What chemicals help define "taste"? As we all know that there are 5 tastes : sweet, sour, bitter, umami, salty.
So what chemicals are there that help form these tastes, in other words how do our tastebuds experience these tastes?

Also can there be more tastes than these basic five. I am not saying that there should be a blend of these to make a taste but rather are there any other undiscovered tastes?

Maybe we haven't tasted them yet?

It's not exactly true that there are 5 "tastes". There are 5 generally accepted classes of gustatory transducers, each comprising many different types. The perception of taste is a product of not only the output of these transducers, but also information from the olfactory system, visual cues, textures, sound, etc.

The gustatory system itself is based (like the olfactory system) on overlapping sensors with partial selectivity. So a given compound, sucrose for instance, may stimulate several sweetness sensors (and probably even some from other classes) to varying degrees, and the particular combination of the responses is what gives sucrose its unique taste.

The types of molecules that stimulate the different sensors is not known clearly, but we can make some broad generalizations:

Sour - These sensors are thought to be primarily ion channels sensitive to pH—acidic solutions produce a sour taste.

Salty - Also thought to be mostly ion channels sensitive to small metal cations. Different ions have somewhat difference response. ($\ce{Na+}$ doesn't have the same response as $\ce{K+}$) There is some debate about the actual mechanism though.

The rest are g-protein coupled receptors. They change shape when they bind a molecule which causes a chemical signalling cascade.

Sweet - Not known precisely, but many compounds containing hydroxyl or carbonyl groups trigger these receptors.

Umami - Mostly triggered by amino acids, particularly glutamate. Inosinate and guanylate are known to enhance the effect of glutamate.

Bitter - This one is rather strange. A very wide variety of compounds stimulate these receptors, even with very different structural features. This taste is thought to have evolved to protect us against toxic compounds as many poisonous alkaloids and such taste bitter. Denatonium is an example of a bitter compound (one of the most bitter known).

As for whether any other classes of transducers exist, it's entirely possible. Some researchers have identified candidates for receptors for things like fats. Also, there some receptors like TRPV1 (which responds to capsaicin) are known that respond to compounds found in food, but aren't usually considered taste receptors.


Pseudoelementary reactions in the Briggs-Rauscher oscillatory reaction mechanism In this paper on the Briggs-Rauscher mechanism (JACS, 1982, paywalled), Noyes and Furrow propose an explanation of the oscillatory behaviour of the reaction based off of about eleven reactions from a selection of processes they consider "might contribute". The authors, however, call these processes "pseudoelementary reactions" and some of them, such as
\begin{equation*}
\ce{IO3- + HIO2 + 2H+ <=> 2IO2 + H2O}
\end{equation*}
seem to involve the collision of four molecules at the same time, which I was always taught is not very plausible. Many of these reactions are trimolecular, like
\begin{equation*}
\ce{HIO2 + I- + H+ -> 2HIO}
\end{equation*}
which I also find strange (these two reactions that I picked out are specifically reactions they say postulate that are important for the oscillatory behaviour).

I had never come across such a thing and I am nor really sure what to make of this. Are these reactions not actually elementary? What do they mean when they say the reactions are pseudoelementary? Are reactions involving the collision of four or three molecules more common than I thought? When they write down the equations for these reactions they write them as if the stoichiometric coefficients were the order with respect to each reagent, which seems to imply that the reactions are somehow elementary, which further confuses me. For instance, for the first reaction (which they call I3 in the paper) they explicitly write
\begin{equation*}
\frac{\mathrm{d}[\ce{HIO2}]}{\mathrm{d}t} = k_{I3} [\ce{H+}]^2 [\ce{IO3-}][\ce{I-}]
\end{equation*}

What do they mean when they say the reactions are pseudoelementary?

Here is a portion of the first page:

So for the reaction (I added an H+ to the product to balance it)
$$\ce{IO3- + HIO2 + 2H+ <=> 2IO2 + H2O + H+},$$

they assume that the two protons attach in a fast reversible step (not sure to which of the reactants), and then there is a bimolecular reaction between the two iodine-containing intermediates X and Y:

$$\ce{IO3- + HIO2 + 2H+ <=>[fast] X + Y -> 2IO2 + H2O + H+},$$

The rate law for this set of reactions taken together is the one given in the question.

  Are reactions involving the collision of four or three molecules more common than I thought?

No, they are rare (and slow). They could have written out the separate elementary reactions, but they probably also did not know where the protons attach and in what order, and they are not claiming to get the mechanism right. They just want to show a plausible set of reactions and rate constants that leads to the observed oscillations.

  When they write down the equations for these reactions they write them as if the stoichiometric coefficients were the order with respect to each reagent, which seems to imply that the reactions are somehow elementary, which further confuses me.

It does not imply that. From a rate law alone you can't distinguish between mechanisms. Take these two mechanisms for example:

$$\ce{ A + H+ <=>[fast] AH+}$$
$$\ce{ AH+ ->[slow] B + H+}$$

on the one hand and
$$\ce{ A + H+ -> B}$$

on the other hand. Both will be first order in $A$ and in $\ce{H+}$.


Solubility involving precipitation Question
  How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$.
  Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$
  Other given values are $$K_{\text{sp}_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{\mathrm{f}_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$
  Answer: $4.4\rm\,g$

My effort:
 Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.

$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [\ce{Ag}] = 1.8 \times 10 -9$$

Then I placed that value in formation eq.

$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$

$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$

$$x = 0.0288\,\rm M$$

and tried to find out $\ce{AgNO3}$ mass by using that by

$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$

That value doesn't seem to be the answer. What did I do wrong?

Step 1: Find equilibrium concentration for Ag+

As you correctly showed in the first step, we need to find the maximum solubility equilibrium concentration of $\ce{[Ag+]_{eq}}$ that is still allowed. Your calculation is correct, the result is
$$ \ce{[Ag+]_{eq}} = \frac{K_\text{sp}}{\ce{[Cl- ]}} = 1.8\cdot 10^{-9}$$

Step 2: Find equilibrium concentration for [Ag(NH3)2]+

We now need to know how much of the $\ce{AgNO3}$ we put into the solution actually ends up in the complex $\ce{[Ag(NH3)2]+}$, which is not in danger of precipitating. The formation equilibrium is according to the reaction equation
$$ \ce{Ag+ + 2NH3 <=> [Ag(NH3)2]+} \,,$$
and reads
$$ K_\text{f} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+]_{eq}}\ce{[NH3]^2}} \,.$$

Substituting $x=\ce{[Ag(NH3)2+]}$ we quickly reach
$$ K_\text{f} = \frac{x}{\ce{[Ag+]_{eq}}(1-2x)^2} \,.$$
This is the step where you went wrong: You also need to consider that the concentration of $\ce{NH3}$ is changed by the formation of the complex. In principle we should also have taken this into account in the section above, but there the error is many more magnitudes smaller than here.

Solving the above gives $x = \ce{[Ag(NH3)2+]_{eq}} = 0.0258942$.

Step 3: Calculate the total amount of Ag+ in the solution

Armed with the knowledge from the two previous sections, this step is pretty trivial. As the contribution of free silver ions is many orders of magnitude smaller than the concentration of the complex, we can safely ignore it and approximate:
$$ \ce{[Ag+]_{tot}} = \ce{[Ag+]_{eq}} + \ce{[Ag(NH3)2+]_{eq}} \approx 0.02589 $$

Step 4: Calculate the amount of AgNO3 to add

If we assume that the activity coefficient is 1, we can directly infer the concentration
$$ c_{\ce{Ag^+_{tot}}} = \ce{[Ag+]_{tot}} ~ \mathrm{mol\, L^{-1}} \,.$$

From the concentration we can calculate the amount of $\ce{AgNO3}$ to add, as you did correctly,
$$ \begin{align}
m_{\ce{AgNO3}} &= c_{\ce{Ag^+_{tot}}} V M_{\ce{AgNO3}} \\
&= 0.02589 ~ \mathrm{mol\, L^{-1}} \cdot 1~\mathrm{L} \cdot 169.87 ~ \mathrm{g\, mol^{-1}} \\
&= 4.398~\mathrm{g}\,,
\end{align}$$
which, after rounding to two significant digits, leads directly to the result your solution says you should have gotten:

$$ m_{\ce{AgNO3}} = 4.4~\mathrm{g} $$


Shea Stearin Extraction by Dry Fractionation I have been able fractionate palm oil and extract palm stearin using the following method:

Melt the palm oil (over 80 °C)
Keep the melted oil at 33.5 °C for 24 hours (using an immersion circulated water bath)

At this point I was able observe crystals being formed. I assume they were palm stearin crystals.

Centrifuge to separate the crystals to obtain palm stearin.

I was able fractionate almost 30 % crystals by weight.

However the same process did not really hold true while processing the shea butter to fractionate shea stearin. After Step 2, it was still in liquid form and there was no visible crystal formation. I'm further considering seeding (with crystals) or playing with the time and temperature variations.

How can I modify my approach to promote stearin crystal formation in shea butter?

Shea butter is a complex combination of esters:

(Wikipedia). The overall ratio of oleic to stearic acids is about 2/1, but statistically, there will a little of the high melting stearic, more of the 2 stearic to 1 oleic; much more of the 2 oleic to 1 stearic, and a lot of all oleic ester. The melting point of shea butter is low (body temperature) and so if it does not precipitate out the tristearin even at room temperature, you will have to do something different.

A fractional crystallization involving higher temperatures would use a poor solvent for the tristearin (mp ~ 72C), like ethanol. The oleic esters will be more soluble. Make a 20-30% solution in hot ethanol (be careful!!) and let it cool. Experimenting with similar alcohols (propyl, isopropyl, and perhaps with a bit of water added (~5-10%) could give better results.

Oiling out is a common problem in trying to crystallize organic compounds.


Why wash a sample before titration? In the awwa notebook for $\ce{Na2SiF6}$ determination, why it is required to wash the product sample in an alcoholised $\ce{KCl}$ solution before titration?

Your protocol gives you the answer: to make sure your sediment is neutral. After all, you titrate with NaOH afterwards, so any wrong starting point in terms of pH would skew your result.


How to name this quintuply-bonded chromium dimer? Depicted below is a chromium compound with a metal–metal quintuple bond. It boasts two identical bridging ligands, derivatives of 1,3-diphenylbenzene. The aim of this question is to figure out

the proper name.

Proper means without losing information. When answering, application of the 2005 Red Book and 2013 Blue Book is desired.

  Effort

  Identify central atom, derive ending: chromium
  Naming unsubstituted ligand, derive ending: 1,3-diphenylbenzene $\longrightarrow$ 1,3-(diphenyl)phenyl
  Allow for chromium's probably higher priority $\longrightarrow$ 2,6-(diphenyl)phenyl
  Add in the isopropyl groups $\longrightarrow$ 2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl
  Look for stereoisomers of ligand: probably none
  Count central atoms: two
  Identify and denote ligand connectivity to central atoms: 1κ$C^1$,2κ$C^7$
  Exploit symmetry, count ligands, prefix di already used: two $\longrightarrow$ bis
  Combine: bis(2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κ$C^1$,2κ$C^7$)chromium
  Account for the bridging nature of ligands: add μ
  Notice metal–metal bonding: add $(Cr$—$Cr)$
  Probably (***) count number of bonds between metals: five $\longrightarrow$ $(5\ Cr$—$Cr)$
  Combine:

  bis(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κ$C^1$,2κ$C^7$)chromium$(5\ Cr$—$Cr)$

  Change parentheses to allow for two chromium atoms: add []

Final proposed name (***):

bis[(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κ$C^1$,2κ$C^7$)chromium]$(5\ Cr$—$Cr)$

Is this correct? Where have I gone wrong?

*** Point 12 conflicts with the Red Book. See IR-10.2.3.5 which says

  For the purpose of nomenclature, no distinction is made between different metal–metal
  bond orders.

This is at odds with imposed criterium of preserving information, hence was ignored. ***

Assembling the chemical formula will most likely be covered in a follow-up question.

It was a great effort. However, in the proposed name

  bis[(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κC1,2κC7)chromium](5 Cr—Cr)

I can see several issues:

In you point 12: Forcing the metal-metal bond multiplicty in the name is interesting, however I wouldn't go against the rules. See the example with the multiple (triple) Cr-Cr bond, in the rules:

  IR-10.2.5.1 The eta (η) convention
  (…)
  Example 26.

  (μ-2η4-buta-1,3-diene-1,4-diyl-1κ2C1,C4)carbonyl-1κC-bis[(η5-cyclopentadienyl)chromium](Cr—Cr)

not "…(3 Cr—Cr)".

The purpose of the rule seems to be to just specify (only) the connectivity, i.e. the count of such and such metal-metal connections.

  IR-9.2.5.6 Trinuclear and larger structures
  (…)
   Example 6.

  octacarbonyl-1κ4C,24κC-bis(triphenylphosphane-3κP)-triangulo-diironplatinum(Fe—Fe)(2 Fe—Pt)

One could speculate that they, to denote the bond order as well, intend to use a multiplied Em dash character variants, which are non-existent presently.

Another motivation might have been that the bond nature / bond-order is often disputed in organometallics.
Your name can denote a different isomer (though maybe impossible), with the two ligands oriented the same way, not say antiparallel, or head-tail, but head-head. I.e. one of the ligand flipped, and connected to chromiums the other way.

From the rules:

  IR-10.2.5.1 The eta (η) convention
  (…)
  Example 24.

  (μ-1η5-cyclopenta-2,4-diene-1,1-diyl-2κC)(μ-2η5-cyclopenta-2,4-diene-1,1-diyl-1κC)bis[(η5-cyclopentadienyl)hydridotungsten]

not bis(μ-1η5-cyclopenta-2,4-diene-1,1-diyl-2κC)bis[(η5-cyclopentadienyl)hydridotungsten].

I.e. repeating the ligands name, with the 1/2 metal locants swapped. Which, however, is obviously inconvenient, maybe the rules will be improved in the future to somehow support relative locants, or so.
2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl part (you construction in points 3, 4) is weird.

The standalone ligand residue would be named according to the organic chemistry nomenclature:

2,6-bis[2,6-di(1-methylethyl)phenyl]phenyl, more correctly
2,6-bis[2,6-di(propan-2-yl)phenyl]phenyl

Inorganic nomenclature does not have so strict "preferred name" concept. Table IR-10.1 lists isopropyl as an acceptable alternative name for the propan-2-yl. So I would go with 2,6-diisopropylphenyl instead of 2,6-di(1-methylethyl)phenyl or , to not complicate the name further.
Problematic locant for the kappa notation in the ligand 2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κC1,2κC7. Maybe a prime symbol / apostrophe could be used like in biphenyls, terphenyls..

I think it should be rewritten to more verbose, but unambiguous 2-(2,6-diisopropylphenyl-1κC1)-6-(2,6-diisopropylphenyl)phenyl-2κC1 (both locant superscripts after κC's are now 1, but its not a problem).

And one note:

Even though the rules state

  IR-9.2.4.3 Comparison of the eta and kappa conventions

  The eta convention (Section IR-10.2.5.1) is applied in cases where contiguous donor atoms
  within a given ligand are involved in bonding to a central atom. Thus, it is used only when
  there is more than one ligating atom, and the term η1 is not used.

in the literature(web) for the structure with a more complicated ligand

featuring the similar 2,6-di(iPr)Ph···Cr bond, the following shortcut is used

  Ar′Cr(μ2:η1,η3-N3(1-Ad))CrAr′

which contains the η1 (and, it interestingly relates to the NNN part of the ligand on the right, not to the phenyl). But I'm not gonna use it in the proposed name.

So my candidate name is
[μ-2-(2,6-diisopropylphenyl-1κC1)-6-(2,6-diisopropylphenyl)phenyl-2κC1][μ-2-(2,6-diisopropylphenyl-2κC1)-6-(2,6-diisopropylphenyl)phenyl-1κC1]dichromium(Cr—Cr)

Finally, lets mention the paper in which this interesting compound was described: Nguyen, T.; Sutton, A. D., Brynda, M.; Fettinger, J. C.; Long, G. J.; Power, P. P. Synthesis of a Stable Compound with Fivefold Bonding Between Two Chromium(I) Centers. Science 2005, 310 (5749), 844–847.


What is [U-14C]-glucose? In Helen Tweeddale's article,[1] they labelled metabolites of E. coli with [U-$^{14}$C]glucose. What sort of glucose is it? I understand they used isotope 14, which is radioactive but what does the U stand for?

Helen Tweeddale, Lucinda Notley-McRobb, and Thomas Ferenci, J Bacteriol. 1998, 180(19), 5109–5116. PMID: 9748443 (pdf)

According to the IUPAC organic chemistry nomenclature 2013, section P-83 ISOTOPICALLY LABELED COMPOUNDS, it has to do with uniformly labeled compounds:

  P-83.5 GENERAL AND UNIFORM LABELING

  P-83.5.1 In the name of a selectively labeled compound in which all positions of the designated element are labeled, but not necessarily in the same isotopic ratio, the symbol ‘G’ is used in place of locants to indicate a ‘general’ labeling.

  Examples:

  Mixture of isotopically substituted compounds (selective labeling) $\ce{CH3-CH2-CH2-^{14}COOH}$, $\ce{CH3-CH2-^{14}CH2-COOH}$, $\ce{CH3-^{14}CH2-CH2-COOH}$, $\ce{^{14}CH3-CH2-CH2-^{14}COOH}$, etc., when added to $\ce{CH3-CH2-CH2-COOH}$, is designated as [G-14C]butanoic acid.
  D-Glucose in which all six positions are labeled with 14C, but not necessarily uniformly, is designated as
  D-[G-14C]glucose.

  P-83.5.2 In the name of a selectively labeled compound in which all positions of the
  designated element are labeled in the same isotopic ratio, the symbol ‘U’ is used in place of locants to denote ‘uniform’ labeling.

  Examples:

  Mixture of isotopically substituted compounds (uniform labeling) $\ce{CH3-CH2-CH2-^{14}COOH}$, $\ce{CH3-CH2-^{14}CH2-COOH}$, $\ce{CH3-^{14}CH2-CH2-COOH}$ and $\ce{^{14}CH3-CH2-CH2-COOH}$, in equal amounts, when added to $\ce{CH3-CH2-CH2-COOH}$, is designated as [U-14C]butanoic acid.
  D-Glucose in which 14C is equally distributed among the six positions is designated as
  D-[U-14C]glucose.

  Note: In the case of radioactive nuclides, ‘same isotopic ratio’ means ‘same specific radioactivity’.

(Please note that the D- stereodescriptor should be in smallcaps)


Do alkali metals not form diatomic molecules? I know hydrogen forms a diatomic molecule $\ce{H2}$, where the electronic configuration of hydrogen is $\ce{1s}$.

But why doesn't lithium also form a diatomic molecule? Its electronic structure is $\ce{1s^2 2s}$, so can't two lithium atoms come together, share their outer electron and form $\ce{Li2}$? Same with $\ce{Na}$, $\ce{K}$ etc.

Am I missing something obvious?

Diatomic molecules of alkali metals are detected in the gas phase. However, it's so happens that the bond in them is very weak and at the temperature the alkali metal vapors develop only a few percents of the metal in the vapors exists as diatomic molecules. It so happens, that metal bonding allows to achieve an overall more energetically favored state.

Outer orbitals of the alkali metal atoms are very diffuse, so their bonds are weak. Only Lithium, the smallest of the alkali metals, has practically meaningful covalent chemistry to speak of.


Addition of HOBr across C=C Source: Carey Advanced Organic Chemistry B, problem 4.1 b

Looking for interesting problems about addition to C=C, I came across this in the back of Carey. It looks like a simple addition across the alkene, but using HOBr which is something I've never came across in the lab (and perhaps not seen used as an example since I was an undergrad...).

The most obvious way to dissociate HOBr is to lose a proton, leaving the OBr anion behind, Wikipedia confirms this:

  [...] hypobromous acid partially dissociates into the hypobromite anion OBr− and the cation H+.

So far so good. And based on this fact, you might expect the product to be as drawn below. However the original literature points towards a different product, also shown below.

Obviously, the product I've drawn as expected looks a little odd, with the unstable Br-O bond present. However, I cannot rationalise how the actual product has formed.

  Ref.: JOC, 1968, 33, 3953.
  The original literature from which the problem was taken

Three possibilities:

The HOBr doesn't dissociate as I'd imagined. There is the possibility of the Br forming the bromonium ion which is then opened by hydroxide, however to get the product as isolated, you have to attack in the anti-Markovnikov sense (i.e. the product would look like the one formed if you'd attacked a primary carbocation with hydroxide).
The product forms as I drew it, but rearranges to give the product they isolate. This is what I consider to be the most likely explanation, but I'm unable to see how this happens.
The secondary carbocation forms upon protonation, but this rearranges. Again, struggling to see how this would work as the secondary cation is the most stable one that can form in this system, as there is no possibility of a tertiary carbocation.

Any suggestions to point in the right direction? Even the name of a rearrangement would be helpful.

The 'expected' product comes from analogy of addition of strong acids like HBr or acid (often sulfuric acid) catalyzed hydration. However, The pKa of HOBr is 8.7 (ref), much lower than those conditions, so it's not reasonable to make that analogy. Put another way, the base in those reactions is an alkene, which is very weak. Only a very strong acid will protonate an alkene.

In some quick SciFinder searching, I haven't found any overwhelmingly convincing evidence for the regioselectivity observed in the product, but it does seem reasonable to me by a mechanism similar to what Aditya Dev suggests.

I look at HOBr as an oxygen with a leaving group (analogous to Br-Br). Attack by the alkene on the oxygen releases bromide and gives the tertiary carbocation. Kick back by the oxygen closes a protonated epoxide, which is analogous to the bromonium. In the bromonium case, these are considered resonance structures. I don't know if that's definitely the case here (oxygen is much smaller than bromine), but I've written it that way to continue the analogy. In any case, protonated epoxides are typically attacked at the site best able to stabilize positive charge, which is the tertiary carbon. The resulting bromohydrin is the 'observed' product.


How do I calculate the normality of NaBrO3? Find the mass of $\ce{NaBrO3}$ required to prepare $\pu{150 ml}$ of $\pu{0.75 N}$ solution based on the reaction
  $$\ce{BrO3- + 6 H+ + 6e- -> Br- + 3 H2O}$$

My analysis:

$\pu{150 ml}$ solution of $\pu{0.75 N}$ means $0.1125$ equivalents of $\ce{NaBrO3}$ are present in the solution. Since, according to the given equation, n-factor (of equivalence) of $\ce{NaBrO3}$ is $6$. So $0.1125$ equivalents means $0.01875$ moles of $\ce{NaBrO3}$. The molecular mass of $\ce{NaBrO3}$ is $151$. Thus, I get the mass as $\pu{2.83 g}$ approximately. Am I correct?

  Since the book says the answer should be $\pu{1.42 g}$

To me your answer is correct. It seems that the book result have a '2' dividing the final result. Could you post all the text of the book exercise? Two possibilities, the result from the book is wrong (could be) or there is a mistake in one of the values (75 ml instead 150, 0.375 N instead 0.75 N...).


Are precipitation and crystallization both analogous between Chemistry and Meteorology? A comment under the Space SE question Is there really precipitation on Mars? Does frost count? brings up the use of the term "precipitation" in chemistry and we wonder if precipitation in weather and in chemistry are in fact analogous insofar as they both start as small particles suspended in the 3D medium (atmosphere or solution) and only fall to the ground or the bottom of the container if/when sufficiently large to make headway against Brownian or other motion.
In chemistry, on Earth in gravity, if a material from solution conformally, over all surfaces vertical and horizontal (e.g. the sides as well as the bottom of a container) is that still sometimes considered precipitation? Or is that crystallization from solution, and analogous to frost formation but not snow or rain?
Images of frost formation: http://www.its.caltech.edu/~atomic/snowcrystals/frost/frost.htm

In chemistry, on Earth in gravity, if a material from solution conformally, over all surfaces vertical and horizontal (e.g. the sides as well as the bottom of a container) is that still sometimes considered precipitation? Or is that crystallization from solution, and analogous to frost formation but not snow or rain?

Precipitation, as typically used an understood in chemistry, is really a very very old terminology (~ several hundred years). It refers to the formation of a solid by adding or mixing chemical reagents or even by cooling. There is no such requirement that the precipitate has to settle to the bottom. For example, colloids may not settle for ages. Precipitation involves crystallization, but not all precipitates are crystals (organic substances can be precipitated out by adding suitable solvents to a solution).
Now there are special ways to generate precipitates in solution. There was an experiment which we taught in classical analysis. The experiment goes like this. Add urea to an iron salt. No precipitate forms and there are no seeds to begin with. Let it boil slowly. With the passage of time, the entire solution becomes basic and it would uniformly produce precipitate iron hydroxide.  Since it was a in situ chemical decomposition of urea to produce iron hydroxides in solution, this process occurs very uniformly in the solution i.e., there is no localization where two reagent were being mixed.  This yielded uniformly sized particles. Such a process is called homogeneous precipitation in solution. Contrast this situation where we would add NaOH solution to iron salts. It will be impossible to avoid localized mixing.
Meterologists use the term precipitation as well, but as per the unabridged Oxford Dictionary (can't provide a link as it is by subscription), the full word is atmospheric precipitation.
In case you are interested, the etymology is

Etymology: < classical Latin praecipitāt-, past participial stem (see
-ate suffix3) of praecipitāre to throw or cause to fall headlong, to ruin, destroy, to fall headlong, to suffer ruin, come to grief, to
hasten, to rush, in post-classical Latin also to cause to be deposited
as a solid from a liquid solution (a1490, 1652 in British sources) <
praecipit- , praeceps (adjective) headlong, sheer (see precipe n.).
Compare Middle French, French precipiter , précipiter precipit v.


Basification of acid solution. pH calculation. How? This is my first post.

In order to understand basification I've created a pretty simple example as I currently understand it mainly because.. I really don't understand all of it.

Example:

Chemicals = Lye & Vinegar

Reaction  = NaOH + C2H4O2 = H2O + NaC2H3O2

Since the molar ratio of the equation is 1:1...

                Molar Mass          It takes...
Lye     NaOH    39.99710928  = 40   Grams of Lye to NEUTRALIZE,
Acetic  C2H402  60.05196     = 60   Grams of Acetic Acid.
                               *....a Weight Ratio of 2:3

Now to the part I don't understand...

1 liter = 1000 grams

At 5% standard acidity, a liter of vinegar has exactly

1000 Grams x 0.05 = 50   Grams of acetic acid in it.

Thus, at the above calculated Weight Ratio of 2:3...

50 Grams of acetic acid x 2/3 = 33.33   Grams of Lye to NEUTRALIZE,
                                50      Grams of Acetic Acid.
                                *....proportional Weight Ratio of 2:3

So far...

We have neutralized the 50 grams of acetic acid in the solution to pH7.

Question:

What would be the pH of the remainder 950ml of water in the bottle of vinegar? Is it the obvious? pH7?

Is my goal only to neutralize the acetic acid in the solution because the remainder is obviously just water and naturally at a of pH7?

You've made some assumptions that we can deal with, and some that need adjustment.

1) 1 liter of water weighs 1000 grams. Correct. However, 1 liter of 5% acetic acid weighs 1005 grams. No big deal; 1 liter will still contain 50 grams of acetic acid.

2) As the commenter noted, neutralizing the acetic acid gives sodium acetate, a weak base with pH ~9. You can use phenolphthalein as an indicator because it changes color at this pH.

3) The liter of neutralized acid will be homogeneous, so the 50 grams of acid and 33.33 grams of sodium hydroxide will be distributed throughout; any sample of the solution will have the same pH.

(Although it is interesting to do an acid/base titration with phenolphthalein. As you approach the endpoint, a drop of base will splash red, then fade to colorless. The next drop will splash a larger red cloud, then fade; etc. Finally, you might get a huge red cloud and think you are done, but you are just shy of the last bit of neutralization and the solution turns colorless again. The next drop flashes totally red.  The reason I mention this is that you could say that for those few milliseconds when the solution is partially red and partially colorless, the pH is different in different locations, but this is not a steady state, so you mix it till you can get a repeatable measurement.)

4) What you measure when you stick a pH meter into the solution is the concentration of the hydrogen ion (hydronium ion) that results from the water molecules reacting with everything else all over the solution. The acetic acid is a reactant in the original solution that increases the hydrogen ion quantity, but the sodium hydroxide reduces it, so you have made a solution of sodium acetate, pH ~9.


Could a magnet pull oxygen out of the air? I read that the $\ce{O2}$ molecule is paramagnetic, so I'm wondering: could a strong magnet pull the $\ce{O2}$ to one part of a room – enough to cause breathing problems for the organisms in the room?

(I'm not a professional chemist, though I took some college chemistry.)

I'm a physicist, so apologies if the answer below is in a foreign language;  but this was too interesting of a problem to pass up.  I'm going to focus on a particular question:  If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region?  The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field.  Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point:  $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility."  Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$.  But potential energy is given by $\vec{F} = - \vec{\nabla} U$;  which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor:
$$
\rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT}
$$
In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample.  To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$;  in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be?  The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$; and so we have1
$$
\chi_\text{mol} B^2 \gtrsim \mu_0 RT
$$
and so we must have
$$
B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}.
$$
If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$;  and plugging in the numbers, we get a requirement for a magnetic field of
$$
B \gtrsim \pu{258 T}.
$$
This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines.  Even at $\pu{91 Kelvin}$ (just above the boiling point of oxygen), you would need a magnetic field of almost $\pu{150 T}$; still well out of range.

1 I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules.  A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation;  and if the gas gets sufficiently dense, then all bets are off.


Reference for electronegativities of different metal oxidation states A long time ago I was researching the effect of the self regulatory response in Fe and Co. I found that my results made sense based on the idea of the electronegativity of the ions considered. I found a  webpage that listed the relationship between the different electronegativities for different oxidation states of Fe and Co.

My problem is I didn't save the webpage and I don't know where I found it. I am not a chemist so I don't know in which kind of books I could find a list of values of electronegativities for different oxidized ions.

It basically said that the relationship between the electronegativities was something like

$$\ce{Co^2+} < \ce{Fe^2+} < \ce{Fe^3+} < \ce{Co^3+}$$

Note: I don't remember it well, so what I wrote could be lies. The important thing was that there was a flip of order in electronegativities for Fe and Co when changing oxidation state.

Does anyone know where I could find that kind of information? I tried googling, but I am not finding the webpage and don't remember the terms I used to find it in the first place.

I never understood the why of the order either, and I have seen chemist webpages that tell you which ion is more electronegative by just looking at it. If someone could explain that too I would appreciate it.

Edit:

I forgot to add that it was in octahedral complexes. I was able to find this pdf online (page 32) but it doesn't state a reference.

Pearson conveniently lists cumulative experimental data in the 1988 paper [1], referrring to the earlier work of Moore [2].
Selected values of $I$ (ionization potential), $A$ (electron affinity), $χ$ (absolute electronegativity – probably, that's what you are looking for) and $η$ (absolute hardness) for iron and cobalt cations are:

Table I. Experimental Parameters for Monatomic Cations (eV)
$$
\begin{array}{lcccc}
\hline
\text{ion} & I & A & χ & η \\
\hline
\ce{Fe^2+} & 30.65 & 16.18 & 23.42 & 7.24 \\
\ce{Fe^3+} & 54.8 & 30.65 & 42.73 & 12.08 \\
\ce{Co^2+} & 33.50 & 17.06 & 25.28 & 8.22 \\
\ce{Co^3+} & 51.3 & 33.50 & 42.4 & 8.9 \\
\hline
\end{array}
$$

So it looks like the relation is a bit different:
$$\ce{Fe^2+} < \ce{Co^2+} < \ce{Co^3+} < \ce{Fe^3+}$$
Complete table as a screenshot:

References

Pearson, R. G. Absolute Electronegativity and Hardness: Application to Inorganic Chemistry. Inorganic Chemistry 1988, 27 (4), 734–740. https://doi.org/10.1021/ic00277a030.
Moore, C. E. "Ionization Potentials and Ionization Limits"; Natl. Stand. Ref. Data Ser. (U.S. Natl. Bur. Stand.); 1970, NSRDS-NBS 34. (NIST - PDF)


Uses of ruthenium-106? I read a news article recently about a cloud of ruthenium-106 being released over Europe. The article mentioned one of the uses for this specific isotope is in fuel cells for satellites. I tried to do some research but i couldn't find information on a fuel cell that used this specific isotope. What kinds of fuel cell or power generator use ruthenium-106?

Expanding @Mithoron's comment into an answer, $\ce{^{106}Ru}$ is used in radioisotope thermoelectric generators that power satellites. Plutonium-238, curium-244 and strontium-90 are the most often cited candidate isotopes for this purpose but other isotopes such as polonium-210, promethium-147, caesium-137, cerium-144, cobalt-60, curium-242, americium-241 and thulium isotopes have also been studied.
Apart from being an energy source, it also finds its application in medicine. $\ce{^{106}Ru}$ is used in cancer radiation therapy, especially to treat eye and skin tumors.


aluminum as cathode electrode Aluminum really wants to and will form aluminum oxide. Upon research, aluminum oxide is fairly unreactive. I wanted to ask if I had aluminum anode which would quickly form aluminum oxide (which is unreactive), will  that inhibit the aluminum from being oxidized? Basically, will aluminum metal loose electrons and become an aluminum ion?

Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions.
As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere.
In an aqueous solution, if the potential is high enough, $\ce{OH-}$ and $\ce{O^2-}$ are generated at the anode, which can then react with the aluminum to produce aluminum oxide.
$\ce{Al^3+}$ can also be generated directly.
The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the $\ce{Al^3+}$ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode.
In this way, coatings thicker than the normal passivation in air can be produced.
However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made).
This is the normal behaviour in aqueous solutions at near-neutral pH (5–7). [1]

However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface.
One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed $\ce{OH-}$, leaving pores in the oxide layer through which the ions can travel and continue to react.
These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.


Which has stronger hydrogen bonds: water or ice? As ice is the solid form of water and it has more hydrogen bonds than water, because it's oxygen atoms are precisely tetrahedrally positioned and each oxygen is hydrogen bonded by four neighbouring oxygen atoms.

This changes the density of ice by expanding it, but does it change the strength of hydrogen bonds of ice? Or more precisely, which one has strong hydrogen bonds, water or ice?

The quick answer is that ice has stronger hydrogen bonds than liquid water on average.$\require{mediawiki-texvc}%for angstrom unit$

As has already been brought up, this question is not easy to answer comprehensively as it may very well depend on temperature and pressure in ways that are not immediately obvious. That being said, I will answer this question for the most obvious case which is looking at hydrogen bond strength when water is just barely a liquid (just above $\pu{0~^\circ C}$) and when water is just barely frozen (just below $\pu{0~^\circ C}$). Also, as you may or may not know, ice freezes into many different forms which all have different lattice arrangements. I cannot speak to unusual forms of ice such as the one with a cubic lattice, but what I say is immediately applicable to forms like proton-ordered hexagonal ice (I think this is ice IX). Regular ice is called ice Ih.

It is worth looking at the very comprehensive reference [1], which discusses details of water from the dimer all the way to the many forms of ice which can form. The important points for us are only the following structural values, however:

In ice Ih, the average $\ce{O-H}$ bond length is $\pu{1.01 \AA}$ and the $\ce{HOH}$ bond angle is $109.5^\circ$. The trend which has been observed experimentally is that in going from the gas-phase water monomer, through the clusters, to the liquid, and finally to ice, the bond angle slowly increases to this tetrahedral value. Additionally, the average bond length in the liquid is shorter than in the solid, but you will have a hard time finding reliable reported values of this number from experiment because it is very difficult to determine the position of hydrogen atoms from scattering experiments in non-solid phases (this is mostly because of an ambiguity as to which oxygen atoms a hydrogen should be associated with in the scattering data). Nonetheless, there are reliable average $\ce{O-O}$ distances from scattering experiments.

In the liquid, the average $\ce{O-O}$ distance is $\pu{2.85 \AA}$ while in ice Ih it shrinks to $\pu{2.74 \AA}$. Intuitively, it is very hard to understand how this would happen if it is not simply due to increased hydrogen bond strength.

We can look at an even simpler case though and see that increased hydrogen bond strength causes the $\ce{O-O}$ distance to decrease in ice. This example is a comparison of the water dimer and water trimer. In the water dimer, the $\ce{O-O}$ distance is $\pu{2.976 \AA}$. In the water trimer, the $\ce{O-O}$ distance is about $\pu{2.82 \AA}$. This decrease in distance and the corresponding red-shift in vibrational frequency are given physical meaning in the Ref. [2].

Ref. [2] gets at a very important point for us, one which Ref. [1] also uses to explain the shrinking in $\ce{O-O}$ distances in ice. This is that hydrogen-bonding in water highly cooperative. What do I mean by this? This is exactly the topic of Ref. [2]. The key idea is this: when you allow three water molecules to mutually hydrogen bond, as in the ring shape of the water trimer, the increase in attractive interaction strength is more than simply three times the interaction strength of a water dimer. By attractive interaction strength, I mean that if you subtract out the initial decrease in binding energy which comes from distorting the water monomers. Again, all of this is laid out in detail in Ref. [2], but we learn the general rule that many-body interactions in water are very important.

So, finally, I feel we have said enough to explain why the knowledge that a shrinking $\ce{O-O}$ distance and widening $\ce{HOH}$ angle tells us that ice has stronger hydrogen bonds than liquid water. In the first case, the decreasing $\ce{O-O}$ distance means that every pair of monomers experiences greater nuclear repulsion. This must be at least offset by a greater hydrogen bond strength in order for this geometry to be favorable. Part of the way this repulsion is offset is that as the system cools and forms a solid, the hydrogen-bonds are allowed to become more ideal. This is why the bond angle widens. The tetrahedral arrangement of atoms means that the $\ce{OH--H}$ hydrogen-bond angle is much closer to 180 degrees on average than in the liquid where it frequently deviates fairly significantly.

There is one last point to make. It can be estimated both from accurate simulations and from experiment what is the average dipole moment of a water monomer in bulk liquid and in ice. It is found that the dipole increases going from the liquid to ice. This additionally tells us that the average charge on each hydrogen atom must have increased from in the liquid. We are guaranteed this because we know the bond angle increased, which should generally decrease the dipole moment (a linear configuration has no dipole). Instead, the bond angle increases, so we get more ideal hydrogen bonds, and the charge increases, so the hydrogen bonds would be stronger even if they were no more ideal than in the liquid!

Hopefully this rather long-winded answer helps you understand how we can take the evidence at hand to arrive at the conclusion that ice has stronger hydrogen bonds than liquid water.

Now to the question of different forms of ice and different temperatures and pressures. I would think that pretty much across the board, every form of ice will follow the same trends we have seen here and hence have stronger hydrogen bonds. One strange case is cubic ice, which has $\ce{OHO}$ angles somewhere close to 90 degrees. This is very strange, but should lead to a large increase in the dipole of each molecule and probably increase in positive charge on each hydrogen atom, so even then the hydrogen bonds should be stronger than in the liquid.

There may be temperatures and pressures where some of these arguments break down, but the $\ce{O-O}$ distance should smoothly decrease to that in the liquid and the angle should close to that in the liquid as the pressure increases, so one would expect the hydrogen bond strength to decrease to that in the liquid and never overshoot it.

References:

Ludwig, R. Water: From Clusters to the Bulk. Angew. Chem. Int. Ed. 2001, 40 (10), 1808-1827. DOI: 10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1.
Xantheas, S. S. Cooperativity and hydrogen bonding network in water clusters. Chemical Physics 2000, 258 (2-3), 225-231. DOI: 10.1016/S0301-0104(00)00189-0.


How to find the combining valence of an anion when it exhibits variable valency? I'm doing an online course and can't for the life of me figure out how to write chemical formulas from the chemical name.

Take the example of calcium chloride.

I know the the valence for calcium is +2 because my periodic table shows me this, but I cannot figure out the valence of chlorine, (or any other element) with multiple valences. My periodic table lists chlorine as having a possible valence of -1, +1, +3, +5, or +7. How am I to know which one is to be used?

First of all, sorry for the (very!) long post!

Determining the Number of Valence Electrons

The number of valence electrons of atom can be determined by the group in which the atom is in. For example, Na is in group 1A, and thus has one valence electron (1+ charge). Chlorine is in group 7A and thus has 7 valence electrons (1- charge). A quick (but very simplified) way of determining if an element has a positive or negative charge is to look again at its group. Groups 1A, 2A, and 3A are positive; while groups 5A, 6A, and 7A are negative. An element in group 4A can be both.

Ionic Compounds:

If the charges are the same, no "balancing" is necessary:

if Na (1+) were to bond with Cl (1-) it would form NaCl.

If the Charges are different, a quick way to determine the formula unit is to assign each element the charge of the other:

if B (3+) were to bind with Cl (1-) it would form BCl3: Boron get the 1 from Chlorine; Chlorine gets the 3 from Boron.

Covalent Bonding

Covalent Bonds are a little trickier and has more contributing factors. It seems like you're having trouble with Lewis-Dot diagrams so that's what I'll address. In order to correctly draw the Lewis-Dot diagram of a compound, you have to determine the formal charge of each atom. For demonstration purposes I'm going to use CH4.

The formal charge of an atom can be determined by:

Number of Valence Electrons - non bonding electrons - number of bonds

So in our example, the formal charge of Carbon is :

4 Valence electrons - 0 non bonding electrons - 4 bonds = formal charge of 0.

Then Hydrogen:

1 Valence electron - 0 non bonding electrons - 1 bond = formal charge of 0.

So if we add the formal charged together we get 0, which makes sense because the charge of the compound is 0.

In general, the more compounds that can have a formal charge of 0, the more likely it is that that particular structure will occur, as it will be more stable. If the compound is a poly atomic ion, the more electronegative atom should get the negative formal charge (electronegativity increases as you move up the group and to the right in a period).

One more significant detail is the ability of some atoms to expand the octet, which can occur if the element is in the third period or beyond. In this case, the charge of the atom can be made 0 (or whatever appropriate) if more bonds are made than typically seen.

i.e. in ICl4-, Iodine, though typically only making one bond, bonds to four Chlorine atoms. In this case formal charge = 7 valence electrons - 4 bonds - 4 non bonding electrons = 1-, the charge of the polyatomic ion. While Cl is more electronegative, only I can expand its octet, thus I has the negative formal charge. The 4 non bonding electrons on Iodine were determined by counting the total number of valence electrons (7 from I + 4*7 for Cl + 1 from the charge = 36) and adjusting accordingly.

There are other exceptions such as free radicals and atoms that can have less than a full octet, but the "rules" above should be sufficent guides for basic Lewis Dot structures.

Hope that covered everything :)


What should be the molarity of HCl and H2SO4 acid so that they can be considered safe to touch? I know $\ce{HCl}$ and $\ce{H2SO4}$ acid of concentration 17 M is dangerous and corrosive, but I wonder can we dilute these acids to the extent that they become safe to touch?

If yes, then what should be the molarity of that safe acid?

According to the Globally Harmonised System (GHS), hydrochloric acid with a concentration of c ≥ 25 % and sulfuric acid with a concentration of c ≥ 15 % are labelled "causes severe skin burns and eye damage" (H314) and are classified based on standard animal test data in skin corrosion Sub-category 1B ("Corrosive responses in at least one animal following exposure > 3 min and ≤ 1 h and observations ≤ 14 days"). "Skin corrosion refers to the production of irreversible damage to skin; namely, visible necrosis through the epidermis and into the dermis occuring after exposure to a substance or mixture."

Hydrochloric acid with a concentration of 10 % ≤ c < 25 % and sulfuric acid with a concentration of 5 % ≤ c < 15 % are labelled "causes skin irritation" (H315) and are classified based on standard animal test data in skin irritation Category 2. "A substance is irritant to skin when it produces rversible damage to the skin following its application for up to 4 hours."

Concentrations well below these values may be considered safe for the exposure of skin to the substance.

Note that the natural skin surface pH is below 5.


What is the compression work? A cylinder contains 2 kg of steam at 200 kPa and 200 $^\circ$C. Now steam is compressed at constant pressure until its temperature rises to 250 $^\circ$C. What is the work done during this compression process?

To calculate work at constant pressure, you need the following:

$$w=-P \Delta V$$

where $w$ is work, $P$ is pressure and $\Delta V$ is the change in volume.

You will need to solve some gas laws relations to get both $P$ and $\Delta V$ in terms of the single variable $T$.

Algebraically, it will simplify to $P\Delta V = nR\Delta T$, assuming ideality.

If you need to assume a non-ideal gas (which is a good assumption for water at high pressure). If we use the van der Waals equation, things get dicey and calculus is needed. I'm hoping you don't need calculus:

$$d[(P-\frac{n^2a}{V^2})(V-nb)]=d(nRT)$$

$$d(PV-\frac{n^2a}{V}-nbP+\frac{n^3ab}{V^2})=d(nRT)$$

$$PdV+VdP+\frac{n^2a}{V^2}dV-nbdP-\frac{2n^3ab}{V^3}dV=nRdT+RTdn$$

Since $n$ and $P$ are constant $dP=0$ and $dn=0$

$$PdV+\frac{n^2a}{V^2}dV-\frac{2n^3ab}{V^3}dV=nRdT$$

$$\int_{V_i}^{V_f}(P+\frac{n^2a}{V^2}-\frac{2n^3ab}{V^3})dV=\int_{T_i}^{T_f}nRdT$$


Disproportionation of 2 KO₂ + 2 H₂O → 2 KOH + H₂O₂ + O₂ How is the following reaction a disproportionation reaction?

$$\ce{2 K\overset{-1/2}{O}_2 + 2 H2\overset{-2}{O} ->2 KOH + H2\overset{-1}{O}_2 + \overset{0}{O}_2}$$

In this, the OS are $-1/2$, $-2$, $-1$, $0$, respectively, but in disproportionation  the element$ O $both oxidizes and reduces. How do we know that this happens here? $\ce{KOH}$ may get oxidized, and $\ce{H2O}$ get reduced. How do we know the same compound has undergone both?

The disproportionation happens to elements, not to compounds.

In this particular case, to the part of oxygen,

$$  \overset{\ce{2 KO2}}{ 4 \times -0.5}=  \overset{\ce{H2O2}} {2\times -1} +\overset{\ce{O2}}{ 2\times 0}$$

while the other part keeps the oxidation state constant:

$$  \overset{\ce{2 H2O}} {2 \times -2}=  \overset{\ce{2 KOH}} {2\times -2}$$

$\ce{H2O}$ does not get directly involved in disproportionation.

$\ce{KO2}$ oxidizes and at the same time reduces  itself.

This is the nature of isproportionation, which happens , if the particular compound of the element in the intermediate oxidation state has higher Gibbs energy then particular compounds of this element in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred.

Typical cases are $\ce{2 Cu^{I}->Cu^{II} + Cu^{0}}$ and $\ce{2 Hg^{I}->Hg^{II} + Hg^{0}}$


I am looking for any substance that would be unstable or dangerous when it is cooler or not on heat, and is stable when it is being boiled I am a first responder and one of our protocols when entering any unknown lab environment is to turn off any burners. Most of the time, we try to not disturb the environment as much as possible so a colleague of mine and myself were curious as to any substance that would actually be more dangerous after turning off said burner. Thank you.

Well, in theory, a few compounds could become explosive when crystallizing out of a solution (which is promoted by low temperatures) though I do not think that "turning off burners" would be enough. The "best" candidate I can imagine would be "azidoazide azide" which can detonate for... just any reason! But nobody uses it, just because of that.


Can dissolution reactions be endothermic? Is it possible for a dissolution reaction of the form

$$\ce{X(s) ->[solvent] X (solv)}$$

or

$$\ce{M_mX_x (s) ->[solvent] m M^x+ (solv) + x X^m- (solv)}$$

to be endothermic, i.e. $\Delta_\mathrm r H > 0$?

You will find the answer in probably any textbook on chemical thermodynamics, or in a thermochemistry chapter in most physical chemistry textbooks, or in Wikipedia's enthalpy of dissolution page:

  Dissolution can be viewed as occurring in three steps:

  Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.
  Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding
  Forming solvent-solute attractions (exothermic), in solvation.

  The value of the enthalpy of solution is the sum of these individual steps. Dissolving ammonium nitrate in water is endothermic. The energy released by solvation of the ammonium ions and nitrate ions is less that the energy absorbed in breaking up the ammonium nitrate ionic lattice and the attractions between water molecules. Dissolving potassium hydroxide is exothermic, as more energy is released during solvation than is used in breaking up the solute and solvent.

So, it can have any sign. The linked page also gives data for common compounds in water, some of them endothermic, some of them exothermic. Ammonium nitrate, $\ce{NH4NO3}$, is a good example of a salt which dissolves in water in an endothermic reaction: if you dissolve some $\ce{NH4NO3}$ in water, the drop in the temperature will be noticeable.

Regarding why the substance is soluble why its dissolution is endothermic, you have to remember that the reaction takes place if $\Delta_\mathrm r G$ is favourable (i.e. negative), and $\Delta_\mathrm r G = \Delta_\mathrm r H - T \Delta_\mathrm r S$. Overall the free energy must be negative for dissolution to occur (on a thermodynamic basis; kinetics are another issue), not the enthalpy.


In the elementary steps of a reaction, why do we assume that the coefficients of a reactant correspond to its order? Recently, I was taught that the coefficients of the reactants in an elementary step of a reaction indicates that reactant's order. Why is this the case? I thought that the order of a reactant was not related to its coefficient.

For example, in the following termolecular reaction mechanism

$$\ce{A + 2B -> C + 2D}$$

the rate law is $k[\ce{A}][\ce{B}]^2$. Why is this the case?

EDIT: Also, for equilibrium reactions, do we always assume the coefficients of the reactants to provide the order for those reactants? What about for equilibrium reactions involving multiple steps?

If a reaction has just one step, it's named an elementary reaction. In this reaction all of reactant directly participate on reaction and effectively collide with each other. So, the order of reaction corresponds to its coefficient, but if a reaction is not a elementary reaction, there isn't any reason for colliding all reactants with each other. For example, consider the following overall reaction:

$$\ce{A + B + D -> F}$$

Assume the corresponding mechanism is constituted of two elementary reactions:

$$\ce{A + B -> C}$$
$$\ce{C + D -> F}$$

As shown above, there is no need to collide A with D or A with C ... Then, in these type of reactions coefficients are not corresponded to order of reaction .


How do thermodynamics and kinetics control this reaction? The answer is A, but how is that? I know B is an intermediate, so that after short time it will be bigger than C, but what does "thermodynamic control" and "kinetic control" mean?

Initially no B and C exist. At short times as $k_1 > k_2$ production of B is favoured over C. (The initial amounts of B and C are so small that back reactions can be ignored wrt. forwards ones.) At long times  equilibrium constants matter and as the equilibrium constant $K_2$ is greatest this is favoured to produce C over B.


Product of Williamson Synthesis I read that, and was wondering that in the first reaction shown they have used a tertiary butoxide where ether is formed because there is a primary alkyl halide but in the second reaction, they have used a methoxide. The Alkene formed is due to the tertiary alkyl halide my doubt here, is that is there any reason for taking a tertiary butoxide in one and methoxide in the other? Does the product also depend on the attacking species carrying the O-?

Update:

What I am trying to say, is that in the book, they first gave 4th reaction and if they were to follow the similar reaction then they should have mentioned something like 2nd reaction but they rather changed the tert-butoxide to a methoxide. Is there a reason behind it? And what would be the products for the other two reactions? Would it be a mixed product with no component in a majority?

In Williamson ether synthesis, the reactive species is an alkoxide, which can act as a strong nucleophile or base. When reacted with an alkyl halide, if the alkoxide acts as a nucleophile in an SN2 reaction, the ether product is formed. If the alkoxide acts as a base in an E2 reaction, then an alkene is formed.

So when considering a the reactivity of alkoxides, there is a competition between SN2 and E2. Which pathway will dominate is determined by several factors, and the steric nature of the nucleophile and electrophile are critical.

With respect to the alkyl halide, we know that SN2 reaction is favored by sterically unhindered electrophiles. So the trend in rates is methyl > primary > secondary >> tertiary (to the point where this is generally impossible). Steric hindrance of the nucleophile will also slow the SN2 reaction. The E2 reaction is relatively insensitive to sterics, so if there is a competition between SN2 and E2, a bulky nucleophile/base will favor E2. That is why often potassium tert-butoxide is used to perform an E2 reaction.

Let's analyze each of the reactions presented.

1) A primary alkoxide (sterically unhindered) is reacted with a tertiary bromide (very hindered). SN2 reaction is extremely slow for tertiary halides, and nine beta-hydrogens are available to deprotonate from the halide, so E2 reaction dominates, and we would expect the alkene is formed.

2) A tertiary alkoxide (very sterically hindered) is reacted with a tertiary bromide (very hindered). Again, SN2 reaction is extremely slow for tertiary halides, and nine beta-hydrogens are available to deprotonate from the halide, so E2 reaction dominates, we would expect that the alkene is formed.

3) A primary alkoxide (sterically unhindered) is reacted with methyl bromide. Not only is methyl bromide very unhindered, but there are no beta-hydrogens available to participate in an E2 reaction. SN2 is the only viable pathway, so the ether will be the only product formed.

4) A tertiary alkoxide (very sterically hindered) is reacted with methyl bromide. Again, not only is methyl bromide very unhindered, but there are no beta-hydrogens available to participate in an E2 reaction. SN2 is the only viable pathway, so the ether will be the only product formed.


How does the rate constant change with the change of the temperature? In class, we discussed the following multiple-choice question.

The rate expression for a reaction is:
$$rate = k[X][Y]$$
Which statement is correct?
A. As the temperature increases, the rate constant decreases.
B. The rate constant increases with increased temperature, but eventually reaches a constant value.
C. As the temperature increases, the rate constant increases.
D. The rate constant is not affected by a change in temperature.

Considering the Arrhenius equation given below, the rate constant is dependent on the absolute temperature of the reactants and activation energy of the reaction.
$$k = Ae^{-\frac{E_\mathrm{A}}{RT}}$$
Taking the following limit as the absolute temperature approaches infinity reveals that the rate constant converges to the Arrhenius constant at very high temperatures. This is also evident when graphing the Arrhenius equation. Therefore, we conclude that option B is correct.
$$\lim_{T \to \infty} Ae^{-\frac{E_\mathrm{A}}{RT}} = A\lim_{T \to \infty} e^{-\frac{E_\mathrm{A}}{RT}} = A$$
However, when checking the mark-scheme, option C is listed as the correct answer. Is there a mistake in my reasoning?

Answer B is technically correct although C could also be accepted. Assuming $E_A$ to be constant, at low temperatures because of the inverse $T$ in the exponential $\exp(-E_A/(R\cdot \text{small number}) \equiv \exp(-\text{big})$ which is a small number. At high temperatures $\exp(-E_A/(R\cdot \text{big number})\equiv \exp(-\text{small})$ is a big number, so the rate constant increases with temperature.

At temperature such that the exponential  $\to 1$ the rate constant is $A$ for all practical purposes. Usually this limit is not reached unless $E_A$ is very small and then as $A$ is also a function of temperature this ($A$) term becomes now important. Generally, however, for the vast majority of reactions the exponential term is the most important.


Is water with oxygen-17 safe to drink? Is it safe to drink and what else can it be used for?

The isotope 17 of oxygen is NMR-Active, so it can be useful if isotopic labelling (eg: in the experimental study of a reaction mechanism) is needed.

Aside from that, since Oxygen-17 exists in nature, you are currently breathing it, and you do drink it.

Now, the question is: could we survive by breathing and drinking only Oxygen-17?

I don't think that anyone has ever answered that question, for economical reasons (from a fast search, a gram can be bought for around 500$!).

But, if I have to take a guess, I would say yes, since I don't think that the nuclear and mass properties of such an "heavy" atom would interfere with biochemical processes (unlike hydrogen's isotopes, the mass of oxygen and its isotopes is so large that they are not involved in quantum tunneling mechanism, and from this point of view their behaviour should be quite similar).


How can you explain the solubility of BeF2? I know that Be has a tendency to create bonds with an increase covalent character so I can tell that BeF2 is more covalent than let's say MgF2. So why is BeF2 soluble in water while MgF2 is barely soluble ? Shouldn't the ionic compound be more soluble ?
My guess is that the lattice enthalpy of BeF2 is not as great as the enthalpies of the other fluorides so it can be break down with much more ease and the energy gained from hydration outweighs the energy needed to separate the ions.

It seems that for BeF2 the hydration enthalpy of Be2+ is really high (due to Be2+ being a small cation and having a high charge density) so the energy required to break the lattice is smaller than that gained from hydration of Be2+ while for something like MgF2 the enthalpy of the lattice is greater than the energy you would get from hydration of Mg2+. In general when dealing with compounds of the group II with F- because of the small radius of F- the lattice enthalpy is great and only the hydration energy released by the hydration of Be2+ outweigh it thus making BeF2 soluble while all the other fluorides formed with the group II elements are quite insoluble in water


Why is the van der Waals coefficient b equal to four times the volume of the particle? In the van der Waals equation of state

$$\left[p + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$

the coefficient $b$ is supposed to represent the volume occupied by the particles. Why then is it equal to four times the volume and not simply the volume?

While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically.
In deriving the VDW equation, the particles are still assumed to be hard spheres, but this assumption is corrected for with the parameter $a$.
The hard sphere approximation forbids that two particles penetrate each other's radii. Thus, we find that two spheres in closest contact are surrounded by a sphere of radius $2r$ (or the diameter of one of the original spheres).

(source: nyu.edu)
Thus, the volume excluded by the particles from the larger sphere surrounding the two spheres shown is
$$b' = \frac{4}{3} \pi d^3 = 8 \cdot \frac{4}{3} \pi r^3$$
Thus, the excluded volume per particle $b$ is $b'/2$ or,
$$b=4 \cdot \frac{4}{3}\pi r^3$$
which, as you point out, is four times the volume of a single particle.
The interesting thing about this is that it does not represent the actual value of $b$ for any given atom, but represents the upper bound of $b$ for any given atom. What I mean by that is, $b$ could very well be correct by calculating four times the volume, but often experiment will show that it less than the calculated value of $b$ because atoms are not hard spheres.
For instance, using helium, which is the closest we'll get to a hard sphere:
$$b_{\ce{He},\mathrm{calc}} = 4 \cdot \frac{4}{3} \pi (140\ \mathrm{pm})^3 \cdot N_\mathrm{A} = 0.02767\ \mathrm{L\ mol^{-1}}$$
while,
$$b_{\ce{He},\mathrm{exp}} = 0.0238\ \mathrm{L\ mol^{-1}}$$
So, the experimental value of $b$ is indeed smaller, but the calculated value gives a rough idea.


E2 reaction - strong base In E2 reaction, there is more than one product. According to Zaitsev’s rule – “The alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.” (Wikipedia)

Now, my question is, how can the base effect the percentage of the minor and major products? Does a stronger base give us more major product and vice versa?

The particular question I’m trying to solve is:
In the E2 reaction of 2-bromo-2,3-dimethyl butane, which base would give more of the minor product and which of the major product?

It will be more difficult for sterically hindered bases to remove the proton of the β-carbon having the fewest hydrogen substituents, since this carbon atom has more bulky groups around it compared to the the others. So base 3, being the less hindered base, will provide a higher percentage of the major Zaitsev’s product and base 2 will provide a higher percentage of the minor product compared to the other bases.


Determine reaction order and reaction rate In a solution of $\ce{Fe^3+}$ the concentration was measured to be:
$$\begin{array}{c|c}
t/\pu{min} & \ce{[Fe^3+]}/\pu{\mu M} \\ \hline
10\ & 238\\
20 & 227\\
40\ & 206\\
60\ & 187 \\
80\ & 169\\
100\ & 154 \\
\hline
\end{array}$$
The decrease in $\ce{Fe^3+}$ concentration is due to the hydrolysis of the ion and precipitation of hydroxide. Calculate the reaction order and reaction rate.

I am having trouble understanding how to tackle this problem. I began by find the reaction formula:
$$ \ce{Fe^{3+} + H_2O <=> Fe(OH)^{2+} + H^+} $$
So this is a second order reaction. I then did:
$$\begin{array}{c|c|c}
\text{time} & \ce{[Fe^3+]} & \text{H2O} & \ce{[Fe(OH)^2+]} & \ce{[H+]}\\ \hline
t=0 & a\ & a & - & -\\
t=t & a-x & b-x & x & x\\
t = \infty & - & b-a & a & a \\
\hline
\end{array}$$
But I am unsure if I should include water and the hydrogen ion, and also how I should continue if this is correct. Can someone please push me in the right direction here. Thank you!
EDIT
First order equation:
$$ \ln \frac{a}{a-x} = kt$$
Is a then equal to $238$ and $a-x$ is the initial value ($238$) minus the next value? e.g., $238-227 = 11$.
If I plot:
$$\begin{array}{c|c|c}
t & \ln \left (\frac{a}{a-x}\right) \\ \hline
10 & 0\\
20 & 3.07\\
40 & 2.43\\
60 & 2.53\\
80 & 2.58 \\
100 & 2.76\\
\hline
\end{array}$$
I get that the slope is $0.0168$ and that would be the rate of the reaction?

The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the  next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned equation: $$\ln[A] = \ln[A]_0 − kt$$ Calculate the value of $\mathrm{\ln[\ce{Fe^{3+}}]_0}$ by eliminating k. Thereafter, you can graph the above mentioned equation $\ln[\ce{Fe^{3+}}] \text{ vs time t}$ with $\ln[\ce{Fe^{3+}}]_0$ as the y-intercept. The slope will yield the value of the rate constant.

[...]that would be the rate of the reaction?

For some reason unbeknownst to me, a lot of people tend to confuse between rate and rate constant. The slope will yield the rate constant. Next,
$$\textrm{rate}=k[\textrm A]^{a} \quad \text{}$$ where $a$ is the order of the reaction with respect to reactant $\ce{A}$.
As a side-note, since this is a pseudo-first order reaction, the rate law equation, initially this: $$\text{Rate} = k^{'}[\ce{Fe^{3+}}][\ce{H2O}]  $$
now appears as this:$$\text{Rate}=k[\ce{Fe^{3+}}]$$
the variable $'k'$ in the second equation is the slope that you obtained from the graph. Here $k=k^{'}[\ce{H2O}]_o$
For more about pseudo-first order reactions you can check one of my previous answers.


CO₂ molecule π bonds Why does $\ce{CO2}$ make two π bonds? Why can't it make four σ bonds? Also, why do the π bonds not involve for repulsion?

Recall the definition of σ-bonds. They are caused by the head-on overlapping of the orbitals.

$s$ orbitals always, and only form σ-bonds while the $p$ orbitals in the direction of bond axis (which is $p_x$), also form σ-bonds.

The other two $p$ bonds cannot form σ-bonds as they are perpendicular to the bond axis, and cannot overlap head-on. In such cases, π-bonds are formed, by the sideways overlapping of the orbitals.

The $p$ bonds not perpendicular to the bond axis involve in π-bond formation ($p_y$ and $p_z$).


What is the modern hardware/supermarket term for 'laundry bluing'? I want to make this amazing paper Crystal Tree with my kids. One of the ingredients is laundry bluing. It is described thus:

  The magic crystals that are left behind are a combination of the Mrs. Stewart’s Bluing and the table salt. The solution that you created is supersaturated by the bluing and salt that you add to the water. The bluing is a colloid, with many tiny particles suspending themselves within the water. It’s just like when you shake up a snow globe, except the particles of bluing are much smaller than the snow. As the bluing and salt water make their way up the tree, the water begins evaporating. These means there is less water able to support the bluing particles and dissolved salt. This evaporation allows the salt and bluing particles to crystallize, resulting in your beautiful tree.

No one in my supermarket has heard of laundry bluing. Is it possible it is hazardous and has been banned in some states/countries? Perhaps it has another name? Perhaps this has a chemical name?

My question is: What is the modern hardware store/supermarket term for 'laundry bluing'?

The active ingredient is prussian blue which you can make yourself from potassium ferrocyanide and an Fe(III) salt.


Why is a comma used to separate hydrogen gas and platinum when using cell notation? I have come across many online articles and books where I see the following:

  Gaseous electrode

  Gases such as $\ce{H2}$ and $\ce{Cl2}$ can act as anode and as cathode by loosing electrons ans by accepting electrons, respectively. For example,

  $\ce{Pt,H2 | H+}$ (acts as oxidation potential)
  or
  \begin{align}
\ce{0.5 H2 (g)(\pu{1 atm}) &-> H+ (aq) + e-} \\
E_\mathrm{cell} &= E_\mathrm{cell}^\circ -\frac{0.059}{1}\log{[\ce{H+}]} \\
 &= 0 + 0.059\mathrm{pH}
\end{align}

Aren't the two species in different phases? If so, aren't they supposed to be separated by a '|' and not by a comma?

Since $\ce{Pt}$ is not a reactant and only a catalyst, one only wants to write $\ce{H_2}$ on the reactant side. When it is specified, since it is not involved in the reaction, the | is not warranted since Pt doesn't show up in the reaction as the comment by TAR86 pointed out.

Generally, the electrode material is not entered into the shorhand notation since the material of the electrode doesn't change the thermodynamic potential. In this case, the choice of $\ce{Pt}$ is critical and that is why it is specified. The potential of the electrode would be affected by the details of the measurement device with any other electrode material that is a worse catalyst.


Will water allow salts to be dissolved in methanol (that are otherwise poorly soluble)? If we make a mixture of water, methanol and a salt like $\ce{FeCl2}$, $\ce{NaCl}$ etc, will the ability of water and methanol to mix allow $\ce{FeCl2}$ and $\ce{NaCl}$ to be a part of this mixture and form a solution as opposed to when $\ce{NaCl}$ and $\ce{FeCl2}$ are just added to methanol in which they are poorly soluble?

Or will we see layers forming (which I don't see happening because of the strong tendency for water and methanol to mix)? Or will we just see a methanol-water layer with some salt dissolved in it and some salt left at the bottom (even though they are soluble in water)?
Thanks in advance

This is a highly complicated question, and the answer can't be derived by logic alone.

Maybe the liquid phase will remain mixed, and just dissolve some salt (more than would dissolve in methanol, but less than would in pure water). Or maybe it will separate into water-rich and alcohol-rich layers, with nearly all dissolved salt staying in the former. This may be considered an extreme case of salting out.

I know for sure that separation does occur in water/ethanol/$\ce{CaCl2}$ system, even though ethanol by itself is miscible with water in any ratio, as is methanol. Then again, $\ce{CaCl2}$ is much better soluble in water than $\ce{NaCl}$, so the latter might not be strong enough to produce the same effect.

The bottom line is that you can't tell for sure until you try it. Chemistry is an experimental science, after all.


How are cells rechargable? I have learnt that you can test if a reaction is thermodynamically feasible by testing if the cell voltage $E_\mathrm{cell}$ is greater than $0$. Surely, if a reaction is feasible, then the $E_\mathrm{cell}$ of the reverse reaction is negative, and so not thermodynamically feasible; in which case, how can a cell be recharged? Taking the following example:

$$
\begin{align}
\ce{Li+ + CoO2 + e- &<=> LiCoO2} &\quad &\pu{+0.60 V}\tag{1a}\\
\ce{Li+ + e- &<=> Li} &\quad &\pu{-3.00 V}\tag{1b}\\
\end{align}
$$

$$
\begin{align}
\text{In use:} &\quad &\ce{CoO2 + Li &<=> LiCoO2} &&\tag{2a}\\
\text{Charging:} &\quad &\ce{LiCoO2 &<=> CoO2 + Li} &\quad \mathrm{EMF}&= \pu{+3.60 V}\tag{2b}\\
\end{align}
$$

Is it the case that putting an EMF against the EMF makes the reaction feasible? In this case, could any reaction be made thermodynamically feasible by putting a potential difference across it?

Firstly, I would replace the word "feasible" with "favorable". I'm also going to replace $E_{cell}$ with $E_{OCV}$, where the open circuit voltage (OCV) is the potential of the cell without any applied electric potential. So you're right in that if the $E_{OCV}$ is positive, the net reaction is thermodynamically favorable: it will occur spontaneously if allowed to and will produce energy that can be used by the load connected to the battery (i.e. the cell can do work).

You're correct in saying that the reverse reaction is thermodynamically unfavorable, but that doesn't mean the reaction won't happen, just that you need to do work on the battery in order for it to happen. In most cases, this means applying a bias potential ($E_{cell}$) that is more positive than the $E_{OCV}$. When that happens, the current flowing across this bias potential takes energy into the cell and recharges it.

However, not every reaction will proceed in this way. In many cases, if a high potential is applied, another reaction will occur. If you have water in your battery and apply a high potential, you may electrolyze water, splitting into elemental hydrogen and oxygen. This will then reduce the usable capacity and/or cause rupture of the battery.

Lastly, all of this ignores the role of kinetics. Just because a reaction is thermodynamically favorable doesn't mean that it happens at a rate that useful. If you have a cell with $E_{OCV} = 3.60V$, applying $3.601V$ will begin to charge the cell but at such a slow rate than you'll never notice. You generally need to apply at least a few tens of millivolts of over-potential to a lithium ion battery in order to get any appreciable current. So in the case on non-rechargeable (chemically non-reversible) batteries, the over potential needed to reverse the reaction will be big enough that you will instead hit the potential for some other, undesirable reaction.


What is the correct nomenclature for Glucuronyl C5-epimerase? I see people writing "Glucuronyl C5-epimerase", "Glucuronyl C-5-epimerase", "Glucuronyl-C-5-epimerase" and "Glucuronyl C5 epimerase", which one is correct? Is there a naming convention?

I have not found this particular enzyme in Enzyme Nomenclature. However, following the Recommendations of the Nomenclature Committee of the International Union of Biochemistry and Molecular Biology, the name should be glucuronyl 5-epimerase. See also this sublist for some analogous names of other epimerases.


Protonation of a compound having alcohol and ethereal linkage I approached the problem this way:

The ethereal oxygen gets protonated(due to greater + I , the electron density on it is higher, hence it is more likely to behave as a lewis base).

Consequently the product formed is pent-2-en-1,5-diol

Apparently that is the wrong answer, could anyone help me understand what exactly happens here and why my answer is wrong?

The alcohol group of your educt (tetrahydrofurfuryl alcohol) is protonated, and after subsequent loss of water, the resulting primary carbocation can rearrange to a more stable 6-membered ring cation, which is stabilized by resonance. Initial protonation at the ether oxygen is therefore disfavored, as it will not result in the formation of a similarly stable carbocation (with positive charge adjacent to oxygen). After final deprotonation, 2,3-dihydropyran is obtained.


Why does ice cream not go back to its original form upon refreezing? I came from a birthday party and someone gave me a sundae. I'm already full so I just decided to eat it at home. After arriving at home the sundae melted. So I left it in the freezer overnight. When I looked at my sundae it just formed ice crystals. It didn't went back to its original form. Why is that?. Why is it when I do this to ordinary ice cream it goes back to it's original form?

Ice cream is a frozen emulsion mixed with air. When the ice cream melted, the air escaped. When you re-froze it, it formed a solid mass.


I am seeing Doublet of Doublets peaks when I should see just a doublet The following is a tutorial question I need to do:

I initially thought the answer was isobutyl formate (CAS:542-55-2) but what confuses me is the dd peak at $\pu{3.9-4.0 ppm}$. To my knowledge, isobutyl formate should give just a doublet at that region. What am I missing out?

Your question is how can the CH2 be split into a dd when the -OCHO shows no triplett, right?

Well, the second splitting at the $\ce{CH2}$ is very small, and the $\ce{CHO}$ looks like it is very nearly split. That´s not just a broad peak, it has a rather clear shape imo. If you do just a bit of zero filling and apodisation, the triplett might well (should) appear.

But generally, you are right. Every splitting must have its counterpart.

Zero filling: Adding blanks to the end of your FID. Artificially increases the resolution, i.e. point density of your spectrum.

Apodisation: Changing the gain of your FID towards the end, by multiplying it with a decreasing or increasing function. The former gives you a less noisy spectrum, the latter a better resolution (sharper peaks actually), each at the cost of the other. The mathematical term for that is "convolution".


Best ways to compress and store tons of CO2? We are looking at doing planet changing work, and removing CO2 from the atmosphere. The issue is storing the CO2. Please let us know the best ways to compress millions of tons of CO2, as we have limited storage space. For example, how much could we compress a ton of CO2 gas? What size container would a ton of CO2 fit in, when fully compressed? Also, would turning tons of CO2 into liquid be more space effective? How much space would a ton of CO2 gas that has been turned into liquid take up?

Poor carbon dioxide has been unnecessarily defamed as a greenhouse effect. This is not only gas in the atmosphere which is an infrared absorber. Guess what, water vapor is a major culprit as well along with methane (recall cows digestive system). Nature has a very delicate balance, carbon dioxide is plant's food as well.

I recall some people doing PhDs on carbon dioxide remediation in chemical engineering. One of options is to pump carbon dioxide into deep rocks with a hope that it will convert into carbonates, mainly calcium carbonate in deep layers. One can then study all types of phase diagram of carbon dioxide etc.

One green solution is a massive drive for the plantation of trees. This option is far better than more expensive options.


Can you theoretically hold more of a volume of water the colder it is? I have basic knowledge of chemistry and how it all works (will be taking the class next year) but I understand density and mass. Since the hotter a liquid, in this case, the molecules are more everywhere and not contained so to say. In a way, a $\ce{200 cm^3}$ cup will hold less water if say brought up to a temp of 90°C (194°F) and will hold more water if brought to a temp of 2°C(35.6°F) due to the molecules being more contained and not bouncing everywhere. Does this make sense or did I underestimate my understanding of this whole thing?

*I am talking about mass not volume.

Water reaches its maximum density at about $\pu{4 ^\circ C}$.  As its cooled from higher temperatures the density decreases as you would expect as the molecules move more slowly and they can thus "settle" in closer together.

But, as it approaches the freezing point, the molecules are moving slowly enough that they start combining together into a more crystalline, ice-like conformation.  It's still completely liquid, but is starting to organize into the ice structure, and ice is about $\pu{7\%}$ less dense than liquid water.

So in the case of water, you can "fit" the most water into a given volume at about  $\pu{4 ^\circ C}$.


Complementary base pairing In the molecular structure of a DNA it is seen that purines (adenine and guanine) bind with pyrimidines (cytosine, uracil and thymine) through hydrogen bonds, and this is always between a purine and a pyrimidine. Can't there ever be hydrogen bonds between 2 purines or between 2 pyrimidines?

Can't there ever be hydrogen bonds between 2 purines or between 2 pyrimidines?

Yes, it is possible to have hydrogen-bonded base pairs (and triplets and quadruplexes) beyond the canonical Watson-Crick base pairs A:T and G:C. You can explore these in 3D on the DSSR-Jmol site for example. If you look at the 1ehz structure (a tRNA from yeast), which loads automatically on DSSR-Jmol, and click on non-canonical base pairs, you will see some examples. Below is an A:A pair with two symmetric hydrogen bonds from that structure:


Vaporized sodium hydroxide? Some time ago, I was making a $\ce{NaOH}$ solution. The solution I had made was much too concentrated, and the heat released in the dissolution of the ions caused the water to begin steaming. Inadvertently, I inhaled some of this steam, and instantly my nose began to sting, but only for an instant. I can only assume I had inhaled some $\ce{NaOH}$, but how is this possible? How can ions be present in this steam, yet the process of distillation produces steam without ions?

When a liquid is evaporating or even boiling, minute drops of the liquid are formed by the mechanical disintegration of the surface of the liquid (bubble bursting, splashing or foaming). Some drops are carried away from the heated liquid by the steam (by convective and vapour flow).

The concentration of non-volatile dissolved matter in the entrained drops approximately corresponds to the concentration in the bulk liquid. Therefore, if the bulk liquid contains $\ce{NaOH}$, the entrained drops also contain $\ce{NaOH}$.

A quantitative parameter that describes this transport is the airborne release fraction (ARF). The ARF increases with temperature of the liquid and air velocity above the liquid. Measured values for liquids that are evaporating but not boiling range from $< 10^{-8}$ to $3\times10^{-5}$ (Mishima, J., L.C. Schwendiman and C.A. Radasch. November 1968. Plutonium Release Studies IV. Fractional Release From Heating Plutonium Nitrate Solutions in Flowing Air Stream, BNWL-931, Pacific Northwest Laboratory).

In principle, distillation and similar technical processes have the same problem. That is one reason why distillation shows only a limited decontamination factor (DF) even for non-volatile dissolved matter.


How to identify a redox reaction? Which balanced equation represents a redox reaction?
  \begin{align}
&\mathrm{A.} &\ce{AgNO3 + NaCl &-> AgCl + NaNO3} \\
&\mathrm{B.} &\ce{BaCl2 + K2CO3 &-> BaCO3 + 2KCl} \\
&\mathrm{C.} &\ce{CuO + CO &-> Cu + CO2} \\
&\mathrm{D.} &\ce{HCl + KOH &-> KCl + H2O}
\end{align}

I don't understand how to identify a redox reaction. The answer key shows the correct answer as D, but I'm confused as to why.

D shows an acid-base reaction. Are all acid-base reactions redox reactions and why aren't the other answer choices correct?

You are confused because the answer key is wrong.
The way to figure it out is to assign oxidation numbers to each species following the rules. Oxidation states are simply a convenient way to keep track of electrons without having to draw out Lewis structures, count electrons pairs and so on. The rules help make that happen.

Notice the position of the sign relative to when you indicate charge. It is not the same.

The correct answer is C though: $$\ce{CuO + CO -> Cu + CO2}$$
with oxidation states
$$\ce{\overset{(+2)}{Cu}\overset{(-2)}{O}
    + \overset{(+2)}{C}\overset{(-2)}{O}
   -> \overset{(0)}{Cu}
    + \overset{(+4)}{C}\overset{(2\times(-2))}{O2}}$$
Notice Cu went from +2 to 0. It gained 2 electrons to do that, thus it was reduced. What gave up the electrons? The other carbon, because it went from +2 to +4: its oxidation state increased by 2, thus it lost 2 electrons.

Recall OIL RIG: Oxidation is loss of electrons and reduction is gain of electrons.
Whatever is oxidized increases its oxidation state, and whatever is reduced decreases it.


Correlation Diagram Doesn't Match Selection Rules? From Miessler's Inorganic Chemistry, 5th edition, there is this correlation diagram for $d^{2}$ in a octahedral field.

On the left hand side, if it's not clear, it goes (from lowest to highest):

$^{3}F, ^{1}D, ^{3}P, ^{1}G, ^{1}S$.

But by Hund's rules mentioned even in the same chapter, one would expect the $^{3}P$ to show up right after the $^{3}F$ because of their higher spin multiplicities. And even after that, G should come before D. (the ordering for that being s,p,d,f,g in reverse for same spin multiplicities). I don't know of any reason for why the $^1D$ term is just above the $^3F$ term in this graph though. Is this an error or does anyone have an explanation?

No, it's not an error. Take a look at the Tanabe Sugano diagram for $d^2$ (from Wikipedia):

You can see that again, the $^1D$ state is below the $^3P$ state. Why? If we were considering a $p^2$ configuration, the $^3P$ is indeed lower in energy than $^1D$. And again anyone might predict that $^3P$ is lower for $d^2$ too, for the exact reasons you give.

We know this result from spectroscopy. The difference is because of the Racah parameters for the different terms (i.e., the electrostatic repulsion between electrons in the same orbital. The Wikipedia article had a nice link to how you'd compute the difference in this case.

Now of course a $^3F \rightarrow ^1D$ transition is not an allowed transition, which is shown in the textbook on the next page.

The moral of the story is that Hunds' rules and the basics will always get you the lowest energy term symbols for multi-electron atoms, but the details can be tricky for higher-level states.


What is the product when Lithium solid is dropped in water? I am working on reaction predictions for the AP Chemistry and I need some help with this problem. On a practice quiz I tried to use the rule of "Metal + Water = Metal Oxide" so I answered with $\ce{2Li + O^{2-} -> Li2OH}$. I got zero points for this answer meaning both reactants and products were wrong. What is the correct answer and how would you get to the correct answer?

I think the major error you made is applying what you call the "Metal + Water = Metal Oxide" rule.  This may apply to some metals, but not all.  Specifically, Lithium is a Group 1 metal, the so-called Alkali Metals.  This means that Group 1 metals tend not to form oxides with water, but rather form hydroxides with water.  That is, the metal ions bond with hydroxide ion, $\ce{OH-}$, which has a -1 charge.  If you have ever seen this reaction (lots of YouTube videos if you have not done the reaction yourself), you will see that lithium in water produces lots of gas.  And in some videos, you will see that this gas can be set on fire!  Given the elements involved, that's a good indication that the gas is hydrogen, a very flammable gas.

So the most likely equation for lithium in water forms an hydroxide of lithium along with hydrogen gas.  Is that enough information for you to propose a different equation than what you answered?


What does solubility limit mean in practical terms? I am not really a chemist guy so please bear with me. Glycerol monolaurate has an upper limit of solubility in water of 20 μg/mL, which is enough for my experiment. Does this mean that no matter how much glycerol monolaurate I add to water the concentration will always stay at 20 μg/mL? Glycerol monolaurate is in powder form so does this mean that the powder would flow on top of water and the concentration of 20 μg/mL glycerol monolaurate would be in the water below?

Thanks.

Q1) If 20 micrograms / ml is the upper solubility limit and you added less than 20 micrograms the solution could be less concentrated but a concentration higher than 20 micrograms / ml H2O can't be achieved.

Q2) No more than 20 micrograms of glycerol monolaurate will dissolve in 1 ml of water so if more than 20 micrograms was added the remaining undissolved powder would stay separated in some way (eg. precipitate) which could be filtered (as @MaxW has stated), or a colloidal suspension could also form. If more than 20 micrograms / ml were to be used the supernatant (GML/H2O layer) could be estimated to be 20 micrograms/ml) after filtration.


Ammonia and hydroxide ion - Lewis basicity Which is the stronger Lewis base, and why?

I'm thinking that the answer is "it depends." With regard to the hydrogen proton, which hydroxide ion is obviously the stronger Lewis base. However, with regard to other, more complex Lewis bases, the distinction may not be so clear, as hydroxide ion has several lone pairs (three) and these can weaken the stability of Lewis adducts through lone pair/"bonding pair" (the pair that forms the coordinate covalent bond) repulsions. Ammonia only has one lone pair so there isn't an issue of lone pair/"bonding pair" repulsion.

With respect to Brønsted basicity, definitely hydroxide is the stronger base. With respect to Lewis basicity, I think the same scenario holds. As you said, hydroxide has three lone pairs, whereas ammonia has only one lone pair. As far as the energetics of these situations are concerned, the magnitude of lone pair to lone pair repulsion is far greater than that of lone pair to bond repulsion. Thus, it would make sense for hydroxide to grab the proton instead of ammonia.


Bond enthalpy of hydroxide Why does it require more energy to break the first O-H bond in water than is required to break the second O-H bond?

The Masterton Hurley (6th ed.) textbook has a bond dissociation energy of +499 kJ/mol for breaking water into a hydrogen atom and the hydroxyl radical OH. The bond dissociation energy for breaking the hydroxyl radical into another hydrogen atom and an oxygen atom is given as +428 kJ/mol.

You could make the argument that you are starting out with a situation where all electrons are paired and end up with two radicals in the first step, whereas in the second step you already start with unpaired electrons (the hydroxyl radical has seven valence electrons).

In the title of your question, you mention the hydroxide ion. This is not a relevant species because bond energies usually refer to homolytic bond cleavage, with one electron each remaining with the two fragments. It would require heterolytic bond cleavage to go from water to the hydroxide ion (water reacting as a Bronsted acid). This reaction merely requires +55 kJ/mol in aqueous solution.

Some sources contribute to the confusion between hydroxide ion and hydroxyl radical in this context. For example, Boundless states:

  For instance, the HO-H bond in a water molecule requires 493 kJ/mol to break and generate the hydroxide ion (OH–). Breaking the O-H bond in the hydroxide ion requires an additional 424 kJ/mol. Therefore, the bond energy of the covalent O-H bonds in water is reported to be the average of the two values, or 458.9 kJ/mol. These energy values (493 and 424 kJ/mol) required to break successive O-H bonds in the water molecule are called ‘ bond dissociation energies,’ and they are different from the bond energy. The bond energy is the average of the bond dissociation energies in a molecule.

They incorrectly refer to the hydroxide ion (OH-) and should be referring to the hydroxyl radical (OH).


Why did I get these inconsistent experimental results in silanization? I used a 50% wt mixture of Tetraethylorthosilicate + Kerosene for silanization of some glass plates with different treatment times (from 1 sec to 35 sec). However, the contact angle(CA) that I got does not seem consistent as it does not always increase as I increase the treatment time (when treatment time increases, the CA fluctuates).

What might be the reason? and whether I should change the method or the solution(s)?

P.S. I cleaned the glass with acetone in a sonicator. Then I put the glass into solution and after specific time put it into water to clean the solution from surface and then clean the remaining oil (from kerosene) by rubbing a napkin to the glass plate. Then I measured the CA.

Rubbing is a very inconsistent process, and will introduce a lot of variation. I would suggest using a volatile non-polar solvent such as acetone, hexanes, or ethyl acetate to remove residual kerosene and allowing to air dry.


How to calculate whether a redox reaction is spontanuous? I have some difficulties to apply the basic concepts of calculating whether a reaction is spontanuous or not.

given:

$$\ce{H_2SO_3(aq) + 2Mn(s) +4H^+(aq) -> S(s) + 2Mn^{2+}(aq) +3H_2 O(l)}$$

and those tables containing the stadnard reduction potentials:

The formula to apply is the following: $E_{cell} = E(reduction)_{cathode} - E(oxidation)_{anode} $

$E_{cell} = 0.20 - (+1.23)$
$E_{cell} = -1.03$

The answer is that the reaction occurs sopontaniously. However this is not what I calculated, according to my calculations it doesn't. What am I doing wrong?

You need to approch the question in a systematic manner.  The first step is to write the reduction and oxidation half reactions:

Reduction Half Reaction: (cathode)

$\ce{H2SO3 + 4 (H+) + 4 e- -> S(s) + 3 H2O}$

Oxidation Half Reaction: (anode)

$\ce{Mn -> (Mn^2+) + 2 e-}$

Now find the reduction potentials for both half reactions (only one was on the tables you provided, but a different set of tables might have both).  Remember they are reduction half potentials so to find the reduction potential of the oxidation half reaction you must reverse the reaction direction:

$\ce{(Mn^2+) + 2 e- -> Mn}$

The reduction happens at the cathode, and oxidation happens at the anode.

$$\ce{E_{cell} = E_{cat} - E_{an}}$$

$\ce{E_{cat} = +0.45 V}$

$\ce{E_{an} = -1.18 V}$

$\ce{E_{cell} = (+0.45) - (-1.18)}$

$\ce{E_{cell} = +1.63 V (spontaneous)}$

Just to clarify again, the cathode's reduction potential (+0.45) was NOT on the tables you provided.  I had to search the internet for a different set of tables which did have the cathode's half reaction reduction potential.