New answers tagged physical-chemistry
3
Have you ever heard of explosions? Explosions are chemical reactions of solid or liquid substances which produce a huge amount of gas in less than one second. For example, nitroglycerin is a liquid with the formula $\ce{C3H5N3O9}$, which makes dynamite. One mole of nitroglycerin weighs $227$ g and has a volume $142$ mL. The explosion corresponds to an ...
3
It seems like you're wondering how the products can have a greater volume than the reactants without violating conservation of mass.
So: In a chemical reaction, the number of atoms is conserved. That preserves conservation of mass. But, as you can see from your balanced chemical equation, the number of molecules isn't conserved: There are more molecules on ...
4
Since AC can be capacitively coupled to the solution, metallic electrodes don't need to contact the solution -- they can be outside an insulating container, e.g., glass or PTFE, avoiding introducing metal ions.
At moderate AC frequencies, ions won't migrate an appreciable distance in one direction before returning to that position on the opposite half of ...
2
[OP] As I have read that every reaction(endothermic or exothermic)requires activation energy.
Yes, and this activation energy is for a single set of reactants reacting to form product.
[OP] So is it correct to say that even exothermic reactions absorb heat?
No. Heat is a bulk transfer of thermal energy from one object to the next.
If you search for ...
2
Exothermic/endothermic (or exergonic/endergonic, if you equally account for changes in entropy) only compares initial state prior, and final state after the reaction.
So the height to overcome that this reaction happens (the activation energy) is irrelevant. It equally is irrelevant if this energy was invested by heat (e.g., Bunsen burner/fire, as in ...
2
Every exothermic reaction releases net heat - by definition.
Every reaction absorbs energy, e.g. thermal energy from heat, for its activation, but the amount of released energy is for exothermic reactions bigger than absorbed energy.
4
«How well a mark shows up» is an ambiguous description. Yet,
Does softness of a metal correlate to how easy material abrades? If you refer to Mohs' scale, yes. Modern test kits contain pens with synthetic alloys to offer reproducible access to Mohs' levels of hardness
(credit, commercial source)
like it is mostly known in art for silver, the silverpoint. ...
1
Ionic compounds as a general rule have greater bond strength than covalently bound molecules. This makes them more heat resistant, therefore it seems likely that a binary explosive consisting of an oxidizer salt and a fuel would be the way to go.
Since you don't want the explosive to change phases during the mission, since it could leak or unmix, you'd want ...
1
You overthink it.
It is just matter of eventual multiplication or division with respective stoichiometric coefficients.
If there is an elementary reaction
$$\ce{a A -> b B}$$
then for respective reaction, appearance and disappearance rates:
$$\begin{align}
R_\mathrm{r} &= k_\mathrm{r}[A]^a\\
R_\mathrm{a,B} = \frac {\mathrm{d[B]}}{\mathrm{d[t]}}=bR_\...
0
Well, consider this argument on the undoubtedly actual expanded system, which would commence with the action of strong UV on a Chlorine gas in presence of Methane, and seemingly would proceed based on random kinetics, as follows:
\begin{align}
\ce{Cl2 + UV &-> Cl^. + Cl^.}\\
\ce{CH4 + Cl^. &-> CH3^. + HCl}\\
\ce{CH3^. + Cl2 &-> CH3Cl + ...
1
Your interpretation is interesting. As you state, the free radical $\ce{Cl·}$ is regenerated at the end of the reaction. It behaves like Vanadium(V) which oxidizes $\ce{SO2}$ into $\ce{SO3}$. And once reduced into Vanadium(IV), it gets reoxidized by $\ce{O2}$ into Vanadium(V), according to the two equations $$\ce{V2O5 + 2 SO2 -> 2 VO2 + 2 SO3}$$
$$\ce{2 ...
7
The answer is a definitive affirmative yes.
You don't need to reach out for proteins, organic dyes are known to suffer from solvatochromism in function of the polarity of the solvent they are dissolved. It either may be a bathochromic shift (or, red shift) of the UV-Vis absorption recorded, or a hypsochromic shift (or blue shift). In large proteins, you ...
1
As far as I can remember, I was taught in 6th grade that warm water would dissolve sugar with greater ease than cool water.
That is true. Sometimes, the sugar in honey crystallizes. If you carefully heat up the honey, you can get it back to a liquid. So there is experimental evidence that sugar is more soluble as you heat up the solution.
In 11th grade, I ...
2
I propose this simple graphic answer as a complement to the more thorough replies above (comments welcome to confirm validity of my answer).
Trace pH Vs % [HA] and [A-]. 0% means you only have the acid [HA] and 100% you only have [A-].
Addition of acid or base, to the buffer solution, affects the ratio [A-]/[AH] and consequently, pH. This has been explained ...
0
[This question was asked long time ago, but it appeared on the list today. Let me give it a shot.]
It would make more sense to compare the eutectic point with the azeotropic point for a heteroazeotrope (not for a typically seen homoazeotrope). You can see the similarity between them in the following two phase diagrams, both showing three phases at the ...
3
We can start by watching two videos showing how sugar and salt are dissolved in water: https://www.youtube.com/watch?v=fwjvwoFHTbg and https://www.youtube.com/watch?v=14iv71ZTRTo. Particularly, the second video is fun to watch, where molecular dynamics simulation is used to show how water molecule solvates an ion at the corner of a NaCl particle.
To ...
1
Here's the answer if anyone is wondering.
$$m_{X_\mathrm{4}O_\mathrm{6}} = \frac{16*6*100}{43.7} = 219.7g$$
$$m_{X_\mathrm{4}} = 219.7 - 96 = 123.7g$$
$$m_X = \frac{123.7g}{4} = 30.99g$$
$$n_X = \frac{43.7\%}{30.99} = 1.443\ \ \ \ \ \ \ \ \ \ \ \ n_O = \frac{56.3\%}{16} = 3.519$$
$$\frac{1.443}{1.443}\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3.519}{1.443}$$
$$1\ \ \ \...
2
I had read about the DLVO theory and colloids a long time ago, so in case I make a mistake, feel free to correct me.
The secondary minimum is not deep enough to hold the particles for a long enough time to cause flocculation. Basically, in room temperature, colloid particles should have enough energy to get out of that secondary minimum and drift away from ...
1
It is overcomplicating the answer to assume it has anything to do with diffusion
Framing this as a "diffusion" problem is a distraction which is why it is hard to see the key features if you start with the assumption diffusion has anything to do with it.
The basic setup is simple. The starting position is two cylinders of gas with the same mass of ...
3
This problem can be thought of as a linear combination of atomic orbitals $\phi_-$ and $\phi_+$ to molecular orbital $\phi$ with broken symmetry (i.e. LCAO-MO and $c_1 \neq c_2$). The answer to it can be figured out as follows.
As stated in the conditions, the normalized atomic orbitals are $\phi_-$ and $\phi_+$ for the left and right intervals centered at $-...
0
Okay, so I want to provide another answer just to elaborate on what was said.
$\ce{E^\circ}$ refers to the cell potential at standard conditions (By the way, it is not STP. STP is for gases primarily). So if we were doing an experiment at 50°C instead of 25°C, then we would use the Nernst to find the cell potential at the specific temperature (E). The ...
4
Air cannot be liquified at temperature naturally occuring on Earth.
More exactly, oxygen cannot be liquified above -118.6°C, nitrogen above -146.9°C.
Things are little more complicated as air is a mixture, but you can get the picture.
At extreme pressures like tens or hundreds gigapascals, air would solidify.
2
It should be clear to you even just from the chart. Think of drawing the third p=f(V) curve above the upper one, for higher temperature. There would be no edges, pressure would continuously increase as volume would decrease. The parabolic-like curve tells you properties of liquid and gaseous phase converge to each other for raising temperature. See also ...
0
If the concentration $\ce{[Cu^{2+}]}$ is $0.01 M$, the potential of the copper electrode becomes $\pu{E_{Cu/Cu^{2+}} = 0.34 V - 2· 0.0295 V = 0.28 V}$. If this copper electrode is coupled with a zinc electrode where $\ce{[Zn^{2+}] = 1 M}$, the potential of the cell becomes $\pu{0.76 V + 0.28 V = 1.04 V}$
Now the effect of temperature change can be calculated ...
7
You are unclear if you are confused by $n$ itself, or by the particular notation $n(\ce{Na2CO3})$
$n$ is widely accepted to denote amount of substance, expressed in moles, as @Loong , IUPAC and Wikipedia say.
If $n$ means the amount of the particular substance like $\ce{He}$ or $\ce{H2O}$, it is often written in index form like $n_{\ce{He}}$ or $n_{\ce{H2O}}$...
13
Yes, the standardized quantity symbol according to ISO 80000-9:2009 Quantities and units — Part 9: Physical chemistry and molecular physics as well as the recommended quantity symbol according to IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book) for amount of substance is $n$.
The quantity “amount of substance” shall not be called “...
1
The short answer is: the Lennard-Jones potential is a completely heuristic model, and the short-range part of the potential is not physically meaningful except in the sense that it becomes very repulsive very quickly. Pauli repulsion is known, also heuristically, to decay as an exponential, so using a very steep polynomial works reasonably well. As to ...
Top 50 recent answers are included