Balancing a reaction where dichromate ion reacts with ethanol to give Cr(III) and carbon dioxide I have only started to learn balancing redox reactions, and a recent question has me confused.
The question is :

Balance in both acidic medium and basic medium:
$$\ce{Cr2O7^2- + C2H5OH -> Cr^3+ + CO2}$$

I usually look for symmetries between the reactant and product sides of a reaction to pair reactants with the corresponding products to form the half reactions, as per the first steps of balancing a redox reaction. I'm having a difficult time doing that here; what am I supposed to pair Cr with? I would personally pair it with the $\ce{Cr2O7^2-}$ molecule but that's just by process of elimination because I would naturally pair $\ce{C2H5OH}$ with $\ce{CO2}$ due to the stronger resemblance they have with each other.
Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:
$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
In a basic solution:
$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$
It seems like in the answer for the reaction in acidic medium, the left side of the reaction has a charge of +14 and the right side has a charge of +12.
However,  in the basic medium the answer has the left side with a total charge of -2 and the right side -4. How can these answers be correct if the charge isn't balanced?
How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.

Let's get the simple question out of the way first:

How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.

You can not balance it then. Also, you wouldn't be given such a reaction and is always a typo when they've said that it is a redox reaction.

Now, onto the bit more complicated question.

Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:
$$\require{cancel}\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
In a basic solution:
$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$
[sic] It seems like in the answer for the reaction in acidic medium, the left side of the reaction has a charge of +14 and the right side has a charge of +12.
However,  in the basic medium the answer has the left side with a total charge of -2 and the right side -4. How can these answers be correct if the charge isn't balanced?

A simple explanation would be that the excess charges that you feel are there are actually not. Try breaking the given reaction in the answer into each atom with their own oxidation state. I'll do the one for acidic medium, and the other will be up to the reader.
$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
Here, we start from the reactant side, we assume the following for simplicity:

$\ce{C-H}$ bond leads to positive charge on $\ce{H}$ and a negative charge on $\ce{C}$
$\ce{Cr-O}$ bond leads to positive charge on $\ce{Cr}$ and a negative charge on $\ce{O}$
$\ce{C-O}$ bond leads to positive charge on $\ce{C}$ and a negative charge on $\ce{O}$

\begin{array}{lc} \hline
&\text{Reactants} &\text{Products} \\ \hline
\ce{Cr} & \ce{Cr^6+} & \ce{Cr^3+} \\
\ce{O} & \ce{O^2-} & \ce{O^2-} \\
\ce{H} & \ce{H+} & \ce{H+} \\
\ce{C} & \ce{C-,C^3-}  & \ce{C^4+} \\ \hline
\end{array}
Now, we split the actual equation into this form:
$$\ce{4Cr^6+ + \cancel{22\mathrm H^+} + C- + C^3- +\cancel{15\mathrm O^{2-}} -> 4Cr^3+ + 2C^4+ + \cancel{15\mathrm O^{2-}} + \cancel{22\mathrm H^+}}$$
Which simplifies the reaction to:
$$\ce{4Cr^6+ +C- +C^3- -> 4Cr^3+ +2C^4+}$$
Adding the charges on both sides we get:
\begin{align}
\text{Charge on reactant}=(4\times 6 + (-1) + (-3)) = 20 \\
\text{Charge on products}=(4\times 3 + 2\times4)= 20
\end{align}
Therefore the charge on both sides are equal ...
Note: The reaction in basic medium uses $\ce{CrO4^2-}$ and not $\ce{Cr2O7^2-}$
$$\ce{4CrO4^2- + 7H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 20OH-}$$


How easy it is for MoS2 to become oxidized I made I high pressurized $\ce{MoS2}$ pellet, using a 6 ton pressurizer. I used very thin powder to create the pellet.

I am using this pellet on beam, with some nuclear techniques. In my spectra I can see oxygen.

How easy it is for $\ce{MoS2}$ to become oxidized?

It seems that the problem has been examined even 50 years ago, probably in order to determine if/how oxidation would alter the the properties of $\ce{MoS2}$ as a lubricant.

Here is an old NASA document from 1949, and in Surface Oxidation of Molybdenum Disulfide from 1955, the authors state in the abstract, that the oxidation is limited to the outmost surface layer, which then passivates and protects the rest of the batch.


What is the difference between fire and warm body? A burning body has flames (Candle flame)whereas a hot or warm body (warm utensil or body being hot but doesn’t burn )is just very hot. Keeping hand near a burning body makes you feel hot but not exactly near a hot body.A burning body has flames which are yellow,then blue and purple for some outside in colour.
What exactly is a burning body and how does it differ from a warm-body?
Please tell how can I improve my question if not suitable ?

You're really asking about the difference between a physical change and a chemical change.*
In a physical change, the chemical identity of the substance doesn't change.  In a chemical change, it does.
Suppose you take cube of sucrose (table sugar) at room temperature.  If you heat it up, it will become warmer.  But it will still be, chemically, sucrose.  It will just be sucrose at a higher temperature.  Since the chemical identity did not change, it is a physical change:
$$\ce{C12H22O11(s, room\,T)  -> C12H22O11(s, warm\,T)}$$
[$\ce{C12H22O11}$ is the chemical formula for sucrose; "s" means solid, "l" means liquid, "g" means gas.]
If you warm it more (to $170 \,^\circ \text{C}$) it will melt.  Melting is a physical change, because after the cube melts, it still consists of sucrose molecules; though these will now be in the liquid state:
$$\ce{C12H22O11(s)  -> C12H22O11(l)}$$
But suppose you instead flow enough energy into it to break the bonds that hold the sucrose together, causing it to combust, in the present of oxygen, into water and carbon dioxide.  That is what happens when it burns.  This is a chemical change because, as it is happening, the sucrose and oxygen are being changed into new chemicals: water and carbon dioxide:
$$\ce{C12H22O11(s) + 12 O2(g) -> 12CO2(g) + 11 H2O(l)}$$
This chemical change releases energy, which (in this case) takes the form of heat and light (which is why it feels hot, and you can see a flame).
In sum, a warming body is undergoing a physical change.  A burning body is undergoing a chemical change.
[*Yes, some might take issue with this distinction, but it is used in all first-year chemistry textbooks I've seen, indicating there is a consensus among educators in this field that this is a useful distinction for beginning students to learn.]


How to calculate the rate of consumption of aluminium with hydrochloric acid? An experiment is done to determine the rate of the following reaction:
  $$\ce{2 Al (s) + 6 HCl (aq) -> 3 H2 (g) + 2 AlCl3 (aq)}$$

  The following data is collected:
  \begin{array}{rr}
\text{time}/\pu{s} & (\text{Mass of beaker + contents})/\pu{g}\\\hline
   0.0 & 270.230\\
  30.0 & 270.200\\
  60.0 & 270.170 \end{array}

  Calculate the rate of consumption of $\ce{Al}$ in $\pu{g/min}$.

The answer is $\pu{0.534 g/min}$, but I don't know how to get there.

Since one of the products is a gas $\ce{H2}$, the decrease in system mass must reflect the escape of hydrogen. Thus, after thirty seconds, the system has lost $\pu{0.030 g}$ of hydrogen, which is $\pu{0.01488 mol}$ of $\ce{H2}$.

Thus, the rate $r$ of hydrogen production is:

$$
r (\ce{H2})
= \frac{\Delta n(\ce{H2})}{\Delta t}
= \frac{\pu{0.01488 mol}}{\pu{30 s}}
= \pu{4.96E-4 mol//s}$$

Now, how can you use this rate and the stoichiometry of the reaction to determine the rate of aluminum consumption?

   $\ce{2 Al (s) + 6 HCl (aq) -> 3 H2 (g) + 2 AlCl3 (aq)}$
 For every three moles of hydrogen produced two moles of aluminium are consumed, therefore $r(\ce{Al}) = -\frac{2}{3}r(\ce{H2}) = \pu{3.31E-4 mol//s}$.


Doesn't the isothermal irreversible expansion/compression of an ideal gas law violate the ideal gas law? Let's say a cylinder has some ideal gas. We use a piston to suddenly expand the gas in isothermal condition from $1\ \mathrm{atm}$, $20\ \mathrm L$ to $0.5\ \mathrm{atm}$, $50\ \mathrm L$.

Since the gas is ideal, $pV = nRT$ implying that $nRT = 20\ \mathrm{L\ atm}$ initially. Later $pV$ changes to $25\ \mathrm{L\ atm}$. However, $nRT$ has to continue to be $20\ \mathrm{L\ atm}$ as the number of moles and the temperature remain constant.

Doesn't this violate the ideal gas equation, as $pV$ and $nRT$ are no longer equal. Does this mean that the irreversible expansion or compression of an ideal gas can not be isothermal unless the number of moles change (through some chemical reaction)?

The ideal gas law describes the behavior of an ideal gas exclusively for thermodynamic equilibrium states.  During an irreversible expansion or compression, the gas passes through non-thermodynamic-equilibrium states, except at the very beginning and at the very end.  In these irreversible non-equilibrium states, the ideal gas law does not correctly describe the gas behavior.  For rapid expansions and compressions, the force that the gas exerts on the piston depends not only on the gas volume, but also on the rate of change of gas volume.  So yes, in irreversible/expansion of an ideal gas, the ideal gas law is violated.

That said, if we can control the force between the gas and the piston by some external means (say by using an automatic control system in conjunction with a flush mounted pressure transducer in the inner piston face), we can calculate the amount of work that the gas does on the piston (without trying to resort to the ideal gas law) even for an irreversible expansion/compression.  This then would enable us to use the first law of thermodynamics to calculate the final state of the gas (at least for an adiabatic process).


Development of experiments to optimize HPLC-MS procedure using 2D central composite full factorial design I have recently been studying development of experiments and how to statistically plan/optimize an experiment for x number of variables based on y number of responses.

To begin with, I am just going to use a 2D central composite full factorial design (CCD) to make sure I don't screw up bigger designs later. My two factors are 'load pH' and 'load conductivity'. My min/max values will be 6.5 - 7.5 and 5.5 - 6.5 respectively. My alpha value = sqrt(2) and I have scaled my model so far to (from min to max):

$$
-\sqrt{2}, -1, 0, +1, \sqrt{2}
$$

In a 2D CCD where 9 experiments are needed (depending on number of centre points), what relevance are the -1 and +1 values? To clarify, if my 2D array is effectively a 3x3 grid then the x and y axis will respectively read $-\sqrt{2}, 0, +\sqrt{2}$. What is the purpose of -1/+1?

FCC is a variant of a CCD design. It's usual to run your 2 factorial design first and establish which factors are relevant.  In your case, it won't matter so much but if you have 5 or more factors, it does.  The 2 factorial establishes your range with an upper and lower limit (+1 and -1). When you move to CCD,  you now have options. Using alpha of root 2 gives your "response surface design" orthogonality and rotatability, meaning, loosely, it is easier to analyse for variance in your model. The reason for varying alpha usually comes down to operational reasons e.g if a negative time or negative mass would result.  This is when a FCC might be used instead. If you ignored your coded +1 and -1, in effect your root 2 values would become your new +1 and - 1. The rationale for this is when a formula is generated to model each response,  the one using coded factors (set at +1 or -1) will indicate the relative influence of each factor, which the actual factor values will not do. The formula using coded values will not base it on your alpha value.


What would be the effect of the addition of an inert gas to a reaction at equilibrium? Why does a dissociation reaction shift to the right with the addition of an inert gas?

Dissociation obviously increases the number of moles.
The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.

There are two cases on which equilibrium depends. These are:

Addition of an inert gas at constant volume:
When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.

Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

Addition of an inert gas at constant pressure:
When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Consider the following reaction in equilibrium:
$$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$
The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants.

Please read about the Effect of adding an Inert Gas


What is the molecular structure of polymetatelluric acid? Telluric Acid, $\ce{Te(OH)6}$, dehydrates to form polymetatelluric acid: $\ce{(H2TeO4)10}$. However, I have been unable to locate the exact structure of polymetatelluric acid, and it does not seem entirely obvious.

Are the tellurium centers connected directly to each other or are they separated by oxygen? Are these polymeric bonds single bonds or double bonds or some mixture of the two? My best guess is that the $\ce{Te}$ centers are separated by oxygens with three other oxygens surreounding each $\ce{Te}$ center, two of which are $\ce{-OH}$ groups and the third being a $\ce{=O}$ group: this would make the oxygens between the $\ce{Te}$ centers alternating single and then double bonds — possibly conjugated on some level. Is there any chance polymetatelluic acid has direct $\ce{Te-Te}$ bonds?

It would be ideal to check out x-ray analysis data, but it seems like polymetatelluric acid is a rather amorphous substance and does not form suitable crystals. As of now, I haven't discovered anything relevant neither in ICSD, nor in CCDC.

Another method of structural analysis would be vibrational spectroscopy. There is an article "Ultrarotspektren von Tellursäuren, Telluraten und Antimonaten" [1, p. 165] (in German) containing an extensive research of various $\ce{Te}$-containing substances performed with infrared and Raman-spectroscopy (I reduced the table to the frequencies for polymetatelluric acid exclusively):

  \begin{array}{ll}
\hline
\ce{(H2TeO4)_x} &\text{Zuordnung} \\
\hline
450~\mathrm{(m, b)} & \delta~(\ce{TeO}) \\
\hline
600~\mathrm{(st, b, Sch)} & \nu~(\ce{TeO}) \\
720~\mathrm{(sst, b)} & \\
600~\mathrm{(sst, b, Sch)} & \\
\hline
1085~\mathrm{(mst, b)} & \delta~(\ce{TeOH}) \\
\hline
1618~\mathrm{(s, b)} & \delta~(\ce{H2O}) \\
\hline
2360~\mathrm{(m)} & 2\delta~(\ce{TeOH}) \\
\hline
3200~\mathrm{(sst, b)} & \nu~(\ce{TeOH}) \\
3360~\mathrm{(sst, b, Sch)} & \nu~(\ce{H2O}) \\
\hline
\end{array}

  Polymetatellursäure $\ce{(H2TeO4)_x}$ ist eine feste amorphe Substanz. In ihrem UR-Spektrum sind alle Banden sehr breit, wie man es bei amorphen Substanzen meist antrifft. Das Tellur ist in dieser Verbindung wahrscheinlich wie im $\ce{Te(OH)6}$ sechsfach koordiniert, wie die annähernd gleiche Lage der $\ce{TeO}$-Valenz- und Deformationsschwingungen in beiden Verbindungen zeigt. Hieraus folgt, daß $\ce{TeOTe}$-Bindungen vorhanden sein müssen. Bei Viererkoordination würden die Valenzschwingungen höher liegen, wie es in den vergleichbaren Molekeln $\ce{IO}$, und $\ce{H5IO6}$ der Fall ist.[...]

  Im Gegensatz zum $\ce{Te(OH)6}$ tritt im Spektrum des $\ce{(H2TeO4)_x}$ eine Frequenz bei $\pu{1618 cm-1}$ auf, die nur von freiem Wasser herrühren kann. Ebenso beobachtet man neben der $\ce{OH}$-Valenzschwingung der $\ce{TeOH}$-Gruppen ($\pu{3200 cm-1}$) die Bande $\pu{3360 cm-1}$ des freien Wassers. Aus den Intensitäten dieser Banden ist zu entnehmen, daß $\ce{H2O}$ und $\ce{TeOH}$-Gruppen in größenordnungsmäßig gleicher Menge vorhanden sind. Eine stöchiometrische Konstitutionsformel wird man dieser Substanz wohl nicht zuordnen können.

Briefly, the IR-spectra bands of $\ce{[H2TeO4}]_x$ bands are very broad, suggesting amorphous non-stoichiometric compound (as Mithoron suggested) with polymeric structure where $\ce{Te}$ atom is 6-coordinated. Bands were only assigned for $\ce{Te-O}$, $\ce{Te-OH}$ bonds and $\ce{H2O}$. IR-spectra do not show the presence of $\ce{Te-Te}$ bonds. Most likely there are $\ce{[TeO]6}$ units with edge- or corner-sharing, linked together in a chain-alike fashion.

Reference

Siebert, H. Z. anorg. allg. Chem. 1959, 301 (3–4), 161–170. DOI: 10.1002/zaac.19593010305 (in German).


Chemical Equilibria - Fraction Bound equation based on Kd, [R], [L], [RL], is actually an average? Let's say 150 stem cells express exactly 150 copies of the same mebrane antigen X. While all differentiated cells express exactly 15 copies of X. To the cell mixture you add a large amount of 0.2 μM solution of antibody to X. The $K_D=$ 4 μM.
If you experimentally analyze the number of antibodies bound per stem cell, you will find that it is not constant.  Explain why, even for cases where the number of antigens expressed is the same, the binding follows a binomial distribution?

I found the fraction bound to be $\frac{1}{\frac{K_D}{\left[L\right]}+1\:}=f_{bound}$

I scaled the [L] contribution by the number of antigens expressed per cell, so for the differentiated case, I scaled 0.2 μM by 15, and for the normal case, I scaled 0.2 μM by 150.
the differentiated case has a lower $f_{bound}$ because it has less ligands.

To describe the binomial distribution for binding part

the antigens exist in the environment of the beaker (or wherever the reaction is taking place) so they can exchange antigens? resulting in some cells $f_{bound}$ being lower. How exactly the interaction takes place occurs stochastically.  <--- want help clarifying this part

When a reaction is at equilibrium, there will be stochastic fluctuations. These are usually not measurable (for example, if we have more than a mole of products and reactants in our reaction mixtures).

In this specific case, the fluctuations are easy to measure because the total number of antigen molecules per cell is countable, and you are separating cells (I presume) before analyzing the number of antibodies bound.

If you expect a fluctuation of $\sqrt{N}$ for N objects, that is significant for $N= 150$ but not for $N = \pu{6.022e23}$ when expressed relative to $N$.


Activities and molar concentrations I would be really happy if someone would clarify the term "chemical activity" and its implications for me, and also how it is related to the concept of molar concentration. My textbook fails at explaining this clearly and, sadly, the Internet is filled with vague definitions and scientific verbiage that's too difficult for me to grasp. Also, why are activities included in various formulae, and what are their uses?

I never quite actually understood what are activities and how do they differ from molar concentrations, nor how are they used. In addition, when working with the Nernst equation, why are we allowed to compute the activities of the reactants and the equilibrium concentrations of the products? This is the stuff that really baffled me and I'd be highly appreciative if you could clarify it.

Most thermodynamic equations are derived assuming a perfect gas or ideal solution. However, for real a solution the chemical potential $\mu_i=\mu^0_i +RT\ln(x_i)$ may not be accurate for the solvent where $\mu^0_i$ is the chemical potential of the liquid solvent at 1 atm and $x_i$ is the mole fraction. The solute may, similarly, not follow ideal behaviour.  The activity is introduced to allow us to use the familiar equations but now as $\mu=\mu^0_{liq} +RT\ln(\gamma x_i)$ where $\gamma$ is the activity coefficient thus $\mu_i=\mu^0_{liq} +RT\ln(a_i)$ where $a=\gamma x$. Thus we can think of activity as effective concentration or effective pressure (now called fugasity) relative to its standard state.  The activity coefficient is unity in the ideal case. The activity has to be measured experimentally because $\gamma$ depends on exactly what intermolecular interactions exits as any temperature and pressure and cannot be easily calculated.  As an example, naphthalene dissolved in benzene behaves almost ideally and the activity coefficient $\gamma = 1$ but when dissolved in hexane this is not the case and $\gamma=2.5$. A positive deviation from Raoult's law indicates that the vapour pressure is larger than that expected from perfect behaviour and so the activity is larger than the mole fraction and $\gamma >1$.


why are Ca(OH)₂ and Al(OH)₃ less soluble than Na(OH) in water Even though it has more $\ce{OH}$ groups and still it is very weak in terms of dissolving?

please clarify to me in terms of polar and non polar solubility, how to determine whether a compound is soluble or not in water.

$\ce{OH}$ is small anion and bonds effectively with small cations. Thus, in general, the bigger (and less specific charge) of the cation, the higher is solubility of the corresponding base in polar solvents, meaning that $\ce{Ca(OH)_2}$ and $\ce{Al(OH)_3}$ having much smaller and more charged cations are almost covalent in nature and thus, do not dissociate.

In general, solubility of salt in water is determined by energy balance between breaking of the ionic bonds and hydratation of anions formed. Thus, the most soluble salts are usually formed by small and big ions, that cannot bond effectively, but at least one of them can be hydrated effectively.

Please, note: solubility for molecular compounds, like sugars and some non-dissociating compounds that looks like salts (like $\ce{Cu(CH_3COO)_2*H_2O}$) is an entirely different matter and determined by relative strength of intermolecular interactions (read: the molecule have to be able to form hydrogen bonds)


AgI and AgCl reduction potentials The reduction potential for $\ce{AgI}$ is $-0.15\ \mathrm{V}$ and for $\ce{AgCl}$ it’s $0.22\ \mathrm{V}$. As $nFE =-\Delta{G} = T\Delta{S} - \Delta{H}$ this must imply that $\ce{AgI}$ is a stronger bond in order for it to be negative.

I would think $\ce{AgI}$ is weaker due to having a much larger ionic radii for $\ce{I-}$ than $\ce{Cl-}$ and the lack of covalence due to a smaller $\Delta\chi$

I think that it must be something to do with the size of the $\ce{I-}$ mean that it causes significant ordering of the water molecules in the solution and it's this entropic effect that is actually the key idea as otherwize I can't rationalise it at all.

The answer is hard-soft acid/base theory (or HSAB theory). Essentially, some Lewis acids and bases are hard and some are soft, with a few borderline cases.

  Hard acids and bases tend to have

Small atomic/ionic radius
Highly charges (or highly positive/negative oxidation state)
High electronegativity for bases and low electronegativity for acids
Low polarizability

  Soft acids and bases tend to have

Large atomic/ionic radius
Low or zero charge / oxidation state
Moderate electronegativity
High polarizability

According to HSAB theory, matches that are hard-hard and soft-soft form strong interactions, where hard-soft matches are weaker interactions.

$\ce{Ag+}$ is a soft Lewis acid. $\ce{Cl-}$ is a hard Lewis base, while $\ce{I-}$ is a soft Lewis base. Therefore, we would predict that $\ce{Ag+}$ and $\ce{I-}$ will have have a stronger attraction than than $\ce{Ag+}$ and $\ce{Cl-}$.

We can see this difference in attraction in the solubilities of $\ce{AgCl}$ and $\ce{AgI}$ in water. $\ce{AgCl}$ is much more soluble in water than $\ce{AgI}$ (realizing that both are nearly insoluble). See the solubility constant data:

$\ce{AgCl}\ \ \ k_{\text{sp}}=1.8\times 10^{-10}$
$\ce{AgI}\ \ \ \ \ \ k_{\text{sp}}=8.5\times 10^{-17}$


When there is an increase in temperature, how is the pH of the ocean affected? (Question is related to ocean acidification) This is my current understanding:
Ocean acidification is the result of atmospheric $\ce{CO2}$  dissolving in the ocean's water. When this occurs the $\mathrm{pH}$ of the ocean decreases (from the increase in $\ce{H3O+}$ ions produced from the dissociation of $\ce{H2CO3}$).
When there is an increase in temperature, the solubility of carbon dioxide decreases which results in a lower concentration of carbonic acid (the formation of carbonic acid is from the reaction of $\ce{CO2}$ with $\ce{H2O}$ which is exothermic, so the reverse reaction is favoured).
So since the concentration of carbonic acid is lower, the concentration of $\ce{H3O+}$ ions decreases (lower concentration of acid dissociates in the water).
Therefore, the $\mathrm{pH}$ of the ocean increases as there are fewer $\ce{H3O+}$ ions present in the water.

Is my understanding correct, or have I made a mistake in my explanation?
If you can help me out I would be very grateful :)

You have to compare the size of these counteracting effects.

The temperature increase slightly decreases the acidification caused by increased carbon dioxide partial pressure.

The shift of reaction constant is minimal.

From the Wikipedia carbon dioxide data page, Water solubility section, we can see for temperature  15 resp 17 $^\circ$C V/V solubility 1.019 resp. 0.956. That is 6.2% lower solubility.

Increased partial $\ce{CO2}$ pressure from 0.03 to 0.04% gives 33%, what is significantly more than by 6.2% decreased solubility.

Aside of that, higher $\ce{CO2}$ concentration and lower sea water $\mathrm{pH}$ are already observed.

This is directly affecting carbonate related equilibriums, particularly in context of forming of mollusk shells.

Some mollusks are already observed  having troubles in forming aragonite - the more soluble form of calcium carbonate.


A problem regarding solubility and concentration A solution is $\pu{0.10 M}$ $\ce{Ba(NO3)2}$ and $\pu{0.10 M}$ $\ce{Sr(NO3)2}$. If solid $\ce{Na2CrO4}$ is added to the solution, what is the $\ce{[Ba+]}$ when $\ce{SrCrO4}$ begins to precipitate?  $\mathrm{K_{sp}}=\pu{1.2 \times 10^{-10}} \text{ for } \ce{BaCrO4}$
$\mathrm{K_{sp}}=\pu{3.5 \times 10^{-5}} \text{ for } \ce{SrCrO4}$

The solution given in my textbook is:

$\mathrm{K_{sp1}}=\ce{[Sr^2+][CrO4^2-]}$ $$\ce{[CrO4^2-]}=\frac{3.5\times 10^{-5}}{0.1}=3.5\times 10^{-4} \pu{M}\tag{1}$$ $\mathrm{K_{sp2}}=\ce{[Ba^2+][CrO4^2-]}$ $\ce{[CrO4^2-]_{total}}\approx\ce{[CrO4^2-]}\text{ from } \ce{SrCrO4}$ $$\ce{[Ba^2+}]=\frac{1.2\times 10^{-10}}{3.5\times 10^{-4}}=3.4\times 10^{-7}\pu{M}$$

First of all, in equation $(1)$ why did they put $\ce{[Sr^2+]}$ same as the initial concentration (before the addition of $\ce{Na2CrO4}$)? When $\ce{Na2CrO4}$ is added,  $\ce{[Sr^2+]}$ will react with $\ce{CrO4^2-}$ (obtained from $\ce{Na2CrO4}$) to form precipitate. The concentration of $\ce{[Sr^2+]}$ will undoubtedly change until the equilibrium is established.
This given solution is totally confusing, I would appreciate if you could tell me the proper way of solving this problem. I tried this on my own, but after calculating the concentration of $\ce{CrO4^2-}$, I couldn't make sense out of this question.

It is supposed addition of solid $\ce{Na2CrO4}$ would cause negligible volume change, so we can afford to consider the same $[\ce{Sr^2+}]$ as the initial one.
The task is rather theoretical, supposing the time point when $\ce{[Sr^2+][CrO4^2-]}$ just reached $K_\mathrm{sp,\ce{SrCrO4}}$, while $K_\mathrm{sp,\ce{SrCrO4}}$ and $\ce{[Sr^2+]}$ are given.
The proper computation procedure is already presented in you solved task. What  else is unclear ?


Chemical bonding and dipole moment We know that ammonia undergoes amine inversion. Why doesn't the dipole moment decrease in ammonia, since the direction of the dipole changes to the opposite direction every time there's inversion?

Shouldn't the effective dipole moment become zero?

Ivan Neretin is correct in stating that no particle in a (nondegenerate) state can have a permanent electric dipole moment in the laboratory frame. In fact, a permanent dipole moment would require a violation of time (T) and parity (P) symmetry. This is because any such dipole should be directed along the angular momentum vector of the molecule and time-reversal and parity operators have a different effect on the rotational angular momentum and direction of the dipole.

What chemists call a permanent dipole moment refers to the dipole moment in the molecular point group to which the molecule belongs. Only when a polar molecule is subjected to an electric field (e.g. an EM wave or DC field), this field will mix rotational states of opposite parity in such a way that the molecule "orients" itself in the field. You can only speak of the inversion of ammonia in the presence of an electric field (the eigenstates of the Hamiltonian are time independent in the absence of a field or other perturbing force).

Although degenerate states could in principle have a permanent dipole moment, it was shown by Klemperer et al. (J. Phys. Chem. 1993,97, 2413) that the degeneracy of these states is removed by higher-order terms in the Hamiltonian.


Confusion about the electrolyte in an electrochemical cell So the cathode is magnesium and the anode copper. The question asks why silver nitrate would not be a good electrolyte. I am confused. Also is the anode always inert in these cells?

As Ed said, that both copper and magnesium will reduce silver ions to elemental silver (you have to justify why in your answer. Hint: Look at the electrochemical series and compare the position of Mg, Cu, and Ag half cells. Another fundamental aspect is that one cannot apply Nernst equation if the metal dips in anything else besides its ions.

For instance, think little dip deeper, can we calculate the potential of copper electrode dipping NaCl solution (wrt to some reference cell?). In this case sodium ions will not react with copper or magnesium.


3D Printing a molecular orbital Before I begin I know that what I am asking will deform the M.O but for what I am doing this for that is okay if it is only around the atoms and bonds.

My question is in JMOL is there anyway to display the isosurface of the molecular orbital and have it leave a gap around the molecule. I am going to a conference soon and need to print a model of a cyclic molecule. I would like to be able to "mount" the MO to the molecule using tiny magnets at certain points so I may dislodge the two phases and show the structure of the molecule. Yes using two different models (one with MOs and one without) is possible but not preferred.

I have explored isosurface command and different syntax and have not been able to find anything that may be of help.

What I ended up doing was using jmol to produce the different phases of the MO and then I used TinkerCAD and made the molecule itself a "hole" structure. This created the space to actually put the molecule inside the orbital.


Why are signals of carbon and hydrogen atoms (in NMR spectra) split by coupling to 19F? Why signals of carbon and hydrogen atoms (in NMR spectra) are split by coupling to $\ce{^{19}F}$? Some spectra books only states that

  $\ce{^{19}F}$ (natural abundance 100%) has a spin quantum number I of ½. The signals of carbon and H atoms up to a distance of about four bonds are split by coupling to $\ce{^{19}F}$.

but there is no explanation. Could someone please explain why $\ce{^{19}F}$ can couple with H or C and split their signals.

You always have a coupling if you have to NMR-active nuclei connected, there is nothing unusual about fluor here. But in the most common and simplest NMR experiments on small organic molecules you usually don't encounter such heteronuclear couplings.

The first important aspect is that many common nuclei in organic molecules are present mostly in an NMR-inactive isotope. Carbon and nitrogen for example are only present in the NMR-active isotopes 13C and 15N in tiny amounts. So you'll be able to see small satellite peaks sometimes, but no complete splitting. If only 1% of the nuclei is NMR-active, only 1% of the signal will be affected by the coupling.

The second aspect is that some experiments like 13C 1D experiments are typically recorded with decoupling. There is a coupling between 1H and 13C, but it is intentionally suppressed as it would complicate the spectra.


Ionisation vs Dissociation This is a statement given in my book:

A strong electrolyte is completely ionised at all dilutions but not completely dissociated.

But the book also uses the terms "Ionisation" and "Dissociation" interchangeably in other places.
When I further tried to search about this topic I found out there is equation which governs dilution of strong electrolyte-
$\lambda_m = \lambda_o + a\sqrt C$
So is it right to say that it will only get completely dissociated at infinite dilution (which is not practically possible)?
Hence my final questions are:

Are ionisation and dissociation the same or different?

If they are not the same then why are they different?

Also the correct explanation of the above statement given in my book?

Will this change for a weak electrolyte?

Also please consider that I am in high school. So please try to explain in simple terms so that I am able to clear my confusion.
Please also tell whether I can assume both the terms to be same or not at my current high school level.

The parameter a in front of √C is empirical. It varies from solute to solute.
And yes, the dissociation is complete at infinite dilution.
Dissociation is the separation of ions when dissolving a salt in water. If NaCl is dissociated, it yields two independent ions, one Na+ and one Cl−
Ionization is what happens when an atom or a molecule looses one electron, without being modified or broken. If $\ce{NaCl}$ is dissociated, it gives two independent ions, one $\ce{Na+}$ and one $\ce{Cl−}$
If $\ce{H2O}$ is ionized, it gives an ion $\ce{H2O}$+ and one electron, but the atoms of the molecule are not separated from one another.


Why can Cu have an oxidation number of +2? The electronic configuration for $\ce{Cu}$, adjusted for Hund's rule, is:

$[\ce{Ar}] 3d^{10} 4s^1$

So, shouldn't $\ce{Cu}$ have an oxidation number of $+1$? whereby it gives off its outermost 4s electron? Why can it also have a $+2$ charge?

Atomic copper has the electron configuration $\ce{[Ar] 3d^{10} 4s^1}$.  By removing one electron and producing $\ce{Cu^{+1}}$, an inert gas configuration $\ce{[Ar] 3d^{10} 4s^0}$ is produced.  While it does take a lot more energy to remove the second electron from copper (first IP=745 kJ/mol, second IP=1,958 kJ/mol), if this energy can be offset by the energy gained through bond formation and lattice energy (or hydration energy) then $\ce{Cu^{+2}}$ compounds will form.  In fact $\ce{Cu^{+2}}$ is the most common oxidation state of copper so the energetics must generally work out that the energy gained by forming more than one bond to copper and gaining additional lattice (or hydration) stabilization, more than offsets the energy cost of removing that second electron.  Removal of additional electrons can also occur with copper to form $\ce{Cu^{+3}}$ and $\ce{Cu^{+4}}$ compounds, for example, $\ce{K3CuF6}$ and $\ce{Cs2CuF6}$.


What type of test can I do to show that water is one of the products in a neutralisation reaction? I've completed an acid base titration experiment of $\ce{HCl}$ and $\ce{NaOH}$. This forms salt + water

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

Indicator was used and we evaporated the water to get the salt.

I've been asked to design a test to show that water was the other product of this reaction. I'm not sure what to do here. I'm thinking maybe a distillation type test but not sure where to go with this question …

I'm a middle school student and would like an answer that is of a middle to high school level if possible. Could you provide some examples of tests I could do, and explain how they work?

EDIT -

I really appreciate the answers guys/gals. I'm not going to pretend like they did not go over my head though!

I think the issue might be the wording of the question in the textbook. What I've written above is exactly how it was asked in the textbook but I get the feeling that the expected answer is much simpler considering the level of chemistry I'm dealing with. Even though I don't expect you to read all of this I've gone ahead and scanned the book page anyway. Below is the experiment I completed, and question 4 in the discussion section which is highlighted is what I am trying to answer.

As was written in comments, it is nigh impossible to detect the formation of more water in an aquaeous solution. Unless you coloured the oxygen atoms in $\ce{NaOH}$ pink or purple you would have no clue which water molecule is new and which isn’t.

Say you had a $1~\mathrm{M}$ solution of both acid and base and used $25~\mathrm{ml}$ each; you would create $25~\mathrm{mmol}~\ce{H2O}$ in the reaction or approximately $0.45~\mathrm{g}$. That’s just about $1~\%$ of the total water added, so you may be able to detect a slight increase in weight if you distilled away the water.

Do not try to get your hopes high, though.


Galvanic cell structure and function I was thinking about galvanic cells the other afternoon and I was wondering the situation
if a chemical engineer designs a galvanic cell using copper and zinc as electrodes, what would the structure, function and output voltage of the cell be, and how might you possibly increase the voltage?

If by voltage, you mean electric potential, then just analyze the nernst equation.

The standard reduction potential for both half-reactions is known, so add those to get the standard cell potential. Then realize that changing the concentrations of reactions and products (hence, the reaction quotient) can change the actual cell potential (the log term in the equation).


How quickly does ethanol evaporate? Suppose I open the drinks cabinet, and pour some of the strong stuff into a saucepan.

Now the boiling point of ethanol is lower than water, so if I boil the mixture for a while, all the alcohol will surely evaporate long before the water does.

But how quickly does the alcohol evaporate? Is it all gone by the time the water starts to boil? Or does some remain unless you boil the mixture for a long time? (How long?) Or maybe the alcohol is gone by the time the mixture starts to get slightly warm?

Certainly water and ethanol and all mixtures thereof will completely evaporate without boiling given enough time.

Water and ethanol form a postive azeotrope that is 95.6% ethanol and 4.4% water and boils at 78.2 degrees C.

If you boil a mixture of water-ethanol that has less than 95.6% ethanol, the concentration of water in the liquid phase will increase, but it will only asymptotically approach pure water as all the liquid boils away.  There will not be a time when all the ethanol is gone and you are left with a finite amount of pure water.

If you boil a mixture of water-ethanol that has more than 95.6% ethanol, the concentration of ethanol is the liquid phase will increase.


Anti-Markovnikov addition of HCl The following reaction has been troubling me for a while now.

Since peroxides won't give the anti-Markovnikov product when used with HCl, I cannot think of any reagents that would result in the product shown in the image above. There doesn't seem to be any Wagner–Meerwein rearrangement either. This left me wondering, what reagent(s) should be used to form the product shown above.

One way to do it would be do first do a hydroboration-oxidation reaction, which leads to the anti-Markovnikov addition of $\ce{H2O}$. So you will get an $\ce{-OH}$ group instead of $\ce{-Cl}$. Then, we can replace the $\ce{-OH}$ group with $\ce{-Cl}$ by treating it with $\ce{PCl3}$ or $\ce{PCl5}$.


Biodegrability of straight chain and branched chain detergents From NCERT Chemistry: Textbook for class XII, Part II [1, p. 461]:

Main problem that appears in the use of detergents is that if their hydrocarbon chain is highly branched, then bacteria cannot degrade this easily. Slow degradation of detergents leads to their accumulation. Effluents containing such detergents reach the rivers, ponds, etc. These persist in water even after sewage treatment and cause foaming in rivers, ponds and streams and their water gets polluted.
These days the branching of the hydrocarbon chain is controlled and kept to the minimum. Unbranched chains can be biodegraded more easily and hence pollution is prevented.

Why are straight chain detergents generally biodegradable while branched chain detergents are non biodegradable? Basically, how does branching affect biodegradation?
Is there any more detail/ better explanation than this or explanation from any other view point?
References

NCERT. Chemistry: Textbook for Class XII. Part II, Reprint: August 2019 Bhadrapada 1941; M. Siraj Anwar, Shveta Uppal, Arun Chitkara, Bibash Kumar Das, R.N. Bhardwaj, Mukesh Gaur, Series Eds.; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-716-7. (PDF)

According to Scott and Jones1:

... problems arise with branched alkyl chains, a side chain methyl group or a gem-dimethyl-branched chain cannot undergo β-oxidation by microorganisms and must be degraded by loss of one carbon atom at a time (α-oxidation).

* text by the authors, highlighting and hyperlinking by me.
A study by Whyte et al2 on the degradation of alkanes by Rhodococcus yielded similar findings:

After 28 days of growth at 5°C in MSM containing 0.1% diesel fuel, Rhodococcus sp. strain Q15 degraded almost all the n-alkanes (C10 to C21) and, to a lesser degree, many of the branched alkanes found in diesel fuel.

P.s: both of these articles are open access, and explain the biodegradation process quite well. You can refer to them directly if you need more detail or specific examples.
References:

Scott, Matthew J., and Malcolm N. Jones. “The Biodegradation of Surfactants in the Environment.” Biochimica et Biophysica Acta (BBA) - Biomembranes, vol. 1508, no. 1–2, 2000, pp. 235–51. doi:10.1016/S0304-4157(00)00013-7.
Whyte, L. G., et al. “Biodegradation of Variable-Chain-Length Alkanes at Low Temperatures by a Psychrotrophic Rhodococcus Sp.” Applied and Environmental Microbiology, vol. 64, no. 7, 1998, pp. 2578–84. PMID: 9647833


Emission Spectrum of Hydrogen and Helium? Why do hydrogen and helium have different emission spectra? I thought that emission spectra were the result of electrons 'jumping' from one energy level to the next - since hydrogen and helium both have electrons on n=1, l=0 and only there, why are their emission spectra different?

We have solved the Schrödinger equation for hydrogen-like atoms such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, and $\ce{B^4+}$. The energy levels are neatly arranged:

$$E_n = - \frac {E_0} {n^2}$$

where:

$E_n$ is the energy of the $n$-th energy level
$n$ is a positive integer ($1$, $2$, $\dots$)
$E_0 = 13.6~\mathrm{eV}$

This is possible because we only need to consider two particles: the nucleus which has a positive charge and the electron which has a negative charge. We only need to consider the interaction between those two particles. Also, the nucleus is not moving, which simplifies some calculations

However, for higher atoms such as helium, where there are more than one electron, the Schrödinger equation contains two terms for the two nucleus-electron attractions and a term for the electron-electron repulsion. The term for the electron-electron repulsion makes things difficult because both electrons can be moved. That term makes the equation impossible to solve analytically.

For hydrogen, the energy of an energy level is determined solely by its principal quantum number. That means, the $2s$ orbital has the same energy as the $2p$ orbital (ignoring the virtual particle effect).

This is not true in higher atoms such as helium. The energy levels of helium can be seen here:

(The energy levels of hydrogen is on the right of the diagram for comparison.)

Here, the azimuthal quantum number has an effect on the energy of the energy levels, which makes for more possible differences between energy levels, which makes for more lines as well as different on the emission spectrum.


Can alkyl-substituted phenols be reduced to alkyl-substituted benzenes by zinc dust? I know that phenol can be reduced to benzene by distilling it with zinc dust. But I want to know if that concept can be expanded. E.g., take chavicol or 4-(prop-2-en-1-yl)phenol; could you still remove the hydroxy group by distilling it with zinc dust?

This is the basic principle of functional groups. Organic chemists quickly noticed historically, that some compounds reacted in the same way when presented with the same chemicals; thus allowing them to deduce that it was only a certain functional group (‘radical’ in historic terminology) that was reacting while the remainder of the molecule remained untouched.

Unless conditions or reagents are known to also attack other functional groups, they will typically be unharmed in a reaction and whichever features your molecule has, it will still react in the desired manner. For example, if phenol gets reduced to benzene by distillation with zinc dust, we can be pretty sure that cresols (e.g. paracresol, 4-methylphenol) will react in the same way, only that the final product will be toluene rather than benzene.

As I am not aware of zinc dust reacting with a double bond — the only other functional group present in chavicol — I am willing to say that it will indeed lose the hydroxy group and be reduced to allylbenzene as you desire.


Sodium Silicide (NaSi) Synthesis, Reversability I'm trying to understand a) the reaction that would produce $\ce{NaSi}$ from $\ce{Na}$ and $\ce{Si}$, as well as how to reverse the reaction of $\ce{NaSi}$ with $\ce{H2O}$. The product of the latter would be $\ce{Na2Si2O}$ (aq), and I want to understand the reaction which would transform it back to $\ce{NaSi}$ (including heat inputs).

Where could I find this information? Is there an obvious repository of chemistry information where I could look this up?

Thanks you!

The reaction of $\ce{NaSi}$ with water is both highly enthalpically and entropically favored—the reaction produces 5 molecules of $\ce{H2}$ gas for every 2 $\ce{NaSi}$ molecules that react, and the enthalpy of reaction is –175 $\mathrm{kJ\ mol^{-1}}$. The reverse reaction would require ludicrous amounts of energy and is likely not even kinetically feasible.

It would be especially pointless to attempt the reverse reaction when $\ce{NaSi}$ is generated easily enough from $\ce{Na}$ and silica gel. Liquid sodium-potassium alloy absorbed onto silica gel results in a fine black powder that spontaneously ignites in humid air. Heating $\ce{Na}$ and silica gel to 165ºC with continual agitation produces a shelf stable, yet still highly reactive $\ce{NaSi}$. When slowly heated to 400ºC, air stable $\ce{NaSi}$ may be produced, though with less reducing power than those produced via the other two methods. It will, however, still react with water to produce $\ce{H2}$ and $\ce{ Na2Si2O5}$.$^{[1]}$

$^{[1]}$ Alkali Metals Plus Silica Gel:  Powerful Reducing Agents and Convenient Hydrogen Sources


Is it safe to use lab-grade calcium nitrate as a fertilizer? I want to use lab-grade calcium nitrate in my hydroponic system (as a fertilizer) to grow lettuce, is it safe?

Usually you can buy chemicals in different purity grades.
If you buy something that has the grade purum >95% will be your desired ingredient. However there is not usually any information about what the other <5% will consist of.

Personally I would never ingest (or use in food production) anything that is not food grade. Food grade does not necessarily mean very pure, it just means that there is nothing in the other x% that is potentially harmful. Look out for an FCC number, that is the Food Chemical Codex, when you order it and you should be fine.


If graphite's just a huge polymer of benzene why isn't it an aromatic compound (polymer?) My teacher had once told me that graphite was just an insanely huge polymer of benzene (Those weren't his exact words but the gist). Today While we were studying aromaticity in organic compounds we steadily advanced from benzene to napthalene the annulenes and so on. What really stuck me was how as the number of  benzene rings increased the stuff tended towards graphite (atleast thats what the structure seemed to indicate). So the main question is why isn't graphite aromatic if its just a big collection of benzene molecules

Graphite/graphene is aromatic. The first image below is a common crystallographic representation of graphite and does not account for covalent bonding.  The second one shows the molecular structure of graphite as a kekule representation. Graphite is made of planar sheets of carbon atoms in 6-atoms rings with conjugated double bonds. This also satisfies the 4n+2 π electrons criteria needed within the aromatic ring for graphite to be aromatic. This aromaticity is why graphite is not only stable to high temperatures but also a more thermodynamically favorable allotrope than diamond under typical conditions.

Crystallogrphic Representation of Graphite.

Kekule Representation of Graphite.


Is the effectiveness of chemical catalysts affected by temperature the same way enzymes are? Enzymes do not need high temperatures to function while metal catalysts tend to need a bit of an energy kick to get going.

Enzymes will denature, or break, if heated up too far beyond their optimum temperature (for most around 40 degrees, although some bacterial enzymes can work at 100 degrees).

Do chemical catalysts also have an ideal operating temperature?
I am not asking if the rate of the reaction itself speeds up when a chemical catalyst is used. Rather, if the chemical catalyst becomes more effective at certain temperatures.

This is a somewhat complex question to answer because even your ‘simple’ enzyme example isn’t as simple as it looks. Let me discuss enzyme optimum temperatures first.

For any reaction that is exergonic — and that explicitly includes enzyme catalysis — an increase in temperature raises the reaction rate. This is due to at least two microscopic features:

higher temperature means more and faster movement thus a higher probability of collisions
each collision has a greater chance of being energetic enough to supply the required activation energy

This is condensed in van ’t Hoff’s postulate that an increase of the reaction temperature by $10~\mathrm{^\circ C}$ leads to an increase of the reaction rate by a factor of $2$ to $4$. Therefore, we would initially expect all enzymatic processes to be faster with increased temperatures.

However, the enzyme itself is a highly fragile system that is held together a good deal by noncovalent interactions. These can break apart under elevated temperatures causing the well-balanced enzymatic structure to break down (‘denature’)  into a seemingly random protein chain. This also breaks the catalytic centre meaning that the enzyme can no longer catalyse efficiently. Taking these two effects together, we arrive at the optimum curve of enzyme catalysis.

Chemical catalysts come in a number of shapes and varieties. Consider for example:

the catalyst (mainly platinum as I heard at school) in your car’s exhaust
the complex catalytic oxide system in the Haber-Bosch process
transition metal catalysts such as $\ce{[Pd(PPh3)4]}$
acid catalysis, i.e. dissolved $\ce{H+}$ is the catalyst
organic catalysis
…

Most of these will have some equivalent of a denaturation temperature; e.g. once platinum metal (or palladium, if on charcoal for typical organic hydrogenations) melts, it can no longer act as a catalyst for the reaction in the way it could in solid state. Thus, there should be a ‘denaturation equivalent temperature’ for most chemical catalyst systems at some point. Some compounds — the Haber-Bosch catalysts jump to mind — will, however, probably melt at such high temperatures that you would need to live on the sun’s surface for that to happen. Therefore, we could say that no limit exists for all intents and purposes.

Catalysts such as tetrakis(triphenylphosphane)palladium(0) need to dissociate at least one ligand to be catalytically active. An increase in temperature should enhance that dissociation. Therefore, many homogenously palladium-catalysed reactions are indeed sped up by a moderate increase in temperature. However, side reactions also speed up.

Organocatalysts and simple proton catalysis should be among the most temperature-insensitive — the former because all bonds should be covalent, the latter because it is just a reactive, activating species that cannot be poisoned. However, these two require a medium in which they act; the boiling point of the solvent is the highest possible temperature for their reactions.

Thus basically: yes, all chemical catalysts have some maximum useable temperature somewhere, but for most it does not matter for practical intents and purposes. Also, there are no simple models to explain this like in the enzyme case, hence it isn’t typically taught.


Why are hydrides of transition metals called hydrogen-deficient? My textbook states:

Metallic hydrides are formed by many d- and f-block elements. Unlike saline hydrides, they are almost always non-stoichiometric, being deficient in hydrogen, for example, LaH$_{2.87}$, YbH$_{2.55}$, etc.

After some research online, I found that non-stoichiometric compounds are so called because they don't follow the law of definite proportions, that is, "their elemental composition cannot be represented by a ratio of well-defined natural numbers."
From what I understand, some transition metals form hydrogen bonds with H atoms that fill interstitial spaces between metal atoms in crystal lattices. However, what does "hydrogen-deficient" mean in this context? Apologies if this is an elementary question, but I could not find an answer online.

As you note, many hydride phases have the hydrogen occupying interstitial sites within a metal phase, not unlike many carbide/nitride/oxide phases as well. In theory this could range from no hydrogen at all up to full occupancy (and perhaps even double occupancy).
Since you mentioned lanthanum hydride, I pulled up a Calphad analysis for the H-La-Ni system, focusing on the H-La part. Hydrogen has almost no solubility in dhcp-La, yet shows a wide solubility range in both fcc-La and bcc-La phases.
Now for the hydride phase, where I'll quote from section 3.4 in the paper above.

The binary LaH$_{x}$ hydride has an fcc structure... Usually, La atoms occupy the 4a sites, while hydrogen atoms occupy the tetrahedral 8c and octahedral 4b sites. Experimental determinations using using several different techniques ... show that hydrogen atoms occupy tetrahedral sites first, for concentration up to $H/M = 1.95$, and only above this concentration value octahedral sites begin to be occupied.

Their calculated diagram is:

So, hydrogen happily fills in to interstitial spaces, but prefers some more than others. Eventually (perhaps strain, perhaps getting too close to each other) other sites become more desirable for the additional hydrogen. In any event, often a quite broad range of compositions are stable, and perhaps (depending on the metal) you never make it to the ideal stoichiometric composition.


Why does Hydrogen Peroxide decompose? I understand that Hydrogen Peroxide decomposes into oxygen and water over time. But why exactly does it do this? Could someone give me an explanation down to the molecular level as to why and how this happens. And also, why does MnO2 catalyse this.

One could say that hydrogen peroxide decomposes because it is unstable. But that would be cheating because this is simply restating the fact that we observe it to decompose in fancier language.

We could use some statistics about the reaction, like stating that the decomposition is thermodynamically favourable (this doesn't explain why a catalyst works but it helps to explain why the reaction happens). But, again, this is just a sophisticated way to repackage the observed fact that the reaction is observed to happen.

However, if we think what is happening at the molecular level, we can get some extra insight.

All molecules above absolute zero have some kinetic energy (that is they are bouncing around and banging into each other with a range of speeds, some slow, some fast). Temperature summarises the average amount of energy present in the compound (but the individual molecules have a range of energies, some higher, some slower than the average).

In a solution of hydrogen peroxide (or in the pure liquid, though I don't recommend ever handling the pure liquid) the individual molecules will be bouncing off each other a lot. Each collision will exchange some energy between the colliding molecules. Sometimes a Collision will involve enough energy that one or more of the O-O or OH bonds will rupture. This will generate some reactive species. Then, sometimes, the reactive species will reform in a way that causes water and oxygen to be formed (the details of the reaction are probably fairly complicated). The products oxygen and water have much stronger bonds than the hydrogen peroxide (or at least eh O-O bond in it) so the probability that they will react to reform hydrogen peroxide is very low.

The probability that peroxide will decompose this way is still pretty low, though, at least in dilute and very pure water solutions. But many substances (including some present in slightly impure water) will catalyse the reaction. The presence of manganese dioxide, for example, provides a different route to the same products (that is pretty much the definition of a catalyst). If a peroxide molecule bumps into a manganese dioxide surface, there is a much higher probability that a reaction will happen than if it hits another peroxide molecule (it probably transfers an electron from the manganese oxidising it and this can be very fast and also can precipitate more reactions that break up the peroxide and, ultimately, yields water, oxygen and leaves manganese dioxide). Human blood contains an enzyme that is designed to do the same job as peroxides and related active species are generated by other biological reactions and you really, really don't want them to build up in the blood. Catalase uses oxidation reactions involving iron atoms to do an even better job of facilitating the reaction than manganese dioxide.

So the detail of the reaction is all about molecules bumping into each other and causing bonds to break and reform in a different way. The overall reaction releases energy (which is to say that the new bonds formed are stronger then the ones that get broken and the reverse reactions are therefore a lot less likely to happen). Catalysts proved an easier route to the products than the reaction where peroxide molecules just bump into each other.


test tube cracked when added glycerin to potassium permanganate I am not expert in chemistry.I learned about combustion in reaction when added glycerine to potassium permagnet.I tried it at home, I took potassium permagnet in test tube and when I added glycerine to it test tube just CRACKED as combustion started.I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..

I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...

Any Suggestion..

I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...

In case of doubt you should not. Anyway, I recommend to avoid doing it with any significant amount. It may and will spread around and even if not it produces manganese (IV) oxide smoke that settles down nearby and is quite hard to wash out.

  I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..

Potassium permanganate is a strong oxidizer, glycerol is a reducer. Mixed together they quickly start to react. The exact nature of the products is debatable, but the mixture self-ignites if warm. The thermal shock is enough to crack glass. This is a common weakness of glass and ceramics.

Furthermore, in case of bad luck it may produce enough gases to spread the mix around. Consequently, safety precautions must be always employed (incombustible table, safety goggles etc). A movie how it may go http://www.youtube.com/watch?v=6F4cuQkKt2U . The reaction has some quirks, specifically if the glycerol is cold, it may not start immediately, so a rather well mixing is achievable. Now, once it warms up to room temperature, it suddenly bursts in flame, potentially delivering itself to nearby surfaces, including your face.


Distillation Heating Mantle Setup In my lab courses, we were taught to use textbooks off the shelf as elevating props to adjust the heights of heating mantles to the right height during distillations, but I was wondering if there was a safer and better way to do so? I recently used a different type of heating mantle that was like a nest and charred the cover of a textbook as well as a heat glove that I put on the second time to prevent further damage to the book.

You need to use something called a lab jack ... just search google to see what they look like.  They are quite cheap too.

You should tell whoever told you to use books about lab jacks, too.


Reaction of one mole of (PO4)3- with one mole of H2CrO4 I am looking for the endproducts of a reaction of one mole of $\ce{PO4^3-}$ with one mole of $\ce{H2CrO4}$ in one litre of water.

First, the following reaction can happen:

$$\ce{PO4^3- + H2CrO4 <=> HPO4^2- + HCrO4-}$$

Then the following reactions may happen:

\begin{align}
\ce{PO4^3- + HCrO4- &<=> HPO4^2- + CrO4^2-} \\
\ce{HPO4^2- + HCrO4- &<=> H2PO4^- + CrO4^2-} \\
\ce{HPO4^2- + H2CrO4 &<=> H2PO4^- + HCrO4^-} \\
\ce{H2PO4^- + H2CrO4 &<=> H3PO4 + HCrO4^-}
\end{align}

Is there a way to determine how many moles there will be of each end product?

Note: this is not homework, rather just a thing I'm interested in. However, I have barely an idea where to start.

To answer this question, you should stop thinking about the acids and bases reacting with each other, but rather they react with water exclusively.
Firstly, the phosphoric acid and its conjugate bases will have the following reactions with water

$$\ce{H3PO4 + H2O <=> H2PO4^- + H3O^+}$$

$$\ce{H2PO4^- + H2O <=> HPO4^2- + H3O^+}$$

$$\ce{HPO4^2- + H2O <=> PO4^3- + H3O^+}$$

Then there is the chromic acid and its conjugate bases

$$\ce{H2CrO4 + H2O <=> HCrO4^- + H3O^+}$$

$$\ce{HCrO4^- + H2O <=> CrO4^2- + H3O^+}$$

So now you have five equlibrium reactions.  You have eight unknowns, $[\ce{H3PO4}]$, $[\ce{H2PO4^-}]$, $[\ce{HPO4^2-}]$, $[\ce{PO4^3-}]$, $[\ce{H2CrO4}]$, $[\ce{HCrO4^-}]$, $[\ce{CrO4^2-}]$, and $[\ce{H3O^+}]$.

The Ka expressions from the above reactions give you five equations.  Then you have the initial concentrations of phosphate and chromic acids, which will give you two more, then the fact that the number of protons is conserved will give you the eighth and final equation.


Clathrate Compounds with Quinone I prepared a compound "$\ce{SO2}$-Quinolclathrate" by dissolving hydroquinone and sodium sulfite with hydrochloric acid in warm water.

After this, I tested a little sample of the product on a flame and it vaporised, following which I smelled the gas. I am supposed to define this gas and my prediction is that this could be $\ce{SO2}$(with orange flame).

Clathrate compounds include a trapping system with guest molecules (held "guest" in cages) and the host (hydroquinone here).
$\ce{SO2}$ is guest in the cages of hydroquinone and the gas is held due to crystal lattice forces. Does this mean there are no bonds?

What can be the reason for $\ce{SO2}$ vaporizing?

Additionally, the product was added to starch/iodine solution to find out if it is soluble in that solution and it is so. What can be the reason of this solubility.

Palin, D. E.; Powell, H. M. "Hydrogen Bond Linking of Quinol Molecules," Nature  156, 334 (1946), and DOI: 10.1039/DF9694800181

White aqueous starch/iod$ide$ tests for oxidizing agents by turning brown, liberated iodine and ionic oligomers intercalated within amylose starch (linear).  Amylopetcin (branched) gives red.  The mix is purple-brown.  Purple-brown aqueous starch/iod$ine$ tests for reducing agents by the reverse.


Role of calcium oxide in soda lime decarboxylation Why do we use calcium oxide $\ce{CaO}$ in soda lime decarboxylation?
$$\ce{\underset{sodium acetate}{CH3COONa} + \underset{sodium hydroxide}{NaOH} ->[\Delta][CaO] \underset{methane}{CH4} + \underset{sodium carbonate}{Na2CO3}}$$
It seems as if it is dehydrating agent to keep NaOH dry), but why is the dehydrating agent required if the heat would be enough to dehydrate the container as the temperature is very high (expected to be over 100 °C)?

Calcium oxide is not necessary for carrying out this reaction. It only prevents NaOH from touching the glass of the container where the reaction is done. If a solid mixture of sodium acetate and NaOH is heated to high temperature, some NaOH will attack the glass before reacting with acetate, and make the glass melt down. If NaOH is intimately mixed with CaO, this trouble is avoided. The glassware is saved.


Do alcohols react with alkyl halides? I wonder if alkyl halides react with alcohols or not. I searched on Google but could only find reaction of alcohols with hydrogen halides

I am assuming that you want the alcohol to react with the $\ce{OH}$ group and the alkyl halides to react with the $\ce{Cl}$ group.

Both groups like to withdraw electrons, so they tend to be negative, and if we take them away, the resulting ions would be positive so they cannot react.

This leaves us to the possibility of the $\ce{H+}$ being cleaved off from the $\ce{OH}$, producing an alkoxide anion, which can react with the cation resulting from the cleaving off of the halide.

However, to effectively cleave off an $\ce{H+}$, we would need an acidic alcohol, such as phenol, which does react with alkyl halides (taken from here):

  In the presence of a base, such as potassium carbonate ($\ce{K2CO3}$), and heat, phenol reacts with an alkyl halide to form an ether.

For example:

$$\ce{\underset{phenol}{PhOH} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + HCl}$$

where $\ce{Ph}$ denotes the phenyl group.

The reaction proceeds through the ionization of phenol:

$$\ce{\underset{phenol}{PhOH} ->[K2CO3] \underset{phenoxide}{PhO-}}$$

The next step is an SN$2$ step:

$$\ce{\underset{phenoxide}{PhO-} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + Cl-}$$

For other alcohols, they are generally not acidic, so you would need the corresponding alkoxide by reacting with sodium:

$$\ce{\underset{alcohol}{R1-OH} ->[Na] \underset{alkoxide}{R1-O-} ->[R2Cl] \underset{ether}{R1-O-R2}}$$

This, as the comments have pointed out, is called the Williamson ether synthesis.


f-electrons in chemistry of heavy transition metals In one lecture at a recent MCR-X conference I was puzzled by a side-note "f-hole is critical to describe DoS of $\ce{IrO2}$ correctly." The context was DFT periodic plane-wave calculations with pseudization of the core electrons.
This implies that f-electrons are somewhat involved into chemistry of Ir. More precisely, it seems that polarizability of f-shell affects subsequent d-shell in somewhat dramatic way.
Could someone point at relevant articles, preferably discussing the matter from both ab initio and experimental perspective? Bonus points for generalization of the matter: d-electrons affect chemistry of heavy p-elements in no less way.

"f-hole is critical to describe DoS of IrO2 correctly."

[...]

This implies, that f-electrons are somewhat involved into chemistry of Ir.

Perhaps naïvely, but I would dispute that the second statement follows from the first. While f-electrons are generally not critical for the chemical reactivity of molecules or materials, their inclusion indeed is critical for a proper description of the Density of States. The magnetic and optical properties, notably, can be derived from the DoS and depend directly from f electrons.


Why can't some salts be used in the salt bridge of specific voltaic cells? I had an exam question in reference to a galvanic cell with $\ce{Pb{(s)}}$ as the anode in $\ce{Pb^2+}$ solution and platinum cathode in a solution of  $\ce{Fe^2+}$ and  $\ce{Fe^3+}$:

  Identify an appropriate electrolyte to use in the salt bridge.

I gave $\ce{KNO3}$ which was correct. However, the marking criteria specified that sulphate and chloride salts were not usable.

Why are these salts not usable? Is there any way to tell which salts would be usable for other galvanic cells?

The main reason would be that sulphates and chlorides can form precipitates in solution. Potassium nitrate essentially never forms in precipitates in solution by going off the solubility rules. The lead in the cell would form precipitates with the sulphate and chloride, and thus it would be ineffective as a salt bridge.


Disparity between LeChatelier's Principle and Solutions Let's say that there is a solution
$\ce{Na+(aq) + Cl-(aq) <=> NaCl(s)}$
If the above solution is at equilibrium we can say that it is saturated and hence no more Sodium Chloride will dissolve. However, according to LeChatelier's Principle, if we add more Sodium Chloride it will just dissolve. Is there another rule that governs how this kind of dissolution behaves?

You misinterpret the principle. It does not say if you add more salt, more salt dissolves. More generally, the principle focuses on changes of intensive properties, that do not change with the system scaling.
The principle says if the thermodynamic activity of the solid had increased by adding more salt, more salt would have dissolved to decrease this activity. But by adding more salt, the salt activity remains constant.
You can also consider  chemical potential   $\mu = \frac{dG}{dn}$  instead of activity, as they are closely related.
By other words, thinking more salt dissolves by adding more salt to saturated solution is equally wrong as thinking more salt crystalizes by adding more of saturated solution. As the salt activity in saturated solution or in solid state does not depend on their amount, which is extensive property.


How much CO2 could you remove from the atmosphere with 2 trillion USD? I know it's possible to capture $\ce{CO2}$ with various chemical reactions. For example NASA's space shuttle had some kind of regenerative $\ce{CO2}$ scrubber. But how expensive is it? Could a huge number of these devices, or something like it, significantly reduce the atmosphere's $\ce{CO2}$ level?

I realize this is probably not practical, but it's an interesting thought experiment. To make my question more specific: can anyone show some rough calculations of how much $\ce{CO2}$ you could remove if you had, say, 2 trillion USD to spend? To simplify, assume that energy is at current prices, but clean. (i.e., if the device requires electricity, assume it's coming from a nuclear power plant, solar farm, etc.).

This is very hard to answer precisely, as there are many different carbon capture strategies, and economics at the scale required is quite different from our normal understanding. However, I'd love to see some attempts to at least get order of magnitude estimates, or sources with more in-depth analyses.

Here is an implementation of carbon capture and storage which is useful to consider for its simplicity rather than its real-world applicability, to give a sense of scale. One way to remove anthropogenic $\ce{CO2}$ is to do "inverse combustion", more specifically:

$$\ce{CO2(g) -> C(s) + O2(g)}\ \ \ \ \ \ \ \mathrm{\Delta H=+390\ kJ/mol}$$

Assume this process can be done with perfect efficiency, and that the only energy expense in the process is to drive the reaction forwards (that is, zero energy consumed in transportation, collection, construction, etc). According to this source, the amount of anthropogenic $\ce{CO2}$ emissions from 1750 to 2008 has totalled about $1250\times 10^9\ \mathrm{t_{CO_2}}$. Suppose you wish to remove all this carbon dioxide (only about half is in the atmosphere, the rest is trapped in the ocean or land) using the above process. This would require about $\mathrm{10^{22}\ J}$ of energy, which Wolfram Alpha suggests is about 50% more energy than can be retrieved from combustion of all global proven oil reserves in 2003. Put another way, this is about 20 times the world energy consumption in 2012, or 150 times world electrical energy production. This would put the cost of this process in the range of hundreds of trillions of US dollars, meaning two trillion USD barely makes a dent.

Is there some other process where two trillion USD is close to enough? I seriously doubt it.

Edit: Several comments and answers mention biological sequestration, which is a legitimate carbon capture strategy. I did not consider it, however, because its costs are far more complicated to calculate. My intention with this answer was to find a quick and comparatively simple way to attach an energy cost to carbon sequestration. Whether the monetary cost even makes sense at this scale (how do you define "monetary cost" when it's larger than world GDP?), I don't know.

But here's another amusing comparison, which should somewhat temper hopes that biological sequestration will be a magic bullet. The amount of anthropogenic carbon released between 1750 and 2008, ~350 billion metric tons of carbon, is comparable to a significant amount of the biomass on Earth (all eukaryotic life contains approximately 560 billion metric tons of carbon, multicellular life is a fraction of this). Thus, biological sequestration of the majority of anthropogenic carbon would be broadly equivalent to sacrificing all eukaryotic life on Earth (or ~10-30% of all living organisms by mass) in order to collect and bury carbon, then seeding the Earth back to its current biological state.

Over 250 years of constant stimulation to produce as much energy/goods from fossil fuels, we have released a lot of carbon.


How is red oxide of copper converted into black oxide & vice versa? I can't really find any information about it.
Any help would be appreciated.

Red Copper oxide is Copper(I) oxide Cu2O. Black Copper oxide is Copper(II) oxide CuO. Copper(I) oxide is oxidised to Copper(II) oxide in moist air.

Copper(I) oxide is oxidised to Cu(II) by dissolving in dilute nitric acid which can be thermally decomposed to Copper (II) oxide by heating at 180C.

Copper(II) oxide can be reduced back to Copper metal by Hydrogen, Carbon Monoxide or Carbon. This may then be heated with oxygen to make Copper(I) oxide. Copper(I) oxide may also be made by reaction of solutions of Copper(II) salts with Sulfur Dioxide.

Source - Wikipedia entries for Copper(I) oxide Cu2O and Copper(II) oxide CuO


What is the product formed in the reaction of dimedone and benzaldehyde? I'm having a hard time deducing the structure of the product formed when dimedone and benzaldehyde react in the presence of catalytic piperidine. I think that a Knoevenagel reaction will get me to this product:

but the answer says that the product has a molar mass of 368 g/mol. The product above is only 246 g/mol. What else can have occurred?


OFET vs Traditional Transistors What is the advance of OFET's (Organic Field effect transistors) over the traditional CMOS Transistors?

The advantages of OFET's over traditional CMOS transistors all derive from how OFET's can be manufactured.  Traditional transistors are manufactured using vacuum deposition techniques.  These are slow, time consuming, batch processes, hence the higher cost for electronic components produced by such methods.

OFET's can be manufactured using standard coating technologies that have been widely used in industry for a long time, and are well understood.  Therefore, the manufacturing equipment needed is readily available. With this coating equipment uniform layers ranging from sub-micron to tens of microns (you don't need thicker layers for an OFET, but the coating technology would allow you to coat much thicker layers if you wanted to) can be coated at high speed on thin, flexible, web supports such as aluminized mylar. Multiple layers can be coated by making multiple passes down the coating line, into a drying oven and then on to the next coater.  Thousands of OFETS can be obtained from a small, coated, mylar spool. Since the process is continuous and high speed the manufacturing cost of the device is much reduced compared to similar CMOS devices.  Each layer contains an active organic compound dissolved in a low-cost plastic (for example, polycarbonate) binder, so material costs are reduced as well.  Another key advantage is that since organic materials, plastic binders and flexible web materials are used, the final product is flexible and can be easily cut to any desired shape.  This allows the manufacturer to use the product in new applications where flexibility (newspapers) and shape (electronic clocks) might preclude the use of rigid CMOS devices.


Methods to quantify lead(II) concentration in water on the ppm/ppb level While there are several ways to quantify metal concentrations in water, a lot of them are only specific to select metals. For example, there is the orthophenanthroline method for quantification of Fe(II) concentrations via spectroscopy. How can lead(II) concentrations be determined using a similar spectrophotometric method?

If spectroscopy isn’t the most efficient way to quantify these concentrations, is there a better way to do it?

The salt used to release $\ce{Pb^2+}$ ions into solution is $\ce{Pb(NO3)2}.$

Russian interstate standard GOST 18293-72 "Drinking water. Methods for determination of lead, zinc and silver content" (PDF in Russian) suggests to use sulfarsazene (plumbone), which forms orange-yellow complex with lead(II).
Reported method sensitivity (spectrophotometry): $\pu{0.5 μg L-1}.$

A brief review in English [1, p. 315]:

  To determine lead in solid particles of industrial facilities, natural waters, biological objects and salts, sulfarsazene is used (4''-nitrobenzene-1'',4-diazo-amino-1,1-azobenzene-2''-arsono-4-sulfoacids).
  While it is less sensitive than dithizone , it has a number of advantages: it is water-soluble, easy to handle and needs no cyanide as masking agent.
  But, it has insufficient reaction selectivity.
  Manganese, nickel, copper, zinc, mercury, cadmium, iron, zirconium, and other elements interfere with the determination. The methods developed on the basis of sulfarsazene require a preliminary ion separation; for instance, lead is separated by extracting with dithizone solution in carbon tetrachloride …

  Study of the reaction of sulfarsazene with lead ions points to simultaneous participation of the lead cation with triazene and arson groups in complex formation.
  Polarographic investigation confirmed the existence of lead bonds with the o-arsonotriazene group of the reagent, and cleared up the azo-group contribution to the formation of the complex which may be due to coordinative bonding with lead atoms.
  The role of the nitrogroup as an electrophilic substituent that improves the mobility of the hydrogen atom of the aminogroup was experimentally confirmed.
  The structure of the resultant compound is as follows:

GOST 18293-72 also includes polarographic method for lead detection using the dropping mercury electrode.
Reported method sensitivity: $\pu{10 μg L-1}.$

References

Roeva, N. N.; Savvin, S. B. Spectrophotometric Determination of Minute Lead Contents of Natural and Industrial Objects. Reviews in Analytical Chemistry 1986, 8 (4). DOI: 10.1515/REVAC.1986.8.4.313.


How to determine the packing geometry of compounds with contradicting coordination number and radius-ratio values? Determine the packing geometry of the following compounds by using the radius ratio rules. For your convenience assume a coordination number of six for both cation and anion.
Be specific in the your answer, "body-centered cubic with the anion occupying the corners of the cell [1 atom] and the cation occupying the central hole [1 atom].

Below is the table we are given, and we're are choosing between geometries of cubic, body-centered, or face-centered.

The compound I'm on is $\ce{LiBr}$
I did $\frac{r^+}{r^-}=\frac{79}{182}=0.43$ for the Radius-Ratio value.
But from here would I choose square planar because it's closer to 0.414 or do I choose octahedral because we're assuming coordination number 6? And how would I determine the locations of the cation/anion?
My best guess is cubic geometry because the cation is too large to fit tetrahedral holes and octahedral holes, with $\ce{Li}$ and $\ce{Br}$ taking occupying complementary corners.

I'm making the assumption that the coordination number 6 is used as a reference to the atomic radii.
I've currently adjusted my answer based off of other examples I could find to

FCC with OH holes occupied by $\ce{Li^+}$ [4 atoms] and the $\ce{Br^-}$ forms the FCC Lattice [4 atoms (by stoichiometry}] for a 100% occupancy rate.

But I still have an issue as to how to conceptually approach the problem.

You've calculated your radius ratio value and you know you're between 0.414 and 0.732, so cubic is off the table, based on your chart.

You're looking at either square planar or octahedral. Your chart also says that ionic compounds do not use square planar packing and the coordination number would have to be 4. LiBr is an ionic compound (like NaCl) with a coordination number of 6, therefore you're looking at octahedral.

From there you can determine your "specific" answer.


What is the mechanism of photosynthesis? I just want to know what particle in plants can change $\ce{H2O}$ into $\ce{O2}$ and what $\ce{CO2}$ does in this conversion system. What if we just use $\ce{H2O}$ and light to convert it into $\ce{O2}$?  If light is what we need to convert $\ce{CO2}$ into glucose and $\ce{H2O}$ into oxygen, why can't scientists make a machine to do that?

I'm definitely not going to outline light harvesting in the light harvesting complexes (LHC), the role of the quinones in electron transfer, etc.

But let me give a short answer on your last question:-

The molecular aggregates, such as the LHCs, involved in the different steps of photosynthesis are not optimized for duration. In contrast to man-made machines, they don't have to! Remember that all these fragile molecular assemblies are part of a living system. If parts break, they will resynthesized and replaced.

While it is partly possible to mimic photosynthesis in the lab, it is the self-repair mechanisms that we are unable to replicate in artificial systems.


Effervescent in mineral spirits? I'm shooting video of paint moving in mineral spirits.  To make for a little more action, is there a household granular/solid I could add to the mineral spirits that would gently effervesce ala alka-seltzer in water?

A fish-tank bubbler with needle valve or regulator would not only provide bubbles but be controllable and would not need to be replenished, as would chemicals.

N.B. However you make bubbles in mineral spirits, it would help the highly-flammable liquid to evaporate, creating a potential explosion and fire hazard. You might get better images with less-miscible mineral or silicone oil, with less danger.


Why does ammonia "clean" scratches? I read that ammonia can be used to clean scratches of certain materials, for example from wedding rings. I was thinking about how this would work but I actually don't know. I suspect that the question I might have to ask is "what is a scratch" and that thus the reason for my misunderstanding comes from some ambiguity in what a scratch might be in the English language.

This is how I would phrase my understanding of a scratch: A scratch on a material is a place on which parts of the surface are removed, creating a "cut" in the surface. Thus there are less particles in one place than the surrounding.

Either way, that is pretty much how I imagine it. If this is the case, why does ammonia "remove" the scratch? Does the ammonia-water solution fill the crevice left and attach itself to the surrounding material, for example the gold. If the ammonia-water solution is transparent you would just see the gold underlying the scratch and it would seem to have "filled" the scratch. But for that to happen the ammonia would have to become solid, which does not happen at room temperatures.

Perhaps the reason why a deep cut can not be "cleaned" with the solution, whilst the small cuts can, is that the word "scratch" is used for two different physical phenomena.

I'm hoping to get a deeper understanding of what is actually happening to the material being cleaned by the water-ammonia solution. Or, perhaps, a link to a paper or other resource explaining what happens.

Links:

https://www.leaf.tv/articles/how-to-get-scratches-out-of-a-gold-ring/
http://www.wikihow.com/Remove-Scratches-from-Glass

This seems more like folklore than a tested method. However, gold is extremely soft (malleable), and it must be alloyed with other metals to make it hard enough to survive wear as jewelry. Simply rubbing a gold item with a cloth, with or without "ammonia" (ammonium hydroxide, $\ce{NH4OH}$), is enough to polish the gold, i.e. relocate some gold particles to fill in the scratch.

Household ammonia does not solidify, nor will it fill scratches. In fact, it simply evaporates into the air. It is an effective cleaner, though, and would help remove anything adhered to the gold, which might appear like a scratch.

For the same reason, ammonium hydroxide is an effective glass cleaner, but don't expect it to remove any scratches. The suggestion at your second reference to use toothpaste, though, is a good idea, particularly for those brands containing silica.

BTW, "milk of magnesia", $\ce{Mg(OH)2}$, is an effective mildly abrasive scratch remover for some plastics, if rubbed very well with a soft cloth.


How do the pKa and pH indicate to the proper environment of protonation? I have been reading an abstract of a paper for a couple of minutes (reference below) but I cannot understand a piece of it. The abstract:

  The ionization state and $\mathrm pK_\mathrm a$ of the inhibitor 6,8-dimethyl-N5-deazapterin bound to the recombinant human dihydrofolate reductase (rhH2folate reductase) complex with NADPH was determined by a spectrofluorimetric method. The excitation spectra for bound ligand as a function of pH from 6.1 to 9.7 indicated it was the same cationic form as for unbound ligand, which is protonated on N3. However, the lower limit for the $\mathrm pK_\mathrm a$ of the bound form was determined to be 9.1, a value about pH 2.5 higher than that for free ligand, indicating that ligand bound to the enzyme is protonated at neutral pH. The excitation spectra for bound ligand as a function of pH were generated by computer simulation by employing corrections for the pH dependence of the proportion of bound ligand (variable $K_\mathrm d$; ligand-dissociation constant) and taking account of the different $\mathrm pK_\mathrm a$ values for bound and unbound forms. A plot of $K_\mathrm d$ values against pH showed a bell-shaped curve indicating that 6,8-dimethyl-N5-deazapterin bound to rhH2folate reductase. NADPH to form a ternary complex of ionised enzyme with protonated ligand and/or protonated enzyme with unprotonated ligand; the spectrofluorimetric results are consistent with the first alternative.

I made the not-understandable part (for me) in bold. Could some one help me to understand how do the previous sentences help to indicate to this sentence? Maybe I am missing something in the previous sentences.

Jeong, S.-S.; Gready, J. E. Ionization state and pka of pterin-analogue ligands bound to dihydrofolate reductase. Eur. J. Biochem. 1994, 221 (3), 1055–1062. DOI: 10.1111/j.1432-1033.1994.tb18824.x.

The sentence immediately before the bolded one indicates that the $\mathrm pK_\mathrm a$ value of the ligand bound to the enzyme was determined to be $9.1$. It also indicates that the $\mathrm pK_\mathrm a$ value o the free ligand is 2.5 logarithmic units less, i.e. $\mathrm pK_\mathrm a(\ce{Lig}) \approx 6.6$.

The human cell is well-buffered to ensure that the pH stays within a very narrow range. There are enough buffering compounds around, that we never need to use any Henderson-Hasselbalch equation or other difficult calculation methods to determine the protonation state of a given compound. We can simply examine its $\mathrm pK_\mathrm a$ value; if that is lower than the cell’s pH value, the compound is deprotonated. In all other cases, it is not.

The $\mathrm pK_\mathrm a$ value of the free ligand indicates it should be deprotonated in the cell. However, upon binding to the protein with the $\mathrm pK_\mathrm a$ value rising by $2.5$  logarithmic units, the ligand suddenly becomes protonated. This is what the bolded sentence implies.


Prove Law of Chemical Equivalence A somewhat infamous law (to me at least), that states:

$$a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$$

for any and every balanced chemical equation -

$$a.A+b.B = c.C+d.D$$

where $i$ denotes the stoichiometric coefficient(/no. of moles) of the chemical species $I$ and $I_{eq}$ denotes the number of equivalents in 1 mole(/formula mass...?) of the chemical species $I$.

Am I wrong about any of the facts I've presented above? Please mention any errors you spot.

Assuming that I did fine, my simple question is,

  What's the mathematical proof for this law?

I tried surfing online, but all I came across (with my not-so-deep searches) were verbal explanations of the law.

Is there no mathematical proof to this law whatsoever? I'll be satisfied to just find a proof, even if I don't understand the complicated math going into it (for I'm an eleventh grader).

Edit to provide an example:

$$\ce{2NaOH + H2SO4  -> Na2SO4 + 2H2O}$$

For A, $a \cdot A_{eq} = 2 \ (moles) × 1 \ (equiv.\ per\ mole) = 2$

For B, $b \cdot B_{eq} = 1 \ (moles) × 2 \ (equiv.\ per\ mole) = 2$

And this follows for other species, hence showing $a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$

An example

Here is a balanced equation:

$$\ce{2 H2 + O2 -> 2 H2O}$$

Both sides of the equation show four atoms of hydrogen and two atoms of oxygen, so it is balanced. The coefficients are $\nu_{\ce{H2}} = 2$, $\nu_{\ce{O2}} = 1$ and $\nu_{\ce{H2O}} = 2$.

If 20 moles of $\ce{H2}$ and 10 moles of $\ce{O2}$ are used up in a reaction forming water (you can figure out 20 moles are formed), I can write:

$$\frac{n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{n_{\ce{O2}}}{\nu_{\ce{O2}}},$$ where $n$ is the amount that is used up in the reaction.

I can not write:

$$n_{\ce{H2}} \cdot \nu_{\ce{H2}} = n_{\ce{H2}} \cdot \nu_{\ce{H2}},$$

as you will find when substituting the values from the example.

I should also not write:

$$2 \cdot \ce{H2} = 1 \cdot \ce{O2} $$

Hydrogen atoms are not oxygen atoms, no matter what we multiply it (nuclear reactions aside)

What about $I$ and $i$?

If you call the stoichiometric coefficients $i$ instead of $\nu$, and you want the equation below to be true:

$$i_{\ce{H2}} \cdot I_{\ce{H2}} =  i_{\ce{O2}} \cdot I_{\ce{O2}},$$

you have to make sure that $I$ is proportional to $\frac{1}{i}$, i.e. define them by

$$I_{\ce{H2}} = \frac{ \mathrm{const} }{ i_{\ce{H2}} } $$

$$I_{\ce{O2}} = \frac{ \mathrm{const} }{ i_{\ce{O2}} }$$

If you look at definitions of "number of equivalents" (a concept that is outdated), you will find that this is the case. Once that is in place, it is a matter of showing that for any non-zero $a$ or $b$,

$$\mathrm{const} \cdot \frac{a}{a} = \mathrm{const} \cdot \frac{b}{b} $$

Source: https://www.askiitians.com/iit-jee-chemistry/physical-chemistry/stoichiometry-and-redox-reactions/law-of-equivalents.aspx


Does carbon monoiodide (CI) exist? Many of my students get chlorine $\ce{Cl}$, confused with carbon monoiodide $\ce{CI}$, despite my promises to use serif fonts whenever possible and to never give them carbon monoiodide in a question.

Everything I know about chemical bonding tells me carbon monoiodide should not exist ($\ce{CI4}$, sure), but weird things happen sometimes. My only results from internet searches for carbon monoiodide bring up a set of Quizlet flash cards.

Will my students prove me wrong if I tell them that carbon monoiodide does not exist (except possibly in extreme laboratory conditions).

Provided that mellitic anhydride ($\ce{C12O9}$, empirical formula $\ce{C4O3}$) is considered one of carbon oxides, then e.g. hexaiodobenzene ($\ce{C6I6}$, empirical formula $\ce{CI}$) could be considered a form of carbon(I) iodide, formally. But please rather don't teach that in regular lectures.


Is it possible to remove salt from food without changing the flavor? I love soup, even a thick soup for dinner, yet many soup products have 400 to 1000 mg of sodium in them. Is there a simple chemical procedure any consumer could do to take the sodium out of the product without 'hurting' the flavor?

If one takes some, say, Campbell's soup and heats it to boiling, then freezes it, then heats it again, then freezes it etc., would this extract some sodium?

Short answer: no, there is no easy (or even moderately hard) procedure, and certainly not boiling and freezing.

The thing is that water dissolves sodium ions better than just about anything else, so almost everything will come out of solution before the sodium does. You're much better off with a product that starts with less salt to begin with.


How does changing the concentrations of intermediate products/reactants change the rate law? A = CH2(CN)2
B = CH(CN)2-
C = BrCH(CN)2
The rate law above is derived from applying the steady state approximation.

I am trying to work out how to modify the rate law to accomodate the following conditions:

1) [Br2] is low relative to [H+]

2) [Br2] is high relative to [H+]

And in the 2nd case how this affects the RDS and order of reaction.

It seems to me like these conditions would just affect the rate coefficient, not actually change the rate law.

EDIT:

$$
\frac{d [CH(CN)_{2}^{-}]}{dt} = k_{1}[CH_{2}(CN)_{2}] - k2[CH(CN)_{2}^{-}][Br_{2}] - K_{-1}[CH(CN)_{2}^{-}][H^{+}]
$$

$$
Applying\ Steady\ State\ Approximation:
$$
$$
k_{1}[CH_{2}(CN)_{2}] - k2[CH(CN)_{2}^{-}][Br_{2}] - K_{-1}[CH(CN)_{2}^{-}][H^{+}] = 0
$$

$$
k2[CH(CN)_{2}^{-}][Br_{2}] + K_{-1}[CH(CN)_{2}^{-}][H^{+}] = k_{1}[CH_{2}(CN)_{2}]
$$
$$
Dividing\ by\ [CH(CN)_{2}^{-}]:
$$

$$
k2[Br_{2}] + K_{-1}[H^{+}] = \frac{k_{1}[CH_{2}(CN)_{2}]}{[CH(CN)_{2}^{-}]}
$$

$$
[CH(CN)_{2}^{-}](k2[Br_{2}] + K_{-1}[H^{+}]) = k_{1}[CH_{2}(CN)_{2}]
$$

$$
[CH(CN)_{2}^{-}] = \frac{k_{1}[CH_{2}(CN)_{2}]}{(k2[Br_{2}] + K_{-1}[H^{+}])}
$$

$$
\frac{d [BrCH(CN)_{2}]}{dt} = {k_{2}[Br_{2}][CH(CN)_{2}^{-}]}{}
$$

$$Rate\ Equation:$$

$$
\frac{d [BrCH(CN)_{2}]}{dt} = \frac{k_{2}[Br_{2}]k_{1}[CH_{2}(CN)_{2}]}{{k_{2}[Br_{2}]}+k_{-1}[H^{+}]}
$$

So if $ [Br_{2}]$ is low compared to $ [H^{+}]$ then I would remove $ [Br_{2}]$ from the rate equation giving:

$$
\frac{d [BrCH(CN)_{2}]}{dt} = \frac{k_{1}[CH_{2}(CN)_{2}]}{k_{-1}[H^{+}]}
$$

If $ [H^{+}]$ is low compared to $ [Br_{2}]$ then I would remove $ [H^{+}]$ from the rate equation giving:

$$
\frac{d [BrCH(CN)_{2}]}{dt} = k_{1}[CH_{2}(CN)_{2}]
$$

So in this case the reaction is 1st order and the RDS no longer proceeds backwards with rate coefficient $k_{-1}$.

Is this right or have I made a mistake somewhere?

The steady state approximation allows us to assume that the concentration of $\ce{B}$ does not change.

$$\dfrac{d[\ce{B}]}{dt}=0$$

Thus the rates of production and consumption of $\ce{CH(CN)2-}$ must be equal.

  $\ce{CH(CN)2-}$ can be produced by one pathway:

$$\ce{CH2(CN)2 ->[k_1] CH(CN)2- + H+} \\ \mathrm{rate}=k_1[\ce{A}]$$

  $\ce{CH(CN)2-}$ can be consumed by two pathways:

$$\ce{CH(CN)2- + Br2 ->[k_2] BrCH(CN)2 + Br-} \\
\ce{CH(CN)2- + H+ ->[k_{-1}] CH2(CN)2} \\
\mathrm{rate}=k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}]$$

  Thus:

$$\dfrac{d[\ce{CH(CN)2-}]}{dt}=k_1[\ce{CH2(CN)2}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$

Now you can can solve for $[\ce{CH(CN)2-}]$ and substitute into the rate law equation for the second step:

$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=k_2[\ce{CH(CN)2-}][\ce{Br2}]$$

You should get a final rate law that contains $[\ce{CH2(CN)2}],\ [\ce{Br2}],\ \&\ [\ce{H+}]$, and then you can answer your questions.

  Update based on revised question.

Your new rate late is correct:

$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=\dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]+k_{-1}[\ce{H+}]}$$

Your interpretation of the simplification you can do is correct for one case and almost correct for the other.

  $[\ce{H+}] \ll [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\ll k_2[\ce{Br}]$ so

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_2[\ce{Br2}] \\
\therefore \
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]}\approx k_1[\ce{CH2(CN)2}$$

which is as you figured.

    $[\ce{H+}] \gg [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\gg k_2[\ce{Br}]$ so

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_{-1}[\ce{H+}] \\
\therefore \
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_{-1}[\ce{H+}]}$$

Note that this approximation does not remove the $k_2[\ce{Br2}]$ term in the numerator.