text:
Why are the relative masses of isotopes not close to whole numbers?
Understandably, the relative atomic masses of isotopes are often not close to whole numbers as they're adjusted for isotope abudandancy eg. $A_r(Cl)\approx35.45$. However, wouldn't one expect the relative atomic masses of individual isotopes to be rather close to whole numbers, given that they're merely the sum of the relative atomic masses of the protons, neutrons, and electrons within the atom and that the masses of protons and neutrons $\approx1$?
I.E, we have $Rb-85$, which has a literature $u=84.91$. Attempting to calculate $u$ given proton, neutron and electron count:
$$ A_r(p^+) = 37\times 1.0073\approx 37.2701\space Da $$
$$ A_r(n^0)=48\times 1.0087\approx 48.4176\space Da $$
$$ A_r(e^-)=37\times 5.4858\times 10^{-4}\approx 0.02030\space Da$$
$$37.2701+48.4176+0.02030=85.708$$
The answer I get is $85.708$, whereas something close to $84.91$ is expected.

Do you consider the total nucleus mass is less than sum of masses of protons and neutrons by the nuclein bound energy?

E.g. Mass of 4He is less than mass of 2p + 2n by about 0.7%, AFAIK. And as average bonding energy per differ from  the one for 12C,  nuclei masses in Da differ from integer values. Another minor factor is various n/p ratio.

Please note that the dalton (unity symbol $\mathrm{Da}$) is not a unit of the quantity relative atomic mass (quantity symbol $A_\mathrm r$). Relative atomic mass is a quantity of dimension one (a so-called dimensionless quantity). Also please note that $u=84.91$ doesn't make any sense at all.

Interesting facts: The lowest mass per nucleon has 56Fe, while the highest binding energy per nucleon has 62Ni.

Because the energy involved in nuclear forces holding nuclei together are so large they affect the net mass of the nucleus
The underlying reason is Einstein's famous $E = mc^2$.
The forces that hold nuclei together are very strong, as they have to be to stop charged protons from flying apart because of their electrostatic repulsion. And those really strong nuclear forces involve a lot of energy, which has to come from somewhere.
When multiple protons and neutrons are combined into a stable nucleus (which is stable because the binding forces are strong enough to outweigh the electrostatic repulsion) the result has a much lower energy than the isolated protons and neutrons. And this can be a lot of energy (enough to make a star shine brightly or to create an enormously destructive fusion bomb).
That energy, using Einstein's famous equation, constitutes a significant proportion of the mass of the resulting nucleus so a combination of neutrons and protons into a stable nucleus will have less mass than the sum of the masses of isolated protons and neutrons.  In the case you calculate that difference is a little under 1% of the isolated masses of protons and neutrons in the Rb nucleus.
The exact mass defect depends on the specific nucleus which complicates things for chemists who tend to reference atomic masses versus specific isotopes where it is easy to measure (once $\ce{^16O}$, more recently $\ce{^12C}$). But, if you understand the principle, the non-integer masses of isotopes is easy to understand.
And that history of how chemists measure relative nuclear mass also explains why the mass of isolated nuclei are not exact integers (they are measured relative to a specific isotope which has a lower mass than the sum of its isolated components).

This explanation better suits why 2+2 is not equal to 4 for atomic masses, however there is no fundamental explanation as to why the mass of a single proton or a neutron is not an integer. In short, we do not know why atomic masses are non-integers. The Einstein's equations is perhaps only a small part of the story. The reason is more or less circular- C-12 has been historically set to an integer. I know it has been changed very marginally but...

@achem That is simply an obvious result of measuring the mass relative to a particular isotope.

@AChem The title of the question mentions integers, but the body of the question is more about the non-additivity. Trivially, 1.0073 looks more like an integer than e.g. 1.0073 * 100 = 100.73. BTW the molar mass of 12C is 11.999 999 9958(36) g/mol according to CODATA2018.

@Karsten, I agree but the student should also ask why the masses and protons are non-integers. These numbers are all circularly defined but they are also self-consistent.

Carbon -12 was chosen to keep the atomic weight of the natural abundance of oxygen as close to 16. as possible and choosing an isotope for the physical scale. This prevented previous chemical data from becoming not so good. The improvements in mass spectrometry meant all the physical masses were being redone anyway. The chemical scale was based on the mix of oxygen set to 16. There is nothing circular about defining a standard mass.

@jimchmst This is correct. The underlying reason why the numbers we get for isotope and nucleon masses is the mass deficit caused by nuclear binding energy. The "circular definition" isn't circular: it is simple about choosing a point (specific isotope) to which other masses are referenced. The fact we need to choose a reference point is a side effect of the mass deficit issue, not the cause of non-integer masses.

In the world of atomic masses, everything is "weighed" with reference to carbon-12. Unfortunately, this is a circular approach, in the sense that carbon-12 is arbitrarily set to 12 as an exact integer. Today, there is a very small change due to the refining of the definitions of mass/ length in terms of fundamental constants.
There is no reason as to why the mass of a single isotope be an integer. Recall, carbon-12, it is arbitrary. Yes, the count of protons and neutrons is an integer but not their masses. Two centuries ago, atomic weights were all different in the time of Berzelius because they had other standards.
To add insult to injury in the nuclear world, 1+1 is not equal to 2. It is always slightly less. This is one of the fascinating mysteries of nature. The missing mass is accounted by the binding energy of nucleons (protons+neutrons) in the nucleus.

One can even state that each proton and each neutron loses a small fraction of its mass when joining to make a nucleus. This mass loss is transformed into energy by Einstein's law : $\pu{E = mc^2}$


text:
Handling of tautomerism in QSAR modelling
This article from 2009:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2776169/
mentions the potential impact of tautomerism on QSAR modelling, and concludes that more work is needed to address it.
12 years later, what is the consensus in the cheminformatics community regarding this (i.e. regarding how tautomerism should be handled when doing QSAR modelling)? Or in fact, is there a consensus at all?
I found only scattered later contributions on the topic.
This one:
https://www.simulations-plus.com/assets/RFraczkiewicz-QSAR_Modeling_Independent_of_Input_Tautomer-GCCh-2015.pdf
seems to argue that the solution is to create tautomer-independent descriptors.
This one:
https://link.springer.com/article/10.1007/s00044-013-0776-0
seems to be more focused on the analysis of the problem than on suggesting solutions.
[But I have not read it yet, I will see if I can get it].
Several other papers argue that the solution is to convert all molecules to a 'standard' tautomeric form, and they even provide large sets of transformations to apply.
Personally, I have no trouble believing that 'some' approach based on special descriptors or standard tautomer forms may work better than most on specific cases (specific molecular properties to predict), but I am wondering if any such approaches are theoretically/fundamentally justified, or just approximations.
Take for instance logD_7.4. You can measure this property by shaking a compound in a biphasic mixture of water and 2-octanol buffered at pH 7.4, taking samples from the two layers and finding the concentrations of compound in them. logD_7.4 is then the log10 of the ratio of the 2-octanol / water concentrations.
However, if the compound has 2 possible (and rapidly interconverting) tautomeric forms, each with its own concentration in the two layers, in fact the concentrations we measure are the sums of the concentrations of the two tautomers (assuming the analytical method does not discriminate the tautomers). Then 3 constants are necessary to describe the existing equilibria, and the 'experimental' logD_7.4 is in fact a combination of these constants. We are measuring the logD_7.4 of a mixture, not of an individual chemical species.
If this is true, how can representing one single tautomer and associating it to one single measurement referred to mixture be a correct way to handle this situation?
But OK, this is only my view; I'd be glad to be wrong, because that would make our modelling work much easier :)

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text:
Can large, sticky, sodium polyacrylate crystals be coated in a powder to facilitate handling?
I am separating a larger container of sodium polyacrylate crystals into smaller containers for student experiments. Commonly known as "Deco Cubes" and used by florists in decorations and centerpieces, they are about 1/8" cubed. They are sticky which makes them hard to handle. Is there a water resistant powder that I can coat these with, to facilitate handling, that can be washed off just before hydrating?

They are not crystals.

It's hard to know what would be best without knowing what state they are in initially and how you are going to use them.
Assuming they are dried to begin with then they are likely sticky due to absorption of humidity from the air. Try drying them in an oven or vacuum desiccator (if you have one) and then storing with silica gel in a sealed container (the little packets you find in electronics boxes) to keep them dry.
If that doesn't work, try coating with a very fine water-soluble powder that  won't ruin your future experiment. Icing sugar might be a good idea but be warned that the resulting gels may end up being the perfect place for bacteria to grow due to the wet, sugary environment so this might be a bad idea depending on how long you plan to use them for.

I called the manufacturer and like you said, the product is absorbing moisture from the environment and becoming sticky. However, they had nothing to add for handling. They just deal with it. I tried adding corn starch, but after 3 days, the results were iffy. Looked like the moisture seeped into the corn starch and bacterial growth could become an issue. I have access to a vacuum chamber so I will try that next.

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text:
Can 5% sulfuric acid and 3% hydrogen peroxide form piranha solution?
I'd like to use 5% sulfuric acid and 3% hydrogen peroxide as an etching solution for copper. (In the past I've used dilute hydrochloric acid and hydrogen peroxide but the generation of chlorine gas has always made me a little nervous.)
I'm planning to use 200ml 5% sulfuric acid and add a 25ml portion of 3% hydrogen peroxide to it whenever the etching stops so that the concentration of hydrogen peroxide in the solution stays low. This solution should not be able to generate any gases other than oxygen (which is harmless as long as I don't store it in a sealed container).
What I'm not sure about, however, is whether this solution will be able to dissolve organic material. As far as I know, acidic piranha solution depends on the dehydrating properties of concentrated sulfuric acid to form the strongly oxidizing Caro's acid and oxygen radicals. I'm not sure whether this reaction can take place in my case. I'd like to avoid dealing with piranha solution when possible.
I've not been able to find any resources on the effects of very dilute piranha solutions (or whether dilute solutions of sulfuric acid and hydrogen peroxide act like piranha at all). The only thing I could find was instructions on how to dispose of piranha, which recommend dilution with water followed by neutralization with a base, for example this page from Princeton.

Your final solution will be $4.45$% sulfuric acid and $0.33$% hydrogen peroxide. These values are rather low, and will not have any effect on organic material. On the other hand, I suspect these low concentrations may not be able to act on copper metal.

@Maurice Thanks! I've done it the same way with hydrochloric acid (5% acid + small portions of H2O2) and it does etch, it's just quite slow and requires constant agitation. Etching speed is far less important to me than the fact that the solution doesn't generate any nasty gases and can be stored.

Hydrochloric acid has the advantage that it reacts and forms complex ions containing chlorine atoms, like $\ce{[CuCl2]^-}$ or $\ce{[CuCl4]^{2-}}$. This helps accelerate the reaction. Sulfate ions do not make such complexes. So the reaction with copper may be slower.

To directly address just the title - no, it cannot. By definition, it is mixture of concentrated sulphuric acid ( like >=95%) and  H2O2 30% ( ratio in formulations differ)

I guess even vinegar could be used for etching instead of sulphuric acid, as copper dissolves slowly in vinegar too, oxidized by aerial oxygen. If you leave copper in vinegar overnight, solution gets blue. H2O2 is better that air.

@Poutnik Maybe vinegar and air would be called “guppie solution”? ;-)

Small amounts of chlorine are not a big hazard. Your drinking water probably contains a small amount of it.

First, per a educational source

copper does not react with dilute sulphuric acid

As such, you may wish to follow a path employed for the commercial leaching of copper ore. The reaction is accomplished with copper metal in aqueous ammonia, optionally an ammonium salt (carbonate), an electrolyte (like $\ce{NaCl}$) and air (or, dilute $\ce{H2O2}$ works just fine).
The underlying chemistry was debated for decades but is now resolved to consist of a mixed reaction involving some spontaneous electrochemistry (and so, for the associated electrolyte, I use a pinch of sea salt). Here is a review article from ResearchGate. Also, this article, "Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia".
The cited provided chemistry is even a little more complex (and unfortunately, in my opinion, does require an exposition due to safety issues involving implementation and handling) with a mention of a minor secondary radical related reactions resulting in an unstable nitrite ($\ce{NH4NO2}$). My take is that the $\ce{O2/Cu}$ reaction with $\ce{NH3}$ is, in effect, a metal/air battery. The latter can introduce solvated electrons into the mix (as sourced from copper, so it is correct to claim that copper is catalytic to the oxidation of ammonia to nitrite from $\ce{O2/H2O2}$). The battery cell provided $\ce{e-(aq)}$ with $\ce{H+}$ produces the hydrogen atom radical, which can react with $\ce{NH3}$ forming $\ce{•NH2}$ radical, which can further interact with $\ce{O2/H2O2}$ as follows:
$\ce{•NH2 + O2 → NH2O2•}$ (aminylperoxyl radical unstable) → $\ce{NO + H2O }$
As a supporting reference on the above, see "On the aqueous reactions of the aminyl radical with molecular oxygen and the superoxide anion", particularly Page 24, Table 2.1.
where the $\ce{NO}$ is a precursor to nitrite, and in particular, a small amount of unwanted $\ce{NH4NO2}$. The latter has the annoying property (based on a lowing of the pH following ammonia consumption) of suddenly decomposing releasing volumes of $\ce{N2}$ gas, which apparently is sufficient so as to cause leakage and even breakage of sealed reaction vessels (as I have experienced).
Consequentially, don't tightly seal your suitably large reaction vessel (and I would recommend avoiding any skin contact with the solution due to also potentially toxic ammonium nitrite here (unclear as to whether its toxicity extends to a skin ingestion pathway).
Note, the ascribed electrochemical reaction has an inception period (waiting time).
Have problems with ammonia?  An alternate electrochemical path is Copper metal, lemon juice, dilute $\ce{H2O2}$, a piece of carbon (large surface area) and some sea salt. Heat (especially for a few minutes in a microwave, if possible, which eliminates the inception period) and you will have a colorful solution of copper citrate/ascorbate.

Thanks a lot for the elaborate explanation! Nitrous oxides, nitrites and ammonia are something I'd like to avoid even more than trace amounts of chlorine, so that's pretty much off the table. I just did a test etch with dilute H2SO4 and it does etch when I add a small amount of table salt (as you suggested in your citric acid method). The problem with that is of course that it generates some chlorine again, although way less than what I got when using dilute HCl. I'd imagine that the citric acid (lemon juice) method will also produce small quantities of chlorine as the chloride gets oxidized.

Copper metal, lemon juice, dilute H2O2, a piece of carbon (large surface area) and some sea salt is electrochemical in nature. Basically, as carbon is very noble (a good cathode) and Copper is not as much (so it is the anode here), the Cu undergoes galvanic corrosion releasing Copper ions. The H2O2 also captures electrons forming radicals which assist. In a microwave, the process is surprisingly fast. and never any chlorine.

per a educational source I had a little chuckle and then started crying... the site you have linked is an insult to actual educational sources.


text:
Can a gas expand and change its volume at STP by doing work on its surrounding?
I came across this question:
1 g of H2 gas expands at STP to occupy double of its original volume. The work done during the process is?
So can a gas expand and change its volume at STP by doing work on its surrounding ie without changing pressure and temperature conditions?

for given temperature and pressure, only one volume is possible

Is the question wrong then? thank you for your reply

The above wording is very poor - better would be "1g of H2 gas at STP expands to occupy double its original volume ...". Even better would be to put "initially" before STP.

If the volume of $\ce{H2}$ doubles, its final pressure is divided by $2$. And the gas is not at STP any more. But this is probably what the author of the problem wants.
Now this process can be done in a  reversible or in a irreversible way. If this is reversible, the pressure outside and inside are maintained constantly at the same value, and the temperature is maintained at $273$ K. In this case the work is the integral of $pdV = nRT ~\mathrm{log}2 = 0.5  \mathrm{mol} · 8.314~ \mathrm{J·K^{-1} mol^{-1}}·273~ \mathrm{K} ·0.301 =  341.6 ~ \mathrm{J}$. If the work is done irreversibly, there are a lot of ways of carrying it out.

The ideal gas law (I'm assuming you're treating the gas as ideal) says $pV=nRT \implies V = \frac{nRT}{p}$. Since $R$ is a constant, the only way $V$ can change is if you change at least one of $n$, $T$, or $p$.
My guess is that the intent of the questioner is that $n$ doesn't change. So, to double $V$, you need $\frac{T}{p}$ to double, which could be achieved either with a doubling of $T$, a halving of $p$, or changes in both that net out to a doubling of $\frac{T}{p}$ (e.g., $p$ doubles and $T$ increases by four-fold). Regardless, when you do this, the gas isn't at STP any more.
To determine the work, you need to know what $p_{external}$ is during the process.  It could be a fixed pressure.  Or it could be done reversibly, in which case $p_{external}$ would vary continulously and be equal to $p_{gas}$.


text:
Combustion Temperature Calculation Using C_p and Heat of Formation
I am simulating the combustion temperature for various compositions of syngas and finding the product temperature. Part one is varying the ratio of CO and H2 from 3:1 to 1:3 in the syngas, burning it with air (21/79 O2/N2), and finding the temperature/composition. Part two is fixing the composition of the syngas and increasing the amount of N2 introduced into the system and again finding the temperature and composition. Lastly, everything is fixed but there is heat loss through the system (~1000 J / mol). Temperature and composition is found. As you can tell, the code between all three parts would be similar. I am stuck on on part that I think is the last step. I have already solved the system of equations described below using sypmy fsolve(). Then I need to calculate the temperature by iterating.
The general equation for the chemical reaction is as follows:
aCO + bH2 + c*(O2 + 79/21 * N2) + dN2 = eCO + fH2 + gO2 + hN2 + jCO2 + kH2O + lNO
From conservation of species:
Carbon: a = e + j
Oxygen: a + 2c = e + 2g + 2*j + k + l
Hydrogen: 2b = 2f + 2*k
Nitrogen: 2*(79/21)c + 2d = 2*h + l
Since there are three compounds, there are three partial pressure equilibrium values called K_p. K_p is a function of temperature and from empirical data is a known constant.
K_p_NO = (X_NO * P) / (sqrt(X_N2*P)*sqrt(X_O2 * P))
K_p_H2O = (X_H2O * P) / (sqrt(P*X_O2)X_H2P)
K_p_CO2 = (X_CO2 * P) / (sqrt(P*X_O2)X_COP)
Where X is the mole fraction. E.G. X_H2O = k/(e+f+g+h+j+k+l)
Please note that in the below code, enthalpy is denoted with a dG instead of a dH.
# There are three parts to this project is split into three parts.
# The First is to calculate the flame temperatures and equilibrium
# composition of syngas and air. Syngas is comprised of mainly
# CO and H2. Sometimes CO2 and CH4 are present, these are ignored.

# Importing Properties. The Specific heat and delta enthalpy of each species is input via empirical_funcs.py.
# The gases used are O2, CO2, H2O, CO, H2, N2, and NO.
# Air is assumed to be 79% Nitrogen Gas and 21% Oxygen
# Air is at 600 K, fuel is at 300 K, pressure is 5 atm.

import matplotlib.pyplot as mpl
import math as m
import numpy as np
import sympy as sp
import empirical_funcs as emp
import scipy
from scipy.optimize import fsolve

# Constants
R = 8.3145  # J / mol / K
dec = 50  # decimal point precision
i = 0
p = 5
# standard heat of formation for elements is zero
h_0_f_CO = -110.541  # kJ/mol
h_0_f_CO2 = -393.546  # kJ/mol
h_0_f_H20 = 241.845  # kJ/mol
h_0_f_NO = 90.297  # kJ/mol
flag = 0
T = 1.7  # Temperature (K)/1000

# PART 1: calculate the equilibrium product composition and adiabatic flame
# temperature for a stoichiometric syngas-air flame fueled only by CO and H2.
# For CO/H2, vary composition from 75/25 to 25/75 by volume.
T_fuel = 0.3
T_air = 0.6
ratio_lower = 25
ratio_upper = 75
increment = 5
T_data = np.zeros((len(range(ratio_lower, ratio_upper, increment)), 1))  # Initialise T_Data array
co_h2_ratio = [*range(ratio_lower, ratio_upper, increment)]
i_len = len(co_h2_ratio)

def solver():
    global T, flag, i
    # aCO + bH2 + c(O2 + 79/21 N2) + dN2 -> eCO + fH2 + gO2 + hN2 + jCO2 + kH20 + lNO
    a = co_h2_ratio[i] / 100
    b = 1 - a
    c = a + b
    d = 0  # For first part

    # Starting temperature is found from mixing reactants, DBL CHECK
    n_cp_air = c * emp.get_cp('O2', T_air) + (79 / 21) * c * emp.get_cp('N2', T_air)
    n_cp_fuel = a * emp.get_cp('CO', T_fuel) + b * emp.get_cp('H2', T_fuel)
    T_reactants = (1000 * T_fuel * n_cp_fuel + 1000 * T_air * n_cp_air) / (n_cp_fuel + n_cp_air)
    T_reactants = T_reactants / 1000
    kp_NO = emp.get_kp('NO', T)
    kp_H2O = emp.get_kp('H20', T)
    kp_CO2 = emp.get_kp('CO2', T)

    # Solving the system of equations
    def equations(vars):
        e, f, g, h, j, k, l = vars
        f1 = e + j - a
        f2 = e + 2 * g + 2 * j + k + l - a - 2 * c
        f3 = f + k - b
        f4 = 2 * h + l - 2 * d - (2 * 79 / 21) * c
        f5 = kp_NO - (l / (sp.sqrt(h)*sp.sqrt(g)))
        f6 = kp_H2O - 1/(k / (f * sp.sqrt((g * p) / (e + f + g + h + j + k + l))))
        f7 = kp_CO2 - 1/(j / (e * sp.sqrt((g * p) / (e + f + g + h + j + k + l))))
        return [f1, f2, f3, f4, f5, f6, f7]
    e, f, g, h, j, k, l = scipy.optimize.fsolve(equations, (0.00004, 0.00004, 0.49, 3.76, 0.25, 0.75, 0.01))

    # e, f, g, h, j, k, l = sp.symbols('e, f, g, h, j, k, l')
    # f1 = e + j - a
    # f2 = e + 2 * g + 2 * j + k + l - a - 2 * c
    # f3 = f + k - b
    # f4 = 2 * h + l - 2 * d - (2 * 79 / 21) * c
    # f5 = kp_NO - (l / (c * (79 / 21) * c))
    # f6 = kp_H2O - (k / (b * sp.sqrt((c * p) / (e + f + g + h + j + k + l))))
    # f7 = kp_CO2 - (j / (a * sp.sqrt((c * p) / (e + f + g + h + j + k + l))))
    # results = sp.solvers.nsolve([f1, f2, f3, f4, f5, f6, f7], [e, f, g, h, j, k, l],
    #                                 [0.0001, 0.00000001, 0.00000000001, 0.000000001, 0.0000001, 0.000000001, 0.000000001])
    #
    # e = round(results[0], dec)
    # f = round(results[1], dec)
    # g = round(results[2], dec)
    # h = round(results[3], dec)
    # j = round(results[4], dec)
    # k = round(results[5], dec)
    # l = round(results[6], dec)

    print('CO = ', e, 'H2 = ', f, 'O2 = ', g, 'N2 = ', h, 'CO2 = ', j, 'H2O = ', k, 'NO = ', l)
    print('a = ', a, 'b = ', b)

    # Reactants
    dG_CO = emp.get_dG('CO', T)
    dG_H2 = emp.get_dG('H2', T)
    dG_O2 = emp.get_dG('O2', T)
    dG_N2 = emp.get_dG('N2', T)

    # Products
    dG_CO2 = emp.get_dG('CO2', T)
    dG_H20 = emp.get_dG('H20', T)
    dG_NO = emp.get_dG('NO', T)

    # Insert equation for components below.
    # Reactants are either consumed or there is some excess/remainder meaning that the Product side is simply summed

    eq_co2 = j*(h_0_f_CO2 + dG_CO2 - emp.get_cp('CO2', T_reactants))
    eq_h2o = k*(h_0_f_H20 + dG_H20 - emp.get_cp('H20', T_reactants))
    eq_no = l*(h_0_f_NO + dG_NO - emp.get_dG('NO', T_reactants))
    eq_co = e*(h_0_f_CO + dG_CO - emp.get_dG('CO', T_reactants))
    eq_h2 = f * (dG_H2 - emp.get_dG('H2', T_reactants))
    eq_o2 = g * (dG_O2 - emp.get_dG('O2', T_reactants))
    eq_n2 = h * (dG_N2 - emp.get_dG('N2', T_reactants))
    print('CO2 = ', eq_co2, 'H2O = ', eq_h2o, 'NO = ', eq_no, 'CO = ', eq_co, 'H2 = ', eq_h2, 'O2 = ', eq_o2, 'N2 = ', eq_n2)
    f8 = eq_co2 + eq_h2o + eq_no + eq_co + eq_h2 + eq_o2 + eq_n2  # should equal zero, if not, iterate T
    print('f8: ', f8)
    print('T =', T)
    if 100 > f8 > -100:
        flag = 1
        print(T)
        T_data[i] = T
        T = 1.7
        i = i + 1
        print('i = ', i)
        flag = 0

        if i == i_len:
            mpl.plot(range(ratio_lower, ratio_upper, increment), T_data)
            mpl.show()
            exit()
        solver()
    else:
        if f8 > 0:
            T = T - 0.02
        else:
            T = T + 0.02
        solver()

solver()

The way I am solving for temperature is by (assuming adiabatic) that
0 = n_products(h_product_formation + cp_product * dT) - n_reactants(h_reactants_formation + cp_reactant * dT)
dT being (T(im trying to find) - T_mixing (I think) The air and fuel are 600 and 300 K respectively. Since I have to iterate to find T, I start with a guess and continue until the above equation is satisfied. However, I am not getting results that align with what gasEq gives me. IS THE ABOVE EQUATION CORRECT? If so, how is it being implemented incorrectly?
Example result:

Welcome to the site, but please try to condense and make this question clearer. As it stands it's rather lengthy and contains information likely not relevant to the question(s) at hand.

You may find useful these links for text formatting ( not to be applied to titles ):
notation ,  text + formula formatting  and
upright vs italic
and Math SE MathJax tutorial.

See Carbon Dioxide - Specific Heat of Gas vs. Temperature and the same for water: Water Vapor - Specific Heat

Do you have a reference for your chemical equation for the composition of syngas?


text:
Where does the energy from combustion come from?
I am aware that combustion reactions are exothermic. The energy stored in the chemical bonds of the products is less than the energy stored in the chemical bonds of the reactants. The difference in energies manifests itself as energy released into the environment. However, this sounds somewhat mystical. How does breaking bonds result in thermal energy, or greater kinetic energy of surrounding molecules? The most satisfying answer that I have seen so far is: the repulsion between like-charge molecules held in a covalent bond is constrained by the covalent bond; once something breaks that covalent bond, the molecules move away from each other, increasing their kinetic energy. Does anyone have further insights on this?

See also: https://chemistry.stackexchange.com/q/119093

The forces between the atomic and sub-atomic particles are not "mystical", just complex. For example, the atom nucleus consists of protons that electrically push each other apart, but at the very small scale, the "strong interaction" will keep these together. If you study this kind of physics, you will be able to compute the energy levels of atoms and molecules.

One way of looking at it: Richard Feynman Fire

Better is to say "Energy released by forming bonds of combustion products is bigger than the energy needed to break bonds of combustion reactants.".
Particularly breaking $\ce{C-C}$, $\ce{C-H}$ and $\ce{O=O}$ bonds needs less energy than is released by forming $\ce{O-H}$ and $\ce{C=O}$ bonds. The nature of the released energy of chemical bonds is electrostatic potential energy being lower than for separated atoms.
In other words, breaking the bond does not release energy, it consumes energy, like if you detach two magnets or two macro objects with the opposite charge. It is the forming of bonds that releases energy, via interaction with other molecules while the bond is forming. (*)
When a bond is being created, the bond gets very high vibration energy(**). It has to pass at least part of this energy somewhere else, otherwise, the bond breaks again. This is frequently done in a mechanical way, when moving atoms in this vibrating bond pushing other atoms or molecules, passing a part of their energy.
Such a bystanding molecule transforms this energy to its electron excitation or some form (translation, vibration, rotation) of molecular kinetic energy.
The increased kinetic energy of molecules after the reaction contributes to the increase of the average kinetic energy of molecules of the mixture being combusted. And, as consequence, increasing temperature as the statistical measure of the average kinetic energy (per degree of freedom) for very large samples of interacting molecules.
E.g. $$\ce{R-CH2-H(slow) + O(slow) -> R-CH2(fast) + O-H(fast)}$$
or
$$\ce{H(slow) + O(slow)  + N2(slow) -> O-H(fast) + N2(fast)}$$
Note that
$$\ce{H(slow) + O(slow) -> O-H(fast) }$$
is not possible due to the momentum conservation law.
Some energy can be released as photons too. Some are emitted by excited molecular bonds ($\ce{C-C}$, $\ce{C-H}$), but most of the radiation in the context of combustion and flame is caused by black body thermal radiation of condensed carbon atom chains, soot or solid particles.

(*) Explanation of quantum chemistry is the topic for thick textbooks, but illustration based on classical analogies would serve well. The proton and the electron in hydrogen atoms, even if in motion, have lower energy than if they were at rest but free. You need to provide energy to separate them, as they attract each other. Similarly as for planets in the Solar system. When two hydrogen atoms are bound in a hydrogen molecule, each electron is around other electrons to repulse each other, but also around another proton to attract each other. The mean squared distance electron-proton is significantly smaller than the mean squared distance electron-electron or proton-proton, so the total energy is even smaller than for separate atoms.
Reaching states with a lower total energy of electrostatically bound systems is the source of chemical energy.

(**) The energy of a hydrogen molecule - as an example - depends on the distance of hydrogen atoms. It decreases with shortening distance, reaches the minimum, and starts to increase again when atoms keep getting closer. As the result, there is pushing together if they are too distant and pushing away if they are too close. As the result, the hydrogen molecule is able to vibrate along the bond, as if both atoms were connected by a spring, trying to keep them at the particular distance = the length of $\ce{H-H}$ bond. It is just a coarse approximation as the behavior is non-linear. If too close, repulsion raises really steeply. If too far, attraction stops raising and starts decreasing toward zero.

I was posting a comment about IR when I saw that the answer has been edited. Fine! Could you add a warning at the final part (the reaction as collision between slow and fast particles) ? As it is, it might induce readers in mixing  statistical ensemble and collisions in well defined systems. (Of course two molecules can react in an quasi vacuum, but then there is no relation to the T). This confusion is the cause of most questions on this topic (although it does not seem the case here, where OP just ask HOWv exothermic reactions heat the surrounding medium) .

@Alchimista Hmm, OK. I have thought it was clear and still think it is correct, but I will reformulate it.

I upvote. I wonder if you get what I mean. Basically that if we burn 1 C and 2 O no need to speak of T. The label slow and fast can induce in treating a chemical reaction as a simple well define collision (which can be).

@Sure. I have originally written "contributes  to increasing temperature", what definitely does not say these few atoms have temperature, but I do agree it may lhave lead to confusion. I have already said multiple times in many questions atoms/molecules do not have temperature.

"The nature of released energy of chemical bonds is electrostatic potential energy being lower than for separated atoms...When a bond is being created, the bond gets very high vibration energy...a bystanding molecule transforms this energy to its electron excitation, or some form (translation, vibration, rotation) of molecular kinetic energy." I am still somewhat confused as to where the electrostatic and vibration energy comes from. How does the combustion of a molecule of methane create kinetic energy of surrounding molecules, explained in terms of electrons, protons, etc.?

You do not want me to explain quantum chemistry in a single answer, do you?

If you imagine the ideal lossless round  hole and ball, then if you put the ball at the edge of the hole, it would start oscillation (=vibration) between the opposite hole edges. Or escapes at the other side.One side is 2 separate atoms, bottom is 2 bound atoms at the right distance, the opposite edge is 2 bound atoms too close,  repulsing each other.

Check if added explanation helps.


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isotopologue distribution, difference among CID (carbon isotopologue distribution) and MID
I am new into LC-MS/MS stable isotope targeted downstream analysis. I've found numerous papers talking about MID (Mass isotopomer distribution) but in some european publications I found CID (carbon isotopologue distribution) and it seems to be the same, the vector of relative isotopologues abundances for a metabolite, could you please tell me if MID==CID ?  Thank you in advance

Have you looked up the definitions of isotopomers and isotopologues? Just like all oranges are fruits, but not all fruits are oranges, isotopologues and isotopomers are different types of isomers but not all isomers are isotopologues and isotopomers.

yes, I understand that isotopomers are indistinguishable in MS, so  $i$ labeled species are grouped in M+i isotopologue., but check for example ' Computational data mining method for isotopomer analysis in the quantitative assessment of metabolic reprogramming': " A metabolic flux ratio between two pathways can be estimated from the 13 C-labeling patterns or the mass isotopomer distribution vector (MDV) . The only vector that one can determine from MS is that of isotopologues , isn't it ?


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Why is it that the enthalpy of hydration equals the difference of solvation enthalpies of anhydrous and hydrous salts
Taking an example from my textbook
At finite dilution of CuSO4, heat is liberated
$$\ce{CuSO4(s) -> CuSO4 (aq)}.$$   ΔH of solvation = -15.8 kcal (anhydrous CuSO4 solvation reaction)
and when it is diluted further to infinity the enthalpy change is positive
$$\ce{CuSO4.5H2O(s) -> CuSO4.5H2O (aq)}.$$   ΔH of solvation = +2.9 kcal (CuSO4 pentahydrate solvation reaction)
1. Why is it that the enthalpy of hydration equals the difference of solvation enthalpies of anhydrous and hydrous salts.
ie
ΔH of hydration = ΔH of solvation of anhydrous salt - ΔH of solvation of hydrous salt.
2. Why does the anhydrous salt get hydrated only when there is excess solvent present? When water interacts with CuSO4 anhydrous should it not hydrate it before dissolving it?

What textbook is it, and what is the exact wording of the question?

See e.g. https://www.lacitycollege.edu/Departments/Chemistry/documents/Chemistry-102-Experiments-Documents/E09ENTROPY page 3 and 4

I've added the paragraph from my textbook. Thank you.

Sorry. Your book is wrong ! $\ce{CuSO4·5H2O(aq)} $ does not exist. When dissolved in water, this substance is transformed into two ions :  $\ce{Cu^{2+}}$ and $\ce{SO4^{2}}$, which may be hydrated. But they are independent from one another. And $\ce{CuSO4·5H2O(aq)}$ does not exist in water

What do hydrated ions mean. Thank you

Be aware of terminology confusion, as hydration in context of ionic compound solutions usually means solvation by water, as hydration is special case of solvation. // Consider that enthalpy change over closed path ( real or formal) must be zero. So 1/salt hydrate loosing crystal water 2/dissolving anhydrous salt 3/ crystallization of salt hydrate has the total enthalpy change exactly zero.

A hydrated salt, like $\ce{CuSO4.5H2O(s)}$, is a solid ionic crystal that incorporates water into its interstices (i.e., into the spaces between the copper and sulfate ions).  If you heat it, this will tend to drive off the water (dehydrate it).  Such dehdrated compounds are hygroscopic, since they will then tend to absorb water back from the air.  By contrast, hydrated ions, like like $\ce{Cu^{2+}(aq)}$ and $\ce{SO4^{2-}(aq))}$, are ions dissolved in water.  They are called "hydrated" both because they are dissolved in water, and because the water forms a solvation shell around them.

@Maurice I didn't read the photo but that could be just a thermodynamic cycle, I suspect, and is perfectly OK doing it.


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What sort of chemical would require a school to be evacuated?
I am doing research for a fictional story. I want to create an accident which would force students to be evacuated from a school. Is there a type of chemical spill that might take place in a chemistry class that would lead to all students being evacuated?

This is a very broad question because there are so many possible examples!

As seen in real life while in grad school, the following incidents required evacuation and visits from the fire department: bromine spill, peroxide formation in diethyl ether and accompanying spark of vacuum pump catching on fire, acrylamide incident, and the mother of all accidents involving sodium metal and ethanol dumped into a sink described here. I was very thankful to be in a theoretical lab during my stint there.

Breaking a mercury barometer and not cleaning it up before the mercury goes into the cracks would do it. Or trying to clean up the mercury with a vacuum cleaner.

Mixing a couple of litres of hypochlorite and ammonia solutions .

The comments provide great examples, but they all were valid in the age of logic and reason. These days, it seems, a spill of dihydrogen monoxide in a chemical lab would cause a complete evacuation and a team in hazmat suits.

Asbestos found in the ceiling or walls

Just open a can of this food item: https://en.wikipedia.org/wiki/Surstr%C3%B6mming. It has actually been used at protests.

I recommend visiting Worldbuilding Stack Exchange for questions like this one.

I am surprised no comments picked the easy choice; mercaptan( ethyl or methyl). The odorant that is added to natural gas. Several drops will stinkup a building as if there is a gas leak and evacuation will be ordered. I suppose if someone immediately confessed to the spill the evacuation might be cancelled. If it is done deliberately one would put it somewhere in the ventilating system. No health risk other than falling down while evacuating.

Given that an answer has been provided and if you continue to insist on committing a crime, a possibly less damaging path for clean up and otherwise  problematic (so that you could try arguing 'harmless prank' upon your arrest) is to ignite a magnesium ribbon inserted into a metal container filled with an ammonium salt. This creates a smoke screen.
My safest choice would be ammonium citrate or ascorbate. However. the smoke itself could still possibly trigger an asthmatic attack which is just one example of what could go very wrong.
Also, please note that anyone injured in an evacuation, precipitated by your actions, may seriously escalate the criminal charge(s), so do re-consider, especially if you cannot afford a good lawyer or do not have a strong affinity for prison life.


text:
A good lab demonstration of this particular optical transform pair?
At about 1:11 the video linked below shows what think is the optical Fourier transform of a sine wave as approximated by a saw blade.
https://www.youtube.com/watch?v=jzmqeRp_tmk
The set-up appears to be a high-quality 4F system with a green laser, two lenses, and a display screen--a good apparatus for demonstrating Fresnel diffraction.* This video is the closest I have found to a simple visual demonstration of the $\sin \leftrightarrow F(sin)$ Fourier transform pair (but it isn't exactly that).
My question is whether it is possible to demonstrate the $\sin\leftrightarrow F(\sin)$ pair optically in the lab.
*Fresnel diffraction here just means the aperture is not small.

For the close votes: I edited the question and found an authoritative answer to this question which is not opinion-based.

Back in the day when X-ray diffraction still was recorded on film, the symmetry relationship was demonstrated with $\pu{35 mm}$ slides with such patterns. Lisensky et al., 1991JChemEduc91 (https://doi.org/10.1021/ed068p91) is one (late) testament to this how to create them, especially when portable Nd:YAG lasers became popular.

See a setup discussed on physics.se: https://physics.stackexchange.com/questions/155216/possibility-of-making-an-experiment-in-a-classroom-to-simulate-dna-diffraction

@Buttonwood: Yes I have done that lab and have some nice images of the X pattern from a helical spring.

Well, this is reminds to photo 51 ...

Although the YouTube videos you linked is nice, one can only wish if they added some explanations too and a complete picture of the set-up. After watching the experiment. All I feel with the sawtooth diffraction is that it is a very broad central maximum. It is not a sin x <-> F(sinx) pair. I don't think such a pair exists in diffraction, because if the were a pure sine FT, the central maximum must be absent. We should get just two spot corresponding to the sine wave frequency. Diffraction could be occuring at the very tips of the sawtooths. However, single slit diffraction (rectangular window) and the sinc function like appearance on a wall (Fourier transform of the rectangular window) is a decent illustration of optical transform pair.
In terms of X-rays, I recalled an demo experiment in an old book using four transmission gratings. You can easily buy transmission diffraction gratings from Edmund Optics. Fortunately, the book is available on Internet Archive. See Nokes, Demonstrations in Modern Physics (Second Edition). You can judge from the yellow pages that how old the book is but sometimes old books have treasures.
The author on page 103-105 shows four diffractions gratings in a concentric tube and each tube can be independently rotated or in other words, each diffraction grating can be crossed as follows:

When you pass collimated beam of visible light, through the four gratings set-up, a beautiful diffraction pattern is formed, very much like a Laue's diffraction pattern of X-rays and crystals.

This is neat. Not sure I see a direct connection to the question but a very interesting set-up.

The diffraction is similar to Laue pattern of X-ray diffraction which is also a Fourier transform of electron density in solids. Can you elaborate your question a little bit more? You have a saw there in the video, which you say, is close to a sine wave shape. Its FT is a displaced delta function.  Optically, how do you see a delta function on the screen?

Right, it's diffraction that gives Fourier images. My question concerns the particular Fourier transform pair. What I expect is that both the sine and the delta function will  be rough approximations of the mathematics. For the delta function, maybe two points of intensity to either side of the center. So in the video we see a saw blade and a series of vertical lines. I expect only two. Is this because there are other frequencies in the saw-sine?  I have no idea. Maybe Young's double slit gives a good approximation for certain slit dimensions. Does this help?

@daniel, I do not think that such an optical pair exists for sin x,->F(sin x). Maybe a single slit diffraction will be equivalent to a sinc function. This could be used to illustrate optical FT concept.

You may be right but then what do you think the saw-blade is showing at 1:11?

I really wish the YouTuber showed the optical path of the whole set up. All I feel is that it is a very broad central maximum. Diffraction could be occuring at the very tips of the sawtooths. However, single slit diffraction (rectangular window) and the sinc function like appearance on a wall (Fourier transform of the rectangular window) is a decent illustration of optical transform pair.

@daniel, I edited the answer.

Thanks--Could you include your assertion that the particular optical pair probably isn't feasible? I don't trust comments to hang around indefinitely..

In Optics at p. 513$^*$ the authors state that Young's double-slit is the practical realization of the Fourier sine-delta pair, noting that the approximations to $\delta$ functions have finite shape which means the transform (sine wave) will not be ideal.
(Can skip)
The practical transform is created by convolving the $\delta$ slits with a rectangle. In the frequency domain this corresponds to multiplication of a sinc and sine function which gives a frequency-modulated sinc. That function is then squared to get the intensity and resembles the usual double slit pattern.
Also, as the distance between slits is decreased to nothing, the double-slit pattern merges continuously into the single-slit pattern shown below (my photo). So the answer to the OP is "yes" but it's not visually obvious.

$^*$Hecht and Ganesan, Optics (Pearson, 2015)


text:
XANES simulation software/code besides FEFF?
I need to generate the XANES region spectra of the Pt L3 edge of a theoretical Pt6 crystal with adsorbed atoms (such as hydrogen, oxygen, etc). It is to generate theoretical references as described in this paper: https://doi.org/10.1039/b927120c
However, based on what I have read, I would need the FEFF8 code but I only have access to FEFF8-lite (free version, limited to EXAFS analysis) and the Demeter package which is for analysis of experimental data, if I'm not mistaken. Is there an alternative software (free) or method to generate or calculate a theoretical XANES spectra?

There might be someone here who can answer, but since this question is largely about finding modeling software, it may be a better fit on Matter Modeling SE

Try FDMNES
http://fdmnes.neel.cnrs.fr/ Team
SINThe FDMNES project is developped in the SIN team, Institut Néel, CNRS, Grenoble, France, by: Yves Joly, Yvonne Soldo-Olivier and Oana Bunau Very important numerical improvements were included by: Sergey Guda, Alexander Guda, Alexander Soldatov, Mikhail Soldatov
Southern Federal University, Rostov-on-Don, Russia An extension for EXAFS, FDMX, is due to: Christopher Chantler and Jay Bourke
Melbourne University, Australia


text:
Do Kp and Kc depend on initial concentration of reactants?
Are $K_P$ and $K_C$ dependent on initial concentration of reactants? So far I had known that $K_P$ and $K_C$ are only dependent on temperature which is explained by thermydnamic approaches perhaps. I didn't learn the derivations of those equilibrium constants. I just learnt that they are the ratio of product of active masses of  respective products and reactants. But I am asking this question because I was stuck at a point while solving a problem regarding $K_P$ and $K_C$
The problem:
In the dissociation reaction of Nitrogen tetraoxide$(N_2O_4)$ $100\alpha$% of the reactant is dissociated at $P atm$ pressure and at $TK$ temperature[suppose volume=$V$]
so as per the problem if the initial concentration of the reactant is $C M$ then concentration of product is $\alpha C M$ at equilibrium
$$N_2O_4<=>2NO_2$$
$$initial:C <=>0 $$
$$equilibrium:C-C\alpha<=>2C\alpha$$
so $$K_C=\frac{[NO_2]^2} {[N_2O_4]}=\frac{4C^2\alpha^2} {C(1-\alpha)}=\frac{4C\alpha^2} {(1-\alpha)}$$
$$K_P=\frac{P_{(NO_2)}^2} {P{(N_2O_4)}^2}=\frac{\frac{4C^2P^2\alpha^2V^2} {CV(1-\alpha)+2CV\alpha}} {\frac{CPV(1-\alpha)} {CV(1-\alpha)+2CV\alpha}}=\frac{4P\alpha^2} {(1-\alpha)}$$
so It can be seen that in the equation of $K_C$ there is $C$ involved in the numerator. since equilibrium constants are not dependent on intial concentration, I suppose that the degree of dissociation($\alpha$) varies depending on initial conditions. But if $\alpha$ changes depending on intial concentration does not that make $K_P$ dependent on intial concentration? Where am I wrong here?
Thank you.

The constants  and  have dimensions, which are in bar and mol/L. So if the pressure or the concentration changes, the parameter  adapts its value so as to obtain the same value of the equilibrium constants.

ohh I messed up by thinking one dimensionally that is considered the change of $\alpha$ only to maintain consistency of $K_C$ but ignored the necessity of maintaining numerical consistency of $K_P$ alongside.

a small point but I think you mean bar and mol/L in your first line.

@porphyrin. Thank you for the comment. I have edited it in my text.

It's not generally the case that Kp and Kc have dimensions.  That depends on the convention used by the textbook or paper.  See https://chemistry.stackexchange.com/questions/107374/what-are-the-units-of-kc-and-kp/107414#107414  Your conclusion—that Kp and Kc don't depend on initial concentrations—is correct, but the reason you give for this is not:  The reason Kp and Kc are independent of initial concentrations is not that they have dimensions, since they are also independent of initial concentrations when the dimensionless convention is used.

@theorist. I agree about your critics. I must also state that the initial question is not clear. First, the data do not specify whether this reaction is carried out at constant volume, or at constant pressure. Furthermore, the value $K_c$ should not be used with gas reactions. Only the $K_p$ constant makes sense, thermodynamically speaking. Equilibrium constants $K_c$ are introduced in thermodynamics with solutes, because it is obvious that their partial pressures cannot be measured.


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When base is added to water it produce OH- ions but why does it not react with H2O as when acid is added to water H+ ions forms H3O+ ions?
Why bases produce OH- ions and why this ions Doesn't react with H2O ? Is these ions are stable ? In the case when acid is added it releases H+ ions which then reacts with water to form hydronium ion but why not hydroxide ion ?

Welcome to Chemistry SE site! When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit compact summary of partial answers/ideas/thoughts you have got until then. It is quite common requirement across all StackExchange sites. It is evidence of effort and gives the hint what you are already aware of. How do I ask a good question.

If you realize H+ - if taken literally - is just a proton/deuteron then you can notice the huge difference between a proton and OH-. Both H3O+ and OH- are further hydrated by H2O molecules via hydrogen bonds.

What happens to the concentration of OH- when an OH- reacts with a water molecule?


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What is the yellow coating seen on the sides of containers storing copper Sulphate?
When storing a solution of copper sulphate, a yellow colored deposit slowly formed on the bottom and sides of the container. What/Why could this be?
I'm using it for simple 8th - grade level reactions. Like electroplating, reacting it with iron etc. I'm storing around 250 ml for around a month or less until used.
I suspect some excess Sulphur but I have no way to prove it.

I would guess if is hydrolyzed/oxidized/precipitated iron(III) oxide/hydroxide from copper sulphate impurities.

What would I have to do to prevent this in future? Purify it somehow? Its not a problem, it doesn't really interfere with reactions, but I have to filter it every time I use it, otherwise the solution is really cloudy.

Provide more context about the purity, purpose, amount or volume, storage interval etc, as the optimal advice may largely depend on that. // Background or context info omitted in too laconic questions has to be frequently clarified later with big effort on all sides.

I unfortunately can't provide info on purity because the amazon listing did not have any info about it and I'm afraid I don't have the equipment to measure its purity. However, I've added the rest as an edit

Small volume of added sulphuric acid to keep low enough pH could help.

Any other alternate?, I'm just doing basic chemistry at home, and I don't have access to sulphuric acid.

More expensive, but safer to handle than sulfuric acid, is sodium bisulfate, NaHSO4. This used to be available in crystal toilet bowl cleaners, but these have been discontinued. Chemical suppliers should have it, though.

What is the container made of?

The one in the picture is made of glass, but I had similar results trying to store it in plastic


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Interpretation of azeotropes from statistical mechanics derivation
In an earlier posted question titled Reason for the formation of azeotropes, @porphyrin presented an elegant derivation from statistical mechanics to interpret partial pressures for the two components of a binary mixture and the azeotropic composition.
The figure of $p_A/p_A^0$, which is normalized partial pressure or activity $\alpha_A$, shows several curves calculated with different $\beta$ values or equivalently $\Delta E/k_B T$ as a function of $x_A$. Most curves in that figure show monotonic changes with mole fraction except for the curves with $\beta$=2 and 3. From realistic experimental data, however, partial pressure curves appear to only show monotonic changes as a function of mole fraction, i.e, the derivative of partial pressure curves won't change sign (it does change sign for the total pressure thereby yielding one point with a derivative of zero at the azeotropic composition). Therefore, the partial pressure curve with $\beta$=2 and at least curve with $\beta$=3 may not be realistic, even though math does produce such curves if large $\Delta E$  is used. By taking the derivative of partial pressure $p_A$ w.r.t. $x_A$, one can find the condition for $p_A$ to show only monotonic changes as a function of mole fraction. It turns out that when $\Delta E/k_B T$ = 2, $x_A$ is located exactly at $1/2 \pm 1/2 \sqrt{1-2/2}$ = 0.5 giving only one derivative value of zero, and when $\Delta E/k_B T$ = 3, $x_A$ is located at $1/2 \pm 1/2 \sqrt{1-2/3}$ = 0.211 and 0.789 giving two derivative values of zero. Therefore, as long as $\Delta E/k_B T$ is less than 2, the partial pressure curves will be showing monotonic changes as a function of mole fraction. An educated guess is that the binary mixtures are miscible if $\Delta E/k_B T$ < 2 and not completely miscible if $\Delta E/k_B T$ > 2, which form homogeneous and heterogeneous azeotropes, respectively.
I am wondering if the derivations given by @porphyrin based on statistical mechanics for the partial pressures are available somewhere in the literature (papers or books) so I can read more to understand and possibly cite the reference instead of the website link in my work?

What is the parameter $\beta$ ? And what is $\Delta E$ ?

As can be seen from the previous post, @porpyrin defined in his derivations $\beta = N_0 z w/RT = zw/k_B T = \Delta E/k_B T$.

Found in Atkins' Physical Chemistry that the critical parameter $\beta=2$ has to do phase separation, and $p_A= p_A^0x_Ae^{\beta x_B^2}$ and $p_B=p_B^0x_Be^{\beta x_A^2}$ are related to the Margules equations $ln\gamma_A=\beta x_B^2$ and $ln\gamma_B=\beta x_A^2$.


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Mass of silver iodide that dissolves in ammonia

What mass of $\ce{AgI}$ will dissolve in $\pu{1.0 L}$ of $\pu{1.0 M}$ $\ce{NH3}?$ Neglect the change in concentration of $\ce{NH3}.$
Given: $K_\mathrm{sp}(\ce{AgI}) = \pu{1.5E-16};$ $K_\mathrm{f}(\ce{[Ag(NH3)2]+}) = \pu{1.6E7}.$ Atomic masses of silver and iodine are $\pu{108 u}$ and $\pu{127 u},$ respectively.

For this question I combined two reactions and wrote
$$K_\mathrm{eq} = K_\mathrm{sp} K_\mathrm{f},$$
but after that I think some data is missing, so I'm stuck here.

Can you elaborate your working?

Please  note that MathJax, as the name implies, is used to typeset in math mode. It makes no sense to enforce \text{…} on everything in math mode. While this issue has been fixed, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. As for the question, it would be nice to see your thoughts beyond generic formula appearing out of nowhere.

There is enough data, try again and more. Provide details of your work. The site policy is to refuse solving such tasks for users unless sufficient explicit effort is provided.

We will show you the way. Try first to see that when $\ce{AgI}$ gets dissolved, most of the silver ions are transformed into $\ce{[Ag(NH3)2]^+}$ ions, so that the concentration in iodide ion is $\ce{[I^-] = [Ag^+] + [Ag(NH3)2]^+}$. Now the residual $\ce{[Ag^+]}$ is negligible, and the concentration of the complex is equal to $\ce{[I^-]}$. As a consequence you can deduce $\ce{[Ag^+]}$ from the solubility product. Draw then the concentration of the complex, and here you are.

I doubt that the solubility/formation product constants are given without dimension. Please cite the source and quote the exercise verbatim. Also, include all your work attempting the problem. In this case, write down the combined reactions.

I believe the above answer is correct. I will show my process:
We have $\ce{K_\mathrm{f} = \frac{[Ag(NH3)2+]}{[NH3]^2[Ag+]}} = \pu{1.5E-16}$
and $\ce{K_\mathrm{sp} = \frac{[Ag(NH3)2+]}{[NH3]^2[Ag+]}} = \pu{1.6E7}$
Set up mass balance: $\ce{[I-] = [Ag+] + [Ag(NH3)2+]}$. Since $\ce{[Ag+]}$ is relatively small, we can assume $\ce{[I-] \approx [Ag(NH3)2+]}$.
That means $\ce{K_\mathrm{sp}\cdot K_\mathrm{f} = \frac{[Ag(NH3)2+]^2}{[NH3]^2} =} \pu{2.4E-9}$.
Now, we can set $\ce{NH3} = 1-x$, and $\ce{[Ag(NH3)2+]} = x$.
Solving for $x$ gives $x = \pu{4.899E-5 M}.$
Based on $\ce{[I-] \approx [Ag(NH3)2+]}$, $\ce{[I-]} = \pu{4.899E-5 M}$.
This correlates with a mass dissolved of $\ce{AgI}$ of $\pu{0.0115g}$.

Here we have to assume that $\ce{[Ag(NH3)2^+] \approx [I-]}$:
\begin{aligned}
\ce{AgI (s) + 2 NH3 &<=> [Ag(NH3)2]+ + I-}\\
K_\mathrm{f} K_\mathrm{sp}
  &= K_\mathrm{eq}
   = \big[\ce{[Ag(NH3)2]+}\big][\ce{I-}]\\
\implies \pu{2.4E-9 mol2//L2}
  &= \big[\ce{[Ag(NH3)2]+}\big]^{2}
   = [\ce{I-}]^{2}
   \qquad\because \ce{[Ag(NH3)2^+] \approx [I-]} \\
\implies [\ce{I-}]
  &\approx \pu{4.89E-5 mol//L} \\
\implies n(\ce{AgI})
  &= \pu{4.89E-5 mol}\\
\implies m(\ce{AgI})
  &= \pu{4.89E-5 mol} \times \pu{235 u}
   = \pu{0.01149 g}
\end{aligned}
A mass of $\pu{1.1E-2 g}$ silver iodide dissolves.

No ! $K_f K_{sp}$ is not eqial to $\ce{[I^-]^2}$, as stated. It is equal to  $\ce{\frac{[I^-]^2}{[Ag^+]}}$

Welcome to Chemistry! On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. I have cleaned up the format as best as I could, added units here and there, but it is unfortunately far from correct.


text:
Potassium from Potassium superoxide
Is it possible to get $\ce K$ from $\ce K{O_2}$ using a synthesis reaction? If yes, can you describe it?

I suppose easier and/or cheaper ways are preferred. KO2 assumes it has (probably) already been K once. It is like to isolate K twice. // Attempts to post the most laconic question ever are at risk they will be closed and most users will not notice them.  How do I ask a good question.

$\ce{KO2}$ is a dangerous and explosive substance. Don't use it. Better destroy it in large amounts of water.

Thermal decomposition should help but it requires very high temperature. It goes through various oxides and ultimately decompose to potassium.
$$\ce{KO2 ->[<400 ^\circ C] non-stoichiometric oxides ->[400-480 ^\circ C] K2O2 ->[500 ^\circ C] K2O ->[560 ^\circ C] K (+oxygen ^)}$$
Do note the reaction is performed in vacuum otherwise it can oxidized back to peroxide/superoxide.
Ref. STUDIES OF THE THERMAL DECOMPOSITION OF THE POTASSIUM OXIDES IN THE TEMPERATURE RANGE 300--480 DEGREES C, RILEY, JOHN F. 1968. (link)


text:
Reaction mechanism of phenylhydrazine with carbonyl
I am currently working on the following problem, but I don’t know how it is solved:

I guess it is a Fischer-Indol synthesis, but I am not sure. This is what I got:

Any help would really be appreciated^^

If your study of chemistry advanced to the level to address questions like the one presented, familiarize yourself with an editor to sketch the reaction schemes, and to export the scheme as a .png you may upload here (limit: 2MB). Some of them are compared here. If your institution does not offer a campus license, e.g., ChemDraw and Marvin offer sketchers with less functionality yet running from a web browser (no installation).

The question you pose is the work of Stork and Dolfini that led to the synthesis of aspidospermine 7 and quebrachamine 9. The aminoketone 1a was condensed with either phenylhydrazine or o-methoxyphenylhydrazine to form the respective arylhydrazones 2a and 2b. Exposure of either hydrazone to hot acetic acid led to a "Fischer indole synthesis" that produced the indolenines 6a and 6b. Reduction of indolenine 6b with LiAlH4 followed by acetylation afforded d,l-aspidospermidine 7, whose stereochemistry was not known at the time. Reduction of indolenine 6a with KBH4 effected reduction of immonium salt 8 (blue arrows) leading to d,l-quebrachamine 9,
One might ask why a normal Fischer indole synthesis did not occur by having the imine double bond of hydrazone 2 tautomerize toward the methylene group of the tricyclic ring moiety. My recollection, having been in this laboratory at the time, was that this issue was a problem when the sequence of reactions was run on the ketolactam 1b. [@Waylander]

G. Stork and J. E. Dolfini, J. Am. Chem. Soc., 1963, 85, 2872.

If you thought that a Fischer indole synthesis was involved, you should have made some effort in that direction.

@user55119 Speculation: The OP did not recognize the tautomeric relationship between compounds 2 and 3.

@user55119 Good answer. I was not aware of this work. Is it known whether the isomeric indole is also formed in the reaction from 1a?

@Waylander: After acetylation the paper states"...gave a crude material from which aspidospermine..." I suspect that some indole was formed. But then again, that was 58 years ago. My memory is good but not that good. ;)

@user55119  and the analytical methods of the time may have struggled to find it if it was present in small amounts

@Waylander: We had tlc and a 60 MHz NMR. Ketolactam 2b formed an indole. LiAlH4 reduction of the lactam would have given an authentic sample. I don't know if that was done. My previous quote, which was  carefully couched by the authors, makes me believe that the indole was in the crude product.

@user55119  I agree, I think it very likely some was formed. We need that keen project student to run the reaction and get an LCMS of the reaction mixture

Fischer Indole for the first step looks possible depending on reaction conditions, but this source states that the indole nucleus is not reduced by LiAlH4, nor do I recall ever seeing it done. Therefore I think what is happening is formation of the phenylhydrazone followed by reduction to the 1,2 di-substituted hydrazine.


text:
How can we recover potassium and phosphorus from ashes?
When plant based solid fuels, such as firewood, coal etc. are burned, their potassium and phosphorus content stay in the solid ashes. Quick search on the net shows ashes contain about 4% potassium and 1-2% phosphorus.
What chemical methods exist to separate potassium-something, something-phosphate out of ashes?
It seems mixing ashes with water will dissolve potassium salts, but still unsure what can we do the get the phosphorus out.
(I find the question interesting because currently both potash and phosphate are mined, and thus non-renewable. So if we want to be sustainable, we will need a way to recover these nutrients from organic waste and ashes.)

Why would you want to separate them from ashes?

Search for they recovery, you should find quite some material. Urine can be a relatively good source for phosphates. Ex https://www.sciencedirect.com/science/article/abs/pii/S1385894719326944

@IvanNeretin Because I think ashes contain a few simple inorganic compounds rather than a lot of complex organic ones. Perhaps it's easier to reason about.

With my comment I wanted also to say that it is done. I do not know the scale of that industry, but recovering for fertiliser is done, at least for phosphates. Just the source are so that the phosphate is already concentrated by living animals and, yes, humans too.

If you want to use these compounds as fertilizers, ashes are good enough as is, without any separation. On the other hand, if you want to do some chemistry, this source is too dirty anyway.

Ashes are made of
$30 - 70$% $\ce{CaCO3}$,
$5 - 12$% $\ce{K2CO3}$,
$5 - 12$% $\ce{Na2CO3}$,
$5 -10$% $\ce{MgCO3}$,
$2 - 15$% $\ce{SiO2}$,
$2 - 5$% $\ce{Ca3(PO4)2}$,
$2 - 5$% $\ce{Mg3(PO4)2}$,
$2 - 3$% $\ce{K2SO4}$,
$< 2$% $\ce{AlPO4}$,
$< 2$% $\ce{FePO4}$,
plus less than $1$% oligoelements. Water will dissolve the sodium and potassium compounds. Adding then acidic solutions will extract the calcium, magnesium and iron compounds. It will not be easy to separate potassium and phosphorus compounds out of these solutions.
Ref.: http://gallica.bnf.fr/ark/12148/bpt6K33845.image.r=Dictionnaire+de+chimie.f181.langFR


text:
What happens inside the orbital when energy provided to electron is not sufficient for it to change orbitals?
While studying the chapter called Atomic Structure, we were introduced with Bohr's model of an Atom. Even though not all of his postulates were right, I believe some were. A doubt arose in the postulate stating energy is only absorbed or emitted when an electron jumps from one orbit to another. My query is, if the energy we supply is not sufficient for it to jump to the next orbit, what would be taking place inside the orbital? Would it absorb that energy at first, then lose it due to the centrifugal attraction and radiate it? Or would it not even absorb it and just let it pass? The way it is written in my textbook makes it sound like the electron just doesn't take the energy, or atleast that's what my interpretation of the manner was. But it's a bit weird, even though everything is in the quantum world. It's like I throw a ball at you, and you only catch it if the speed is X, but you can't measure the speed of the ball. Hence, I want to clarify if this is a misconception I have or another one of the beauties of the quantum world.
Thanks,
Ravi

"Let it pass" is the most adequate description.

In particular let it go given the context. As far you are at Bohr level, or even more advance treatment, let it go is what happens as far the electronic levels are concerned. But, as you suspect, there are other interactions like scattering. (I personally reserve the word Absorption for the above interlevel transitions, but the interactions I have mentioned are often described - as you did - like an absorption fast followed by re-emission, this might be justified if one is to look and formalise the inner mechanism).

"It's like I throw a ball at you, and you only catch it if the speed is X [, otherwise it passes through you]." I haven't heard that metaphor before, but I think it's quite fitting to describe absorption.

@IvanNeretin So , I think this Q is related to emission & absorption spectrum. If I pass a light through an object which it doesn’t absorb , it lets it pass but we do see the light after that .


text:
On the definition of eigenframe
While reading the following article,

Elizaveta A. Suturina, Kevin Mason, Carlos F. G. C. Geraldes, Ilya Kuprov, David Parker, Beyond Bleaney's theory: experimental and theoretical analysis of periodic trends in lanthanide-induced chemical shift, Angewandte Chemie, September 25, 2017

I came across the term eigenframe, which I am unfamiliar with. What is an eigenframe?
Here is the context in which it was used:

In particular, pseudo contact chemical shifts (PCS) induced by lanthanide complexes are commonly described by McConnell's long-range relation (SI units),
$$ \delta^{PCS} = \frac{1}{12\pi r^3}[\chi_{ax}(3\cos^2 \theta -1)+3\chi_{rh}\sin^2\theta \cos2\phi]$$
where $\theta, \phi, r$ are the polar coordinates of the nucleus relative to the lanthanide in the eigenframe of the molar magnetic susceptibility tensor, and $\chi_{ax}$ and $\chi_{rh}$ are axiality and rhombicity of the susceptibility tensor.

I can't find anything definitive but both
https://www.thestudentroom.co.uk/showthread.php?t=4172326
and
https://www.physicsforums.com/threads/eigenframe-definition-please.876035/
suggest that an eigenframe is the basis formed by the eigenvectors of the relevant operator, the molar magnetic susceptibility tensor in this case.


text:
What is the chemical equation of the decomposition of natural cork?
I am currently doing a study project on natural cork wine stoppers and I am investigating the emissions of composting natural cork (C123H182O56N). However, I am struggling with setting up the chemical equation of this decomposition. Anyone knows how to tackle this?

Ultimately, you get carbon dioxide, water and nitrogen or nitrogen oxides. Lots of intermediate steps, of course, and no way to make a sensible list of those. If the decomposition involves fungus, bacteria, etc., then more intermediate steps.

I believe cork is almost entirely composed of lignin and suberin with some cellulose and hemicellulose thrown in. There should be plenty of info available on the biodegradation of each of those substances separately. It will also depend on how good the aeration of your compost pile is because that will affect the amount of aerobic metabolism vs anaerobic

What is the origin of the chemical formula of cork you state? Since it is a (varying) blend of natural products (with lignin and cellulose itself describing polymers of varying chemical composition, too), you probably have to rely on empirical data ($\pu{x kg}$ of cork yield $\pu{xk L}$ of $\ce{CO2}$ over time and temperature. Some landfills monitor the release of gases anyway (example, a brief search) to draw inspiration.

Natural cork is not a chemical compound. The meantioned formula is just the supposed empirical element composition recalculated to the empirical formula. Furthermore, the decomposition is biological process of biodegradable material, with part being reused by living organisms.

Unfortunately, natural cork is not just a simple chemical compound. It is a mixture of suberin, lignin, cellulose and other organic compounds and you can't write a simple decomposition reaction. If you want reaction, google "thermal decomposition of 'the individual components e.g. lignin'".

The thermochemical decomposition of natural cork has been studied previously:

The thermochemical degradation of cork from Quercus suber L. was studied in the temperature range 150°C–450°C in relation to mass loss, chemical composition and the influence on the cellular structure. The degradation of cork is strongly dependent on temperature and mass losses become significant at 200°C (15% of initial dry weight) and increase rapidly for higher temperatures (27% at 250 °C, 49% at 300 °C, 62% at 350 °C) until ashing at 450 °C. The polysaccharides are the most heat sensitive components: at 200 °C, hemicelluloses disappear and cellulose is degraded to a considerable extent. Suberin is more resistant and degradation starts at approx. 250 °C; 300 °C-treated samples only contain 7% suberin. The cellular structure of cork is also significantly influenced by temperature. Upon heating, cells expand and the cell walls stretch, attaining at 250 °C a maximum cell volume increase corresponding to a factor of approximately 2. Above 300 °C, the structure of cell walls is considerably changed and show profound physical damage; in the later stages of pyrolysis, a cellular structure is no longer observed.

Reference:

Pereira, Helena. (1992). The thermochemical degradation of cork. Wood Science and Technology. 26. 259-269. 10.1007/BF00200161
https://www.reddit.com/r/chemhelp/comments/r14w9z/what_is_the_chemical_equation_of_the/


text:
Reaction between hydrobromic acid and sodium borate

Write the reaction for a a sodium borate solution being titrated by hydrobromic acid, assuming sodium borate is represented by $\ce{Na3BO3(aq)}$.

I think this is a simple double replacement reaction:
$$\ce{Na3BO3(aq) + 3 HBr(aq) -> 3 NaBr(aq) + H3BO3(aq)}\tag{R1}$$
Official solution without explanation:
$$\ce{Na3BO3(aq) + HBr(aq) -> 2 NaBr(aq) + NaH2BO3(aq) }\tag{R2}$$
I don't understand how the official reaction was reached. Why should a sodium cation take the place of a hydrogen?

It is rather $\ce{Na[B(OH)4]}$

The whole task seems to me like being made up. Who would use HBr for titration and end titration with alkalic solution ? I also do not believe much in Na3BO3(aq).

Don't forget, in contrast to e.g., $\ce{S}$ in $\ce{H2SO4}$, $\ce{B}$ in $\ce{H3BO3}$ (or $\ce{B(OH)3}$) still has the electron gap characteristic for Lewis acids like $\ce{AlCl3}$.

Poutnik is right. $\ce{Na3BO3}$ does not exist.

@Maurice Googling "trisodium orthoborate, na3bo3" give me 168 hits. The promising results are this and this.

The 2nd link states: "Both NaBO2 and Na2B4O7 were found to react readily with molten Na2CO3 at any Na:B molar larger than 3:1 to form trisodium borate (Na3BO3). The formation of Na3BO3 as a reaction product means that the amount of borate required for the autocausticizing reactions can be substantially lowered than previously suggested."

@NilayGhosh. I was thinking of aqueous chemistry. In the same reference, it is said that $\ce{Na3BO3}$ does exist. But it is immediately decomposed by water into $\ce{NaOH}$ and $\ce{NaBO2}$

Notice I have written Na3BO3(aq) not just Na3BO3.

The borate anions are a series whose charge depends upon pH. A good (and long, but probably still incomplete) reference is https://europepmc.org/article/pmc/6470676.

The request for an equation identifies two starting materials but does not suggest an endpoint, and there are many. The OP gives a complete reaction: react all the possible basic anions with acidic HBr. Good answer. The official solution is a partial neutralization - to what pH? The suggestion of titration needs to say what indicator or what pH is the endpoint, or else it must have been explained in class beforehand.
Or, the question is just a bad one - or a trap.

Wikipedia says at boric acid that Polyborate anions are formed at pH 7–10 if the boron concentration is higher than about 0.025 mol/L. The best known of these is the 'tetraborate' ion, found in the mineral borax.

@James Gaidis. Where have you found these diagrams of Boron stuff versus pH ?

@Maurice: the reference for the diagram is in the third line of my answer.


text:
Smell produced in reaction of sodium hydroxide and aluminium
I was doing an experiment with $\ce{NaOH}$ dissolved in some $\ce{H2O}$ and aluminium. My textbook says that the following reaction takes place
$$\ce{2 NaOH + 2 H2O + 2 Al -> 2 NaAlO2 + 3 H2}$$
It also says that hydrogen gas ($\ce{H2}$) is colorless, odorless, tasteless and highly combustible. But when I did the experiment the gas was kind of having a strong smell (not of rotten eggs) but indeed it was a strong smell which got me coughing really hard.
When I was doing some research about this I found this video with the same results:-
https://www.youtube.com/watch?v=nG0qUBxcvko&ab_channel=ChanKaiWee
( After the 30 second mark)
Can someone please explain why this happens?

Tiny droplets of the solution carried into the air along with the evolved gas. May want to to this reaction in a better-ventilated area or even under a laboratory hood.

so that cough was due to tiny NaOH and Sodium Aluminate Vapor ?

Proper enumeration of  chemical reactions requires just ability to count and compare small integer numbers, what is taught early in elementary school. 3=2 is not true just because it is in the reaction. // Traces of other elements may create volatile smelly hydrides, as the result of intensive reduction process.

it might be because I used tap water which might have some impurities and trace substances but is it enough for that strong of a smell ?

If we can exclude effect of aerosol of hydroxide solution, human nose is quite sensitive for some volatile compounds of heavy elements and these traces can have origin in Al. ( See e.g.  telluric breath, being perceived months after tellurium exposure -  https://episodictable.com/tellurium/  )

Please note that chemical names should in general not be capitalised (unless at the start of a sentence).

Oscar Lanzi's answer addressed the points correctly. This reaction of Al + NaOH pellets is used in drain openers for obvious reasons. The reaction is so exothermic that solution can start boiling. It generates a lot of heat, efferevesence and steam.
The key point is that only chemically pure hydrogen is odorless and colorless, otherwise it is not. The OP mainly smelled the caustic mist that is generated by vigorously evolving hydrogen bubbles which bursting over the solution surface. This reaction should be carried out in a fume hood.

I used to make hydrogen this way, when I was a kid and didn’t have manganese available. I always thought the stinky smell was due to impurities. In any event, it is easy to do an experiment: filter out all droplets with a drying tube, etc., and smell the dry gas.

@EdV, the garlic like stench of hydrogen is usually present when an acid is used with iron or zinc. With NaOH and Al, there is no stench, it is just the caustic spray from hydrogen bubbles that severely irritates the nose. You also reminded of childhood experiments!

Interesting. I know from several experiences that powdered iron, reacting  with either hydrochloric acid or sulfuric acid, produces bad smelling hydrogen. The last time I used the Al and NaOH method, as a high school kid, I underestimated the time required to flush out enough air in the generator, so lighting the hydrogen resulted in a loud bang and a piece of glass tubing embedded into a ceiling beam in my basement lab. I would not have thought glass could do that!

The consensus is that the odor could come from multiple causes:

Droplets of solution: Evolved gas (or even gas being blown over the liquid, as with winds blowing over a lake or ocean) can carry tiny particles of the liquid, which in this case is a caustic solution. In general, any gas-evolving reaction likely involves a toxic or corrosive medium that reacts to produce the gas, so it is good practice to carry out chemical reactions that evolve gas in a laboratory hood or in a well-ventilated setting.

Impurities: Most likely, the aluminum was an alloy whose al-loying elements may also react with the sodium hydroxide or even with the highly active nascent hydrogen that forms transiently during the main reaction. Products of such side reactions may trigger odors even at low levels. These may also be toxic and the same safety precautions noted above apply.

Actually, the reason for the smell is very important as a related reaction (Sodium hydroxide added to an Aluminum soda can with residual amount of a phosphate based drink) is reported in Bretherick's Handbook of Reactive Chemical Hazards, Volume 1, page 23-24, to quote:
"A student mixed aluminium foil and drain cleaner in a soft drink bottle, which started emitting gas. Another student carried the bottle outside and was claimed to have been overcome by the toxic fumes [1]. Most drain cleaners are alkalis, so that aluminium will dissolve to produce hydrogen. The bleach that it is is suggested may have been present will produce no toxic fumes in alkali, and one would be surprised to find arsenic or antimony compounds present. If the collapse was not purely hysterical, the remaining, though remote, possibility would be phosphine. The soft drink the bottle had contained was one of the many perhaps best described as impure dilute phosphoric acid [2]. "
Basically, nascent hydrogen or more precisely, the potentially effects of reducing hydrogen atom radical acting on impurities and anions present.
Associated metal impurities occurring in the case of this question, can also produce very problematic compounds, as further noted in Bretherick's handbook.


text:
Relation of temperature to effusion and diffusion
We were studying Graham's Law today and its dependence on Temperature, Pressure, Area, etc.
However, our teacher said that, for diffusion: $$R \propto \sqrt T$$ and for effusion:$$R\propto1/\sqrt T$$ giving the reason that during effusion, transfer of gases takes place through a very small hole, and so, increasing temperature will cause increased vibration of those gases, thus making them harder to pass through.
However, for diffusion, he said, there is no hole, and so an increase in temperature will cause increased diffusion.
I've tried finding answers for this here, but all I got was that even in effusion, the formula is:
$$r= \frac{PA}{\sqrt{2RTM\pi}}$$
and using formula $PV=nRT$, you can get that $r\propto \sqrt T$ even for effusion.
Can anybody clear this up for me?
Edit: I want to know whose explanation is correct and why, and also what is the (not too complicated) physical reason for this behaviour.

Yeah, I'd seen that answer as well, but it didnt really answer my question. Primarily because whatever he said is way too advanced chemistry for me, and also I dont think what he said in his answer is for an ideal gas.

Also, if $ R\propto \sqrt T$ for both diffusion and effusion, then how is that possible? I mean, whatever our teacher said to us made intuitive sense to me, so if someone could explain that as well, it would be helpful.

What exactly is your question - are you looking for a discussion of whether your teacher's explanation is correct, or an entirely different physical rationalisation of why the dependence is of that form, or a mathematical proof of the dependence, or maybe a proof of the full formulae? As it stands, you wrote a number of statements, but it's not immediately obvious what about them you want "cleared up".

For both an explanation of my teacher's reasoning, and a (not too complicated) physical rationalisation of why the dependence is of that form, as you said.

@stonecraft bros: your teacher has influenced your intuition. For effusion, imagine not a gas, but a liquid: increasing the temperature decreases the viscosity, and the flow thru a hole with the same force will be greater. This viewpoint suggests a different (more accurate) intuition.

I think part of the confusion comes from assuming that these temperature dependences are fixed properties that remain true under any conditions. To explain what I mean, let's start from porphyrin's equations for rate of diffusion/effusion for ideal gases from the linked answer.
$$r_{\text{diff}}\propto\frac{1}{\rho}\sqrt{T}$$
$$r_{\text{eff}}\propto\rho\sqrt{T}$$
If we assume $\rho$ (the density) remains fixed while we change the temperature, then diffusion and effusion both have a $\sqrt{T}$ dependence. This makes sense, as various measures of the "typical" speed of the gas particles are proportional to $\sqrt{T}$ and an increase in speed should increase diffusion and effusion.
However, if we allow the density to change with the temperature, we will get a different temperature dependence for these two processes, since $\rho=\frac{P}{RT}$ for an ideal gas. So if we keep $P$ fixed, $r_{\text{diff}}$ would now have a $T^{3/2}$ dependence while  $r_{\text{eff}}$ would have a $T^{-1/2}$ dependence.
Intuitively, the dependence of these properties on the density also makes sense.  As we make the gas less dense by heating it, the particles have more space to move around before colliding, allowing them to diffuse faster. For effusion, we have essentially the opposite effect, where either increasing the volume or decreasing the number of gas particles decreases the opportunities to escape.
The overarching point is that the temperature dependence is related to the conditions you conduct your experiment with. The choice to keep density/pressure constant or not will change the overall temperature dependence. More generally, whether the system is a gas/liquid/solid will also affect the dependence, along with how sophisticated a model we use to describe these systems.
For your teachers explanation, I could see why they might arrive at the dependences they emphasized (admittedly I'm speculating a bit here). For diffusion, it may make sense to consider the constant density dependence in a situation where all the possible places the gas can go are included as your system (i.e. you aren't considering particles "escaping" via diffusion). Even if you started with a sample where the density was very concentrated, it should become more or less constant once it has spread to fill it's container. For effusion, it makes more sense to look at nonconstant density, as particles are escaping from the system through effusion.
However, I don't think the explanation in terms of vibrations makes sense. If the gas particles are single atoms, they wouldn't have any vibrational modes. Even if they were molecules, the ideal gas law treats them as point particles that don't occupy any volume, so at least within this model, we can't use vibrations to explain the temperature dependence (vibrational effects could play a role in real gas model of diffusion/effusion).


text:
True or false: "If a reaction has a large negative value of ∆G, then it will be a fast reaction."
I recently took a Gen. Chem. 2 exam that contained this question. I answered false, as I knew that thermodynamics and kinetics are separate matters. My professor said the answer is true. I pressed him on it, citing a specific portion of our textbook that stated that the spontaneity of a reaction must not be conflated with the reaction rate. He admitted that it's true that a highly negative ∆G value does not necessarily imply that a reaction will be fast, but he said that that's usually the case, so I should have answered "true". This strikes me as absurd, not only because it's not a reasonable interpretation of a true/false question (they're supposed to be interpreted strictly, and if there's no qualifier like "generally" or "typically", then an if/then proposition should be treated as a "necessarily true" claim), but also because most students probably answered "true" because they did conflate spontaneity with reaction rate, which is something the textbook specifically warned against.
My professor also sent me the following via email:

"Please note that delta G combines all the properties of thermodynamics, which includes kinetics. Delta G has two components. Delta H (enthalpy) and Delta S (entropy). A large negative delta H means the reaction is highly exothermic. A large positive delta S means the product "randomness" is highly favorable, the system wants to happen even if enthalpy isn't high. I know of no system with these two feature that isn't fast. Note: rust (iron + oxygen going to rust) is highly exothermic, but the delta S is negative (solid and gas going to a solid), and that's why the reaction is favorable but slow."

I was surprised by this, as it directly contradicts what our textbook says on the matter. My understanding is that the ∆G equation contains only thermodynamic state variables and simply expresses the magnitude of thermodynamic potential. I see nothing in the equation that could convey information about the rate at which a process happens.
I would greatly appreciate not only theoretical comments on this, but also some specific examples of reactions with "large negative ∆G values" that are not "fast". (The inherent ambiguity of the terms "large" and "fast" here also seems to be a problem.)

This answer to an earlier question about spontaneity should support your case. They give the example of gasoline and oxygen, where combustion even at room temperature should be highly favorable (large positive entropy and large negative enthalpy), but it doesn't occur because the barrier is so high.

I think that "will" is supposed to be more like "would be if activation energy is applied". Such reactions tend to be fast, when they actually happen. For example burning of magnesium will be fast when it actually happens.

Indeed, delta G has nothing to do with kinetics. Ask him if he gave his wife a diamond ring…

The reaction  $\ce{C + O2 -> CO2}$ has a large ∆G value. But this reaction is so slow at room temperature that it seems not to happen at all.  In fact you are right, there is no connexion between kinetics and thermodynamics values.

Some of today's explosives are safe in the way that you can burn them without them exploding. See https://www.youtube.com/watch?v=OOWcTV2nEkU for a demonstration. The energy released by burning or exploding a sample is (approximately) the same, but the reaction speed is very obviously not.

@JonCuster, I agree with the general sentiment of your comment. However, your and the other examples are individual data points and therefore not suitable to exclude any correlation between kinetics and thermodynamics of reactions in general.

I am here to comment that you seem particularly unlucky with your teacher. In the sense that s/he had no idea of why it is as s/he said. Fortunately, @Snijderfrey's answer fixes the issue. In this case a brutal true/false option isn't appropriate.

True or false?: If the valley behind this mountain ridge has much lower altitude than this valley then the ridge is very low.

There is a big difference between a vague general relationship and truth/falsehood. and this is one big reason why your prof is wrong. It is easy to find examples that completely defy his generalisation. Ask him, for example, why we don't catch fire when we breathe.

UPDATE: I wrote two very long emails with detailed arguments regarding this question and the other question I posted the other day. My professor responded with just the following: "Why can't you admit that my response to both questions made sense and are most accurate? Is it really worth this much of your time to attempt to prove a point?" If anyone has advice to give on how to deal with this situation, I'd be grateful to hear it. I think this is unacceptable behavior from anyone, but particularly from an academic. I'm prepared to take it as far as it needs to go.

Thermodynamics is concerned with the differences in state functions and is path independent. Kinetics depends on the mechanism or path. Should there be a relationship? A reversible elementary reaction follows the same path in both directions so the forward and reverse rates depend on the differences in the free energies of products and reactants.  This principle is extended in the study of reactions with similar mechanisms such as free radicals or carbonium ions saying that the transition state has similar energies to the intermediate. How far this can be extended merits discussion not  T-F.

I am trying to say that both you and the prof have valid arguments and shouldn't be in opposition or dogmatic.  Show him that you are thinking about this and not simply quoting a text[s]

The professor's other view (AA batteries lose electrons as they discharge) is more worryingly wrong than this one. This is more about what a true/false question means in academia rather than simply being utterly confused about how electrochemistry works. However, your "next steps" are probably best discussed on https://academia.stackexchange.com/ rather than here. It does concern me that someone as confused as this can be teaching, but it is not unknown.

There is the Bell–Evans–Polanyi principle stating that "the difference in activation energy between two reactions of the same family is proportional to the difference of their enthalpy of reaction" (from the linked Wikipedia page). Looking at the Arrhenius equation, it is clear that this should lead to an increase in reaction rate with an increase in reaction enthalpy. Your professor might have this in mind, although admittedly he does not mention that explicitly.
Acutually, as far as I can tell the work by Bell, Evans, Polanyi and co-workers from the 1930s and 1940s can also be used to state that there is no universal relationship between thermodynamics and kintics valid over all reaction types. However, it is also apparently not true that there never is any relationship, as it is often claimed. As usual, the real world is more complicated than we are taught in basic university courses. I must leave the detailed reading of the papers to you – I could only have a quick glance – but below are some possible sources that I extracted from this paper. As it is often the case, the original papers from back in the days appear to be a rewarding read.
Concerning your question, the entry on Bell–Evans–Polanyi principle in the IUPAC gold book is quite informative:

The linear relation between energy of activation ($E_\mathrm{a}$) and enthalpy of reaction ($\Delta H_\mathrm{r}$) sometimes observed within a series of closely related reactions.

As you can see, this statement is far from making any general claims.

Bell, R. P., Proc R Soc London Ser A 1936, 154, 414.
Bell, R. P.; Lidwell, O. M., Proc R Soc London Ser A 1940, 176, 114.
Ogg, R. A., Jr.; Polanyi, M., Trans Faraday Soc 1935, 31, 604.
Evans, M. G.; Polanyi, M. Trans Faraday Soc 1935, 31, 875.
Evans, M. G.; Polanyi, M. Trans Faraday Soc 1936, 32, 1333.
Evans, M. G.; Polanyi, M. Trans Faraday Soc 1937, 33, 448.
Evans, M. G.; Polanyi, M. Trans Faraday Soc 1938, 34, 11.
Evans, M. G.; Warhurst, E. Trans Faraday Soc 1938, 34, 614.
Evans, M. G., Trans Faraday Soc 1939, 35, 824.
Baughan, E. C.; Evans, M. G.; Polanyi, M., Trans Faraday Soc 1941, 37, 377.
Baughan, E. C.; Polanyi, M., Trans Faraday Soc 1941, 37,648.
Evans, M. G., Trans Faraday Soc 1946, 42, 719.

Thank you for the response, @Snijderfrey. How relevant would you say this is to the general claim about the relationship between ∆G and reaction rates, though? I looked at several textbooks, and several of them explicitly state that kinetics and thermodynamics are different subjects, which seems to be enough to establish that the proposition I quoted is false.

@Logicus, yes, I agree with what you say, and the work by Bell, Evans, Polanyi et al. does not contradict you, either. As far as I can tell, it acutally supports your case. I edited in some extra info you might be interested in. You will have a hard time finding reliable sources saying that the more exergonic a reaction is, the faster it is (in this generality).

@Logicus I would say there is weak correlation, not causality between values of reaction enthalpy and activation energy ( reaction rate ).

This is the complete question at the GenChem 2 level (found posted on Chegg):

The correct answer is C. Of course, there is a relationship between most concepts, but not one where the speed of a reaction would reliably predict the sign of the standard Gibbs energy. Also, fast reactions is a fuzzy concept. A given reaction will be faster or slower depending on reaction conditions. For any fast reaction, you can probably find a faster one, but does that make the other reaction slow?


text:
Experimental evidence for anomalous electronic configurations of chromium and copper
Chromium ($\mathrm{3d^5\,4s^1}$) and copper ($\mathrm{3d^{10}\,4s^1}$) are fairly well-known exceptions to the aufbau principle, often attributed to the stability of the half- or fully-filled 3d subshell (as discussed in this prior question).
However, what experimental evidence is there for this? Is there a spectroscopic method (for example) which can be used to probe electronic configurations?

@ToddMinehardt, There is an article in the Journal of Chemical Education, The Full Story of the Electron Configurations of the Transition Elements, by W. H. Eugen Schwarz. He has theoretical explanations but no experimental discussion. I don't fully understand article either. It is based on "d-orbital" collapse!

Atomic spectroscopy (both emission and absorption) would give you this answer. Any method that involves spin-orbit coupling as well. I didn't pot this as a full answer because I am not sure

X-ray Photoelectron Spectroscopy (XPS) can be used to analyze 3d and 4s orbital energies of these elements. This unique behavior comes from the Aufbau principle, which explains why one electron moves to the d-orbital to fill half of the shell or total of the shell.
I used XPS for my previous research with elemental analysis (i.e. ratio of elements and which oxidation state they are in). I am not an expert in this device but comparing metallic copper's 3d shell binding energy with a different element (let's say Zn) would show that their B.E.'s are pretty close. To the best of my knowledge, lower atomic numbered elements in the 3$^{rd}$ row would not show any 3d peak due to very low B.E. (this is the outermost shell).
I don't know if that is the answer but XPS is a strong candidate for demonstrating this phenomena experimentally.

Thanks, I have also used XPS but I am not an expert. However, I do not think that XPS can be used for determine electron configurations.

I believe this is the correct answer, I will try and get an reference/proof of this soon. We have discussed this question in my lab lol.