Nomenclature for two cyclohexane rings linked by a single bond From my current basic organic chemistry knowledge, I suppose it is 1,1-dimethylcyclohexane-4-methylhexane. I considered one of the hexane ring as parent chain. But I'm looking for the correct answer with appropriate explanation for it.

The relevant sections on ring assemblies of two identical cyclic systems in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually read as follows:

P-28.2.1 Ring assemblies with a single bond junction
Assemblies of two identical cyclic systems joined by a single bond are named by one of two methods:
(1) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding parent hydride enclosed in parentheses, if necessary. Parentheses are used to avoid confusion with von Baeyer names;
(2) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding substituent group (for names of substituent groups, see P-29), enclosed in parentheses, if necessary.
Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
(…)

P-52.2.7.1 Preferred IUPAC names for assemblies of two or more identical cyclic systems joined by a single bond are formed using the names of parent hydrides rather than the names of substituent groups, (…). For two-component assemblies, locants of one ring are unprimed; locants of the second ring are primed. Locants, including the locants 1 and 1', are necessary in preferred IUPAC names to indicate points of attachment of rings or ring systems (see P-28.2.1 ).

Therefore, the preferred IUPAC name for the compound that is given in the question is formed according to Rule (1) as 4,4'-dimethyl-1,1'-bi(cyclohexane).
The corresponding name according to Rule (2) is 4,4'-dimethyl-1,1'-bi(cyclohexyl). This name may be used in general nomenclature.


Why doesn't water play a role in reactions in aqueous reactions? $$\ce{Ba(NO3)2(aq) + Na2CO3(aq)}$$

In the above reaction both the substances are in water, however water plays no role in the reaction. Why is it that although water is present in the reaction, it does not play a reactant role? To clarify, why isn't the above written as:

$$\ce{Ba(NO3)2(aq) + Na2CO3(aq) + H2O}$$

Why does water act as a bystander and not a reactant?

Water is far from a bystander in aqueous solutions. It has a number of features that greatly facilitate chemical reactions, as well as being an important ligand (the aqua ligand) of transition metal complexes.

Here are some (unbalanced) examples:

Water has a rather large dielectric constant and as such is very effective at shielding charges from one another, facilitating the charge separations inherent in the dissolution of ionic compounds. This is why you can easily dissolve sodium chloride in water, but not in hexane. The nature of this shielding is often articulated as several concentric 'solvation shells', where 'shells' of water molecules cluster around an ion with their dipoles approximately aligned towards or away from the ion.

Here's an image from Wikipedia qualitatively depicting the first solvation shell surrounding a sodium ion:

More generally, the large dipole moment of water allows it to partake in hydrogen bonding, where water will weakly associate with other polar molecules or functional groups. This extends to water forming (short lived, highly fluxional) clusters with itself.
Water is also happy to accept (forming hydronium, $\ce{H3O+}$) and donate (forming hydroxide, $\ce{OH-}$) protons in solution. So much so that it will actually do this with itself, in a process called autoionisation. This process makes water molecules significant in a wide range of acid/base and redox equilibria.


The C-13 chemical shift of acyl chlorides vs ketones In lectures, we were taught that ketones and aldehydes tend to have chemical shifts of around 200ppm, while acid derivatives (carboxylic acids esters etc) tend to have values around 160-170 ppm.

The reasoning given by my supervisor was that despite acid derivatives having more electron withdrawing groups, due to conjugation, the net effect was that these additional groups donated electron density to the carbonyl carbon, hence shielding it more than the carbonyl carbons in ketones/aldehydes (i.e. the conjugative effect dominates over the inductive effect). This all makes sense for me except for acyl chlorides.

The reason why acyl chlorides are not adding up for me is that in the very next topic, we were taught that the CO double bond of an acyl chloride is stronger than that of a ketone/aldehyde! Since carbonyl bonds are strengthened by anything that withdraws electrons, I can only conclude that the Cl has a net electron withdrawing effect (i.e. its inductive effect dominates the conjugative effect).

Since the Cl group appears to actually withdraw electron density, going back to NMR, shouldn't the carbonyl carbon in acyl chlorides resonate at a higher frequency than the carbonyl carbon in a ketone/aldehyde?

Halogens $(\ce{Cl},$ $\ce{Br},$ $\ce{I})$ also act as conjugative donors. Not very good ones but you still have to take that into account especially because, as you said, conjugative effect is larger than inductive effect (and it is very different in $\ce{^1H}$- or $\ce{^13C}$-NMR).
The $\ce{^13C}$ chemical shift in $\ce{CH2Cl2}$ is $\pu{53 ppm}$ but the one for $\ce{CH2I2}$ is $\pu{–62 ppm}$ (yes, with a minus sign).


Bends in cis Isomers It’s been said over and over in my biochemistry books that the cis-double bonds between the Cs in the hydrocarbon tails in triglycerides causes a kink. This kink disappears when the bond is isomerized into its trans conformation. What is it about the cis bond that causes the kink?

The best way to demonstrate this kink is with pictures. Usually we have some motivation to try to represent the molecule on a single line. So for oleic acid (cis-octadec-9-en-1-oic acid), we often get pictures like the following:

Elaidic acid, the trans-isomer is also sometimes depicted unhelpfully to save space:

The second structure for elaidic acid is not as misleading as the corresponding structure for oleic acid. A better bond-line structure for oleic acid is:

Better yet, here are some space-filling structures from Wikipedia based on crystal structures of both molecules.

Oleic

Elaidic

And here's $\gamma$-linolenic acid, which has three cis alkenes in it:


Removing aluminum oxide before galvanic current and potential measurements I have an aluminum sample stored about 1 month ago. I am going to couple it for galvanic current and potential measurements. I think the Al sample grew an oxide film due to exposure to air. Would the aluminum oxide film can affect the measurements? Should I remove it first? How?

sure it's going to affect the test so you have to remove the oxide layer
Generally, we adopt the method of alkali corrosion to remove the oxide layer on the surface of the aluminum.The purpose of alkali corrosion is removing the oxide layer on the surface of the aluminum. (Note: Technological process of the method is as follows: under normal temperature, 20% nitric acid; the process parameters is about 0.5 minutes. Pre-degrease--degrease--rinsing--rinsing--alkali corrosion--rinsing--rinsing--bright dipping--rinsing--rinsing--pure rinsing. The purpose of bright dipping is to remove surface residual black ash after alkali corrosion.Temperature:50-60 degrees; time:0.5-1 minutes.The purpose of degreasing is to remove grease on the surface of the aluminum.The process parameters is 0.5-1 minutes. You can use acid to wash it


Gas phase during water sampling I have a question concerning simultaneous sampling of water and gas phase.

The situation is a pipe (diameter somewhat around 15 cm) where a mixture of water and gas is flowing through (two-phase flow). Assuming that the flow is separated and the gas phase is moving on top, what will happen if I open a sampling tap at the bottom of the pipe? The flow is induced by a pump and the pressure in the system is above ambient (20 bar).

My assumption is, that I only sample the water and the gas phase on top will not leave the pipe unless I allow the gas phase to "connect" with the tap by extracting a big amount of fluid.

The problem is maybe comparable to a bottle of sparkling water, where I drill a small hole at the bottom. Will I only sample the fluid phase or will I sample a mixture of both?

I hope somebody has an answer. Thanks in advance!

That's a tricky question.

Given your setup, it follows that you will only be sampling water (not simply "fluid", mind you; fluid is a common term for both gas and liquid). But gases might be more or less soluble in liquids. Chances are that your water is saturated with some gas (say, $\ce{CO2}$) and pressurized with the same gas in the pure form. Then, as soon as you take your sample and put it in a sealed (or worse yet, unsealed) container, it will release some gas, and so the composition of the sample will become different from that of the liquid in the pipe.

Whether or not this issue is relevant to your situation, is up to you to assess.


How to calculate the aqueous solubility of CO2 vs Pressure I would like to make Na2CO3 react with HCl in a closed fixed volume vessel, and accordingly I'd like to understand how ensuing pressure will affect the equilibrium of my reaction (and to which extent my reaction will affect the level of pressure when equilibrium is reached).

I guess that the link between pressure and the reaction mostly has to do with CO2 retention into the solution, which leads me to want to understand how I can calculate CO2 solubility in water vs pressure.

Would anyone care to explain me how I could calculate CO2 solubility vs pressure? And tell me if there's anything big I am missing other than CO2 solubility?

The acidity of HCl is so great compared to H2CO3 that you have no equilibrium between Na2CO3 and HCl; the equilibrium is only between CO2 and water and the vapor space. H2CO3 is one of the forms of CO2 in water, but it is unnecessary to delve into this.

After your reaction, you will have a solution of NaCl in which some of your CO2 is dissolved; the rest of the CO2 will be in the vapor space. We will assume that the solubility of CO2 in this solution is the same as in pure water. Data from CRC Handbook, per 100 cc H2O at 1 atm pressure: 171 cm^3 at 0C, 90 cm^3 at 20C. In grams: 0.348 at 0C, 0.145 at 25C, 0.097 at 40C, 0.058 at 60C.

That's the chemistry. Now the mathematics is more complicated, and I'll leave that up to you. But a few points to help you get started: if you have a container with a volume of 100 cc and put 100 mL of water into it, you could also dissolve 171 cm^3 (0.348 grams) of CO2 into that same volume at 1 atm pressure (OK, maybe a slight volume increase - ignore it for simplicity). If you put only 50 mL H2O in that 100 cc volume and added 171 cm^3 of CO2, the pressure in the vessel would be greater than 1 atm (but less than 1.71 atm, which is the pressure you would have with no water in it at all).


Spin state change during crystallisation for [NiCl2(PPh3)2] $\ce{NiCl2(PPh3)2}$ is a complex that is borderline between tetrahedral(high spin) and square planar(no unpaired electron, high splitting parameter) because $\ce{Cl-}$ is a weak field ligand and $\ce{PPh3}$ is a strong field ligand. Apparently, the tetrahedral and square planar states of the compound are interconvertible. The tetrahedral high spin state is blue, and produced directly by reacting hydrated nickel chloride and triphenylphosphine in alcohol. But when the complex is crystallised out from a cholrinated solvent like dicholoromethane, it converts to the red square planar complex.

Is there any theoretical explanation as to

1) Why the square planar form is favoured in the crystalline state?
2) What role does the chlorinated solvent play in isomerising the complex to its sq. planar form?

And are there any other examples like this?

I made both isomers as a PhD student many years ago. The red diamagnetic form crystallises out when a concentrated CH2Cl2 solution is cooled in a dry ice-acetone bath, and you can quickly filter off the red product. But on leaving the solid at room temperature, it reverts in a couple of days at room temperature to the blue, paramagnetic form. So it's a thermal equilibrium.


Overlapping oxygens in crystal structure -- what's going on? This is a screenshot of  a portion of a metal organic framework (MOF).  These are two oxygens of two NMP solvent molecules.  What I am wondering is why they are so close and overlapping like this.  I checked the CIF file, and occupancy for all atoms in this structure is 1, so it doesn't seem to be the case that there is only one NMP molecule in two equally likely positions.

Could someone explain this?  Thanks

It's a bit hard to answer without having coordinates file (*.res or *.xyz), but the simplest answer that I would expect is that these oxygen atoms belong to two different molecules that are disordered over the same position (i.e. they have only partial occupation on the site that in sum is 1). It's very common for f.e. thf coordinated to alkali-metals.

If you have shelx input for the structure look for the line that defines one of oxygens and see if 6th parameter is 11.0

O1    5   -0.180348    0.245831    0.298215    11.00000    0.02208    0.02526 =
                                               ^^

This means fixed (10) occupancy of 1.

If you have cif-file look for the atom loop

loop_
 _atom_site_label
 _atom_site_type_symbol
 _atom_site_fract_x
 _atom_site_fract_y
 _atom_site_fract_z
 _atom_site_U_iso_or_equiv
 _atom_site_adp_type
 _atom_site_occupancy
 _atom_site_symmetry_multiplicity
 _atom_site_calc_flag
 _atom_site_refinement_flags
 _atom_site_disorder_assembly
 _atom_site_disorder_group
O1 O -0.18035(6) 0.24583(6) 0.29822(5) 0.02794(8) Uani 1 1 d . . .

8th parameter is occupancy (_atom_site_occupancy).

These both examples were for not disordered atoms. If they are disordered then it looks like this:

shelx-ins/res

O1    4    0.297097   -0.198418   -0.064164    10.50000    0.34896    0.34935 =
                                                  ^ occup. 1/2

or

FVAR       0.22777   0.51340   0.51266
...
O1    4    0.551971    0.536866    0.295322    21.00000    0.03100    0.03031 =
                                                ^ occup. bound to 2nd free variable
                                                  here 51%

cif-file:

O1_20 O 0.6673(9) 0.3140(7) 0.1724(3) 0.0434(14) Uani 0.513(3) 1 d PDU C 1
                                                      ^


Determine what the unknown element is A piece of electronic equipment originally contained $\pu{1.250 g}$ of a valuable metal. Corrosion over time has given a compound of formula $\ce{M(OH)3}$, which is isolated but not weighed. This compound is treated with sulfuric acid to give a compound that has beautiful red crystals and is found to have a molecular formula of $\ce{M2(SO4)2}$ and a mass of $\pu{1.860 g}$. What is this valuable metal?

My attempt:
$$ 1.860 ~\mathrm{g} - 1.250~\mathrm{g}= 0.61~\mathrm{g}\, \ce{SO4} $$

I took the $\pu{0.61 g}$ and converted them to moles of $\ce{SO4}$, then I converted 2 moles $\ce{SO4}$ to 2 moles $\ce{M}$ and got $0.0064~\mathrm{M}$.
Then I divided that by $\pu{1.250 g}$.

I got 195.31, so is platinum the correct answer?

The mass of the sulfate is $m_{\ce{SO4}} = 0.610~\mathrm{g}$, the mass of the metal is $m_\text{M} = 1.250~\mathrm{g}$.

With the molar mass of the sulfate anion $M_{\ce{SO4}} = 96.061~\mathrm{g\, mol^{-1}}$ we can calculate the amount of sulfate present in the final compound:
$$ n_{\ce{SO4}} = \frac{m_{\ce{SO4}}}{M_{\ce{SO4}}} = 6.35~\mathrm{mmol} $$

Since the sulfate has the formula $\ce{M2(SO4)2}$ we know that we have the same amount of metal as sulfate: $n_\text{M} = n_{\ce{SO4}}$.

We can now calculate the molar mass of the metal using the information above:
$$ M_\text{M} = \frac{m_\text{M}}{n_\text{M}} = 196.85~\mathrm{g\, mol^{-1}} $$

This number resembles most closely gold with a molar mass of $M_\text{Au} = 196.97~\mathrm{g\, mol^{-1}}$.

Note: It is important that you don't round your numbers while doing the calculations, but round them at the end. So you should keep as much precision as possible in between to minimize rounding errors along the way.

If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.35\ mmol}$ you will get $\mathrm{196.85\ g/mol}$ (gold).
If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.4\ mmol}$ you will get $\mathrm{195.31 g/mol}$ (platinum).


Why doesn't a table sublimate, while ice does? A table does not sublimate, and nor does a spoon. Ice does, however. What is the fundamental difference?

This relates to the difference between how ice, iron spoons and wooden tables stick together.

In wood, most or all of the bonds between the individual atoms and fibre units are covalent; making them very strong. Similarly, a spoon is an alloy of iron, nickel, chrome and carbon, also a very coherent structure.

In ice, on the other hand, the units of the crystal is held together with comparatively weak hydrogen bonds, meaning that not a lot of energy is required for a surface molecule of water to escape, in a process called sublimation.

Usually, once a water molecule escapes the ice, it is, given a relatively constant temperature, very unlikely to rejoin the body of ice; meaning that especially in open air and windy conditions, ice will sublimate at sub-zero temperatures, although as the temperature of the ice drops, this effect becomes steadily less noticeable.

Edit: It is worth noting that given enough time, a table or a spoon can indeed sublimate as well, at any temperature, but the odds against it are utterly staggering.


What is the chemical basis for superglues? There are a variety of different types of glues, some derived from animal or natural materials and some designed by chemists. Some glues need to be activated before they harden and stick (many epoxies, for example) but others seem to stick very quickly without any obvious activation (like Loctite Superglue or The Original Super Glue).

What chemical reactions make these glues work?

This is a very general answer to your question.

According to the Chemistry World podcast transcript, Cyanoacrylate and the article The Chemistry of Superglue (Brubaker, 2000), the main ingredient cyanoacrylate, which is

  a type of plastic known as an acrylic resin. It exists in the tube as single molecules in liquid form. But when they come into contact with water, the molecules react with hydroxide ions to form long polymer chains that set into a hard solid. As there is a thin layer of moisture covering most surfaces, cyanoacrylate can bond them together very effectively.

This also explains why superglue tends to harden very quickly when in contact with skin.

A diagram of cyanoacrylate polymerisation in the presence of water is shown below:

from the University of Rochester, who also state that asides from water, that the presence of

  Weak bases (such as those on the proteins in skin) work even better

for polymerising the cyanoacrylate.


Should I include solvation in quantum chemical calculations of a simplified protein-ligand interactions? I am running a B3LYP-D3/6-31+G geometry optimization of a protein-ligand interaction. Well, not quite, I have simplified the protein to four amino acids surrounding the active site, but I keep the entire ligand (170 atoms total).

I am wondering if I should optmize the geometry in solution, i.e. will including solvation effects make my results (energies) be more realistic? My mind is split.

On the one hand, I am modelling a biological system which usually takes place in aqueous solution. On the other hand, my system is so simplified, that it no longer has much root in reality. By "placing my molecules in water", the water molecules will interact with parts of my system that otherwise would not interact as strongly (there would be much less water inside the protein than outside it, I mean). So, I feel there is no guarantee that my results will improve, and I am not able to find any data indicating one or the other. If I included the entire protein, then the case would be different, perhaps, since at least then that protein would be "solvated correctly".

So, given the simplified system, will solvation improve the resulting geometry, and thus the interaction energy between the protein residues and the ligand, or not?

This is difficult to answer without full knowledge of the system, but some tips:

1) The overall energy of the system is arguably irrelevant.  Rather, the relative difference between different conformations, and perhaps, different ligands is of utility.

2) The level of theory and basis set your using is quite low.  If you have the time and the resources, scale it up.  For biomolecules I've found the minnesota functionals to be most accurate (if insisting on DFT) but I still believe MP2 is king.

3) Referring back to #1, if you solvate, then your comparison should too be solvated.  Perhaps your approach would be to run a solvated and unsolvated model for both your ligand, and perhaps, another known ligand. This could too help convince you of the utility of the solvation model.


What kind of system is required for dynamic equilibrium to be set up? Most websites and books mention a closed system, but technically isn't an isolated system required? If energy can be transferred, then the position of of the equilibrium will continuously change and no equilibrium will be set up, right?

If energy can be transferred, it does not mean it is transferred.

If there is e.g. zero net thermal energy transfer, a system is in thermal equilibrium with the neighborhood, even if it is not isolated.

As Ivan says, an equilibrium is possible even in an open system, if net transfer of particular matter kinds is zero.

The thermodynamic equilibrium consists of thermal, radiative, chemical and mechanical equilibrium.

Note that thermodynamic equilibrium happens at following conditions:

  For a completely isolated system, the entropy $S$ is maximum at thermodynamic equilibrium.
  For a system with controlled constant temperature and volume, the Helmholtz free energy $A$ is minimum at thermodynamic equilibrium.
  For a system with controlled constant temperature and pressure, the Gibbs free energy $G$ is minimum at thermodynamic equilibrium.

$$\begin{align}
G&=H-TS \\
\Delta G&=\Delta H-T\Delta S \\
\end{align}$$

Total entropy of the not isolated system and it's surrounding increases in time.

Entropy of the system alone may decrease with time, if the Gibbs energy of system decreases, as this latter decrease also means
$$\Delta S_{\rm surrounding} +  \Delta S_{\rm system}  \gt 0$$


What happens to the hydroxide anion during the electroreduction of water? The reduction of water at a cathode is represented by the following equation:

Cathode (reduction): $\ce{2 H2O_{(l)} + 2e^{−} → H2_{(g)} + 2 OH^{-}_{(aq)}}$

If this is done in a divided cell, what would happen to the $\ce{OH-}$ anions? The hydrogen gas would accumulate and bubble would form. Would the $\ce{OH-}$ anions just build up in the solution? Do they interact with each other naturally forming $\ce{H2}$ and $\ce{O2}$?

You only show one a half-reaction, reduction at the cathode.  Write the other half-reaction, oxidation at the anode to give oxygen.  A real world water electrolysis cell requires a salt electrolyte for conductivity.  $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.

http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif

If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$.  $\ce{Na2CO3}$ can be interesting.


Why is the specific gravity of 0.9 % saline 1.0046? Medical Normal Saline (0.9% Saline) contains 9 g of salt dissolved in 1000ml of water.

Hence a litre of NS should weigh 1.009 kg.

Hence its specific gravity should be 1.009, but it’s only 1.0046.
(I've measured it, and verified it on-line. It is apparently, a fact.)
I thought I understood SG, it’s a simple concept but I am missing something.

EDIT: after more reading, I learned that dissolving NaCl in water results in a reduction in the original volume. So my question now becomes:
We're dissolving NaCl in pure water at a temperature where 1.000kg mass of water has a volume of 1.000 litres. By my (erroneous) reasoning:
Say the volume decreases by 1 ml, giving a final volume of 0.999 litres.
So the density of the solution would be 1.009 kg/ 0.999 l = 1.010 kg/l
Giving an SG of 1.010 (because water has a density of 1.000 kg/l, as above) If the volume didn't change, it would be 1.009, but it increases because of the reduction in volume.
BUT: The known, published SG of 0.9% w/v NaCl solution is 1.0046.
Can anyone explain why, please? or at least, why the SG is below 1.009, when it should be greater?

0.9% medical saline is defined to contain $\pu{0.90 g}$ sodium chloride per $\pu{1000 ml}$ solution at $\pu{22 °C}$. Its density is $\pu{1.0046 g/cm^3}$ at $\pu{22 °C}$.

The density of water is $\pu{0.9978 g/cm^3}$ at $\pu{22 °C}$.

If the volume would stay constant when you add $\pu{0.90 g}$ sodium chloride to $\pu{1000 ml}$ water, the resulting density would be

$$\pu{0.9978 g/cm^3} + \pu{0.0090 g/cm^3} = \pu{1.0068 g/cm^3}$$

The density of the 0.9% saline shows us that $\pu{1000 ml}$ of it contain

$$\pu{1004.6 g} - \pu{9.0 g} = \pu{995.6 g}$$

or

$$\frac{\pu{995.6 g}}{\pu{0.9978 g/cm^3}} = \pu{997.8 ml}$$

water.

This means that the volume of the liquid phase increases when we add sodium chloride to water, what explains the difference in density from $\pu{1.0068 g/cm^3}$ to $\pu{1.0046 g/cm^3}$.


Identity of a green substance formed during electrolysis reaction? I did a short experiment in which I connected a $1.5\:\mathrm{V}$ battery with 2 paper clips into a bowl of water to which I added salt. After a while, I saw a yellowish-greenish cloud inside the water around one of the paper clips, and when I took it out, the part that was in the water was more black than the rest of the paper clip. After a while, some of it turned into a greenish color.
What caused the paper clip to change its color?

Without further details of your equipment, this might be a possible answer. I assume the electrolyte you added was table salt (NaCl).

What you saw as a yellowish cloud was gaseous chlorine which discharged on the clip you connected to the positive electrode (cathode) of your battery. The reason why you did see this, instead of seeing just colorless bubbles of oxygen, is because of something called "overpotential". It's a joint property of the species in your cell and the material and surface termination of the electrode; in your case, the clip (made of wire steel) has a lower overpotential for the discharge of chloride anions (from the salt you added) than oxygen.

Elemental chlorine is a very reactive oxidant, and you had a thin chlorine film around your clip electrode. The reaction between chlorine and iron in aqueous solution most probably yielded iron (II) chloride tetrahydrate which is the reason you saw the greenish colour of the previously submerged fraction of the clip.


Current in a galvanic cell I was looking through chemistry textbooks to find out how to determine how much current a galvanic cell should generate and what affects that current value. However, I did not find anything: textbooks just talk about the EMF of a cell.

I wanted to know what determines how much current a cell generates. Certainly, the amount of electrons produced externally contributes to the current produced by the cell. However, what about internally? Batteries have internal resistance. Is that true of cells too (it should be because batteries are made up of many cells)? What determines amount of internal resistance: does the type of electrolyte(s) affect it. If a cell uses a weak electrolyte (like acetic or carbonic acid) will it cause more internal resistance then if something like sulfuric acid (strong electrolyte) is used?

What if a cell used weak electrolyte as the cathode electrolyte and the strong electrolyte was the anode electrolyte, will that cause more internal resistance due to there being less ions in the cathode compartment as not all of the electrolyte has been dissociated into ions (thus, the cathode cation cannot be reduced)?

Although voltage can be calculated from the electromotive series, maximum current is harder to predict due to a number of factors, some of which you've mentioned.

Electrolyte conductivity, which varies with components, concentration, temperature, etc. As you state, weak electrolytes are less conductive. Resistance of a composite electrolyte would be the sum of its components.
Cell polarization, which is an effect at the electrode/electrolyte interfaces. The causes are buildup of gas bubbles, concentration gradients that develop in the electrolyte, etc.
Deterioration of the electrodes and/or electrolyte, particularly under high current drain.
Of course, as @MaxW states, a major factor is cell geometry: area of electrodes and separation due to insulators and electrolyte.

For these reasons, though the maximum current can be calculated from the resistances and EMF, in actual use, cells often fall far short of their theoretical potential (no pun intended).

The resistance of a battery is primarily the sum of that of the cells. Since cells are joined with metallic connections, the cell-to-cell resistance is usually negligible (though I've used poorly designed battery holders with high-resistance spring contacts that severely limited the current, and that damaged the plastic holder due to ohmic heating).


Oxidizing properties of sulfuric acid: can it oxidize hydrochloric acid to chlorine? Can $\ce{H2SO4}$ oxidize $\ce{HCl}$ to $\ce{Cl2}$? I know that $\ce{KMnO4}$ can do it, but my question is considering $\ce{H2SO4}$. In sulfuric acid, sulphur is in +6 oxidation state and in $\ce{KMnO4}$, manganese is in +7 oxidation state. But sulphur is more electronegative than manganese, so I guessed that, $\ce{H2SO4}$ should be equally good oxidizing agent as $\ce{KMnO4}$.
However, I got to know that $\ce{KMnO4}$ can oxidising $\ce{HCl}$ to $\ce{Cl2}$ but $\ce{H2SO4}$ can't. Why it is so? I would also like to know which one has more oxidizing power and why?

Concentrated sulfuric acid cannot oxidise chloride or fluoride, see here for a detailed explanation here

Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize fluoride or chloride. In those cases, only the steamy fumes of the hydrogen halide—hydrogen fluoride or hydrogen chloride—are produced. In terms of the halide ions, fluoride and chloride are not strong enough reducing agents to reduce the sulfuric acid. This is not the case for bromides and iodides.


Can we use HPLC to purify an organic reaction product? I know HPLC is a form of column chromatography that uses high pressure to pump a sample mixture/eluent through a column. However, I'm not sure if this technique is being used to purify and isolate organic products? After doing some research, I found that automated flash chromatography systems use LPLC technique rather than HPLC. I can't understand why high pressure can't be used in these automated systems?

For semiprep applications, you need larger columns, because you cannot just increase the concentration (at least not a lot) with a normal one. You can get along with an analytical column using an automated system running over the weekend, but that too fails if you want to scale up by another order of magnitude.

A HPLC pump that can run at 100x the standard flow rate is likely twenty times as expensive. The column will be much more expensive. The larger volume needs totally different safety considerations. And so on ...

As Waylander commented, you can do preparative HPLC at a pinch and save your day (or your dissertation), but to use it as a standard lab procedure is most often un..sustainable. ;)


Corey -House synthesis vs Wurtz reaction In Corey House synthesis the first step is formation of Alkyl Lithium. why can't we react alkyl lithium with alkyl halide to produce alkanes. Rather we choose to prepare Gilman's reagent? Does it have something to do with the reactivity of alkyl lithium?

As $\ce{R-Li}$ is very reactive ( even more than $\ce{RMgX}$ ). To decrease the reactivity of $\ce{R-Li}$ , it is reacted with $\ce{CuX}$ to form Gilman Reagent which is less reactive than $\ce{R-Li}$ to produce alkanes.

If  $\ce{R-Li}$ reacts directly with $\ce{R'-X}$, there will no formation of alkanes, but interchange reaction occurs.

$$\ce{RLi + R'X --> RX + R'Li}$$


When an expansion of a system does work on its surroundings, is the sign for work negative? This question comes to me from trying to explain the answer $1.67\ \mathrm{L}$ of this question:

A cylinder with a moving piston expands from an initial volume of $0.250\ \mathrm{L}$ against an external pressure of $2\ \mathrm{atm}$. The expansion does $288\ \mathrm{J}$ of work on the surroundings. What is the final volume?

It means the change in internal energy is negative, so you are on the right track. Because the system is doing work on the surroundings, that $288\ \mathrm{J}$ is a loss for the system and a gain for the surroundings.

One way to intuitively work your way through this is to think about what the signs on the terms must be in order for the problem to make sense.

You know the final volume is greater than the initial (it's an expansion of the system into the surroundings), so $\Delta\mathrm{V}$ has to  be positive. You know that $w = -\mathrm{p_{ext}}\Delta\mathrm{V}$ is negative, because $\mathrm{p_{ext}}$ is positive. Thus, the sign of $\Delta\mathrm{E}$ is negative.

You want to solve

$$\Delta\mathrm{E} = -\mathrm{p_{ext}}\Delta\mathrm{V}$$

for $\Delta\mathrm{V}$ and add that to your initial volume of $0.25\ \mathrm{L}$.

Solving for $\Delta\mathrm{V}$:

$$-288\,\mathrm{J} = -2\ \mathrm{atm}\times\Delta\mathrm{V}$$

$$\Delta\mathrm{V} = 144\ {\mathrm{J\over atm}}$$

and then converting units

$$\Delta\mathrm{V} =
\left(144\ {\mathrm{J\over atm}}\right)\cdot
\left({\mathrm{L\cdot atm}\over 101.325\ \mathrm{J}}\right) =
1.42\ \mathrm{L}$$

we now solve for the final volume as (spoiler alert):

  $$\mathrm{V_{final}} = 0.25\ \mathrm{L} + 1.42\ \mathrm{L} = 1.67\ \mathrm{L}$$


How exactly does the electrolysis of water work? When we pass electricity through water it breaks into hydrogen and oxygen gas.

$$\ce{2H2O -> 2H2 + O2}$$

What I couldn't understand is how the electrical energy is converted to breaking the bonds of water molecule. I've searched a lot but couldn't find any satisfactory answers on how the energy conversion happens.

Few thoughts :

The fast moving electrons from the battery hit the water molecules breaking them into hydrogen and oxygen ions.
I think above idea is incorrect as pure water is an insulator and blocks passage of electrons. Then it might have to do with the electric field that exists through out the water between the anode and cathode. Does this electric field somehow transfer energy to the water molecules and split them ?
Above idea makes sense to me as we can think of each $\ce{H2O}$ molecule as a dipole $\ce{2H+-O^2-}$, so it can interact with electric field.
I feel all above ideas are silly and there is some sophisticated way to reason this which I don't know yet.

Electrolysis of pure water is very difficult, but adding only a small amount of ions make the process easily achieved. In most places, there are enough minerals in the water  that the ionic strength or conductivity of the water is  great enough for electrolysis to effectively occur without needing to add additional ions to the water

Water dissociates into H+ ions and $\ce{OH–}$ ions; the $\ce{H+}$ ions are attracted to the negative electrode (the cathode) and are converted (reduced) to a hydrogen atom ($\ce{H}$) (i.e. $\ce{e- + H+ -> H}$). This is highly unstable and immediately reacts with another hydrogen atom to produce $\ce{H2}$, molecular hydrogen gas.
At the other electrode (the anode), oxidation occurs.  The $\ce{OH-}$ ions are attracted to the positive electrode where they are oxidized to form oxygen gas ($\ce{O2}$) and hydrogen ions ($\ce{H+}$).  However, if chloride is present, it will oxidize (instead of the $\ce{OH–}$ ions) and form chlorine gas, which will then react with the water to form hypochlorous acid.

The cathode reaction is: $\ce{2H+ + 2e- (cathode) →H2 (g)}$
The  anode  reaction  is:  $\ce{2OH– → 4e-(anode) +O2 + 2H+}$
The overall reaction is: $\ce{2H2O → 2H2 (g) + O2 (g)}$


Nomenclature of carbohydrates In the chapter on Biomolecules in my book, the following points are given regarding carbohydrates.

Numbering the C-chain with consecutive rank
The C-atom present in the aldehyde group of aldoses and the terminal C-atom nearest to the keto-group of the ketoses are numbered as the first carbon or C1.
D- and L- Configuration
i) The D- and L- symbols refer to the configuration of monosaccharides. These do not indicate the dextrorotatory or laevorotatory specific rotations.
ii) In case of a monosaccharide, if the asymmetric C-atom of the highest rank is similar to the asymmetric carbon of D-glyceraldehyde (H-atom on the right,-OH group on the left) then the monosaccharide is of D-configuration. If the H-atom is on the left and -OH group is on the right, then the monosaccharide is of L-configuration.
iii) As per convention, if in a monosaccharide unit (expressed by Fischer projection formula), the -OH group falls on the right of the highest ranked asymmetric carbon, then the monosaccharide has D-configuration and if the hydroxyl group falls on the left, it has L-configuration.
Use of + and - sign
D- and L- symbols are followed by '+' and '-' in brackets. These repesent the dextro- and laevo- specific rotation. There is no relation between optical activity and nature of rotation. Example: Both glucose and fructose have D-configuration, but their optical activities are positive (dextrorotatory) and negative (laevorotatory) respectively.

Points ii and iii seem to be contradictory to me. Can someone please shed some light on this. I noticed that point iii is talking about Fischer projection. May be point ii is talking about some about mode of representation. Can someone please explain the paragraph under Use of + and - sign

It's really strange and contradictory info, I don't know what the ii paragraph is talking about. Referencing the source as “my book” makes is hard to verify and e.g. check if there is some illustratory graphics.

To the last paragraph: The cited information would lead to e.g. “D(+)-glucose”, which is not consistent with typographical details of IUPAC nomenclature of carbohydrates (or IUPAC nomenclature of organic chemistry). The correct name would be (+)-ᴅ-glucose.[1]

  2-Carb-4.5.  Optical rotation

  If the sign of the optical rotation under specified conditions is to be indicated, this is done by adding (+)- or (−)- before the configurational prefix. Racemic forms are indicated by (±)-.

  Examples:

  ᴅ-Glucose or (+)-ᴅ-glucose
  ᴅ-Fructose or (−)-ᴅ-fructose
  ᴅʟ-Glucose or (±)-glucose

Fructose is an example of laevorotatory ᴅ- sugar (therefore is also called levulose).

The stereochemical configuration (especially when it is of one of many chiral atoms, and in sugars, sometimes it's the one that forms the acetal group of the major cyclic form, in some sugars not) is not directly related to the physical optical rotation.

References:

Nomenclature of Carbohydrates (Recommendations 1996), 4.5. Optical rotation


How does one properly do this alkali melt? (Pfleger's indigo synthesis) TL;DR: Needing advice on the best procedure (e.g., preparation of reagents, and the order of addition) for an alkali melt/fusion used in Pfleger's indigo synthesis.

I need some advice on how to carry out the alkali fusion/melt step in Pfleger's industrial synthesis of indigo. SciFinder comes up with nothing but vaguely-worded patents and two horrible documents.

I would love to find a recently-written procedure on how it is made industrially (e.g., indigo for jeans, or however it would be made when ordering it from Sigma-Aldrich), but I can't find anything like that. All original documents (e.g., Pfleger's patent) lack important details.

This scheme (#3) outlines the synthesis that is used industrially; a melt of NaOH, KOH, and NaNH2 forms indoxyl out of phenylglycine, which then spontaneously (in the presence of oxygen) oxidatively couples using a radical mechanism to form indigo. The part that I'm having trouble with is finding a good procedure on how to do the melt.

I've seen arrow-pushing mechanisms that show NaNH2 being used for either deprotonation or for imine formation (depending on the literature source), while another source says that it's merely a dehydrating agent, Le Châteliering the reaction forward.

Since I don't want to waste my NaNH2 by having it react with water, I figured that I should dehydrate my NaOH and KOH first. So far, I have ground NaOH and KOH together and brought the temperature up to about 200 °C, at which point they become a liquid. Putting this on a vacuum, I see a fair amount of water come off, and when I remove the vacuum, the mix solidifies (but is still soft, probably because I haven't removed all of the water).

I'm assuming that I would then add the sodium/potassium salt of phenylglycine, then the NaNH2. In the Nigerian papers, they simply grind 5:5:1 NaOH, KOH, and NaNH2 (seemingly in open air), then add nitrobenzene and reflux. I'm leery about grinding NaNH2 in open air (safety guidelines suggest grinding it under a hydrocarbon solvent, though perhaps I'm being overly cautious?), and I'd expect that the NaNH2 would react with the water in the NaOH and KOH.

Assuming that nobody has a good literature procedure for this, I'm hoping to get some guidance on things like preparation of reagents, and the order of addition. I'm also interested in any other advice that anyone may have.

On academia, I found a reading of Industrial Dyes: Chemistry, Properties, Applications by Wiley (2003) that on first view sheds some light on the processes and chemistry.  Looks like an interesting secondary reference, rather than a Organic Synthesis.  I took some screen photos from there

and

and

however, the concluding bibliography may turn out sour

as many of these references are likely only in German.

Perhaps one may trace either in these, or more recent work (than 1890) citing them provide a more detailed procedure.

Addition: On archive, I found a "medium old" book about dye chemistry by Fierz, initially published in 1949, that comes pretty close to your intended alkaline melt; except omitting sodium amide:

Perhaps other old books dedicated to this topic equally tell more about these secrets, than modern textbooks?


Enantiomerically pure ketone reaction under acidic conditions gives 1:1 mixture of diastereoisomers? My problem is such:

My thinking is that perhaps the explanation has something to do with the carbonyl group being prochiral and having two faces, each side generating one of two diastereoisomers? I'm not sure...

Carbonyls are able to undergo keto-enol tautomerisation, this is accelerated in the presence of acid/base.

This phenomenon means that alpha-chiral carbonyls are generally regarded as configurationally unstable, as they are prone to racemisation, although in the absence of acid/base the extent to which the alpha-chiral carbonyl exists in its enol form is often negligible (the C=O, C-H combination is ever so slightly more favoured than C=C, O-H in a lot of cases).

The enol form of the carbonyl (the middle structure in my diagram below) is planar. When the enol tautomerises back to the ketone form, the proton may be transferred to either face of the planar structure, leading to both diastereomers being formed, as you describe.


Photocurable Polymer with no oxygen functional group or whose the structure is well known I'm experimenting with the PDMS. I want to change the material with no oxygen containing functional group.

Is there a UV curable polymer with no oxygen atom?
Preferably cheap.

Or

Please recommend any UV curable or photoresist with well-known morphological property and molecular (polymer) structure.

I'm not sure if this addresses all your criteria,  but acrylonitrile is polymerisable by light and is oxygen free.


Relation between enthalpy and internal energy for chemical reaction at constant volume Question

The combustion of benzene(l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}.$ Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: $(R = \pu{8.314 J K-1 mol-1})$

Answer

 $\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

 $\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
 $\Delta n_\mathrm g = 6- 7.5 = -1.5$
 $\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
 $\Delta H = \Delta U + \Delta n_\mathrm g RT$
 So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question
When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:
$$\mathrm dH = \mathrm dU +\mathrm d(PV)$$
At constant volume:
$$\mathrm dH = \mathrm dU + V\mathrm dP$$
In the question, heat of combustion is given as $\pu{-3267.6 kJ},$ while solving for constant volume $\mathrm dU$ is given as $\pu{-3267.6 kJ}.$
Why is $\mathrm dH = \mathrm dE$ at constant volume?

why is dH=dE at constant volume?

It isn't, and nowhere in the problem or answer is this implied.
First of all some definitions.
For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume:
$$\Delta U = q_V ~~~~~\text{constant volume} \\
\Delta H = q_p ~~~~~ \text{constant pressure}$$
There is a simple integral form for the relation between internal energy and enthalpy that follows from the definition of the latter:
$$\begin{align} \Delta H &= \Delta (U+pV) \\ &= \Delta U + \Delta (pV) \end{align}$$
This relation always holds (it is a definition).
For this particular problem, the key to the next step is to use the ideal gas law, $pV=n_gRT$. This is allowed only if that equation of state holds for the substances in question (here carbon dioxide and oxygen). Inserting the ideal gas relation gives
$$\begin{align} \Delta H &=  \Delta U + \Delta (n_gRT) \\ &= \Delta U +  RT\Delta n_g\end{align}$$
where T is assumed constant. This is the expression used to find the solution. Clearly it shows that the change in enthalpy and internal energy are different.
You could derive this last expression slightly differently to emphasize the differences in the energy and enthalpy. The enthalpy change includes the heat input required to perform work against a constant pressure atmosphere and isothermally. In the case of an ideal gas, this heat input is
$$q_p-q_V = \Delta q = -w_{pV} = p\Delta V = p \Delta (n_g RT/p) = RT \Delta n_g$$
Note that for an ideal gas the enthalpy is a function of T only, so it's true then that a change in enthalpy is the same under constant V or p, provided you converted equal amounts of substance at same T. The same holds for the energy (since for an ideal gas it is also a function of T only).


The real structure of difluoro(dioxo)xenon (XeO2F2) I have a problem regarding structure of difluoro(dioxo)xenon ($\ce{XeO2F2}$). I have posted a image above. my teacher is saying that the structure B is correct as there is very less repulsion in structure B that is what I too think but many of the sources and books have written the structure A to be correct and me as well as my teacher is in a doubt that why structure A is correct not structure B as there is an angle of 180° between oxygen atoms. If I am wrong please guide me and also please tell me what are factors should I look at for judging a correct shape for a compound.
Any hint will be very very helpful to me .

Xenon is a noble gas and thus does not need to form any bonds to complete its octet. When such a bond is formed, this is known as a hypercoordinate structure: it contains more bonding partners than the most simple of simple Lewis structure theories would allow. Thanks to Linus Pauling, for many years chemists have been trying to invoke d orbitals to explain this bonding situation when the actual case is far more trivial and can be explained with simple octet considerations.

The two oxygen atoms which are depicted as double-bonded in the d orbital model are actually connected by a single bond. If you want, this bond can be formed by oxygen keeping its six valence electrons in three orbitals and using the forth, empty orbital to bond with a single orbital of xenon. Both electrons can be described as originally donated by xenon and a good way to draw it in the Lewis formalism is to use charge separation:

$$\ce{\overset{-}{O}-\overset{2+}{Xe}-\overset{-}{O}}$$

This leaves the two fluorine atoms. They actually form a slightly more complex bond known as a 4-electron-3-centre bond (thus fluorine only bonds to xenon in neutral compounds in multiples of 2). For a deeper explanation of the 4-electron-3-centre bond, see this previous answer of mine. In a Lewis formalism, this can only described with resonance structures as shown below:

$$\ce{F-\overset{+}{Xe}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Xe}-F}$$

This bonding situation means that the fluoride atoms must occupy opposite positions with $\angle(\ce{F-Xe-F}) \approx 180°$. Thus, A is the correct depiction of  $\ce{XeO2F2}$ which was confirmed in a neutron diffraction experiment by Peterson, Willett and Huston.[1]

The paper determines the $\ce{O-Xe-O}$ angle as approximately $105°$. This is somewhat in-between a pure p orbital and an $\mathrm{sp^2}$ description of xenon, $15°$ from either. At first approximation it would, however, be fine to assume unhybridised xenon and derive the bonding situation from there with each oxygen bonded to one of the p orbitals and the two fluorines bonding to the third.

Reference:

[1]: S. W. Peterson, R. D. Willett, J. L. Huston, J. Chem. Phys. 1973, 59, 453–459. DOI: 10.1063/1.1679827.


Better understanding the pyrolysis reaction I'm no chemist, but much of my work revolves around biochar.  I've always understood it with the following rough approximation:

$$
\ce{CH2O (dry plant matter) + heat -> black carbon + H2 + CO + CH4 + hydrocarbons}
$$

I've got two questions:

Is the above a decent approximation?  In what ways is it misleading (as all approximations always are)?
A colleague recently told me that methane is only formed in this reaction when there is water vapor present -- he said that it mixes with CO at high temperature to form the methane.  Is this true?  Will one not get any methane from oven-dried biomass that is exposed to heat in the absence of oxygen?

The presumption here is that the biomass is purely sugars: glucose- and xylose-based starches and cellulose.  The essential components of lignin, which make up about 20% of the biomass, is phenols and propane. You may need to rebalance based on those additional components.


What is the numbering of substituted benzene derivatives according to IUPAC nomenclature? According to the lowest locant rule shouldn't the numbering of the compound 1 start from the carbon at the 4th position, thus giving the IUPAC name as 1,2-dimethyl-4-hydroxybenzene?

For compound 2 they could have prioritized the functional group, but they have prioritized the lowest locant rule instead.

There are several criteria for the numbering of locants. The most important simplified criteria for the numbering for substituted benzene compounds are:

lower locants for the principal characteristic groups, which are expressed as a suffix or implied by the name of the functional parent compound (If there is a choice, the principal characteristic groups are assigned in accordance with the order of seniority of classes of compounds.)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The first compound given in the question has the functional parent compound phenol. The name ‘phenol’ implies the presence of the characteristic group $\ce{-OH}$. Since there are no other characteristic groups and low locants are assigned first to the principal characteristic group, the locant ‘1’ is assigned to the  substituent group $\ce{-OH}$. Therefore, the systematic name for this compound is 3,4-dimethylphenol.

The second compound given in the question does not have any principal characteristic group – neither expressed as a suffix nor implied by the name of a functional parent compound. Therefore, low locants are assigned to the prefixes ‘chloro’, ‘methyl’, and ‘nitro’, all considered together as a set in a series of increasing numerical order. Thus, the systematic name for this compound is 2-chloro-1-methyl-4-nitrobenzene and neither ‘1-chloro-2-methyl-5-nitrobenzene’ nor ‘3-chloro-4-methyl-1-nitrobenzene’ since the locant set ‘1,2,4’ is lower than ‘1,2,5’ or ‘1,3,4’.


What are the active ingredients of shampoo, and how do they work? After seeing one too many TV commercial about the fantastical properties of different high-profile shampoo products all "with documented effect" it suddenly struck me that I never actually researched what the active ingredients of a shampoo really is.

So what are the active ingredients of most shampoos, and how do they work?

Is there is any significant difference between popular brands?

Is there any significant difference between cheap and expensive brands?

Will they actually heal your hair?

Shampoo has a few common components: Surfactants, preservatives and a number of additives meant to give the product desirable properties.
When you wash your hair, you're removing debris and buildup in two steps; the polar components (such as salts and aerosols that land in your hair) are washed away by the water, and the non-polar by the surfactants.
Surfactants are molecules where one side is polar and the other is non-polar. Things with similar polarity like to group together, and the opposite happens with polar/non-polar pairs (I'll call this the 'Like attracts Like rule, it's very important to this particular problem).
These surfactants can group together with  each other to form little spheres known as a micelle if there are enough around. This level is known as the critical micelle concentration.

(source: ekshiksha.org.in)
Because the non-polar parts of the surfactants come together, the center of the micelle is also non-polar and can hold small droplets and pieces of other non-polar substances like the sebum that builds up on your hair naturally over time.
Almost all shampoos use a nearly identical set of surfactants. Common examples are sulfates (ammonium lauryl sulfate, Sodium laureth sulfate, sodium pareth sulfate, sodium lauryl sulfate) and/or polysorbates (PEG20 and PEG80).
There are also additives that thicken it up, keep it from going bad and make that rich wonderful lather that we've all come to know and love.
What you are paying for aside from the above is mainly botanical or synthetic additives meant to soothe or beautify. The best way is to look up these ingredients on a case by case basis and decide if they would benefit you. I would say "be wary"; a large number of ingredients provide virtually zero added benefit for your money. Some work extremely well for certain people; you never know until you try.
Always read your labels! If you don't know what something is, google it!


Cp - Cv for a real gas I am trying found a relation between cp - cv for a real gas. I know how to calculate for an ideal gas, but when I try to do the same for a real gas I stopped at some point and I don't know how to continue.
The EOS used is Van der Waals equation.
If someone could help me, please, I am glad for it.

You need to show that $$\left(\frac{\partial U}{\partial V}\right)_T=-P+T\left(\frac{\partial P}{\partial T}\right)_V$$


What liquid other than water that is safe to put on your skin will provide the most evaporative cooling? Suppose that you are in some extremely hot and humid place where your sweat just doesn't evaporate well enough to cool you down. You can then consider putting a liquid other than water on your skin, e.g. methanol would be an excellent liquid with a decent latent heat and high rate of evaporation at room temperature, but this is poisonous. Ethanol is better, but it will also cause problems as you need to use large quantities.

The question is then what liquid that can be used safely, will provide the best cooling.

There aren't many liquids other than alcohols which you can easily use. Liquid butane, for example, would be a coolant, but it would probably cause frostbite (and is flammable, not to mention from cylinders it stinks).

If you can't use ethanol for any reason, you can use isopropanol (rubbing alcohol): its cooling properties are similar. Note that you should only use isopropanol outdoors, because not only does it smell nasty in high concentrations, it is slightly toxic. Make sure to use the water solutions (~75%), as gels will leave sticky and nasty residue upon evaporation.

Even ethanol is not completely "safe", because it has the potential to cause drunkenness, even in vapor form.


Why don't molecules of ionic compounds exist? My book says:

  As single ions of a metal are not associated in the solid with single ions of a non metal, separate units of ionic compounds do not exist. It is, therefore, wrong to talk of a molecule of an ionic compound.

I know ionic solids exist in form of crystal lattice but why can't we isolate a single molecule of ionic compound?

Ions are electrostatically bound.  One need only separate them with inegrated inverse squared distance work.  If you mix multiple ionic compounds in solvent, a melt, gas phase, the original paired associations are spontaneously scrambled.  Benzene plus xylene does not give you toluene.  "Separating" benzene is not electrostatic attraction.

Quaternized glycine, $\ce{^{+}N(CH3)3-CH2-C(=O)O^{-}}$, is betaine.  It is ionic, but the ions are covalently bound into a molecule.  Where are the discrete formula units in alumina?  You can draw lines to connect closest atoms, but those are not bonds. (Note fractional atom counts for atoms embedded in planes, edges, and corners.)


Why can't I balance this reaction algebraically? $$\ce{aCu + bHNO3   ->  cCu(NO3)2 + dNO + eH2O}$$
  (Source)

I calculated:
\begin{align}
a &= c &(\ce{Cu})\\
b &= 2e &(\ce{H})\\
b &= 2c + d &(\ce{N})\\
3b &= 6c + d + e &(\ce{O})
\end{align}

And I tried substituting $a= 1$, but it couldn't solve the whole equation. How can I solve this algebraically?

Here's how to solve the system in python:

>>> import numpy as np
>>> stoich_mat = np.array([[-1, 0, 1, 0, 0],  # copper balance
                           [0, -1, 0, 0, 2],  # hydrogen
                           [0, -1, 2, 1, 0],  # nitrogen
                           [0, -3, 6, 1, 1],  # oxygen
                           [1, 0, 0, 0, 0]])  # impose constraint that a == val
>>> rhs = np.array([0, 0, 0, 0, 1])  # last number is assumed val of a
>>> np.linalg.solve(stoich_mat, rhs)
array([ 1.        ,  2.66666667,  1.        ,  0.66666667,  1.33333333])

Thus this solution says:

$$\ce{Cu   +   2.67 HNO3   ->  Cu(NO3)2    +   0.67 NO   +    1.33 H2O}$$

Multiplying through by 3:

>>> rhs_times_3 = np.array([0, 0, 0, 0, 3])
>>> np.linalg.solve(stoich_mat, rhs_times_3)
array([ 3.,  8.,  3.,  2.,  4.])

$$\ce{3 Cu   +   8 HNO3   ->  3 Cu(NO3)2    +   2 NO   +    4 H2O}$$

If your question is how computers (or people) can solve systems of linear equations, you can read up on Gaussian elimination and related techniques.


What are "tensides"? I was reading about chemicals that control pore size of polyurethane foams and there it was mentioned that "tensides" decrease pore size. What are tensides? I assume it's the opposite of a surfactant, that is, it increases surface tension, but that is just my guess. I have no idea what it actually is.

I tried to Google it but strangely I just got bunch of "surfactant" results even through I wrote it just tensides. Some German site when translated from German to English changed the word tenside to surfactant, so I don't know if it is different class of chemicals or just a German word for surfactant. From the original article where I saw the word, it was written in a way as to suggest that they both are different from each other.

I don't remember to have seen the term tensid outside the context of the German language.

The term is typically applied to agents that decrease the surface tension (hence tensid) of water. Note that the logic behind the English equivalent surfactant is just the same.

So, in summary tensid and surfactant mean the same.


How does 1,3-dichloro-1,3,5-triazinane-2,4,6-trione (dichloroisocyanuric acid) generate hypochlorous acid when dissolved in water? Can someone draw the reaction that leads to the generation of HOCl when dichloroisocyanuric acid (structure shown below) is dissolved in water?

$\ce{N-Cl}$ bond is polarized toward nitrogen: $\ce{N^--Cl+}$ This means, that any chloramine can produce $\ce{HOCl}$ in water by mean of hydrolysis of $\ce{N-Cl}$ bond.

$$\ce{R2NCl + H2O -> R2NH + HOCl}$$

The mechanism is more subtle matter. Basically, it should be one of $S_N$ mechanisms, but it is hard to choose right one here ($S_N1$ or $S_N2$). The central atom in mechanism here should be $\ce{Cl}$, and $\ce{R2N-}$ is leaving group.


Why are alkyl halides more soluble in alcohol water solution rather than pure water? Why are alkyl halides more soluble in alcohol water solution rather than pure water?

I could not find any satisfactory explanation using Google.

As others have mentioned, it generally comes down to "like dissolves like".  Water is highly polar, low molecular weight alcohols like ethanol a bit less so.  By increasing the ratio of alcohol to water in a solution you are effectively making the solution less polar as a whole.  There are effectively fewer hydrogen-bonding opportunities in an alcohol water solution than in pure water.  In general, alkyl halides tend to fall in the low-to-intermediate polarity range.  Methylene chloride for example is only very slightly soluble in water, which makes it an excellent candidate for performing extractions.  By adding alcohol to water, you are giving the solution as a whole a more non-polar character, and increasing the potential for favorable interactions between the alkyl halide and the solvent.


What is the actual voltage to the standard hydrogen electrode and why is it used? Why is the standard hydrogen electrode chosen to be of 0 voltage, what is its actual voltage and how could that be calculated experimentally?

Also, why is it that we cannot measure the voltage or potentials of the range of electrodes and we must use the standard hydrogen electrode?

The voltage given for any reduction is totally meaningless in the absence of either a comparison to another reduction potential or an oxidation potential. Absolute energies are totally meaningless. Only differences matter.

As usual then, we need references. For example, heat of formation of standard Gibbs energy if formation are set to 0 for elements in their standard forms at standard state.

Likewise, for reduction potentials, we need a reference, and that reference if the standard hydrogen electrode, whose standard reduction potential is 0 by fiat. All other potentials can be determined by comparison.


Equilibrium Constant and Surface Area At equilibrium, the rates of the forward and backward reaction become equal. In a reaction
A(s) giving B(g) + C(g)
Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected. By the time the rate of the backward reaction becomes equal to that of the forward reaction, more of A must have converted into B as compared to how much it did when there was less surface area. Since the activity of a pure solid is unity, it doesn't appear in the expression for K. So K must have increased. The equilibrium constant on the other hand depends on the ∆G standard for a reaction. The standard ∆G is unchanged, so K must be unchanged. How does one resolve these two?

In a reaction A(s) giving B(g) + C(g) Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected.

This is an assumption that you're making and it is thermodynamically and geometrically questionable.  If increasing the surface area of $\ce{A}$ increases the rate at which it sublimes/gasifies to form $\ce{B + C}$, it means that the reactive molecules of $\ce{A}$ are only those at the surface, in which case it is misleading to denote the reactant as $\ce{A(s)}$.  It is really $\ce{A(surface)}$ which is reacting.  Similarly, if the rate of the reverse reaction is not really dependent on the surface area and surface geometry of $\ce{A(s)}$, then it implies that very quickly molecules of $\ce{B}$ and $\ce{C}$ would condense to form a nanocrystallite of $\ce{A}$ that is one molecule big.  These nanocrystallites of $\ce{A}$ would have, collectively, nearly infinite surface area.  So the backward reaction creates surface area under the assumptions you have made.  You have to account for that created area.

Solid-phase reactions are complicated by phenomena such as crystal nucleation and other mechanisms of crystal growth.  Treating solids as uniformly reactive materials with unit activity is inappropriate for understanding such phenomena.  Another way to say it is that the assumption that solids have unit activity and don't appear in expressions for $K$ or $\Delta G$ is an assumption, one that is inapplicable when dealing with reactions whose rates depend on surface area.


Why does Le Chatelier's principle not apply to pure solids and liquids? If we had a homogeneous reaction of pure liquids only, would Le Chatelier's principle still apply ?

Also, wouldn't increasing the amount of solid in a heterogeneous reaction mixture at equillibrium result in more product, therefore indicating a shift in equilibrium?

Le Châtelier does apply to pure solids and liquids - what happens is that pure solids and liquids experience almost no change in concentration[*] in chemical equilibria, so for all intents and purposes they are not affected by Le Châtelier considerations.

Heterogeneous systems

Imagine a system with two phases; for instance, a gas phase and a solid phase, and we have some sort of equilibrium between them; for instance if we put water in a closed container (remember that Le Châtelier only properly applies to closed systems), it reaches an equilibrium with its own vapour:

$\ce{H2O(l) <=> H2O(g)}$

Here we have two phases: a liquid phase and a gas phase. The molecules in both are the same but in different aggregation states, which crucially differ in a fundamental behaviour regarding volume: in the liquid, molecules occupy a fixed volume each in certain conditions. If you have twice the number of molecules, you get twice the volume, and the density remains constant. So, if the equilibrium above is displaced to the left, the concentration of water molecules in the liquid phase doesn't change: the size of the liquid phase does. You simply get more volume of liquid. The same is true about solids.

For the gas it's different: the gas phase will always occupy the whole free volume in the container, and what changes is its pressure (which is to say, its concentration). If you double the number of gas molecules, the volume doesn't change - it still fits the whole container - but their concentration doubles.

This is a simplified picture; formally, the concentration of real solids and liquids is not exactly constant. However, it is much, much less sensitive to conditions than gases or solutes, so for all practical purposes it's a constant. See the next section for a numerical example that also applies here.

Dilute homogeneous systems

When we deal with dilute homogeneous systems (only one phase), the concentration of the solvent can be safely treated as constant because it's so much larger than anything else that changes simply don't impact it.

Imagine that you are applying Le Châtelier to the self-ionisation equilibrium of water; if you have a sample of pure water, a very small fraction of the molecules dissociate per the following reaction:

$\ce{2H2O<=>H3O+ + OH-}$

If you have a pure water in equilibrium and you add $\ce{OH-}$ ions, Le Châtelier predicts that this will displace the equilibrium to the left, and some of the previously dissociated water molecules will associate again to form $\ce{H2O}$.

However, water itself is the medium of the reaction, so its concentration is pretty much constant. Let's calculate some actual figures to get an idea. The self-ionisation constant is $\mathrm{K_w=10^{-14}}$, so for a chemically pure sample of water we'd get these concentrations:

$[\ce{H3O+}] \approx 10^{-7} \mathrm{M}$

$[\ce{OH-}] \approx 10^{-7} \mathrm{M}$

$[\ce{H2O}] \approx 55.55 \mathrm{M}$

Now, imagine we add a lot of $\ce{OH-}$ ions - and we make their concentration a thousand times bigger than before - that will have a big impact on the concentrations of both $\ce{H3O+}$ and $\ce{OH-}$. When we reach equilibrium we'd have

$[\ce{H3O+}] \approx 10^{-10} \mathrm{M}$

$[\ce{OH-}] \approx 10^{-4} \mathrm{M}$

Note that this means that 99.9% of the dissociated molecules have been displaced to the left and formed water. However, this number of new water molecules (around 0.0000000999 M) is simply irrelevant compared with the previous concentration of water (55.55 + 0.0000000999 is still 55.55 for all practical purposes). Even further, as with heterogeneous systems, even if the addition of more water were relevant, its impact would be increasing the volume of the phase, not changing its concentration.

That's why the concentration of water isn't even included in the self-ionisation constant - yes, formally the equilibrium depends on $[\ce{H2O}]$, but since that's pretty much always the same, it's easier to factor it into the constant and write the equilibrium as $\mathrm{K_w=[\ce{H3O+}][\ce{OH-}]}$.

[*] More precisely, activity


How does conjugation affect reactivity? From this topic on MOs of butadiene vs ethene, it is clear that when two ethene molecules are combined to extend the conjugate chain, the HOMO is raised in energy and the LUMO is lowered in energy. This implies that the resultant compound is more reactive towards nucleophiles(Lower LUMO more easily matched in energy with HOMO of Nu), and also towards electrophiles (Higher HOMO more easily matched in energy with LUMO of E).
In "Why chemical reactions happens" by Keeler and Wothers, "The conjugative effect" is mentioned as something that decreases reactivity (see p. 163-165):
"The delocalization of a lone pair from X into the carbonyl π system decreases the reactivity of the carbonyl towards nucleophiles by raising the energy of the π LUMO" .
The second argument seems to contradict my original argument, so what am I missing?

Perhaps extending conjugate system by pi bonds, empty p-orbitals or filled single orbital (LP) etc. have different effects on the reactivity?

I think an extensive answer explaining the effects of conjugation on reactivity(towards nucleophiles and electrophiles), perhaps also touching on the common statement that "presence of conjugation stabilizes a compound", would be very helpful for the site.

EDIT: Seeing as my original question is very broad, i would appreciate if someone could simply explain the apparant contradiction between the previous post and the Keeler&Wothers quote.

The effect of conjugation on electrophilicity and nucleophilicity of chemical species varies depending on the type of conjugation. There is no contradiction between the two cases you have presented in your question because in each case, the conjugation at work is different.

The conjugation which occurs in conjugated polyenes has been well elaborated upon by Philipp (and thus, I shall not elaborate further), in the answer you have linked to. This type of conjugation occurs as a result of interactions between parallel p orbitals in a chain of carbon atoms.

The other sort of conjugation you mentioned reading in the book by Keeler and Wothers is an interaction involving an interaction between the nonbonding molecular orbital, which holds the lone pair, localised on ${X}$ (e.g. $\ce {NH2,Cl, Br}$) and the antibonding $\pi$ molecular obital of the $\ce {C=O}$ bond, which is more localised on $\ce {C}$. This is a $\ce {HOMO-LUMO}$ interaction between the two orbitals and electron density is donated into the antibonding $\pi$ MO from the nonbonding MO on $X$. This interaction produces yet another set of bonding and antibonding molecular orbitals. Since the bonding MO produced is of a lower energy than the nonbonding MO on $X$, the lone pair is stabilised and its reactivity is decreased. Also, since the antibonding MO produced is of a higher energy than the previous antibonding MO of the $\ce {C=O}$ bond, the LUMO is raised in energy as a result, decreasing reactivity of the carbonyl group towards nucleophiles. One of the consequences of this conjugation is the lower reactivity of amides towards nucleophilic addition (Clayden et al., 2012).

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.


Number of Sigma Bonds in this Structure Containing a Ring (This question was on my last gen. chemistry exam.)

I counted 12 sigma bonds, but the answer was 13. I'm taking this course online so there is no explanation given. Could anyone offer a quick explanation on how this structure could give 13 sigma bonds?

There're are 13 sigma bonds there (count the red lines),

You've clearly tried to attempt this yourself, so good on you (and by virtue of that, it's equally clear that you counted the bonds too hastily).

I counted (quickly, because it looked simple enough) and got 12 too at first. Counted a little slower the second time (I was actually going to attempt to prove to you that the number of bonds is 12), turns out the ring is not entirely conjugated, so one carbon atom in it actually has two hydrogen atoms bonded to it. ;-)


Why Benzene is a good solvent even while it is not polar Consider for a moment, ether, THF, benzene and hexane.

Why is the solubility of many organics in these solvents in the following order

THF > ether / benzene > hexane

What makes benzene a good solvent for many solutes when compared with hexane even while both solvents lack a permanent dipole moment.

Benzene is a good solvent for many organic compounds, the term good solvent to me does not mean a nice solvent with a good all round profile.

To me the perfect solvent would be non toxic, not flammable but combustable, very resistant to radiation, inert to nitric acid, be easy to distill, be cheap and easy to obtain. It would also have to be able to dissolve a useful range of different things.

Benzene is more able to dissolve many organic substances than a saturated hydrocarbon such as hexane, this is becuase the pi cloud of the benzene ring is able to interact with the pi clouds of various solutes. For example polystyrene will be solvated better by benzene than by hexane for this reason.

This good ability to dissolve things is why I regard benzene as a good solvent. It is a good solvent to dissolve things but often a bad thing to work with in the lab. Benzene has one special use, as a NMR solvent. One of my favourite tricks is if my student can not understand a proton NMR spectrum due to overlapping peaks. I tell them to repeat the measurement in d6 benzene. Often a lot of the chemical shifts for protons change, with a little luck the resulting spectrum is more easy to understand. Sometimes you need to look at both the chloroform and benzene solution NMR spectra to be able to understand enough of the spectrum to understand what your structure is.

While benzene is an unpopular solvent due to the health issues and flammability, there are plenty of aromatic solvents which have better health and fire profiles. For example Solvesso 150ND is an aromatic kerosene which is marketed by Exxon, this has different solvent properties to the aliphatic saturated kerosene marketed by Statoil under the name of "solvent 70".

This can have some advantages, if I want to dilute the ionic liquid aliquat 336 to 30 % in a diluent and then use this solution to extract metals such as copper, cobalt and iron from aqueous chloride media. Then with solvent 70 as the diluent (diluting agent) then the organic phase is very prone to separate into a layer of ionic liquid with metal complexes as the anions and a diluent rich layer. But when an aromatic solvent such as ethyl benzene or solvesso 150 is used then the organic phase can hold a lot more metal before this unwanted phase splitting occurs.

The aromatic solvent (diluents) tend to be more toxic than the saturated aliphatic (alkane) diluents. If you look at "A comparison of two methods of recovering cobalt from a deep eutectic solvent: Implications for battery recycling" by F.J. Albler et. al., Journal of Cleaner Production, 2017, volume 167, pages 806-814 it contains an assessment of leaving a puddle on the floor of an unventillated room is made for both solvent 70 and ethyl benzene / toluene is made. The solvent 70 level in the air is likely to pose a far smaller threat to health than the aromatics. But aromatics like solvesso 150 are likely to pose a far smaller carcinogenic threat than benzene.

It is important to not regard all alkanes or aromatics as being the same, hexane is a popular solvent in chemistry but it is very toxic. In the body it is converted to hexane2.5-dione which is very bad for you. But the other longer alkanes do not have the same harmful effect. You also need to consider the exposure route. For example a puddle of diesel fuel on the floor of a garage is far less harmful than exposure of a person's hand to a high pressure spray of diesel which injects it into the person's body.


How to read a chromatography calibration curve? I am a high school student who has been studying a bit of chromatography. I came across calibration curve during class and I trying to figure how to read it and use it? I have no clue how to read off the graph nor do I really now the use of it. Can someone please explain what information I can extract from this graph and possibly provide an example if that's fine?

I assume your teacher explained the HPLC separation. If injected a mixture of four ions you will get four peaks. Each peak has an area, which is proportional to the concentration of the substance.

Imagine you wanted to determine the concentration of chloride ions in your tap water. What you would do is that you will prepare several known concentrations of chloride ions using $\ce{NaCl}$ solutions, say $0$ to $\pu{350 ppm}$ (look at your calibration chart). Inject them one by one, and measure the peak area, let us say the second peak is that of chloride ion. The peak shape is Gaussian in real work but you can "estimate" them as triangles. Recall triangle's area is very easy to calculate.
First step: Plot the peak area for concentration. What you get is a calibration curve.
The next step is to find a mathematical equation that fits this data.
Fortunately it is linear in your case of the form $y=mC$ where $y= \text{peak area}$, $m=\text{slope}$, $C=\text{concentration}$ in $\pu{ppm}$.
Now you would inject your tap water in the HPLC, you do not know the concentration ($C_1$) but you do know the peak area.
The last step is called interpolation. You go from the known area, and read the corresponding concentration axis. Viola! Now you know the chloride concentration in your tap water!
Alternatively, you use the equation, $y=0.1312C$, so I know the area of the tap water chloride injection, say, it was $\pu{40.0 units}$, so $C_1 = 40/0.1312 = \pu{304.9 ppm}$.
In typical tap water it is around $\pu{100 ppm}$, I just made up those numbers.
Bonus: Rationalize the units of the peak area on the $y$-axis! Why it is conductivity times minutes?


Why do we use [H⁺] and [OH⁻] in balancing redox reactions instead of using [H₃O⁺] and [OH⁻]? For what reason is it universally agreed upon that we use $$\ce{H+ + OH-}$$ ions in balancing water on either side of the reaction, but not the $$\ce{H3O+}$$ ions?

It is by far not universally agreed upon. My year 10 chemistry teacher taught me to use $\ce{H3O+}$ in balancing. The result is a larger number of water molecules on the other side of the equation but everything else is the same.

Many chemists will prefer $\ce{H+}$ because it is shorter and simpler, that’s all.


Pure substances have only one kind of atom I read in a book recently that no substance is ever totally pure, even if we take great care to separate it from other substances.
And also that pure substances contain only one kind of matter.
But I actually have some doubts related to both of the notes.
I would rather number...

Pure substances contain only one kind of matter? True for elements but what about compounds? Are compounds not pure substances? They are homogenous at composition, right?
No substance is totally pure? Reactive elements are found in combined state, but what about less reactive or maybe radioactive?
As well, can't we obtain any substance with 100% purity?

I did surf the questions but did not get any reasonable explanation. Also, please do correct me where I am wrong.

Most of the issues can be resolved with precise definitions.

Let's start with purity. Purity is relative. In many cases (eg table salt) 99% is pure enough. What this means is that for all practical purposes we don't care about the remaining impurities. Chemists tend to describe something as "pure" when there is less of any contaminant than matters for what they are doing with the chemical. Sometimes purity of very high levels matters. Semiconductors behaviour depends on very precise doping with very small amounts of other substances. The wrong contaminant even at levels of 1 atom per million is a problem. So the industry have worked out good ways of making extremely pure silicon, for example. These techniques have been used to make some of the purest substances ever (the silicon sphere now being used to redefine the Kg is pure to parts per billion atoms.) Strictly speaking it isn't pure in the sense of having absolutely 100% of all its atoms as silicon, but that is as pure as it needs to be for its purpose. So, strictly, nothing is ever absolutely pure, but it doesn't matter.

The word "substance" is a bit vague. In chemistry it could be a compound an not just an element. Purity for a compound means that all the molecules in the substance are the same (to the level of purity that matters). Having more than one type of atom doesn't make it impure as long as all the molecules are the same.

Then there are some homogeneous mixtures that are hard to separate by any simple means, some metal alloys for example. We wouldn't normally describe these as "impure" but chemists might use special terminology to describe them (alloy rather than compound for example).

And as for "Pure substances contain only one kind of matter" it depends what you mean by "matter". Chemists would prefer the term "molecule" to "matter". Matter can be interpreted different ways (does it mean atoms or molecules: it's unclear). "Pure substances contain only one kind of "molecule" would be an unambiguous statement chemists would agree with.


1-5 dicarboxylic acid to lactone with SOCl2? I came across a reaction whereby a 1-5 dicarboxylic acid is converted to lactone with $\ce{SOCl2}$. I was wondering what the mechanism of this was.

I know that $\ce{SOCl2}$ converts carboxylic acids to acyl chlorides. Is it the case that one carboxylic acid undergoes this transformation, then the oxygen from the $\ce{OH}$ of the other carboxylic acid attacks the acyl chloride, making the $\ce{Cl}$ a leaving group?

Is this the case or am I missing something?

The simplest 1,5-dicarboxylic acid is glutaric acid (pentanedioic acid in a more systematical naming). If you react it with $\ce{SOCl2}$, you will indeed intermediately arrive at a compound that has one carboxylic group and one acid chloride group. Then, as you suggest, the other carboxylic acid can attack the acid chloride and $\ce{HCl}$ is eleminated.

However, the resulting product is not a lactone but rather glutaric anhydride; a cyclic acid anhydride. To convert it to a lactone you would need to remove exactly one carbonyl group. Whether this is done before or after the reaction with $\ce{SOCl2}$ does not matter, but before sounds easier to me.


Methods used to compare two spectra in order to determine the effects of irradiation of sample I would like to compare two absorption spectra (or interferograms) and conclude whether between these two there are statistically significant differences at particular wavelength intervals. At the moment, I have data of two experiments that look like this:
    # A tibble: 6 x 5
      t     x1     y1     x2     y2
  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
1 3999. 0.0124 0.0132 0.0122 0.0113
2 3998. 0.0125 0.0130 0.0122 0.0116
3 3997. 0.0122 0.0131 0.0122 0.0113
4 3996. 0.0121 0.0136 0.0122 0.0114
5 3995. 0.0124 0.0139 0.0122 0.0122
6 3994. 0.0125 0.0141 0.0122 0.0129

The first column represents the wavenumber, the x columns represent the absorbance of sample and the y columns represent the absorbance of irradiated sample (before and after).
I was wondering whether I could compare these data (x and y) as time series and if so, what could be the method to quantify the differences, if any, between the samples before and after irradiation. Maybe it's already been done and there is somewhere some information as to how to compare the spectra if the wavenumber is interpreted as time ($x$ axis).
I did the t-test in R and in both experiments the null hypothesis could not be rejected, although for the second experiment (x2, y2) the $p$ value was much lower than for the first. If I average the x and y, and then plot both data, I see that there are visible differences at certain wavelength intervals. But how can I check for sure the differences between spectrums?
Here is a project with similar experiments by Zezell et al. [1]. For statistical analysis they use ANOVA and Tukey's test, but how do I do it for the vectored data?
Reference

Zezell, D. M.; Benetti, C.; Veloso, M. N.; Castro, P. A. A.; Ana, P. A. FTIR Spectroscopy Revealing the Effects of Laser and Ionizing Radiation on Biological Hard Tissues. J. Braz. Chem. Soc 2015. DOI: 10.5935/0103-5053.20150246.

The table below illustrates a possible screen, values computed displayed are rounded to four decimals.  The suggest is to use the observations prior the intended irradiation ($x_1, x_2$) separate from the observations after the intended irradiation ($y_1, y_2$).  Per wavelength

compute for $x$ and $y$ the arithmetical mean value
determine the standard deviation of this sample (e.g., $x_1$ and $x_2$), the one pocket calculators sometimes label with $\sigma(n-1)$.  The table doesn't include it, but you may compute the confidence interval with a reasonable $t$ value.  Because there were only two realizations, the degree of freedoms ($f = n - 1$) equates to 1, and an understandably high $t_{1, 0.95} = 12.71$ to map the interval of

$$\bar{y} - \frac{t \cdot \sigma}{\sqrt{n}} \le \bar{y} \le \bar{y} + \frac{t \cdot \sigma}{\sqrt{n}}$$

subtract the arithmetical mean values from each other to determine the effect of the irradiation.  Expect positive as well as negative values to occur.
compute the standard deviation for this effect.  Because the effect is computed as a difference of $(\bar{x} - \bar{y})$, the error propagation of is a sum of the corresponding standard deviations. Or add the corresponding halfes of the confidence intervals about $x$ and $y$ instead.

I have no hands-on experience with R.

| lambda |     x1 |     y1 |     x2 |     y2 | mean_x | stdev_x | mean_y | stdev_y | mean_x - mean_y | stdev_x + stdev_y |
|--------+--------+--------+--------+--------+--------+---------+--------+---------+-----------------+-------------------|
|   3999 | 0.0124 | 0.0132 | 0.0122 | 0.0113 | 0.0123 |  0.0001 | 0.0123 |  0.0013 |          0.0000 |            0.0015 |
|   3998 | 0.0125 | 0.0130 | 0.0122 | 0.0116 | 0.0123 |  0.0002 | 0.0123 |  0.0010 |          0.0001 |            0.0012 |
|   3997 | 0.0122 | 0.0131 | 0.0122 | 0.0113 | 0.0122 |  0.0000 | 0.0122 |  0.0013 |          0.0000 |            0.0013 |
|   3996 | 0.0121 | 0.0136 | 0.0122 | 0.0114 | 0.0122 |  0.0001 | 0.0125 |  0.0016 |         -0.0003 |            0.0016 |
|   3995 | 0.0124 | 0.0139 | 0.0122 | 0.0122 | 0.0123 |  0.0001 | 0.0130 |  0.0012 |         -0.0007 |            0.0013 |
|   3994 | 0.0125 | 0.0141 | 0.0122 | 0.0129 | 0.0123 |  0.0002 | 0.0135 |  0.0008 |         -0.0011 |            0.0011 |


Why should all the acid be used up when making zinc sulphate crystals? So to make zinc sulphate crystals you add granulated zinc to heated sulphuric acid. However, you have to make sure all the acid is used up and you know so when the zinc stops reacting and producing bubbles "hydrogen". You then filter out the solution and set it on a water bath to crystallise it. The question is : if all the acid is not used up in the reaction, what will happen when I start to evaporate the solution of zinc sulphate?

I don't think you have really explained the experiment correctly. I'd guess that you're using a diluted solution of sulfuric acid.

$\ce{Zn + H2SO4 ->[H2O] ZnSO4 + H2 ^}$

It would be difficult to have exactly the right amounts of zinc and sulfuric acid, so one or the other will be in excess.

If zinc is in excess the solution can be filter to easily remove the
excess Zn.
If Sulfuric acid is in excess then when all the water is gone you get zinc
sulfate which is wet with sulfuric acid. The sulfuric acid won't evaporate
on a water bath like water.


Why does F have a higher ionization energy than O? The book I'm working from says F has a higher ionization energy than O because the nucleus has more protons so there is a higher attraction on the electrons. Isn't that balanced out by the higher number of electrons or is it because the atomic size is smaller?

At first approximation, we need to consider the orbital we are extracting an electron from. Both oxygen and fluorine (and nitrogen, carbon, boron and neon) have a $\mathrm{2p}$ orbital as their highest occupied, so we expect to ionise from that. If you draw a trend of the ionisation energies, you observe that the energy rises from boron to nitrogen, is lower for oxygen and rises from oxygen to neon. Disregarding the slight decrease at oxygen at first approximation we can say that it becomes increasingly hard to ionise an electron from a $\mathrm{2p}$ orbital with higher element number.

This is because essentially the shape and size of the $\mathrm{2p}$ orbitals are (close enough to, at first approximation) equal. So the wavefunction of the electrons we are removing does not change significantly. However, in their vicinity we have a nucleus which changes charge. For boron, we have an effective nuclear charge of $+1$ or $+3$ (depending on whether you count the $\mathrm{2s}$ electrons as shielding or not). For every step we take and every move we make to the right, the effective nuclear charge ($n(\ce{protons}) - n(\ce{core~electrons})$) will increase by one. We need to counter that nuclear charge to remove an electron which, naturally, because increasingly harder the stronger the charge holding it back.

The slight drop in ionisation energy at oxygen warrants a separate explanation. If you observe closer, you realise that the first three p-electrons (boron to nitrogen) are added as α spins while oxygen is the first element to obtain a β spin electron. A single β electron at first approximation has three degenerate orbitals it can populate. This is an unfavourable situation — compare with the Jahn-Teller effect in coordination chemistry that exists to counteract this situation. Thus, the energy required to remove this seemingly extraneous electron is lower than it would be just looking at effective nuclear charge. On the other hand, three α spins are more stable since the system is no longer degenerate. Thus it is slightly harder to remove an electron from nitrogen than it should be.


Are there any other elements that get less dense in their solid state? I was learning about how water, because of its hydrogen bonds, actually gets less dense as it goes into its solid state - I was just wondering, what other elements do this? Are they similar to water?

P.S. I want to make sure that I understand why ice is less dense than water in its liquid state, so please correct me if I am wrong.

In water, the hydrogen bonds have the ability to move around and bounce with other hydrogen bonds. They also have the ability to break and reform. Their ability to move around and bounce around allow them to be more 'closely knit'. But, as they get cooler, they, the molecules, loose their kinetic energy and thus stop moving around and become more framed and rigid. They loose the ability to come closely together. And thus, ice is less dense.

What I don't understand also, is why making water cold makes the hydrogen bonds spread out. Why don't they just freeze in the position they are in, instead of spreading out?

Your terminology is a little off: water isn't an element.

There isn't really a simple single explanation of why some solids are less dense than their associated liquids. The general explanation (which doesn't really explain much) is that the solid has a structure and, sometimes, that structure takes us more space than the average structure that occurs in the liquid.

In water, ice has a definite structure where each oxygen sits at the approximate centre of a tetrahedron with two bonds to hydrogen and two hydrogen bonds to the hydrogens of other water molecules. In liquid water the structure is far more fluid and averages out to be slightly more dense that the (more ordered) solid. Exactly why this is true in this case requires a lot more than could fit into an answer here. If water was frozen extremely quickly you might be able to freeze the structure in the liquid; normal cooling allows enough time for the bonding to become more organised.

But the phenomenon isn't unique to water, though the specifics will be very different for other substances. Among elements Gallium, for example, also expands on freezing (don't store it in glass bottles: see this other answer here).

Other examples of elements are germanium and silicon. Acetic acid also forms a less dense solid. In all cases the substances form a relatively ordered crystal structure which just happens to be less dense than the (less ordered) liquid phases.


Why is this molecule considered a cis isomer of 2-bromobut-2-ene? I know that we have two CH3 groups on one side and restricted rotation around the double bond but aren't we supposed to have two similar groups on the other side as well ? Does this mean that we only need one side of the molecule to have identical groups/atoms in order to call it a cis isomer ?

With the same reasoning as provided here, I recommend that you name this compound (2 E)-2-bromobut-2-ene.

Using (cis/trans) may be useful around an isolated double bond with only two substitutents. While in the example presented by you the two methyl groups are on the same side of the double bond, they are not the only ones to be considered here. Triple and quadruple substituted dienes should be named in accordance to the CIP rules only.

Edit by 02-Jan-2019:
Agreeing with a comment by @Loong, the previously suggested name, "(E)-bromo-2-butene", was replaced by "(2 E)-2-bromobut-2-ene".  This edit removes simultaneously the ambiguity about the position of the bromine, as well as the double bond.  The interested is referred to the illustrated examples of rule R-7.1.2 of the corresponding Blue Book here.


What is "mer", the first ingredient in a type nail polish? The first ingredient in a type of nail polish is "mer". What is this? Is it some kind of chemical? How is it used? How does it work?

In the organic lab, sometimes we would run reactions that produced a tar-like substance in the reaction flask as an undesired by-product.  It was just a gooey polymer, and it happened often enough that this substance was just referred to as "mer", short for polymer.  "Mer" was therefore a fairly common term used in many (most?) labs.
Since "mer" is listed as the first ingredient in nail polish, that indicates that it is the major component in the formulation.  If you google "nail polish", you'll find a Wikipedia article. Its opening paragraph states:

Nail polish consists of an organic polymer with various additives

I suspect they use the term "mer" to refer to whatever polymer is the main proprietary ingredient in their formulation.  The Wikipedia article also notes that

Nail polish consists of a film-forming polymer dissolved in a volatile organic solvent. Nitrocellulose that is dissolved in butyl acetate or ethyl acetate is common.

Once the volatile solvent evaporates, you are left with a hard polymer coating on your fingernail.