How to decrease the percent of ammonia that is converted to the ammonium ion in water? Ammonia is a weak base that reacts with water according to this equation:

NH3(aq)  +   H2O(l)     ⇌     NH4+(aq)  +    OH −(aq)

Will Fe(NO3)3 decrease the percent of ammonia that is converted to the ammonium ion in water?

If ammonia gas is bubbled into pure water, the following reaction takes place:

$$\ce{NH3(aq)  +   H2O(l)   <=>     NH4+(aq)  +    OH−(aq)}\tag{1}$$

How to decrease the percent of ammonia that is converted to the ammonium ion in water?

Le Chatelier's Principle indicates that adding $\ce{OH-}$ would push the equilibrium to the left.

I lifted the following figure from:

  HYDROPHOBIC MEMBRANE TECHNOLOGY FOR AMMONIA EXTRACTION FROM WASTEWATERS by Airton Kunz and Saqib Mukhtar Eng. Agríc. vol.36 no.2 Jaboticabal Mar./Apr. 2016
  https://doi.org/10.1590/1809-4430-Eng.Agric.v36n2p377-386/2016

As can be seen in the figure, an increase in pH favors formation of $\ce{NH3}$ and a decrease in the pH favors $\ce{NH4+}$.

  Since the OP asked only for the percent of $\ce{NH4+}$ to be reduced, there is another way. Dilute the solution.

Will $\ce{Fe(NO3)3}$ decrease the percent of ammonia that is converted to the ammonium ion in water?

Iron (III) forms an relatively insoluble hydroxide as shown by the following equation:

$$\ce{Fe^{3+} + 3OH- <=>[K_{\mathrm{sp}} = 2.79\times10^{−39}] Fe(OH)3(s)}\tag{2}$$

So again according to Le Chatelier's Principle removing $\ce{OH-}$ from reaction (1) favors the reaction moving to the right. Thus the precipitation of $\ce{Fe(OH)3}$ will increase the formation of $\ce{NH4+}$.

  In a comment on the question user Poutnik indicated the reaction $\ce{Fe^{3+} + 3NH3 + 2H2O -> FeO(OH) + 3NH4+}$.  I'm sure there is a study somewhere on $\ce{Fe(OH)3 \pu{ vs } FeO(OH)}$ formation, but I didn't find it in 15 minutes of fooling around. I'd expect that strong $\ce{OH-}$ concentrations and higher temperatures (boiling solution) would favor $\ce{FeO(OH)}$.


How do I determine the equilibrium concentration given Kc and the concentrations of component gases? I feel my method for solving it is correct and I've checked my calculations yet I can't get the right answer.

Nitrogen and oxygen gases may react to form nitrogen monoxide. At 1500 °C, $K_c$ equals 1.0E-5.

$\ce{N2_{(g)} + O2_{(g)} <=> 2 NO_{(g)}}$

If $2.75 \times 10^{-2}$ mol $N_2$ and $2.75 \times 10^{-2}$ mol $O_2$ are sealed in a $0.886 L$ flask at $1500 °C$, what is the concentration of $\ce{NO_{(g)}}$ when equilibrium is established?

Notes: $K_c$ = equilibrium constant

a) 1.38E-7 M
b) 4.35E-5 M
c) 3.85E-5 M
d) 8.70E-5 M
e) 2.30E+4 M

I set up the problem with $(2x)^2$ as a numerator and $(.03104 - x)^2$ as the denominator

I took square root of both sides and did algebra and got $4.9*10^{-5}$

For the record the answer is $8.7 \times 10^{-5}$

For a reaction in the form $\ce{ \mathit{a}\,A + \mathit{b}\,B <=> \mathit{c}\,C}$, the quilibrium constant $K_c$ is given by

$$K_c = \frac{[C]^c}{[A]^a[B]^b}$$

You might want to check yours.


In the Wittig reaction, why do stabilised ylides produce E-alkenes and unstabilised ylides produce Z-alkenes? Wikipedia says that with unstabilised ylides they react very quickly in step 1-2 and step 4-5 is the rate limiting step. But with stabilised ylides step 1-2 is the rate limiting step and this is why stabilised ylides give E-alkenes. But there is no explanation as to why this makes E-alkenes and where in the mechanism it occurs.

I would have thought stabilised and unstabilised ylides would give E-alkenes to reduce steric hinderance but this is not the case. Any help explaining this would be very appreciated. :)

I have given some justification on why unstabilized ylids give Z-alkenes in this answer of mine.

A possible explanation why stabilized ylids show E-selectivity is given in the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on p. 817:


Reaction of 2,4,6-triaminobenzoic acid in a protic medium and heat The given answer is (A). At first I thought that it was a simple decarboxylation (due to high heat).
But that does not seem feasible at all as the ring has no electron withdrawing groups which stabilise the carbanion; infact it has electron donating groups which rules out decarboxylation. What else could be a possible mechanism here?
Edit:In the answer posted below could anyone explain how the hydrolysis of triamminobenzene occurs?

My initial guess for the product was Phloroglucinol, or 1,3,5-trihydroxybenzene1,2

Belskaya et al.3 says that 1,3,5-triaminobenzene can be formed, but it can eventually be hydrolyzed to Phloroglucinol

The sources mention decarboxylation as the mechanism. Note that it is the trihydroxylamine 10 that is getting decarboxylated here. The mechanism becomes clearer if you consider the imino tautomers 6 or 7 with the acid group attached and then dehydrate.
In short, Phloroglucinol should be the final product. If that is not one of the options, then 1,3,5-triaminobenzene is the answer you should select.
References:

“PHLOROGLUCINOL.” Organic Syntheses, vol. 9, 1929, p. 74. doi:10.15227/orgsyn.009.0074
Tewari, K.S and Vishnoi, N.K : A textbook of organic chemistry, 4th edition.
Olga B.Belskaya, Valentin P.Talsi, Roman M.Mironenko, Vladimir A.Rodionov, Sergey V.Sysolyatin, Vladimir A.Likholobov, "Transformation pathways of 2,4,6-trinitrobenzoic acid in the aqueous-phase hydrogenation over Pd/C catalyst", Journal of Molecular Catalysis. doi: 10.1016/j.molcata.2016.04.014


Spin coat of Graphene oxide on glass slides I would like to spin coat Graphene oxide on the glass slides and intend to have a thickness of 100nm/0.1micrometre. May I know what is the amount of Graphene oxide to be poured onto the glass slides, spin time, spin speed and time for it to develop to obtain the optimized and desired thickness of 100nm?

The procedure will vary with the size and shape of the glass, the concentration and viscosity of your graphene oxide dispersion, the solvent you're using, the wettability of your glass slide, and so on. You'll need to figure out the parameters yourself by varying the spinning conditions, measuring the resulting film thicknesses via profilometry/AFM/cross-sectional SEM, and then constructing spin curves (examples).


Separating NaClO from its aqueous form in bleach I was wondering, can I separate the $\ce{NaClO}$ from the $\ce{H2O}$ in bleach. The decomposition point for $\ce{NaClO}$ is $374.14\ \mathrm{K}$, so I'm risking decomposing the compound if I do simple distillation.

What should I do?

PS: I don't have a vacuum pump.

Sodium hypochlorite, $\ce{NaClO}$, is unstable and can't be separated from the water in which you usually find it without it decomposing.

That's the short answer.  To learn more about it check Wikipedia for sodium hypochlorite.


Why does freezing point decrease on adding impurities? It is a well known fact that freezing point decreases on adding impurities. I feel that it should increase on adding impurities.

Reason: I have watched this video on supercooled water. It says that when ice forms, it needs some nucleation points. If water is pure and left undisturbed, then there would be no nucleation points and water won't freeze even below $0^o$C.

So, my reasoning is that when impurities are added, they provide additional nucleation points where water crystals can gather. This would cause water to freeze faster.

But it doesn't happen. Why?

No, but good question. Freezing is a thermodynamic process. Nucleation is a kinetic process.

Kinetic Explanation: At the freezing point of water, water wants to become ice (thermodynamically driven). In order for freezing to initiate and ice to form a temperature dependent minimum number of water molecules must arrange them selves in a crystal or else the particle will dissolve and no freezing will occur.  Impurities or containers with rough surfaces can help arrange and hold molecules which allows freezing to occur with a fewer number of molecules.

Thermodynamic Explanation:  Freezing is a thermodynamic process dictated by $\Delta G_f = 0 = \Delta H_f -T\Delta S_f$.  When water freezes it must dispel the impurities to form an ice crystal (reverse mixing). Since mixing increases entropy, this means that for water to freeze the negative change in entropy ($\Delta S_f$) is larger than that of pure water.  This must be compensated by a lower temperature ($T$) neglecting and differences in ethalphy. The enthalpy ($\Delta H_f$) of mixing will contribute as well to the freezing temperature of water and must be accounted for.  In the case of Sodium chloride though the enthalpy mixing is positive which would favor a higher melting point, but the entropy contribution is large and magnified by the temperature such that entropy dominates and reduces the temperature of freezing.

If you look at the phase diagram of water and salt you will see that the mixing of the salt and water depresses the melting temperature due to the increased change in entropy from mixing water and salt to make brine from ice and salt.


Acetic acid reaction sequence Can anyone tell me where I am doing it wrong for the following reaction?

In the first step, Br should be attached to the alpha carbon.

In second step, the bromide is replaced by cyanide.

I am able to get the third product that is final compound.

Step 1: $\alpha$-bromination
$$\ce{CH3-COOH + Br2/P -> Br-CH2-COOH}$$

Step 2: Nucleophilic Substitution by Sodium Cyanide
$$\ce{Br-CH2-COOH + NaCN -> NaBr + NC-CH2-COOH}$$

Step 3: Acid-catalyzed Conversion of Nitrile to Amide or Acid

(1) First hydration... $$\ce{NC-CH2-COOH + H2O -> H2N-CO-CH2-COOH}$$
(2) Further heating... $$\ce{H2N-CO-CH2-COOH -> HOOC-CH2-COOH + NH4+}$$

Special Notes

Be VERY CAREFUL with Step 2 as you are potentially forming $\ce{HCN}$ which could get everyone in the lab killed if you're not careful.


Changing iron(III) acetate to iron(II) acetate Is there a way simple way to turn iron(III) acetate into iron(II) acetate or vice versa?

Edit (as per answer below): an oxidiser like hydrogen peroxide will turn iron(II) acetate into iron(III) acetate. A reducer (not sure what household product) will do the reverse.

Some background: my partner soaks scrap cast iron in white vinegar for several weeks and then uses the resulting liquid as a mordant for dyeing. According to wikipedia's page on iron acetate, soaking iron in vinegar creates a mix of iron(II) acetate and iron(III) acetate. As I understand it, the oxidation state of the iron used as a mordant effects the colour of the dye on the fabric. As such, it would be nice to be able to experiment with just iron(II) or just iron(III). Is this possible to achieve using household chemicals?

Even more background: The dyeing process is soak fabric in mordant water. Dry fabric. Wrap fabric with leaves and place in boiling water. Unwrap and admire beautiful prints :)

You need a reducing agent that will reduce iron(III) back to iron(II). Actually finely distrubuted iron could be the best reagent to do this, as it will not introduce any contaminants. The reaction taking place is 2 Fe(III) + Fe = 3 Fe(II).

Of course the iron used should not be rusty. I am not sure that you can easily get hold of fine iron powder, but I would give a try boiling the vinegary solution with fresh iron shavings or steel wool and as little contact with air as possible. The color change should be visible to greenish from the red-brown of the iron(III) acetate.

Actually, during dissolution of Fe in acetic acid you should get iron(II) acetate predominantly until the Fe runs out, but usually scrap iron is covered by iron(III) oxides/hydroxides, and these also dissolve in the vinegar. When the iron(II) solution is in contact with oxygen in air, it oxidises rather quickly. It is a good idea to start with a batch that was not left standing for weeks.

Other reducing agents work quicker than iron (e.g. tin(II) salts or zinc powder), but these introduce other products that would interfere with your dyeing.


What is the product of the reaction of 2-methyl-5-phenyl-2-pentene with sulfuric acid? The reaction is 2-methyl-5-phenyl-2-pentene with $\ce{H2SO4 }$

I have tried multiple attempts at this question but am still getting it wrong. I have no idea why any of these answers are wrong.

I know that the chain off the aromatic ring is weak aromatic activator and an ortho-para director. I figured para would be better than ortho because of the size of the chain but obviously both answers were incorrect. I then figured that $\ce{H2SO4}$ is just a strong acid so its basically $\ce{H3O+}$ but adding $\ce{H}$ and $\ce{OH}$ across the double bond in the chain did not work either.

With the help of user SendersReagent in comments, the answer to the question was a simple intramolecular Friedel–Crafts reaction:


How does caesium’s density change with temperature? According to Wikipedia caesium’s density is $1.90\ \mathrm{g/cm^3}$ at $\Theta =20\ \mathrm{^\circ C}$. How does this change when $T$ changes? E.g. will it expand when melting?

Looking at the wikipedia page you can see that the density is $\rho=1.93$ kg/l at room temperature and $\rho=1.843 kg/l$ at it's melting point.

Additional information can be found in this paper, where they experimentally explore the density of liquid Cesium in the vicinity of its melting point. In the range of 302 to 375 K to be exact.

From their experimental data they deduce the following approximate equation for the density as a function of the temperature:
$$\rho= 1829.12- 0.61483 \left(T-T_{melt}\right) $$
where $T_{melt}=301.6$ K.

This shows that the density reduces with temperature. Plotting the information that we have from the paper and from wikipedia yields:

The jump at the melting point is from the mismatch between the paper and wikipedia. In the paper they specify specifically that the density is for the liquid case, on wikipedia it might (I'm not sure) be the density for the solid case at the melting point. If that is the case then the jump in density from solid to liquid is: $1843-1829=14 $kg/m$^3$


What is true about this reaction involving hydroxy carboxlic acid? A sequence of reactions is shown below starting from an optically active compound P.
  $$\Large{\ce{\underset{\ce{C4H8O3}}{P}->[SOCl2]\underset{\ce{C4H6Cl2O}}{Q}->[C2H5NH2]\underset{\ce{C6H12ClNO}}{R}}}$$
  P does not react with 2,4-dinitrophenylhydrazine (2,4-DNP) to form a hydrazone. Q reacts very rapidly with one molar equivalent of ethylamine at low temperature to form R. Select the most appropriate statement(s).
  (A) Q as well as R are optically active.
  (B) Q on heating with an excess of ethylamine in the presence of a base gives a
  basic compound.
  (C) R is a basic compound and can react with an acid to form a salt.
  (D) An optically inactive isomer of P on heating can form compound S $(\ce{C4H6O2})$ which does not react with thionyl chloride.

My approach:

P is hydroxy butanoic acid. In Q the hydroxyl groups gets replaced by chlorine. I have confusion about the structure of R.

I feel Option (A) to be correct. I am doubtful about (D). And have no idea about (B) and (C).

I think your guess for P is correct, we just don't know whether it is 2- or 3-hydroxy-butanoic acid.  Reaction of thionyl chloride with the hydroxyl group in the alkyl chain can proceed with retention or inversion of configuration depending on reaction conditions, but the point is, if P is optically active, then Q will be optically active too (so A seems correct).  Reaction of Q, an acid chloride, with ethyl amine will produce the corresponding ethyl amide, R.  Since amides are basic (they can form salts with acids), I think B and C are true too.  An optically inactive isomer of P would be 4-hydroxy-butanoic acid.  On heating it would dehydrate and form a stable lactone (S) that would not react with thionyl chloride, so I think D is true also.  I just wonder if we're missing something that would allow us to determine if P is the 2- or 3- isomer?

Update:  To be complete and accurate let's update.  Based on the comments below (particularly JA's) I was right about "B" (but for the wrong reason, I thought the amide would be basic enough, but it is not; B is true because, as JA points out, heating with excess amine will convert P into an amino amide (I've labeled it as X) which is basic).  I was wrong about C, R is not basic enough to form a salt.  I still don't know if P is 2- or 3-hydroxy-butanoic acid, but I've added a picture based on the 3-hydroxy isomer to try and summarize the chemistry.


Chemical bonding - lattice energy Question: Why is ionic lattice energy inversely proportional to the radius of the atom?

Most heterogeneous covalent molecules are polar to some extent. The degree of polarity, or the dipole moment, depends on the difference in electronegativity difference between the two atoms. The larger the dipole moment, the higher the ionic character.

What I know: Electronegativity decreases as you go downward in a group, however, the size increases, usually, as you go downwards in a group. Thus, ionic character will increase upon going downward, but the ionic lattice energy will decrease? This seems contradictory. Is this true? And if it is, why is it so?

Considering your doubt, from Born-Lande equation:

  $$E = -\frac{N_\text{A}Mz^+z^- e^2 }{4 \pi \varepsilon_0 r_0}\left(1-\frac{1}{n}\right)$$

  $N_\text{A}$ = Avogadro constant;
      $M$ = Madelung constant, relating to the geometry of the crystal;
      $z^+$ = charge number of cation
      $z^-$ = charge number of anion
      $e$ = elementary charge, $1.6022\times10^{-19}\ \mathrm{C}$
      $ε_0$ = permittivity of free space
      $4πε_0 = 1.112\times10^{-10}\ \mathrm{C^2/(J\cdot m)}$
      $r_0$ = distance to closest ion
      $n$ = Born exponent, typically a number between 5 and 12, determined experimentally
      by measuring the compressibility of the solid, or derived theoretically.

And what you are assuming the radius is in fact the distance between the two atoms.
As the distance increases, the attraction decreases from Coulomb's law :

  $$F=\frac 1{4\pi\varepsilon_0r_0^2}Q_1Q_2$$

and thus there is less energy involved in breaking the lattice, so lattice energy decreases.

As you move down, the size increases and the valence electron in the outermost shell experiences lesser repulsion and thus it can be easily removed, therefore the elements at the bottom are highly electro-positive or conversely very less electro-negative. So when forming bonds, the elements at the bottom easily remove the electron(s) and gain more developed charged rather than the ones on the top who just release it just partially. This charge variation varies the dipole moment $\vec \mu=q\times\vec d$, by altering the partial charge $q$. Thus downwards as $q$ increases, $|\mu|$ also does increase and make the compounds formed by the element more ionic. Also the ionic/covalent character may be varied depending upon the bonding partner, in sense of, elements or other entities.

Also as charges $z^+$, $z^-$ increases, the magnitude of lattice energy does also increase as can be seen clearly from the Born-Lande equation.

$$\begin{array}{r|c|l}
&&\text{Ionic Solid}&\text{Lattice Energy (kJ/mol)}\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&r_0 \text{ i.e. distance}&\ce{NaCl}&786\\
&\text{between ions}&\ce{NaBr}&752\\
&&\ce{NaI}&702\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&z^+,z^- \text{ i.e. charge}&\ce{MgO}&3800\\
&\text{on ions}&\ce{Al2O3}&15900\\
\end{array}$$


How do you work out whether reduction would occur based on reduction potentials I'm having a bit of trouble with standard reduction potentials. Specifically, when you're given a table of half reactions and reduction potentials...is the potential of the reverse reaction (the oxidation potential) just the negative of the reduction potential? So, say you were given the two reactions :

$\ce{Cl2 + 2e- -> 2Cl- }$   $     E = 1.36$V

$\ce{F2 + 2e- -> 2F-  } $   $  E = 2.87$V

and the question asked whether F- would reduce Cl2 to Cl-, so

$\ce{2 F- +  Cl2 -> 2Cl- + F2 }$

...how would you work this out?

First, using observations, since the E value for fluorine's reduction is higher, fluorine more preferentially undergoes reduction compared to chlorine. Furthermore, if you know your periodic table, fluorine being higher in the group more readily gains a valence electron (with fewer electron shells and hence a stronger force of attraction) and would be reduced preferentially and oxidise chloride ions.

Now using the numbers

We know the following formula: $$E^o_{cell}=E^o_{cathode}-E^o_{anode}$$
And also that oxidation is at the anode and reduction at the cathode.

Now, if and only if $E^o_{cell}$ is positive would we have a spontaneous reaction.
All that is left is to consider the two cases and see which would work.

Case 1: $\ce{2 F- +  Cl2 -> 2Cl- + F2 }$

Reduction Equation (Chlorine): $\ce{Cl2 + 2e- -> 2Cl-  }$,  $E^o_{cathode} = 1.36\,V$

Oxidation Equation (Fluorine): $\ce{2F- -> F2 +2e-  }$,  $E^o_{anode} = 2.87\,V$

Then $$E^o_{cell}=1.36\,V-2.87\,V=-1.51\,V\lt0\,V$$

Hence this reaction would  not be spontaneous and not take place in this direction.

Case 2: $\ce{2 Cl- +  F2 -> 2F- + Cl2 }$

Reduction Equation (Fluorine): $\ce{F2 + 2e- -> 2Cl-  }$,  $E^o_{cathode} = 2.87\,V$

Oxidation Equation (Chlorine): $\ce{2Cl- -> Cl2 +2e-  }$,  $E^o_{anode} = 1.36\,V$

Then $$E^o_{cell}=2.87\,V-1.36\,V=1.51\,V\gt0\,V$$

Hence this reaction would be spontaneous and take place in this direction.


Is an isobar the same as an isotope? I am a little bit confused about what an isobar is. Its online definition is that it's an element with the same number of neutrons but a different number of protons from an element $\ce{X}$.

To me, it doesn't make sense from the get-go, because once you change the number of protons the element changes as well so why exactly is it defined as the same element $\ce{X}$ with the same number of neutrons and a different number of protons.
Definition of an isotope: An isotope is an element $\ce{X}$ with the same number of protons and a different number of neutrons.

So to the actual question now. Isn't an isobar just an isotope? Here is an example to clarify what I mean.
If we take for example carbon $\ce{^12C(p:6, n:6)}$ and turn it into an isotope it will be $\ce{^13C(p:6, n:7)}$, and that makes sense, but if we turn it into an isobar it would be $\ce{^13C(p:7, n:6)}$, which doesn't make sense, because it looks exactly like an isotope of nitrogen $\ce{^13N(p:7, n:6)}$.

If the atomic number changes than the element changes as well. So isn't an isobar just an isotope of the following element with a smaller neutron number?

Not quite, an isotope has same number of protons ($ A- N = Z = \mathrm{constant}$), but a different number of neutrons ($\mathrm N$ varies; e.g. $\ce{^3_\color{red}{1}H}$ and $\ce{^2_\color{red}{1}H}$, or $\ce{^235_\color{red}{92}U}$ and $\ce{^238_\color{red}{92}U}$ are isotopes).

An isobar has a fixed number of total nucleons ($Z + N = A = \mathrm{constant}$; e.g. $\ce{^\color{red}{40}_19K}$ and $\ce{^\color{red}{40}_20Ca}$, or $\ce{^\color{red}{3}_2He}$ and $\ce{^\color{red}{3}_1H}$ are isobars). Not nearly as mainstream as isotopes, but isobars are important to consider when doing mass spectroscopy.

Extra fact: For nuclei of the same number of neutrons ($A - Z = N = \mathrm{constant}$), the term is isotones.


Heating organic compound in a stream of nitrogen gas This is related to estimation of percentage of oxygen in an organic compound and my textbook states:

  A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas.

I would like to ask will all oxygen present in the compound evolve as molecular oxygen? Isn't there any possibility that carbon and oxygen of organic compound combine and form CO2 thereby producing error in measurement?

This is a strangely written text. In a practical setting, the way to get oxygen concentration is by sending an organic sample to a lab for an ultimate analysis.  The ultimate analysis chemistry is explained elsewhere.

The highlighted text seems to describe anoxic pyrolysis of organic matter well above 450°C (for coals this temperature may be higher).  Pyrolysis proceeds as follows.

$$\ce{Organics ->[heat]  syngas + tar + char  }$$

If any molecular oxygen were present in the system these two reactions would very quickly happen.
$$\ce{Organics + O_2 ->[heat]  2CO }$$
$$\ce{2H^+ + O_2 ->[heat]  H_2O }$$

The syngas contains elemental oxygen mostly in the form of $CO$ and $H_2O$, however, other gasses like $CO_2$, $H_2S$, and perhaps even other hydrocarbons may be present.  For the sygas to have most of the elemental oxygen as opposed to the char or tar phase high temperatures exceeding 450°C is important.

If the syngas containing the volatilized matter is passed over a bed of hot coal around 900°C then several things happen.

Complex particles break down (hence the lysis in pyrolysis.)
The gas stream picks up elemental carbon from the bed of hot coal.
$$\ce{CO_2 + C ->[heat]  2CO}$$
$$\ce{H_2O + C ->[heat]  H_2 + CO}$$

By this point, if the temperature is hot enough, all but perhaps an insignificant fraction of the elemental oxygen will convert into $CO$.

Testing for $CO$ is done with the iodine pentoxide test.


Material that doesn't melt at 800 deg F, but can be dissolved? I'm a sculptor and I have a chemistry/plastic question. I thought here would be a good place to ask.
I'm looking for a substance (probably a plastic) that has these characteristics:

Doesn't melt at the annealing temperature of Copper (doesn't melt at 700 degrees Fahrenheit)
Can be dissolved/made non-solid/melted by a process that will not affect copper (a specific acid?)
Is available in a wire form
Is relatively flexible

I want to wind long coils of tiny copper wire around a flexible core (the plastic I'm looking for), form that coil into complex shapes, then get rid of the core so as to be left only with the copper wire showing

I understand these requirements are perhaps impossible to meet...Just wondering if someone with good knowledge of plastics maybe has an idea.

Aluminium and Magnesium
react vigorously with mineral acids.
A lot of heat is evolved.
Try iron/steel wire, which is cheaper and
available in any diameter.
Georg


How exactly is energy created by oxidative phosphorylation I understand how the protons and electrons move around through various carriers, shuttles and proteins, but there's this talk of electrons being in "lower in energy" at the end. What does that mean? How is an electron lower in energy?

Also why do they need an electron acceptor? What would happen if there was no $\ce{O2}$ to accept the electrons? Wouldn't they just bind to something else?

How exactly is energy created by oxidative phosphorylation?

Energy is "created" by the syntesis of ATP. The ATP molecule is considered an energy carrier, because it is used in other reactions as a source of energy. The released energy comes from the loss of one or two phosphate gropus to form ADP or AMP.

Remember that energy is not created, but simply transferred. The oxidative phosphorylation is a mechanism to recreate the ATP using the energy from other sources (O2, nutrients, etc).

Also why do they need an electron acceptor? What would happen if there was no O2 to accept the electrons?

I'm not sure I can give you a precise answer, but I'll try. In chemistry reactions, the system is always working towards equilibria. In general, if you take away one reagent of a reaction, there won't be a reaction at all. If there's no oxygen available, ATP won't be formed, and as a consequence the body resorts to other sources of energy, or cells will die.

Wouldn't they just bind to something else?

I don't think so. The electrons aren't free particles moving around the cell, the transfer only happens if all the reagents for the reaction are complete. Enzymes are, generally speaking, very specific, and what could happen is a "saturation" of NADH's which wouldn't be converted back to NAD and hence the cycle would stop.

But there's this talk of electrons being in "lower in energy" at the end. What does that mean? How is an electron lower in energy?

I don't remember hearing or reading that. Can you post some context or source?


Types of organic chemicals? I am looking for a list of classes of synthetic organic chemicals. I'm not very strong in chemistry. The only information I could find are from here which seem to describe SOC's are like pesticides or fuel additives, but I don't think you can call those classes.

There are a lot of organic chemicals[citation needed].

Due to the incomprehensibly vast number of chemical species and variation between them, almost every classification system is domain-specific.

In terms of pure chemistry, the best you can do is usually to organize them according to what functional groups are in play; this will usually reveal something about the fundamental chemical interactions available to the species in question, but this is not necessarily useful or enlightening:

In many applications, such as as antifreeze, fuel or a solvent, methanol and ethanol are (largely) interchangeable: They both have short carbon chains attached to a single alcohol group (-OH), their boiling points, dipole moments, chemical energy and other properties are similar, and they mostly react in the same way on a chemical level.

In other applications, however, they are as  night and day: In biology, methanol is a potent poison, and ethanol is a mostly harmless intoxicant.

Organic chemistry is, in short, notoriously resilient to generally useful classification systems.


Why does diamond have no molecular formula? Why does diamond have no molecular formula, while buckminsterfullerene does have it, despite both being allotropes of the same element?

To quote my answer to Is a diamond a single molecule?:

Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).
Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc.

Another way to think about it, since diamond is a covalent network solid, each diamond is a single giant molecule. Since each diamond is one molecule, there is no universal diamond molecule with the exact same number of carbon atoms every time it occurs.
Fullerenes form distinct molecules with the same number of carbon atoms in each species every time. Thus buckminsterfullerene is always comprised of molecules of $\ce{C_{60}}$.


What is a kg.cal per gram, as a heat of combustion? I see heats of combustion given in kg [dot] calories per gram molecular weight (Handbook of Chemistry and Physics, 1969-1970). I don't know how that makes sense, as I don't know what a kg cal might be.  I would have expected units of energy per mass (calories per gram) or energy per mole.  What is a kg cal, and what is wrong in my thinking?

$\mathrm{kg~cal}$ probably represents the kilogram calorie (especially given that it is an old book). Nowadays it is usually just known as the Calorie, usually abbreviated $\mathrm{kcal}$. It is roughly the amount of energy required to raise the temperature of $1~\mathrm{kg}$ of water by $1~\mathrm{^\circ C}$. Since this is temperature and pressure dependent, there are slight variations in the actual definition depending on the context in which it is used.


Is it more reasonable to use stratification to separate similar compounds rather than using temperature? In a simple reaction between butane and chlorine, it is more likely for the secondary carbons of the butane to lose a hydrogen and gain a chlorine. The carbocation is a more stable intermediate in that situation compared to a primary carbon losing the hydrogen. So if the desired synthesis is 1-chlorobutane as opposed to 2-chlorobutane then it seems reasonable that a chemist might use stratification. The reason that I believe it is less desirable to use temperature as a method of separation is because that, at least in this example, both compounds will move between the gas and liquid phase randomly providing the possibility of still having mixed samples. Is this true, and will stratification work more reliably if purification is the desired result?

Furthermore, I never really learned why mono-chlorinated products form. Will there not also be compounds that have been chlorinated twice? Or will, over time due to entropy, all compounds be mono-chlorinated assuming that the reagents are in a ratio of 1 chlorine atom per molecule of butane?

The answer to your second question:

The halogenation of saturated hydrocarbons occurs by free radical substitution in the presence of UV light or heat and not by formation of carbocations. There's a very clear representation of the reaction mechanism on this Wikipedia page.

Since free radical intermediates are involved, this is a chain reaction and so what you said is correct, a large number of substituted products can be formed depending on the reaction conditions.

For example: If you consider the simple reaction between Methane and Chlorine in the presence of sunlight, Chloromethane, Dichloromethane, Chloroform, Carbon tetrachloride and a small amount of Ethane (by combination of Methyl radicals) are formed depending on the amount of chlorine available.

All these products will be formed even if the reactants are in a 1:1 ratio but they will be present in low concentration since there is less chlorine available.

In the case of higher alkanes like Butane and Pentane, the secondary carbons are substituted faster since the secondary free radical intermediates are more stable (reasons are the same as that for secondary carbocations).


How can the number of molecules and atoms be the same in a gas? A sample of certain mass of gas was collected and it was found that the number of atoms and the number of molecules in the collected sample are equal. You are required to state the conclusion drawn from this observation.

I am not sure of the answer. I think it is related to the quantity of a mole being constant.

The gas was one (or a mixture) of the noble gases, as every other gas is composed of more atoms than molecules.

Example

Helium: $1~\text{mol}~\ce{He}~\text{gas} = 1~\text{mol}~\ce{He}~\text{atoms}$
Nitrogen: $1~\text{mol}~\ce{N2}~\text{gas} = 2~\text{mol}~\ce{N}~\text{atoms}$


Evaporate propylene glycol I tried to make a cannabis infused e-liquid. I made a tincture with marijuana and vegetable glycerin, but it became to thick so a mixed in more propylene glycol than I'd originally planned. Now the e-liquid is to weak, not enough THC per milliliter.

My question is: can I just heat up the e-liquid to 115-123 degrees celsius and let the glycol evaporate over a long time, or is that temperature too low? I don't want to destroy the psychoactive chemicals in the e-liquid.

Best wishes.

Let me reword the question:

Does $\Delta^9$-tetrahydrocannabinol (THC) significantly degrade when its solution in propylene glycol is heated to 115-123 °C?

This was somewhat answered in a study by Michael A. Repka et al. (DOI) on the stability of THC embedded in a  hydroxypropylcellulose polymer matrix. When heating this system to 120 °C for 75-90 min, about 2.7% of the initial THC degraded. With other words, 97.20% remained unchanged.

Heating to higher temperatures did not lead to significantly higher degradation, but among the degradation products, the amount of cannabinol (CBN), the oxidation product of THC, was much higher here.

Technically, heating to ~120 °C seems to be no problem.

You might want to check the reference given, have a look at the articles cited, compare toxicological data of THC and CBN and decide for yourself.


Writing the net reaction for reaction of sodium hypochlorite with hydrochloric acid and potassium bromide I am currently taking a general chemistry course and during a lab they had us do the following procedure:

Place $\pu{2 mL}$ of $5\%$ $ \ce{NaClO} $ solution in a test tube.
Add $10$ drops of cyclohexane.
Add ~$10$ drops $\pu{6 M}\,\ce{HCl}$
In another test tube, add a pinch of solid $\ce{KBr}$
Use a dropping pipette to withdraw the chlorine/cyclohexane layer from above and add it to the test tube containing $\ce{KBr}$

When I did this, the mixture turned dark red and bubbled with a hint of purple to it. In my lab report, my teacher wants me to write the net ionic equation(s) of this reaction, but I have no idea how to do that.
How would you proceed?

I'd like to add some things to what Tanith Rosenbaun said.
First, the point of net ionic equations is not just to write out the equations in terms of ions, but to focus on only those ions that experience a change in charge or valence.  Ions which don't change in charge don't really participate in the reaction and called spectator ions.
Here's the last equation from Rosenbaum's answer:
$$\ce{2H3O+ + 2Cl- + 2Na+ + CO3^2- -> 2H2O + 2Cl- + 2Na+ + CO2}$$
Notice that the sodium and chloride ions are unchanged in this reaction.  So we drop them, and just write out the ions that change:
$$\ce{2H3O+ + CO3^2- -> 2H2O + CO2}$$
This is the net ionic equation: it focuses precisely on what is interesting, which in this case is the neutralization reaction.
Now let's go back to your problem:  People have been asking what you know about it.  You seem to know that you are transferring a solution of chlorine dissolved in cyclohexane.  You got that chlorine from the following reaction:
$$\ce{2HCl + NaClO -> Cl2 + NaCl + H2O}$$
and the chlorine preferentially dissolves in the cyclohexane.  Now it's important to understand that the cyclohexane doesn't play any role in the second part of your experiment, it's just how you transport the chlorine.  (And it's why you need to be careful with acids around bleach: chlorine gas is poisonous!)
The key question is: what happens when you put elemental chlorine in the presence of dissolved potassium bromide?
$\ce{Cl2 + KBr -> ?}$  (Note: this is not balanced yet!)
In particular, do you know what you produced when that solution turned dark red?  Hint: this is a very important property that halogens like chlorine and bromine have towards each other, which is indeed seen best via a net ionic equation.  I'll happily provide the answer if you've tried some stuff and are still stuck.


why is NaClO 'liquid chlorine'? Sodium hypochlorite. $\ce{NaOCl}$.
Appears to be: one part sodium, one part chlorine, one part oxygen.
Five questions:

Couldn't it just as easily be 'liquid sodium' or 'liquid oxygen'?
It's not really any of these though, is it?
It's a compound of all 3, right?
So, doesn't the new molecule have it's own properties?
Or is a compound/molecule(s) properties simply derived from it's constituent elements?

Short answer:
No. No. Yes. Yes. No.
Next time, try searching before posting.
Longer answer:
It is just a sale trick. Chemistry community has no direct influence on used trademarks of commercial products. Some may call whiskey "liquid gold", but nobody expects there is any gold there.
The product is called "liquid chlorine" just because it shares some of characteristics of chlorine dissolved in water/water based solutions. As in both cases , there are some equilibrium concentrations of hypochlorous acid, free chlorine and atomic oxygen.
The "liquid chlorine" is basically a solution of sodium hypochlorite, sodium chloride and sodium hydroxide. There is the answer directly on the photo: "130 g/L available chlorine(Cl) present as sodium hypochlorite". I.e. chlorine is not present in its free form, but in chemically alternative form of hypochlorite, that can release chlorine at different than storing conditions.
Sodium hypochlorite is produced by dissolution of chlorine in solution of sodium hydroxide:
$\ce{Cl2(g) + 2 NaOH(aq) -> H2O + NaCl(aq) + NaClO(aq)}\tag{1}$
If acidified, e.g. by sulphuric acid, the reaction (1) is reversed:
$\ce{H2SO4(aq) + NaCl(aq) + NaClO(aq) -> Na2SO4(aq) + Cl2(g) + H2O}\tag{2}$
If dissolved by about neutral solutions, reaction (2) happens in lower extent, as there occurs also
$$\ce{ClO-(aq) + H2O <=> OH-(aq) + HClO(aq)\tag{3a}}$$
$$\ce{HClO(aq) -> H+(aq) + Cl-(aq) + O(aq)\tag{3b}}$$
The atomic oxygen is reportedly the true desinfecting agens, rather then chlorine itself.


Hydrogen molecule potential energy graph In the above graph, I was confused at the point where the internuclear distance increases and potential energy become zero. Though internuclear distance is very small and potential energy has increased to zero.

Another question that though the internuclear distance at a particular point is constant yet potential energy keeps on increasing.

I think the point where potential energy become zero we can say that attractive forces between electron and nucleus of bonding hydrogen atoms just equals the repulsive force between bonding hydrogen electrons. It's just what I am assuming not sure either its true or not.

Normally, when the atoms are separated by a huge amount (infinity) we take this energy to be zero. Bringing the atoms closer to one another has an attractive potential and the bond has formed on reaching the lowest point on your curve.  Moving the atoms closer together increases the potential energy because here we push electrons closer to one another and this energy increases rapidly  with each small decrease in separation due to repulsion of similar charges. The slope of the curve gives the force on the atoms, so at small separation they experience a large force and repel one another. At large separation, where the slope is positive, there is a smaller attractive force.


Second order reaction with two reactants I'm interested in the second order reaction $\ce{A + B -> P}$. Mass action kinetics is assumed.

Let $[A]_0 = a$ and $[B]_0 = b$, then $[A] = a - x$ and $[B] = b - x$. The rate law becomes
$$-\frac{\mathrm dx}{\mathrm dt} = -k([A]_0 - x)([B]_0 - x)$$

Why is the right-hand side of the differential equation negative? Shouldn't it just be the left side when we look at the decrease of the reactant $\ce{A}$? Or is it because $x$ is equal to the concentration of $\ce{P}$?

Assuming mass action kinetics, we have the following differential equation

$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = -\kappa \, [\ce{A}] [\ce{B}]$$

where $\kappa > 0$ is the rate constant. If we denote

$$x (t) := [\ce{A}]_0 - [\ce{A}] = [\ce{B}]_0 - [\ce{B}]$$

then

$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = \frac{\mathrm d}{\mathrm d t} \left( [\ce{A}]_0 - x \right) = 0 - \dot x = - \dot x = -\kappa \left( [\ce{A}]_0 - x \right) \left( [\ce{B}]_0 - x \right)$$


Is it possible to prevent saponification by modifying free fatty acids? Not big on chemistry so this question might be a bit out there. It is my understanding that a soap molecule is made up of a free fatty acid and lets say a sodium ion. Correct? Would it be possible to modify the free fatty acid (with a solid or liquid substance) in such a way that when it comes in contact with the sodium ion it wouldn't be able to form a soap molecule?

Thanks

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

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


How were silver and gold separated using the salt cementation process? Gold and silver are often extracted from the same ores and are difficult to separate due to their chemical similarity, which prevents the use of techniques such as cupellation. Acid-based methods were discovered during the medieval period but were too expensive for use, so the main method of gold parting (the separation of gold from other metals) until the C16th was by salt cementation, in which sheets or granules of low-purity gold were heated with salt (NaCl) and/or other other compounds.

I would like to understand how this process was actually used. Wikipedia says the silver chloride (AgCl) produced is 'removed' but does not describe how. It also mentions that AgCl is volatile and the vessel used for the process is sealed to prevent its escape, which seems to contradict the previous statement. At the same time, it mentions the process is kept below 1000degC to stop the gold from melting, which is a long way below the boiling point of AgCl at atmopsheric pressure (although I don't know what pressure would be reached inside the vessel or how that would affect the volatility of AgCl). Furthermore, I thought that upon heating AgCl would normally decompose to silver (and chlorine) anyway, which in a simple, sealed vessel would presumably just remix with the gold. Some sort of alembic arrangement, maybe? I'm pretty sure I've misunderstood something fairly basic here.

I'm then further confused by this 1974 paper by Notton (open access) which seems to describe a recreation of what sounds like salt cementation process but then compares it to cementation (meaning it must differ significantly in some way I'm missing), and also sheds no light on the mechanism for recovering silver, although notes that the alumina plug of the sealed container ends up discoloured by silver salts after heating. Were the silver salts absorbed by the walls of the container, as in lead/silver cupellation?

I am interested in historical (pre-C16th) application of this process, rather than modern solutions.

This is an old process

M. Berthelot gives it from Papyrus V from the Leyden Papyrus X, in his Introduction à l'étude de le Chimie, des anciens et du moyen âge (Introduction to the study of Chemistry, of forefathers and the middle age).
The papyrus is at least from the III° century, the process very probably older. From page 14 (my translation):

  "Take piquant vinegar, thicken, take some [lacuna], 8 drachmas of common salt, 2 drachmas of lamellar alum (schist), 4 drachmas of litharge, crush with vinegar for 3 days, separate by decantation and employ. Then add to the vinegar 1 drachma of copperas, half an obole(1) of [lacuna], 3 oboles of chalcite(2), one and a half oboles of sory(3), one silique(4) of common salt, two siliques of Cappadoce salt(5). Make a lamella having two quarters (of an obole?) Submit it to the acion of fire... until the lamella breaks, then take the pieces and view them as refined gold.

  1 drachma = 6 oboles
  copper ore, such as pyrite
  product of alteration of pyrite, which can include both copper sulfate and basic iron sulfate.
  silique = third of an obole, a measure of weight
  a variety of rock salt"

What about the chemistry?

Berthelot cites the process of départ par cémentation (departure by cemetation) from Macquer's Dictionnaire de la Chimie (1778). From the latter, page 312 (still my translation):

  "First prepare a cement composed of 4 parts crushed bricks passed on a sieve, one part green vitriol calcined to red, and one part common salt: mixe the whole very-exactly, and make a firm paste, moistening it with a little water or urine. This cement is called cément royal, because it is used to purify gold, which chemists view as the king of metals.

  On the other hand, reduce the gold one wishes to prepare, to lamellas about as thin as billon pieces: at the bottom of a crucible or cementation pot, place a layer of the cement of thickness the width of a finger: stratify the gold lamellas on this layer: place on top a new layer of cement: thus fill the pot, always placing the gold between two layers of cement; and cover it with a lid sealed with sand and clay. Place the pot in a furnace or oven ; heat progressively, until the pot is a dull red ; maintain this level of heat for about twenty four hours: it is very-essential that the heat not be able to melt the gold. After that, let the pot cool down, and open it to remove the gold, which must be separated thouroughly from the surrounding cement: it also has to be boiled in large quantities pure water several times. One can try this gold on a touchstone or otherwise ; and if it is found insufficiently pure, submit it a second time to the process.

  The vitriolic acid of the brick and of the calcined vitriol, releases the acid from the common salt during this cementation ; and this last one dissolves the silver alloyed to the gold, and seperates it by this mean.

Berthelot has this to say about the chemistry occuring here, (page 16):

  By proceeding thus, the silver and other metals dissolve in the sodium chloride, with the help of the oxidative -- and then chlorinating -- action exerted by the iron oxide derived from the vitriol ; while the gold remains unchanged.

And elsewhere (page 15):

  [...] At a stretch, it could be that sodium chloride, in the presence of basic salts of [ferric oxide] [...], releases chlorine [...].

So what I think happens is: the silver reacts with $\ce{Cl2}$ and maybe $\ce{FeCl2}$ (speculation?, b.p.= 1023°C, vapor pressure at 800°C = 0.1bar) to give $\ce{AgCl}$. Then the $\ce{AgCl}$ evaporates (b.p.= 1547°C, vapor pressure at 800°C = 0.1mbar), condensates on the cement, and dissolves in it. It is unclear whether the silver stays in oxidized form. I would bet yes, but I am no inorganic chemist.

Recovering the silver

Macquer also says (page 313):

  We can see it is essential to wash thouroughly the gold, after the process is completed, to remove all the disolved silver parts, which would otherwise stay mixed with it.

  Concerning this silver, it can subsequently be separated from the cement, by heating it with a sufficient quantity of lead and litharge, and taking the resulting silver containing lead slag and cupelating it to remove the lead."

So apparently you can redissolve the silver in lead, and then remove the lead by normal cupellation.

Food for (even more) thought

Macquer (page 313) also mentions that the process works just as well if you replace common salt by saltpeter. Which is weird because $\ce{AgNO3}$ decomposes at 440°C.

He then adds that you can add both common salt and saltpeter, and not only does the silver still get dissolved, the gold doesn't. Which is also weird because you have all the components of aqua regia, but it doesn't attack the gold. Macquer does acknowledge his low confidence in his sources for this last process.


Why do sodium halides react so differently with sulfuric acid? Why do sodium halides react so differently with sulfuric acid?
\begin{align}
  \ce{NaF + H2SO4 &-> NaHSO4 + HF}
    \tag{1a}\label{NaF}\\
  \ce{NaCl + H2SO4 &-> NaHSO4 + HCl}
    \tag{1b}\label{NaCl}\\
  \ce{2 NaBr + H2SO4 + 2H+ &-> Br2 + SO2 + 2H2O + 2Na+}
    \tag2\label{NaBr}\\
  \ce{8 NaI + H2SO4 + 8H+ &-> 4 I2 + H2S + 4 H2O + 8Na+}
    \tag3\label{NaI}
\end{align}

Conventional explanation: $\ce{NaI}$ is a strong enough reducing agent to reduce the sulfur, and $\ce{NaBr}$ is a little less strong, so sulfur is not reduced as much. Equivalently, $\ce{H2SO4}$ is not as strong an oxidising agent to reduce $\ce{NaF}$ and $\ce{NaCl}$.

My ensuing question: Why is reaction \eqref{NaI} more favourable than \eqref{NaBr} and $(\ref{NaF},\ref{NaCl})$ for sodium iodide, so that it reacts with sulfuric acid like it does. What factors are involved: what is more stable, for instance, (or there is lower Gibbs free energy or something else) about reducing sulfur as much as possible (as $\ce{NaI}$ 'chooses' \eqref{NaI} over $(\ref{NaF},\ref{NaCl},\ref{NaBr})$?

Ditto for sodium bromide. I understand why \eqref{NaI} is impossible, but why is \eqref{NaBr} chosen over $(\ref{NaF},\ref{NaCl})$?

A factor you didn't touch upon which may have some weight in determining the reactions is the acidities of the hydrogen halides formed by reactions of the first type (which is fundamentally a coarse qualitative analysis of the free energy of reaction). In aqueous solutions, the order of acidity is $\ce{HF} \ll \ce{HCl} < \ce{HBr} < \ce{HI}$. While concentrated sulfuric acid is obviously not the same as water, perhaps it is fair to say the relative acidities of the hydrogen halides are approximately the same, given they are both polar protic solvents with very high dielectric constants.

As you go down the family, the acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway. $\ce{HI}$ is somewhat close in acidity to $\ce{H2SO4}$ since neither conjugate base is very coordinating. Meanwhile, as you say, the reducing power of the halides increases, and since sulfur is in a high oxidation number in $\ce{H2SO4}$, redox reaction pathways become more attractive. A detailed analysis of free energies of reactions should agree with the change in the mechanism since it is unlikely these reactions are kinetically controlled.

Edit: Another neat consequence of the acidity strength argument is that it explains why the reactions stop at the bisulfate anion instead of going all the way to the halide sulfates; $\ce{HSO4-}$ is a much, much weaker acid than $\ce{H2SO4}$, so there is no free energy to lose out of consuming it to produce a much stronger acid (in the case of $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ at least). $\ce{HF}$ is similar in acidity to $\ce{HSO4-}$ (in water), so it's not as readily explainable in such a qualitative analysis.


How do you figure out fuel specific impulse non-experimentally? Simple question, probably very complicated answer:

Without building a test engine/rocket and finding out experimentally, that is to say, based on chemical characteristics alone, how can one compute the expected specific impulse of a given fuel?

The relevant property is the temperature of the burnt fuel when it leaves the rocket engine nozzle, i.e. when it stops exchanging impulse with the engine.

That temperature you can convert into a velocity distribution of the gas particles, multiply with their molar mass, and that's it. (somewhat less than that of course, because no all gas particles are flying in the exact same direction. depends on the nozzle type, among other influences)

The temperature you can estimate to some extent from the reaction enthalpy, but of course you have radiative and conductive losses. Which is where this becomes an engineering problem.


Why do steel and iron not passivate the way aluminum does? When aluminum is exposed to oxygen a micro layer or film of oxide appears on the surface and oxygen cannot penetrate any further.

With iron or steel, however, oxygen continues to reach into the metal, rusting it deeper over time. When iron is heated the oxide ("scale") forms much faster and the oxygen can deeply penetrate into the metal.

What is the reason that aluminum and iron are different in this way?

The main reason is nature of the oxide layer formed on the surface. Iron oxides are not remotely like aluminium oxide. They tend to be flaky and porous meaning that they offer no protection to the surface when formed. And iron oxides tend to be lower density and higher volume than the iron they are made from. Aluminium oxide is not at all porous and is very strong.

Bulk aluminium oxide is corundum, which is one of the strongest minerals (it is just about the hardest other than diamond). Corundum occurs as the gemstones sapphire and ruby when it has the right impurities and they are known for their hardness as well as their beauty. It is also unusually dense for a mineral made up of relatively light elements. So even a thin layer provides strong protection.

It is also worth noting that the aluminium oxide layer can be disrupted (e.g. by gallium or mercury), in which case aluminium does the equivalent of "rust", in some cases quite spectacularly.

It is also worth noting that some steels can be made to have a passivation layer which is why the are "stainless". This is achieved by alloying with metals such as chromium that facilitate the creation of a stronger oxide layer on the surface.


How does one actually get the energy from the Kohn–Sham equations? This might be a silly question, but how does one acquire the energy of the system from the Kohn–Sham equations?

$\left[ -\frac{1}{2}\nabla^2 + V_{eN}(\vec{r}) + V_{ee}(\vec{r}) + V_{xc}(\vec{r}) \right]\phi_i(\vec{r})=\epsilon_i\phi_i(\vec{r})$

Where $\nabla$ is the Laplacian operator, $V_{eN}(\vec{r})$ is the electron-nuclei attraction, $V_{ee}(\vec{r})$ is the electron-electron repulsion, $V_{xc}(\vec{r})$ is the exchange-correlation potential, $\phi_i(\vec{r})$ are the one-electron wave functions and $\epsilon_i$ are the orbital energies.

Assuming that we run through this set of equations for each electron, we get a set of orbital energies: $\epsilon_i \cdots \epsilon_N$. How is the overall total energy of the system then calculated?

A common mistake is the thought that the total energy is the sum of all orbital energies $\{\epsilon_i\}$.

From Step #6 of Daniel Crawford's SCF programming project (modified slightly in some places):

  The SCF electronic energy may be computed using the density matrix as:

  $$
E_{\text{elec}} = \sum_{\mu\nu}^{\text{AO}} D_{\mu\nu} (H_{\mu\nu}^{\text{core}} + F_{\mu\nu})
$$

  The total energy is the sum of the electronic energy and the nuclear repulsion energy:

  $$
E_{\text{total}} = E_{\text{elec}} + E_{\text{nuc}},
$$

where the density matrix is defined as (Step #8)

  $$
D_{\mu\nu} = \sum_{m}^{\text{occ. MO}} C_{\mu m} C_{\nu m},
$$

the Fock matrix as (Step #7)

  $$
\begin{align}
F_{\mu\nu} &= H_{\mu\nu}^{\text{core}} + \sum_{\lambda\sigma}^{\text{AO}} D_{\lambda\sigma} \left[ 2(\mu\nu|\lambda\sigma) - (\mu\lambda|\nu\sigma) \right] \\
&= H_{\mu\nu}^{\text{core}} + 2 J_{\mu\nu} - K_{\mu\nu}
,
\end{align}
$$

and the core Hamiltonian as (Step #2)

  $$
H_{\mu\nu}^{\text{core}} = T_{\mu\nu} + V_{\mu\nu}.
$$

I've also introduced the definitions of the Coulomb matrix $J$ and the exchange matrix $K$:

$$
\begin{align}
J_{\mu\nu} &= \sum_{\lambda\sigma}^{\text{AO}} D_{\lambda\sigma} (\mu\nu|\lambda\sigma) \\
K_{\mu\nu} &= \sum_{\lambda\sigma}^{\text{AO}} D_{\lambda\sigma} (\mu\lambda|\nu\sigma) \\
\end{align}
$$

Now, identify each of the terms in the Kohn-Sham equations with the terms from above.

$$
\begin{align}
\hat{T}_{e} &= -\frac{1}{2} \nabla^2 \rightarrow T_{\mu\nu} = \left< \chi_{\mu} \left| \hat{T} \right| \chi_{\nu} \right> \\
\hat{V}_{eN}(\vec{r}) &= \sum_{A}^{\text{nuclei}} \frac{Z_A}{|\vec{r} - \vec{R}_{A}|} \rightarrow V_{\mu\nu} = \left< \chi_{\mu} \left| \hat{V}_{eN} \right| \chi_{\nu} \right> \\
\hat{V}_{ee}(\vec{r}) &\stackrel{?}{\rightarrow} 2 \hat{J} \\
\hat{V}_{\text{XC}}(\vec{r}) &\stackrel{?}{\rightarrow} - \hat{K}
\end{align}
$$

This last part isn't quite correct though. Usually, when looking at the Kohn-Sham equations, one replaces the full electron-electron interaction $\hat{V}_{ee}$ with the sum of the Hartree potential $\hat{V}_{H}$, which gives the Coulomb energy, and the exchange-correlation potential $\hat{V}_{\text{XC}}$, which replaces the exact exchange $\hat{K}$ with a (currently approximate) expression for both the exchange term and the true electron-electron (correlated) interaction.

In terms of how the energy is actually calculated, all quantities from above are the same as in Hartree-Fock theory, except the calculation of the exact exchange integrals during the Fock build is replaced with calculating the exchange-correlation matrix $F^{\text{XC}}$, leading to

  $$
\begin{align}
F_{\mu\nu}^{\alpha} &= H_{\mu\nu}^{\text{core}} + J_{\mu\nu} + F_{\mu\nu}^{\text{XC}\alpha} \\
F_{\mu\nu}^{\beta} &= H_{\mu\nu}^{\text{core}} + J_{\mu\nu} + F_{\mu\nu}^{\text{XC}\beta}
\end{align}
$$

For a density functional approximation (DFA) based on the generalized gradient approximation (GGA), where the functional is dependent on both the density $\rho(\mathbf{r})$ and its gradient $\nabla \rho(\mathbf{r})$,

  $$
\begin{align}
E_{\text{XC}} &= \int f_{GGA}^{\text{DFA}}(\rho_{\alpha},\rho_{\beta},\gamma_{\alpha\alpha},\gamma_{\alpha\beta},\gamma_{\beta\beta}) \, \mathrm{d}\mathbf{r} \\
\gamma_{\alpha\alpha} &= |\nabla \rho_{\alpha}|^{2} \\
\gamma_{\beta\beta} &= |\nabla \rho_{\beta}|^{2} \\
\gamma_{\alpha\beta} &= \nabla \rho_{\alpha} \cdot \nabla \rho_{\beta} \\
\end{align}
$$

  The exchange-correlation parts of the Fock matrices are given by

  $$
F_{\mu\nu}^{\text{XC}\alpha} = \int \left[ \frac{\partial f}{\partial \rho_{\alpha}} \chi_{\mu}\chi_{\nu} + \left( 2\frac{\partial f}{\partial \gamma_{\alpha\alpha}} \nabla\rho_{\alpha} + \frac{\partial f}{\partial \gamma_{\alpha\beta}} \nabla\rho_{\beta} \right) \cdot \nabla(\chi_{\mu}\chi_{\nu}) \right] \mathrm{d}\mathbf{r}
$$

$f^{\text{DFA}}$, $\frac{\partial f^{\text{DFA}}}{\partial \rho}$, and $\frac{f^{\text{DFA}}}{\partial \gamma}$ are unique closed-form expressions for each DFA, and are usually evaluated numerically on an atom-centered grid (ACG) such as a Lebedev grid. This generally requires mapping the set of AOs/basis functions $\{\chi\}$ onto this grid.

References

John A. Pople; Peter M. W. Gill; Benny G. Johnson; Kohn–Sham density-functional theory within a finite basis set. Chem. Phys. Lett. 1992, 199 557-560

$\tiny{\text{As usual, sorry if I'm lazy with notation, being consistent is so difficult...}}$


Why not use all the colours in colorimetry? Colorimetry (and it's cousin visible-UV spectroscopy) is a form of chemical analysis that works by sending visible light of one wavelength through the sample. It then analyses how much of the light was absorbed in that process so you can determine concentration.

What I don't understand is why only one wavelength is used. Apparently, even for Infrared Spectroscopy, only one wavelength is analysed at one time yet the whole spectrum is needed. Why not send in white light (with all the wavelengths at once)? The output light could be split through dispersion and every wavelength could be analysed for its absorbance. Perhaps some wavelengths of light aren't worth measuring or too many frequencies at once cause interference?

Why not send in white light (with all the wavelengths at once)?

This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.

Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!

In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.

Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.


Chiral products from achiral reactants and 'principle of racemic mixture formation' I have studied that if in a reaction system chiral reactants are absent but chiral products are produced; then all enantiomers of respective diastereomers are produced in equal concentration or a racemic mixture is produced.

But what is the reason behind this 'principle of racemic mixture formation'? Is it simply because as the enantiomers of respective diastereomers have no difference in physical or chemical properties, the reaction system shows no preference to any of them?

Because the starting material is achiral and the products are enantiomers, the reaction, by symmetry, must proceed via two enantiomeric transition states. Enantiomers have the same energy, so these two transition states have the same energy, so the rate of product formation is the same for both products. Therefore, they will be produced in equal amounts.

However, there is a complication that was not covered in your statement. There are examples of autocatalytic reactions like the Soai reaction where the product aids in catalysis and a tiny random preference for one enantiomer over the other leads to amplification of that preference. This is akin to how life uses preferentially one enantiomer instead of another, since presumably, we started with achiral materials in the primordial soup.


What is the difference between a coefficient and a subscript in a chemical equation? Observe the following parts of an electron transfer equation:

\begin{align}
\ce{2Li + S &-> Li2S}\tag{1}\\
\ce{2Al + 3Br2 &-> 2AlBr3}\tag{2}
\end{align}

The first equation says $\ce{Li}$ with a small subscript 2, however the second one says Al with a coefficient 2 instead of $\ce{Al2}$.

Can someone please explain this discrepancy and inconsistency I'm seeing? Why is lithium in the first reaction written with a small $2$, while in the second one aluminium is written with a big $2$? Why can't both of them be written in the same way? Does it matter? Is there a difference? How do I tell which one to use?

Why can't both of them be written in the same way? Does it matter? Is there a difference?

Yes, there is a difference, and it does matter! The placement of subscripts and coefficients tells you how the substances are put together on the atomic level.

Let's look at the first equation:

$$
\ce{2Li + S -> Li2S}
$$

This one indicates that two moles of lithium combine with one mole of sulfur to form one mole of lithium sulfide. $\ce{Li2S}$ is called a formula unit or empirical formula, and tells you the smallest whole-number ratio of atoms that make up a "building block" of the compound. In this case, the smallest building block of a lithium sulfide crystal would contain two lithium ions for every one sulfide ion. Here's a picture from wikipedia to demonstrate:

From this, we can learn that:

  Coefficients tell us the number of moles, molecules, or formula units of a species

and

  Subscripts tell us how many atoms or ions there are within a compound

Now if we look at the second equation:

$$
\ce{2Al + 3Br2 -> 2AlBr3}
$$

we can see that in this case, it take 2 moles of aluminum to combine with three moles of bromine to produce two moles of aluminum tribromide. If it were instead $\ce{Al2Br6}$, that would indicate a different formula unit - it would imply a different compound.

So to answer your question:

  Why is Lithium in the first reaction written with a small 2, while in the second one Aluminium is written with a big 2

This is because in the first equation, the product has two lithium ions for every one sulfur in the formula unit, while in the second, two moles of aluminum tribromide are produced for every two moles of aluminum, but the formula unit contains only one aluminum atom and three bromine atoms.

For your final question:

  How do I tell which one to use?

You would have to know what the formula of the product is. In these examples, you can predict it by assuming that they are ionic compounds and balancing charges. You would get the correct answer, even though this assumption is not true for the second equation (it is actually a covalent compound, and you would just get the right answer by luck).


How to determine the empirical formula for an unknown iron oxide? I have to determine the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ dioxygen by mass in it.

I have started as follows:

For our convenience we take $\pu{100 g}$ of that iron oxide:
We have $\pu{69.9 g}$ of iron in it and $\pu{30.1 g}$ dioxygen in it.
We convert into amount of substance of each element:

$\text{amount of substance of iron} = 69.9/55.8 = 1.25$
$\text{amount of substance of dioxygen} = 30.1/32 = 0.94$

Is my method and calculations right?

First assume the mass of an oxide of iron to be $\pu{100 g}$ which implies mass of iron would be $\pu{69.9 g}$ and mass of oxygen would be $\pu{30.1 g}$. Now calculate amount of substance of both the elements by dividing the assumed mass by their atomic number.
Then calculate relative amount of substance of each element by dividing the obtained amount of substance of each element by the smallest of them. (E.g. suppose you've got $\pu{5 mol}$ of iron and $\pu{2 mol}$ of oxygen, divide 5 by 2 and 2 by 2 to get relative amounts of substance of both.)
Now just write relative amount of substances in the form of a ratio and bring the obtained ratio to least simplified form. (Remember the ratio should come in whole numbers.)
After obtaining ratio, write the empirical formula as I did in this example.
(eg. suppose I got this ratio $2:5$ from last example. So the empirical formula will be $\ce{Fe2O5}$ .)

Actually, you can skip the relative amount of substance step when you've got 2 elements in empirical formula, but you will need to perform all the steps when there are more than 2 elements as the answer comes to be different in some cases.


Principles of Thermodynamics A gas expands and does 82.7 J of work on the surroundings. At the same time, the gas loses 103 J of energy to the surroundings as heat. What is the change in the internal energy of the gas?
would this be ΔE = q + w  (-103) + (82.7) = -20.3.   I get confused with the positive and negative charges.

The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is

$$\Delta U = Q + W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur).

Therefore, heat transferred from the surroundings to the system is assigned a positive sign in the equation, whereas heat transferred from the system to the surroundings is assigned a negative sign.

Similarly, work done by the surroundings on the system is assigned a positive sign in the equation, whereas work done by the system on the surroundings is assigned a negative sign.

(Note that some books on physics or engineering still use a non-standard definition when discussing the work done by a thermodynamic heat engine.)


Why is a ketone more nucleophilic than an ester? I guess the ester is a weaker nucleophile because it does have an additional oxygen atom, unlike the ketone, that is pulling electrons from the C-O double bond towards the carbon atom (this happens vice versa too of course with the ester oxygen). This lowers the ionic character of the C-O double bond (smaller $\delta^{+}$ and $\delta^{-}$ charge on the carbon and oxygen atoms) and thus increases the energy of the C-O $\pi^{*}$ bond, making it less nucleophilic.   Is this correct?

The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester.  Look at the resonance structures drawn below.  Both the ketone and ester have a resonance structure that places positive charge on the carbonyl carbon, however you can also draw a third resonance structure for the ester that removes some of the positive charge from the carbonyl carbon and places it on the ester oxygen. This makes the ester carbonyl carbon less positive (more negative, more nucleophilic, less prone to attack by a nucleophile) than the ketone carbonyl carbon (more positive, less nucleophilic, more prone to attack by a nucleophile).  A nucleophile has electrons it would like to share and will react preferentially with centers of low electron density (more positively charged).  Therefor a nucleophile will react faster with a ketone carbonyl than an ester carbonyl.

EDIT

I like Martin's posts, they make me think - and that is always a good thing. Here is where my thinking has taken me:

I believe Martin's approach (basically frontier MO) is fundamentally correct.  I would be more convinced that his actual numbers are meaningful if he were to calculate 1) the carbonyl carbon LUMO coefficient and 2) the HOMO-LUMO separation for the following series of compounds in order to see if his calculated values correlated with this order of reactivity of carbonyl compounds towards nucleophilic attack (we could use water as the nucleophile) - but Martin, I think such calculations might be a poor use of your time and resources.

acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion

My argument based on how resonance affects the charge density at the carbonyl carbon is one approach used to explain ("teach") the above order of reactivity.  See for example page 4 on these UCLA notes.  I was just trying to keep it simple.
The issue I have with Martin's approach is, what should the reader do next week when he wants to know if an acid chloride reacts faster or slower than an ester.  While the resonance approach may not be rigorous, it does yield the correct order of reactivity and allows the user to make predictions while only using pencil and paper.
I am considering further editing of this post to use the resonance approach to explain differences in ground state stabilization of the carbonyl series.  Then, since they all go (more or less) to the same tetrahedral intermediate, one could explain the reactivity order based on differences in Eact.  This is another textbook method used to explain the carbonyl reactivity pattern.
Finally, what is the "correct" answer?  Is it the real truth as nature knows it, but in a form that is difficult for the common individual to use and extend; or is it a rough approximation of the truth that can be easily used?  I think the answer to that question depends upon who is asking the question, who is the audience.


Verification of variants of the Ebullioscopic formula Wikipedia cites the following formula for the ebullioscopic constant $K_{\mathrm{b}}$ regarding the elevation in boiling point of a volatile solvent upon addition of a non volatile solute (https://en.wikipedia.org/wiki/Ebullioscopic_constant):

(All symbols carry their usual meaning, unless specified otherwise)

  $$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}}{\Delta H_{\mathrm{vap}}}$$

But my textbook cites another formula :

  $$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}}{L_{\mathrm{v}}}$$

Where $L_{\mathrm{v}}$ is the latent heat of vaporization. Since there exists the equation $$\Delta H_{\mathrm{vap}} = {mL_{\mathrm{v}}}$$

my textbook further modified its formula to:

  $$K_{\mathrm{b}} = \frac{RT^2_{\mathrm{b}}m}{1000\Delta H_{\mathrm{vap}}}$$

Where $m$ is the mass of solvent in grams.

Now my question is, which one's correct? The only difference between the equation on Wikipedia and the one in my textbook is the mass term $m$. I can't seem to find any reliable references on this :/. So a good reference would be appreciated.

This answer expands upon my first comment:

  I am familiar with the version with $\Delta_{\text{vap}}H$, and in the derivation somewhere I remember that the Gibbs-Helmholtz equation is invoked and that is the origin of this term.

I have tried to perform the derivation I alluded to, and it is given:

Assuming a binary mixture, A denotes the solvent and B denotes the solute. Superscript (*) implies pure solvent without solute (e.g $T^*$ is boiling point of pure solvent). Also, I assume the enthalpy of vaporisation doesn't change over the temperature range we are interested in.At equilibrium the chemical potential obeys:

$$\mu^*_A(g) = \mu^*_A(l) + RT\ln(\chi_A)$$
$$ \ln(\chi_A) = \frac{\mu^*_A(g)-\mu^*_A(l)}{RT} = \frac{\Delta_{\text{vap}}G}{RT}$$

Differentiating with respect to $T$ and invoking the Gibbs-Helmholtz equation

$$\frac{\mathrm{d}\ln(\chi_A)}{\mathrm{d}T} = \frac{\mathrm{d}(\Delta_{\text{vap}}G/T)}{R\mathrm{d}T} = -\frac{\Delta_{\text{vap}}H}{RT^2}$$

Integrating from $\chi_A = 1 \ \text{to an arbitrary value}$

$$\int_{0}^{\ln(\chi_A} \mathrm{d}\ln(\chi_A) = \frac{-1}{R}\int_{T^*}^{T}\frac{\Delta_{\text{vap}}H}{T^2}\mathrm{d}T$$

$$\ln(\chi_A) = \ln(1-\chi_B) = \frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T}-\frac{1}{T^*}\right)$$

For dilute solutions, $\chi_B < 1$ so the approximate expression

$$\chi_B =\frac{\Delta_{\text{vap}}H}{R}\left( \frac{1}{T^*}-\frac{1}{T}\right)$$

Since, $T \approx T^*$
$$\left( \frac{1}{T^*}-\frac{1}{T}\right) = \frac{T-T^*}{TT^*} \approx \frac{\Delta T_b}{T^{*2}}$$

Now, we can rearrange to get the desired expression,

$$ \Delta T_b = \overbrace{\left(\frac{RT^{*2}}{\Delta_{\text{vap}}H}\right)}^{K_b} \chi_B$$

  Moreover, $\Delta_{\text{vap}}H$ and $L_v$ have the same physical meaning, and only the dimensions are different ($\mathrm{kJ \ mol^{−1}}$ vs. $\mathrm{kJ\ kg^{−1}}$  The multiplicative factors that result from the conversion between the two, can just be absorbed into the constant itself,I believe. So I am guessing the tabulated values might differ based on definition. Plus, I think it is an empirical parameter anyway.

Note, that the relation given above is in terms of molar fraction however it is easier to work with molalities when you are doing experiments. The same relation can thus be reformulated

$$\Delta T_b = \tilde{K_b}b$$

and this $\tilde{K_b}$ has the definition which involves $L_v$, I believe. (just look at the dimension in both the expressions and you'll see).


Why don't heavy and light gases separate in the atmosphere? Everyone must have heard that balloons are filled with helium, and the fact associated with it that helium gas is light and light gases always go upward.

There comes a question to mind: if the molar mass of $\ce{CO2}$ is greater than that of $\ce{O2}$ and $\ce{N2}$, then why doesn't $\ce{CO2}$ occupy the lower layer of the atmosphere, since it is heavier than $\ce{O2}$ and $\ce{N2}$, as in the case of balloons, where helium being light rise upwards.

That's because of two reasons. One is entropy, the ultimate force of chaos and disorder. Sure, gases would like to be arranged according to their density, but even above that, they would like to be mixed, because mixing creates a great deal of entropy. If you prevent the mixing, then they would behave just as you expected. Indeed, a balloon filled with $\ce{CO2}$ would drop right to the floor and remain there.

On the other hand, if you allow mixing, light gases wouldn't necessarily go upward. Just pierce that balloon with helium, and... Well, you won't actually see much, but in fact, the helium will disperse in the atmosphere and remain there. True, its contents in the upper layers would be somewhat higher, but only somewhat. It is not like a layer of pure helium floating atop all those $\ce{O2}$ and $\ce{N2}$.

The importance of entropy is by no means limited to gases. Think of all that salt in the oceans. Salt is much more dense than water; wouldn't it just drop to the ocean floor? Well, no, it rather wouldn't.

The other reason is the constant action of winds and currents mentioned by Zhe. They stir the atmosphere (or sea water, for that matter) and make it even more uniform than it might have been otherwise.


Alcohol or Carbonyl - Which is more reactive towards PCl5? We're aware that ketones and alcohols both react with phosphorus pentachloride, to produce geminal dichlorides and alkyl chlorides respectively.

What happens in the compound shown below, when treated with $\ce{PCl5}$? Which group reacts first? (which is to say, what happens if I were to add just one equivalent of $\ce{PCl5}$?)

Please help me out here, I'm not able to qualitatively predict an order.

Reaction of alcohols with PCl5 and PCl3 (for reaction with alcohol)
Mechanism for conversion of ketone to dichloride with phosphorus pentachloride
(for reaction with ketone)
Now as per the question as it is not given in excess so it will react with any one of the alcohol or ketone. It's quite hard to tell which one of them will react first because both of them show totally different kinds of reactions. Like ketones prefer neucleophile attack and alcohols prefer substitution and elimination.  But if you look at the mechanism quite closely you would find all the favourability of the reaction depends on the lone pair donation tendency of oxygen.  Now as you know, in sp³ hybridization state the s -character is nearly 25% and in sp² hybridization state s-character is nearly 33% and as s-character increases electronegativity increases
Why a higher s character increases a carbon atom's electronegativity?
and the tendency to hold the electron pair increases so for ketone it is quite difficult to donate it's electron pair to PCl5. So in that perspective I can say alcohol is quite more reactive than ketone in this case.


Why is liquid oxygen more reactive than gaseous oxygen? Liquid oxygen is somewhat hairy to work with, as, if it comes in contact with essentially anything combustible, it has an unfortunate tendency to detonate violently if looked at sideways.

Gaseous oxygen obviously does not do this, as can be seen from the fact that combustible objects can sit indefinitely in gaseous oxygen without exploding when touched.

This seems backwards, as, at a given pressure, liquid oxygen is colder than gaseous oxygen, and materials are generally more reactive at higher temperatures - thus, intuitively, one would expect liquid oxygen to be less reactive than gaseous oxygen, not more.  Why is this not so?

Since oxidation is generally strongly exothermic, a large part of the difference in reactivity can be explained by kinetics (rather than thermodynamics): the denser liquid brings oxidant (oxygen) into intimate contact with the oxidized substance throughout the reaction. Consumed oxygen is quickly replenished by surrounding liquid oxygen. In the gas the exothermic nature of combustion reactions accelerates reaction with oxygen gas by increasing the collision rate between oxygen molecules and the surface being oxidized. In the liquid this temperature effect is not as important because the oxidized surface is already bathed in oxygen molecules. Also, although the gas might be at a higher T, as assumed in the question, local heating of liquid oxygen around the oxidized surface due to the highly exothermic nature of most oxidation reactions can also add to the faster propagation rate of reactions in the liquid. The importance of thermal effects might also be reduced (if not trumped) by the acceleration of concerted events during the oxidation (such as sequential events at the same or neighboring sites or while in a short-lived activated state), since in the liquid collisions between surface and more than one oxygen molecule at neighboring sites will occur at rates that are orders of magnitude faster than in the gas.

This is admitedly an answer in need of some quantitative support.

Here's a "back-of-the-envelope" justification for the importance of density:

It turns out that if you use a model based on the Maxwell-Boltzmann kinetic theory then the collision frequency with a surface is proportional to $p/T^{1/2}$. Therefore, if you ignore the thermal activation energy for reaction, you can assume that the reaction rate is proportional to $p/T^{1/2} = \rho _m RT^{1/2} = RT^{1/2}/V_m$. This means we will have the same collision frequency for two different set of values of T/V when

$$\left(\frac{V_{m2}}{V_{m1}}\right)^2=\left(\frac{T_{2}}{T_{1}}\right)$$

Therefore, we can counter the effect of a 200 K drop in T by increasing the density (reducing the molar volume) by a factor of ~14. Now at STP the molar volume of a gas is ~20 L/mol, while for a compressed (liquified) gas the molar volume is 100 times smaller, say 0.2 L/mol. Therefore the increase in density more than makes up for the drop in T, at least within this simplified model.

In more general terms, we can explain away much of the loss of the kinetic advantage of higher temperature by the higher density in the liquid. So based on this simplified argument oxygen at a liquid-like density can be more reactive than a much warmer gas.

Note that the activation energy for reaction of $\ce{O2}$ is not negligible due to the relatively stable structure of the oxygen molecule. Therefore kinetic effects involving rapid heating and thermal activation of oxygen (due for instance to local accumulation of thermal energy) are probably also important to explain the reactivity of LOX.


What is the difference between ideal stages and transfer units in a packed bed gas absorber? I am having a hard time grasping what these two concepts are supposed to mean and how they differ. What I know so far is:

In a counter-current gas absorber, if we set the inlet and outlet concentrations of solute in gas and liquid, and also have control on the ratio of gas to liquid flow rates, we can find the outlet concentrations by making a "step" like a staircase (with the number of steps equaling the number of stages) from the operating line to the equilibrium curve.

I understand where this comes from. When we draw a "step", we are equating the two outlet concentrations to be in equilibrium.

Now, for the number of transfer units, we usually find this out by using the integral formula:

$$\text{NTU} = \int_{y_1}^{y_2}{\frac{dy}{y-y*}}$$

where $y$ is the operating line solute concentration in gas phase for a given $x$, i.e. solute concentration in liquid phase, and $y*$ is the equilibrium curve solute concentration for the same $x$.

What I have noticed is that often, the NTU is not an integer, it can also be a fraction. And this fraction, multiplied by the height of transfer unit (HTU), gives the total height of the bed needed for such a transition.

I am confused. I dont understand what NTU is exactly, or better, what a transfer unit is. I would greatly appreciate some guidance.

A transfer unit gives the change of composition of one of the phases equal to the average driving force producing the change
Source: http://www.separationprocesses.com/Absorption/GA_Chp04c.htm

So to phrase it differently, for some ideal zone in the absorber that has a particular value to the average mass transfer driving force, there will be a fixed change in the phases present at that ideal zone.
NTU is somewhat analogous to the total number of theoretical stages if the operating and equilibrium line are straight and parallel.
There is no direct physical interpretation of what a transfer unit. If you design any appreciable size packed bed tower, you will find that the vendors of the packing will use a lot of empirical correlations for their own products to determine separation efficiencies and other related parameters.


When to use Da or u for mass units Following up on the question Units of mass on the atomic scale,
Daltons (Da) and unified mass units (u) are equivalent, but which is used seems to depend on context.  For example, when discussing single atoms one typically sees the mass expressed as u, but when discussing large biomolecules one typically sees Da or kDa.

Is there a rule about whether to use Da or u? Or, is this simply a matter of convention or left to the writer's choice?

It doesn’t matter. Both units are basically identical, much like there is no everyday difference between the units mbar and hPa. There may be situations where one is preferred — e.g. hPa are, in my opinion, rarely used outside weather reports — but they should be understood nonetheless.

If there is anything to be said about the two, it seems using dalton is more popular in biologic contexts. But that wouldn’t be a tendency to base anything on.

Being consistent within one’s lab, within the journal one is publishing in or throughout one’s thesis is far more important.


How do i detect the compounds a plant is made of, automatically? Let's  say I want to take a random plant , and get the chemical formulas of  the compounds he's made of. Is there an automated method for doing so  and how costly is it(roughly) ?

HPLC-MS (high performance liquid chromatography- mass spectrometry), the latter with a suitable database for known small molecule plant components (alkaloids, terpenes, THC, etc.) would do it.  Cost would be three to low four figures/shot.  Saccharides and proteins are different and more expensive problems.

Consider what you are asking.  Mere pokeberry contains astragalin, canthomicrol, caryophyllene, esculentic acid, jaligonic acid, lectins, oleanolic acid, 3-oxo-30-carbomethoxy-23-norolean-12-en-28-oic acid, 6-palmityl-Δ7-stigmasterol-Δ-D-glucoside, 6-palmytityl-α-spinasteryl-6-D-glucoside, phytolaccagenic acid, phytolaccagenin, phytolaccatoxin, phytolaccosides A,B,C,D,E,F,G (esculentoside E), pokeweed mitogen, tannin, triterpene alcohols, α-spinasterol, and α-spinasteryl-β-D-glucoside and a whole bunch more.  One injection is hardly a beginning.  If you do not know how to cook it, it kills you.


Derivative of Gibbs Energy wrt Temperature at constant Internal Energy I came across with the following solution from Sandler's book: $$\begin{eqnarray}\left( \frac{\partial G}{\partial T} \right)_U &=& \frac{\partial (G,U)}{\partial (T,U)} \\ &=& \frac{\partial (G,U)}{\partial (T,V)}\cdot \frac{\partial (T,V)}{\partial (T,U)} \\ &=&\frac{(\partial G/\partial T)_V (\partial U/\partial V)_T-(\partial G/\partial V)_T (\partial U/\partial T)_V}{(\partial U/\partial V)_T} \end{eqnarray}$$

Please help me understand these steps. What happens in line 1? I know that line 2 is by using Chain Rule by then how did line 2 becomes line 3?

The starting point is the two equations $$dG=\left(\frac{\partial G}{\partial T}\right)_VdT+\left(\frac{\partial G}{\partial V}\right)_TdV$$ and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$From the second equation, you determine the partial of V with respect to T and constant U by setting dU=0.  I leave the rest of the math up to you.

Chet


After mixing, which species has the dominant affect on the pH of the solution that remains? A solution is created by mixing $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{HCl}$, $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{CH3COOH}$ and $500\,\mathrm{mL}$ of $1.5\,\mathrm{M}$ $\ce{CsOH}$.

The $K_\mathrm{a}(\ce{CH3COOH}) = 1.8 \times 10^{-5}$

What is the $\mathrm{pH}$ of this solution?

I am stuck between $\ce{HCl}$ and $\ce{CsOH}$ since they are both a strong acid and base (respectively) meaning that they should both have a substantial affect on the $\mathrm{pH}$.

Also when determining what the $\mathrm{pH}$ is, which species would we use in our ICE table?

CsOH is a strong base.

0.5 L X 1.5 M = 0.75 moles of hydroxide.

Similarly, there are 0.5 moles of monoprotic acid.

After mixing there will be 0.25 moles of hydroxide in 1 liter of solution.

In other words [OH-] = 0.25M

pH is approximately 14 + log (0.25) = 13.4


Why is the work done by a gas during free expansion zero and not infinity? Suppose there's a spaceship that has a hole in it. Gas is expanding through the hole from the spaceship to outer space (vacuum). According to pressure volume work ($P\Delta V$) the work done by the gas is infinity. According to my homework assignment, it's $0$. Can someone please explain why?

Edit:

According to Wikipedia:

  $P$ denotes the pressure inside the system and outside the system, against which the system expands; the two pressures are practically equal for
  a reversible process;

How can $P$ denote two different quantities? Can't this rule be applied for situations where the pressures inside and outside the system are different?

The external pressure in space is 0 (well, depends where and not quite, but for the discussion let's say that it is). So $P \cdot \Delta V$ is also 0. Gas expanding from a hole in a spaceship is not an irreversible process. Such processes are very slow and gradual, such that the external pressure has time to equilibrate with the internal pressure.

On the contrary - you have the most irreversible process in here. Gas leaking from a hole in a spaceship into a medium (space) that will never equilibrate with the internal pressure.

Think about it from an intuitive point of view. Work done by an expanding system is there because it moves other things around. In space there is nothing to move, so no work done.


Calculate pKa for lipids I would like to calculate $\mathrm{p}K_\mathrm{a}$ of lipid molecules using some software tools. I have tried ChemAxon and compared with ChemDraw professional 16.0 version they are mostly similar. I think both are in gaseous phase.

Do I have any chance to calculate the $\mathrm{p}K_\mathrm{a}$'s in aqueous phase using solvent parameters? Any software tool suggestion are highly appreciated.

See H++ if you want a rough prediction of $\mathrm{p}K_\mathrm{a}$.

If you want to more accurately calculate the $\mathrm{p}K_\mathrm{a}$ of your molecules, you should run a quantum mechanics calculation with a continuum solvent model (example calculation: [1]).

The general outline is as follows. Run a geometry optimization and vibrational mode analysis on your system in basic and conjugate acidic forms. So, $\ce{BH+}$ and $\ce{B}$, or $\ce{BH}$ and $\ce{B-}$, depending on what your system looks like. This will allow you to extract a free energy of reaction for each of your lipid molecules.

Importantly, these calculations will not give accurate single point free energies, but the relative values ($∆G$) will be. As such, you will need a reference $∆G$ from literature of a similar reaction to "set the 0" of your values. Finally, you can get $\mathrm{p}K_\mathrm{a}$ from $∆G$ using

$$\mathrm{p}K_\mathrm{a} = \frac{∆G^\circ}{2.303RT}$$

Refrences

Liptak, M. D.; Shields, G. C. Accurate $\mathrm{p}K_\mathrm{a}$ Calculations for Carboxylic Acids Using Complete Basis Set and Gaussian-n Models Combined with CPCM Continuum Solvation Methods. Journal of the American Chemical Society 2001, 123 (30), 7314–7319. https://doi.org/10.1021/ja010534f.


What are some examples of a clear, liquid, UV photoactive polymer? Something that is not reactive to lesser wavelengths of light, specifically IR.

Methacrylates make nice starting materials for the lightinduced polymerization.

In methyl methacrylate (MMA), camphorquinone (a typical initiator) has an absorption maximum around $\lambda$ = 473 nm.


Difference Between Betaine & Zwitterion? what is difference btw Betaine & Zwitterion, Both having positive and negative charge in a molecule.

How to categorize Zwitterion and Betaine? or both are same.

From Wikipedia:

  A betaine (BEET-ah-een, /ˈbiːtɑːˌiːn/) in chemistry is any neutral chemical compound with a positively charged cationic functional group [...] which bears no hydrogen atom [...]. A betaine thus may be a specific type of zwitterion.


Interpretation of Figure - biofunctionalization In this paper on page 1985 they say:

  Fluorescence analysis confirmed that enhanced protein binding at higher electron dose depends only to a small extent on the slight increase in the lateral dimensions of the nanodeposits but to a much larger degree on their height (Figure 4E)

  The amount of adsorbed IgG-Cy3 is largely independent of the actual area of the nominal 100 nm islands but scales with the height of the nanodeposits.

I don't understand this yet: why do I see that the fluorescence is proportional to area in this figure?

The amount of adsorbed protein is proportional to the area of the 100 nm islands. Figure 4E proves it. The authors misinterpreted their own figure.


sp2 hybridized carbon versus sp3 hybridized silicon Carbon's most commonly found allotrope is graphite/soot/fullerene/carbon nanotubes which are all $\mathrm{sp^2}$ hybridized forms of it, while silicon being its first periodic twin is mostly found in the $\mathrm{sp^3}$ hybridized form in bulk, while formation of silicene is extremely difficult. While in the case of carbon, diamond is the one which is harder to find, what is the reason for this discrepancy?

Carbon — and all the other elements of the second period, for that matter — is exceptionally good at forming double bonds. A double bond is formed if there is a σ bond and a π bond. Forming a σ bond is not hard: You have two orbitals coming from two different atoms pointing at each other. Forming a π bond is much harder, since the orbitals are pointing orthogonally away from the bond-axis. Thus, the overlap of π bonds is often a lot weaker than for σ bonds.

This gets more important if you consider third-period elements rather than second-period ones. They have a total of eight more core electrons making them much larger in size and the orbitals more diffuse. A σ bond is still formed easily, there will be the point when two diffuse orbitals are close enough to interact in a σ fashion. To arrive at favourable interactions for a π bond would require the atoms to come much closer than what would be comfortable; unfavourable interactions between the core electrons would already be a thing.

In fact, I was taught in my Concepts of Inorganic Chemistry class, that disilene $\ce{Si2H4}$, the heavier homologue of ethene, does not form a classical π bond. Instead, it can be understood as a par of two singlett silene fragments $\ce{SiH2}$ which approach each other so that the empty p-orbital of one silene can interact with the lone pair in an $\mathrm{sp^2}$ orbital of the other silene fragment. Obviously, the symmetry is lower and the bond is much weaker.

So the reason behind it is that $\ce{Si=Si}$ double bonds are hard to create and this gets even harder for heavier homologues while $\ce{C=C}$ double bonds are reasonably stable and easy to generate.