Answer:
to find the number of electrons you need the charge of the element.
Elements and compounds are both examples of pure substances. Compounds are substances that are made up of more than one type of atom. Elements are the simplest substances made up of only one type of atom. i hope that helps:)
Hope this might help ask me if u have any doubts
Answer:
1) The power needed to process 50 ton/hr is 135.4 HP.
2) The void fraction of the bed is 0.37.
Explanation:
1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).
We assume the units of Ei are kWh/t.
The equation that relates this parameters and the power is (size of particles in μm):
![W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t](https://tex.z-dn.net/?f=W%3DEi%2A%28%5Cfrac%7B10%7D%7B%5Csqrt%7BP80%7D%7D%20-%5Cfrac%7B10%7D%7B%5Csqrt%7BF80%7D%7D%20%29%5C%5C%5C%5CW%3D9.45%2A%28%5Cfrac%7B10%7D%7B%5Csqrt%7B3175%5Cmu%20m%7D%7D%20-%5Cfrac%7B10%7D%7B%5Csqrt%7B76200mm%7D%7D%20%29%5C%5C%5C%5C%5C%5CW%3D9.45%2A%280.1774%2B0.0362%29%3D2.019%20kWh%2Ft)
The power needed to process 50 ton/hor is
![P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP](https://tex.z-dn.net/?f=P%3D2.0194%5Cfrac%7BkWh%7D%7BTon%7D%2A%5Cfrac%7B50Ton%7D%7Bh%7D%2A%5Cfrac%7B1.341HP%7D%7B1kW%7D%3D%20%20%20135.4%20%5C%2C%20HP)
2) The density of the packed bed can be expressed as
![\rho=f_v*\rho_v+f_s*\rho_s](https://tex.z-dn.net/?f=%5Crho%3Df_v%2A%5Crho_v%2Bf_s%2A%5Crho_s)
being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).
Then we can rearrange
![\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37](https://tex.z-dn.net/?f=%5Crho%3Df_v%2A%5Crho_v%2Bf_s%2A%5Crho_s%5C%5C%5C%5C%5Crho%3Df_v%2A0%2B%281-f_v%29%2A%5Crho_s%5C%5C%5C%5C%5Crho%2F%5Crho_s%3D1-f_v%5C%5C%5C%5Cf_v%3D1-%5Crho%2F%5Crho_s%3D1-990%2F1570%3D1-0.63%3D0.37)
The void fraction of the bed is 0.37.