text:
Moles of Sodium Peroxoborate

The question involves the addition of $\pu{0.15 g}$ of an already prepared sample of sodium peroxoborate which is added to $\pu{100 mL}$ of $\pu{1 M}$ $\ce{H2SO4}$ and $\pu{2 g}$ of $\ce{KI}$. The solution is left to stand for 10 minutes and is then titrated against standardized sodium thiosulfate. Starch indicator was used and a titration figure of $\pu{17.1 mL}$  was obtained. Find the moles of sodium peroxoborate.

Sodium peroxoborate, sulfuric acid and potassium iodide react as follows:
$$\ce{Na2[B2(OH)4(O2)2] + 4 KI + 3 H2SO4 -> Na2SO4 + 2 K2SO4 + B2O3 + 5 H2O + 2 I2}$$
Iodide reacts with thiosulfate as follows :
$$\ce{I2 + 2 [S2O3]^2- -> 2 I- + [S4O6]^2-}$$
I'm not sure how to use the titration value to find the result of this answer. I'm not looking for the solution I just need a point in the direction that will help me at least get a start on the problem.
What about this:
$$2 \, \text{mol} \, \ce{S2O3 ->} 1 \, \text{mol}  \, \ce{I}$$
when we made the sodium thiosulfate I calculated the moles to be $0.096 \, \text{mol/L}$. Hence
$$0.096 \, \text{mol/L} \, \, \ce{S2O3->} 0.048 \, \text{mol/L}  \, \, \ce{I}$$.
the relationship between iodine and peroxoborate is as follows:
$$4 \, \text{mol} \, \ce{I->} 1 \, \text{mol}  \, \ce{[B_2(O2)2(OH)4]}$$
Hence
$$0.048 \, \text{mol/L} \, \, \ce{I->} 0.012 \, \text{mol/L}  \, \, \ce{[B_2(O2)2(OH)4]}$$.
But I didn't use the titration value of thiosulfate in this solution?

Well, you can use the amount of thiosulfate to compute the amount of iodine. You can use the amount of iodine to compute the amount of peroxyborate. There are stoichiometric relationships between all of those.

Can I say $\text{moles} = \frac{2}{166}=0.012 , $ mol $\ce{KI}$

Because I'm struggling to see why I  can't just say $\frac{0.15}{307.62}= 4.88 \times 10^{-4}$ mol where $307.62$ is the $M_r$ of peroxoborate.

How many moles of thiosulfate react with 1 mole of iodine? How many moles of peroxoborate react with 1 mole of iodine? You can't solve these types of problems without understanding the stoichiometric relationships. Nowhere in your solution do these ratios come into play...

This problem is out of whack. You stated "standardized sodium thiosulfate" but then you don't give a concentration for the thiosulfate. Just knowing how many ml of solution was used doesn't define the number of moles of thiosulfate. The sulfuric acid and KI would be in excess so knowing how much of those doesn't help. The only way to solve with the information given is to use the weight of the sodium peroxoborate. IS THAT THE POINT OF THE PROBLEM!?!

I edited the question, I'm still unsure as to whether the solution is correct or not.

Amounts correlate directly, not concentrations. You are missing the core point if stoichiometry, which is that the balanced equation provided ratios between amount of substance, e.g., number of moles. That ratio does not apply to mass or concentration.

But I don't have the amount of moles of thiosulfate, all I have then is the titration value.


text:
Analytical solution for kinetics of bimolecular reaction
Consider two chemicals, $\ce{A}$ and $\ce{B}$ that react with each other to make $\ce{C}$ with a reaction rate $k$. The reaction can be expressed as $$\ce{A + B->C}$$ The equation expressing the rate of the reactions can be expressed as $$\frac{d[\ce{A}]}{dt}=\frac{d[\ce{B}]}{dt}=-\frac{d[\ce{C}]}{dt}=-k[\ce{A}][\ce{B}]$$
I can separate this equation to make a system of differential equations.$$\frac{d[\ce{A}]}{dt}=-k[\ce{A}][\ce{B}]$$ $$\frac{d[\ce{B}]}{dt}=-k[\ce{A}][\ce{B}]$$
With these two equations, I note that they are similar and will only work with one of these equations for the time being. Therefore, we can write one of these equations as $$\frac{d\ln([\ce{A}])}{dt}=-k[\ce{B}]$$ and by taking another derivative $$\frac{d^2 \ln([\ce{A}])}{dt}=-k\frac{[d\ce{B}]}{dt}=-k\frac{[d\ce{A}]}{dt}$$
I solved this equation using Wolfram Alpha (QED) $$[\ce{A}](t)=\frac{c_1 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}$$ Therefore the rate of reaction can is $$[\ce{A}]'(t)=\frac{c_1^2 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}-\frac{k c_1^2 \exp^2[c_1(t+c_2)]}{(k \exp[c_1(t+c_2)]-1)^2}$$
I observed that the rate of change can be written as $$[\ce{A}]'(t)= c_1 [\ce{A}](t)-k [\ce{A}](t)^2$$ so that $c_1$ may be solved, given the initial conditions of $[\ce{A}](0)$ and $[\ce{A}]'(0)$ such that $$c_1=\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}$$
Substituting the definition of $c_1$ into the equation of $[A](t)$ and $[A]'(t)$ an equation for $c_2$ can be found.
$$c_2=\frac{1}{c_1} \ln(1-\frac{c_1 k}{[\ce{A}](0)}) $$$$
c_2= \frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2}  \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k) $$
Using the equations for $c_1$ and $c_2$ an explicit equation for $[\ce{A}](t)$ can be found.
$$ [\ce{A}](t)=\frac{\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)} \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2}  \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]}{k \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2}  \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]-1} $$
Side note: since $[\ce{A}]'(0)=[\ce{B}]'(0)= -k[\ce{A}](0)[\ce{B}](0)$ then $c_1$ can be rewritten as $$c_1=\frac{-k[\ce{A}](0)[\ce{B}](0)+k[\ce{A}](0)^2}{[\ce{A}](0)}=k([\ce{A}](0)-[\ce{B}](0))$$
this simplifies $c_2$ to
$$c_2=\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)})$$
which simplifes the equation for $[\ce{A}](t)$ to
$$[\ce{A}](t)=\frac{k([\ce{A}](0)-[\ce{B}](0)) \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]}{k \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]-1}$$
with a similar equation for $[\ce{B}](t)$
My quesion is: Is this a valid mathematical model for a bimolecular reaction? If not, what is commonly used?

I think it would be easier to take $[\ce A]_t = [\ce A]_0 - x$ and $[\ce B]_t = [\ce B]_0 - x$ (this must be true because of the stoichiometry of the reaction), then find an expression for $\mathrm dx/\mathrm dt$, then find $x(t)$.

At first glance, I see nothing wrong with this. I haven't checked your computations, but I believe they should be alright. @orthocresol's suggestion is also a valid, and probably easier approach to the problem.

If you assume that $[A] \gg [B]$, do you get a pseudo-first-order (i.e. single-exponential) expression?  This could be a good check on your work.

Except that you've made it look absurdly complex for no reason at all, it seems OK. Expand $c_1(t+c_2)$ as $c_1t+c_1c_2$, and everything will simplify a great deal.

Assuming that the bimolecular chemical reaction $\ce{A + B ->[\kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs
$$\begin{array}{rl} \dot a &= - \kappa \, a \, b\\ \dot b &= - \kappa \, a \, b\end{array}$$
where $\kappa > 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain
$$a (t) - b (t) = a_0 - b_0$$
where $a_0 > 0$ and $b_0 > 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0  - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows
$$\dot a = - \kappa \, a \, \left( a - (a_0  - b_0) \right)$$
which can be rewritten in the form
$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$
Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion
$$\left( \frac{1}{a - (a_0  - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0  - b_0) \, \mathrm d t$$
Integrating, we obtain
$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a_0 - (a_0  - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0  - b_0) \, t$$
which can be rewritten as follows
$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0  - b_0) \, t$$
Exponentiating both sides, we obtain
$$\frac{a (t) - (a_0  - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)$$
and, eventually, we obtain
$$\boxed{\begin{array}{rl} &\\ a (t) &= \dfrac{a_0  - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\\\ b (t) &= \dfrac{(a_0  - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\ & \end{array}}$$
Taking the limit,
$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 & \text{if } a_0 > b_0\\\\ 0 & \text{if } a_0 < b_0\end{cases}$$
$$\\$$
$$\lim_{t \to \infty} b (t) = \begin{cases} 0 & \text{if } a_0 > b_0\\\\  b_0 - a_0 & \text{if } a_0 < b_0\end{cases}$$

What if $a_0 = b_0$?
Previously, we assumed that $a_0 \neq b_0$. If $a_0 = b_0$, then
$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$
becomes
$$-\frac{\mathrm d a}{a^2} = \kappa \, \mathrm d t$$
Integrating, we obtain
$$\frac{1}{a (t)} - \frac{1}{a_0} = \kappa \, t$$
and, eventually, we obtain
$$\boxed{ a (t) = \frac{a_0}{1 + a_0 \, \kappa \, t} = b (t)} $$
In this case, both reactants are eventually exhausted
$$\lim_{t \to \infty} a (t) = \lim_{t \to \infty} b (t) = 0$$

You need to enforce the boundary conditions of your problem to evaluate the integral. I think that might be the material point here.
But let's start at the beginning: You have the rate equation
\begin{align}
    - \frac{\mathrm{d}c_{\mathrm{A}}}{\mathrm{d}t} = k c_{\mathrm{A}} c_{\mathrm{B}} \ .
\end{align}
Now, introduce the conversion $x = c_{\mathrm{A}0} - c_{\mathrm{A}} = c_{\mathrm{B}0} - c_{\mathrm{B}}$. This means that an infinitesimal change in the conversion can be written as $\mathrm{d} x = -\mathrm{d} c_{\mathrm{A}} = -\mathrm{d} c_{\mathrm{A}}$. Substituting this into the rate equation yields
\begin{align}
    \frac{\mathrm{d}x}{\mathrm{d}t} = k ( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x ) \ .
\end{align}
To solve this differential equation it is sufficient to separate the variables and integrate both sides of the equation. And this is the point where the boundary conditions enter the game. To evaluate the integrals fully, i.e. without leaving a constant of integration $C$, you need to use some limits for your integrals for which you know the values of you variables beforehand. So the question is: What is known beforehand? You know that at the start of the reaction, and we will define this point in time as $t=0$, none of you reactants have reacted with one another. So, $c_{\mathrm{A}}(t\!=\!0) = c_{\mathrm{A}0}$ or in terms of the conversion $x(t\!=\!0) = 0$. This will set the lower limit of the integrals. For the upper limit you can use the value that you actually want to calculate, i.e. the concentration $c_{\mathrm{A}}(t)$ or conversion $x(t)$ at some time $t$. Using all this leads to
\begin{align}
    \int \limits_{x(t=0) = 0}^{x(t)} \frac{1}{( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x )}\mathrm{d}x = \int \limits_{t=0}^{t} k \, \mathrm{d}t \ .
\end{align}
For the integral on the left-hand side you can use the hint already given to (although calculating it yourself isn't that difficult if you use partial-fraction decomposition) but you can leave out the constant $C$ as this will be fixed by setting explicit integration limits. The integral on the right-hand side is no problem.
\begin{align}
    \int \limits_{x(t=0)}^{x(t)} \frac{1}{( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x )}\mathrm{d}x &= \int \limits_{0}^{t} k \, \mathrm{d}t \\
    \left[ \frac{1}{ c_{\mathrm{B}0} - c_{\mathrm{A}0} } \ln\left( \frac{c_{\mathrm{B}0} - x }{c_{\mathrm{A}0} - x} \right) \right]^{x(t)}_{0} &= \Bigl[ k t \Bigr]^{t}_{0} \\
    \frac{1}{ c_{\mathrm{B}0} - c_{\mathrm{A}0} } \ln\left( \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - x) }{c_{\mathrm{B}0} (c_{\mathrm{A}0} - x)} \right) = k t
\end{align}
Exponentiating this gives nearly the result that you are after:
\begin{align}
    \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - x) }{c_{\mathrm{B}0} (c_{\mathrm{A}0} - x)} = \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr)
\end{align}
Now, the only work that is left is to make the back-substitution $x(t) = c_{\mathrm{A}0} - c_{\mathrm{A}}(t)$ and solve the resulting equation for $c_{\mathrm{A}}(t)$:
\begin{align}
    \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - c_{\mathrm{A}0} + c_{\mathrm{A}}(t)) }{c_{\mathrm{B}0} c_{\mathrm{A}}(t)} &= \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \\
    c_{\mathrm{A}0} (c_{\mathrm{B}0} - c_{\mathrm{A}0} + c_{\mathrm{A}}(t)) &= c_{\mathrm{B}0} c_{\mathrm{A}}(t) \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \\
    c_{\mathrm{A}0} c_{\mathrm{B}0} - c_{\mathrm{A}0}^{2} &= c_{\mathrm{A}}(t) \left( c_{\mathrm{B}0} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - c_{\mathrm{A}0} \right) \\
    c_{\mathrm{B}0} - c_{\mathrm{A}0} &= c_{\mathrm{A}}(t) \left( \frac{c_{\mathrm{B}0}}{c_{\mathrm{A}0}} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - 1 \right) \\
   \Rightarrow \qquad  c_{\mathrm{A}}(t) &= \frac{c_{\mathrm{B}0} - c_{\mathrm{A}0}}{\frac{c_{\mathrm{B}0}}{c_{\mathrm{A}0}} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - 1}
\end{align}

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text:
Nonsubstituted Cycloalkene HBr Reaction
If I have a cyclohexene with no substituents and add HBr, how do I know what side of the former double bond to place the Br anion? I think I understand Markovnikov's rule, but in my chemistry book they always show one carbon specifically that the Br attaches to and I seem to always pick the other. Does it matter or are they equally likely? Is there some way to decide which carbon is more substituted if there are no other groups attached to it and it's just a carbon ring?
Hopefully I've worded this in a way that makes sense, thanks for any help!

Cyclohexene is very symmetric. At first approximation, you can draw two planes of symmetry: one containing the double bond and one orthogonal, bisecting the double bond.
These two planes of symmetry mean, that both sides of the double bond are equal as are both carbon atoms. Therefore, the hydrogen atom protonating the double bond in $\ce{HBr}$ addition cannot sense any difference between either of the four possible attack pathways and chooses one at random. The bromide then has to choose the other carbon, but can again choose the plane from which it attacks (‘above’ or ‘below’ — in inverted commas because they are equivalent!) at random.
Even after the reaction, the product bromocyclohexane is highly symmetric. You can again draw a plane of symmetry through it; this time, the plane must include the $\ce{C-Br}$ bond and also bisect the carbon atom on the other side of the ring. The presence of a plane of symmetry means that the molecule cannot exist in enantiomeric forms, and since you cannot say on which side the hydrogen was previously added (they, too, are all identical), you only get one possible product.

We might consider a more complicated case and decide to add $\ce{DBr}$ instead of $\ce{HBr}$. In this case, we would indeed observe more than one type of product since a deuterium atom cannot be mapped onto a hydrogen atom by symmetry. However, we would obtain all symmetry-equivalent products at equal ratios because of the high symmetry of the starting compound.

In general, if you can find planes of symmetry then both sides are equally likely to react and you cannot predict a preference.

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text:
Why isn't the hydroxyl group of citric acid deprotonated while the carboxyl groups are?
My AP chemistry book "Chemistry: the Central Science 12th edition" says that one molecule of citric acid generates 3 H+(aq) ions when dissolved in water. Looking at the structural formula:

There are three carboxyl groups. However, there are four instances of an -O-H link. Why don't all of these break?
My guess is that the carboxyl groups can form resonance groups and thus obey the octet rule.

You are correct in saying that the carboxyl groups will have resonance structures after being deprotonated. Thus, the carboxyl groups are fairly acidic and will deprotonate in water. On the other hand, deprotonation of the hydroxyl group will not lead to any resonance stabilization, such that the hydroxyl group will be much less acidic than the carboxyl groups. In fact, the $\mathrm{p}K_\mathrm{a}$ of alcohols is usually in the range 16-20, which means that the alcoholic hydrogen will not deprotonate in water. It can deprotonate, however, in other solvents.

How fitting that an alcohol answers this question ;)


text:
Why does only a single azeotropic composition exist for a mixture of two given substances?
In the image below, which represents a generic composition vs. temperature plot for a mixture, there is a single intersection point between the vapor and liquid curves. This is due to the fact that the composition corresponding to this intersection point has an extremely low boiling point due to negative deviation from Raoult's law. If this is the case, why is it not possible for the vapor curve to "hug" the liquid curve along an entire section, rather than just at one point? Similarly, why does the vapor curve have to intersect liquid curve at all? Would it be possible for the minimum point of the vapor curve to be located just above the minimum point of the liquid curve?


text:
Why can the electron gain process be endothermic or exothermic whereas the ionistation process is always endothermic?
I know that this may seem a very silly question to some of you but please help me.I am extremely confused.
PS:Sorry if you accidentally stumbled to this question.I can't find the appropriate tags

This is a good question, and there is nothing silly about it. See, when you ionize something, you are left with a positive ion and negative electron, and they are attracted to each other. On the other hand, electron gain is about adding an electron to a neutral atom; they may or may not be attracted.


text:
Confusion about definition of Bond Enthalpy
My Chemistry textbook defines bond enthalpy as "amount of energy required to break one mole of bonds of a particular type between two atoms in gaseous state".
Can someone confirm to me whether

a particular type

is related to bond order, to the atoms involved, both, or something else?
Also, should not the bond enthalpy of all diatomic molecules be large as one mole of bonds to be broken. But this does not happens. Why?

Well, it is pretty large.

Bond energies are usually given for each bond-type. You can see this table of values, for example (that's a great site). In order to clasify them we could say that particular type means clasify the bond according to the atoms and to bond-order. So you have C=O, C-O, N=C, N-C, and so on.
If you check the site, look that those values are positive, meaning that the energy -as you said- is the necessary to break the bond. If you need to estimate the difference in enthalpy of formation of:
$$\ce{A<=>B}$$
you will need to change sign of bond-enthalpy values (because usually entalphy is of formation).
The energy is calculated per mol, as you say. And it is a very big value, bigger than enthalpy of vibration, rotation or translation.

So by different type we mean different kind of atoms and bond orders and not single,double,triple bonds etc?Thanks

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text:
What are good representatives for the concepts of endergonic and exergonic?
We must demonstrate two reactions: endergonic and exergonic. I'm wondering if thefollowing reactions are accurate.

Exergonic: I have sodium bicarbonate and vinegar. Would it be catabolic, spontaneous, and $-\Delta G$ as well?
Sodium bicarbonate and vinegar is your basic science-fair volcano reaction. You have the baking soda in a container of some sort and pour the vinegar. The formula would be something like
$$\ce{Bicarbonate of soda + vinegar -> Carbonic acid + Water + Carbon dioxide gas}.$$

Endergonic: I have ammonium chloride and water. Would it be anabolic, nonspontaneous, and $+\Delta G$?
Ammonium chloride would be dissolved into the water. The formula would be something like
$$\ce{NH4Cl + water -> NH4+ + Cl-}$$

I've just barely touched on the $G$ value thing. I don't exactly know what's going on there. What I mean I guess is that if it's $+\Delta G$, the enthalpy increases, and if it's $-\Delta G$, the enthalpy decreases. To be more precise, we're on the Gibbs free energy lesson at the moment.
Endergonic means energy is absorbed during the reaction. Exergonic means that energy is released during the reaction. Is energy released or absorbed during a vinegar and baking soda reaction? Is energy released or absorbed when dissolving ammonium chloride and water?

I think you should describe the processes better. Do you mean disolution of salts or what? When you give G values, always specify the process, i.e, the initial state and final state.

@HernanMiraola I edited my post a bit. Thank you

Please do not use edit statements, they carry no information whatsoever for anyone who encounters the question for the first time. Use the [edit] function to improve your post, but still let it read consistently as one post (as if you were writing it for the first time). For those who are interested in the becoming of the post, there is an edit history available.

As you say you are starting with Gibbs free energy, just pay attention to the $\Delta G$ values. Remember that $\Delta G= \Delta H -T\Delta S$ so it doesn't depends only on enthalpy but also on entropy. A process where $\Delta G<0$ is exergonic and in the opposite case it is endergonic.
About the examples we can say that:

Salts that disolve in water are exergonic processes, $\Delta G_\mathrm{disol}<0$.
Sodium bicarbonate will disolve in vinegar to produce $\ce{CO2}$ (because of the high concentration of acetic acid), it is also exergonic.


text:
2017 Nobel Prize; What does Cryo-electron microscopy data actually look like?
It was just announced that the 2017 Nobel Prize for Chemistry has been

awarded to Jacques Dubochet, Joachim Frank and Richard Henderson "for developing cryo-electron microscopy for the high-resolution structure determination of biomolecules in solution".

What does the data that Cryo-electron microscopy produces when examining the structure of "biomolecules in solution" actually look like? Is it electron transmission images of the molecules, or a diffraction pattern produced by single molecules, or something different? Are the molecules still in liquid solution when they are measured, or are they actually frozen by the time the data is collected?
Including an example image or diffraction pattern or some representation of the type of data obtained from a biomolecule in solution would be greatly appreciated!

From this Nobel Prize tweet:

From what I've seen (admittedly not that much), the first picture on the Wikipedia page is actually more-or-less the raw data for CryoEM.

@chipbuster I'm trying to understand if  the Nobel Prize announcement's quote "high-resolution structure determination of biomolecules in solution" really means in solution, or actually frozen. Also, is high resolution molecular structure information really contained in those images? Comparing to this and this I'm still having a hard time understanding how this Scanning TEM image provides high resolution data.

@uhoh it is Frozen but the idea is that the freezing process takes place much faster than the time scale of reorientation. For macromolecules like proteins, these bulk motions are on the order of microseconds. So once they are frozen, the spectroscopy essentially samples a snapshot of the natural environment which for large N is sufficient to sample any static property.

@jheindel I see, so even though the water molecules are frozen, it's a similar-enough environment to liquid molecules. But it allows longer exposure times at lower electron beam currents because the molecules are not moving as much. Thanks!

This is a summary of the account by Peter Brzezinski, Professor of Biochemistry, Stockholm University, Member of the Nobel Committee for Chemistry from the Scientific Background document.

[Brzezinski:] Jacques Dubochet developed methods for preparation of samples for cryo-EM studies of biomolecules in water. Joachim Frank developed methods for structural determination of biomolecules from analyses of ensembles of particles in solution. Richard Henderson demonstrated that it is possible to obtain atomic resolution structures of biomolecules using cryo-EM.

To answer the questions of the OP, I will use some of the excellent figures in the document.

[OP:] What does the data that Cryo-electron microscopy produces when examining the structure of "biomolecules in solution" actually look like? Is it electron transmission images of the molecules, or a diffraction pattern produced by single molecules, or something different?

For larger particles such as viruses, it looks like the actual objects. Here is an example:

As you can see, the virus particles are not oriented (not a crystal, different molecules rotated differently). Other studies used two-dimensional or three-dimensional crystals. No diffraction images are used (in contrast to single molecule X-ray diffraction using free electron lasers). When the objects are large and have high symmetry, orientation of the molecules is easier than for small objects lacking symmetry. Thus, success in structure determination started with highly symmetric virus particles (pioneered by Henderson) and progressed to smaller objects requiring more powerful algorithms for orienting images (developed by Frank).

[OP:] Are the molecules still in liquid solution when they are measured, or are they actually frozen by the time the data is collected?

Samples are in vitrious ice. For the technique developed by Dubochet, no cryo-protectant is necessary (different from cryo-cooling crystals for X-ray diffraction):

Advances in detectors
The image posted by the OP mentions advances in detector technology. Traditionally, electron microscopy uses film or CCD detectors. In both cases, signal from electrons are "converted" into visible light, introducing noise. The new type of detectors are "constructed from Monolithic Active Pixel Sensors, based on Complementary Metal Oxide Semiconductors, known as CMOS technology". They directly detect the electrons, giving images with less noise.
Disclaimer
For brevity, I skimmed over the contributions of the laureates and did not mention any of the many other contributors. The article I reference above is excellent; I recommend it as a starting point to a more in-depth study of the topic.


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3D Printing a molecular orbital
Before I begin I know that what I am asking will deform the M.O but for what I am doing this for that is okay if it is only around the atoms and bonds.
My question is in JMOL is there anyway to display the isosurface of the molecular orbital and have it leave a gap around the molecule. I am going to a conference soon and need to print a model of a cyclic molecule. I would like to be able to "mount" the MO to the molecule using tiny magnets at certain points so I may dislodge the two phases and show the structure of the molecule. Yes using two different models (one with MOs and one without) is possible but not preferred.
I have explored isosurface command and different syntax and have not been able to find anything that may be of help.

Perhaps someone will actually know how to do this in JMOL, but if klunky hacks are acceptable, you could export the isosurface and then try to edit it in a program like MeshMixer (free, but unfortunately not on Linux)

Kind of the direction I might have to go. I would use TinkerCAD or Meshlab. Increase size of atoms and bonds, then delete it, leaving a cavity, then replace the molecule.

I've used VMD in the past to make 3D-printed molecular models: https://www.youtube.com/watch?v=8hdmBRxCrnw . Never done it with orbitals, though. I know VMD can visualize orbitals but whether converting JMOL format to VMD is easy, is another question.

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


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Why don't Gringard Reagents react with carbonyls at the alpha position?
According to many texts, the reason why Gringard reagents cannot be used with alcohols is because the Gringard reagent is a strong base, which would deprotonate the alcohol. ($p\pu{K}_b$ of Gringard $\approx 50$).
However, carbonyls also have a fairly acidic hydrogen at the alpha position ($p\pu{K}_a$ of enol $\approx 20$), so why does the Gringard reaction react with the electrophilic carbon, rather than as an acid-base reaction with the alpha position to form an enolate?

It actually happens sometimes and you end up with the enolate being formed in appreciable amounts. But it depends on other adjacent substituents that can withdraw or donate electron density to the alpha carbon.
If you draw the mechanism for the alpha attack you can see that the enolate can react further with the aldehyde and produce aldol condensation adducts. Furthermore, the carbon atom of the aldehyde is way more electrophilic than the hydrogen atom bound to the alpha carbon, that is the main reason why the enolate end up being formed in very small amounts and sometimes you don't even detect it.


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Born habor Cycles (specifically of MgO)
So on an exam paper they have said to find out the enthalpy of formation for MgO is : ∆H = 148 + 736 + 1450 + 496/2 - 141 + 798 – 3791
 = –552 kJ mol–1 , however if you draw out the born habor cycle you can see that the second enthalpy of electron affinity for O- is endothermic, so the arrow points up, meaning it should be 148 + 736 + 1450 + 496/2 - 141 -798 -3781
Please explain which is right
Thanks

Wow, that's a lot of numbers. It would be nice of you to tell us what does each of them mean.

The answer key is correct. You're trying to determine $\Delta H$ for:
$$\ce{Mg(s) + 1/2 O2(g) -> MgO(s)}$$
Since enthalpy is a state function, we can take any path between reactants and products and the net change in enthalpy will be the same.
This sequence is deduced from your numbers and the reaction. All units are $\mathrm{kJ\ mol}^{-1}$.
Vaporization energy for magnesium:
$$\ce{Mg(s) + 1/2 O2(g) -> Mg(g) + 1/2 O2(g), \Delta H = +148}$$
First ionization energy for magnesium:
$$\ce{Mg(g) + 1/2 O2(g) -> Mg+(g) + e- + 1/2 O2(g), \Delta H = +736}$$
Second ionization energy for magnesium:
$$\ce{Mg+(g) + e- + 1/2 O2(g) -> Mg^{2+}(g) + 2e- + 1/2 O2(g), \Delta H = +1450}$$
Bond dissociation energy for dioxygen (that's where there's division by 2):
$$\ce{Mg^{2+}(g) + 2e- + 1/2 O2(g) -> Mg^{2+}(g) + 2e- + O(g)}, \Delta H = +\frac{496}{2}$$
First electron affinity of oxygen:
$$\ce{Mg^{2+}(g) + 2e- + O(g) -> Mg^{2+}(g) + e- + O-(g)}, \Delta H = -141$$
Second electron affinity of oxygen. This one is positive:
$$\ce{Mg^{2+}(g) + e- + O-(g) -> Mg^{2+}(g) + O^{2-}(g)}, \Delta H = +798$$
There are two steps left but only 1 number, so I'm not sure what's going on there, unless you're forming magnesium oxide in the gaseous state:
$$\ce{Mg^{2+}(g) + O^{2-}(g) -> MgO(g)}, \Delta H = ?$$
$$\ce{MgO(g) -> MgO(s)}, \Delta H = ?$$
The key point is this: If you take all of the equations I wrote and you add them up and cancel out equivalent species on both sides, you get the original equation at the top which is the reaction for formation of magnesium oxide. The change in enthalpy is the simple the sum of all the deltas for the components of the cycle. The contribution from the second electron affinity is therefore positive.
It's really easy to get the signs mixed up if you're just focusing on the procedure instead of thinking about what the sign means. Here the positive sign means that energy must be put into the system to achieve the reaction, hence endothermic.


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Which is the suitable process for ball milling?
Why do we mainly choose the wet ball milling process in spite of dry ball milling process? Both processes are used to make a homogeneous mixture of different compounds.

Wet ball milling is much more effective due to Rehbinder effect producing a disjoining pressure.  Adsorption layers from surfactants that arise in microcracks significantly reduce the energy expenditure necessary to crush solid materials as a result of a decrease in their free surface energy upon contact with a medium containing substances capable of adsorption. Effect is observed in solids and porous materials, e.g. metals, ionic, covalent and molecular mono- and polycrystalline substances, glasses, polymers, both partially crystallized and amorphous.
This reduces milling time and wear of the working surface, at the same time improving homogeneity of the grinding fraction. The same effect is also used at the lab scale when, for example, metal oxides are dispersed in a mortar much more effectively when acetone is added. Sometimes after such treatment the reaction even takes place in a solid phase.


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Movement of solid particles in colloids
A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular forces. Where a solid is the same, but the atoms have much less freedom of movement.
Even if the water is still, its molecules are still moving against each other.
My question is: If you take a solid metal and create a fine dust and mix it with water, will the water molecules move the solid powder all the time?

You ask about Brownian motion ?

The short answer to this question is: yes.
The movement of particles in suspension like the powder in water would be a perfect example of Brownian motion as Mithoron already said. The particles are being kicked around by the water molecules, which are in constant motion as you already said.
This is also true for small airborne particles that collide with fast moving gas molecules.


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What is the correct way to notate conformation preferences of polysaccharides?
In some webpages (1 or 2) are found the way to notate the conformational preferences of monosaccharides such as furanose and pyranose.
But, it lacks to give a brief description on how to do it with polysaccharides.
For a given example:

Would it be: $^4C_1\ \phi\ \psi\ \omega\ ^4C_1$ ?
Or
Would it be mixed with Newman projections (3, 4 or 5): $^4C_1\ [a,s,t,g],\ [a,s,t,g],\ [a,s,t,g]\ ^4C_1$ ?
Or
What is the proper way to notate?

If I understood the question correctly, your first notation looks correct and less ambiguous.
Common polysaccharides (glycogen, starch, and cellulose are composed of pyranoside subunits (having six-membered rings)). Such molecules can be represented as a series of rigid pyranose rings connected by an oxygen atom bridging two carbon atoms (the glycosidic bond).

There is, in principle, free rotation about both $\ce{C–O}$ bonds
  linking the residues , but rotation about each bond is limited by
  steric hindrance by substituents. The three dimensional structures of
  these molecules can be described in terms of the dihedral angles,ϕ and
  Ψ, about the glycosidic bond.

The bulkiness of the pyranose ring and its substituents, and electronic effects at the anomeric carbon, place constraints on the angles ϕ and Ψ thus certain conformations are much more stable than others, as can be shown on a map of energy as a function of  ϕ and Ψ:
(a)

(a) When the relative energy (Ʃ) is plotted for each value of  ϕ and Ψ with isoenergy (“same energy”) contours drawn at intervals of 1 kcal/mol above the minimum energy state, the result is a map of preferred conformations.
A map of favored conformations for oligosaccharides and polysaccharides.
The torsion angles  ϕ and Ψ which define the spatial relationship between adjacent rings, can in principle have any value from 0 to 360 degrees.
In fact, some of the torsion angles would give conformations that are sterically hindered, whereas others give conformations that maximize hydrogen bonding.
(b)

(b) Two energetic extremes for the disaccharide Gal(β-3)Gal; these values fall on the energy diagram (a) as shown by the red and blue dots. The red dot indicates the least favored conformation, the blue dot the most favored conformation.
Conformation at the glycosidic bonds of cellulose, amylose, and dextran.

The polymers are depicted as rigid pyranose rings joined by glycosidic bonds, with free rotation about these bonds. Note that in dextran there is also free rotation about the bond between C-5 and C-6 (torsion angle ѡ (omega)).
So the notation considering the glycosidic linkage will be $\ce{Glu(ϕ)-(β1-4)-Glu(Ψ)}$ or another detailed way of illustrating this is $\ce{C1(ϕ)-O4(Ψ)-C4}$ for cellulose.

There are other ways of illustrating polyasacharides notations, and their side groups e.g hydroxyl group (in newman projection)
For the precise specification of the orientation of a polyatomic ring substituent, it is necessary to specify the torsion angle about the exocyclic bond.

The reference atom in the ring is the carbon atom with the number one
  lower than that of the substituted carbon, unless substitution is on
  the anomeric carbon, when the ring oxygen is the reference atom. The
  reference atom in the exocyclic –CH2X group is X. The exocyclic
  torsion angle is denoted by χ(chi), followed by the atoms to which it
  refers e.g χ(C1-C2-O2-H).

References

Lehninger Principles of Biochemistry 5th ed: Polysaccharides
Symbols for Specifying the Conformation of Polysaccharide Chains

Thank you but your answer did not response the question. Along your answer, you showed the conformational space, but you did not show, in any part of your response, a treatment related to the main question, on how to notate each conformer. At least, did you read the links that I shared? Thank you again

I tried to give an example from the diagrams drawn just before the underlined paragraph "So the notation will be......" Is it different from what you're looking for? It considers the anomeric form of polysaccharides with repeating Glu units

No, that notation is not related to conformational studies. It only relates the type of bond between two monosaccharides.

Ok thanks I will update if I get something,  along the lines


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Reaction between copper II sulfate and ammonium
The reaction between hexaaquacopper(II) and ammonium reacts as follows
$$\ce{[Cu(H2O)6]^2+ + 4NH3 -> [Cu(NH3)4(H2O)2]^2+ + 4H2O}$$
Apparently it's known from physical measurements that the hexaamine complex $\ce{[Cu(NH3)6]^2+}$ is not formed? Why is this?
Edit: This isn't a homework question by the way, I'm genuinely curious seeing as the hexaamine complex is formed when ammonium is reacted with complex cobalt ions.

"Simply speaking, the missing $\ce{[Cu(NH3)6]2+}$ in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites."

The six ligand positions are not equivalent. I think the best answer is this. https://chemistry.stackexchange.com/a/29172/22102


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Why does some compound as given in the image below change from low spin to high spin above a particular transition temperature?

Magnetic moment sudden increase at a particular temperature is explained by CFT. But why does it happen at all?

Related: https://en.wikipedia.org/wiki/Magnetochemistry#Paramagnetism


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What is an example of an endergonic reaction?
I am doing a project where I must find both an endergonic and exergonic reaction. The latter seems to be abundant, but the endergonic reactions seem to be purely biological. I have to present the reaction. The only thing I can think of that is endergonic (that isn't biological) is ice melting. Is there some kind of chemical reaction that's endergonic that anyone knows about? I have access to laboratory chemicals if I need to use them. If all else fails, I can do ice melting, but that sure will be pretty boring.
Thanks!

If you take an exergonic (spontaneous) reaction, the reverse reaction is endergonic (nonspontaneous). Ice melting is not good enough. Below the freezing point, ice melting is endergonic, but above freezing, it is exergonic.

You can look examples by yourself. Type on google any reactions and analize Gibbs Free energy (if they occur at constant T and P as in the body). Here you have a very good link http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php. Good luck!


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Decomposition of AsH3
In German problems olympiad 2014, the solution says that decomposition of arsine is not a redox reaction. Why would that be? Shouldn't H turn from -1 to 0 and As from 3+ to 0?
$$\ce{2AsH3 -> 2As + 3H2}$$

If they really said so, they probably implied that with electronegativity difference that small we may just as well consider both oxidation states 0 from the start, because they are arbitrary anyway.

Often in chemistry "facts" depend on perspective. You're right that essentially H goes from -1 to 0 and As from 3+ to 0. But in general redox reaction involve two different chemical species. In this case it is better to think of the reaction as a decomposition reaction rather than a redox reaction.

Decomposition reactions are autoredox reaction (Disproportionation ).Are they not ?


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are Butyrate and n-butyrate the same compound
in the biomedical literature I see references to both butyrate and n-butyrate which seem to be used interchangeably. Are they the same compound? What is the reasoning behind this nomenclature?

They are about as same as Smith and John Smith.

Neither are compounds. Both are names for the conjugate base of n-butanoic acid, but n-butyrate is more specific in that it specifies the straight chain connectivity explicitly. In principle, butyrate could refer to a branched version, but typically, when the name is ambiguous like this, it defaults to the straight chain variant.

I guess this answer answers the actual question; however, I don’t think that all statements are correct: Butyrate actually is a compound, “butyrate” cannot refer to a branched isomer, the name “butyrate” is not ambiguous, and generally an ambiguous name does not default to any particular variant. Anyway, it might be worth mentioning that “butyrate” is in accordance with IUPAC nomenclature (although it is not the preferred name) whereas “n-butyrate” is not.

@Loong "Butyrate" is an actual compound? Or are you talking about a compound word?


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What is the difference between an analytical standard and a certified reference material?
I am analysing the analytical standard market, but every time I try to understand what an analytical standard is, I only get the definition of a certified reference material. Is there a difference?

An analytical standard (as I understand it) is a defined amount of analytical reagent (EDTA, phosphate, iron, cer, etc.) that you can flush into a calibrated flask that you then fill up to the marking where it has it's defined volume, and you have a solution with defined "guaranteed" concentration, e.g. for titration. Does that fit to what you think it is?

I don't think you find any universally recognized definitions for the two terms. In essence an "analytical standard" has to be certified by  some organization/company. So to me the terms are essentially the same. Exactly how you use the standard would depend on what the standard sample was. You can have a certified weight, a set of steel standards for XRF, or KHP for acid/base titrations.

An analytical standard is a compound of suitable purity and known concentration to be used as a calibration standard for an assay.
A certified reference material is a material which has been certified by some trusted organisation to be of a consistent quality and composition. Normally, but not always, the concentrations of the major components of a CRM have been carefully measured. For example, the US-NIST produce hundreds of reference materials for everything from limestone to baking chocolate to dried human lung tissue. Labs analysing these materials use the reference material as a quality control to check that they are consistent with other labs.


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Can minerals form in the subsurface?
Can gypsum form in the subsurface? If there is another source of sulfate and maybe limestone dissolution happening, if the ionic strength is high enough can gypsum crystals form in groundwater?

yes, gypsum can form from solution in the ground. See: https://en.wikipedia.org/wiki/Gypsum


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Weak buffers and accuracy when mixing acid and base

What is the weakest buffer possible - that is, what has the lowest buffer capacity possible?

I would just assume that the lower the concentrations of $[\ce{A-}]$ and $[\ce{HA}]$, the lower the capacity is, but it seems like the question is looking for a specific solution?
Also, how precise do you have to be when mixing a strong acid and base to get a pH value of 7 and not 6? since $[\ce{H+}] = 10^{-7}$ and  $[\ce{OH-}] = 10^{-7}$ at $\mathrm{pH} = 7$, while $[\ce{H+}] = 10^{-6}$ and $[\ce{OH-}] = 10^{-8}$ at $\mathrm{pH} = 6$, does this imply you can at most be off by a factor of $10^{-2}$ mol/L?

What is the weakest buffer possible - that is, what has the lowest buffer capacity possible?

You've conflated two different aspects of a buffer. First there is the $K_\mathrm{a}$, or $\mathrm{p}K_\mathrm{a}$ for the species.
$$K_\mathrm{a} = \frac{\ce{[H^+][A^-]}}{\ce{[HA]}}$$
$$\mathrm{p}K_\mathrm{a} = -\log{K_\mathrm{a}}$$
So weaker acids would have a high $\mathrm{p}K_\mathrm{a}$, close to 7 or greater.
Then there is the buffer capacity which is a function of the absolute concentrations of the various species ($\ce{[H^+]}$, $\ce{[A^-]}$, and $\ce{[HA]}$). So if there is relatively much more $\ce{[H^+]}$ or $\ce{[A^-]}$ in the solution than acid or base being added, then the pH will change by only a small amounts.

Also, how precise do you have to be when mixing a strong acid and base to get a pH value of 7 and not 6?

If the strong acid is $10^{-6}$ greater than the strong base then the pH will be 6.
$\mathrm{pH} = -\log\ce{[H^+]}$


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How does this reaction occur?
I was doing homework for my chemistry class, and I came across a question asking to complete and balance the following acid-equation:
$$\ce{Cu(OH)2 (s) + HClO4 (aq) -> ?}$$
At first I thought it was an acid-base reaction, but clearly $\ce{Cu(OH)2}$ is a precipitate. Thus, it could not be an acid-base reaction, and I was under the impression there was no reaction at all.
However, it turns out that the answer was to the question was (courtesy of the back of the textbook):
$$\ce{Cu(OH)2 (s) + 2 HClO4 (aq) -> Cu(ClO4)2 (aq) + 2 H2O (l)}$$
How is this possible? The answer to the question was exactly the same as an acid-base reaction, despite a precipitate instead of a base in the reactants.
Anyone know why this is the answer, or if it is even correct?

What makes you think it cannot be an acid-base reaction because one of the reactants is in solid state? $\ce{Ca(OH)2}$ is also poorly soluble in water, but I don't think you'd doubt it reacts with $\ce{HCl}$ reagrdless.

Since its not soluble, how can it produce OH- in a solution?

Why it should produce $\ce{OH-}$ in solution? The reaction takes place at the interface.

If no $\ce{OH-}$ is produced, how is it a base?

What is it then, if not a base?

I was under the impression it was just a normal ionic compound. If it was a base wouldn't it produce $\ce{OH-}$?

An acid or base can react with a solid. For instance HCl will dissolve $\ce{CaCO3}$ with much fizzing of $\ce{CO2}$ gas.
You also miss the point that $\ce{Cu(OH)2}$ has a solubility product. So there is a small amount of $\ce{Cu^{2+}}$ and $\ce{OH^-}$ in solution.
$\ce{K_{sp} = [Cu^{2+}][OH^-]^2}$
The perchlorate anion is a powerful oxidizing agent, but since $\ce{Cu^{2+}}$ is fully oxidized no redox reaction occurs.

Oh ok, thanks. That clears up my confusion a lot, but since $\ce{Cu(OH)2}$ is a weak base (I'm assuming it is), shouldn't the neutralization of it with perchloric acid be in the form $\ce{HA+B->A-+HB+}$, rather than forming water?

@Frank you have essentially the right equation in your question $$\ce{Cu(OH)2 (s) + 2 HClO4 (aq) -> Cu(ClO4)2 (aq) + 2 H2O (l)}$$ The only quibble I'd have is that the $$\ce{Cu(ClO4)2 (aq)}$$ is solvated , so it would be $$\ce{Cu^{2+} + 2ClO4^-}$$ in solution. Also I think the copper ion would coordinated with water molecules rather than the perchlorate anion.


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Is the Fischer Esterification mechanism very simplified?
Example mechanism:

Is that very simplified, or is that pretty much what is very likely happening at the molecular/atomic level when you do the reaction in real life?  (Of course besides the free $\ce{H+}$, that is obviously hydronium, protonated alcohols, or the acid itself.)

You can write 8 mechanisms for both esterification and deesterification. As long as the mechanism is reasonable, it has a right to live.

That is the wrong question to ask. Mechanisms are putative pathways. Accepted mechanisms are mechanisms that we have not found evidence to refute. There is no way to determine whether a mechanism is the true mechanism.

This isn't exactly actual. There are methods that let analyse mechanisms directly.

You have a window but that doesn't rule out things that you aren't measuring. Your putative mechanism is consistent perhaps but not guaranteed to be correct. @Mithoron

What Zhe has said is undoubtedly correct.
However, the currently accepted mechanism that is taught in lectures typically represents exactly our understanding of how a mechanism would work. There may be non-productive side steps, there may be backwards and forwards shuffling, but the effective pathway that a molecule takes will — under the assumption that the mechanism is correct at all — happen in pretty much exactly that order.


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What molecule is this?
I have tried some only searches but I couldn't find the molecule name. Can someone identify this for me?

Possible duplicate of What is this formula representing?

This is caffeine.
By the way, you can draw structures in ChemSpider (or related websites) for somewhat simple structural identification.


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Exothermic and Endothermic reactions stability
We Know that when a system becomes stable,it gives out energy.
In case of exo reaction,heat is liberated resulting in energy loss and thereby stabilizing the system.
In case of endo reaction,heat is taken in,thereby increasing the energy of the system thus making the system unstable.
So why would a endothermic reaction occur to make the reactants even more unstable..(we know that reaction occurs to make the reactants stable)

Enthalpy of reaction isn't the only criteria used for determining the spontaneity of a reaction. For example, water evaporates from the ocean even though it is an endothermic process. But the process of evaporation increases the entropy of water to that extent where it compensates the +ve enthalpy of the process and thus, make the process spontaneous.

TL;DR
System spontaneously tries to reach most stable state. This depends on not only enthalpy, but entropy and hence, Gibbs free energy.
"Why is water a solid below $\pu{0^\circ C}$, a vapour above $\pu{100^\circ C}$ and a liquid between these two states, all under atmospheric pressure conditions?"

A substance prefers to exist in the most stable form it can - let us consider this generally true, and consider a relatable example. This is a qualitative explanation.
Consider a system which contains pure $\ce{H2O}$ - just one component and one phase. Suppose you are able to maintain it at any desired pressure and temperature. You set it at atmospheric pressure. You can vary the temperature.
You know that the stable state for $\ce{H2O}$ below $\pu{0^\circ C}$ is solid. Suppose you take not-solid $\ce{H2O}$ (i.e. water or steam) in your system and bring it to atmospheric pressure and a temperature below $\pu{0^\circ C}$. You will need to cool the system, meaning that you will need to take the evolved heat away. It is becoming more stable at atmospheric pressure and a temperature below zero by giving heat out.
You know that the stable state for $\ce{H2O}$ above $\pu{100^\circ C}$ is gas. Suppose you take not-gaseous $\ce{H2O}$ (i.e. ice or water) in your system and bring it to atmospheric pressure and a temperature above $\pu{100^\circ C}$. You will need to heat the system, meaning that you will need to provide heat to the system. It is becoming more stable at atmospheric pressure and a temperature above zero by absorbing heat.
You need to know about another thermodynamic property - entropy, represented using $S$. A system is considered stable when it does not generate any more entropy. The heat you mentioned in your question is called enthalpy, represented using $H$.
It is found that the Gibbs free energy, which is defined as $G=H-TS$, always decreases while approaching the stable state. When it eventually 'reaches' the stable state, i.e. the equilibrium condition, it stops decreasing. So, at a constant pressure and temperature, $\Delta G=\Delta H-T\cdot \Delta S$. If a system is approaching a stable state, its value of $G$ is decreasing. So, $\Delta G$ is negative. So, as a system goes to a stable state, it has to maintain $(\Delta H-T\cdot\Delta S)<0$. The heat you mentioned in the question covers only the $\Delta H$ term.
When you factor in the positive $\Delta S$ due to melting of ice and vaporisation of water, everything will fall in line.

We Know that when a system becomes stable,it gives out energy. It depends on what you mean by stable, e.g. a  green algal cell is 'stable', but it is absorbing light energy all the time.
In case of endo reaction,heat is taken in,thereby increasing the energy of the system thus making the system unstable. This is not true for spontaneous physical reactions.
Secondly I can not think of any spontaneous chemical reaction that is endothermic - you are not taking into account whether heat energy input is driving the reaction or the high potential energy of the reactants is driving the reaction
So why would a endothermic reaction occur to make the reactants even more unstable (we know that reaction occurs to make the reactants stable)?
Reactions occur to make the entropy of the whole system less - not the reactants. that is where you have to be careful.

$$\ce{2H2(unstable) + O2(unstable) -> 2H2O(stable) + heat}$$
$$\ce{CaCO3(stable) + heat -> CaO(unstable) + CO2(stable)}$$
The above describes a hot hydrogen fuel flame that is causing the calcium carbonate to slake.
You have to consider the whole system not just one reaction. At the beginning you have 2x-unstable, at the end you have only 1x-unstable. The amount of instability is less.


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Can I safely use gelatin to mix with 25%-35% peroxide to make a gel?
Can I safely use gelatin to mix with 25% - 35% peroxide to make a gel
or should I use glycerin?

What are you trying to achieve?

Hydrogen peroxide? You've got to be kidding me...

This sounds like a particularly dangerous case of amateur chemistry...

@mithoron it's the 30% part that is particularly frightening...

Not to tell if it was another peroxide...

That depends. There is no way to handle 30%+ hydrogen peroxide completely safe, it is one of those ever interesting and laboratory-equipment-destroying substances. In your case, judging by the casual way you pose the question, I think it would be interesting in a stack-exchange-user-detonating way as well.
Please reconsider whatever you are attempting. The answer is no and no.


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Is it possible to have Volume(solution) < Volume(solvent before adding solute)?
I understand the answer to the question "Is it possible to have Volume(solution) < Volume(solvent) + Volume(solute)?" is "yes" -- but it is possible that adding a solute causes the solvent to contract so much that the volume of the solution is less than the initial volume of the solvent?
An ideal answer would include either a specific example of such a case, or an explanation of why it is impossible.
If the above is not possible, is it possible for the volume of the solution to equal the initial volume of the solvent?

@Zhe I would count that only if the total volume of the mixture were less than the initial volume of one of the components (either the water or the alcohol, but not the sum of the two).

I remember encountering this question elsewhere. The expert opinion was that there is no law forbidding this, but a convincing example remained elusive.

@IvanNeretin I would accept that as an answer if you can locate and cite that expert opinion.

I will not post this as an answer because it does not meet my own standards. The question is really good, it deserves a more comprehensive handling .

Sorry, I misread your question. I'm not sure how this contraction is going to take place. Liquids are generally incompressible since there is not much intermolecular space. So what you're asking is essentially to compress an essentially incompressible material and add more matter into that mix...

@Zhe Liquid hydrogen has density of about 0.07. Liquid methane is 0.4, and 1/4 of its mass is hydrogen. So the concentration of hydrogen in it is greater than in pure liquid hydrogen. That's how, for example, an incompressible can be compressed.

While that's technically true, we're not dissolving carbon into liquid hydrogen to produce liquid methane... @IvanNeretin

@Zhe That's why this is not an answer.

Gosh, sure. According to various on-line info (of unknown quality) a 2% by wt. solution of NaOH in water has a density of 1.0207 at 20°C. This means 100 g of it has a volume of 97.97 ml. At 20°C water has a density of 0.9982 meaning that 98 grams of water has a volume of 98.18 ml.

For record-keeping purposes, here is 1 such online source for the 2% NaOH solution (which claims its data comes from "Perry's Chemical Engineers' Handbook"): http://www.handymath.com/cgi-bin/naohtble3.cgi?submit=Entry


text:
Are aspartic acid and glutamic acid at physiological pH present in zwitterionic form?
I am wondering whether aspartic acid and glutamic acid at physiological pH are present in zwitterionic form? In my textbook it says that all aminoacids at  $\textrm{p}H$ of 7.4 or 7 are in zwitterion form. I wonder if $\textrm{p}K_\textrm{a}$ values of side chains of aspartic acid is 3.65. There are also arginine, lysine, cysteine and tyrosine that  will not have overall neutral charge at $\textrm{p}H = 7$.

What's unclear about it? -NH2 is protonated, both -COOH are deprotonated, so here's your zwitterion.

How can it be zwitterion if definition of zwitterion is that it has separated charges but it is overall neutral  .

Related on Biology SE: https://biology.stackexchange.com/q/23824. I believe the issue might be the meaning of present. It may not be the dominant form of the amino acid, but all of them should have some population of the zwitterion at biological pH.

Then again, probably the guys who wrote the textbook still considered this a zwitterion despite the overall negative charge, or they forgot that some aminoacids have an additional acidic group in the side chain, or they just did not care.

The IUPAC goldbook defines a zwitterion as being overall neutral. Any other definition is thus contrary to the standard definition.


text:
Why does the optical rotation prefix differ for glyceraldehyde and glyceric acid when both rotate polarised light in the same direction?
When both glyceraldehyde and glyceric acid rotate the plane-polarised light in the same direction, why is glyceraldehyde assigned as D(+) and glyceric acid assigned as D(-)?
The first image shows the chemical conversion of D(+)-glyceraldehyde to D(-)-glyceric acid. The second image shows the rule for assigning absolute configuration.

Well, there are only two configurations, no matter what do you call them.

ᴅ-(+)-Glyceraldehyde and ᴅ-(−)-glyceric acid do not rotate polarised light in the same direction.

@Loong  kindly elaborate . I have assigned priorities to groups around the carbons in both the compounds . Kindly correct if I have assigned in a wrong way

D and L are determined by arbitrary rules that you've referenced. (+) and (-) specify the net optical rotation of plane polarized light by a sample. There is no way for a human to determine (+) and (-) just by looking at a molecule. It is fixed, but it is whatever it is.
There is no relationship between the D/L and +/- designation across all substances. However, for a given substance the relationship is fixed.

Thank u for your answer. But as far as I know D,L is different from d,l.  D,L refers to Relative configuration whereas d,l or +,-  or R,Srefers to absolute configuration. I want to know why both glyceraldehyde and glyceric acid are assigned different absolute configurations when both rotate plane polarised light in same direction or is there a difference between +,- and R,S. kindly correct me if I am wrong

The D/L designation is fully irrelevant, as Zhe’s answer points out. The same is true for R/S or any other minor standards that are based on the molecular structure alone. Whether a molecule rotates polarised light to the right ($+$) or to the left ($-$) can be determined only by experiment.
There is a clear discrepancy between your two statements:

D-(+)-glyceraldehyde and D-(−)-glyceric acid

and

rotate the plane of light in the same direction

Both cannot be true; either the naming was incorrect and D-glyceric acid should be called D-(+)-glyceric acid or the measurement result was in fact the opposite, i.e. the polarisation is opposite. The answer by Zhe and Loong’s comment assume the latter. However, checking Sigma-Aldrich’s webpage, I get told that the optical activity of D-glyceric acid is positive by the sign convention. That would mean that it should be correctly labelled as D-(+)-glyceric acid. The caveat being that the disodium salt was measured.
The dispute will only be settled by confirmation experiment or by literature citations. In any case, one statement will fall. Either the designation is correct; then they polarise in different directions. Or the result you state is correct, then glyceric acid’s designation is wrong.

I think as your said, only R and S nomenclature of a compound can be assigned theoretically by looking at the compound. But, + and- can be assigned only after experimentally determining the compound's optical activity using a polarimeter. This is what I have understood. Thank you your answer

@Shakti I didn’t explicitly mention it but all stereochemical descriptors aside from ($+$) and ($-$) are determined based on the structure. Whether it is (R / S), (aR / aS), d/l, D/L (if there even is a difference between the two), syn/anti, cis/trans or (E / Z). If I forgot to mention one, it was probably oversight. So none of these correlate in any way with the direction of rotation of polarised light, and there is no way to predict the polarisation based only on knowing the structure.

Thank you Jan......so basically these configurations are only to say the spatial arrangement of groups around a chiral carbon but cannot determine the optical rotation . Am I right ?

@Shakti Yes. You cannot even infer that the optical rotation will be the same as in a sample compound after a reaction that does not touch a single chiral centre. Source: it happened to me after opening a lactone. (I have not seen any examples but I do not deem it impossible for the sign of optical rotation to change upon something as simple as alcohol methylation without that inducing any severe conformational change.)


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Solubility of Alkali Metal Chlorides
Why does Potassium Chloride less soluble than Sodium Chloride in water, isn't it supposed to be other way round? Since Potassium Chloride would be more ionic than Sodium Chloride?

So what? MgO is pretty ionic, yet insoluble at all.

Well MgO does react, even with moisture from air (moderately strong base).  The hydroxide has limited solubility but still greater than most other metal hydroxides.

When dissolving, the ionic crystal’s lattice enthalpy must be broken while at the same time the hydration shell is generated around the cations and anions. The latter process is typically exergonic while the former is strongly endergonic.
How well a salt dissolves depends not only on the strength of solvation but also on the strength of the crystal lattice. If the lattice is strong, the solubility will be lower because there is an increased tendency to reform the crystal lattice.

Sodium ion has smaller size than potassium ion.So sodium ion can more easily form a hydration sphere around it than potassium.So NaCl is more soluble than KCl.


text:
Which has higher lattice energy: MgO or LiF?
Which has higher lattice energy: MgO or LiF ?

A compound has higher lattice energy if it's ions have smaller size and greater charge

Li cation is smaller than Mg cation => +1 for (LiF)
Fluoride anion is smaller than oxide anion => +1 for (LiF)
Mg cation has greater charge than Li cation => +1 for (MgO)
Oxide anion has greater charge than fluoride anion => +1 for (MgO)

Therefore, 'charge' factor favour MgO while 'size' factor favour LiF.
So, which one will have higher lattice energy?

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/lattice.html may help

$\ce {MgO}$ ($\approx 3800$ $\pu{kJ mol^{-1}}$) has higher lattice energy than $\ce {LiF}$ ($\approx 1045$ $\pu{kJ mol^{-1}}$) mainly because of the greater charge on  $\ce{Mg^2^{+}}$ ion and $\ce {O^{2-}}$ as lattice energy is directly proportional to the charges of the combing atoms.
Your reasoning isn't incorrect but remember that $\ce{Li}$ and $\ce{Mg}$ show diagonal relationship so the size of their ions has only slight difference in magnitude. Similarly, oxygen and fluorine are placed next to each other in the periodic table so their anions don't have a large difference in size.
Thus, the "charge factor" dominates.

according to me MgO has more lattice energy that LiF.
The lattice energy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
More ionic is a compound, stronger would be the ionic bond and more would be the lattice enthalpy (Ref).


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Could you make a helium atom from hydrogen gas?
Hydrogen gas is $\ce{H2}$, and contains 2 electrons and 2 protons. My question is, with sufficient pressure, could you fire two neutrons at the molecule and have them form a single Helium atom?

The sun does this all the time. It is hard to convert H2 gas to helium. That is the gist of fusion reactors which my someday be a power source here on earth.

https://chemistry.stackexchange.com/questions/82648/can-helium-gas-be-formed-from-deuterium-gas

Not really H2 @MaxW.  Not even atomic hydrogen.  Hot enough to have unadulterated protons in the soup.

I'm voting to close this question as off-topic because its about physics, not chemistry

The hydrogen nuclei are spaced out by about $\pu{70pm}$, if I guessed/remembered the number correctly. To generate a helium nucleus, this distance has to be reduced to almost $\pu{0pm}$ against the repulsion of like charges. So if you just bombarded a hydrogen molecule with neutrons from one side, nothing would happen since the two would not fuse.
It may work if you manage to shoot both neutrons into different hydrogen atoms in such a manner that both hydrogens are displaced towards the other. But this type of collision can be considered more of a thought experiment than anything practical. (Even just hitting a single molecule with two neutrons at the same time is hard enough.)

The process is nuclear fusion which takes place in sun where H is combined to form high energy and He is produce   which produce light too.You can do it but at rare instance when neutron is bombarded it may form neutron when protons and electron like in neutron star for more detail
https://youtu.be/Cb8NX3HiS4U


text:
How should one check an instant-equilibration algorithm for conservation?
We have a system of equations that describes the equilibration between several phases $A, B, ...$.  If we define the total concentration $T = A + B + ...$, then $\partial_t T=0$ must hold.
We are further assuming equilibrium at all times, so we have devised an algorithm for this that transforms $A\rightarrow A'$, $B\rightarrow B'$ etc.  Finally, the total concentration must be conserved, so we must check if $T'=T$.  But should one also check if $\partial_t T'=0$, i.e. the conservation equation with the equilibrated variables?
Does it make sense to test both $T'-T=0$ and $\partial_t T'=0$?

I would think one of those conditions would always be sufficient. If you have shown that $T'-T=0$ for all $t$ and that $\partial_tT=0$, then  $\partial_tT'=0$ must be true by default. That is $\partial_tT=0$ shows that $T$ is a constant and if $T'-\text{constant}=0$ for all $t$, then $T'$ is also a constant.

Clear answer. Indeed, if for all $t$, $\partial_t T=0$ and $T' = T$, then $\partial_t T'=\partial_t T=0$. The converse is not true, so it appears that $T'=T$ is a stronger condition than $\partial T'=0$.


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Why does a carbocation undergo covalent bonding with, say, chloride, rather than ionic bonding, like what occurs between Na+ and Cl-?
I've been trying to look for an answer for a while but I can't find one. Why does a carbocation accept an electron pair from a nucleophile and form a covalent bond rather than just accepting one electron like what happens between a sodium ion and a chloride ion in the formation of sodium chloride?

Why do two chlorine atoms not form an ionic bond with $\ce{Cl+}$ and $\ce{Cl-}$?

check out https://chemistry.stackexchange.com/a/17072/9961 and maybe https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond

Chemistry 001, octet rule?

Some carbocations do form true ionic compounds.  Many are known in which the positive charge is incorporated into an aromatic ring (e.g., cyclopropenyl, tropyl) or conjugated with multiple aromatic rings (e.g., triphenylmethyl).  But in most carbocations you have a compact, relatively low-energy vacant orbital that will overlap to form a bond with an electron pair from, say, a chloride ion.  Alkali metals don't have such compact valence orbitals and thus do not form a strong covalent bond even with the positive charge all on one atom.

You've reminded me that it's possible to make pure hydrocarbons which are actually ionic salts. Amazingly, these substances reversibly dissociate in organic solvents and their solutions conduct electricity!

I remember seeing those.  They are awesome.  Might they be better approaches to true, full-fledged ions than things like sodium chloride?

@OscarLanzi No, if you want as much ionic character as possible then check Non-coordinating anions.


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Hydration Isomerism in Cr(III) and Co(III)
Cr(III) and according to a paper I found also Co(III) are famous for the property called hydration isomerism. I already asked about how changes in the environment of a complex, for example temperature causes a ligand substitution but Cr(III) seems to be treated differently here. I guess it's somehow basically the same reaction but Cr(III) often adapts to changes quite slowly.
I once made a Cr(III) salt, which was green in solution and it took a week till the color suddenly changed to purple. On the other hand if I dissolve some anhydrous CrCl3 and wait for a day the purple color will change to green. This exchange of ligands and counter-ions or solvent seems to be a bit slower for Cr(III) and of course some reactions don't even happen for other metals. The chlorocuprate(II) would only form at elevated temperatures when in equilibrium with the aqua-complex.
But perhaps this temperature is just lowered for Cr(III) and room temperature is enough to cause the change here. Because according to a paper I found is this property of hydration isomerism something very unique in Cr(III) and Co(III) or d3 and d6 ions.
But then I again I don't see that much of a difference to other changes in ligands. Most anhydrous compounds will change from a situation where the counter-ions are coordinated to the metal to an aqua-complex. They do it faster than Cr(III) and they do the opposite reaction but still I don't see that much difference here.
The problem is that I cannot find any good explanation about what makes Cr(III) so special. Does this have to do with the d6 configuration? I know Cr(III) tends to act as an acid as well forming polynuclear compounds, so perhaps the Cr-OH2 bond is weaker than the Cr-Cl bond or something like that?
To summarize it and formulate is as a question, because it doesn't really matter if we give this property a name or not, what makes Cr(III) so special, that it will exchange its aqua ligands for chloride ligands when dissolved in water? Or the other way around? Because there are also reports where, if you dissolve the aqueous CrCl3 in water the green solution will turn purple and in my case the purple solution turned green. And with other Cr(III) complexes I watched transitions in both directions as well.

I fail to follow. Hexachlorideferrate(III) is easily formed and noticeable by its yellow colour. Tetrachloridocobaltate(II) likewise with its blue colour. Where did you get the idea that you are elaborating?

related https://chemistry.stackexchange.com/questions/83469/why-does-heating-transform-the-aquacomples-of-slightly-acidified-cobaltii-chlo

Possible duplicate of Why does chromium(III) have the highest tendency to form complexes, comparatively, among Mn(III), Cr(III), V(III) and Ti(III)?

This question^ is still unclear and should be edited accordingly to answer if it's supposed to make a duplicate of this here.


text:
Understanding Galvanic Cells
I am an electronics guy and not into chemistry so please forgive my ignorance.
A Galvanic Cell is meant to act like a pump, separating positive and negative charges, and thereby creating an EMF. Obviously energy has to be expended in this process.
All my reading tells me that this occurs on account of redox reactions.
If I have understood right, redox reactions happen because of the chemical properties of elements and their atomic structures, and on account of the need for some elements to acquire or get rid of electrons, basically in order to attain octet 'stability'.
Now I presume that this urge for 'stability' is based on the natural process whereby things trend towards their lowest energy states.
Now if this is so, how is it that in redox reactions, charges get forced apart, which requires the expenditure of energy to accomplish. Where, how and why does this energy come from ?
How do we get this free gift of energy from nature just because a chemical element strives towards energy 'stability' by filling an octet.
Obviously I am missing something.
Can someone patiently explain what happens in a redox reaction, and why ?
I request you to use simple examples for electrode and electrolyte materials.

Sorry but the answer is contained in your question. The energy needed to separate charges is provided by the substances striving to get into a more stable state. In this respect, it is like to boil water using the energy provided by burning substances...


text:
Relative humidity above various mixtures of table salt and water
I heard that if I saturate some water with NaCl and put it in a closed box, the relative humidity near the water surface should be around 75 % at room temperature.
I am using this to calibrate some electronic sensors measuring relative humidity.
If a saturated solution gives 75 % RH, how can i know which RH more diluted mixtures should result in?
--
My setup is a small 1l plastic container where in I put a mixture of table salt (iodized) and tap water (about 150 ml). The sensor is hanging a few cm above the water surface and measures for a day or so.
The solutions I use are:

Fully saturated water (Plenty of undissolved salt crystals)
75 % saturated water (3 parts saturated, 1 part pure water)
66 % saturated water (2 parts saturated, 1 part pure water)
50 % saturated water (1 part saturated, 1 part pure water)

I am also monitoring the temperature in the container, and it is usually swinging a few degrees below 20 °C.

What do you mean by "part"? Mass, volume?

I mean volume..

Your information on relative humidity (RH) above saturated $\ce{NaCl}$ is valid within about 0.5% over the range of 0 °C to 80 °C. For other standard humidity solutions, such as $\ce{MgCl2}$ or $\ce{K2CO3}$ the U. S. National Institute of Standards & Technology provides a convenient reference.
However, though it is possible to adjust RH by varying concentration, there are, "special  problems...  because  their  concentrations  must  be  measured  and controlled." If you need only a few accurate RH points, it would be simpler to use saturated solutions of some alternate salts.
BTW, it may take some time for RH to equilibriate because of stratification. It would help to keep the solution on a slow magnetic stirrer, moving not only the liquid but causing eddies in the air above it.

I don't need very accurate calibration.
Most interesting for me is to be able to detect 90% RH, and 5% accuracy would be fine.

For ideal behaviour the vapour pressure above the sodium chloride solution is given by Raoult's law as
$$p_w = p_w^{\star}~x_w \tag{1}$$
where $p_w^{\star}$ is the vapour pressure of pure water, and $x_w$ the mole fraction of water in the solution.
Therefore the relative humidity
$$\mathrm{RH} = \frac{p_w}{p_w^{\star}} = x_w \tag{2}$$
equals the mole fraction of water in the solution. Note that this is independent of temperature.
The molar mass of water is $18.02~\pu{g mol^{-1}}$, that of sodium chloride is $58.44~\pu{g mol^{-1}}$.
The solubility of sodium chloride at $20\pu{°C}$ is $359.2~\pu{g l^{-1}}$.
The density of saturated sodium chloride solution at $20\pu{°C}$ is $1.200~\pu{g cm^{-1}}$.
From this it can be calculated that $1$ litre of saturated sodium chloride solution at $20\pu{°C}$ contains $n_w = 46.65~\pu{mol}$ water and $n_s = 6.146~\pu{mol}$ sodium chloride.
Thus the relative humidity above the solution is
$$\mathrm{RH} = x_w = \frac{n_w}{n_w + 2 \cdot n_s} = 0.791 \tag{3}$$
The factor $2$ in the denominator of equation $3$ comes from the assumption that $\ce{NaCl}$ as a strong electrolyte is completely dissociated in aqueous solution.
The actual behaviour of sodium chloride is not ideal, so the real value for the relative humidity may differ from the calculated one.
Since the mole fraction of water increases for more diluted solutions it follows that you can only prepare relative humidity in the range from $100$ to $79.1$ percent using sodium chloride solution.
The values for diluted solutions can be calculated using equation $3$. However, if you need more accurate values, you had to use tabulated ones.


text:
Bonding in PCl3
Why is there a need to produce 4 equivalent sp3 hybrids? Wouldn't the 3s2 orbital be just fine on its own since it's spherical? Wouldn't it be easier to just be closer to the nucleus and not get involved in the 3p business? What would change if there were just three 3p orbitals?

Yes, it works as you say.

related https://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h


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Unpaired electrons from magnetic moment

A complex formulated as $\ce{Cs_x[NbCl6]}$ has spin-only magnetic moment $\mu_{s.o}$ of approximately 2 B.M. From this value calculate how many unpaired electrons are present.

Spin-only magnetic moment is related to the number of unpaired electrons by the formula
$$\mu_{s.o}= \sqrt{n(n+1)}$$ where $n$ is the number of unpaired electrons.
I let $\sqrt{n(n+1} = 2$ and solved for $n$. I got two values one $+$ and one $-$. Do I just take the positive value and round it to the nearest whole number. In this case I got $n_+= \sqrt{5} - 1 \approx 1.2$.
Does this mean there is $1$ unpaired electron?

Yes for both questions.

@a-cyclohexane-molecule Would you be willing to expand and make your comment an answer?


text:
Confusion between electronegativity and electron affinity
Electronegativity is a chemical property that says how well an atom can attract electrons towards itself.
The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.
However,fluorine is taken as the most electronegative element but chlorine is accepted as the element with most electron affinity.
But shouldn't the atom which attracts the electron the strongest also release the most energy?

They have almost nothing in common.

https://chemistry.stackexchange.com/questions/6901/why-does-chlorine-have-a-higher-electron-affinity-than-fluorine?rq=1

I wouldn't shoot this down so quickly - consider, e.g. Mulliken electronegativity, defined as the average of the electron affinity and ionisation energy. If you think about it, electronegativity depends not only on how strongly an atom pulls other electrons to itself, but also how willing it is to give up its own electrons. The former can be somewhat represented by the EA, the latter by the IE.
One point that hasn't been mentioned is that electronegativity is not a physical parameter which can be measured. It is a human construct, which is often helpful in explaining chemical reactivity, but at the end of the day it is not something which can be derived from physics. Therefore, there is no real reason to expect that there should be any one-to-one correlation between the EN and the EA (unless the EN scale is explicitly defined to follow the EA). For example, if you use the more common Pauling scale (defined by bond dissociation energies), there's no real reason why the trend in EN should match the trend in EA exactly.
Broadly speaking, the trends match (because of what you've said) - e.g. both EN and EA generally increase going across a period. However, the small details like chlorine vs. fluorine won't necessarily match.

You should keep in mind that electron affinity is a property of atoms, while electronegativity is a property of atoms when they form bonds with other atoms.
Fluorine is indeed the most electronegative element, which means that if there is a bond between an atom of fluorine and an atom of another element, then the fluorine atom will attract the electrons of the bond towards itself.
When it comes to electron affinity, chlorine has a lower value than fluorine because fluorine is small, so when you add an electron to it, the increase in attraction of the electrons to the nucleus leads to more energy being released when an electron is added. Chlorine, on the other hand, has an increased atomic radius so the distance from the electron to protons increases and the electrons have less of an attraction to the nucleus meaning that when an electron is added lesser energy will be released due to the lesser attractions.

this question is good even I had the same question and I searched web and found that it already has an answer here although the answer is bit long but it's coomprehensive. The answer has mainly to do with sum of electron affinity and ionization energy of the atom being considered.


text:
How close are we to building an actual nanobot?
I want to know how far current nanotechnology is from creating an actual real-world nanorobot? Also, does such a thing actually exist, or is it just a nice name for an umbrella term to show off to the general public different fields of science and engineering which might not be actually linked?
P.S. I understand that the term nanorobot may be taken into consideration because it may mean different things to different people, that's why when I use it I mean something bigger than a molecule but smaller than a biological cell which could replicate and do some "stuff" like the things described in Eric Drexler's book Engines of Creation or in foresight's institute nanotechnology roadmap. I guess they will be somewhat the size of biological organelles and just like them be composed of various molecular complexes some capable of self assembly. I want to know what is the current state of research into the creation of such structures and how far are we from building them.
P. S. 2. And, please, guys can you not use only Wikipedia for your answers. I want to know what serious research is being done and where it's published in, not just a general introduction to the field. I'll be very thankful if you could provide some sound material published in peer reviewed journals or at least a list of keywords I could use for searching for articles related to the creation of real nanorobots or a list of journals where such research can be found. If you could do at least that you will have my gratitude.

You described the "nanostuff" quite accurately. "Real nanotechnology" is... biology :D

I just want to have the answer to a question about the current state of nanotechnology to its "ultimate" (if I could even use this word) goal of building something we can call a nanorobots.  think this section of SE communities is the best place to ask since there isn't one on nanotechnology and I couldn't find the question asked so clear anywhere on this site. Do you think here is the right place and this is the best way to ask it?

I would like to see your questions on a forum with a lot of discussion, but Stack Exchanges site aren't forums. Closest you can get may be Worldbuilding. SE, where are various speculative and open ended questions.

I'm starting to see the limitations of this community, however, it's still the best community I had encountered so far on the Internet in terms of people who actually do serious science and aren't "bible mumblers"or "new age priest" trying to bring on some very twisted form of pseudoscience they use to further their ideology. I think I have something interesting to ask and offer however I have no job in science and can't get publish from the place I'm currently in, so my best option is asking questions here @Mithoron. What else do I have?

Actually your question isn't exactly clear. Various people talking about "nanobots" usually don't know what they are talking about. "Real nanobots" are for example ribosomes

Let us continue this discussion in chat.

Yes, nanotechnology does exist. And yes, this is an umbrella term. Before it came along, same researchers did the same stuff under different names. As for nanorobots: still pretty far.

Come on guys, why are you doing this? What's wrong with that question? There is nothing unchemical or broad here! I just want to know what is the current state of research on nanotechnology we are in? How is that not  right for Chemistry SE???

@YordanYordanov it is a debate/discussion type question with no objectively correct answer. First, nanorobot may mean many things, if you want, nanorobots already exist, second, it asks about the future and how it will unfold.

Yordan I came across your questions here and in biology SE. You are really aiming at some life or life mimicking stuff. Therefore, although I encourage you to keep curiosity and fantasy alive, you cannot be surprised if you find this site limited. It might be, but not because one cannot answer your questions which are always interesting but broad and, as it was clear to me in your (Ill posed) Q about single molecule and equilibrium or in the Q about single molecule life, at the border life/no life. Faced to a question like the current one I can only say, like others, "umbrella - term" .

@Alchimista I think you really would like to see this question of mine-https://academia.stackexchange.com/questions/91075/if-i-start-a-blog-to-spread-my-ideas-in-academia-could-it-be-successful . It says much more about my true intentions, than what I have written so far in either the chemistry or the biology SE sections.


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Conductivity measurements of demineralized water - Unexpected results. Looking for explanation
I'm trying to measure the conductivity change of demineralized water over a few days. But I get results I can't explain.
Setup
This is the circuit and cell I'm using:

Where $R_x$ is the conductivity cell, R is $1797 \Omega$.
The right picture shows the cell ($R_x$). The metal plates shown (barely visible) are made of stainless steel.
Evaluation
On ch 1 and ch 2 I measure the voltage by consecutively taking a sample of $10^4$ points over 4 ms on each channel then I calculate the RMS.
Using ohms law I calculate the resistance of the cell:
$R_x = R \frac{Ch_1-Ch_2}{Ch_2}$
The applied current is a sine curve of 0.5V with varying frequency between 1, 5, 10, 50 and 100 kHz. It is turned on for about 2-3 seconds during the reading of the two channels.
For each frequency I make 10 measurements and take the mean of those.
These measurements are then repeated every 30 min. Shown in the graphs below.
(Sample 1-3: I changed the water and repeated the measurement to see if it's caused by some residue in the cell from previous measurements, although I cleaned it thoroughly)

Why do the curves look like this?
If it's caused by polarization of the ions in the solution wouldn't the resistance increase with time?
I would be happy if someone could point me in the right direction finding a theory explaining this.

CO2 dissolve in water. This lead to increased ionic concentration so increased conductivity. Water evaporate too. This also lead to increased ionic concentration and thus conductivity.

@Alchimista Would it remove this effect or reduce it below the sensitivity of my measurement if I put the cell in a plastic bag then flood it with $N_2$ and afterwards fill it with water?

If you are looking for a boring almost flat curve in your graph then do it. Jokes a part, yes. More you hinder the contact between water and atmosphere more constant will be the conductivity you measure. Better you do it more constant conductivity. You can bubble the water too but prior of measuring. Then stop else you improve evaporation. Also avoid a dry stream of N2. Ideally the N2 should function as a static layer. But also seriously, what you have seen is the only interesting outcome... Else, ideally, you have a conductivity stable for ever. In case of constant conc only T effect is left

After everything is in place keep little inlet outlet for N2 and cabling, and just go with a minimal overflow


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Hardness of water using EDTA calculation issue
I have been asked to work out the hardness of water in German degrees and this is what I did:
I got a sample of water I didn’t measure how much I put into my volumetric flask perhaps it was $\pu{8.5ml}$ but I didn’t think I would need it, but I do use it which made me believe I was wrong.
The sample was made up to $\pu{100ml}$. $\pu{10ml}$ of this dilution was used as the analyte along with buffer and the indicator.
I Titrated $\pu{9.07ml}$ of EDTA with a concentration of $\pu{0.0185mol/dm^3}$
So the calculation I did was:
$$\begin{align} n(\ce{edta})&= (9.07\times0.0185)/1000= 0.000167795\\
n(\ce{Ca^2+}) &= n(\ce{edta})\end{align}$$
So then I find the mass of Ca:
$$m(\ce{Ca^2+}) = 40\times0.000167795= \pu{0.0067188g}$$
Now my issue:
So $\pu{0.0067188g}$ of Ca in the $\pu{10ml}$ therefore in the $\pu{100ml}$ there will be $10$ times that, as it was a dilution.
So $\pu{0.067188g}$ which is $\pu{67.118mg}$.
So there is $\pu{67.118mg}$ of Ca in my $\pu{8.5ml}$ sample.
Now a $1^\circ\ \mathrm{dH}$ is $\pu{10mg}$ of Ca per $\pu{1000ml}$.
So is there $\pu{7896.2mg}$ in $\pu{1000ml}$ of sample —
therefore having $789.6^\circ\ \mathrm{dH}$ of hardness?

Please always, always, always, always use units. An amount is not equal to a simple number like in your first calculation. Likewise, the multiplication of two simple numbers does not give a value in grams like in your second calculation.
That said, as far as I can tell your calculations add up. (It took me a few tries since you juggle with units so liberally — another reason to use them consistently.) So maybe you were not given tap water, maybe you grossly misestimated the volume you added to the flask initially or maybe your titration was wrong. What exactly happened is nothing we can answer.


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Why is enol content higher in non-polar solvents?

Apparently, for keto-enol tautomerism, the enol content of a given carbonyl compound increases when in a non-polar solvent.
What is the reason for this?

From my brief search in some textbooks, the matter seems to be much more complicated than your question suggests. Solvent definitely has an important effect on tautomerism. However, this is dependent not only on the polarity, but also on the structure of the carbonyl compound, hydrogen bonding, etc. A rather involved account is given in J. Org. Chem. 1985, 50, 1216; slightly more accessible is Reichardt's Solvent Effects in Organic Chemistry, 3rd ed. (p 106ff.).

Seconding orthocresol. Also, I note that the close votes are given for ‘unclear’ which doesn’t make sense at all as this question is sufficiently clear.

Naive guess: higher dipole moment in carbonyl group compared to enol. Probably more complicated though as suggested in comments above. Enol=hbond donor while keto=hbond acceptor so it would be reasonable to assume that e.g. keto favored in hdonor solvent and enol favored in hacceptor solvent. Just some ideas.

I doubt I am not good as others ,but comparing this thing with cyclic structure is completely unfair because that includes the strain as well as ortho para effect @orthocresol what I feel is the main and strong reason is the hydrogen bonding

@logicalx2 Keto group is a poorer H bond donor ($\beta = 5.3$) than alcohol which is also a donor ($\beta = 5.8$)

The type of solvent used is generally not major factor in determining the enol content formed in an keto-enol tautomerism. Generally it is overshadowed by other factors like aromaticity and hydrogen bonding.
But this being said doesn't mean that solvent has no role to play in the enol content. As you mentioned a polar medium prefers the keto form whereas a non-polar medium prefers the enol form.
This is because in a polar solvent, the lone pairs will be already involved in hydrogen bonding with the solvent, making them less available to hydrogen bond with the enol form and thus taking away from the stability of the enol form.
Whereas in a non-polar medium there is no hydrogen bonding with the solvent and thus it is preferable for the molecule to attain an enol form to gain stability via hydrogen bonding.
Sometimes this can cause drastic effects. For example, in benzene, the enol form of 2,4-pentanedione predominates in a 94:6 ratio over the keto form, whereas the numbers almost reverse completely in water.
To quote from Solvent effects on the tautomeric equilibrium of 2,4-pentanedione[1]:

For solvents such as $\ce{CCl4}$ and cyclohexane the intramolecular bond of the enol form persists and bulk solvent effects account for the equilibrium enol–keto content. In solvents such as DMSO, disruption of the intramolecular bond occurs and the percentage of enol falls due to unfavorable entropy changes. The enol intramolecular bond is disrupted by the solvents water and methanol. Enol hydrogen bond formation through self-association and with the solvent accounts for the entropy changes upon enolization in these solvents.

Reference:
(1)   Spencer, J. N.; Holmboe, E. S.; Kirshenbaum, M. R.; Firth, D. W.; Pinto, P. B. Solvent Effects on the Tautomeric Equilibrium of 2,4-Pentanedione. Can. J. Chem. 1982, 60 (10), 1178–1182.


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What happens to minerals once you boil off liquid in a water and fat mixture?
Let us say you are cooking a curry.  As water evaporates the minerals will become more and more concentrated in the curry's water but they will not deposit in the fat which would be at the top layer of the liquid. Regarding the minerals not depositing in the fat whilst water exists, is this correct?
Once you have evaporated all of the water which I imagine happens before any fat starts evaporating, would the minerals then deposit in the fat?
Is there a way to know by observation that water has evaporated and you are left with minerals and fat?

I don't know what you mean by "deposit". Mineral content of tap water is minimal, and you can find it as solid precipitate once you boil water off. Minerals generally don't dissolve in fats or oils, so you find the minerals as white solid in the pan, as a powder or stuck on the wall

Because dietary minerals are predominantly water soluble and not significantly soluble in fats, your first statement is correct; the minerals will not dissolve in the fats as you boil away the water.  This is true even after all the water has boiled off, so the answer to your second question is no. As the water boils away, the minerals will precipitate out until no water remains. At that point, you will predominantly have the minerals at the bottom of the pan, covered in the fats. It is unlikely any of the minerals will have a low enough density to float on top of the fats. With mixing, you may be able to keep a suspension of the minerals within the fats even though they don't technically dissolve.
It will probably be difficult to tell exactly when all of the water is gone. There should be a difference in the appearance of the way the stuff boils once the water is gone and it's just fats boiling, but this could be a subtle change. You might be able to tell when the water is gone by monitoring the temperature of the mixture. It should stay close to the boiling point of water until the water is gone and then increase when you just have the fats boiling.
I'm far from a chef, so I hope I haven't made any improper assumptions on that front. Don't hesitate to ask for clarifications in the comments.

@James Wilson. I am a chemist not a cook, but I cannot recall any cooked food preparation in which one should evaporate ALL the water. I mean, even after barbecue there will be some water leaking at the surface of the meat. If the matter of your question is not loosing minerals, even in the case you concentrate so much that the salts precipitate, they will be finally dispersed (or could be finely disperded by spoon etc) in your fatty sauce or sticked to the aliments... Something like that. No worry much.