text:
What is the electrolytic bath used in gold-plating of copper articles?
The answer is $AuCl_3 + NaCN$
Now I understand that for electroplating, we need gold to be deposited on the copper article, and the electrolyte must enable that deposition. But I don’t know what is the use of $NaCN$ in the process.
Also, if copper is the cathode, then pure gold should be the anode. But the oxidation potential of gold is very low, then how does it get dissolved in the solution.
Can I get an explanation on this process would work?

There are several cyanide processes for gold plating. Just do some searching and you will find them.

@EdV I don’t know how appropriate it is to ask, but can I get an upvote? I really need it. It isn’t ethical to do so, but I need one.

I can ethically upvote if you flesh out the question text. Look up the cyanide process for gold extraction. Once you see that, you will know why cyanide is used. Fair enough? By the way, why would anyone actually need an upvote? There is nothing in the real world riding on stack exchange rep, after all.

@EdV You see, I am on a question ban here, for a few bad questions 6 months ago. Now I agree, it’s totally my fault, but I was a new user, and SE is a really harsh place for such people. But I know how to use the site properly now (I be over 2000 rep on Math), but the system won’t lift the ban until I get an upvote (at least, that’s what I could make out from the guidelines). So yes, it’s kind of important, or else it will ban me for another 6 months


text:
Understanding the use of Le Chatelier's principle in the industrial production of ammonia
I understand why the reaction
$$\ce{N2 + 3 H2 -> 2 NH3} \qquad \Delta H = \pu{-92.2 kJ mol-1}$$
isn't carried out in the lowest possible temperature (even though it seems that it should be as it favors the right side of the reaction). As the rate of reaction would be very slow, the amount of $\ce{NH3}$ produced would be lesser than the amount that would've been produced if the reaction was carried out at the optimum temperature in a given time span.
Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

There is an optimal temperature, for given production line setting, where the effective yield is optimal, as reaction thermodynamics and kinetics have the opposite effects, wrt temperature.

Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

The rate constant would be high, but that does not mean that the net forward rate is high. If you start without product, the initial rate would be high, but would drop to zero very quickly as equilibrium is reached. The equilibrium concentration of product would be very low at high temperature, as you said, so the yield would be very low.
If you had a very cheap and highly efficient process to remove product directly from the reaction vessel, you could run the reaction at high temperature. However, the product removal is slow and expensive (cool down the reaction mixture, remove ammonia, and reheat the remaining reactant to feed back into the reactor). Because this is costly, the reaction is run at a compromise temperature where at equilibrium, about 15% of the reaction mixture is product.

I think the simplest way to conceptualize this would be to pretend the heat produced by the reaction is a "product" (it's not really correct to say that, but it may help you understand it). Since the reaction is exothermic, you could rewrite the reaction as: N2 + 3H2 ⟶ 2NH3 + Heat
If you raised the temperature, you'd add heat, which is a "product".  As per Le Chatelier's Principle, increasing the concentration of a product shifts the reaction equilibrium left, which is why the reactants predominate at a higher temperature.
If you're familiar with Gibbs Free Energy, I think that would be a better (and more accurate) way to look at it. The forward reaction converts four moles of gas (3 H2 + 1 N2) to two moles of gas (2 NH3). This reduces the total entropy of the system.
Now, we know that: ΔG = ΔH - TΔS
We also know that since entropy is lowered by the forward reaction, ΔS < 0, which makes the term -(TΔS) positive. Lastly, we know a reaction will be spontaneous only if ΔG < 0.
If the temperature were very high, the (TΔS) term would also be very high, making ΔG positive. For the reaction to be thermodynamically possible, the temperature needs to be very low, otherwise the decrease in entropy would offset the favorability added by the loss of enthalpy. This is why there's an "optimal temperature", and why you can't just try and maximize the temperature during ammonia synthesis.
There may be some other ways to look at it, but that's how I would conceptualize it.

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text:
Limiting molar conductivity of multiple solutes
Suppose there is a water solution at $\pu{25 °C}$ with $\ce{Na+},$ $\ce{Mg^2+}$ cations and $\ce{Cl-},$ $\ce{HCO3-}$ anions. Concentrations $[\ce{Na+}],$ $[\ce{Mg^2+}],$ $[\ce{Cl^-}]$ and $[\ce{HCO3-}]$ are known.
I cannot understand how to apply the Kohlrausch's law of the independent migration of ions here:
$$\Lambda_\mathrm{m}^0 = \nu_+\lambda_+ + \nu_-\lambda_-$$
How this formula will transform for the solution with many kinds of ions?
Edit:
Electroneutrality condition holds, so:
$$[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]=[\ce{Cl^-}]+[\ce{HCO3-}]$$
I've found the table with the limiting ionic conductivities $\lambda$ of each individual ion in water at $\pu{25 °C}$:
$\lambda_{\ce{Na+}} = \pu{50 \cdot 10^{-4} S m2 mol-1}$
$\lambda_{\ce{Mg^2+}} = \pu{106 \cdot 10^{-4} S m2 mol-1}$
$\lambda_{\ce{Cl-}} = \pu{76 \cdot 10^{-4} S m2 mol-1}$
$\lambda_{\ce{HCO3-}} = \pu{45 \cdot 10^{-4} S m2 mol-1}$
But I don't know what values for $\nu$ I should take.

I've found a answer for my question from this question.
The way to compute the conductivity of electrolyte with multiple ion types is given from the work Pawlowicz, Rich, ( 2008), Calculating the conductivity of natural waters, Limnol. Oceanogr. Methods, 6, doi:10.4319/lom.2008.6.489.
For general case consider the system, which consist of $N_+$ number of cation types, $N_-$ number of anion types and $N_{types} = N_+ + N_-$ total number of types, fully dissolved in solvent. This system can be viewed as a weighted sum of all possible pairwise combinations between cation and anion types. Then the conductivity $\sigma,\pu{[S m-1]}$ of the solution will be determined by:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} \Lambda_{eqm, \, ij}}$$
where $c_i^{\pm}$ - molar concentration of corresponding ion type, $\pu{[mol m-3]}=\pu{[mM]}$;
$z_i$ - valency of corresponding atom;
$c_i^{\pm} \cdot z_i$ - equivalent ionic concentration of corresponding ion type $\pu{[mol m-3]}=\pu{[mM]}$;
$\Lambda_{eqm, \, ij}$ - equivalent molar conductivity of binary subsystem of cation type $i$ and anion type $j$, $\pu{[S m^2 mol^{-1}]}$.
Equivalent ionic concentration $C_{eq}$ is defined as:
$$C_{eq} = \sum_i^{N_+}c_i^{+} \cdot z_i = \sum_j^{N_-}c_j^{-} \cdot z_j = \frac{1}{2}\sum_k^{N_{types}}c_k^{\pm} \cdot z_k$$
In general a binary solute dissociates according to formula:
$$\ce{A_{\nu^+}B_{\nu^-} -> {\nu^+}A^{z^+ +} + {\nu^-}B^{{z^-}-}}$$
where $\nu^+$ and $\nu^-$ are coprime numbers and represent the moles of ions  for 1 mole of solute. It follows:
$$\nu^+ = z^-$$
$$\nu^- = z^+$$
Equivalent molar conductivity $\Lambda_{eqm}$ of single binary electrolyte is defined as:
$$\Lambda_{eqm} = \frac{\sigma}{c\nu_+z^+} = \frac{\sigma}{c\nu_-z^-}$$
For a binary solute, dissolved in water at normal pressure, $\pu{25 °C}$ and infinitely diluted, the law of the independent migration of ions can be written to find a limiting equivalent molar conductivity $\Lambda_{eqm}^{0}$:
$$\Lambda_{eqm}^{0} = (\frac{\nu_+}{z^+ } \lambda_{+}^{0} + \frac{\nu_-}{z^- }\lambda_{-}^{0}) = (\frac{z^-}{z^+ } \lambda_{+}^{0} + \frac{z^+}{z^- }\lambda_{-}^{0})$$
where $\lambda^{0}$ - limiting ionic conductivity of the ion, $\pu{[S m^2 mol^{-1}]}$.
Here interaction between ions is ignored.
Change of equivalent molar conductivity from changing of the solute equivalent concentration is determined by Debye-Hückel-Onsager equation of the form:
$$\Lambda_{eqm} = \Lambda_{eqm}^0 (1-A \sqrt{I}) - B \sqrt{I} =  \Lambda_{eqm}^0 K_A - K_B$$
where $A, \, B$ - Debye–Hückel–Onsager coefficients;
$I$ - stoichiometric ionic strength;
$$I = \frac{1}{2} \sum_{k=1}^{N_{types}}c_i z_i^2$$
$$ A = \frac{z^2eF^2}{3 \pi \eta}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
$$ B = \frac{qz^3eF}{24 \pi \varepsilon RT}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
where $\eta$ - viscosity of solvent, $\pu{[Pa s]}$;
$\varepsilon$ - dielectric permittivity of solvent;
$q$ - coefficient depends on $z^+/z^-$ of binary solute.
The expression for the conductivity of electrolyte (solvent - water at $\pu{25 °C}$ and normal pressure, ion pairng reduction factor is ignored ($\alpha_{ij}=1$ for all $i,j$)) will be:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} (\frac{z_j^-}{z_i^+} \lambda_i^0 \, K_{A, \, ij} + \frac{z_i^+}{z_j^- } \lambda_j^0 \, K_{A, \, ij} - K_{B, \,ij})}$$
In my case I have 2 cations and 2 anions, their pairwise combination give 4 different solutes, which are assumed to be completely dissociate in water. Relation between solute molar concentration and its ions molar concentrations can be established.

$\ce{NaCl -> Na+ + Cl-}$
$\ce{NaHCO3 -> Na+ + HCO3-}$
$\ce{MgCl2 -> Mg^2+ + 2Cl-}$
$\ce{Mg(HCO3)2 -> Mg^2+ + 2HCO3-}$

Conductivity of my electrolyte will be:
\begin{align*}
\sigma &= \frac{1}{[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]} \cdot \\
           &\cdot \Bigl(\ce{[Na+]}  \ce{[Cl-]}       K_{A(\ce{NaCl})}        ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{NaCl})}}{K_{A(\ce{NaCl})}}      )+\\
           &+             \ce{[Na+]}  \ce{[HCO3-]}  K_{A(\ce{NaHCO3})}  ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{HCO3-}}^0 - \frac{K_{B(\ce{NaHCO3})}}{K_{A(\ce{NaHCO3})}})+\\
           &+             2\ce{[Mg^2+]}  \ce{[Cl-]}       K_{A(\ce{MgCl2})}        ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{MgCl2})}}{K_{A(\ce{MgCl2})} }      )+\\
           &+             2\ce{[Mg^2+]}  \ce{[HCO3-]}  K_{A(\ce{Mg(HCO3)2})}  ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{HCO3-}}^0 - \frac{ K_{B(\ce{Mg(HCO3)2})}} {K_{A(\ce{Mg(HCO3)2})}})
           \Bigr)
\end{align*}
All the values here depend from given concentration of ions or from values found in tables. Hope it is correct, please, correct me if I'm wrong.

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text:
Cause of m/z = 56 peak in 4-methyl-1-pentanol
This is the mass spectrum for 4-methylpentan-1-ol:

From its structural formula, what could be the fragmentation which results in a peak at $m/z = 56?$
I was considering a few possibilities; could it be $\ce{C3H4O+}?$

Unfortunately I did not find a high resolution spectrum of this compound, which would have quickly answered your question.
Nevertheless, ionized alcohols and even more primary alcohols have a main fragmentation pathway which is the loss of water to yield an ionized alkene. In this case, the corresponding alkene would be 4-methyl-1-pentene, at m/z 84. The peak is present although rather small in the EI spectrum. From there, it is quite obvious, when one looks at the fragmentation spectrum of 4-methyl-1-pentene (NIST Webbook spectrum), that it shares a lot of similarities with the 4-methylpentan-1-ol EI spectrum. (See for instance peaks at m/z 69, 56, 43, 42, 41.) From there, it is quite likely that the m/z 56 peak is ionized isobutene ($\ce{C_4H_8}^{\bullet +}$) arising from a retro-ene type rearrangement.


text:
Why is Electrochemistry often the method of choice for determining Ksp values?
I don't understand the relationship they have with one another. Please, someone explain.

Have a look at this excellent answer: https://chemistry.stackexchange.com/a/118123/79678. Then search around the sire a bit: lots of good stuff here!

Perhaps a simple example will illustrate how a solubility product constant, $\ce{K_{sp}}$, may be determined via measurement of an appropriate voltaic cell potential. Consider the estimation of the $\ce{K_{sp}}$ value for the following equilibrium:
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq) \quad K = K_{sp}} \tag{1}$$
This equilibrium lies far to the left because AgCl has very poor solubility in water and, as a direct consequence, the concentrations of silver ions and chloride ions are very low. Therefore, the $\ce{K_{sp}}$ value is small, i.e., much less than unity, and a direct measurement of the $\ce{K_{sp}}$ value is difficult due to the low solubility of AgCl.
However, we have the following two half-reactions, with their standard reduction potentials (from standard tables):
$$\ce{Ag+ +  e−  <=> Ag(s)} \quad E^\circ_\mathrm{red} = \pu{+0.7996 V} \tag{2}$$
$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−} \quad  E^\circ_\mathrm{red} = \pu{+0.22233 V} \tag{3}$$
Half-reaction (2) has the more positive standard reduction potential, hence more negative $\ce{\Delta G}$, so it will be the spontaneous reduction half-reaction at the cathode in a voltaic cell that can be constructed. Reversing half-reaction (3) yields the oxidation half-reaction at the anode in the constructed voltaic cell:
$$\ce{Ag(s) + Cl− <=> AgCl(s) + e−}  \tag{4}$$
So the voltaic cell notation is
$$\ce{Ag(s)|Cl^-(aq;\pu{1 M})||Ag^+(aq;\pu{1 M})|Ag(s)}$$
and the net cell reaction is obtained by adding half-reactions (2) and (4), resulting in
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s) \quad K = 1/K_{sp}} \tag{5}$$
This equilibrium lies far to the right because it is simply the reverse of half-reaction (1). Therefore, the $\ce{K}$ value is large. Indeed, it is simply the reciprocal of half-reaction (1)'s $\ce{K_{sp}}$ value. This is consistent with the fact that mixing aqueous solutions containing silver ions and chloride ions immediately results in spontaneous formation of AgCl solid that precipitates out of solution.
The Nernst equation, where $\ce{Q}$ is the reaction quotient, is
$$ {E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{n F} \ln(Q) } \tag{6}$$
The natural logarithm term is basically a correction term that takes into account the fact that voltaic cells are usually not at standard state, i.e., they are not dead yet. This is accomplished using $\ce{Q}$. At standard state, $\ce{Q} = 1$, and at equilibrium, $\ce{Q} = K$, where $\ce{K}$ is the equilibrium constant. In the present case, $\ce{K}$ pertains to equation (5).
Now, at equilibrium (and ignoring units), $\ce{\Delta G_{cell} = 0, E_{cell} = 0 and Q = K}$, so
$$E^\circ_\mathrm{cell} = \frac{R T}{n F} \ln(K) = {E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{+0.7996 V} -(\pu{+0.22233 V}) = \pu{+0.5773 V}} \tag{7}$$
Note that the standard cell potential is defined as
$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag{8}$$
With $\pu{T = 298.15 K}$ and $\pu{n = 1}$, $\pu{RT/nF \approx 0.02569 V}$, so equation (7) can be solved for $\ce{K}$:
$$\mathrm{K = e^{nfE^\circ _{cell} /RT} \approx e^{+0.5773V/+0.02569V} = 5.74 \times 10^{9} \tag{9}}$$
Hence $$\pu{K_{sp} = 1/K = 1.74 \times 10^{-10}} \tag{10}$$
which is close to typically quoted values in the literature.

I think I am going to need simpler words. I don't think I understand, sorry.

If either answer is acceptable or helpful to you, please consider giving it the green checkmark. It encourages people to put some thought and time into crafting answers that are correct, understandable and helpful to future questioners. Thanks for considering this!

The standard half cell potential, $\mathrm{E^0_{red}}$, is for a 1.000 molar solution for the ion being reduced. For other concentrations the Nernst Equation gives the half-cell potential, $\mathrm{E_{red}}$.
$$\mathrm{E_{red}} = \mathrm{E^0_{red}} - \dfrac{\mathrm{RT}}{\mathrm{zF}}\ln{\dfrac{a_{\mathrm{Red}}}{a_\mathrm{Ox}}}$$
Note that the Nernst equation should use activities, $a$, but for low quantities of a solute the concentrations are often be used.


text:
Creating a buffer with with a target pH and given weak base?
I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride salt.
I used the Henderson-Hasselbalch equation to figure out the relative amounts of $\ce{NH3}$ and $\ce{NH4+}$ I need:
$$
\textrm{pH} = \textrm{pKa} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\textrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89
$$
So for every mole of $\ce{NH4+}$ I need $0.89$ moles of $\ce{NH3}$ to reach the desired pH of 9.20. I know that the buffer must withstand $0.004$ moles of $\ce{NaOH}/\ce{HCl}$, so I can set up the inequality
$$
8.70 < \textrm{pKa} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} < 9.70 \\
-0.55 < \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} < 0.45 \\
0.28 < \frac{[\ce{NH3}]}{[\ce{NH4+}]} < 2.82
$$
So I know the relative ratios I need to be within after the strong base and acid, but from here, how do I proceed with my $0.1\textrm{M}$ ammonia solution and ammonium chloride to get the desired ratio and buffer capacity?

Your initial calculations using Henderson-Hasselbalch equation is correct:
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\mathrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\log \frac{[\ce{NH3}]}{[\ce{NH4+}]} = -0.05\\
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89
$$
Yet, since you have only $\pu{0.10 M}$ ammonia solution, you can't exceed $\ce{[NH3]}$ by $\pu{0.10 M}$. Thus, to prepare $\pu{100 mL}$ of ammonia buffer with $\mathrm{pH}=9.20$, you have to add $m$ grams of $\ce{NH4Cl}$ to $\pu{100 mL}$ of $\pu{0.10 M}$ ammonia solution where:
$$ m = \frac{\pu{0.10 molL-1}}{0.89}\times \pu{0.10 L}\times \pu{53.49 gmol-1}=\pu{0.601 g}$$
Assuming that the volume has not changed when you add $\pu{0.601 g}$ of $\ce{NH4Cl}$ to $\pu{100 mL}$ of ammonia solution, you have gotten the buffer you needed with $\mathrm{pH}=9.20$. Now, you have to check the buffer capacity by adding $\pu{20 mL}$ of either $\pu{0.20 M}$ solutions of $\ce{NaOH}$ or $\ce{HCl}$.
Let's add $\pu{20 mL}$ of $\pu{0.20 M}$ solution of $\ce{NaOH}$ first. The following reaction would take place:
$$\ce{NH4Cl + NaOH -> NH3 + NaCl + H2O}$$
Thus, $0.20 \times \pu{0.020 mol}= \pu{0.004 mol}$ of $\ce{NH4Cl}$ will convert to $\pu{0.004 mol}$ of $\ce{NH3}$. The new volume of solution is $\pu{(100+20) mL}=\pu{120 mL}$.
The new $\ce{[NH3]} = \dfrac{\pu{(0.1\times 0.1+0.004) mol}}{\pu{0.120 L}}=\pu{0.117 molL-1}$
Similarly, The new $\ce{[NH4Cl]} = \dfrac{\pu{(\frac{0.1\times 0.1}{0.89} -0.004) mol}}{\pu{0.120 L}}=\pu{0.060 molL-1}$.
Therefore, the new pH would be:
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
= 9.25 + \log \frac{0.117}{0.060} = 9.54 \lt 9.70
$$
Thus, this buffer stands for the higher $\mathrm{pH}$ limit required. You can also check the lower limit by similar method.

Late edition:
My friend MaxW insists in elsewhere that this is the case of $\ce{NH4Cl}$ solution to use the buffer (opposes to what I did with solid). Max correctly calculated volumes of two $\pu{0.10 M}$ solutions would be $\pu{52.91 mL}$ of $\ce{NH4Cl}$ and $\pu{47.09 mL}$ of $\ce{NH3}$. If Max was correct, this should still stand the required buffer capacity. Let's check:
The new $\ce{[NH3]} = \dfrac{\pu{(0.1\times 0.04709 + 0.004) mol}}{\pu{0.120 L}}=\pu{0.0726 molL-1}$
Similarly, The new $\ce{[NH4Cl]} = \dfrac{\pu{(0.1\times 0.05291-0.004) mol}}{\pu{0.120 L}}=\pu{0.0108 molL-1}$.
Therefore, the new pH would be:
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
= 9.25 + \log \frac{0.0726}{0.0108} = 10.08 \gt 9.70
$$
Thus, this buffer would not stand for the higher $\mathrm{pH}$ limit required.

Yep, I just finished my calculations after some fiddling and reading-I came out with exactly 0.601 grams of salt as you mentioned. I had to assume that since NH3 was a weak acid that [NH3] = 0.10M and since NH4+ is mostly from the salt and not from dissociation that [NH4+] = same concentration as if salt were dissolved, then worked backwards as you wrote.

This answer can't be entirely correct. The question asks the solver to use 0.1 molar solution of ammonia, and 0.1 molar solutions of ammonium chloride to make 100 ml of the buffer. You'd have to be able to add some pure water too...

@MaxW: Sure, there are a lot of possibilities. The question did not pinpoint use of ammonium solution. I operated with given data and OP didn't object to it.

Hey all, my bad. I failed to see it could be ambiguous. I mean NH4Cl salt. I'll edit the question.

The question asks:

I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride.

User Mathew Mahindaratne has provided an answer, but did not answer the question. The question supposes that 0.1 molar solutions of ammonia and ammonium chloride are to be used. Mathew used ammonium chloride salt.
EDIT User Mathew Mahindaratne did in fact answer the question correctly. The OP edited the question and confirmed that a salt of $\ce{NH4Cl}$ was being used.
To me there is an added constraint.
$$\mathrm{V_\ce{NH3}} + \mathrm{V_\ce{NH4Cl}} = \pu{100 ml}\tag{1}$$
From Mathew's answer
$$\dfrac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89\tag{2}$$
If we neglect the amount of $\ce{NH4OH}$ in the ammonia solution, then the concentrations of ammonia and ammonium are equal so:
$$\dfrac{\mathrm{V_\ce{NH3}}}{\mathrm{V_\ce{NH4+}}} = 0.89\tag{3}$$
Substituting into equation 1 yields:
$$
0.89*\mathrm{V_\ce{NH4Cl}} + \mathrm{V_\ce{NH4Cl}} = \pu{100 ml} \\
\mathrm{V_\ce{NH4Cl}} = \dfrac{\pu{100 ml}}{1.89} = \pu{52.91 ml}\tag{4}
$$
So $\pu{47.09 ml}$ of the ammonia solutioon must also be used.

My bad: I phrased the question wrong. I am using ammonium chloride salt.


text:
Electrode potentials vs interfacial potential differences
Between an electrode and an electrolyte there exists an interfacial potential difference.
Many authors give the electrode potential to be the electric potential of the electrode relative to the electrolyte. I have a few problems with this, and was wondering whether anyone could help.
We define the electrode potential of SHE to be 0V. It is impossible to measure the interfacial potential difference, however nevertheless one certainly exists between the reference electrode and the solution. Are we saying that we define the interfacial potential difference of the SHE to be zero, and then have electrode potentials of other couples as interfacial potential differences measured relative to this interfacial potential difference?
To my mind, this cannot be right. Don't we just set our zero reference (i.e. one end of our voltmeter) of electric potential at the electrode of the SHE, and measure electric potentials of other electrodes relative to this?
Sergio Trasatti wrote this regarding electrode potentials,

Equation (4.7) shows that the "electrode potential", as obtained in
  practice, does not measure the electric potential drop between the
  bulk of the metal and the bulk of the solution, but it is a more
  complex quantity which includes both electrical and chemical
  contributions.

Why do so many authors conflate interfacial potential differences and electrode potentials? Or am I missing something here? Thank you!

Comments are not for extended discussion; this conversation has been moved to chat.

The comment by the author was very useful. "
@M.Farooq I have now obtained a copy and have read through the chapter, and it has indeed given me a better understanding (though some parts I couldn't fully appreciate since I haven't formally studied electrochemical energy yet!). On another note, I have found a truly outstanding explanation by Professor Richard G. Compton of Oxford University on "Electrode Potentials", which clears up almost every doubt I had. Pages 18-22 are especially illuminating in the context of my original question.

IUPAC defines the electrode potential of an electrode to be the EMF of a cell in which the electrode in question is on the right hand side of the line representation of the cell and a SHE is on the left hand side.
As such, the electrode potential of the $M^+/M$ couple is
$E^o_{M^+/M} = \phi_{M} - \phi_{R}$
where $R$ refers to the reference electrode, in this case the SHE. This is equivalent to
$E^o_{M^+/M} = (\phi_{M}-\phi_{S}) - (\phi_{R}-\phi_{S})$
Where $\phi_{S}$ is the potential of the solution. Essentially, this means that a relative electrode potential really means
$E^o_{M^+/M} = \Delta \phi_{M^+/M} - \Delta \phi_{H^+/H_2}$
That is, the potential difference of the electrode in question and the solution relative to the potential difference between the SHE electrode and the solution.
It doesn't help the cause that electrochemists refer to potential differences when they use the word potential, however this is just a case of getting to grips with the terminology used in the field!


text:
Finding Symmetry number of a molecule

An isotopic variant of the molecule SF$_6$ is S$^{19}$F$_4^{18}$F$_2$, with the two $^{18}$F nuclei oriented axially, so that the $^{18}$F-S-$^{18}$F angle is 180°and the four $^{19}$F nuclei form a square in a plane perpendicular to the $^{18}$F-S-$^{18}$F axis. All of the bond lengths are equal.  In the figure, the regions surrounding the $^{18}$F nuclei are shown in blue, and the regions around the $^{19}$F nuclei are shown in green.Two of the moments of inertia of S$^{19}$F$_4^{18}$F$_2$ are identical, and the third is slightly larger: I$_A$= I$_B$ = $2.9942 ×10^{-45}$ kg m$^2$ and I$_C$= $3.0749×10^{-45}$ kg m$^2$. Wikipedia describes SF$_6$ as an “extremely potent greenhouse gas.”
What is the symmetry number of S$^{19}$F$_4^{18}$F$_2$?
I have considered it and think there is one 4-fold axis, and two 2 fold axes, but still unsure if that is right

Welcome to Chemistry Stack Exchange! Please edit your question to show your efforts at solving the problem

I made one edit, but haven't really been able to get anywhere as I am very confused!

One way to think about it is: Start from SF$_6$ without any isotope labeling, and see what point group you get. What symmetry operations do you lose when you change the two axial fluorines to $^{18}$F?

SF6 has three 4-fold axes (counting for 3 each) -> 9, four 3-fold axes (each counting for 2) -> 8, and six 2-fold axes (each counting for 1) -> 6, plus the identity so the symmetry number is 24

Not even close.

I am more confused with the addition of the isotope, so is there one 4-fold, two 2-fold, and I'm not sure how the 3-fold would work here..

See https://www.chemtube3d.com/octahedral-sulfur-hexafluoride-sf6-symmetry/ for exploring the O$_h$ symmetry of the untagged (all $^{19}$F SF$_6$. Which operations do you lose when you do the replacement I mentioned above?

If you assume that your molecule is of the form $\ce{AB4C2}$ then one form is that shown in your figure and belongs to the  $D_{4h}$ point group, the other has the two blue atoms at 90 deg to one another and belongs to $C_{2v}$


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How concentrated does the analyte solution need to be in order to perform a (chelatometric) titration
I’m currently investigating a method to chemically pretreat cellulosic waste as bisorbents for enhanced uptake of lead in water. I intend to measure the initial and final concentration of water to record the change in concentration of aqueous lead via a chelatometric titration.
In my experiment, I’ll be performing filtration in the batch mode at three different initial concentrations (5 mg/L, 20 mg/L, 50 mg/L). To visualize the magnitude of these concentrations, I’ve converted the units to the more conventional mol/L (M).
Initial concentrations of water samples I’ll be testing on in mol/L (M)
My question is, are these initial concentrations too low to be conducting a titration on. If so, what would be a good initial concentration of lead to be working with (considering after filtration, the concentration of lead will diminish significantly). If these initial (and very small post filtration) concentrations are measurable for a titration, what is a good  concentration for the EDTA titrant. Should the titrant concentration be significantly less than the analyte concentration?
All the best!

Sorry to tell you a bad news. Classical methods, like titrimetry or gravimetry do not work very well at low concentrations. Failure would be a strong word. The beauty of EDTA is that it forms 1:1 complex with most metals. Pb is not an exception. If your lead solution is 0.000001 M, your EDTA solution should be 0.000001 M. Such concentrations are never recommended in classical methods.
For ppm level, spectrophotometric methods can work quite well. Go to Google Scholar and search the methods in Google Scholar

If I were to perform a chelatometric titration, what concentrations would work well?

About 0.01 M or 0.001 M, but rarely lower than that.

Do you have atomic absorption spectrophotometer?

I don’t unfortunately. Where might I be able to access one? I’m very interested.

Most chemistry departments of universities will have an atomic absorption spectrophotometer.

I see. Is the procedure like ICP Mass Spectrometry where one analyzes a spectra to determine concentrations?

ICP relies on atomic emission (very expensive instrument), ICP-MS is even more expensive. Atomic absorption spectrophotometer relies on atomic absorption (recall Beer's law). It is a much more affordable instrument.

That makes sense. Thank you for the heads up!


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Network covalent bonds during melting
What happens to the intramolecular bonds when a network covalent substance is melted? I understand that in covalent molecular substances, the intermolecular bonds weaken between the molecules once molten. My confusion is with what happens to the bonds between network covalent substances when melted.

Do the intramolecular bonds break, leaving only the individual atoms (carbon for example in graphite or diamond).
Do the intramolecular bonds weaken
The basis for this query is when I was asked to predict the electrical conductivity of graphite and silicon dioxide in molten states (Graphite sublimates, but under pressure is liquid). I started by researching the atomic and bonding structure of the elements when they are melted but could find very little information.

TLDR: What happens to covalent network bonds when they melt
Thankyou

The nature of liquid carbon can't be explained in a few words, and besides, isn't even known all that well.

@IvanNeretin thankyou, I rabbit holed a bit on a highschool assignment. I became very curious about it. Thankyou for pointing me towards liquid carbon, that should have more results than "help! Network covalent melting does what!?!?"

Here are some previous related questions: (1) (2)

Silica is melting at a rather high temperature, because the whole crystal is made of a huge molecule, where each atom is attached to the whole structure by covalent bonds. All silicium atoms are attached by four sigma bonds to different Oxygen atoms. And all Oxygen atoms are attached by two sigma bonds to two different Si atoms. So the whole structure is rigid, and it is difficult to break all the bonds Si-O simultaneously.

" intramolecular bonds break, leaving only the individual atoms (carbon for example in graphite or diamond)" - that like never happens. Even in gas phase there are usually small molecules.

@Mithoron oh ok, so how are molecules formed in a molten network covalent substance?

Some bonds are weakened, some get broken even before melting, some with further heating. That's complicated and highly case dependant.

@Mithoron ah alright, I guess I was trying to pigeon hole a complex question


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Is there a difference between hand soap and body soap with killing viruses?
With the Coronavirus in full swing around the globe, I wanted an answer to this, particularly since my dad has just been diagnosed with Leukaemia.
We can't find hand wash anywhere but instead I just bought some Radox shower gel and put that in a hand soap bottle.
Is there a significant difference between hand soap and shower gel when it comes to killing off germs/viruses? Or is the case that both are detergents and it's all around marketing?

No significant difference, both will disrupt the lipid bilayer of virus particles

@Waylander are you saying the alcohol component in sanitizers is not needed? That common detergent 'soaps' will kill virus?

Alcohol based sanitisers and soaps are two different things. With a sanitiser you need an alcohol concentration of at least 60% and a long contact time. Common detergents (e.g. hand soap) do 2 things, they break up the lipid bilayer of virus particles and they facilitiate the mechanical removal of the virus from a surface (provided you rinse it off well). A soap is not a sanitiser and a sanitiser is not a soap, you use them under different circumstances.

@MarsJarsGuitars-n-Chars  Since you ask, yes soap does kill viruses  https://www.vox.com/science-and-health/2020/3/11/21173187/coronavirus-covid-19-hand-washing-sanitizer-compared-soap-is-dope

@Waylander Awesome. Thanks. Makes good sense. Wonder why this is not being stressed in news reports, etc.. The mechanical action I knew about. Strange this is not more common knowledge.

@mars wouldn't really matter, I think. .You can feel alcohol, it burns and cools as it evaporates. That feeling is what makes people relieved and sure of their cleanliness. Can't get that with soap. The feels trump the reels always...


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Why does electrode potential depend upon enthalpy of hydration and sublimation?
It's given in text that "Electrode potential among other things depends upon: enthalpy of sublimation, ionization enthalpy and hydration enthalpy."
I could understand why it would depend upon ionization enthalpy as less is the IE easily we could remove the electron thereby reducing the other reactant. But I couldn't figure out why the other two enthalpy comes. But after some thinking I thought maybe we are adding the enthalpy of the three with sign and substituting in $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ and obtaining $\Delta G^\circ$ to substitute in $\Delta G^\circ = -n\mathcal{F}E^\circ$ cell to find E°cell value.
I couldn't think further clearly after this. Could you help me? (I'm trying to understand why lithium because of its high hydration enthalpy overweighs its high ionization enthalpy and becomes the strongest reducing agent in aqueous solution)

You might find this helpful: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810. It addresses some of the issues you raise, but is not for total novices.

In a thought experiment, you can substitute the electrochemical half reaction by three steps:

Removing atoms from the electrode into the gas phase (sublimation)
Removing electrons from the atoms (ionization)
Solvating the resulting ions

If you have quantitative descriptions of these processes, you get a quantitative description of what is actually going on. This is because many thermodynamic functions (such a enthalpy) are functions of state, i.e. they don't depend on the path from A to B, just on the state of A and B.
I would have said these quantities are related rather than one depends on the other. There is really no causality here.

Copper has the positive standard electrode potential because when we use copper as an electrode. copper would not convert into copper + 2 ion because the energy which is needed to convert copper to the copper + 2 iron is not full filled by the hydration energy, which is ivolve when we emerged copper into the electrolytic solution.
so that's why copper will not converted into copper + 2 because of this less hydration energy it does not came into the copper + 2 ion it means copper will not give electron and converted into positive + 2 ion, as it not give electron so it will not have negative electrode potential because only such elements which can donate electrons for example Lithium or sodium or potassium can have (-)electorde pottential.thus the only option which is left for the copper is for having the positive standard electrode potential as it could not give electrons and converted into positive ion, (only due to low hydration energy)
the root behind this is due to the low hydration energy. Only due to this low hydration energy electron does not convert into copper + 2 and due to which it does not give electrons and due to which it does not have negative  electrode potential and due to which the copper has the positive standard electrode potential.

First, I present an example, which provides a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:

$\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$

Reference: one of many possible links .
This produces a pH effect, which per a source on the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?', the answer employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:

As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.

If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.

Next, with respect to the impact of sublimation (and relatedly, ionization), I note that this can be energy (and especially temperature) related.  Quoting a reference:

As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space. ... Continue to supply even more energy and more electrons will be removed from the atom.

On the connection of temperature to the electrode potential, I cite an answer from The Student Room:

Increasing the temperature AFFECTS electrode potentials, not increases them.
The electrode potential is a measure of the extent of a redox process - negative and the equilibrium lies relatively to the LHS (compared to the standard hydrogen electrode)
example: $\ce{Zn(2+)(aq) + 2e- -> Zn(s)}$
The process is not quite as simple as it appears. To move from right to left the equilibrium has to have the zinc atomised and then ionised and then the ions solvated.
Each of these processes is either exo or endothermic and is affected by the temperature.
The equilibrium between the ions and the pure metal is also affected by the temperature.
If the equilibrium is exothermic overall from left to right then an increase in the temperature will make the equilibrium move more to the left hand side making the electrode potential more negative.
The reverse is also true - if the equilibrium is endothermic overall from left to right then increasing the temperature makes the electrode potential more positive.

Note, both provided answers, are in effect, very situational, with, for example, is the removal of atoms via sublimation an exothermic or endothermic process.


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Photoelectric effect question

My attempt:- 1)At 0.68e15 Hz frequency of light,Metal A gives 7.2eV Kinetic Energy(KE).
2)At 1e15 Hz frequency of light, Metal B, gives 6eV Kinetic Energy.
3) At 1.1e15 Hz frequency of light, Metal C gives 5.5eV Kinetic Energy.
4)At1.2e15 Hz frequency of light, Metal D gives 5eV Kinetic energy.
Now $KE_{max}=(\frac{hc}{\lambda}-\phi)$ where c= speed of light, $\lambda$=wavelength of light and $\phi$= work function of metal.
Now $7.2eV=(4.14e-15eVs \times 0.68e15 Hz- \phi)$ for metal A.The wavelength of light is 441 nm. Likewise we can compute wavelength of light incident on Metal B, C and D.
But we don't know the work functions of Metal A,B, C and D.
So, how to answer this question?

For each of the 4 metals, the work function energy is given by the abscissa intercepts, using E = h $\nu$. Those are the minimum energies, for each metal, to just get electrons free, but with no kinetic energy. Should be easy now!

I just did what I said in my other comment and the answer is B. Use the information given to compute the work function energy, in Joules, then use the energy of a photon expression in my other comment!

@EdV, Yes, my computed answer also is in agreement with your answer.

Excellent! Now I don’t have to use my analogy using prisons (metals), prisoners (incarcerated electrons), a sack of money tossed over the prison wall (absorbed incident photon energy), bribing the guard to escape (paying the work function energy), and having extra money for moving around after escaping (kinetic energy of the electron). ;-)


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Electron arrangement in Nickel
I'm struggling to understand why the element Nickel can be written like so: [Ar]3d84s2, but cannot be written as [Ar]3d10.
I know that the d-orbital can 'house' 10 electrons, and that according to the Aufbau principle, electrons must be 'housed' in the vacant orbital which has the lowest energy, so the first way I presented seems wrong to me.
I'm sure that there's some confusion on my end here, or rather, I'm missing something that might seem obvious, but I really don't get it.

I believe that the 4s orbital is of lower energy than the 3d orbital, yet where writing out the electron configuration, you put 4s after 3d. You can look at the periodic table as a guide for this, i.e. period 4 column 1 and 2 is 4s and period 4 column 3 to 12 is 3d.

Nickel is an element that exists and it has its actual electron configuration. Aufbau, on the other hand, is just a rough guideline with plenty of exceptions. Elements do not care about Aufbau. At best, Aufbau is like parlay in the first Pirates of the Caribbean movie: a rule that is not such a great rule.

The available energy levels are arranged in the following order, going from the lowest level to the highest :
$1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p,$  etc. So the $4s$ level is filled before the $3d$.

Particularly once you throw in $d$ and $f$ electrons, the neat, simple ordering of energy levels is no longer applicable. A check of electron configurations across the transition elements shows that quite clearly.

Thank you! What's an easy way to keep in mind how the energy levels are arranged?


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The burning candle - rising water experiment
This is a classic experiment.
On a plate with a little water place a candle and light it. Place a drinking glass over it. The light goes out while sucking up the water.
At least two things happen:

The oxygen O₂ is changed to CO₂
The air is heated up a few degrees
Water is sucked up

But then I tried to change the experiment with a vase (big glass) and opening upwards and place the candle at the bottom. Then lit the candle and place plastic over the opening. The result was that the light went out but it didn't suck down the plastic. So it there is no vacuum in the glass?
A strange thing is that the plastic was pulsing like it was breathing.
It can be seen here in my video https://youtu.be/aolz8Y27Lsk
Why is the experiment so different from the other?

Obviously the plastic didn't completely seal the vase.

Your chemical description is incomplete.

There must be a leak somewhere, because the gas volume does not change during the combustion. And the volume must change during the combustion. It must increase or decrease, as explained by Karsten Theis

The seal on your vessel has to be perfect for this to work. An inverted glass vessel achieves this; it is very, very unlikely that a plastic seal will do so.

Changes in volume
The chemical reaction for burning a candle is something like this:
$$\ce{C25H52 + 38 O2(g) -> 25 CO2(g) + 26 H2O(g)}$$
For every 38 dioxygen molecules used, you are making 25 carbon dioxide molecules and 26 water molecules (which start out as a gas, but will condense once they reach an area of lower temperature such as the glass surface). So once the flame is out and the water has condensed, the volume should be less than that of the fresh air (20 % of the volume initially are oxygen. Its volume will be replaced by that of carbon dioxide, so the volume will go from 100% to 25/38 * 20% + 80% = 93%). While the candle is burning, however, the temperature is higher, and some of the water will be in the gas phase. Also, while the air was already hotter than room temperature before the container was covered, the temperature will rise as soon as the influx of cooler (fresh) air is cut off.
How hot will it get?
A tea light produces heat at about 30 J/s, and (with a molar heat of combustion of about 15,000 kJ / mol) used up $\pu{80 \mu mol}$ oxygen per second. This means you need about 10 mL of fresh air each second the candle burns. The container has a volume of about 400 mL, so the candle should burn <40 seconds. In the video, it burned about 2 minutes (not bad for an estimate). Using up all the oxygen in the container would give off about 1.2 kJ, which is sufficient to raise the temperature by 1000 degrees Celsius if it were isolated. As it is, most of the heat flows into the container.
What is different when using water?
Water has a higher heat capacity, so it is more efficient in cooling down the gas (and the container). As a consequence, the water level will rise appreciably. As MaxW mentions in the comments, the plastic probably did not seal perfectly. The pulsing could be intermittent flow of gas, or temperature fluctuations because of mixing.

So in the next experiment I will

put water at the bottom to increase coolant
light the candle and cover immediately
use more heat resistant plastic
use better and more rubber bands.
Goal: To see the plastic cover be sucked in the container (will that happen?)
(Sorry for making such a lousy experiment)

@hschou Instead of a candle, you could use some steel wool soaked in vinegar. It will use up the oxygen without making carbon dioxide (the oxygen ends up being rust). This experiment removes two complications - the carbon dioxide and the large changes in temperature. But I encourage you to try your 2nd experimental design as well - the first try is always lousy, and you only learn by repeating experiments.

Why do most people want to believe that on burning fuel little or no carbon monoxide is created?
Yes, it could be true, mainly CO2 is formed, but read these qualifying comments from a source:

A properly designed, adjusted, and maintained gas flame produces only small amounts of carbon monoxide, with 400 parts per million (ppm) being the maximum allowed in flue products.

And, per the same source:

Incomplete combustion occurs because of:

Insufficient mixing of air and fuel.
Insufficient air supply to the flame.
Insufficient time to burn.
Cooling of the flame temperature before combustion is complete.

Also, per Wikipedia, to quote:

In the presence of oxygen, including atmospheric concentrations, carbon monoxide burns with a blue flame, producing carbon dioxide.[10]

So, many combustion reactions perhaps should be expressed in steps including:
$\ce{2 CO + O2 -> 2 CO2}$
So, perhaps more likely, the presence of CO is more of a reality than many would want to believe.
In the case of your experiment, leakage (from possible thermal exposure damage and/or a poor seal) along with incomplete combustion may be some route causes (as carbon monoxide is not very soluble in water, 27.6 mg/L  as compared to CO2 with 1,450 mg/L at 25 °C).


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How can I calculate Kp given Gibb's free energy?
I need to solve a problem but I can't seem to figure out how to move from the given values to Kp. Here's the question:
2SO2(g) + O2(g) ⇌  2SO3(g)
ΔrH° = −200 kJ.mol-1 and ΔrS° = −187 J.mol-1.K-1 at 25 °C.
Calculate the temperature at which Kp = 1.
I suppose I'm supposed to calculate ΔG first, and then Kc, but then I'm pretty much stuck.

Using the following relations:
$\Delta{G}^⦵ = \Delta{H}^⦵ -T\Delta{S}^⦵$
$K_P = e^{\frac{-\Delta{G}^⦵}{RT}}$
How would you find the temperature at which $K_P = 1$?

 $K_P = 1$ when $\frac{-\Delta{G}^⦵}{RT} = 0$, which is when $\Delta{G}^⦵ = 0$

Given what $\Delta{G}^⦵$ must equal, how could you find T?

 $\Delta{G}^⦵ = 0 = \Delta{H}^⦵ -T\Delta{S}^⦵ \therefore T = \frac{\Delta{H}^⦵}{\Delta{S}^⦵}$


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Why Molecular Solids can be treated as Monoatomic in Reactions
I'm trying to figure out what is the precise rule for the formula of elemental substances in reactions. I'm afraid I can't word the question properly, so I'll try to give some examples.
1)Oxidation of some metal, e.g. Calcium
$2Ca+O_2 \rightarrow 2CaO$
Since Calcium is a metal I understand that each atom reacts independently and the formula is plain $Ca$
2)Oxidation of some molecular gas, e.g. Hydrogen
$2H_2+O_2 \rightarrow 2H_2O$
Now atomic hydrogen in gas form is not what one usually expects, but rather the diatomic gas, so the formula must be $H_2$.
3)Oxidation of Sulfur
$S+O_2\rightarrow SO_2$
Now, here I get confused. I just copied the reaction from a textbook, but everywhere I looked in the internet says that this is the correct formula. On the other hand the textbook says that the most common allotrope of sulfur has formula $S_8$. And now I wonder why the reaction should not involve molecular sulfur and be
$S_8 +8O_2\rightarrow 8SO_2$.
I do understand that either way one would get the correct mass proportions, but I'm sot sure if that wold be the case in other reactions or with other substances.
So, I get that for metals the formula for the elemental substance that appears in reactions is just the atom. I get that for carbon and other network covalent materials the same applies. And I get that for molecular gases you need the diatomic formula, because each molecule participates as a whole. What puzzles me is what is the rationale beyond plain atoms appearing in molecular solids like sulfur, phosphorous, etc.
I apologize for not being able to state the problem in a more succinct fashion, feel free to edit the question or ask for further clarifications.

;/ I've a feeling you wouldn't make more ambiguous title even if you tried. As far as your q. is concerned, it's just that specific form of sulfur is largely irrelevant for reaction.

@Mithoron Yes, I reckon that the title is not good. I would appreciate any suggestion to improve it

Kinda better now.

Generally, other than the seven diatomic gases at STP ($\ce{H2}$, $\ce{O2}$, $\ce{N2}$, $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$... though  $\ce{Br2}$ & $\ce{I2}$ have fairly low vapor pressure at room temperature), for convenience, other elements are considered as if they reacted as individual atoms, though, as you state, that is not actually the case. Phosphorus, for example, can be in allotropes with formulas of $\ce{P2}$, $\ce{P4}$, or even a graphitic form, but in a reaction, it's usually considered to be monatomic (unless one must take into account the energy to break the bonds of a specific form).
BTW, iodine vapor is diatomic only at moderate temperatures. From Wikipedia: "1% of a sample of gaseous iodine at atmospheric pressure is dissociated into iodine atoms at 575 °C." Furthermore, at STP, iodine solid behaves as a semimetal, so perhaps it shouldn't be considered diatomic... but it's easier to just say "all the halogens are diatomic", ignoring iodine and fleeting astatine.
In brief, then, this is the conventional list.

This is indeed a clear statement of the rule, but do you have an argument for why it doesn't affect stoichometric calculations to just ignore the molecular nature of the substance and treat it as monoatomic?

As for stoichometric calculations, for solids and liquids, they're based on the mass based on the chemical formula for solids and liquids. Since gases are traditionally measured volumetrically, knowledge of the number of particles is required.

Sulfur is made of $S_8$ molecules when cooled down from gaseous or liquid state. But when it is obtained by a chemical reaction at room temperature, it is a mixture of many allotropes, like $S_n$, where n is not well defined and may vary.
Also, by cooling abruptly liquid sulfur in carbon disulfide and extracting the soluble part, the dissolved sulfur is $S_4$ The residue $S_6$.
Anyway, whatever its formula, it will react the same way with $O_2$ or any other reagent.
That is why usually sulfur is described simply by the formula S.
Reference (in French) : Clement Duval, Soufre, Presses Universitaires de France, Paris 1967.

Thanks, this contains relevant information. But is the presence of various allotropes in a mixture what allows one to treat it as monoatomic or either way it will act as if it were monoatomic?

Different allotropes of sulfur will react in the same way with oxygen, chlorine or any other reagent, producing the same products, with the same rate law.

@Maurice, the rate of reactions do depend on allotropic form, e.g  white phosphorous reacts readily in air, with luminescence, though black and red forms are relatively inert. Ozone reacts readily with saturated hydrocarbons at STP, but O2 does not.


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Auto-ionization equilibrium of water shifted with acid-base addition?
Adding acids or bases to water, so that either pH or pOH decreases independently of the other, will that affect the auto-ionization of water? For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?

@ed v I am thinking about what happens when a tiny amount of ammonia is added to the water.  Ammonia takes away hydrogen ions rather than putting in hydroxide ions, so it actually increases the extent of water autoionization as the water chases the hydrogen ion ansorption.

The autoionization with ammonia is the hydroxide ions which came entirely from water molecules, and there are more of those than in pure water.  There are fewer solvated hydrogen ions, true, because the ammonia consumed some of them, whereas it didn't react with the hydroxide ions from the water.

I think you misread what I mean. I'm thinking of the auto-ionization independently from the reactions with other acids and bases (water itself can be viewed as an amfolyte, capable of giving away both a proton and a hydroxide ion. ) Then, if the addition of other acids or bases will tend to increase or decrease the auto-ionization.

Addition of acids or bases to pure water will always decrease the auto-ionization.

@Rafiki  As acidobasic equilibrium reactions of acids and bases share the common ions with the equilibrium reaction of water autodissociation, you cannot consider them separately.  Each equilibrium honors its equilibrium constant. Shift in ion concentrations due one equilibrium affects all other equilibriums sharing this ion.

It depends how you write the reaction.

For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?

If we call the base $\ce{B}$ and the conjugate acid $\ce{BH+}$, you could write:
$$\ce{B + H3O+ <=> BH+ + H2O}$$
This would mean that you just lost some hydronium ions, and water will ionize a bit more to attain equilibrium with its ions.
Or you could write it like this:
$$\ce{B + H2O <=> BH+ + OH-}$$
This would mean that you just gained some hydroxide ions, and water will ionize a bit less to attain equilibrium with its ions.
What about adding acid?
Same story:
$$\ce{AH + H2O <=> A- + H3O+}$$
or
$$\ce{AH + OH- <=> A- + H2O}$$
Which one is correct?
Either one is fine. You will arrive at the same equilibrium concentrations if you consider one of the acid/base reactions and water auto-dissociation. Depending on what is the major species and what is the minor species, you can choose one or the other. Or if you can't decide, you can be creative and write:
$$\ce{AH + 1/2 OH- <=> A- + 1/2 H3O+}$$


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Is there a quick way to approximate a logarithm in Nernst equation?
The problem is as follows:

In a galvanic cell the cathode is an $Ag^{+}(1.00\,M)/Ag_{(s)}$
  half-cell. The anode is a standard hydrogen electrode immersed in a
  buffer solution containing $0.10\,M$ benzoic acid $(C_6H_5COOH)$ and
  $0.050\,M$ of sodium benzoate $(C_6H_5COO^{-}Na^{+})$. The measured
  cell voltage is $1.030\,V$. What is the $pK_a$ of benzoic acid?.

What I did to solve this problem was to find the potential for the cell involving the standard hydrogen electrode. At first I was confused because there are three "elements" featured in the problem. One being the silver electrode, the other the standard hydrogen electrode and the other a buffer solution, so I didn't know how to proceed from there.
Then I noticed that to get the constant of equilibrium I only require the concentration of $[H^{+}]$ ions as,
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
Therefore to obtain those protons I did this:
The half equations in the cell are:
$\begin{array}{cc}
Ag^{+}+1e^{-}\rightarrow Ag_{(s)}&E^{0}=0.7999\,V\\
H^{+}+1e^{-}\rightarrow \frac{1}{2}H_{2(g)}&E^{0}=0.0000\,V\\
\end{array}$
Hence the overall reaction for this process would be:
$E^{0}_{cell}=E_{cathode}-E_{anode}=0.7999-0.0000=0.7999\,V$
Which is for:
$Ag^{+}+\frac{1}{2}H_{2}+\rightarrow Ag_{(s)} + H^{+}$
Hence:
$E_{cell}=E^{0}-\frac{0.0592}{n}\log\frac{[H^{+}]}{[Ag^{+}]p^{\frac{1}{2}}_{H_{2(g)}}}$
Since it indicates that the cell potential is $1.030\,V$ then:
$1.030=0.7999-\frac{0.0592}{1}\log\frac{[H^{+}]}{[1](1)^{\frac{1}{2}}}$
Solving this I'm getting:
$[H^{+}]=0.000125306\,M$
Now all that's left is to plug in this value in the equation to get the equilibrium constant:
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
$[C_6H_5COO^{-}]=0.05\,M$ and $[C_6H_5COOH]=0.1\,M$
Hence:
$K_a=\frac{(0.05)(1.25306\times 10^{-4})}{(0.1)}=6.2653\times 10^{-5}$
Therefore the $pKa$ of benzoic acid would be:
$pKa=-\log Ka=-\log\left(6.2653\times 10^{-5}\right)=4.20306$
Which does seem to be within the value of benzoic acid which I have on different references. But the problem with this method it is that it required the use of logarithm.
Given this situation, does it exist an approximation or anything that can be done right of the bat to get an idea where that value would be?. Does it exist another method which I could use?.

You can involve Czebysev orthogonal polynomial serie as logarithm approximation.. :-)

How did you come to  $[\ce{C6H5COO-}]=[\ce{H+}]=\pu{0.05 M}$ ?  It is approximately valid for the acid solution, but not for the solution of the acid and its salt. Rather  $[\ce{C6H5COO-}]=[\ce{H+}]+[\ce{Na+}=[\ce{H+}]+\pu{0.05 M}$

@Poutnik, good answer! Use an orthogonal polynomial if you don't like logs or Taylor series expansion. The question is why avoid logs to begin with?

@M. Farooq  I remember the golden age of the "true programmers", 8bit computers and Z80 Assembler language. I owned the commented source code of ROM content of the legendary Sinclair ZX-Spectrum and noticed Czebyshev orthogonal polynomials ( COP ) were the way how it approximated functions like EXP, LN, SIN and ArcTan. For those not familiar, A polynomial function based on COPs minimizes the maximal approximation error for given function and an approximating polynomial based on them, for the given polynomial order.

0
@Chris Steinbeck Bell You may use few useful points for rough numerical approximation: Log ( A . 10^N ) = Log ( A ) + N, where A is +1.0 .. +10. Log(1)=0 Log(2)~=0.30 Log(3)~=0.48 Log(5)~=0.70 Log(10)=1

@Poutnick I am sorry I made a mistake in the typing. It should had been only the concentration of the salt $C_{6}H_{5}COO^{-}$ equal to $0.05,M$  As you can see in the following steps I did calculated the $[H^+]$ from the cell reaction from silver and the SHE. Are we okay with this?. By the way I am aware of the Chebyshev's polynomials but I don't have enough experience with them to use them as an approximation for logarithm function. Can this be done by hand?

@M.Farooq The thing is what if I don't have access to a calculator and I must find the result or an approximation to a reasonable accuracy. Can this be achieved without using a logarithm?. The previous comments mentioned a golden age of programming. But as already mentioned, what if I dont have access to such technology? By the way I am familiar with the golden age of programming but for me the only language which made me feel comfortable was BASIC.

@Poutnik Thanks for those log shortcuts. It seems than other than the hat rule for logarithms and change of base and product or division and memorizing the values for some integers there isnt much that can be done, which of course would be to use the polynomials you mentioned.

@ChrisSteinbeckBell, I am not a programmer either, but calculator is in eveyone's reach.  I have studied in a time in school which taught us how to use printed tables to calculate log. It was fun looking up mantissa and characteristics. Anyway, you can use Taylor series expansion https://www.efunda.com/math/taylor_series/logarithmic.cfm to approximate log function.

Everybody, who can post to Chemistry SE site, has available plenty of free software based calculators. For Windows, Linux, Android. Not sure about Mac and iPhone, but I bet there as well. There are online web based calculators as well.

@M. Farooq Taylor serie for logarithm has quite slow convergence.

I still do not understand why the original poster is avoiding a log. Certainly, he wishes to avoid log "button" on a calculator, so go for the slow but algebraic version.

@M.Farooq I don't know if you are getting my idea. What I intended to ask is if there was a method to solve these kinds of problems which involve logarithms by inspection or by a rough and still reasonable approximation with the least steps. Something you can do perhaps mentally or with short lines by hand. Needless to say that there are plenty of tools, but I'm aiming at an abreviated method of computation. By the way I don't have a personal aversion to calculators or anything along those lines, but rather as a recommendation of what to do when for some reason a calculator isn't available.

@M.Farooq And this extends to a cell phone or whatsoever. But sure It seems that the only methods mentioned are either Taylor's expansion, or memorization of some log values, or having to consult with a table. Are we right on this?.

@Poutnik It might not seem obvious but of course you can use any free tool available when you're in front of a computer. But the intention of this question was what approach can be recommended when is not available. And as I mentioned in the lines from above, the best is to either memorize values or consult with a table or if you have more space and time, to try out Taylor expansion. But in all and all Nernst equation requires a logarithm and you can't avoid this unless the concentrations are near to the values of the base either as a base 10 logarithm or an euler number log if you go

@Poutnik by the natural logarithm route, which of course was already simplified by using the $\frac{0.0592}{n}$. I'm not sure if this way was I better understood. :)

From your comments it seems that you are looking for an approximation of a log. I wish you clarified that in the main question without mentioning calculators. It seemed you just wanted to avoid a calculator for some unknown reasons. As Poutnik states, anyone who can post here, will certainly have access to computers and hence the ability to calculate logs.
A first-hand approximation is $ln (1+x)$= $x-x^2/2+x^3/3-x^4/4+...$
Also see https://www.quora.com/How-can-I-calculate-ln-x-without-using-a-calculator
Use you use a simple ln to log conversion factor in the Nernst equation.
So whatever number you get after dividing the concentrations, subtract "one" from it. This is your $x$. Since you are assuming that a calculator or computer is not available, all you need is a paper an pencil to evaluate the right hand side.
No need to memorize anything for logs.
If you can afford a slide rule, https://en.wikipedia.org/wiki/Slide_rule, it can also help as a mechanical calculator. I never used them but Youtube has videos on it.

@M.Farroq As Poutnik's mentioned in the end this depends on how far are you going with the approximation as the function has slow convergence?. The thing here is that you can use a computer or any technological tool but if later you cannot have access to it for some reason does it exist another way to aleviate this caveat?. From your answer its a yes. By the way i do have a japanese slide rule, but the thing here was to do a quick calculation on the go that you can do in few minutes or something without compromising the answer too much.

But as I indicated above, galvanic cells in non standard conditions must require resort to Nernst equation and solely to that reason one way or another you might have to deal with them. If you are allowed to modify or set the experiment or whatsoever, perhaps the best choice without avoiding much fuss is to select concentrations which cancel or simplify the logarithm. :)

At least give Taylor series a try. I do understand your point now. Sorry there is no simple short cut.

It would benefit the answer to note that concentration of $[H^{+}]$ in these cases are found from SHE (standard hydrogen electrode) and any other electrodes which this is interacting. Am I right with this?.

No, the hydrogen ion concentration is determined via pH of each half cell. What I do, is that I apply Nernst equation on individual half cells rather write the entire cell reaction for one Nernst equation.

In my calculations I did used the SHE to get the concentration of $[H^{+}]$ ions. Applying the Nernst equation to individual half cells in the end is subtracted cathode minus the anode and those logarithms will multiply and end up as a fraction, which would account for the same as adding the two half reactions. In the end it ends the same. But yes to get the $pH$ depends on the concentration of those ions for each half cell. Does my approach exhibit any flaw in this question?. Sorry, I just want to get my concepts correct!.

I think your concepts are alright.

Note that the mentioned Taylor serie is for the natural logarithm, not decadic logarithm. Be aware that depending on context and domain customs, natural logarithm is ln or log, decadic log or log10.

About Czebyshev approximation of ln ( 1 + x ), see https://math.stackexchange.com/questions/2447075/chebyshev-expansion-of-log1-x

Padé Approximation for ln(1+x) provides very interesting trade off between simplicity and accuracy ( See also Wikipedia - Padé approximant ):

$$P\{ \ln( 1+x ) \} = \frac{x(6+x)}{6+4x}$$
$\ln(1) = 0$, $\ln(2) = 0.7$,  $e_\mathrm{max} = 0.00685$, $e_\mathrm{max, rel} \lt 1\% $, $e_\mathrm{RMS} = 0.00258$
This is already a good and fast approximation to $\ln(1+x)$ and in many applications like realtime displays can be used as a basis for logarithmic scaling with a few minor modifications for integer arithmetic.
Where greater accuracy is required the expression can be optimised by least squares fitting of the coefficients over the range 0-1. This yields:
$$p'\{ \ln( 1+x ) \} = \frac{x \cdot (6 + 0.7662\cdot x)}{5.9897 + 3.7658 \cdot x}$$
$\ln(1) = 0$, $\ln(2) = 0.69358$, $e_\mathrm{max} = 4.3E-4$, $e_\mathrm{max, rel} \lt 0.1\% $, $e_\mathrm{RMS} = 1.5E-4$
This is about as good as you can get with this simple formula, and these coefficients may be scaled up to suitable integers for use.
Much higher accuracy is possible by starting from a better series:
Define: $y = x/(2+x)$,  then $\ln{ \frac{1+y}{1-y} } = 2y + 2y^3/3 + 2y^5/5 + ... $
$\ln(1) = 0$, $\ln(2) = 0.69300$, $e_\mathrm{max} = -0.00014$
From which the Pade approximation yields:
$$P\{ \ln \frac{1+y}{1-y} \} = \frac{2y*(15 - 4y^2 )}{15 - 9y^2}$$
$\ln(1) = 0$, $\ln(2) = 0.693122 $, $e_\mathrm{max} = -0.000025$
Again this expression can also have its coefficients tweaked to improve accuracy over a narrow range still further.


text:
Effect of the counterion on the colour emitted by a metalic salt
Could someone please enlighten me on what would be the theoretical effect of a counter ion on the colour of the flame emitted by a metallic salt in a flame test?
Eg. NaCl - what would be the effect of the non-metal on the general colour of the flame of the compound?
I am looking into specific groups of metallic salts based on 3 metals: Sodium, Potassium and Copper II.
I have searched throughout the literature but have not found an answer yet.

One anion giving colour is Borate

Nice answer here: https://chemistry.stackexchange.com/a/117691/79678. Another colorful anion would be cuprate ion.

Recall the flame emission experiment is basically an atomic emission experiment. We don't say it is a compound emission experiment. The job of the flame/plsama or any high temperature medium is to decompose the compound into the constituent atoms or even ions for an experiment. Once a dilute metal salt solution is aspirated into the flame, the compound decomposes into constituents and then to the atoms. For example, you start with NaBr, NaF or NaI or NaNO3, in the flame you will get one and only emission at 589 nm from the sodium atoms. All of these compounds will make the flame bright yellow. So what happened to the non-metallic anion? Provided the flame temperature is high, one should "see" emission from non-metallic atoms as well. Our eyes cannot see the emission bc it in the deep UV region.
Sometimes the anion interferes with the atomization process such as phosphate for calcium. It forms very thermally stable compounds in the flame. You will see the signal of calcium go down quickly if phosphate ion is present.
Forget about copper as atomic emission, I guess you are thinking about the green flame of copper compounds. This only happens in low temperature flames. That color originates from molecular compounds of copper in the flame, and the anion may play a role. You need a very high temperature flame and copper emits in the UV.

You can put in the flame any compound of Sodium, Potassium or Copper. The flame will be colored. But the emitted light will probably be highest with chlorides, because chlorides are more volatile than sulfates, carbonates or any other compounds. In order for the flame to be colored, the substance must first be volatilized, then decomposed into its atoms, then excited by the flame up to an electronic level. And then, on the way back, light is emitted. The first step, the volatilization, is easy with chlorides (maybe also with other halogenides - I don't know). With copper, it seems to  be more complicated, because the light can be emitted from copper metal and also from volatile chlorides like $CuCl$, before their decomposition. It depends on the temperature.


text:
Comparison of lattice energy
Which compound has largest amount of lattice energy among $\ce{AlF3}$ and $\ce{AlN}$. I think it should be $\ce{AlF3}$ because lattice energy in inversely proportional to radius and size of flourine is quite small as compared to nitrogen. Talking about charge then both have charge of magnitude 3 but my textbook says $\ce{AlN}$ has higher lattice energy could you explain

Have a look at https://chemistry.stackexchange.com/questions/104805/comparing-the-lattice-energy-of-different-compounds/104864#104864 FWIW I don't think this is a very good question.

In my opinion the difference in charge always prevail over atomic radius. Additionally, according to the Born-Lande equation says the lattice energy is proportional to the charges on each ion, not the charge difference. The fluoride ion has a 1- charge which is smaller than the nitride which is 3-.
Anyways these ionic solids have different structures and thus different Madelung constants, it does contribute to the lattice energy difference as well.
You can also refer to the Kapustinskii equation (search on Wikipedia). If you can assume that the radius of the nitride and fluoride ions are roughly the same (in fact the nitride is larger, but as I said the difference in charge always prevail over atomic radius), then the lattice energy is proportional to the total number of ions in the empirical formula and the absolute value of the ion charges. For AlF3, it is 4*3*1=12 and AlN is 2*3*3=18. So the aluminium nitride should has larger lattice energy.


text:
What is "dithreitol"?
The article by Hu et al. [1] mentions a compound "dithreitol" in the Saliva Proteome Analysis section (emphasis mine):

For the "shotgun" approach, saliva samples (1 mL) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA).
  Individual fractions were treated with 10 mM dithreitol for 30 min and then 50 mM iodoacetamide for 30 min.

PubChem doesn't reveal any details as to what this compound exactly is and suggests alternative spelling, e.g. D-threitol.
Google Books search results in several mentions of this compound in the literature, but I didn't find any further details except that it's probably a sulfhydryl reagent.
This makes me think that "dithreitol" could be another name for dithiothreitol, but I didn't manage to find any solid references to support this proposition.
To sum it up: what is dithreitol, exactly, and what chemical and structural formulas does it posses?
References

Hu, S.; Li, Y.; Wang, J.; Xie, Y.; Tjon, K.; Wolinsky, L.; Loo, R. R. O.; Loo, J. A.; Wong, D. T. Human Saliva Proteome and Transcriptome. J Dent Res 2006, 85 (12), 1129–1133. DOI: 10.1177/154405910608501212.

In some papers, like this https://pubmed.ncbi.nlm.nih.gov/3139612/, are mentioned urea-dithreitol-sodium lauryl sulphate (UDS) and urea-dithiothreitol (UDT). If you google "uds urea lauryl", you get many results which define UDS as the ingredients above plus dithiothreitol. So I guess that is just a kinky name for dithiothreitol.

@The_Vinz If you found an article that explicitly states that dithreitol≡dithiothreitol, please feel free to post an answer.

@andselisk, It is most likely a spelling mistake which is not uncommon. If SciFinder does not list it (as written) then it is not a standard name. I got zero results.

@M.Farooq I agree it could be a typo if the term is encountered in a single place or two. Here, we are dealing with several dozens (if not hundreds) entries for the textbooks (Google Books) and papers (Google Scholar). This is weird.

@andselisk, Could you locate the original method of the procedure used by the authors? It is a chain mistake and other people blindly quoted the same name for dithiothreitol. The same thing happens for references. Believe it or not there are so many (mostly minor) mistakes in the published works that one cannot imagine. My boss used to say "Paper never refused an ink"...meaning you can write whatever you want, the paper won't say anything.

It's probably a typo- there is no other good explanation: (1) dithreitol could be a misstyped version of "D-threitol" (2) dithreitol could be a synonym (redundant) of threitol (3) dithreitol could refer to a covalent dimer of threitol, say an ether, or  longer polyol. Given that the context in which the compound appears is often as a buffer additive, often in texts in which the purported compound acts as a (thiol) reducing agent, the best explanation is that "dithreitol" is a typographical error where DTT should have stood.

DTT is very commonly used. It is interesting that the (puported) error arises so often, but some typos should not seem entirely strange, given the frequency with which DTT is reported used, the ease with which the typo can be committed, and the ease with which it could be overlooked or ignored as insignificant.

For the "shotgun" approach, saliva samples ($\pu{1 mL}$) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA). Individual fractions were treated with $\pu{10 mM}$ dithreitol for $\pu{30 min}$, and then $\pu{50 mM}$ iodoacetamide for $\pu{30 min}$.

This reaction sequence in Ref.1 has a clear intention: First, break disulfide bonds in saliva proteins using a thiol exchanging reagent and then trap the recently formed thiol groups by iodoacetamide (Ref.2). According to Ref.2:

Alkylation of cysteine thiols with iodo, bromo or chloro substituted acetic acid or acetamide is a classic approach that has been exploited since the 1930′s.

Thus, it is safe to assume that this second reaction is intended to trap thiol group as quickly as possible (The relative reaction rates between glutathione and these halogenated acetates are $100:60:1$ for iodo-, bromo- and chloro-acetates respectively; Ref.3).
Also keep in mind that disulfide reduction to corresponding thiol is accomplished primarily by thiol exchange type reagents such as dithiothreitol (DTT) or 2-mercaptoethanol (ME). Thus, it is also safe to assume that dithreitol mentioned in Ref.1 is definitely a thiol exchanging reagent, possibly DTT. Therefore, I think it is a chain mistake as M. Farooq suggested in his comment.
References:

S. Hu, Y. Li, J. Wang, Y. Xie, K. Tjon, L. Wolinsky, R. R. O. Loo, J. A. Loo, D. T. Wong, "Human Saliva Proteome and Transcriptome," J. Dent. Res. 2006, 85(12), 1129–1133 (DOI: 10.1177/154405910608501212).
Jakob R. Winther, Colin Thorpe, "Quantification of Thiols and Disulfides," Biochimica et Biophysica Acta 2014, 1840(2), 838-846 (https://doi.org/10.1016/j.bbagen.2013.03.031).
Frank Dickens, "CLII. Interaction of halogenacetates and SH compounds. The reaction of halogenacetic acids with glutathione and cysteine. The mechanism of iodoacetate poisoning of glyoxalase," Biochem J. 1933, 27(4), 1141–1151 (doi: 10.1042/bj0271141).

One of the publication in andselisk's search even had the CAS number, "dithreitol, 3483-12-3". Another publication had "dithio dithreitol". Many publications have something like "dithreitol (DTT)". It reminds me of mis-citations of https://www.nature.com/articles/227680a0.


text:
What exactly is temperature?
I've read at many places that temperature is the average kinetic energy of particles present in an object. I just don't intuitively get how kinetic energy is connected with temperature. And how is heat connected with temperature then? Then what exactly is temperature? All of the descriptions given online are very confusing.

Try to study hyperphysics-temperature and Wikipedia - temperature and links leading from there.

I'm voting to close this question as off-topic because it's outside of scope of the site.

@Mithoron I'm a little confused why it's out of the scope of this site. Doesn't this fall nicely within stat mech and thermo? Not to mention, the relationship between temperature and average kinetic energy is fundamental to vibrations and physical chemistry.

@jezzo Isn't temperature a question for [physics.se], not [chemistry.se]?

@CJDennis it depends on where you draw the line. For example, there is a thermodynamic definition of temperature. I'd say thermodynamics falls within both physics and chemistry.

@jezzo https://www.xkcd.com/435/

hahaha love it. Guess we can only apply it then, no asking questions! ;)

Temperature is not THE kinetic energy of particles. It is only proportional to this kinetic energy, and the proportionality constant is k, the Boltzmann constant.

@Maurice The proportionality constant is not k, but rather k times f(T), where f is a function of temperature. It is easier if you say it is proportional to the energy in one translational degree of freedom, and the proportionality constant is k/2.

Temperature vs kinetic energy

[OP:] I've read at many places that temperature is the average kinetic energy of particles present in an object.

Temperature has to do with the average kinetic energy of particles, but to say the two concepts are the same is incorrect. What is correct is that if the particles in two mono-atomic gas samples have the same average kinetic energy, they will have the same temperature. For samples that are not mono-atomic gases, see What are the degrees of freedom that define the temperature of an ionic solid (such as sodium azide)?.

[OP:] I just don't intuitively get how kinetic energy is connected with temperature.

If you have a gas in a container (such as the air in a room), gas molecules will collide with the walls. If the walls are colder than the gas (such as a cold window pane in the winter), these collisions will slow down the gas particles on average, decreasing the temperature of the gas. If the walls are hotter than the gas (such as a window pane in the summer), these collisions will speed up the gas particles on average, increasing the temperature of the gas. Because energy (and momentum for elastic collisions) is conserved, changes in the temperature of the gas will be reflected in opposite changes in the temperature of the walls (the magnitude of change will not be the same, it depends on the heat capacities).
Heat vs temperature

[OP:] And how is heat connected with temperature then?

Heat is the transfer of thermal energy. If nothing else is going on, heat transferred from sample A to sample B will go along with a drop in temperature of A and a raise in temperature of B. See also: https://chemistry.stackexchange.com/a/112057
Definition of temperature

[OP:] Then what exactly is temperature? All of the descriptions given online are very confusing.

In the simplest terms, it is what you measure after you put a thermometer in thermal contact with the sample. The sensing part of the thermometer (mercury or alcohol bulb, thermocouple, etc) has to reach the same temperature as the sample. The sample should be much bigger than the sensor so that bringing them into contact does not significantly change the temperature of the sample. The temperature measured by the thermometer is equal to the temperature of the sample because they are at thermal equilibrium (heat exchange is zero), and the thermometer has some property that changes with temperature (such as the volume of alcohol) in order to sense its temperature. See also: Temperature measurement
The quantitative definition of temperature is given in the official definition of its SI unit Kelvin:

The kelvin, symbol $K$, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $\pu{1.380649e−23}$ when expressed in the unit $\pu{J K−1}$, which is equal to $\pu{kg m2 s−2 K−1}$, where the kilogram, metre and second are defined in terms of $h$, $c$ and $Δν_{Cs}$.

This definition requires a lot of physical chemistry to understand. However, it is sometimes formulated as:

One kelvin is equal to a change in the thermodynamic temperature $T$ that results in a change of thermal energy $kT$ by $\pu{1.380 649e−23 J}$.

So if the thermal energy (average per particle, not stated above) goes up, the temperature goes up.

Small caveat here "If nothing else is going on, heat transferred from sample A to sample B will go along with a drop in temperature of A and a raise in temperature of B." The temperature of B will not change if B is undergoing a phase transition.

@M.Farooq Yes, that would be something else going on - but it is good to have an example. If there is a reaction going on in A, that might be why the temperature is higher in the first place, so A might not decrease in temperature until the reaction is done/

Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing).
Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic energy of the particles in the body. As to why this is, I'd ask: what happens at absolute 0?

 At absolute 0, heat has transferred out of the system so much so that you cannot lower the energy of the system any more.

(Note for the knowledgeable reader: there is indeed still a quantum mechanical phenomena -- zero point energy-- that prevents some energy from leaving the molecule, but that's a conversation for another time).
From a thermodynamic definition, temperature is the description of how the internal energy changes with entropy for a closed (no matter flows in or out) system of constant volume:
$T=\left(\frac{\partial{U}}{\partial{S}}\right)_{N,V}$
By increasing the entropy of the system by a fixed amount, the temperature of the system tells me by how much the internal energy will increase. Now, this is not a very useful form, as you cannot directly increase the entropy of a body (you must add energy and then let the entropy indirectly increase). It's much more useful to consider the inverse temperature:
$\frac{1}{T}=\left(\frac{\partial{S}}{\partial{U}}\right)_{N,V}$
At absolute zero, everything is in the lowest energy state. Any small transfer of energy to the system will result in a large increase in entropy. But this was only a small change, so the internal energy doesn't increase by much. Compare to a system at room temperature, where I must transfer much more energy to the system to achieve the same magnitude of increase in entropy.

I find the temperature argument as a circular definition. Experimentally, how do we increase entropy for a closed system with constant volume? It is unfortunately not your issue, it is thermodynamics itself- full circular definitions.

And yet it's a function of state. I think temperature is an especially tricky notion, one that we'd be better off thinking about in terms of thermodynamic beta. Unlike entropy, you could directly change U by changing the potential, or in cases where you can't change the potential, there is at least a more clear mapping from transferring heat to the system and seeing U increase.

Yes, temperature and heat have bothered the greatest minds & physicists. What annoys me in thermodynamics is the way it is taught to students and the authors never tell you how that mathematical notion is practically implemented. For example, if you say "you could directly change U by changing the potential", the next question comes to my mind is how is that experimentally achieved.

One could vary the external electric field, for example. I agree it's a shame to miss the finer points of the discussion of temperature, but that much is possible.

@M.Farooq To reiterate though, I completely agree that temperature is circularly defined. What do you think of my previous comment though?

You can directly increase the entropy of a system without changing its internal energy.  Just let it expand adiabatically into a vacuum.  Or consider an isolated system containing two pure gases divided by a partition.  If you remove the partition, thus allowing the gases to mix, you are increasing the entropy of the system without changing its internal energy.

@theorist but in the first case V isn’t constant. In the second, N, V isn’t constant

A better term than "circular" is "self-consistent" :-)

This answer is "circular" because it requires a thermodynamic description of entropy, which in turn requires a definition of T .....

@jezzo.  In the second case if you define the system as the two gases together, both N and V are constant.  But let me offer a more obvious example.  Suppose you prepare a system consisting of A+B, and isolate it.  After a time, A+B spontaneously react to form C+D.  Here neither U, V, or  N has changed.  But S has increased.  The resolution is that you need a more general formulation for closed systems, which is that U = U(S, V, {n$_i$}).  I.e, if your only restrictions are that the system is closed & has const. V, it is not sufficient to hold N const., you need to hold all the n$_i$'s const.

Temperature is related to kinetic energy, but it can't be simply equated to the average kinetic energy of the system.  As I wrote in response to another answer,  different systems can have different average kinetic energies/particle, but the same temperature. E.g., at the same temperature the avg. kinetic/energy particle of a diatomic gas is greater than that of a monatomic gas, because the monatomic gas particles have only translational kinetic energy, while the diatomic particles will have the same average translational kinetic energy, but rotational and vibrational kinetic energy as well.
What you can say, however, is that temperature is a measure of the average kinetic energy per available degree of freedom, $\langle H_{kin,DOF}\rangle$:
$$\langle H_{kin,DOF}\rangle = 1/2 N k_B T/f,$$
where $f$ is the fractional availability of the degree of freedom.
Thus we can write:
$$T = \frac{2\langle H_{kin,DOF}\rangle}{N k_B f}$$
Consider again a monatomic vs. diatomic gas.  At low temperatures, the vibrational degrees of freedom might not be entirely available, but the translational degrees of freedom will be (hence, for each of the three translational degrees of freedom, $f=1$). At the same temperature, the average KE/particle of the two gases will be different.  However, for both gases, the average KE per particle per translational degree of freedom will be the same!

It looks like you are invoking the equipartition theorem, but what is N? If N is the number of independent particles, isn't the total number of degrees of freedom for a rotationally and translationally invariant system generally f=3N-5?

@BuckThorn (1) N is the number of independent particles; (2) Note that I've not used f to mean the the number of degrees of freedom, I've used it to indicate the fractional accessibility of any single degree of freedom. (3) Regarding the equipartition theorem: My answer is actually designed to be a more general description that, while keeping the answer relatively simple, also accommodates systems in which the equipartiton theorem fails (because that partiucular degree of freedom isn't fully accessible, because of quantum-mechanical efffects). [I may expand my ans. to make this more explicit.]

@BuckThorn Didn't understand your last comment:  "This is probably the answer that gets closest to a useful description of temperature. What is your other answer (link ?)" Looks like you perhaps didn't finish editing it (?).

Never mind my last comment, after review it looks like I misread a sentence in your answer.

In thermodynamics, the basis for a definition of temperature is provided by the $0^{\text{th}}$ Law: two bodies independently in thermal equilibrium with a third body are in thermal equilibrium with one another. Thermal equilibrium allows the definition of temperature: two bodies in thermal equilibrium are said to be at the same "temperature".
The $0^{\text{th}}$ Law is useful because it provides a means to determine whether two systems that are not in thermal contact would be in thermal equilibrium if placed in contact. This transitive property of thermal equilibrium provides a method of ranking systems, for instance by bringing them separately into thermal contact with a reference body. A thermometer represents such a reference body. A practical measure of temperature (a scale) can be provided by an observable intensive property defines the state of the thermometer. If the thermometer indicates the same "temperature" when in contact with two different bodies then those two bodies are in thermal equilibrium (or would be if placed in thermal contact).
Once a temperature scale is set by using the state of a thermometer as a reference, the next question to answer is, what happens when two bodies at different temperatures are placed in thermal contact? To answer this we invoke the 1st law. The first law defines the additivity of different forms of energy (heat and work). Heat is the change in the internal energy of objects when they equilibrate in thermal contact, absent work. The 1st law also provides a way to measure "heat" by relating it to an equivalent amount of work. For instance, electrical work can be used to change the state of a substance in a rigid adiabatic container. The amount of work is equivalent to the heat that would result in the same temperature change of the substance if the energy transfer had been performed thermally and in the absence of work:
$$\begin{align} \Delta U &= w \tag{adiabatic} \\ &= q\tag{diathermal, rigid}\end{align}$$
Note finally that there is a thermodynamic temperature scale which is based on the 2nd law (as a corollary of the properties of entropy), but in practice we rely on scales whose limiting (ideal) behavior approaches that of the thermodynamic scale.

Temperature is, like entropy and unlike energy, an absolute property, in that it has a "null" point, an absolute zero. But it can be more easily understood in view of this answer as a "relative" measure that becomes evident when bodies are brought into thermal contact, without need to reference the internal microscopic workings of the system. Here kinetic energy is an extra-thermodynamic microscopic property. However it is difficult to explain heat transfer without a kinetic model. Therefore kinetic energy is fundamental to an understanding of heat and temperature.

More accurately one would say that heat is the change in internal energy unaccounted by work. If there is no work, all of the change in internal energy is due to heat.

I like to think of Temperature as a measure of the LEVEL of energy, but not the QUANTITY of energy.
An analogy with water tank levels:
A small tank (0.1m dia) with high level (1m)
A large tank (1m dia) with a small level (0.1m)
It's obvious which has a higher level and which direction the flow would be if you connect the tanks together.
However the large tank has a larger quantity (78 litres) compared to the small tank (7.8 litres), so it has a greater capacity.
Similarly with Temperature - compare a boiling kettle and a huge iceberg. One has a higher level of energy, the other has a higher quantity of energy.

Temperature is the average kinetic energy of the particles making up a system. That's it, and it is correct. Any other definition, and there are many of this page, are either equivalent or incorrect. What's the problem?

The problem is that different systems can have different average kinetic energies/particle, but the same temperature.  E.g., at the same temperature the avg.  kinetic/energy particle of a diatomic gas is greater than that of a monatomic gas, because the monatomic gas particles have only translational kinetic energy, while the diatomic particles will have the same average translational kinetic energy, but rotational and vibrational kinetic energy as well.

So what? Doesn't change the definition.

Temperature is not the average kinetic energy of the particles making up the system, because two different systems can have the same temperature, yet different average kinetic energies.  I.e., what you wrote in your answer is incorrect.

Temperature and energy have different dimensions, so they can't be equal to one another.

Dave, I think you are right. But to avoid disputes you should add "translational" kinetic energy. $\bar{E},=,\frac{3}{2}k_\text{B} T_\text{k}$ considers only the translation part. $T=\left(\frac{\partial{U}}{\partial{S}}\right)_{N,V}$ with a fixed $\partial{U}$ added to a diatomic gas, $\partial{S}$ distributes to a translational component and a component internal to the molecule, thus leading to a lower $T$ compared to a monoatomic gas ($\partial{S}$ being bigger than what accounts for temperature).

@RolandPuntaier:  Systems can have a well-defined temperature without having any microscopic translational kinetic energy at all.  The canonical examples of this are magnetic systems.

@RolandPuntaier And there are also more commonplace examples. E.g., a covalent network solid (such as a diamond) will have no microscopic translational kinetic energy.

@michael-seifert Yes, how the energy is transferred can depend, but if it increases the temperature of the target system, then it increases the translational energy, unless you define a different kind of temperature. In-place vibrations in crystal I consider translational.