# Tag Info

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Yes there is validity to the finding that two atoms of hydrogen form molecular hydrogen at very slow rates. Review Gould, R.J., and Salpeter, Edwin E., Astrophysical Journal, Vol 138, pg. 393. It is publicly available at: http://articles.adsabs.harvard.edu/pdf/1963ApJ...138..393G The exact answer to your question is found on page 138 of that work, under ...

0

Potential difference and EMF are equal if the galvanic cell does not produce any currant. When the cell works and produces even a small currant $I$, the potential of the cell decreases a bit, because of the internal resistance $R_i$ of the cell. If $E_o$ is the the EMP, the potential $E'$ measured between the plus and the minus poles of the working cell ...

1

We know from the first Newton motion law, that the net force acting on an object in rest must be zero. The forces acting on the piston are gravity and gas pressure: $$\vec F_g + \vec F_\mathrm{p,down} + \vec F_\mathrm{p,up}=\vec 0 \tag{1}$$ If $V$ is the given bottom gas volume, $V_0$ is the total gas volume, $n$ is the molar amount of each of gases, the gas ...

1

Whatman number their filters according to the type of material e.g. glass fibre, paper, wet strengthened, hardened, ashless paper and also the porosity e.g. 6 micron, 2.5 - 5 micron. These are also related qualitatively to speed, e.g. fast, medium. So 54 is fast speed (because it's larger pore size) and hardened paper. In this case, 41 is a fast (20 micron) ...

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yes, there are several errors in your calculation. Germicidal lamps specify an effective power output in UVC. This lamp https://www.osram.de/ecat/PURITEC%20HNS%20UV-C-UV-C%20lamps%20for%20purification-Ultraviolettstrahler-Industrie-Spezialbeleuchtung/de/de/GPS01_1028570/ZMP_1022255/ consumes 16W of electricity but has a UVC output of 4W. You need to ...

3

You kind of had the right idea, but not quite. First of all, they gave you the molar heat capacity at constant pressure, so you didn't have to express it in terms of the degrees of freedom. Secondly, the change in entropy is not Q/T, it is $\int{\frac{dQ_{rev}}{T}}$. Third, for a constant pressure process, $dH=dQ=nC_pdT$ This should be enough information to ...

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I think your methodology is a bit wrong. When the frozen brine melts into the water, the water won't be 100% water anymore. So, the specific heat capacity of the water changes, making the change in temperature different. Maybe that's why you have a 25% drop when compared to water. You can follow the steps below only if you have a heat source that generates a ...

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A1: There is no displacement of the wall, forces are in equilibrium with zero net force. There are initiated stronger vibrations on warmer wall side by energetic hot air molecules. And weaker vibrations on colder wall side by energy leaching cold water molecules . A2a: If the wall initial temperature was the cold water temperature, then yes. Otherwise no, as ...

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If you have an initial estimate for the pressure, $p_0$, then you can quickly solve iteritively for the pressure that matches the desired entropy $s=s^*$ using modified Newton's method: $$p_{n+1}=p_n-\frac{s(p_n,T)-s*}{\left(\frac{\partial s}{\partial p}\right)_T}$$The partial derivative of s with respect to p can be obtained by finite difference at the ...

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J is the total angular momentum. L is the orbital angular momentum and S is the intrinsic total electron spin angular momentum. J = L + S L = $\sqrt{L\left(L+1\right)}\frac{h}{2\pi }$ --> L can be n-1 where n is the Principal quantum number S = $\sqrt{S\left(S+1\right)}\frac{h}{2\pi }$ --> S is an integer or half an odd integer, depending on whether ...

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Note that there is the law of mass and energy conservation, but there is no law about volume conservation. Molecules of matter are in eternal motion. Molecules of gases move freely by flying between collisions. Nitrogen or oxygen molecules of air have an average speed of a supersonic fighter, colliding at rate typically 10 billions collisions per second, ...

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The rate is determined by the slowest step, as you correctly assumed. The transition state of C->P is assumed to occur significantly faster than the first step in Organic Chemistry at least. Unless you have an energy diagram for the reaction or any Le Chat, I think you have it right as rate = k[A][B].

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Note that pH effect is reaction specific. If a reaction is e.g. $\ce{A + B + H+ <=> C + D+}$ , or $\ce{A + B- <=> C + D + OH−}$, then lowering pH supports the spontaneity. If OTOH, the reaction is $\ce{A + B + OH- <=> C + D−}$ , or $\ce{A + B+ <=> C + D + H+}$, then lowering pH goes against the spontaneity. It can be generalized by ...

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Suppose you want to study the equilibrium of the reaction of a metal like zinc $\ce{Zn}$ with water to produce $\ce{ZnO(s)}$ and $\ce{H2}$ :$$\ce{Zn(s) + H2O(g) <=> ZnO(s) + H2(g)}$$ At high temperature, the equilibrium constant may be written $$K_\mathrm{1}=\frac{[\ce{ZnO}]·p\ce{(H2)}}{[\ce{Zn}]·p\ce{(H2O)}}$$ In this expression three parameters are ...

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Aspartame 4 calories per gram source Sucrose 3.9 calories per gram source But you need far less aspartame than sucrose to get the same level of perceived sweetness

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Simply put, it is because we don't have complete or near complete understanding of the forces that drive chemical reactions, every few atoms added to the structure of the compounds will add new forces and layers of complexity that we haven't accounted for in our simple 300 years of chemistry knowledge. You can sense this when you learn the theories show ...

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I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M). While $\ce{H2SO4}$ is a strong acid, $\ce{HSO4-}$ is not ($\mathrm{p}K_\mathrm{a}$ around 2). So your pH should be about zero, and you might not need to use the Hammett acidity at all.

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Let's be positive. The edited proposal has merit. You can pass a current thru a solution of a carboxylic acid and measure it with an ammeter (a more sensitive milliammeter would be better). How about this: check for conductivity. Commercial conductivity meters use AC current and an AC ammeter. Use two inert, identical electrodes for this, and use DC if you ...

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Yes, square brackets are used to denote a coordination sphere, but they may sometimes be used incorrectly. Ideally you want to check the crystal structure, luckily there is one : The $\ce{Cu(II)}$ atom has square-planar geometry with two bidentate oxalate ligands. Each $\ce{K+}$ cation is coordinated by eight oxygen atoms from one bidentate and two ...

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Such questions troubled physicists in the early 20th century. Their inability to add more than one electron to the planetary Bohr-model was one of the reasons that quantum mechanics was developed. Nevertheless, continuing to ask this type of questions may help us sharpen our thinking and develop our intuition, as far as it goes (you might also take a look at ...

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A simplified explanation would be to think about a balance between minimising potential and kinetic energy. While an electron sticking to the nucleus would decrease potential energy due to electrostatic attraction, the spiked and extremely localised wave function would be high in kinetic energy (correlated with the curvature of the wavefunction). I do ...

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To add to the plethora of recent answers: it's not that the electron cannot "crash" into the nucleus, it certainly can, the likelihood depending on the isotope in question. The process of fusion of proton and electron to form a neutron is called electron capture. None other than Luis Alvarez studied such processes early in his career. You might ...

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This is only a comment on the answer by Pallas, which I upvoted: Fig. 3-4. The wave function and probability distribution as functions of r for the n = 1 level of the H atom. The functions and the radius r are in atomic units https://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html I visualize the electron as a ball of cotton candy (spun sugar), ...

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The planetary view of a a small electron orbiting a nucleus does indeed not give a satisfactory explanation; Bohr simply postulated stable orbits in his original model. In QM, you could argue that the bound electron is the wave function , and therefore much larger than the nucleus, despite being much lighter. The closest thing to sticking together in this ...

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