction I am not sure where to start with this problem. I know that $\small q = mc(T_f-T_i)$, but that seems like it would not help here. I think I need to balance the equation, but I am not sure how the rate factors into the problem. Anyway, if anyone could help me find the relevant equations/method for solving this problem, it would be greatly appreciated:

A natural gas mixture is burned in a furnace at a power-generating station at a rate of 13.0 mol per minute.

If the fuel consists of 9.3 mole $\small\ce{CH4}$, 3.1 mole $\small\ce{C2H6}$, 0.40 mol $\small\ce{C3H8}$, and 0.20 mole $\small\ce{C4H10}$, what mass of $\small\ce{CO2} (g)$ is produced per minute?
How much heat is released per minute?

You know the amounts of the four gases burned in a minute.  Just write combustion reactions for each of these and work out the heat produced by each reaction as written. That's your answer to the second part.  Adding up the carbon dioxide produced in each of the four combustion reactions is the answer to the first part.


Inductive effect and bond strength What is the relation between the inductive effect on bond length and strength.

how would one rank the bond strength of

$\ce{CH3-H}$, $\ce{CH3CH2-H}$, and $\ce{(CH3)2CH-H}$

I know that alkyl groups donate electron density by induction, but how does electron density in a bond affect its length or strength?

Just as hyperconjugation stabilizes carbocations and radicals by sharing the electron density of adjacent $\ce{C-H}$ bonds with the unoccupied/partially occupied orbital, hyperconjugation destabilizes carbanions. So more substitution means less acidic and more ionizable. I had a hard time finding a single resource listing bond lengths, but they seem to correlate with bond-dissociation energy. The wikipedia page for methane states that its $\ce{C-H}$ bond length is 108.7 pm and an answer by Jan says ethane’s is 109.4 pm.

$$ \small
\begin{array}{lcc}
\hline
\text{Hydrocarbon} & \mathrm{p}K_\mathrm{a} ^{[1]} & \text{BDE (kcal)}^{[2]}\\
\hline
\ce{CH3-H} & 48 & 104\\
\ce{CH3CH2-H} & - & 98\\
\ce{(CH3)2CH-H} & 51 &  95\\
\ce{(CH3)3C-H} & 53 & 93\\
\hline
\end{array}
$$

 $[1]$  Evan’s pKa Table
$[2]$ Wikipedia: Carbon-Hydrogen Bond


Why does a mixture of nitrous acid with concentrated nitric acid cause nitration of benzene instead of nitrosation? A question came up in our worksheet to predict the reaction product of nitrous acid and concentrated nitric acid with benzene.

I quickly recognized it to be similar to the nitration reaction. Following this ChemGuide page, which gives the reaction of sulphonation as:

  $$\ce{HNO3 + 2H2SO4 -> NO2+ + 2HSO4- + H3O+}$$

I wrote this reaction:

$$\ce{HNO2 + 2HNO3 -> NO+ + 2NO3- + H3O+}\label{a}\tag{1}$$

and thought that this electrophilic $\ce{NO+}$ would then perform EAS on the benzene to give nitrosobenzene (aka nitrosation reaction).

However, the answer given in my worksheet is nitrobenzene instead. I am unsure if the following reaction (for the generation of the electrophile $\ce{NO2+}$):

$$\ce{HNO3 + 2HNO2 -> NO2+ + NO2- + H3O+}\label{b}\tag{2}$$

is even valid. To me, reaction $\ref{a}$ seems more favorable because it involves the formation of the more stable $\ce{NO3-}$. Then why is reaction $\ref{b}$ favored instead? Why is nitrobenzene the product of this reaction, and not nitrosobenzene?

You were absolutely correct when you said that nitrosation takes place.

What actually happens is that after nitrosation occurs, concentrated $\ce{HNO3}$ oxidises the nitroso group to form nitrobenzene.

Source
A Guidebook To Mechanism In Organic Chemistry; Chapter 6, section 6.2;Peter Sykes


Metallic Aluminum and Dry Ice We see $\ce{2Mg + CO2 -> 2MgO + C}$ , quite often in lab experiments.

Under which conditions can $\ce{4Al + 3CO2-> 2Al2O3 + 3C}$ be used - if chemically feasible?

Based on this (very old) paper - http://link.springer.com/article/10.1007%2FBF00740367#page-1 - the aluminum will ignite if you heat it past the melting point of aluminum oxide, 2060C.


Why is a hydrocarbon tail non-polar? Here it says that the hydrocarbon tail is non-polar. I understand that the $\ce{H}$ on top of the $\ce{C}$ and on the bottom have dipoles that cancel, but at the very end, with the $\ce{CH3}$ group, there is no opposing vector to cancel with that dipole. Why then is the hydrocarbon tail still non-polar?

The hydrocarbon tail is considered nonpolar mainly due to

The Pauling electronegativity of Carbon is 2.55 and Hydrogen is 2.2. Due to the small difference, .35, and the definition of a polar bond, C-H bonds are considered nonpolar.

Indeed, the molecule methane, $\ce{CH4}$ has a dipole moment of 0 D (debyes). I think part of your confusion arises from the picture you provided. Skeletal drawings are 2-d in nature and do not convey information about 3-d. For instance, each carbon has it's hydrogens arranged in a tetrahedral structure, with bond angles 109.5 degrees. The symmetry of this cancels all the vectors contributing to polarity.

If you look at the picture of methane above, and consider the top most hydrogen as the next carbon in the hydrocarbon tail, you can see that in reality, the bottom 3 hydrogens do experience a net dipole moment towards the carbon, and this increased by having another carbon further up in the tail. Relatively speaking, this distance is rather large from carbon 1 to carbon 2. We can say that the vector that arises from this is negligible, and indeed, there is a net dipole moment anyways towards the oxygen at the polar head.


How is the hydration state of a chemical related to its adsorption of water? I am trying to understand adsorption and hydration states better, and I was thinking about the different hydration states of calcium sulfate - anhydrous ($\ce{CaSO4}$), hemihydrate ($\ce{CaSO4*1/2H2O}$), or dihydrate ($\ce{CaSO4*2H2O}$).
I would assume that $\ce{CaSO4}$ and $\ce{CaSO4*1/2H2O}$ would adsorb more water than $\ce{CaSO4*2H2O}$, and I am suspecting that $\ce{CaSO4*2H2O}$ barely adsorbs any water at all since it is "saturated" in a way. But I cannot find any information on this. In fact, I can't seem to find any sources mentioning adsorption in the context of hydration, which is surprising to me - I would intuitively assume that these phenomenons are closely related.
I would therefore appreciate an explanation of the relation of adsorption and hydration. Basically, does a lower hydration state generally imply stronger adsorption? Is adsorption in fact the dominant process by which a chemical changes its hydration state? And how does desorption work for hydrated chemicals?

Surprisingly enough, hemihydrate reacts quickly with water to produce dehydrate (gypsum). It is due to the fact that hemihydrate is produced by heating gypsum to a high temperature. This heating produces a structure with plenty of holes or voids corresponding to the previous positions of the water molecules. Water molecules can easily fill these holes if hemihydrate is mixed with water. Anhydrous calcium sulfate is a mineral (anhydrite) that does not react easily with water. The structure of anhydrous calcium sulfate is compact and does not contain any holes between its atoms to include easily water molecules.


How does the cyclopropyl group influence conjugation and aromaticity? While answering the question Why isn't cyclobutane planar?, I recollected that cyclopropane and its derivatives behave like double-bond compounds, e.g. a cyclopropyl group can potentially be conjugated with neighboring $\ce{C=C}$ bonds and affect aromaticity, like in spiro[2.4]hepta‐4,6‐diene which has pronounced aromatic character (I remember this example from an early edition of March's Organic Chemistry):

Cyclopropyl group is generally known to be a poor $\pi$-electron acceptor, but a very good $\pi$-electron donor [1]. So, I'm wondering whether the cyclopropyl conjugation in general may result in altered aromatic properties, and to what extent; also, on how Hückel's $4n + 2$ rule is applicable here.

I've picked up existing cyclopropane derivatives which I think mimic electronic configuration of well-known aromatic and non-aromatic compounds, assuming each cyclopropyl group acts as a double bond:

[3]Rotane $\ce{C9H12}$ [2] (1) vs. benzene (1a);
Dispiro(2.2.2.2)deca-4,9-diene $\ce{C10H12}$ [3] (2) vs. cyclooctatetraene (2a);
9,10-bis(Spirocyclopropyl)-9,10-dihydroanthracene $\ce{C18H16}$ [4] (3) vs. dibenzo(a,e)cyclo-octatetraene (3a).

Crystal structure of (1), obviously planar triangle (1a is planar as well):

Crystal structure of (2), planar $\ce{C6}$ ring (2a is not planar):

Crystal structure of (3), all three $\ce{C6}$ rings are planar and are within a single plane (3a is not planar):

Provided with these three examples of planar systems for illustration (you can also pick up the systems of your interest), I would like to state a couple of questions:

Is there always a conjugation between a cyclopropyl group and an adjacent double bond? In other words, do cyclopropane's "bent" bonds always have extensive $\pi$-character and can therefore act as an alkene?
1a. Are electronic structures 1 and 1a; 2 and 2a; 3 and 3a similar, and to what extent?
Can we use Hückel's $4n + 2$ and $4n$ rules for cyclopropyl-containing monocyclic rings? If not, how to estimate aromaticity/anti-aromaticity in such systems?

Bibliography

de Meijere, A. Angew. Chem. Int. Ed. Engl. 1979, 18 (11), 809–826. DOI 10.1002/anie.197908093.
Boese, R.; Miebach, T.; De Meijere, A. J. Am. Chem. Soc. 1991, 113 (5), 1743–1748. DOI 10.1021/ja00005a043.
Haumann, T.; Boese, R.; Kozhushkov, S. I.; Rauch, K.; de Meijere, A. Liebigs Ann./Recl., 1997 (10), 2047–2053. DOI 10.1002/jlac.199719971007.
Rosenfeld, S.; White Tingle, C.; Jasinski, J. P.; Whittum, J. E.; Woudenberg, R. C.; Van Epp, J. J. Am. Chem. Soc. 1994, 116 (26), 12049–12050. DOI 10.1021/ja00105a054.

About the molecule that you remember cited in your textbook, there's a study in ref.1. According to these authors:

  The delocalization of the C1-C2 and Cl-C3 electrons of spiro[2.4]hepta-4,6-diene (1a) and partial rehybridization at C1, C2, and C3
  result in a contributing structure which is perhaps best represented as a $\pi$ complex (lb) of cyclopentadienylidene and ethylene.

From NMR studies they concluded that:

  The comparison of the predicted downfield shift
  of approximately 6.6 ppm for the complex and the
  observed shift of approximately 1.0 ppm leads to the
  the conclusion that the $\pi$ complex may contribute as
  much as 15 % to the total electronic structure of 1.

The most recent reference I have found is from 1997(ref.2). They studied these structures:

  In the series of the investigated compounds 1-5, spi- ro[2.4]hepta-4,6-diene (1) shows the strongest  influence of cyclopropyl  conjugation on molecular structure in the group of doubly  unsaturated   cyclopropyl  hydrocarbons

The conjugation properties of cyclopropane are subject of debate and other references don't support this. You can find a blog about this on the royal society of chemistry website(http://my.rsc.org/blogs/84/1009).

I have found another paper in ref.3 and according to this:

  In one chemical attempt to test the transmission
  of conjugation by cyclopropane rings, Cannon, et al.were unable to alkylate or acylate the methyl group in diethyl 2-methylcyclopropane-l, 1-dicarboxylate and concluded that transmission of conjugative effects in the transition state for carbanion formation are small compared with analogous, unsaturated compounds.

Also:

  Trachtenberg and Odian concluded from their review
  of the various conflicting reports that a cyclopropane
  ring is incapable of transmitting conjugation in the
  ground state but may be able to do so in some excited
  states. It is likely that, during chemical reactions,
  such transmission occurs only in a transition state,
  i.e., after bond breaking of the ring has commenced.

From what I have read, even assuming a conjugate effect, this would be much less evident than a C-C double bond.

About your questions:
Is there always a conjugation between a cyclopropyl group and an adjacent double bond? In other words, do cyclopropane's "bent" bonds always have extensive π-character and can, therefore, act as an alkene?
 Partially, the effect is not comparable to a true double bond. This hypothesized conjugation effect was found to be stronger in piro[2.4]hepta‐4,6‐diene(1) compared to the other molecules studied in ref.2(molecules 2 to 6 in the second picture).

1a. Are electronic structures 1 and 1a; 2 and 2a; 3 and 3a similar, and to what extent?
Considering that the supposed conjugation contribution of the cyclopropane ring is much less than a C-C double bond, I think they differ substantially. Unfortunately, I can't find any spectra or other additional data supporting this.

Can we use Hückel's 4n+2 and 4n rules for cyclopropyl-containing monocyclic rings? If not, how to estimate aromaticity/anti-aromaticity in such system?The Huckel rule is not valid in these complex cases. The Hückel's rule is based on calculations using the Hückel method and therefore has a limited validity. Here you need advanced MO calculations and as you have seen from all the references mentioned in such systems there's a lot of debate.

References:
1) Aromaticity via cyclopropyl conjugation. Electronic structure of spiro[2.4]hepta-4,6-diene, J. Am. Chem. Soc.197092154736-4738 https://pubs.acs.org/doi/abs/10.1021/ja00718a047

2) Structural Aspects of Cyclopropyl Conjugation: Experimental Studies and Ab Initio Calculations.  https://doi.org/10.1002/jlac.199719971007

3)Transmission of conjugation by the cyclopropane ring, Stewart JM, Pagenkopf GK, The Journal of Organic Chemistry, 1969. https://pubs.acs.org/doi/pdf/10.1021/jo00838a003?rand=vimssm6d


The sign of enthalpy of formation of magnesium oxide I'm currently doing a lab to calculate the enthalpy of formation for $\ce{MgO}$. However at the moment me and my lab partner are having a disagreement. We've both calculated and agreed upon the same heat released in the reactions ($\pu{4.2\frac {kJ}{mol}}$ (reaction of $\ce{MgO +2HCl})$ and $\pu{7.5\frac {kJ}{mol}}$ (reaction of $\ce{Mg +2HCl}$)). Where we disagree is over how to calculate the enthalpy of reaction.

My argument:

Enthalpy is the amount of energy put into the reaction, what we measured is the heat released thus the equation should be:
$$\Delta H=-\frac{Q}{n}$$
Also, this would make sense considering our reactions were blatantly exothermic, and using this equation provides negative values.

Her argument:

The equation is simple and the teacher agreed with her results:
$$\Delta H=\frac{Q}{n}$$

My question: Who is correct?

I guess you've been answered in the comments already, but anyway. You are correct. In thermodynamics, enthalpy is the energy supplied to the system at constant pressure. If a reaction is accompanied by the release of heat, i.e. if it is exothermic, its $\Delta H$ is negative. If $x$ kJ is released into the calorimeter upon dissolving 1 mol of substance, resulting in the rise of calorimeter temperature, then $\Delta H_{diss} = -x$ kJ/mol.

As for the heat, depends on the type of the calorimeter, i.e. on the direction in which the heat flows. For conventional "coffee cup" calorimeter, where $Q=mC\Delta T$, $Q$ is the energy released, so $Q = -\Delta H$. For some power-compensated calorimeter, where you vary the heater power so as to maintain the temperature, $Q$ would be equal to the energy supplied. Since you have to supply excess energy ($Q>0$) to offset the endothermic reaction ($\Delta H>0$), $Q = \Delta H$.

I've heard that a long time ago, at least in some countries, there were two systems of signs for enthalpies, so-called "thermodynamic" and "thermochemical," with reversed signs in the latter. However, I haven't seen "thermochemical" one in real life, and it seems to be long since abandoned. It is not even mentioned in modern textbooks on thermodynamics. So only the thermodynamic sign system, where $\Delta H>0$ for endothermic processes, remains.


How to find the number of stereoisomers of a tetra-substituted allenes, exhibiting both axial and central chirality? I need a reliable method for determining the number of diastereomers and enantiomeric pairs of the following series of compounds bearing in mind that the substituent groups A are chiral on their own (exhibiting central chirality).

There are two planes of interest bisecting the allene sistem, the one constains the left A groups, the other is orthogonal to it. If two A groups are to be reflected through one plane (so it becomes a symmetry plane) their configuration must be different. That is all regarding the central chirality, but I don't know how to include axial chirality in this description (so I don't count one stereoisomer twice).

I think that these are the four diastereomers. Instead of A, I wrote the configuration of A (R/S). Now, as R and S are connected with a mirror plane, diastereomers 2 and 4 shouldn't exhibit axial chirality, as R and S can be regarded as same substituents on one side of the allene system (because of the mirror plane). That's why the third diastereomer exhibits axial chirality (P/M). I am not sure whether my reasoning is correct. The rest beats me. The answer should be 4 diastereomers and 3 enantiomeric pairs (last page), but how?

Axial and planar are just adjectives we can use to describe what type of chirality a molecule possesses.  Usually, just saying the molecule is chiral is good enough.  Being planar or axially chiral does not create any new enantiomers or diastereomers compared to if we just said the molecule is chiral.

Let's analyze your problem using standard methods - no special techniques just because the molecule happens to be axially chiral.

I get the same 4 diastereomers that you've drawn above (good job!).  When analyzing allene stereochemistry I've always found it convenient to analyze them using Newman projections.  I've redrawn the 4 diastereomers below using Newman projections.

In the top diastereomer, all 4 chiral substituents are the same; the molecule has a non-superimposable mirror image (enantiomer).  Therefore this diastereomer is chiral.

In the second diastereomer, 3 chiral substituents are the same, one is different; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.

In the third diastereomer, two chiral substituents with different chirality are attached to the same carbon; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.

In the fourth diastereomer, two chiral substituents with the same chirality are attached to the same carbon.  This diastereomer belongs to the interesting point group $\ce{S_4}$.  I say interesting because this point group does not contain a $\ce{\sigma}$ plane, but is still nondissymmetric (achiral).  Look at my Newman projection for this diastereomer and picture it lying behind the plane of the screen.  Reflect it in the plane of the screen and then rotate it 90° (n=4, 360/4 = 90°) and you have recreated the original molecule. This molecule is achiral (all molecules belong to point group $\mathrm{S}_n$ [n is an even integer] are achiral).

So what do we wind up with?  4 diastereomers, 3 of which are chiral (enantiomeric pairs) and one is achiral.


Acidity order of organic compouds Here is the question:

We have to find the order of their acidity.
Here is my thought process:
We check the acidity order by finding which conjugate base is more stable. Thus we remove $\ce{H+}$ from all of them. One question I have here is that in (IV), from which position should I remove $\ce{H+}$? Will it be the position from where resonance can take place or from the middlemost carbon?
In (I), (II), and (III), resonance can take place in all of them after removing $\ce{H+}$. Resonance will be more in (II) and (III) than in (I).
Conclusion:
(II) and (III) are more stable than (I). I am uncertain about the stability between (II) and (III) and about (IV).
Please tell me if I have been wrong somewhere and how I could proceed further...

Okay, so if we talk about the acidic strength of a compound, consider adding a base to a solution of all the above. Practically, a reaction is bound to take place-if possible-due to thermodynamic reasons. Hence, reaction in IV will take place since the base will try to find the most acidic Hydrogen, which is why one should consider the removal of H atom from which resonance can take place.

Now for the order of acidic strength, I, II and III are individually more acidic than IV, due to extensive conjugation in all three but the last. This consequently makes the conjugate bases stable and acids stronger. Now for the order among these, notice that Cross-conjugation takes place in III, as we have COO- group. Hence, III will consequently be less stable and a weaker acid.

The order and reason stated by you for II and I are correct, and hence the order for acidic strength is:

II>I>III>IV

i.e. option A.


Polar compounds that doesn't dissolve in water? I'm studying for my exam introduction to chemistry. In the course, there's an exercise where we have to explain why something dissolves in water and why other compounds doesn't.

For me, if the substance is polar, than it dissolves in water because H2O is polar to. (and non-polar dissolves in non-polar).

The first example is AluminumOxide. In the course, it's mentioned that this doesn't dissolve in water. They say that the forces are to high? But it's a polar compound? How do you know if a polar compound will dissolve?

You may accidentally have chosen an awkward example because I'm not sure that $\ce{Al2O3}$ is ever properly molecular. It is usually crystalline and may even be in the form of nano-crystals, in which case there will be considerable lattice binding energy to overcome. In its doped form, with the odd Cr$^{3+}$ ion, it is the mineral Ruby which is clearly not soluble in water.

In the general case polarity is not the only consideration for solubility. There is entropy change of mixing which is always favourable when two species are mixed and also enthalpy change when placing molecules of one type into the other. The enthalpy is where the polarity part comes in.

In water in particular there are other  entropy  changes to consider because water is so hydrogen bonded. The H-bonding structure is broken up when a solute is added and water has  then to 'bridge the gap' as it were round the solute. This forms a more structured water arrangement than before adding the solute and so entropy is reduced. This is primarily responsible for insolubility in water.

We measure the process by considering the free energy change as $\Delta G = \Delta H - T\Delta S$ where $\Delta H$ is the change in enthalpy, $\Delta S$ that of the entropy. When the process if favourable $\Delta G$ will be negative.

Surprisingly adding, for example, non-polar  cyclohexane to water has a change in enthalpy ($\Delta H$) that is close to zero, whereas for a polar molecule $\Delta H$ will be negative. The H-bonding bridging with cyclohexane will make the change in entropy negative so making the free energy positive, which is unfavourable.

With a polar molecule in water H bonding bridging is not necessary to such an extent as for a non-polar molecule and the change in entropy is smaller and so with its negative $\Delta H$ the polar molecule has a negative $\Delta G$ and can dissolve.


What is the derivation for the expression of depression in freezing point? In my textbook, it is written that the expression for depression in freezing point is
$ΔT_{f}=K_{f}m$
But how did this expression come? There is literally no derivation given in my textbook. Can someone please rigorously prove this expression? I am so confused. Please help. (There is also a graph given in my textbook for the derivation; I am not sure if it is a triple phase diagram or not so I did not understand the derivation; Please someone explain).

To derive it, you have to know how an equilibrium depends on concentration and on temperature.
At the freezing point, pure liquid and the solid are at equilibrium. If you lower to concentration of the liquid by adding solute, you disturb the equilibrium. If you lower the temperature by the right amount, you are back at equilibrium.
The relationship you get is:
$$x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\ \Delta T$$
This is from a General Chemistry text that also does not derive it. With some algebra, you can express $K_f$ in terms of the enthalpy of fusion comparing this and the equation given by the OP.


Acidic, Basic, and Neutral Medium So I'm new to this concept and I can't find any proper explanation to this. What does it mean for a reaction to occur in an acidic, basic, or neutral medium?

Why do we call it an acidic medium instead of acidic solution?

This is just a guess, but redox reactions can take place in solution, on surfaces, in the vapor phase, underground, in pastes found in batteries etc. So to have a single word instead of listing all the different environments, we just say the redox reaction takes place in a "medium" that is acidic, neutral or basic.

There are similar collective terms for other situations in chemistry. If you want to talk about reactants, product, catalysts and spectators at the same time, you can refer to them as "species". If you want to talk about atoms, molecules and ions at the same time, you can refer to them as "particles".


Are the collisions between the real gas particles perfectly elastic? Well my question is simple, if two real gas particles are colliding (head on collision) then will the kinetic energy will be conserved i.e. will it be a perfectly elastic collisions?

Well, what is an inelastic collision, really? Suppose you have two balls made of steel; they collide, then fly away with some lasting deformation, so some energy is lost. With molecules, it is not quite like that. You can't leave a dent on a molecule. It has certain discrete energy levels, and that's it. You either excite the molecule to one of these levels, or you don't excite it at all.

To sum it up, some collisions of molecules are perfectly elastic, and others are combined with excitation of some rotational or (more probable at higher temperatures) vibrational mode in one of the molecules, or maybe in both.

Noble gases which have no molecules and hence neither rotational nor vibrational modes may enjoy perfectly elastic collisions up to pretty high temperatures.


How do we know the concentration of hydroxide ions in this derivation? In the derivation of the pH of a solution of a salt prepared from a weak acid and strong base, I used two equilibrium reactions:

  $$\ce{ CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)} \qquad k_{eq} = K_a \\
    \ce{ 2H2O(l) <=> H3O+(aq) + OH-(aq) } \qquad k_{eq} = K_w$$

Reversing the first equation and adding it to the second gave me the "hydrolysis reaction":
$$ \ce{ CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq)} \qquad k_{eq} = \frac{K_w}{K_a} $$
My textbook uses this third equation to derive the pH of the salt solution by asserting that the equilibrium concentrations of the reactants and products of the third reaction are:
$$ \ce{[CH3COO-(aq)]} = C-C\alpha \\
   \ce{[CH3COOH(aq)]} = C\alpha \\
   \ce{[OH-(aq)]} = C\alpha $$

What I am confused about is, why is the equilibrium concentration of hydroxide ions $C\alpha$, and is only governed by the third reaction? Why
does the second equilibrium not contribute any hydroxide ion concentration? That is to say, the derivation assumes the water equilibrium does not exist, and all the hydroxide ion concentration comes only from the third reaction.

Also, even if we suppose some $x$ concentration of hydroxide ions and use it in the 'equilibrium constant expressions' of both the second and third reactions, we have more unknowns then conditions. So that's a problem too.

All in all, I need to know that, why, inspite of hydroxide ion participating in two equilibria, is the hydroxide ion concentration decided by only one of the reactions. And even if approximation is involved, if we try to be accurate, why do we have fewer conditions than variables?

Thank you.

In a solution of acetic ions, say from sodium acetate, some acetate will react with water to produce acetic acid and hydroxide anion (your "hydrolysis" equation"):

$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$

Now, let's say you dissolved 1.00 moles of sodium acetate in 1.00 L of water. For the sake of convenience, you would describe this solution as $\mathrm{1.00\ M\ }\ce{CH3COONa}$. However the actual concentration of acetate ion is less than $\mathrm{1.00\ M}$ because of the reaction with water.

Enter $C$, the "total concentration of acetate". This concentration is the sum of all concentrations of species that are related to or derived from acetate. In this case the relevant species are the acetate ion and acetic acid, so $C = [\ce{CH3COO-}]+[\ce{CH3COOH}]$.

What is $\alpha$? $\alpha$ represents that percent dissociation or percent ionization of the acetate ion. Therefore, $C\alpha$ represents the concentration of acetate that has ionization (reacted with water). In this case, since every acetate that reacts with water produces acetic acid, $C\alpha = [\ce{CH3COOH}]$.

How does this derivation lead to $C\alpha = [\ce{OH-}]$? You need to look back at the reaction equation for the process. For each acetate ion that reacts with water, one acetic acid and one hydroxide ion are produced. Based on stoichiometry, the concentrations of acetatic acid and hydroxide ion must be the same.

How do we determine $[\ce{OH-}]$? We plug the three expressions for the three species into the law of mass action and solve for $\alpha$ (we know $C$):

$$K=\dfrac{K_w}{K_a} = \dfrac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}=\dfrac{(C\alpha)^2}{C-C\alpha}$$

Now, if there was some additional hydroxide $x$ already present, say because the pH was fixed at 12.00, how do we handle that? The expressions for the concentrations of acetate and acetic acid are still valid. $C$ is still the total concentration of acetate + acetic acid. $C\alpha$ is still the concentration of acetic acid. $C-C\alpha$ is still the concentration of acetate ion. However, the concentration of hydroxide ion is different in this case. There is some concentration from the initial conditions $x$, and there is some increase from the ionization of acetate ion $C\alpha$.

If the initial hydorixde concentration is from the autoionization of water, we can account for that:

$$\begin{aligned}
[\ce{CH3COO-}] &= C-C\alpha \\
[\ce{CH3COOH}] &= C\alpha \\
[\ce{OH-}] &= 1\times 10^{-7} + C\alpha \end{aligned}$$

However, except for the most dilute solutions, the contribution from the autoionization of water is usually orders of magtitude smaller than $C\alpha$ and gets lost in the insignificant figures. Let's compare the same calculation for our $1\ \mathrm{M}$ acetate solution. Note that you get the same outcome regardless if you include the hydroxide concentration form $K_w$.

$$\begin{array}{|c|c|c|} \hline
 & \text{without autoionization} & \text{with autoionization}\\ \hline
C & \mathrm{1.00\ M} & \mathrm{1.00\ M} \\
[\ce{CH3COO-}] & C-C\alpha = 1-\alpha & C-C\alpha = 1-\alpha \\
[\ce{CH3COOH}] & C\alpha = \alpha & C\alpha = \alpha \\
[\ce{OH-}] & C\alpha = \alpha & 10^{-7} + C\alpha = 10^{-7} + \alpha \\
K & \dfrac{K_w}{K_a} = \dfrac{10^{-14}}{1.74\times 10^{-5}} & \dfrac{10^{-14}}{1.74\times 10^{-5}}\\
K & \dfrac{\alpha^2}{1-\alpha} & \dfrac{(10^{-7} + \alpha)(\alpha)}{1-\alpha}\\
\alpha & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\
[\ce{CH3COO-}] &  1.00-2.40 \times 10^{-5} \approx 1 & 1.00-2.39 \times 10^{-5} \approx 1 \\
[\ce{CH3COOH}] & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\
[\ce{OH-}] & 2.40 \times 10^{-5} & 10^{-7} + 2.39 \times 10^{-5} = 2.40 \times 10^{-5}  \\ \hline
\end{array}$$


Why does green ferrous sulfate solution change color to yellow upon addition of hydrochloric acid? I was demonstrating the formation of $\ce{Fe(OH)2}$ precipitate at school when the light green $\ce{FeSO4}$ I was using reacted with dil. $\ce{HCl}$ turned yellow. This only happened with dil. $\ce{HCl}$ and not any other acid available at the lab.
[EDIT- I did not wash the reaction and after some time, the ferrous sulphate reacted with nitric acid turning a dark reddish brown, but the colour had vanished the next day.]
I have no idea what this is, but I have a feeling that this is iron being oxidized to $\ce{Fe^3+}$ in solution. Someone please explain what was happening here.

Dry ferrous sulfate heptahydrate is green(ish) and is expected to make a green(ish) aqueous solution.

Anhydrous ferrous chloride is described as green to yellow, and the dehydrate and tetra hydrate are green to blue-green (CRC Handbook).

Now color is broad and continuous (see the visible spectrum),

not digital, like how many fingers am I holding up. This set of curves relating ferrous to ferric comes from https://www.sciencedirect.com/science/article/abs/pii/S0143720816309263 There's a continuity, not a change that can be described as this color to that color.

There is also an effect called dichromatism (not dichroism) in which the perceived color of a material that has some depth will change depending on the depth/thickness or concentration observed (https://en.wikipedia.org/wiki/Dichromatism).
In addition, ferrous ion picks up oxygen easily and becomes ferric (ferric chloride is deep red in concentrated solution (~20% or more), but quite definitely yellow in more dilute solutions (~5%)). The reaction 2Fe$^2$$^+$ + 0.5 O$_2$ + 2H$^+$ --> 2Fe$^3$$^+$ + H$_2$O is favored with lower pH. So your supposition that the (greenish) ferrous ion in solution is picking up oxygen and producing some yellowish ferric ion is almost certainly true. As ferric is formed green will diminish and yellow will increase; the spectral curve will just slide over from the green to the yellow.
However, because there is no quantification mentioned, everything is ishy. Greenish, yellowish, pHish, and yellow and green are both in the green and in the yellow solutions. You also don't mention specifically what other acids you tried.
To explain this to a class, you need some numbers. What you could do is get the visible absorption curve of fresh ferrous sulfate of known concentration (do it at two concentrations, at least), and then get the spectra after addition of (at least) two levels of known concentration HCl. Versus time, because the oxidation will probably take a few minutes in order to get enough exposure to air/oxygen. That will suggest how much of the color change is due to oxidation.
But you should also take the known concentrations of ferrous sulfate and add sodium chloride at the same molar concentrations as the HCl, to see how much color difference occurs because of chloride complexation while keeping the pH essentially constant. These solutions will yellow in time also, but slower than the acidified solutions.
These are some of the variables in your experiment, but how to express this is a short time will be up to you.


Why would hydrogen not be generated at the cathode when electrolysing zinc sulfate? Describe how zinc metal can be obtained from zinc sulfate solution by electrolysis.
  A labelled diagram is acceptable. Include all the products of this electrolysis. The electrolysis is similar to that of copper (II) sulfate solution with inert electrodes.

The answer key says that zinc would be formed at the cathode. However, zinc is more reactive than hydrogen, so logically hydrogen should be given off at the cathode, not zinc.

As I understand it, the ions involved in this electrolysis are $\ce{Zn^2+}$, $\ce{H+}$, $\ce{OH-}$ and $\ce{SO4^2-}$.

Also, why can possible cathodes in this reaction be carbon, platinum OR zinc while the anode can only be carbon or platinum? What allows the cathode to be zinc?

In an electrolytic cell we are using the electrical energy from an outside power source to drive a chemical system against the direction it would spontaneously move.  In this case, we can take a solution of zinc ions and force them to become zinc metal.  As long as we put the zinc ions in an environment where they can be reduced, they will be.

Consider the half reaction that occurs when zinc metal is produced:
$\ce{Zn^{2+}(aq) + 2e- -> Zn(s)}$

This is a reduction reaction, and reductions always occur at the cathode, whether the cell is galvanic or electrolytic.

The other half reaction involves hydroxide being converted into water and oxygen.
$\ce{4 OH- -> 2 H2O + O2 + 4e-}$

This is an oxidation reaction, and oxidations always occur at the anode.

Platinum and carbon are nonreactive in this system, so they can be used for either electrode.  Zinc can be used as a cathode because that's where it is being deposited, so we're just adding additional metallic zinc onto our already existing zinc electrode.

With the idea that oxidation occurs at the anode in mind, what would happen if we used a zinc anode?


Is diamond transparent to X-rays? Why is diamond said to be transparent to X-rays? Does this statement mean that X-rays can pass through diamond? If that is the case, then how does one explain this phenomenon?

There are x-rays (Li K-$\alpha$ is 50 eV) and there are x-rays (Uranium K-$\alpha$ 98 keV).

In reality all electromagnetic radiation tries to pass through everything. Let $\rm I_0$ be the intensity of the beam at the surface and $\rm I_x$ be the fraction that gets through to depth $x$, then:

$\rm I_x = \rm I_0 e^{-\mu\rho x}$

where:

$\mu$ is the mass absorption coefficient, units = $\rm{cm}^2/\rm g$
$\rho$ is the density of the material, units $\rm g/\rm{cm}^3$
$x$ depth of radiation, units = $\rm{cm}$

As a gross generality as the energy of the radiation increases (wavelength shorter) the transparency of matter increases. So a thin slice of carbon, be it graphite or diamond,  will stop the vast major of Li K-alpha x-rays, but very few U K-alpha x-rays.

Note with some rearrangements and variable substitution this gives the Beer-Lambert Law used in UV-Vis spectroscopy.


How to translate the mobile phase from thin layer chromatography to column? I recently was told that once you find a good solvent mixture for separation on thin layer chromatography (TLC), you should only use 10% of the more polar solvent (for silica) when you run your column. I tried it and it worked great!

Since I'd never heard that before, is that a standard rule of thumb, or did I just get lucky?

I never heard about the "10% rule" and I never applied it. Transferring TLC (silica on aluminium plates, Merck) results to preparative column chromatography on similar silica usually worked without changes, although the situation is a bit different in the column: the upper part of the TLC plate is rather dry when letting it run, while the silica in the column is not. Note that I was never a friend of the dry packing method and always prefered the slurry method to prepare the columns.

Anyway, I never thought about increasing the polarity.

Bad separation typically was a result of column overload, bad packing or cracking of the stationary phase, e.g., when the diethylether evaporates in the column.

+1 For citing the classic article on flash chromatography!

I can really recommend the technique and I used it ... more than once ;)


Atmospheric Humidity Effects on Non-water Evaporation Does evaporation of non-water liquids into the atmosphere vary with atmospheric humidity?

Yes, because though the liquid phase initially contains no water, water molecules may enter the liquid phase from the gas phase.  The more hygroscopic the liquid is, the more the evaporation rate will be affected by humidity.  If the liquid is very hygroscopic and has a low vapor pressure, the initial evaporation rate may even be negative in humid conditions, as water enters the liquid phase at a rate greater than liquid enters the gas phase.


analytical chemistry standard deviation Analysis of a bottle of $100$mg of vitamin C, yields an average vitamin content of $99.8\pm0.3$ mg. Assuming Gaussian statistics, which of the following is true and the answer is $68\%$ of tablets contains between $99.5$ and $100.1$ mg of vitamin C how is that?

You should look into what is meant by Gaussian statistics or normal distribution. In the case of a normally distributed random variable, approximately 68% of the set should be within one standard deviation of the mean. Particularly look at the section entitled "Standard deviation and coverage".


pH of bleach solution slowly decreasing? I work within a science department in a school. We are demonstrating Universal Indicator and something odd has happened. Our bleach solution (~2.5% Sodium hypochlorite) turns blue when UI is added but within about 1 min it goes green and then after 1-2 mins it goes yellow. Overnight it goes clear but when more UI is added it is yellow. Why would this happen? The bleach smells like bleach and has been used in the last week for the luminol reaction which worked.

Looks like your bleach bleaches your indicator, just like it does pretty much everything else. After all, we don't call it "bleach" for nothing.

Sure, pH may be in fact slowly decreasing, and the bleach would eventually decompose and lose its properties, but these processes do not typically occur in a matter of minutes.

As to the mechanism, it is complicated, as are indicators themselves (usually, large organic molecules). One might think of a huge conjugated system of double bonds that produces color; all of a sudden, an oxidant appears and starts randomly snapping bonds in half. Soon the conjugated system is gone, and so is the color.


transpiration of trees and the cooling of the air The change from the liquid state of water to the water vapor requires a large latent heat to loosen the hydrogen bonds. The process of transpiration in the plant involves exactly this change of state. Here, the water evaporates from the lower leaf surface and the leaf gives up some of its heat. In this way, the leaf is prevented from overheating that would result from exposure to sunlight. Sweating in animals involves this same latent heat aspect of water.

If trees use transpiration to decrease their temperature, shouldn't the air around them  be warmer? The leaves are cooled as water evaporates, and the air should be warmer.

I always experience the opposite as I approach a forest. I'm not sure if the effect could be attributed to the reduced irradiation due to foliage.

You seem to misunderstand the thermodynamics of evaporation: "Evaporation is an endothermic process, in that heat is absorbed during evaporation." It appears you are confusing evaporative cooling with a heat pump. In a heat pump, the compression half of the cycle changes the vapor back to liquid, which does make things warmer.

Consider boiling water at sea level: Water liquid at 100°C becomes water vapor at 100°C, requiring ~540 cal/g energy to change state from liquid to vapor at the same temperature. At lower temperatures, as water evaporates, thermal energy is withdrawn from the liquid, cooling it below ambient temperature. "Swamp coolers" work on that principle.


Why does MgCl2 lower the freezing temperature, and how does it do so? I am working on a project in my chemistry class, and even though we've already finished the lab, I'm still confused as to why and how $\ce{MgCl2}$ melts ice/ lowers the freezing temperature of water. Can anybody tell me why this happens?

The $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$ ions disrupt hydrogen bonding, preventing ice from forming a lattice structure. Therefor, the water must be colder so those weak hydrogen bonds will be acting on water molecules with less kinetic energy (i.e., colder), finally allowing them to create a crystal around the dissolved "impurities" $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$.

This is a basic colligative property. Its more effective for electrolytes because they dissociate, creating more lattice disruptions. As a general resource, see: Ebbing, Darrell D., General Chemistry, 3rd ed, Houghton Mifflin, 1990.


How should I identify my leaving group in SN1 reactions if there might be more than one option? Before I can even begin to contemplate the products of a given SN1 reaction, I need to be able to determine what my leaving group will be, which should be a species with a strong conjugate acid, and as I understand it the species should be relatively stable once it has left the substrate. Please correct me if I am making any mistakes in my understanding.

A hydroxide ion will be a terrible leaving group, since its conjugate acid is water, and hydroxide is not very stable by itself.

So when I am asked this question "Draw the major organic product(s) for this SN1 reaction:

How do I know what will be the best leaving group? Hydroxide is not very good, so perhaps $\ce{CH_3OH}$ will work better? How do I know it is not $\ce{CH_3CH_2OH}$ instead?

Is this only one decent option, and I am over thinking this?

This case (and similar cases) you need to think about the following:

When a leaving group leaves, only one bond is broken. By this logic, $\ce{OH}$ is a leaving group - it can leaving by the breaking of the $\ce{C} -\ce{O}$ bond. The alcohols you mention, $\ce{CH3OH}$ and $\ce{CH3CH2OH}$ require breaking two bonds to leave: the the $\ce{C} -\ce{O}$ bond and a $\ce{C} -\ce{C}$ bond.
What is that acid $\ce{H-Br}$ doing? When you are given a strong acid as a reagent, very commonly the first step in the reaction's mechanism is a proton transfer (an acid-base reaction). Proton transfers are very rapid; they often happen before anything else has a chance to. In this case, the proton transfer is also exothermic (or exergonic if you prefer) - the equilibrium constant is on the order of $10^8$ in favor of the products. What does the product of the initial proton transfer look like? Is there a better looking leaving group on that structure?


What would the pH change be if I drop one gram of sodium hydroxide into a pH 8.1 solution I have an experiment where I drop 1 gram of lab-concentration Sodium Hydroxide into a litre sea water (pH 8.1). What should be the expected change in pH? Would this change be constant? i.e if I was to drop 2 grams of NaOH into the sea water, would the change double?

If we know that the pH is 13 (let's take it as 13 for the simplicity), then we can back-calculate the concentration of sodium hydroxide in this solution from $\ce{14=pH + pOH}$. This gives us pOH=1, thus the concentration of sodium hydroxide of 0.1M. Since 0.1M sodium hydroxide is hardly a very concentrated solution, we can safely assume that 1 gram of the solution is going to be roughly 1 mL. Thus, with each 1 gram of the lab solution you're adding about 0.0001 moles of $\ce{NaOH}$ into your 1 liter of sea water giving you $1*10^{-4}$M solution with the pH = 10.

Using the same trick to calculate the concentration of $\ce{OH-}$ in your sea water as we did for the stock sodium hydroxide solution, we get about $1.25*10^{-6}$M, which is negligibly small compared to $1*10^{-4}$M solution that you get after adding sodium hydroxide to it. Thus, after 1 gram of sodium hydroxide in a liter of sea water your pH should go from 8.1 to about 10. If you add 2 grams, you'll get a $2*10^{-4}$M solution, which yields pH of about 10.3.


Can suspensions be separated by centrifuge or by separatory funnel? I want to know if we could separate suspensions whose particle sizes are more than $\pu{100 nm}$ like mud by a normal centrifuge?

Could we use a separatory funnel for these situations?

If you are indeed dealing with a proper suspension (not a colloid and not a solution) then centrifugation should always work to separate the components. It may be that you need a very high rotation speed or a long time, but you will manage to get the solid particles to the bottom of the tube and the clear liquid above.

A separatory funnel is a bad idea for a number of reasons. For one, you don’t know how long it will take for the components to separate. For two, you will probably have difficulties removing the solid portions; they might clog the opening or they might attach together not intending to go through at all. Finally, you will always get some liquid together with the solid fraction in a separatory funnel while you should arrive at a much better dryness with centrifugation.


What is a good way for a laymen to purify alcohol to near 200 proof? I'm making herbal and medicinal mushroom tinctures and extracts.  Normally I'll use everclear but it isn't cheap and 95% alcohol.  Also I don't know what that other 5% is.  Is it just water?

Is there a way I can reduce alcohol to near 100% without a still?  I've heard poring it over epsom salts through a few layers of coffee filters will remove some of the non alcohol contents.  Is this true?

Distillation of ethanol yields 95% ethanol and 5% water. That is the best one can do unless you don't want to drink it afterwards (it will have a little benzene, a known carcinogen). The Epsom salts may remove some water, but I don't think it will remove all the water, not to mention that some of the Epson salt will dissolve into solution.


What's the Acid Ionization Constant of the Hydrogen Proton? What is the $\ce{K_{a}}$ value for $\ce{H^+}$. I understand that the hydrogen proton doesn't stick around in solution by itself for very long. Nonetheless, does it have a  $\ce{K_{a}}$ value? Can it have a  $\ce{K_{a}}$? I think it can have one but quantifying it might be hard. Its  $\ce{K_{a}}$ value I think would correspond to this reaction:

$\ce{H^+ + H_2O ->H_3^+O}$

However this doesn't look like a conventional acid/base reaction; we don't have the conjugate base of $\ce{H^+}$ here, unless the definition of a conjugate base isn't what I think it is (acid missing one proton).

So my questions are:

1) Does the above mentioned $\ce{K_{a}}$ value exist.

2) Is my definition of a conjugate base too narrow?

The equillibrium between a proton plus water molecule and a hydronium ion is better considered using the concept of proton affinity.

\begin{aligned}
\ce{H^+(g) &-> H^+(aq)} \\
\Delta E &= −1530~\mathrm{kJ/mol}
\end{aligned}

You could roughly convert to an equillibrium constant using

$$\Delta G = -RT \ln(K)$$


Is sodium percarbonate artificial? If I add sodium percarbonate to my natural laundry soap recipe, can I still claim "No Artificial Anything"?
I know it's used in many "natural" products, but is it considered an "artificial ingredient"?
I don't want to make any false claims with my product.

thank you!

Yes, it is artificial. Sodium carbonate is a substance that occurs in nature, even though any of it is most likely manufactured.

Sodium percarbonate, on the other hand, does not occur in nature and it has to be manufactured. It is an oxidiser, and oxidisers don't occur in nature normally. That's a problem for you, because oxidisers (aka bleach) are desired in laundry applications.


Compare reactivity of carbocations and stability of carbanions, radicals, and carbocations For ranking the carbocations in the image above,
I thought of taking the approach that the 1st molecule, the secondary carbocation was the most reactive because it has no electron donating group, then the carbocation with a terminal NH2 because nitrogen has an electron donating group, then the tertiary carbocation, and lastly the last carbocation has resonance making it very stable and less reactive. Is this the correct train of thought?

Order of stability: 3 > 2 > 1 > 4
I would say that the third structure is the most stable (least reactive) as an octet-complete resonance structure of it exists which is much more stable that the initial carbocation. This is basically what you said, the carbocation is stabilized due to the electron donation tendency of the nitrogen atom in this case.

Then structure 2 then 1 since 2 is tertiary while 1 is secondary. More the alkyl groups, more the positive inductive effect stabilizing the carbocation.
I would say that the fourth structure is the least stable. If you try to draw the resonance structure in which oxygen ends up donating electrons stabilizing the carbocation, you'll notice that the oxygen ends up with a positive charge which is even more unstable than the initial carbocation. Resonance does not help here.

(the structure above should have a positive charge on the oxygen; I wasn't able to draw that using the software available)
Do note that resonance structures don't actually exist and there are better ways to explain more or less the same thing. I've stuck with resonance since you've mentioned that in the question.


Is it correct to write Na2B2O4 as 2NaBO2? My teacher wrote the reaction for heating of borax like this:
$$\ce{Na2B4(OH)4 .8H2O->Na2B4O7 + 10H2O}$$
$$\ce{Na2B4O7->B2O3  +Na2B2O4}$$
And told us that $\ce{Na2B2O4}$ can be written as $\ce{2NaBO2}$,is this correct?
If yes then why?

No, $\ce{Na2B2O4}$ can not be written as $\ce{2NaBO2}$. $\ce{Na2B2O4}$ means 1 mole of $\ce{Na2B2O4}$ and $\ce{2NaBO2}$ means there are 2 moles of $\ce{NaBO2}$. Both are different compounds and have different molar mass.

Actually in real $\ce{2NaBO2}$ should be written as $\ce{(NaBO2)2}$. Now, both $\ce{Na2B2O4}$ and $\ce{(NaBO2)2}$ are same.


how to speed up Ferric Chloride and copper reaction im doing pcb etching process (for those who don't know pcb is that thing with electronics on it lmao) using Ferric Chloride and it's taking forever

is there any chemicals i can find that will speed up this process (could it also be found at home, salt or acid for instance)

thanks in advance

Chemicals are probably not the good way to go. What is the temperature of the solution? Increasing the temperature speeds up the reaction. Btw: take the proper safety precautions when etching pcb's!


What easily available chemicals do I use to turn oak light grey? I have 10 square meters of dry and freshly sanded oak that I would like to artificially age/weather to the colour of the above photo. Once treated the oak is to be used for a bedroom interior. I was wondering if there are any chemists out there who may know how to do this.

Graying is symptomatic of bleaching (oxidation) caused by air and sunshine. You can use chemical bleach for a similar effect.
See for instance: https://www.wikihow.com/Bleach-Wood.
There are a lot of guides online on the subject. This one seems pretty useful, discussing different bleaches, their safe handling, and treatment steps. Even the US Department of Agriculture has a guide on the subject.
The appropriate bleach for your application appears to be the "two-part (A/B) wood bleach" in which a mix of peroxide and sodium hydroxide reacts with the wood. Commercial kits are available. The USDA guide has the following to say:

One of the most powerful, satisfactory, and widely used bleaches in recent years is based on  a  strong  solution  of  hydrogen  peroxide  (30  or  35  percent  strength).  The  wood  is  first coated evenly with a solution of sodium hydroxide (caustic soda, or ordinary lye) containing pounds of the solid per 10 gallons of solution.  After this has dried for about 30 minutes, an even  coat  of  the  hydrogen  peroxide  is  applied.  The  sodium  hydroxide  applied  as  the  first coat  renders  the  hydrogen  peroxide  unstable,  whereupon  it  decomposes  and  liberates  a  large amount of oxygen in a chemically active condition, which does the bleaching. The liquid on the  surface  should  be  allowed  sufficient  time  to  dry,  during  which  time  the  bleaching  occurs; then  the  wood  should  be  rinsed  thoroughly  with  cold  water.  To  neutralize  any  remaining traces  of  sodium  hydroxide,  the  surface  should  then  be  washed  or  sprayed  with  a  solution containing 1 pound of acetic acid (or 1 pound of oxalic acid) per 10 gallons of water, allowed to stand for 15 minutes, then rinsed thoroughly with water and allowed to dry. After the wood is sanded lightly to remove raised grain the new finish may be applied.

See also the section labeled "precautions".
Regarding the chemistry of the bleaching process, Wikipedia provides the following summary:

Hydrogen peroxide
Using hydrogen peroxide to delignify chemical pulp requires more vigorous conditions than for brightening mechanical pulp. Both pH and temperature are higher when treating chemical pulp. The chemistry is very similar to that involved in oxygen delignification, in terms of the radical species involved and the products produced.
Oxygen
Oxygen exists as a ground state triplet state, which is relatively unreactive and needs free radicals or very electron-rich substrates such as deprotonated lignin phenolic groups. The production of these phenoxide groups requires that delignification with oxygen be carried out under very basic conditions (pH >12). The reactions involved are primarily single electron (radical) reactions.[14][15] Oxygen opens rings and cleaves sidechains giving a complex mixture of small oxygenated molecules. [...] While the radical reactions are largely responsible for delignification, they are detrimental to cellulose. Oxygen-based radicals, especially hydroxyl radicals, HO•, can oxidize hydroxyl groups in the cellulose chains to ketones, and under the strongly basic conditions used in oxygen delignification, these compounds undergo reverse aldol reactions leading to cleavage of cellulose chains.

Lignin is a polyphenolic compound that lends wood much of its color, so it is the main target of bleaching. The cellulose is best left alone, so a relatively mild bleaching protocol is best for wood.
Another important factor affecting the aged appearance, evident in the posted image, is more difficult to replicate with a chemical treatment. That is water loss, which is responsible for cracking. A quick surface treatment might not provide the highly aged appearance in the picture$^\dagger$.
Update (and somewhat off-topic)
$\dagger$ Oak is apparently among the most difficult woods to dry without it developing cracks. If the wood is already dry then it is unlikely to crack, but if it is fresh and is allowed to dry quickly (for instance by setting in the sun for a few weeks) than it is almost certain to crack.


Why is the plateau on a heating curve in equilibrium? From the interval of 100 to 140 minutes, why is the substance in equilibrium?

I believe it's because at this interval, freezing and melting begin to occur at the same rate and, when two opposing reactions occur at the same rate, equilibrium is present. However, since the potential energy is decreasing (since the substance is freezing), the equilibrium favors the exothermic reaction causing the substance to eventually become solid in the interval of 140 to 160 minutes.

You are right about the equilibrium existing because both freezing and melting occur at the same rate.
Imagine ice and water kept in a thermos flask under normal atmospheric pressure. You will observe that there is no change in the mass of both the phases.
But there is actually a lot of activity going on. Some molecules of ice enter the liquid phase and some molecules of water enter the ice phase. As both these processes occur at the same rate you don't see any change in the mass of both the phases.
In fact such equilibriums exist for all pairs of solids in their own liquids and gases above their own liquids at particular temperatures.
In your case I think there is a mechanism to extract heat from the system which shifts the equilibrium towards liquid changing to solid ie freezing.


Calculate limiting molar conductivity of BaCl2 How can I calculate limiting molar conductivity of BaCl2 in water at 25 degree Celsius?
I have got a formula for calculating limiting molar conductivity .

limiting molar conductivity  =  V+ A+ +
V- A-
V+  =  number of cations per formula unit of the
electrolyte
V-  =  number of anions per formula unit of the
electrolyte
A+ = molar conductivity of the cations
A- = molar conductivity of the anions

Here I figured out, V+ = 1 and V- = 2
How can I solve the question and tell me  if there is any information that is not present above to solve the question.

I'll answer the second part of your question first: you have one equation with three unknowns, so you need the value's of two more unknowns, namely the molar conductivities of Ca++ and Cl-.  They are readily available from the CRC Handbook of Chemistry and Physics.  Every chemist and chemistry prof and library has one.  You could also find those values from online resources, just be sure to check with reliable sources.  An internet search will turn up many.  Once you have those two values you just plug them in to your equation and you are done.  You are correct with your V values.


What forces keep a cotton fiber together? According to Wikipedia, a single cellulose molecule is:

  a linear chain of several hundred to many thousands of β(1→4) linked
  D-glucose units.

which means a cellulose molecule is made of ~1μm of glucose units.

According to Trotman, 1984,

  Each cotton fibre is a unicellular hair [...] Each fibre, however long
  or short it may be, consists of one single complete vegetable cell
  [...] Average fibre length 13-51 mm.

  (6th ed., pages 27-33)

  (source: Trotman, page 31)

So if cellulose molecules are only ~1μm in length, what forces keep the cellulose molecules together inside the S1,S2,S3 structures throughout the fiber? How ~1μm units end up giving cotton its mechanical properties?

A glucose molecule is roughly 1 nanometer long (ref).  There are roughly 9,000-15,000 glucose molecules in a cellulose fiber (ref, see Table 5).  If we use 10,000 as the number of glucose units, then the length of a cellulose fiber will be
$$10^{-9}~ \mathrm{m/molecule} \cdot 10^4 ~\mathrm{molecules/fiber}=10^{-5}~\mathrm{m/fiber} $$
or ~10 microns/fiber.

The mechanical strength of the fiber comes from two sources.

Structural: As your diagram illustrates there are many walls in the cellulose fiber.  All of these walls reinforce one another and build up the mechanical stability of the fiber. Each wall is composed of highly-organized cellulose chains.  "Highly-organized" is another way of saying "crystalline".  Cotton has an average crystallinity around 75%.  This allows the cellulose fibers to pack closely together further increasing mechanical stability.  The cellulose fibers in the different walls are orientated in different directions, again serving to increase the fiber strength.
Chemical: Here is a drawing of 2 glucose molecules from a cellulose chain.

Note that each glucose molecule has 3 equatorial hydroxyl groups oriented out and away from the polymer chain.  Because of the high crystallinity mentioned above, the cellulose chains are positioned very close to one another.  This means that thousands of hydrogen bonds can form between hydroxyl groups on different chains.  As a rule of thumb, a hydrogen bond might have a strength around 5 kcal/m; if we have thousands of these bonds holding the cellulose chains together, then we have the equivalent of hundreds of normal chemical bonds holding the chains together!  These hydrogen bonds account for much of the stability of the cotton fiber.


When calculating a dihedral angle for a chemical bond, how is the direction defined? What are the most common convention? I understand the geometry and calculation of a dihedral angle, discussed in this question. It is not clear how the direction is defined for a chemical bond.  On the wiki page for dihedral angles, it states that, "the dihedral angle $\phi$ is the counterclockwise angle..." between the planes $P_{123}$ and $P_{234}$ (where $P_{123}$ is defined as the plane containing atoms 1, 2, and 3 in the diagram below, and likewise for $P_{234}$).

$\hspace{45 mm}$

This counterclockwise definition means that the direction of the dihedral angle is defined using the left-hand rule about the line connecting atom 2 to atom 3.  The left-hand rule is unusual in mathematics and geometry.  Alternatively, this example may be defining the directionality of $\phi$ simply so the result is a positive angle less than $180^\circ$.

I have not been able to find any chemistry standards on this.  Is there a common convention?

The IUPAC Gold Book states that,

  In a Newman projection the torsion angle is the angle (having an absolute value between 0° and 180°) between bonds to two specified (fiducial) groups, one from the atom nearer (proximal) to the observer and the other from the further (distal) atom. The torsion angle between groups A and D is then considered to be positive if the bond A-B is rotated in a clockwise direction through less than 180° in order that it may eclipse the bond C-D: a negative torsion angle requires rotation in the opposite sense.

This implies a range of -180 to 180, which is by far the most typical range used, but I certainly have seen reports ranging from 0..360, e.g., 225 degree dihedrals.


Why does (2z²-x²-y², x²-y²) have -1 trace in the character table of Td for C3 rotation? I'm trying to understand the irreducible rep for $(2z^2-x^2-y^2, x^2-y^2)$ in the $T_\mathrm{d}$ point group. Specifically, for $C_3$ rotations, a trace of $-1$ is listed in the character table for this group. For any matrix, one can rotate the coordinates (basis for the matrix) - i.e. an orthogonal transformation, yielding a matrix with different entries, but the trace will remain the same. So in the case of $120^\circ$ rotations around an axis through a vertex and opposite face of the tetrahedron, one can rotate the coordinates so that the $z$-axis aligns with rotation axis making the calculation of the trace easier.

Any rotation in this basis will leave $2z^2-x^2-y^2$ unchanged, ie. a $1$ on the diagonal. However, the projection of a rotated $x^2-y^2$ back to its own basis will yield $(\cos 2\theta)(x^2-y^2)$. The trace will then be $1 + \cos 2\theta$.
Can anyone help me where I went wrong?

I think I know now.

Choose rotation axis that makes equal angles with the $x$-, $y$-, and $z$-axes so that the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose a $C_3$ rotation that permutes $x$ to $z$, $z$ to $y$, and $y$ to $x$. Rotation in either direction will give the same trace with this $C_3$ rotation. With this rotation (permutation), $(2z^2-x^2-y^2,x^2-y^2)$ becomes $(2y^2-z^2-x^2,z^2-x^2)$. Multiplication of basis function vector a by $2\times 2$ matrix in this representation yields

$$\begin{pmatrix}
  -1/2 & -3/2\\
  1/2 & -1/2
\end{pmatrix}
\begin{pmatrix}
  2z^2-x^2-y^2 \\
  x^2-y^2
\end{pmatrix}
=
\begin{pmatrix}
  2y^2-z^2-x^2\\
  z^2-x^2
\end{pmatrix}
$$

The trace of this matrix is $-1$ as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

The tricky thing about this example was that I needed to use a rotation in a three-dimensional vector space and project it into a rotation in a two-dimensional vector space.


Why is the helium wavefunction a linear combination? the general solution for the wave function of Helium is a linear combination of the two possible states:

$$
\text{state 1:} \quad \psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2}) \\
\text{state 2:} \quad \psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1})
$$

(where $\vec{r}_{1}/\vec{r}_{2}$ denotes electron 1/2.)

But aren't the products of these wavefunction the same? Why then make a linear combination; it would be just like same function +/- same function?

$$
\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})
=
\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ?
$$

First of all, note that
$$
\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})
\neq
\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ,
$$
since same functions in these products ($\psi_a$ and $\psi_b$) have different arguments ($\vec{r}_{1}$ and $\vec{r}_{2}$) to the left and to the right of the inequality sign.

Now, to the question, why do we have a linear combination. For a system of two indistinguishable particles (say, electrons) the obvious requirement is that there should be no observable difference between the system in state $\psi(1, 2)$ and the system in state $\psi(2, 1)$,
\begin{equation}
    |\psi(1, 2)|^{2} = |\psi(2, 1)|^{2} \, ,
\end{equation}
where $1$ and $2$ stand for coordinates of particles.
This implies that
\begin{equation*}
    \psi(1, 2) = \pm \psi(2, 1) \, .
\end{equation*}

And in order to satisfy this equality in what is often called the orbital approximation, where the state of a many electron system $\psi$ is represented as a product of a single-particle states $\psi_i$, the wave function $\psi(1, 2)$ should be actually written as not just a product of $\psi_1$ and $\psi_2$ (none of which, neither $\psi_1(1) \psi_2(2)$ nor $\psi_2(1) \psi_1(2)$, is symmetric or antisymmetric), but rather as the following linear combination of these products,
\begin{equation*}
    \psi(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) \pm \psi_1(2) \psi_2(1) \Big) \, ,
\end{equation*}
where $1 / \sqrt{2}$ is the normalization factor.

The choice of the sign in the expressions above is not arbitrary, but rather determined by the particles spin $s$, namely,
\begin{equation*}
    \psi(1, 2) = (-1)^{2s} \psi(2, 1).
\end{equation*}
Particles with integer spins, called bosons (photons, alpha particles), are described by a symmetric wave function
\begin{equation*}
    \psi_{\mathrm{s}}(1, 2)
    =
    \frac{1}{\sqrt{2}}
    \Big(
        \psi_1(1) \psi_2(2)
        +
        \psi_1(2) \psi_2(1)
    \Big) \, ,
\end{equation*}
while the particles with half-integer spins, called fermions (electrons, protons, neutrons), are described by the anti-symmetric wave function
\begin{equation*}
    \psi_{\mathrm{a}}(1, 2)
    =
    \frac{1}{\sqrt{2}}
    \Big(
        \psi_1(1) \psi_2(2)
        -
        \psi_1(2) \psi_2(1)
    \Big) \, .
\end{equation*}


Hyperconjugation and the stability of alkenes Clearly, this issue has been touched on here and here. However, I would like to raise a new point on the issue.

I have always thought that stabilisation of the $\ce {C=C}$ $\pi$ bond via hyperconjugation is through an interaction between a $\ce {C-C}$ or $\ce {C-H}$ $\sigma$ MO and the $\pi^*$ MO. This is also the explanation presented by orthocresol and Philipp in the posts linked above. However, the explanation given by Fleming (2009) is somewhat different as can be seen from the diagram below, taken from p. 74

There seems to be an internal contradiction in the text. On the previous page, Fleming does mention about the interaction between the $\sigma$ MO and the $\pi^*$ MO yet the diagram shows an interaction between a $\sigma$ MO and the $\pi$ MO. Not sure if it's a misinterpretation on my part...

On p. 73, he wrote:

  One factor is the hyperconjugative stabilisation of the $\ce {C=C}$ $\pi$ bond by the alkyl groups. Fig. 2.11 shows the interaction of the orbitals of a $\sigma$ bond with the orbitals of a $\pi$ bond, which is similar to the interaction of two $\pi$ bonds in butadiene (Fig. 1.32). Although the $\sigma$-bonding orbital and the $\pi^*$-antibonding orbital are further apart than $\psi_2$ and $\psi_3^*$ in butadiene, they are just close enough to mix in a bonding sense effectively to lower the energies of $\psi_1$ and $\psi_2$ in Fig. 2.11, making the drop in energy $\ce {E1}$ a little greater than the rise inenergy $\ce {E2}$.

Reference

Fleming, I. Molecular Orbitals and Organic Chemical Reactions. John Wiley & Sons, Ltd. United Kingdom, 2009.

Ordinarily without the contribution of the π* orbital, we would expect the interaction of two filled orbitals to be destabilising, in the sense that $E_2 > E_1$ (as drawn in the diagram). This is the same rationale as why simple MO theory predicts the helium dimer to be unstable with respect to dissociation, for example.

It might be slightly unclear, but my interpretation of the text is that the empty π* orbital contributes a little bit to both MOs $\psi_1$ and $\psi_2$. This leads to a tiny stabilisation of both of these two MOs, such that the destabilisation $E_2$ is a bit smaller and the stabilisation $E_1$ a bit larger. Although it is not very obvious, you can see this from the figure.

It could probably have been made clearer by drawing extra dotted lines from the π* orbital to those MOs. My guess as to why Fleming hasn't done that is because he wanted to use dotted lines to illustrate the most significant orbital interactions (those between orbitals of similar energy). Note how he emphasises in the text that the σ and π* orbitals are not very close in energy (which implies that the interaction is small).


What happens during oxidative and hydrolytic rancidity of fats? I understand that when lipid molecules - triacylglycerol - are hydrolysed, the fatty acids 'break' off from the glycerol backbone. And when fatty acids oxidise, these in turn can break into smaller end products. Does hydrolysis always happen first? I have made this image - would it be a fair representation of what happens when oils turn rancid?

I am currently finishing my Ph.D. in rancidity, and therefore I know quite a bit about it. The picture is not a fair representation of what happens, since free fatty acids should be present, though you just show the triacylglycerols. Which rancidity happens first will depend on the matrix and on the storage condition of the oil. Usually, but not always, oxidative rancidity happens first.


Gas chromatography/mass spectrometry: Can I use peak ratios and molecular mass to derive quantity? Can derive the quantity of a substance from a GC/MS report if I know:

the ratio to another substance in the data,
the quantity of that second substance, and
the molecular mass of both substances.

The site was very dense and vague regarding how GC/MS test work, but fairly upfront regarding how they derive their ratios.

Here are the numbers (slightly tweaked, but more or less in proportion):

Substance A:

Quantity: $300 \:\pu{mg}$ (Know already, not part of GC/MS report)
Mol. Mass: $500 \:\pu{g/mol}$ (found in external source)
Peak Proportion: 3

Substance X:

Quantity: Unknown
Mol. Mass: $425 \:\pu{g/mol}$ (again, external source)
Peak Proportion: 1

To be clear, the ratio of 1:3 means that substance X has a horizontal peak $\frac13$ of substance A.  I'm not clear on what the x-axis is actually a measurement of, which is part of why I'm at a loss. The site does make it very explicit that the ratios are not directly proportional  to mass ratio (so if it found a ratio of 3:1 of glucose to arsenic, this doesn't mean the substance is 75% glucose and 25% arsenic, only that the glucose "peaks" 3 times higher, which I've taken to mean "3 times the oomph", but that may be incorrect, as well).

So, if I know that the molecular mass($\rm M$) of substance X is 0.85 of substance Y, can I derive the actual mass of substance X using the formula:

\begin{equation}\mathrm{qty}_X=\frac{\mathrm{qty}_A\times(M_X/M_A)}{\mathrm{peak}_A/\mathrm{peak}_X}\end{equation}

with data being:

\begin{equation}\rm qty_X=\frac{300\:\pu{mg}\times(425/500)}{3/1}\end{equation}

Which simplifies to:

\begin{equation}\rm qty_X=(300\:\pu{mg}\times 0.85)/3\end{equation}

and finally the result of $85 \:\mathrm{mg}$.

So I guess in the end there are 3 questions:

Is this even how GC/MS results work?
If so, is my assumption to derive the mystery quantity using the molecular mass correct?
Is the math itself in order? (I'm specifically worried that I should invert either the mass ratio or the peak ratio or both).

Of course, if the answer to the first questions is no, then my true question is: can I derive the quantity of substance X with the given data, and if so, what would be the right approach?

If anyone is curious for some context, I need to know the actual quantity of substance X as I know that, by mass, it has a threshold between harmless and toxic, so just knowing substance X is $\frac13$ "peak" of substance A doesn't let me know if I should let my dog/child/self ingest it.

Is this even how GC/MS results work?

As cbeleites said the method you described is a proper technique but not likely to be appropriate given the information you cited.

In GC/MS you should have two sets of information. The first is the GC Total Ion Chromatograph (TIC) which will have time as the x-axis and response (abundance) as the y-axis. For each retention time on the TIC there will be a corresponding Mass Spectra (MS). In the MS the x-axis is m/z (ion) and y-axis is also response (abundance).

Different compounds have different responses so if you inject the exact same amount of two different compounds you could get a much larger response from one compared to the other. For example tramadol gives much higher response compared to hydrocodone. This is why you want your internal standard to be structurally similar to your analyte.

Quantification of a compound is often done by running 3 - 5 calibrators at known concentrations in order to make a calibration curve. Once an acceptable calibration curve is made the sample along with controls can be fit onto the curve to calculate the amount of the compound of interest in the sample and controls. If the controls are correct then the sample value can be used.

To make a calibration curve you need an internal standard in each calibrator, control and sample. You can then make the curve either with ion ratios between paired internal standard and analyte ions OR if your internal standard response is consistent in all the calibrators, controls and sample you can use the GC peak response.

In conclusion if you only have one GC/MS run with the two dissimilar compounds in it you will not be able to easily calculate the quantity of the second compound based on the GC response of the first unless you have additional information that was not listed in this question.


Will borosilicate glass absorb 808nm laser light? Will borosilicate glass absorb 808nm laser light? If it instead refracts, then what is the refractive index? Also, will borosilicate reflect lower frequencies of light like those used in non contact temp. measurements?

Borosilicate glasses are a family that has many members with differing properties. You can look up the provider for your glass, and check whether it is transparent in the near infrared. I think it would be in most cases (see here for one example).


What is the effect of adding HCl on a certain solution at equilibrium? If we have
  $$\ce{CH3COOH + H2O <=> CH3COO- + H+}$$
   what is the effect on acetate ions when adding each HCl and NaOH to the solution?  (State why according to Le Chatelier's principle)

Le Chatelier's principle states that if any of conditions of a chemical reaction at equilibrium changes, the reaction will shift to a direction to oppose that change and restore the equilibrium.

I guess adding $\ce{HCl}$ will increase hydronium ion concentration so the reaction will be shifted to the left side, accordingly acetate ions concentration will decrease. If we add $\ce{NaOH}$, this will increase water concentration so the reaction will be shifred to the right side accordingly increasing the concentration of acetate ions.

You are absolutely correct. Adding $\ce{HCl}$ adds protons to the mix, which you can imagine as applying pressure on the right-hand side, which means that the reaction will shift backwards slightly and the acetate concentration will decrease. For $\ce{NaOH}$, you also need to consider the underlying equation:

$$\ce{H2O <=> H+ + OH-}$$

which will be shifted to the left by additional hydroxides. This means that the overall proton concentration decreases and you can imagine that as lifting a weight from the right-hand side: The reaction will shift forwards.


How to determine which aqueous solution has the largest freezing point depression Question

This is more of a conceptual question. Let's say that we are given different aqueous solutions and we are asked to choose the one with the highest freezing point depression, which do we choose? Is there a pattern? do we choose the one with salts inside of it?

Example question

  Which aqueous solution exhibits the largest freezing point depression?
  $$\begin{align}(\mathrm{A})&\  1.0\ \mathrm{M}\ \ce{KBr} & (\mathrm{C})&\ 0.5\ \mathrm{M}\ \ce{MgCl2} \\
(\mathrm{B})&\ 0.75\ \mathrm{M}\ \ce{C6H12O6}& (\mathrm{D}) &\ 0.25\ \mathrm{M}\ \ce{Ga2(SO4)3}\end{align}$$

now this question is a bit different because it has concentrations, so do we choose the one with the highest concentrations or the one with salts?

BTW

The answer is A)

But my hypothesis is that we only look at the ones that contains salts/ions inside of them and then the one with the highest concentration. Please let me know if this is correct or not

Freezing point depression is a function of the amount of dissolved substance — but substance must be treated as distinct particles rather than compound. This is an important distinction because ionic compounds, when they dissolve in water, typically separate into hydrated anions and hydrated cations — two distinct species as far as freezing point depression is concerned.

Thus, for ionic compounds you first need to determine how many individual ions are generated and then multiply that by the overall concentration. In your examples:

$1~\mathrm{M}\ \ce{KCl}$. One cation, one anion. Thus, there are two moles of dissolved particles per mole of compound; giving $2~\mathrm{M}$ effective particle concentration.
$0.75~\mathrm{M}\ \ce{C6H21O6}$. This compound is not ionic. Thus, it does not dissociate and one mole dissolved gives one mole of particles. $0.75~\mathrm{M}$ effective particle concentration.
$0.5~\mathrm{M}\ \ce{MgCl2}$. This again is ionic but we get two moles of anions per mol of compound in addition to one mole of cations. Thus three moles of ions per mole of compound. Effective particle concentration $1.5~\mathrm{M}$.
$0.25~\mathrm{M}\ \ce{Ga2(SO4)3}$. Here, we have a total of two moles of cations and three of anions per mole of compound. Thus, the effective particle concentration is five times the original or $1.25~\mathrm{M}$.

The actual formula for freezing point depression is based on a constant factor of degrees per dissolved mole; and this analysis was used to confirm that ionic compounds separate into ions and to determine molecular weight in the past.


What does 2 vol (w/v) mean? I came across a protocol saying:

  Homogenize 50 g powdered plant material (accurately weighed and recorded) in 2
  vol (w/v) acidified methanol.

What does 2 vol (w/v) mean?

Weight per volume percent. It's basically the density.

$$ w/v = \frac{mass_{solute}}{volume_{solution}} * 100 $$

Usually it includes units too, but you didn't specify. I'm guessing it means grams per liter?

For more, see http://www.ausetute.com.au/wtvol.html


Using acidic detergents If detergents were acidic rather than basic, can we still achieve a clean surface? What would be more effective? Strong acid or a weak acid?
How would the result of cleaning be different if an acid of pH 4, for example, is used?
Is it possible to achieve the same result with an acid and a base, or even if they are both present in a detergent?

Many commercial detergents are composed of pH-neutral surfactants like alkyl polyethylene oxides plus alkaline additives like sodium silicate or carbonate or borax, which help dissolve greasy materials better than acids.

There are some acidic detergents, like citric acid cleaners for stainless steel and other metals and toilet bowl cleaners. The acidic function is paramount, and the reduced surface tension is secondary, useful in providing better wetting.

It is instructive to consider what cleaning really means, and the individual steps that go on. First, water is the primary cleaning agent. It is the universal solvent, but not quite good enough or fast enough. So we add surface tension lowering agents - big improvement: oily materials are wetted and can be suspended better as the first step in removal. Alkaline additives then combine with polar surfaces to further displace oily materials, improving cleaning action. Acids don't help with oily materials much but can remove some contaminants better.

Dispersants for solids, like mud or clay, are typically anionic polymers, usually alkaline because the anion contains groups that will adsorb on polar surfaces and disperse the particles by electrostatic repulsion, assisting their removal.

There are classes of surfactants where the oleophilic group is anionic, like sulfonates or phosphonates, and these can be acidic if the cation is hydrogen or alkaline if an alkali metal, or nearly neutral if the cations are mixed. Another class of surfactants, called cationic, has the oleophilic group attached to nitrogen, and is typically somewhat acidic because the anion is the anion of a strong acid like HCl.

So you could optimize your cleaner by selecting a pH of water that cleans best; then add a specific surfactant to improve removal of your specific dirt - or the reverse. And consider interactions and whatever other additives might help.


Why and how carboxylate ion has greater stability than carboxylic acid? I found it written in my textbook. I think it's a misprint, I can't think of any logic behind this statement.

I've only learned that ion is less stable specie so carboxylate ion cannot be more stable than carboxylic acid.

First of all, it would have been clearer if it's a misprint or not if you had attached a screenshot of the wording in your textbook (you still can).

At any rate, the subject of stability of carboxylic acids and carboxylate anions usually arises when discussing the level of acidity or in other words, the strength of the acid. It is pointed out in most organic chemistry books (where electron delocalization is explained) that electron delocalization or resonance is a concept that contributes to a certain structure's stability. That is to say, when a negative charge can be delocalized over several atoms (as is the case with a carboxylate group), instead of being localized on a single atom (as is the case with an alkoxide group), it is considered more stable energetically. (see here, section B - Resonance).

The relative stability of the carboxylate anion means that a carboxylic acid will favor (again, due to energetic considerations) the release of a proton in a basic environment (this is important, since in an acidic environment, the equilibrium will shift towards the acid, and much less carboxylate will exist in the solution), since the conjugate base (the carboxylate anion) is relatively stable, and therefore can exist in the solution in this form.

 (Taken from Organic Chemistry, by Paula Yurkanis Bruice, any edition really...)

To sum up, Carboxylic acids and their conjugate carboxylate anions are not (and cannot be) compared in terms of stability. Under the right conditions, i.e. in a basic solution, a carboxylate anion will be favored over a carboxylic acid because:

a. There is a base to take a proton (be it a hydroxide ion, an aminic group or any other H+ acceptor).

b. The carboxylic group will release the proton rather easily, because what's left (the carboxylate anion) is relatively stable due to electron delocalization.


Is every substance in equilibrium with its vapours? In one of my previous questions, I got an answer from user Jan:

The smell of a compound as we perceive it usually depends on which receptors in the nose will bind the molecule (or not).
These receptor, being proteins, are usually rather specific for certain shapes, not for bonding patterns. So chlorobenzene might fit into similar receptors as toluene which would lead to at least partly a similar smell, but that is it.

This means that a substance must be in the vapour phase in order to smell it, which brings me to my question:

Do all substance (at least those which have a characteristic smell) maintain an equilibrium with their vapours (if the substance is in closed container)?

But if it's true for those solids, then it must be true for all other solids, as well as nonvolatile liquids that do not have a characteristic smell to humans. Is this not the case? Also, if this substance is placed in open air, will it evaporate totally?

If they were maintaining an equilibrium with their vapour you wouldn't be smelling it.

In theory any equilibrium in the case of vapor pressure occurs in a closed container, in reality a vapor pressure equilibrium is rarely if ever reached.


Why is the product of the reaction "Bi + O2" Bi2O3 instead of Bi2O5? I am stuck on a homework question that states:

  A piece of solid bismuth is heated strongly in oxygen.

The answer key states the answer as $\ce{Bi + O2 → Bi2O3}$. However, it is to my understanding that the metal Bismuth has a charge of +5 because of its location on the periodic table. Therefore, $\ce{(+5)[charge of Bi]*2 + (-2)[charge of oxygen]*5 = 0}$ giving me $\ce{Bi2O5}$. Please advise.

Many elements have multiple valencies. The most common oxidation state of Bi is +3. Bismuth (V) is also possible. Read more here on Bismuth oxides.


Something about the Roothaan equation Why we cannot solve Roothaan equation
$\mathbf{F} \mathbf{C} = \mathbf{S} \mathbf{C} \mathbf{\epsilon}$
by just move S matrix to the left, as
$\mathbf{S}^{-1} \mathbf{F} \mathbf{C} = \mathbf{C} \mathbf{\epsilon}$ as a new matrix $\mathbf{F}'$, and then solve the eigenvalue problem
$\mathbf{F}' \mathbf{C} = \mathbf{C} \mathbf{\epsilon}$
why we cannot do that?

We could. Mathematically there is nothing wrong with that - the overlap matrix is positive definite and therefore must have an inverse, and so your manipulation is perfectly correct mathematically. The problem is that $\mathbf{S}^{-1} \mathbf{F}$ is not a symmetric (or in the more general case Hermitian) matrix. Computationally it is MUCH more difficult to find accurate eigenvalues and eigenvectors of a general non-symmetric matrix than a symmetric one. Thus, when solving the problem on a computer, if one can cast one's algorithm in terms of diagonalisation of a symmetric rather than a non-symmetric matrix you are likely to get more correct answers, quicker. And this is what the standard manipulations achieve.


How to Test Auto-Ionization of Ammonia I was wondering how we could test that the auto-ionization of liquid ammonia actually occurs experimentally. The chemical equation is described below:
$\ce{NH_3 <=> NH_4^+ + NH_2^-}$
I was thinking if there might be a substance that reacts with $\ce{NH4+}$ to form a noticeable precipitate, but I couldn’t find anything. Maybe we could test pH? But do pH tests work in liquid ammonia, and what would it tell us?
Any tips would be appreciated!
Thank you!

No, that's not how it all works on more than one level.
First, the amount of $\ce{NH4+}$ out there is small. And by "small", I mean really, really small - about as small as you can imagine, only smaller. Many times smaller, I must reiterate. If some substance was to react with it (which is certainly not outside the realm of possible) and form a product in equally small amount, you won't be able to see it.
I intentionally avoid any numeric estimates here. You must have seen them already. They are too far outside the everyday experience to make an impression. Well, you may look up the auto-ionization constant of ammonia and see how it relates to that of water (which itself is not big). Guess what? It is many orders of magnitude smaller.
Now comes the second point, which kind of negates the first one. If something reacts with that $\ce{NH4+}$ (which, I repeat, is totally possible), then the auto-ionization reaction will produce more $\ce{NH4+}$ to replace the loss, and then more and more, probably until no more $\ce{NH3}$ is left. How so? Well, imagine a crowded theater emptying through one door. Obviously all those people can't fit in the door at once. Not even half of them can. But they don't have to. They just walk through it one by one, and before long, the theater is empty. Much like that, all $\ce{NH3}$ molecules will go through the form of $\ce{NH4+}$ (or $\ce{NH2-}$) to whatever products are there.
In a way, this will prove that the auto-ionization was there in the first place, but tell us nothing about its extent. To measure that, we need other methods; conductometry comes to mind.
As for pH indicators for water, they are definitely no good for liquid ammonia. They all will be far off-scale, much like your cooking thermometer in the middle of Antarctica. To work in liquid ammonia, one needs to develop a whole new set of indicators. Some seems to be known already, but I have no firsthand experience with them, and Google is the same over there as it is over here.
So it goes.


Element Song, updated with 118 names? Tom Lehrer’s The Elements song (listen) mentions all elements up to Nobelium, and ends with the appology

  These are the only ones of which the news has come to Harvard,
  And there may be many others, but they haven't been discovered.

Now that’s a bit dated. We can’t ignore trans-uranic elements when every bedroom in developed countries contains Americium-241, and now that official names have been ratified for all elements through period 7, I can hope that an updated song will be available.

Are there any such songs, whether a revision of the original or based on a new melody, or any known efforts underway to complete one?

The NEW Periodic Table Song (Updated)
AsapSCIENCE
https://www.youtube.com/watch?v=VgVQKCcfwnU


Why is C4 of butynone electrophilic? A question asks whether C4 in 3-butyne-2-one can act as a nucleophile, electrophile, or acid (it can be more than one). The answer is that it can be both electrophilic and acidic.

The acidity makes sense, it's a terminal alkyne, sp orbitals stabilize the negative charge on the conjugate base.

The electrophilicity I don't quite understand. I can write a resonance structure that puts some positive charge on C4, analogous to an alpha-beta unsaturated carbonyl compound, but it looks bizarre. I know ketenes are a real thing, but is  this resonance structure a real contributor? Is there another explanation for the electrophilicity?

Google searches for alkynone, ynones, etc. yielded mostly peer-reviewed literature that's above my understanding.

Yes, that second resonance structure, the one on the right, is a significant contributor to the overall description of the molecule.  That resonance looks a little different because the p orbitals on the alpha and beta carbons are on $\ce{sp}$ hybridized carbons, rather than the $\ce{sp^2}$ hybridized carbon more commonly seen in enones.   As this second resonance structure explains, alpha-beta unsaturated enones and ynones are both subject to attack by nucleophiles at the electrophilic beta carbon.

Resonance structures, like the one pictured below, explain the reactivity seen in enamines and yneamines (the triple-bond analogue); however, now the positive charge is on the nitrogen and a negative charge on the beta carbon.  Therefore these nitrogen compounds are subject to attack by an electrophile at the nucleophilic beta carbon.


Is alcohol based sanitizer flammable? After applying sanitizers can they catch fire, if yes, then after what time is it safe to use the hands, say in the kitchen.
Also one more question is that if they are flammable then the flammable component is alcohol. However as soon as we apply the sanitizer the alcohol evaporates? Then how can they remain flammable on our hands?

Firstly, we should realise that sanitizers should not be considered as pure alcohols, alcohols are indeed a major component of it but many other important components are also present. A typical sanitizer contains these things -

Isopropyl alcohol/ Ethyl alcohol, Glycerin, Water, Carbomer(stabilizer) and some other components mostly related to cosmetics.

Now, i will answer your questions one by one

After applying sanitizers, can they catch fire?

Yes, there is 60%-70% alcohol content which is highly flammable. Organic compounds like alcohols, alkanes etc have very high enthalpy of combustion ($\Delta_\mathrm{c}H^\circ$ = -2000 kJ/mol for Isopropyl alcohol) and they can evaporate very easily due to the heat of our hands($\Delta_\mathrm{vap}H^\circ$= 40kJ/mol for isopropyl alcohol).

As soon as we apply alcohol, it evaporates so why is it still flammable?

This is actually wrong, as I earlier mentioned, that sanitizers are not only alcohols, in fact, the 'carbomer' present acts as a stabilizer to emulsify/dissolving all the components in it. These stabilizers prevent the alcohol to evaporate immediately and allow it to disinfect our hands fully. Also, after applying sanitizers, many of you might feel hands becoming sticky it is because of the presence of Glycerin and other cosmetic products used in it which act like a moisturizer. So generally it takes about 30-40 seconds for all(mostly) the alcohol to evaporate and sustaining a risk to be inflammable for a long time.
A thing to keep in mind is that the residue (mostly Glycerin) is also flammable(not that much, it's flammability rating is one) so after the alcohol has evaporated, you aren't fully safe from catching fire.
If your are working in kitchen, there is no actual need of using sanitizers as you can always wash hands in sink with soap and water.
I hope this answer cleared some of your doubts, stay safe :).


Why can't a reverse fission reaction be automatically classified as a nuclear fusion reaction? $$\ce{_2^4He + _13^27Al -> _15^30P + _0^1n}$$

Will this reaction qualify as a nuclear fusion reaction? The answer to this question in the test says that it won't, yet it seems to be the reverse of a fission reaction.
Why would it not be considered fusion?

The Wikipedia article on Nuclear Fusion starts off:

  In nuclear physics, nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons). The difference in mass between the reactants and products is manifested as the release of large amounts of energy.  ...

$\ce{^4He}$ and $\ce{^{27}Al}$ are certainly different than $\ce{^{30}P}$.

The masses of the products however are more than the masses of the reactants

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\text{4-He} & 4.&\hspace{-1em}00260325415 \\
\text{27-Al} & 26.&\hspace{-1em}98153863\\ \hline
\text{total} & 30.&\hspace{-1em}98414188\\
\end{array}$$

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\text{30-P} & 29.&\hspace{-1em}9783138 \\
\text{n}   & 1.&\hspace{-1em}00866491588 \\ \hline
\text{total} & 30.&\hspace{-1em}9869787 \\
\end{array}$$

So it would seem that the definition used would require the products to have less mass than the reactants. This seems like a weird definition to me...