Titration of acid with base or base with acid: can I always use the same indicator? If you are doing a titration with $\ce{NaOH}$ and $\ce{HCl}$, does it matter which one goes in the burette, and will this change the outcome of whether the solution in the beaker will turn pink due to phenolphthalein?

When phenolphthalein is used as an indicator, NaOH is taken in a burette because then the colour change is from colourless to pink. And this time, it will be easier to observe.

On the other hand, if HCl is taken in a burette, colour change will be from pink to colourless which can't be observed in a single drop.

When NaOH is titrated with HCl using methyl orange as an indicator, HCl is taken in a burette because then the colour at the end point changes from yellow to red which is easier to observe rather than that from red to yellow.


Why chlorine acts as a strong field ligand in tetrachloridoplatinate(II)? In most of the cases I have learnted, halogens act as a weak field ligand. However, I recently learned somewhere on the web that the strength of a ligand also depends upon the size of the atom. A classic example is tetrachloridoplatinate(II): $\ce{[Pt(Cl)4]^2-}$.

Why is it so, and how can I predict the hybridization correctly in such exceptions, if I come across them for the first time?

I would like to add an answer stating a few exceptions which would come handy to you in future:

$\ce{Ni^{2+}}$ has a $3d^8$ configuration. In this case strong field ligands ($\ce{C}$ or $\ce{N}$ donors) will make the complex square planar. In all other cases, the complex will be tetrahedral.
In the case of $4d^8$ and $5d^8$ configuration and any ligand the hybridisation will be that of a square planar complex.
With $\ce{Co^{3+}}$, nearly all ligands behave as strong field ligands ($dsp^2$ hybridisation). Notable exceptions to this rule being $\ce{[CoF6]^{3-}}$ and $\ce{[Co(H2O)3F3]}$.
In case of bulky ligands (for instance $\ce{PPh3}$) present with $3d^8$ configuration, the ligands cannot get adjusted in the same plane leading to a tetrahedral shape ($sp^3$ configuration).


Plasma, Fourth State of Matter, and Magnetic Fields and Electric Fields Plasma is the fourth state of matter consisting of positive atoms, and freed negative electrons.  Does plasma have an electric field, and magnetic field?

There is not necessarily a preexisting magnetic or electric field in a plasma, since the charges can be randomly distributed. Think of plasma as similar to a metallic state conductor -- a sea of mobile electrons (and, of course, ions in plasma, so perhaps it better approximates a semiconductor, with electrons and holes). A plasma in thermal equilibrium, not exposed to an external electromagnetic field, should be uncharged.

That said, unbalances occur for a number of reasons. For example, electrons are thousands of times less massive and therefore more mobile than ions, so charge separation occurs, leading to a plasma potential.. Lasers can generate a charged plasma, accelerating particles with the wakefield effect.


What does 'preconcentration techniques' mean? In the scope of the analysis of organic compounds in column-switching liquid chromatography, what does the following statement mean?

  sample preconcentration techniques with high enrichment factor(EFs)

The word  preconcentration is not understandable for me.
Is it a technique for extracting the components?

Preconcentration means to increase the concentration of a sample prior to analysis or detection. For example you can do preconcentration for an non-volatile organic sample with evaporation its solvent.

An operation (process) as the result of which microcomponents are transferred from the sample of larger mass into the sample of smaller mass, so that the concentration of the microcomponents is increased. Examples include the decrease in solvent volume during distillation or evaporation, and the transfer of microcomponents from an aqueous solution into a smaller volume of organic solvent by extraction.(IUPAC)

There are lots of methods for preconcentration: solvent evaporation, using a different phase (ex: a solid or liquid that could absorbs sample from the solvent) and ...


Why is Standard heat of reaction equal to difference between standard heats of formation of products and reactants? The knowledge of standard heat of formation of various substances can be used to calculate the heat of reactions under standard conditions. The standard heat of any reaction is equal to the difference between the standard heat of formation of the products and the reactants

Why is that so? I'd prefer physical derivation for this:
$$\Delta H^\circ = \sum\Delta H^\circ_{products} -  \sum\Delta H^\circ_{reactants}$$

It is the consequence of the Hess's law and that is the consequence of the energy conservation law. Overall it says the enthalpy change does not depend on the path, but only on the start and final point.
Comparing to the real reaction path, you can formally get there also

via the path of conversion reactants to pure elements in their standard form
and then via forming products from those elements.

If the result were not the same, you could create perpetuum mobile switching both states via 2 different paths, gaining the nonzero net energy.


Should the liquid be included in Kc expression? Should the concentration of liquid water be included in the $K_\mathrm{c}$ expression for the reaction
$$\ce{2 NO2_{(g)} + 7 H2_{(g)} -> 2 NH3_{(g)} + 4 H2O_{(l)}}$$
if water is liquid and all the other compounds are gasses?

In the equation that you stated initially, water would be omitted from the equilibrium constant expression.

Your second question (about esterification that you posted in a comment to one of the answers) is extremely interesting.

I have given it some thought: in the standard case of esterification water should be included  because it is responsible for causing the reverse process ( hydrolysis of ester formed).

But in the case of zeolites, which by virtue of their structure, probably act as shape selective catalysts..ester hydrolysis by the water would be affected. (in some cases, I have read that selectivity towards ester formation is nearly 100%) In a case such as this, one can safely remove water from the equilibrium constant expression.

(I am not be a 100% correct, please excuse me, since I am just a student myself. Perhaps, someone else can provide a better/clearer explanation)


What is the name of the compound Sc(HCO2)3? I would like to know what the name of the compound $\ce{Sc(HCO2)3}$ is. I've been told that it is not scandium hydrogen carbonite and that it is not scandium formate. Any hints or suggestions about how I can solve this problem?

The formula you have is definitely scandium formate. There is one of four reasons why someone (like your instructor or your online homework service) might tell you that scandium formate is wrong.

Someone wants you to call this compound scandium(III) formate because scandium can have multiple oxidation states.
Someone wants you to call this compound scandium methanoate (or scandium(III) methanoate) since methanoate is another name for the formate ion.
You are correct and your instructor is wrong (or your assignment software has the wrong answer keyed in).
There is a typo in the formula you were given.


Factorizing Slater determinant into product of spin-up and spin-down slater determinant I am reading the book on MONTE CARLO METHODS IN AB INITIO QUANTUM CHEMISTRY by Hammond and Lester. (Chapter 5-Variational Trial Functions, Section 5.3-Hartree Fock and Beyond, Sub-Section 5.3.3 Hartree Fock and Correlation Energies)In section 5.3.3, the book says the following,

"In Monte Carlo, we may factor the determinant into electron spin components, i.e., $\psi_{D}=\psi_{\alpha}\psi_{\beta}$. This factorization is justified because all possible permutations of this term that are present in the full antisymmetric $\psi$  are operationally equivalent: each term simply corresponds to a relabeling of the electrons. Expectation values, and even local values of any observable, are unchanged."

I couldn't understand how these two are equal. Is it really possible to factorize this into two slater determinants? How does one justify that this is true?

When exchange effects are not taken into consideration, the simplest possible description of a many-body wave function is simply the Hartree product, i.e. the product of individual-particle wave functions. This is reasonable for bosons, but electrons are fermions, which means that the exchange of any two must flip the sign of the wave function.
Slater determinants are used to antisymmetrize the wave function so that fermion antisymmetry (Pauli principle) is obeyed. Since this is only relevant for electrons of the same spin, it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product. It is like up- and down-spin wave functions occupy different subspaces and interact via the electron–electron correlation term in the Hamiltonian.


What do the horizontal lines in a P-V phase diagram mean? I have been studying phase diagrams of pure substances, and for a substance such as carbon dioxide for example, it's diagram looks something like this:

The horizontal lines in the liquid phase puzzle me. It seems if we shrink the specific volume by a certain amount, the pressure change is much much less. This doesnt seem to be what is observed in reality (liquids have a very high bulk moduli and are generally considered incompressible), which would require a large pressure change for a less change in specific volume. The solid phase has vertical lines, which seems to be true, but the liquid portion baffles me.

Can someone please help me understand what I'm missing here?

the graph is trying to show three things (pressure, density and temperature)  in 2D.

The yellow portion shows the super critical fluid of $\ce{CO2}$.

The blue portion liquid $\ce{CO2}$

The light green portion shows the gas phase for $\ce{CO2}$.

The horizontal lines are called "tie lines." The two phases at the ends of the tie lines are in equilibrium.

If you follow the line for say for 12 $^\circ$C, from the degree notation there is at first one phases, a gas phase.

Now when you reach the dotted curve there are two phases a gas phase and a liquid phase which are joined by a horizontal tie line.

The whole thing is a lot easier to figure out with a more traditional phase diagram.

Found above on page:
https://www.engineeringtoolbox.com/CO2-carbon-dioxide-properties-d_2017.html


Why are acids characterized by giving hydrogen and bases characterized by receiving it when acid and base reactions are double displacement reactions? One example of this is $\ce{HCl}$ and $\ce{KOH}$. Why would the acid not just give the hydrogen ion in $\ce{HCl}$ to form $\ce{Cl-}$ and $\ce{KOH2+}$?

$\ce{KOH}$ does not exist as a molecule.  In water it is completely dissociated as $\ce{K+}$ and $\ce{OH-}$.  It does not make sense to write $\ce{KOH_2^+}$ for this reason.


Why do the d orbitals have these notations? Why do the d orbitals have the following notations: $xy, yz, xz, z^2$ and $ x^2-y^2$? What do they represent in their wave-functions?

The notation is shorthand for the hydrogen atom $\ell = 2$ wavefunctions in real form, in cartesian coordinates.  In other words, the Schrodinger equation is solved in spherical coordinates, in terms of spherical harmonics, which involve complex numbers; however, linear combinations of degenerate (equal energy) solutions are also solutions.  This permits the wavefunction solutions to be expressed using real numbers only. These real solutions can be transformed to cartesian coordinates.

$z^2$ represents that the wavefunction is proportional to $(3z^2-r^2)/r^2$.

$xy$ represents that the wavefunction is proportional to $xy/r^2$.

$yz$ represents that the wavefunction is proportional to $yz/r^2$.

$xz$ represents that the wavefunction is proportional to $xz/r^2$.

$x^2 - y^2$ represents that the wavefunction is proportional to $(x^2 - y^2)/r^2$.

See this Table of Real Spherical Harmonic Functions for the a list of functions in both spherical coordinates and cartesian coordinates, for s through f orbitals.


How to calculate a reaction's DeltaH using a MM forcefield calculator software? I want to calculate $\Delta H$ for hydrogenation of benzene and producing cyclohexane using MM+ Force Field Energy. I am using Hyperchem for energy calculation and optimization. I calculated the energy for cyclohexane and then I calculated the energy for benzene. But the difference of two energies is totally false. The calculated energy is $8~\mathrm{kcal/mol}$ however the real value is $-49~\mathrm{kcal/mol}$.
Is it a true way to calculate $\Delta H$? If not, how can I do it ?

What you are forgetting is that the reaction will also involve hydrogen.
$\ce{C6H6 + 3 H2 -> C6H12}$
In principle, the $\Delta H$ for $\ce{H2}$ should be 0, but in practice, that may not be true with a force field. The energies from molecular mechanics are seldom accurate enough to compute reaction energies. They are parameterized on a set of molecules, but the errors are usually large, both systematic and random errors.


Dispersal of pyrethroid insecticide by vaporization - role of petroleum distillate in distribution in air I live in Indonesia where there are many mosquitoes and products sold for their control.
One such product is this one:

The product contains 22ml of liquid, the contents of which likely vary slightly by jursidiction
https://www.whatsinsidescjohnson.com/ph/en/brands/Baygon/baygon__liquid_starter

Prallethrin - 1.316% (Wikipedia indicates 1.6% in typical products in India https://en.wikipedia.org/wiki/Prallethrin)
Petroleum Distillate (carrier) - Enables the delivery of a product to a surface.
Butylated hydroxytoluene (BHT) - preservative
Fragrances

They sell a liquid for use in a sprayer like this:

Unlike the plug-in which releases very small amounts of insecticide continuously, a sprayer would be sprayed manually perhaps once or twice a day. The product is sold in quantities of around 300-700ml.
Active ingredients vary by jurisdiction: in the Philippines
https://www.whatsinsidescjohnson.com/ph/en/brands/Baygon/baygon__multi_insect_spray_waterbased_refill

Water
Transfluthrin / Cyfluthrin
Isopropanol "wetting agent" - Helps a formula spread across a surface.
Benzyl Alcohol - anti microbial preservative
Sodium Lauryl Sulfate - detergent
Acetyl Tributyl Citrate - Helps ensure even distribution of a product's ingredients.

The Indonesian version is:
https://www.whatsinsidescjohnson.com/id/id/brands/baygon/baygon__liquid_spray_floral

Water
Cybermethrin, imiprothrin, prallethrin
Propylene Glycol Butyl Ether and PEG-60 Hydrogenated Castor Oil instead of the Acetyl Tributyl Citrate
Methylisothiazolinone & Benzisothiazolinone instead of the benzyl alcohol
Ethoxylated Alcohol in place of the SLS

Another company sells a similar liquid product to the Baygon liquid

This is sold in two formulations, the regular and the water-based which notes

'tanpa minyak tanah' (literally 'without ground oil', i.e. paraffin - this is not a very technical term and obviously means that the alternative product contains some kind of petroleum distillate)
Both regular and water-based formulations contain the same active ingredients: 1.14% prallethrin, 0.29% cyfluthrin.
I presume that there are particular properties of prallethrin that make it better for vaporization with an electric plug-in device (which I guess is not close to boiling - this should be evaporation) than certain other pyrethroids, but the question is:

if one were to refill the plug-in device with the 'spray'-type 1.14% prallethrin, 0.29% cyfluthrin petroleum-based product, would it be reasonable to expect it to have similar dispersal effectiveness to the 1.32% prallethrin petroleum-based 'vaporizer' product
given the existence of 'water-based' and 'petroleum-based' products with the same insecticides, is the 'water-based' less effective at dispersal, and does it have any advantage beyond presumably some sort of antipathy towards hydrocarbon vapours for health reasons.

In short:  the petroleum most likely is used as a solvent.
Small detour:  tablets of asperin contain acetylsalicylic acid as the intended active compound.  But the tablets contain other chemical compounds as well, for example to retain the active ingredient in «good shape» (e.g., stabilizers, antioxidants, etc.) and not at last, fillers.  Because, to moderate the wanted action of aspirin in the body, you possibly need a smaller mass of the pure compound, than your balance at home allows you to measure. Thus, the active ingredient is diluted; you equally could say, it is dissolved in the solid matrix of the tablet.
Now to the pyrethroids in question.  Similar to perfume, only a relatively small number of the molecules per cubic foot / cubic metre of air suffices to trigger the action; here, against the insects without triggering nausea / headache etc. to you.  (It is a question of balancing the effects, and the cost producing the materials vs. return of investment for the company, too.)  Thus, the pyrethroids are distributed as diluted.  Given the structures of two of the compounds you mention (they differ)

(prallethrin, source)

(cyfluthrin source
there are no, e.g. hydroxy-groups which could render the substances well soluble in water (hydrophilic).  Instead, they are more likely to dissolve in apolar solvents, such as petroleum is, then called «carrier».  In the other recipe, the PEG and propylene glycol butyl ether may function as the slowly evaporating solvent (mixture).
(Possibly, the list of ingredients was rearranged for this question; often, the containers list them in decreasing relative amount to the total content.)  The list of ingredients from the Philippines still may indicate lots of water in the bottle because simultaneously there are other compounds which may mediate between the polar water and the less polar pyrethroids to yield a solution; especially isopropanol (an organic solvent miscible with water), and the detergent, sodium lauryl sulfate may increase the solubility here.
It may be, that overall the recipe using both prallethrin and a little of cyfluthrin is more efficient or / and is less costly to produce than using the other recipe of prallethrin  alone.  It may be local legislation setting an upper threshold for one, or the other compound differently in one country, to the other; which may be to protect you directly as a consumer, or the environment in general (they don't decompose well in normal wastewater treatments and thus may enter the food chain).  Because the evaporation of an aqueous solution may well differ from one based on petroleum, it is not evident that one may be used to refill the other in the electric nebulizer.


How to check percentage yields in a given reaction? For a general reaction , if it is asked to predict and not given by theoretically, how would we determine the percentage yield of the products based on their mechanism , stability , path taken for the reaction etc. ?
For example , where my main problem lies is  :

According to Solomon and Frhyle it is said that chlorine is highly unselective. But in a problem I was asked percentage yields of the two products:

So how do I go about determining the percentage yields in this reaction ?

They are determined experimentally. There isn't any possible way you can determine percentage yield by looking at a reaction (unless there is only 1 possible product).

What the text is trying to explain to you is that when comparing analogous reactions of $\ce{Cl2}$ and $\ce{Br2}$, there will be higher percentage yields of the more highly substituted (and stable) product in the $\ce{Br2}$ reactions. For example, in the reaction between propane and $\ce{Cl2}$, the yields of 1-chloropropane and 2-chloropropane are 45% and 55% respectively. In a reaction between propane and $\ce{Br2}$, the yields of 1-bromopropane and 2-bromopropane are 3% and 97% respectively.

The exact reason for these selectivities has to do with the differences in energies in the transition states for the formation of the free radicals involved in the mechanism for these reactions. If you're interested in learning more you can see for yourself here.


Different statements of second law of thermodynamics This may sound trivial but I am having a hard time linking different statements of second law. What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy. Also from second law, entropy of any isolated system never decreases. How are the two statements similar? Please use simple layman terms, I am simply trying to grasp the idea.

Please use simple layman terms, I am simply trying to grasp the idea.

Thermodynamics is a rigorous field, which is one of the things that makes it so powerful. Explaining something in simple layman terms will make it less powerful and more mysterious. One way around this is to pick and choose some parts that are necessary to answer the question and ask you to suspend your disbelief while following the argument. In this case, we need to know that in a ideal heat engine, the change in entropy of the reservoirs is:
$$ \Delta S_\mathrm{reservoir} = \frac{Q_\mathrm{rev}}{T}$$
Also, the overall change in entropy will never be negative (and for an ideal heat engine, it will be zero).

What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy.

You can figure that out because if you extract heat from a reservoir, the entropy of it increases. Your process has to increase the entropy of something else by at least the same amount, so you can't just turn thermal energy into work.

Also from second law, entropy of any isolated system never decreases. How are the two statements similar?

If it never decreases, it is zero at best. You can apply this general statement to the specific case of an ideal heat engine. In this process, you have to heat up the cold reservoir to match the entropy decrease of taking heat out of the hot reservoir. Because we are dividing by $T$, and it is different for the two reservoirs, we can extract some of the energy in form of work without breaking the second law. The higher the temperature difference, the more work we can extract for a given amount of heat transfer.


4 compounds in equilibrium with one another - determine their equilibrium concentrations I know that if I have two compounds in equilibrium and I know $\mathrm{\Delta G}$, then I can use $$\mathrm{\Delta G=-RTlnK}$$
to determine the concentrations of the two components at equilibrium.  I have 4 compounds that are all in equilibrium with one another
\begin{aligned}
\mathrm{\Delta G(1,4)} ~&= ~~~3.41~ \mathrm{kcal/mol}\\
\mathrm{\Delta G(3,1)}~&=-2.21~ \mathrm{kcal/mol}\\
\mathrm{\Delta G(2,3)}~&=-0.46~ \mathrm{kcal/mol}\\
\end{aligned}

I applied the equation sequentially and found that
\begin{aligned}%
[1]~&=~0.34\%\\
[2]~&=~29.1\%\\
[3]~&=~13.5\%\\
[4]~&=~57\%\\
\end{aligned}

Are those the correct concentrations? Is there an "easy" way to do it?

In case anyone else travels this road...

Using $\mathrm{\Delta G~=~-RTlnK_{eq}}$ and the 3 $\mathrm{\Delta G's}$ supplied above, we find that
\begin{aligned}
\mathrm{K^{1/4}_{eq}}~&=~0.003\\
\mathrm{K^{3/1}_{eq}}~&=~40.749\\
\mathrm{K^{2/3}_{eq}}~&=~2.163\\
\end{aligned}
We also know that the concentrations of the 4 species must total to 1

$${[1] + [2] + [3] + [4] = 1}$$

so we have the following 4 equations and 4 unknowns
\begin{aligned}%
[1]~ &=~~~ 0.003 ~[4]\\
[3]~ &=~ 40.749 ~[1]\\
[2]~ &=~~~ 2.163 ~[3]\\
[1] + [2] + [3] + [4]~ &=~ 1\\
\end{aligned}
solving for the 4 concentrations yields
\begin{aligned}%
[1]~ &=~~~ 0.23\%\\
[2]~ &=~ 20.27\%\\
[3]~ &=~~~ 9.37\%\\
[4]~ &=~ 70.13\%\\
\end{aligned}


What is the reaction order of a reaction? Here's is what I understand so far and the following is referenced directly from UNSW CHEM1061 lecture notes by Professor Pall Thordarson:

order of the reaction SHOULD be equal to the number of molecules of that reactant involved in each collision ( however this is not always the case)
it is only possible to predict the order of a reaction is it occurs in a single step (elementary reactions)
reaction order is experimentally derived from rate equation

so I understand all of the above but what I fail to fully comprehend is that actual DEFINITION of 'reaction order' what is it in it's entirety, what does it's value represent if it only sometimes correlates with molecularity (no. of molecules which react in an elementary reaction) ? If someone can answer this question could they please provide a qualitative answer that explains the value conceptually? Or is it just a value to relate concentration to reaction rate? that helps us work out more information?

Very Shortly, Rate of reaction is directly proportional to product of the active masses of reactants raised to some power which is Order of the Reaction.

Consider an elementry chemical reaction
$$\ce{2A(g) + B(s) -> C }$$
As reaction is elementry, stoichiometric coefficient and order will be same. The order of the reaction will be 2 (as the other component is solid, therefore will not affect concentrations). The rate law for reaction will be given as
$${R = K[A]^2}$$
where K is reaction constant.
[A] is concentration of A.
If the order of the reaction would be 0, then Rate would be equal to reaction constant itself as the rate law would be given as
$${R = K[A]^0}$$

This explains the maximum use of Order of the reaction and also what order of reaction is.
Edit: for more wide explaination on this, see wikipedia. (Same explaination is provided but a widely expanded.)


Naming compounds with multiple -COX functional groups I know that this compound is named propanoyl chloride.

However I am not sure what to do when here are multiple -COX groups and they are of the highest seniority. There are two cases, when they are the same:

And when they are not:

For the first case I think it is called propanedioyl chloride, but I do not know what to do for the second. Is one of them cited as a prefix and if so which one?

Systematic names for acyl halides from suffix acids are formed according to Subsection P-65.5.1.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

  P-65.5.1.1 Acid halides in which hydroxy groups of all acid groups expressed as the suffix denoting the principal characteristic group (carboxylic, sulfonic, sulfinic, selenonic, etc. acids) have been replaced by halogen atoms (F, Cl, Br, and I) are named by citing the name of the acyl group (…) followed by the name(s) of the specific class(es) as a separate word(s), in alphabetical order, each preceded by a multiplicative prefix, as needed.

Therefore, the preferred IUPAC name (PIN) for the examples given the question is ‘propanoyl chloride’, ‘propanedioyl dichloride’, and ‘propanedioyl chloride fluoride’, respectively.

However, note that various carboxylic acids were often given trivial names (e.g. reminiscent of their animal or vegetable origin), which are retained for general nomenclature (including functionalization).


Why does potential chemical energy decrease with increasing oxidation of carbon? I saw an interesting spectrum in my biology class the other day. On the left end was methane - the least oxidized form of carbon - and the most energetic form of carbon (at least relative to what else was on the slide). On the right end was carbon dioxide - the most oxidized form of carbon - and the form with the lowest energy.

Now, I do agree with the slide - we can combust methane but we can't combust carbon dioxide - at least I don't think we can.

However, is there an intuitive explanation of why increasing oxidation state correlates with lower potential energy, at least for carbon?

Does increasing oxidation state correlate with lower potential energy for all elements?

Why does it seem that oxygen forms rather strong bonds with other elements in many cases?

is there an intuitive explanation of why increasing oxidation state
  correlates with lower potential energy, at least for carbon?

The heat of formation of methane, carbon dioxide and carbon tetrafluoride are -74.9, -393.1 and -927.2 kJ/mol respectively (the trend remains unchanged even if you divide by MW and compare per unit mass).  As we shift electron density from electropositive elements to electronegative elements we stabilize electrons, their energy is lowered.

  Does increasing oxidation state correlate with lower potential energy
  for all elements?

I haven't checked extensively, but the trend seems to hold up.  I'd expect it to hold up.

$\ce{BeH_2}$, $\ce{BeF_2}$  127.2, -796.7 kJ/mol

$\ce{CaO}$, $\ce{CaF_2~~}$      39.1, -789.1 kJ/mol

$\ce{NaH}$, $\ce{NaF~~}$      126.2, -288.0 kJ/mol

  Why does it seem that oxygen forms rather strong bonds with other
  elements in many cases?

I'm not sure what you mean by "strong bonds".  The C-O bond in carbon dioxide is much more reactive than the C-H bond in methane.  Perhaps you're alluding to molecular stability which takes us back to the top, e.g. compounds with polar bonds typically have stabilized electrons, electrons in lower energy orbitals, as discussed above.


Determine the molecular formula of the compound using percentage composition I’ve attempted to solve this problem with percent composition, but it says the answer is wrong. Is it a $pV=nRT$ problem, and if so, can I use the units as they are given?

  A $\pu{0.02867 g}$ sample of gas occupies $\pu{10.0 ml}$ at $\pu{290.0 K}$ and $\pu{1.10 atm}$. Upon further analysis, the compound is found to be $38.734 \%~ \ce{C}$ and $61.266\%~\ce{F}$. What is the molecular formula of the compound?

I keep on getting $\ce{CF}$ for the answer via percent composition.

You're actually on the right track. Looking at the percent composition, you've correctly identified that the ratio of $\ce{C}$ to $\ce{F}$ atoms is 1:1, however, you cannot assume that the formula is just $\ce{CF}$ (which isn't a known compound), it could be any compound with that ratio, $\ce{C2F2}$, $\ce{C3F3}$, $\ce{C4F4}$, etc.

The way to narrow it down is to use the gas data. You are right that you need to use the ideal gas law. If you look at the equation, you have everything except the number of moles, so you're interested in finding $n$:

$$n=\frac{PV}{RT}$$

As for the units, that depends on what units you have in $R$. If you have $R$ in $\mathrm{J/(mol\cdot K)}$, then $P$ needs to be in $\mathrm{Pa}$ and $V$ in $\mathrm{m^3}$, if it's in $\mathrm{L\cdot atm/(mol\cdot K)}$, then $P$ should be in $\mathrm{atm}$ and $V$ in $\mathrm{L}$. You can either convert what you have to match $R$ or convert $R$ to match your variables; it doesn't really matter.

Once you have the number of moles, you can calculate the molar mass of the compound from the sample mass:
$$m_\mathrm M=\frac{m_\mathrm s}{n}$$

Then, since we know the ratio of $\ce{C}$ to $\ce{F}$ is 1:1, the molar mass will be some multiple of the mass of $\ce{CF}$ which is $ 31.0091\ \mathrm{g/mol}$, so just divide the molar mass by that and you'll know $n$ in $\ce{C_nF_n}$.

Also, while not a sure way if fake compounds are possible answers, not all compounds with formula $\ce{C_nF_n}$ are known ($\ce{CF}$, for instance), so you can narrow down your choice of answer.


Plotting the data using Beer-Lambert law For the spectroscopy lab we had a stock dye ($\pu{1.00E-4 M}$ Erioglaucine Blue) that we had to dilute. We had to dilute $\pu{1 mL},$ $\pu{3 mL},$ $\pu{5 mL},$ $\pu{8 mL},$ and $\pu{10 mL}$ in $\pu{100 mL}$ of water and measure their absorbance $A.$ I obtained the following data:
$$
\begin{array}{ccc}
\hline
V/\pu{mL} & A & λ/\pu{nm} \\
\hline
1 & 0.127 & 629.0 \\
3 & 0.342 & 626.2 \\
5 & 0.550 & 627.6 \\
8 & 0.850 & 628.3 \\
10 & 1.064 & 626.9 \\
\hline
\end{array}$$
How do I get the concentration in $\pu{mol L^-1}$ in order to make the graph? If I could get one example from this data set I should be able to calculate the rest for myself. I just need a head start.

You need to be aware of the dilution formula in order to plot concentration on the x-axis and absorbance on the y-axis:
$C_iV_i = C_fV_f$
So your initial concentration is your stock concentration and your initial volume is listed in the table. Your final volume is $\pu{100 mL}$. Can you determine $C_f$? With that you would be able to calculate all 5 different concentrations.
Keep in mind, that you need to fix the wavelength of measurement. Since the spectrum is very broad, $\pu{1 nm}$ will not make a huge difference. You may not have perfect linearity.


Where does the characteristic peak for phtalic acid diethyl ester at m/z 149 originate? Which fragmentation is possible at this m/z value for phthalic acid diethyl ester?

I tried McLafferty but it didn't solve my problem.

First of all, you should indicate if you consider the fragmentation in electron ionization or in other ionization conditions. Considering it is the electron ionization conditions and that you refer to this spectrum, one can postulate for a reasonable mechanism, which can be derived from what is known on the fragmentation of benzoic acid and esters. I am not aware of any complete study of phtalate ester fragmentation, but I did not carry a complete litterature search.

In the figure below is a proposal for the mechanism leading to the formation of two possible structures for the $m/z$ 149 ion.

If one consider that is fragments as do benzoate esters, through radical induced $\alpha$-bond cleavage leading to the loss of $\ce{CH3CH2O^.}$ followed by standard decarbonylation of the acylium ion. For phtalate esters, the question at this point is whether the structure will remain as I or might undergo isomerisation to more stable structures.

I have provided a few alternative structures: simple addition of the neighbouring carbonyl would lead to structure II which although it has some constraints, allows a good stabilization of the positive charge both in the ester and by the aromatic ring. Other structures can be postulated, such as III after addition of the ethanoyl oxygen to the aromatic cycle. This four membered ring can reopen to lead to the acylium structure IV which can in turn undergo decarbonylation to form the ion $m/z$ 121, also present in the spectrum. So there is some indirect evidence that these structures might be involved in the fragmentation process.

Finally, in order to explain the formation of ions at $m/z$ 105 and 77, one can also propose a last isomer for ion $m/z$ 149, structure V, formed after 1,5 hydride transfer. This one can quite easily lead to loss of acetaldehyde, eventually followed by decarbonylation to lead to ion $m/z$ 77.

If phtalates present specific reactivity, the other option is that ion A at $m/z$ 177 can cyclize to structure B which is a very good candidate for ethene loss leading also to an ion at $m/z$ 149.

The question then becomes which of these two $m/z$ 149 is the actual ion. These two isobaric ions could be resolved by high resolution experiments, but after looking in publicly available databases, I did not find any high resolution spectra. Deuterium labelling of the ethyl groups could also prove handy, but once again I did not find any spectra of diethylphtalate labelled on the ethyl groups. (The usual diethylphtalate used in quantitation experiments seems to be the $d_{4}$ isotopomer, with deuterium on the aromatic ring.) Finally one can also have a look at isomers and related ions to see how they fragment. And indeed if one takes the dipropylphtalate ester, the ion at $m/z$ 149 is present (produced after propanoyl loss followed by propene loss) and there is no strong ion at $m/z 163$ which would tend to be a good indication of ethene loss for diethylphtalate. But, on the other hand, both the meta and para isomers of diethylphtalate have a much weaker ion at $m/z$ 149, which cannot be accounted so easily by an equivalent to intermediate B. So I would probably favor the concurrent occurence of both pathways, with a ratio that still needs to be measured.

In addition to the answer above, it should have been quite immediate that a McLafferty rearrangement was not an option: since there are no nitrogen, the nitrogen rule allows to conclude directly that the odd mass of the ion at m/z 149 indicates that it is an even electron cation, as opposed to radical cations produced by McLafferty rearrangements.


How exactly does the reaction between alkyl halides and silver nitrite occur? $$\ce{R-X + AgNO_2 -> RNO_2}$$
Now depending on the $\ce{R}$ and solvent the reaction can occur via $\mathrm{S_N1}$ or $\mathrm{S_N2}$.
When the reaction proceeds via $\mathrm{S_N1}$ the product formed is $\ce{R-O-N=O}$ whereas for $\mathrm{S_N2}$ it is $\ce{R-NO_2}$.
Why is this so? According to me as the $\ce{Ag-O}$ bond is covalent the lone pairs on nitrogen must attack in both the cases forming the product $\ce{R-NO_2}$.

The concept of hard and soft acids and bases (HSAB) proved to be
  useful for rationalizing stability constants of metal complexes. Its application
  to organic reactions, particularly ambident reactivity, has led
  to exotic blossoms. By attempting to rationalize all the observed
  regioselectivities by favorable soft–soft and hard–hard as well as
  unfavorable hard–soft interactions, older treatments of ambident
  reactivity, which correctly differentiated between thermodynamic and
  kinetic control as well as between different coordination states of ionic
  substrates, have been replaced. By ignoring conflicting experimental
  results and even referring to untraceable experimental data, the HSAB
  treatment of ambident reactivity has gained undeserved popularity. In
  this Review we demonstrate that the HSAB as well as the related
  Klopman–Salem model do not even correctly predict the behavior of
  the prototypes of ambident nucleophiles and, therefore, are rather
  misleading instead of useful guides. An alternative treatment of
  ambident reactivity based on Marcus theory will be presented.

  — H. Mayr’s, M. Breugst’s and A. Ofial’s introductory remark in their review pulished in Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.

Within the review, in the section on the nitrite ion, the authors give two competing effects that control product distribution.

For electrophiles with an electrophilicity parameter $E < -3$ (the bis-(4-methoxyphenyl)carbenium ion has $E = 0$, lower values are more stabilised carbocations such as bis-(4-dimethylaminophenyl)carbenium) the reactions proceed at equilibrium under thermodynamic control and the product yields are equivalent to their relative stabilities. Since nitro compounds are more stable than nitrite esters, they predominate.

$$\ce{R-ONO <=> R+ + NO2- <=>> R-NO2}$$

For electrophiles with a parameter $E > 0$, the reactions are essentially diffusion controlled, i.e. the free electrophile will capture the first nucleophile it can find. This includes tertiary alkyl carbocations ($E \approx 7{-}8$). The reactions do not go through classical transition states, and any means of describing them via frontier orbitals must fail. That explicitly includes describing them according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. In these reactions, kinetic control determines that the $\ce{O}$-attack is slightly more likely with $k_\ce{O} / k_\ce{N} \approx 3.4$. (This corresponds to computational models that show a slightly smaller activation barrier for the oxygen attack.)

Reality is more complex than these simple models imply especially for the diffusion controlled reaction. While the attack of oxygen is favoured kinetically, the attack of nitrogen results in less required reordering of the molecule (the two $\ce{N-O}$ bonds would still be identical while they become $\ce{O-N=O}$ in the nitrite ester) which reduces the expected favourism for oxygen. Effects such as solvent and counterion slightly affect the product distribution, and the highest ratio of nitromethane : methyl nitrite can be achieved using $\ce{Bu4N+ NO2-}$ and $\ce{MeI}$ in $\ce{CDCl3}$ (a product ratio of $70\,:\,30$ in favour of nitromethane). Note that $\ce{AgNO2}$ and $\ce{MeI}$ in DMSO yield more methyl nitrate ($54\,:\,46$) although the same electrophile is used. The authors conclude the nitrite section with the statement:

  It should be noted, however, that the selectivities
  shown in [the table presented in the review] are also not related to the hardness of the electrophiles.

Note

Throughout this answer, electrophilicity parameters according to Mayr’s equation are used:

$$\lg k_{20~\mathrm{^\circ C}} = s(E + N)$$

$E$ is a nucleophile-independent electrophilicity parameter while $s$ and $N$ are electrophile-independent nucleophilicity parameters.

Bibliography

A paper related solely to studying the nitrite anion: A. A. Tishkov, U. Schmidhammer, S. Roth, E. Riedle, H. Mayr, Angew. Chem. Int. Ed. 2005, 44, 4623. DOI: 10.1002/anie.200501274.

The review mentioned in the opening, concerning all sorts of ambient nucleophiles often explained via the HSAB concept: H. Mayr, M. Breugst, A. Ofial, Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.


Generation of singlet oxygen by irradiation What is the mechanism involved in the generation of singlet oxygen when ground state triplet oxygen is irradiated in the presence of rose bengal?

Rose bengal (pictured below) is a member of the fluorescein family.

(image source: wikipedia)

There are 3 steps in the process where rose bengal sensitizes the formation of singlet oxygen..

\begin{gather}
\ce{RB ->[$h\nu$]\  ^1RB}\tag1\\
\ce{^1RB ->T[ISC]\  ^3RB}\tag2\\
\ce{^3RB + ^3O2 ->T[sensitization]\  ^1RB + ^1O2}\tag3\\
\end{gather}

In the first step, ground state rose bengal (a ground state singlet) absorbs light producing an excited singlet state.  Because of the heavy iodine atoms in the molecule intersystem crossing (ISC) to the excited triplet state rapidly occurs (heavy atom effect). In the presence of ground state triplet dioxygen, a process called "sensitization" occurs. In this step the excited triplet rose bengal transfers its energy to the dioxygen promoting it to the excited singlet state while the rose bengal is deactivated and returns to its single ground state.  In this sensitization step, 2 triplets are converted to 2 singlets so, importantly, spin is conserved.


Regarding enthalpy of a reaction Enthalpy of a reaction $$\ce{2H2 (g) + O2 (g)-> 2H2O(g)} \Delta H_1$$

$$\ce{2H2 (g) + O2 (g)-> 2H2O(l)} \Delta H_2$$

Now I was asked to compare the enthalpy.

If I subtract reaction 2 from 1, I get $$\ce{2H2O(g)-> 2H2O(l)} \Delta H_2 - \Delta H_1$$

Since this is exothermic enthalpy is negative and $$ \Delta H_2 - \Delta H_1 < 0 $$ or $$ \Delta H_2 < \Delta H_1 $$

But if I consider

Enthalpy of $\ce{2H2O(l)}$ is less than enthalpy of $\ce{2H2O(g)}$ therefore more energy will be released in the formation of $\ce{2H2O(l)}$ and thus $$ \Delta H_2 > \Delta H_1 $$

So which one is it, where am I going wrong ??

You should be careful to distinguish between the magnitude of enthalpy (or some other change in energy) and the sign of the property:

the magnitude tells you how large the change was (yes, self-evidently)
the sign tells you in which direction the change occurred:

a) $\text{system}\rightarrow \text{surroundings}$, $\Delta H<0$, or exothermic
b) $\text{surroundings}\rightarrow \text{system}$, $\Delta H>0$, or endothermic

This is the typical convention used (not the only one).

In the case of the condensation of a gas, the magnitude can vary depending on the strength of intermolecular interactions in the liquid compared to the gas. The sign is negative because condensation at a constant temperature requires dissipation of heat to the surroundings.

  So which one is it, where am I going wrong ??

That the enthalpy of $\ce{H2O(l)}$ is less than the enthalpy of $\ce{H2O(g)}$ means in the scenario you present that
$$\Delta H_2 < \Delta H_1$$
not the other way around. You could also start from the fact that condensation is exothermic (or vaporization endothermic) so that $H_g-H_l=-\Delta_{cond}H =\Delta_{vap}H>0$$^\dagger$, from which $$\begin{align}H_g&>H_l\\ H_g-H_{reagents}&>H_l-H_{reagents}\\ \Delta H_1&>\Delta H_2\end{align}$$

$\dagger$ (Writing "$H_g$" is a bit "dangerous" because it suggests there is an absolute enthalpy scale (rather than relative) but you can assume that here you are using a common reference state for water).


How do I figure out the composition of a material given specific densities? The Problem
A waste material consists of sand (SG 2.7), a chemical salt (SG 2.1), and a dense plastic (SG 1.4). 200 g of a representative sample from the material is found to have a density of 2300 kg/m3. The salt is dissolved out of the sample which, after drying, now weighs 170 g. What is the composition of the waste material?
My Attempt

Based on the given information, we can derive:

densitysand = 2700 kg/m3
densitysalt = 2100 kg/m3
densityplastic = 1400 kg/m3
masssalt = 30 g = 0.03 kg
proportionsalt = 0.03 kg / 0.2 kg = 0.15

To try and calculate the individual mass of sand and plastic in the sample, I used the density = mass / volume equation to first figure out the volume using the density and mass of the total sample.

densitytotal = masstotal / volume =
volume = masstotal / densitytotal = 0.2 kg / 2300 kg/m3 = 0.00008696 m3
masssand = 2700 kg/m3 * 0.0000869 m3 = 0.2348 kg
massplastic = 1400 kg/m3 * 0.0000869 m3 = 0.1217 kg
But I don't think this step was the right choice since the mass of the sand came out to be larger than the mass of the sample. Was I correct in trying to figure out the value of the mass of sand and plastic or was that the wrong approach to determine the material's composition? If it's the first case, how should I find out the mass values?

Your mistake is that you have calculated the mass of a sample with the same volume as ours consisting entirely of sand, and a sample with the same volume as ours consisting entirely of sand.
volume = mass_total / density_total = 0.2 kg / 2300 kg/m^3 = 0.00008696 m^3
gives the initial volume of the sample before the salt is removed, however by multiplying this by the densities of sand and plastic, we calculate the mass of a 0.0000869 m^3 sample of sand and a 0.0000869 m^3 sample of plastic.
First we need to find the new volume and density of the sample after the salt has been removed.  This can be done by calculating the volume of salt removed:
volume_salt=0.03kg_salt/2100 kg/m^3=0.00001429m^3
By subtracting this volume from the initial volume you already calculated:
0.00008696m^3-0.00001429m^3=0.00007267m^3
dividing the new mass by this gives:
0.170kg/0.00007267m^3=2,339kg/m^3
Now we need to use this to calculate the proportion.  We can do this by creating and solving a system of equations.  We know that some mass of sand plus some mass of plastic gives us a 170 gram mixture, and that some volume of sand plus some volume of plastic gives us a mixture with a volume of 0.00007267m^3.
xkg_sand+ykg_plastic=0.170kg
am^3_sand+bm^3_plastic=0.00007267m^3
Now, we know that the volume and mass of each material are related by their densities, and because of this it is possible to replace the terms in one equation with the terms in the other by relating them through density, so if we decide to solve for mass, our volume equation becomes:
(xkg_sand/(2700 kg/m^3))+(ykg_plastic/(1400 kg/m^3))=0.00007267m^3
Because the x and y in these equations are the same, we can solve a system, which yields:
x=0.1418kg
y=0.0282kg
Dividing these by our starting mass gives:
proportion_sand=0.709
proportion_plastic=0.141
These can be rounded to match your answer if needed, however these proportions combined with your calculated salt proportion adds up to 1.


Is it customary to express heat loss as a negative number? In my chemistry class, we had a problem akin to the following:

  If an object with mass bla and specific heat bla is placed into a bla g calorimeter with an initial temperature of bla and a specific heat of bla, and the temperature of the calorimeter rises to bla at the end of the experiment, how much heat was lost by the object?

I performed the calculations and got an answer of, let's say, $8 J$. When I submitted this answer, I was told that it was wrong, and the correct answer would have been $-8 J$. Wouldn't this response to the above question mean that the object gained heat?

Your assumptions are correct, and from the wording of the question, the "correct" answer doesn't make sense, unless like you said, the object gained heat, which is possible; people have negative profits all the time. Sign conventions are not flawed in chemistry, but you just have to be careful when wording a question. To answer the title of your question, yes it is not only customary, it is necessary to express heat loss as a negative number, but the question still has to be phrased correctly--using the term "lost" serves as a negation in this use of language, the same way the unary operator in -1 serves as negation for a number. If heat loss were not expressed negatively, it wouldn't make sense because the system of an exothermic reaction must lose energy for its surroundings to gain it.


How does the diffusion coefficient of a species affect the limiting current for a RDE voltammagram? And how do I find the diffusion coefficient from a Levich plot?

I would recommend you take a look at the book Electrochemical Methods fundamentals and applications.

For the Koutecý-Levich equation (totally irreversible one-step, one-electron reaction):

$\Large \frac{1}{i}=\frac{1}{i_K}+\frac{1}{i_{l,c}}=\frac{1}{i_K}+\frac{1}{0.62nFAD^{2/3}\omega^{1/2}\nu^{-1/6}C^*_O} $

The higher diffusion coefficient is, the higher is the current.

The diffusion coefficient can be computed from the Koutecý-Levich equation if you know all other coefficients.


What happens when salt water is electrolyzed with a copper anode and a copper cathode? I heard that copper oxide is formed on the cathode and copper chloride is formed on the anode, is this correct?
I don't have much knowledge of chemistry, so I can't find out much by searching on Google.

The chemistry here can be explained from the associated products of the electrolysis of brine. To quote an education source:

Three important chemicals, NaOH, Cl2, H2, can be obtained by electrolyzing an aqueous NaCl solution (brine). This forms the basis of the chlor-alkali industry. The diaphragm cell (also called a Hooker cell)...

As such, depending on the cell design, expect also hypochlorous acid:
$\ce{Cl2 (g) + H2O (l) <-> HCl (aq) + HOCl(aq) }$
and hypochlorite with OH- as a side product of the electrolysis of the brine:
$\ce{ OH- + HOCl -> H2O + ClO-}$
However, some of the $\ce{HOCl}$ can also engage in a Fenton-type reaction (example here) with $\ce{Cu}$ (and, to a lesser extent, $\ce{HOCl}$ acting on $\ce{Cu2O}$ resulting in some possible $\ce{CuO}$ or even a basic chloride formation) generally depicted as follows:
$\ce{Cu/Cu+ + HOCl -> Cu+/Cu++ + .OHCl-}$
where the latter radical intermediate can further break down with products dependent on the pH level (here, likely over pH 7 from $\ce{OH-}$ created in the parallel electrolysis reaction) and the chloride concentration. Source: See reactions listed in Supplement Table S1 available here by clicking on Supplement File to obtain a download.
Support: With metal spoons serving as electrodes for the electrolysis of a saltwater solution, there is clear discoloration of one of the spoon electrodes (visible at 5 minutes and 20 seconds into this science experiment video).
Bottomline in time, expect similar discoloration of a copper electrode as well from the associated electrolysis products.


Do the mechanism (SN1 or SN2 ) which reaction follow depends more upon intermediate stability than on substrate and nucleophilic? For eg : would the reaction of 3-methylbut-2-en-1-ol with HBr follow SN1 mechanism due to stability of your intermediate carbocation stability  or SN2 due to the compound being primary alcohol and good nucleophilic. Also please correct me if I got something wrong.

It will follow SN1 as the intermediate carbocation formed is resonance stabilised.


Is this what an orbital means? In this picture, there is a $1s$ and a $2s$ orbital. Are the orbitals only the dark gray circles ? Or are the light gray circles (which are surrounding the dark gray ones on both sides) also included in the orbitals? Other than these two types of circles , there are the nodes which never include any electrons. And since people define an orbital as a region where electrons are most likely to be found, this must mean that the orbitals are only the gray circles because if they included the light gray circles too, then the probability of finding electrons in an orbital would be 100%.

This picture is quite bad, since there are patterns inside the gray areas which shouldn't be there but the gray areas are the orbitals. You could think of it like writing down the position of an electron over and over again and every dot is a position where you found the electron. More dots means it was in that area more often (=higher probability of finding it there).

An electron, theoretically, can be everywhere except the nodes. So an electron from an atom right in front of you could be lightyears away. The chance is very, very, very low for that to happen but it is possible. So for having a 100% chance of finding the electron you have to look in the whole universe, of course it has to be somewhere. That's not very useful at all, since every electrons can be found somewhere in the universe so we don't work with 100% chance but rather with 95% ( or 99% or 99.9%,..) since that's already a probability of finding the electron very, very close to the core of the atom and it is there most of the time.


Am I visualizing/understanding atomic form factor correctly? I'm having difficulty visualizing the concept of atomic form factor or atomic scattering factor as it has been presented in a chemistry course on X-ray crystallography that I'm taking. At the moment, I visualize it as such:

(Please note that the question marks in the diagram above are included to communicate my confusion.)
Am I visualizing this correctly? Is there any other way for me to interpret these diagrams? The course doesn't dive into the mathematics of atomic form factor (and neither does our accompanying text), so I don't have much to go on in developing a mental image of the phenomenon at play here--and I strongly suspect that the mental image that I do have is wrong since I can't find any evidence to support it.

In your illustration, there seems to be a little «almost step».  But more importantly, if displaying a heavy scatterer (iron, Fe) next to a weaker one (C), you should put a bit more emphasis on the descent for iron.  This is one of the reasons why the atomic position of the heavier atoms are easier to attribute in the Fourier map, than the lighter ones.
Take a look at IUCr's illustration from their freely accessible on-line dictionary:

(credit)
Maybe your school has the printed or / and electronic version of the IT (the International Tables of Crystallography) which goes into further detail (e.g., listing $f$ and $f'$ for the elements, obviously dependent on the wavelength of radiation used during the diffraction experiment).
Because currently many schools' libraries may restrict access and borrow the books in person, note that IUCr moderates a dedicated section of teaching pamphlets, too.  Topics include e.g., crystal packing and symmetry.


Why doesn't nitric acid have a positive charge? From the diagrams I've seen $\ce{HNO3}$ has a dative covalent bond so why doesn't it have a positive charge?

One side of the $\ce{HNO3}$ molecule does have a positive charge, but this is neutralized by the negative charge resulting from the Oxygen atom with only one bond to the nitrogen. These charges cancel leaving an overall neutral molecule.


In what order should the compounds containing Ru and Os, or Mo and W be named? According to IUPAC recommendations for naming of inorganic compounds, the more electropositive element is written first with a suitable prefix, and the more electronegative element is written at the last with a suffix -de and a prefix. For example, $\ce{H2O}$ (Water) can be called dihydrogen monooxide. [this is an modified statement extracted from the wiki encyclopedia-IUPAC nomenclature of inorganic chemistry]
  In the above example, $\ce{H}$ (Pauling scale = 2.20 or Allen scale = 2.300) is less electronegative than $\ce{O}$ (Pauling scale = 3.44 or Allen scale = 3.610). So, $\ce{H}$ is placed first, followed by $\ce{O}$.

If suppose we were needed to name the compounds containing

$\ce{Ru}$ (Pauling scale = 2.2) and $\ce{Os}$ (Pauling scale = 2.2)
$\ce{Mo}$ (Allen scale = 1.47) and $\ce{W}$ (Allen scale = 1.47)

, in what order should we write the elements?

If we use pauling scale as referance, we would face difficulty in naming $\ce{Ru}$ and $\ce{Os}$, if we use allen scale, we would be in trouble with respect to the elements like $\ce{Mo}$ and $\ce{W}$, etc, as we can see that elements have same electronegativity value. I don't know whether the compounds with the above mentioned elements exist or not, but if we suppose, assume that, they exist, in what order should they be named?

When IUPAC rules refer to electronegativity (and electropositivity), they don't mean actual electronegativity measured by one or another scale. IUPAC decided, for nomenclature's sake, on a formal ordering of electronegativity, which goes like this:

This is Table VI, on page 260 of IUPAC's very own 2005 Red Book, aka Nomenclature of inorganic chemistry. The cases in which it is used are detailed in IR-2.15.3.1 (page 42).

The explanation of the changes from earlier IUPAC recommendations is in IR-1.6.3 of the Red Book:

  The element sequence of Table VI
  In Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990 (Ref. 11), the position of oxygen in certain element sequences was treated as an exception. Such exceptions have been removed and the element sequence of Table VI is now strictly adhered to. In particular, oxygen is treated as the electropositive component relative to any halogen for constructing compositional names (Section IR-5.2) and corresponding formulae (Section IR-4.4.3) for binary compounds. This results in, for example, the formula O2Cl and the name dioxygen chloride rather than the formula ClO2 and the name chlorine dioxide.

  In Ref. 11, the formulae for intermetallic compounds were also subject to an exceptional rule although no guidance was given for naming such compounds, and the term ‘intermetallic compound’ was not defined. The problem is to define the term ‘metal’. Therefore, no attempt is now made to make a separate prescription for either the formulae or the names of intermetallic compounds. It is stressed, however, that the present recommendations allow some flexibility regarding formulae and compositional names of ternary, quaternary, etc. compounds. Several ordering principles are often equally acceptable (see Sections IR-4.4.2 and IR-4.4.3).

  The element sequence of Table VI is also adhered to when ordering central atoms in polynuclear compounds for the purpose of constructing additive names (see Section IR-1.6.6).

For more on this topic, you've got to go read (and upvote) this excellent answer by Manishearth, which is where I first learnt that useful and intriguing nugget of IUPAC idiosyncrasy!


Rate of Reaction for Elementary Reactions The rate law for an elementary step of a reaction of the form $2A \Rightarrow$ Products is given by $rate=k[A]^2$, while in the form $A \Rightarrow$ products is given by $k=[A]$. Is there some sort of intuition as to why the the reaction with $2A$ is quadratic, instead of double the one with $A$? Essentially, why is the equation $k[A]^2$ instead of $2k[A]$?

Intuitively, a reaction between A and B to give C will go twice as fast when you double the concentration of A (because collisions between A and B will occur twice as frequently). Similarly, doubling the concentration of B will also double the reaction rate. But what about the specific case where A and B are the same substance (A + A -> C)? When you double the concentration of A, the rate is quadrupled; when you triple the concentration of A, the rate goes up by a factor of 9, etc.


A thermoset or a thermoplastic? How do you know if a polymer is a thermoset or a thermoplastic? For example:

hydroxybutanoic acid
2-amino-3- hydroxybutanoic acid

(I am not sure if these are correct in English)

I know what thermosets and thermoplastics are, and their properties, but I don't know how to see if polymers of a certain molecule is a thermoset or a thermoplastic by looking at the structural formula.

Some background, since most readers would probably need it.

A thermoplastic is a polymer that becomes pliable/moldable upon heating, and then returns to a solid (keeping its new shape) when cooled. Thermoplastics are generally high molecular weight polymers with very few or no crosslinks. Polymer chains are held together by intermolecular forces (most often van der waals) and physical entanglement, both of which can be overcome with addition of heat.

A thermoset is a polymer that is irreversibly cured, usually through covalent crosslinking (during or after polymerization). The covalent crosslinks hold the chains together in such a way that the chains cannot be separated by the addition of (small amounts of) kinetic energy. Thermosets tend to dramatically fail after the addition of lots of heat, following which they do not return to their original state. It is possible for the cumulative forces from many strong non-covalent interactions (ionic, hydrogen-bonding, and even pi stacking) to make a polymer behave like a thermoset. For example, proteins are thermosets.

  Can you tell if a monomer will produce a thermoplastic or themoset based on its structure?

Yes. All polymers produced from only mono- (addition polymerizations) or bi- (condensation polymerizations) functional monomers will always produce thermoplastics. All of the hydroxybutanoic acids (e.g. $\gamma$-hydroxybutanoic acid) are difunctional monomers for condensation polymerizations. Their polymers will be thermoplastics.

Unless...

The addition of tri- or tetra- functional monomers produces crosslinks, which always produces thermosets. For example, glycerol, malic acid, or lysine could be trifunctional monomers in condensation polymerizations. 2-Amino-3-hydroxybutanoic acid (threonine) is a trifunctional monomer. It has three functional groups (alcohol, amine, carboxylic acid), that can all react in a condensation polymerization. Any synthetic polymer containing threonine is likely to contain crosslinks and be a thermoset.

Or...

Many thermoplastic polymers can be converted into thermosets by curing them afterwards. Some curing is thermal, some is irradiative (espccially using UV), and some is chemical. A prime example is polyisoprene (natural latex rubber), which is a thermoplastic (with a very low melting point). It is also an elastomer - it will stretch and return to its original shape (although melting it erases the "memory"). Rubber is converted into a thermoset by vulcanization to make it more useful while still having the elastomeric properties.


A basic question about tetramethylammonium hydroxide --strong or weak? This is a follow-up to the question here.  There is an answer claiming that tetramethylammonium hydroxide (TMAH) is a suitable strong base that can be used instead of the usual alkalies to reach pH 14.  However, Wikipedia says that TMAH is only a weak base.

But, when I state this I get comments that other sources document TMAH as a strong base.  The other sources, however, just mention it in passing and do not offer direct proof.  Can anyone provide a reference that actually does settle this question?  Thanks!

Note: At the time the question was posed Wikipedia did make the weak base claim. It has since been edited, and it now appears that the edit is correct.  See the answers.

TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society 87 (1905): 955-961:

  Practically the preparation of tetramethylammonium hydroxide
  from its salts resolves itself into a question of solubility as follows.
  In general, the equation

  $$\ce{NMe4X + MOH = NMe4OH + MX}$$

  will represent a real action proceeding nearly to completion if M, X,
  and the solvent are so chosen that all the substances except MX shall
  be soluble, or at least that MX shall be much less soluble than either
  of the original reacting substances. This principle was applied by
  Hofmann in his preparations with water as solvent. For M and X
  he chose either the pair Ag,I or Ba,SO4, in both of which cases MX is
  practically insoluble in water.

  It is clear that if the general application of the above principle is
  justifiable, tetramethylammonium hydroxide may be prepared from a
  tetramethylammonium salt by means of potassium hydroxide if we so choose X and the solvent that of the substances represented in the equation

  $$\ce{NMe4X + KOH = NMe4OH + KX}$$

  all shall be soluble except KX.

...

  An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.

So by two independent methods it is shown that tetramethylammonium hydroxide  is much stronger than pKb=4.2.  Firstly, the halide cannot not be converted to the hydroxide by adding metal hydroxide, unless an insoluble halide is produced to force the equilibrium forward.  Seconding, it is shown by velocity of saponification to be only slightly weaker then NaOH.

See also Essential Principles of Organic Chemistry (1956)

  It follows that a quaternary ammonium hydroxide is completely dissociated and should be a strong base, as is sodium hydroxide. The experimental facts are entirely consistent with this view. Although trimethylamine in water solution gives only a small concentration of hydroxide ion, tetramethylammonium hydroxide gives a full molar equivalent of hydroxide ion in solution.


Calcium chloride hydrolysis I was wondering about $ \ce{CaCl2} $ hydrolysis.
The most primitive answer would have been:
$ \ce{Ca(OH)2} $ does not dilute that great, therefore it is not strong enough to have the $ \ce{Ca^2+} $s not hydrolise. Thus:
$$
\ce{Ca^2+ + H2O <<=> CaOH^+ + H+}
$$
Yet, as we can see in this answer - $ \mathrm{pK_B^{II}} $ for $ \ce{Ca(OH)2} $ is $ 1.4 $, which is not bad. Therefore, some reckon the diluted part of $ \ce{Ca(OH)2} $ to be a pretty strong base, making hydrolysis impossible. To make things worse, the cation hydrolysis is known to be harder to trace rather than anion hydrolysis.
Still, I have come across some articles on $ \ce{CaCl2} $ hydrolysis (like this one - a top Google search line; yet, this one first shows that $ \mathrm{\Delta G_{900^oC}} $ is $ \mathrm{90.951 >  0} $).
So, is it safe to say that at $ \mathrm{\approx 25^oC} $ $\ce{CaCl2}$ does not undergo hydrolysis?

Just calculating from the head by the simplifying formula $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{(\frac {[\ce{A-}]}{[\ce{HA}]})}$, $\mathrm{pH}$ of $\pu{1 M } \ce{CaCl2}$ is approximately 1/2( 12.6 - log 1)=6.3.
So concentration $\ce{[CaOH+]}$ is about million times smaller then $\ce{[Ca^2+]}$. Hydrolysis happens, but is fully negligible. Note that the real $\mathrm{pH}$ depends on impurities and dissolved $\ce{CO2}$.
About the calculation of respective fractions, we can use the formula for gradual dissociation of multiprotic acid:
If we formally suppose $$\ce{H2A <=>  HA- + H+ <=> A^2- + 2 H+}$$ then respective molar fractions for particular acid forms are
$$x_{\ce{H2A}} = x_{\ce{Ca^2+}} =\frac {[\ce{H+}]^2}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{HA-}} = x_{\ce{Ca(OH)+}} =\frac {K_\mathrm{a1} \cdot [\ce{H+}]}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{A^2-}} = x_{\ce{Ca(OH)2}} =\frac {K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
where $K_\mathrm{a1} = K_\mathrm{w} / K_{\mathrm{b1},\ce{Ca(OH)2}}$ and $K_\mathrm{a2} = K_\mathrm{w} / K_{\mathrm{b2},\ce{Ca(OH)2}}$


How is a car battery recharged? When recharging a lead-acid battery, how are the terminals of the battery connected to the voltage source. I mean, do we connect the positive terminal of the battery to the negative terminal of the source and vice versa? Or does the positive go with the positive and the negative with the negative? Also, are the electrodes reversed while the battery is being recharged?

The voltage on the terminals is different by 12 V (± a few depending on the charge state), you can think of this as the positive terminal being at +12 V and the negative terminal at being 0 V. The polarity of the voltage does NOT change.

On the current, things are a bit different. During discharge, the current flows from the positive to the negative.

During charge, the current flows out of the positive terminal of the charger into the positive terminal of the battery (and out of the negative terminal of the battery into the negative terminal of the charger).


Energy of Eclipsed Interactions I am asked to identify the highest and lowest energy conformations for the following bond-line structure (I think it is 2,2-dimethylbutane). I drew a Newman projection (for all 6 conformations) and was able to identify the lowest energy conformation correctly but the textbook solution for the highest energy conformation does not match what I drew. It seems as though the textbook solution is missing a methyl group, and has 3 $\ce{H}$ bonds instead. Can someone please clarify?

You are correct in drawing this. The "answer" molecule depicted by your book is different from the one it first depicted, so it cannot be right.

You correctly depicted the high energy conformation of the molecule as a rotation of the low energy that shows the methyl groups aligned. This is high energy because of the intramolecular repulsion of the hydrogen atoms and the electron clouds.


Is there an A value for the carbaldehyde and vinyl groups? A methylene ($\ce{=CH_2}$) group ist sterically less hindered than a methyl ($\ce{CH_3}$) group.

I'm wondering if the carbonyl ($\ce{C=O}$) group is sterically more or less larger than a methylene ($\ce{=CH_2}$) group.

Bond length: ($\ce{C=O}$) is shorter than ($\ce{C=C}$) [1]
Diaxial i.a.: Me//H is larger than Me//n(O) (0.9 vs. 0.4 kcal/mol; no ref. for the 2nd value)

While the shorter distance makes the carbonyl more sterically, the lone pairs are less sterically than the Hydrogens of methylene. I think this effect is larger and hence a carbonyl to be more sterically hindered than a methylene group.

The A-values gives the energetical difference of a monosubstituted cyclohexane between the conformation in which the substituent is placed axial (1,3-diaxial interactions, hence higher in energy) versus equatorial.

To reformulate above problem of the sizes I have to set following question:

Comparing the axial conformations of vinylcyclohexane and cyclohexancarbaldehyde: Which molecule is more strained?

Why I am interested in above question?

If we don't assume protonation of a ketone's oxygen, I'd like to know if I have to place the carbonyl axial or equatorial when R=Me.

For R=long alkyl chain it makes sense to place R equatorial, but I'm not sure if a methyl group is also bigger than a carbonyl.

Because I think methylene and carbonyl are almost of the same size from above arguments, and I'm knowing that methyl is bigger than methylene, it sounds logical to me to place the methyl equatorial.

http://evans.rc.fas.harvard.edu/pdf/evans95.pdf

Comparing the axial conformations of vinylcyclohexane and
  cyclohexancarbaldehyde: Which molecule is more strained?

As you point out, "A-values" can be used to get some idea of the relative "size" of a substituent in the cyclohexane system.  More specifically, the A-value reflects both the steric size of the substituent and the length of the bond connecting the substituent to the cyclohexane ring (the longer the bond, the further away the substituent and steric interactions are therefore reduced).

The A-value is just the free energy difference between the axial and equatorial conformers. Since destabilizing steric interactions can occur in the axial conformation, the larger the substituent, less axial and more equatorial conformer will be present.

(image source)

As a consequence, the "larger" the substituent the larger the A-value. The A-value for a methyl group is 1.8, using

$$\mathrm{\Delta G = -RT*ln(K)}$$

we can solve for K and find that the axial/equatorial ratio for methylcyclohexane is ~ 5/95 (as expected, the less sterically strained equatorial conformer predominates) at room temperature.

(image source)

Here is a link to a nice table of A-values.  The A-value for a vinyl group ($\ce{-HC=CH2}$) is 1.6, while it is only 0.7 for the aldehyde group ($\ce{-CHO}$). Hence the aldehyde group is the smaller of the two. Or said differently, axial vinyl-cyclohexane would be more strained than axial cyclohexane-carbaldehyde


What is the difference between electric current and electric charge? Are electric current and electric charge the same thing? I'm just getting back into relearning chemistry, so a more in-depth answer about how electricity works in general would be wonderful.

Edit: Thank you all for clarifying that this belongs to the Physics SE. That being said, I guess this question doesn't have much relevance to me anymore. I'm honestly not sure if I should delete the question or not, so I'll leave that up to the mods to decide. Thanks again.

Your question is very broad. The first thing. No they are not the same. Simple explanation for these terms can be:
Charge: Its the "amount" of unbalanced electricity in a body.  Its usually measured in the form of electrons (either excess or deficiency). Its units is coulomb.
Current: It is the "flow" of electrons from higher electric potential towards the lower. Its unit is ampere.

To visualize how they are not the same is by considering two situations (without going into any calculations). Two bodies (A & B) are connected via some copper wire. A has 1 million Coulombs charge & the second one has 1million & 1 coulombs charge. Now individually these charges are enormous, but as the difference between the two is only 1 coulomb, its a very low difference in charge, so a little current will flow between them.

Now if you change the bodies' charge. A has 100 Coulombs & B has 10 .. much more current will flow between the bodies, because there is a way bigger potential difference.

ps. and yes its a physics questions.


Reaction of hydrazine and hydroxylamines with carboxylic acids Carboxylic acids and their derivatives on reaction with hydroxylamine $\ce{NH2OH}$ and hydrazine $\ce{N2H4}$ give hydroxamic acids and acid hydrazides, respectively, through acyl nucleophilic subsitution.

Why don't they give the same reaction as aldehydes and ketones (i.e. removal of $\ce{H2O}$) giving formation of oximes as the acids also have oxo ($\ce{=O}$) group?

For simplicity, I will consider the reaction of acetic acid with ammonia, because the same principles apply. The two alternatives are:
$$
\underbrace{\ce{CH3C(O)NH2}}_{\mathrm{amide}} + \ce{H2O} \leftarrow \ce{CH3C(O)OH} + \ce{NH3}\rightarrow \underbrace{\ce{CH3C(NH)OH}}_{\mathrm{imidic \ acid}} + \ce{H2O}
$$
where the group connected via a double bond has been set in parentheses. The two products are connected by protonation/deprotonation in a tautomeric equilibrium, so the question boils down to thermodynamic stability. For most cases, we expect the C-O double bond/C-N single bond combination to be more stable than the C-N double bond/C-O single bond combination.


Why are condensation reactions endergonic? My lecture notes state that

  Condensation reactions, like all biosynthetic reactions, are endergonic.

I am unsure as to why this would be the case. I thought bond formation was an exothermic reaction, so you would get a negative enthalpy change from the bond being formed. I appreciate that you wold have a decrease in entropy when going from two small molecules to one larger molecule, but would this really counter-balance the negative enthalpy change every time? Perhaps the small molecule produced in the biosynthetic reaction, e.g. $\ce{H2O}$ for condensation reactions, is important here?

In a nutshell, I am just confused as to what ensures that all biosynthetic reactions (and condensation reactions) are always endergonic?

One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to synthesise lipids. It can be broken down to:

$$\ce{R-O-H + H'-O'-C(=O)-R' -> R-O-C(=O)-R' + H'-O'-H}$$

With each important bond highlighted and the oxygen and hydrogen atoms made distinguishable by primes. A simplistic way of deriving standard reaction enthalpies is to use Hess’ law and add up the bond dissociation enthalpies of bonds that are broken and subtract that from the bond dissociation enthalpies of bonds that are formed. Doing that in this case gives:

Broken:

$\ce{C-O'}$
$\ce{O-H}$

Formed:

$\ce{C-O}$
$\ce{O'-H}$

Since the bonds being formed and those being broken are identical, at first approximation $\Delta_\mathrm{r} H^0 \approx 0$, meaning the reaction depends entirely on the entropic term.

On the entropic side of things, you are creating one huge, dissolved entity out of two not-as-huge dissolved entities. Making an exact call here is difficult, as it depends on a number of things, most notably how large the hydration sphere is for all three of these compounds. If each of them has a hydration sphere of approximately the same diameter, the product side will have more entropy since less water molecules are needed for an identical hydration sphere (akin to the hydrophobic effect). The opposite effect is also possible, though.

I would like to challenge the ‘all condensations are …’ It seems to simplistic to me. Many condensation reactions are probably endergonic but likewise I expect a few to be overall exergonic. The cell just takes no chances and uses an $\ce{ATP}$-dependent pathway in all cases to ensure the outcome it desires.


Help understanding the pressure's effect on equilibrium and Le Chatelier's principle So let's say we have a closed system in the following reaction:

$\ce{N2O4 -> 2 NO2}$

The only way to increase the system's pressure without adding any mole of gas and keeping temperature constant is reducing the system's volume. Once we increase the pressure by reducing the total volume, formation of $\ce{N2O4}$ is observed.

In many text books, the explanation for this is that 2 moles of $\ce{NO2}$ occupy a bigger volume than 1 mole of N2O4, thus the equilibrium is restablished when more $\ce{N2O4}$ is formed. What doesn't make sense for me is that both the substances are contained in the same volume, so how 1 mole of $\ce{N2O4}$ would occupy less volume than 2 moles of $\ce{NO2}$? It only makes sense to think about partial pressures, not partial volumes, right?

I'm trying to use entropy and energy microstates to solve this problem, because it seems more fundamental than this mantra in the general chemistry textbooks. From quantum mechanics, I know (can't explain, just know; need further studying on this subject) that translational energy states become closer when a gas expands in a bigger volume, so I'm assuming that in the reaction above, with a bigger pressure (and thus a smaller volume) these energy states will be more spaced between each other. With energy states more far apart, why are there more possible energy microstates when $\ce{N2O4}$ is formed?

I can't get past this! Why does this implies in the formation of the substances with less gaseous coeficients? How are the energy states of reactants and products and their changes on pressure related with that?

Edit: to make myself clearer, I'll use this example from this textbook I'm reading:

Chem1 Virtual Textbook, Stephen Lower, link

In the endothermic reaction H–H (g) ⟶ 2 H (g), you only get hydrogen atoms at very high temperatures, because you'll have a higher distribution of populated states on the 2 H side, thus you favor the products at higher temperatures for endothermic reactions. I can completely understand the effects of themperature in Le Chatelier's principle, and I'm trying to understand an analogous thought for changes in pressure in the microscopic point of view.

What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.

In order to understand this a little better you can have a look at the ideal gas law:
$$
p = \frac{nRT}{V}
$$
As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG < 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.


Why are indicators used only in the form of dilute solutions? Why are indicators used only in the form of dilute solutions? I get that indicators are a form of weak acids themselves, and the extent of their dissociation varies with the pH of the solution they are in.

The most simple and general answer is: because we can! However, it may require explanation.

Indicators will always affect the observable of the solution they are measuring. For example, an acid-base indicator is an acid or a base itself (depending on the pH of the solution). It will take at least one indicator-equivalent of acid/base to protonate/deprotonate the indicator and induce the colour change. (Some indicators may even require two or more equivalents.) Until that full equivalent is supplied, we cannot witness the new colour. Thus, we want the indicator concentration to be as minute as possible because that will give us the lowest indicator error possible.

All other properties and effects of indicators are secondary to the major point above. For example, indicators are typically chosen to be very intensely coloured so that the colour may be observed even at very low concentrations; this allows us to reduce indicator loading and error and thereby increase accuracy.
Synthesising these indicators may be time-consuming and they may be expensive. This further strengthens our wish to use as little as possible.
The colour change of an indicator is ideally abrupt and strong, again meaning we can reduce the loading a lot.
And so on and so forth.

Thus, because we can use indicators at these low concentrations, we do.


How many kilograms of graphite are consumed while producing 4 kg of Al through Hall-Heroult process? How many kilograms of graphite are consumed while producing 4 kg of $\ce{Al}$ through Hall-Heroult process?

This looks like a simple stoichiometry problem:
$$\ce{4Al^3+ + 12e- -> 4Al} \; \; \; \; \; \text{(at cathode)} \\
\ce{3C ->[O2] 3CO2 + 12e^-} \; \; \; \; \; \text{(at anode)}$$
For every $4$ moles of $\ce{Al}$ consumed, $3$ moles of $C$ are consumed. The mass of graphite consumed is thus:
$$M = \frac{4000}{27} \times \frac{3}{4} \times 12 = \pu{1.33 kg}$$
However, my textbook mentions the answer to be $\pu{2 kg}$, along with the following explanation:

for every kilogram of $\ce{Al}$ produced, approximately $\pu{0.5 kg}$ of graphite is consumed.

Why is this so? Could anyone provide a valid reference?

According to Aluminium Production1:

The production of 1 ton of aluminum typically requires 420 kg of carbon, 1920 kg $\ce{Al2O3}$, and 16 kg $\ce{AlF3}$ ....
The theoretical carbon consumption is only $\pu{333 kg C/t} \space \ce{Al}$. The excess carbon consumption is mainly due to air oxidation of the hot anodes in the hooded cells. The reaction occurs preferentially in the pitch binder matrix and leads to the physical loss of coke particles to generate dust that floats on top of the electrolyte.

Theoretically, your answer holds good. However, it fails practically. It's a stupid tricky question by the problem setters, who are apparently testing your knowledge of metallurgy rather than stoichiometry.
References:

Tabereaux, Alton T., and Ray D. Peterson. “Aluminum Production.” Treatise on Process Metallurgy, Elsevier, 2014, pp. 839–917. doi:10.1016/B978-0-08-096988-6.00023-7


What results from the reaction of 5-bromopentan-1-ol and NaOH? In our chemistry class, we noted three molecules that result from the reaction:

pentane-(1,5)-diol

This one is simple to understand, simply a nucleophilic substitution between $\ce{Br^-}$ and $\ce{OH^-}$.

(1,5)-epoxypentane

This is the main result, resulting from the intramolecular esterification of the two $\ce{OH^-}$ groups of the previous molecule. This is wrong, please see the explanation
by  ringo

4-penten-1-ol

This one, I don't understand. How is this a possible result? Could anyone help me understand? Thank you for your time.

The three reactions are:

An $\mathrm{S_N2}$:

Deprotonation followed by an intramolecular Williamson ether synthesis ($\mathrm{S_N2}$):

An $\mathrm{E2}$:


Oxidation of guanine with singlet oxygen I'm studying the oxidation of guanine using singlet oxygen and one step of the reaction is not very clear to me.

My problem is with the first step: it is a 4+2 Diels–Alder cycloaddition, and Fukui's rules suggest that this kind of reaction doesn't happen photochemically, only thermally.

However, since the reaction is between guanine and singlet oxygen, the oxygen has been photochemically excited, and I don't think it would happen thermally with oxygen at the ground state because otherwise we would have constant oxidation.

So what exactly is happening there?

You're right that ground state $\ce{^3O2}$ doesn't undergo this reaction.

There is no direct excitation of oxygen either.

The key in these processes is the electronic excitation of tetraphenyl porphyrine (TPP). The excited singlet state of the dye undergoes intersystem crossing to a triplet state. The latter interacts with ground state oxygen and generates singlet oxygen.

$$\ce{TPP ->[h\nu] ^1TPP^*}$$
$$\ce{^1TPP^* ->[ISC] ^3TPP^*}$$
$$\ce{^3TPP^* + ^3O2->[sens] TPP + ^1O2}$$

Singlet oxygen then acts like the typical dienophile in a Diels-Alder reaction.


Iodine ion vs caesium ion size An iodine ion is the larger of the two, but why if they have the same amount of electrons? Also, $\ce{Li}$ is larger than $\ce{Li+}$, don't more electrons mean larger radius?

Iodine is having 53 electrons and cesium is having 55 electrons in general. Iodine ion can be made by adding or by removing electrons from it. Here you added two electrons to iodine and made it an iodine ion in which number of electrons of iodine ion became equal to the number of electrons of cesium but the composition of nucleus(number of protons) is different in both iodine ion and cesium ion. Obviously cesium is having more number of protons than iodine, so cesium nucleus can hold it's electrons more closer in cesium than iodine nucleus in an iodine ion which results in larger radius of iodine ion than cesium..
Same number of electrons doesn't mean that their radii are same. Same logic is applicable for lithium ion and lithium.. Lithium ion is having less electrons and more protons compared to lithium.. So radius of lithium is larger than radius of lithium ion


Density of saturated KBr Google search shows the density of saturated $\ce{KBr}$ is $\mathrm{3.119\ g/ml}$ at $\mathrm{25\ ^\circ C}$.

We have a solution of saturated $\ce{KBr}$ and density is about $\mathrm{1.5\ g/ml}$ when weighed on a calibrated analytical balance.
How do we account for this discrepancy?

According to “Aqueous Solubility of Inorganic Compounds at Various Temperatures”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the solubility (expressed as mass fraction) of potassium bromide $(\ce{KBr})$ in water is $39.4\ \%$ (at $T = 20\ \mathrm{^\circ C}$) and $40.4\ \%$ (at $T = 25\ \mathrm{^\circ C}$).

According to “Concentrative Properties of Aqueous Solutions: Density, Refractive Index, Freezing Point Depression, and Viscosity”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the properties of a potassium bromide solution of with a mass fraction of $w_{\ce{KBr}} = 40.0\ \%$ at $T = 20\ \mathrm{^\circ C}$ are as follows.

molality: $b_{\ce{KBr}} = 5.602\ \mathrm{mol\ kg^{-1}}$
concentration: $c_{\ce{KBr}} = 4.620\ \mathrm{mol\ l^{-1}}$
density: $\rho = 1.3746\ \mathrm{g\ cm^{-3}}$


Chemiluminescence - Determining Temperature from Luminol Imaging I'm working on an imaging project and I have no background in chemistry or chemiluminescence for that matter. This is a Computer Science project.

While most of the details for it have been worked out, I'm having difficulty figuring out the scope and possibility of an important requirement.

Given a luminol image, I will develop a heat-map. For now the heat map is built based upon the intensity of the light.

However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?

I would assume that this chemical would burn differently at various temperature and would have a specific signature of light emitted depending on it its temperature but I'm having difficulty pinpointing whether this is true, possible and backed by research (or science).

The closest parallel I've found is combustion analysis. The heat release rate is calculated by measuring the amount of light radiated from a flame at various wavelengths, since certain chemicals give off radiations at specific wavelengths.

Can anyone confirm this, suggest alternatives or point me int the right direction?

However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?

Probably not! The energy released from the excited state of the 3-aminophthalate dianion formed in the oxidation of luminol (or rather its dianion) by hydrogen peroxide in alkaline is released in form of light. There's little to no heat released in this chemoluminescent reaction.


Is the quantum number $L$ a combination of the azimuthal and magnetic quantum numbers $(\ell, m_\ell)$? Somehow, I am getting confused about this...
In the $J=L+S$ equation about total angular momentum, which of the four quantum numbers used to describe the electrons and their states is included in the $L$?
And does the $J$ include the first, principal quantum number?

The values of $L$ come from the individual $\ell$ values, according to the Clebsch-Gordan series:
$$L = |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2.$$
where the dots indicate all "in-between" values with a step size of 1. The values of $S$ come from the individual $s$ values, in a similar fashion. Note that it is $s$, not $m_s$.
Finally, the values of $J$ come from $L$ and $S$:
$$J = |L - S|, \ldots, L + S.$$
The principal quantum number $n$ does not play any role in this, since it has no direct relation with angular momentum.

Usually it helps to see an example. Consider the case of the ground-state carbon atom; the $\mathrm{1s}$ and $\mathrm{2s}$ orbitals can be neglected since those are closed-shell, so we only need the two highest-energy electrons, i.e. a $\mathrm{p^2}$ configuration.
We therefore have two electrons, both having $\ell = 1$ (p-orbital) and $s = 1/2$ (because they are electrons, and all electrons are spin-$1/2$). The permissible values of $L$ are
\begin{align}
L &= |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2 \\
  &= | 1 - 1 |, \ldots, 1 + 1 \\
  &= 0 , \ldots, 2 \\
  &= 0, 1, 2
\end{align}
The permissible values of $S$ are
\begin{align}
S &= |s_1 - s_2|, \ldots , s_1 + s_2 \\
  &= | 1/2 - 1/2 |, \ldots, 1/2 + 1/2 \\
  &= 0 , \ldots, 1 \\
  &= 0, 1
\end{align}
The permissible values of $J$ depend on what $L$ and $S$ are, and must be evaluated on a case-by-case basis. So far, we have identified three possible values of $L$ and two possible values of $S$. Consider the case where $L = S = 0$: here,
\begin{align}
J &= |L - S|, \ldots , L + S \\
  &= |0 - 0|, \ldots, 0 + 0 \\
  &= 0 , \ldots, 0 \\
  &= 0
\end{align}
So when $L = S = 0$, the only permissible value of $J$ is also $0$. If you've seen term symbols before, the corresponding term symbol is denoted as $^1\!S_0$; if you haven't, then all you need to know is that this is a chemist's "shorthand" way of writing $(J, L, S) = (0, 0, 0)$, which is the combination we just found.
You will need to work through this for the other five combinations of $L$ and $S$. This will give you a full series of term symbols, except that there's a catch. The Pauli exclusion principle forbids some of those six combinations: to be precise, the combinations $(L, S) = (2, 1)$, $(1, 0)$, and $(0, 1)$ are not permissible, which leaves only three valid combinations, one of which I've already demonstrated. See Pauli-forbidden term symbols for atomic carbon for more details.

Addendum. As for the $m_\ell$ and $m_s$ quantum numbers: they do have a use, but in a rather more subtle way. I don't expect this very brief summary to be useful, but I'll try; feel free to ignore it at this stage of your learning if it doesn't make much sense.
Much like how (for one electron) $\ell = 1$ implies $m_\ell = -1, 0, 1$, for the entire atom each value of $L$ is associated with a range of $M_L$ values: for example, $L = 1$ implies $M_L = -1, 0, 1$. Therefore, the $L = 1$ term is actually not one single electronic state, but rather a collection of three different values of $M_L$, which is why we call it a "term" rather than a state. Likewise, the $L = 2$ term actually corresponds to five different values of $M_L$. How does this relate to $m_\ell$? Well, it turns out that $M_L$ is the sum of the individual electron $m_\ell$ values. A similar situation exists with $S$, $M_S$, and $m_s$. However, I think that's a topic for another question, if necessary.


What does it mean by 'the residue was taken up in solvent' in synthesis protocol? The below is a synthetic protocol I need to perform:

N-trifluoro acetoxy succinimide Synthesis
Trifluoroacetic acid anhydride was added to N-hydroxysuccinimide at 0 °C. The reaction was allowed to warm up to r.t. over 24 h. Then toluene (50 mL) was added and the volatile was removed in vacuo. The residue was taken up in DCM (15 mL) and the remaining solvent was removed in vacuo.

I don't quite get the last sentence. What does it exactly mean by 'taking up a residue in solvent'? Does it mean I just need to add 15 mL DCM to the reacting mixture?
Also, the second bit also gets me confused — if I put my mixture under vacuum, wouldn't all the solvents get sucked up? what would be the point of adding DCM if I am placing my mixture under vacuum immediately after adding DCM?

You add the DCM to the material left after removing the toluene and then evaporate it off under vacuum. The reason is to ensure that all the trifluoroacetic acid is removed as this would adversely affect the next reaction, and TFA is readily soluble in DCM.