How to calculate Gibbs free energy from pKa? Is it possible to calculate Gibbs free energy ($ΔG = -RT\ln K_\mathrm{eq}$) of an acid dissociation reaction using the $\mathrm{p}K_\mathrm{a}$? If so, how?

To find the change in Gibbs free energy $\Delta G$ for the reaction of an acid, you can simply use $K_\mathrm{a}$ in the formula you gave in place of $K_\mathrm{eq}$ ($K_\mathrm{a}$ is an equilibrium constant).


Best way to raise pH without alkali metal? I have a process that requires pH~14-14.5 to work. I’ve been achieving this with concentrated NaOH solutions, but the Na cations are later causing side reactions which are detrimental.  Is there any organic base/non-alkali base that has the solubility in water and pKa to reach these pH values?

Tetramethylammonium hydroxide

This is a strong base like NaOH.

See Behavior of Polyethylene Oxide Based Nonionic Surfactants
in Silicon Processing Using Alkaline Solutions J. Electrochem. Soc. 1995   volume 142, 621-627:

  alkaline cleaning solutions based on quarternary ammonium hydroxides such as choline ($pK_b = 0.1$) and tetramethyl ammonium hydroxide ($pK_b = 0$)

and Ceramic Nanomaterials and Nanotechnology II which says:

  In this research, tetramethylammonium hydroxide (TMAH) was used, instead of KOH, to adjust pH of the precursor solution to 14.0

Also, TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society (1905):

  An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.


Finding the sum of coefficients of the reaction What is the sum of the coefficients $(a + b + c)$ in the following reaction?
$$\ce{x Zn + y HNO3} -> {a Zn(NO3)2 + b H2O + c NH4NO3}$$

Here if the equation is balanced we get 1 mol of $\ce{Zn}$ and 4 mol of $\ce{HNO3}$. As  $x + y = a + b + c$ we get sum of coefficients $a + b + c = 5$. According to me, this is the correct answer, but in the actual answer key the answer is 8.
I may be wrong, but can someone help me clarify the right answer?

Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me):

$$\ce{$x$ \overset{0}{Zn} + y H\overset{+5}{N}O3 -> $a$ \overset{+2}{Zn}(NO3)2 + b H2O + $c$ \overset{-3}{N}H4NO3}$$

$$
\begin{align}
\ce{\overset{+5}{N} + 8 e- &→ \overset{-3}{N}} \tag{red}\\
\ce{\overset{0}{Zn} &→ \overset{+2}{Zn} + 2 e-} &|\cdot 4 \tag{ox}\\
\hline
\ce{4 \overset{0}{Zn} + \overset{+5}{N} &→ 4 \overset{+2}{Zn} + \overset{-3}{N}} \tag{redox}
\end{align}
$$

So that the complete balanced equation is

$$\ce{4 Zn + 10 HNO3 -> 4 Zn(NO3)2 + 3 H2O + NH4NO3}$$

leaving $a = 4$, $b = 3$, $c = 1$ and the sum $a + b + c = 4 + 3 + 1 = 8$.


Can ICE tables give two solutions? What does this mean physically? The following question comes from General Chemistry: Principles and Modern Applications, 11th ed. by Pettruci.

  Exercise 15-33. A mixture consisting of 0.150 mol H2 and 0.150 mol I2 is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H2, I2, and HI?
  $$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$

First, I found the concentrations of H2 and I2:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\
&=0.0461538\text{ M}
\end{aligned}
$$

Setting up the ICE table:

$$
\begin{array}{|l|c c c c c|}
\hline
&\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\
\hline
\text{Initial}&0.0461538&&0.0461538&&0 \\
\text{Change}&-x&&-x&&+2x \\
\text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
 \\
50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\
0.327009&=9.0852x&&&0.327009&=5.0852x \\
x&=0.0359936&&&x&=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\
&=0.0101602\text{ M}&&&&=-0.0181522\text{ M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.

My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this?

Edit: To reiterate, I am aware that one of the solutions is to be eliminated for being "non-physical". My question is whether it is possible for both solutions to be physical.

In other words, is there a scenario where I could end up with two possible [solutions]?

No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range of possible equilibrium constant), and it is always unique (because for a given set of starting values, every value of Q is unique - follows from the monotonic variation).

The only exception is a process where none of the concentration vary with x (when all species are pure solids or liquids). In this case, Q takes a single value (except for running out of reactant or product). If that value is equal to K, x can take a range of values. If Q is different from K, there is no solution, and there won't be any equilibrium (i.e. the reaction would go to completion).


Zinc evaporation/toxicity at room temperature I have a (slightly embarrassing) question regarding zinc powder.

I have recently broken an old mercury thermometer, and read that I can handle the mercury with powdered zinc - so I've sprinkled some on my floor. Too much probably. It turned out that unfortunately the powder is then quite hard to get out of the particular material of my floor (wood). I got out as much as possible, but I think there's still some left. So now I'm wondering if there is some danger if having all that zinc laying around.

Thanks anybody for the answer :)

Zinc will amalgamate with mercury, but whether that reduces the vapor pressure of mercury very much is debatable. At best, the increased volume of zinc amalgam and the ability to amalgamate with fine droplets of mercury would make it easier to clean up the mercury as amalgam.

It might be better to spread some sulfur dust (flowers of sulfur) over the area. The sulfur should react with the mercury and tie it up as sulfide, with very low vapor pressure. Sulfur powder (and possibly the HgS) might also be easier to clean up than zinc dust, because sulfur is more like ordinary dust than a dense metal dust.

BTW, I love mercury thermometers. And I hate it when I break them!


How to get meniscus down to the calibration mark with a volumetric pipette I am having trouble taking aliquots using the 10 ml volumetric pipette.

After rinsing, I draw the solution up past the calibration mark and stopper the pipette with my index finger. Then I slowly rotate the pipette and try to let a tiny bit of air inside to get the meniscus down to the calibration mark. But every time, the solution just shoots down, or goes just below the calibration mark.

How can I improve my technique?

If you have to stopper the pipette with your finger, it just takes practice. I believe the proper way to do it is to just roll your finger to the side very slowly, but I've heard several strategies including rotating the pipette like you said, and rocking/sliding your finger back and forth.

It is a lot easier if you have access to pipette bulbs like these, which let you squeeze the side arm to expel liquid smoothly:

These are easy to use too - you just roll it down to expel the liquid:

Most labs should have either of these, especially university ones. Maybe less common in high schools.


Ice Crystal Matrix measurements of H2O molecules? I want a mathematically generated 3D STL copy of this image, however it's represent algorithmically.

If I do all the H2O's as individual pieces with female connections prepared for chopped bike spokes, are all the H2O's the same and what angles do they have?

Is there a written array of positions and degrees of rotation of the water molecules of a regular ice matrix?

You can produce STL (and VRML) files directly from Mercury - see https://www.ccdc.cam.ac.uk/Community/blog/post-56/

I'm not sure how best you'd go about including female sockets for bike spoke segments, but be aware that for some 3D printing types you're charged for the total volume of material used, so you might be better off trying to create a model with multiple molecules in a smaller area with less empty space.


How to dissolve calcium sulfate fast? I would like to dissolve $\ce{CaSO4}$ in a hardened condition (e.g. gypsum board) as fast as possible. I have seen in posts about $\ce{CaSO3}$ that vinegar and sulfamic acid are able to dissolve it.

However, I would appreciate leads to reactants that are a little more active. Basically I am looking for a drop of something that would instantly create a hole.

Unfortunately calcium sulphate, gypsum, $\ce{CaSO4}$, is depressingly stable and nothing non-deadly will dissolve it.

The other posts you have seen are about calcium carbonate, chalk/limestone/marble, $\ce{CaCO3}$, which acids will dissolve, although people do not in practice use that fact to make holes in it. One of the first difficulties would be to get the application of a liquid focused enough to make a hole-shaped hole.

If you ever did want to dissolve marble, I'd use hydrochloric acid, $\ce{HCl}$, because calcium chloride is nice and soluble. But check beforehand that it doesn't dissolve the gloves you intend to use.


Value of mixing ratio and rate constant in Chapman mechanisn In the Chapman mechanism (which tells about the activities of ozone in the stratosphere), there is a equation for the lifetime ($\tau_{\ce{O}}$) of an oxygen atom, $\ce{O}$ against conversion to ozone, $\ce{O3}$, through a certain reaction is given as
$$\tau_{\ce{O}} = \frac{1}{k_2 C_{\ce{O_2}}n_a^2},$$
where $k_2$ is the rate constant of a reaction involving consumption of $\ce{O2}$ to form $\ce{O3}$,
$$\ce{O + O2 + M -> O3 + M},$$
 $\ce{M}$ is a third body present during the reaction. $C_{\ce{O2}}$ is the mixing ratio of $\ce{O2}$ (which is given as $0.21~\mathrm{mol/mol}$), $n_a$ is the air number density.
Source: Daniel J. Jacob: Introduction to atmospheric Chemistry. 1999, To be published. Chapter 10.

I am confused regarding this mixing ratio of $\ce{O2}$. Shouldn't be this value in the stratosphere different from that in troposphere?

Also, regarding $k_2$, nothing is mentioned about how and where this was measured. I mean if the value of $k_2$ is derived through laboratory experiments, then how reliable is this value for the reactions happening in the stratosphere?

The entire stratosphere lies within the "homosphere", where the composition of gases is dominated by mixing processes and the atmosphere is thus homogenous.  This reaches up to about 100 km, whereas the stratosphere doesn't go any higher than about 50 km.


Continuously replenished reaction Substrate ($m$) concentration in a reaction scheme is governed by the equation $$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}$$ where $c_1$ is the initial enzyme concentration and each $r_i$ is the reaction rate of the reactions present within this reaction scheme (which are positive). Now additionally suppose the substrate is being continuously replenished by a reaction $$\ce{A->[r_5]S,}$$ where $A$ has constant concentration $a_0$. Modify $dm/dt$ and show there are two distinct positive equilibria $0<m_1<m_2$ if $r_5<r_5^C$ where $r_5^C$ is a critical value to be determined.

With this information we have

$$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}+r_5a_0.$$

Equilibria when $dm/dt=0$. We rearrange this to produce the quadratic

$$r_2r_4r_5a_0 + r_1r_4(r_5a_0 - r_2c_1)m + r_1r_3r_5a_0m^2=0$$

So the equilibria occur at

$$m_{1,2} = \dfrac{r_1r_4(r_2c_1-r_5a_0)\pm\sqrt{r_1^2r_4^2(r_5a_0-r_2c_1)^2-4r_1r_2r_3r_4r_5^2a_0^2}}{2r_1r_3r_5a_0}$$

Now I need to justify these are positive and find the critical value for $r_5$ but I have no idea how to continue here.

The first step is to set the discriminant equal to zero, and thereby solve for the critical value of $r_5$. When I did this, I got $$r_5a_0=\frac{r_2c_1}{1+2\sqrt{\frac{r_2r_3}{r_1r_4}}}$$.  The next step is to substitute this into the equation for m, and find the two  corresponding equal values of m.


Is the heat for constant pressure the same as for constant volume in the formation of nitric oxide? In the reaction $\ce{N2 (g) + O2(g) -> 2NO (g)}$
my textbook[1]  says $q_v$ and $q_p$ are not the same, without any explanation.

I know that
\begin{align}
\Delta U &= q_p + w\\
\Delta U &= q_p - P\Delta V\\
\Delta U &= q_p - Δn_\mathrm{gases}RT
\end{align}

I also know that
$$\Delta U = q_v.$$

because under constant volume, work done ($-P\Delta V$) is zero, since there's no change in volume.

In the reaction mentioned,
$$n_\mathrm{gases} = 0,$$
so shouldn't
$$\Delta U = q_p$$
and therefore
$$q_p = q_v?$$

Here's a picture of question 143 in Chapter 7 of [1]. The answers say a) and d) are false:

Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette: General Chemistry: Principles and Modern Applications, 11th edition. Published by Pearson (February 23rd 2016)

Kevin is right. In that reaction, the number of moles does not change. So carrying it at constant pressure or at constant volume does not make any difference. And $q_v$ = $q_p$


Do molecular orbital diagrams exist for ionically bonded compounds and molecules? Do molecular orbital diagrams exist for ionically bonded compounds and molecules?

Molecular Orbital theory is a simplified approach to describe chemical bonding based on symmetries and orbital overlap. In principle, it makes no assumptions on the type of bonding, i.e., covalent, ionic, Van der Waals. This distinction is more a construct to understand the strength of the bond.

In case of a covalent bond, the electrons that make up the bond are shared by both nuclei, whereas in an ionic bond the bonding originates mostly from the Coulomb interaction, that is, the electrons are mostly located on a single nucleus. In MO theory, this can be understood in terms of the interaction strength. When the atomic orbitals that make up the bond have similar energy (and same symmetry), there will be a strong interaction and the resulting MO will have character of both AOs. In other words, the electrons will be shared more or less equally by both atoms. When there is a large difference between the AOs, the resulting bonding MO will have mostly the character of the AO which is lowest in energy and as such the electrons in this MO stay mostly around this atom. This resembles an ionic bond.


Electrolysis - What is the basis for the discharge rules? At the cathode, the one that is discharged is the less reactive one (compared to hydrogen) so why at the anode is the one that’s discharged more reactive (less stable) ?

The redox potential is a measure of tendency of  redox system to exist rather in oxidized form ( negative potential values) or reduced form ( positive values ).

Therefore the higher the standard redox potential is, the stronger oxidation effect and weaker reduction effect have the respective oxidized/ reduced forms of the redox system.

And vice versa.

Particularly, on the opposite sites of the standard redox potential table are
$$\begin{align}\ce{
Li &<=>> Li+ + e- \\
F2 + 2e- &<=>> 2 F-  \\}
\end{align}$$

The former is very strong reducting agent, while latter is very strong oxidating agent.

I will illustrate it on anode case, with the cathode case it is the same, but with the reversed potential order.

When the anode potential is externally forced to raise up, if first start to oxidate the reducted form of the redox system with lowest redox potential.

If the current is already saturated by diffusion rate of redox reagents, futher increase of potential does not increase the current until it reaches potential region of next easiest to oxidize redox system.

If the redox systems have similar values, the next system start to react on the electrode before the system

The result is, the curve current versus anode potential has a typical look of the  rounded stepwise  voltammetry curve.

See also Voltammetry,  Polarography and example of the cathodic polarographic curve.


Determining the redox half-reactions of single displacement reactions For the following double displacement half reaction, the half reactions can be determined easily using the oxidation numbers.
$$\ce{MnO4- {(aq)} + IO3- {(aq)} -> MnO2 {(s)} + IO4- {(aq)}}$$
We know this is the reduction half reaction because the oxidation number of $\ce{Mn}$ changes from +7 to +2:

$$\ce{MnO4- {(aq)} -> MnO2 {(s)}}$$
and we know this is the oxidation half reaction because the oxidation number of $\ce{I}$ changes from +5 to +7:
$$\ce{IO3- {(aq)} -> IO4- {(aq)}}$$

However, for the following reactions, this strategy cannot be used. So how do we determine the half reaction equations for the following reactions:

  $\ce{Mn(OH)2 {(s)} + H2O2 {(aq)} ->T[Base] Mn(OH)3 {(s)}}$
  $\phantom{iiiiiiiiiiiiiiiii}\ce{MnO4^2- {(aq)} ->[Acid] MnO2 {(s)} + MnO4- {(aq)}}$
  $\phantom{iiiiiiiii}\ce{IO3- {(aq)} + I- {(aq)} ->T[Acid] I3 {(aq)}}$
  $\phantom{iiiiiiiiii}\ce{P {(s)} + PO3^3- {(aq)} ->T[Base] HPO3^2- {(aq)}}$

Let's take the first equation in the question by example:
$$\ce{Mn(OH)2 (s) + H2O2 (aq) -> Mn(OH)3- (s)}$$
In this equation ion $\ce{Mn(II)}$ is oxidised into $\ce{Mn(III)}$, and oxygen in $\ce{H2O2}$ (oxidation number -1) is reduced into ion hydroxide (oxidation number -2):
$$
\begin{align}
\ce{Mn^{2+} &-> Mn^{3+} +e^- \tag 1}\\
\ce{H_2O2 (aq) + 2e^- &-> 2OH^- \tag 2}
\end{align}
$$
If we multiply the first equation by 2 and add the equations $(1)$ and $(2)$:
$$\phantom{iiii}\ce{2Mn^{2+} + H_2O2 (aq) -> 2 Mn^{3+} + 2OH^- \tag 3}$$
By adding four ions hydroxide to the two sides of equation $(3)$:
$$\ce{2Mn^{2+} + H_2O2 (aq) +4OH^- -> Mn^{3+} + 6OH^- }\phantom{iiiiiiiiii}$$
By rearranging the equation $(3)$, we find:
$$\ce{2Mn(OH)2 (s) + H2O2 (aq) -> 2 Mn(OH)3 (s) \tag 4}\phantom{iiiiiii}$$
We do the same for the other equations.


Why is methoxy group an electron donating group? I am studying acidity of carboxylic acids and I found that benzoic Acid is more acidic than p-methoxybenzoic acid. It is written in my book that this happens because methoxy group is an electron donating group. I can understand why electron donating groups can reduce acidity of an carboxylic acid but I can't understand why methoxy group acts as an electron donating group?

The lone pair on the oxygen delocalizes into the conjugated aromatic ring.

Edit (based on Jan's suggestion):

Notice that the answer is actually a lot more complicated. A methoxy substituent actually impacts the ring electronics via two competing effects. The oxygen's lone pair is well-placed to delocalize and increase electron density within the ring's conjugated system. This allows delocalization to better stabilize positive charges. So the methoxy is electron-donating from a resonance perspective.

On the other hand, because oxygen is quite electronegative, the methoxy group is electron-withdrawing in an inductive sense via the $\sigma$ bonds.

Take a look at a linear free energy relationship (LFER) plot. For acidity of benzoic acids, the methoxy group has a positive $\sigma_{\mathrm{meta}}$ value but a negative $\sigma_{\mathrm{para}}$ value.

The meta $\sigma$ value measures the effect when there is no resonance structure that delocalizes effectively to the carboxylic acid, and its positive value tells us that the methoxy is electron-withdrawing in that position. The negative para $\sigma$ value tells us that when there is effective conjugation into the carboxylic acid, the methoxy is electron-donating. Notice that it is still electron-withdrawing inductively, but the resonance donation is much stronger.


How to create black precipitate in clear liquid I am not a chemistry person at all, so my apologies if my question seems dumb.

I am working on the set for a short film and in one of the scenes we would like the main actor to look at a sample of water, drop something into it and watch a black precipitate form. Clouding the water.

I have looked online to see what reaction would yield the desired result, however, my chemistry knowledge is very basic.

Could somebody here help me out and let me know what I could use? Safely.
I understand that we need to handle all chemicals with care.

I can think of iron(II) sulfide($\ce{FeS}$). There are many ways to make iron(II) sulfide but one reaction procedure will suit your purpose. It uses iron(II) sulfate and lime sulfur(calcium polysulfide) to form a clear black precipitate of iron(II) sulfide with a clear supernatent. See this e-book excerpt.

You can use other sulfides as well. See this picture (full picture here):

From left to right: silver sulfide ($\ce{Ag2S}$, black), lead sulfide ($\ce{PbS}$, black), nickel sulfide ($\ce{NiS}$, black-brown), copper sulfide ($\ce{CuS}$, dark indigo). These are formed by reacting $\ce{Na2S}$ with corresponding metal ions. If you want to fully make your solution black, then your options would be to form silver sulfide and nickel sulfide. You could also make your solution black by forming elemental mercury solution but that would be too toxic for your purpose.

Other ways include adding iodide to excess amount of bismuth(III) nitrate  in dilute nitric acid but be careful not to overdo as excess concentration of iodide will make solution orange-red(See here).

If you do not want any reaction to happen then just dump manganese(IV) dioxide in water. They are black and insoluble.


Can Fuel Exhaust Be Recycled When the spark plug of a car ignites the pressurized air/fuel mixture, could you instead send the exhaust to another system within the car to be Recycled?

I think it's a great idea that could drastically lower pollution, that is if it's possible.

One of the major components of exhaust is carbon dioxide, produced by the combustion of hydrocarbons - if in complete combustion, follows the general reaction:

$$\ce{C_{x}H_{y} + O2 ->CO2 + H2O}$$

Other possible emissions include $\ce{NO_{x}}$, particulate matter, carbon monoxide, VOCs etc. The amounts of these emissions vary from car to car (due to a lot of factors, e.g. age, wear-and-tear etc). According to a self study by Audi, in the document Motor Vehicle Exhaust Emissions, on page 6, they determined that asides from $\ce{N2}$, the main emissions were $\ce{CO2}$ and $\ce{H2O}$.

For this answer, the focus will be on the possibility of recycling the $\ce{CO2}$ emissions (and linking to the $\ce{H2O}$ emissions).

This very question is the subject of the Scientific American article Reverse Combustion: Can $\ce{CO2}$ Be Turned Back into Fuel?, the articles details that there are several research groups looking at this problem.

A couple of examples of ideas and experiments that have been put forward include:

  In the 1990s a graduate student named Lin Chao at Princeton University decided to bubble carbon dioxide into an electrochemical cell. Using cathodes made from the element palladium and a catalyst known as pyridinium—a garden variety organic chemical that is a by-product of oil refining—he discovered that applying an electric current would assemble methanol from the $\ce{CO2}$.

Inspired by the natural process that removes $\ce{CO2}$ from the atmosphere - photosynthesis, continued research has sought to emulate this - essentially, the current research is effectively looking at a possible solution by taking

  $\ce{CO2}$, water, sunlight and an appropriate catalyst and generate an alcoholic fuel

The research reported in the Scientific American article and the National Geographic article Carbon Recycling: Mining the Air for Fuel go along the lines of extracting the excess $\ce{CO2}$ from the air rather than directly from the car's engine or exhaust. The reason for this is the current practicality, with the National Geographic reporting that

  The Sandia prototype’s solar collector has an area of about 20 square meters (215 square feet) for a reactor the size of a beer keg, Stechel says. About 300,000 acres (121,400 hectares) of mirrors would be required to collect enough sunshine to make the equivalent of 1 million barrels of oil per day, she says. (The world currently consumes about 86 million barrels per day of petroleum and other liquid fuels, including biofuels.)

The research shows that, yes it is theoretically possible and practically possible on a larger scale to recycle exhaust gases.


Space filling model for nitrogen molecule I want to make space filling model for nitrogen molecule.

I've learned that van der Waals diameter is to be taken for atom radius.

Since van der Waals radius for nitrogen is $\pu{155 pm}$ and bond length is $\pu{110 pm}$, does this mean that nitrogen molecule space filling model is two blue spheres of diameter $\pu{155 pm}$ and center to center distance $\pu{110 pm}$?

One of the reasons for this question is that samples of space filling model for this molecule (see for example Wikipedia) do not agree with these data; there's about a 11% difference between the length/width ratio of the drawn molecule, and what it should be based on the van der Waals radius and bond length.

\begin{array}{cccccc}
  \displaystyle\frac{d}{2r}
    &\sim& \displaystyle\frac{\pu{1000 pix}-\pu{716pix}}{\pu{716 pix}}
    &\sim& 0.397\\
  \displaystyle\frac{d}{2r}
    &\sim& \displaystyle\frac{\pu{110 pm}}{2\times\pu{155 pm}}
    &\sim& 0.355
\end{array}

Yes, two spheres with a radius of 155 pm and the distance between the center is 110 pm.

The axis in that picture is not aligned, that's why you get an error. Here's a picture with aligned axis and van der vaals spheres produced in Avogadro:

The corresponding *.xyz:

2

N   -0.55   0.0 0.0
N   0.55    0.0 0.0

I'm also unable to follow your math here. height of the molecule should be $2\times 155=310$ and length should be $155\times 2+110=420$ so height to length ratio should be $0.73$, you got $0.72$ which is pretty close.


Coupling constant (Jab) - Why more unpaired electrons means a smaller coupling? I hope this is not a "homework" question, but I'm having a bad time trying to figure this out.
Available literature proposes 3 equations to calculate the coupling constant during a Broken-Symmetry approach:

\begin{align*}
J(1) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}^{2}} \\
J(2) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}\left ( S_{max}+1 \right )} \\
J(3) = \frac{-\left ( E_{HS}-E_{BS} \right )}{\left \langle S_{HS}^{2} \right \rangle-\left \langle S_{BS}^{2} \right \rangle}
\end{align*}

I'm intrigued by the fact that, from the equations, the more the system have unpaired electrons, the minor will be $J_{ab}$. Why does this happen? Doesn't more unpaired electrons increase magnetic momenta (and an increase in magnetic coupling))?

I hope I have understood your question correctly.

For illustrative purposes, I will restrict myself to the Hartree-Fock picture. In a broken spin system, e.g., UHF, the spin state is not a good quantum number. That is to say that the UHF wavefunction need not be an eigenfunction of $S^2$ or $S_z$. For example two complexes with a single unpaired electron each will combine three singlet states (two ionic states and one open-shell) and three triplet states ($m_s\in \{-1,0,1\}$).

The open-shell singlet ($|\alpha\phi_1 \beta\phi_2|-|\alpha\phi_2 \beta\phi_1|$) and the $m_s=0$ triplet ($|\alpha \phi_1 \beta \phi_2|+|\alpha\phi_2 \beta\phi_1|$) states are not single Slater determinants. The singlet states need not be of the same energy, whereas the triplet states have to be degenerate for a Hamiltonian without explicitly spin-dependent components such as spin-orbit coupling. Hence, you have to accommodate for the multiple counting.

For higher spins, the degeneracy grows with $S_{max}$ and must be corrected accordingly. This does not imply that coupling necessarily decreases with higher spins.


H1 NMR Spectra of inorganic complex, Cu4OCl6(2-mepy)4 I'm having difficult interpreting the following H1 NMR I have been given for Cu4OCl6(2-mepy)4.

The NMR spectra obtained with integration trace beneath. I'm not too sure how to rationalise the 4.5:1 ratio between the peaks nor the cause of the small broad peak at 115ppm (well to the left of the spectra).

I'm also not too sure why the H1 NMR range is so vast. I appreciate being bonded to metals the electron density is changed across the relevant proton environments but by that much?

Thanks a lot

What you have there is a typical example of an H NMR spectrum of a paramagnetic compound. The formal oxidation state of the Cu here is Cu(II) resulting in d9 configuration. This means that there will be an unpaired d electron making the Cu atoms paramagnetic.

The lone electron in the Cu atoms disturbs the magnetic field the Hs' feel resulting in extreme chemical shifts and also provide a very efficient relaxation mechanism resulting in extreme peak broadening (in some cases you cant even see the resonances at all). The highest paramagnetic shift I have encountered so far was around 250 ppm for an Fe(II) complex I made. In another occasion we had resonances at 1200 ppm!

As for the peak broadening I had peaks of ν1/2 of a few thousands of Hz. The integration wont always make much sense with paramagnetic compounds either.


Why does clearance increase after phenytoin overdose? I am finding it hard to envision how an overdose of phenytoin ultimately leads to an increase of the clearance of phenytoin after the enzymes that metabolize phenytoin stop getting saturated.

I know it has to do with zero order and first order kinetics, but I just can't put it in perspective.

Would appreciate if someone could help out.

Phenytoin is primarily eliminated by hepatic metabolism (>95%).
  Hepatic metabolism is mainly via the CYP2C9 enzyme system with a
  smaller amount metabolized by CYP2C19. About 5% of a phenytoin dose is
  recovered in the urine as unchanged drug. Phenytoin follows
  Michaelis-Menten or saturable pharmacokinetics.

This is the type of nonlinear pharmacokinetics that occurs when the number of drug molecules overwhelms or saturates the enzyme’s ability to metabolize the drug. When this occurs, steady-state drug serum concentrations increase in a disproportionate manner after a dosage increase:

In this case the rate of drug removal is described by the classic Michaelis-Menten relationship that is used for all enzyme systems: rate of metabolism =$\ce{(Vmax ⋅ C) / (Km + C)}$, where Vmax is the maximum rate of metabolism in mg/d, C is the phenytoin concentration in mg/L, $\ce{Km}$ is the substrate concentration in mg/L, and where the rate of metabolism = $\ce{Vmax /2}$.

The clinical implication of Michaelis-Menten pharmacokinetics is that the clearance of phenytoin is not a constant as it is with linear pharmacokinetics, but is concentration- or dose-dependent. As the dose or concentration of phenytoin increases, the clearance rate (Cl) decreases as the enzyme approaches saturable conditions: $\ce{Cl = Vmax / (Km + C)}$. This is the reason concentrations increase disproportionately after a phenytoin dosage increase.

However, since clearance is dose or concentration-dependent, half-life also changes with phenytoin dosage or concentration changes. As doses or concentrations increase for a drug that follows Michaelis-Menten pharmacokinetics, clearance decreases and half-life becomes longer for the drug.

  For drugs such as phenytoin with a low hepatic extraction ratio
  (≤30%), the numeric value of liver blood flow is much greater than the
  product of unbound fraction of drug in the blood and the intrinsic
  clearance of the compound (LBF >> fB)

  Plasma protein binding
  displacement drug interactions cause major pharmacokinetic alterations
  but are not clinically significant because the pharmacologic effect of
  the drug does not change.

Because the clearance of the drug is

  dependent on the fraction of unbound drug in the blood and intrinsic clearance for a low hepatic extraction ratio agent, a
  decrease in plasma protein binding and increase in unbound fraction
  will increase clearance.

Currently there is no substantive evidence of phenytoin auto-inducing its own metabolism (the theory of phenytoin displacing itself from plasma proteins really isn’t convincing at all but perhaps in a drug interaction with a drug which is predominantly plasma bound). The reason being that the information regarding phenytoin auto-induction is sparse and conflicting. The results of this study aren’t convincing enough when taken in perspective of its ability to auto-induce itself after an overdose.

References

Applied Clinical Pharmacokinetics: A. Bauer, Phenytoin: Basic Clinical Pharmacokinetic Parameters


When I burn isopropyl alcohol (IPA), it burns orange. But when I burn ethyl alcohol, it burns totally blue. Why is this? I burned them in a small aluminium tray. While IPA is burning orange, it produced the smell of soot, but while ethanol is burning blue there isn't any smell.

Also, ethanol made the tray really cold when I poured it on the tray before burning it, compared to isopropyl alcohol. How can this be explained?

Note: IPA concentration is 99%, Ethanol is 97%

IPA has a different carbon:hydrogen ratio than ethanol. There is more incomplete combustion occurring with IPA, hence the smoky orange flame and smell of soot. Ethanol combusts more completely, leading to a blue (soot-free) flame and no smell.

In response to your second question, ethanol likely has a lower latent heat of vaporisation than IPA, resulting in it evaporating rapidly. It takes a lot of heat energy away from the tray when doing this, leading to the tray cooling. A similar effect can be observed if you accidentally get some types of solvent on your hand and feel a sudden coldness as they vaporise, taking heat from your skin.


Zinc and copper salt bridge replacement If I were to replace $\ce{Zn}$ with $\ce{Al}$ in  $\ce{Zn|Zn^2+ || Cu^2+|Cu}$. Would it produce the same results?

No, it would not.

$\ce{Al}$ is normally covered by oxide layer which effectively renders it inert to most things. Your cell would stop dead, as if the electrodes were made out of wood.

Then again, you might want to know whether it is possible to remove the said layer. Yes, it can be done, but then $\ce{Al}$ turns out to be just too active. It reacts with water quite readily (albeit not as spectacularly as $\ce{Na}$). Merry little bubbles of $\ce{H2}$ will pop up, flakes of $\ce{Al(OH)3}$ will drop to the bottom, and in a matter of an hour or so your electrode will be gone. While it lasts, the cell will produce some voltage; as Dr.J. correctly noted, it will be even greater than that of the normal $\ce{Cu/Zn}$ cell.

So it goes.


Can you in reality give neon a charge of +8? First off, I am currently confused about why neon can even be ionized at all. But since it can be ionized, this is the energy required to give a mole of neon a charge of +8: 207,390,000 joules!! Or ~0.5 GJ! Giving one-fifth of a kilogram of neon the maximum charge would be like the energy from 50 armor-piercing rounds of the ISU-150 assault tank, or approximately one Tour-de-France! Charging up one kilo of neon would be basically either throwing a lightning bolt or melting a tonne of steel.

My question is, what insidious chemical or procedure could rip away all of neon's electrons like that? Or is it just simply impossible to do?

I suspect that it is possible to do (otherwise how would they be able to calculate the energy expenditure), but how is the question.

Of course you can take all the electrons off an atom - it is then called "fully stripped" in atomic physics. You don't need to do it to an entire mole, mind you. In accelerators one would send energetic neon ions through a background gas or a thin foil, and the interactions will result in various charge states coming out, up to and including fully stripped.


What is meant by the "probability density" of finding an electron? Could someone please explain to me in layman's terms what probability density of finding an electron means, just as probability means chances of finding an electron. With due respect, please don't answer the formula or the distribution curves as I understand and can plot those, I just need the definition.

In quantum mechanics we often need to find the probability at some position $x$. However, $x$ is a continuous variable with an infinite number of values so it does not make sense to ask what is the probability of being at some exact position , say $x=1/4$, as this will be vanishingly small. Instead we imagine that the probability is that of being at position x to $x + \mathrm{d}x$ where $\mathrm{d}x$ has some very small value. If $f(x)\,\mathrm{d}x$ is that probability then $f(x)$ is the probability density which must be a real non-negative number. It turns out that quantum mechanics postulates that the probability density is given in terms of the wavefunction is $|\psi|^2$.


How to find out the energy gained from hydrolysis of Dimethylsulfoniopropionate? I am trying to find out what the energy gain from hydrolysis of DMSP (Dimethylsulfoniopropionate) is, versus the energy gain from oxidation of a sugar or carb (e.g. glucose) in a terrestrial atmosphere.

To be clear, I am talking about the 'cleavage of chemical bonds by the addition of water' to DMSP.

This question is motivated by the issue of whether heterotrophic animals living in a reducing atmosphere of an exoplanet which has no free atmospheric oxygen might use DMSP or some similar compound for energy storage in preference to carbohydrates.

DMSP was mentioned by the paper linked below, which suggests that ocean algae get energy from this process and so could alien metabolisms which don't have access to O2. The author states that the energy available from hydrolysis of DMSP is the same in a reducing or oxidising atmosphere but doesn't state what that energy gain actually is in j/mol.

I realise the actual energy storage molecular species used by an actual organism on such a planet is probably going to be something else, but this seems like a good proxy.

"Photosynthesis in Hydrogen-Dominated Atmospheres", William Bains, Sara Seager and Andras Zsom

When the literature refers to "hydrolysis" of DMSP, it is referring to the following cleavage reaction, even though no water is involved in the overall reaction:

DMSP --> DMS + acrylate + H+

See for example at page 156 of Biological and Environmental Chemistry of DMSP and Related Sulfonium Compounds: "hydrolysis of DMSP to DMS and acrylic acid".

For the above reaction:

$\Delta G = -11.8$ kcal/mol

Source: MetaCyc Reaction: 4.4.1.3

This means 11.8 kilocalories (49.4 kilojoules) of energy are released for each mole of DMSP that reacts.


The definition of n-factor, what was it when it was introduced? I realize that n-factor is an outdated concept however I am interested in its formal definition when it was introduced.
We have many definitions now on how to calculate n-factor depending on the type of reaction and compounds involved.

For acids, n-factor is defined as the number of H+ ions replaced by 1 mole of acid in a reaction.
For bases, n-factor is defined as the number of OH– ions replaced by 1 mole of base in a reaction.
The n-factor for such salts is defined as the total moles of catioinic/anionic charge replaced in 1 mole of the salt
The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt.

However, none of these definitions explain what n-factor actually is. Is there a formal definition for n-factor?

Instead of asking what n-factor really was, a better question would have been what an equivalent was (But that would have been a duplicate)
According to the above question, the formal definition of an equivalent is as follows:

An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

Therefore given any chemical reaction, one equivalent of each reactant reacts to form one equivalent of each product, irrespective of the stoichiometric coefficients
$$\ce{aA + bB -> cC + dD}$$
In all such scenarios, the n-factor of the compound is defined as,
$$\textit{n}\text{-factor} = \frac{M_\mathrm{molar}}{M_\mathrm{equiv}}$$
However, why are there so many different definitions for calculating the same thing. This is because an equivalent doesn't have one fixed definition, rather depending upon the situation, the definition of an equivalent changes and hence the n-factor definition changes as well.
Therefore there is no formal definition per se for the n-factor, simply because it was never formally introduced. As M.Farooq pointed out in the comments,

I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books. The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions.


According to the Molecular Orbital Theory, what makes a compound aromatic? For example, benzene has 6 $\pi$ electrons. Which means the p orbitals will be occupied, and only the bonding molecular orbitals.
But 1,3-cyclopentadiene has 4 $\pi$ electrons, which means 2 of the 3 molecular bonding orbitals will be occupied. Why does this make it non-aromatic?

There's 4 criteria for aromaticity:
1. The compound must be cyclic. As in a ring structure of some kind.
2. The compound must be planar. All the atoms making up the ring must be on the same plane. The hybridization of bonds causes this which is why benzene is planar and cyclohexane is not.
3. The compound must be conjugated. Therefore it must alternate between single bonds, double bonds, lone pairs, or empty orbitals. For example, two double bonds in a row mean it is not conjugated. But a single bond connected to an atom with an empty orbital still satisfies the rules of conjugation. Another example, benzene alternates single bonds and double bonds, so it is conjugated.
4. Must satisfy Huckel's rule. So the number of pi electrons must fit in the equation 4n+2, with n being a whole number or 0. So 6=4n+2 works but 4=4n+2 would mean n=1/2 so that does not work.


Does hyperconjugation into sigma-star explain methylamine's increased basicity relative to ammonia? The inductive effect with the polarisation of the C-N bond obviously plays a role in stabilising the positive charge of the conjugate acid. But given the similar electronegativities for C and H, CH3 group should donate as much electron density to N as H does (roughly), if only the bonding atoms are considered. And if we're considering atoms other than the two bonding atoms in their contribution to stability, is it not a question then of hyperconjugation?

This question seems to deny hyperconjugation plays a role in stabilising ammonium derivatives. However, it takes a very narrow view of hyperconjugation and focuses only on electron density donation into bonding orbitals. But if more substituted double bonds are more stable because of the donation of alkyl groups into pi-star orbitals, antibonding orbitals must be viable recipients. And similarly in methylammonium, we have sigma-star orbitals on the positive nitrogen which the C-H bonds of the methyl should be able to donate into, destabilising the N-H bond in methylammonium but overall stabilising the system.

So, my question is does hyperconjugation from sigma C-H bonds into sigma-star orbitals of sp3 centres occur, and then from that - does hyperconjugation explain methylammonia's strength (or alkylammonias' strength, more generally) relative to ammonia?

Yes, I once thought about this interesting phenomenon also. There is no good reason why hyperconjugative interactions cannot take place between the $\ce {N}$ lone pair orbital and the $\ce {C-H}$ $\sigma$. However, the interaction may be less favoured if you consider the geometry, compared to the the usual hyperconjugation to alkene $\pi$ systems or to carbocation $\ce {p}$ orbitals. Although geometry does not help to favour this interaction as much, the change of atom from $\ce {C}$ in the usual case to $\ce {N}$ does help to decrease the energy gap for the interaction since $\ce {N}$ is more electronegative than $\ce {C}$ and thus, its atomic orbitals would all be of a lower energy compared to $\ce {C}$.

Raising the energy of the HOMO of a general nucleophile would increase its nucleophilicity by decreasing the energy difference between its HOMO and the LUMO of the electrophile. In the same way, by this filled orbital-filled orbital interaction, the HOMO of the methylamine is now raised, allowing it to more easily interact with acids. Thus, $\ce {MO}$ theory can be used to explain the increasing basicity of amines with increasing alkyl substitution.


Why is aniline less basic than methylamine? The reason given for the question is that in aniline the lone pair on N atom is delocalized throughout the benzene ring, making it less available for donation.

But the conjugate acid of aniline should be more stable given the fact that the positive charge on N atom is also delocalized throughout the ring, hence making it more stable than the conjugate acid of methylamine which has +I effect. On that account, shouldn't aniline be more basic?

N in aniline is forced into an $\ce{sp^2}$ configuration because only then can its p orbital interact to form aromatic MO’s. By contrast, methylamine has its N in the usual $\ce{sp^3}$ configuration. An $\ce{sp^2}$ lone pair has a great contribution from an s AO than an $\ce{sp^3}$ lone pair (33% vs 25%), and because an s AO is lower in energy than a p AO (due to a region of electronic density closer to the nucleus), it is also lower in energy. Therefore, the HOMO of aniline has a poorer energy match with the H 1s LUMO than the HOMO of methylamine, and the resonance energy stabilizing the protonated adduct is smaller. Hence, aniline is less basic.


Are polyhydroquinones considered to be polyphenols? Looking at the Wikipedia article for polyphenol, I was struck by how the monomer units shown are all substituted meta or ortho for the hydroxyl groups.

https://en.wikipedia.org/wiki/Polyphenol#Chemical_structure_and_synthesis

I assumed, naively, that polyhydroquinones would be considered polyphenols, but I am uncertain now. From what I can see, there is no IUPAC definition of polyphenols, so I am interested to know what commonly accepted criteria (and what contentious aspects, if any) are there for polyphenols.

The WBBSH definition states that polyphenols are defined as:

generally moderately water-soluble compounds
with molecular weight of 500–4000 Da
with >12 phenolic hydroxyl groups
with 5–7 aromatic rings per 1000 Da

This is the most widely accepted definition. However, due to the increase in research, there is a proposed Quideau definition stating

“The term "polyphenol" should be used to define compounds exclusively derived from the shikimate/phenylpropanoid and/or the polyketide pathway, featuring more than one phenolic unit and deprived of nitrogen-based functions.”

However, this is not an official definition (yet)


Endothermic reaction absorbs heat from environment I read on my school book about chemistry the reaction:

$$\ce{H2O + NH4Cl}$$

with $T(\ce{H2O}) = \pu{25 °C}.$

The temperature of the final solution is $\pu{20 °C}$.
I know that it is an endothermic reaction, and the initial solution absorbs temperature from the environment.

I don't understand why the temperature of the final solution is lower than the initial solution, I think it should not be lower because  all the heat that it needs comes from the environment.

An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released
into the surroundings, resulting in an overall negative quantity for the heat of reaction (-ΔE ).
An endothermic reaction
occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat.
Endothermic reactions result in an overall positive heat of reaction (+ΔE ).
Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (ΔE ), the sum of
all potential and kinetic energies.
ΔH is determined by the system, not the surrounding environment in a reaction.
A system
that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the
products is lower than the enthalpy of the reactants of the system.
The enthalpies of these reactions are less than zero, and are therefore exothermic reactions.
A system of reactants that absorbs
heat from the surroundings in an endothermic reaction has a positive , because the enthalpy of the products is higher than
the enthalpy of the reactants of the system.
Because the enthalpies of these reactions are greater than zero, they are endothermic reactions.
EDIT ( See this )
I understand your confusion. See, in endothermic reaction the product will be "colder"....this might seem counterintuitive as in endothermic reaction heat is absorbed by the system form the environment. Therefore the heat absorbed must increase the temperature right?
No! See temperature shows the direction of heat flow. If the temperature of the system increased ( assuming that previously the system and environment were in thermal equilibrium ~ same temperature ) this would mean that heat should flow from system to the environment as the system has a higher temperature than environment.
The following example makes everything clear:
$$T_{(env)}= 25°$$
$$A + B _{(25°)}  = C _{(20°)}  ...... (1)$$
$$A' + B' _{(25°)}  = C' _{(30°)} ..... (2)$$
Compare the temperature of system and environment in the above rxn.
Therefore in rxn. 1 heat flows from environment to system ( from 25° to ~20°).
This heat energy is used to break bonds and stuff.( Endothermic )
And in rxn. 2 heat flows from system to environment ( from 30° to 25°).( Exothermic )
Also see this practical video:
https://youtu.be/NHOHKH3808c


What is an example of an isopropyl alcohol resistant caulk? What is a caulk (or bonding adhesive) that can be exposed permanently submerged in an isopropyl alcohol (>91%) bath?

The idea came from this previous question:
Silicone vs Alcohol

It seems from one of the answers there the alcohol would be resistant but the adhesion properties may be compromised.

The more readily available the product, the better.

Not an answer, but a strategy to get one:  I would start by googling 'alcohol-resistant caulk' or 'alcohol-resistant sealant'.  This will give you hits for chemical companies that produce such products.  For instance,

https://www.azom.com/article.aspx?ArticleID=5854
claims to have an epoxy ("UV18MED") that is alcohol-resistant and can be used as a sealant.

https://bowersindustrial.com/product-category/concrete-sealers-caulking/chemical-resistant-caulk/
claims to have caulks that are resistant to alcohol.

Since each of these companies are experts on the properties and performance of their respective products, you should call them, describe your specific application, and ask if they have anything that would work reliably.  That's the best way to be assured the product will do what you want.

One problem you might run into is that these may be B2B companies, i.e., ones that sell in quantities designed for businesses rather than consumers.  If that's a problem, try calling some manufacturers of consumer adhesives and asking if they have anything.

I know this is more work than simply posting a question on CSE (and there is of course nothing wrong with posting such a question), but it's only by speaking directly to the manufacturer that you can be sure the product is designed to work in your application (unless you have a product label that indicates this).

While you're speaking to them, you might also want to ask about leaching -- whether components of their product might dissolve into the alcohol; you would need to judge for yourself how much leaching your system could tolerate.  You'll also need to let them know what kind of mechanical stresses and temperature changes the product must tolerate (e.g., an adhesive that is sealing an aquarium and also holding it together needs to be a lot stronger than a caulk that is merely sealing the top of a sink). The more specific details you can give them about your system, the better.


Charge to mass ratio of an electron The charge to mass ratio of an electron is -1.7588×10^8 coulomb/g, and is fixed. My textbook, however says that as the speed of an electron increases, the value of charge to mass ratio increases. How is this possible? Shouldn't the mass and charge of an electron always be the same? How is the ratio changing?

The electric charge on an electron is fixed, regardless of its speed. The apparent mass of an electron INCREASES with its speed (relative to the observer). So charge/mass will decrease as mass → ∞. The major problem with these statements is the use of "mass" to mean "apparent mass" or "observed mass". More and more (and probably the majority now days) of the Physics community reserve the term "mass" to mean "rest mass" (which is the mass observed at zero velocity relative to the observer). This is a bit more complicated than it seems: consider a gas. The center-of-mass velocity may be zero (relative) but obviously each molecule has its own velocity not dependent on center-of-mass. Fortunately, in the range of temperatures chemists are concerned with, relativistic velocities aren't seen, so no correction is necessary between a molecule's rest mass and its apparent mass. Apparent mass, observed mass, relativistic mass, dynamic mass, are (usually) all different ways of expressing the fact that the energy of an object affects its mass; E= mc² and that energy is the sum of potential and kinetic terms, where the kinetic term is ½mv². Hence the mass you will observe depends on how fast the object is traveling relative to you (or your observational equipment). The Special Relativity equation for mass is m = m0 ÷ √( 1 - (v/c)²) where v is the particle's speed (magnitude of its velocity), c is the speed of light, and m0 is its mass when v=0, i.e. its rest mass. As you can see, as v approaches c, the denominator approaches 0 and so the observed mass gets larger without limit.

I should also note that the generalization I made, that chemists need not "worry" about relativistic velocities, isn't entirely correct. For instance, the color of gold is due to relativistic effects (otherwise it would be silver just like the other more common metals). In general, atoms do not have relativistic velocities, but larger atoms have electrons with relativistic momentum, which can be thought of as relativistic speed. There's no reason why atoms can't have relativistic speeds, other than the obvious problem of how we'd contain and observe them. If there wasn't any "chemistry" at relativistic speeds, we'd have fusion power already - confinement of plasma is "hard" because it interacts with the container.


Confused about weak/strong acids, conjugated acid-base pairs and buffers I learned that

a) the conjugate base to a weak acid is a strong base (and vice versa).

b) a buffer consists of a weak acid (base) and its conjugate base (acid).

However, this explanation of buffers says the following:

"[...] a weak acid is one that only rarely dissociates in water [...]. Likewise, since the conjugate base is a weak base, [...]"

which seems to stand in conflict with my assumption a) above. So, what is correct?

Furthermore, if b) is correct, isn't any solution of a weak acid solution a buffer, since any weak acid in water makes an equilibrium of the form

$HA \text{ (weak acid)} \leftrightarrow H^+ + A^- \text{ (conjugate base)}$

and is thus a solution of a weak acid and its base?

Your statement a) isn't always true.

Water dissociation is represented by:

$$\ce{H2O + H2O <-> H3O+ + OH-}$$

$$ K_\mathrm w=[\ce{H3O+}]\cdot [\ce{OH-}] = 1\times 10^{-14}\ (\textrm{at}\ 25^\circ ~\mathrm C) $$

Note 1: We don't write the $\ce{H_2O}$ activity, since it can usually be rounded to 1 and the ions activities can be rounded to their concentrations.

Note 2: This value was obtained experimentally, considering the concentration of $\ce{H_3O^+}$ and $\ce{OH^-}$ in the medium was the same and measuring the ionization of water.

On a weak acid dissociation:

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

$$K_\mathrm a=\frac{[\ce{A^-}]\cdot [\ce{H3O^+}]}{[\ce{HA}]}$$

On its conjugate base:

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

$$K_\mathrm b=\frac{[\ce{HA}]\cdot [\ce{OH^-}]}{[\ce{A^-}]}$$

If we 'add' both reactions we expect that both equilibriums will happen in the mixture, so we have:

$$\ce{HA + H2O <-> A- + H3O+}\ \ \ \ (K_\mathrm a)$$

$$\ce{+}$$

$$\ce{A- + H2O <-> HA + OH-}\ \ \ \ (K_\mathrm b)$$

$$\ce{============================}$$

$$\ce{H2O + H2O <-> H3O+ + OH-}\ \ \ \ (K_\mathrm w)$$

$$K_\mathrm b\cdot K_\mathrm a=[\ce{H3O+}] \cdot [\ce{OH-}] \cdot ([\ce{A-}]\cdot [\ce{HA}])/([\ce{A-}] \cdot [\ce{HA}]) = [\ce{H_3O^+}]\cdot [\ce{OH^⁻}] = K_\mathrm w$$

$$K_\mathrm b=K_\mathrm w/K_\mathrm a~~~~~~ K_\mathrm b=10^{-14}/K_\mathrm a\ \ \textrm{at}\ 25^\circ ~\mathrm C$$

Another way to write this:

\begin{align}-\log(K_\mathrm b)&=-\log(10^{-14}K_\mathrm a) \\ \implies \mathrm pK_\mathrm b &= -\log(10^{⁻14}) - (-\log(K_\mathrm a))\\ \implies \mathrm pK_\mathrm a+\mathrm pK_\mathrm b &=14\;.\end{align}

This is the relation between a conjugate base strength and its acid strength. A very strong acid has a weak conjugate base, but a weak acid doesn't necessarily have a very strong base. Say you have a $\mathrm pK_\mathrm a=5$, which is a weak acid, with $K_\mathrm a=1\times 10^{-5}$. The conjugate base would have a $\mathrm pK_\mathrm b=14-5=9$ or a $K_\mathrm b=1\times 10^{-9}$, which is not a strong base.

However, if we have a strong acid, like $\ce{HCl}$ with a $\mathrm pK_\mathrm a=-6.3$ and $K_\mathrm a=10^{6.3}$. Its conjugate base would have a $\mathrm pK_\mathrm b=14-(-6.3)=20.3$ and $K_\mathrm b=10^{-20.3}$ which is a really weak base.

Hopefully I didn't make it even more confusing for you! It's mostly an reaction equilibrium issue.


making patinas for copper Copper nitrate, calcium carbonate and ammonium chloride mixed will patina copper green. I need to know

If calcium carbonate should be a heavy powder or a reagent.
If calcium chloride is a dehydrate or reagent.
If ammonium chloride should be granular or granular reagent.

These mix with water in equal parts. I`m not a chemist, and this is for a sculpture with a recipe from an old metal techniques book.

Nothing about your process is quantitative in nature, nor does it require high-purity reagents.  So most definitely do not invest in reagent-grade anything.  To answer your questions in order:

1) The "heavy powder" calcium carbonate will be sufficient.  It may be "clumpy", in which case you will want to physically break it up prior to mixing it in with the water.  You should wear rubber gloves and a dust mask if you need to do this.  The only purpose of this compound in your procedure is to raise the pH.

2)  You definitely do not need anhydrous or dried calcium chloride. You are not trying to quantitatively measure out this compound and you are just going to put it in water anyway, so hydration is not an issue for you at all.  The purpose of this reagent (generic term, not referring to any "grade" or purity) is to provide chloride ions, that at high pH will form a layer of cupric chloride, which is green when it absorbs water from the air to form cupric chloride dihydrate. Again, this will be a product of your procedure and the degree of hydration of the calcium chloride is not relevant. As in step one, this compound may be clumpy and if so you should break it up as described there.

3) The phrase "should be granular or granular reagent?" is the key. As stated above, there is no need for reagent grade materials for your process, so if your options are "granular" vs. "granular reagent" then just get the granular, not the granular reagent ;)  The role of this compound is to both raise the pH and to provide chloride for the purpose described in step 2.

Again, it may be tempting to think you will get a better patina by investing in expensive reagents, but for your process that would simply be a waste of your money to buy reagent-grade anything if there is a lower grade alternative.


What is the minimum distance of Rubidium and Iod ions in a centered cube? I have a problem to visualize the following assignment:

What is the minimum distance between $\ce{Rb^+}$ and $\ce{I^-}$ ions if radius of  $\ce{Rb^+}$ is ${1.49\cdot10^{-10}}~\mathrm{m}$ and radius of $\ce{I^-}$ is ${2.17\cdot10^{-10}}~\mathrm{m}$ if they are arranged in centered cube?

The distance between $\ce{Rb^{+}}$ and $\ce{I^{-}}$ in the fcc lattice would be $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$


Does mercury amalgamate with Brass or Steel? Does mercury amalgamate with Brass & Steel ?

Details: I am designing a micro pump to dispense mercury in micro quantities ($\mathrm{300-700}\ \pu{mg}$). I am planning to use 2 check valves with syringe to suck mercury from a mercury reservoir and push it for dispensing.
The problem is that all the high quality check valves are mostly Brass & I am not sure if mercury amalgamates with it.

Brass contains zinc and copper.  Both of those materials will readily amalgamate with mercury and thus it is not recommended to allow brass to contact mercury.

Steel will not form an amalgam and is perfectly fine for use with mercury. In fact is Approved as a container for mercury in transport.


Why does phosphorus tribromide act as a Lewis acid (electron acceptor)? This question is purely out of curiosity and to help me remember the "why" of a mechanism to help me remember the "what/how".

In organic chemistry we recently studied the  Sn2 Substitution of Primary Alcohol with PBr3:

My question is, why is it favorable for PBr3 to act as a Lewis acid (i.e. accept electrons from the oxygen to form a bond) when it already has a full octet and no formal charge?

The why of a mechanism is always the best approach.

The reasons for this attack are:

$\ce{P}$ can expand its octet due to presence of vacant $d$-orbitals.
The $\ce{P-Br}$ bond is polar due to the electronegativity difference between the atoms and hence P develops partial positive charge favouring a nucleophilic attack on it.
Looking at the molecule as a whole, $\ce{Br}$ is a good leaving group so $\ce{P}$ doesn't mind a nucleophilic attack on it by the lone pair of oxygen.
"Nature loves stability" and  $\ce{P-O}$ bond is stronger than $\ce{P-Br}$ so that might be the driving force for this reaction mechanism.


Which of these can undergo decarboxylation reaction readily Which of these carboxylic acids can undergo decarboxylation readily?

According to what I know beta-keto acids undergo decarboxylation easily and readily due to formation of a stable intermediate. So my general answer was option (A) however the answer key states that (A),(B) and (D) are correct. I am not able to understand, based on my current theory what is the reason for (B) and (D) to undergo decarboxylation readily.

You have said that  β−ketoacids undergo decarboxylation easily and readily due to formation of a stable intermediate. That intermediate is the carbanion on the alpha carbon. So, any species that can stabilize that carbanion can undergoe decarboxylation easily. In (B), the carbanion formed is stabilized by the double bond beside it so it undergoes decarboxylation easily.
(D) is a special reaction as referenced to in this answer.


Is time a measurement of Entropy? Is time a measurement of entropy? Is it because time is increasing forever?

Well after reading about the Second Law of Thermodynamics and working with Entropy, I began to realize that if time can be interpreted as a measure of Entropy, since it is always increasing.

  Since almost everything decays with time, I thought that time could be
  a measure of Entropy. Since it is achieving the lowest energy state
  possible.

Thanks

Time and entropy has not much to do with each other.

From practical point of view: Equilibrium systems can hang around for infinite time, and their entropy does not change at all - so time would be an impractical measure.

From conceptual point of view: Entropy comes from statistical behavior of the system of many bodies. Time is unrelated to statistics; it exist even in single body systems.


Can a volume of gas be measured by change in PSI? I had a visit from an air conditioner repairman who brought with him a tank of refrigerant. He measured the system and determined that the pressure was about 20 psi too low. So he added refrigerant to the system until it was right.

When it came time to pay him, he said, "well I had to raise your system by 20 psi so it's going to be [20 x price for 1 psi]."

The price was fine, but I am still wondering: would it have taken the same amount of gas to raise my system from 0 to 20 psi as it would to go from 80 to 100, say?

Or, in other words, is it absurd to charge by pressure delta instead of some more reliable measure of what was dispensed from the tank?

Volume is inversely proportional to pressure only when:

1) the number of moles of gas is held constant. We can say this is true in your situation because the repairman measured the pressure in the tank and I don’t think a significant amount of gas disappeared from the tank in the time it took for him to measure the pressure.

2) The temperature is held constant. Again, I don’t think temperature dropped 20 kelvin in the time it took for him to measure the pressure in the tank.

This follows from the ideal gas law equation – a good approximation most of the time:

$PV=nRT$.

So I say your repairman still has his wits about him and isn’t trying to rip anyone off.


What is the shape of [Cr(NO)₄]? I was learning about crystal field theory recently. I am quite confused about predicting the shape of $\ce{[Cr(NO)4]}$ complex because I am not able to conclude whether it will be square planar or tetrahedral.

How to predict the shape of $\ce{[Cr(NO)4]}$ using crystal field theory?

Will $\ce{NO}$ act as a 3-e- donor here?

Will all the four $\ce{NO}$ that are in coordination with $\ce{Cr}$ be of the type nitrosyl or will three of them be nitroso and one nitrosyl (because $\ce{Cr^3+}$ is one of the common oxidation state of $\ce{Cr})?$

Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $\ce{Cr(NO)4}$.

The NO ligand can be rendered as a three-electron donor, if we take
it as uncharged.  It is actually similar to CO, really, in that it
interacts through its $\pi^*$ orbitals as well as its $\sigma$
orbital; but NO unlike CO has an electron to offer from its $\pi^*$.
That plus the $\sigma$ electrons makes three.
We then may identify the central metal atom as having 18 valence
electrons.  When this happens with multiple ligands we typically see
an arrangement with highest order symmetry, thus tetrahedral for
four-coordination.  This gives the best possible bonding to the
entire set of valence orbitals involved in the 18-electron
structure.  This is seen in the tetrahedral four-coordinationn of
$\ce{Ni(CO)4}$, and in the octahedral six-coordination of the series
$\ce{[V(CO)6]-, Cr(CO)6, [Mn(CO)6]+, [Fe(CO)6]^{2+}}$.  So, we should
expect $\ce{Cr(NO)4}$ to be tetrahedral.
Heberhold and
Razavi
found evidence for a tetrahedral structure in $\ce{Cr(NO)4}$ from
its IR spectrum.


Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes? Octahedral complexes have greater splitting in the d orbitals. Is it because octahedral complexes have more atoms and thus more interactions?

The answer is, as @orthocresol mentioned, due to two parts.
Initially, if we are assuming the ligands are point charges (crystal field theory) then in octahedral complexes, the ligands are more closely associated with the d orbitals, as shown in this diagram:
But, in tetrahedral complexes, the ligands are not so closely associated with the d orbitals, as shown in this diagram:
By associating the d orbitals seen in the first diagram, with the tetrahedral point charges in the second diagram, you can see how close the point charges are to the d orbitals in the octahedral case compared to tetrahedral.
Also as otho mentioned, octahedral has 6 ligands interacting with 5 d orbitals, whereas tetrahedral has only 4 ligands interacting with 5 d orbitals. Octahedral has the more negative charge due to more electrons, therefore more repulsion.


Confused with 'discharge potential',preferential discharge theory My book says that the ions which have least value of discharge potential are deposited first in the concerned electrode  because the 'require' the least energy to do so.I am confused by 'require' , do ions upon discharging gain energy or lose energy , are ions in general not less stable than neutral counterparts?further,suppose there are two contenders for the same electrode, say sodium ion and hydrogen ion during alkaline electrolysis of water , what is the need of adding the base , what actually is it doing and how does it improve electrolysis ?

I can answer your first two questions.

  I am confused by 'require', do ions upon discharging gain energy or lose energy?

Ions lose free energy during discharging, as you are referring to the spontaneous process of transferring charge at the electrodes with conversion of some of the ions into neutral species (in general you are speaking about the process described in the next paragraph).

For a process to proceed spontaneously, the free energy change for the system (combined half-cells, say) has to be negative. The Nernst equation tells us that the free energy difference is given by $$\Delta G = -nFE$$ where E is the total cell potential. This value must be negative for the cell to discharge, which means E is required to be positive. This is what is meant by "require" (language is used in funny ways, ions don't really require anything). You have more choices than just discharging the cell: you can charge it (creating more of some of the ions). Unless the potential is such to cause discharge you may actually produce more of the ions that are disfavored under spontaneous (discharging) conditions.

  are ions in general not less stable than neutral counterparts?

You should not attempt to generalize, as this depends on the context (details of the reaction in question). An ion may be the most stable state of an atom in one context (an oxidizing environment, say) whereas it will be unstable in another (a corresponding reducing environment). Moreover, you can apply a charging potential and thereby invert the sense of stability, rendering one species more stable than the other.


Reactions that occur at the anode electrode of electrolysis dilute H2SO4 How to know that reaction that occur at the anode electrode is $\ce{2H2O -> 4H+ + O2}$?

Why the reaction is not the oxidation of  $\ce{SO4^{2-}}$?

This problem does not give the E0 of $\ce{SO4^{2-}}$ too.

How to know that reaction that occur at the anode electrode ...

There are two kinds of anions: oxidizable (ones that contain central element in lower oxidation states), like $\ce{ClO3-}$, $\ce{Cl-}$, $\ce{NO2-}$, and unoxidaziable, where central element is in hightest oxidation state ($\ce{ClO4-}$, $\ce{NO3-}$. Fluoride obviously is unoxidizable as well.  As a rule of thumb, it is usually assumed that unoxidizable anions do not react in electrolysis. A rationalization for it is that in such cases an oxygen should be oxidized, and oxygen in water reacts faster. For the sake of basic chemistry course it is enough.

However, it is not entirely true: electrolysis of cold solution of ammonium sulfate in sulfuric acid results in formation of ammonium persulfate $\ce{(NH4)2S2O8}$, where anion has structure $\ce{ (O3SOOSO3)^{2-}}$ Similiarly, electrolysis of very cold solutions of carbonates produces various percarbonates. A synthetic method exists, that is based on electrolysis of carboxylic acid. The anions on oxidation form $\ce{RCOO\cdot}$, that further loose $\ce{CO2}$ and resulting radicals pairs, resulting in net reaction $\ce{2RCOO- \rightarrow R2 + 2CO2 + 2 e-}$ However, in general, formation of peroxocompounds requires low temperatures, usually below zero.

The exact mechanism of the process highly depends on the nature of the electrodes, but likely includes formation and transformation of surface particles on the electrode, making reasoning difficult and electrode-specific. Since commonly anodes used have oxide coating, it can be more-or-less safely assumed, that the process includes dissociative adsorption of water with subsequent oxidation of adsorbed (hydro)oxide atoms.


Mechanism of the oxidation of alcohols with KMnO4 Many oxidising agents, like chromate, dichromate, iodine in $\ce{NaOH}$ etc. seem to work via ester formation and elimination. For example, chromic acid will react with the $\ce{-OH}$ of alcohol to form a chromate ester, and then this will undergo an E2-like elimination, with a proton abstracted from carbon and loss of leaving group from oxygen.

However it seems that manganate works via a different mechanism. I think this because here:
http://www.masterorganicchemistry.com/2015/05/21/demystifying-alcohol-oxidations/

It says in a footnote at the bottom:

  "..main exception you’ll encounter is $\ce{KMnO4}$, which likely proceeds through a C-H abstraction/internal return type mechanism followed by collapse of the hydrate to give the new carbonyl."

I don't quite understand this brief summary of the mechanism, could someone expand on it? I don't quite understand what "internal return type mechanism" is refering to - I kind of understand it refers to something a little like the "intimate ion pair" in the $\mathrm{S_Ni}$ mechanism, but haven't been able to find online a more expanded mechanism - many places seem to show it like the case of chromic acid above, and so I am slightly confused as to what mechanism it really follows.

$\ce{KMnO4/OH-}$ Oxidation

$\ce{H2CrO4}$ Oxidation

The two main differences are

$\ce{H2CrO4}$ Oxidation occurs in acid, $\ce{KMnO4}$ oxidation occurs in base
The source of the nucleophile in the removal of the metal ester. In $\ce{KMnO4}$ oxidation, the $\ce{Mn}$ ester itself extracts the $\ce{H+}$ from the alcohol carbon, while in $\ce{H2CrO4}$ oxidation the nucleophile is the solvent.


Relationship between Productivity and Yield I am trying to formulate an optimization problem. Can you help me understand the definition of productivity and yield? Why is there a trade-off between them during the reaction?

Thanks in advance.

Depending on your application and the specifics, to keep it simple. . .

Yield: The amount of product obtained in a chemical reaction
Productivity: The rate of speed a product can be obtained in a chemical reaction

For the compromise, as vague example, perhaps the more of the product you want to yield, the more time it may take thus a longer productivity you'll have.


Find out of the friend is getting a good deal I have a strange chemistry question that I've never came across before, and I have no idea about how to solve it.

  Your friend comes across a good deal to purchase a gold ring. She asks you for advice for you to test the ring. The ring has a mass of 4.54 g. When the ring is heated with 94.8 J of energy, temperature rises from 23.0 °C to 47.5 °C. The gold specific heat capacity is 0.1291 J/g. Is she getting a good deal?

So I did the equation to solve for Q which is 14.35 J, and I have all the numbers the (considering they are given) but I have no idea how to actually calculate how much of this ring is pure gold.

I came here as a last resort.

The jewelry gold usually comes historically rated in grades 14 or 18 karats, where 24 means pure gold. The rest is mostly silver.

Find out specific heat capacity for the complementary metals like silver or copper, so you could roughly estimate via interpolation.

Another method would be pyknometry, determining the ring density by differential weighting the mass of displaced water - old good method of Archimedes. Gold has almost double density, compared to other expected metals.

Edit:  list of densities:

Gold 19.3.
Silver 10.5.

If linearly interpolated by mass fraction:.

14 karat gold  15.63.
18 karat gold  17.1


Where does this equation for the activity in the gas phase equilibrium between N₂, H₂ and NH₃ come from? I have a question with Equilibrium Constant.

I understand that for a reaction in equilibrium $aA +bB \rightleftharpoons cC +dD$
$$K_c= \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
and $K_p$ is just using the partial pressures.

I'm looking through my PChem1 tutorials and I saw that they wrote for

$$\ce{1/2 N2 + 3/2 H2 <=> NH_3}$$
$$K=\frac{a(NH_3)}{a(N_2)^\frac12a(H_2)^\frac32}$$

I believe that a represents activity. However what i don't get is what follows:
$$K=\frac{a(NH_3)}{a(N_2)^\frac12a(H_2)^\frac32}=\frac{\left(\frac{n(NH_3)RT}{p_oV}\right)}{\left(\frac{n(N_2)RT}{p_0V}\right)^\frac12\left(\frac{n(H_2)RT}{p_oV}\right)^\frac32}=\frac{n(NH_3)}{n(N_2)^\frac12n(H_2)^\frac32}\left(\frac{p_oV}{RT}\right)$$

I cant' figure out how $a(NH_3)=\frac{n(NH_3)RT}{p_oV}$

It obviously has something to do with the ideal gas equation $pV=nRT$ but I don't get how.

additional context:
The question asks for the equilibrium composition starting with 1 mol each of $\ce{N_2}$, $\ce{H_2}$ and $\ce{NH_3}$ in a 1L reaction vessel at 298K

P.s. I get that I am to use extent of reaction to solve this.

The activity of a molecule $\ce{M}$ as a function of its concentration in the solution can be written as

\begin{equation}
  a_{\ce{M}} = \frac{\gamma_{\ce{M}} c_{\ce{M}}}{c^0} \ ,
\end{equation}

where $c^0$ is the standard concentration and $\gamma_{\ce{M}}$ is the (dimensionless) activity coefficient (see also: here and Wikipedia).
If you have a sufficiently dilute solution so that non-ideal contributions become very small, the activity coefficient can be approximated by 1, i.e. $\gamma_{\ce{M}} \approx 1$.
So, what you're left with for the activity under these conditions is

\begin{equation}
  a_{\ce{M}} \approx \frac{c_{\ce{M}}}{c^0} \ ,
\end{equation}

Now, the concentration is defined by $c = \frac{n}{V}$ and with the ideal gas law $\frac{n}{V} = \frac{p}{RT}$ you get:

\begin{align}
  a_{\ce{M}} &\approx \frac{c_{\ce{M}}}{c^0} = \frac{\frac{n_{\ce{M}}}{V}}{\frac{n^0}{V}} \\
  &= \frac{\frac{n_{\ce{M}}}{V}}{\frac{p^{0}}{RT}} = \frac{n_{\ce{M}} R T}{p^{0} V}
\end{align}

where $p^{0}$ is the standard pressure.


Can I solubilize lemon essence in water somehow? I have a water based solution (95% water, the rest are salts) to which I want to add a lemon fragrance. I bought "water based lemon essence", but upon putting a mere drop in about 500 mL, the solution turned from crystal clear to milky off-white.
I researched and found that lemon essence is apparently one of many essential oils (citral, citronellal, cintronellol). None of which are water-soluble. That's why the result is an emulsion, and it's not translucent. (It mixed very well and it's very stable, that's not my problem)
I was wondering then if it's possible to give lemon fragrance to a water solution without turning it into a milky emulsion.

Some hints are provided by Wikipedia:Lemon liqueur:

To produce the Lemon liqueur requires sugar, water, lemon zest,
mixing in organic salutes  liquor, and time to mature. Lemon zest is soaked in high proof neutral spirits to extract from it the lemon oil (an essential oil). The extraction is then diluted with simple syrup.

In other words, lemon zest is drawn into water by mixing in water-soluble organic compounds, in this case sugar and (ethyl) alcohol.  If you want it to be nonalcoholic, maybe just the sugar will do.
During late autumn I make a cranberry jam by boiling the cranberries with oranges, sugar and water.  In the presence of the sugar and organic material from the oranges and cranberries and with the application of heat, orange zest is extracted into the jam.  The jam is too dark and thick to see whether the zest remains in solution when the jam is cooled, but the extraction of the zest is evident by taste.


Confusion about hybridization terminology I was just introduced to the concept of orbital hybridization. I believe I understand the idea behind it, but there is some accompanying terminology that prevents me from achieving a greater understanding.

In the following image:

I see that one of the electrons in the $2s$ orbital of carbon is promoted to one of the $2p$ orbitals. I start to get confused when hybridization occurs, simply because there are four $\ce{sp^3}$ hybrid orbitals, as also shown in this diagram:

I learned that the name $\ce{sp^3}$ comes from the combination of one s-orbital and three p-orbitals ($\ce{s + p + p + p}$), and this makes sense to me. What doesn't make sense is why each individual hybrid orbital is called an $\ce{sp^3}$ orbital. When the $\ce{2s}$ orbital, for example, becomes an $\ce{sp^3}$ orbital, why is it referred to by its combination with p-orbitals ($\ce{sp^3}$) when it was just an s-orbital before?

Could someone please offer a simple explanation that would help clear this up? Why is each individual hybrid orbital called an $\ce{sp^3}$ orbital? It seems like all four orbitals together should be called $\ce{sp^3}$, because they all make up the combination defined by $\ce{sp^3}$ ($\ce{s + p + p + p}$).

The number of hybridized orbitals is equal to the number of original orbitals. The name of the hybridized orbitals (e.g., $sp^{3}$) means just that the ratio of the contributions of $s$ to $p$ orbitals is $1:3$. We do not write $s^{\frac{1}{4}}p^{\frac{3}{4}}$.


Why would permeability for krypton be higher than argon? So, I've been looking over data on permeabilities and there's something that bothers me.  Permeability data on the noble gases is somewhat hard to find, but the Parker O-Ring Handbook has them for a number of materials:

http://www.parker.com/literature/O-Ring%20Division%20Literature/ORD%205700.pdf

(as an example material to choose in the list, say, PTFE)

What I don't get is... why is the permeability rate for krypton usually higher than that of argon for most materials? ?  Their chemical properties are basically identical, krypton is just a larger atom.  By all standards one would expect it to permeate slower.  Xenon permeates slower than both of them, as expected.  So what's up with krypton?

As noted in some of the comments above, a paper that directly compares noble gas permeation through several materials is S.J. Schowalter et al., Permeability of noble gases through Kapton, butyl, nylon, and "Silver Shield", in Nucl. Instr. Methods A 615 267-271 (2010) . They note that $K = Db$ where $K$ is the permeability, $D$ is the diffusivity, and $b$ is the solubility of the noble gas in the material. For the noble gases, the diffusivity monotonically decreases from He to Xe in all materials. However, the solubility has an odder behavior, with Ne being the oddball with lower solubilities than He or Ar on either side of it.

The net result is that the permeation can end up not being monotonic in nature across the noble gases. In particular, for butyl, $K$ is highest for He, drops for Ne, increases for Ar, drops a bit for Kr, then continues lower for Xe.

Lots of interesting data in the paper, a nice direct comparison of the noble gases and several materials in the same aparatus under the same conditions .


Why are more easily oxidized elements more reactive than less easily oxidized elements? Why do the more easily oxidized elements displace the less easily oxidized elements?

Question inspired by Activity Series of Metals

Ease of oxidation is how easily it loses electrons. Electrons are the basis of chemical reactions. If chemical A is more easily oxidized than B, then B is a relative oxidizer to A, so in terms of the activity series, the elements that are more easily oxidized can react with more chemicals, since they will be able to act as reducing agents to more chemicals.


Can metals have a net negative charge Normal metals like sodium or Calcium have a positive charge as $\ce{Na}^+$ or $\ce{Ca}^{2+}$.
Transition metals have a loot of variable oxidation states.

Yesterday I read about Iridium in Wikipedia and it stated that

  a +9 oxidation state has been found in $\ce{IrO_4^+}$

which is astounding for me...

So i was thinking can metals have a net negative charged ion also?

According to Google :

Since it is written that metals USUALLY have positive charges which means there may be conditions or compounds when they have negative charges too. Is it possible?

Also can Noble gasses form ions too?

These species usually do not exist in nature, but they can be synthesized.

Silver has been reduced in liquid ammonia to give $\ce{Ag-}$.
A lot of anionic metal carbonyl complexes $\ce{M(CO)_{n}^{m-}}$ have been synthesized:

-1

$\ce{[V(CO)6]-}$, $\ce{[Nb(CO)6]-}$, $\ce{[Ta(CO)6]-}$, $\ce{[Mn(CO)5]-}$, $\ce{[Ir(CO)4]-}$, $\ce{[Co(CO)4]-}$, $\ce{[Rh(CO)4]-}$

-2

$\ce{[Ti(CO)6]^{2-}}$, $\ce{[Zr(CO)6]^{2-}}$, $\ce{[Hf(CO)6]^{2-}}$, $\ce{[Fe(CO)4]^{2-}}$, $\ce{[Ru(CO)4]^{2-}}$, $\ce{[Os(CO)4]^{2-}}$

-3

$\ce{[V(CO)5]^{3-}}$, $\ce{[Nb(CO)5]^{3-}}$, $\ce{[Ta(CO)5]^{3-}}$, $\ce{[Mn(CO)4]^{3-}}$, $\ce{[Re(CO)4]^{3-}}$

-4

$\ce{[Cr(CO)4]^{4-}}$, $\ce{[Mo(CO)4]^{4-}}$

There's a review by John E. Ellis: Highly Reduced Metal Carbonyl Anions: Synthesis, Characteri's zation, and Chemical Properties on these compounds.

Edit

While Christian Aichinger's comment concerning $\pi$-backbonding in metal carbonyls is correct, it is important to realize that $\ce{CO}$ is a neutral ligand!

With other words: The Mond process: $\ce{Ni(s) + 4CO -> Ni(CO)4}$, using nickel powder and carbon monoxide, is not a redox reaction and the oxidation state of the metal still is $0$ in the metal carbonyl.


Why is kerosene preferred over other non-polar solvents to store sodium in? I think we all heard this again and again in our early schooling days:

  Sodium is always stored under kerosene so that it doesn't react with air, and corrode itself.

But lately I've been wondering, why kerosene? Why not benzene? Why not cyclohexane? Why not $\ce{CS2}$? What is that makes kerosene preferred?

Is kerosene the cheapest non-polar liquid available? Or is there any particular reason that I'm not aware of?

Kerosene is a fraction from the distillation of petroleum, the one with boiling point 180° to 230°, containing hydrocarbons from $\ce{C_11}$ to $\ce{C_12}$. In reaction to what has been said in the commentaries (below the question), I note that mineral oil is a cloudy expression whereas kerosene seems to be the precise name of the more or less broader base material of this cloud.

The best means to store alkali metals, which are all strong reductants, is an alkane because they are inert (as indicated by the name “paraffin”) and cannot be reduced.$^1$ The reason why one does not take the lower boiling fraction (gasoline 40° to 180°, $\ce{C_6}$ to $\ce{C_10}$), cyclohexane (b.p. 81°) or hexane (b.p. 69°) is obvious: it would be too evaporable, which is dangerous since its evaporation when someone forgets to close the bottle sufficiently, would lead to spontaneous ignition of the alkali metal (especially in the case of potassium) by contact with moist air. (Additionally, cyclohexane would anyway be unnecessary expensive.) The reason why the higher boiling fraction (light gas oil, 230 to 305°, $\ce{C_13}$ to $\ce{C_17}$) is not used is that this is much more greasy than kerosene.

It is strange that in the question benzene and carbon disulfide are proposed since they are both more expensive, very toxic and their b.p. is very low. — In the case of carbon disulfide this would be a paradigm of a very dangerous choice for three reasons: 1. concerning the boiling point (46°, extremely fiery), 2. concerning its extreme toxicity and 3. because it is not inert against reductants.

Footnote:

$^1$ In the case of lithium, which is much less reactive than the other alkali metals, the container should ideally be filled level with the top since it is swimming on the kerosene. But anyway the kerosene stays over as a protecting film on the lithium, which obstructs its reaction with moist air, even if it is not submerged.


Apparent paradox regarding spontaneity Consider the following analysis ( rev and irrev respectively refer to a reversible and an irreversible path, between the same initial and final states):
$dU_{rev}=dq_{rev}+dw_{rev}$,and,$dU_{irrev}=dq_{irrev}+dw_{irrev}$.
$dU_{rev}=dU_{irrev}$. Therefore, $dq_{rev}+dw_{rev}=dq_{irrev}+dw_{irrev}$. Rearranging,
$$dq_{irrev}-dq_{rev}=dw_{rev}-dw_{irrev}\tag{E01}$$
Now: If the volume increases, then $dw_{rev}$ and $dw_{irrev}$ are negative: with $|dw_{rev}|>|dw_{irrev}|$. Thus $dw_{rev}-dw_{irrev}<0$.
On the other hand, if the volume decreases, then $dw_{rev}$ and $dw_{irrev}$ are positive: with $|dw_{rev}|<|dw_{irrev}|$. Thus, again, $dw_{rev}-dw_{irrev}<0$.
We can thus conclude, in general, $dw_{rev}-dw_{irrev}<0$. E01 thus becomes:
$$dq_{irrev}-dq_{rev}<0\tag{E02}$$
Now the Gibbs free energy (G) is defined as $G=H-TS$. Thus, $dG=dH-TdS-SdT$. At constant pressure and Temperature, $dH=dq$ and $dT=0$. Thus,
$$dG=dq-TdS= dq-T(dq_{rev}/T)=dq-dq_{rev}\tag{E03}$$
If the process is reversible , then E03 becomes $dG=0$. If the process is irreversible, E02 implies $dG<0$. Combining these together, we get $dG \leq0$.
This analysis ensures that $dG$ is never positive...... Which seems to suggest that (if we take $dG$ as the criteria for spontaneity) every process is spontaneous.
The logical conclusion is that there's something in my analysis that isnt general: it holds for spontaneous processes only. What is it? Is it that the first-law requires some modification to be general?

The problem is that you are starting the derivation from conditions that ensure reversibility or spontaneity, but you are not considering conditions that lead to the third possibility.
The following expression from your post is generally true:
$dq-dq_{rev}=dw_{rev}-dw $
I've dropped the "irreversible" labels and assume that $dq$ and $dw$ refer to some process that you have dreamed up.
Now the condition that leads to non-spontaneous processes must be
$dw_{rev}-dw>0$
since you have shown that the other conditions are for spontaneous processes, which leads to
$dq−dq_{rev}>0$
and therefore
$dG>0$
which is what you are looking for.
If you forget about $dG$ for a second and focus on  how you can get to the equation $dq−dq_{rev}>0$ (since you already showed the next step to an inequality in terms of $dG$), you see that to arrive at this equation you have to say that the system can do more work in a process you dreamed up than in a reversible process  ($dw_{rev}-dw>0$), which violates the 2nd law. To carry out a process in which $dG>0$, you have to correct this violation and make up the work difference by having the surroundings do additional (more) work on the system than was assumed in the original analysis.
By the way, you should be careful using PV-work in your examples, as for an ideal gas in the absence of reactions, Boyle's law says that the pressure will change during an isothermal expansion and Charles' law says that the temperature changes in an isobaric process. In neither case would the Gibbs free energy provide the criterion for spontaneity.


Theoretical model for the hydrogen bond I am curious how, from the perspective of quantum chemistry, the hydrogen bond is modeled. In particular I am interested in using such a model to estimate the amount of energy released per $\ce{H2O}$ molecule, if say two of them approach each other e.g. in case of condensation. Any hint for precise literature would be great!

Please go through papers by Prof. Ralf Ludwig. You will certainly gain a lucid understanding of the subject both at experimental end as well as computational end. He advocates use of low frequency vibrational modes such as Far IR along with quantum chemistry for studying hydrogen bonding and allied weak interactions in very simplistic yet the most accurate way.

References:

"Hydrogen Bonding in protic ionic liquids: Reminiscent of Water." K. Fumino, A. Wulf, R. Ludwig., Angew. Chem., 2009, 121, 3230-3233. Angew. Chem. Int. Ed., 2009, 48, 3184-3186.
"Advancing into Water’s “No Man’s Land”: Two Liquid States?" D. Paschek, R. Ludwig, Angew. Chem., 2014, 126 (44), 11888-1890. Angew. Chem. Int. Ed., 2014, 53 (44), 11699–11701.


When SiO2 changes state, what forces/bonds are being overcome/broken? When silicon dioxide changes state, are the covalent bonds being broken (ie. the intramolecular forces) or is it the intermolecular forces?

$\ce{SiO_2}$ is a network covalent compound that has an extremely high melting and boiling point, because many silicon-oxygen bonds have to be broken in order for it to achieve the necessary freedom. To clarify, SiO2, which has a tetrahedra network lattice formation, shows that each silicon is actually bonded to 4 oxygens; each oxygen is bonded to 2 silicon. These excess bonds aside from the ones of SiO2 are broken.


How does the pH of fresh glacial meltwater compare to marine seawater? How does the pH of fresh glacial meltwater (at the time of melting, so it has not travelled downstream for example) compare to marine seawater? I assume that glacial ice is more neutral since it cannot uptake as much alkalis into its structure. Any insight would be helpful.

In general, the pH of seawater can range from 7.5 to 8.4 depending on where you are in the world, this is due to temperature differences, shipping activity, pollution, and several other factors. Reefs are quite sensitive to pH and are usually found in bodies of water with a pH ranging from 8.0 - 8.4. Glacial water should be nearly neutral as most glaciers were formed thousands of years ago when pollution levels were relatively low. It is quite possible that the glacier would actually be very slightly acidic due to the carbon dioxide that would have been trapped in the ice when it formed bubbling through the water creating a very weak carbonic acid. Scientists actually measure the levels of carbon dioxide in the ice to get an idea of what the Earths atmosphere was like thousands of years ago.

Other users please feel free to correct me if I am incorrect in my answer,  hopefully you can find this useful.


What is the use/importance of the equilibrium constant Kc I understand what the equilibrium constant is, however when reading in my textbook it has occurred to me that I do not understand what it's actual use/importance is. One specification in my syllabus is "Understand the importance of the numerical value of Kc".

Thank you for any help.

I do not understand why this question has been put on hold. How can I demonstrate that I have put in an effort into understanding this question when I literally cannot find the answer on the internet or in any of my four textbooks? I have familiarised myself with Kc, however nowhere can I find it's importance, hence the reason that I ask the question.

$K$ is the value of $Q$ at equilibrium. The expression for $Q$ and $K$ are the same however $Q$ describes the state of a system and $K$ describes the system at equilibrium. Le Chatelier's principle tells us that a system that has $Q\neq K$ will approach $K$. When $Q$ is not $K,$ the system is not at equilibrium.

K tells you where an equilibrium lays. For a given reaction, if $K$ is larger than 1 than the equilibrium is towards the right. If $K$ is less than one then the equilibrium is towards the left.
   For a simple equilibrium such as acid dissociation, the forward reaction's equilibrium constant is

$$K = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
so you can compare the strengths of acids by comparing the equilibrium constant. Since the $K$ of nitric acid is larger than the $K$ of hydrofluoric acid we know that it is a stronger acid.

Hope this helps!