23

For steel wool the combustion reaction is roughly: $$\require{mhchem} \ce{2 Fe (s) + 3/2 O2 (g) -> Fe2O3 (s)}$$ So the object 'absorbs' (and chemically binds) air oxygen and thus gains weight.


3

All oxidation reactions will result in a greater mass of oxide than of the original material, due to the added oxygen in the oxide. Oxygen has mass, after all. The important part to note here though is that because of the mass of the original element(s) and the structure, the majority of the oxide falls to the surface, so its mass can be measured fairly ...


2

Because any reaction involving oxygen not requiring heat has already happened. So, all the reactions that have not already happened (and so you can observe them) require heat.


2

Simply, no. You have a model of a 2D diffusive process, but most flows are not dominated by diffusion. What weather phenomena are you hoping to model with this? There is a reason that one of the uses of supercomputers is for weather forecasting, the flow is incredibly complicated! Atmospheric flows are typically turbulent, advection dominated and 3D, as well ...


1

DNA computing is essentially mechanical computing using macromolecules. The computers made in this way are slower and less scalable than the electronic chips: DNA self-assembly and other similar processes are essentially chemical reactions, much slower than switching electric current. DNA is a molecule that has teh size bigger than the logical gates of ...


1

Sodium is not a blackbody. Rather the spectrum is determined by all the possible transitions between energy levels of it's electrons. Some of those levels are degenerate to varying degrees. If there are a lot of possible transitions with the same energy, the spectrum of the material will have a higher intensity for those energies, or wavelengths, ...


1

These two examples are of different origins. The black body spectrum is the results of two factors: Bloltzmann distribuition $e^{-\beta \hbar \omega}$ and the available number of modes between $\nu$ and $\nu + d\nu$, $N(\nu)$. The intensity of spectrum of frequency $\nu$, $$ I(\nu) = N(\nu) \sum_{j=1}^\infty j e^{-j\beta h\nu} = \frac{N(\nu)}{ e^{\beta h \...


1

The answer here is that the water wets the dust grains. The wetting process involves weak chemical bonding between the water droplet surface and the dust grain surface. In the ideal case, the droplet "swallows" the dust grains it touches and because the droplet is far far bigger than the typical dust grain, gravity pulls the droplet down through ...


1

You can't just jump from a diffusion problem for one variable and one equation to weather prediction. As you aspire to do this in 2-D, you will already have to solve at least the mass, and two momentum equations simultaneously, together with the advection problem and pressure gradients. The mathematical character of your equations that you solve changes (...


1

I suspect there is some confusion here because you are considering the entropy of the total system. There is nothing wrong with as long as the total quantities and the quantities of the separate phases are not confused. In your last equation \begin{equation} n_1\,d\mu_1 + n_2\,d\mu_2 = -S\,dT +V\,dP \end{equation} You should have \begin{equation} \left(\frac{...


1

Why [must $S/n_1$] be equal to $S_1$ (molar)[?] By definition. $S$ depending on a variety of other parameters doesn't preclude such a definition. We can conclude that the molar entropy of species 1 in this system also depends on those parameters. why we are able to [replace] $\Delta S=S_2−S_1$ with $\frac{1}{T}ΔH$[?] $\Delta G = 0$ during a phase ...


1

I take care of older people's houses. When I cleaned their windows, the water would not distribute itself over the glass in a smooth way. Many small disconnected patches were visible. After I had cleaned them, these patches had grown in size significantly. Bt they were still disconnected. That was a sign that I didn't clean them "to the bottom".


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