Any free tool for Calculating Gasteiger-Hückel charges for ligands? I want to know if there is any free open source tool available for Gasteiger-Hückel charge calculation? I know it is available on Sylbyl but is there any other(free) way to calculate these charges for small molecules?

I was actually trying this paper, if any one is interested,

How to improve docking accuracy of AutoDock4.2: a case study using different electrostatic potentials.

Hou X1, Du J, Zhang J, Du L, Fang H, Li M.X et al.

The method is not fully published according to Gasteiger.

There is an online server for PETRA run by the Gasteiger group, and in principal, it's available from Molecular Networks. I note that PETRA is no longer listed on the products page, but it's probably part of some other product now.

The open source CDK package has an implementation of the PEPE method, although I don't know if it has been validated against Gasteiger's code.


Possible products from anion analyses What are possible products from these two anion analyses?

The nitrite test, reacting diphenylamine with a nitrite:

$$
\ce{(C6H5)2NH + NO2- -> ?}
$$

The sulphite test, reacting potassium permanganate with a sulphite:

$$
\ce{KMnO4 + SO3^{2-} -> ?}
$$

Under acidic conditions, a blue colour is observed. This results from the formation of N,N'-diphenylbenzidine. In the course of the reaction, diphenylamine is oxidized to a quinoid radical which subsequently adds to another molecule of diphenylamine.
Under slightly acidic conditions, bleaching of the solution is observed. Sulfite is oxidized to sulfate while $\ce{MnO4-}$ is reduced to $\ce{Mn^{2+}}$.


What's the chemical formula of "Everitt's salt"? According to my school textbook (which is notorious, for being in possession of several errors), a certain "Everitt's salt" is a complex that results from the reduction of Prussian Blue.

Everitt's salt (according to our inglorious textbook)  has the chemical formula $\ce{K2[Fe(CN)6]}$... which I straightaway dismissed as nonsense, because it would mean it's got $\ce{Fe^{4+}}$ (ferryl ion). From my (very limited) knowledge of the subject, the ferryl ion doesn't look like something that would occur in any reasonably stable compound; since Everitt's salt is, well, a "salt" (and since salts can usually be crystallized into stable solids).

Google wasn't much help in this regard,

  I've no idea how to interpret the formula that is supposedly that of Everitt's salt

Questions:

1) Is there really something called "Everitt's salt"? Is it produced by the reduction of Prussian Blue?

2) What's its chemical formula?

According to various sources Everitt's Salt has the formula $\ce{K2Fe[Fe(CN)6]}$, where both $\ce{Fe}$ are in the $+2$ Oxidation State. Everitt described it as a yellow precipitate, resulting from the reaction:

$$\ce{2 K4Fe(CN)6(aq) + 6 H2SO4(aq) \to 6 HCN(g) + 6 KHSO4(aq) + K2Fe[Fe(CN)6](s)}$$

Acc. Everitt, the wet salt can be oxidised to Prussian Blue by the action of air (oxygen).

Source.


Explain volume contraction in mixtures of alcohol and water Could someone explain why volume contraction occurs when you mix an alcohol such as ethanol with water in relation hydrogen bonding and the dipole-dipole forces?

Liquid water can be thought of as having an average configuration that tends to be an open, almost crystalline structure, somewhat like that for hexagonal ice crystals $\mathrm{^{[1,2,3]}}$.  This structure has been confirmed by both x-ray and neutron diffraction experiments.  Below is a depiction of the hexagonal structure of ice for reference:

Source: Boundless. “The Structure and Properties of Water.” Boundless Chemistry Boundless, 26 May. 2016. Retrieved 06 Feb. 2017 from boundless.com

This phenomenon of the ice-like structure of liquid water is also illustrated in the observation that at temperatures below $\pu{4^oC}$, the density of water decreases with decreasing temperature as it becomes more ice-like prior to actual solid crystalline ice formation.

So, if we envision this average crystal-like configuration for liquid water, then two effects come into play regarding the volume contraction observed when water is mixed with ethanol.

Firstly, some fraction of the hydrocarbon tails of the ethanol molecules is able to fit within this lattice, filling some of the voids within this water structure.

Secondly, the hydroxyl moiety of ethanol engages in hydrogen bonding with water.  As ethanol only has a single hydrogen bond donor, however, the crystal-like structure of the liquid water cannot be fully maintained, and thus it somewhat collapses.

Also, please take a look at these answers regarding the general concept of volume contraction of liquid mixtures, not specific to the case of ethanol and water.

1 http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000003174/2_Chap1.pdf;hosts=
2 http://crystal.med.upenn.edu/sharp-lab-pdfs/sharp_EncLifeSci.pdf
3 "Water structure and changes in thermal stability of the system poly(ethylene oxide)–water", J. Chem. Soc., Faraday Trans. 1, 1981,77, 2053-2077


Why are super absorbent polymers not made from polysaccharide-rich wheat straw and chaff? What are the qualities in a substance that chemists look for when making a super absorbent polymer?

If I am right, polysaccharides are one of them.

Until the 1920's, polysaccharides were a prime choice for producing absorbent materials. However, their water absorbancy was limited to about 10:1 by mass (i.e. 1g of materiel could absorb 10g of water).  This metric, along with absorbancy by volume, are the properties of interest when designing a super absorbent polymer (SAP).

Subsequently, for several decades hydrogels dominated the SAP landscape. According to this Wikipedia article:

  "Water-absorbing polymers, which are classified as hydrogels when
  cross-linked, absorb aqueous solutions through hydrogen bonding
  with water molecules. A SAP's ability to absorb water is a factor of
  the ionic concentration of the aqueous solution. In deionized and
  distilled water, a SAP may absorb 300 times its weight (from 30 to 60
  times its own volume) and can become up to 99.9% liquid...
  [...]
  Superabsorbent polymers are now commonly made from the polymerization
  of acrylic acid blended with sodium hydroxide in the presence of an
  initiator to form a poly-acrylic acid sodium salt (sometimes referred
  to as sodium polyacrylate). This polymer is the most common type of
  SAP made in the world today.

  Other materials are also used to make a superabsorbent polymer, such
  as polyacrylamide copolymer, ethylene maleic anhydride copolymer,
  cross-linked carboxymethylcellulose, polyvinyl alcohol copolymers,
  cross-linked polyethylene oxide, and starch grafted copolymer of
  polyacrylonitrile to name a few.

There is also ongoing research into improving the ability of natural products to perform as SAP's.  According to the same Wikipedia article quoted above:

  "Today´s research has proved the ability of natural materials, e.g.
  Polysaccharides and Proteins, to perform super absorbent properties in
  pure Water and saline solution (0,9%wt.) within the same range as
  synthetic polyacrylates do in current applications."

Summary, TL;DR:
Until the 1920's, polysaccharides were the prime choice for producing absorbent materials. Primarily due to their relatively low absorbancy (~10:1 by weight, and lost water when pressure was applied), they were subsequently replaced with hydrogels (absorbancies of 50:1 to 300:1, and maintained water when pressure was applied). However, by the 1960's, polysaccharide based super adsorbent polymers (SAPs) saw a resurgence in use as researchers (primarily at the US Department of Agriculture) were able to re-engineer polysacharide based SAPs to have absorbancies comparable to that of the synthetic hydrogels.

The qualities that make a SAP desirable are primarily it's absorbancy and ability to maintain absorbancy under pressure, but also cost and availability of materials. It is the latter of these that drives the research to develop better techniques for the use of polysacharide based SAP's.


How to figure out number of atoms in a molecule? e.g. Phosphorus exists as P4 We know that Phosphorus exists as P4 but how can I figure out the number of atoms in a molecule for other elements?

This is the question that requires the use of P4. I would never have known Phosphorus is tetra-atomic and that the equation will involve this. Is there a way to figure that out?

You are asking how to identify the standard state chemical structure of an element; this is obtained from empirical data.

The standard state is defined as the most stable allotrope of the element under standard conditions. For example, graphite is the standard state for carbon. Phosphorus, however, is an exception: while white phosphorus is not the most stable allotrope, it is the easiest to characterize, and is defined as the standard state for phosphorus.

The chemical formula for each element at standard conditions is that of the standard state. Except for the gases $\ce{H2, N2, O2, F2, Cl2, Br2, I2}$, and the solids $\ce{P4, S8}$, all other elements exist in monatomic form in the standard state.


Single Point Energy in Gaussian software I used Gaussian software to determine the single point energy (SPE) of 3 molecules namely ethane, ethene and phenol.

The SPE for the molecules are -79.83, -78.59 and for phenol it is quite large at -307. However, I am not able to explain the trend.

I am thinking does SPE have anything to do with the molecular weight of the molecule for example, larger mass = more negative SPE?? I cant think of a better explanation. Can someone please help me out?

thank you

The energies you're seeing are those resulting from solving for the eigenfunctions of the Coulomb Hamiltonian under the Born-Oppenheimer approximation, where the Hamiltonian can be broken down into the following 5 components:

The most important thing to take away from these expressions is that they all contain summations over nuclei, electrons, or both; combined with the presence of the set of nuclear charges $\{Z\}$, there is a non-linear dependence of the molecular energy on more than just the nuclear masses.

This is one reason why it usually isn't instructive to compare absolute energies of unrelated molecules.


Why is CH4 a gas at room temperature but CH3Cl a liquid? I cannot figure out a question on my review sheet. The full question is:

  Methane ($\ce{CH4}$) is a gas at room temperature but chloroform ($\ce{CH3Cl}$) is a liquid. Explain why changing just one $\ce{H}$ atom for a $\ce{Cl}$ atom changes the compound's state at room temp.

I thought about it, and I think it does have to do with $\ce{CH4}$ is non polar, so it does not tend to stick to each other (gas state), while $\ce{CH3Cl}$ is polar, so it does stick, such as like $\ce{H2O}$, which is liquid and is cohesive, so $\ce{CH3Cl}$ would be as well. Is anyone able to tell me if I am on the right path?

Yes, I think that is the right answer expected for this question.

Three factors to consider.

(1) Most importantly there four intermolecular forces which effect BP

Ionic bonds  > Ion-dipole interactions > H bonding >  dipole-dipole interactions > Van der Waals dispersion forces.

(2) Increasing the number of carbon atoms (e.g. n-alkanes) increases the BP

(3) Branching on carbon chain lowers the BP.

The answer being given though is a bit of hand-waving that oversimplifies the situation. Dipole moment alone can't explain the overall trends. Van der Waals dispersion forces must be considered also.

For instance with increasing substitution $\ce{CH2Cl2}$, and $\ce{CHCl3}$ both have lower dipole moments than $\ce{CH3Cl}$ but higher BP's. $\ce{CCl4}$ has no dipole moment like methane, but has the highest BP of all.

                                          Bond Lengths
           Dipole     B.P.       Mol.
           Moment     (°C)       Wt.       C-H     C-X
 CH4         0        −161.49    16.04     108.7    ---
 CH3Cl       1.9       −23.8     50.49     111      178.3
 CH2Cl2      1.6        39.6     84.93     106.8    177.2
 CHCl3       1.15       61.15    119.37    107.3    176.7
 CCl4        0          76.72    153.81    -----    176.6

Also look at the halomethanes as a series. $\ce{CH3F}$ has a smaller dipole moment than $\ce{CH3Cl}$ because the bond length is shorter.

           Dipole   B.P      Mol.      Electro-  C-X     C-H
           Moment   (°C)     Wt.       Negat.    Length  Length
 CH3F       1.85    −78.4    34.03     4.0       138.5   109.5
 CH3Cl      1.87    −23.8    50.49     3.0       178.4
 CH3Br      1.81      4.0    94.94     2.8       192.9
 CH3I       1.62      42     141.94    2.5       213.9

The trend for molecular weight looks good, but the deuterated form of methane, $\ce{CD4}$, has the same (or at least nearly the same) boiling point as $\ce{CH4}$.


Can silica gel beads cause silicosis if crushed? Can silica gel beads (which I believe are amorphous) when crushed to dust cause silicosis?

No, it won't. Silica beads dust may cause irritation to the skin and eyes. Unlike silica crystalline,  synthetic amorphous silica gel is  indurated, and so does not cause silicosis.


TL;DR: PM6 only works for elements up to fluoride in GAMESS.

I have played around with the GAMESS version installed at our cluster, the ancient 5 DEC 2014 (R1) and I came to the conclusion that there must be a 'bug' in the implementation of PM6 with GAMESS. (I reproduced the problem with 20 APR 2017 (R1); I have also downloaded 30 Jun 2019 (R1), but I won't try it as I think the result is the same.) When I tried it with Gaussian 16 Rev. B.01 on the PM6 level of theory it converged fine.

Actually, I have looked at the documentation, and it seems to be officially not supported (from INTRO.DOC):

Capabilities

   [...]
   3. Calculates semi-empirical MNDO, AM1, or PM3 models
      using RHF, UHF, ROHF, or GVB wavefunctions.
   [...]

Or from INPUT.DOC:

        * * * semiempirical basis sets * * *

GBASIS = MNDO - selects MNDO model Hamiltonian
       = AM1  - selects AM1 model Hamiltonian
       = PM3  - selects PM3 model Hamiltonian
       = RM1  - selects RM1 model Hamiltonian
       = DFTB - selects tight binding Hamiltonian

I am pretty sure the error is associated with sulfur. I have tried a few molecules, among them $\ce{H3C-NH2}$, $\ce{(H3C)2NH}$, $\ce{(H3C)2CO}$ (see input below), urea, which converged fine. Whenever I tried a molecule containing sulfur the SCF seemed to be oscillating. I have tested $\ce{CS2}$, $\ce{SO2}$, $\ce{(H3C)2CS}$, and (obviously) thiourea. (With PM3 everything worked fine.)

My best guess is that they simply didn't finish implementing it. You could issue a support ticket on GitHub: issues. I found an article by (the here known) Jan Jensen from 2013: Project status: PM6 and PCM in GAMESS. And then from 2014: PM6, DFTB, GAMESS, enzyme design, protein-ligand binding, and protein structure validation. I stopped there.

I'd recommend using a different program, the obvious choice would be MOPAC (http://openmopac.net/). Gaussian has it implemented and a couple others probably too. Or you switch to a completely different method like xtb, HF-3c (in ORCA), DFTB (e.g. DFTB+), etc.. Depending on your system size, you can probably also get DFT based calculations done.

Here is also something related: GAMESS crash course?

 $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE MAXIT=100 MULT=1 COORD=UNIQUE
    NOSYM=1 $END
 $BASIS GBASIS=PM6 $END
 $DATA
 C3H6O
C1
CARBON      6.0      0.156332162        -0.059231571         0.417752882
OXYGEN      8.0      1.200569512        -0.535604279         0.831319560
CARBON      6.0     -0.037914597         1.466530485         0.340822109
HYDROGEN    1.0      0.279389785         1.860652999        -0.631337255
HYDROGEN    1.0     -1.089141584         1.742544046         0.481697509
HYDROGEN    1.0      0.545018653         1.982387227         1.112307247
CARBON      6.0     -1.053925162        -0.882111362        -0.061565447
HYDROGEN    1.0     -0.987243509        -1.102851552        -1.132983096
HYDROGEN    1.0     -1.121341433        -1.838731138         0.468944868
HYDROGEN    1.0     -1.993938204        -0.344384630         0.106761952
 $END


Understanding how mass spectroscopy works I’m trying to get a deeper understanding of how mass spectroscopy works. Most tutorials and textbooks I’ve encountered omit certain details about the process and I’m hoping someone out there who understands the process can fill in the gaps. I’ll start by explaining the way I currently understand the process and then list where I start to get confused.

1 First a particular molecule is bombarded with a beam of electrons (ionization step). This step works to free an electron from the molecule under analysis which generates a cation.

Note: It’s important to remember that the only reason why this molecule was ionized is to make use of a magnetic field to bend its direction. Only ions will be affected by this magnetic field which allows us to pinpoint the molecule in question.

2 Next, these ions are accelerated by an electric field toward a magnetic field that bends the moving ions by a certain amount.

  Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields.

3 The ions that bend will travel around a tube by a certain degree and hit a detector. The amount of bending will tell us about the mass of a particular molecule. Heavier molecules will bend less and lighter molecules will bend more.

  Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.

Any help understanding this concept would be appreciated.

Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields.

Electric fields accelerate charged particles (linear acceleration). Magnetic fields deflect them (angular acceleration).

Charged particles respond to electric fields and magnetic fields differently. Electric fields accelerate charged particles linearly in the direction of the flux. For example, if an electric field exists between two plates, one with a buildup of positive charge and the other with a buildup of negative charge, then positively charged particles will accelerate from the positive part toward the negative plate. The acceleration of the particle ($\vec{a}$) depends on the strength of the field ($\vec{E}$), the mass of the particle ($m$), and the charge on the particle ($q$). Since the acceleration is a scalar multiple of the electric field, they have the same direction.

$$\vec{F}=q\vec{E}$$
$$\vec{F}=m\vec{a}$$
$$q\vec{E}=m\vec{a}$$
$$\vec{a}=\frac{q}{m}\vec{E}$$

Magnetic fields are orthogonal to electric fields. Magnetic fields alter the angular acceleration of charged particles (i.e., they deflect them). Thus, charged particles have a circular trajectory in a magnetic field. The radius of the circular path ($r$) is dependent on the strength of the magnetic field ($\vec{B}$), the charge of the particle ($q$), the mass of the particle ($m$), and the velocity of particle ($\vec{v}$). The radius of curvature is proportion to the mass to charge ration ($m/q$).

$$|\vec{F}|=m\frac{|\vec{v}|^2}{r}$$
$$|\vec{F}|=|q\vec{v}||\vec{B}|$$
$$m\frac{|\vec{v}|^2}{r}=|q\vec{v}||\vec{B}|$$
$$r=\frac{m}{q} \frac{|\vec{v}|}{|\vec{B}|}$$

The acceleration of the particle depends on the velocity, magnetic field, radius, and mass. Since acceleration in a magnetic field is a scalar multiple of the cross product of velocity and field strength, the direction of acceleration is always changing.
$$\vec{F}=q(\vec{v}\times\vec{B})$$
$$m\vec{a}=q(\vec{v}\times\vec{B})$$
$$\vec{a}=\frac{q}{m}(\vec{v}\times\vec{B})$$

  Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.

Now that we know the relationship between radius of curvature and mass-to-charge ratio, this question is easy. The detector counts the number of particles that hit it. That is all that it can do. That is all that it needs to do. The hard work of separating the ions by m/q is done by the mass analyzer, which is the magnetic field in curved tube. The magnetic field scans (quickly) over a range of $\vec{B}$ to alter the curvature of the ions by their m/q. Since the tube has a fixed curvature, only the ions with the correct m/q will pass to the detector. All others will hit the walls of the tube since their radius of deflection would be too large or too small. Thus, since the instrument controls the strength of the magnetic field, and physics determines what m/q will have the correct deflection at that field strength, there is no need to calibrate the detector.

More complex mass analyzers exist, for example the quadrupole mass analyzer. Instead of the typical bipolar magnetic field, it generates a quadrupolar magnetic field. Ions are thrown all over the place to hit a spherical detector. The physics is more complex, but the trajectory once again depends on m/q, so the detector only has to count hits in certain spots.


What software was used to create these organic reaction scheme figures? I am aware that software such as ChemDraw, ChemDoodle or MarvinSketch can be used to draw skeletal formulas and the like. But what can I use to draw (in a similarly professional way) reaction schemes, including the skeletal formulas, arrows, reagents/conditions above the arrows etc. such as

$\mathrm\LaTeX$ to the rescue
I reproduced these reaction schemes with $\mathrm\LaTeX$ using the following particular packages:

chemfig for drawing 2D chemical structures (note that I enabled sans-serif font for the labels like the original did);
chemmacros for the typesetting of chemical formulas and chemistry-related notations;
siunitx for handling physical quantities (here, temperatures);
stix for loading STIX fonts, which in my opinion are the best choice for technical writing due to canonical and coherent appearance of the glyphs and great coverage for special symbols.

All of this is free for any type of use, works on all major operating systems used in production and the source code is available.
Result

Source code
% !TEX program = pdflatex

\documentclass{report}
\usepackage[T1]{fontenc}
\usepackage{stix}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathtools}
\usepackage{siunitx}
\usepackage{chemmacros}
\chemsetup{
    modules = all,
}

\newcommand{\bndlen}{1.7em}
\newcommand{\bndlenshort}{0.5}
\newcommand{\bndlenshorter}{0.75}
\newcommand{\bndlenlonger}{1.5}
\newcommand{\bndlenlong}{1.75}
\newcommand{\lnwidth}{0.8pt}
\newcommand{\cramwid}{0.3em}

\usepackage{chemfig}
\setchemfig{
    angle increment = 15,
    atom sep = \bndlen,
    bond offset = 1pt,
    double bond sep = 3pt,
    cram width = \cramwid,
    compound sep = 5.0em,
    arrow coeff = 1,
    arrow offset = 12pt,
    scheme debug = false,
    bond join = true,
    chemfig style={line width=\lnwidth},
    arrow style={line width=\lnwidth},
}
\renewcommand*\printatom[1]{\ensuremath{\mathsf{#1}}}

\begin{document}

\schemestart
    \chemfig{MeO-[::30]*6(-=-(-~--[::-60]CHO)=-=)}
    \arrow{->[\parbox{5cm}{\centering
        i) \ch{Hg^2+}, dil.~\ch{H2SO4}\\
        ii) \ch{AgNO3}, \ch{NH4OH}
    }][
        iii) \ch{Zn}-\ch{Hg}, conc.~\ch{HCl}]}[,2.5]
    \textbf{Q}
    \arrow{->[\parbox{5cm}{\centering
        i)~\ch{SOCl2}\\
        \hphantom{i)~}pyridine
    }][
        ii) \ch{AlCl3}]}[,1.5]
    \textbf{R}
    \arrow{->[\parbox{5cm}{\centering
        \ch{Zn}-\ch{Hg}\\
        conc.~\ch{HCl}
    }]}[,1.5]
    \textbf{S}
\schemestop

\bigskip

\schemestart
\chemfig{(-[::120]*5(---=-))-~-(-[::-60]*6(---=--))}
\arrow{->[
        i) \ch{H2}, \ch{Pd}-\ch{BaSO4}, quinoline
    ][
    ii) dil.~\ch{KMnO4}~(excess), \SI{273}{\kelvin}]}[,3.5]
\textbf{P}
\schemestop

\bigskip

\schemestart
    \chemfig{H_3C-[::-30](-[::-60]Br)-[::60]-[::-60]Br}
    \arrow{->[\parbox{5cm}{\centering
        i) alc.~\ch{KOH}\\
        ii) \ch{NaNH2}
    }][
        iii) red hot iron tube, \SI{873}{\kelvin}]}[,3]
\schemestop

\end{document}


Nernst equation calculations I am stuck on 2 problems and would be grateful if anyone decided to give me a proper explanation as to how to solve them. I am not sure how to use the given information to correctly solve what is being asked. I have included the answers so its not a matter of academic integrity, I am genuinely lost.

  Given the equation:
  $$\ce{2X^3+ + 3 Y -> 3 Y^2+ + 2X}$$
  The standard reduction potential for the X cation is $\pu{0.498 V}$.
  The standard reduction potential for the Y cation is $\pu{0.187 V}$.
  Using the Nernst equation, calculate the $\log$ of the equilibrium constant $K$ for the above reaction.

The answer is $\log K = 31.52$.

Using the answer, I attempted to work backwards:

$$0.685 = \frac{0.0592}{n} \times 31.52 \implies n = 2.724 ,$$

which I don't understand since $n$ is supposed to be the number of electrons.

  Given the equation:
  $$\ce{2 X+ + Y -> Y^2+ + 2 X}$$
  The equilibrium constant for the reaction is $5.912 \cdot 10^{13}$.
  You measure the concentration of $\ce{Y^2+}$ to be $\pu{0.0674 M}$.
  Calculate molar concentration of $\ce{X^+}$.

The answer is $\pu{33.8 nM}$. This one I'm not even sure.

For (1), I think you made the mistake of using a different value for overall cell potential when you worked backwards. Because if you use the correct overall cell potential, you get $K = 31.52$, your $\log K$ would be different if you used $0.685$ (which would make $\log K = 69.43$). Using the correct $E_\mathrm{cell}$ will get you the correct value of $n$ that you're looking for.

To solve for (2), think of what $K$ stands for. It's the ratio of something. Rearrange the ratio to solve for the unknown. It might seem confusing since you might think of an electrochemical solution, but that isn't always the case.


The effects of H2o2 (hydrogen peroxide) with oils I have some 35% $\ce{H2O2}$ that I've been cutting down to 3% to create a product. Cheaper than buying pre cut solution from store.

However, I've been thinking about trying to try it out with producing a new product which would induce mixing of the $\ce{H2O2}$ with pure oils and essential oils. I am not sure if the $\ce{H2O2}$ would break down the oils and render them useless in the product or not.

I select $\ce{H2O2}$ cause I want it to be able to kill germs to help kill a particular smell. But I want the oils to maintain their properties in which they keep there defined smells. So pretty much the $\ce{H2O2}$ is a catalyst to make to oils more liquidity able to spray and also helping hide original scent.

Oils will oxidize (and burn, too). With dilute H2O2, it's not a rapid reaction (in laymans terms)... some oils (like linseed) will oxidize more rapidly than others (like mineral oil).

Essential oils are not really oils... they are more like phenolic and terpenic extracts.


Combustion of saturated hydrocarbons vs combustion of unsaturated hydrocarbons I read in my science book that saturated hydrocarbons burn with a blue flame (indicates complete combustion) and unsaturated hydrocarbons burn with a yellow flame (indicates incomplete combustion) due to the highier percentage of carbon in unsaturated hydrocarbons as compared to saturated hydrocarbons. So I think the reason why unsaturated hydrocarbons don't burn completely is the presence of double and triple bonds in unsaturated hydrocarbons which are more stronger than single bonds because of which they are difficult to break.  There was a question that :-

Why is a mixture of ethyne and oxygen used for welding?

And the answer was(as written in the book) :

So that ethyne burns completely without producing soot but according to me the issue is not about availability of oxygen.
As I saw in the reactions of combustion, unsaturated hydrocarbons require less oxygen to burn than saturated ones. For example :-
$$\ce{C2H6 + 7/2 O2 = 2CO2 + 3H2O}$$
And
$$\ce{C2H2 + 5/2 O2 = 2 CO2 + H2O}$$
Here we can see that unsaturated hydrocarbons require less oxygen as compared to saturated hydrocarbons.
So my queation is :-

Is the answer given in the book correct. If yes, why?
$$Thank you$$

You are viewing it from the wrong perspective.
In the case of normal burning, Generally, the oxygen is the limiting reagent (therefore some of the hydrocarbons left behind ).
Let's take 1 mole of Oxygen.
So you have to take  0.285 mol of ethane (0.57 mol carbon )  and 0.4 mol of ethyne (0.8 mol of carbon).
It's quite clear ethyne has more carbon content for a given amount of oxygen. As if some hydrocarbon get left, Ethyne will give more carbon content.

https://www.youtube.com/watch?v=FJi4ODPQSE0

(Note: If concentration of oxygen is very high and enough temperature is given then both will burn with Blue Flame, as it is done in case of welding
https://www.youtube.com/watch?v=hKMDxag8yAo )

Now In the case of welding, welding is done in presence of excess of Oxygen, so there is no question of limiting reagent of oxygen.
What matters is Auto Ignition point, Ethyne has much lower auto ignition temperature that is 305 °C as compared to ethane which has 515 °C.

Of course, you pointed out above that for, given 1 mol of ethyne and ethane, ethyne need less oxygen to be burned, so it can be taken in quite concentrated form.

You can take maximum 80% Ethyne (concentration in air), it will burn.
Whereas ethane can't be burn as its concentration exceeds 12.5% (in air).

More concentrated heat implies Higher can be temperature achieved.


How to rotate a chiral carbon I know that to correctly assign R,S priorities I have to rotate the above figure until the H is projecting into the page (away from me). However, I don't know how to rotate the H and the rest of the molecule when the H is in the plane of the page.

When I imagine tilting the molecule back I get a see-saw shaped molecule ... obviously incorrect.

Imagine viewing it from right side:

H is upwards and CH3CH2 is downwards.
OH is leftwards and CH3 rightwards.
H and CH3CH2 are backwards(from right) so they must be on vertical line of Fischer projection and similarly for OH and CH3.

Alternatively:

Put the dotted wedge CH3 in any vertical position let it be bottom.
Then imagine all other groups to be planar, we see in clockwise order:

CH3CH2 , H , OH

Draw them similarly on the Fischer projection in clockwise sense.!

Note that both are (S)


Why would a ~1 cm thick layer of argon be a significantly poorer conductor of heat than air? In yesterday's new Periodic Video, Argon (new) - Periodic Table of Videos, after about 07 min 00 sec, Sir Martyn Poliakoff says:

The final, and I have to admit perhaps slightly boring application of argon, is in double glazing (of windows) to keep our houses warm. Because argon has a heavy atom, the atoms don’t move very fast in the gas phase, and therefore they’re bad at conducting heat (emphasis added).
So if you have double glazing (that’s two sheets of glass in your window) if you put argon between the glass, it is much more effective at insulating the inside of the house, keeping cold out, than if you use air, because air has a much higher conductivity of heat.
So, if you look out of the window, you may find that you are looking through argon gas.

At a given temperature or kinetic energy per molecule, velocity will roughly scale with the inverse square-root of the mass, so I estimate that the argon atoms would be moving
$$\sqrt{\frac{28}{40}} \approx 0.84 $$
only 16% slower than nitrogen molecules would be moving.
Why then would a trapped ~1 cm thick layer of argon gas be a significantly poorer conductor of heat than nitrogen gas or air?
I'm wondering if it has more to do with the fewer internal degrees of freedom of a monatomic gas, than it does with the mass of the molecule. There would be a similar collision rate, but each collision of an atom of Ar could transfer less energy to the glass surface than a molecule of N2 could transfer.

The thermal conductivity coefficients of some gasses in units of W/(m•K) are helium 0.142, argon 0.016, air 0.026, methane 0.03, propane 0.015, bromine 0.004, and steam 0.018. These are all rather similar except for helium and bromine.

Using the kinetic theory of gasses the thermal conductivity coefficient is given by $ \kappa \sim n\bar c\lambda C_V$ where $n$ is concentration, $\bar c$ the mean speed, $\lambda$ the mean free path and $C_V$ the heat capacity. Using definitions $n\lambda \sim 1/\sigma^2$ where $\sigma$ is the diameter of a gas atom or molecule; $\bar c \sim 1/\sqrt{m}$ and the heat capacity is $C_V \sim sk_\mathrm B$ where $k_\mathrm B$ is Boltzmann constant and $s$ is a multiplier depending on the type of species, atom, diatomic molecule, etc.

The conductivity coefficient at constant temperature is therefore $\displaystyle \kappa \sim \frac{sk_\mathrm B}{\sigma^2\sqrt{m}} $.

Thus we can see that helium will have high $\kappa$ as it has a small diameter and mass, and bromine will have a small value of $\kappa$; large diameter and mass. The other gasses are similar to one another because although the heat capacity increases between an atom and molecule and increases further as the molecule has more vibrational modes, the diameter and mass also increase when the molecule has more atoms and these terms cancel out to some extent.

So it is not just about speed, but also size, mass and number of vibrational modes. Using argon over dry air would seem to give a small advantage, but it may not be cost effective, i.e. heat saved vs. initial cost and energy needed to use argon.

(The molecular heat capacity increases above that for translational motion with the number of squared terms in the energy, potential and kinetic, each term contributes $k_\mathrm BT/2$ to the energy and so $k_\mathrm B/2$ to the heat capacity. Each vibration populated thus contributes $2k_\mathrm B/2$ and rotations the same for a linear molecule, but $3k_\mathrm B/2$ for a not linear one.)


Calculating cell potential First of all, is the problem even worded correctly? From how the problem is worded, it seems that copper is being oxidized as well as iodide ion. This is what I am taught:

Second of all, I'm not getting 0.24 volts as the cell potential. I'm getting 0.19 volts.

Here is my work. What am I doing wrong?

In electrochemistry you always write the half reaction with reduced species in the left and oxidized species on the right. This is done for consistency purposes. Species with higher potential will reduce and species with lower potential will be oxidize.

I$_2$ + 2e- ----> 2 I$^-$   +0.54     //will reduce (go from left to right), has positive charge
Cu$^{2+}$ + 2e- ---> Cu  +0.34       //will oxidize(go from right to left), has negative charge

Next, you should use Nernst equation to each half reaction. You will get values E$_{Cu^{2+}/Cu}$ and E$_{I_2/I^{2+}}$. Now subtract one from another to get the voltage.


If atoms are mostly empty space then how are we able to interact with things? Note: This question has come out of curiosity

Q: We all have read that atoms are mostly empty, they are generally compared to a football stadium in which a pea is kept in the center (nucleus) and flies buzzing in stands (electrons). Then why is it that we can interact with things?
It would be of great help if anyone can provide me with any article because I can't find any.

When we think about the structure of atoms and molecules we need to use a range of simplifying analogies as the objects we are talking about are too small to see and behave according to quantum rules we are unfamiliar with in the macroscopic world. None of the analogies we use are perfect and may mislead unless they are brought back to reality by recognising their limitations an checking their implications again the mathematics we know that does describe their behaviour.

That is the problem here. On one analogy atoms are built from nuclei containing protons and neutrons surrounded by electrons orbiting a long way away. This analogy is useful for explaining what happens when, for example, alpha particles are fired at a gold film a few layers of atoms thick. This is the famous Rutherford scattering experiment first done in 1909. This experiment led to the "solar system" analogy for what atoms look like. But really it only explains what they look like from the point of view of fast alpha particles. The mass of atoms is concentrated in the small nucleus and it is the nucleus that alpha particles bounce off, occasionally.

But that is only one, imperfect, way to describe what atoms and molecules look like. It doesn't describe how individual atoms join to form molecules, for example. For that we have to look at how electrons behave. Here a different range of analogies or mathematical descriptions are needed. The solar-system view of electron behaviour is deeply inadequate to describe what electrons actually do. And it is a very poor description of where electrons are. It is better to think of electrons as occupying 'clouds' of probability than being at a single point in space. Those clouds have a range of shapes from simple spheres to multi-lobed, multi-layer things in atoms with many electrons. Many of these have some probability of the electron being anywhere inside the cloud even, in some cases, right into the nucleus. In this analogy atoms are not empty at all but are filled with some probabilistic amounts of electron. No more than two electrons can occupy any orbital which imposes some strong constraints on what can happen when the clouds of two atoms or molecules meet.

Almost all the interactions we experience in normal macroscopic life involve those clouds of electrons interacting with each other. There are a range of different ways they can do this, but at the heart of things they all consist of electrostatic interactions at the atomic or molecular level where electron clouds push (or pull) each other. The low density of electron clouds doesn't matter since the quantum mechanical rules determine what can happen when the clouds meet. Normal interactions in the real world are just the macroscopic impact of these interactions.


What difference between pi bond and bent bond? Is it possible to replace sigma-pi bond model and use only bent bond instead? Between sigma-pi bond and 2 bent bonds, is it actually equiavalence? Or there is specific molecule that require to be bent bonds and cannot exist as pi bond and vice versa? I felt like we actually don't need pi bond model at all if we consider it all to be bent bond
If it not, then how can we consider which chemical result will be bent bond, and which will form pi bond?

In the famous words of George Box, "All models are wrong. Some are useful."
The bent bond and sigma/pi descriptions are two apparently distinct models for representing the underlying structure of electron density of molecules. Both are oversimplifications, but the key question (as with any model) is which is better for accounting for all of the known molecular structures and predicting the results of observations we have not yet made.
There are examples that support both. Bent-bonds have been cited famously as better predicting the conformation of propene, but the sigma/pi gives more intuitive results for reactions like the Diels-Alder. As in the case of the VB vs MO theory debate, both models can be elaborated with greater complexity to the point that they give indistinguishable results, but the sigma/pi approach has largely taken over because in its basic form it is more intuitive and gives accurate predictions in most cases.
The general concept of a bent bond remains useful for highly strained systems such as cyclopropanes, where the electron density of a single bond does not appear to align with the axis connecting the atoms.
To answer your final question,

If it not, then how can we consider which chemical result will
be bent bond, and which will form pi bond?

it is rare that the sigma/pi model will fail you in modeling a molecule with a double bond, and since it is the prevailing model pedagogically, you would do best to use it preferentially and only invoke bent bonds for strained single-bond systems.


Kinetic energy of an ideal gas I've heard that the kinetic energy of any different gases are the same

  All gases at a given temperature have the same average kinetic energy.

But if we have 2 moles of hydrogen and 4 moles of helium in a container at a specific temperature will the kinetic energies become the same?
I think it cannot be as the n is different..

As the kinetic energy is
3nrt/2

This might be a small question but this has been a conflict between my friends..

The answer is in front of you. $E_v$ is the Total translational  kinetic energy not the Average kinetic energy. If you want average translational kinetic energy per mol this is given by: $E_v/n$ or $3/2RT$.


Why is silicon used in computer chips? Most computer chips contain or are made of silicon. Why silicon? Why not copper or tellurium or some such?

Most electron components of computer chips and boards are semiconductor devices like transistors and diodes.
Silicon is, from the Wikipedia article on semiconductor devices:

  By far, silicon (Si) is the most widely used material in semiconductor devices. Its combination of low raw material cost, relatively simple processing, and a useful temperature range make it currently the best compromise among the various competing materials

Silicon is the second most common element in the earth's crust after oxygen, and present in a variety of silicate minerals.

Electronics grade silicon can be produced easily by molten salt electrolysis of these minerals followed by purification by zone refining.

Copper is a conductor, and therefor not suitable for semiconductor devices. Tellurium is one of the rarest elements in the earth's crust.


Factors that influence the kinetics of an irreversible exothermic reaction I'm trying to understand what are the main factors that influence the reaction rate of an irreversible exothermic reaction.

I think these could be the main factors:

chemical nature of the reagents and their concentration;
nature of the solution (homogeneous or heterogeneous)
presence of catalysts
temperature.

(Let me know if I missed some factors or added one or two that had nothing to do with what I'm looking for.)

But I struggle understanding how the temperature influences an exothermic reaction. I can't use Le Chatelier's principle since the reaction is not reversible, so how do I know what is the effect of (for example) an increased temperature on such reaction?

I'm prone to think that an increased temperature will bring more products, but then again I'm not so sure since is an irreversible reaction.

I assume by kinetics you mean the rate of the reaction (i.e. how fast it happens), in which case the factors you mention are fine.

Regarding the temperature: the rate constant and therefore the reaction rate at a given concentration always increases with temperature as you can see in the Arrhenius equation. The fact the the reaction is exothermic means that it releases heat which, in the case where no adequate temperature control is applied (such as an ice bath), will increase the temperature of the solution and therefore the rate of the reaction releasing even more heat and so on, resulting in a runaway exotherm. This is the reason why exothermic reactions must be well temperature controlled especially when performed in large scale.

If your reaction was reversible, then heating it would not only accelerate the reaction (both forwards and backwards at equilibrium) but also shift the equilibrium towards the reactants.


Hydrochloric acid vs hydrogen chloride? If one of them is an aqueous solution, in this case hydrochloric acid, why doesn't it have a different chemical formula ? How would I explain the reaction between hydrogen and chlorine without the use of water to make it acidic ?

Hydrochloric acid is an aqueous solution of hydrogen chloride.


How would one express pemoline in terms of phenethylamine? Pemoline is a psychostimulant that is no longer used in most countries. Similarly to the psychostimulants amfetamine and methylphenidate I can clearly see a phenethylamine backbone to this compound. I know it is also an oxazoline, but I am writing an article on my Wiki about substituted phenethylamines and their pharmacologic and structural properties so I would like to know what its chemical name would be in terms of its phenethylamine backbone. If you want an example of what I mean, see the current revision of my Wiki article on said drug [1] and look at the chemical names I have written. For example for methylphenidate the chemical name is N,α-butyl-β-methoxycarbonylphenethylamine.

My chemistry knowledge is only first-year university chemistry level, hence why I need your help as I don't know all the adjective forms of functional groups.

Below is the 2D structure of pemoline if you would like it in order to deduce its name in terms of phenethylamine.

I would like to know what its chemical name would be in terms of its phenethylamine backbone.

Artificial numbering and naming schemes that only exist in ones own cloud cuckoo land are pointless.

IUPAC nomenclature exists for a reason.

Besides the IUPAC nomenclature, other naming schemes - based on established (trivial) names exists and are widely used - for good reason and with acceptance by the scientific community. Typical examples are found in the field of natural products chemistry (steroids, alkaloids, etc.)

It is however crucial to understand that in these cases, prefixes and/or suffixes denote a variation of the underlying core structure, such as

additional substituents
lack of methyl groups
an elongated side chain, etc.

In the current case, the central structure is a completely different heterocyclus, so markedly different from the original phenethylamine that it would not make any sense to apply a phenethylamine-based name here.

Neither would one try to apply such a naming scheme to the morphine skeleton - despite the fact that one might spot a six-membered ring with $\ce{CH2-CH2-NCH3}$ side chain here.

This does not mean that a member of the phenethylamine family is not among the ancestors of morphine in the biosynthesis! In fact, reticuline, a morphine precursor, most likely is a grandchild of dopamine.

But again, at a certain point structures have become so markedly different that a valid nomenclature (= valid for a particular compound class) no longer is applicable.

Trying to describe similarities between structures is not wrong, but trying to apply an inappropriate naming scheme is not helpful here.

Similarities between molecular entities should be expressed otherwise.
For sequences (DNA, RNA, etc.), Hamming and Levenshtein distances can be used. For the title molecule and the whole dopamine/amphetamine class, I'd consider the comparison of their fingerprints (Tanimoto coefficients).


Finding new pH after NaOH added to buffer solution I have a buffer containing 0.2 M of the acid $\ce{HA}$, and 0.15 M of its conjugate base $\ce{A-}$, with a pH of 3.35. I need to find the pH after 0.0015 mol of $\ce{NaOH}$ is added to 0.5 L of the solution.

I started by converting everything to moles, writing out an equation for $\ce{HA}$ and $\ce{OH-}$ reacting to make $\ce{H2O}$ and $\ce{A-}$, getting 0.0985 mol of $\ce{HA}$ left after all the $\ce{OH-}$ was used up, and 0.015 mol of $\ce{A-}$ produced.

At that point I got a bit stuck. I tried doing things a bunch of different ways after that, such as trying to calculate $\mathrm{p}K\mathrm{_a}$ by using the Henderson-Hasselbalch equation, but I'm not getting anywhere near the answer. I calculated a $\mathrm{p}K\mathrm{_a}$ of 3.47. I added the amount of $\ce{A-}$ made when $\ce{HA}$ reacts with $\ce{OH-}$ to the original moles of $\ce{A-}$ in the buffer (I calculated 0.18 M $\ce{A-}$), and tried putting that, the leftover moles of $\ce{HA}$, and my $\mathrm{p}K\mathrm{_a}$ value into the equation, but that wasn't right either.

What am I doing wrong?

Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$

Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$

Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$


How to separate HCl and NaOH from salt and water solution? I mean, how can we reverse the neutralization reaction of $\ce{HCl}$ and $\ce{NaOH}$ to get back $\ce{HCl}$ and $\ce{NaOH}$.

I want to extract $\ce{HCl}$ and $\ce{NaOH}$ from a mixture of common salt and distilled water in equal proportions at home.

Basically, how can I make this reaction happen?

$$\ce{H2O + NaCl -> HCl + NaOH}$$

What temperature or pressure conditions are required?
Also, would the extracted components be pure? (I know common salt contains iodine and even distilled water is not pure, that's why I am asking this.)

Unless time travel is an option, you could

Electrolyze the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases.
Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation  at $\lambda$ < 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the chain reaction to yield $\ce{HCl}$ gas.

$$\ce{Cl2 ->[h\nu] 2 Cl*}$$

$$\ce{Cl* +\  H2 -> HCl + H*}$$
$$\ce{H* +\  Cl2 -> HCl + Cl*}$$

EDIT 1

Unless you know exactly about the risk of both processes, don't do it!
This video gives a nice impression of how violently hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect!

EDIT 2

The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can not simply revert the neutralisation reaction.

The neutralisation is irreversible.

In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the entropy changes.


Why does a 1:1 mixture of Polyvinyl Alcohol (PVA):H2O decompose/carbonize during heating? I bought a powder of Polyvinyl Alcohol (PVA from Sigma Aldrich). I mixed the powder with water in a mixture 1:1 to place a droplet on a hot plate. To figure out the melting temperature of the PVA, I heated it, meanwhile the water should be evaporated. According to Wikipedia or other chemical databases, partially hydrolysed grades of PVA melt between $150-180\ ^\circ C$
.

My questions:

Why did my mixture decomposed/carbonized above $120\ ^\circ C$
(its color changed from transparent to brown, so I guess it decomposed)?
The mixture did not melt between $150-180\ ^\circ C$
. Why?
Did I make any general mistake at all?

I experienced similar phenomena in my experiments.
In my case, I put PVA pallets (M.W. is 84000~12500, 87~89% hydrolyze degree) in the oven at 120∘C. After over 12 hours, the color became yellowish (partially brown). But after dissolving the pallets in the water, there was no difference between control sample(without heat treatment).

There are several papers about formation of polyene in PVA chain under acid catalyst. And, almost commericial PVA product has always small amount of acetic acid which hydrolyzed from PVAc spontaneously. Therefore partial polyene-nization (semicarbonization) was possible with heating under melting temperature.

Also, I made acidic PVA film by solution casting (10% PVA, pH 2 by adding HCl). Then, the color of PVA film was changed to black at 120∘C in the oven.

So color change of PVA is common while experiments. Maybe it is related with carbonization.

maybe polyenenization or carbonization.
I remembered that PVA becomes viscous near 150-180∘C. Measuring Tm of polymer exactly is not easy. Especially hot plate is not suitable tool to analysis physical properties of polymers. Maybe due to cooling by air, melting of PVA is indistinguishable to glass-transition. I recommand to use DMA instead of your experiment.
To measure Tm, you don't have to mix PVA with water. DSC or DMA is better.


What is the nucleophile in alkaline hydrolysis of a haloalkane? We know that haloalkanes are able to undergo $\mathrm{S_N1}$ or $\mathrm{S_N2}$ in alkaline hydrolysis, and the hydroxide ion is the nucleophile responsible for attack. But when we talk about hydrolysis, isn't it a reaction with water? Where does the hydroxide ion that attacks the haloalkane really come from; the water, or the alkaline catalyst?

Consider the reaction of a haloalkane with aqueous sodium hydroxide to produce the alcohol plus sodium chloride. The net reaction consumes one hydroxide ion for each alcohol molecule produced, so it is not really correct to refer to it as a catalyst.

(1) If this reaction occurs via the $\mathrm{S_N2}$ mechanism, a one-step displacement mechanism, you are correct to say that hydroxide (being a much better nucleophile than neutral water) is the nucleophile attacking carbon.

(2) If the reaction occurs via the $\mathrm{S_N1}$ mechanism (note that which mechanism is preferred depends on the structure of the haloalkane, and in some cases a haloalkane can react via both mechanisms), you have a two-step process in which the haloalkane initially dissociates (a slow step) to form a carbocation plus halide ion. The carbocation then reacts very rapidly with whatever nucleophile is handy to form the alcohol. Water is likely more abundant than hydroxide, and is a strong enough nucleophile to react instantly with a carbocation, so most of the product will be formed that way, with release of $\ce{H+}$, rather than by combination of the carbocation directly with hydroxide. The $\mathrm{S_N1}$ reaction of a haloalkane in this manner with excess water or other polar solvent is often referred to as a "solvolysis" reaction.


Could anyone suggest some catchy chemistry experiments or demonstrations, to be performed in front of a young audience? I'm wanting to volunteer at some of our local children centers, and do some experiments to interest them in Science/Chemistry. One cool, flashy experiment is to light my hand on fire - and not have to go to the hospital after.

The general scheme I've found for it (I've not tried it - and the instructions aren't .edu or other '95% correct' sites, so if you know of any corrections for this please chime in) is:

50 - 50 EtOH and Water, plus a small amount of salt for color. Dip your hands in, light them on fire, the EtOH burns coldly enough that the water doesn't evaporate and your hand stays safe and insulated.

Of course, I'll test it on small scale on paper in lab first.

However, I'm initially hesitant to play with fire around children. For obvious reasons.

Does anyone have anything similarly flashy or attention grabbing that they've used for a similar purpose with children? Obtaining restricted materials isn't an issue, but low-cost is preferable.

Note: I know this stackexchange frowns upon questions that don't have a specific answer, with the rules stating that discussion type questions are non-preferred - but this question isn't for the meta site, and I trust this community of educated scientists to have better advice on experiment safety than say, the educator stackexchange, so forgive me if this doesn't fit the general site mold.

To talk to kids about chemistry I don't think you just need some exotic experiments.

I visited the Sterling Hill Mining Museum and one of founders had a neat gimmick that got kids attention. He asked them to name some product that had nothing to do with mining. He'd then rattle off a number of things that are mined to make that product. So for a truck the oil and gas are from an oil well. The steel comes from an iron mine and the copper in the electrical wires comes from a copper mine. So have the kids try to name a product that doesn't use chemistry.

Another thing that has always bugged me is the perception of the general public about "chemicals." It seems that the public believes that only things created in some chemical processing plant are chemicals. Well table salt is a chemical. Sugar is a chemical. A strawberry is a bag of water with a zillion chemicals in it. Take a vitamin tablet? The tablet is full of chemicals that our bodies need. Our bodies run on chemistry. So chemicals are all around us.


Why doesn't liquid propane in a closed coke bottle burst? I was watching this YouTube video in which the original experiment was to make a small rocket out of liquid propane and Coke. When that failed, the person doing the experiment decided to try pouring liquid propane into the Coke bottle and closing the cap on it to see what would happen.

I would have expected (and I believe he did too) that the bottle eventually would build up enough pressure to explode. However, that did not happen and along with that not happening, the liquid appeared to warm up to about air temperature and stopped boiling. When he opened the lid of the bottle, it quickly depressurized, the liquid began boiling and returned to its normal, cold state.

My guess is that once the propane vapors filled the bottle and expanded sufficiently, it put enough pressure on the liquid to keep it from boiling. So my question here is, is my guess correct and, along those lines, if it had been warmer outside, would that bottle have built up enough pressure to explode?

Note: Even if my guess is correct, I want to see the hows and whys of it in an answer.

Yes, based on what we can see in the video, your guess appears to be correct: as the propane-filled bottle warmed up, just enough propane evaporated to keep the pressure inside the bottle equal to the equilibrium vapor pressure of the liquid propane.

According to the video, the ambient temperature outside at the time it was recorded was "about 45 °F", or about 7 °C.  Using the formula given here, I calculate the vapor pressure of propane at that temperature to be about 4400 mmHg, or about 590 kPa or about 5.9 bar.

Meanwhile, according to this page, the pressure inside a warm can or bottle of Coke can reach at least 380 kPa, or about two thirds of the vapor pressure of propane on a cold day.  As the bottles are certainly designed with a considerable safety margin, to make sure that they won't burst even if handled carelessly or slightly damaged, it's not surprising that they can easily withstand the pressure of the propane in the video.

BTW, this is exactly how aerosol spray cans work: they contain a mixture of the liquid being sprayed and a propellant substance (quite often propane) that has a boiling point at 1 atm only slightly below room temperature (or, equivalently, that has an equilibrium vapor pressure only slightly above 1 atm at room temperature).  Thus, as the can is drained, the partial boiling of the propellant maintains the pressure inside the can at the propellant's vapor pressure, which is high enough to propel the spray out of the nozzle, but not so high that it would require an excessively sturdy and expensive can to contain it.

As for what would happen at higher temperatures, at room temperature (i.e. 25 °C), the vapor pressure of propane would be about 7100 mmHg or 950 kPa (according to the formula, or about 7600 mmHg or 1000 kPa according to the table, which seems to be taken from a different source).  According to this random forum post, the small ½ liter Coke bottles used in the video can apparently withstand at least 180 psi, or 1250 kPa, so the propane-filled bottle probably wouldn't burst even at room temperature (unless it happened to be damaged or otherwise particularly weak).  If the temperature was raised to, say, 45 °C (113 °F, a very hot day), the vapor pressure of the propane would rise further to about 1500 kPa, which just might be enough to make the bottle fail.  Also, elevated temperatures will soften and weaken the plastic somewhat, making failure more likely.

In any case, in my personal experience, the weakest point of such bottles seems to be the relatively thin and weak cap, which would likely fail at some point before the bottle itself did.  I wouldn't be surprised if that was by design, to make the typical failure mode relatively safe and predictable.


Actual color of colloids I have read that the color of colloid depends upon the wavelength of light scattered by the dispersed phase and also in the sense in which the receiver receives the light.
If all of this depends on wavelength scattered and in the sense we see, then what is the actual color of a colloid?

Colloidal solutions scatter light of different wavelenghts in different directions, and that's also dependent on the size(!) of the colloidal particles.
To see any scattering at all, there only has to be a difference in the refractive index of pure solution and dissolved particles.
The individual colour (i.e. absorption spectrum) of the particles is irrelevant to the scattering, but of course it additionally changes the colour that you see. Absorbed light does not get scattered, obviously.


How is solid PVA filament (for 3D printing) synthesized? Polyvinyl alcohol (PVA; also referred to as: PVOH, PVAl) filament--the sacrificial support polymer filament used in 3D printing, how is that made?  All the reactions I've seen that make PVA create a gel-like substance that could be used as a glue or perhaps a film.  But it's not solid.  (See image with beaker.)

The reactions I've seen use an alcohol (ethanol or methanol) and an alkaline catalyst (sodium hydroxide) which are combined with polyvinyl acetate.  Here's the reaction that was used to make the beaker's gel PVA:

Another preparation can be seen in this video.

Is there some kind of coagulation/further reaction I must do to the gel to solidify its structure, or is the trick to do something during the first reaction, such as varying the temperature or pressure?

To clarify, I don't actually need/want it in the typical extruded filament form that would be needed for a 3D printer.  A simple solid blob of the stuff would suffice. I'm very new to materials science and only a novice in chemistry, but I'd love any answers you wonderful folk could provide, even if they are dense scholarly articles. Thank you for your time!

All the reactions I've seen that make PVA (also called PVOH, PVAl), create a gel-like substance that could be used as a glue or perhaps a film.

Yes, the reaction you describe is done in a solvent (water, ethanol, methanol, etc...) so that no acetate groups are unaccessable to reaction in a crystalline phase. A gel is formed because when poly(vinylacetate) is fully hydrolyzed, it is less soluble, typical white glue has about $12\%$ of the acetate groups unhydrolyzed to reduce crystallinity and keep solubility high.  The polyvinyl alcohol in filament is just dried out solid PVA, whereas the gel has water in a polymer matrix. All they have to do is melt and extrude the dry PVA powder into filament.

  I don't actually need/want it in the typical extruded filament form that would be needed for a 3D printer.  A simple solid blob of the stuff would suffice.

Side note: you can buy PVA powder online with less mess than self synthesis or depending on your application, you might be able to just use dried out white glue.


Activated Carbon - Desorption in sunlight? I have some activated charcoal pads or small "pillows" that can be used to adsorb odors and chemical fumes from different materials.

The instructions say to put them out in sunlight to reactivate them once they have been used for a while.

What is the mechanism whereby sunlight is able to reactivate the charcoal?

I tried googling but could not find a definitive answer. I ended up assuming it must be just heat that forces the gasses trapped in the charcoal to come loose and evaporate, thereby restoring some of the adsorption potential. Is that the mechanism or what is the sorcery going on here?

Your guess is right: the mechanism of charcoal regeneration is undoubtably temperature-induced desorption. Gas molecules are adsorbed on the large surface area of activated carbon by a mechanism of physisorption (or physical sorption) caused by weak nonspecific interactions (van der Waals forces) between gas molecules and the carbon surface. Unlike chemisorption, this process is reversible and, since it involves only weak interactions, it can actually be reversed by relatively small increases of temperature.

Wikipedia gives a specific example:

  The activated carbon can adsorb a large amount of methanol vapours in ambient temperature and desorb it at a higher temperature (around 100 degrees Celsius). During the daytime, the sunshine irradiates the collector, so the collector is heated up and the methanol is desorbed from the activated carbon. In desorption, the liquid methanol adsorbed in the charcoal heats up and vaporizes.

You will surely notice the resemblance between this text and your case.


Molar mass of a protein from osmotic pressure The height of a column of liquid that can be supported by a given pressure is inversely proportional to its density. An aqueous solution
  of $0.010 g$ of a protein in $10 mL$ of water at $20^oC$ shows a rise of
  $5.22 cm$ in a mercury tube. Assume the density of the solution to be $0.998 g/cm^3$ and the density of the mercury to be
  $13.6 g/cm^3$. (a) What is the molar mass of the protein?

The answer key says the answer is $4.8*10^3 g/mol$, but I got $3.5*10^2 g/mol$. Below is my reasoning:

Osmotic pressure = pressure to raise mercury, so $MRT = rho * g * h$

$\frac{(0.010/M)}{(10.010/0.998/1000)} * 8.3145 * 293 * 10^3$ (pascals) = $13.6/1000*(100)^3 * 9.8 * 0.0522$ (pascals). Solving this equation for $M$ gets $3.49*10^2$, instead of 3.8*10^3.

Can someone please point out where I am wrong?

Please do not point this out as a duplicate, as I have already referenced the question here. I used their approach but still cannot get the correct answer.

Osmotic pressure of a solute in water is given by the formula

$$\Pi = i\rm{MRT}\tag{1}$$

Where:
$\Pi$ is the osmotic pressure
$i$ is the Van't Hoff factor
$\rm{M}$ is the molarity of the solute (mol/L)
$\rm{R}$ is the universal gas constant 62.36 ($\rm{L}\space \rm{Torr}\space$ $\rm{K}^{−1}$ $\rm{mol}^{−1}$)
$\rm{T}$ is the absolute temperature ($^\circ\rm{K}$)

We'll assume $i =1$.

Let's use $\rm{M_W}$ for the molecular weight of the protein. So the moles of the protein is equal to $\rm{g/M_W}$.

$$\dfrac{0.010\space\rm{g}/\rm{M_W}}{0.010\space\rm{L}}\times 62.36\space\rm{L}\space \rm{Torr}\space\rm{K}^{−1}\space\rm{mol}^{−1}\times293\space\rm{K}  = 52.2 \space \rm{Torr}\tag{2}$$

$$\rm{M_W} = \dfrac{0.010\times62.36\times293}{0.010\times52.2}\space\rm{g/mol}= 350\space\rm{g/mol}\tag{3} $$

Your mistake was believing the book answer.

  If you need to check this sort of answer, then choose R with convenient units. It makes solving the problem much easier.


Why does mercury not form amalgam with iron? I am learning about solutions and while giving an example of a liquid (solute) and a solid (solvent), my teacher told us that mercury forms an amalgam with all the metals except iron.

I want to know why this is so?

Iron does not form an amalgam readily, hence the use of iron electrodes in mercury switches.  Apparently, iron does form alloys with mercury under specific conditions; see http://www.madsci.org/posts/archives/2011-04/1304143502.Ch.r.html.

By contrast, mercury spilled on gold (e.g. gold-alloy jewelry) soaks in immediately, discoloring the gold, and migrates further into the bulk metal in a few hours, thus restoring the color.


Relationship between enthalpy change and endo/exothermic reactions Something that confused me, is there a relationship between enthalpy change and endo/exothermic? Does region "x" represent the enthalpy change?
What represents enthalpy change, kinectic energy, activation energy and potential energy on the graph below?
But why must, for an endothermic reaction, the value of activation energy be higher than the enthalpy change? Then, how about the backwards in equillibrium, if the forward reaction is exothermic, is it the same?
And why is, for an exothermic reaction, the value of the activation energy equal to the enthalpy change? Is that a must? Can the activation be energy greater than the enthalpy change?

Yes the region x represents Enthalpy change.The relation between Enthalpy change and endothermic reaction  is that if a reaction is endothermic then Enthalpy change is +ve and if the reaction is exothermic then the Enthalpy change is -ve.

In an endothermic reaction the amount of energy you supply is lesser that the amount of energy released when product is formed so the amount of energy you supply to transform the reactants to activated complex will be more than the energy released when the activated complex transforms to product.

In the graph you have provided the reaction in forward direction is exothermic and in backward direction is endothermic. Whichever direction you take  activation energy will always be more than the Enthalpy change because Enthalpy change is the difference in the Enthalpy of the products and the reactants whereas the activation energy is the difference in the Enthalpy of the activated complex and the reactants. The Enthalpy of the activated complex is always more than the Enthalpy of the products and the reactants so both these quantities cannot be equal.

For both exothermic as well as endothermic reactions it not a necessary condition for the activation energy to be equal to the Enthalpy change as activation energy is always greater than the Enthalpy change.


Judging orders in a series of compounds My teacher frequently gives us questions where we have to arrange the given binary compounds on basis of their melting points or solubilities. He did give us some instructions in order to judge this order. But they don't seem to work for any questions. I asked a senior student as to how to tackle this problem, and he simply said inorganic predictions are tough to make, wo just stick to the data and make appropriate relations between the given compounds. But the problem is we don't get data in tests and have to judge the compounds solely on the basis of logic. But I have no clue as to how to figure out this logic. So is it really true that all the orders are simply unpredictable without their actual data? Is every order just not following the standard rules to figure out these orders? It would really help if someone is able to guide me how to approach such questions without knowing the related data and maybe even share some appropriate further readings. I tried looking on Wikipedia and google, but everywhere I go, they quote the experimental data. I even tried looking at some answers here on stackexchange, but the people in comments tend to confuse me even more.(some have even written the obvious answer-"stick to experimental data").

Don't worry too much about it, this educational exercise will be over when the exam is over. In real life nobody can predict the melting point or solubility in water just by looking at the structure of a given molecule. There are software like ACD, ChemDraw and SciFinder database which have predicted melting points and boiling points of most known compounds but those values are often off. Sometimes the values are correct. This is why only the experimental values of are considered reliable. Maybe in future, our understanding will become so much better that melting/boiling points and solubilities in water would be accurately predictable.
All your teacher wants is to remember some general trends and periodic properties, which should be in your mind (in other words memorize, just like 2+2 is always 4). Take the example of halogens (diatomic molecules). As you move down the group the melting point increases. Fluorine has the lowest melting point and iodine has the highest.
In the alkali metals, the melting point decreases down the group. All you have to keep in mind the melting point trends down in the groups as well as across the periodic table for simple molecules. Similarly, look for the trends in alkali halides, hydrogen halides, other hydrides of the oxygen group.
Read this article on Is there a trend in melting points on the periodic table? https://antoine.frostburg.edu/chem/senese/101/periodic/faq/melting-point-trend.shtml


Will the sulfur-iodine cycle and Haber Method work at lower pressures and temperatures? I have a year-long research project next year in my senior year of high school, in which we have to make some product or prototype to solve some issue.

My interest was in the Liquid Fluoride Thorium Reactor but considering that I do not have the access I need, I opted to invest in the heat it produces.

One use as proposed by Kirk Sorensen proposed using the heat from the reactor to fuel the sulfur-iodine cycle to split water into $\ce{H2}$ and $\ce{O2}$.  Then use those combined with carbon collected from the waste of coal and fossil fuel plants and nitrogen from the atmosphere to make methanol dimethyl-ether and ammonia.

My teacher and I have ruled out the first two, but decided that we might be able to make ammonia.  My question is whether it would be feasible and practical to use the sulfur-iodine cycle in combination with the Haber process to split water and make ammonia at a much lower temperature and pressure to ensure safety.

Yes this is possible but you must remember it is a reversible reaction so if more pressure creates ammonia, applying less will cause the reaction to reverse creating some of the reactants again.


If carbon dioxide is non polar, why does it react with water? I’ve read in many places that carbon dioxide forms carbonic acid with water since the carbon is partially positive and thus the oxygen bonds with it. But isn’t carbon dioxide in its entirety non polar since the 180 angle removes any dipole moment, so how would oxygen be attracted to the carbon?

Is it that individually each atom has partial charge but not collectively? Or am I wrong in some way?

I can only repeat myself here: Polarity is an ill-defined concept that has a nice potential for confusion.
In most cases, when specifying a molecule as polar, one is colloquially referring to the presence of a dipole moment, i.e. one actually categorises the molecule as dipolar. As described by ron in "Why is carbon dioxide nonpolar?", $\ce{CO2}$ has no dipole moment, it is therefore not dipolar, or colloquially it is not polar.
However, $\ce{CO2}$ has two very dipolar bonds, and a significant quadrupole moment. If one were to extend the nomenclature, one would say the molecule is quadrupolar. However, this may lead to complications down the line.
On the other hand, as I have written in the linked question, toluene is often considered an unpolar/non-polar solvent, which is not really true considering it has a small dipole moment.
There are a couple of things one can predict with the concept of polarity, and fortunately, the more complex the molecules become, the better the approximation becomes. It is small highly symmetric molecules, which break these approximations.
Related reading:

Some more extended reading by others and myself can be found in this question: What are dipole moments in a molecule supposed to act upon?

And to further the confusion: Is the triiodide ion polar?

Where to draw the line? What is practical and what not? Why is CO practically nonpolar? What is the dipole moment direction in the nitrosonium ion?


Can the alkyl-oxygen bond in lactones be cleaved in an SN2 reaction? I understand that nucleophiles typically attack the electrophilic carbonyl carbon of the lactone and open the ring through a nucleophilic acyl substitution. Has there ever been the case where the nucleophiles open the lactone ring by attacking the carbon adjacent to the oxygen (not the carbonyl oxygen) via a SN2 reaction instead? Is there a preferred pathway for certain nucleophiles to take and why so?

Yes, it is possible. If you consider the chemistry of a four-membered lactone (a β-lactone), it is able to act as an alkylating agent, as nucleophiles can attack the C–O bond in an SN2 fashion.

One example is in the synthesis of a series of β-substituted amino acids (J. Am. Chem. Soc. 1985, 107 (24), 7105–7109). An intramolecular Mitsunobu reaction of Cbz-protected serine 1 forms the β-lactone 2 which is then opened by a variety of nucleophiles. Here, benzylthiolate is used as an example, giving N-Cbz-S-benzylcysteine 3 as the product:

The authors note that:

  Only relatively “hard” nucleophiles like ammonia and methoxide attack the carbonyl to give acyl-oxygen cleavage. In earlier reports serine β-lactones were used mainly for synthesis of seryl peptides by attack of amines at the carbonyl. However, with ammonia as the nucleophile, simply altering the solvent [from MeCN to THF] inverts the 3:1 amide/amine ratio.

The following article has a wide selection of more examples (see references 6–8): J. Org. Chem. 2002, 67 (14), 4680–4683.


Why does gas particle velocity affect rate of effusion? I understand why smaller particles have more velocity, but I don't understand what velocity has to do with rate of effusion:

My reasoning is thus:

Pressure is the number of impacts of particles in a given period of time.
If He and Ar are both in balloons at, say, 1.5 atm, both gasses have the same average Kinetic energy but He is moving faster (because it's smaller). Simple enough.
If both gasses are subject to the same pressure, they have the same number of impacts over a given area over a given period in time.
This should mean that each gas has an equal number of particles approaching a hole over a given period of time.
If a hole is big enough for both particles to fit through, the rate of effusion should be the same, since the same number of particles are approaching the exit hole.

For an analogy to explain my thinking ... if you have two lanes of cars going through a checkpoint, one has 30 cars per minute at 50 miles per hour and one has 30 cars per minute at 30 miles per hour, the resulting number of cars through the checkpoint should still be 30 cars per minute in each lane, regardless of their velocity.

My book (and professor, and Wikipedia...) say that it is the higher velocity of He particles which cause the faster rate of effusion; however, from my reasoning, that doesn't make sense except with small enough holes where the size of the particles would have an effect, i.e. the holes are very small in size and He could fit through but Ar could not, or there are simply more holes that He could fit through.

But Graham's law is referring to pinholes, which should be large enough for either atom to fit through.

So why does velocity affect rate of effusion?

Your point 1 is mistaken (or incomplete)
$$P=\frac{F}{A}=\frac1A\frac{\Delta p}{\Delta t}=\frac1A\times\frac{2nmv}{\Delta t}$$

This is assuming all particles hit perpendicularly (you can always modify this for all colissions by taking the average component of velocity in the perpendicular direction).

So, pressure is proportional to number of impacts, mass, and velocity. Pressure is not simply "number of impacts", rather it is a combination of number of impacts with other stuff. Since the energies are equal, we can say that $v\propto m^{-\frac12}$, and we get $P\propto n m^{\frac12}$. P is also the same, so $n\propto\frac1{\sqrt m}\propto v$.

Thus, when avg KE and pressure are the same, rate of diffusion is inversely proportional to square root of mass.


Combination of several unit symbols when finding new pressure from ideal gas law As a car is driven, the tires heat up and the volume and the pressure change. Calculate the new pressure of a 35.0 L tire initially at 235 kPa that expands to 36.5 L when it heats up from 15.0 °C to 43.0 °C. (4 marks)

$P_1 = \pu{235 kPa}$
$P_2 = ?$
$V_1 = \pu{35.0 L}$
$V_2 = \pu{36.5 L}$
$T_1 = \pu{15.0 °C} = 15.0 + 273 = \pu{288 K}$
$T_2 = \pu{43.0 °C} = 43.0 + 273 = \pu{316 K}$

The number of moles of gas in the balloon is not changing, so $n_1 = n_2$.

$$P_1V_1/n_1T_1 = P_2V_2/n_2T_2$$

(since $n_1 = n_2$ we can cancel them from both sides)

$$P_1V_1/T_1 = P_2V_2/T_2$$

(cross multuple to allow for isolation of $P_2$)

$$P_1V_1T_2 = P_2V_2T_1$$

(divide both sides by $V_2T_1$ to isolate $P_2$)

$$P_2 = P_1V_1T_2/V_2T_1$$

(now plug in values)

$$\begin{align}
P_2 &= (\pu{235 kPa})(\pu{35.0 L})(\pu{316 K})/(\pu{36.5 L})(\pu{288 K})\\
  &= 2 599 100 / 10 512 \\
  &= \pu{247 kPa}
\end{align}$$

The new pressure of the tire would be 247 kPa.

I like keeping track of the units throughout solving, but not sure how to do that with three units...

Can I write $P_2 = 2599100\cdot\pu{kPa/L/K}/10512\cdot\pu{L/K}$ (L and K cancel out and leave me with kPa), or is it recommended not to record the units in that step and only write the final unit with the final answer?

Section 5.1 Unit symbols in SI Brochure [1, p. 130] has an answer:

  In forming products and quotients of unit symbols the normal rules of algebraic multiplication or division apply. Multiplication must be indicated by a space or a half-high (centred) dot (⋅), since otherwise some prefixes could be misinterpreted as a unit symbol. Division is indicated by a horizontal line, by a solidus (oblique stroke, /) or by negative exponents. When several unit symbols are combined, care should be
  taken to avoid ambiguities, for example by using brackets or negative exponents. A solidus must not be used more than once in a given expression without brackets to remove ambiguities.

  $\pu{m kg/(s3 A)}$,
  or $\pu{m kg s–3 A–1}$,
  but not $\pu{m kg/s3/A}$,
  nor $\pu{m kg/s3 A}$.

I personally prefer to use powers of the units — subjectively, this makes the expressions clearer and also reserves a slash "/" for separating the symbol of the quantity from the units on plots and in the table headers. However, in your case you don't need any of that. According to the ideal gas law

$$pV = nRT$$

and

$$n_iR = \text{const} \quad \implies \quad \frac{p_iV_i}{T_i} = \text{const},$$

so you have a simple proportion

$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \implies \quad p_2 = \frac{V_1T_2}{V_2T_1}p_1.$$

Obviously, there is no need to juggle around with the units as all of them except for the pressure units are cancelled out (note how I also grouped homogeneous quantities within the fraction one under another, and wrote pressure separately for clarity):

$$p_2 = \frac{V_1T_2}{V_2T_1}p_1 = \frac{\pu{35.0 L}\cdot\pu{316 K}}{\pu{36.5 L}\cdot\pu{288 K}}\cdot\pu{235 kPa} = \pu{247 kPa}.$$

And yes, generally it is considered a good practice not to do any intermediate calculations unless you are explicitly asked so, and solve the problem algebraically in the first place.

References

BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 8th ed.; Bureau international des poids et mesures: Sèvres, 2006.


Why is pyrimidine less basic than pyridine? From Wikipedia's article on pyrimidine:

  Because of the decreased basicity compared to pyridine, electrophilic substitution of pyrimidine is less facile.

But why is pyrimidine less basic than pyridine? Pyrimidine has two $\mathrm{sp^2}$-hybridised lone pairs available for protonation, compared with pyridine's one.

It is not the number of lone pairs that in any way explains basicity. Take a random sugar and it will have ten times the number of lone pairs (albeit on oxygen, not on nitrogen) without being significantly basic.

The problem is electronics. Nitrogen, being a rather electronegative atom, is able to draw the π electrons towards it well — to the extent that pyridine derivatives are termed extremely electron poor and thus very slow in electrophilic substitution reactions.

Going from pyridine to pyrimidine we need to double that. Thus, there is a perceived severe shortage of electrons in that ring since both nitrogens are trying to draw them towards themselves.

Now assume you protonate that species. We now have two nitrogens of which one is positively charged. Immediately, that positively charged nitrogen multiplies its electron-withdrawing force. Equally immediately, the other nitrogen feels electron density leaving its vicinity. The whole system is much less favourable than the pyridine system with only a single nitrogen.

Rephrasing that argument: Basicity is increased when electron-donating neighbours increase the electron density (that equals a partial negative charge) on an atom. In pyrimidine’s case, we have an electron-withdrawing neighbour that reduces the electron density (giving a partial positive charge) and making protonation (i.e. more positive charge) less favourable.


Does humidity increase oxygen levels in the air we breathe? I'm testing a new form of meditation and it requires certain breathing exercises in order to generate a specific effect on the nervous system; I have noted significant differences in the results being generated based on the setting in which i practice the exercises... here is some data, hopefully you can help me put all of these pieces together, id like to base this off of solid data so i can present my findings for further evaluation.

QUESTIONS:Why do the effects of the technique increase in different settings?

Am I breathing in more O2 based on the environment i do the exercise?
Does high humidity increase the levels of oxygen in the air?
Or are other factors such as temperature & pressure contributing to the effects?

SETTING:
Steam Room:
- 100% humidity, temperature
- HIGH heat
- Pressurized steam room

Normal Room Temp Setting:
- "Normal" Dry Colorado Air (Assumed approx 78% nitrogen, 21%o2, 1% misc)
- Normal home setting

-the "effects" are intensified when performed in the sauna as compared to normal, dry Colorado air, i am trying to pin point why this might be.

Hopefully this is clear, thanks.

As to the answer in the title, no, increased humidity does not increase oxygen levels in the air.  In fact it does the exact opposite, since the air in each breath will have some fraction displaced by water molecules instead of the normal nitrogen and oxygen mix we breathe.  It's a relatively small effect, with water making up at most a few percent of the air under high humidity/high temperature conditions that are safe for people to inhabit for reasonable periods of time.  That would mean that oxygen content would be decreased by about 1% in your steam room (again, this assumes you're not doing something dangerous with the temperatures and humidity.)

As to your other questions, you've got serious methodological issues to deal with.  Your first goal in finding the reason for the significant differences you've noticed should be verifying whether those significant differences actually exist, or whether they are based on things like confirmation bias or the simple effects of being in a boiling hot room vs outdoors in Colorado.  How you do that is really more of a psychology experiment, as has been suggested by the comments above, and would be better suited for the Cognitive Sciences Stack Exchange.


Why is periodicity seen in these certain properties? I missed my lesson on periodicity so had to teach myself, and have always forgotten to ask my teacher to explain to me why these trends are seen, which, unfortunately, the textbooks don't.

Density: the density of an element tends to increase across a period up to group 3, and then starts to decrease and become very low at group 5 or so. Why is this?

Melting and boiling point: they tend to rise up to about group 3 or 4, before falling rapidly after that. I would suggest that this is because group 4 elements tend to have giant covalent structures, so also a higher boiling point, whereas on the right hand side of a period they are simple molecular structures, so have a low boiling point. Am I correct?

Atomic size: atomic size always decreases along a period from left to right. My suggestion is that this is because there are more electrons and more protons as the group increases, so there becomes a stronger attraction between them, pulling the electrons closer to the nucleus. Am I right?

So the only one I have no idea about really is density, and the other two I just need clarification on.

For 2. I would add, on top of what you already have, that the elements start as metallic elements, hence we have metallic bonding. Across the period, the number of electrons increase, meaning there are more delocalised electrons among the metal cations so that electrostatic attraction increases along the period. For period 2, that's the trend.

For period 3 however, it is a bit more tricky because of the transition metal elements and I think Nicolau explains it better than I could in this answer. But I don't think that you will be required to know those for your level, but it's still interesting for your own knowledge. :)

For 3. I would add that the number of electron shells containing electrons remains constant. If it increased instead, the atomic radius would increase as well.

And you could use 2. and 3. to explain 1:

First, we observe that the atomic mass increases along the period, and that the atomic radius decreases. For giant structures, this generally means that the density increases ($\rho = \dfrac{m}{V}$ and while $m$ increases, $V$ is decreasing slightly, so $\rho$ increases)

When we hit the simple molecular structures, the density sharply falls again, because of the average distance between the atoms of the element has drastically risen, so the volume increases. With $m$ increasing gradually and a sharp rise in $V$, you get a sharp decrease in $\rho$.

These observations change down the groups though. For instance, Astatine is a solid, so you won't expect a huge fall in density across period 6 around group 5. But for the purpose of your syllabus coverage (I'm assuming High School), those should be enough.


How to dissolve bronze at room temperature Is there any way to dissolve about 10 kg of bronze at room temperature?

If yes, which substance would be the best to use and how would the reaction look like?

Supposing that the bronze is only 78% copper and 12% tin (and there are no other metals). Nitric acid will dissolve the copper... but this reaction is exothermic, which means it will heat up.

At boiling temperature. Nitric acid will also react with tin to make tin oxide which is soluble. However, if you kept it cool/room temperature with an ice bath, then you would have leftover tin which did not react.

Tin will react with aqua regia to form tin (IV) chloride, but copper is slower to react with aqua regia... so it would be better to use nitric acid first (to dissolve the copper) and then add hydrochloric acid to react with and dissolve the tin. Once again, the reaction between tin and hydrochloric acid (or aqua regia) will proceed much faster with heat, but it will happen at room temperature.

Aqua regia = 1 mole of Nitric acid and 3 moles of hydrochloric acid).


Should I provide a citation for atomic masses? I am writing my first lab report ever, and the handbook I was given did not say anything in particular about this issue. I will use the atomic masses for my calculations and I am not sure if I should cite the source from which I obtained the data for those figures.

Should I cite periodic tables? If yes, how?

You can’t cite ‘a’ or ‘the periodic table’ because neither are published books. What you can do is state that you took the atomic masses as given by <book> or <source>.

I personally would not do that and I have not yet seen a research article that includes citations for what is considered a well-determined atomic property such as atomic mass or even electronegativity — even though the values for the latter vary greatly depending on source and method used. (Papers that discuss the difference between different definitions being the obvious exception.) Whether you are required to do that depends on what your TA or professor thinks. So go and ask them.


How are the π bonds arranged in Osmium Tetroxide? The structure of $\ce{OsO4}$ has puzzled me greatly for a while. I know that it has a tetrahedral geometry.

I am quite piqued at the thought of how the $π$ bonds were arranged in the molecule.

My knowledge is very limited, and I know that $π$ bonds are formed due to the sideways overlap of $p$ orbitals.

While, in a regular unhybridized atom, the 3 $p$ orbitals are arranged orthogonal to each other ($p_x$, $p_y$, $p_z$), I am unsure on how they are arranged in $\ce{OsO4}$. Is it something like this?

Or is it that some other orbitals may be involved?

Your conception of $\pi$-bonds is a bit too restrictive. A $\pi$-bond is a bond in which there is one node along the internuclear axis. So, yes, two side-on $p$-orbitals do form a $\pi$ bond because there is one node along the internuclear axis (the one separating the bottom lobe from the top lobe). A $\sigma$-bond is a bond with no nodes (i.e. cylindrically symmetric), a $\delta$-bond has two nodes, etc. It's important to note that these symmetry labels work best in molecules where each bond axis either exactly lines up with or is exactly orthogonal to all the others (like diatomics and octahedral metal complexes) but we can work in other molecules with a little more effort.

Now, in osmium, the valence orbitals most involved in bonding are actually the $d$-orbitals. Unfortunately, because the molecule has tetrahedral symmetry and the bond axes are not all orthogonal or parallel, it's less easy to see how the oxygen and osmium orbitals come together to form $\pi$ bonds, and we need to resort to MO theory to get a qualitatively correct answer. If you don't know MO theory, the most basic way of approaching the following is to think of all the ligands and metal orbitals combining (much like the simple $p-p$ overlap in your question) and we label each combination by a label that has to do with the symmetries (rotations and reflections) of the combination of atomic orbitals.

Let's start with a qualitative MO diagram from this answer:

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs. Orbital energies here are qualitative (and in fact the order is wrong here for $\ce{OsO4}$ which has occupied orbitals energies $t_1>t_2>a_1>t_2>e$).

The oxygen $p$ orbitals that point toward the metal and will make up the $\sigma$ bonds form the $a+t_2$ orbitals. The oxygen $p$ orbitals that make up the $\pi$ bonds are the $e+t_1+t_2$ orbitals. The osmium $d$ orbitals are $e+t_2$. Since only orbitals of the same symmetry can mix, the $e$ orbitals are purely $\pi$-bonding while the $t_2$ orbitals are a mix of $\pi$ and $\sigma$ bonding.

So your picture is sort of correct for the oxygens but they have to be combined in specific ways in order to overlap with the osmium. Here is a picture of one of the $\pi$ bonds in this molecule:

You can see the $p$ orbitals on each oxygen, the node along the bond axis and the osmium $d$ orbital at the center taking part.

Here is a purely $\sigma$ orbital:

In this case, the Os $s$ orbital participates with the oxygen $p$ orbitals pointing towards the center.

And here are two views of one of those weird $t_2$ orbitals, which is a mix of both $\sigma$ and $\pi$ in tetrahedral symmetry:

You can see that the oxygen $p$ orbitals clearly have a node near the O, but as you get towards the metal center the node bends off-axis. Meanwhile, near the metal, the nodes are not oriented near the bond axes at all. Very confusing!

In tetrahedral symmetry, some orbitals are clearly $\pi$-type, some are clearly $\sigma$-type, and some are a mix and therefore not a well-defined $\sigma$ or $\pi$.

EDIT: The careful reader will note that I ignored the $t_1$ orbitals. These are $p$ orbitals on the the oxygens that have no osmium orbitals to bond with and so are non-bonding. Here is a picture of one of these non bonding orbitals.

You can clearly see all the oxygen $p$ orbitals, but note that the osmium center has no orbital.


Does a technical umbrella term exist that covers the different kinds of biomolecules? I want to write a sentence along the lines of:

  There are four ____ in biochemistry: proteins, lipids, carbohydrates and nucleic acids.

Does a specific technical term exist that I can use to fill the void? Or am I relatively free in my choice?

I don’t know what your background is. Maybe it is organic chemistry, where the term functional group is used in a very specific way and using what would seem to be a synonym instead is wrong. Maybe it is biology (taxonomy) where it makes a big difference if a ‘group’ is an order, a family, a genus or something else.

The typical primary metabolites in biochemistry — carbohydrates, peptides/proteins, nucleic acids and lipids — do not have an overarching, widely accepted technical umbrella term. You can call them groups, classes, kinds or even types and be well understood.

If I had to suggest something, I would go with classes of primary metabolites. But I wouldn’t call that a strict technical term like functional group (organic chemistry) or species (taxonomy).