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Experimental approach
Here are 100 marbles in a polyethylene bag:
If a marble touches the bag, it is counted as being on the surface. I did not count, and the bag was not perfectly round. A balloon might be better.
Theoretical approach
This works for large $N,$ and the result is as Ivan predicted (from scaling arguments in 2D and 3D, I would imagine). We ...
8
Note that the classical idea of an atomic nucleus as a set of nucleon balls is very wrong, similarly as another classical idea of an atom as a planetary system of orbiting and spinning electron balls.
I suppose you have had bad luck until now as this task is in opinion of my non-math brain resistant to both general theoretical ( geometry, topology ) and ...
6
Here I used a numerical approach roughly equivalent to the analytic approach in Poutnik's answer. The procedure is to generate the smallest enclosing sphere containing within its radius 100 hexagonally close-packed beads, and then determine the number of beads in a second smaller concentric enclosing sphere. If r=1 is the radius of a bead, R=5.074 is the ...
4
Have you used anhydrous $\ce{CaCl2}$, dihydrate $\ce{CaCl2 . 2 H2O}$ or hexahydrate $\ce{CaCl2 . 6 H2O}$ ( 3 most common forms) ? It is a big difference in the resulting thermal effect.
Hexahydrate causes cooling down of the solution while being dissolved. If ice is used instead of water, as the mixture hexahydrate : ice 2 : 1, it forms the famous ...
2
As the unperturbed Hamiltonian is Hermitian it follows that
$$\int{ψ^{(0)*}_0}\hat {H}^{(0)}{ψ^{(2)}_0} dτ = \left[\int{ψ^{(2)*}_0}\hat {H}^{(0)}{ψ^{(0)}_0} dτ\right]^*=\left[\int{ψ^{(2)*}_0}E^{(0)}_0{ψ^{(0)}_0} dτ\right]^*=E^{(0)}_0\int{ψ^{(0)*}_0}{ψ^{(2)}_0} dτ$$
by the definition of Hermiticity and the fact that all eigenvalues of a Hermitian operator ...
1
Although the question has been already answered, I will like to add a bit more of detail.
The expression you are quoting come from the thermodynamic formulation of the TST that expresses the dependence of the rate constant of temperature as
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta G^\ddagger_0}{RT}\right)$$
and can be partitioned in an entropic ...
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