New answers tagged computational-chemistry
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When exchange effects are not taken into consideration, the simplest possible description of a many-body wave function is simply the Hartree product, i.e. the product of individual-particle wave functions. This is reasonable for bosons, but electrons are fermions, which means that the exchange of any two must flip the sign of the wave function.
Slater ...
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As Poutnik pointed out, if you boil a 98% ethanol solution (2% H$_2$O), then you will lose a few % of ethanol until you get to ~95.5% ethanol; then the rest of the solution will boil off as if it were a pure compound, for the composition will not change. Well, this is strictly true at the high temperature of boiling at 1 atm pressure. At about 0.1 atm, the ...
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I don't think that any realistic modeling is possible. There are a lot of details which would made such a system chaotic. For the liquid side of the gas/liquid system consider the differences in evaporating from a thin layer in a very large diameter vessel to evaporation from a extremely long tube of the mixture.
For the gas side how is the diffusion away ...
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Is the particle in a box concept analogous to an electron in an orbital?
No, but there is an analogy to the $\pi$ systems of dye molecules, and (for the 3D box) to the band structure of nanodots. Some physical chemistry courses have a lab that explore this relationship, e.g. at Saarland university (sorry, in German, but the figures and the math are ...
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I decided to give a help here since OP has given enough effort to solve this problem. First, OP needs to know the molarity of a solution is defined as the amount of solute (in this case $\ce{NaBr}$) in $\pu{mol}$ per $\pu{1.0 L}$ of solution. The first part of problem states:
If $\pu{2.60 g}$ of $\ce{NaBr}$ is dissolved in enough water to make $\pu{160.0 mL}...
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It comes to be $\pu{252 mL}$ precisely:
$$V = \frac{\pu{2.6 g}}{\pu{103 g}} \times \frac{1}{0.1} \times \pu{1000 mL} = \pu{252.4 mL}$$
$\text{Molar Mass of NaBr} = \pu{103 g mol-1}$.
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