What is the significance of "magic numbers" of atoms in nanoparticles? In gold nano particles, as the synthesis method improves to form smaller and smaller nanoparticles, they tend to congregate in magic numbers.

I want to know if these magic numbers occur in other metals besides gold, and whether they contribute to stability, reactivity, catalysis, increased surface area, and their lattice structure? In short, what are the main differences between magic and non-magic nanoparticles?

As indicated in the comment above, "magic numbers" (i.e., atom counts with much higher stability) have been known in atomic clusters of many types for a long time.

Each magic number has a specific geometry. Some are non-spherical, so it's hard to make generalizations about surface area or lattice structure.

A very famous non-gold example is buckminsterfullerene ($\ce{C60}$). Multiple groups had seen the signature of 720 Daltons in mass spectra of soot for years before the official discovery. There are other magic numbers for $\ce{C_n}$ clusters.

In nanoparticles and nanoclusters, this is an active area of research, but the main difference is stability. The so-called magic-numbers are simply more stable. (Case in point, the thiol-termated gold nanoparticles were created by heating up the samples until only the most stable ones survived.)

While I'm not an expert, I suspect the increased stability likely means less reactivity, lower catalytic activity (i.e., the species is more stable, less reactive so less likely to activate).

I'll give a comparison with Hückel's 4n+2 $\pi$ electron rule. We know that benzene is aromatic and has increased stability relative to the 5-membered and 7-membered conjugated hydrocarbon rings. It's less reactive and thus requires specific reaction chemistry.


What is the relationship between the constant of equilibrium (Kc) and the position of equilibrium? I understand that the constant of equilibrium is a ratio of the concentration of the products to the reactants and that this value allows one to estimate whether the position of equilibrium lies to the right (in which there will be more product at equilibrium) or lies to the left (where there will be more reactant at equilibrium). I'm also aware that Kc is affected only by a change in temperature and not a change in concentration, pressure etc.

I know the position of equilibrium is affected by changes in concentration, pressure and temperature (unlike Kc which is only affected by change in temperature). I understand Le Chatelier's Principle and how the position of equilibrium shifts in order to minimise the change, and that this maintains the equilibrium constant.

My problem arises when trying to relate the two. The position of equilibrium shifts so that the equilibrium constant will be maintained however I don't understand how this works or why!
When we say that the position of equilibriums shifts (to the right for example), what does this actually mean? I know that the amount of product will increase but what is actually happening. Is the rate of the forwards reaction increasing? In other words, what actually is the position of equilibrium? If Kc is the ratio of product concentration to reactant concentration then what actually is the position of equilibrium?

Thank you in advance!

What you are probably referring to is the degree of dissociation $\alpha$ when you ask about the position of equilibrium. At a given temperature the equilibrium constant is just that, constant, but as the pressure changes, say for a gaseous reaction then $\alpha$ changes.An example shows this.

In the gas phase reaction $\ce{A2 <=> 2A}$ with partial pressures $p_{A_2}$ and $p_{A}$ respectively, the  equilibrium constant is $\displaystyle K_p=\frac{p_A^2}{P_{A_2}}$ and is defined at standard conditions of 1 atm.  If $P$ is the total pressure then the two partial pressures are $\displaystyle p_A=\frac{2\alpha}{1+\alpha}P$ and $\displaystyle p_{A_2}=\frac{1-\alpha}{1+\alpha}P$ where and  so after some simplification $\displaystyle K_p=\frac{4\alpha^2}{1-\alpha^2}P$.

Thus, as the equilibrium constant is constant at any total pressure $P$, the degree of dissociation must change as the pressure changes to compensate, thus the 'position of equilibrium' changes.

[Notes: The amount of A$_2$ reacted is $1-\alpha$ and the amount of A produced is $2\alpha$, thus the mole fractions of each are $(1-\alpha)/(1+\alpha)$ and  $(2\alpha)/(1+\alpha)$ and so the partial pressure are as given above. The total amount needed for the mole fraction is $1-\alpha+2\alpha=1+\alpha$. You can derive related expressions in terms of concentrations.]


What is the rationale for calling something as an 'inorganic [organic compound]'? While studying the main groups of p-block, I have seen many compounds being labeled as an analog to some popular organic compound. For example, borazine is termed as 'inorganic benzene', while boron nitride has been called 'inorganic graphite'.
Now, my question is, whether these monikers (if that's what they may be called?) reflect just the structural likeness, the chemical similarities or, perhaps, both?
Moreover, are there some other examples which illustrate this?

I would not call graphite organic, but there is no clear-cut way of defining organic and inorganic compounds.
To your question, the pairs of compounds are isoelectronic. That means that if we assume that molecular orbitals (MOs) arise from atomic orbitals, the corresponding MOs in the two compounds have the same occupancy. In these cases it arises because the number of electrons in the dimer BN is the same as in the dimer CC. Another example of isoelectronic compounds would be N$_2$ and CO. Being isoelectronic, they do share some properties, e.g. both N$_2$ and CO have a triple bond. If only the valence electrons are identically occupied, we call them isovalent, which would include e.g. H$_2$O and H$_2$S.
The bonding of benzene and borazine differs in that not all sites are equivalent, but the form of the MOs is broadly similar and the same applies to boron nitride. Similarity in MOs is then reflected in similarity in reactivity. (Even though they are not that similar in reactivity.)


Calculation of concentration at equilibrium for extraction The equation for equilibrium of solvent-solvent extraction is:
$$\ce{PhOH(toluene) <=> PhOH (aq)}$$
An aqueous solution contain $\pu{1.5 \times 10^{-2} M}$ phenol and is shaken with the same volume of toluene. Determine the equilibrium concentration of $\ce{PhOH(toluene)}$. $K_\mathrm{D} = 14$

I have done lots of similar questions and each time I think I've gotten the method of how to do these types of questions it seems like the method doesn't apply anymore. In this question, I am really confused because I initially solved it and got the right answer. But when taking a closer look, I noticed that my calculations were wrong, and did it again in the correct way (or at least what I think is the correct way), but I got the wrong answer.
My attempt at solving the problem
Given that the aqueous solution had the concentration $1.5 \times 10^{-2}$, I made the following table, where $n_a$ is what I want to calculate.
$$\begin{array}{c|c|c|}
 & \text{PhOH(toluene} & ⇌ & \text{PhOH (aq)} \\ \hline
\text{t=0} & - & &\ \pu{1.5 \times 10^{-2} M}\\ \hline
\text{n equilibrium} & n_a & &n_0-n_a \\ \hline
\text{c equilibrium} & \frac{n_a}{V} & & \frac{n_0-n_a}{V}
\end{array}$$
My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator? and the aqueous phase in the denominator? Meaning that if my reaction is written as above, do I still have the organic phase divided by the aqueous phase or do I follow the rule of products over reactants. If I follow the rule of products over reactants, I get the correct answer, but if I take org/aq which is the formula our professor refers to, then I get the wrong answer.
Which way is the correct way to solve a problem like this? All help is truly appreciated!

I'd like first to answer your question:

My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator and the aqueous phase in the denominator?

The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio ($K_\mathrm{D}$) as follows (also see the Goldbook):

Partition Ratio ($K_\mathrm{D}$): The ratio of the concentration of a substance in a single definite form, $\mathrm{A}$, in the extract to its concentration in the same form in the other phase at equilibrium, e.g. for an aqueous/organic system.
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}$$

Thus, if you have used equal volumes of organic and aqueous phases, the equation you can set up to solve is:
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}= \frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{[\mathrm{PhOH}]_\mathrm{aq, initial}-\mathrm{PhOH}]_\mathrm{org,eq}} \\ =\frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{\pu{1.5 \times 10^{-2} M}-[\mathrm{PhOH}]_\mathrm{org,eq}}= 14 \tag{1}$$
When you solve the equation $(1)$ for $[\mathrm{PhOH}]_\mathrm{org,eq}$, you get your answer:
$$[\mathrm{PhOH}]_\mathrm{org,eq} =\frac{\pu{1.5 \times 10^{-2} M} \times 14}{15}= \pu{1.4 \times 10^{-2} M} $$
Hence, $[\mathrm{PhOH}]_\mathrm{aq,eq} = \pu{1.5 \times 10^{-2} M} - \pu{1.4 \times 10^{-2} M} =  \pu{1.0 \times 10^{-3} M}$
Note: If you perform extraction in different volumes of aqueous and organic phases, you should read this Wikipedia article to get help for solution.
References:

N. M. Rice, H. M. N. H. Irving, M. A. Leonard, "Nomenclature for liquid-liquid distribution (solvent extraction) (IUPAC Recommendations 1993)," Pure & App. Chem. 1993, 65(11), 2373-2396 (DOI: https://doi.org/10.1351/pac199365112373).


How many moles of the substance are in 2.5 g of lead? How many moles of the substance are in 2.50 g of lead?

What I did:

Atomic mass of Pb: 207.2 amu (or g?)

Applying unitary method, 1 mol of Pb -> 207.2 g; so, x -> 2.5 g

$x = \frac{2.5}{207.2}= \pu{0.0121 mol}~\text{of}~\ce{Pb}$

Shouldn't it be like this?:

1 mol of Pb -> $(\pu{207.2 amu} \times1.66\times(10^{-24}))~\pu{g}$ so x -> $\pu{2.5 g}$

Do not mix amu with mass of mol of substance! In your last formula you've forgot about Avagadro number.

One Pb atom weighs $\pu{207.2amu}$ or $\pu{207.2amu} \cdot \pu{1.66\times10^{-24}g}$.

But one $\text{mol}$ of Pb weighs $=\pu{207.2amu}\cdot \pu{1.66\times10^{-24}g}\cdot6.02\cdot10^{23}=\pu{207.2g}$


Nitrogen and an unknown gas is leaked into a common vessel... The gaseous mixture contains 0.4 moles of nitrogen. Molecular mass of the other gas is? For 10 minutes each, from two identical holes, nitrogen and an unknown gas is leaked into a common vessel of 3L capacity at 300K. The resulting pressure is 4.15 bar. If the gaseous mixture contains 0.4 moles of nitrogen, then molecular mass of the unknown gas is?

attempt: According to Graham's law of effusion, $$\frac{r_1}{r_2}=\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$$ where $V_i$ is the volume of gas effused. Let $V_1$ and $M_1$ be volume and molecular mass of nitrogen. $$V_1=22.4\times0.4=8.96;M_1=28$$
Substituting, $${V_2}=8.96\sqrt{\frac{28}{M_2}}$$
and $V_1+V_2=3$, so $$8.96(1+\sqrt{\frac{28}{M_2}})=3$$
This is not possible since $\sqrt{x}\ge0$.
What am I doing wrong?

Calculate total moles using the ideal gas law and solving for $n$:

$$n = {4.15\;{\rm bar}\cdot 3\;{\rm L}\over {0.0831\;{\rm L}\,{\rm bar}\,{\rm K}^{-1}\,{\rm mol}^{-1}}\cdot 300\;{\rm K}} = 0.5\;{\rm mol}$$

Given 0.4 mol $\ce{N2}$, moles of unknown gas = 0.5 mol - 0.4 mol = 0.1 mol.

Rate of effusion of $\ce{N2} = r_{\ce{N2}} = 0.4\;{\rm mol}/10\;{\rm min} = 0.04\;{\rm mol}\cdot{\rm min}^{-1}$.

Rate of effusion of unknown ${\rm X} = r_{\rm X} = 0.1\;{\rm mol}/10\;{\rm min} = 0.01\;{\rm mol}\cdot{\rm min}^{-1}$.

Using Graham's law of diffusion:

$${r_{\ce{N2}}\over r_{\rm X}} = \sqrt{M_{\rm X}\over M_{\ce{N2}}}$$

we solve for $M_{\rm X}$:

$$M_{\rm X} = {M_{\ce{N2}}\times (r_{\ce{N2}})^{2}\over (r_{\rm X})^{2}}$$

and plugging in numbers we get the molecular mass of the unknown gas:

$$M_{\rm X} = {(28\,{\rm g}\cdot{\rm mol}^{-1})\times (0.04)^{2}\over (0.01)^{2}} = 448\,{{\rm g}\over{\rm mol}}$$

where the units on the effusion rates have been omitted for clarity and because they cancel.


Which compound is responsible for the borax bead test? When borax is heated, the reaction that takes place is :

$$\ce{Na_2B_4O_7.10H_2O \ce{->} 2NaBO_2 +  B_2O_3 + 10H_2O}$$
And the compounds that are responsible for the colors are $\ce{M_x(BO_2)_y}$ salts, like $\ce{Co(BO_2)_2}$. My question is, does this $\ce{BO_2}$ come from $\ce{NaBO_2}$ or from $\ce{B_2O_3}$ or do both of them react?

A little chemical detective work gives a plausible answer.

Say the metal is cobalt and it's oxidation state is +2.  Then, with the cobalt calcined to the oxide, you have two possible acid-base reactions:

$\ce{CoO + 2 NaBO2 -> Co(BO2)2 + Na2O}$

$\ce{CoO + B2O3 -> Co(BO2)2}$

Both reactions look good, but upon further review the first one gives something we recognize as a powerful base, more likely to be a reactant than a product.  Moreover,  a quick check here reveals that boric oxide has a low enough melting point to liquefy in the flame, favoring the second reaction kinetically.

So the second reaction above is really the only plausible one.


Equilibrium rate constants forward and backwards Correct me If I am wrong, given a reaction at equilibrium, if we say add more concentration of products to the reaction, the rate going backwards would increase to try to get back to equilibrium.

I was wondering if the rate forward stays the same or slows down?

In simple cases, the rate is equal to the rate constant "k" times the concentration.  The forward rate constant and the backward rate constant don't change, but the rate of change (rate constant times concentration) will change as the concentration of a reactant or product changes.


hygroscopicity vs solubility Phenols are hygroscopic and form a red liquid, upon contact with air. Yet, Phenol’s solubility in water is said to be moderate. ( Tilted to the low solubility side) This is explained with the benzene ring being non polar.
Then how can it be hygroscopic? Are solubility and hygroscopicity even related at all, then?

Hygroscopicity is measure of how much the molecule likes to bond  molecules of water, by hydrating/covering charged part of the molecule with molecule/s of water, which relates to it having some polar or charged functional groups, like hydroxyl  group for phenols.

Solubility is about how molecule likes to be surrounded by water, so in this case the structure of whole molecule makes a difference.
Since water is a polar solvent it likes polar solutes. If a molecule despite having one or more polar groups has, for example, a huge part which is not polar, like a carbon chain, then it will not be soluble even though it may be hygroscopic.


Why does raising the surrounding pressure raise the vapor pressure of a liquid? What’s the qualitative explanation behind the fact that an isothermal increase in surrounding pressure raises the vapor pressure of a liquid? The quantitative expression relating surrounding pressure and fugacity is well-known, but I’d like a qualitative one.

The answer lies in that the chemical potential of the liquid increases with pressure.

Quantitatively, the chemical potential changes with a change in pressure from $p_1$ to $p_2$ as follows:
$$\Delta\mu(\textrm{l})=\int_{p_1}^{p_2}V_mdp$$

A qualitative explanation is possible by linking the change in pressure to the concept of chemical potential. The chemical potential difference is the amount of non-pV work (the "chemical" work) which the system can perform. In essence, increasing the pressure means more of the liquid has the potential to transfer into the gas state (the chemical work in question).   You might visualize the transferred molecules as being "squeezed out" of the liquid by the increasing pressure applied by their neighbours.


Cis/trans stereochemistry dictating pKa Why is trans-4-phenylcyclohexane-1-carboxylic acid a stronger acid than the cis isomer?

In the cis-isomer the carboxyl group is axial; and, given the 1,4 distance from the phenyl, why would solvation be any more difficult?

The bulky phenyl group will preferentially occupy an equatorial position. Therefore, the trans acid has the carboxyl group equatorial and the cis acid has it axial, as you have rightly pointed out.

The difference in acidity is to do with the resulting solvation of the carboxylate group. The negatively charged carboxylate group is more strongly solvated by water molecules, and has a larger effective steric bulk, than the neutral, protonated form.

So, there is an increase in the steric bulk of the group upon deprotonation. Such a process would be more favoured if the group is equatorial, i.e. the trans isomer is more acidic.

Of course, in the absence of pKa values, such an explanation is only theoretical in nature. Solvation is a very difficult concept to quantify objectively.


Largest molecule that isn't polymer? What is the largest molecule (synthetic or natural) that isn't a polymer -- let's say, bigger than 3000 dalton?

An old German book about chemical records (Quadbeck-Seeger, H.-J., Ed. Chemie Rekorde; Wiley-VCH: Weinheim, 1997) gives the Cullinan Diamond as the largest found natural molecule and a synthetic diamond of 38.4 carat as the largest made artificial molecule.


Change in internal energy for the heating and vaporisation of water What is the value of change in internal energy at $\pu{1atm}$ in the process $$\ce{H2O(l,$323\ \mathrm K$)->H2O(g,$423\ \mathrm K$)}$$
  given $C_{\mathrm m,V}(\ce{H2O,l})=\pu{75.0 J K^{-1} mol^{-1}}$ and $C_{\mathrm m,p}(\ce{H2O,g})=\pu{33.314 J K^{-1} mol^{-1}}$

  a. $\pu{42.91 kJ/mol}$
  b. $\pu{43086 kJ/mol}$
  c. $\pu{42.6 kJ/mol}$
  d. $\pu{49.6 kJ/mol}$

I tried to use the First Law here. For chemical reaction I found $\Delta n_\mathrm g$ to be $1$ and the heat of the reaction $\Delta H$ is given as $\pu{40.7 kJ mol-1}$. However, I am not able to use the specific heat data given in the question. What am I doing wrong here?

I am getting stuck in phase change reactions, particularly in using $C_p$ and $C_V$ here. I know the First Law will work, but how do we use $C_p$ and $C_V$ in such reactions?

The easiest way to do this problem is to first determine the  change in enthalpy for this constant pressure process.  Do you think you can do that?  Then determine the change in PV between the initial and final states.


What is the simplest way to create a precipitation reaction at home? What is the easiest way to create a precipitation reaction at home, preferably with easily retrievable chemicals?
I need this because I'm making a video for my film education.

Here's one easy way :

Make a solution of salt ($\ce{NaCl}$) in water, and another solution of Silver Nitrate ($\ce{AgNO_3}$), which you can buy easily) in water.

Upon mixing the two, $\ce{AgCl}$ (Silver chloride, a white compound) will instantly precipitate.


Last step of nicotine extraction using DCM After using hydro-distillation method with distilled water and NaOH, separating the extract with dichloromethane, distillate it at 40 °C/104 °F, I'm left with some yellow pale oil.

I'm just wondering if traces of DCM could still be detected in the final product or if I need more steps to purify the extracted alkaloids ?

If you are worried only with DCM, then you should be fine since it is quite volatile. To be sure you have no DCM left in your nicotine leave it for a few days in a closed recipient filled with some dried silica gel or calcium chloride (here in my country those are easily found, even for domestic use), and if you are in a proper laboratory just leave it in the desiccator with the vacuum on for a few hours.

However, nicotine is hard to isolate since other alkaloids from tobacco are also extracted in the most commonly used methods. I don't know which extraction methodology you used but just for comparison, here are two papers with some information for isolating nicotine:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3263647/
https://www.ncbi.nlm.nih.gov/pubmed/19135499


What is the range of sizes for molecular systems? Molecular systems - are in a simple count systems made up of molecules. But almost every system, surrounding us are molecular than. Considering the molecular systems in terms of chemistry compels us, firstly, impose dimensional constraints on the molecular system (which?), molecular system could not be macroscopic (How big?), otherwise you'll have to talk about it as a solid body; secondly, energies for such systems are limited (<1 eV); in third place we will be interested in properties of such system as whole, but not as sum of its parts properties.

So the main question is raised: What is the range of sizes for molecular systems?

Well apportioned, mixed, and cured epoxy glue is a single covalent molecule.  Single crystal white diamond is a single covalently-bonded molecule.  On the small end, a dihydrogen molecule, is 0.7414 Å (ortho or para, your choice to trim the decimal).

The largest human chromosome is ~220 million base pairs, or about 1.5x10^11 g/mol.


Time to effervescent an effervescent tablet Say I have an effervescent tablet and I put in water.  I thought that if I break the tablet in two halves it would dissolve quicker than with the whole table (without breaking), but it actually takes the same time.

What would be the scientific reason of this?

Dissolution in general

Dissolution is normally a function of several aspects of the solid and liquid and one of them is the interfacial area between the solid and the liquid. A common relation for the dissolution rate which is not always exact, but shows all trends/dependencies very well is the following:

$$\frac{\mathrm dm}{\mathrm dt}=A \frac{D}{d}\left(C_\mathrm s-C_\mathrm b\right) $$
where $A$ is the interfacial area, $D$ is the diffusion coefficient of your solid molecules in the liquid, $d$ is the thickness of the solute boundary layer and $C_\mathrm s$ and $C_\mathrm b$ are the concentrations of the solid molecules at surface and in the bulk (this term is the driving force).

Dissolution of a whole or broken tablet

To make a simple calculation for the whole vs. the broken tablet let us do the math.  Assuming that we have a disc-like tablet the surface area is 2 times the frontal area $+$ the area of the sides.

$$A=2 A_\mathrm f+A_\mathrm s $$
For the whole tablet
$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t$

For the broken tablet
$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t  + 2 h D_\mathrm t$

Here I assume that the tablet is cleanly broken in the exact middle of the disc. So you can already see that the surface area is bigger for the broken tablet, but how much will depend on the specifics of the tablet. Rewriting the area we can see how the different aspects of the broken tablet will influence the area:
$$A=\pi D_\mathrm t \left[\frac{D_\mathrm t}{2}+h\left(1+\frac{2}{\pi}\right)\right] $$

From this equation it is clear that if $D_\mathrm t \gg h$ the height term doesn't do much and therefore it will not matter much that the tablet is broken.

For a final check: let's assume some typical numbers now. A typical effervescent tablet is about $D_\mathrm t=2\ \mathrm{cm}$ in diameter and $h=3\ \mathrm{mm}$ thick. This will result in the following surface areas for the broken and the whole tablet: $A_\text{whole}=8.2\ \mathrm{cm^2}$ and $A_\text{broken}=9.4\ \mathrm{cm^2}$.

This means that the difference in dissolution rate will only be about $20\ \%$ which is probably hard to notice due to other effects that disturb the comparison. One such effect is that $d$ depends on the amount of stirring which will be different already when you drop the tablets in the glass in a slightly different way (which is unavoidable, because you have 1 broken and 1 whole tablet).


What is the standard molar weight of Carbon, and why? The mole is defined as the amount of a chemical substance which contains as many representative particles, e.g., atoms, molecules, ions, electrons, or photons, as there are atoms in 12 grams of carbon-12 (12C), the isotope of carbon with relative atomic mass 12 (from Wikipedia).

However, when I go to websites such as http://www.webqc.org/molecular-weight-of-C.html or http://www.convertunits.com/molarmass/Carbon and I ask what is the molar weight of Carbon, I get answers such as:

  Molar mass of C is 12.01070 ± 0.00080 g/mol

When I asked on a forum why this isn't precisely 12, I was told that typically carbon is a mixture of isotopes which have relative atomic masses of 12, 13 and 14. With the ratio of each, the average value comes out to 12.0107.

But where does that number come from?  Is it by definition?  Is it by convention?  Is it a practical number?  Where do the error bars come from?  For instance, I would expect to see different isotope distributions at different geographic locations (or what about the moon, asteroids, Jupiter, etc.).   I tried googling for "standard carbon isotope ratio" but didn't find anything definitive.  Of course then I start to wonder how molar weights of other elements are defined, with varying isotopic combinations, and I have the same question.

As you stated the molar mass is a weighted average based on isotopic abundance, specifically abundance on Earth. We can obtain the molar mass of carbon by averaging the mass of $\ce{^{12}C}$ (about $98.938\%$ abundance) and $\ce{^{13}C}$ (about $1.078\%$ abundance). $$(\mathrm{mass\space of\space \ce{^{12}C}})*(.9889)+(\mathrm{mass\space of\space \ce{^{13}C}})*(.0111)=(12)*(.9898)+(13.00335)*(.0111)=12.0206\text{g}$$ This deviates from your value above because the values I used for the abundance aren't very precise.

In terms of where these abundances comes from and how the error bars are determined, the short answer is that its complicated. The long answers is here in a technical report by IUPAC, particularly sections 1.2-1.4.


Acid-Base Titration, why does for a weak acid indicator, [HA] = [A-] I am reading about indicators and their behaviour as weak acids.

I know that the end-point is when exactly all of the H+ ions from an acid have neutralised the OH- ions of the base.

The book later says

  At the end point of a titration, the indicator contains equal concentrations of HA and A- and the colour will be inbetween the two extreme colours e.g. for methyl orange, the colour at its end point is orange.

It continues to say

  For an indicator HA ⇌ $H^+$ + $A^-$
  At the end-point, [HA] = [$A^-$] and $$K_a = \frac{[H^+][A^-]}{[HA]} = [H^+]$$

I am confused why this is true. why is the expression [HA] = [$A^-$] valid?

In an Acid titration, we want the end-point to be as close to the equivalence point as possible.

Different indicators have different end-points (they will change colour at different pH values).

When an end-point is reached, the colour of the indicator will change with respect to the [$H^+$] concentration of the solution it is measuring.

As [$H^+$] increases, there will be a point when [HA] = [$A^-$] and depending on the indicator used by the person doing the titration, this would be close to the calculated equivalence-point.

Note, the difference in volume between the end-point and equivalence-point is very very small, hence we can approximate the end-point as the same as the equivalence-point in any experiment.


Finding molarities of individual reagents given the chemical equation and total concentration in a container I am given a table that gives me the different total pressures at certain times. The decomposition of phosphine is the reaction:

$$\ce{4 PH3(g) <=> P4(g) + 6 H2(g)}$$

The time/pressure values are:

\begin{array}{rr}
\mathrm{time\ [min]} & \mathrm{pressure\ [Torr]} \\
\hline
0 & 100 \\
40 & 150\\
80 & 167\\
100 & 172\\
\end{array}

This reaction takes place at 950K. I can then find the molarity of the total gas mixture at each time using $$\frac{n}{V}=\frac{P}{RT}$$

I need to know how to get the pressure of the $\ce{P4}$ for each individual time. How would one go about doing this?

The assumption you need to make is that at time zero, only $\ce{PH3}$ is present.  So relative to the moles at time zero, or $N_0$, you should find $\Delta N = N(t) - N_0$, the total new moles formed by the reaction since it began.

The reaction converts 4 moles of gas to 7.

So pressure increases mean the reaction is moving to the right.  Every mole of pressure increase must be $\frac{6}{7}$ due to $\ce{H_2}$ and $\frac{1}{7}$ due to $\ce{P4}$ formation.

So just find the total new moles of gas and multiply by $\frac{1}{7}$.


What chemical should I use to thin UV cure epoxy? The epoxy that I am working with is UV curable with a cycloaliphatic epoxy resin base. I am trying to reduce the viscosity. Whatever I add to thin the epoxy, it needs to be able to evaporate off so that when I cure it, it's just the epoxy remaining. I have used solvents like acetone to thin 2 part thermal cure epoxies, but I am not sure if this is appropriate for a UV cure epoxy or which solvent would be best to use.

Adding solvents that need to evaporate, to your epoxy, can cause pinholing. You need a integratable solvent. Styrene monomer works well. You can get it from a fibreglass supply or by pyrolizing poly-styrene.

The monomer will combine with the epoxy so it won't need to evaporate out of the epoxy.


Does a sulfuric acid/nitric acid mixture dissolve PTFE (ie. teflon)? In our lab, we commonly use a mixture of sulfuric acid and nitric acid to clean borosilicate glassware.  I'd like to clean some PTFE pieces in there, too, but not sure if they will dissolve or not.

Does anyone have any first-hand experience with this, and know whether or not a sulfuric acid/nitric acid mixture dissolve PTFE?

I can't claim to have any personal experience with that sulfuric acid/nitric acid cleaning mixture specifically, but, as far as I'm aware, PTFE is generally inert towards acids. As a testament to that, it is apparently the material used for storing fluoroantimonic acid. I found a chemical resistance chart confirming its stability against strong acids. It is evidently resistant to concentrated sulfuric and nitric acids, as well as aqua regia (a concentrated mixture of nitric acid and hydrochloric acid). I have used the last of those for cleaning PTFE lab equipment before with no problems.

Having said all that, it would be wise to consult the handling and safety data for your particular labware to be certain.


Is there a way to experimentally measure entropy? I've been doing thermodynamic chemistry, and recently focusing on Gibbs Free Energy. Whilst doing calculations using, $$\Delta G = \Delta H - T \Delta S$$ I have been given a table of values for $\Delta H$ and $\Delta S$, and a temperature to work with. I was wondering , as the unit for entropy is $\mathrm{J/K}$ or $\mathrm{J\ K^{-1}}$ how exactly one would experimentally measure entropy/change in entropy, or can you?

The most common way of measuring $\Delta S^\circ$ for a chemical reaction is probably by making a van't Hoff plot. You measure the equilibrium constant $K$ at different temperatures and plot $\ln K$ vs $T^{-1}$.  The $y$-intercept = $R\Delta S^\circ$ and the slope = $-R\Delta H^\circ$.

Another option is to measure $\Delta H^\circ$ by calorimetry and measure $K$ by some other means. Then compute $\Delta G^\circ$ from $K$ and solve for $\Delta S^\circ$


How is temperature constant in enthalpy change formula? I have learnt that temperature sort of indicates the internal energy of a system, or at least this is my intuition. So if temperature is constant, then internal energy is constant as well.
But in the derivation of the following formula
$$\begin{align}{
{\Delta H = \Delta U + \Delta n_g RT}}
  \end{align} $$
T doesn't have a delta, which means it's treated like a constant value, but then wouldn't this make $\Delta U = 0$? In fact most questions in my textbook have the statement "at constant pressure and 298/300 K", but $\Delta U$ isn't 0 despite the temperature being constant.
Am I missing/neglecting something here?

The internal energy of a system is in general not only a function of temperature. This means that holding the temperature constant does not guarantee that the energy will be constant. The statement that the energy can be expressed as a function of only the temperature applies to the particular case of ideal gases.
Assuming that gases are ideal simplifies expressions for the change in internal energy and enthalpy. For ideal gases at constant temperature and composition ($\Delta T=0$ and $\Delta n=0$):
$$\Delta U = 0\\\Delta H = 0 \tag{1}$$
These expressions are not general. They apply to ideal gases at constant temperature and composition.
In general,
$$\Delta H = \Delta U + \Delta (pV) \tag{2}$$
For an ideal gas at constant temperature,  $ \Delta (pV)= \Delta (nRT)= RT\Delta n$
so that
$$\Delta H = \Delta U + RT\Delta n \tag{3}$$
It is also worth remembering that at constant pressure $w=-p \Delta V$ so that  equation (2) becomes $\Delta H = q_p$ (in general for any substance, provided p is constant and only pV work is done). But what is $\Delta U$?
We can show from the ideal gas law that it is only a function of T for a constant amount of a pure ideal gas, and, from the kinetic theory of gases, one can also show that for one mole,
$$U_{m,i} = f_iRT$$
where $f_i$ is a constant that depends only on the type of gas (not on thermodynamic variables). Then
$$\Delta U = \sum_i U_{m,i}\Delta n_i = RT \sum_i f_i \Delta n_i $$
for a mixture of ideal gases.


A dehydrating agent acting as an hydrating agent? We know that $\ce{P4O10}$ is a dehydrating reagent and I might list on 100's of reaction of it's dehydrating nature, mainly orgainc elimination reaction and few inorganic oxoacid dehydrations, but it is acting like a hydrating agent, cf. these:
$$\ce{PCl5 +P4O10->POCl3}$$

  "Phosphorus pentachloride also forms $\ce{POCl3}$ by reaction with water, but this reaction is less easily controlled..."

Reaction with water is not always a hydration reaction.

A hydration reaction is one in which all of the atoms of the water molecule get added to the reactant to form a product:

$$\ce{MX + H2O -> MH2XO}$$

$$\ce{C2H4 + H2O}\overset{\ce{H+}}{\longrightarrow}\ce{C2H6O}$$

$$\ce{P4O10 + 6H2O -> 4H3PO4}$$

The reaction of water with phosphorous pentachloride is more properly a substitution or partial metathesis reaction. The hydrogen atoms in water are not incorporated into the product:

$$\ce{PCl5 + H2O -> POCl3 + 2HCl}$$

This reaction is difficult to control because $\ce{POCl3}$ is reactive with water also,
and under the right conditions can yield phosphoric acid or even $\ce{P4O10}$.

$$\ce{POCl3 + 3H2O -> H3PO4 + 3HCl}$$
$$\ce{4POCl3 + 6H2O -> P4O10 + 12HCl}$$

The reaction of $\ce{P4O10}$ and $\ce{PCl5}$ is thus also a partial metathesis. It is more obvious when you balance the equation. The oxygen and chlorine atoms are rearranging between the phosphorous atoms.

$$\ce{P4O10 + 6PCl5 -> 10POCl3}$$


Interaction between energy and entropy in highly elastic polymers? Consider the interaction of energy and entropy in the highly elastic materials of an ideal polymeric network.

Now Gibb's free energy cannot be used directly $\Delta H = \Delta G + T \Delta S$ where $\Delta S$ and $\Delta H=U+pV$ for the internal energy $U$ such that

$S(R) = -k_{B} \frac{ 3R^{2} }{ 2N l^{2} } + const$
$U_{eff}(R) = k_{B} T \frac{ 3R^{2} }{ 2 N l^{2} } + const $

of which I am unsure.

I think the Gibbs free energy may have some nonlinear behaviour with the material such as some correlations between different terms and more passive energy terms due to elesticity -- this may be described by things such as fugacity.

How can you describe the interaction of energy and entropy in highly elastic polymers?

If I remember correctly, the relationships you have presented are related to the configurational entropy of the polymer chains between cross links of the polymer network.  In these equations, R is the spatial distance between cross links, N is the number of chain segments between cross links, and l is the length of each chain segment.  The smaller the value of R, the greater the number of configurations that the chain can exhibit (i.e., the greater the entropy).  Thus, I believe that there should be a minus sign in your equation for S, because, as R increases, the fewer the number of configurations that the chain can exhibit, and thus the lower the entropy.  The parameter you call U is, I believe the Helmholtz free energy, and is a measure of the stored elastic energy of the polymer network.  As the R gets longer, the polymer chains have been stretched more, and more elastic energy is stored in the chains.  Conceptually, the polymer chains are like springs between the cross links.


Confusion on syn and anti I came across the following compound syn acetaldoxime
However why is it syn because the lone pair having lowest priority is anti to H which is lowest priority on other side.
Please clarify

In an oxime, the syn or anti configuration is determined by the position of the lone pair on nitrogen with respect to the alkyl or aryl group. That is, syn configuration is one in which the lone pair on nitrogen and the alkyl or aryl group are in the same side, while, in anti configuration, they are on opposite sides.


Is there an inexpensive solvent I can use to dissolve thermal paste without scrubbing? I'm removing large voltage regulators from circuit boards which are often coated with a generous amount of thermal grease. I've found that 70% isopropyl alcohol works great for softening this and removing it with some light scrubbing.

The problem I have is that, without a lot of additional effort and scrubbing, the surfaces will often retain a film from the grease. There are also areas which I may have missed or were unable to reach when scrubbing.

This type of grease is often electrically conductive and can cause a short circuit if accidentally brushed over the pins of a chip. It's also just plain annoying when it gets on your hands.

Is there an inexpensive solvent that I can use to dissolve the thermal paste without having to scrub the components? I felt that it would be more appropriate to ask this here on the chemistry stack exchange as I'm also interested in understanding how other solvents would work better for this.

I tried leaving the components in a bag of 70% isopropyl alcohol but this confirms that it only softens it. The thermal paste don't actually free itself from the surface without some mechanical force.

See here for a photo where the surface looks clean but you can see where I scraped my finger across it through the bag and it removed more of the grease.

The color of the component becomes a dull grey due to the grease becoming embedded in the pores of the plastic which is what it's supposed to do. I would like to know if there is a way to simply drop these into a chemical and have the grease dissolve away on its own without damaging the package.

First the sad part - there is no magic compound that will dissolve thermal paste like a champ and leave your electronics alone. If that was no "sine non qua" as they say, we could have suggested a varied assortement of highly aggressive halogenated solvents which we would mix with something else for the extra bang, some hydrogen peroxide perhaps, or the good old fuming nitric acid. Things would happen, rapidly.

As for all-round no-nonsense solvents, isopropyl alcohol is a big favourite, it is fairly environmentally friendly especially if you consider your hands and your tabletop as part of the environment. And of course it solvates a wide range of icky substances that needs removal. As a small step up the agressive solvent ladder would be to consider acetone, "citrus solvent" (basically oils and acid of lemons, it kicks ass) or acetic acid based solvents. But none of them will magically clean the surface, you'll have to use some elbow grease. Or thumb grease, I guess.


Radioactive Decay Process Please bear with my long winded description.

It is classical to model radioactive decay of some particles $A \rightarrow A^\ast$ by the differential equation
 \begin{align}
 \frac{\mathrm dN_A}{\mathrm dt}=-\lambda N_A
 \end{align}
where $N_A$ refers to the number of particles in state $A$ and $\lambda$ is the rate of decay. If we initially start with
$N_0$ number of particles in state $A$ then by solving the above differential equation we see that
\begin{align}
N_A(t) = N_0\mathrm e^{-\lambda t}
\end{align}
gives us an explicit number of particles in state $A$ as a function of time $t$. In particular, we also see that
\begin{align}
\lambda = \frac{\ln 2}{t_{1/2}}
\end{align}
where $t_{1/2}$ is the half-life. However, the model is completely deterministic, meaning no probability required.

However, quoting Wikipedia

  Radioactive decay is a stochastic (i.e. random) process at the level of single atoms, in that, according to quantum theory, it is impossible to predict when a particular atom will decay, regardless of how long the atom has existed.

But collectively, the mean number of particles in state $A$ is well modeled by the above deterministic model. However, the
difficulty here is to articulate the connection between the deterministic model and the stochastic model in a rigorous manner.

Let $T$ be a continuous positive random variable (RV) that denotes the lifetime of a single (microscopic) particle in state $A$. It is natural to make the assumption that in the stochastic model the decay of a particle in state $A$ is independent of the time that the particle spends in state $A$. With this assumption and a little bit of math (which is not important for the discussion), one can show that $T$ has an exponential distribution with a parameter $\lambda$, i.e. the
  PDF of $T$ is given by
  \begin{align}
  p_\lambda (t) =
  \begin{cases}
  \lambda\mathrm e^{-\lambda t} & \text{ if } t \geq 0,\\
  0 & \text{ if } t<0.
  \end{cases}
  \end{align}
  and the cumulative distribution function (CDF) is given by
  \begin{align}
  P(t\geq T) = 1-\mathrm e^{-\lambda t}
  \end{align}
  for $t \geq 0$ and 0 otherwise. In particular, we see that the probability of the particle in the decayed state $A^\ast$ at time $t$
  is precisely $1-\mathrm e^{-\lambda t}$ and the probability in of the particle in the state $A$ is $\mathrm e^{-\lambda t}$. Hence at each fixed time
  $t$ we have a Bernoulli distribution. If we make another assumption that a collection of the radioactive particles decay independently, then we see that given $N_0$ particles at time $t$ the probability that $N$ particles survives and $N_0-N$ particles decay is given by
  the binomial distribution
  \begin{align}
  P(N_t=N) = \frac{N_0!}{N!(N_0-N)!}(1-\mathrm e^{-\lambda t})^{N_0-N}\mathrm e^{- \lambda N t}
  \end{align}
  By standard computations, we also see that the mean of the binomial distribution is $\operatorname{E}[N_t]= N_0\mathrm e^{-\lambda t}$ and the variance is $\operatorname{Var}[N_t]= N_0\mathrm e^{-\lambda t}(1-\mathrm e^{-\lambda t})$ which leads to the relative standard deviation
  \begin{align}
 \frac{\sqrt{\operatorname{Var}[N_t]}}{\operatorname{E}[N_t]} = \frac{1}{\sqrt{N_0}}\left(\mathrm e^{\lambda t}- 1\right)^{1/2}.
  \end{align}

I have two questions.

  Question 1: Looking at the relative standard deviation, it seems to me that for very large time $t$ the deterministic model seems to be a poor approximation of the stochastic model since the relative standard deviation grows exponentially in time. Is my assertion valid?

I actually disagree with the above calculation but have no valid argument against it.

  Question 2: My interest in the above analysis comes from my desired to give a rigorous justification for using Monte-Carlo simulation of radioactive decay to arrive at the above deterministic model. However, I'm stuck after the above analysis. Could someone provide me with some reference or explanation to rigorous justify the connection?

The average disintegration rate is given by $\bar n=\sum_{n=0}^{N_0} nP(n)$ where $P$ is the binomial distribution. If $M$ is the average number of events in time $t$ then $\displaystyle M=N_0(1-e^{-\lambda t})$. For times smaller than the lifetime $1/\lambda$  the $\displaystyle e^{-\lambda t} \to 1-\lambda t$ then $M=N_0\lambda t$ and as the disintegration rate is $M/t$ this leads to $\displaystyle \frac{dN}{dt}=-\lambda N$. Thus provided the measuring time-bins are small compared to the lifetime this rate law should hold.

As the decay proceeds the number of counts becomes small, perhaps $10^6$ at short times to just 1 or 2 at long times then of course the signal to noise becomes huge, zero counts in some time bins for example and 1 or 2 counts in others. (This becomes an issue in analysing data, see Anal. Chem., 2003, 75 (16), pp 4182–4187).

The expected variance is, as you quote, $V=\sigma^2=N_0e^{-\lambda t}(1-e^{-\lambda t})$ and can be rewritten as $V=Me^{-\lambda t}$ where $M$ is the mean number of events recorded in time $t$. As the time bins are normally short so that $\lambda t <<1$ then $\sigma=\sqrt{M}$ so that $\displaystyle \frac{\sigma}{M}=\frac{1}{\sqrt{M}}$


Why do we say that a transition state that resembles a stable species in free energy also resembles it in structure? My textbook states," According to the Hammond-Leffler postulate, the transition- state structure for a step that is uphill in energy should show a strong resemblance to the structure of the product of that step."
Also stating,"One way that the postulate can be stated is to say that structure of a transition state resembles the stable species that is nearest to it in free energy. For example, in a highly endergonic step the transition state lies close to the products in free energy,and we assume,therefore,that it resembles the products of that step in structure as well. "
(Source:Solomons' Organic Chemistry
Book by Craig B. Fryhle, Scott A. Snyder, and T. W. Graham Solomons)

EDIT for a different approach:

The energy of a quantum mechanical system of nuclei and electrons is governed by the positions of the pieces. So, we're looking at the energy associated with the wave functions of all the pieces.

For two species that are similar in structure, the idea is that those two structures lead to relatively similar solutions of the wave functions. If the wave functions are similar, the energies of the particles described by those wave functions must be similar as well.


Should the mean energy be written in angular brackets or with an overline? How would you write mean energy? I came up with two different ways:

$\displaystyle\langle G_{\textrm{f}}^\circ(j) \rangle$ with the MathJax code \langle G_{\textrm{f}}^\circ(j) \rangle
$\displaystyle\overline{G}_{\textrm{f}}^\circ(j)$ with the MathJax code \overline{G}_{\textrm{f}}^\circ(j)
Or maybe another way?

I am looking for the proper use of this notation in the scientific literature.

Both the IUPAC 2008 Green Book (section 4.2) as well as the ACS Style Guide (specifically Chapter 11) agree that both methods of representing the mean value are acceptable. There might be some style guides / international standards that state otherwise, but for the average chemist these two are generally sufficient.

If ever in doubt, just state the meaning in text.


Which of the following are true based on this comparison? In the following chart, the horizontal line is time and vertical line is temperature. $K$ and $L$ are solid matters that are placed in two identical pots within the same room and are being heated up with identical heaters.

(Translation: Sıcaklık = temperature, Zaman = time, dak (dakika) = minute, maddesi = substance)

Which ones of the following are true?

A-) If $K$ and $L$ are different matters, they have the same mass.

B-) If masses of $K$ and $L$ are different, they can be the same matter.

C-) If masses of $K$ and $L$ are equal, they are different matters.

I chose "only $B$" which turned out to be wrong while the correct answer is "$B$ and $C$" and in the explanation it said "You picked the wrong answer because you think different matters can't have the same boiling/melting point." which is true because they can't, right?

A is not true - no such connection exists.
B is true (and given the same temperature indicating the phase transition it is in fact likely that K and L are the same matter).
C is true and can be explained as follows: if K and L have the same mass, they must have different specific heat capacities, defined as:

$$
c = \frac{\text{the amount of energy to be added to raise the temperature by 1°C}}{\text{mass}}
$$

This is a property that is unique for each matter - same as the melting and boiling point. It can be said to be a function of the matter. Note that the $T(t)$ graphs appear to start at the same temperature, but it takes longer to heat L than K to 85°C although the same heating is applied.


¹¹B NMR in a borosilicate tube I'm just wondering, if one were to run a $\ce{^{11}B}$ NMR on a boron containing sample in a borosilicate tube, how could we separate the spectra of the tube and the sample?

Apart from using a different tube - Perhaps through a pulse sequence like BIRD for $\ce{^1H/^13C}$?

Or decoupling - which I've only just started to learn about and don't really understand?

The borosilicate tube gives a very broad signal, the signals from the boron compounds you're measuring should be much sharper. If your samples are concentrated enough and have reasonably sharp lines, that should be sufficient to distinguish them from the broad background signal from the NMR tube.

You could of course also measure an identical NMR tube with the same amount of solvent, but without your boron sample. Comparing this with your real sample should make it easy to identify which signals are background from the tube.

And of course you could use an NMR tube made from a different material.


Is it possible for a compound to be soluble in a mixture of solvents but not the individual solvents? Some weeks ago one of my friend needed to use turpentine. As a good but ignorant chemist on that point I did some research about it. I found (at the beginning) in the French wikipedia page of turpentine :

  "L'essence de térébenthine est un mélange de molécules variées,
  comprenant en particulier des terpéniques, des acides et des alcools.
  Ce liquide est insoluble dans l'eau, dans l'alcool absolu et dans
  l'éther, cependant il est légèrement soluble dans un mélange
  eau-alcool."

Which translated in English gives,

  "Turpentine is a mixture of different molecules, particularly terpenes, acids, and alcohols. This liquid is insoluble in water,
  absolute alcohol, and ether, but it is slightly soluble in a mixture of
  water-alcohol."

I found no reference about that. Is it really possible? It sounds particularly strange to me. And obviously if that is possible, how?

The short answer is: yes, this is possible. Unfortunately, solubility is a fairly complex phenomenon to explain simply. Let's start with some examples where solubility is higher in a binary mixture than either solvent alone.

For a solid-liquid-liquid example: phenanthrene-cyclohexane-diiodomethane.[1]
For a liquid-liquid-liquid example like the one you were asking about: trifluoromethane-tetrafluoromethane-ethane at low temperatures and high pressures.[2]

In the Hildebrand Theory of solutions, there is a quantity called the "internal pressure", loosely how much these compounds want to break apart. In this model, this is usually something like the molar energy of vaporization divided by the molar volume. If the internal pressure of the solid is between the internal pressures of the liquids (as in this case), the solid is more soluble in the mixture than in either one individually. Another way to asking your question is about phase separation in ternary mixtures. You can find many examples of curved lines in ternary phase diagrams which are due to the ternary mixture not being the sum of its parts.

Qualitatively, you could imagine that some solvent solvates some parts of the molecule well and other poorly and vice versa for the other solvent. Hence, together they solvate better than each individually.

Gordon, L. J.; Scott, R. L. J. Am. Chem. Soc. 1952, 74 (16), 4138–4140. DOI: 10.1021/ja01136a054
Paas, R.; Peter, K. H.; Schneider, G. M. J. Chem. Thermodyn. 1976, 8 (8), 741-747. DOI: 10.1016/0021-9614(76)90053-7


Memory foam chemicals from pillow spoiled bed and rendered it useless I have recently purchased a memory foam pillow and in two days i ve experienced breathing problems and stomach problems and eyes irritation due to the harsh toxins in it and I returned the memory foam. But, The memory foam has touched the walls next to the bed and the plywood on the bed. Because of which i m unable to use the plywood which is still causing the harsh effects of the memory foam pillow whenever i m spending time on it. I recently bought another plywood it picked the same toxins ( may be from the wall) and i m again facing the same problem. Need suggestion for a way to clean these chemicals away from plywood and wall ?

I would suggest that perhaps it wasn't the pillow but another issue altogether.

If in two days your pillow outgassed enough to saturate the wall and the plywood, a couple days should also be sufficient for those to also release any trapped gasses. As I doubt they have any properties better at trapping gases than the foam in your former pillow.

Just a couple thoughts:

Do you have a carbon monoxide monitor in your home? A faulty furnace or water heater could be causing some of your problems.

Have you opened a window to provide better airflow?

Have you febreezed the area? The active ingredient in febreeze, cyclodextrin, does trap unwanted chemicals and gasses.

Or another chemical trap, is activated charcoal.


Penta-valent ion in water? I would like to have a solution with a pentavalent positive ion. I need something that, preferably in water:  $\ce{XY -> X^5+ + Y^5-}$. What can XY be?

Even the tetravalent cations (most of them) are heavily hydrolyzed and barely exist in water solutions. Pentavalent cation is even more of a challenge. Protactinium might be a possibility, but you can't have it anyway.

Speaking of anions, there is not even a trivalent single-atom anion that would not hydrolyze instantly and completely upon touching water.

On the other hand, polyatomic anions (and cations, for that matter) can have pretty much any charge you want.


How to test for methanol? I wanted to make some ethanol from baker's yeast and I used sucrose. I've made it and put it in a glass jar. I left it on a warm place (near the radiator) until the yeast died and after they died I distilled the yellowish liquid using simple distillation. When I smelled the product in the receiving flask I was confused. It resembled ethanol a little. The smell was like sweet and medium-intense ethanol. I think that it might be methanol. If there are any reagents or tests other than smelling that I can use at home, please do tell.

Sorry @ JavaScriptCoder, but I've to agree with aventurin. I'd say idoform test should have been done in a certified laboratory because of safety concern. However, I think fractional distillation might be helpful for this identification (or simply verification). OP can do this at home since he/she has done simple distillation at home. Ethanol/water binary system give an azeotrope, which boils at $\pu{78.1 ^\circ C}$ and gives $95\%$ ethanol. Methanol, on the other hand, make no azeotrope with water, and boils at $\pu{64.7 ^\circ C}$. The temperature range of $\pu{13.4 ^\circ C}$ need only about ≤20 theoretical plates to separate. If you get first distillate below, say, $\pu{67 ^\circ C}$, then you got methanol or something other than ethanol. So, don't drink it.

For more information, read the Wikipedia page having Azeotrope tables.


Difficulty understanding redox in terms of hydrogen and oxygen transfer? I have come across 4 ways of defining oxidation:

loss of electrons
increase in oxidation number
gain of oxygen
loss of hydrogen

From these definitions, I have an idea of what I think 'oxidation' and 'reduction' really refer to, but I would appreciate if someone could verify or tell me if I am going wrong.

I think oxidation and reduction simply refer to the degree to which the electrons of an atom 'dissociate' or 'distance' themselves from the nucleus of an atom. We define this extent of distancing to be 0 when an atom is bonded to another atom of the same element.

Thus oxidation really is 'loss of electrons', and reduction is 'gain of electrons', but the electrons do not have to be completely gained or lost. It is more like, oxidation is the valence electron shell moving further from the nucleus and reduction is the valence electron shell coming in closer to the nucleus. I.e. in a chemical reaction when the atoms surrounding a particular atom, say A1, are changed, the terms 'oxidation' and 'reduction' give a measure as to whether the atoms surrounding A1 have an overall effect of being more electronegative than the atoms that were bonded to A1 previously (in which case A1 has been oxidised) or less electronegative (A1 has been reduced).

Then the increase in oxidation number coincides with these definitions.

When it comes to REDOX in terms of hydrogen and oxygen transfer, I think this is only valid for organic chemistry. In general, loss of gain of hydrogen is to a carbon atom or an oxygen atom, or maybe nitrogen and sulfur too. As these are more electronegative than hydrogen, a gain in hydrogen means that these atoms can attract more electrons towards themselves, whereas a loss of hydrogen means that they lose electrons that they could be attracting towards themselves.

In terms of oxygen, in most of these organic systems it is gain of oxygen to carbon, sometimes nitrogen, which are less electronegative than oxygen. So by the same idea as above with hydrogen, a gain in oxygen corresponds to decrease of electron density being held around the atom, i.e. oxidation.

You are quite right, those rules of thumb are not always valid.

To be precise, the "adding hydrogen = reduction" rule only works for elements more electronegative than hydrogen, and the "adding oxygen = oxidation" rule only works for elements less electronegative than oxygen. For example, sodium is less electronegative than hydrogen, and as such adding hydrogen to sodium ($\ce{Na -> NaH}$) is in fact an oxidation of sodium from 0 to +1.

However, I wouldn't be so quick to limit their application to "organic chemistry". The elements that are less electronegative than hydrogen are just the metals, and the only element more electronegative than oxygen is fluorine.

Therefore, adding oxygen to chlorine is an oxidation, although the oxides or oxoanions of chlorine are classified under inorganic chemistry. Likewise, nearly the entirety of the p-block obeys these rules (except for Group 13). Think about germanium, arsenic, etc.. those are undoubtedly not "organic chemistry".


Why is the binding angle of HOF lesser than that of H2O, despite the lone pairs and size of the flourine? Why is the binding angle of HOF(101°) lesser than that of H2O?(104.5°) Since fluorine has unpaired electrons wouldn't it repel hydrogen more strongly than the corresponding hydrogen of H2O? Also, the size of fluorine is bigger than hydrogen, which also should have made the repulsion greater.

In $X_2O$ molecules, such as $H_2O$, $F_2O$, $Cl_2O$, the oxygen is at the center of a tetrahedron and therefore theoretically the X-O-X bond angle should be 109.5°. However, this is never the case and it's always a balance between the repulsion of the two lone pairs of the oxygen(they repels more strongly than the bond pairs) and the steric repulsion of the X atoms.

Among the series H, F, Cl, hydrogen is the least electronegative and fluorine the most.
F is very electronegative and thus the electron density of the O-F bond near O is less than that in O-H.
If the electron density is farther away from the oxygen, then the 2 lone pairs of O squeeze the two bonding pairs closer together and the F-O-F bond angle will be less than the H-O-H angle.
Based on the electronegativity, you should expect that the Cl-O-Cl bond angle is less than H-O-H but more than F-O-F. This is not the case and the reason is that chlorine is big enough so that the steric repulsion prevails.

In your case you are considering the molecule HOF:
According to the logic above, you may expect that its bond angle is less than $H_2O$ but more than $F_2O$. Instead, H-O-F bond angle is 97° so even less than $F_2O$!

The reason for this unexpected small angle is that the bond length O-F and O-H are very different and their electron density(bond pairs) don't interact much and therefore they repel even less than in the F-O-F case(see picture above, where electron density is represented in blue).


Hydrolysis under basic conditions Which route actually occurs? Is it really true that the carboxylic acid never forms under basic conditions?

Why not just form the -COOH and deprotonate it afterward?? Or is there a higher activation every barrier to the second route?

I understand that carboxylic acids don't stick around long in basic solutions but that doesn't necessarily mean they don't exist in basic solutions at all.

I find the top mechanism unlikely for two reasons.

1) In the top mechanism, the proton removed in the first step should be way less acidic than a typical hydroxyl since the intermediate is already negatively charged. Going to a di-anion should be very far uphill, and therefore, slow. If that di-anion was ever formed, the next step would be exceedingly fast

2) In the bottom mechanism, the first step should be fast. It's intramolecular and entropically favored. The second step is instantaneous.

I can't imagine any text book includes the top mechanism. Check your book for saponification of esters.


Could we possibly discover new extraterrestrial elements and minerals? So, after dissolving a good portion of my brain by watching a ton of sci-fi, I started thinking about all of the strange new materials that we will supposedly have in the future. For example, the ever-present "di-lithium" in Star Trek. But, as I started thinking about it, I thought it was a little ridiculous (I know science-FICTION) that all these exotic elements could exist. No matter where a species comes from in the galaxy or universe, they would be working with the exact same Table of Elements as us. Furthermore, so long as they are similar to human-life, i.e. they breathe oxygen, need water, etc., and therefore live on similar planets, wouldn't the compounds be the same as well?

Let's say we are 250 million years into the future and have discovered superluminal space-flight at 100,000,000c. Say we cruise around to inhabitable planets throughout the galaxy and universe. Upon landing on these new worlds, the elements and compounds we find would be the same that occur on Earth, right? Would we still find diamonds, rubies, uranium, iron, etc.? Or is it possible that there are combinations of the elements yet unknown? Is it for certain that we have catalogued every element and isotope that occurs naturally? I understand that new compounds could be formed, but as far as the base elements and the naturally occurring minerals, have we found them all?

One thing which can be said with absolute certainty is that no species  in the Universe (or even all their combined efforts) will even come close to exhausting the daunting enormity of chemical space. When you drag combinatorics into a problem, you can easily stumble on massive numbers, and this is the case here; it has been roughly calculated that there are on the order of $10^{60}$ "chemically reasonable" distinct molecules below a molar mass of $\mathrm{500\ g\ mol^{-1}}$! This means that almost all chemical substances which can exist, never will exist, be it from natural or artificial production.

No matter how far chemistry develops, we will always be interested in investigating the chemical composition of new samples, and will always find many new compounds doing so. Most of the time we'll find the same of what we do on Earth; there's no escaping the fact that $\ce{N_2}$ and $\ce{Al2O3}$ are very stable substances, for example, and can exist in a wide range of conditions. But there will always be something new for us waiting out there. Organic substances are tremendously varied thanks to the concatenability of carbon atoms. One might think inorganic substances are far less variable, but just the class of silicate minerals is enormous, and many polymorphs of the same substances can exist.

Having said that, at least one area of chemistry seems far closer to being exhausted; the elements themselves. There is no possibility of us having "missed" any element among the first 112, and we're still filling in the gaps at higher atomic numbers, but each heavier element discovered involves ever more complicated procedures and lower yields. Much of the experimentally determined properties of elements above $\mathrm{Z=104}$ come from the analysis of a tiny amount of atoms, as little as tens of them. Each new element is significantly more difficult to synthesize, and is less stable, which is hampering progress. We don't yet know very well how to get past $\mathrm{Z=122}$ or so even in principle, and are unsure how accessible the ultraheavy elements will prove to be. There is still great academic interest in several aspects of the elements and their isotopes, but current research deals with the production of very shortlived species, which have limited applicability.

In all, we will always find something new to explore, but whether we will always find something strange depends on what you mean by the word. There are still many, many substances to find which will display impressive biochemical properties, for example, and many lethal or crippling diseases today might still become curable in the future. But there is zero chance of finding a substance which makes magic real, which unlocks faster-than-light travel, which has endochronic properties or which can be used to make overunity engines and perpetual motion devices. Ultimately, Chemistry is a form of applied quantum electrodynamics, a subsection of Physics, and quantum electrodynamics is a very developed theory with incredible precision. Thus, there is little room for surprises, and they likely will not come from a chemical system unless it is very specially tailored, rather than randomly found in nature.


What is the point of using molality when we have molarity? Molality and molarity are both concentration terms. Given molarity's popularity molality seems rather antiquated.  Why is molality still discused in modern chemistry classes?

One of the key attractions of molality is that changing the temperature of a solution does not change the molality, while it may change the molarity. This is because the volume of the solution changes as a result of expansion or contraction of the solvent upon changing the temperature, and thus the molarity changes, since $$M=\frac{n}{V}.$$

The downside, of course, is that the amount of solute required to reach a given molality depends on the nature of the solvent; ambiguity may also arise when a mixed solvent system (e.g. a water-ethanol mixture) is used, as the choice of which compound is the solute changes the molality of the resultant solution.1 Molarity avoids this problem by requiring only the volume of solvent.

As for why it's introduced in school, that's a pedagogical question beyond the scope of this answer, but both molarity and molality have their uses in the laboratory.

1 Consider, for example, a solvent made from 95% water and 5% ethanol. Molality could be determined with respect to only the water, or the combined water-ethanol solution, leading to different calculated concentrations.


Electron Configuration of an element with no. of electrons > 18 Let's say that we have an element, say $\ce{Ca}$.
According to law, the last shell should contain only upto 8 e-.
Hence, electronic configuration of $\ce{Ca}$ would be 2,8,8,2.

But, according to the 2n2 rule, the M-shell can hold upto 18 e-.
So, doesn't this mean that the e- configuration of $\ce{Ca}$ can also be written as 2,8,9,1 or say, 2,8,7,3 or something?

But it's not so... Why?

But, according to the 2n2 rule, the M-shell can hold upto 18 e-. So, doesn't this mean that the e- configuration of Ca can also be written as 2,8,9,1 or say, 2,8,7,3 or something?

Experimentally, the ground state electron configuration is 2, 8, 8, 2, so there is no other correct way to write it.  2,8,9,1 and 2,8,7,3 would be incorrect.

Really the only way to know is through experiment or possibly very high level calculations that include relativity and quantum electrodynamics.  As atomic number increases, the problem becomes more and more complex.  It isn't necessarily true that one configuration accurately describes the ground state.

For example for Lawrencium (element 103), for which no experimental determination has been made, you will see some periodic tables that say it has a 7p electron (8 electrons in the P shell and 3 electrons in the Q shell) and some that say it has a 6d electron (9 electrons in the P shell and 2 electrons in the Q shell).

See The low-lying level structure of atomic lawrencium (Z=103): energies and absorption rates to appreciate how difficult it is to really calculate such information.


Why is quickly cooling a a hot pan bad for it? I've heard that really quickly cooling a hot pan is bad and reduces the lifetime of the pan. I'm wondering, from a chemical perspective, why this would be true. My thought is that as a pan is heated up, there may be slight rearrangements in the lattice of the metal, which, if allowed to cool, would likely reset into the correct portion of the lattice. On the other hand, if quickly cooled, perhaps these atoms don't make it back into the correct place, and this warps the pan.

As a secondary question to this idea of warping a pan, why would reheating the pan not cause it to return to the original shape? Seemingly, that original shape is a lower energy structure of the lattice, so with enough energy to overcome barriers, why would it not "unwarp"?

Specifically it depends on material the pan is made of. Just like how heated glass can crack once immediately expose to a source of the opposite temperature. For example, a cast iron skillet has the potential to crack when at a high temperature it is exposed to a much lower temperature. To answer your second question, the reason why reheating the pan wouldn't return it to its original shape is first, you would have melt the metals to reform it and second of all if a crack has occurred in the situation, it means the elastic limit has been reached. The elastic limit is the greatest stress something can undergo and return to its original shape. Beyond the elastic limit, it is impossible to return something to its original shape. I hope this helps!


Does anything like this covalent concerted bond exchange reaction exist? I'm wondering if any reaction pathway exists (not involving enzymes) to do the following in aqueous solvent at temperatures < ~100°C:

To clarify, it's important however that the reaction is concerted, and that the W - X and Y - Z bonds are not broken until both the W-Y and X-Z bonds are formed.  Bonds here must be covalent, but otherwise can be of any type.  Any kind of catalyst can be used.  Of course, it would be nice to know of an efficient reaction with minimally exotic conditions or reagents (i.e. it would be nice to avoid heavy metals for oxidative-addition type reactions), but I won't push my luck.

Does something like this exist?

$$\ce{W=X + Y=Z ->[\mathrm{cat}] W=Y + X=Z}$$ is known as olefin metathesis, achieved using Grubbs/Hoyveda/Fürstner catalysts.

Water-soluble Ru catalysts have been prepared in the Grubbs group (DOI).


Why can't carbon and nitrogen be detected in ICP-MS and ICP-OES techniques? Why can't carbon and nitrogen be detected in ICP-MS and ICP-OES techniques?

My book says this technique is efficient for detecting multiple elements, with exception of halogens and carbon. I found somewhere that ICP is an open system, therefore it doesn't allow carbon detection. But why is that exactly?

As you say, air is allowed to enter the plasma. Air consists primarily of $\ce{N2}$ and $\ce{O2}$, and would clearly interfere with any oxygen or nitrogen in the sample. The problem is not that ICP-MS or -OES cannot detect them, the problem is air.


Why don't B, C, Si, have ionic forms? Also, why can some elements such as $\ce{Fe}$ have different ionic forms such as $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$? What determines each form?

Boron: borates, borides.  $\ce{MgB2}$ is the highest temperature BCS superconductor by a large margin.

Carbon: carbonates, carbides, hexacoordinate carbon in complexes, organolithiums, Grignard reagents, carbenium ions ($\ce{Ph3C+}$).  $\ce{Al4C3}$ hydrolyzes to methane.

Silicon: silicates, aluminosilicates, silicides.  Cook up silica gel in ethylene glycol plus potassium hydroxide (NOT in glass) to get pentacoordinate siliconate,

http://patentimages.storage.googleapis.com/WO1994020507A1/imgf000014_0001.png

You asked the second half in the Physics Stack.


London dispersion forces and distance relationship Why is the London dispersion force inversely related to distance but not to the second power but the sixth power? Why is there not an inverse square relationship as there is for Coulombic forces?

Is there something non-Coulombic about London dispersion forces?

London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.

Some electrostatic (Coulombic) relationships:

Two Point Charges: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)
Point Charge and Permanent Dipole: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)
Two Permanent Dipoles: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)

The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules rotate then it falls off to $1/r^6$.

In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.

So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:

$\mu_2^* = \alpha_2 E$

where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:

$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$

The potential energy is given by:

$V = \alpha_2 E^2 / 2$

$V \propto r^{-6}$

Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:

$V = \frac{C}{r^6}$


Why does superglue lose strength at high temperatures? I'm not a chemist. When I superglue my fingers together, by accident, I put them under warm water to peel off the superglue. Why does superglue become less "sticky" at higher temperatures? Is it the water or the temperature?

It is the temperature because as you probably know all kinds of substances get denatured at high temperatures. Another reason could the that the rate of reaction increases after running hot water, therefore taking no effort to take it off. Hope the answer helped!


What is the pressure belonging to in Boyle's Law? Just a hour ago, I was reading the great and highly interesting, Boyle's Law. In my textbook, there is given Boyle's Law and leaving some pages, there is stated Dalton's Law of Partial Pressure.

Boyle's Law is stated as, "As so long as the mass of the gas and temperature is constant, the product of pressure by volume of the gas remains constant."

Experiment of U-tube is well known. I am not going to copy it here but I would directly explain my problem. From U-tube experiment, I grasped that the pressure, mentioned there, is one that is applied and not of the gas. However, the difference stays vague in the law stated earlier that whether this is the pressure applied by external agency or the pressure applied by gas molecules or gas.

Later, I read there the Dalton's Law of partial pressure. There is a primary equation concerning to partial pressure of a gas viz.,

$$\frac {P_{gas}}{P_{mixture}}=\frac{m_{gas}}{M_{mixture}}$$

It is clearly mentioned there that this pressure either $P_{gas}$ or $P_{mixture}$ is the pressure applied by gas/mixture and not the pressure applied by external agency. This is even more clear from the relevant experiment.

The problems start when I begin to analyze the derivation for the equation above. Author of my textbook hasn't given any derivation for it but my teacher let us did that.

Derivation:

Consider  the mixture of gases A, B and C.

Let the number of moles of gas    $A=n_A$

Similarly, number of moles of gas $B=n_B$

And that of C is given by $n_C$

Total number of moles=$n_T=n_A+n_B+n_C$

We know that, $$P=\frac{nRT}{V}$$

Therefore pressure applied by gas A is given by, $\color{red}{P_A=\frac{n_ART}{V}}$

For gas B, $\color{red}{P_B=\frac{n_BRT}{V}}$

And for gas C, $\color{red}{P_C=\frac{n_CRT}{V}}$

For mixture of A, B and C, $\color{red}{P_T=\frac{n_TRT}{V}}$

Diving $P_A$, $P_B$ and $P_C$ individually by $P_T$ we get,

$$\frac{P_A}{P_T}=\frac{n_A}{n_T}$$

$$\frac{P_B}{P_T}=\frac{n_B}{n_T}$$

$$\frac{P_C}{P_T}=\frac{n_C}{n_T}$$

  The red marked statements are where my previous concepts were denounced. I was thinking that the derivation maybe wrong but I am not sure so I want you people to judge my concepts and my teacher as well and inform me.

Thanks.

Your derivation of Dalton's Law of partial pressure is fine.

I think you are confused about how pressure is "applied" to a gas, and what that does to the internal pressure of a gas.

Imagine that you have a "sphere" of gas surrounded by a "force field" (like in Star Trek) in empty space, then the force field containing the gas is dropped. Since there is nothing to confine the gas it just expands into infinite space.

A lot of ideal gas models, like the Wikipedia article on Boyle's Law, show the gas as being contained in a cylinder with a piston pushing on the gas to create the volume and pressure of the system. The external force on the piston is counterbalanced by the pressure of the gas times the surface area of the piston. The pressure of the gas is the same on all surfaces of the container. The gas doesn't just "press back" on the piston.

Pressure and volume are inversely related for the same system. So at the same temperature,

$PV = k$

Thus if the pressure increases, then for the same amount of gas the volume must decrease to keep $k$ a constant.


Is there any difference between table salt and common salt? Iv'e been reading about $\ce{NaCl}$ and this doubt crossed my mind. Are they anything other than just household terms?

They are both the same. If you look at this wiki page, it uses common salt and table salt as the same name. It's mainly just household terms, like people calling hydrochloric acid muriatic acid etc.


What are physical properties and chemical properties? As the question title says, I would like to know what the physical and chemical properties are. If possible, please include examples. Since it seems to be related, I would also like to know what physical and chemical changes are. I believe that things such as crushing a can is a physical change, but I am not sure what would constitute a chemical change (perhaps doing something that rearranges the molecules?)

Edit: I originally wanted to ask if cooking an egg constitutes as a chemical or physical change, but I thought that that would be closed as too vague. If possible, please explain which one it is too. Thanks!

This is a tricky question.

I have to concur with Paul - it's difficult to give a truly hard and fast definition of the distinction between a chemical change and a physical change as there is a fairly smooth continuum between 'physical' and 'chemical' interactions. However, a good rule of thumb is that chemical changes involve changes in electronic structure, usually in the form of chemical bonding or redox. The problem is that traditionally 'physical' properties may have big implications for chemistry - I'll give some examples.

The Van der Waals and dipole-dipole interactions are mediated respectively by 'instantaneous' and permanent polarisations of electron density with respect to positive charges. Dipole-dipole interactions such as those found in water are quite significant and give rise, for instance, to water's frankly freaky liquid range as well as water's ability to mediate a wide range of chemical processes. These interactions don't involve significant changes in electronic structure however they have chemical implications in, for example, making a choice between liquid water or liquid nitrogen as a solvent.

The atomic mass (and hence the isotope) of an element is typically considered a physical property, however it has implications for reaction kinetics. This can be illustrated (expensively) by giving living organisms nothing but heavy water to drink. They will eventually experience toxicity due to the slowing of key biochemical reactions.

The magnetic properties of an atomic nucleus are typically considered to be physical, however they can be exploited in the observation that e.g. different isotopes of uranium have slightly differing hyperfine splittings and hence different ionisation energies. This allows isotopes to be enriched by selective ionisation.

Regarding the egg:

The dramatic changes that occur when you cook an egg are the result of denaturation of the proteins in the egg. Proteins are long chains of amino acids that take on their characteristic active conformations through a process called 'folding', so called because that's what it looks like (if you've heard of Folding@Home, that simulates the folding of proteins). If you raise the temperature much higher than physiological temperature, the protein degrades through unfolding into a different conformation. This involves the Van der Waals and dipole-dipole interactions holding together the secondary/tertiary/quaternary structure of the protein being broken en mass as well as the formation and/or scission of covalent bonds.


Aminoethyl or ethylamine? I was trying to name the above compound which happens to be 3-(2-aminoethyl)-2-methylcyclohexan-1-ol.

On my first attempt, I had named it 3-(ethyl-2-amine)-2-methylcyclohexan-1-ol.

Which one is correct?

The IUPAC rules use "aminoethyl" for such a group. This is copied from the IUPAC "blue book" which can be found online here

Rule 811.3 - An -NH2 group, when not the principal group, is named by the prefix "amino-".

Your original name is very close to what would be used if the N were attached to the ring and the ethyl attached to the N. That would be "3-ethylamino-2-methyl. . ." according to this rule:

811.4 - Radicals RNH-, R2N-, and are named as substituted amino groups by changing "ine'' of the amine to "ino".


Calculating change in enthalpy of combustion of 1.12g of hexane I will calculate the change in enthalpy of combustion of 1.12g of hexane.

1) Calculate the energy transferred to 200g of water using the equation: J = mass of water in grams (200g) x specific heat capacity of water (4.18 J/gK) x temperature increase in Kelvin (24C = 297.15K). This results in the amount of energy transferred to the water of a total of 248,417.4J or 248.4kJ.

2) Calculate number of moles of hexane burnt using n=m/M. Weight of hexane before experiment = 222.07g. Weight of hexane after experiment = 220.92g. Therefore 1.15g of hexane was burnt during the experiment. 1.15g of hexane = 0.013mol of hexane.

3) Find enthalpy change of reaction. Combustion of 1mol of hexane produces (by my calculations) -4194 kJ/mol.  Therefore, 0.013mol will produce (0.013/-4194) -3.1x10-6 kJ/mol.

So this is the procedure that I used to find out the answer! Was it correct?
Because, apparently, you can use the values of steps 1 and 2 to find out the enthalpy change?

I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:

1.) 0.2 kg Water was icreased by 24 °C:
U = 0.2 kg * 4187 J/(kgK) * 24 K = 20,100 J = 20.100 kJ

2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol

3.) 1.15 g caused the temperature increase of 24 °C:
20.100 kJ / 0.013 mol = 1,510 kJ/mol

So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane.


Is paper aging(oxidation) and paper burning(rapid oxidation) the same thing? I woke up today with a question of what is fire; looked up the definition and the first response was that fire was rapid oxidation.

I was curious what the difference was between rapid oxidation and regular oxidation that we have all around us as seen with metals rusting, book pages yellowing, artificial aging, paint drying, apples browning due to air exposure.

I cannot find an answer to what the difference between rapid and regular oxidation is, other than the speed of the reaction, and the visibility of the reaction when it becomes fast enough.

I am curious whether there is any difference at all between metals "rusting", paper "burning"/"aging", paint "drying", or if these are all different names for the same process at different rates.

Is it reasonable to say that there is fire everywhere there is oxidation occuring, but it's invisible until the reaction is fast enough?

Rapid oxidation- burning - leads to two main products carbon dioxide and water. You cannot keep any records if paper is burned.
In case of slow oxidation some of paper components are oxidized which changes the colour of paper, but just that. Old manuscripts can be stored for hundred of years if no none burns them.


What is lithium pyroborate? This component is mentioned in an article called "Novel geopolymer materials containing borate and phosphate structural units".
I've never heard about this before, and a google search didn't really give an answer it seems.

Lithium pyroborate ($\ce{Li2B4O7}$) is the salt of lithium ($\ce{Li}$) and pyroboric acid ($\ce{H2B4O7}$).

Nomenclature

The prefix ortho- designates an acid with the maximum number of hydroxyl ($\ce{OH}$) groups [reference]. For boron ($\ce{B}$), orthoboric acid is $\ce{H3BO3}$ (also written as $\ce{B(OH)3}$ to show that there are $3$ hydroxyl groups).

The prefix pyro- here seems to not match with the usual usage, so I will not be explaining the usual usage here. One common feature between this and the usual usage is that their molecular formulas are multiples of the ortho acid's molecular formulas, minus some water molecules:

$$\ce{4H3BO3 -> H2B4O7 + 5H2O}$$

Gallery

Some proposed structures of lithium pyroborate:

Some proposed structures of pyroboric acid:

The above are modified from two proposed structures of sodium pyroborate, with the first one referenced here and the second one referenced here.


Reactions of alcohol with sulfuric acid I knew two chemical reactions of alcohol with sulfuric acid

$\ce{CH3CH2OH + H2SO4 -> CH2CH2}$
Here product is having a double bond (ethene) and this reaction happens at 443 K temperature.
$\ce{CH3CH2OH + H2SO4 -> C2H5OC2H5}$
Here product is ether an happens at 413 K temperature.

But today I came across another reaction.

$\ce{C2H5OH + H2SO4 -> C2H5OSO2OH + H2O}$

Now my question is when will this third reaction happen? Any specific temperature?

The book from which I found this reaction doesn't describe anything. Just says: Alcohols react with inorganic acids (except halogen acids) to form inorganic esters.

This article on Wikipedia says your compound, ethyl sulphate, can be synthesized, by adding sulfuric acid drop-wise to boiling ethanol and temperature should be kept below 413K which agrees with your data. I found also an article according to which authors conducted esterification at about 300K, but with starting presence of anhydrous sodium sulfate.

Also proportion of reagents should be right, if you add more than equimolar amount of acid, removal of leftover is problematic, as sulfuric acid has high boiling point and there's no suitable selective extractant.


water vapor vs air (in a balloon) I poured some water in a balloon, inflated, and tied it up.  Over a period of couple weeks, all water in the balloon is gone.  But the balloon is still at about 2/3 of original inflated size.

I thought the the air molecules are much smaller than water vapor.  Why the water is all vaporized?

The air in the balloon will be saturated (with water vapour) and the air outside not. As a result there will be a vapour density (or partial pressure) gradient between the inside and outside of the balloon that will drive the mass flow of water vapour through the balloon and to the outside.


Is dissolution of H2S in water exothermic? While doing certain laboratory tests for qualitative identification of cations, I found that on passing $\ce{H2S}$ in the solution, the temperature of the solution increased considerably. When the temperature decreased due to addition of $\ce{NH4Cl}$, the solution again attained room temperature in passing $\ce{H2S}$.

This rouses the question: is the dissolution of $\ce{H2S}$ in water exothermic? Or does it react with other components of solution to make an exothermic reaction?

The Wikipedia article stated only this:

  Hydrogen sulphide is slightly soluble in water and acts as a weak acid (pKa = 6.9 in 0.01–0.1 mol/litre solutions at 18 °C), giving the hydrosulfide ion $\ce{HS^−}$. Hydrogen sulphide and its solutions are colourless.

The full experiment can be found here.

Koschel et al. (2013, p. 14486) report that

  dissolution of gaseous hydrogen sulfide in aqueous phase is exothermic at conditions where $\ce{H_2S}$ is gaseous, i.e., at a temperature of 323 K and pressure 2 MPa, the saturation pressure $p_\text{sat}(\mathrm{H_2S})$ being 3.5 MPa. The dissolution process is endothermic at the medium and highest temperature where hydrogen sulfide is liquid and supercritical, respectively, the critical parameters of $\ce{H2S}$ being $T_c = 373.4$ and $p_c = 8.94$ MPa.


Where does the energy of solar radiation hitting the atmosphere go? I was reading about solar radiation, and there was a part that says that the atmosphere absorbs most of the radiation emitted by the sun. I wonder if when the atmosphere absorbs the energy, where does it go since energy doesn't disappear or that's what I've been taught in high school. I'm guessing it transforms to heat. Correct me if I'm wrong, and a explanation would be very appreciated.

It is broadly true that some solar radiation ends up heating the atmosphere. But it isn't true that the atmosphere absorbs most of the heat emitted by the sun.

Broadly there are several ways that radiation is absorbed:

Some of it (but not much of it) is absorbed directly by certain molecules in the atmosphere (carbon dioxide and water vapour are biggies), directly heating them.
Some of it is reflected by clouds and ice and other parts of the earth and reemitted to space (quite  a lot of it)
Some radiation is absorbed by plants and used to drive photosynthesis (important but not a lot of absorbed radiation)
A lot of it is absorbed by the surface of the earth (land and sea) which is the part that contributes most of the warming effect of sunlight (of course this warmer sea and earth then warms the atmosphere above it either directly be contact or by being absorbed before it is transmitted back into space, though a lot of it is emitted to space).

So, in the end, a lot of the radiation ends up as heat, but mostly via heating of earth and sea not by directly heating the atmosphere. On the other hand, the atmosphere does end up much warmer as a result. And the absorption by the atmosphere does make the difference between the earth being very cold or very habitable as a planet. The details of this are very complicated which is why people build very big computer models when they want to understand global warming which depends of the complicated details of how the atmosphere absorbs fractions of the heat.


Energy of orbitals under a crystal field It is shown in octahedral crystal field splitting that both the $e_g$ orbitals have equal energy. But how can that be possible ?

The $d_{x^2-y^2}$ is directed along both the axes (x and y) whereas the $d_{z^2}$ is directed along only the z-axis, shouldn't the later have less energy than the former ?

They are equivalent in energy because symmetry says so, and because the calculations give that as an answer. Okay, that wasn’t exactly satisfactory, was it?

The next complicated answer on the road from over-simplified to reality is that the $\mathrm{d_{z^2}}$ extends to both ligands, too: Do not forget the ‘ring’ in the $xy$ plane. By extending towards the ligands on the $z$ axis further than the $\mathrm{d_{x^2 - y^2}}$ orbitals do to their ligands, it is less stabilised than it looks, and the interaction with the equatorial ligands also destabilises.

The really sophisticated and probably most correct answer is that we generally have a wrong view of those two orbitals. They are mathematically equivalent and would actually form a group of three: $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2 - x^2}}$ and $\mathrm{d_{y^2 - z^2}}$ — but because we can only have two, wee need to add two together and commonly take the two that give ‘$\mathrm{d_{z^2}}$’. In some images of molecular orbitals you can see that they are much more identical than they initially look.


What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base? Which of the following statements best characterizes the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?

  A) Both the strong acid and the strong base are completely dissociated in solution whereas the weak acid dissociation equilibrium constant is small enough that there is a measurable amount of the un-dissociated acid present in aqueous solution.

  B) The equivalence point of a titration of a strong acid with a strong base may be observed with an indicator whereas the equivalence point of a titration of a weak acid with a strong base can only be observed with a pH meter.

  C) The titration of a strong acid with a strong base is a fast reaction whereas the titration of a weak acid with a strong base is generally a very slow reaction.

This website seems to indicate that choice C is correct.

For B, all I know is that the pH at the equivalence point does not equal 7.00 (pH > 7.00) for the weak acid titration.

For A, I know the first part of that sentence is true. Strong acids and bases are characterized by completely dissociating whereas weak acids do not dissociate to the same extent.  But is it true that there's a "measurable amount" of the un-dissociated acid present in aqueous solutions?

The answer is A.

Let's look at a weak acid: acetic acid. It has a $K_a$ of $1.8\cdot  10^{-5}$. From this you can calculate how much it will dissociate in water. Let's say we have a concentration of $1~\mathrm{M}$:

\begin{align}
K_a &= \frac{[\ce{CH3COO^{-}}][\ce{H+}]}{[\ce{CH3COOH}]}\\
1.8 \cdot 10^{-5} &= \frac{x \cdot x}{1~\mathrm{M} - x}\\
x &= 4.233 \cdot 10^{-3}\\
\end{align}

This means, at equilibrium, $[\ce{CH3COOH}] = 1~\mathrm{M}- 4.233 \cdot 10^{-3} = 0.995~\mathrm{M}$. As you can see, by far the largest part is not dissociated!

B is not true, you can still observe the equivalence point. Have a look at the following picture, that shows the titration curve for weak / strong acid titrated with strong base. When choosing an indicator for colorimetric titration select one so that the pH jump at the equivalence point contains the interval $\mathrm{p}K_a\pm1$. Phenolphthalein has a $\mathrm{p}K_a \approx 9$, so to decide if it is a suitable indicator you would check if the pH jumps from 8 to 10 at the equivalence point. It does:

C is not true either. If it was a very slow reaction, you could not titrate accurately, as you would have to wait a long time between every drop of $\ce{NaOH}$.