What is the meaning of n-Pr and i-Pr? Apologies in advance, I have a background in physics, so this question might seem awfully simple.

Anyway, I am interested in the chiral spin selectivity effect in organic molecules, and have read a paper in which they synthesize helicene molecules with side groups R = n-Pr, i-Pr with absolutely no reference only a slight hint as to what it is an abbreviation of in either the paper or supplementary information.  I thus assume that it must be some very common notation, and my guess is that Pr refers to some kind of propane/propene/propanol, and n and i refers to some normal and altered structure, respectively.  But I don't know.

I would be grateful if someone could shed some light on it for me.

Propyl is a simple hydrocarbon with three carbons in it. Propane is $\ce{CH3CH2CH3}$.

Propyl is the unit formed by attaching that chain to something else (which implies replacing one of the carbons with another bond). There are two ways to do this: one attaches the propyl unit using the end carbon on the chain (giving $\ce{X-CH2CH2CH3}$); the other involves attaching the middle carbon to something else (giving $\ce{CH3CHXCH3}$). The first is called (at least in older terminology) normal-propyl, n-propyl or $\ce{n-Pr}$; the second is called iso-propyl, i-propyl or $\ce{i-Pr}$. These reduce the amount of space required to write the structure.

In your compound helicene I'd guess that the propyls hanging off the rings are there to make the compounds far more soluble in organic solvents as flat aromatic structures tend to be fairly insoluble due to the way the pack in crystals.


is there any hydrophobic material to be used on paper cups? Is there any bio degradable hydrophobic material that is not poisonous and easy to spray up on paper cups?

The paper cups I have seen already have wax on the inside to prevent water from soaking the paper.


Which compound should I avoid in deodorants to avoid "sticky, unwashable" feeling? I tried some personal deodorants to avoid the smell of sweat, both in stick, no gas, gas sprays. Some can be washed away easily with body soap on the next day, some require either hair shampoo (because it is significantly harsher than body soap and washes much better) or to wait for two days for the effect to disappear by itself (whether the compounds break down or the compounds get transferred to the shirt is untested).

I don't like to have on me a product I cannot wash away, so I would like to know which compounds in the deo's I should check to avoid the sticky/unwashable effect.

The answer should list or indicate the compounds or rules in a way that people with little knowledge of chemistry can understand, like "the name comprises this string" or something like that. Generic classes which include compounds with a significantly different name is regarded as too complex.

Trying each product may work of course, but then quite some flacons would end up wasted or I would need to use products I don't like for a long time (one flacon lasts months, usually). Also, the composition may change.

Regrettably, most deodorants work as antiperspirants,  by clogging the pores of apocrine glands. These include salts of aluminum and zirconium, which hydrolyze (break up and combine with water). One example is aluminum zirconium tetrachlorohydrex glycosine complex, which works by forming a sticky, colloidal gel with water (and sweat). It's like the tale of the scorpion and the frog -- its their nature to act that way.

A bad alternative is deodorants with antibacterial compounds such as triclosan, which don't block sweating, but kill some odor-causing bacteria. Some of these antibacterials are endocrine disruptors, and all promote a population of antibacterial-resistant germs on the skin.

Traditionally, personal deodorants were just masking scents, and in 16th century Europe, flowers were worn to cover the stench... personal and environmental.


Do we need to check for all kinds of symmetry in a compound to declare it achiral? I know that a compound is achiral if it contains any one of the following kind of symmetry.
1.Plane of symmetry
2.Centre of symmetry
3.Normal axis of symmetry
4.Alternate axis of symmetry
It is difficult to write, for many compounds, its mirror image and then check using visualisation that if the image and compound are same or not, so we check for these 4 symmetries in molecule. My question is that for any compound do we need to check for all the symmetry elements one by one till we get a symmetry possessed by the molecule to declare it achiral, for example, does there exist any compound which has only alternate axis of symmetry and not any other kind of symmetry, so that any one will have to check all the symmetries one by one, till he reaches alternate axis of symmetry, to declare it achiral?
I know compound which have centre of symmetry, but not plane of symmetry, but  I don't know about a compound which has only alternate axis of symmetry and not any other kind of symmetry. Most of the achiral compounds that I have come across have plane of symmetry or centre of symmetry. So, do we need to check for all axis of symmetry or only centre and plane of symmetry, or any other shorter combination?

We need to check all 4 symmetries to determine whether a compound is optically active or not.
Consider the following compound. It has only Alternating Axis of Symmetry

Note that the rings are in perpendicular planes. The following diagrams illustrate the top view of the molecule.The stereochemistry of the methyl groups can be:

$S_4$ axis of symmetry is seen only in the following diastereomer where the methyl groups are anti to each other in each ring and the view looks like a Nazi cross from the top.


Why don't unstable odd electron species dimerize so that they become stable? Why doesn't unstable odd electron compound $\ce{NO}$ dimerize to $\ce{N2O2}$? Why doesn't this structure of ozonide dimerize? But why then does $\ce{BH3}$ dimerizes to $\ce{B2H6}$, and $\ce{AlCl3}$ to $\ce{Al2Cl6}$, both with something like bridged bonds?

Odd electron species do dimerize.

  $\ce{NO}$

Nitric oxide dimerizes, but only at low temperature (and probably high pressure}.

  $\ce{NO2}$

Nitrogen dioxide does dimerize. In fact, this is a well known property of $\ce{NO2}$. $\ce{NO2}$ (orange-brown) is in equilibrium with $\ce{N2O4}$ (colorless).

$$\ce{2NO2 <=> N2O4}$$
The position of this equilibrium is dependent on temperature and pressure (like all gas equilibria), and so it makes a nice demonstration of gas properties.

  $\ce{O3}$

Ozone does not dimerize. Dimerization would lead to $\ce{O6}$, which likely is a 6-membered ring of all oxygen atoms. While this structure violates no rules, the oxygen-oxygen single bond is not stable. This instability of the peroxide bond makes it both useful and dangerous.

  So why does $\ce{NO2}$ dimerize readily, $\ce{NO}$ dimerize under duress, and $\ce{O3}$ not dimerize at all?

The extra electron in the nitrogen oxides is found in an antibonding orbital. Conversion to the dimer gets that electron into a bonding orbital. There are no antibonding electrons in $\ce{O3}$.

  $\ce{BH3}$

Borane dimerizes because the boron atom is electron deficient - it has an empty p orbital. This species is so electron deficient that it will form bonding interactions with nearly any electron pairs, including those already in sigma bonds. (More properly the B-H-B bonds are three-center-two-electron bonds). Aluminum compounds dimerize the same way.

  So why does $\ce{BH3}$ dimerize and $\ce{O3}$ does not?

Ozone has no electron deficient atoms. Count the electrons. All atoms have an octet, and while one has a positive formal charge, the compound as a whole is not electron deficient.


Mechanism for bromoetherification of olefinic alcohol Propose a mechanism for the following reaction showing the relevant stereochemistry of intermediates and the final product.

I am confused as to what is going on here. I guess there's some sort of neighboring group participation after the formation of the bromonium intermediate. From the molecular formula of the product, I deduce the mechanism seems to proceed via an addition and subsequent elimination. But then again I'm not very sure.

As you already know, treatment of an alkene with aqueous bromine leads to halohydrin formation. This results when the intermediate bromonium ion is trapped by the solvent, water:

Generally, water is a lousy nucleophile, but this process can happen because there is simply so much water around. In our case, the alcohol in the same molecule acts as the nucleophile, instead of water which may or may not be present. Alcohols aren't great nucleophiles, but what this has going for it is that the process (called bromoetherification, because you form a bromo+ether) is intramolecular.

As you can see in the scheme above, there are two possibilities in the formation of the bromonium ion: either on the top face of the alkene, or on the bottom face. There is probably a slight bias because of the presence of the substituent on the ring, but it turns out that this probably does not matter.

When the alcohol attacks the bromonium, it is an $\mathrm{S_N2}$ process, which requires the oxygen to attack carbon from behind the C–Br bond. Now, if the bromonium ion forms on the top face, this can't happen, because it's tough for the oxygen to loop all the way to below the plane of the alkene. Only if the bromonium ion is formed on the bottom face can the oxygen attack. Thankfully, bromonium formation is pretty much reversible, so even if the bromonium ion is initially formed on the top face, it can revert to starting material.

Extra

This bromoetherification reaction doesn't seem to have been carried out on this particular substrate - perhaps it's too simple - but it is a thing. Here's a more impressive example (Org. Lett. 2004, 6, 2961):

It also has a somewhat more popular cousin, called bromolactonisation, where the nucleophile is a carboxylate instead of an alcohol. Here's an example, taken from E. J. Corey's prostaglandin synthesis. The halogen used is different ($\ce{KI3}$ acts as a source of $\ce{I2}$, so this is an iodolactonisation) and the nucleophile is different, but the mechanism is generally the same - except that $\ce{NaHCO3}$ deprotonates the carboxylic acid before it attacks the bromonium. Can you work it out?

See also: Facial selectivity of iodonium formation in an iodolactonisation for a related question on iodolactonisation.


The relationships between n , T , P , V for a gas? The symbols :

T : temperature //
V : volume //
P : pressure //
n : number of mole

what it is clearly known about the relationship between one another is :

P is inversely proportional to V ( from Boyle's law : PV=constant )

V is directly proportional to T ( from Charles's law : V/T=constant )

but the other ones I am not sure about them, So i might say :

P is inversely proportional to T

T is inversely proportional to n

p is directly proportional to n

What do you think?

All statements like "P is inversely proportional to V" should be qualified with "other variables being equal". With this qualification you can read off all the proportionalities from $pV=NRT$.  Any two variables on the same side are inversely proportional; any two on opposite sides are directly proportional.  So $p$ is directly proportional to $T$.  Your other assertions are correct.


Proof of Dynamic Nature of Equilibrium Here is how my textbook proves that dissolution of solid in liquid is dynamic in nature. It considers the example of a saturated solution of sugar. It says that,

  Though the solution is in equilibrium, the process of dissolution of sugar and crystallization of sugar takes place continuously and also at the same rate. This can be confirmed with the help of radioactive sugar. If we drop some radioactive sugar into saturated solution of non-radioactive sugar, then after some time radioactivity is observed both in the solution and in the solid sugar. Initially there were no radioactive sugar molecules in the solution but due to dynamic nature of equilibrium, there is exchange between the radioactive and non-radioactive sugar molecules between the two phases.

It also proves the dynamic nature of chemical equilibrium in a similar way.

From this proof, my argument is that dynamic nature of equilibrium is proved only by adding another substance. This disturbs the equilibrium. It is possible that the saturated solution was in static equilibrium but addition of this substance has led to the dynamic nature.
I do not find it logical to prove that it is dynamic by the process of disturbing it.

Is this proof invalid or am I going wrong in my arguments?

You are technically correct in saying that radioactive sugar is another species in that it is possible to distinguish it from 'normal' sugar, although chemically identical. But think about it this away. Some of the added radioactive sugar solution (which could be micromolar or less) finds its way into the solid sugar at the bottom of the flask, but some remains in solution so that (dynamic) equilibrium is reached with respect to the radioactive sugar. You could do it the other way round, add solid radioactive sugar and it will appear in the solution.  What is so special about the radioactive sugar that it can do this and 'normal' sugar cannot? Well, there is nothing special so it is entirely reasonable to suppose that the normal sugar does this also. By inference, other molecules also do this as we know of no experimental reason why sugar is peculiar among all the other types of molecules known.

If you are still not convinced an experiment could be done with a coloured salt such as copper sulfate, so only one species. Make a saturated solution in which there will be solid in equilibrium with water, which will be blue due to the copper ions. Next, filter off the water and retain the solid and (gently) add more water, initially this will be clear but becomes increasingly blue as the sulfate dissolves. You will see equilibrium being established again as the sulfate dissolves, and provided there is some solid left equilibrium solid to solution will be achieved.

There is an experimental technique called temperature jump that is used specifically to study the time it takes to 'relax' back to equilibria and so obtain rate constants for the process. In this method, a short duration laser pulse is used to rapidly heat a solution (typically by $\ll 1$ C) and change its equilibrium a little. The return to equilibrium is followed by optical/spectroscopic methods so nothing has to be taken from the solution. Many reactions have been studies this way e.g. $\ce{H3O+ + OH^- <=> 2H2O}$.


While electrolyzing concentrated aqueous sodium chloride, why is it that chlorine is discharged but not sodium? Hydrogen and hydroxide both exceed sodium and chlorine in terms of reduction and oxidation potential respectively.
While electrolyzing a concentrated solution of aqueous NaCl, it is known that chlorine is discharged at the anode (contaminated with traces of oxygen). However, at the cathode, hydrogen is charged in place of sodium. Why is this so? I have read many explanations, but they never provide the reasoning.
I suspect these particular properties of factoring in preferential discharge:

Atomic mass (and consequently weight).
Magnitude of oxidation & reduction potentials.

Let's take a look at a table of standard electrode potentials and find the relevant half-reactions:

Reductions (there is also a half-reaction for water reduction that doesn't involve a proton at -0.83 V, but for the sake of simplicity, we can ignore it):

$$\begin{align*}
&\ce{Na+ + e- <=> Na} & &E°=-2.71\ \mathrm{V}\\
&\ce{2H+ + 2e- <=> H2} & &E°=0\ \mathrm{V}
\end{align*}$$

Oxidations (reduction reactions reversed):

$$\begin{align*}
&\ce{2H2O <=> O2 + 4H+ + 4e-} & &E°=-1.229\ \mathrm{V}\\
&\ce{2Cl- <=> Cl2 + 2e-} & &E°=-1.36\ \mathrm{V}
\end{align*}$$

Now, if you know the concentrations of the species involved, you can use the Nernst equation to figure out the actual potentials, but we can reason around it without knowing the exact numbers.

The big thing to notice is how close the potentials of each of the reductions and oxidations are to one another—the reduction reactions are separated by almost 3 V, but the oxidation reactions are very similar. This means that for reduction, it is far less energetically costly to produce hydrogen gas than to reduce sodium, and in fact, if any sodium were produced, it would instantly react with the water to produce hydrogen gas, so if there is any water around, that half-reaction will be favoured. In the real world, the exact potential hydrogen is produced depends a great deal on the electrode used (see overpotential), but it's more or less impossible to electrodeposit metals with much lower reduction potentials than zinc in aqueous solutions. To actually produce metallic sodium, electrolysis of molten NaCl without any water present is the most common method.

Looking at the oxidations, the story's quite different. It doesn't take much more energy to oxidize chloride than it does water (and if the chloride concentration is high, the oxidation potential will go up). This means that (subject to the effects of the electrodes on overpotential), some amount of both half-reactions will occur and because chlorine is a gas and therefore leaves the solution and the oxidations potentials are so close (leaving little energy to overcome kinetic barriers), the chlorine produced will not significantly oxidize the water.


Reactions on alpha tetralone to synthesize naphthalene Alpha-Tetralone is reacted by $1.\ce{Zn(Hg),HCl}\ \ 2.\ce{ Pd/C}$ What is the product and mechanism that is followed for the 2nd reaction?

A book by SN Sanyal says that Naphthalene will be produced . I know that in the 1st reaction, clemenson reduction is followed thereby reducing the Carbonyl group to $\ce{CH2}$ group .Then how the reaction proceeds is not clear.

I may be wrong of the 1st step also. Please correct me.

Moinak Banerjee: Your concern about the mechanism of the conversion of tetrahydronaphthalene (tetralin) 1 to naphthalene 5 has been addressed in a comment by Waylander. It is important to recognize that the process is a 4-electron oxidation of tetralin 1 and a four electron reduction of the hydrogen that is liberated as 2 equivalents of hydrogen gas. Ideally, the reaction is catalytic in palladium (Pdo). Mechanistically, the reaction likely proceeds by oxidative addition of Pdo into the labile benzylic C-H bond of tetralin 1 to afford Pd(II) species 2, which, in turn, isomerizes to π-complex 3. These steps are shown as reversible because complex 3 is the likely intermediate in the hydrogenation of 1,2-dihydronaphthalene 4 to tetralin 1. Collapse of complex 3 regenerates the catalyst Pdo and an equivalent of hydrogen gas. A repetition of these steps at either the benzylic or allylic C-H bonds of 1,2-dihydronaphthalene 4 leads to naphthalene 5 and a second equivalent of hydrogen gas. These reactions are driven to the right by the loss of hydrogen gas in an open vessel. --continued--

This reaction played a critical role in early attempts to synthesize estrone since it served to reduce the number of asymmetric carbons, thereby facilitating stereochemical analysis. An example of this process is illustrated in the sequence 6 --> 8 from the work of Bachmann and Chemerda1. The double bond in tetracycle 6 is isomerized by Pdo to the tetrasubstituted position in isomer 72. The conversion to the naphthalene 8 follows the same steps as in the case of 1,2-dihydronaphthalene 4.3

1) W. E. Bachmann and J. M. Chemerda, J. Am. Chem. Soc., 1948, 70, 1468.
2) The double bond migration involves a similar C-H insertion and subsequent π-allyl complex. But that is another story.
3) See also: J. Heer and K. Miescher, Helv. Chim. Acta, 1949, 32, 1572.


Understanding Mesomeric Effect I am getting confused between the type of mesomeric effect (+M or -M) shown by compounds containing $\ce{-COX}$ where $\ce{X}$ is an electronegative element.

Examples: $\ce{-CONHCH3}$, $\ce{-COOCH3}$, $\ce{-COOH}$.

According to me, all these compounds must show both a +M as well as a -M effect. Since the group (e.g. $\ce{-COX}$) can accept as well as donate $\pi$ electrons (as in $\ce{CH_2=CH-COOH}$).

However, the book says these compounds show only -M effect. Please explain me the reason behind it.

In the above example, isn't the group showing +M effect, since it's donating the electrons?

Following up on Permeakra's comment, the relative importance of resonance structures is typically determined by:

1) Maximize atoms that fulfill the octet rule.
2) Electronegative atoms have negative formal charges and avoid positive charges.
3) Minimize charge separation.

Carboxylic acid derivatives (esters, amides, etc.) have resonance structures that break #2, but since all atoms fulfill the octet rule, they are important contributors to the overall electronic picture of the compounds. Hopefully the image below is legible.

In general, resonance (mesomeric) structures that contain an electronegative atom that both breaks the octet rule AND bears a formal positive charge are not reasonable resonance structures and do not contribute to the overall electronic picture of the compound. The resonance structure of the carboxylic acid drawn in the original question falls into this category as the 'top' oxygen both breaks the octet rule and contains a positive charge.


Will strong concentrated acid destroy litmus paper and pH paper if they are dipped in acid? I am curious to know what will happen if I have litmus paper and pH paper, and I dip them into strong concentrated acid like conc. $\ce{H2SO4}$ to test its acidity or basicity. Will acid destroy papers? if YES, then how will I determine the acidity? do I need something else to carry my experiment?

Liquid pH testing solutions or drops are also available. Other indicators like phenolphthalein, methyl orange , methyl red can be used if you expect your test solution to fall in the pH range of the respective indicators. Also it will be wise to take a small sample of the solution to test it rather than testing the whole thing.


How to precipitate Chloroplatinic acid? For the precipitation reaction, I have to react $\ce{H2PtCl6}$ with $\ce{(NH4)2CO3}$ while maintaining the $\ce{pH}$ in the range of 7-8 at 60 °C. But the problem is Chloroplatinic acid is not precipitating. Instead, the acid evaporated by making holes in the aluminium foil which were not present before.
Can you please point out what I'm doing wrong?

Theoretical reaction:

$$\ce{(NH4)2CO3 (aq) + H2PtCl6 (aq) -> (NH4)2PtCl6 (s) + H2O (aq) + CO2 (g)}$$

PS: Aluminium foil was just used for proper heating

I suspect that you could be forming an aqueous solution of $\ce{(NH4)2PtCl6}$.

I would like to make an observation as a person who has done platinum group metal chemistry, the anionic chloro complexes of PGMs like $\ce{PdCl4^2-}$ and $\ce{PtCl4^2-}$ are very toxic. They can induce a nasty allergy to PGMs. I would suggest that you do not work at home with these substances.

Also in my experience at least with $\ce{PdCl4^2-}$, I have always found that if you want to bring it out of solution then cation methathesis with a potassium salt can bring the substance out of solution as a brown solid. I suspect for the hobby chemist that low sodium salt would be the best reagent for doing this transformation.

For the lab chemist I would suggest using tetramethyl ammonium chloride or even tetraphenyl phosphonium chloride


What material are heat bags made of? I just bought a heat bag from store and it has special property.

It is made up of transparent plastic bag and transparent unknown liquid inside when sold. There is a tiny circular metal piece in there. If you bend the metal piece, crystalization will occur and release quite a lot of heat. The crystalization will spread out and change all the liquid into crystals.

When the hardened plastic bag cool down, you need to put it into boiling water and the crystals will dissolve and become liquid again.

What is more interesting, I tried to heat it up but not letting all the crystal dissolve, take it out of the boiling water and assume it is ready for the next use. However, the half-dissolved liquid/crystal turns into all crystals again (without me bending the metal piece). I am guessing the transformation was not complete and the process was reverted.

Did the Japanese invent this material?

The bag contains a supersaturated solution of some salt with a highly negative enthalpy of crystallisation, which could be something like sodium acetate ($\ce{Na^{+}CH3CO2^{-}}$). The dissolved salt would really like to come out of solution but lacks a nucleation site. Bending the metal disk produces a transient site where crystallisation can begin, and from there are chain reaction results, causing all of the solute to crystallise and release the heat associated with this crystallisation.

Your guess regarding the half-dissolved crystals is correct - the crystals that have not dissolved act as seed crystals for further crystal growth. You need to dissolve all of the crystals to prevent the process from spontaneously reversing.

Despite the popularity of this kind of reaction for modern self-heating food/drink packaging in Japan, I suspect the discovery of this kind of reaction is lost to antiquity, as for instance sodium acetate can be made by simply mixing salt and vinegar. I recall reading that the highly exothermic reaction of calcium oxide and water was used in self-heating cans of food back in World War I.


Could metal hydride storage tanks for hydrogen store helium as well, given their repective atomic radii? With atomic radii very similar, could helium dissolve into metallic foam about as well as hydrogen, thus making metal hydrite hydrogen tanks suitable for helium as well?

Hydrogen does not just insert itself physically into the metal.  Although only a few metals form stoichiometric hydrides with predominantly ionic character, most of them (and all those involved in hydrogen/hydride storage) form some type of bond, generally involving saturation of the hydrogen 1s orbital with ionic or polar covalent bonding.  Helium simply does not have such capability for the same reasons it does not bond with much else, and it is generally not significantly soluble in metals.

An introduction to various methods to store hydrogen is given by Wikipedia.  The
US Department of Energy reviews some proposed materials here.


Dry paint becoming lubricant with heat I am undertaking a physics experiment and have come up with a chemistry problem. Could paint perhaps become a lubricant under mild temperature. (It would take a while to explain my experiment, but an object sliding down a ramp is increasing velocity with increasing mass, this should not happen if if coefficient of friction stays constant). I am wondering if anyone has information about paints lubrication vs temperature, or anything similar. The coefficient of friction is the same before and after the experiment, so I am wondering if paint changes its coefficient of friction or lubrication when temperature increases.

I'm sorry if this questions doesn't make sense, if anyone can point me towards anything paint and friction related, thankyou.

If you take "paint" in the very broad meaning of substances used to treat surfaces:

wood can be waxed, and waxes melt at not too elevated temperature and can then behave like oils.

I'd expect the behaviour of waxes here as very prone to surprises, since they can also become sticky when some compounds are liquid while others are solid.


Why is chlorine and not fluorine labelled as the most reactive halogen for halogenation? The reactivity of halogenation in sunlight is as follows:

$$\ce{F2}>\ce{Cl2}>\ce{Br2}>\ce{I2}$$

So why does we often label chlorine as most reactive halogen towards halogenation reaction?

The reactivity of halogens in sunlight

OK, this tells us that we are dealing with a free radical reaction.  Sunlight breaks the halogen bond ($\ce{X-X}$) to generate halogen free radicals.

$$\ce{X2 ->C[{h\nu}]\  2X.}$$

In the next step the halogen radical abstracts a hydrogen atom from any hydrocarbon present to generate $\ce{HX}$ and a hydrocarbon free radical (note that multiple hydrocarbon radicals may be generated depending on the structure of our hydrocarbon).

$$\ce{X. + R-H -> HX + R.}$$

Finally, the hydrocarbon radical can react with the halogen free radical to produce a halogenated product.

$$\ce{R. + X. -> R-X}$$

Let's now consider the energetics of these steps.  The first step, generation of the halogen free radical, uses the sunlight to break the $\ce{X-X}$ bond, so no additional energy is required for this step. Then we break a $\ce{C-H}$ bond (so we must add energy) and make an $\ce{H-X}$ bond and a $\ce{C-X}$ bond (so energy is given off).  The various bond strengths are contained in the following Table.

\begin{array}{|c|c|c|c|c|c|} \hline
\ce{C-H}~ \text{Bond Strength} & \ce{H-X} & \text{Bond Strength}& \ce{C-X} & \text{Bond Strength} & \text{Overall}  \\ \text{(kcal/mol)}\ & & \text{(kcal/mol)} & & \text{(kcal/mol)} & \text{(kcal/mol)} \\ \hline
\ 99 & \ce{H-F} & -135 & \ce{C-F} & -116 & -152\\ \hline
\ 99 & \ce{H-Cl} & -103 & \ce{C-Cl} & -81 & -85\\ \hline
\ 99 & \ce{H-Br} & -88 & \ce{C-Br} & -68 & -57\\ \hline
\ 99 & \ce{H-I} & -71 & \ce{C-I} & -51 & -23\\ \hline
\end{array}

(bond strengths taken from Bill Reusch's table)

We can clearly see that all of these free radical reactions will occur (e.g. they are exothermic) in the order you suggested, with fluorine being, by far, the most reactive.

$$\ce{F2 >> Cl2 > Br2 > I2}$$

Saying that chlorine is the most reactive is incorrect.

Note: 99 kcal/mol was used as a typical $\ce{C-H}$ bond strength.  We could refine our model by using more exact $\ce{C-H}$ bond strengths (see this Wikipedia table) and calculate the relative rates of halogenation at the various carbon atoms in a complex hydrocarbon.


Chemical potential and it's implications in an open system Take the fundamental equation for the Gibbs free energy of a system:

$$dG=VdP-SdT+\sum_i\mu_idn_i$$

Thus, the chemical potential can be defined in terms of the Gibbs free energy as follows:

$$\mu_i=(\frac{\partial G}{\partial n_i})_{P,T}$$

Now, assuming that the chemical potential is positive (i'm not sure if that is always true) if I add $dn$ of component $i$, $dG>0$ which means (given that the Gibbs free energy is the capacity of a system to do non PV work) I have enabled the system to do more work. Consequently, I have done work on the system by adding more substance to it because i think I have increased it's potential - like raising a mass to a greater height.

It makes sense to me that adding a substance to a system would increase it's capacity to do work however, it doesn't make sense to me that work must be done to add moles of a substance. It's more clear when I think about the converse...

If I remove a small amount of the substance I reduce it's capacity for work and the system must therefore give out energy. This is confusing to me because surely i'd have to do work to remove a portion of a substance from a system - breaking the intramolecular forces and so on. However, for an ideal gas, there aren't any intramolecular interactions so unless the chemical potential for an ideal gas is zero (which I assume it's not) I must be missing something?

Too long for a comment, so here goes.

It's quite subtle and I'm really not sure how to explain in words, but the gist is that the chemical potential does not reflect a real, physical process of adding a substance that can be carried out. If that were the case, then the chemical potential would be the difference in the Gibbs free energies of the two systems:

System 1: $n$ moles of substance and $\mathrm{d}n$ moles of substance, separated by some imaginary barrier
System 2: $n + \mathrm{d}n$ moles of substance

and measuring the chemical potential would simply mean removing the imaginary barrier, allowing intermolecular attractions to form, etc. But that's not the case.

The chemical potential is the difference between these two systems:

System 1: $n$ moles of substance
System 2: $n + \mathrm{d}n$ moles of substance

That's all there is to it. Where do the extra $\mathrm{d}n$ moles come from? It doesn't come from anywhere. There's no real physical process going from System 1 to 2 here, so don't think of it as one.

Since Gibbs free energy is an extensive property, the $\mathrm{d}n$ moles of substance carry their own Gibbs free energy, just like they carry their own internal energy. That is why the chemical potential is positive - yes, the formation of intermolecular forces, etc. will affect the exact value of the chemical potential, and in a sense that is why the chemical potential is a partial derivative - but remember just simply having more of the substance means that there is more Gibbs free energy.


What is the chemical reaction in a home-made high-bouncer ball? I took the kids to the Science Museum in Canberra (Questacon) and one of the toys we brought back was a Home-Made High Bouncing Ball.

Now I did it with the kids and it was amazing. You pour the powder in the mould, hold it underwater for a minute. Dry it in the mould for two minutes and open the mould and let it dry.

The end result was like the high-bouncer balls we used to play with as kids. (Lots of fun).

My question is: What is the chemical reaction in a home-made high-bouncer ball? (I'd also love a picture I can show my kids - but that is not essential).

I have google it and got the link which explains the process but don't show the organic reactions(might be because they are complex)

  This activity demonstrates an interesting chemical reaction, primarily between the borax and the glue. The borax acts as a “cross-linker” to the polymer molecules in the glue – basically it creates chains of molecules that stay together when you pick them up. The cornstarch helps to bind the molecules together so that they hold their shape better. - See more at: Sciencebob


The logic for hybridisation formula Hybridisation for a molecule is given by:
$$\frac{1}{2}(V + H - C + A)$$
Where,

V = Number of valance electrons in central atom
H = Number of surrounding monovalent atoms
C = Cationic charge
A = Anionic charge

If this expression is 2 then its $sp$ hybridised, 3 then $sp^2$ hybridised and so on.

However, I am not able to understand why this works.

Since you understand why it works for monovalent surrounding atoms, I will assume   that we have a molecule $\ce{MA_{$m$} B_{$n$}}$ where $\ce{M}$ is the central atom, $\ce{A}$ represents the monovalent atoms which may or may not be identical, and $\ce{B}$ represents not necessarily identical atoms which are polyvalent. First up, there are some assumptions to be made:

All "surrounding atoms" are directly bonded to the central atom $\ce{M}$, and with no other atom;
All atoms form as many covalent bonds as is indicated by valency;

Now say the $n\ \ce{B}$ atoms are absent. Let the steric number be $x$,
Then, $x=0.5\times (V+m-C+A)$
Now add the $\ce{B}$ atoms. Now the central atom forms $n$ more sigma bonds, but if the sigma bonds were not present, these would exist as lone pairs, and hence already counted in $V$. Take for example, $\ce{PCl_3}$ and $\ce{POCl_3}$. For $\ce{PCl_3}$ you would get $0.5\times (5+3)=4$ and for $\ce{POCl_3}$ you would get $0.5\times (5+3)=4$. The reason the $\ce{O}$ made no difference is because phosphorus uses two of its electrons to bond with both of which are nothing but valence electrons of phosphorus, only present as a lone pair in one case and a sigma bond in the other. Also, this formula does not work when you have triple bonds, say for $\ce{HCN}$.
As to how it works for mono valent atoms being the only surrounding atoms, it is simple.The number of sigma bonds is the number of mono valent atoms, and thus by adding that to the number of lone pairs we get the hybridization. Say you have $ n $ lone pairs and $s$ sigma bonds.You can see that $V=2n+s$, and that $s=H$. So $0.5*(H+V)=n+s$, as required.


Why is ethanol written as C2H5OH not C2H6O? This is balanced equation of fermentation of glucose:

$$\ce{C6H12O6(aq) -> 2C2H5OH(aq) + 2CO2(g)}$$

I know that the result of glucose fermentation is ethanol ($\ce{C2H6O}$). What represented in the aforementioned equation is slightly different. I would like to know what $\ce{C2H5OH}$ is?

What is the OH functional group doing here? Does it make ethanol acidic? Does it mean that ethanol is diluted with water?

It is better to write $\ce{C_2H_6O}$ as $\ce{C_2H_5OH}$, because there is more than one compound with the same overall formula, and emphasising the hydroxyl group $\ce{-OH}$, which is what characterizes an alcohol makes it ethanol and nothing else.  So when you ask, "What is OH functional group doing here", it is exactly that the product is an alcohol, characterized by a hydroxyl group. And yes, alcohols are acidic in the right situations, given that their hydroxyl groups do produce acidic hydrogen ions, and yes, that is why it's so soluble in water.


D3d point group matrices I am trying to construct all the matrices for the point group D3d operations.   I find that the matrices of E, i, S6, inverse of S6, C3, and inverse of C3 are easy,  but I do not know how to construct the matrices of the three C2s and the three dihedral reflections.   When I construct these matrices,  they do not look like those of usual reflections or rotations.   I cannot find any resources about finding these more complicated matrices.   Any help would be appreciated.

One of the usual setup for the coordination system would be that $x$ axis coincident with a two-fold axis. If we use this setup, we can easily write out the matrix for rotation around this 2-fold axis:
$$\begin{bmatrix}
1&0&0\\
0&-1&0\\
0&0&-1
\end{bmatrix}.$$

Then you can combine this with the transform matrix of the 3-fold axis to get the transform matrices of the other two 2-fold axes. That's just matrix multiplication.

In this setup, $yz$ is one of the reflection plane and the corresponding matrix is
$$\begin{bmatrix}
-1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}.$$

Note that this is just the combination of the previous matrix with inversion. You can apply the 3-fold axis rotation to it (once again, matrix multiplication) to obtain the matrix with regard to other two reflection planes.


Why are double bonds at more substituted carbon more stable? That $\ce{C=C}$ bonds are more stable at more substituted carbon  atoms is a common argument used throughout Clayden et al., but is never actually explained (at least not in the first 21 chapters). They do explain however, why the HOMO is increased in energy for more substituted double bonds, through the argument of electron donation by means of hyperconjugation of neighboring groups, increasing the electrophilicity of the $\ce{C=C}$ bond.

I find this somewhat counterintuitive. Usually when something is more reactive, it is less stable (I understand that of course by kinetic pathways this does not always hold, e.g. with $\ce{I-}$ as nucleophile).

How can both observations be incorporated in a consistent explanation in this case?

Let's start by providing some useful background information that will help answer your question.

1,3-Butadiene is more stable than 2 separate ethylene molecules due to the extended conjugation possible in the diene.  We can illustrate this by drawing simple resonance structures for the diene that can't be drawn for ethylene.

Huckel calculations show the same thing, 1,3-butadiene is more stable than 2 individual ethylenes.  Here is the Huckel MO diagram for ethylene.  The two electrons in ethylene have a net stabilization of 2$\beta$ compared to an electron in an isolated p-orbital.  4 electrons in 2 ethylenes would have an overall stabilization energy of 4$\beta$

Here is the Huckel diagram for 1,3-butadiene.  The 4 electrons have a net stabilization of [(2 x 0.62) + (2 x 1.62)] = 4.48$\beta$.

Even though the HOMO in 1,3-butadiene is higher in energy then the HOMO for ethylene, overall 1,3-butadiene is more stable than 2 isolated ethylenes by 0.48$\beta$.  This is because the MO below the HOMO is stabilized much more (0.62 $\beta$ compared to ethylene).

Now let's return to your question.  Hyperconjugation is a reasonable way to explain why more highly substituted double bonds are more stable. Just like in 1,3-butadiene, when we delocalize electrons over a larger number of atoms, which is what hyperconjugation does, an overall stabilization of the molecule should result.  And, as the hyperconjugated resonance structure shows, electron density will shift from the alkyl group to the double bond.

Even though the HOMO may be raised in energy, (at least some of) the other occupied MO's below the HOMO will be lowered in energy (just like in 1,3-butadiene) and overall a net stabilization of the molecule will result.


Use of ethanol versus DMSO as a solvent My professor has had the reaction shown in DMSO and ethanol as solvents but made clear these are used for different circumstances. I did not understand what the difference in circumstances are. Does it have something to do with being polar and aprotic and what alkene is being reacted with the cyanide?

The alkene being reacted with hydrocyanic acid here goes by benzylideneacetone, or 4-phenylbut-3-en-2-one in the language of the IUPACians. But you shouldn't view this molecule as an alkene, you should view it as an "enone", or $\alpha,\beta$-unsaturated carbonyl.

Enones are notable for having "regioselective" reactions where nucleophiles add to either the beta-carbon (via 1,4-addition, vinylogy chemistry) or the carbonyl carbon (via 1,2-addition, standard carbonyl chemistry) regions of the molecule.

In your question, you see we have a 1,4-addition of $\ce{H+}$ and $\ce{CN-}$. (After cyanide attacks the 4-position, the proton adds to O, the "1-position", and the enol intermediate then tautomerizes to a ketone.)

The choice of solvent can have some effect on influencing whether 1,2- or 1,4- addition dominates. Aprotic solvents can encourage 1,4-addition (conjugate addition) to be favored over 1,2-addition (carbonyl addition) (Source).

Your ethanol is a polar, protic solvent because the hydroxyl is capable of hydrogen bonding or donating the proton (its Bronsted-Lowry acidity is about that of water, pKa 15.7 vs 16). DMSO is a polar, aprotic solvent because it does not contain hydrogens that can H-bond nor are (appreciably) acidic. Use of DMSO can encourage 1,4-addition (source).

Simply changing the protic-aprotic nature of the solvent may not always be enough to effectively toggle the dominant regiochemistry between 1,2- and 1,4- addition motifs. It may depend on the other reaction conditions (such as temperature), the structural qualities of the particular enone substrate at hand, and the nature of the nucleophile. In the Hard-Soft Acid-Base Model, "hard" Lewis bases tend to add directly to the carbonyl (1,2) whereas "soft" Lewis bases tend to perform conjugate addition (1,4). For nucleophiles (Lewis bases) not on either extreme end of the hard-soft spectrum, changing the solvent type can sometimes be enough to change the regioselectivity.

I recommend you ask your professor whether there is a rule of thumb he wants you to use to determine the solvent's effect (such as "polar aprotic always gives 1,4"), or whether you should memorize specific examples (because there are many exceptions).


Lead-Acid Battery: Standard Electrode Potential I'm making a lead-acid battery for an investigation. Since I only had sulfuric acid and lead electrodes, I decided to firstly run 2 lead electrodes through electrolysis in a sulfuric acid electrolyte. This would create the following half equations:
$$\ce{4H+ + 4e- -> 2H2}$$
$$\ce{Pb +2H2O->PbO2 +4H+ +4e-}$$
At this stage, I still need to perform some theoretical calculations of the required potential to run this electrolysis. Obviously, the standard reduction potential of the hydrogen ion is 0V. But I can't find the potential for the 2nd reaction. Could someone provide me with a source for it?

Your reactions are wrong.

$$\ce{Pb(s) + HSO^−_4(aq) -> PbSO4(s) + H+(aq) + 2e−}$$
$$\ce{PbO2(s) + HSO^−4(aq) + 3H+(aq) + 2e− -> PbSO_4(s) + 2H2O(l)}$$

Lead acid batteries are fairly reversible, hence their use as rechargable batteries. Anything that produces a gas, like in your reduction reaction, is unlikely to be reversible.


Explaining differences in rate of propagation based on monomer structure The values given below have been taken from my lecture hand out, and they weren't cited so I don't know where my instructor got them from

The $k_\mathrm{p}$ (rate constant for propagation in radical polymerisation ) step for the following monomers is given below:

$$\begin{array}{cc}
\hline
\text{Monomer} & k_\mathrm{p}~/~\mathrm{mol^{-1}~s^{-1}} \\ \hline
\text{Styrene} & 106 \\
\text{Vinyl ester} & 3000\mathrm{-}4000 \\
\text{Ethylene} &  16 \\ \hline
\end{array}$$

And the instructor explained the difference between the first two as follows:

  the benzyl radical is stabilised by conjugation, however the radical produced in a vinyl ester lacks stabilisation by conjugation and thus reacts more rapidly.

Seems like a fair explanation, however in the same table the $k_\mathrm{p}$ for ethylene is given as $16~\mathrm{mol^{-1} s^{-1}}$, the slowest propagation step of the three. Thus his explanation is not consistent with this observation I believe, because similar to the vinyl ester radical this one too is not stabilised by conjugation and should be similarly eager to react.

My instructor did not offer a satisfactory explanation and I was wondering if people could weigh in on this here.

The argument presented by your instructor is based on one factor: the stability of the radical terminus of the polymer. But for bimolecular polymerization reactions we need to consider 4 things:

Stability of the radical
Stability of the monomer
Stability of the transition state
The probability of the radical and the monomers being correctly oriented

The first three items affect the activation energy ($E_\mathrm{a}$), while the final item affects the frequency factor ($A$), or pre-exponential factor in the Arrhenius equation:

$$ k = A \exp{\left(\frac{-E_\mathrm{a}}{RT}\right)} $$

For styrene, vinyl acetate (a vinyl ester), and ethylene, I have found the following values for radical polymerization processes.[1,2] These have been calculated to provide an apples-to-apples comparison:

$$\begin{array}{cc}
\hline
\text{Monomer} & E_\mathrm{a}~/~\mathrm{kJ~mol^{-1}} & A~/~\mathrm{L~mol^{-1}~s^{-1}} & T~/~^\circ\mathrm{C} & k_\mathrm{p}~/~\mathrm{L ~mol^{-1}~s^{-1}} \\ \hline
\text{Styrene} & 32.5 & 4.3 \times 10^7 & 60 & 341  \\
\text{Vinyl acetate} & 20.4 & 1.5 \times 10^7 & 60 & 9460 \\
\text{Ethylene} & 27.7\mathrm{-}32.8 & 2.0 \times 10^3 \mathrm{-} 3.0 \times 10^4 & 60 & 0.1\mathrm{-}0.2 \\ \hline
\end{array}$$

Note:

  "the propagation rate coefficient for most monomers is strongly pressure-dependent with large negative activation volumes; thus an increase in pressure leads to an increased $k_\mathrm{p}$ value."[1]

This is particularly relevant for the ethylene example, which is done at $100\ \mathrm{bar}$. At $60\ ^\circ\mathrm{C}$, this should be about $0.15\ \mathrm{g~mL^{-1}}$,[3] which is a factor of 7 less dense than the styrene and vinyl acetate systems.

With those caveats noted, the argument of the vinyl ester having a lower activation energy ($E_\mathrm{a}$) due to the reactants being less stable relative to the transition state seems valid. However, the data for ethylene shows that the pre-exponential factor ($A$) is likely responsible for the lesser rate of propagation displayed by ethylene.

Sources:

Krzysztof Matyjaszewski & Thomas P. Davis, Handbook of Radical Polymerization, 2003, p 199.
Etienne Grau, Jean-Pierre Broyer, Christophe Boisson, Roger Spitz, and Vincent Monteil, Macromolecules 2009, 42, 7279–7281. DOI: 10.1021/ma901622u
Thomas, W. & Zander, M. Int J Thermophys (1980) 1: 383. DOI: 10.1007/BF00516565


Is the equilibrium constant in the expression based on pressure or concentration? $$\Delta G^\circ=-RT\ln K$$

I have always assumed it to be $K_p$ because $\Delta G^\circ$ represents standard free energy change and the standard state for a gas is 1 bar. So I thought we must express the activities of gases as ratios of partial pressures to $\pu{1 bar}$. But while solving sums, I found my answer to match only if I consider $K$ as $K_c$. Where am I wrong? Or am I missing something trivial?

Neither. It is exactly what is given there $K$, the standard equilibrium constant, and the equation given is its definition,
$$\Delta G^\circ=-RT\ln K.\tag{1}\label{def}$$

A general equilibrium constant for any reaction
$$\ce{\nu_A A + \nu_B B + \nu_C C + \cdots
   <=>\nu_{A'} A' + \nu_{B'} B' + \nu_{C'} C' + \cdots  }$$
is given by
$$K_x = \frac{ x_{\ce{A'}}^{\ce{\nu_{A'}}}\cdot x_{\ce{B'}}^{\ce{\nu_{B'}}}\cdot
               x_{\ce{C'}}^{\ce{\nu_{C'}}}\cdots}{
               x_{\ce{A}}^{\ce{\nu_A}}\cdot x_{\ce{B}}^{\ce{\nu_B}}\cdot
               x_{\ce{C}}^{\ce{\nu_C}}\cdots},$$
or more general, gathering all components in $\mathbb{B}=\{\ce{A, B, C, \dots, A', B', C', \dots }\}$
$$K_x = \prod_{B~\in~\mathbb{B}} x_B^{\nu_B}.\tag2$$

If you insert that into \eqref{def}, then $x$ becomes the chemical potential of the component. You can derive the link to the activity of the component (see here), and you can link the activity to concentrations, pressures, etc.

Given the right boundary conditions, e.g. ideal dilution, $K$ essentially becomes $K_c$, etc.

$$\mu(i) \propto a(i) \propto c(i) \implies K \propto K_c\tag3$$

Obviously, this is a very sloppy generalisation, and you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which.


Does Intercalation and Deintercalation of Lithium ions occurs in Half Cell? The important feature of Lithium ion batteries is the ability to  intercalate and deintercalate lithium ions in  anode and cathode . However when we use  half cell, using lithium metal foil as anode, intercalation and deintercalation process occurs in this case too. How the lithium ion moves and what process cause the current to flow in discharge?

When you're using a lithium metal anode there is no intercalation really, just reduction of the lithium ions to lithium metal:

$$\ce{Li^+ + e^- -> Li^0_{(s)}} $$

The cathode reaction is still the same.


Finding molar mass of uranium fluoride using Graham’s law of diffusion Problem

A sample of uranium fluoride is found to effuses at the rate of $\pu{17.7 mg/h}.$ Under comparable conditions, gaseous iodine effuses at the rate of $\pu{15.0 mg/h}.$ What is the molar mass of the uranium fluoride?

Answer

 $\pu{354 g mol^-1}$

My approach
Perhaps I’m still having issues with significant figures, but I am getting the wrong answer. I know to use Graham’s law of diffusion, which leaves me with:
$$\frac{\pu{17.7 mg h^-1}}{\pu{15.0 mg h^-1}} = \sqrt\frac{\pu{253.8 g mol^-1}}{x}$$
I get $x = \pu{181.70 g mol^-1}.$ I know that the molecular mass of uranium fluoride should be less than iodine, so the answer makes no sense to me.
What am I doing wrong?

Graham's law of diffusion deals with the molar effusion rate rather than mass effusion rate.
This can be confirmed by the description of Graham's Law on Wikipedia.
An excerpt from the aforementioned page,

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$
where:
$r_1$ is the rate of effusion for the first gas (volume or amount per unit time)
$r_2$ is the rate of effusion for the second gas
$M_1$ is the molar mass of gas 1
$M_2$ is the molar mass of gas 2

Now the mathematical part,
$$\begin{align}
r_{\ce{UF6}} &= \pu{17.7 mg/h} = \frac{\pu{17.7 mmol/h}}{M_{\ce{UF6}}} \\
r_{\ce{I2}} &= \pu{15 mg/h} = \frac{\pu{15 mmol/h}}{M_{\ce{I2}}}
\end{align}$$
By Graham's Law,
$$\begin{align}
\frac{r_{\ce{UF6}}}{r_{\ce{I2}}} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{M_{\ce{UF6}}} \cdot \frac{M_{\ce{I2}}}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \cdot \frac{M_{\ce{UF6}}}{M_{\ce{I2}}} \\
\frac{17.7}{15} &= \sqrt{\frac{M_{\ce{UF6}}}{M_{\ce{I2}}}}\\
M_{\ce{UF6}} &= \left(\frac{17.7}{15}\right)^2 M_{\ce{I2}}\\
  &= \pu{353.39 g mol^-1}
\end{align}$$


Why won't water freeze if you put ice in it, despite ice being frozen? Just now I got some room temperature water and added some ice to cool it, and it just came to me, like the scientist I wish I was ... why doesn't frozen (freezing point) ice added to higher temperature liquid compound force the water molecules to solidify?

Why won't water freeze if you put ice in it?

It will, even at room temperature. You just need a big enough, cold enough ice cube.

Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably because it has frozen. Repeat, but immediately turn the tray upside down. You'll get a wet floor. You can conclude that, given enough time, the water will be frozen by the ice. To the whys and hows:

From the second law of thermodynamics we know that:

  An isolated system, if not already in its state of thermodynamic equilibrium, spontaneously evolves towards it.

And

  Heat cannot spontaneously flow from a colder location to a hotter location.

So if you put ice and water together in a cup, heat will flow from the water to the ice until they reach thermal equilibrium, which will be a temperature between the water's initial temperature and the ice's initial temperature. If the temperature of the ice was low enough or if there was enough ice for the water to keep giving away heat, then the water will freeze. That's considering our cup is an isolated system.

Why don't you usually see that happening? Well, regular ice cubes are probably around −5 °C, and there usually is more water than ice in a cup, so before the water has a chance to reach 0 °C all the ice has already melted. Besides that, the ice is receiving heat from the room, unless you're in a place where's below 0 °C. In that case, the water WILL eventually freeze, but mostly because of the environment.


How do we explain mutations, in the deterministic world of chemical reactions? I'm not a chemist, neither a molecular biologist. But I'm fascinated with science.

The way I understood, DNA is a macro-molecule which is made of atoms. And in reproduction, either mitosis or meiosis then mating, or in RNA creation mutation can happen. And mutation means a letter molecule in the chain of DNA can become another letter.

From the other side I know that by combining a known set of reactants, we get the same products, ALL the times. In other words, chemical reactions are deterministic and no matter how much we try, H2 + O results in water, and not in solphuric acid for example.

At this point I'm stuck. And I can't find an answer. How mutation happens in DNA while it's the outcome of chemical reactions and those reactions are deterministic?

The topic is much larger than a scope for a single answer. But staying at general, low level chemical principles, it can be pinned down to several mechanisms leading to possible mutations:

Chemical reactions are not deterministic. Generally, on molecular level, there is less than 100% probability the desired and expected reaction happens, and more than 0%  probability it does not happen, or other reaction happens.
A molecule may react with other molecule in more than one way, with alternative reactions having their probability to form  "by-products".
A molecule may react with similar, but still native molecule, that it is not supposed to react with to form desired product.
A molecule may react with a molecule that should not be present, coming e.g. from pollution, medical treatment or poisoning.
Some foreign molecules may not react directly, but can affect DNA replication or synthesis of RNA or proteins.
A molecule may be broken by ionizating irradiation like UV, X-ray or  $\gamma$ photons, high energy subatomic particles or ions from the natural body radioactivity ($\ce{^{14}C}$, $\ce{^{40}K}$), from natural radioactive background ($\ce{U}$, $\ce{Ra}$, $\ce{Rn}$), from pollution (radioctive elements from coal burning ), medical treatment, radiation incidents or even poisoning.


Why do we use helium in balloons? While I was looking at the periodic table today, I realised that there were gases that were much lighter than helium such as hydrogen. If hydrogen is lighter than helium, why do we insist on using helium in balloons?

As other answers have noted, the only gas lighter than helium is hydrogen, which has some flammability issues that make it more difficult to handle safely than helium.

Also, in practice, hydrogen is not significantly "lighter" than helium.  While the molecular mass (and thus, per the ideal gas law, the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the difference between the density of the gas inside the balloon and the air outside.

The density of air at STP is about $\rho_{\ce{air}}=\pu{1.2754 kg m-3}$ , while the densities of hydrogen and helium gas are $\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$ and $\rho_{\ce{He}}=\pu{0.1786 kg m-3}$  respectively.  The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences  $\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$  and $\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$.  Thus, helium is only about $7.5\%$ less buoyant in air than hydrogen.

Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant.  For example, if you wished to go ballooning on Jupiter, which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply sink, and even a pure hydrogen balloon (at ambient temperature) would not lift much weight.  Of course, you could always just fill the balloon with ambient Jovian air and heat it up to produce a hot hydrogen balloon (not to be confused with a Rozière balloon, which are used on Earth and have separate chambers for hot air and hydrogen / helium).

Ps. A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about $\pu{2 Da}$, while a helium atom has two protons and two neutrons, for a total mass of about $\pu{4 Da}$.

Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about  $\pu{32 Da}$ (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to  $\pu{28 Da}$ (one proton and one neutron per atom less than oxygen).  Thus, the average molecular mass of air should be between $28$ and  $\pu{32 Da}$; in fact, since air is about three quarters nitrogen, it's about  $\pu{29 Da}$, and so the buoyancies of hydrogen and helium in air are proportional to $29 - 2 = 27$ and $29 - 4 = 25$ respectively.  Thus, hydrogen should be about $\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$ more buoyant than helium, or, in other words, helium should be about $\frac{2}{27} \approx 7.5\%$ less buoyant than hydrogen.

Pps. To summarize some of the comments below, there are other possible lifting gases as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices.

For example, methane (molecular mass  $\approx \pu{16 Da}$) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas.  However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's more dangerous by others (total energy content per volume).  In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium.

A much less flammable choice would be water vapor which, with a molecular mass of  $\approx \pu{18 Da}$, is only slightly less buoyant than methane at the same temperature and pressure.  The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all.  This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster — whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses.

Despite these drawbacks, hot steam balloons have certainly been suggested, studied and tried in the past — alas, not always particularly successfully (although, apparently, there have been much more successful attempts as well).  There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope.  So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas.

Other potential lifting gases, with molecular mass similar to methane and water, include ammonia and neon.  Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium.

Ammonia, on the other hand, while much less flammable than methane, is rather toxic and corrosive (not to mention really stinky, which, given its other properties, is probably a good thing).  I don't think I'd like to fly in an ammonia balloon, but apparently, some people do!  It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form.

Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second.  If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical.  Then again, so would hydrogen — its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.


How do we get rid of dangerous explosive acetone peroxide and nitroglycerine? Imagine the police (or army or whoever's job is that) found out that some maniac placed a lot of primary explosives under a building. They have evacuated the building and deactivated the trigger.

What now? You have a barrel full of Acetone peroxide (terrorist's favourite explosive). You move it - it might blow.

Even if it doesn't - can you react acetone peroxide slowly, to avoid explosion?

What do factories do with this compound formed as a side waste?

Could you answer the same for nitroglycerine?

I don't know if it's practical to dissolve or otherwise stabilize acetone peroxide, but I could see absorbing nitroglycerine with sawdust or trapping it in gelatine, rendering it more stable, which is how dynamite is made. Certainly if it can be done safely, detonating in place is be the preferred solution for dealing with unstable explosives. It's done frequently with unexploded munitions if they're deemed too dangerous to move, e.g. WWII bomb in Munich. I also visited a potash mine once. Old dynamite tends to "sweat" nitroglycerine, but the explosives lockers are designed such that the contents can be detonated in place without causing damage, if this happens as it's safer than attempting to remove them.


How to calculate the concentration of conjugate acid from given pH and pKa values? Preface: Buffer solution (acid-base buffer). I am provided with a weak base, which I will designate B. $\mathrm{p}K_\mathrm{a}$ for $\ce{B}$'s conjugate acid, which I will designate $\ce{BH}$, is $8.1$, and its mole weight (sic) is $121.1$. I'm assuming the latter is the molar mass, though I don't know how that helps me solve this problem.

  Problem: $\pu{0.1 M}$ ($\pu{mol/L}$) of $\ce{B}$ was dissolved in $\pu{1.0 L}$ water, then titrated to $\mathrm{pH}$ $8.1$ and $7.5$. I am to assume no change in volume.

The last statement sounds like an odd assumption to me, how do you titrate something without [significantly] altering the volume? I always assumed titration is only done by adding a liquid to another liquid, is this not correct?

I am to find the concentration of $\ce{BH}$ at $\mathrm{pH}$ $8.1$ and $7.5$.

This question looked really easy to me at the first glance, but after a few hours I realized I had no idea on how to go about this.

I've tried calculating the equilibrium concentrations and going from there. This just ends up in $x \neq x$ in my calculations (where $x$ is the concentration of some substance), which obviously doesn't hold.

I've used the acid dissociation-, base dissociation- and Henderson–Hasselbalch equations, but nothing seems to do the trick. I might not be the brightest person to have grazed the surface of the planet, but I'm confident that I've got a firm grasp on elementary arithmetic.

The answer is provided, and should be $\pu{0.05 M}$ at $\mathrm{pH} = 8.1$.

I would highly appreciate somebody showing or explaining the thought process to me.

To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand.

Since we are working with dissolved $\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation:

$$\ce{B + H2O<=>BH +OH-}$$

So the equilibrium expression of the reaction is given by the equation:

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

If the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid is 8.1, then by mathematical manipulation:

$$\mathrm{p}K_\mathrm{a}=-\log K_\mathrm{a}$$

$$-\log K_\mathrm{a}=8.1$$

$$\log K_\mathrm{a}=-8.1$$

$$K_\mathrm{a}=10^{-8.1}\approx7.94\times10^{-9}$$

Since $\mathrm{p}K_\mathrm{w}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, the $\mathrm{p}K_\mathrm{b}$ of the base should be, and since $\mathrm{p}K_\mathrm{w}=1\times10^{-14}$:

$$1\times10^{-14}=7.94\times10^{-9}\cdot K_\mathrm{b}$$

$$K_\mathrm{b}=\frac{1\times10^{-14}}{7.94\times10^{-9}}\approx1.26\times10^{-6}$$

For the same reason that $1\times10^{-14}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, $\mathrm{14=pH+pOH}$. Given that the pH of the water is 8.1 for our first calculation:

$$14=8.1+\mathrm{p}\ce{OH}$$

$$\mathrm{pOH}=5.9$$
and
$$-\log{[\ce{OH-}]}=5.9$$

$$\ce{[OH^{-}]}=\mathrm{10^{-5.9}}\approx1.26\times10^{-6}$$

Since we know that any $\ce{BH}$ that is made must come from the reaction of 1 molecule of $\ce{B}$, then:

$$[\ce{B}]=0.1-x~~~~[\ce{BH}]=x$$

Using the knowledge we have gained from the problem, we can now solve (finally!):

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

$$1.26\times10^{-6}=\frac{[x]\cdot[1.26\times10^{-6}]}{[0.1-x]}$$

$$x=0.1-x$$

$$2x=0.1$$

$$x=0.05$$

So:

$$[\ce{B}]=0.1-0.05=0.05~\mathrm{M}$$

Try using this process to solve for the $\mathrm{pH}$ of 7.5.


Tip of my tongue: what is the name of this process? I remember back in Chemistry class, we did this thing where we would drop some liquid into a solvent. The moment it touched the solvent, it would turn into a gel sphere with the solvent inside it, and you could take these little spheres out and they'd essentially be capsules of solvent. What is the name of this process? The closest thing it reminds me of is Prince Rupert's Drop.

The general term for the process would be "encapsulation".

The Wikipedia article on micro-encapsulation lists different processes by which such encapsulation is achieved. Two which sound similar to your described process are interfacial polycondensation and interfacial cross-linking:

  In interfacial polycondensation, the two reactants in a polycondensation meet at an interface and react rapidly. ... Condensed polymer walls form instantaneously at the interface of the emulsion droplets.

Alternatively, the process your class did might have been ionotropic gelation:

  Ionotropic gelation occurs when units of uric acid in the chains of the polymer alginate crosslink with multivalent cations.


How does equalizing voltage for lead acid battery depend on the temperature? I am using advanced lead-carbon batteries (these ones) in an off-grid solar system. The spec specifies equalizing voltage 14.1 V for 12 V block @ 25 °C (77 °F). I keep the batteries in an outdoor shed, and the temperature where I live tends to hang around 0 °C (32 °F) these days.

So, for lead acid chemistry, would the equalizing voltage go up or down with temperature going down? I feel like it would go up (more voltage needed to do the same job at lower temperatures). How does this reasoning stand? What are the underlying scientific considerations?

Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge.
$$\ce{Pb <=> Pb^2+ + 2 e-}$$
At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$.
$$\ce{Pb^4+ + 2 e+ <=> Pb^2+}$$
For a classical lead–acid battery, the overall cell reaction is approximately
$$\ce{Pb + PbO2 + 2H+ + 2 HSO4- <=> 2 PbSO4 + 2 H2O}$$
As long as $\ce{Pb}$, $\ce{PbO2}$, and $\ce{PbSO4}$ are available at the electrodes, the equilibrium cell voltage depends only on the acid concentration (that's why equilibtrium cell voltage can be estimated based on measured acid density).
The dependence of the equilibrium voltage on concentration is given by the Nernst equation:
$$U_\text{cell}=\left(1.931+0.0592\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
This equation applies to one cell; a battery, however, has six cells in a row. Thus the total voltage is
$$U_\text{battery}=6\left(1.931+\frac{RT}{ZF}\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
where
$R$ is the gas constant,
$T$ is temperature,
$z$ is the number of electrons transferred in the cell reaction, and
$F$ is the Faraday constant.

We are only interested in the dependence of voltage $U$ on temperature $T$. All the other parameters may be taken as constants. Thus, our equation can be simplified to
$$U=\left(11.586+T\cdot k\right)\ \mathrm V$$
where $k$ is a constant.

The exact parameter values for a classical lead–acid battery might differ a bit from a modern battery with lead–carbon electrodes; however, we can calibrate our model using the data given by the manufacturer, i.e. $U=14.1\ \mathrm V$ at $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$
$$14.1\ \mathrm V=\left(11.586+298.15\ \mathrm K\cdot k\right)\ \mathrm V$$
and we find
$$k=0.008432\ \mathrm{K^{-1}}$$
Therefore, our complete equation is
$$U=\left(11.586+T\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V$$
For a new temperature of $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$, we get an estimate of
$$\begin{align}U&=\left(11.586+273.15\ \mathrm K\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V\\
&\approx13.9\ \mathrm V\end{align}$$


Which of these reactants will not generate this racemic amine? I understand that in E there is no electrophile the amine can attack, however, I do not understand why A is an answer as well. The amine in A can perform an intramolecular attack on the aldehyde and form the product below. Why can it not?

Also, how does the oxygen leave anyway?

This is an example of reductive amination, which generally proceeds through two main steps: (i) condensation reaction between the amine and carbonyl to generate an imine; and (ii) reduction of the imine to the corresponding amine.

Clearly (e) can't do this reaction since there is no carbonyl. The reason why (a) is also an answer comes down to the wording of the question - it specifically states that the product is racemic. The stereocentre in (a) is (R) in the starting material, and is not involved in the reaction so the product will exist as a single (R) enantiomer.


Why is the boiling point of heavy water more than that of normal water? In class we learnt that the London forces become stronger as the relative molecular mass increases. Not just as in organic chemistry but in things like the halogens.

However, as I understand, the London forces are just temporary dipoles forming indicating that this would have nothing to do with the molecular mass but instead the number of electrons because there will be more electrons to form the dipole. This would mean that the trend would be stronger London Forces with more electrons instead of increasing molecular mass.

How can that explain the slightly higher boiling points of heavy water (deuterium oxide ~101˚C) than normal water?

The hydrogen bonds in deuterium oxide are slightly stronger than those in water. This is due to a quantum mechanical effect; the bonding interaction has a lower zero point energy due to the greater mass of the deuterium atom. It, therefore, requires more energy to excite the bonding electrons from the ground level to the dissociation point, and a higher boiling point is observed.

You can find a good explanation of the physics behind these effects on this page.


What is the geometric configuration of the four fluorine atoms during the synthesis of xenon tetrafluoride? Neil Bartlett (1932–2008) first synthesized $\ce{XeF4}$ (and $\ce{XeF6}$) in 1962.  In the synthesis, a nickel chamber is used, and heated to 400°C, causing the formation of $\ce{NiF4}$, which Wikipedia claims is not part of the reaction.

Due to the nature of the noble gas, xenon, one might imagine that the fluorines would need to  could be "forced" onto the $\ce{Xe}$ atom at the same time, otherwise with each addition, the molecule would be even less energetically likely.  I was trying to ascertain in what geometric form the four fluorine atoms would configure themselves to make this addition possible.

I had first assumed that the nickel was being used as a catalyst, and the fluorines could be plucked from its surface accordingly, but this does not jive with the description of the synthesis above.  According to the response below, one synthesis of the molecule does rely on forming fluoride radicals on a catalyst, but it's mysterious that the synthesis above does not mention this.

Assuming there is no catalyst, how are the fluorines at 400°C oriented to allow the addition of 2 molecules at presumably the same time?

F - F
.   .
.   .
F - F

(and is that even possible?)

The answer appears to be right there in the linked synthesis details - the nickel is non-catalytic and is only there to generate a passivated surface that won't degrade in the high-temperature fluorine atmosphere.

[Edit]

The plot thickens! this paper suggests that xenon may in fact abstract fluorine off the passivated nickel fluoride surface due to observed zeroth order kinetics, which lends some serious credence to jonsca's original inference of Ni being catalytic. This will take some more investigation...


Does adding ice to sparkling mineral water extend the 'fizziness' of the solution before the drink goes flat? To my understanding the carbonated beverage loses its $\ce{CO2}$ content as gas bubbles which gives the drink its 'fizziness'. The loss of $\ce{CO2}$  occurs due to the low solubility of carbonic acid formed during carbonation resulting in an excess of unreacted $\ce{CO2(g)}$ which is trapped in the container when bottled and escapes when opened:

$$\ce{H2O(aq) + CO2(g) -> H2CO3(aq) + CO2(g)}$$

Temperature and volume of the solvent play a critical role in the solubility of matter.

Would the decrease in temperature and slight increase in solvent caused by the ice cube being added to a carbonated beverage exposed to air cause the solution to remain 'fizzy' for longer than the same solution at RT°C with no solvent increase?

Let me lightly gloss over the issue of nucleation sites and ice. While one observes bubbles forming on ice in the glass, they continue to form on the glass as well.

Instead, one could well ask the impact of temperature on release of the carbon dioxide, and thermodynamically that does come down to the solubility of carbon dioxide in water. Since this is of great importance to the soda industry, the study goes back some ways. R. Wieb and V.L. Gaddy, "The solubility of carbon dioxide in water at various temperatures from 12 to 40 degress and at pressures to 500 atmospheres - Critical phenomena', Journal of the American Chemical Society 62 815-817 (1940) shows measured solubility curves close to the temperatures of interest. One should note the increase in solubility as the temperature is dropped from room temperature down towards zero C.

While the carbon dioxide is supersaturated in the soda water, the driving force relates to the difference of the supersaturation to the solubility limit. As the temperature decreases, the thermodynamics favor retention of more carbon dioxide, which will then come out in your (warmer) mouth, giving you that 'fizz'.

Further, ignoring nucleation sites on the ice being much better than on the glass, as the temperature is decreased, the kinetics of nucleation will be at least partially suppressed as well.

Plus, I like my soda (and beer) cold, not warm!


Finding the concentration from a dilution This must be a very basic question but I have gotten confused over it.
How would I find the concentration of each component of the reaction for each experiment from this data? I am guessing it can’t be the concentrations given, is that for some stock solution?

Solving one out of the four would be good enough to understand the concept behind this. Therefore, I shall explain the first experiment and the concentrations of the components in the solution produced.
In experiment 1 all four are added in equal quantities. Finding the total volume, we get the total volume
$$V_\text{tot} = \pu{4 ml}.$$
Assuming the amount of $i$th solute to be denoted by $n_i,$ its molarity $c_i$ can be found as
$$c_i = \frac{n_i}{V_\mathrm{tot}}.$$
Now, we add $\pu{1 ml}$ of $\pu{2.0 M}$ acetone. This is equivalent to saying we add $1/1000$ of $\pu{2 mol}$ of acetone. Therefore we add $\pu{2 mmol}$ (millimoles) of acetone.
This means that we have $\pu{2 mmol}$ of acetone in $\pu{4 mL}$ of water. Using the formula for molarity we get that acetone is in $\pu{0.5 M}$ in solution.
Hope you can follow on for the other parts as well, the concept remains the same.


Need to find series of equation products from given reagents I can balance equations but only if I already know both the sides. I am trying to find the following products:

$$\begin{aligned}
\ce{Na2S2O5 + O2 ->}\ ?\\
\ce{Na2S2O5 + KNO3 ->}\ ?\\
\ce{Na2S2O5 + KNO3 + Fe2O3 ->}\ ?\\
\ce{Na2S2O5 + KNO3 + C12H22O11 ->}\ ?\\
\ce{KNO3 + C12H22O11 ->}\ ?\\
\ce{KNO3 + C12H22O11 + Fe2O3 ->}\ ?
\end{aligned}$$

I'm guessing that the sodium metabisulfate will produce sulfur dioxide and sodium oxide, but that would just be plain decomposition, so I'm not sure where the oxygen would fit. I know that $\ce{KNO3 + C12H22O11}$ produces carbon dioxide and water, but I'm not sure where the nitrogen and potassium go.

Of course once I know the products, balancing and calculating specific energy is simple, if a touch laborious.

I will assume that all of these reactions are happening in solution. Since we've established that all these reactions are redox reactions, we know the general structure of each reaction--- $$\text{reducing agent} + \text{oxidizing agent} \to \text{oxidized thing} + \text{reduced thing}$$ ---and we generally won't have to worry about other things (like what you've mentioned in your edit) happening. I will work out one reaction, and leave you with the rest. Let's look at reaction 4.

We can ignore $\ce{Na+}$ and $\ce{K+}$, because we know from prior experience that these are relatively inert spectator ions that won't participate in most reactions. Then we have the three species $\ce{S2O5^2-}$, $\ce{NO3^-}$, and $\ce{C12H22O11}$.

$\ce{N}$ has its maximum oxidation state of $+5$ in $\ce{NO3^-}$, so we expect that $\ce{NO3^-}$ will be a good oxidizing agent. $\ce{NO3^-}$ is commonly reduced to $\ce{NO2^-}$ in solution.
On the other hand,  $\ce{S}$ has an oxidation state of $+4$ in $\ce{S2O5^2-}$ and a maximum oxidation state of $+6$, so we might expect $\ce{S2O5^2-}$ to be a reducing agent. Fully oxidized $\ce{S}$ is usually present as $\ce{SO4^2-}$ in solution.
Finally, $\ce{C}$ also has its maximum oxidation state of $\ce{+4}$ in $\ce{C12H22O11}$, so $\ce{C12H22O11}$ is likely to also be a reducing agent. Fully oxidized $\ce{C}$ is commonly present as $\ce{CO2}$.

This gets us the three half-reactions
$$\begin{aligned} \ce{NO3^- + H2O + 2e- &-> NO2^- + 2OH-}\\ \ce{S2O5^2- + 6OH- &-> 2SO4^2- + 3H2O + 4e^-}\\ \ce{C12H22O11 + 48OH- &-> 12CO2 + 35H2O + 48e-} \end{aligned}$$ where we've also assumed basic conditions and hence added $\ce{H2O}$ and $\ce{OH-}$ as required to balance out charges and atoms on both sides of each equation. Now we simply need to take linear combinations of these half-reactions to cancel out electrons on both sides, and we're done.

In response to your comment, everyone starts out bad at chemistry. You'll get better.


Can I change glass transition temperature (Tg) and impact resistance with additives? I have three different HDPE polymers with different melting index but they don't have any elasticity. Physical form of polymers is granules. Can I change glass transition temperature (Tg) and impact resistance with additives?

The glass transition temperature depends on the cooling rate and molecular weight distribution and could be influenced by additives:

Most liquid plasticizers are low molecular weight organic materials with a glass transition temperature in the range of $125-225^0\text{C}$. When this kind of additives is added to an organic polymer, the result is a weight averaging of the glass transition temperature between that of the polymer and that of the plasticiser if they are miscible. The reduction of the the glass transition temperature of a polymer by the plasticiser is usually about   $5^0\text{C}$ per weight percent of plasticiser.

Impact modifiers ( a kind of additives) can affect the impact resistance of a polymer. Impact modifiers are key additives for
increasing flexibility and impact strength
to meet physical property requirements of
rigid polymers. All impact modifiers are elastomeric or
rubbery in nature, with a lower modulus
than the host polymer. The dispersed
rubber phase acts to absorb or dissipate the
energy of impact in order to stop craze or
crack propagation.


Boiling point comparison between cyclohexane derivatives I am trying to order these according to boiling point. I think that B is the highest because it has hydrogen bonding. I am not sure how to differentiate A and C though. They both have about the same molar mass. The only thing I can think is that C has a double bond. Does this double bond make the boiling point of C higher than A because it is stronger than a single bond? Or perhaps the double bond is shorter which decreases branching in C making the boiling point lower than A?

It's not the double bond in C that is significant here, it's the oxygen. The carbon-oxygen bond is polar and will make C more "electrically sticky" with its neighbors compared to A. Since the $\ce{O-H}$ bond in B is even more polar than the ketone in C, its boiling point will be even higher. Read more about it on this page.

(If you'd like to skip to the relevant section, on Google anyway, search "ketone vs alcohol boiling point".)


can i use an isobutane stove (like jetboil) or butane blowtorch with borosilicate glass? I don't have a bunsen burner, but have a camping stove (similar to https://www.amazon.com/Stansport-183-B-Isobutane-Backpack-Stove/dp/B004EPWM5S) and a blow torch.  I'd like to do some distillation with things like salt water and juice, and am looking at a distillation set (https://www.amazon.com/Distillation-Apparatus-Equipment-Chemistry-Glassware/dp/B07R8JFK97/). Is it safe to use this with a blowtorch or camping stove?

While this would have been considered normal in the past (I used such a burner when I was a kid), today bunsen burners are considered outdated as heat sources.
Of course it will still work fine fine, but our view on risk has changed.
Normally one wants to avoid having any source of ignition in the lab, and burners are also difficult to control.

That being said, a burner like this one  will make for a very nice burner with a focused flame. This type of burner works really well when you need an open flame, like for melting salts onto magnesia rods, sterilizing inoculation spatulas, and showing how metals change flame colors.

You can even buy a portable bunsen burner that is pretty much the same:

One big downside is that the burner will transform into a flame thrower if is falls over.
Also, you will use a lot of gas. If you like working with an open flame, I would recommend to buy a small propane tank with a pressure regulator and a regular bunsen burner, a setup like this:

Just get the smalles propane tank feasible, like a 2 or 5kg tank, just to keep the total amount of flammable material as low as possible.
Any store that sells gas supplies will be able to set you up for little money, just buy a bunsen burner for propane.

Doing a distillation of salt water will work, but it will be difficult to control and you can easily shatter the flask by uneven heating. Maybe that is a good lerning experience!

Distilling anything flammable, such as alcohol, on an open flame would be reckless.

Some general advice:

Never leave a burner unattended, not for the shortest moment
Remove anything flammable nearby. Work outdoors if possible.
In any distillation setup, always be able to remove the source of heat very quickly. Don't use the burner as a support, so that you can take out the burner without your setup falling apart. That is especially important for a burner with an attached gas tank.
Make sure the gas tank does not get hot.
Always be prepared for any flask to break at any time. What would happen if the boiling liquid would be released. Would you get injured?
Wear safety glasses
Wear non flammable clothing that is closed at the neck. A traditional lab coat is not a good protection for your chest. Buy something like this
Learn about superheating and use boiling stones


Why doesn't phosphorus form as many oxides as nitrogen? Nitrogen has oxides like $\ce{N2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{N2O3}$, $\ce{N2O5}$, but as far as I know, phosphorus has probably only $\ce{P2O3}$ and $\ce{P2O5}$.
Why does nitrogen have so many oxides in the first place?
Also why doesn't phosphorus have the analogous oxides?

It is generally due to the larger size of phosphorus. Nitrogen and oxygen both are members of 2nd period.Hence the p$\pi$ - p$\pi$ bond formed between nitrogen and oxygen has larger area of overlap resulting in shorter bond length and increased stability. Phosphorus is a member of 3rd period so when 3p orbital of phosphorus overlaps with 2p orbital of oxygen, due to difference in energy and size the area of overlap decreases which leads to lesser stability.Hence it forms less oxides.


Determining percentage of copper in brass Source: an IB Chemistry past paper.

I looked it up and found out that brass is an alloy of copper and zinc - however, I don't see zinc in any of the equations, and therefore any brass being separated into copper and zinc, which I had originally thought would be the method to determine the percentage of copper in brass (by looking at the coefficients of the compounds in the resulting chemical equation).

My question is how do these reactions help determine the percentage of copper in brass?

Well the question would had more to it. The question would have probably gone something like this. A sample of brass was dissolved with excess nitric acid. The $\ce{Cu^{2+}}$ ions are then reduced with excess iodide ions. Iodine is then titrated with thiosulfate (giving its concentration and volume used). Find the percentage of copper in brass.

To do this question, calculate the amount of moles of thiosulfate used, using its concentration and volume. Then using equation 3, you can see the number of moles of iodine is half the number moles of thiosulfate. Then using equation 2, you can see that the number of moles of $\ce{Cu^{2+}}$ ions are twice as the number of moles of iodine and hence are equal to the number of moles of iodine. Finally, using equation 1, the number of moles of copper is equal to the number of moles of $\ce{Cu^{2+}}$ ions and hence are equal to the number of moles of thiosulfate.

Now that we know the number of moles of copper, you can multiply by its molar mass to get its mass and then find the percentage of copper in brass.


Are we using an assumption in calculating the ratio of first state excited to unexcited chemical species using the Boltmann distribution? In calculating the ratio of first state excited to unexcited chemical species using the Boltzmann distribution, are we assuming that we only have species in the unexcited and first state excited states? It seems to me that this must be the case, because there's no part of the equation that accounts for the population of e.g. the second excited state.

The expression for the Boltzmann equation is:

$\eta_{i} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}$,

wherein $\eta_{i}$ the probability is to encounter the ensemble in state $i$ is in the ensemble, $g_{i}$ the degeneracy of state $i$, and $\epsilon_{i}$ its energy (I reckon the other variables are familiar to you).

To calculate the ratio between two states, you can simply divide the above expression for two different states by which you obtain:

$\dfrac{\eta_{i}}{\eta_{j}} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)} \cdot \dfrac{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)}$

The expression above shows that because states $i$ and $j$ are of the same ensemble, you do not have to use (or even know) the energy levels of the other states in that ensemble if you are only interested in the probability ratio between the two states.


At what position in the distillation apparatus does the temperature match the boiling point of the distilled substance? A student is separating $\ce{CHCl3}$ (bp = $\pu{61 ^{\circ}C}$) from $\ce{CHCl2CHCl2}$ (bp = $\pu{146 ^{\circ}C}$) by distillation. She has just begun to collect the first distillate in the receiving flask. At what position in the apparatus will the temperature be $\pu{61 ^{\circ}C}$?

The answer to this questions is B.
I narrowed down the options to either A or B, because I know that the temperature at A would be slightly higher than $\pu{61 ^{\circ}C}$, but lower than $\pu{146 ^{\circ}C}$, and the temperature at B would be slightly cooler than the temperature at A. C and D would be much lower in comparison.
What makes the answer to this questions B rather than A? I think that the two temperatures should be somewhat close to each other, but is there some intricacy to this that I am missing?

The temperature in area A can well exceed B all right, especially if the distillation adapter is long or if there is a fractioning column or another condenser before the three-way adapter.
However, no target component is going to make it through the condenser unless its vapors reach the side arm (the one with the joint to condenser).

Apparently, it's only possible if the temperature of that area coincides with the boiling point of the target, that's why the thermocouple or thermometer probe is installed exactly at that area (B).

As you rightfully suggested the temperatures in areas C and D are arbitrary: it only should be above the m.p. to prevent clogging and less than b.p. so that the target is not evaporating away.

P. S. there is a vacuum adapter shown on the picture (area C), so there also might be distillation at lower pressure, which would also lower the b.p. of the mix constituents.
Since the question doesn't mention its usage, atmospheric pressure is assumed.


can I mix. trisodium phosphate, potassium hydroxide, and sodium hydroxide together? Can I mix trisodium phosphate (tsp), potassium hydroxide, and sodium hydroxide ( or sodium hypochlorite ) and not cause a bad reaction?  And will this make for a good degreaser and cleaning chemical for me ?  I own a pressure washing business.  I do fast food resturanta, gas stations

I advise against using potassium or sodium hydroxide in homemade cleaning agents. You could easily end up with a mixture that could severely burn eyes (up to blindness), skin or lungs.

Also, grease from fast food restaurants is fundamentally different from gas station grease. The former consists most probably of fat while the latter is a mineral oil product.

Your home-made degreaser, with a hydroxide concentration that could be handled without complete personal protective equipment, might or might not have some effect on fat due to saponification but for removing mineral oil based grease I would not expect it to be effective.


In sourdough bread baking what do the wild yeast and bacteria not eat during dough fermentation and why? I bake bread using a sourdough starter. The starter is a culture of lactobacillus and some kind of yeast. They apparently eat the sugars or carbohydrates in the dough during fermentation but when I look at the nutritional info for fully baked sourdough bread I see 56gm of carbohydrates per 100gm and my question is, why do these organisms stop eating and what is it exactly that they leave behind? They seem to consume something in the dough then they stop as if they have run out of food at which point you need to bake the bread else it becomes "over-proved". Why do they stop eating and what is actually left over when the fully fermented dough goes in the oven?

Flours contain a diverse array of carbohydrates which vary depending on the specific type of flour (wheat, rye, teff, etc). There will usually be a lot of starch (itself having variable amounts of different types of branching), some small mono and oligosaccharides (again of varying types) and "fiber", which is a broad term for polysaccharides that are more resistant to digestion.
To add to the variables, the specific types of bacteria and yeast in a sourdough fermentation can vary in their ability to digest the different types of carbohydrates.
So one part of the answer to your question is that yes, there are carbohydrates that cannot be digested, but the specific molecules that are not digested will depend on the flour and the microbes. The cause of the undigestibility is either 1) the monosaccharides are linked in a way that they cannot be cut apart or 2) the microbes lack the metabolic capability to digest the monosaccharide.
But generally in a bread dough, the digestible carbohydrates have not been completely consumed, so some of the remaining carbohydrate is digestible. The reason it is not used up is that the baker stops the fermentation before it reaches that point so that the dough still has a good consistency.
It is also true that the fermentation activity slows down considerably before this point, even though it does not stop. There are two reasons for this. 1) the pH drops, which slows the metabolism of the bacteria and 2) the easily digestible carbs get used up first, so the later fermentation rate can be affected by the slow rate of breakdown of the more difficult to use carbohydrates. Ultimately, the dynamics of the fermentation are determined by the choice of flour, the microbial population and the conditions (temp, water content, whether and how often the baker works the dough, etc).


Difference between nickel catalyst and Raney nickel catalyst Many times I have seen the catalyst as Raney nickel instead of nickel catalyst. Raney nickel catalyst is developed by Murray Raney, and I think Raney nickel catalyst and nickel catalyst are two different things, because various sources use both names.

How do they differ from each other?

Yes, the Raney nickel catalyst is different from simple, fine nickel powder.  Raney nickel is prepared in two steps.  First, the nickel is alloyed with aluminum and then in second step much of the aluminum is removed through a leaching process with sodium hydroxide. As the aluminum is removed two things happen.  The resulting nickel-aluminum alloy becomes very textured and porous with extremely high surface area (sponge-like).  Also hydrogen gas is liberated during the leaching process and some of it remains adsorbed on the alloy surface.  This is the active catalyst that is ready for use.

$$\ce{2 Al + 2 NaOH + 6 H2O -> 2 Na[Al(OH)4] + 3 H2}$$

If you'd like to read more about the initial alloying process or the activation (leaching) step, see this Wikipedia article.

As an aside, Murray Raney discovered this catalyst (US1628190) in the late 1920's and he was the sole assignee.  I'm guessing that W. R. Grace bought the patent rights from him since they hold the trademark on use of the term "Raney nickel"